3^ CORNELL UNIVERSITY LIBRARY MATHEMATICS uate uue Cornell University Library QA 371. B32 Differential equations. 3 1924 001 518 152 »»«> Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924001518152 LONGMANS' MODERN MATHEMATICAL SERIES General Editors P. Abbott, B.A., C. S. Jackson, M.A. F. S. Macaulay, M.A., D.Sc. DIFFERENTIAL EQUATIONS LONGMANS' MODERN MATHEMATICAL SERIES. General Editors : P. Abbott, B.A., and F. S. Macaulay, M.A.,D.Sc. Crown 8vo. A SCHOOL COURSE IN GEOMETRY (including the Elements of Trigonometry and Mensuration and an Introduction to the Methods of Co-ordinate Geometry). By W. J. Dobbs, M.A., sometime Foundation Scholar of St. John's College, Cambridge. With 361 Diagrams. 3s. 6d. With Answers, 4s. ; Answers separately, 6d. THE TEACHING OF ALGEBRA (including Trigonometry). ByT. Percy Nunn, M.A., D.Sc, Professor of Education in the University of London. With Diagrams. 7s. 6d. EXERCISES IN ALGEBRA (including Trigonometry). By T. Percy Nunn, M.A., D.Sc. Part I., 3s. 6d. With Answers, 4s. Part I., Section I., is. 6d. ; Sections II. and III. 2s. 6d. Part II., 6s. With Answers, 6s. 6d. EXERCISES IN ARITHMETIC AND MENSURATION. By P. Abbott, B.A., Head of the Mathematical Department, The Poly- technic, Regent Street, London, W. With Diagrams. 3s. 6d. With Answers, 4s. 6d. In 2 Sections, 2s. each. Answers separately, is. THE GROUNDWORK OF ARITHMETIC. By Margaret Punnett, B.A., Vice-Principal, London Day Training College, Southampton Row, London, W.C. With Coloured and other Diagrams. 3s. 6d. *** A Handbook for Teachers. EXERCISES IN ARITHMETIC. By Margaret Punnett, B.A. Book I., 4d. Book II., 6d. Book III., 6d. PROJECTIVE GEOMETRY. By G. B. Mathews, M.A., F.R.S., Lecturer in Pure Mathematics, University College of North Wales, Bangor. With Diagrams. 5s. SLIDE-RULE NOTES. By Colonel H. C. Dunlop, late Professor of Gunnery, Ordnance College, Woolwich, and C. S. Jackson, M.A. With Diagrams. 2s. 6d. net. RULER AND COMPASSES. By Hilda P. Hudson, M.A., Sc.D. With Diagrams. 6s. net. THE ELEMENTS OF NON-EUCLIDEAN PLANE GEOMETRY AND TRIGONOMETRY. By H. S. Carslaw, Sc.D. (Camb.), D.Sc. (Glasg.), Professor of Mathematics in the University of Sydney; formerly Fellow of Emmanuel College, Cambridge. With Diagrams. 5s. net. INFINITESIMAL CALCULUS. By F. S. Carey, M.A., Professor of Mathematics in the University of Liverpool. In two Sections. 8vo. Section I. 6s. net. Section II. 10s. 6d. net. DIFFERENTIAL EQUATIONS. By H. Bateman, Ph.D., M.A., Lecturer in Applied Mathematics, Johns Hopkins University! Baltimore, formerly Fellow of Trinity College, Cambridge, and Reader in Mathematical Physics in the University of Manchester 8vo. ELEMENTARY NUMERICAL TRIGONOMETRY. By P. Abbott, B.A., Head of the Mathematical Department, The Polytechnic, Regent Street, London, W. Crown 8vo. LONGMANS, GREEN, & CO., LONDON, NEW YORK, BOMBAY, CALCUTTA, AND MADRAS. ^Longmans' /iBo&ern /IftatbemattcaL Series DIFFERENTIAL EQUATIONS BY H. BATEMAN, Ph.D., M.A. LECTURER IN APPLIED MATHEMATICS, JOHNS HOPKINS UNIVERSITY, BALTIMORE FORMERLY FELLOW OF TRINITY COLLEGE, CAMBRIDGE, AND READER IN MATHEMATICAL PHYSICS IN THE UNIVERSITY OF MANCHESTER LONGMANS, GREEN AND CO. 39 PATERNOSTER ROW, LONDON FOURTH AVENUE & 30TH STREET, NEW YORK BOMBAY, CALCUTTA, AND MADRAS %* . m/n PREFACE The subject of Differential Equations has grown so rapidly in recent years that it is difficult to do justice to all branches of the subject in a single volume. In writing this book I have endeavoured to supply some elementary material suitable for the needs of students who are studying the subject for the first time, and also some more advanced work which may be useful to men who are interested more in physical mathematics than in the developments of differential geometry and the theory of functions. The chapters on partial differential equations have consequently been devoted almost entirely, to the discussion of linear equations. The order in which the material has been arranged is slightly different from that which is usually adopted. Instead of beginning with the standard forms of equations which can be solved very easily, I have devoted the second chapter to the method of inte- grating factors and the third to the method of transformation. A reader who is not already familiar with the subject will find it worth while to postpone the reading of §§ 13-17 and §§ 32-35 until he has mastered the more elementary work. Chapters VII and VIII contain a few results which I think are new. In Chapter VI the treatment of special solutions bears some resemblance to a treatment which has just recently been sketched by Prof. Hill in an abstract of a memoir which will soon be published in the Proceedings of the London Mathematical Society. My work is independent of Prof. Hill's, for the manuscript was sent to the publishers at the beginning of 1915. Example 3, p. 143, was, however, added last September when marking the proof sheets, but these had been returned to the printers before I received my copy of Prof. Hill's abstract. Prof. Hill has evidently discussed the matter much more thoroughly than I have done, and I hope that some of my readers will consult his memoir. vi PREFACE I have naturally to acknowledge my indebtedness to previous books on differential equations, particularly to the works of Forsyth, Boole, Cohen, Wolsey Johnson and others too numerous to mention. I have also found Whittaker's Analysis, Carslaw's Fourier Series and Integrals and Bromwich's Infinite Series of great assistance in the preparation of Chapter IX. Some articles have, been partly borrowed from Darboux' TMorie of Surfaces, and the treatises on analysis by the great French writers Jordan, Picard, Goursat and others have been invaluable to me. I also wish to express my gratitude to my teachers at Cambridge, among whom I may mention Dr. Glaisher, who first roused my interest in the subject, Dr. Forsyth and Prof. Whittaker, who developed it by their lectures and writings. Some of their memoirs have been a source of inspiration to me in my research work. I have great pleasure in thanking the officers and staff of the Glasgow University Press for the admirable work they have done in connection with the printing, and I am very grateful to the Editors of this Series of Mathematical Books for many useful suggestions. HARRY BATEMAN. November, 1916. CONTENTS CHAPTER I DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS SECTION PAGE 1. Introduction - - 1 2. The nature of the solution of an ordinary differential equation 2 3. Supplementary conditions - 4 4. Equations in which the variables are separated 10 5. Continuous and discontinuous solutions 13 6. A type of Green's function 15 CHAPTER II INTEGRATING FACTORS 7. Linear equations of the first order 17 8. Equations reducible to linear equations 21 9. Integrating factors of two linear equations 22 10. Systems of linear equations with constant coefficients 24 11. Linear equations of the second order with constant co- efficients - 28 12. Symbolical methods 33 13. Discontinuous solutions of linear equations with constant coefficients 38 14. A system of linear equations occurring in the theory of radio-active transformations 41 15. A system of linear equations occurring in the theory of probability 45 16. General theory of a system of linear equations - 46 17. Equations of the first order and of the first degree - 48 18. Integrating factors of linear equations of the second order 53 19. Equations of various types - - 56 PAGE viii CONTENTS CHAPTER III TRANSFORMATIONS SECTION 20. Preliminary remarks - - 60 21. Homogeneous equations of the first type - - - 60 22. Equations which can be reduced to homogeneous equations 63 23. Clairaut's equation 63 24. Equations of the first order which can be solved for either x or y - 64 25. Equations of the second order which do not contain x explicitly - 65 26. Equations of the second order which do not involve y explicitly - 66 27. Equations of the second order which may be solved by differentiation 66 28. Equations of Lagrange's type 67 29. Riccati's equation 68 30. The linear differential equation of the second order - 71. 31. Equations reducible to linear equations with constant coefficients 73 32. Reduction of an equation of Laplace's type to a canonical form 75 33. Homogeneous equations of the second type 80 34. The principle of duality and the theory of contact trans- formations 81 35. Jacobi's equation 83 CHAPTER IV GEOMETRICAL APPLICATIONS 36. The differential equation of a family of algebraic curves 87 37. Singular solutions of differential equations - 88 38. The c-discriminant 92 39. The p-discriminant - 94 40. Differential equations which can be resolved into factors 97 41. Orthogonal trajectories 9g 42. Orthogonal trajectories of a family of circles 100 CHAPTER V DIFFERENTIAL EQUATIONS WITH PARTICULAR SOLUTIONS OF A SPECIFIED TYPE 43. Euler's equation ^03 44. Linear differential equations with simple solutions 110 CONTENTS ix CHAPTER VI PARTIAL DIFFERENTIAL EQUATIONS SECT,0N 45. Functional determinants and their properties - H6 46. Solution of the equation J = 1 119 47. The derivation of a partial differential equation from a given primitive 121 48. Geometrical representation of the different types of primi- tives ' 125 49. The characteristics of a partial differential equation of the first order . . " 131 50. Lagrange's linear equation - 134 51. Special solutions of Lagrange's equation 140 52. The homogeneous linear equation 143 53. Simple types of soluble partial differential equations. Standard forms 144 54. The principle of duality 148 55. Equations with three independent variables 150 CHAPTER VII TOTAL DIFFERENTIAL EQUATIONS 56. The condition of integrability 153 57. Transformation of variables 154 58. Invariance of the condition of integrability 161 59. First method of solving the equation A = 161 60. Canonical form for A, second method of solution 164 61. Homogeneous equations - 167 CHAPTER VIII PARTIAL DIFFERENTIAL EQUATIONS OF THE SECOND ORDER 62. The homogeneous linear partial differential equation of the second order 1 69 63. Laplace's method 173 64. D'Alembert's equation 175 65. The equation of wave-propagation in three dimensions. Maxwell's equations - - .... 179 x CONTENTS SECTION PAGE 65a. The fundamental equations of the theory of -electrons 186 66. Laplace's equation 193 67. The equation of the conduction of heat 211 68. Harmonic equations 216 69. The condition that a given equation can be reduced to the harmonic form - - 217 CHAPTER IX INTEGRATION IN SERIES 70. Cauchy's existence theorem for an equation of the first order 223 71. Runge's method of approximate solution 227 72. The representation of solutions of linear differential equa- tions by means of power series 230 73. The hypergeometric equation 232 74. A limiting case-of the hypergeometric equation 236 75. Bessel's equation 237 76. Legendre's equation 240 77. Two convergence theorems 242 78. The method of successive approximations 245 79. The integration of linear partial differential equations by means of infinite series ■ 253 CHAPTER X THE SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS BY MEANS OF DEFINITE INTEGRALS 80. Laplace's transformation 260 81. Equations of Pfaff's type 263 82. Euler's transformation 265 83. The solution of partial differential equations by means of definite integrals 268 84. The Green's function of a linear differential equation with assigned boundary conditions - 275 85. Riemann's method 280 CONTENTS xi CHAPTER XI THE MECHANICAL INTEGRATION OF DIFFERENTIAL EQUATIONS SECTION PAGE 86. The reduction of a differential equation to a convenient form - 287 87. Mechanisms for the solution of equations of Pascal's type - 288 88. Mechanisms for integrating the product of two functions - 290 89. Pascal's mechanism for the solution of a linear equation of the first order - 293 90. The solution of a linear differential equation of the second order - 294 Miscellaneous Examples ... - - 296 Index - - 304 CHAPTER I DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS §1. Introduction. When a mathematician attempts to describe natural phenomena with the aid of symbols, he frequently finds that he is unable to express the relation between two variable quantities x, t directly in the form ¥(x, t) = 0. The invention of the differential calculus has, however, led to a method by which the difficulty may sometimes be overcome. It is often easy to write down a relation between x, t and some of the derivatives of x with regard to t. Such a relation is a particular type of differential equation, and must be regarded as an auxiliary equation which expresses some but not all of the properties of the desired relation. In general, a differential equation must be understood to mean a relation involving a derivative or derivatives of a function ; it may involve the derivatives of other functions, and it may involve partial derivatives if there are several independent variables. If, however, the only derivatives which occur are partial derivatives, it is customary to call the equation a partial differential equation. A differential equation which does not involve partial derivatives is frequently called an ordinary differential equation. Thus F / dx d 2 x\ is an ordinary differential equation, and »(■"■■■»£) -° < 2 » a partial differential equation. In the first case t is regarded as the independent variable and x as the dependent variable ; in the second case, z is the dependent variable, x and y the independent variables. By making a slight change in the form in which a differential equation is presented, it is possible to make any one of the variables appear as the dependent variable ; it is generally convenient to choose the dependent variable in such a way that the equation takes the simplest possible form. Thus the equation 2 DIFFERENTIAL EQUATIONS may also be written in the form dj 2 + \dy) f® " ° but the first form is simpler and easier to deal with. § 2. The nature of the solution of an ordinary differential equation. A relation between the dependent and independent variables is said to be a solution or integral of a differential equation if, when the dependent variable and its various derivatives are calculated from the relation and substituted in the differential equation, the latter becomes an identity ; in other words, it is satisfied for all values of the independent variable. Thus x = sin t is a solution of the differential equation d 2 x dfi +X = ° d 2 x because when we calculate -^ from the given relation we find that d 2 x — = -smt = -x and with this value of -^ the differential equation is satisfied '■ identically. M A differential equation of type _/ dx d 2 x d n x\ n ,„, i in which the highest derivative which occurs is of order n, is called ' a differential equation of the n th order. If the equation can be thrown, . into such a form that P is a polynomial of the m th degree in the d n x quantity — . it is called a differential equation of the m th degree. Thus (^Y^ + ^0 \dfiJ dt dt 2 is a differential equation of the second order and of the third degree. A differential equation generally possesses an infinite number of solutions, as might be inferred from an examination of a simple equation, & which is satisfied by a relation of type x =\f(t)dt +c where c is an arbitrary constant. An equation will be regarded as solved when an expression is obtained for the most general function which satisfies it, even if this expression involves integrals which AND THEIR SOLUTIONS 3 cannot be evaluated by the ordinary methods of the integral calculus. In solving a differential equation of type (3), the chief aim is to obtain a class of relations of type f(x, t) =. which is sufficiently general to include all possible solutions of the equation. It has been found by experience that if a solution involving n arbitrary constants can be found, it is generally possible to derive any particular solution from it either by making a special choice of the arbitrary constant or by some other artifice. Such a solution is called a general solution or primitive of the differential equation. A solution involving n arbitrary constants may fail to be general if it can be expressed in terms of a fewer number of arbitrary con- stants ; for instance, the primitive x 2 + y 2 = a 2 + 6 2 appears to involve two arbitrary constants a and b, but if we write a a + b z = c 2 , an expression is obtained in which only one arbitrary constant, c, appears. When a primitive involving n arbitrary constants is given, a differential equation of the n th order can generally be derived from it by differentiating the given relation n times with regard to the , independent variable x and then eliminating the n arbitrary con- stants from the n equations so obtained, making use of the original equation. No general method of performing this elimination can be given at this stage, but in a large number of cases a simple method of performing the elimination readily suggests itself. Example 1. Find the differential equation corresponding to the primitive x + a where a and b are arbitrary constants. First solve for a, and differentiate so as to eliminate a ; we get yx + by - x = a ... s^ + y+6^-1-0 dx a ax Solving for 6 and writing j- = p, we get i y z. 2. _ x = b P P p z dx p 2 dx 4 DIFFERENTIAL EQUATIONS The differential equation is thus Example 2. Find the differential equation corresponding to the primitive C 2 X ~ c (x 2 + xy + y) +y{x + y) = In this case the simplest plan is to differentiate the equation as it stands. Writing -p = p, we get c 2 -c(2x + xp + y +p) + (xp + y + 2yp) = Eliminating c 2 , we find that c(x 2 + x 2 p +px - y) = x*p + 2xyp - y 2 Also c 2 {x 2 p + 2xyp - y 2 ) = c [p(x 3 + 2x 2 y + xy 2 + y 2 ) + y 2 - 2xy 2 - x 2 y - y 3 ] :. (x*p + 2xyp - y 2 ) 2 = \$x(x + 1) + x 2 - yf x [p (x 3 + 2x 2 y + xy 2 + y 2 ) + y 2 - 2xy 2 - x 2 y - «/ 3 ] or {p + l)ipx- y)(2x 2 y + 2xy 2 - z 4 - 2x*y - x 2 y 2 - y 2 ) = The differential equation is thus £+0(-£-»)-<> A solution involving the proper number of arbitrary constants may fail to give all possible solutions if the differential equation can be resolved into factors. For instance, the equation fdy ^\(dy dx *~J\dx ) is satisfied by y = x + c, and is of the first order. Consequently this solution involves the proper number of arbitrary constants, but there are solutions of type y = 2x + a which are not, included in the general relation y = x + c and cannot be derived from it. Some simple examples on the formation of differential equations are given on p. 16. § 3. Supplementary conditions. We have already seen that a differential equation does not specify completely the relation between two variables x and t. To complete the specification it is necessary to add certain supplementary conditions which are gene- 1 rally indicated by the circumstances in the problem to be solved.:? Sometimes the value of x is known for a particular value of t, and' this suffices to determine the unknown constant. Sometimes the fact that x is real when t is real becomes of importance, while the condition that x is finite for all real positive values of t frequently AND THEIR SOLUTIONS 5 occurs and leads to the determination of a constant. It may happen that there is more than one way of choosing the arbitrary constants in the solution so as to make x a real function of t, and in this case the real relations to which the differential equation is equivalent form a number of classes instead of only one. Again, it may happen that there is only one real solution of a differential equation or none at all. The following examples will illustrate the different points : 1°. A general solution of the equation dv -~ = cos t at is given by y = sin t + c If, however, a supplementary condition states that y = when ( = 0, the constant c is completely determined, and we have y = sin t. /dx \ 2 2°. The equation {-^ - 1 J + (x - t) 2 = possesses only one real solution, namely x = t. /dx\ 2 3°. The equation {^j + {x - t) z = does not possess a real solution. 4°. The equation t -^ + x 2 = possesses the general solution x = , , but the condition x = when t = is not sufficient to determine the unknown constant c, for the condition is always satisfied. 5°. If a is a positive constant, the equation ~ = a 2 - x 2 dt possesses two distinct types of real solutions. A solution is given by f dx i dx , t = I — 1 + const., and the integral on the right-hand side has different real forms according as x 2 is greater or less than a 2 . We may write t - t = ^log^^ - a a 2 . If x = a when t = 0, we must have t = - co , and both equations give x = a. If we use the symbol \x - a\ to denote the modulus or absolute value of a; - a, the two real solutions may be embodied in the single formula 1 , Is + al '-'•-S l0g JF^| In formulating a differential equation corresponding to a given problem, the dependent and independent variables should be chosen so that the differential equation is as simple as possible ; care should also be taken to include all the conditions and restrictions peculiar to the problem in the supplementary conditions ; note should be made, for instance, of the range of values of the independent variable (or some other quantity) for which the problem is a possible one. Some examples will now be given for the sake of illustration. EXAMPLES. 1. A smooth board with a rough curved edge is placed on a smooth table with its curved edge resting' against a straight ledge of length 2a. A light string, carrying a weight W at each end, passes round a nail fixed in the board at C and over the ends of the ledge, the weights being suspended in mid air. Find the form of the curved edge so that the board may be in neutral equilibrium. AND THEIR SOLUTIONS 1 Let A, B be the ends of the ledge, P the point of contact and M the middle point of AB. For equilibrium, the resultant of the two equal tensions at C must pass through P and make with the normal at P to AB an angle which is not greater than the angle of friction. The A A resultant bisects the angle at C, and so we must have ACP = PCB. Let MP = s ; then the equilibrium will be neutral if the board is still in equilibrium after it is rolled a small distance ds along AB. The con- dition for rolling is that s should be always equal to the length of the arc of the edge measured from some fixed point V which comes in contact with M. Let us take s as the independent variable, and r = CP as the dependent variable. If the angle CPB be denoted by b, vc, ... denote the masses of the original substances at the beginning of the reaction and Ix, y.x, vx the amounts of these substances that have been transformed at time t, the law of mass action states that if the velocity of reaction in the reverse direction may be neglected, -j-(Xa;) = KX(a - x)\x{b - x)v(c - x) ... where K is a constant. This equation may be written in the form -j- = k(a - x)(b - x)(c - x) ... where k is a constant. EXAMPLES t 1. The equation of a unimolecular chemical reaction is dx , . where a is the initial amount of the original substance, x is the amount * For further information the reader is referred to J. W. Mellor's Chemical Statics and Dynamics. t Most of these have been taken from Lamb's Infinitesimal Calctdus. 12 DIFFERENTIAL EQUATIONS transformed at the end of time t and k is a positive constant. Obtain an expression for x, using the supplementary condition that x = when t = 0. 2. Discuss the equations -£ = k(a - x){b - x) dx -^- = k(a - xY dt v ' regarding them as equations of bimolecular chemical reactions. 3. A particle falls through a resisting medium in which the resistance is proportional to the square of the velocity. Show that the equation of motion is g„ where k is a constant and g is the acceleration due to gravity, and is also supposed to be constant. Integrate this equation, and show that as t increases without limit the velocity v tends to the value VI' 4. The distance x from the axis of a thick cylindrical tube of metal is related to the internal pressure p as indicated in the equation dp a -2p dx x where a is a constant. Hence show that p = Ja + Car" 2 , where C is a constant to be determined by the supplementary condition. 5. The equation of rectilinear motion of a particle under an attractive force varying inversely as the square of the distance from a fixed point is dv _ (X dx x 2 where v is the velocity of the particle when it is at a distance x from the fixed point and [j. is a positive constant. Prove that if the particle starts from rest at a distance 2o from the fixed point, the motion is represented by the equations a:=2ocos 2 t = a*|T*(2e + sin 26) 6. If in the last example the particle has an initial velocity u when at a distance c from the fixed point, the equations defining the motion have different forms according as u % is greater, equal to, or less than 2(x/c. 7. The form of a heavy chain of uniform density hanging from two fixed points is determined by the condition that the centre of gravity of any portion is vertically above the point of intersection of the tangents at the two ends. If s denote the length of an arc of the chain measured from a point at which the tangent is horizontal, the form of the chain is determined by the differential equation i^i-s^S AND THEIR SOLUTIONS 13 Hence show that the form of the chain is given by the equations s = c sinh - y = c cosh - + a c c where c and a are constants. 8. In a suspension bridge with uniform horizontal load the form of the chain is determined by the condition that any two tangents to the curve intersect on the vertical line bisecting the chord of contact. Prove that the curve formed by the chain must be a parabola with its axis vertical. 9. Find a curve in which the tangent and normal at each point form a triangle of constant area with a fixed line. 10. Find a curve in which the projection of the normal on the radius- vector from the origin is of constant length. (Hint. Obtain a differential equation in which the coordinate x and the angle made by the radius vector with the axis of x are variables. The normal is supposed to be terminated by the axis of x and the curve.) 11. A rotating fly-wheel is being stopped by a fluid resistance which decreases the angular momentum at a rate Fco, where t y 3. A ray of light passes through a medium in which v, the velocity of light, depends only on the distance y from a fixed plane. If ^ is the angle which the tangent to the ray makes with the fixed plane, the law of refraction is v sec ijj = const. = k, say. Prove that if v increases with y and surpasses the value k, the differential equation for the path of the ray should be written in the form dy dx = + Vk 2 - v 2 for -M has different signs on the two sides of the vertex of the ray. dx 4. Find the differential equations * corresponding to the following primitives, a, b and c being arbitrary constants : (x - c)(y - c) = 1 (x - a) 2 + (y - 6) 2 = 1 y 2 = iax + b y = (c - x) 3 y = a sin (bx + c) x 3 + y 3 + Zaxy = a 3 y= ex - c 3 (y - b) 2 = kax 5. The primitive y - ae bx + c leads to a differential equation of the second order, while the primitive y = a bx leads to an equation of the first order. Explain why the order is less than the apparent number of arbitrary constants in these cases. * Of course innumerable differential equations can be obtained from a given primitive by eliminating the constants, but in this example the equation of lowest order is required in each case. CHAPTER II INTEGRATING FACTORS § 7. Linear Equations of the First Order. Let us consider the differential equation a„ l + p ^ = Q w in which P and Q are any integrable functions of x. This equation is said to be linear, because each term is at most of the first degree in the function y and its derivative ; it is said to be of the first order, because the first derivative of y is the highest derivative occurring in the equation. To solve the equation we shall endeavour to find an integrating factor v. This is defined to be a function of x such that •£ + *)-£<■»> « If a suitable function v(x) can be found, the differential equation may be written in the form ^vy) = Qv (3) any may be integrated at once, giving vy = \Qvdx + C (4) where C is an arbitrary constant. To find a suitable value of v, we equate the coefficients of y in (2), and obtain a v ¥- = P*> ax Writing this equation in the form ldv = p vdx and integrating, we obtain logw = YPdx' + A 18 DIFFERENTIAL EQUATIONS where A is a constant. Since a particular integrating factor will serve our purpose, we may put A = 0, and write v = exp (M The general value of v is obtained by multiplying this value by an arbitrary constant ; if we use this general value in (4), we simply obtain the same result as if we used the particular value. Substituting the particular value of v in (4), we obtain y = C exp ( - \ P dx) + exp ( - JP dx) J Q exp ( JP dx) dx This is the general solution of the equation. The constant C can usually be determined with the aid of a supplementary condition. If, for instance, we are given that y = b when x = a and v(x) is not infinite or indeterminate when x = a, the value of C may be deter- mined uniquely by putting x = a in equation (4). The complete solution may be written in the form v(x)y - v(a)b = P Q{t)v(t) dt j a If a maximum or minimum value of y is known to be a given real quantity h, there may be several possible values for the constant C. ■ dy In fact, if we put y = h, -p = in equation (1), we obtain an equation P{x)h = Q(ar) which may determine several real values of x or none at all. If x = a is one of the values of x for which the equation is satisfied, an appropriate constant C may be obtained by putting x = a, y = h in equation (4) . If we are given the mean value of y for a certain range of values of x, the constant C can be determined uniquely, provided the mean value of - is not zero. In the alternative case the supplementary condition is incompatible with the differential equation, unless the assigned mean value M is also the mean value of the quantity \\<* dx When this condition is satisfied, the constant C may be chosen arbitrarily, and so every solution of the differential equation satisfies the supplementary condition. Thus, when the mean value of - is v zero, the problem of finding an appropriate solution is poristic ; in other words, there is either no solution at all or an infinite number of solutions. INTEGRATING FACTORS 19 If we are given the mean value ofy 2 over a certain range, it follows from equation (4) that there are generally two values of C which may or may not be real. Hence, in this case, the differential equation possesses an appropriate real solution only when a certain inequality is satisfied. It should be noticed that the difference equation y n+l + p n y n = q n , in which n is a positive integer and p n , q n given functions of n, may be solved in a similar way. An integrating factor u n must now be chosen so that the equation may be written in the form %2/n+i - M n-i2/ n = u rtln- This requires that u n _ t = - u^ n . Solving this equation, we get u n = ( - 1)" 5 p where u is an arbitrary P1P2 ••■Pn constant which may be taken to be unity. The solution of the original equation in its new form is obtained at once by addition of the equations in which n has successive integral values, and we obtain u n y n+1 - u a y 1 = u^ + u 2 q 2 + ... u n q n where y 1 is an arbitrary constant. EXAMPLES. 1. If * is the strength of the electric current flowing along a uniform wire at time t, L the coefficient of self-induction of the wire, R the resistance and E the electromotive force generating the current, we have di L - + Ri = E at (The first term is analogous to mass x acceleration and the second to a frictional force proportional to the velocity.) We shall integrate this equation on the supposition that E = E sin at and that E , co, R,'L are constants. dv The equation determining an integrating factor v is L — = ~Rv, hence we may take iu v= e L Writing the equation in the form — (Lie L ) = E e L sin at and integrating, we get _ . y T _ /R sin at - u>L cos u>t\ y- L*e Ij = LE„l — ' _ ■ — - e 1. '"' R 2 + co^ 2 where C is an arbitrary constant. To simplify this result, we write 01L = S sin 9, R = S cos 9, AL = CS, where A is a new arbitrary con- stant. Hence, finally, m Si = E sin ( E 2 . Show also that when t is very large, 2S 2 a 2 is approxi- mately equal to E 2 , and is thus practically independent of the initial conditions. 4. Prove that equation (1) may be solved by writing y = uw, and determining w so that the coefficient of u vanishes. 5. Prove that the values of x for which a solution of equation (1) can become infinite or indeterminate are independent of the supple- mentary conditions. 6. Prove that equation (1) can only possess one solution which is a continuous function of x and has a given value when x — 0, except in the case when v is infinite or indeterminate for x = 0. 7. In the differential equation the constant X is unknown. Prove that it may be determined uniquely if it is known that at time t, x has just half its initial value when t = 0. 8. Show that equation (1) is transformed into another equation of the same type by the substitutions x=¥(i) y = t)G(5) + H(|) where F, G, H are arbitrary functions, and i), \ are regarded as the new dependent and independent variables respectively. 9. Solve the equations x — - y = x n dx dy dx~ Xy=X INTEGRATING FACTORS 21 10. A radioactive substance Aj is changing into another radioactive substance A 2 , while A 2 is changing into a stable substance A 3 . If P lf P 2 , P a are the amounts of the substances present at time t, the following differential equations are satisfied : dPl - XP dP °-xp -XP dp a m - x ^ 2 where X^ X 2 , X 3 are constants peculiar to the substances A t , A 2 , A s respectively. If when t = 0, P x = N, P 2 = 0, P 3 = 0, the amounts of the different substances present at time t are given by the equations P t = Ne'- A i* P 2 = mi [e-Xxt _ e -A 2 (] X 2 - Xj P 3 =-^-[-X 2 e-^+X l6 -M + X 2 - \] A 2 - /.j § 8. Equations reducible to Linear Equations. A given differential equation may sometimes be transformed into a linear equation by a change of variables. For instance, in the case of the equation * where P and Q are functions of x only, a suitable change of variables is „ . dy dv The transformed equation is ^ + (1 _»)Pw = (l-n)Q dx and the method of § 7 may be used to obtain the solution. The equation £ + Pe" = Q may be reduced to the linear form by putting z = e-» The equation for z is then — - + Qz = P ax * This is sometimes called Bernoulli's equation. 22 DIFFERENTIAL EQUATIONS EXAMPLES. 1. A particle describes a curve in a vertical plane under the action of gravity, and meets a resistance which acts in a direction opposite to that in which the particle is moving at each instant, and varies as some function f(v) of the particle's velocity v at this instant. Prove that the path of the particle is determined by the equations ldv , , f(v) - 3-j + Cot lb + J \ ' = v dtp g sm ijj gt=* Jd * f VI Jsm sin ib where g is the acceleration due to gravity and i/ is the angle which the tangent to the path makes with the vertical. Integrate the first equation in the case when f(v) = kv n , k being a positive constant. dy 2. Solve the equations -^ + xy = y % dy ~ + y tan x = y 2 dx " a § 9. Integrating factors of two linear equations. Let us consider the two linear equations dx dt -ax- by dy Tt -bx + ay (1) where a and b are functions of t. Multiplying the first equation by 2x, the second by 2y and adding, we get j t (x* +2/ 2) =2a(x* +«/ 2 ) . . . . (2) Hence x 2 + y % = A 2 exp M 2a dt J where A is an arbitrary constant. Again, if we multiply the second equation by x, the first equation by y and subtract, we get 4 - '% - '<*■ + »•> Hence tan- 1 ^ = j bdt + e (3) INTEGRATING FACTORS 23 where s is an arbitrary constant. Solving equations (2) and"(3) for x and y, we find that x = AeJ" cos H6 eft + e) - y = AeJ sin (\bdt + e) (4) If a = and 6 is a constant, the solution is simply x = A cos (bt + e) y = A sin (bt + e) . . (5) Let us now consider the equations !+«*+%=/] (6) d f t -bx + ay=g\ where a, b, f, g are functions of t. To solve these equations, we shall endeavour to find two functions x , y , such that the equations j t {xx + yy ) = fx + gyA j t {xy - yx ) = fy ~ 9 x o ) are a direct consequence of (6). Substituting the values of -=-, -j- given by (6) in (7), we find that the last pair of equations are satisfied identically if ^ x -^ = ax - by dy n , -j£ = bx + ay Solving these equations by the method already explained and choosing special values of the arbitrary constants, we may write x = e) a 'cos (ficftj #o = e sin (& eft Substituting in equations (7) and integrating, we obtain the values of x and y. If we use the notation -H J cos(j*& + Wo)# + A] + y* [|(/y - gx )dt + B V = 2/o* [ {(/*« + WoW + A ] - V [fCfro - !7*o)# + B where A and B are arbitrary constants. x * = e x 24 DIFFERENTIAL EQUATIONS EXAMPLES. 1. A charged particle starts from rest in the plane x = at time 1=0, and is acted upon by a uniform electric field of strength X, parallel to the axis of x, and by a uniform magnetic force H, parallel to the axis of z. Assuming that the equations of motion are d 2 x dy m -r-r = eX - He -f dt 2 dt d 2 V rj dx to ~ = eH — dt' dt where the mass m and the electric charge e are constants, prove that the position of the particle at time t is given by the equations m~X. ( /e TT to X fe TT . (e, TT \) y=- m \~Ht- sm -Hf e H z [to \to J) 2. Solve the equations given in the last example for the case in which X = a cos toi H = b cos est and a, b, to are constants. 3. A rigid lamina moves in its own plane so that its angular velocity co and the coordinates (5, tj) of the centre of instantaneous rotation are given functions of t. Write down and solve the differential equations satisfied by the coordinates of an arbitrary point of the lamina. 4. Solve equations (1) and (6) by writing x + iy = z, where i = y/ - 1. 5. Solve the difference equations x n+i = a n x„ - b^„ y n+1 = b n x n + a^y n where a„, b n are given functions of n. § 10. Systems of linear equations with constant coefficients. Let us consider the equations dx Tt =ax + by dy , Jt =lx + my (1) where a, b, I, m are constants. If we multiply the second equation by an arbitrary constant X and add to the first, we obtain j t {x + \y) = (a + tyx + (b + -km)y .... (2) Now choose X so that b + Xm = X(a + XZ) / 3 \ INTEGRATING FACTORS 25 and write z = x + \y. We then have the following equation for z: j t = {a + Vk)z (4) This gives us z =-Ce (a+U)i where C is an arbitrary constant. There are generally two distinct values of X for which equation (3) is satisfied, let us call these X x and X 2 ; then we have x +\ 1 y = C 1 e( a +' x i)' x + \y = G 2 e<- a + l ^ t where G x and C 2 are arbitrary constants. If x = x , y = y when t = 0, we have C^ = x + X^,,, C 2 = x + X 2 i/ . The values of x and y are now completely determined, and we have x(\ - XJ = X 2 (z + X l2/o )e(«+^ - X^^o + \, yo )e<»+^ Sfta - ^) = (*o + X l2 / )e("+^)' - (x + X 2 y )e<«+^< J ( j If the equation for X has two equal roots, we use the equation x +\y = Ge ( - a+lK > t to simplify the second of equations (1). We then find that y satisfies the equation du "f=(m- Vk)y + ZCe<«+^> ( Now, since equation (3) has two equal roots, its derivative with regard to X must vanish ; hence m = a + 2tk The preceding equation may consequently be written in the form ^ = {a + Vk)y + ICe^+W or d L\ye-^+ l ^ t ] = IQ at Hence ye -W)t = ICt + A where A is an arbitrary constant. If x = x , y = y when t = 0, we have C = x + ly A = y Hence, finally, x = [*„ - Xl(x + \y Q )t\ e<»+») If the roots of equation (3) are not real, that is, if (m - a) 2 + 4bl < we may proceed as follows. Write y = xv x + £y = u where x and £ are constants. (6) 26 DIFFERENTIAL EQUATIONS The differential equations satisfied by u and v are then ^ = (a + ll)u + x(b +lm -la - l 2 V)v az dv I . 7} -, — = -u + (m - lt,)v dt x We now choose I and x, so that a + y, = m - 11 x(6 + lm - la - IH) = - l - the differential equations for u and v are then of the type considered in § 9. Solving for I and x, we get I 2 211 = m - a 4 -g = - 46Z - (m- a) 2 = + 4o> 2 , say. Writing a + m = 2cr, our differential equations take the forms du — - = GU - COW at dv -j? = cow + av dt and the solution is consequently u = Ae°" ( cosco(i - t ) v = Ae^'sinco^ - t ) where A and t are arbitrary constants. The corresponding expressions for x and y are now found to be y = Axe^'sinco^ - t ) \ Ae* , . in - a . . . cosco(£ - t ) s smu(( - t ) (?) If we wish to solve the equations dx di dy -j t = ax + by + f(t) dt =lx + my + g(t) we may adopt a similar method. If the roots of equation (3) are real and distinct, we write z 1 = x + X-^y, z 2 = x + \$, and obtain dz -^ = (a + X x Q Zl + f(t) + X lS f(«) dz -^ = (a + A 2 Z)z 2 + f(t) + \£{t) These equations can be solved by the method of § 7, and the INTEGRATING FACTORS 27 values of x and y derived from the expressions obtained for z 1 and z 2 . When the roots of equation (3) are not real, we write x + t,y - u y = v.v as before, and obtain du -^ = au - cow + f(t) + lg(t) dv 1 /% j t = com + ot + -C/(<) These equations can be solved by the method of § 9. When equation (3) has two equal roots, we write z = x + Xy, and obtain ^ Tt = (a + Xl)z + f(t) + \g(t) -£ = {a + \l)y +lz + g(t) The first equation can be solved by the method of § 7, and when the value of z so found is substituted in the second equation, this new equation can be solved by the same method. EXAMPLES. 1. If in equations (1) x = for t = and y = for t = t, the func- tions x and y are always zero unless X^i 1 " = X 2 6 JA 2 T When this condition is satisfied, the value of y can be chosen arbitrarily. 2. The magnitudes x and y of the electric currents flowing in two circuits acted on by electromotive forces f(t), g(l) respectively, satisfy the differential equations where L, N are the coefficients of self-induction, R, S the resistances and M the mutual inductance of the two circuits. Integrate the equations on the supposition that L, M, N, R, S are positive constants. 3. Solve the difference equations x n+l = ax n + by n +/„ 2/«+i = lx n + m y n + ffn where a, b, I, m are constants and /„, g n given functions of the positive integer n. 28 DIFFERENTIAL EQUATIONS § 11. Linear equations of the second order with constant coefficients. Let us consider the differential equation d 2 x dx , W =m lTt +lx in which I and m are constants. Putting- -5- = y, we obtain the two linear equations dx di = y dy , ~dt = lx + my Treating these by the methods of § 10, we see that if the equation 17? = Xm + 1 has two distinct roots X 1; X 2 , and we write a x = Vk lt oc 2 = tk 2 > the solution is given by the formula lx(x 2 - aj) = a. 2 {lx + a 1 2/ )e°i i - a 2 (Za; + a 2 y )e^' This is the solution corresponding to the supplementary con- ditions dx "o -j t = «/o when * = ° It should be noticed that the general solution is of the form x = A 1 e a i* + A 2 e°^ where A lf A 2 are arbitrary constants and a. v a 2 are the roots of the equation a 2 = ma + Z When this equation has equal roots, so that oq = a 2 = a, we see from equation (6) of § 10 that lx = [lx - a(lx + v.y )t] e°- 1 When the roots of the equation for a are not real, it follows from (7) that the solution is of the form x = e' 7 * [B cos att + C sin coi] where B and C are arbitrary constants, since A and t were arbitrary. The constants a and w are given by the equations 2g = m 4co 2 + m 2 + 41 = di or dec The equation -=-^ = m -=- + fo + /(£) may be replaced by a pair of linear equations by writing dx dt =y ^ = lx +my + f(t) and the methods of § 10 are again applicable. INTEGRATING FACTORS 29 To illustrate the method of obtaining the solution, let us first of all consider the equation d 2 x dx -^2 ~ K t «2) ^ + «l«2^ = f(t) This may be replaced by the pair of linear equations dx dt =y dn J- = K + a 2 )y - a. t a 2 x + f(t) The equation for a is now a 2 - «(a x + a 2 ) + a x a 2 = (a - oc^a - a 2 ) = 0. Writing y - ol-^x — z lt y - «. 2 x = z % , we obtain the two linear equations dz II = ^ + M Hence y - a. x x = z 1 = A^" 2 ' + e^' le -0 "' f(t)dt y - a 2 x = z 2 = A 2 e a >' + e" 1 ' e~° l( /W <* where A 2 and A 2 are arbitrary constants. Solving for x, we find that (a x - a 2 )a; = A 2 e«i ( - A^ + e^ \e-^f{t)dt - e«*'J e~°-*f{t)dt In this solution the part which involves the function /(<) is called the particular integral, the term A 2 e a i ( - A^' is called the com- plementary function. When <*! = a 2 = a, the quickest way of solving the equation is to make use of the fact that e _a * and fe- ai are integrating factors; we thus obtain d dt\ d, [e-«* J - ae-«a;] = e-7(«) ai = A x + [ e- aT f{i)dT te-*j - (at + l)xe-« l = A 2 + f e~" t/(t)*c whence e_a 'd/ _ aa;e 30 DIFFERENTIAL EQUATIONS Solving for x, we get x = e a '(Aji - A 2 ) + e at \ e- ar {t - t)/(t)cZt where A 1 and A 3 are arbitrary constants. In particular, if f(t) — Ae a< where A is a constant, we find that x = e^Ajt - A 2 + JA« 2 ) Let us next consider the equation of forced damped vibrations sin&(£ - t)F(t)cZt Jo INTEGRATING FACTORS 31 Let us now consider the special case in which ¥(t) is a particular solution of the equation d?x _ dx w + 2a^ + { a* + b>)x=0 We may then write ¥(t) = Ee-" ( sin (bt + z) where E and s are constants. The general formula for x then gives E x = -j-e' at 1 sin b(t - Jo t) sin (bt + z)dt + Ae" at cos bt - Be -«j s . Now 2sinb(t - t) sin (6t + s) = = cos (bt - 26t - z) - cos (6< + < hence it appears that x = - E* „, 26 e C0S (bt + =) + Aje"* ; cos bt - Bje-" 1 1 sin bt where A v B x are arbitrary constants differing in value from A and B. If a = 0, so that the differential equation reduces to -p- + b 2 x = E sin (bt + e) E« we have x = - -^- cos (6i + e) + A x cos 6i - B x sin 6i The presence of the term - -^ cos (bt + z) indicates that the effect of the force E sin (bt + e) becomes more and more pronounced as t increases. In this case the period of the force P(i) is the same as the period of the free motion when ~F(t) = 0. Let us next con- sider the case in which the period of the force differs from that of the free motion, but the damping coefficient a is the same. In other words, let -p(t) = Ee-" 62 - p t 32 DIFFERENTIAL EQUATIONS Substituting in the formula :<: --- ®e- at [ sin b(t - -v) sin (yr + e)cfc + A.e~ at cosbt - Be at smbt b we find that E x = e~ at sin {pt + s) + A^" 1 " cos 6< + B^""' sin &Z 6 2 - p 2 where A 1 and B x are arbitrary constants. EXAMPLES. 1 . Solve the equations d 2 x dx -— - 5-r- + 6x = e 2t dt* dt d 2 x „dx „ . -r- + 3 — + 2x = e~ l cos t dt 2 dt — - + b 2 x = sin 2 £6« d 2 x , dx 2. Solve the equations — — = ax + by dF =lx+my on the supposition that a, b, I, m are constants, and that (m - a) 2 + 46Z > 3. Solve the equation — 2a -j- + a. 2 x =/ (t) by putting x = e M y. At * dt 4. Solve the difference equation x n+2 + ax n+1 + bx n = by putting x n + l = y n . Hence show that the solution can be expressed in the form ^ = K%X n + A ^n where Xj and X a are the roots of the equation X 2 + ak + b = 0. Deter- mine the form of the solution when o 2 = 46. 5. Prove that the equation y n +i - 2acos (tyn+i + « 2 2/ n = is satisfied by y n = Aa re cos (nft + e) where A and s are arbitrary, constants. 6. Solve the difference equation Vn+z + a Vn+i + by n =/„. INTEGRATING FACTORS 33 § 12. Symbolical methods. Most of the results of the preceding article may be obtained by a symbolical method in which the operator D = ^- is treated as if it satisfied the ordinary laws of Algebra. For instance, the equation d X (lor -jjp ~ (*i + «a) fa + «i««* = /W •■••(!) may be written in the form (D - ai )(D -a 2 )s=/(<) (2) To solve the equation, we must find a value for the expression (D - Jp -J *> < 3 > Resolving the operator into partial fractions, we replace the foregoing expression by Now the expression (D - a.^j- x f(t) must be regarded as the solution of the equation (D - xfo = /(*) and has consequently the value x x = A^'i* + e"i* le-°i'/(*)d< where A x is an arbitrary constant. Evaluating the second term in (4) in a similar way, we finally obtain (a 2 - xjx = A^i'- A 2 e«« t + c»»' [e- a i'/(0^ - e" 2 ' U"" 1 '/^^ This agrees with our former expression; consequently the sym- bolical method gives the correct result. In using the symbolical method, the following rules are sometimes useful : p(D) e atf( {) = e at F(D + k) ^ F(D 2 ) sin (pt + e) = F( - p 2 ) sin (p* + e) When the function F is a rational integral function of its argument the preceding equations have an intelligible meaning. It is easy to verify the equations for the case in which F(D) = D™, n being a positive integer ; consequently we may conclude that the equations hold when F(D) is a polynomial in D. To extend the equations to the case in which F(D) is the ratio of two polynomials in D, we must understand the equations to mean that the right-hand side is one value of the operation on the left-hand side. When f(t) is a polynomial a value of F(D)/(£) may be calculated by expanding F(I>) in ascending powers of D and allowing each term to operate 34 DIFFERENTIAL EQUATIONS on f(t). After a certain term the result of each operation is zero, and so F(D)/(«) is obtained as the sum of a finite number of terms. As a further illustration of the symbolical method, let us take the equation m v || + b 2 x = E sin (pt + e) A particular integral is given by 1 D 2 + E E sin [pt + e) sin (pt + s) b 2 - p 2 ' To obtain the complete solution, we must add the complementary function, that is the general solution of the equation E Hence x = ^ s sin (pt + s) + A cos bt + B sin bt o 2 - p 2 where A and B are arbitrary constants. This agrees with a result obtained by a former method. If p = b, the symbolical method requires modification.* The symbolical method gives us a rapid method of evaluating certain integrals. For instance, if we wish to integrate e at sin pt, we write 1 . . , 1 . , D - a =r e at sin pt = e* 1 ~ sin pt = e at =rj ; sin pt D D+a D 2 - a? , a - D . , , a sin pt - p cos pt = e a( -5 5 sm pt = e at . , , a. 2 + p 2 a 2 + p 2 It should be noticed that in the preceding analysis we have replaced D 2 by - p 2 in the operator acting on sin pt. This is an application of the following rule, which is justified by the fact that it gives correct results : Rule. In calculating a particular value of F(D) sin (pt + s), we may simplify the operator by replacing D 2 by — p 2 wherever an even power of D occurs. Let us now find a particular integral of the equation of forced oscillations (IT (I'Y -p + 2a -j + \j?x = Ee _w sin (jpt + s) *The correct result may be obtained by finding the limiting value of the above expression for x asp-*b. If we differentiate the numerator and denominator of the fraction — 2 _ 2 with regard to p and then put p = b, we obtain the limiting value ~P Et - =r cos (bt + s) INTEGRATING FACTORS 35 wherein a, a, E, X, p, s are real constants. We write = Ee " W (D - X) 2 + 2affl - X) + ^ Sin ^ + £) = Ee ' Ai X^-2aX + ^-^ + 2D(a-X) sin ^ + s) - Jie (X 2 - 2aX + a* - p*y - 4D 2 (a - X) 2 n {P + S) _ n (X 2 - 2«X + a 2 - p 2 ) sin {pi + e) - 2y(» - X) cos (pf + e) ~ (X 2 - 2aX + a 2 - y 2 ) 2 + 4p 2 (a - X) 2 To obtain the complete solution, we must add the complementary function, which, if \j? > a 2 , has the form Ae~ at cos bt + Be~ a( sin bt where (jl 2 = a 2 + 6 2 . The case in which X = 0, p = u, a > is of special interest, for then the solution is E x = - » — cos (at + e) + Ae" a 'cos bt + Be""' sin bt where A and B are constants which depend on the initial conditions. As t -> oo , the last two terms become negligible and the amplitude E of the forced oscillation depends ultimately on the quantity = — , which is large when the damping coefficient a is small. It should be noticed that the amplitude varies directly as the period of the impressed force. To solve the equation (D - a) 2 x = f(t) by the symbolical method, we write (D -a e"*5i[e-*/(0]. = C'Ta +Bt + [ {t - T)e- aT /(T)cZf The symbolical method may be used with advantage in the solution of systems of linear equations with constant coefficients. Let us consider the following example in the theory of electric currents. Currents are generated in two coupled circuits by a 36 DIFFERENTIAL EQUATIONS variable electromotive force f(t) acting on the first circuit. If L, N are the coefficients of self-induction of the two circuits, R, S the resistances and M the coefficient of self-induction, the currents x and y in the two circuits satisfy the differential equations Writing these in the forms (LD +R)a; + MDy =/(«) MDz + (ND + S) y = we may satisfy the last equation by putting x = (ND + S) u y = - MDw Substituting in the first equation, we get [(LD + R)(ND + S) - WD*]u =/(*) Since (JKN + SL) 2 - 4RS (LN - M)« = (RN - SL) 2 + 4RSM 2 and is positive, the complementary function in the solution of the last equation is of the form Ae~ at + Be~P l , where a and [3 are real and positive if, as in the actual case, LN > M 2 . When u has been found, x and y may be derived at once by the equations given above. The symbolical method may also be used to solve a differential equation of the n th order with constant coefficients. Let us suppose that the equation is represented symbolically by (D - ai )(D -« 2 )...(D -cc n )x=f(t) If the constants )x=M (5) where F(D) is a polynomial in D, whose roots are not necessarily all real and distinct. Let us assume that F(D) = (D - ai )*i(D - a 2 )^ ... [(D - 1;)* + yf] ... . (6) where the quantities a. v l v a 2 , X 2 , ... ?, 7) are all real. INTEGRATING FACTORS 37 When we express in partial fractions, we have terms of types 1 ID + m CD - aj" (D - ?)• + y)2 hence, if we wish to calculate — — - fit), we must find the values of P(D) fl> + » (D - Q« + T) 2 ' and m -i— r-/(0 The former may be regarded as the solution of the equation [(D-5) 2 +*]«]«=/(*) and has already been found ; the latter may be replaced by and may be calculated by direct integration . When a particular integral of equation (5) has been found, the complementary function must be added to it. The complementary function may be obtained by adding together the complementary functions corresponding to the different factors on the right-hand side of (6). It is, in fact, e a »*[A + A. x t + ... A^it*"-- 1 ] + ... + e*'[Pcos7)i + Qsinvj«] + ... where A ... A Xl _i, P, Q ... are arbitrary constants. The case in which a factor of type [(D - £) 2 + yf] s appears in F(D) is rather difficult, and will not be discussed here. Perhaps the simplest method is to resolve it into complex factors and work with complex quantities : the preceding method may then be used. For a fuller account of the symbolical method, the reader is referred to Forsyth's Differential Equations. EXAMPLES. 1. Solve the equations d l x — — h 4x = t 2 + sin I df d*v d 2 v —^ + 3 -r4 + 2« = x 2 cos x dx 1 dx 2 dx 2 dx — - - 2 -^ + V = x 3 + e x sin x dx 2 dx a d z x , — - - x = e dt a 38 DIFFERENTIAL EQUATIONS g—2 + -/w _ + „*=/(*) 2. Show that the equation d'y , may be replaced by two equations t where the A's and B's are constants. Since x is continuous, we must have (Ag _ A j cos ^ + ( Bj _ B j sin |J , T = -57 + Xx = when t = x, we must have - [x(A 2 - A x ) sin [xt + [x(B 2 - B x ) cos [xt + X(A 1 cos [xt + B x sin [xt) = Solving for A 2 and B 2 , we find that A 2 = A 1 h — sin [xt(A x cos [xt + B x sin jxt) B 2 = Bj cos [ix(A x cos [xt + B x sin jxt) The values of A x and B x must be determined from the initial conditions, and when these are known the above equations will determine A 2 and B 2 . If A„, B„ denote the values of the constants just before the n th impulse, we find in a similar way that A„ +1 = A„ + - sin w[xt(A„ cos W[xx_ + B„ sin m;xt) B n+1 = B n C0S n V-*(-K cos W* + B « sitl n \y*) 40 DIFFERENTIAL EQUATIONS If t is half the period of a free vibration under the normal spring, we have [it = tz, and consequently sin wjjit = 0. The preceding equations now become An+l — •"■„ B„+i = B„ X A once that -^•n+l = -"-I B n+1 = B l -^A t 1* If B 1 and Aj have opposite signs, the impulses at once magnify the amplitude of the initial vibration, and we get a vibration whose amplitude increases continually with n. If B x and A x have the same sign, the impulses may at first diminish the amplitude of the initial vibration, but as soon as — A x becomes greater than B x , the V- amplitude of the vibration increases continually with n. If A x = 0, the impulses do not produce any effect. The foregoing theory throws some light upon an experiment due to Melde, in which a fine string is maintained in transverse vibration by connecting one of its extremities with the vibrating prong of a massive tuning-fork, the direction of motion of the point of attachment being parallel to the length of the string. The other end of the string is fixed. The effect of the motion is to render the tension of the string periodically variable ; and at first sight there is nothing to cause the string to depart from its equilibrium condition of straightness. It is known, however, that under these circumstances the equilibrium may become unstable, and that the string may settle down into a state of permanent and vigorous vibration, whose period is double that of the fork (compare Rayleigh's Sound, Vol. I. p. 82). If t is any integral multiple of a half period, the theory is exactly the same. This corresponds to a result obtained by A. Stephenson in a discussion of equation (1). A complete discussion of the case in which t is arbitrary is difficult; the case in which t is a quarter period is simpler and will now be discussed. We now have \n = -=, and the equations take the form •A-2W+1 — A 2n B 2re+1 — B 2 „ - A 2n -A-271+2 = A 2 „ +1 + — -B 2 „ +1 B 2n+2 = B 2m+1 Hence -^2n+l = -A-2n-l + 7 B 2n _i B 2b+1 = B 2n- 1 ^2n+l INTEGRATING FACTORS 41 To solve these equations we multiply the second by £ and add to the first ; then A a „ +1 (l + &■) + £B 2n+1 = A^, + (~ + 1;)b. Now choose £ so that 2n-l i.e. so that £ 2 +-i + l=0 {A ' If a = - + £ where £ is a root of the preceding equation, we have t \a + 1 = and A 2n+1 + crB 2 „ +1 = - G 2 {A 2n _ t + ctB 2 „_!) or A 2n+1 + aB 2nU = ( - l)"a 2n (A 1 + CT BJ If ct 1( or 2 are the two values of cr, we have a^ = 1 ; hence if a x is real and numerically less than 1, ct 2 is real and numerically greater than 1. Hence, if a x and cr 2 are real, at least one of them is numeri- cally greater than 1 , and the equations A 2n+ i + S!B 2re+1 = ( - lW^ + 0^0 A 2n+1 + a 2 B 2ra+1 = ( - l^aWAj. + a.Bj) indicate that the amplitude of the vibration is large when n is very large. The two values of cr are real when X 2 > fi. 2 ; consequently, if this condition is satisfied, the effect of the sequence of impulses is to produce ultimately a vibration of continually increasing amplitude. The impulses thus have a cumulative effect. If X 2 < [x 2 , the values of a are of the form e ie , e' ie respectively, where 6 is real and i = V- 1. In this case the values of A, Y 2«+l and B 2n+1 remain finite as n increases and fluctuate between certain upper and lower limits. The impulses do not produce a vibration of continually increasing amplitude. § 14. A system, of linear equations occurring in the theory of radio- active transformations. Let us suppose that a radio-active sub- stance A is changing into A lt that A x is changing into A 2 , A 2 into A 3 , and so on. Let P , P 1( ... P„ be the amounts of A , A 1 , ... A n present at time t. Then, from the law of mass-action, the rate at which A s is transformed into A s+1 is proportional to P s , while the rate at which A s-1 is being transformed into A, is proportional to P s _i. The total rate at which the quantity P s is changing is thus X^jPj,! - X S P S , where the quantities X^j, X s are characteristic 42 DIFFERENTIAL EQUATIONS constants for the substances A^, A s respectively. We thus have the system of equations * ~dt - ~ X ° F ° ni = x ° p ° ■" XiPi w " XiPi ~ XaI>2 d^n _ y p _ > p If the m th product P n is stable, we have \ t = 0, and the equations S ive d s (P + Pi+...P.) =0 hence the total mass remains constant, as it should do. It will be convenient, however, to consider the general case in which X n + 0. To calculate the values of P , Pj, P 2 , ... , we observe that e^ is an integrating factor of the first equation, e* 1 ' an integrating factor of the second equation, and so on. Let us first assume as our supplementary conditions P = N , P 2 = 0, P 2 = 0, ... P M = for t = The first equation gives P = C e~ A °', and since P = N for t = 0, we have C = N , and so P = N e-* The second equation may be written in the form j t (eWPJ = X eW = X N e<^-^ hence e^'Pj. = N ° X ° [e^-^ + CJ Xj - X Since P 1 = when t = 0, we have G t = - 1, and so P x = J^ [e -v _ e -M] A i - A o The second equation may now be written in the form ■j (e**'P 2 ) = X^Pj = ]Sr ° X ° Xl [e^-^' - e^- x >>'] at X : - X and gives e A 2(p = XqX^q^-^' XqX^q^-^' 2 ( A i - A o)( A 2 - A o) + (X - X X )(X 2 - XJ + ° 2 * Cf. E. Rutherford, Radio-activity, 2nd edition, p. 330. INTEGRATING FACTORS 43 Since P„ = when t = 0, we find that 0, = NoXoXi (X 2 - X )(X 2 - XJ H At; — + *o)(* 2 - x„) (X - X X )(X 2 - X x ) + o~ y<^> i - *J (*0 - W We shall now prove by induction that P. - ( - D-N X X 1 ...X m _ 1 [^ + f±- + ...f^\ where / m (X) = (X - X )(X - XJ ... (X - X m ) ' and/ m '(X) denotes the derivative of / m (X) with regard to X. Assuming that the law holds for a particular value of m, we have -£(ewp m+l) = x m p me w = ( _ i)^N x x 1 ...x m g e( 7;;:"; ) ' Now, since / m 'P0 = (\ - X )(X S - Xj) ... (X, - X,.^, - X 5+1 ) ... (X. - XJ we have (X s - X m+1 )/ m '(X s ) = f m+1 (k s ) where / ml (X) = (X - X )(X - X x ) ... (X - X m+1 ) Hence e^*P m+1 = ( - l^IWi ... X m [I; e .^ +1 " A f + C, L-s=0 /m+l\ A s,) Putting £ = 0, we get m "I S T 7 — oTT + Cm = ° s=0 Jm+l\ A s) but from a well-known theorem in the theory of equations m+l J Hence G m = /m+l(Xs) 1 = m+l /m+l(Xm+l) and our formula now gives P*i = ( " IJ^NoXoXj.-.X,, £ ^ , s=0 ,/m+l\ A s,l Thus we have proved that if our formula holds for a given value of m, it holds for the next greater value of m ; but we have already shown that the formula holds for m = 0, 1, 2 ; hence it holds for all positive integral values of m. The solution of the general case when the supplementary con- ditions are P =N , Pi = N 1 ,...P n =N n for * -0 44 DIFFERENTIAL EQUATIONS is obtained by adding together the solutions for the separate cases P =N , P 1 = 0, P 2 = 0,...P„ = P =0, P x =N 1> P 2 = 0,...P„ =0 P = 0, P x = 0, . . . P„ = N n Now it should be observed that if P is zero initially, it is always zero ; if both P and P x are zero initially, they are always zero, and so on. Writing F r (X) = (X -X r )(X -\ +1 )... (X -XJ we find that N X X 1 ...X f ^ 1 £ 1 f 7 — - N^ -X m _ i ;S 1 f 77 f T + - ( " l)"N M e-\. 7=1 -t ! (A,) The case in which there is initially radio-active equilibrium is of some interest. We then have the relations X N = 1 1 N 1 = ... X n N„ = a, say To sum the series, we make use of the relations X 3 (X 1 X 2 ... X m _ t ) _ X Xi ... X m _j X x • 11 X m _] Whence X, F '(X S )- P '(X S ) F,'(X.) X S (X 2 ... X m _ t ) _ >■!■.. X m _ t X 2 ... X m _ t F X '(X.) " F/(X S ) + F 2 '(X S ) x s x s F S '(X S ) ~ F S '(X S ) X X X 2 ... X m _j x 2 . ■■ ^m-1 j _ X X t ... X m _] 1 '(X.) + -J ' F '(X S ) consequently it follows that the coefficient of e~V in the expression for P m is XpX^-X^ a s F '(X s ) Hence, finally, P m = ( - lJ-oXoXi ... X m _ 3 . 2 „ s=o A s J - l A si EXAMPLES. 1. Prove that in the case when there is originally only one substance present, r™ Jo and in the case when P = N , Pj = N 1; ... P m = N m for t = 0, O P m *=N„+N 1 + ...N m What is the physical interpretation of this equation ? INTEGRATING FACTORS 45 2. If there is initially only one substance present and Q m =l e-«P m dt (as>0) prove that Q m = V-i-V-i Hence show that the formula for P m may be obtained by trans- forming this expression for Q m into partial fractions. § 15. A system of linear equations occurring in the theory of prob- ability. If in the last section 1 = X 1 = ... X„, the solution takes a different form. If the initial conditions are P = N , P x =0, P 2 =0, ...P„ =0 the value of P is the same as before, but the equation to de- termine P 2 is j f (eW 2 ) = A oe vp o = .-. e^'P, = X N < X N Again, | (eVP,) = X e^'P 1 = X «N t Proceeding in this way, we find that P«=N e-^M^ (1) The solution for the case in which P = N , P x = N 1( ... P M = N„ may be found by superposition as before, and we have P„_-e- W [ Ho M-" +N ,M^ !+ ...N.] A system of linear equations, in which the quantities X are all equal, occurs in the solution of the following problem in the theory of probability. Consider a large number of intervals and a large number of objects which are distributed at random among the intervals. Let P„(a;) be the chance that an interval chosen at random contains exactly n objects when the average number of objects in an interval is x. To find P„(a;), we shall suppose that the number of objects is in- creased by one, and that the number of intervals is a very large number x. The chance that the new object goes into the chosen interval is -, the chance that it does not is ( 1 — ). The chance y. \ x/ P„ (x + - J of having n objects in an interval chosen at random when the average number has increased to x + - owing to the addition of the new object is the sum of two chances : 46 DIFFERENTIAL EQUATIONS 1°. The chance that the interval originally contained n - 1 objects and that the new object is included in the interval. This chance is i - P_i(«) 2°. The chance that the interval originally contained n objects and that the new object did not enter this particular interval. This chance is / 1\ (l ->.(«) Hence we have the equation P. Since - is very small, we may write •.(*+3-(i-S^(*)+^-i(*) ay small, we may write P " V + x) " P "^ = x~dx + higher P° wers of x "« + "n-l hence we have approximately da; When x = 0, we have P = 1, P x = P g = ... P„ = ; accord- ingly our result (1) gives Poisson's formula P« = -r e ~* EXAMPLES. 1. Prove that the probable value of n is x. 2. Prove that the probable value of the square of the deviation (n - a;)* is a;. § 16. General theory of a system of linear equations. Let us con- sider the system of linear equations : dx, , -j£ + a ll x 1 + a 12 x 2 + ... a ln x n = f t {t) -^ + a il x 1 + . . . . a 2n x n = f 2 {t) r + «»i*i + . . . . a„ A = f n {t) dt wherein each of the quantities a TS is a function of t. INTEGRATING FACTORS 47 If we multiply the equations by y v y 2 , ... y n respectively and add, the resulting equation may be written in the form ^I>i2/i + x 2 y 2 + ... x n y n ] = yj^t) + ... yj n (t) . (1) provided -J- 1 = a^ + a 21 y 2 + ... a nl y n dy z -£f = «i2«/i + a 22 j/ 2 + ... a n2 y n dt HrMi + a 2n y 2 + nVn Let us suppose that these equations can be solved completely, and let us denote the set of solutions corresponding to the supple- mentary conditions «/i = 0. y% = °. Vr = 1. ••• Vn = when t = by 2/ri. Ur2 ■■■ t/m- Let this be done for each value of r ; then when t = 0, y TS is zero if r is different from s, and unity if r is equal to s. The initial value of the determinant A = 2/n> !/i2> ••• 2/i« J/gl> #22" ••■ I/in y»i. y«z. y„ is consequently unity. Differentiating the determinant with regard to t, we obtain the sum of a number of determinants, each of which is derived from A by differentiating one column. On substituting the values of the derivatives, we find that the determinant in which the s th column contains the derivatives is equal to a ss A. Hence - 'a lt + a 22 + ... a nn )& dt Integrating, we get „]("" + "a + •■■ "*nn)dt If the quantities a llt ... a nn are finite, the value of A is different from zero ; consequently, if we know the values of the n quantities jV z r = x 1 y rl + x 2 y r2 ... x n y rn we may solve for x x , x 2 ... x n . The values of the quantities z r may, of course, be obtained by solving the n equations of type (1) . Let us now consider the case when the coefficients a rs are all constants ; we may then write the equations for the y's in the sym- bolical form (0n _ J) )y 1 + a21 y 2 + ... a nl y n = a 122/l + («22 - D )?2 + - a n2Vn = nVx + a 2n y 2 + ■••(«««- D )y« = o 48 DIFFEKENTIAL EQUATIONS Let us solve the first n - 1 equations just as if D were an algebraic quantity ; we then obtain P 1 (D)-Pp0--F„(D) - M ' Say Putting yi = F^DK y 2 = F a (D)«, ... y B = F„(D)« in the w th equation, we obtain F(D)w = 0, where F(D) denotes the operator ! - D, a 21 , a 31 , .., a nl 2> ^22 ~ ^> *32> ■" ®«2 F(D) = «!»> a nn ~ D Let us now assume that P(D) - (X x - D)(X 2 - D) ... (X„ - D) where the constants X 1; X 2 , ... X„ may or may not be all different ; we then have (Xi _ j))^ _ D) ^ _ D)w = This equation may be solved directly by the method of § 12, or it may be replaced by the system of linear equations """1 dt - XjUj = du 2 ~dl - X 2 M 2 = U 1 du 3 dt - X 3 M 3 = « 2 du dl " X„M = «*„-! which is very similar to the system occurring in the theory of radio- active transformations. The solution is obtained as before by using the integrating factors e~ Al ', e' x ^, ... . § 17. Equations of the first order and of the first degree. A differ- ential equation of the first order and of the first degree is of the f0rm |=/M (1) where the function /is single- valued for some domain of the variables x and y. We shall often find it convenient to write the equation in the symmetrical form Mcfo + Ntfy = (2) where M and N are functions of x and y. Definition. The equation (2) is said to be exact when a function m exists such that the equation has the form du , du , INTEGRATING FACTORS 49 The equation may then be integrated as it stands, and the solution ls u(x, y) = c where c is an arbitrary constant. For instance, if M is a function of x only and N a function of y onty, the equation is exact, and the solution is f f IMcfo + Wdy = c It is clear that an equation of type (2) is not generally exact, for the relations du _ du _ dx fry { > imply that M and N are connected by the relation 3N _ dhi 3M dx dx dy dy provided M, N, -— and — — are continuous for some domain of the ox dy variables x and y. Conversely, if the conditions just mentioned are satisfied and -— = -=— , we can show that a function u exists for which equations ox oy (3) are satisfied. ^ Let us first of all integrate the equation ■=- = M ; regarding y as a constant, we get = JM(a, y) da + Y where Y, the constant of integration, is independent of x, but may be considered as a function of y. We shall endeavour to choose this function of y in such a way that the equation ^- = N is satisfied. Differentiating the previous equation with regard to y, we see that Y will satisfy the requirements if N = ^JM(«, 2 /)& + ¥ Now, since — is a continuous function of its arguments,* we dy may differentiate the integral under the integral sign by the rule of Leibnitz. This gives N= j_M(a,y)«fa +¥ = ^ a N(a,y)da + ^ since by hypothesis ^-M(a, y) = grN(a, y) * See Goursat-Hedrick, Mathematical Analysis, Vol. I. p. 193. B.B. D • 50 DIFFERENTIAL EQUATIONS Now — N(ffl, y) is a continuous function of a ; hence the integral may be evaluated, and we have dY Tt(x, y) = N(as, y) - N(*„ «/) + -^ where x is the lower limit of the integral and is independent of y. The function Y may now be obtained by solving the equation dy- = N <*" *> which is certainly soluble if x belongs to the domain under con- sideration, for the function N is then a continuous function of y, and may consequently be integrated with regard to y. When the equation (2) is not exact, it seems natural to try and make it exact by using a multiplier [i(x, y). This idea occurred to Clairaut (1739), and was developed by Euler, who obtained the partial differential equation |(^)=|W W for the determination of [i. This equation is, of course, the con- dition that y.(Mdx + ~Ndy) should be exact. If fji is one integrating factor and u(x, y) = c a solution of the differential equation (2), it is easy to see that [if(u) is also an inte- grating factor where / is an arbitrary function, for the equation is a consequence of (3) and (4). By using different integrating factors, the solution of equation (2) is obtained in different equivalent forms. To understand the nature of this equivalence, we remark in the first place that if u(x, y) = c is a solution of (2), then F(w) = F(c) is also a solution where F is an arbitrary function. Let us now see if there is a more general solution of the form F(m, x) = a where a is a constant. Differentiating with regard to x, we obtain 3F /du dudy\ 9F _ du \dx dy dx) dx The quantitv within brackets vanishes, since u satisfies equation SF (2) ; consequently ^- = ; in other words, F depends only on u. Sometimes the fact that two solutions of a differential equation INTEGRATING FACTORS 51 are equivalent leads to interesting analytical results. As an illus- tration, let us consider the exact equation ydx + xdy = which possesses the solution u = xy = c. Since - = — is an inte- grating factor, we obtain the equation u xy dx dy — + — = x y whose solution log x + log y = a must be equivalent to xy = c. We may therefore write log x + log y = ¥(xy) Putting?/ = 1, we get ¥(x) = log x, and consequently log x + log y = log (ny) It should be remarked that if two relations u(x, y) = a v(x, y) = b lead to the same differential equation of type (2) , the relations du j du , ■=- dx + •=- dy = ox dy dv j dv j n ^-dx + -^-dy = dx dy * must be equivalent, and so 3m dv du dv dx dy dy dx Conversely, if this relation holds, the two primitives lead to the same differential equation, and it follows from the above reasoning that v is a function of u. It is generally impracticable to solve the partial differential equation for the determination of an integrating factor, but Euler has shown that the equation may be used with advantage to find when there is a multiplier of a specified type. For instance, u can be a function of some assigned quantity z depending on x and y if aM 9N dy dx -. T dz ^ r dz N— - M^- ax dy is a function of z. This is seen at once by putting \i = 9(2) in the partial differential equation (4). The cases in which z = x, y, xy, - respectively are of chief interest. For further details the reader is referred to Boole's Differential Equations, Chap. IV. Euler's partial differential equation is of some importance in thermodynamics. It is known that in certain heat exchanges 52 DIFFERENTIAL EQUATIONS between two substances, the heat energy Q, which produces a small change in the state of one substance, can be represented by a simple formula of type Q = l v dv + y v dd where is the absolute temperature and v some other variable, such as the volume of the substance. It is known, moreover, that _1 is a multiplier or integrating factor of the expression for Q. Euler's partial differential equation consequently leads to an important relation between the derivatives of the quantities l v and y„. Another relation is obtained by using the first law of thermodynamics, which states that if p is the pressure of the substance Q - pdv is the increase of the internal energy, and so is an exact differential. EXAMPLES. 1. Prove that the following equations are exact : (x % + y z - 1) dx + 2xy dy = (x 2 - 3y) dx + (y 2 - 3x) dy = 2. Show that the relation x + y = c(l - xy) leads to the differential equation (1 + x 2 ) dy + (1 + y 2 ) dx = Hence obtain the addition theorem for the function tan x. 3. Show that the relation ckVi - y % + ys/i - x 2 = c leads to the differential equation \A - x 2 dy + Vl - y* dx= Hence obtain the addition theorem for the function sin x. 4. Solve the equations • dy »)-0 (x 1 + j/ 2 + 2ax) dx + 2ay dy where a is a constant. 5. The equation Mdx+ N dy = possesses an integrating factor of the form n = x n F f^\ if ^ ;=— - nNx ay ox J Ma; + Ny is a function of ^. Denoting this function by 9(-Y show that the function F is obtained by solving the equation dF x* + y* + a (x — - y) (fe + 9WF = where z = - ■ INTEGRATING FACTORS 53 solution of t + y* = Q(<») 6. ' If P(a;) = y is a particular solution of the equation dy dx ;~ s \ Fdx is an integrating factor (Lacroix). 7. It is known that the equation (a; cos y - y sin y)dy + (x sin y + y cos y)dx = possesses an integrating factor which is a function of x only ; find this integrating factor, and hence solve the equation. 8. Solve the equations [y(x 3 + y 2 ) - xy]dx + x 2 dy - {x 2 + y 2 + 1) dy + xy dx = § 18. Integrating factors of linear equations of the second order. A linear differential equation of the second order may be written in the form d i u du ' a{ ^ dx* + (K) dx + °® U = ^ • ' • • W or in the abbreviated form L^w) = where L x denotes the operator by means of which the expression on the left-hand side of equation (1) is derived from u. A quantity v is called an integrating factor of the expression L z (w) if we have identically du v-LM - ^ va j- + pu where p is some function which is independent of u. Comparing the two sides of this equation, we find that we must have d . , dp Tx (va) + p=bv Tx =cv Eliminating the unknown function p, we find that v must satisfy the equation _ d 2 g hx{v) S tf W - d^ hv) +cv = ° ■ ■ ■ (2) which is said to be adjoint to L x (u) = 0. When u and v are arbitrary functions of x, we have the identity vL x (u) - uL x {v) = ^ R(m, v) .... (3) where B(«, v) = o(t,g - «g) + «t,(ft - g) It is clear from this relation that the expression L x (u) bears the same relation to L x (v) as L x (v) does to L^w). The quantit}' R is called the bilinear concomitant, and L a (w), L^v) are said to be adjoint differential expressions. It is clear that a function w which satisfies the equation l> x {u) = is an integrating factor of li x (v). 54 DIFFERENTIAL EQUATIONS When an integrating factor or solution of equation (2) is known, the equation (1) may be integrated. If we multiply by this factor v, a first integration gives R(w, v) = \vf dx + A where A is an arbitrary constant. The last equation may be regarded as a linear equation of the first order for the determination of u. Multiplying throughout by the integrating factor 1 \ b a II) — — — P. 1 « dx and integrating, we get avwu = I w dx I vf dx + A I w dx + B where B is an arbitrary constant. The solution is thus of the form u = A Ul {x) + Bu 2 {x) + — [w dx j v{Z)f(l) d\ where u-^x) and u 2 (x) are particular solutions of the equation L,(«) = 0. The general solution of this last equation is, in fact, obtained by putting/ = ; it is u= AUi{x) + Bu2{x) It is customary to say that the solution of equation (1) is made up of two parts, viz. the complementary function Au x (x) + B?* 2 (a;) and a particular integral involving the function /. It should be noticed that the general solution of equation (1) involves two arbi- trary constants. The two expressions L(w), L(v) are identical if d 2 a db _ da dx 2 dx dx i.e. if -T- = b. In this case the equation L(m) = is said to be self -adjoint. A self-adjoint equation may be written in the form T , , d ( du\ L ^^Tx\ a d-x) +CU -° and relation (3) then becomes V ^ {u) - Uh * {v) ^i[ a ( V fx- U t)_ The general expression L x {«) of the second order may be made self-adjoint by multiplying it by a suitable factor z. This must be chosen so that , -j-(az) = bz dx y ' INTEGRATING FACTORS 55 The relation (3) may be extended to the case of a linear differential expression of the n th order, such as T , . d n u d n ~hi Since , d n ~ 3 u av d n u d T d^u d , . d n ~ 2 u d 2 , y d^=dxl aV d^ ~ dx^ dx^ 2 + d^ {av) dx n ~ it follows that z * {v) = ( ~ l)n ^ {av) + ( - 1)n_1 £i (a ^ + - +a - v A function v, which satisfies the equation li x {v) = 0, is an inte- grating factor of the equation M M ) = /(*) Multiplying this equation by v and integrating, we obtain B,(u, v) = \vfdx + C x where G 1 is an arbitrary constant. The new equation may be regarded as a linear differential equation of the (n - l) th order for the determination of u, and can be treated in a similar way if a suitable integrating factor can be found. When n operations of this type are performed, an expression involving n arbitrary constants is finally obtained for u. This expression can be regarded as the sum of two expressions, a par- ticular integral depending on the value of the function /, and a complementary function involving two arbitrary constants and independent of the form of /. This complementary function is a solution of the equation L x (m) = 0. EXAMPLES. (1 ^"IJi (1.1 J, 1. If u(x) is a solution of the equation a-=— „ + 6 — + cu= 0, then r h dx* dx U -dx v = - e J & is a solution of the adjoint equation. 2 - If w«>-[s + *m][s + * w ]"-° prove that the adjoint equation is ^W-[^-ff(«)][5-p(-)]-o This means that if the operator ~L S can be broken up into factors, the adjoint operator L consists of the product of the adjoint factors + A„w= 56 DIFFERENTIAL EQUATIONS in the reverse order. Extend this theorem to the case in which there are n factors. 3. If L x (m) and L^u) are two adjoint expressions, the expression L„L a («) is self-adjoint, and the expression L„^I*(«) is self -adjoint except for a difference in sign (Darboux). 4. Prove that an equation of type dtf> L ^"J + d^ 1 " 1 L dtf^\ + '" das L ' 55_ is self -adjoint (Bertrand and Hesse). § 19. Equations of various types. A differential equation of the is said to be ezaci when we have identically J \dx*' dx' y,X ) ~ dx g \dx' J ' In this case the equation is a ^s< integral ; if this equation of the first order can be solved, the original equation can also be solved. Sometimes an equation may be made exact by multiplying throughout by an integrating factor ; for instance, the equation g + ^siny=0 (1) may be made exact by multiplying throughout by ■—-, the resulting equation gives on integration = c l fdyY 2\dx) -V- 2 cos y In this equation the variables are separated, and so the integration can be carried a step further. The differential equation (1) is the equation of motion of a point mass which moves in a circle under gravity. In fact, if is the angle which the radius to the moving point makes with the vertical, the tangential acceleration is a0 where a is the radius of the circle. Equating this to the resolved part of the acceleration due to gravity, viz. - g sin 0, we obtain the equation ai I = - g sin which is of the form (1) except that and t replace y and x. INTEGRATING FACTORS > 57 The method of integrating factors is particularly useful in the case of the differential equations occurring in dynamics, and can be employed when a system of differential equations is to be solved. Let us consider, for instance, the equations d?x _dR dt f adr " " an x* ~ J T 2 "" = J r[2r 2 (R + c) - a 2 ] + where b is an arbitrary constant. Writing y* = x* tan

= tan6 dr r VY ; 1 + tan 0/(tan 0) dr _ 1 +tan0/(tan0) r /(tan 0) - tan flD The solution is evidently of the form r = Ae G W where A is an arbitrary constant. The differential equation evidently corresponds to a class of curves which are similar and similarly situated, the point being the centre of similitude. Sometimes the relation between and i[i cannot be expressed very simply in the form (2) , but may be written in the form tan = F(tan asa new variable. Differ- entiating the equation y = x¥(jp) with regard to x, we get 62 DIFFERENTIAL EQUATIONS whence — - V '^ dp WhenCe x ~ p - F(p) The solution is now obtained by eliminating p from the equations log x = I — - ip - + C y = xF(p) h - F(P) wherein C is an arbitrary constant. We may, however, retain p and regard the last two equations as equations giving a parametric representation of the coordinates of points on the curve in terms of the parameter p. If the relation between and 9 is of the form tan 6 = /[tan 9] = /[tan (9 - 0)] and the resulting equation cannot be easily solved for -^- or -, it is convenient to use polar coordinates. The differential equation is then T dQl tan0=/[rgy = /(tan 9 ) Differentiating with respect to r, we get sec 2 -j- = /'(tan 9) sec 2

= 26. (4) tan = a + b tan 6. (2) (J; = 77 - 6. (5) tan iji = a + b tan 9. (3) 9 = a. (6) tan 6 = a + b tan 9. 3. Prove that the equation ?(!)<**+ 4>(^)dy+ x" x (^){xdy- ydx) = may be solved by making the substitution y - vx and using the method of §8 (G. Boole). TRANSFORMATIONS 63 § 22. Equations which can be reduced to homogeneous equations. If a and b are constants, the equation i-'(Hi) "» can evidently be reduced to the homogeneous form by putting x' = x + a y' = y + b This substitution is successful in the case of the equation (Ix + my + n)dx + (Xa; + \xy + v)dy = . . (2) where I, m, n, X, [A, v are constants. The transformed equation will be homogeneous if a and b are chosen, so that la + mb = n Xa + \xb = v These equations fail to determine a and b when - = — . In this X y. case we write X = hi, y. = km, and take Ix + my as a new variable t. Since Idx + mdy = dt, the differential equation becomes m(t + n) dx + (kt + v)(dt -Idx) =0 dx kt + v dt ~ l(kt + v) - m(t + n) The integration can now be effected. EXAMPLES. 1. Prove that the differential equation (2) can also be solved by writing it in the form dx - Q^y dx - Q 2 dy (X + OjQx + ((jl + % 1 m)y + v + 6jn (X + Q 2 l)x + (ft + 8 2 wi)2/ + v + 6 2 »i and choosing X , 2 so that they satisfy the equation Z6 2 + (X + m) 6 + [i = When does this method break down ? 2. Solve the equations (x - y - 1) dx + (x - y + 1) dy = (a: - 32/ + 2)daj + (3x - y + &)dy = 3. Solve the equation dy m 1 m2 x - a dx (m-i - m 2 ) 2 y wherein a, m 1: m 2 axe constants. § 23. Clairaut's equation. The equation y = px + f(jp) (1) in which p = -~ may be solved by differentiation. The resulting equation is x ,j n [x + rm £ = o 64 DIFFERENTIAL EQUATIONS The transformed equation is thus ■—■ = 0, and this gives p = c ax where c is a constant. Substituting in (1), we obtain the general *»*«*«>» y =cx + /(c) (2) A singular solution is obtained by putting £ + /'(p) = 0, and eliminating p between this equation and (1). This solution does not involve an arbitrary constant, when interpreted geometrically it represents the envelope of the family of lines given by (2). The theory of singular solutions will be discussed more fully in Chapter IV. Equation (1) is usually called Clairaut's equation if it is written in the more general form F(y - px, p) = the general solution is F(y - ex, c) = § 24. Equations of the first order which can be solved for either x or y. When an equation can be written in the form y = v(*,p) (i) it may sometimes be solved by differentiation. The resulting equation is 9F 9F dp dx dp dx and can be regarded as a differential equation for p. If this new equation can be solved, giving f(3>,x, o) = (2) where c is a constant, the solution of the original equation may be obtained by eliminating^ between (1) and (2). When an equation can be written in the form x = ?{y,i>) (3) we must naturally differentiate with regard to y. The resulting equation is 1 _ SF d¥ dp p ~ dy dp dy If this equation in p and y can be solved, giving fip.y.c) = (4) a solution of the original equation is obtained by eliminating p between (3) and (4). EXAMPLES. ]. Solve the equations y = P i f(%) y = P n f(z) * = P n f{y) x + yp = ap 2 y = f(xp) TRANSFORMATIONS 65 2. Solve the equation y = xf(p) + , a, b) be the solution of this equation, then p is determined as a function of <\i, and x, y may be expressed in terms of <|; with the aid of the equations p cos <\> d<\> y = I p sin <\i dty 66 DIFFERENTIAL EQUATIONS EXAMPLES. 1. Solve the equations d 2 v where X, \x, n are arbitrary constants. § 26. Equations of the second order which do not involve y explicitly. The equation / du d 2 y\ may be replaced by an equation of the first order by writing j- = p. If the equation . , . can be solved and p expressed in the form p = f{x, a) where a is a constant, a solution of the original equation is given by y = I f(x, a)dx + c and involves two arbitrary constants a and c. An equation of the present type can evidently be reduced to an equation of the preceding type by taking x as the dependent variable. EXAMPLES. . o , ,, *.- dy d*y fd*y 1. Solve the equation -^ = x-^ - -r-^ u dx dx 2 \dx and show that in addition to the general solution involving two arbi- trary constants there is a singular solution involving one arbitrary constant. 2. Solve the equations (x + p) - - = p dx X d x* -~- + v % - xp = dx § 27. Equations of the second order which may be solved by differ- entiation. It is hardly possible to write down the general form of a differential equation of the second order which can be solved by differentiation. A simple example will be sufficient to illustrate the method. TRANSFORMATIONS 67 Consider the equation 2 d 2 y_(dy\ 2 _ Sdy _ d*yl * V dx 2 \dx) - J ldx X dx*\ where / is an arbitrary function. Differentiating with regard to x, we obtain ^_ x *V r [ m] [dxdx* JKy '\ where f{y), F(6) are arbitrary functions. § 28. Equations of Lagrange's type. The equation considered in the last paragraph is of the form «=/(») (1) where the equations -^ = 0, j- = contain a common factor and are satisfied in virtue of a single differential equation, which, of course, does not depend upon the form of/. Equations which can be thrown into such a form are said to be of Lagrange's type. 68 DIFFERENTIAL EQUATIONS It is instructive to ascertain how an equation of Lagrange's type arises from a given primitive. For simplicity we shall consider the case of an equation of the first order. Let the given primitive be 9 (x, y, a, b) = (2) where the constants a, b are connected by a relation a=f(b) We shall commence by discarding this relation and treating the constants a and b in (2) as independent. Under this understanding we may eliminate a and b in turn from the two equations * = ° dx + *>ty=° Let us suppose that the resulting equations can be written in the form u ( x> y^ = a an( j v(x, y, p) = b . . . (3) respectively. When we differentiate these equations, two differ- ential equations of the second order are obtained. If these are equivalent, equation (1) is of Lagrange's type, and its solution is obtained by eliminating^ from the equations (3). The two differential equations of the second order can generally be expected to be equivalent, since they both have (2) as a primitive. EXAMPLES. Solve the equations y 2 (l + p 2 ) = f(x + yp) yp = xf{y* - xyp) yp - x=f{y 2 -y 2 p z ) y - 2xp = f{xp*) F [y - G(p) + pG'{p), x + G'(p)] = y - xp = f y - xp y* + p J \l + x*p where p = — and/, ~F, G are arbitrary functions. § 29. Biccati's equation. An equation of type g=iy + Qy+R (1) where P, Q, R are functions of x is now generally known as Riccati's equation after Count Riccati, who studied the particular equation ^ + by 2 = ex'" (2) where 6 and c are constants. TRANSFORMATIONS 69 If one solution y =u(x) of equation (1) is known, the equation may be integrated by writing y = u + - and regarding «asa new dependent variable. The transformed equation is Tx + 2(Pu + Q)v + P = and may be solved by the method of § 7. Since the equation is linear, the general solution is of the form v = cf(x) + g(x) where c is an arbitrary constant and /(a), g(x) are independent of c. The complete solution of Riccati's equation is now seen to be of the form 1 _ cF(a) + G(x) V U cf{x) + g(x) cf(x) + g{x) ' ' ' (3) If F, G, /, g are single-valued functions of x, y is also a single- valued function of x, and if we are given that y = «/ when x = x , there is only one value of c, and so y is determined uniquely. If the functions F, G, f,g are not all single- valued functions of x, for instance, if f(x) = Vx z - 1, y is generally a multiform function of x, but its branch points, i.e. the points at which two values of the function become the same, depend only on the positions of the branch points of the functions F, G, /, g and not on the arbitrary constant c, and are consequently the same for all solutions of the equation. In the present case x = 1 and x = - 1 are branch points of f(x), and are consequently branch points of y unless the radical Va; 2 - 1 disappears when the expression (3) is simplified,. At any rate, we may conclude that a solution of Riccati's equation has fixed branch points. This is not true for every differential equation of the first order, for instance, the general solution of the equation *i - ' is y = Vx - c This is a two- valued function with a branch point at x = c. The position of the branch point thus varies with c, and is consequently different for different solutions of the equation. In other words, x = c is a movable branch point. Euler has shown that Riccati's equation may be transformed into a linear differential equation of the second order by making the substitution 1 ^ z y = ~ Vzdx The resulting equation is dx 2 \ """ dxJdx p£-(pQ+£}£h ; •,,■; .-0 70 DIFFERENTIAL EQUATIONS In particular, equation (2) can be replaced by the equation -=-s + bcx m z = dx 2 An important property of Riccati's equation is that the cross ratio of four particular solutions is independent of x. We easily find from (3) that if y lt y 2 , y 3 , y t are the solutions corresponding to the constants c lt c 2 , c 3 , c 4 , (2/ 2 - «/ 3 )(«/i - y*) (c s - C 3 )(Ci-C 4 ) (2/3 - yd(y-z - 2/4) (c s - cjfa - c 4 ) This result is due to E. Weyr and E. Picard. EXAMPLES. 1. A gas is being ionised by an ionising agent, and at time t there are n % free positive ions and w 2 free negative ions per cubic centimetre. < If q ions per cubic centimetre are produced every second by the agent, and the number of recombinations per second is proportional to n^, we have the equations ^ n dn. Solve these equations on the supposition that a and q are constants. 2. "When can the equation be satisfied by all the solutions of an equation of type ^ = ay 2 + by + c ? Obtain the necessary and sufficient relation between the functions P, Q, R, S. 3. The equation dy -j- + f{x) sin y + g (x) cos y + h (x) = can be reduced to a Riccatian equation by means of the substitution z = tan ^. 2 4. Integrate the equations p + y +y ^o dx x f-y +y >=o dx x TRANSFORMATIONS 71 5. Integrate the equation „dy x dx ~ v x *y + x * = ° by making use of the fact that y = x is a particular solution. 6. Prove that equation (1) is transformed into another Riccatian equation by the substitution _ AY + B V ~ CY + D where A, B, C, D are functions of x. 7. Prove that if A, B, C, p, q, r are arbitrary functions of t and P + m? + n 1 = 1, the equations dl — = l(Al + Brn + Cm) + qn - rm - A dm -3- = m(Al + Bm + Cn) + rl - pn - B dn — = n(Al + Bm + Cn) + pm - ql - C €bt can be reduced to two Riccatian equations in s and a by putting s(l + n) = I + vm, a(l + n) = I — im. § 30. The linear differential equation of the second order. If the differential equation L» = a{x) ^ + 6(a:) ^ + c(x)u =f(x) . . (1) is transformed by the substitution u = zw and 2 taken as new dependent variable, the resulting equation is d*z aW dx* /_ dw , \dz ( dhv ,dw \ ... + \ 2a Tx + bw )Tx + ( a dx~* +b dx +CW ) Z "■«*) The function w is at our disposal if we choose it so that h x (w) — ; the new equation may be written in the form 6?z /2 dw b\dz _/ dx 2 \w dx a J dx aw dz This is a linear equation of the first order for -=-, and an integrating - dx). Integrating, we obtain '__e Ja = \-we ia + c ax J a The value of z may now be obtained by integration; consequently, if one solution of the equation l> x (w) = is known, equation (1) can be solved. When a solution of the equation L x (w) = cannot be readily obtained, it is sometimes useful to reduce (1) to a normal 72 DIFFERENTIAL EQUATIONS form in which the second term is missing. This may be done by choosing w so that ^w 2a ^- + bw = dx Substituting this value of w, we obtain the equatioi} — + Iz = — dx 2 aw T- C -— — — — — — a 4a 2 2a dx 2a? dx This quantity I is the same for all linear differential equations which can be derived from (1) by a substitution of the form u = zv. For ' if ,dH u ,dz , a dx-> +b dx +CZ = g is such an equation, we must have ■v / ,.i » ^" i ii dh) ,dv la = av W = 2a t — v bv \c = a -^-z + b-j- + cv dx dx 2 dx where X is some function of x ; r _ C &'* 1 db' JS_aW__c b*_ ±db _b_da_j ~ a' 4a' 2 2a' dx' + 2a' 2 dx ~ a~ 4a 2 ~ 2adx + 2a z dx ~ I is on this account called the invariant for substitutions of type u = vz. We may also get rid of the second term in an equation by making a change of independent variable. Let us first of all reduce the equation to the self-adjoint form d ( du\ d ( du\ _,, , by writing va = p, v6 = -p, vc = q, v/ = F, where \is a solution of |(v«)=v6 Cdx Now define the new variable t by the equation t = I — , and express pq, Fp in terms of t. The resulting equation is "* & d 2 u dt2 +pqu=Fp and the coefficient of u is C 2 f *cte pq = -e a a TRANSFORMATIONS 73 EXAMPLES. 1. Reduce the equation c.dhi du x' - r — + x — + nhi = dx % dx d 2 u to the canonical form -r- + n 2 u = dt 2 and hence solve the equation when n is a constant 2. Reduce the equation d'u 2m + 1 du dx 2 x dx d ^ +z -^K = o to the form dz 2 3. Reduce the equation (l_a..)^_2«^+n(n+ 1)«=0 k 'da; 8 dx v ' d 2 w n(n +1) to the form -— + =-— u = di 2 cosh 2 * and show that it may also be reduced to a canonical form by writing „-„*-. in S + IM "-° sr" +J w=° prove that the product uv = y satisfies the differential equation d fl fd'y m dy dP)l _ dsLQ{^ +2P ^ +2/ ^/J +Q2/=0 where P = I+J, Q = I-J (Schafheitlin) . § 31. Equations reducible to linear equations with constant co- efficients. If we transform the equation , . d 2 u ,, .du , ., > a ^d^ + b{x) Tx +c{x)u=fix) by a substitution of type t = ' = -r-. The new equation can be expressed as T dx 2 T dx a linear equation with constant coefficients if a=<>■ ■ <*> g + 2(e + fx) d £ + (px* + 2qx + r)y = . . (3) where e, /, p, q, r are constants. If I + 0, equation (1) can be reduced to a particular case of (2) by writing a + Ix = x' , while if 1 = 0, equation (1) is a particular case of equation (3). *Orelle's Journal, Bd. 51 (1856), p. 123. t Compendium der hoheren Analysis : Brunswick (1879), Vol. II. p. 523. %Math. Ann., Bd. 3S (1891), pp. 225, 241. tAmer. Journ., Vol. 15 (1893). |l The second equation was discussed by Liouville, Journ. de VEcole Polytechnique, Cah. 21, p. 185. 76 DIFFERENTIAL EQUATIONS If in (2) we put* x = yjx', y = vx' K sP*y', we get -an equation of the same type as (2), but with new constants e', /', p', q', r' denned by the equations e' = e + X /' = x/ + [x p' = X(X - 1) + 2eX + p q' = v.q + xX/ + [ie + Xfi, r' = x 2 r + 2xf/./ + \i? It follows from these equations that p' + e' - e' 2 = p + e - e 2 = A say q' - e'f = x{q - ef) = xB say r' - f 2 = x 2 (r - / 2 ) = x 2 C say The quantities A, B, C are in this account called invariants of B 2 equation (2) ; the quantities A and -^ are absolute invariants. We shall write =^ = I. B 2 The second term in the new equation will be missing if e' = 0, /' = 0, i.e. if X = - e, \i = - x/. Equation (2) is thus reduced by the substitution x = xz, y = z _e e~ xte v to the form d 2 ?; /A 2xB 2r A . Hence, by a proper choice of x, we obtain the following canonical forms : I. B+0. C*0 ^ + (^ + ? + l) v = dz 2 \z 2 z / II. B,0. 0=0 g + (^ 2 > = III. B = 0, C^O g + ^ + i) w=0 IV. B=0, C=0 ^ + ^ = az 2 z J The values of the constants e, f, p, q, r for an equation derived from (1) by the method indicated are as follows : „ bl - am „, m . cl - an n 2*=— p— 2/=^ ^=0 2=—^- '«ji To reduce an equation of type (2) to the form (1), we must first of all transform it into another equation of type (2) with p' = 0, and to do this we must choose X so that X(X - 1) + 2eX + p = 0. Multiply- ing the resulting equation by x, we obtain an equation of type (1). The equation t-| + -^ + (1 --;)«/= dx 2 xdx \ x 2 J is one of the most important equations of type (2) ; it is usually called Bessel's equation. To reduce it to the form (1), we must * e is used here to denote the base for natural logarithms. TRANSFORMATIONS 77 choose X so that X 2 - w 2 = 0. An appropriate substitution is thus y = x n z, and the resulting equation is dH 2n + 1 dz dx z x dx The differential equation (3) is transformed by the substitution x = kx' + X y = ef"'+s«'y x ' ± . . (4) into an equation of the same type, but with new constants e', /', p' , q', r' given by the equations e' = x(e + /X) + fx /' = x 2 / + v p' = >&p + 2vx 2 / + v 2 , q' = y. 3 (q + \p) + fxx 2 / + vx(e + /X) + [xv r' = x 2 (#X 2 + 2qk + r) + 2fxx(pX + q) + fx 2 + v Let P = p - P Q = q - ef R = r - e 2 - / S = PR - Q 2 then it is easy to see that if P', Q', R', S' are defined in a similar way, PV 2 + 2QV + R' = x 2 (Px 3 + 2Qx + R). P' = x*P S' = x 6 S To reduce equation (3) to the normal form, we must choose X, u, v, x so that e' = 0, /' = 0. Dropping the dashes, we find that the new equation is ,72,, U + (Pz 2 + 2Q* + R)*/ = Now put x = ctz + (3 ; then by a proper choice of a and (3 we can obtain the following normal forms : I. P + g + ( * 2 W% = o i = g II. P = 0, S + d 2 j/ 1 /Scherk-Lobatto> di 2 ~ 3 Si/ V equation > II. P = 0, S = g + 2 /=0, ifR*0 P = 0, Q = 0, R = J?=0 or- Ln equation of type (3) can be reduced to an equation of type (1) by choosing v in the substitution (4) so that p' = 0. We shall now show that equation (1) can in all cases be reduced to the canonical form d 2 u . .du _ s d^ + ^- s) dS' au = where a and y are constants. This is Weiler's canonical form. 1°. If I + 0, m + 0, we write y = e kx r, I the new equation is then (a +lx)j£ + [2Xs + b + x(2U + m)] j± + [aX 2 + 6X + c + x(lk 2 + mX + «)]■/] = 78 DIFFERENTIAL EQUATIONS Let us choose X so that t>? + mk + n = ; if this equation has two unequal roots, the new differential equation is of the form {a + lx) d5 + {h + hx) £ + jri = ° The substitution a + Ix = v£ reduces this to the form rd\ ( Ih - ak kvAdy] jv _ *d%* \ Z 2 pVdt, Z 2 ^ ~ Choosing v so that kv = - Z 2 , we obtain an equation of the required type. If a, b, c, I, m, n, x, y are real, the quantities 5, f\ are certainly real when the roots of ZX 2 + mk + n = are real. When the equation 17? + mk + n = has equal roots, 2XZ + m = 0, and the equation for yj is of the form Putting a + Ix = v!; 2 , we get r (2h ,\dri 4yk - The substitution yj = e^w now gives K d 2 u (2h , _Wm TA 1 /4v£ 1\ „"| Choosing v so that 16v£ + Z 2 = and writing £, = - s, we obtain an equation of the required form. If a, b, c, Z, m, n, x, y are real, \ and y) are real when a + Ix am? - 2blm + 4cZ 2 is positive. 2°. Z = 0, m + 0. Put «/ = Y)£ ™; the equation for yj is then of the form «ft, l a4 + \ l + 2)T{ +( " y] where « = St = ~ ~ — + — 2k 2m 2m? 2m? Writing i; = - s, we obtain an equation of the required type. In this case, if a, b, c, m, n, x, y are real, s, yj, co are also real. TRANSFORMATIONS 79 3°. I = 0, m = 0. We may get rid of the second term by writing y = 7je 2a • the new equation is then ^J + (ft + fa)T] = where A = 1 -^ k = - a 4a 2 a Now write A + kx = v^ 7 where v is a constant ; the new equation is then ^ ld 4 V 3 where X = =--,. Making the further substitution v) = e rf w, where p is a constant, we obtain *Z? + fa + $% + [ A EXAMPLES. 1. Prove that the normal form of the equation taking \ and i\ as new variables. We easily find that g = *>-» [(p - l)tp - 2)7) + (p - 3)(tj - pi) | + ( ,-^)(!)\(,-^g TRANSFORMATIONS 81 When these expressions are substituted in the differential equation, the same power of x occurs as a factor of each term. Removing this factor, we are left with a differential equation whose order is one less than that of the original equation. If the original equation is homogeneous for all values of m and n, it must have the form »[»-£. -S.*3.~]-o where F is a homogeneous function of its arguments. If the equation is of the second order, it is transformed by the above substitution into a homogeneous equation of the first order, which may be solved by the method of § 21. If the equation is of the third order, it is transformed into a homogeneous equation of the second order of grade 1. When m = 0, the preceding method fails. In this case we make the substitution t uds y = eJ and take u as new dependent variable. The resulting equation is of order s — 1 when the original equation is of order s. A differ- ential equation which is homogeneous for all values of m and n may be treated by this method. EXAMPLES 1. Solve the equations x'y -.— ; = x * dx* ™S-»[-2-'] § 34. The principle of duality and the theory of contact transfor- mations. A relation between three variables x, y, p, such as

=/ 3 (X,Y,P) . (1) the transformed relation $(X, Y, P) = will be a differential equation of the first order at the same time as 9 (x, y, p) = if the relation dY - PdX = is a consequence of dy - p dx = 0. This condition is certainly satisfied if the transformation (1) is of such a nature that dY - FdX = p(dy - pdx) (2) where p is a quantity which is neither zero nor infinite except perhaps 82 DIFFERENTIAL EQUATIONS for certain exceptional values of x, y, p. Such a transformation is called a contact transformation. Since dy - p dx = d(y - px) + xdp we can evidently satisfy (2) by writing Y = px - y X =p P = x PX - Y = y . (3) Hence a differential equation q, r 1( r 2 , r are constants, the differential equation (2) is transformed into a similar equation with a different set of constants. In fact, since the tangent at a point (x, y) of any curve y is transformed by the substitution (3) into the tangent at (X, Y) to the corresponding curve T, we must have a relation of type Y - Y - g(X - X ) = v[y - 2/o " %(x - *„)] (4) where (X , Y ) corresponds to (x , y Q ) and \i is some function of x , y, x o> y<» which is different from zero. Substituting the values of X , Y in terms of x , y given by (3) and equating coefficients, we find that dx ) q 2 dX - Pi dY - r 2 {YdX - XdY) dy Pl dY - q x dX + 7Jf dX - X dY) ydx — x dy (5) pdY - qdX + r(YdX - XdY) Substituting for x, y, dx, dy in (2), a differential equation of the same type as (2) is obtained. When the substitution (3) is the same as (1), the differential equation (2) may be written in the f orm Ydx - Xdy = ydx - x dy TRANSFORMATIONS 85 and it follows at once from (5) that this is equivalent to an equation exactly like (2), but in the coordinates (X, Y). We shall now endeavour to choose the substitution (3) so that equation (2) is reduced to a simple form. It is clear that the sub- stitution (1), under which (2) is invariant, will be transformed into the substitution under which the new equation is invariant. Let us choose the transformation (3) so that the substitution corre- sponding to (1) is simply XX v _ yY X i — V x x *i-V where X, (x, v are constants ; then the new differential equation is [xYdX - XXriY - v(YdX - XdY) = or (H- - v ) Y~ + ( v »?-o The solution is thus of the form x^'Y" - * = const, or (p +p ± x +p$y- v {q + q ± x + q 2 y)"- K (r + r x x + r 2 y) K -» = const ... (6) To determine the constants p, q, r, etc., we must identify the P + Pi x i + Prf/i X p + p x x + p$ substitution r + r 1 x 1 + rrf/! q + q 1 x 1 + q s y 1 v r + r x x + r 2 y [i. q + q x x + q$ v r + r-jX + r 2 y r + r x x x + r^y x with (1). This gives p x {l x x + l& + I) + p?,{m x x + m 2 y + m) + p(n x x + n$ + n) = ®k(p x x + pgj + p) q x (l x x + 1$ +1) + q 2 (mjX + m$ + m) + q{n x x + n$ + n) = ^{q x x + q& + q) r x {l x x + 1$ + T) + r^m-p + m$ + m) + r{n x x + ngj + n) = ^{r-jX + r$ + r) where is some constant. The first equation gives p 1 (Z 1 - GX) + p 2 m 1 + pn x = Pih + Pa(™2 ~ 0>O + P n -i = ° pj + p z m + p(n - 6X) = hence OX must be a root of the equation in a, l x - 0. Taking logarithms of both sides of (6) and using the values p t = m 1 n i - m 2 n 1 + m x X p 2 = n-J> 2 - n i\ + n 2^ p = l 1 m 2 - l 2 m 1 - \(l ± + m 2 ) + X 2 we find that the coefficient of e on the left-hand side is ^ ~ v {n-.x + um + 2a - L - m 2 ) q +q x x + q& - log (q + qjX + q 2 y) + log (r + r x x + r 2 y) Equating this to a constant, we obtain the solution ±llL±m exp { ([X _ v) »x» + ~tff + 2ti. - t, - m, | = const + q x x + q^ K q + 1\X + q<& ) EXAMPLES. 1. Solve the equations (x + 1) dx - (3y - 1) dy - (5x - 9y + 6){ydx - xdy) = (x + 1) dx - (3y - 1) dy - (Ix - 9y + 8) (y dx - x dy) - 2. Prove that if the transformation (1) leaves the angle between any two lines unaltered, the curve C is an equiangular spiral. Consider the case in which the transformation consists of a rotation round a point 4 followed by a magnification about this point. After finishing this chapter the reader will find it worth while to try some of the miscellaneous examples at the end of the book. CHAPTER IV GEOMETRICAL APPLICATIONS § 36. The differential equation of a family of algebraic curves. Let us consider a family of algebraic curves denned by the equation /(*. V. c) = (1) wherein c is an arbitrary parameter. At a point (x, y) of one of these curves, the direction of the tangent is specified by the value of p = ~f-, and this is given by the equation £+»£-• < 2 > We shall suppose that /is a single- valued function of its arguments, so that the last equation determines just one value of p when x, y and c are given ; it fails to determine p, however, when ~ and -J- are both zero. dx d & If the parameter c be eliminated between (1) and (2), a differential equation of type ^{ x> y> p ) = (3) is obtained, and this is regarded as the differential equation of the family of curves represented by equation (1). If / is a polynomial of the n th degree in c, there will generally be n values of c when the point (x, y) is known ; consequently there are generally n real or imaginary curves of type (1) through a point (x, y), and so there are generally n real or imaginary values of p, some of which may be equal. We can then expect cp to be a poly- nomial of the w th degree in p, whose coefficients are rational func- tions of x and y. It is not true, however, that every differential equation of the n th degree with rational coefficients corresponds in this way to a family of algebraic curves ; for instance, the linear equation du corresponds to the family y = ce^, which is not algebraic. The differential equations of the n th order with rational co- efficients which correspond to families of algebraic curves are of a special type, but they are of considerable interest, because the 88 DIFFERENTIAL EQUATIONS differential equation can be used to define the family of curves, and certain loci connected with the family may be derived at once from the differential equation without finding its general solution. EXAMPLES. 1. Find the differential equation of the family of circles (x - e) 2 + y* = 1 We have the relation 2(x - c) + 2yp = ; /. 2/ 2 (l + P") = 1 2. Find the differential equations of the following families of curves : 2/ 2 - c 2 = x - c (y - c) 3 = x 2 y* = (x - c) 3 ic 2 w 2 + — - — =1 w 3 + c 3 x 3 = 3ca;« a + c p + c 3. Prove that the points of contact of the tangents from the point (x , y„) to the curves of the family is known the values of ~, ~, ... are determined uniquely by the equations dx dx dy dtydy 3? 3? dx p dx*dp + dx + dy P dtydj, 9*9 _S>_ ^p jpy 9 2 9 dx* dp + dx" + p dxdy + P dy* + dx" 3~xlp~ W/ dp" + P dx* dy~dp GEOMETRICAL APPLICATIONS 8» We are not justified, however, in jumping to the conclusion that the values of y, ^|, ^J, ... which are associated in this Way with the value of x are sufficient to define uniquely a solution of the differ- ential equation ; indeed, it is not certain from a priori considerations that they define a function at all. Points of this kind are usually settled by the so-called existence theorems for differential equations which are too difficult to be considered at this stage. At present we shall be concerned only with the case in which the equations (2> fail to determine the values of ~, -=-|, etc., when x, y and p are given. This occurs when -^- = in virtue of the value which has been given to p, that is, a value determined by the equation Now the equations 9=0, ^ = imply that for the chosen values of a; and y the equation

v) = ° there is generally no singular solution, and the ^-discriminant simply gives the locus of cusps of integral curves.* It is easy to see, in fact, that a singular solution only exists when the function 9 satisfies a certain condition, for if the p-discriminant satisfies the differential equation, the value of -^ for the eliminant of the equations y,p) = =2 = should be equal to one of the values of p given by 9 = 0. Now the value of -^ is given by the equation 9? tydy = Q dx dy dx hence, in order that a singular solution may exist, the function 9 must satisfy the equation ■ ¥ \z. V . -| =0 dyl *Oomptes Rendus, t. 70, (1870). Bulletin des Sciences MaMmatiques (1873), p. 158. GEOMETRICAL APPLICATIONS 91 This equation may be satisfied either identically or in virtue of P) = 0. If we eliminate p between the last equation and cp(x, y, p) =0 ; in other words, if we eliminate p from the equations dx +P dy =0 (x,y,p)=Q an equation is obtained which includes the singular solution when it exists. The equation also gives the locus of points of inflexion on integral curves, because the equation -? + « l$ = o is satisfied d 2 y dx r dy when ^ = 0. The first example of a singular solution was given by Leibnitz in 1694. Brook Taylor showed in 1715 that a singular solution can be deduced directly from the differential equation in the case of the equation 9(3. V> P) = 0- + % 2 )p z - ^xyp + y 2 - 1 = This equation is of Clairaut's type, and so its general solution is /(*, V, c) = (1 + x 2 )c 2 - 2xyc +y 2 -l=0 The ^-discriminant is (1 + x 2 )(y 2 - 1) - x 2 y 2 = or y 2 — x 2 = 1 Also =| = 2xp 2 - 2yp ^ = 2y - 2xp and so -^ + p ^ = ox * dy Hence y 2 - x 2 = 1 is a solution of the differential equation, and it cannot be derived from f(x, y, c) = by giving a special value to c ; consequently it is a singular solution. EXAMPLES. 1. Find the singular solutions of the following differential equations : a 2 [x + y + p(y - a;)] 2 = (xp - y) 2 (x* + y*) p 2 (2x* + 1) + p(x* + 2xy + y 2 + 2) + 2if + 1 = p 3 - 4xyp + 8y 2 = Show that in the last example y = is included in the general integral and is also a true ' singular solution ' in the sense that it is included in the envelope (Cauchy). 2. If a differential equation 3 <£= — + P— = 'BX + BY 92 DIFFERENTIAL EQUATIONS § 38. The c-discriminant. The envelope of a family of integral curves fix, y, c) = may also be obtained by forming the c-dis- criminant, i.e. by eliminating c from the equations f(x, y, c) « ^f(x, y, c) = . . . (1) To see this, let us suppose that (x, y) is a point for which the two equations can be satisfied, and that an appropriate value of c, depending on x and y, has been derived from these equations. Call this value of c, c(x, y) ; then, if we calculate -^ from the equation f[x, y, c{x, y)]=0 (2) we et df_ d[dy -discriminants is shown in the following table : c-discriminant. Envelope 1 Particular curve 1 Node-locus 2 Cusp-locus 3 p-discriminant. Envelope 1 Cusp-locus 1 Tac-locus 2 Particular curve 3 EXAMPLES. 1. Let us consider the equation y y - x=\, p = 1 ; hence y - x = ^ is a singular solution. 96 DIFFERENTIAL EQUATIONS 2. In the case of the equation xp 2 - (x- l) 2 = the jj-discriminant is x(x - l) 2 = 0. A family of integral curves is given by the equation 9^ + c y _ 43.(3. _ 3)2 and the c-discriminant is x(x - 3) 2 = 0. In this case x = is the envelope or singular solution, x = 1 a tac-locus and x = 3 the node- locus. 3. Discuss the p- and c-discriminants of the equations 2/2(1 + p 2 ) = 1 j/ 2 + (a; - c) 2 = 1 (x + yp) 2 + 4{y - xpY - o 2 (l + p 2 ){x 2 + y 2 ) 2 2 2 1 (x + ycy + (y - xcY = 0^(1 + c 2 Y (3i/ + 2) 2 £> 2 = 4(1 + y) (x - c) 2 = y 2 + y 3 The solution of each equation is given in the same line. 4. Prove that in the case of the equation (2a; - p) 2 + x(y - x 2 ) 2 (2x - p) + (y - x 2 ) 3 = the particular curve y = x 2 appears as a factor of the p-discriminant. Taking y(\ + cx j _ x i + cx 3 + c a as a family of integral curves, show that y - x 2 = belongs to this family. What is the geometrical meaning of the factor x 2 - 4 of the ■c-discriminant ? (Petrovitch.) d 2 f 5. If =-j = at all points of the node-locus of the family of curves j(x, y, c) = 0, the node-locus N is also an envelope and N' a factor of d 2 f the discriminant. If —^ = at all points of a cusp-locus C, this is an envelope and C* is a factor of the discriminant. (M. J. M. Hill.) 6. Prove by the principle of duality, or otherwise, that the result of eliminating p from the equations 2 {x, y), ... p = (];„(*, y) . (4) can be put respectively in the forms 9ii x - y) = c i 9i{ x > y) = % ••■ 9n{ x > y) = c » the result obtained by substituting the values of c lt c 2 , ... c„ in the equation (c _ c ^ c _ ^ ... ( c _ «g = can be regarded as a primitive of (1). We thus obtain the solution (c - !7i)(e - <7 2 ) .- (c - sr„) = . . . . (5) The jj-discriminant of equation (1) is (+i - s implies that a curve of type g r = c r has the same tangent as a curve of type g 3 = c s . Notice that the factor (<\i r - t^s) appears twice in the discriminant, as it should do. The c-discriminant of equation (5) is (ffi ~ 9z) 2 (9x ~ 9 S ) S ... (<7i ~ 9n) 2 (g* ~ 9z)* - (ft " 9nY - (9n-l ~ 9«Y = and represents a locus of double points of curves represented by (5) ; it is, in fact, the locus of points where a curve of type g r = c crosses a curve of type g 3 = c. If a curve of type g r = c has double points for all values of c, the locus of these double points is apparently not given by the c-discriminant, when it is written in the above form. EXAMPLES. 1. The differential equation (dy\ \dxj 2 . 1 = sin 2 - X is satisfied by -I sin - dt + c x t y= - r • i , 1 sin - at + c t and y PI • l = I sin - Jol * dt + c 3 where c 1 , c 2 , c 3 are arbitrary constants. (O. Perron.) 2. Solve the equations : p 2 2/ 2 - x" = pW - Spxy + 2y 2 = p 2 + p(x + y) + xy = §41. Orthogonal trajectories. Let cp(x, y, p) = be the differential equation of a family of curves f(x, y, c) = 0, which is such that n curves pass through each point (x, y). We shall suppose that cp is a polynomial of the n th degree in p, and that there is usually one value of p for each curve through (x, y). Let us now consider the problem of finding a second family of curves g(x, y,a) =0 defined by a differential equation V—(s) ! dQ_ I _1_ _ 1_ 0r d~r~'\b~ 2 r 2 The variables are now separated, and the solution is given by e=a+ rgv^^" 2 In this example the first system of curves consists of the tangents to the circle x 2 + y 2 = b 2 , while the orthogonal trajectories are involutes of this circle. It should be noticed that in this case the curves of the first family have a real envelope, which is the locus of points at which the curves of the second family come to a full stop. 100 DIFFERENTIAL EQUATIONS 2. If the differential equation of a family of curves in polar coordi- nates is fdfy «. f I — , r, 1 = the differential equation of the orthogonal trajectories is f I , r, 8 I = 3. If m + m> = F(a; + it/), where i = V - 1, prove that the curves t> = const, are the orthogonal trajectories of the curves u = const. 4. Find the differential equation satisfied by the family of confocal conies a 2 -+-^-=1 c c - X where X is a constant, and show that the orthogonal trajectories also belong to this family. 3" /1\ 5. If r* = x a + y 2 , prove that the curves =— - = c are cut ortho- d n+i r 3x«\rJ gonally by the curves . n+1 — a. (Stefan.) 6. Find the orthogonal trajectories of a system of coaxial parabolas of latus rectum 4a. 7. Find the orthogonal trajectories of a system of similar coaxial ellipses. § 42. Orthogonal trajectories of a family of circles* Let the family of circles be given by the equation \x-l{cW+[y -v)(c)P = [p(c)P. ... (1) where c is the variable parameter. Since the tangent at a point of an orthogonal trajectory passes through the centre of the corre- sponding circle through this point, the differential equation of the orthogonal trajectories is dx = d y , 2) x - £(c) y - y)(c) where c is given by (1). Let us now write 1 — t 2 2t X = 1(C) + P (C) j-j-^ V = V)W + P(C) Y^fi ( 3 ) and take c and t as new variables. Differentiating, we find that — - p , 1 - < 2 dt U dc ~ ^ + p 1 + t* P dc (1 + ty dy _ , , 2t dt 2(1 - P) dc ~ r > + p 1 +fi + p dc (1 + *«)* * See Darboux, TMorie des Surfaces, t. 1, p. 113 ; L. Ballif, L'Enseignement mathimalique (1915), p. 215 ; C. Cailler, Ibid. p. 223. GEOMETRICAL APPLICATIONS 101 Multiply the first of these equations by y - yj, the second by x - £, and subtract ; then, after making use of (2) and (3), we get P (c)|;=e(c)«l-K(c)[i -< 2 ] This is a particular case of Riccati's equation : consequently we may conclude that the cross ratio of the four values of t for four orthogonal trajectories is independent of c. The geometrical mean- ing of t is that it is the tangent of the angle which the line joining the two points {xy) ( ? _ p , ,,) makes with the axis of x. When c is given, these points are both on the circle (1), and the latter is independent of t ; hence the cross ratio of four values of t is equal to the cross ratio of the four corre- sponding points on the circle, by which we mean the cross ratio of the four lines joining these points to an arbitrary point on the circumference. Hence we have the following theorem : Each circle of the system (1) is cut in the same cross ratio by four orthogonal tra- jectories. EXAMPLES. 1. Prove that the orthogonal trajectories of the family of circles x 1 + y 2 - 2cx = b 2 , where 6 is a constant, are given by the equations _ <(1 - a 2 b 2 ) _ t 2 + a 2 b 2 X ~ o(l + t 2 ) V ~ a{\ + t 2 ) 1 + a 2 b 2 or by the equation x* + y 2 - 2ky + b 2 = 0, where 2fc = 2. Prove that the orthogonal trajectories of the family of spheres [x - 5(c)] 2 +[y - 71(c)] 2 +[z- C(c)] 2 = [p(c)]* are obtained by writing , , u + v _. . . , . v - u uv - 1 * =P(C) TT^ +?(C) 2/=tp(c) ITW +Yl(c) z=p(c) TTW +f(c) where i = \/ - 1, and u and v satisfy the Riccatian equations ^=?(5'-*y)--S'-^;'W) do 2p p *p dv v 2 ,_. . „ v „ 1 , r , . ,. respectively. (Darboux.) 3. Find the orthogonal trajectories of the family of circles which touch two given straight lines. 4. Prove that when the orthogonal trajectories are known for the 102 DIFFERENTIAL EQUATIONS system of circles (1), they can also be found for a system in which 5(c), r)(c) and p(c) are replaced byj£ (e), '"]o( c )> Po( c )- w h ere 5o'(c) = /< c )5'(c) Y) '(o) = /(c)r)'(c) Po (c) = /(c)p(c) /(c) being an arbitrary function. (V. Rouquet.) 5. Prove that the problem of finding the orthogonal trajectories of a family of circles can be reduced to that of finding the orthogonal trajec- tories of a family of equal circles. (L. Ballif.) 6. Find a curve T on which it is necessary to roll a line L in order that a point of a plane rigidly attached to L may describe a given curve C. Prove that T may be regarded as the evolute of an orthogonal trajectory of a family of equal circles whose centres lie on C. (L. Ballif.) CHAPTER V DIFFERENTIAL EQUATIONS WITH PARTICULAR SOLUTIONS OF A SPECIFIED TYPE § 43. Euler's equation. Since there is no general method of solving differential equations which is always applicable, and since there are many types of differential equations whose integrals cannot be expressed in a simple manner in terms of known functions, it is useful to have a list of equations whose solutions can be ex- pressed in a simple form. The natural method of forming such a list is to start with a given type of primitive and find a corresponding differential equation. If we start with a bilinear relation between x and y, axy + bx + cy + d = (1) in which a, b, c, d are constants, we obtain (ax + c) ^- + ay + b = (2) and so a particular differential equation corresponding to the given primitive is dx i dy _ ax + c ay + b an equation in which the variables are separated. The complete solution is given by (1), wherein d is the variable parameter. If a, b, c, d are linear functions of a parameter X, so that a = a + a x X b = b + b^k c = c + c x X d = d + d x X the differential equation obtained by eliminating X from (1) and M is , dy , , * dy , (a

i)( c o«/ + d o) 104 DIFFERENTIAL EQUATIONS The general solution of (4) is furnished hy (1), wherein X is the variable parameter. It is possible, moreover, to choose the con- stants a , b , c , d , a lt b x , c ls d x so that ¥(x) and G(y) are arbitrarily assigned quadratic functions with equal Hessians. Notice that F(a;) = G{x) if b = c. Next let us consider a biquadratic relation between x and y. X 0? / 2 + 2X lV + X 2 = (5) in which X = a x z + 2a,jX + a 2 X x = b x 2 + 2b x x + b 2 -A. ft == ^Vp^ ' £G"]p£ "T ^2 and a , a x , a 2 , b , b lt 6 2 , c , c lt c 2 are constants depending on a para- meter X. We find that (X y + XJ-£ + {a x + ajy* + 2{b x + bjy + c B x + c x = that is (X y + XJ ^ + Y x + Y x = . . . . (6) where Y = a «/ 2 + 2b y + c Y x = 0,$* + 2b x y + Ci Y 2 = a$ 2 + 2b$ + c 2 Squaring (6) and noticing that (5) can also be written in the form Y z 2 + 2Yia; + Y 2 = we obtain the relation (X x « - X X 2 )(|) 2 = Y> 2 - Y TJ, Hence, if F(z) = X x 2 - X X 2 G(y) = Y x 2 - Y Y 2 the relation (5) leads to the differential equation dx du . ,_. —t= + . * = 7) vw) vw) It should be noticed that if b = a lt c = a 2 , c 1 = 6 2 , F(x) = G(x), and the differential equation is symmetrical in x and y. In order that the relation (5) may represent a general solution of equation (7), the differential equation must be independent of X after it has been multiplied throughout by an appropriate function of X. We must, therefore, make the constants a , b x , etc., depend on X in such a way that X, 2 - X X 2 = /(^(X) Y, 2 - Y Y 2 = g(y) 9 (X) where / and g are independent of X. If the relation between x and y is a symmetrical one, it is only necessary to satisfy one of these SOLUTIONS OF A SPECIFIED TYPE 105 equations, and then the other is a natural consequence. Let us assume that Rx) = px * + qxS + rx 2 + sx + t then we have to satisfy the equations V - «o c o = P When X^Owe must write a, b, h + X, / + \v, g + Aw, c + 27Jc for a, b, h, f, g, c respectively, but the differential equation is un- altered for 2U{v + x) 2 - 2\v(v + x)(ux + k) + 2ku(v + x) 2 x - 2~hx(v + x)(ux + k) = Hence, if f(x) and g(y) are defined by (21) and (22), the solution of the differential equation (23) is given by (20). If we arrange (20) according to powers of X, it takes the form (be -f 2 )(u+ y) 2 + (ac - g 2 )(v + x) 2 + [ab - h?)(k - xy) 2 + 2(gh - af)(v + x){k - xy) + 2(hf - bg)(u + y)(k - xy) + 2{fg - ch){u + y)(v + x) + 2\[(bk - fv)(u + y) 2 + (ak - gu)(v + x) 2 - h(k - xy) 2 + (hu - av + g)(v + x)(k + xy) + (vh - bu + f)(u + y)(k - xy) + (uf + vg - c - 2hk) (u + y)(v + x)] - 7?(xy + ux + vy + k) 2 = EXAMPLES. 1. Show that the relation (x a - i/ 2 ) 2 - 2X 2 [a; 2 + i/ 2 - 2(1 + jfe 2 )zy + kWy^x* + y*)] + X 4 (l - k*x t y 2 )* = is a solution of the differential equation dx dv , + . V(l - x*)(l - k'x*) V 1( - 2/ 2 )(l - fry) Show also that the solution can be expressed in the form xVl - yWl - k'y 2 + yVl - a 2 Vl - k 2 x 2 = X(l - k 2 x 2 y 2 ) 2. If the function f(u) is defined by the property that x = @{u) when u = I (4i 3 - g t t - gjf^dt (g 2 , g 3 constants) show that p(« + v) + P («) + f(v) =l [ P ^:l'^ ] 110 DIFFERENTIAL EQUATIONS 3. If cp (x), *\i(x) are polynomials of the second degree, the equation dx dy \/(y) is satisfied by vv lrKyl ' vi/; ; v ' = const. (Laguerre.) 4. The equation dx ^ dy Va 4 + 2aa; 2 + 1 Vy* + 2ay* + 1 is satisfied by v — J - + -Hf — + — = 1 + X 1 - X a - X Prove that if x and y are regarded as the parameters of the isotropic generators of the sphere 5 2 + tj 2 + £ 2 = 1, so that r . xj/ + 1 xy - 1 „ re + 2/ 5 = * ?) = (, = x - y x - y x - y the integral curves of the differential equation are confocal spher< >- conies. (E. Turriere.) 5. Prove that equation (23) can be made symmetrical in x and y l by the substitution yyi + uy 1 + vy + k= Notice that this is determined by the circle (15). 6. When Euler's differential equation can be reduced to the sym- metrical form this can generally be done in four different ways. 7. Prove that when Euler's differential equation is rationalised the p-discriminant and the X-discriminant both give the singular solution f(x)g(y) = 0, which represents the envelope of the family of confocal bicircular quartics. § 44. Linear differential equations with simple solutions. No method of expressing the solution of the general linear differential equation of the second order in finite terms has yet been discovered. We shall therefore give a brief outline of the properties of some well-known equations which possess particular solutions of a simple type, for we have seen that when a particular solution is known the general solution can be found by integration (§ 30). If a, b, X, [jt, are constants, the function y = (x - a) x {x - by satisfies a number of linear differential equations of the second order of which (x - a){x - 6)g - [(X - l)(x - 6) + (fx - l)(x - a)] d £ - (X + y.)y = (1) SOLUTIONS OF A SPECIFIED TYPE 111 is the most interesting. Differentiating this equation v times with regard to x and using Leibnitz's theorem, we obtain (x _ a)(x - 6)g? - [(X - v - l)(x - b) + (y. - v - l)(x - o)]g + [v 2 - v(X + (i. - 1) - X - (x]« = . (2) where u = — = — (a; - a) K (x - bY dx" dx" In particular, if a = 0, b = 1, we find that the equation *(* ~ *)£ + [> + * -X-(2v + 2-X-^]g - (v + l)(v - X - ti)w = . . . . (3) is satisfied by the function <*" JW — x K (l - x\ dx" v ' The differential equation u = -=— rsf- y (l - xy-"- 1 . . (5) * (1 " * } S + [Y ~ (a + P + ^£ - «P» « . (4) is called the hypergeometric equation, it becomes identical with (3) when a=v +l (3 = v - X - (i y = v + 1 - X We may conclude then that, if a is a positive integer, the hyper- geometric equation is satisfied by the function dT 1 ' dx* If we put u = x l ~ y v, equation (4) becomes - (a - y + 1)@ - y + 1)« = This is also a hypergeometric equation, and is consequently satisfied by j«-y when a - y + 1 is a positive integer. Hence it follows that under the same condition equation (4) is satisfied by M=a; i-vi!2 a; -i(i - x) -fi .... (6) dx°- y If \x\ < 1, we can expand (1 - x)' ? by the binomial theorem and differentiate the product term by term ; if then - y is not a negative integer, we obtain u = (a - l)(a - 2) - Y[l + j^* + ! . 2 y( y + 1) x + 112 DIFFERENTIAL EQUATIONS The series within square brackets is called the hypergeometric Junction and is usually denoted by the symbol F(«, p, y, x). It will be shown in Chapter IX that the series representing F(a, p, y, *) satisfies equation (4) for unrestricted values of a, p, y. When the expression (5) is evaluated in a similar way, we find that , -i dx a ~ l = (a -y)(a -y - 1)... (2 - y) . x 1 ^ x F(a - y + 1, p - y + 1, 2 - y, x) hence x l ' y F(a - y + 1, P - y + 1, 2 - y, «) is a solution of the hypergeometric equation. Similarly, by making the substitution u = (1 - a;) 7 """^, we can prove that when y - a is a positive integer u = (1 - a)*" "-' J^r *-(l - as)"" 1 • • (7) is a solution of (4), and that when 1 - a. is a positive integer u = £-in _ x y—* *_1 x y— \\ _ x f-y . . (g) is a solution. Hence it follows that in these cases x 1 -^! - zy— p V{l - a, 1 - p, 2 - y, x) and (1 - xy~ a ^ F(y - a, y - p, y, x) are solutions of the hypergeometric equation (4). It is clear that a and p can be interchanged in any of the above results ; hence we may conclude that a solution of equation (4) can be expressed by means of a formula analogous to one of those already given if any one of the quantities «, p, 1 - a, 1 - p, y - a, y-p, a-y + 1, p-y + lisa positive integer. If in equation (4) we put * x = - and make p -> oo , we obtain the equation P d 2 u , . du /n . S ^ + fr- S >^- aM= ° ■ ■ • • (9) which has already been chosen as the canonical form of an equation of Laplace's type. Since Lim (i -ff"-* P^V 1 pj _e -we may make the following deductions from the preceding results. * This device was used by Kummer in his well-known paper on the hypergeometric series, Crelle's Journal, Bd. 15 (1836), pp. 39-83, 127-172. See especially p. 138. It will be unnecessary to make a rigorous examination of the limiting processes, as the results can easily be verified. SOLUTIONS OF A SPECIFIED TYPE 113 1°. If a is a positive integer, a solution of (9) is given by U = I^ ll>a ~ V] (10) 2°. If a - y + 1 is a positive integer, a solution of (9) is given by « = ^J^-V] (li) 3°. If y - oc is a positive integer, a solution of (9) is given by 7Y-a-l u = e'-^^ls-e-'] (12) 4°. If 1 - a is a positive integer, a solution of (9) is given by u = s 1 -v e '' — \_s*- a - V s ] (13) The limiting form of the series oJ3s «(q + l)p(p + 1) fay 1 . y P 1 . 2 y (y + 1) \p/ '" ■n/ \ ■■ a a ( a + 1) ., /-i^\ g (*'T-«>° 1+ r^* + i.2 Y ( Y + i) * + -- (14) It will be shown in Chapter IX. that this series satisfies equation (9) for unrestricted values of a and y- It should be noticed that if 1 - a is a positive integer the series terminates ; it may also be deduced from 4° that in this case the equation possesses a solution which is a polynomial of degree - a in s. If in (2) we write a = 1, b = - 1, X = (A = v = w, we obtain Legendre's equation ^-^^- 2x tx + n{n + 1)u ^° ■ ■ (15) which is satisfied by the function u = i% l{x2 ~ m (16) when n is a positive integer. This function is usually denoted by 2" n\ P„(aj), and is a polynomial of the n tb degree. Let us next consider Bessel's equation *"S + x di + ( * 2 ~ n * )y = ° • ■ • ( 17 ) which is reduced to the form X d* + {2n + l) dx +XU = °- ■ ■ ■ (18) ■ by the substitution y = x n u. To reduce this to Weiler's canonical form for an equation of Laplace's type, we write u = ve^ x = |i£ where i = V - 1 B.E. H 114 DIFFERENTIAL EQUATIONS > The new equation is 5g + (2» + l-5)g-(»+«f=0 . . (19) and is of the type (9) if a = n + \, y = 2n + 1. It follows then from 1° and 3° that if m + | is a positive integer, the differential equation is satisfied by the two expressions d m " = ^LT" 1 - 1 ^] • • • • (20) » = ^[r m -v*]. . . (2i) where m = w - J. We thus find that Bessel's equation is satisfied (aT TO -V**) ■ • (22) by the function 7„, ^ -•- _ a , -m-l -2w\ dx n Let us write K^+iC*) = (- lf2-'%/|e* ^+* *l (a r»-V to ) . (23) when m is a positive integer ; then, if n = m + \, we have the result that Bessel's equation is satisfied by the functions JL, n+i (ix) K m+i { - w) A solution of Bessel's equation for this case can also be obtained in the following way. If we write x = Vs in (18), the equation becomes dhl du \ . .... S l? + ^ n + 1 ^s + l u = ■ ■ ■ (24) Differentiating this equation m times with regard to s and putting v = -^ — , we find that v satisfies the equation ds m ^ (11) fll) 1 a j-s + (n + m + l)^- + 7 d = . . . . (25) as 2 v ' ds 4 which is of the same type as (24). Now put n = - \, then (18) d 2 u becomes -^— + u = and is satisfied by u = cos a; and w = sin x ; hence it follows that with this value of n equation (24) is satisfied by cos Vs and sin Vs, and (25) by d m — d m ^(cos V*) and ^(sin Vs) This means that the last two expressions are solutions of the equation ^ n «fo 1 A SOLUTIONS OF A SPECIFIED TYP.E 115 consequently, if n + f is a positive integer, two solutions of Bessel's equation are dm gm a! "^ (c0sVs) and zn d^( sinVs ) where m = n + f , s = x 2 . In the usual notation 2 cZ m+1 J m+i (x) = ( - l) m+1 ^(2x)™+i^^(cos Vs) 2 d"^ 1 ¥„+»(*) -( - l) ro+1 ;^(2zr + *^(sin V~s) When w = m + |, the functions J n (x) and Y„(a;) are independent solutions of Bessel's equation. When n = \, we have the two independent solutions cos x s j n x — y=r- and — r=- Va; Va; EXAMPLES. 1. Prove that V n (x) = f(»+ 1, - n, 1, — *A 2. If o is a constant and p + 2 a positive integer, the equation d 2 u p{p + 1) is satisfied by u^^(l^(^±^y (Gaskin.) 3. Prove that the roots of the polynomial V n (x) are all real and lie in the interval ( - 1, +1). 4. If 1 - a is a positive integer, the roots of the polynomial F(a, y, s) are all real and positive. (Gegenbauer) . 5. Prove that if m + 1 is a positive integer, 1 d m+1 ds m+1 where s = x*. 6. Prove that the equation (x - af(x - by j^ + cy= is satisfied by y = A(x - a) m {x - b) n + B{x - a) n (x - b) m where A and B are arbitrary constants and m and n are the roots of the equation fi p2 - p + (^T6Ji=° < Stokes -> CHAPTER VI PARTIAL] DIFFERENTIAL EQUATIONS § 45. Functional determinants and their properties. Before we can commence a treatment of partial differential equations a preliminary theorem is needed which will enable us to ascertain when n functions u lt u 2 , ... u n of the n independent variables x v x 2 , ... x n are con- nected by a relation of type F(m 1( u 2 , ... u„) =0 ....... (1) which does not involve x v x 2 , ... x n explicitly. If such a relation exists, we have the n relations 9F du x 9«j dx x 9F 3m 2 H 9w 2 dx x + . 9F du n ' du n dx x = 9F 9wj du j dx 2 9F 9m 2 9m 2 Sx 2 + . 9F du n " du n dx 2 = 9F du x du j dx n 9F du 2 du 2 dx n + . 9F du n " du n dx n = which are obtained by differentiating (1) with regard to x v x 2 , ... x„ respectively. If now we eliminate ■=— , ■=— , ... -= — from these equa- tions, we find that 9m j du 2 du„ dx x dx x dx x 9m j 9m 2 du n dx 2 dx 2 ' dx 2 9m j du 2 du n dx n dx n ' dx n 9m j' 9m 2 9m„ = Let us denote the determinant on the left-hand side by the symbol J, and call it the Jacobian or functional determinant of the PARTIAL DIFFERENTIAL EQUATIONS 117 functions u lt w 2 , ... u n with respect to the variables x x , x 2 , ... x n . It will be convenient to use the notation _ _ d(u v u 2 , ... u n ) "( x lr x 2> ■■■ x n) The vanishing of J is seen to be a necessary condition for the existence of a relation of type (1) ; we must next show that it is a sufficient condition. The following property of the functional deter- minant is needed for this purpose. // J is the Jacobian of u^ u 2 , ... u„ when considered as functions of x 1( x 2 , ... x„ and J 1 is the Jacobian of x lf x 2 , ... x„ when these quantities are considered as functions of the n independent variables Yk v 2> ••• y»> then JJ X is the Jacobian of u x , u 2 , ... u„ when these quantities are considered as functions of y lF y 2 , ... y„. In other words > d{u lt u 2 , ... u n ) d(x t , x. lt ... x n ) = d{u lt u 2 , ... u n ) d{x lt x 2 , ... x n ) * 3(2/!, y 2 , ... y n ) d(y lt y 2 , ... y n ) This is proved by multiplying the first two determinants together by the ordinary rule and making use of the n 2 relations of type du T _ du r dx 1 du r dx 2 du T dx„ In particular, we have 3(2/i. V*, - y«) 9(^i. x 2> ■■■ x n) 1. . . . (4) d{x v x z , ... x n ) d{yi,y 2 ,...y^ Let us now suppose that the Jacobian J of the n quantities u lt u 2 , ... u n with respect to x lt x 2 , ... x n vanishes identically, we shall endeavour to prove that in consequence of this the quantities u t , u 2 , ... u„ are connected by a relation of type (1). If the n - 1 functions u lt u 2 , ... u n _ 1 be not independent of one another, then the proposition to be proved is at once granted ; we may therefore suppose them to be independent of one another. Between the n equations expressing the n functions u we can eliminate * n - 1 of the variables ; if the remaining variable, say x n , be not eliminated thereby, the result may be written in the form u„ = u 2> - w »-2. x n) (8) we could deduce from (7) and (8) that there was a relation of type (6), and this possibility has already been considered. Proceeding in this way step by step we find eventually that either there is a relation between u v u 2 , ... u n or ^-1 = 0. Arrang- ox x ing the w's and x's in different orders and proceeding as before, we see that the alternative to there being a relation between u v w 2> ... u n is that all the quantities of type -=-£ should vanish. This, however, ox s is impossible unless all the w's are constants ; hence there must be a relation between u t , u 2 , ... u n . Now consider a system of equations m x = a v u 2 = a 2 , ... u„ = a„ . . . . (9) where the a's are constants. If the Jacobian of the w's vanishes identically, we can say that the equations are inconsistent with one another unless the a's satisfy a certain relation, and when this PARTIAL DIFFERENTIAL EQUATIONS 119 Telation is satisfied the equations are not independent. To prove this, we remark that since J = there is a relation of the form F(w 1; u 2 , ... u n ) = •consequently the equations (9) can only be satisfied simultaneously when the condition F(a lt a 2 , ... a„) = is satisfied. If the relation between the u's is written in the form the condition is simply /(0, 0, ... 0) = ; hence it follows that when the condition is satisfied, the relation u n = a n is a consequence of the other relations u 1 = a x , u 2 = a 2 , ... u n _ 1 = a n _ 1 . Note. If the equation J = is not satisfied identically, but only in virtue of some or all of the equations u 1 = a lt u 2 = a 2 , ... u n = a n , we cannot infer that there is a relation between w 1( u 2 , ... u n , and so the preceding conclusions are no longer valid. Example. If u x = x£ + x 2 2 - 1, u 2 = a; 1 cosa + a; 2 sina - 1, -where a is a constant, we have T du 1 du 2 du x du 2 . . J = ^-i -^ -3—3-^ = 2 («! sin a - x 2 cos a) dx 1 ox 2 ax 2 ox 1 In this case J is not identically zero, but it is zero when u x = 0, it 2 = 0, for (x x sin a - x 2 cos a) 2 = x x * + x 2 2 - (x x cos a + x 2 sin a) 2 = u x + 1 - (u 2 + l) 2 We cannot, however, conclude that the equation u 2 = is a -consequence of u x = 0. §46. Solution of the equation J = 1. Gauss has shown that two functions u, v satisfying the equation d(u, v) d(x, y) may be obtained by writing 1 l?(x, u) y=- If { x, u) -where F is an arbitrary function, and solving for u and v. These equations show, in fact, that vdu - ydx is an exact differential ; hence / g M \ g M \ V Yx- y ) dx+V Ty dy is also an exact differential, and the condition dx\ dyJ dy\ dx y ) -which must consequently be satisfied takes the required form du dv du dv _ dx dy dy dx 120 DIFFERENTIAL EQUATIONS If [u, v), (x, y) are regarded as the rectangular coordinates of two corresponding points P, Q in two planes W, Z respectively, and this correspondence can be defined in such a way that just one real point Q corresponds to a given real point P, then as P describes a closed curve C, Q will describe a closed curve T, provided F is a single- valued function of x and u. The area of the curve T will, moreover, be equal to the area of the curve C if each of them is a simple oval. To prove this we remark that the areas of C and T are represented by the integrals f , f , \vdu \y ax taken round C and V respectively. Their difference is equal to \aT and this vanishes, for when P is given Q is determined uniquely ; consequently x and u are determined uniquely, and so F returns to its initial value after P and Q have made complete circuits round the curves C, F respectively. It should be remarked that when J = 1 it follows that vdu - ydx is an exact differential d¥, and so the solution of the equation J = 1 necessarily has the form given above. The more general equation 8 K v ) _ f(x u) d(x,y) -t {X - y) may be satisfied in a similar way by writing 2 r ?i v = -faFi*. u ) \f( x . y) d v = - g^fo M ) and solving for u and v. EXAMPLES. 1. A correspondence is established between two points P and Q by- making a tangent from P to an oval curve C equal and parallel to a tangent from Q to an oval curve T. Prove that in this correspondence areas are unaltered. Show also that the correspondence can be made one to one by supposing that each oval curve is described by a point in a definite direction and considering only one-half of the tangent at each point, namely, that part which would be described by the moving point if it left the curve. o c i a j.- 8 (w, v > w ) ft \ 2. Solve the equation ^ = f(x, y, z) d{x, y, z) 3. A point which moves in such a way that the quantities f(x, y, z, t), g(x, y, z, t) and h(x, y, z, t) remain constant will always lie on the same moving surface of the family k(x, y, z, t) — const, if the Jacobian d(f, g, h, h) d{x, y, z, t) PARTIAL DIFFERENTIAL EQUATIONS 121 vanishes for each position of the moving point. In this example x, y, z denote the co-ordinates of a point and t the time. 4. A family of moving curves given by the equations f(x, y, z, t) = a, g(x, y, z, t) = b, where a and b are constants which vary from curve to curve, is such that each member continually intersects each one of a singly infinite set of moving curves belonging to a second family 9 (as, y, «,*) = « x{x, y, z, t) = (3 Prove that the Jacobian 3 (/> 9> 9. y> z > P. which must be satisfied either identically or in virtue of (1). If a is constant, equations (4) can be replaced by the single relation I=° < 8 ') Eliminating a and b from equations (1) and (8) with the aid of the relation b = <\>{a), we obtain a relation G(x, y, z) = (9) which can be regarded as a possible primitive of the partial dif- ferential equation (3). If we write the relation between o and b in the more general form F(a, b) = . ... (10) we find from (4) that d }?' Y! = (11) v ' o{a, b) and the equation G(x, y, z) = may be regarded as the result of eliminating a and b from equations (1), (10), (11). To distinguish between the different types of primitives we shall call (1) a complete integral, (7) a singular integral and (9) a general integral. When a rule for finding all possible primitives of a partial dif- ferential equation has been found, we shall say that the complete solution has been obtained. The solution given by equation (7) is called a singular integral for a reason which will appear later ; it " should be noticed that it does not contain any arbitrary constants or arbitrary functions. PARTIAL DIFFERENTIAL EQUATIONS 123 In this book we shall discuss only those methods of solution which are applicable to partial differential equations of the first order or to linear partial differential equations of the second order. It will be instructive, however, to give some examples on the for- mation of partial differential equations in which the equations are not necessarily of the types just mentioned. Some examples illus- trating the derivation of systems of partial differential equations are also included. Sometimes the partial differential equation of a family of surfaces defined by some geometrical property is required. To obtain the required equation a certain amount of geometrical knowledge is frequently necessary ; in this book only a few simple cases will be considered. As an illustration let us find the partial differential equation of the family of all developable surfaces. We shall define a developable surface as the envelope of a singly infinite system of planes, that is, a system in which the coordinates of each member are functions of a single parameter. Let the equation of one of the planes be \x + 7)«/ + *z + t = (12) where £, vj, £, t are functions of the parameter a. The envelope of the system is obtained by eliminating a between this equation and its derivative with regard to a, viz. 3£ dr\ 3C, 3t a /10N &' + £ y + T* Z + 3a ' ° • • ■ • (13) Let us write Bz _ dz _ cPz _ _S%_ = - 3R sV + *U + pT = -tr- oy ^Y+~U+iS+rT=^- (Massau.) da: oj/ ox dy 8. Prove that the partial differential equations of the last example are equivalent to the two equations 1 dw 1 dw - ■=- = - v- = « - WW y> da; g dy ~i a« 1 dv~ _g 82/ p dx_ _ \du 1 3w g dy p dx u-= — (Lecornu.) pq 9. Prove that the function V = F((o) it may happen indeed that the equations / = 6=«|,(o) g +( {/(a)!=0 do not lead to a solution of the partial differential equation at all. To illustrate this let us consider the equation / = (y + bz) 2 - (x + a) 3 = which can be regarded as a complete integral of the partial dif- ferential equation 11q\qy - z) + 8p 3 = If we put 6=0 and form the a-discriminant, we get simply y % = hence in this case the method does not provide us with a solution of the differential equation. The surface y = is a locus of cuspidal curves. If an integral surface F = is an envelope of a doubly infinite system of integral surfaces belonging to the family f(x, y, z, a, b) = 0, there will generally be an infinite number of these surfaces through a point (x, y, z) of the envelope which have the same tangent plane at this point. Let us select a limited number of these by making a an arbitrarily chosen function of x and y, then if F = is not a particular surface of the family (1), we may eliminate z from the equations / = 0, 128 DIFFERENTIAL EQUATIONS P = and express b as a function of x and y. Moreover, b will not be simply a function of a, consequently equations (4) which state that the tangent planes at the point (x, y, z) to the surfaces / = 0, F = are the same can only be satisfied if da db Consequently an envelope of a doubly infinite system of surfaces belonging to the family (1) is obtained when it exists by eliminating a and b from the equations / = o |=0 |=0 J da db The eliminant from these equations gives the general envelope of the family (1) when such an envelope exists, and this envelope may consist of either one or a number of surfaces, each of which provides us with a singular solution of the partial differential equation, assuming as we have done that an enveloping surface does not belong to the family (1). The eliminant may also contain other factors which do not generally provide us with solutions of the partial differential equa- tion. The loci of conical points, double curves, cuspidal curves and more complicated singularities may appear as factors. For a full discussion of the different possibilities the reader is referred to a memoir by Prof. M. J. M. Hill.* The isoclinal surfaces which correspond to the isoclinal curves of § 39 are considered by H. Levy.t •Some interesting theorems are given by Darboux.J It should be remarked that the preceding method may fail to give a singular solution when the function / is not a single-valued func- tion of its arguments. Let us consider, for instance, the case when f{x, y, z, a, b) = y + 2(z - x - yf + a{2y - z) + b = The equations J- = ^ = / = ^ da db J clearly cannot be satisfied simultaneously. The associated partial differential equation {1 + (z - x - yfyp + q = 2 however, possesses the solution z = x + y, which is not included in the above complete integral and is not given by the general integral y + 2(2 - x - yf = F{2y - z) It should be noticed, however, that if we rationalise the complete integral and write it in the form [y + a{2y - z) + b] 2 = 4(z - x - y) * Phil. Trans. A. (]892), pp. 141-278. iProc. Roy. Soc. Edinburgh, Vol. 32 (1911-12), p. 150. J Mimoirts prtsentts a VAcadimie des Sciences, t. 27. PARTIAL DIFFERENTIAL EQUATIONS 129 the method indicates that z = x + y is a singular integral, for it is clearly an envelope of integral surfaces. , Summing up our results we can say that when supposition A is valid, a solution of the partial differential equation

y> z) when p and q are expressed in terms of x, y and z with the aid of the equations ^i = 0, ■— = 0, the partial differential equation ^ ^ '{a + yj Sa), where < t\ < 1 ; hence f a (x, y,z,a+Q Sa, b + Sb) + J 2 / dbdy_ T9 2 / 5 2 / " l_3a; 2 + P 3a- 3z. tfa; + 3 2 / + p- 32/ 3 d * _dxdy + 1 3 2 / ' 3a; 3z. da? 3x 3 3 2 / 3a; dy dy dz. )*/+( 3 2 / 3a; 3z 3*A 3z 2 > 1 l'J)Sz + ** + (Hx + *wdz) Sa + (wL + *Mz) 6b -° < 3 > 3 2 / + 3 2 / \dxdy dx dz. )to + ( 32/ 3»/ 2 3 2 / I* ♦ ( 3a dy + 1 dy dz. da dz) V )sy + (■ 36 3?/ ^36 3z/ 3 2 / dydz + 1 dz Sb = (4) 3 2 J\„ PARTIAL DIFFERENTIAL EQUATIONS 133 Eliminating Sp and Sq from (1), (3) and (4), we obtain an equation which must be a consequence of (2) ; consequently dp\dx 2 dxdz J dq\dxdy dxdz J dxdz dx ftp/ 8 2 / dp \dx dy ay dz ay dydz/ ^ dq\dy 2 dydz dpKdxdz + p dz 2 ) + dq\dydz + q dz 2 J dz dz ~ dz dpxdadx dadz/ fV\ + d s(J* ! dq\dady ftp/ 9 2 / 3p \36 3a; + ^) 3q>/ 3 2 / + a? Va& 3y + ? 5a fe/ da 9 2 A =1 Sf dbdzJ db Eliminating the unknown quantity X from these equations with the aid of the relations h^ dl db df + Qf = dx dz d l + q d i- dy q dz we find that _dadx + p £♦* J 2 / dadz ' 3 2 / .da dy d 2 f dxdz \n \ 92 / ui ■> Vfl&P + q + p 2 >L. m >t + M»t dadz a 2 /] 9? dz 2 J dp db dy db dzj dq 9

„ Sep d

/dv dv\dx nz) =0 d(f>/du du\dy du\ q dz) + 89 /dv dv \dy q dz)=° The eliminant (du du\/dv \dx +P di)\dy dv\ (du = y-dy + du\/dv q dzATx + dv P dz. can be written in the form Vp + i 3q =• R (2) PARTIAL DIFFERENTIAL EQUATIONS 135 where P-X^4 Q=x|^4 R = X#^ o(y, z) d{z, x) d(x, y) and X is an arbitrary multiplier. If X is independent of p and q, the resulting partial differential equation is said to be linear because the left-hand side is a polynomial of the first degree in p and q. Now suppose that a partial differential equation of this type is given. To solve it we must endeavour to find a suitable pair of functions u and v. Let a and b be arbitrary constants ; then the equations u = a, v = b should give particular solutions of the partial differential equation. We may regard them as the equations of two particular integral surfaces. Now let a point (x, y, z) move along the curve of intersection of these two surfaces ; then the increments dx, dy, dz are connected by the equations du du du ^-dx + -=-dy + -=-dz = ox dy oz dv , dv , dv , ^-dx + =-dy + -^-dz = ox dy oz consequently the curve is given by Lagrange's auxiliary equations dx _ dy _dz P ~ Q ~ R W The curve is, in fact, a characteristic of the partial differential equation. To determine a suitable pair of functions u and v we have simply to obtain two independent integrals of the equations (3). These may be regarded as two simultaneous equations for determining x and y as functions of z. In solving these equations it should be noted that when one solution u = a has been obtained we may use this result to eliminate one of the variables and obtain an ordinary differential equation connecting the other two. If v = b is a solution of this last equation, the general integral of the partial differential equation is given by (1). To verify that the partial differential equation is satisfied, we have simply to retrace the steps whereby (3) was obtained. In connection with a system of differential equations of type (3) there is an important theorem due to Jacobi which is sometimes useful. Let us write X = ^ ; then we have MP =|^4 MQ = g^? MR = |^4 d(y, z) 3(2, A S(x, y) consequently M satisfies the partial differential equation 1(MP) + |(MQ) + !(MR) = 136 DIFFERENTIAL EQUATIONS Let us suppose that M and N are two distinct solutions of this equation, and write N = M0 ; then !(M6P) + |(MBQ) + |(MBR) - and consequently MpJ + MQ^ + MR^ = On substituting for MP, MQ, MR, this equation takes the form 9(6, u, v) _ d{x, y, z) This equation is satisfied identically and not in virtue of the equations u = a, v = b ; hence 9 must be a function of u and v. Quantities such as M and N are called multipliers ; the theorem which has just been proved indicates that the ratio of any two multipliers is a solution of the partial differential equation (1) and of the system of equations (3). An important property of a multiplier is that when one integral of the equations (3) has been obtained and when any multiplier is known, another integral can be obtained by quadratures only. Let the integral which has been obtained by u(x, y,z) = a then, pursuing the method which has already been sketched, let one of the variables (say z) be eliminated with the aid of this equation. If v(x, y, z) = b is a second solution of the equations (3), the modified form of the equation Qdx = Vdy will be satisfied by the relation G(x, y, a) = b, which is obtained by eliminating z from the equations u = a, v = b. Now dv _ dG 3G du dx dx da dx dv _ dG 3G du dy dy da dy dv dG du dz da dz consequently M p _ 3{u, v) _ dG du d(y, z) dy dz MQ = f- V \ = f \ U d{z, x) dx dz Hence 7/ ~ dG 7 9G , dG = ^—dx + -=-dy dx dy = H(Q %'A< ••• x n< 2 ) MR =1^1^ J' d{x v x 2 is called a multiplier, it satisfies the partial differential equation J- (MP,) + ^(MP 2 ) + ... ^-(MPJ + | (MR) - The ratio of two different multipliers is a solution of the partial differential equation (II) and of the system of differential equations (III). If u l = ocjl, u 2 = + |h + |(p»> + 1 = ° is satisfied. Example. Consider the equation {y 2 + z 2 - x 2 )~ - 2xy~ + 2xz = Lagrange's auxiliary equations are dx dy dz y 2 + z 2 - x 2 - 2xy - 2xz A first solution is derived from the relation — = — , and may be y z written in the form z = ay. Modifying the differential equations with the aid of this relation, we find that dx dy y 2 {\ + a 2 ) - x 2 - 2xy This is a homogeneous equation ; to solve it we write x = ty : the equation then becomes 2tdt dy . + -^ = 1 + a 2 +t 2 y :. y{\ + a 2 + t 2 ) =b or x 2 + y 2 (l + a 2 ) = by x 2 + y 2 + z 2 Thus a = - b = y y and so a general solution of the partial differential equation is given by /z x 2 + y 2 + z 2 \ = V y ) where F is an arbitrary function. , dz dz In the following examples p and q are written for ^, ^- respec- . , , dz dz dz y tively, while p v p 2 ,p 3 , ... denote ^-, ^-, ^-, ... . EXAMPLES. 1. Solve the equations ap + bq = 1 p{x - a) + q{y - b) = z - c py - qx = where a, 6, c are constants, and interpret the results geometrically. 2. Solve the equations p + zq = py - go; = 1 xzp + 2/zg = xy x 2 p - xyq + y" = 140 DIFFERENTIAL EQUATIONS 3. Solve the equation (a;, + x z + z)p 1 + (x 3 + x x + z)p 2 + {x 1 + x 2 + z)p 3 = x x + x 2 + x, 4. Particles are projected from a given point with various velocities in various directions ; the particles move under gravity without resist- ance from the air. Prove that if a moving surface t = f{x, y, z) always contains the same particles, / must satisfy the partial differential ■!♦'!♦«•- «■>!-' where g is the acceleration due to gravity. Show also that M = f~ l is a multiplier of Lagrange's auxiliary equations, and that the density of the swarm of particles can only be a function of type ^ j 3 \t t t -" , 5. Solve the equations p + q = - xp + zq + y = p q \ xp + yq =nz — I — = - x y z x 2 p + y 2 q = z 2 (x + y)p + (y - x)q = z - px - qy = ayx 2 + y 2 + z 2 where a and n are constants. § 51. Special solutions of Lagrange's equation. If u and v are not both single- valued functions of x y, z, the equation z + x ] = The solution may be obtained directly from the differential equation if we write this in the form cp = (p + l) 2 (z - x) - q* = The equations = au + bv + c leads only to relations involving x,y,z and not <\i ; hence there are no proper special integrals of the equation. § 53. Simple types of soluble partial differential equations. Standard forms. It is natural to inquire what form a partial differential equation must have in order that a complete integral may repre- sent a family of oo 2 planes. There are evidently two forms which the complete integral can have. Standard I. Let the complete integral be z = ax + by + c where c is a constant and a and b are connected by the relation a ) 5 + »4 dz a; + a2/ + 6= J^^) = F(z ' a) ' say the solution of which is ,, 1 (y,q) a complete integral may be obtained by putting each side of the equation equal to a constant a. Solving the equations 9=0, 9 = a for p and q respectively, we get p = Q x (x, a) q = 6 2 (2/, a) Substituting in the equation dz = pdx + qdy and integrating, we obtain the complete integral |6 1 (x, a) dx + |6 2 («/, a) dy + b wherein b is an arbitrary constant. The preceding methods are all particular cases of the following general method due to Charpit. Let f{x, y, z, p,q) =0 be the given partial differential equation ; we shall endeavour to find a second relation between x, y, z, p, q, viz. : = const. If, how- ever, another integral J?(x, y, z, p, q) = can be obtained, 9 = F ( x > y> z > p> q) will be a solution of the partial differential equation for 2 + 9 2 W^ 2 + 2/ 2 can be reduced to a soluble form by the substitutions X = log (a; 2 + 2/ 2 ) Y = tan" 1 ^ Z = . Z 5 v * ' a; -yV + 2/ 2 3. Solve the equations px + qy = pq q = px + p" Vp + Vq = 2x q = (z + px) 2 (p» + g 2 )j/ = gz p(g 2 + 1) + (6 - z)g = xq= yp + xe" +yl § 54. 2%e principle of duality. Contact transformations. We can evidently obtain a relation of type dz - pdx - qdy = (x(dZ - PdX - QdY) . . (1) which is characteristic of a contact transformation, by writing pdx = d(pa;) - xdp qdy = d ( Eliminating x, y, z between the equations (5), (6), (7), we obtain a relation between X, Y, Z, which is a solution of (4). To obtain the formulae of transformation of the second derivatives, we notice that dp - r dx - sdy = dX. - r dP - s dQ dq - sdx - tdy = dY - sdV - tdQ PARTIAL DIFFERENTIAL EQUATIONS 149 Now, if r B 2 z ~ dx 2 s = a 2 « dxdy < = d 2 z By* R 3 2 Z ~ ax 2 S = d 2 Z 3X3Y T = a 2 z W 2 we have dp = = r dx + sdy rfg = s dx + i dy dP - = B,dX + SdY dQ = SdX + TdY Substituting in the preceding equations, we get rR + sS = 1 rS + sT = sR + ts =0 «S + *T = 1 ■ • T _ S ~ R ~ rt S Another transformation is obtained by noticing that dz - pdx - qdy = d(z - px) + xdp - qdy Hence we may put (i = l Z = z - px X = p P = - x Y = y Q = q The partial differential equation (3) is in this case transformed int0 q, px + QY - Z, X, Y) = 3. Reduce the equations F(« - px - qy, x, y) = F(z - px, x, q) = to standard forms. 4. Solve the equations pq + xy = q(z - px) = x pq + xy — 1 (2 - px - qy)(px + qy) + pq = 150 DIFFERENTIAL EQUATIONS 5. A family of circles is defined by a relation p = /(5, v)) between the radius p and the coordinates (£, t\) of the centre. A circle (p , £„, tj ) is drawn to cut the three circles (p. I, -n) (p + ||«*5. 5 + d5, -n) (p + §*■** 5, n + *i) orthogonally. Prove that and deduce that the following transformation is a contact transfor- mation : X = a; - zp Y=y - zq Z = zVp* + q 2 - I P=- , y Q= " 3 Vp 2 + g 2 - l Vp* + g 2 - l 6. Prove that if a is a constant, the transformation a P , r a? Vi + p 2 + g 2 VTT p 2 Z = z .- P=p Q = 9 Vl + P 2 + 3 2 is a contact transformation. What is the geometrical interpretation of the transformation ? § 55. Equations with three independent variables. When there are several independent variables, there are also standard forms of equations which can be solved by simple means. For instance, a complete integral of the equation . '(££©- ■ ■ ™ may be obtained by writing V = ax + by + cz + d (2) where F(a, b, c) = (3) To obtain the general integral we make d a function of a, b, c and eliminate a, b, c from the equations _ dc dd dd do --= X + Zjr- + — + — — da da dc da „ do dd dd dc Q=y+Z db + db + d-cdb =— —^ da dc da 9F d¥dc db + dc db and the equations already given. PARTIAL DIFFERENTIAL EQUATIONS 151 When F is a homogeneous function of its arguments, /(V) is a solution of the partial differential equation when V is a solution. In this case solutions of the partial differential equation may be obtained in a number of ways, which may be described as follows : 1°. Let a, b, c be functions of V which satisfy the equation (3) ; then if V is defined by the equation (2), it satisfies equation (1). The quantity d may be an arbitrary function of V. 2°. Let F(£, 7], t) = be interpreted as the condition that the line \x + r\y + t, = may touch a curve given by the parametric equations x = j (tt) y = g ^ then, since the tangent at the point with parameter a is V(«) - y/'(«) = /(%'(") - /'(«)!7(a) the relation F(£, tj, £) = is satisfied when I = <7» -n = - /'(«) I = /'(«)(7(a) " /(«)!7'(«) Now write a; = zf(a.) + u y = zg(a) + v then dx --= dzf(x) + zf'(*x) da. + du dy = dzg(v.) + zg'(a.) dcr. + dv and so (7'(a) dr - /'(a) % + dz[/'(a)(7(a) - /(a)?' (a)] = (6), where x = iz cos a. + u y = iz sin a + v a = F(w, v) cos a dw + sin a dv = [i, d6 2. Deduce from the preceding result that if is denned by the equation ^ _ ?(0) j 2 + |- y _ ^j, + [ z _ ^ 6) ] a = where !;, tq, C are arbitrary functions, the expression V = $(6) satisfies the partial differential equation. 3. Obtain a general solution of the equation dV\ 2 /dVy , /9Vy fdV dx) \dy) \dz) \dt CHAPTER VII TOTAL DIFFERENTIAL EQUATIONS § 56. The condition of integrability. If a particle is moving on a surface whose equation is F(x, y, z) = 0, the component velocities (dx dy dz\ r , . , . , , . , , , \1f' iff' ~dt) particle are subject to the relation dFdx dFdy dFdz dx dt dy dt dz dt ~ ■ ■ ■ \ I which expresses that the direction in which the particle is moving is at right angles to the normal to the surface. This equation may be written in the abbreviated form 9F, 3F, 3F J ^-dx + -=-dy + -=-dz = . . . . (2) ox ay dz v ' Now let us take the more general case in which the direction of motion of the particle is subject to the condition Fdx + Qdy + Rdz = (3) where P, Q, R are functions of x, y, z but not of t. It is natural to ask if this condition implies that the particle always lies on a surface of the family ~F(x, y, z) = const, whatever may be the initial position of the particle and its initial direction of motion. The answer is, that this is not necessarily the case, for if we assume that equation (3) is simply a multiple of equation (2), so that (4) 3F n 9F p dF = *Tz ■ ■ te multiplier, we find that 3R dQ _ d[i 3F d\x 9F' dy dz dy dz dz dy 3P 9R _ 3fjL 9F 3|x, dF dz dx dz dx dx dz 3Q dP _ dn SF d(x 9F dy dx dx dy dy dx (5) 154 DIFFERENTIAL EQUATIONS Hence the quantities P, Q, R must satisfy the condition Denoting the three quantities on the left-hand side of (5) by £, yj, K respectively, the condition takes the form P? + Qt) + SZ = This condition is necessary, and it will be shown presently that it is also sufficient ; in other words, when the condition (6) is satisfied, the differential form (3) can be thrown into the form (2). We shall find it convenient in the following analysis to denote the differential form on the left-hand side of (3) by the symbol A. EXAMPLES. 1. Prove that the condition of integrability is satisfied in the following cases : (y - z)dx + (z - x) dy + (x - y) dz = e x dx + (e z - e x ) dy + (e* - e x ) dz= (y 2 + yz) dx + (xz + z 2 ) dy + (y 2 - xy) dz = 2. Prove that if P dx + Qdy + Kdz + Sdt can be reduced to the form U du, the equations S5 + Qy - Rp = S£ + P|3 - Qa = Stj + Ra - Py = P£ + Qt) + R^ = are satisfied where 3R 3Q dy dz ap 3r .dz dx _ dQ ap dx dy _ as ap dx dt R 9S dQ P dy dt as 3R Y ~ d(y, z) v d(z, x) v ' 3P 3P hence d(P, »)=-=- tZ(z, x) - j-d(x, y) d(Q, V) = d -§d(x, y) - d ^d(y, z) d( * n ' ^ = ~dy d{y ' ^ ~ 'dlc d{ ' Z ' ^ The symbolical derivative of P dx + Qdy + R dz is thus fdR dQ\ , /3P 3B\ /5Q 3P\ ,. . Notice that if Udw = Ada; + Bdy + Gdz = JJ^dx + V^dy + V^dz dx dy dz its symbolical derivative is the same whether it is calculated from the expression on the left-hand side or its expression on the right- TOTAL DIFFERENTIAL EQUATIONS 157 hand side, for in one case the derivative is d(U, u) , and in the other case it is d(z, x) d(x, y) Ldx\ dy) dy\ dxJ _ - iff* *> + §|f *■*> + &-> which is equal to d(\J, u) . It should be remarked that if d(\J, u) is identically zero the three Jacobians in the preceding expression are zero, and so it follows that there is a relation between U and u which does not contain x, y and z. The vanishing of the first Jacobian indicates in fact that either u is a constant or that there is a relation of the form U = f(u, x). The other Jacobians then take the forms df du df du dx dz dx dy respectively, and indicate that either / is independent of x or that u is a function of x only. In both cases U is a function of u only. If we form the symbolical product of du, dv, dw, replacing these quantities by du , 3m, du 7 dv j dv j dv , ^-dx + ^-dy + -=-dz =-dx + -^-dy + =-dz dx dy dz dx dy dz dw , dw 7 dw -, ^-dx + ^r-dy + -=-dz dx dy dz respectively, we find that In particular d(u, v,w)j, . d{u, v, w) = -=7 f d(x, y, z) K d{x, y,z) d{F, y, z) = -^d{x, y, z) d(F, z, x) = -^d{x, y, z) 3F d(F, x, y) = -fad{x, y, z) hence the symbolical derivative of F d{y, z) + Gd(z,x) + Kd(x,y) is /d¥ dG 9HN,. . fe + W + ^ (W) In particular, if we find the symbolical derivative of the expression -*■»-$$*•" + "^fc*> + "^>^ in two ways — first, by using the expression on the left-hand side, 158 DIFFERENTIAL EQUATIONS and secondly, by using the expression on the right-hand side — we obtain in one case d(w, u, v) , and in the other case {My d(u, v)~ 9 r diu, v] dy L d(z, x) d_ dz diu, v)~ i — - d(x, y). \d(x, y, z) or d(w, u, v) d{x, y, z) d(x, y, z) which is equal to d(w, u, v). We come then to the following conclusion : In performing the operations of symbolical multiplication and differentiation with expressions of type Pdx + Qdy + Rdz Fd(y, z) + Gd(z, x) + Hd(x, y) the results are the same if we replace these expressions by their equivalents when the variables are transformed to u, v, w, provided the increments du, dv, dw are multiplied according to the same rules as dx, dy, dz. The foregoing analysis may be extended without difficulty to the case in which there are n independent variables ; thus, when n = 4, the symbolical product of A x dx + B x dy + C-^dz + D-^dt and A 2 dx + B 2 dy + 2 dz + ~D 2 dt is (B X C 2 - BA)%. z) + (CA - CjAJdfc x) + (AjBjj - k 2 B^d{x, y) + (A 1 D 2 - A g D 1 )c?(a;, t) + (B X D 2 - B^DJdfo, t) + (dD, - CJ)Jd{x, t) and the symbolical derivative of P dx + Qdy + B,dz + S dt is 3R 3Q\ /aP -Wj-Tz) d ^-^ + \Tz te)d(z, x) fdQ 3P\ + u- - + 3Q %, + as dx as as aj? a< 3R' "a* d(a;, t) d(z, t) ^as V3y dt . The symbolical product of 1% z) + Gd(z, z) + Hc%, */) + Ld{x, t) + M<%, t) + Nd(z, i) and P dx + Q cZ«/ + R dz + S c?i is (PS + QN - RM)c%, z, «) + (GS + RL - PN)<% x, t) + (HS + PM - QL)<%, !/, «) + (PF + QG + RH)d(x, «/, z) while the symbolical derivative of the first expression is fd~F 3Q 3R\,. A /3G 3L + \ dt + dx /a_F \dt dz ^)d(z,x,t) W d{x,y,t) + \Tz + dy- + dz:) d{x > y ' z) TOTAL DIFFERENTIAL EQUATIONS 159 The symbolical product of Ad(y, z, t) + Bd(z, x, t) + Cd(x, y, t) - Dd(x, y, z) and P dx + Q dy + R dz + S dt is (AP + BQ + CR + DS) d{x, y, z, t) while the symbolical derivative of the first expression is /9A 3B 9C am,. \Tx + Ty + dz + 'dl) d{x ' y ' Z ' t) References, A. C. Dixon, Phil. Trans. A., Vol. 195 (1899), p. 509. E. Cartan, Ann. de VEcole Normale (1899). Th. de Donder, Rend. Palermo, t. 15 (1901). G. Ricci and T. Levi Civita, Math. Ann., Bd. 54 (1901). J. E. Wright, Quadratic Differential Forms, Cambridge (1908). R. Hargreaves, Gamb. Phil. Trans. (1908). H. Bateman, Proc. London Math. Soc., Ser. 2, Vol. 8 (1910). EXAMPLES. 1. If in a transformation of variables from (x, y, z, t) to (x r , y', z', t), = P dx + Q dy + R dz + S dt is transformed into F'dx' + Q'dy' + R'cfe' + S' dt' prove that the forms of the following expressions are unaltered : \d{y, z) + t\d{z, x) + X,d{x, y) + a.d(x, t) + $d{y, t) + yd(z, t) Ad(y, z, t) + Bd(z, x, t) + Cd{x, y, t) + T>d(x, y, z) (&+ tjP + Z,f)d(x, y, z, t) 7m. tic where 9R 3Q dy dz dP 3R dz dx 3Q ap dx dy _ as ap dx dt as 3Q P " dy dt as 3R Y " dz dt a = s? + Q Y - rp c = st; + P|3 - Qa B = Svi + Ra - Py D = P? + Qt) + BZ, Hence show that if the reduced form for Q, is M du + N dv, u and v satisfy the partial differential equation dx dy dz at 2. Prove that with the preceding notation, 3A 3B 9C 3D , „ 3a; 9j/ a« a« 160 DIFFERENTIAL EQUATIONS 3. Prove that when u, v and w are functions of x, y, z and t, ,. v d(u, v) „ , d(u, v) ,. , d(u, v) ,. , + d p^-d(x, t) + d S^\d{y, t) + ^diz, t) d(x, t) ' d(y, t) a d(z, t) 7/ > S(u, v, w) ,, , 9(m, «, w) ,, d(M > * w > = -k^ d{y ' z - t] + ib^r (z ' "• t] d(u, v, w) ,, , dlu, v, w) ,, + w^) d(x - y - t)+ h^J d{x ' y - z) 4. Prove that if d(u, v, w) is identically zero there is a relation between u, v, and w, and that if d(u, v) is identically zero there is a relation between u and u. 5. If P, Q and R are continuous functions of x, y and z, the expression P d{y, z) + Q d(z, x) + R d(a;, 2/) can be reduced to the form Wd(u, v). When can the expression be reduced to the form d(u, v) ? 6. If 5, 7), £, a, p, y are continuous functions of x, y, z, t, the expression £ c%, z) + 7] d(z, *) + X, d(x, y) + a d(a;, «) + p <%, t) + y d(z, t) can be reduced to the form Wd(u, v) + Pd{u, t) + Qd(v, t) + Rd(w, t) Hence show that if £+ ? p _ $ _ t - 2? ft» dy ^ dz v dz By dt 9y _ Sa _ 37) ^ = 3a _ 3p _ 9^ 3* & dt r " dy dz dt (The results of § 59 are needed in this example.) 8. In the last example, ifp = ? = r = s=0 and £a + 7$ + £y= 0, the appropriate canonical form is d(X, Y). (C. Meray, A. C. Dixon.) TOTAL DIFFERENTIAL EQUATIONS 161 § 58. Invariance of the condition of integrability. If we start with the equation J?dx + Qdy + B,dz = Vdu + Ydv + Wdw we deduce by symbolical differentiation that /9R 0Q\ /3P 3R\ /3Q 3P\ W " ^n y ' z)+ \ih- &><*• *> + (a? - *><*• y) /aw av\,. . /su aw\,, , /av 3u\., 4 Multiplying this equation symbolically by the preceding one, we obtain fdR _3Q\ , n f dF dB !) + R f ?Q _ ?PY [K|-f)^( 3z 3a;/ \dx dy/ d(x, y, z) = fu f— - — ") + v (— - — W w (— - ^?Yli(« L Va« 3w/ \3w 3m/ \du dv/J ' ' ' ' Hence, if the condition of integrability is satisfied in the case of the equation -?dx + Qdy + Rdz = it is also satisfied in the case of the transformed equation U du + V dv + W dw = If we use I and J to denote the quantities within square brackets, we find from the relation j, x 3(u, v, w) ,. d{u > v ' w) = W7yT^ d{x ' y,z) that I =J |WJ d(x, y, z) Hence I is transformed into a similar expression multiplied by the Jacobian of the transformation. I is consequently an invariant. § 59. First method of solving the equation A = 0. If by a change of variables A can be reduced to the form U du + V dv, the con- dition of integrability becomes simply vf-uf-o aw ow ® or .,: I/.; Hence = depends only on u and v. This means that U du + V dv can be made an exact differential dF by multiplying it by a suitable integrating factor. If now we write A = EdF and differentiate symbolically, using the notation of § 56, we find that ld{y, z) + 7] d{z, x) + £ d(x, y) = d(E, F). 162 DIFFERENTIAL EQUATIONS If I, 7) and Z, are zero, we have d(E, F) = 0, and this implies that E is a function of F ; hence in this case A is an exact differential. Conversely, if A is an exact differential, £, 7] and X, are zero. To show that A can be reduced * to the form U du + V dv, we proceed as follows : Integrate the equation P dx + Q dy = on the supposition that z is a constant. We shall suppose that this can be done with the aid of an integrating factor, so that The integral thus obtained may be written in the form F(z, y, z) = C where C must be regarded as a function of z. When x, y and z vary, we have aF aF aF dF = -=-dx + -=-dy + -^-dz dx By oz consequently, we may write (3TJT a "CT \ j-dx + -=-dy) + Rdz = IdF +(b, - Xj-)& 3F Putting u = F, U = X, v = z, V = R - 1-^, our object is accomplished. This method can evidently be used to integrate the equation A = in certain cases. To illustrate the method, let us consider the equation 2xz dx + 2yz dy + (z 2 - x ? - - y 2 ) dz = The equation 2xz dx + 2yz dy = is satisfied by F = x 2 + y 2 = C or by z = Using the first solution, we find that the differential equation may be written in the form zdF + (z 2 - ~F)dz = This becomes a homogeneous equation of the first type when we write s 2 instead of F. The natural substitution is therefore F = zH ; the equation then takes the form (1 + t)dz + zdt = or z(l + t) = c * It is tacitly assumed here that an ordinary differential equation of the first order possesses a solution, a proposition which we have not yet proved. It will be shown, however, in Chapter IX, that if certain reasonable conditions are satisfied, a solution undoubtedly exists. TOTAL DIFFERENTIAL EQUATIONS 163 Thus z (l + x ^±t) =c or a; 2 + «/ 2 + z* = cz The particular solution z = is obtained by putting c = oo . When a total differential equation J?dx + Qdy + Rdz = is reduced to the form dF(x, y, z) = it does not follow that all possible solutions are included in the formula F(a . ( y >z ) =c So much depends, in fact, on the nature of the function F. Let us suppose, for instance, that E(x, y,z) = x + Vl - yz = c The corresponding total differentia] equation is 2Vl - yz and when it is written in the form 2Vl - yz dx - ydz - zdy = it is seen that it possesses the special solution yz = 1 which is not given by the general integral x + Vl - yz = c. If we rationalise this last equation, we obtain 1 - yz = (c - x) 2 and it is seen that 1 - yz = is an envelope of the singular surfaces. Thus yz = 1 is in this case a singular solution. In the case of the equation (y + y 2 x - yz)dx + (x + x 2 y - xz)dy - dz = the general integral is xy + log (xy - z) - c but there is a special integral z = xy. If we write the general integral in the form xy - z = ae~ xy where a is a new constant, the " special integral " can be derived from the general integral by putting a = 0. The integral z - xy = is consequently not a singular solution. For further information on the subject of singular solutions, the reader is referred to papers by Guldberg, Comptes Eendus, Dec. 26 (1898), and E. O. Lovett, Ibid. t. 129 (1899). EXAMPLES. 1. Solve the equation y dx - xdy + (a; 2 - y 2 ) dz = and discuss the nature of the solutions x - y = x + y = 164 DIFFERENTIAL EQUATIONS 2. Solve the equations (ny - mz) dx + (Iz - nx) dy + (mx - ly) dz = x(y 2 + z 2 ) dx - y{z 2 + x 2 ) dy - z(x 2 - y 2 ) dz = y(y + z) dx + z{x + z) dy + y{y - x) dz = (y 2 + yz + z 2 ) dx + (z 2 + zx + x 2 ) dy + (x 2 + xy + y 2 ) tfe =0 zdx - xdy + {xz + x log x) dz = (y + a) 2 dx + z dy - (y + a) dz = (2xz + z 3 ) dx + 2yz dy - 2(x 2 + y 2 + a 2 ) dz = §60. Canonical form for A. Second method of solution. We shall now show that a change of variables can be made which will reduce A to the canonical form * A = dw + udv Let us suppose that this has been accomplished ; then the form A - dw can be made an exact differential dv by dividing it by a suitable factor u, and consequently its invariant must vanish. This condition gives the partial differential equation for the determination of w. Treating this by Lagrange's method, we have the auxiliary equations dx dy dz dtv T = ^T = T = £» + iQ + CR If one integral a.(x, y, z) = a of these equations can be found, we can proceed as follows : Write v = a. and regard x, y, z as connected by the relation a.(x, y, z) = a ; then dv = and A reduces to an exact differential of a function w. Now, when z is eliminated with the aid of the relation a.(x, y, z) = a, the expression A can be thrown into the form A = dG{x> y> a) Let us substitute a — a.(x, y, z) and write G(x, y, a) = w ; the function u may then be found by writing _ dw dv dx dx When a.(x, y, z) = a, $(x, y, z) = b, two independent solutions of -the equations dx _ dy _ dz * This is a particular case of a general theorem concerning the reduction of Xidx-i + X 2 dx 2 + ... X n dx„ to a canonical form. See Forsyth's Differential Equations, p. 327 ; Theory of Differential Equations, Vol. I. The present method is sometimes called Bertrand's method. TOTAL DIFFERENTIAL EQUATIONS 165 are known, we may write ? =Y ^iP) v-Z-K® r_ v %.P) Now |5 + JJ + £ _ o era 5y 9a consequently _y ' K ' " = B(x, y, z) and so y is a function of a and p. Let us write y = /(a, p). « = « « =-J/(«,p)dpl and so E = 8(M ' v ) 9 ( a - P) _ g ( M - «) 3(«.P) 3(11, z) B(y,z) d(u, v) d(u, v) 8(z, x) s d{x, y) The expression A - udv is now seen to be an exact differential, dw, for dy\ n ~ U dz) ~ dz\ Q - U dy)=^-dKz) and similar equations. It is easily seen that if w = g(x, y, z) is one possible value of w, another possible value of w is w = $(w, v) + g(x, y, z) where P is an arbitrary function. For if A = dg + udv, we also have g<£ g A = d(g + #) + udv --^-du - -^— dv = dw + VdV When the expression A has been thrown into the form dw + udv = there are clearly three ways in which the equation A = may be satisfied : \°. v = const. w = const. 2°. w = const. « = 3°. *(,,,») -0 u d ± = d ± T aw dv When the invariant of A is zero, either of the preceding methods may be used to find the solution, which is clearly of the form H(«, v) = const, for the vanishing of the invariant implies that u is a function of v and w. 166 DIFFERENTIAL EQUATIONS EXAMPLES. 1. Solve the equation A = yzdx - zxdy - y'dz fiT (I'M (I Z The auxiliary equations — = -2 = — - are in this case 5 7] <= dx dy dz x - 2y y - 2z and two independent solutions are X a. = y 2 z = a p = j/V = b Now d(*. P) therefore /(a, p) = ^, and so w = - l/(a, p) dp = - log p 9(z, a) 1 P' We now have di« = A - uda = yzdx - zx dy - y'dz + (-+ 2 log 1/ ) (2j/z dy + y'dz) = d[a;2/z + zy 2 (2 log t/ - 1)] = d[a log P - a] Thus A = d[a log p - a] - log fda = a-^ - da P A solution of the equation A = is consequently given by - = const, or ze~" = const. P 2. Reduce the expression (ax + by + cz) dx + (Ix + my + nz) dy + (px + qy + rz) dz to the canonical form, a, b, c, I, m, n, p, q, r being constants. 3. Solve the equations yz dx + zxdy + dz = (y - z)dx + (z - x) dy + (x - y)dz = x(y - \)(z - 1) dx + y(z - l)(x - 1) dy + z(x - \)(y - 1) dz 4. Prove that two differential forms A = Fdx+ Qdy+ Rdz A' = P' dx + Q' dy + R' dz can be reduced to the forms A=Vdu+Vdv A'=XJ'du+V'dv respectively, where u = const., v = const, are two independent solu- tions of the equations dx dy dz QR' - Q'R RP' - R'P PQ' - P'Q TOTAL DIFFERENTIAL EQUATIONS 167 5. Show that, in the last example, we may write XU- ) - ( pp- +qq . + bk .)(p| + q| + b|) XU' = (PP' + QQ' + RR')(p'|? + Q'|? + B,'j) xv.,,p-. + q-. + E -,(p| + q| + r |) - (PP' + QQ' + BR')(p-^ + Q'g + B'g) where X is some function of x, y, z. 6. Prove that, with the notation of the last two examples, the functions 6= U *=V are solutions of the partial differential equation (QR' - Q'R)^ + (RP' - R ' p )|^ + ( p Q' - P 'Q)^ = PT + QY + B.%' - 6(Pr + Qr)' + RC + P'5 + Q'y) + B.%) + e j (P5 + Qrj + bz) . ., 9R' 3Q' ot „ where 5 = ^5 5—. etc. § 61. Homogeneous equations. If in the equation A = 0, P, Q, R are homogeneous and of the same degree in x, y, z, the equation A = can be written in the form * + f(*D«, + gG?;D*-o Putting y = sx, z = tx, the equation becomes dx[l + sF{s, t) + tG(s, t)] + zF(s, *) ds + xG{s, t)dt=0 dx ~Fds + Gdt or — + i ri its = ° a; 1 + sF + «G Now, if the invariant of the first equation vanishes, the invariant of this equation will also vanish, and this means that the preceding equation can be integrated as it stands. 168 DIFFERENTIAL EQUATIONS Example. Let us solve the equation (a; 2 - y 2 - z 2 ) dx + 2xy dy + 2xz dz = by this method. Putting y = xs, z = xt, we get dx „ sds + tdt 1- 2 = x ^ 1 + s 2 +& .'. a;(l + s 2 + t 2 ) = c or a; 2 + z/ 2 + z 2 = ca; EXAMPLES. 1. Solve the equations 2yz dx + 2zx dy - xydz = (2x + y + z)(y + z)dx+ (2y + z + x)(z + x) dy + (2z + x + y)(x + y) dz = (y - z) dx + (z - x) dy + (x - y) dz = 3 (2y - 3z) (8yz - x 2 ) dx + 12a; (a; 2 + 4z 2 ) dy - 2x (9a 2 + \§y l ) dz = 2. Solve the equation eta + (e*~ x - \)dy+ [ev- x - 1) dz = CHAPTER VIII PARTIAL DIFFERENTIAL EQUATIONS OF THE SECOND ORDER § 62. The homogeneous linear partial differential equation of the second order. The general theory of partial differential equations of the second order is beyond the scope of the present treatise ; there are, however, special types of equations of frequent occurrence which can be treated briefly by elementary methods, and these will now be considered. If A, B, C, D, E, F are functions of x and y only, the equation dx 2 dx dy C— Ti- dy 2 dx dz + E~ + Fz = . (1) dy is called a homogeneous linear partial differential equation of the second order. In attempting to solve this equation the first step is to reduce it to a canonical form ; we accordingly make the substitution * = M. v))' v = ?& i) (2) and take \, tj as new independent variables. The transformed equation is d 2 z B 2 z W^ + Fz 07) (3) where A B' ~ A \dxJ a?5v) 9 ^ + 2B^p + dx dy dx dx \dx dy c ' - A (l) 2 + 2B^^ + dx dy dldji dy dx, c (I)' dy By D' A dx 2 E' = A 3a; 2 + 2B + 2B E dy dy 2 dx dy J dx dy dy 2 dx d 2 Y) dxdy (4) 170 DIFFERENTIAL EQUATIONS If B 2 - AC and A are different from zero, the quadratic equation AX 2 - 2BX + C = (5) has two different roots X 1( X 2 , and we may write i/W . 2B^2 ?2 + fift' - A& + X ^ f 9cp + X ^ A W + 2B dx-dy + L \dy) - A \dx + ^)\Tx + **%) Let \{x, y) and r)(x, y) satisfy the equations dx 1 dy h + x ^ = dx 2 dy respectively ; then ,, ' ^. is not zero, and so we may regard Z, and 7) as independent variables. Let the transformation I = 5(*> 2/) >) = t#, «/) be equivalent to (-2) ; then for the new equation (3), A' = C = 0, while B' =/= 0. Dividing by 2B', equation (3) takes the canonical form S 2 z dz L a« ap^ + a aI + 6 ^ +C2 = • • • • < 6 > When the roots of equation (5) are both real, equation (1) is said to be hyperbolic ; when the roots are both imaginary, the equation (1) is said to be elliptic ; when the equation (5) has equal roots, equation (1) is said to be parabolic. These terms were used in this connection by P. du Bois-Reymond. The transformation was given by Laplace in 1773. The last case will now be discussed ; it evidently occurs when B 2 = AC. Writing Am 2 + 2Buv + Cw 2 s A(u + Xv) 2 we have A' = A( -~ + 1~\ --(i-l)d-l) dy) where the functions £,{x, y) and y^x, y) are at present arbitrary. Now let 7) be chosen so that fa + x|a - o ox dy and I so that p + xp =/= o °- -*(£♦*£)■ EQUATIONS OF THE SECOND ORDER 171 then B' = C = 0, while A' =f= 0. Dividing by A', the transformed equation (3) takes the canonical form d 2 z dz dz . ,_. W + a M + % +cz = (7) If A = and B =j= 0, we interchange x and y and proceed as before ; if both A and B are zero, the differential equation (1) is already in the canonical form. In discussing the properties of partial differential equations of the second order, it is convenient to write = 3*2 _ d 2 z d*z dz dz T dx* S ~ dxdy dy* P ~ dx q ~ dy If we are given a series of elements (x, y, z, p, q) depending on a parameter and satisfying the relation dz = p dx + q dy, the second derivatives r, s, t are connected by the relations dp = rdx + sdy dq = sdx + tdy These equations, together with the partial differential equation (1), generally enable us to determine the values of r, s, t uniquely ; for if we eliminate r and t respectively, we get s{Ady* - 2Bdxdy + Cdx*) = Adpdy + Cdqdx + (Dp + Ey + Fa) dx dyj This equation determines s, and the other two equations determine r and t when s is known. If, however, the equation Ady* - 2B dxdy + Gdx 2 = .... (9) is satisfied, there is no finite value of s unless we also have Adpdy + Cdqdx + (Dp + E^ + Fz)dxdy = When this last equation holds, the equation (8) does not determine s at all, and oo' sets of values of r, s, t consistent with the given series of elements (x, y, z, p, q). The equation (9) is called the differential equation of the char- acteristics of equation (1). If 0(a;, y) = constant is consistent with (9), we have ^-dx + -^-dy = 0, and so ox dy a©V»SS + o(£)--.... ( i', This is another form for the equation of the characteristics ; it will be seen that the functions £, y\, which were used in obtaining the canonical form (6), are solutions of this partial differential equation of the characteristics. ,}» 172 DIFFERENTIAL EQUATIONS If we regard (x, y) as the coordinates of a point in a plane, the equation (9) can be regarded as the differential equation of a family of curves ; the equation (9) evidently specifies two possible directions for the tangent to one of these curves. Let the curves defined by the differential equation (9) be called characteristic curves, or simply characteristics ; then, by comparing equation (9) with (5), we see that a hyperbolic equation has real characteristics, an elliptic equation imaginary characteristics, and a parabolic equation coincident characteristics. It should be noticed that if the partial differential equation (1) possesses a solution of the form z=Y/(6) (11) where 0, y are particular functions of x, y and /an arbitrary function, must satisfy the partial differential equation (10). In general, however, the equation (1) does not possess a solution of this form. To see this, let us consider the equation in the canonical form (6), and write z = Y /(5) (12) The differential equation will then be satisfied if a i.e. if .„ ' + a~ + b^- + cv = oc, 9t) ai, 07) ' 57) + ^ = ° (13) Differentiating the last equation with regard to \, we find that hence the two equation (13) can only coexist if ab = c - ^ (14) This, then, is the condition that the differential equation (6) should possess a solution of type (12). When this condition is satisfied, the partial differential equation may be written in the form K! +az ) +6 (l) +a2 )= o Integrating this equation by the method of § 7, we get dz , , - f 6 d£\ ^ + az = tyh)e J EQUATIONS OF THE SECOND ORDER 173 This equation may now be solved by a second application of the method of § 7, and we get z =/(5) e -H + e-HQ^cfy.expjjadT) - bd^ We evidently have y = e J a ' § 63. Laplace's method. The preceding method may be generalised by writing equation (6) in the forms -^ + bz x - hz = ■ z x = ~ + az . . (15) -^ + az_, - kz = 2, = ~ + bz . . (16) where A = == + a& - c « = 7T- + ao - c 07) If either A = or k = 0, the equation may be integrated by the preceding method ; if not, it is sometimes useful to form the partial differential equations satisfied by z t and z_ v These are 9 2z i 3z, , dz x - + «! -=^ + &i -5- + c^ =■• . . (17) a^ * a? * a?) a 2 z_i az_i , az_i n , 10 , ^ +a -^ +6 - 1 ^ +c - lZ - 1 = - • (18) respectively, where ai = a ~ ay/ log ^ 6l = b da db , 3 „ 7l Cl = c " 3| + a^ " V° s } a_ x = a 6_! = b - -^ (log A) _a_ ae db da a ,. ,. c - 1 = c -a^ + al- a aI (l0gA;) Writing t Sa, , . 36, , hl = ~a~t + ai1 ~ Cl 1 = lhi + 0l1 ~ C] t 3a_, , , db_, a 2 we have A x = 2A - 4 - ~-~- (log h) k t = h a 2 A_ x = k k_ x = 2k - h - gFg-( lo S *) 174 DIFFERENTIAL EQUATIONS If either h 1 or k_ 1 is zero, one of the new partial differential equations can be solved by the preceding method. If neither is zero, the transformation may be repeated, and this process continued a number of times until a soluble equation is obtained, or until it is realised that the method is unsuitable for the equation in hand. It should be noticed that if equation (17) is transformed by P uttin g to, , g| + t> 1 z 1 = z we have z' = hz The equation satisfied by z' is thus derivable from (6) by a simple transformation of the 'dependent variable, and will be regarded as equivalent to (6), because the quantities h', k' for the new equation are equal to h, k respectively. In fact, if we make the substitution Z =X(5,T,)8' the equation satisfied by z' is 9V ,dz' ,,dz' ,,, n where a' = a + ^- (log X) V C = c b + ^(logX) + a^(logX) + 6^(logX) + -g p ^ Thus ^' = -_ + a 'b' - c' = h k' = % + a'b' - c' =k 07] The quantities h', k' are on this account called the invariants of the differential equation, and the operation by which h_ v £_! are derived from h, k is regarded as the inverse of the operation by which h lt k-L are derived from h, k. For further information on this subject the reader is referred to Darboux, TMorie des Surfaces, Vol. II. EXAMPLES. 1. Solve the equations {x- y)s + p - q= x{r - t) + 2p = 2. Show that the equation dp 2 dg- EQUATIONS OF THE SECOND ORDER 175 can be reduced to the form ._ d"v 3 - y fdv dv\ by the substitution 2a ^ 2a ^ u = pv i, = p 2 + q ri = p 2 - q Y - 1 y - ! Hence show that when y = 1-4, » = (5 + i)) V'(5) - 6(5 + 7)) «p'(5) + 12q,(5) + (5 + »))» = q-, where y is regarded as a function of x and t. Assuming that -^\ is continuous along the element m question, we may write *Crsm+)-*(T D-T gfa when quantities of the second order are neglected. We thus obtain the partial differential equation ^-c«^ (1) dt 2 ' dx* w T where c 2 = — is a positive quantity independent of x and t . Equation (1) is usually called d'Alembert's equation ; it was the first partial differential equation to be considered, and is of historical interest on account of the discussion which took place as to the nature of the general solution. To reduce equation (1) to the canonical form, we must solve the differential equation of the characteristics, namely dx* - cHP = (2) Two independent solutions of this equation are x - ct = const, and x + ct = const, respectively. Let us write a = x - ct, (3 = x + ct ; then equation (1) is reduced to the canonical form sfe- (3 > and its general integral is seen to be y = P(a) + G(P) (4) where F and G are arbitrary functions. Substituting the values of a and (J, we get y = Y(x + ct) + G{x - ct) (5) The supplementary conditions suggested by the physical problem may take various forms. If the position and transverse velocity of each portion of the string is known at time t = 0, the supplementary conditions are y = /(*) |f = 9(x) when t = . . . (6) The value of y given by (5) will satisfy the requirements if f(x) = F(z) + G(*) g{x) = 6E'(z) - cG'{x) EQUATIONS OF THE SECOND ORDER 177 >t equation, we obtain jVfQdS-Ffc) -G(x) +A Integrating the last equation, we obtain 1 c where A is some constant whose value is not needed We now find that 1 «:+ct 2F(z + ct) = f(x + ct) + - g{l) d\ - A c Jo 2G(x - ct) = /(* -\ct) - ir Ct g(l)dl + A i r i c x+et I Hence y = -[/(* + ct) + f(x - ct) + - 1 j(l) d£j . (7) This formula gives the required solution when the second deriva- tive of /(£) and the first derivative of g(Z,) exist, and are finite for £ = x - ct + e and \ = x + ct - e, where e is an arbitrarily small positive quantity. Now let us consider the case when f(x) = g(x) = for x > a and x and x - ct> a, the value of g(Z,) appearing in the integral in formula (7) is zero for the whole range of integration. .'.«/= when x > ct If t > and a x and «! > x — ct while a; + ct < a, we have 1 rx+ct If <> 0, a x > a; - ci and a; + c<> a, we have 1 f a Finally, if a; + ct < a 1( we have y = 0. It follows from this result that two waves travel out from the region of initial disturbance in opposite directions with velocity c. Before a wave reaches a point x there is no disturbance. Each wave B.E. M + 9(-ct)=o\ (8) 178 DIFFERENTIAL EQUATIONS is of thickness a - a v Some points are influenced by both waves and others by only one. When the supplementary conditions state that y = for x = 0, and for x = I whatever value t may have, and that V = f(m) % = ?(*) for t = 0, 0["3Y _ dZ _ lfa dzLdz dx c dtj cdtLdz dy cdt_ ^ aja d**. d*a. _ T_d*a. ~ dx 2 + dy 2 + dz* c 2 dt 2 Hence a satisfies equation (1), and in a similar way it can be shown that p, y, X, Y, Z satisfy equation (1). Now let us endeavour to satisfy equations (2) by expres- sions of type a = a /(w, v) p = p /(«, ») y = y /(w, «) X = X /( M , «) Y = YJ(u, v) Z = Z„/(«, «) where a , p , y , X , Y , Z , m, v are certain functions of x, y, z, t and / is an arbitrary function of u and v. Substituting in the first of equations (2), we get rs To ap idx i dn du du 1 a«i ; Lay " "& ~ c "aTJ + a«L Yo a«/ Po dz ~ e A ° ad dfV dv dv 1 ^ a«~| . + ib^-P»a- z -c x »a7j =0 Since / is arbitrary, we must have du du 1 __ du ^"Ty-^Tz -c x "¥ =0 dv dv 1 3w Y< % " Po a^ " c x »a7 =0 ay_ _ ap_o _ i ax_ ay az c a< The first two equations give = dj^v) d(u, v) _ d{u,v) while the third equation gives a(x, u, v) d(y, z, t) = and a, ff, 7 are proportional to the components of the magnetic force, the multiplier being \V/k, where k is the specific inductive capacity and /* the permeability of the medium. Also c 2 = — EQUATIONS OF THE SECOND ORDER 181 Treating the other equations in a similar way, we find that we must have Z = 3(w, v) 'WJ) d(u, v) d(x, t) V _ ,. d(w. v) d(u, v) Yo - Cfi, 3(«, v) Y cv 8(m ' v ) 3(2, t) and a number of similar equations. Comparing these with the previous equations, we easily find that v = X, and that ,?M v .,d(u,v) d{u,v)) d(u, v) \w %> 2) x n - cX d(u, v) d(u,v) %, 2) «* 3(x, t) ' d(y. t) To Bf^JV) _ 3(2, oc) d(u, v) Y =-d|^ = ^ (3) 3(2. *) ^ 3(j/, <) - cX d(w, a) 3(«, v) d(x, y) r 3(z, J Comparing these equations, we find that X 2 + \j? = 0. .". \x = iX (i = V - ]). There are also four equations of type 3(X, u, v) _ 3(X, m, v) _ 3(X, w, p) _ 3(X, «, w) 3(2/, 2, t) ~ d{z, x, t) ~ d(x, y, t) ~ d(x, y, z) = which indicate that X is a function of u and v ; we may without loss of generality take it to be unity. We now have the result that formulae (3) give a set of solutions of equations (2), provided the functions u and v satisfy the relations 3(w, v) . B(u, v) 8(x, t) 3(w, v) B(u, v) = ic %. 2) d(u, v) 3(2, x) . d(u, v) o[x, y)J (4) To solve these equations we take u, v, x, y as new independent variables and write the equations in the form B(u, v, y, z) . d(u, v, x, t) d(x, t, y, z) " d(y, z, x, t) 3(w, v, z, x) . B(u, v, y, t) B(y, t, 2, x) ' 3(2, x, y, t) B(u, v, x, y) . 3(u, v, z, I) 3(2, t, x, y) ic B(x, y, 2, t) 182 DIFFERENTIAL EQUATIONS 3^x 11 % t\ Now multiply both sides of each equation by rr- 1 — '- — '—r ; then o\u, v, x, y) the theorem relating to the multiplication of Jacobians gives d(u, v, y, z) . d(u, v, x, t) d(u, v, x, y) d(u, v, x, y) d(u, v, z, a;) . d(u, v, y, t) d{u, v, x, y) ~ d{u, v, x, y) d(u, v, x, y) . d(u, v, z, t) d(u, v, x, y) ~ d{u, v, x, y) Evaluating each functional determinant, we find that the trans- formed equations'are (5) dz . dt dx dy dz . dt dy ~ - ic _ ox l = ic d[z ' l) d(x, y) The first two equations > give d / •3 , Tx {z - ct) = = % {Z - ct) A(z +ct)=- *l(z + ct) Solving these equations by Lagrange's method, we get z - ct = "F(x - iy, u,v) z + ct = G(x + iy, u, v) dz Now 2~ = F x + Gj *I-°- F, 2^-i(G 1 -P l ) 2c| = i(F 1 + G 1 ) where F x denotes the derivative of F with regard to its first argument ; hence d(z, t) AiC; = (Gi - FJ" - (Fa + GJ» = - 4F 1 G 1 . d(x, y) :. F.G, = - 1. This equation can only be satisfied if both F x and G 1 are inde- pendent of x and y ; consequently we may write z - ct = - I - i{y - 7))] z - X, + c{t - t) = - g[a; - l + i(y - v))] where £, 7), £, t are functions of u and v such that £ - ct = 6(5 - «)) + cp S + ct = - q(£ + irj) + ^ If we eliminate 6, we obtain the equation ( X - If + (y - ,)« + ( Z _ Ql = C »(J _ T )l Two independent functions of u and v will evidently serve our purpose as well as u and v, and it is usually convenient to write u = t v = 6 If, then, 8 and t are defined by the equations (x - iy + {y - t[) % + (* - I? = c«(< - t) 2 | z -£ -c(« -t) = 6[* - 5 ~i(y -*))] /' • * ' where £, v), £ are arbitrary functions of 6 and t, we can say that the expressions ^ =/{e,T) 3(M)- z = - c/(6, t) 3M will furnish a set of solutions of equations (2) , and will all be solutions of equation (1). To obtain a real set of solutions, we may retain only the real parts of the above expressions. In the particular case when 5, v), t\ are functions of i only, we have y--n=M Ty *-Z=M Tz g[* - I - % - *))] - * + n| = o ^[* - 5 - % - v))] - 1 + N^ = where M = g(* - 5) + |j(y - tj) + |j(z - S) - c 2 (< - t) (8) and N = |-c-6(|-^) Hence Mfa; - E - *(y - 7>)1 ^7 ( = „ r = ti(z - S) - {y - ij) 184 DIFFERENTIAL EQUATIONS but z - I = |(e - g)(a; - ?) - ^(6 + J)(y - 7j) .-. iQ(z - Q - (y - 7)) - |(8 2 - l)[x - I - i(y - v))] and so Ir4 = i( e2 - 1 ) o(«/, z) 2M V Writing g{Q, r) = (0 2 - l)/(8, t), we have the result that the function 1 V=gfir(6,T) (9) satisfies the partial differential equation (]). When E, = yj = t, = 0, equations (7) and (8) give t = t - - 9 = -Z-I-L M = - cr c x - ly where r 2 = a; 2 + j/ 2 + z 2 Hence the function V . \g(t - I, ±Zl.) . ... (10) r J \ c x - tyJ v ' satisfies the partial differential equation (1), g being an arbitrary function. This result may be generalised by replacing x, y, z by x - \, y - tl,z - X, respectively, where \,t\, t, are arbitrary constants ; this is, in fact, what is obtained when we take £(t), tj(t), £(t) to be constant instead of zero. We may also change the sign of r in the expression (10) and obtain a new solution. The particular solutions V = \ g (t - and V = I,( T ) are of considerable importance in mathematical physics. EXAMPLES. 1. If u is defined by the equation xl(u) + ym(u) + zn(u) - ctp[u) = q{u) where P + m? + n* = p* prove that V = f(u) satisfies the equation 3 2 V 3 2 V 9 2 V 1 9 2 V and that dx' + dy 2 + dz 1 c 2 dt a £(«) dl dm, dn dp dq du du du du du is also a solution, I, m, n, q, f, g being arbitrary functions of u. (A. R. Fdrsyth.) EQUATIONS OF THE SECOND ORDER 185 2. If 6 is a constant and / an arbitrary function, the function V = f\x cos Q + y sin + iz, x sin 6 - y cos - ci] satisfies the differential equation 3 2 V g 2 V 9 2 V _ 1 g 2 V dx* + dy* + 9z 2 ~ c*W 3. Prove that two functions u, v which satisfy equations (4) of § 65 also satisfy the equations /3w\ 2 /Buy (faY_ 1 (fa\ z \dx) + \dy) + \dz~J ~ ~^\di) /dvy /dvy (fay _ ^(fay \dx) + \dy) + \Fz) ~ ~^\di) du dv du dv du dv 1 du dv dx dx By dy dz dz c 2 dt dt 4. Prove that if u and v satisfy equations (4) of § 65, the ratios of thejacobians g( M- „) d(u, v) d{u, v) d{y, z) d{z, x) d{x, y) are functions of u and v. 5. Prove that equations (2) may be satisfied by writing _?5_^? a- — -— -f^f_!5: ~ dy dz dz dx dx dy X -— -— y=- — --— z=-— -- — dx c dt dy c dt dz c dt where F, G, H, V are solutions of the equation (1) and are connected by the relation g F g G g H j gy dx dy dz c dt 6. If «=a + iX t)=S + iY w = y + iZ * = V - 1 Maxwell's equations can be written in the form dw dv idu dv du idw dy dz~ c dt dx dy c dt du dw i dv dv, dv dw _ dz dx~ c dt dx dy dz (H. Weber and L. Silberstein.) 7. Prove that the last set of equations can be satisfied by writing _ Y— - —\- -— — U ~ l \fa dz) ~ cdt dx - 7— _ ^\- l^f ^ ~ \dz dx) ~ c dt dy dG dF\ 19H dA + ^r 3- dy) c dt dz 186 -where DIFFERENTIAL EQUATIONS F = 19L cdt + ./9N 9M\ dz) 9K da; G = 19M ./9L o dt \dz 9N\ 9K 9?/ H = 13N Tdi + ./3M 9L\ _ ~9 + (y - y )* + (z - z a y - c*(* - *„)* satisfies equation (1). 9. If in Maxwell's equations a. = (3 = y = 0, so that there is no magnetic force, X, Y, Z are independent of t, and X dx + Y dy + Z dz is a complete differential - d#. Hence show that the potential # in an electrostatic field satisfies Laplace's equation g 2 $ 92$ d*$ dx* + dy* + lk*~ 10. Prove that, with the notation of Ex. 5, the conditions are satisfied by F = ^ G=^ H-%» V = ^ M M M M where M and t are the quantities denned above. When 5' 2 + V 2 + £' 2 l(x, y, z, t) = const. J These equations may be regarded as integrals of the system of differential equations dx dy dz , — = — = — = dt u v w consequently . x) kv _ 3(9> j?£ x) kw = *(?. h x) kU - d(y, z, t) kV - d(z, x, t) m B(x, y, t) where k = - jj&4^ d (x, y, z) 188 DIFFERENTIAL EQUATIONS Now equations (A) give the relation and can only be solved when this equation is satisfied. Substituting the above values of u, v, w, the relation takes the form 9(co, cp, <\i, x) = d{x, y, z, t) where w = |. Hence | is a function of cp, y, ^, and so we may write r 9(a;, ?/, z) T T *•' If

X) the preceding equation can be written in the form _ d Wi> 9i- Xi) d(x, y, z) Let us take these functions cp x , ip lf y^ for our functions cp, <\i, x, which specify the paths of the particles of electricity ; then we may put G = 1 . Now let us endeavour to find a solution of equations (A) which gives another solution when a, B, y, X, Y, Z, p are multiplied by the same arbitrary function of two parameters p, q, and u, v, w are left unaltered. Substituting the expressions a x = aF(p, q) Si = pP(p, g) Y i = Y^fo ?) X x = XF(p, 3) Y x = YT(p, g) Z x = ZF(p, g) Pl = p F(p, g) in the equations 9jj _ 9^1 _ _1 9X t p^w gp x 9 Kl _ 1 3Z X pjW 92/ 92 c 9£ c dx dy ~ c dt c d^ a Y i _ 1 9Y t p_i« aXi 3Yj az_i 9« 9x c 9< + c 9x + 9?/ + 9z Pl and proceeding as in § 65, we find that we must have *d(z,t) X - cX ^) P A 9(2/,f) Y ~ CX 9(27^) T 9(2,0 z - cX ^) EQUATIONS OF THE SECOND ORDER 189 3(X, p, q) dCk, v, q) pw = c d(y, z, * d(z, x, i) 3(X, P, q) _ ,3(X, p, g) 3(*. V. 9 H " 3(*. 2/, «) Comparing these expressions for p, u, v, w with those obtained previously, we see that the two sets may be made equivalent by writing p==9 ^tj, cX = x Hence, when the paths of the elements of electricity are given by equations of type (B), a set of solutions of equations (A) is obtained by writing „ _ X d() 3( v , 4) c 3Q/, *) A 3(z, sb) .. _ X 3(9» j) 7 ., 3 (9, ji) T - 7 a/, v\ z = - II 3(M) " * 3(^,2/) A more general solution is obtained by adding to these expressions for a, [3, y, X, Y, Z, the corresponding expressions for a set of solutions of equations A when p = 0. Since = 8 (?. <\>> x) 3 (a;, «/, z) we may make p zero by taking ^ to be a function of 9 and <\i. Hence we may add to the above expressions quantities of type « = ?/*.. «%f *--?*(*+.>%$ where 9 S and uv, are arbitrary functions of x, y, z, t and / S (9 S , ty s ) is an arbitrary function of cp s and <\> s . The summation indicates that a number of different solutions of the type under consideration are added together.* The particular solution given by equations II is of special interest for several reasons. In the first place, the equations cp(x, y, z, t) = const. <\i(x, y, z, t) = const, may be regarded as the equations of a line of electric force, i.e. a curve which moves in such a way that the tangent at any point * A much simpler set of solutions of equations A, when p = 0, is obtained by writing X 3R 3Q 1 3P as dy 9z «3( ~dx Y _3P ft 1 3Q , 3S p ~ c dt dy _3P dy 13R as 7 c dt dz where P, Q, R, S are arbitrary functions of x, y, z and t. 190 DIFFERENTIAL EQUATIONS of it at any instant is in the direction of the electric intensity at that point. To see this, we keep t constant and vary x, y, z; then the direction of the tangent of the curves is given by ■^-dx + ~dy + -^-dz = dx dy dz ^ dx + -£ dy + =i d z = o dx dy dz , dx dy dz andso X=Y=Z This proves the theorem. Secondly, if we combine II with (B), it is easy to see that the relations coc = vZ - wY c(3 = wX - uZ cy = wY - vX are satisfied. These are the relations adopted in Sir Joseph Thom- son's representation of an electromagnetic field by means of a single set of moving tubes of force.* It should be noticed also that satisfy certain conditions which have not yet been expressed in a simple form. If we regard (x, y, z, id) as rectangular coordinates in a space of four dimensions, the four equations imply that the surfaces 6 = const., t = const, are orthogonal to the surfaces cp = const., ij; = const. In particular, the conditions are satisfied if the varieties = const., t = const., cp = const., ip = const, form a mutually orthogonal system. EXAMPLES. 1. Prove that if t and are defined by the equations (x - If +(y - 7)) 2 + (z - K) 2 = c 2 (* - r) 2 f(x - I) + g(y -y ) ) +h (z-q+k l\x - I) + v(y - n) + C'(* - - c 2 (* - t) where %, tj, £, /, g, h, k are arbitrary functions of -r and primes denote differentiations with respect to t ; then 9 and <\> will satisfy the four equations at the end of § 65a if they are taken to be the real and imagi- nary parts of the expression - 5 + i(y - 71) '[^ C + e{t - t) where E[a, t] = const, is a solution of the Riccatian equation 2h Tx = K# + ^ ~ o( ^' + c)][a(/ " ig) + h] - [o(5' - •ti') + ? - e][f +ig - oh] 2. Let the expressions I be formed with the values of 8 and t given in the last example, and let 5, vj, £ be regarded as the coordinates at time t of a moving point E. Prove that if particles are shot out from the different positions of E, and travel along straight lines with the velocity of light, they will form at each instant a line of electric force if the direction cosines (I, m, n) of the direction of projection at time r satisfy the differential equations h^- =/(c - IE,' - mrt' - nC/) + {V - cl)[fl + gm + hn) (AT k-r~ =g(c - l£' - mr( - n£') + W - cm)(fl + gm + hn) k^-=h(c - IV - OT7)' - nZ') + (?' - cn)(fl + gm + hn) 192 DIFFERENTIAL EQUATIONS Prove that the solution of these equations may be made to depend on that of the Riccatian equation considered in the last example. 3. If in the last example /(0, -r) is a constant, the velocity of the electricity is the same as if particles of electricity were shot out in all directions from the moving point E and travelled along straight lines with the velocity of light. (The case when k = c 2 - Z,' 2 - t{ 2 - XJ 2 and is positive is of special interest, because a, (3, y, X, Y, Z, p, u, v, w are all finite except at the singular point E, and there is a constant electric charge associated with the moving point E, so that equal quantities of positive and negative electricity are shot out from it at each instant.) 4. Prove that if, in the last example, f=E," g = rf ft = C" h = c 2 - I' 2 - V)' 2 - S' 2 the quantity p and, consequently, also pu, pv, pw are zero. The solution then becomes identical with that of Ex. 10 of § 65, and so represents a point charge of electricity. (R. Hargreaves.) 5. Prove that the fundamental equations may be solved by writing 3H 36 1 dF _5V dy dz c dt dx 3F 9H _ 1 9G _dV dz dx c dt dy _ 9G 9F J_ 1 3H dv dx dy ' c dt ' dz where F, Q, H, V are solutions of the equations d 2 F 5 2 F 9 2 F 1 9 2 F P u dx 2 dy 2 dz 2 c 2 dt 2 c = d 2 G d 2 G d 2 G 1 d 2 G pv dx 2 dy 2 dz 2 c 2 dt 2 c = d 2 H 3 2 H S 2 H 1 d 2 TI pw dx 2 dy 2 dz 2 c 2 dt 2 c = d 2 V 5 2 V d 2 V 1 S 2 V dx 2 + dy 2 + dz 2 1? dt 2 + p = 3F 3G 9H l av dx dy dz c dt 6. Prove that if the fundamental equations possess a solution of *yP e a = ao/(T) P = Po/(t) Y = To/(t) X = X /(t) Y = Yo/(t) Z = ZJ(t) P = Po/(t) + pJ'(t) representing a series of waves or independent pulses, and /(-r) is arbi- trary on account of the independence of the separate pulses originating from the source, the phase t satisfies the partial differential equation ? V> t, and the summation extends over a number of different points of type P, b.e. N 194 DIFFERENTIAL EQUATIONS which may be reduced to the canonical form da dp by writing x + iy = a, x - iy = p. The complete solution of (3) is consequently V = f(x + iy) + g(x - iy) (4) where / and g are arbitrary functions. It is evident that if u + iv = f(x + iy), u - iv = f(x - iy), then u and v are solutions of Laplace's equation ; they also satisfy the equations du = dv du = _ dv dx dy By dx and are called conjugate functions. Laplace's equation (1) evidently possesses particular solutions of the form v = u{x) v{y) w(z) (5) for if we substitute this expression for V, we obtain , ■> / % / v T 1 d 2 u 1 d 2 v 1 dhol . «(*)%) w(2 )|_-_ +__+__ J =0 Hence, if we write ld?u 10 ld 2 v , Idho udx 2 vdy 2 wdz i where I 2 + m 2 + n 2 = 0, equation (1) will be satisfied. The pre- ceding equations may be satisfied by taking u = e lx v = e my w = e ns I, m, n being constants. Hence V = e. lx+mi/+nz ....... (6) is a particular solution of the required type. The usefulness of particular solutions of equation (1) is increased by the fact that since the equation is linear, the sum of any number of solutions is also a solution. We have, in fact, V 2 (U + V + W + ...) = V 2 U + V 2 V + V 2 W + ... which is zero if U, V, W ... are solutions of Laplace's equation. To generalise the solution (5), we may consider the problem of finding solutions of Laplace's equation of type V = w(a) t>(p) w(y) (7) where a, (3, y are certain functions of x, y, z which are supposed to be given. The natural method of solving this problem is to take a, (3, y as new independent variables and to find the new form of Laplace's equation. The transformation may be effected as follows. EQUATIONS OF THE SECOND OKDER 195 Let dx, dy, dz ; Sx, Sy, Sz be two independent sets of increments of the variables x, y, z ; dx, d$, dy ; Sat., (5(3, Sy, the corresponding sets of increments of the variables ct, (3, y ; then, since OlC ox ox ox ox ox dx = tJ* + sp d P + ^ Sx = rJ x + ^ + aA etc - we find that * dxSx + dySy + dzSz = a da. Set + b dfi (5(3 + c dy Sy \ , a > + f(d$8y + dyS$) + g{dySct + Sydct) + h(daS$ + Sctdfi)) { > where a, b, c, f, g, h are functions of a, (3, y. If now we have a relation of type ld(y, z) + md(z, x) + nd(x, y) \ = P d(P. Y) + 1 %> a ) + r d(ct, (3) j (yj we may deduce by multiplication, using the symbolical method of § 57, that (ISx + mSy + n Sz) d(x, y, z) = [(ap + hq + gr) So. + (hp + bq + fr) (5(3 + igp +fy + or) $i\ <%, (3, y) Now write d(x, y, z) = J d(ct, (3, y) ; then the above equation is equivalent to the identical relation 3V» 3V„ l 3V. 9V„ dV„ a 3V„ to & + a7 % + IT* " *** + W S ? + af*r av m = ->£]♦[•«■♦» -t??]—- <"> When m = this reduces to Legendre's equation. The equation for u is satisfied by u = r n and u = r _(,l+1) ; hence we have par- ticular solutions of Laplace's equation of the types r n cos (w«p + e)P„ m (rj) r~ n_1 cos (nap + s)P„ m («r) where v = P n m (cr) is a solution * of equation (17). A comparison of these two solutions suggests the following theorem : IfV = F(r, 0, 9) is a solution of Laplace's equation, then the function V = -F( -, 0, 9) is also a solution of Laplace's equation. * This equation will be solved by the method of infinite series in Chapter IX. EQUATIONS OF THE SECOND ORDER 199 This is easily proved ; for if we write r = - in equation (16), it becomes s slnQ ( s *w) + m\ sm% m) + cosec( W - ° 3

) is a solution of equation (18). This may be verified by a direct comparison of the differential equations, to be satisfied by the function / in the two cases. In the general case when p 2 =/= m 2 , the substitution sin 2 = \ reduces equation (21) to the form F = ... e . d 2 ~F ,. OB . dF TO 2 f> 2 _»(» + *) 4 ^ 4(1 _ £)J m p Putting F = £ T (1 - ^) 2 G, we obtain the hypergeometricec l{\ - ?)g + [y - (a + p + m^jr - «PG = where y = to + 1 oc = n + 1 + i(m + p) p = J(to + #) Jl Hence, if F(a, (3, y, £) denote a solution of the hypergeometric equation, we have solutions of equation (18) of type V = x 2n cos (to

+ -q) sin m cos* 6 F(a, p, y, sin 2 0) This result is due to Stieltjes. Let us next consider the differential equation 32-y g2y g2y U + w + ^ + kW = °- ■ ■ ■ ^ which is derived from the equation of wave propagation by assuming that V is of the form V = F(.r, y, z) cos kc(t - l ) 202 DIFFERENTIAL EQUATIONS where k and t are constants, and F is a function to be deter- mined. Transforming to polar coordinates, we obtain Fr( r '' ays dr/ + &VV i d / . + "^~o an I slrj sm 6 30 \ •S) 1 + sin 2 e S 2 V 9

= C05 «Tz = ° cos a da.} dy (32) while if we differentiate (26) with regard to z, we get 30 3 2 30 3 2 + v 30 3 2 (33) 3a; 3a; dz ' dy dy dz dz dz 2 Equation (29) is now seen to be a consequence of (32) and (33), provided cos a =j= 0. Let us now write then jj, V = y - ix cos a W = Z - 13. • sin a . . (64:) 3(0, v, w) d(x, y, z) - i cos a sin a - i cos a + ix sin a 3a 3a; 3a 1 + ix sin a ^r- 3a - ix cos a ^r- 3a ix sin a ■=- dz - i sin a - ix cos a 3a dx , . 3a 1 - za; cos a ■=- 3z and the determinant is seen to vanish. 204 DIFFERENTIAL EQUATIONS Similarly, it can be shown that 9(oc, v, w) = d{x, y, z) in virtue of (27) and (29). We must conclude then that there is either a relation between v and w, or that there are relations express- ing a and 6 as functions of v and w. Now [idQ = dy cos a +'dz sin a - i dx = dv cos a + dw sin a hence we have the following general method of finding a function which satisfies equation (26) and is not independent of either x or y. Let v and w be defined by the equations (34), wherein a, v, w are connected by a relation not involving x, y, z ; then, if be defined by the equation dy cos a + dw sin « = ^ d0 it is a solution of (26). Conversely , any solution of (26) which involves x and y can be obtained in this way. If is a function of a, we have d(v cos « + w sin a) = y. cZ0 + (w cos a - v sin a) da = X da, say hence v cos a + w sin a is a function of a, and there is a relation of the form ^ cos a + z sin a - ias = '(a) - s cos a + y sin a . Again, -S * ™ (1)' - » ^d 2 a dU/daV da .da M^-s + -3— I a-) + smoc-_- = - sina^- dy 1 da XdyJ dy dy ^d*a dM (da\^ da da M &i + fel&; - cosa fe= cosa ^ Addin, g up, and remembering that (da\ 2 fda\ 2 /day n y + y + y -° we have 5 2 oc S 2 « 3 2 oc da; 2 + dz/ 2 + M Hence the conditions (25) and (26) are satisfied by y = 1, a = 0, and we have the result that V = /(a) is a solution of Laplace's equation, / being an arbitrary function. EQUATIONS OF THE SECOND ORDER 205 3V . Since ^— is also a solution of Laplace's equation and /'(a) can also be regarded as an arbitrary function, we have the result that w = ™ = - i-f ( «) = £M dx M J w M is a solution of Laplace's equation, g being an arbitrary function. We conclude, then, that if = a, where a is defined by (35), we m8 y take , . 6(a) where a(a) and 6(a) are arbitrary functions of a. We shall show presently that this is the most general value of y. Let us next consider the case in which a is a constant ; 6 is then a function of y cos a + z sin a - ix, and it is easy to verify that V = /(0) is a solution of Laplace's equation. A second solution is derived from this by differentiating with regard to the parameter a ; thus a second solution is ^— = (y cos a - y sin a)/ (y cos a + z sin a - ix) When 6 has the value y cos a + z sin a - ix, an appropriate value of y is consequently y = a(0) + 6(6) (z cos a - «/ sin a) where a(0) and 6(0) are arbitrary functions of 6. It will be shown presently that this is the most general value of y. The case in which is independent of x or y leads to a class of potential functions which may be derived from the previous one by a simple interchange of variables and a special choice of a. We have, in fact, = y + iz, and it is easy to see that in this case V = y/(0) satisfies Laplace's equation if y = a(0) + a* (6) where a(0) and 6(0) are arbitrary functions. Let us finally consider the case in which depends on x and is not simply a function of a ; we may then write equations (34) in the form y = ix cos a + /(a, 0) z = ix sin a + g{tx., 0) (37) The equation df cos a + dg sin a = y. dQ . . . . (38) implies that J- cos a + ~ sin a = da da and so we may write J- = i£ sin a ^ = - »^ cos a . . . (39) 206 DIFFERENTIAL EQUATIONS We now require the new form of Laplace's equation when x, a, 6 are taken as independent variables. Let us write Sf ae = ig cos e dg ia sin e then it is easily seen that dx Sx + dy Sy + dz Sz = - {x - lfda.8a. - aHQSQ - ocos (a - e){dx8Q + SxdQ) + a(x - £,) sin (a - s)(da.SQ + SxdQ) The determinant of the coefficients is in this case - {x - Q* a{x - I) sin (a - s) g(x - £) sin (a - e) -a 2 - a cos (a - s) - a cos (a - s) = + a\x - i;) 2 cos 2 (a - s) The new form of Laplace's equation is consequently l\- doL\_x - \ 9 cos (a ,9V . , e) =— + cr sin (a act m 96 [(*-aH+4[ os M« x 9V / ^ 9V o(x - £) cos (a - s) ^— = (40) Let us now substitute V = y/(0) in this equation and equate to zero the coefficient of /'(6). The resulting equation is (x 5)£ + £[y(*-5)] = o dx dx and this implies that y 2 (x - £) depends only on a and 0. Hence we may put y = y/(9) = {x _ ?) _^ (a> Q) When this last expression for V is substituted in the differential equation (40), it is found that there is one term with (x - £)~2 as a factor and no other term which could possibly cancel it. Equating this term to zero, we find that we must have a cos (a - e)$(a, 8) (J^Y = o (41) Now ) y 2 z =sin a cos 01 where y 2 = 1 - sin a sin to and k is a constant. A solution of Laplace's equation is then given by V = yF(a, B), provided F satisfies the partial differential equation . . S 2 F _, sin 2 a + k* cos 2 a 9 2 F „ „ 3F 3 sm 2a -g-j + 4 ^— 3-— + 2 cos 2a 3- - - sin 2a . F = da 2 sm 2a 9B 2 da. 4 (V. Amaldi.) 7. Prove that if we put x = p cos 9 2/ = p sin 9 p = oM sinh a z = oM sin i); where a is a constant and M _1 = cosh a - cos ty, Laplace's equation becomes £(m sinh .*) + £( M sinh .*) + M cosech <^ = EQUATIONS OF THE SECOND ORDER 211 Show also that this equation is satisfied by u = M _i cos n(ty - <]>„) cos m(

Putting 9 = e'"i\i, we get •3 + »-o2-c + »)+-« Comparing this with the equation d?Y , . % + *-<> •••(') possesses the polynomial solution av + » + s ^ - ***: - ° which is of the same form as (3) ; hence the theorem is proved. Let us next suppose that the function V in equation (1) is inde- pendent of z ; we then have the equation av _ /dW d*y\ dt -X W + dy*J (11) Let x = a cos tp y = p sin cp s = ~— 4:Xt Equation (11) may then be written sv _ /aw i av i_ aw \ a< \3p 2 p dp p 2 atp 2 / and this is satisfied by V = G(t, s, 9) .- 3 2 G .. ,3G 3G 1 3 2 G . ,,„. lf s W + {l+s) ^- t Tt + TsW = ° • • (12) Now write G = f cos (my ■)• e)w(s) ; the preceding equation is then satisfied if d?u ,, .du ( ra 2 \ This equation is transformed by the substitution u = s p v into dh> , a is dv I 4p 2 - m 2 \ Let us take # = ± J»» ; then the term in - disappears, and we obtain dh) , n .dv , , . . 8-r\ + ( s ± m + l)j" + (± i m _ v v = ° Writing s = - a, we obtain the equation satisfied by the function V = F(± \m - v ; 1 ± m ; a) Hence equation (11) is satisfied by m V = f(£f F(| - v ; 1 + m ; - ^) cos (imp + e) that is, by V = t n p m F ( - n ; 1 + m ; - ^-J cos (my + e) . (13) 214 DIFFERENTIAL EQUATIONS When 1 - n and 1 - m are positive integers, this is a polynomial in x, y and t. Let us now write t x = - -, s 1 = - s as before ; then equation (12) becomes 3 2 G „ ,3G 3G 1 3 2 G n and this is satisfied by G = ^e s >H if 3 2 H .. ,3H 3H 1 3 2 H . which is of the same form as (12). Hence we have the following theorem due to Brill : If V = ^(x, y, t) is a solution of (11), the function 1 _^±Z 2 /x v 1\ V = " t e ** * (-, |, - -) . . . . (14) is also a solution. In particular, we deduce from (13) that the function V = p^r m ~ n ~ 1 e' itK F ( - n ; 1 + m ; -£-] cos (mtp + e) (15) is a solution of equation (11). Again, if ^ is independent of t, a solution of (11) is given by ■*■ = K x + w) + y( x - iy) where / and g are arbitrary functions ; hence we may deduce that the function satisfies equation (11). In the general case, when V depends on x, y, z and t, we may make the substitution r 2 x = r sin cos cp y = r sin 8 sin cp z = r cos s = — 4xt The transformed equation 1 3 / „ 3V\ 1 3 / . „ 3V\ 1 3 2 V 1 3V V= 7 e / . 3V\ 1 3 / . fi 3V\ r37J + ^in0 30l Sme 30J r 2 dr V 3r / r 2 sin 30 V 30/ r 2 sin 2 3cp 2 y. dt is satisfied by V = G(s, <)0(8)*( M + *• - Q cos ( m is also a solution. In particular, if we take / to be a constant, we obtain the solution 3 r*_ V = t~*e~**t (19) while we may deduce from formula (17) that the function V = r»t- in+ ' +i) ~F( - x, n + f, £-) e~ 4 " ( cos (mcp + e)P n m (cos6) is a solution of (1), where x = v - - is an arbitrary constant. LI References. H. S. Carslaw's Fourier Series and Integrals. E. Cunningham, Proc. Boy. Soc. Vol. 81 (1908), p. 310. J. Brill, Messenger of Mathematics (1891). P. Appell, Liouville's Journal, Ser. 4, t. 8 (1894). EXAMPLES. 1. Prove that if V = f(x, t, t) is a solution of the equation d 2 V _ 3 2 V dx*~ WFt the function V=r i f(j, t - ^, - -) 216 DIFFERENTIAL EQUATIONS is also a solution, and that if V = f(x, y, t, t) is a solution of the equation 9 2 V 9JV _ gay dx 2 + dy*~ dti\ 1 (x y x 2 + y 2 1 the function V =-/(--, t J \t t U t is also a solution. 2. Obtain corresponding theorems for the equations 3 2 V 9JV d 2 V _ dx Y+ dy 2 + ~dz 2 ~ d 2 V d 2 V _ 1 9 g V d 2 V d 2 V d 2 V _ 1 d 2 V 3a; 2 + 92/ a + 3z 2 ~ c 2 9« 2 respectively. § 68. Harmonic equations* A partial differential equation is said to be harmonic when it possesses an infinite number of particular solutions, each of which is the product of a number of functions of one variable, the variable being different for each factor. Thus the equation gs z g2 z is a harmonic equation, because it possesses an infinite number of' particular solutions of the type 2 = sin ~kx . sin \y where X is an arbitrary parameter. An equation of type d 2 z -j^jy- = [(x - y)]« ■ (1) will be regarded as the canonical form of a harmonic linear partial differential equation of the second order in two variables. It possesses an infinite number of harmonic solutions of the form ^ = /(* + y) -g(z - y) (2) where f(s) and g(t) satisfy the ordinary linear differential equations g - [*(•) + X]/ in which X is an arbitrary parameter. * The analysis in §§ 68 and 69 is taken from Darboux's TMorie des Surfaces, t. 2. EQUATIONS OP THE SECOND ORDER 217 Many differential equations which occur in mathematical physics can be reduced to the canonical form which has just been given. The equations in question are of the type P(«) gp- + Q( s ) -gj + R ( s ) w = L(0 -fa + M(t) -^ + N(Qu> (3) To reduce to the canonical form, we write w = zu(Q vfa), where u, v, £, and tj are determined by the equations t>A^Y , du _ d 2 £ _ e£ n \dtJ dy] dt 2 " dt The equation then becomes d*z [1 d 2 u 2 /dw\ 2 "1 _ d*z JTldtv _ 2 fdv\ 2 ~| dl 2 + Z Vu dZ? vAdl) + K J - 9^2 + z \_ v d7] 2 v z\ dri ) + KJ The substitution £ = £+7), */ = £-■/] now reduces the equation to the canonical form. Example. To reduce d 2 w n dw d 2 w m dw W + s a7 = W + l~di n in to the canonical form, we write w = s ~*t 2 z. The equation then becomes ^ w(w _ 2) dH m(m _ 2) •z = 9s 2 4s 2 dy 2 -.«.-„(=♦«)♦«.-,,(«-«).. Differentiating twice with respect to x and twice with respect to y, we find that ^x d*Y dx* = ~dnf EQUATIONS OP THE SECOND ORDER 219 Hence X = ax* + bx 3 + ex 2 + dx + e Y ~ ay* + by 3 + cy 2 + dy + e the coefficients being the same in each case, since X = Y when x =y. By taking different biquadratics, we may obtain various trans- formations of the differential equation (5). In the first place, if we put x = 4z Y = 4y x = x' 2 y = y' 2 the transformed equation is d 2 z Vm{m - 1) m(m - 1)1 dVdy' ~]_{z' + y') 2 " (x' - y') 2 \ Z ' ' ' (7) This result tells us that the equation (6) is transformed by the substitution t = s 2 + a 2 t = 2sa into the form — -* + — ~ = —^ + — _z as 2 s as og 2 a da Secondly, if we put X = 4a; (1 - x) Y = 4y(l - y) x = sin 2 x' y = sin 2 y' the equation becomes d 2 z [~ m(m - 1) m(m - 1) ~| dx' By' Lsin 2 (x' + y') sin 2 (x' - y')j and consequently possesses harmonic solutions of type 2 = f¥ + y')f(x' - y') where f(u) satisfies the equation ^[=^4- • . (8) du 2 L sin 2 u y v ' When X = 0, this is transformed by the substitution cot u = i\i into Legendre's equation k^-O + m(m - 1)/ = When m is an integer, a solution of (8) can be expressed in finite terms. This is seen most easily by using the following theorem, which is due to Darboux : * If we know how to solve the differential equation % = t?(t) + h]y (9) for all values of the parameter h, and y = v(t) is a solution when h has the value h v the general solution of the equation S = [v^) + h- hl >.... (I 0, * Oomptes Rendus, t. 94 (1882), p. 1456 ; Thiorie des Surfaces, t. 2, p. 196. 220 DIFFERENTIAL EQUATIONS ■ • u dy y dv , in is given by z = d± v dt ichere y is the general solution of equation (9). This is easily verified by direct substitution. Let us now apply this theorem to the equation S-* < 12 > When h = - 1, this is satisfied by y = sin t . Let us take this for our function v ; then we may conclude that the general solution of the equation d%z [l2 -, ^=Lsm^ +/l > (13) is z = AVhe^' - BVhe- v * ( - cot *[Ae v *< + Be WKt ] where A and B are arbitrary constants. For particular values of h, equation (13) possesses the solutions sin 2 t, sin 2 t cos t ; consequently the theorem may be applied again. It will be convenient, however, to consider the more general equation d?y _ V m(m - 1) n(n - 1) 1 , dP - y l sin 2 * + cos 2 * +h y • ■ (4) When h = - (m + n) 2 , this equation possesses the particular solution y = sinm f cos „ t = ^ say Employing this solution, we obtain the new equation d?z dP or dH _ J m(m + 1) n(n + 1) "I d* 2 ~ Z l sm 2 * + cos 2 * + M • • (15) This is of the same type as (14), but with m + 1, n + I written in place of m and n. It should be noticed that if n = 0, so that the term — - — r— - is zero in (14), the term — 5 — - is zero in (15) ; cos 2 * v " cos 2 * hence the solution of the equation dH _ dP ~ Z ,a(v, : 1) + q sin 2 * may be derived by successive steps when m is an integer. If we put n = 1 in (14), the term n ^ n ~ ' in (14) is zero, but the term n(n + 1) . .,„ . cosH cos 2 t m (15) is not zero. It is clear, then, that if both m and n are integers, the general solution of (14) can be obtained with the aid of Darboux's theorem. EQUATIONS OF THE SECOND ORDER 221 If in equation (14) we write y = sin m t cos" t u[t) and s = sin 2 t, we find that u satisfies the hypergeometric equation «(i - s ) -£2 + [y - (« + P + !) 5 ] ^ " a P" = ° where 2a = m + n + V - h 2$=m+n-V-h 2y = 2m + 1 Hence we have the result that the solution of the hypergeometric equation can be expressed in finite terms when are integers. T and a + p EXAMPLES. 1. Prove that Laplace's equation possesses solutions of the form V = P m cos m 9 F m (s)F m (o) where s a + c z = iz, 2sa = p and F m (s) is a solution of the equation d 2 F 2w+ 1 dF (is 2 s cfe Hence show that there are harmonic functions of type V= J TO (.s)J m ( P) =0 and that it is possible to solve for p either in one or a number of ways, so that the differential equation can be written in the form l= F ^) W The function F(x, y) will generally be a many-valued function of x and y, but we shall suppose that when x and y are limited by certain inequalities it is possible to assign just one value to ¥(x, y). To fix ideas we shall assume that a point (x , y ) with real or complex coordinates can be found such that F(x, y) can be expanded in a power series F(*, y) = 2 2 A mt „(x - x )™(y - y )" ... (2) which converges uniformly for \x - x \ t) + C where C is an arbitrary constant. The condition z = z , when x = x , is satisfied by taking C = b, and then we have -z = 6 [l-Vl + ^log(l Now the function on the right-hand side can be developed in a power series of type (9) for sufficiently small values of \x - x \. This is seen most easily by reverting the equation *-s = a [l-exp{-^ + ^}]. . (10) and using a known theorem on the reversion of series. Let us assume, then, that the power series which is obtained in this way converges for \x - x \ n INTEGRATION IN SERIES 227 we use the value of y x to calculate the value y 2 of y for x = x 2 , provided \x 2 - x^-ca^ where a x is some positive number. In a similar way we can proceed from x 2 to x s , from x s to a; 4 , and so on. If the quantities x lt x 2 , x 3 ,... are suitably chosen, it may be possible in this way to calculate the value of y for x = x v . In practice, this method is very laborious, especially as the suc- cessive increments of x must generally be small in order that there may be no doubt about the convergence. In actual calculations it is customary therefore to use certain methods of approximation which will now be described. § 71. Runge's method of approximate solution. In the method of approximation invented by Runge * the value of y for x = x p is calculated by successive steps in which intermediate values x lt x 2 , ... Xp-i oi x are used just as in the process of continuation, but instead of calculating y t , y 2 , ... with the aid of power series, certain expressions are used which are similar in form to those used in Simpson's method of approximate integration. The approximate expression for y x is, in fact, vs = vo + m + w + p") • • • • (ii) where P = h F(s , y ) p' = h F(x + \%, y + £P) x x = x + h P" = h ¥{x +h,r i ) r, =y +h ¥(x + h, y + p). When the series tfi - Vo = c i h + c z h2 + c 3^ 3 + (12) converges, the corresponding expansion for y x * - y can be shown to agree with it in the first three terms. To prove this we must show that the first three derivatives of y x * are equal to those of y x when h = 0. Now it follows at once from equation (2) that F(*o. 2/o) = A 00 ^- = A 10 Wf> = A 01 dx~} ~ 2A *° dx~dio ~ u 9y„ 2 ~ ° 2 where F stands for F(a; , y ) ; hence, when h = 0, \dhJo~ 00 (d*p\ _3F#3F_ wAr dx~ + dhWo~ 10 01 °° " 10 * 01 ~ 2 * Math. Ann. Bd. 46. This method and certain graphical methods of approxima, tion are described in C. Runge's Graphical Methods, New York (1912), p. 120. 228 DIFFERENTIAL EQUATIONS (*&) - 3 U»A _d 1 3 2 F 1 dp g 2 F .4 9a; 2 2 d/i dx dy — f [-^-20 + C l-A-ll + C l A 02 ] + 4 W 9 a F ' f^R = 2 TM + ^ *3' V dh 2 / o L9x 3?/ dfe. m) - 3 rs + 2(A 10 + A 0lCl ) = 4c 2 9 2 F cZy] f^FAfyy 9F^5~ \ dh 3 J o~ " Ldx 2 ^ dx dy dh dy 2 \dhJ dy dh 2 _ = 6[A 20 + Cj.Au + c i 2A 02 + 2cgA J Consequently, if we make use of the relations (4), we find that V dh Jo V dh 2 2c, and this means that the first three derivatives of y^* agree with those of 2/j when h = 0. It follows from this result that the quantity «/j* will often give a good approximation for y x whether the series (12) converges or not. In fact, if y x = f(h), we may write 2/i - 2/o = c x h + c 2 h 2 + c,ft» + J/""W where 6 is some positive quantity less than 1, the last term repre- senting the remainder in Maclaurin's expansion of f(h). Other formulae of approximation have been given by Heun * and Kutta.f The latter has compared the results obtained by a number of methods of approximation by considering the case of the equation dy y - x ., - /(*. y) dx y + x when the supplementary condition states that y = 1 when x The accurate solution, as found by the method of §21, is 0. log (x 2 + y 2 ) - 2 tan y The different methods to be compared may be described as follows : / = y /' = 2{x 2 y y + x /'" = - ./" 4(z 2 + y 2 )(2y - x) (y + x) z (y + xf 20(z 2 + y 2 ){3y 2 - 2xy + x 2 ) {y + x) 7 Ay=fA*+r^ + r { -^+r {Ax)i (Taylor.) 2! ' J 3! ' J 4! 1 Zeitschrift fur Math. u. Phys. Bd. 45. -\Ibid. Bd. 46. INTEGRATION IN SERIES 229 3° 3°. 4°. A'" 5°. A' = if(x, y)Ax A" = J/(x + |Ax, y + A') A* A'" = |/(x + \Ax, y + A' + A")Ax A"" = J/(a! + fAx, 2/ + A' + A" + A'")Ax Ay = A' + A" + A'" + A"" (Euler.) A' = f(x, y)Ax A" = /(x + %Ax, y + iA')Ax A'" = /(x + f Ax, y + f A")Ax A' + 3A'" Ay (Heun.) A' = f{x, V)Ax A" = f{x + AAx, y + $A')Ax -- f(x + Ax, y + A')Ax A"" = f{x + Ax, y + A'")Ax A' + 4A" + A"" Ay 6 (Runge.) Ax A'\ . = .fl« + — , y + ^-)Ax A' =f(x,y)Ax A" =f(x , , ., ., A'" = f{x + fAx, y + A" - £A')Ax A"" = /(x + Ax, y + A'" - A" + A')Ax A' + 3A" + 3A'" + A"" , rr , Ay = = (Kutta.. Let the value of y be required for x = 1. The interval (0, 1) will be divided up into three smaller intervals in which Ax = 0'2, 03, and - 5 respectively. The value of Ay must then be calculated for each of these intervals by the various methods of approximation. If Ajy, A%y, A s y are the calculated values of the three increments, the approximate value of y for x = 1 is y x = 1 + A ± y + A^ + A& The values of Ajy, Aay, A$, as calculated by the different methods, are shown in the following table. The last column contains the values of Ay calculated by taking Ax = 1 : Aiy Aay AsV Aoy Taylor 1666667 •3368533 •4936913 Euler - 1754353 •3573505 •5367900 62242 Heun- •1680250 •3395806 •4990390 51613 Runge ■1678487 ■3393690 •4991167 52381 Kutta T678449 •3392158 •4982940 49914 True value - 1678417 •3392094 •4982784 49828 Some applications of the method of approximation are given in R. Emden's Gaskugeln, Leipzig (1906). W. J. Harrison {Cambr. Phil. Trans., Vol. XXII (1913), p. 39) has recently used Runge's method to solve the equation dp A/, B N x ) and x 1 ~y F(q + 1-y»P + 1-Y» 2 - Y > x ) > consequently we have two solutions of the hypergeometric equation which are usually independent, but which become identical when y is a positive integer. Let us now put y = n + e where n is a positive integer, and consider the limiting value of the expression Its r(a + g)r(p + p) tV T(y +p)T(l +p) X _ y. T(a + 1 - y + r)r(P + 1 - y + r) x i- v+ r~ & T(2 - Y + r)T(l + r) when e -> 0. Since the expression satisfies the hypergeometric equation when s is not zero, we can expect its limiting value for s = to be also a solution of the hypergeometric equation. Now, since T(l - e) =(- e)(- e - 1)...(- e - n + p + l)r(- £ - n + p + 1) we have lim^r; rr = ( - l) n ' p (n - » - 1)! n>p e ^o er( - s - n + p + 1) Also - I> + P )T® + P) _ r(q + F - s)r(p + y - e) "I I> + p + s)T(l +p) T(l +p - s)I> + p) J is easily seen to be equal to r(q + j»)r(p + p) r(n + p)T{\ + p) i> + p) r f (p + y) r> + p) r'(i + P n + T(x +p) r(p + p) i> + p) r(i + p) J where r'(z) denotes the derivative of T(z). lim- e->0S x [log 234 DIFFERENTIAL EQUATIONS Again, it follows from the relation r(a + p) = (a + p - l)...aT(a) that h. + p) „ i i 9 + ...i + m r(a + p) a + p - 1 a+p-2 a r(a) hence, we finally obtain as the limiting value of our expression, r(n) * (a - p ' n • x) L log x + r(a) + r(p) r(n) r(i) J + F^a, (3 ; w ; x) where r £f r(» + p)r(i + p) =E i + o + '" a + p - 1 p 1 111 1 P + J3 - 1 1 "\p TO "w+^-l- r(a + 1 - n - r)r(P + 1 - n - r) i I>-r-l) r r=0 S(-!)' ! T(l + r) When y and 1 - a are positive integers, we may obtain two independent solutions of the hypergeometric equation in the following way. We have already seen that the equation possesses the particular solution and can be derived from the differential equation Xl-*)£ + [«-Y+2 + (P-«-3)*]£ | (2) + (p - a - 1)m = Oj which is satisfied by u = sfl~ a ~ 1 (l - a;)' 3- ? by a suitable process of transformation. If, then, we find a second solution of this last equation, a second solution of the hypergeometric equation can be derived from it by the transformation. Putting u = a;v- a - 1 (l - tff—iv, and using the method of §44, we find that a second solution of the differential equation (2) is u = xv-»-i(i _ x y-y — x sa ,, hence a second solution of the hypergeometric equation is y 2 = a;i-v(i _ x )v-*-P. d ~° I ' At ' %y (1 _ x y-y[ *1 dx- a L v ' Jzv-^i - x)^-y +l INTEGRATION IN SERIES 235 Expanding the integrand in ascending powers of x and integrating, we see that this solution is of the form «/ 2 = A log x . F(oc, (3 ; y ; x) + ]£ b r x r r=l-y where A and b r are constants. The case in which y - 1 is a negative integer may be discussed by making the substitution y = x l ~yw, when it is found that w satisfies a hypergeometric equation in which a, (3, y are replaced by oc + 1 - y, (3 + 1 - y, 2 - y respectively. Since 2 - y is now a positive integer, the preceding analysis is applicable. Other expressions for the solutions of the hypergeometric equation may be obtained by making a change in the independent variable. If we write 1 - x in place of x, we find that the differential equation is also satisfied by y a = F(oc, p;a + p-y + l;l-a;) y 4 = (1 - x)^- a -^F(y -a,y-p;y-a-p + l;l-:B) while if we make the substitution x = — , we find that the equation is satisfied by the functions x «/ 5 = x ' aV (a, a-Y + i;«-P + i;-) % =z-"F(p.p -y + l;(3-« + l;^ By combining these transformations, we can also obtain series 11 x of powers of 1 , -z and respectively. The problem of finding the relations between the different solutions which are obtained in this way is of some interest. For a discussion of this problem the reader is referred to Forsyth's Differential Equations and to a memoir by Dr. E. W. Barnes, Proc. London Math. Soc. Ser. 2, Vol. 6 (1908), p. 141. EXAMPLES. 1. Prove that F(a, p ; r ; x) = (1 - x)y- a - p F(y - a, y - (3 ; y ; *) F(a, p ; Y ; x) = (1 - x)~^U, Y - a, Y , — i- 2. Show that the hypergeometric equation can be reduced to the normal form d'v v dx a+ 4 X 2 1 - [i 2 , 1 + (i. 2 - X 2 - v 2 x 2 (1 - a;) 2 a;(l - x) where X 2 =(l- Y ) 2 |x 2 =(a-p) 2 v 2 = ( Y - a - p) = 236 DIFFERENTIAL EQUATIONS 3. Verify that F( - n, (3 ; p ; - x) = (1 + x) n 2F( - \n, - \n + i ; i ; a 2 ) = (1 + x)» + (1 sm ! i = 2 2 2 J nx n , n 3 . , \ sinra; F ( 1 + -s. 1 - s: 5 t: 5 sm * = — . 1 — — * — s 0111 .*/ I =^ ; 2 2 2 / nsmxcosa; „ /n + 1 I - n 1 . , \ cos nx F ( , ; - ; sin's 2 ' 2 ' 2 ' / cos x n + 1 n ; n + 1 ; *) = 2»{1 + Vl 4~ m 2 2 4. Obtain the recurrence formulae F(a, p + 1 ; y ; x) - F(a, p ; y ; a) = — F(a + 1, p + 1 ; y + 1 ; ») F(a+ 1,P+ 1; y; x) - F(a, P; y; x) = ^F(a+ 1, p+ 1; y + 1 ; a) ^-F(ce, P ; y ; x) = ^F(a + 1, p + 1 ; y + 1 ; x) dx y § 74. 4 limiting case of the hypergeometric equation. If we put x = - in the hypergeometric equation and make |3 ->» , we obtain the equation d 2 y , , dy , . which may be written in the forms &(» + y - I)?/ - z(» + a)?/ = where »■ = z -j-. This equation possesses the solutions 2/i = F(a ; y ; z) «/ 2 = ^-»F(« - T + i ; 2 - y ; 2) where F(a ; y ; z) denotes Rummer's function . a oc( y 2 , ... y n ; x) = x + ai-a, -Om g + «i(ai + l)« a (a a + 1) ... a m (a m + 1) ^ + l-Yi-Ya-Y* 1-2-Y 1 (Yi+ 1)Y 2 (Y2+ 1)-Y„(Y»+ !) and by n other series of type, i/! = a^Tfaj - Yj + 1, a 2 - Tl + 1, ... a w - Yl + 1 ; 2 - Yi. Yz - Yi + 1. ••. Y„ - Yi + 1 : x ) If m j> n, the series converge for all values of x, but if m = n + 1 the series converge, for |oj| <1. If m > n + 1, the series diverge. (Pochhammer.) § 75. Bessel's equation. Bessel's equation X *% + X % + {X * - V * )V = ° • • • • W may be written in the form (& 2 - v 2 )«/ + x 2 y = The indicia! equation is now p 2 - v 2 = 0, and so there are two solutions _ ^ ( - l) f g»+" «/i - 2^2*+T I T(r + v + 1) " ° v{X) y ( - l)^ 2 '-" _ T M 2/2 ~ P -^2*—r!r(r - v + 1) " " W These series converge for all finite values of x. When n is an integer, we have 3_ n {x) = (- l) n J n (x), and so the two solutions already obtained are not independent. To obtain a second solution in this case, we choose a number v = n + s very nearly equal to n and consider the limiting value of i[J„(x) - (- l)"J-»] as s -> 0. This limit, which is usually denoted by the symbol izY n (x) can be written in the form « Y -w-[|i j 'wL- ( - l)m K J -' (a,, L 238 DIFFERENTIAL EQUATIONS Making use of the relations lirn^ *— — n = (- l)-*(» -P -1)1 ,_>'<> er(- s - to + p + 1) d (x\ 2r +" (x\ ir + v , (x\ n\a) =y iog y sy =-y iog y i im ir kJ? { _ n. LzJT! I „"" 9vLr! T(r + v + 1) v ; (r + n)! T(r + n - v + 1)J ( - l)^ 1 [ T'(r + n + 1) r'(r + 1) 1 _ r! (r + to)! Lr(r + to + 1) + T(r + 1) J r '( r + J ) i i * r 'd) ,TW 1 r'(i) where y = - p.-j. , we finally obtain ^ (to + r)\r\ v ' v ,J \2/ When v is not an integer, it is convenient to define the function Y v (x) by the equation Y„(a;) = cosec V7r[cos vtc . J v (x) - J_„(a;)] then, as v -> to, Y„(a;) h* Y„(a;). In all cases J„(x) and Y v (x) may be taken as two independent solutions of equation (1). The function K„(x), which has been used previously, can be defined by the equation K,(«) = -fe~[J,(a) -»X(*)] The function e 2 J„(ia;) is usually denoted by the symbol I„(a;) ; it satisfies the differential equation * 2 S + *% ~ v + *>* = ° If we write w = e~"I v (x), we find that w satisfies the equation d 2 d x 2 -r— 2 (e x w) + x-j-(e x w) - (x 2 + v 2 )e x w = ,d 2 w dw 'dx~ 2+{2x2+X ^ or x 2 -r^ + (2a; 2 + x )~ + (x - v 2 )w INTEGRATION IN SERIES 239 Writing this equation in the form (a 2 - v 2 )w + x(2a + l)w = and applying the method of § 72, we see that there are two power series satisfying the equation. One starts with x v and the other with x~ v ; it is clear, then, that if v is positive, the former must be a constant multiple of e~ x I v {x). Denoting the power series by QO ~S\ a n x v+n , we have the recurrence formula [(v + nf - v 2 ]a„ + (2v + 2n - l)a n _ 1 = Hence it follows that e~ x I v (x) t" r <~+~i)L , 2v + 1 x + (2v + l)(2v + 3) 2-r(v + 1)L 1(1 + 2v) 1.2. (2v + l)(2v +2) "J There is an important relation between J r {x) and Y v (x), which is easily proved. Let us write y = J y (x), z = Y (x) ; then, if we multiply the equations dx[ dx L dx. by z and y respectively and subtract, we get d f / dy dz\ l'( Z dx\_"\ dx dx) _ dy dz A .„. .'. 3 ^ - y-r = - ( 2 ) ax ax x where A is some constant To determine the value of the constant, we make use of the formula z = Y„(a;) = cosec V7i[cos v7uT„(a;) - J-„(x)] and the expressions for J v (x), J- V (x). The coefficient of - on the left-hand side of (2) is consequently cosec V7E r n . " r(i + v)r(i-v) [v + ^ = A Now r(i + v) = vr(v) and r»r(l - v) an V7C 2 therefore A '= dz dz 2 consequently */t X T~ ~ — dx dy tw? This relation still holds when v is an integer. 240 DIFFERENTIAL EQUATIONS EXAMPLES. 1. Prove that the functions J„(a;), Y v (x) satisfy the recurrence relations o v J„ +1 (z) + J v+1 {x) = — J„(x) <2J„ v d^ = x Jv{x) ~ J "+ l(:B) dJ " T I \ V T / \ § = K J »-.(*) - Jh-i<*)] ^f = *[J W («) + -W*) - 2J,(*)J 2. Prove that the function F n (x) = e~ x I n {x) satisfies the equation ^=J[F„_ 1 + F n+1 -2F n j 3. Prove that the equation dH — - + a 2 a; m v = ax 2 is satisfied by v= xi[AJ n (az) + BY„(oz)] where z = 2nx 2n - 4. Prove that j (x) cos * - « y ("DT(v+2n + « „ J„(*) cos x - ^ S (2^)TT(2vT2^Tl) ( } J,(.) S .n,= LI j (^TI)Tr(27T2^T2) (2x) + 2 5. Prove that Y„_ 1 (a;)J > ,(a;) - Y v (x)3 v _ 1 (x) = — § 76. Legendre's equation. If we put x = - in Legendre's equation „, d 2 M „ du it takes the form 2 _ IN dz 2 dz , , 1. 2 When m + \ is an integer, there is only one power series in - which satisfies the equation ; a second solution for this case may be derived from the second solution of the hypergeometric equation when y is an integer. It should be noticed that Legendre's equation is transformed into a hypergeometric equation by putting 1 - x = 2s ; hence the function P(m + 1, - n; 1 ; — - — J satisfies Legendre's equation. EXAMPLES. 1. Prove that Legendre's equation is satisfied by the coefficient of a n in the expansion of (1 - 2ax + a 2 ) -1 in ascending powers of a. Hence show that this coefficient is P„(a>). 2. Prove that T? n (z) and Q„(x) satisfy the recurrence formulae of type nP„ - 2(w - l)asP B _! + (n - 1)P„_ 2 = PI - aPUx = mP„_, *K - K-i = riP„ where primes denote differentiations with respect to x. b.e, Q 242 DIFFERENTIAL EQUATIONS § 77. Two convergence theorems. The method of integration by means of power series can be applied to differential equations of a much more general character than those which have already been considered, but the recurrence formulae for determining the co- efficients are rather complicated. There are, however, two general convergence theorems which will be given here for the case of a linear differential equation of the second order. If P = p + pjx + p 2 x 2 + ... , Q = q + q x x + q 2 x* + ... (1) are two power series which converge for \x\ < R, the differential equation d 2 y dy has solutions of the type y = A + A-^x + A 2 x 2 + ... , where A and A x are arbitrary, and the other coefficients are determined uniquely in terms of them by the recurrence formula n(n - 1)A„ = (n - 1)A„_^ + (n - 2)A n _ 2 p 1 + ... A l2 )„_ a \ + A„_ 2 g- + ... Ai?„_ 3 + A q n _J It follows that if the values of y and -j- are given for x = 0, there is one and only one solution y which can be represented by a power series of the above type. To establish the convergence of the series for y, we remark that since the series for P and Q converge for \x\ < R, we can find a positive number M such that i i M , , M \Pn\<- n Wn\<^+ 1 where r is any pre-assigned positive number less than R. This is a particular case of the theorem proved in § 70 for the case of a power series in two variables. These inequalities imply that \{n - s)p s _ 1 + q s _ 2 \ <(n - s + 1)^ We now compare the recurrence formula (3) with a second one, M nT , JV[_ M lfn-2 + "Qyn- which is such that the coefficient of B„_ s is positive and numerically greater than the modulus of the coefficient of A„_ s , and possesses the useful property that the limiting value of the ratio -^^ is easily estimated. We have, in fact, B n(n - l)B n -\{n - l)(n - 2)1^ = nB^M and so lim -^~± = r n(n - 1)B„ = nB^M + (» - 1)B„_ 2 ^ + ... 21^ + B - , INTEGRATION IN SERIES 243 It follows from this result that the series 2B„x" converges if \x\ S and define B p = | A^ | for p < S, we shall have |A„| 1, there are oo' solutions which satisfy the supplementary conditions, for then both y 2 and ~ vanish for x = ; consequently, any constant multiple of y 2 may be added to a solution which is known to satisfy the supplementary conditions, and a new solution is obtained which satisfies the requirements. The case in which a logarithmic term appears in y 2 can.be dis- cussed in a similar way. EXAMPLES. 1. Prove that an equation which possesses the two linearly inde- pendent solutions y = u x (x), y = u 2 (x) is given by d*y dy dx* dx V d 2 u, du, i i It, dx" dx d 2 u 2 du 2 — - u 2 dx 2 dx = INTEGRATION IN SERIES 245 2. Prove that a linear differential equation of the second order, which possesses two linearly independent solutions of the form y = [exp Kz- 1 + ... a n ar n )] [ 2 b m x m ] is necessarily of the form where Pj(a;), P 2 (k), P,(a;) denote power series of type Xc m x m . 771=0 3. Obtain solutions of the following equations in the form of power series : d*y dy dx 2 dx d*y 3 d 2 y dx 3 x dx 2 f _ in 2 - I N, dy 4 \ x 2 ) dx x § 78. The method of successive approximations.* Let us consider the two differential equations g=A(w) £=/.<*.¥.«> • • (1) where f^x, y, z), f 2 (x, y, z) are functions which can be represented by power series /i(*. «/.*) = 2 A ™.».> - a ) m (y - b ) n ( z - c ) p m, n,p which converge absolutely and uniformly in the region \x - a\ y. *)M *{ W -y\ + V - A) For we have and the absolute values of the integrands never exceed k ; hence the theorem follows. * Cf. E. Picard, TraiU d' Analyse, t. 2, 2nd ed. (1905), p. 340. For a system of linear differential equations the method has been developed in a very convenient form by H. F. Baker : Proc. London, Math. Soc. Ser. 1, Vol. 35 (1902), p. 337 ; Ser. 2, Vol. 2 (1904), p. 293 ; Phil. Trans. A., Vol. 216 (1916), p. 129. 246 DIFFERENTIAL EQUATIONS We shall now endeavour to determine y and z as functions of x, so that y = b and z = c when x = a. We commence with the approximate values y = b, z = c, then determine functions y v z x with the aid of the equations -jfc = A(*. 2/0. «o) ^ = /■(*. y . z o) and the initial conditions y x = b, z 1 = c when x = a. We next determine functions y 2 , z 2 with the aid of the equations and the initial conditions y 2 = b, z 2 = c. Continuing in this way, we obtain a series of successive approximations (y , z ), (y v zj, (y 2 , z 2 ), ... (y n , z n ) ... . We must now show that as to-* 00, the functions y n , z n tend to limiting functions y, z, which satisfy the differential equations. „ Let p be the smaller of the two positive quantities r, — ) and let us limit the variable x to the region \x - a\ < p. "*■ We have y 1 - b = A(5, b, c) i\ J a where the path of integration is the straight line joining the points a and x in the complex plane. The length of this line is \x - a\, and since \fi{^,b, c) | < M, we have \y 1 - 6|oo , this limit is the sum of the infinite series b + u 1 + u 2 f ... We shall show that this series converges absolutely and uniformly for \x - a\ < p. We have M n = f Ul( x . Vn-V Z »-l) - fli*. Un-t. Z n- 2 )] dx Now \fi{K, Vn-l,Z«-l) - fi{%,yn-2,Zn-2)\^k{\y n _ 1 - y n _ 2 \ + |z b _ x - Z„_ 2 \} hence \f x {x, y lt z x ) - /^x, y , z )\ < *(!%! + | v x [) < 2&M|a; - a| 2! dx INTEGRATION IN SERIES 247 and so \u 2 \<2kM.( X ~^\x - a\d\x - a\ = 2m \ x ~ a ? Jo * ! c - ., , , , 2kM\x - a\ 2 Similarly, | v 2 | < L_ L Again, l/l(*. 2/2- Z 2 ) - /l(*. 2/l- Zl) I ^ *K| M 2 | + N) ^ ~ 9|, and since u s = f [/^z, y a , z a ) - / x (x, y lf zj] , . (2A) 1 MP*-»I. .... , {2k)m\x - a\ 3 we have |w 3 | < I |x - a| 2 rf|a; - a\ = %- - Z I Jo o ! Continuing in this way, we find that . (2i)«- 1 M|a; - o|» , . 9 „ s Krzl <-^— ^ ^7 L (71 = 1,2,3,...) (2k) n ~ 1 Mp n Since the series with the general term - — - — j — — converges, it follows that the series n ' b + u 1 + u 2 + u 3 + ... converges absolutely and uniformly for \x - a| ^ p. Hence y n tends to a limit y and similarly z n tends to a limit z, and these functions y and z are continuous functions of x. It now follows from the equation y n = & + \Jl( X ' Vn-l, z n -i) *» that y = b + I f^x, y, z) dx and, similarly, we can prove that z = c + /afc, V> z ) dx Hence y and z satisfy the differential equations and the specified boundary conditions. Some particular cases of the foregoing theorem are of special importance. 1°. If /i(a;, y, z) is independent of z, we have a method of solving the equation g^ 2°. If/i(a;, y, z),f 2 (x, y, z) are linear functions of y and z, we. obtain two linear equations % = A i(% + B i( x ) z + Cxfc) dz fa = A 2 (% + B 2 (z)z + C,(x) 248 DIFFERENTIAL EQUATIONS 3°. In particular, if f^x, y, z) = z, we obtain the equation dx2 - h \*. y, d j which becomes a linear equation of the second order when f^x, y, z) is a linear function of y and z. The above method of approximation is closely related to a graphical method of solution which has been developed by J. Massau* and others. f If the differential equation is^> = fix, y), the first step is to draw the isoclinals or curves along which / has a constant value, and to associate with each curve a line y = fx drawn in a direction making an angle whose tangent is / with the axis of x. This correspondence between straight lines and isoclinals is inde- pendent of the system of rectangular coordinates, and gives a convenient graphical representation of the differential equation. Now let a system of isoclinals a, b, c,d,e,... be b cde drawn at small intervals , and let the first of these pass through the point P, whose coordinates (x , y ) represent the initial state of affairs. Through P we draw a line in the direction associated with a to a point Q somewhere be- tween a and b. Through Q we draw a line in the direction associated with b to a point R somewhere between b and c, and so on. The curve touching this broken line at its points of intersection with a, b, c, ... respectively is a first approximation. J To find a better approximation we introduce a rectangular system of coordinates x, y, laying the axis of x somewhat in the mean direction of the broken line. Let y t be the function of x corresponding to the first approximation. The second approximation y 2 is then obtained as an integral curve of f(x, t/ x ) ; in other words, 2/2 = 2/o+ /(*» 2/i) dx To find y 2 we choose the new axis of y at unit distance from the old origin through which the rays were drawn, and associate the ordinates of the points in which the rays meet this axis with the abscissa of the points in which the broken line PQR... meets the corresponding curves a, b, c, ... . Plotting the points with these ordinates and abscissae as coordinates, we obtain a series of points on a curve Y=Mft) * ' Memoire sur l'integration graphique et ses applications,' Ann. de V Assoc, des ing. sortis des icoles spdciaks de Gand (1878), (1884), (1886), (1887), (1890). f See M. d'Ocagne, Calcul graphique et nomographic (1908), p. 149 ; C. Runge, Graphical Methods, New York (1912), p. 120. I This curve need not be drawn. INTEGRATION IN SERIES 249 This curve is to be integrated graphically, beginning at the point associated with P. Adopting the method of approximate integration in which the function /is assumed to have a constant value in a series of intervals each of which contains one of x's for points that have been plotted, we obtain a second broken line, which is treated in the same way as the first, and so on. We thus obtain a series of functions y 1 ,y i ,...y n ,..., and according to the theory of the method of successive approximations they would converge to the true solution if the integration were exact instead of only approximate. For a fuller account of the method and an estimate of the possible error, the reader is referred to Runge's book. There is another method of approximation which is particularly useful * when we wish to establish the existence of a solution of the equation -, 5 -**.»> (1) in the case when the variables are real and f{x, y) is not restricted as much as in § 70. The following presentation of the method is taken from the lectures given by G. A. Bliss at the Princeton Colloquium t and from Picard's Traitd d' Analyse, t. 2, p. 322. Let us suppose that the function /(a;, y) is continuous in the interior of a certain region R of the ;n/-plane, and such that the quotient /to y') - /to y) , 9 , T^y (2) is finite when {x, y) and (x, y') lie in any closed region whose points are all interior to R. This restriction was introduced by Lipschitz. A so-called Cauchy polygon for the equation (1) through a point (£, y\) interior to R is defined by means of equations of the form V\ = t\ +/& t])( x i - I) 2/2=2/1+ /to. yd to: - x d y = y n -i + /to-i. y n -i)( x - x n -i) The division points E, < x x < x 2 < ... may be taken for convenience at equal intervals 8 from each other. Any value x > \ will lie on one of the intervals x n _ 1 x n , and the polygon will either be well-defined for all such values, or else there will be a constant [3 such that for every x in the interval \ ^ x < (3 the points of the polygon are interior to R, while for x = (3 the corresponding point {x, y) will be a point of the boundary of R. The polygon { defined by the equations above may be denoted by *See E. Picard, Comptes Bendus, Vol. 128 (1899), p. 1363; P. Painleve, Bull, de la Soc. Math, de France, Vol. 27 (1899), p. 151. t The Princeton Colloquium Lectures on Mathematics, New York (1913). %i.e. the function y = P^a;). 250 DIFFERENTIAL EQUATIONS Pj^a;), and the analogous one when the division points are distant d/2 n ~ 1 from each other by P n (x). A common interval £ — x< a for two functions P(x), Q(x) with respect to any region R may be defined as one over which both are interior to R, and one such that on any ordinate of the interval all the points between {x, P(x)} and {x, Q{x)} are also interior points of R. Consider now a closed region R x interior to R and containing the point (£, 7)), and let in and k be two constants greater respectively than the absolute values of fix, y) and the quotient (2) in the region R r If I > is given in advance, the partitions for any two polygons P(x), Q(x) through (£, tj) can be taken so small that |P(a;) - Q(a;)|<|{e*l*-*l - 1} for all values of x in any common interval of P(x) and Q(x) with respect to R r For at any point (x, y), where y = P(a;), the equation P' = f(x, P) + {fix n _ v y n _ r ) - fix, P)} = fix, P) + p is satisfied by the forward and backward derivatives of the polygon P. On account of the continuity of fix, y) there exists for any I a constant jx such that the inequalities \x - x'\ < (j. \y - y'\ arbitrarily, and I so small that | e ta, _ l| < s Then |P n .(aO - P„(x)| < |{e te ' - 1} < £ provided that the intervals d/2 n '~ 1 and d/2 n ~ 1 are each less than the constant [j. corresponding to I. Hence the sequence ~P n {x) converges uniformly to a continuous function y{x) on the interval t, < x < £ + a v The equations P»(*) = *1 + pY(*) dx = 7) + f{M P«) + P»}^ 252 DIFFERENTIAL EQUATIONS hold for every n, and the sequences {f(x, PJ} and {p„} approach uniformly the limits/{ x, y(x)} and zero, respectively. Hence 2/0*0 = 7) + f{x,y(x)}dx from which it follows by differentiation that y(x) is a solution of the differential equation. It is easy to show by means of the inequality (3) that there is only one continuous solution y = y(x) of the differential equation (1) in the region R and passing through (£, 7)) . For suppose there were another, Y(a;), distinct from y(x) at a value x' > i*. There would then be a value £j < x' at which yi^j) = Y^), and such that the two solutions would be distinct throughout the interval £i < x < x' . In a neigh- bourhood of the point of intersection (& t , yjj) interior to R, a relation d(J d ~ V) | = l/fc Y) - f{x, y)\ < k\Y - y\ would be satisfied, and hence, from the inequality (3) |Y-j/|<0 This contradicts the hypothesis that y(x) and Y(x) are distinct throughout the interval \ x < x < x± . It should be mentioned that Bendixon has shown * that if y = )) are con- tinuous in all three of their arguments, and if the function f{x, y) has continuous first derivatives with respect to x and y in the interior of the region R, then 9 and

- 1. dy ,- 2. The differential equation ■=- = y V possesses a solution y(x) = \{x - x + 2yV ) 2 x = x o ~ 2 VVo = x^x - 2yV *Bull. de la Hoc. Math, de France, Vol. 24 (1896), p. 220. See also G. A. Bliss, loc. cit., p. 95. t Math. Ann. Bd. 76(1915), p. 471. INTEGRATION IN SERIES 253 which satisfies the conditions y = 2/o when x = -^o y = when x = (O. Perron.) § 79. The integration of linear partial differential equations by means of infinite series. In physical mathematics it is frequently necessary to find a solution of a harmonic linear partial differential equation L,(«) = M ( (w) (1) under the conditions that u satisfies certain supplementary equations or boundary conditions for all values of t within a certain range, and certain other supplementary conditions for all values of x within a certain interval. We shall find it convenient to call these two sets of supplementary conditions X and T respectively. The method which has proved most successful in the treatment of problems of this kind is based on the fact that the sum of any number of particular solutions of a linear differential equation of the above type is generally a solution ofe the quation. Now, in the present case the partial differential equation possesses an infinite number of particular solutions of type u = u.{x)${t) where ol(x) and $(t) satisfy the ordinary linear differential equations L^oc) + Xa = M t ((3) + X(3 = X being an arbitrary constant. Let us suppose that a series of constants X 1; X 2 , ••• and a corresponding series of functions a 1 (a;) ) Pi(0 '• <*i{ x )> P2C) >' ••■ can b e found such that each of the functions a. n (x) satisfies the conditions X, and each of the functions |3 M (<) satisfies all but one of the conditions T. Then, if «. n (x), (3„(£) satisfy the equations L.(a) + X„a = M,(P) + X„p = . (2) respectively, the series „ « = S ±flJ?)$n® (3) will satisfy the partial differential equation (1), the conditions X and all but one of the conditions T whatever may be the values of the constant coefficients A n . The series (3) will represent the com- plete solution of the problem if the coefficients A n can be chosen in a unique manner so that the last of the conditions T is satisfied, and so that the series converges in a suitable manner if m is infinite. The last condition which has to be satisfied is generally of the form u = f(x) when t = and a -->g + »— if the conditions X state that a is to be finite when x = 1 and x = - 1, the constant X must have a value such as s(s + 1), where s is an integer. These values are all real and infinite in number. 2°. It is of some interest to ascertain the different possible types of conditions X for which p (a r -r- 2 - On-p-) is zero at the limits a and b. A useful discussion of this question is given in a paper by D. Hilbert.t In the case of Legendre's equation, the conditions stated above may be used, provided a = - 1,6 =+ 1. 3° The nature of the convergence of a series of type 2c„oc„(a;) must be ascertained. The question of the possibility of expanding a given function f(x) in a series of this type must be answered, and finally the series (3) must be studied with the object of finding if it represents a continuous function of t for t = 0. These questions are too difficult to be studied here in their full generality, so we shall limit the present discussion to that of a simple case which will illustrate the method. For the more elaborate studies the reader is referred to some of the original memoirs.} Let us consider the partial differential equation d*v dv , n ,, w = m (°<*<9 and the supplementary conditions (X) v = for x = and x = I (t> 0) (T) v is finite for t = + oo v = f(x) for t = (0 < x < I) * Proceedings of the 5th International Congress of Mathematicians, Cambridge (1912), Vol. 1, p. 162. See also G. D. Birkhoff, Trans. Amer. Math. Soc, Vol. 10, 1909, p. 259. f Oottinger Nachrichten (1904). See also G. D. Birkhoff, I.e. % See, for instance, A. Kneser, Math. Ann., Vol. 58 (1903), p. 81 ; Vol. 60 (1905), p. 402; Vol. 63 (1907), p. 477; D. Hilbert, loc. cit. ; M. Mason, Trans. Amer. Math. Soc, Vol. 8 (1907), p. 431 ; A. C. Dixon, Proc. London Math. Soc, Ser. 2, Vol. 3, p. 83 ; Vol. 4, p. 411 ; E. W. Hobson, Ibid., Vol. 6, p. 349 ; Vol. 7, pp. 24, 359; J. Mercer, Phil. Trans., A. Vol. 211 (1912), p. Ill; A. C. Dixon, Ibid., p. 411 ; Haar, Math. Ann., Vol. 69 (1910), p. 331 ; Vol. 71 (1911), p. 38. 256 DIFFERENTIAL EQUATIONS The equation -5-5 + Xv = possesses a solution v = A sin (a;\A) + B cos (a;-\A) , which satisfies the conditions X if Zy'X = mz, where n is an integer ; for then it is sufficient to take B = 0. The equation -3- + Xv = possesses a solution u = Ce~ w , which satisfies the first of conditions T if the real part of X is not negative. The value X = raV/Z 2 certainly satisfies this requirement. We must now consider the infinite series A„e l sin -3- n=l ' If this represents a continuous function of t for i = 0, it will satisfy the second of conditions T if /(») = 2 A «sin^ Writing v- n \ x ) = sm -J - «,(!) = sin — p we see that the condition a r - T - 5 - a„ -r- 1 =0 for a; = and a; = Z r dx n dx is satisfied ; hence the coefficients in the expansion may be deter- mined with the aid of the formula P . n-KX . nra, , I sm —j— sin —p ax = (n =f= r) 1 Jo * * We also find by direct integration that I sin 2 -^=— da; = £Z jo I Now Fourier's theorem * tells us that if f(x) is a single-valued function of x in the interval 0-0 Let us assume that f(x) satisfies the conditions of this theorem ; * then, since the series is convergent, we may conclude that each of its terms, at any rate after some finite value of n, is numerically less than some finite positive quantity M. If, moreover, we exclude the points of discontinuity of f(x) at which the series does not converge by small intervals, we may choose M so that it is inde- pendent of x, and make the finite value of n to which we have just referred also independent of x. We now have . . mzx A„sin — 2s A„sin^e * converges absolutely and uniformly for any range of x which does not include any of the excluded intervals and for any positive interval of t. Putting s = 0, 1 in succession, we see that the series oo _«V( A„sin— j— e ' tt 2 ^. 2 . nnx -^ converge uniformly ; hence the second series may be derived from the first either by differentiating it term by term with regard to t or by differentiating it twice term by term with regard to x.\ This * These are usually called Dirichlet's conditions. t The cosine series which is derived by one differentiation is convergent, because it can be derived from a uniformly convergent series (viz. the second series) by integra- tion with regard to x. We may prove by an application of the preceding method that the cosine series is uniformly convergent. B.E. R 258 DIFFERENTIAL EQUATIONS indicates that the function v represented by the infinite series satisfies the partial differential equation -j-j = .-, except possibly in the excluded intervals. In the problems of mathematical physics the function f(x) does not generally become infinite, and so we shall assume that there are no excluded intervals. We have now to prove that as t -> '0 our function v tends to the limit f(x). To do this, let us write . , mzx J™* a n = A n sin -j- c n = e * pR n (t) = a n c n + a n+1 c n+1 + ... o, n +p+i c n+p+i p r„ = u n + a n+ i + ••• a n+p -i Then it is easy to see that pRnW = l r n{ c n ~ c n+l) + 2 r n( c n+l _ c n+i) m + ••• p-i?n[ c n+p-2 ~ c n+p-l) + P r n c n+p-l But 2 a n is convergent and c„ > c n+1 > c n+2 > c n+p _ 1 ; therefore 11=1 y n \ ... and IA.WI 0, and the uniformity of the convergence extends up to and includes the point t = 0. Hence v is a continuous function of i for < t , and so we may conclude that lim v(x, t) = S]A n sin - 1 - = f(x) t->0 ,1=1 ' The function v defined by the infinite series thus satisfies all the conditions of the problem. The general method of solving harmonic partial differential equations which has just been described may be extended to partial differential equations of the form L x (u) + My{u) + N*(«) = where ~L X , My, ~N Z are linear differential operators. The particular solutions from which the required solution is built up are now of the tyP e u = A{x)B{y)G(z) where L^A) + XA = Mj,(B) + [j,B = N*(C) + vC = and X + (j. + v = INTEGRATION IN SERIES 259 This method has already been used in the case of Laplace's equation d * u d z u d 2 u dx* dy z dz* and can also be applied when the equation is transformed to polar coordinates. We may, in fact, generalise the particular solutions of § 66 by writing «, «, u = 2 2> cos (mcp + s M , „)A m , „ P„»>(cos 6) n=-oo m=Q where s m>n and A^,, are constants at our disposal. In many cases the conditions of the physical problem indicate that A m _ „ is zero either when n is negative or when it is not negative. EXAMPLES. 1. Prove that the equation 4["-'>wK["->??] is satisfied by a series of type V=SA„P»P„(v) where P„([x) is Legendre's polynomial. Show also that if V = /((/,) when v = 1, the coefficients A„ may be derived with the aid of the formula r+i P m ((*)P» dp = (m + n) and deduce this formula from Legendre's equation. 2. Prove that the preceding differential equation is satisfied by Hence obtain an expansion for this function in a series of Legendre polynomials. 3. If H„(x) = F(a + n ; - n ; c ; x), where n is zero or a positive integer and c is positive, we have for a~>c — 1 J H m (a;)H„(a;)a: c (l - x) a -°dx =0 (m + n) When m = n the value of the integral is 1 r(n+l)[r(c)]T(a + n- C +l) o + 2w r(a + w)r(c + n) ' CHAPTER X THE SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS BY MEANS OF DEFINITE INTEGRALS § 80. Laplace's transformation. The method which will now be explained can often be employed with advantage when the dif- ferential equation we wish to solve has the form v a_ d m u L» ^Ss/m=0 • • • (1) m=0 »=0 ""«' and is particularly successful when q = 1. The success of the method depends on the possibility of finding an integrating factor of the expression miv)^t±a min t- d ' n dP 1 which is obtained from the left-hand side of (1) by replacing x n by -zr~ and -= — by t m - If o = 1, an integrating factor can be found dt n dx m J * oo at once by the method of § 7. In the general case an integrating factor is a solution of the equation M. t (w) = 0, adjoint to M s (v) = 0. This equation is defined as in § 18 by the identity wM t (v) - vM t (w) = -5-R(v,w) m*m -i;i;(-i) n « m , n ^(^ m=0 n=0 v and in the present case we have _ 'Wnn % dP d n v d n z, ,, d ^ , . WherC Z ltin ~ V d¥ { ~ ^ ^ dt V ^ _ . . d n -h) _ dz d n -h) d 2 z d n ~ 3 v SOLUTION BY DEFINITE INTEGRALS 261 After these preliminaries let us endeavour to satisfy the differential equation by a definite integral of the type first used by Euler, \e xt w{t)dt (2) where the path of integration is at present unspecified. Assuming that the definite integral can be differentiated under the integral sign a suitable number of times by the rule of Leibnitz, we have r L» = }L x {e xt )w(t)dt It should now be observed that we have identically ~L x {e xt ) = M.t{e xt ) hence LJw) = \M t {e xt )w{t)dt Now let w(t) be chosen so that Mt(w) = ; then the expression under the integral sign is an exact differential, and can be written in the form j Jt n { e«, W ) If, then, the limits of the integral (2) are chosen so that the quantity R vanishes at both limits or takes the same value at both limits, the definite integral will represent a solution of the differential equation (1). The equation M. t (w) = is called the Laplace transformed equation of Jj x {u) = 0. The relation between the two equations is such that if a solution of one can be obtained, then a solution of the other can generally be derived from it by Laplace's method. Petzval * has shown, in fact, that if e~ xt is used in place of e*', then L^w) = is the Laplace transformed equation of Mt(w) = 0. To verify this we have simply to prove that L x (e-* ( ) = Et(e- Xt ) — d m Now L» = 2 2 ( - l) m « m>n ^ (x n u) hence the preceding equation holds if d m d n and this can be shown to be an identity by using Leibnitz's theorem. It follows from Petzval's result, that if the first equation can be satisfied by an integral of the form u{x) = I e xt w(t) dt * Integration der linearen Differentialgleichungen, p. 472. 262 DIFFERENTIAL EQUATIONS the equation M t (w) = can often be satisfied by an integral of the form rh w(t) = I e' xt u{x) dx id To illustrate the use of Laplace's method, let us consider the equation _ . . d?u . .du ,„. M«) = X Jtf + (P + 1 + X ) fa + P u = ° • • ( 3 ) In this case Mt{v) = t(t + 1)-^ + [p + (p +q)t]v Mt(w) = t{t + 1)^ -[p-l+(p+q- 2)t]w hence we may take w = #> _1 (1 + t)^- 1 and then uM t (v) = -^[^(l + l) q v] The quantity ^(1 + t)ie xt vanishes when t = and when t = - 1, provided p > 0, q > ; it also vanishes when t = - go if x > 0. Writing - t in place of t, we obtain the two solutions u x {x) = e-»*fP-i(l - t)i-^dt ° p > 0, q > 0, z > u 2 {x) = e- a *^- 1 (l - i)?- 1 ^ Jo Sometimes it is convenient to use a definite integral of the type I sin(xt + z)w{t)dt in place of an integral containing the exponential function ; in this case we seek a relation of the form li x [sm(xt + e) = M. t [sin(xt + ej] For instance, if L» = x-^ + (2v + 1)^ + xu = . . (4) we have Mt(v) = (I 2 - 1) ^ + (2v + 1) Iv £ i = 2 + E An integrating factor of M t {v) is easily found to be w{t) = {t 2 - iy-i and we have wM«[cos (xt + e)] = -|-[(Z 2 - l)"+icos (a* + e)] SOLUTION BY DEFINITE INTEGRALS 263 If v > - \, the quantity within square brackets is zero when t =± 1 ; hence one solution of the differential equation (4) is given by f +i u = sin(xt + e)(l - py-idt where, s is an arbitrary constant. This solution is of the form u = sin e I cos xt (1 - t 2 )"' i dt and is finite when x = 0. Another definite integral satisfying the differential equation may be obtained in the following way. Let the identity L^wsina*] = ^-[(fi - iy+icosxt] (Jib be integrated twice with regard to x between and x ; then, if V(ar) = w(t) sin xt, we get xV(x) + (2v - 1) [ X Y{xj) dx x + [ X dx x i\v(x„) dx 2 \ JO JO J0 I /g\ Now, if - -| < v < -|, the quantity within square brackets is zero when t = 1 and also when t - oo ; hence it follows that the function p» u = sin xt(t 2 - iy-idt (6) satisfies the equation xu(x) + (2v - 1) I u(x-^j dx t + j dx x I x 2 u(x 2 ) dx 2 = (7) Jo Jo Jo Differentiating twice, we find that u satisfies the differential equation (4). It should be noticed that in this case it is not per- missible to differentiate the integral under the integral sign, but the integrations under the integral sign which aie needed when the expression (6) is substituted in (7) are quite justifiable. It is now easy to verify with the aid of (5) that the definite integral (6) satisfies equation (7). § 81. Equations of F 'faff 's type- A differential equation which can be written in the symbolical form M-i) - <41 u = .... (8) is said to be of Pfaff's type, in honour of Pfaff's investigations ; the equation appears to have been first discussed by Euler. A solution of a differential equation of this type can often be expressed as a definite integral of the form u = PK(xt)w{t)dl (9) J T 264 DIFFERENTIAL EQUATIONS The method depends on the fact that the function K(xt) = K(z) satisfies the partial differential equation K Now , H«e) + g (-e)]« - J-»[°('ff) + <- H (4)> hence the function w{t) should be chosen so that the quantity under the integral sign is an exact differential ; in other words, w must be a solution of the differential equation adjoint to In the case of the hypergeometric equation fi 11 (in x{x - l)-3-£ + {{a + b + \)x - c}-j-+ abu = (10) we choose ~K.(xt) so that x i (x - 1)^-2" + {{a + b + l)x 2 - cx}-^- + abxK = t(l-t)^ + {d-ct) Tt This means that z{z - i)d S + & a + b + i ) z - d ^ + abK = ° hence we may take K(xt) = F(o, b, d, xt) An integrating factor w(t) is given by w = t d ~ 1 {\ - £) c_ 0, c> d, we may take t = 0, t 2 = 1, and we may con- clude that the integral f ^-!(1 - ty-^Fia, b, d, xt) dt Jo satisfies the hypergeometric equation (10). We have, in fact, the formula f l*-i(l - ty-^F(a, b, d, xt) dt = IW(c -) is simply dv P(t)% + Q(t)v and the function w{t) must be chosen so that it is an integrating factor of this expression. To illustrate the method, let us consider Legendre's equation d u nil M«) - (i - *") ^ - 2x fa + n ( n + i)« = o In this case we have P(a;) =1 - x 2 , aP'(i) + Q(x) = 2x, a( " + 2 ^ P"(») + (a + 1)Q» = n(n + 1), .-. Q(.r) = 2(1 + a)x, - a (a + 1) + 2 (a + l) 2 = n(n + 1). The last equation is satisfied by either a=w-lora=-n-2. Now, an integrating factor of the expression (1 -«»)*? + 2(1 + *)t.v is given by the equation ^[w(l - < 2 )] = 2(1 + oc)i . w Hence w = (1 - J 2 )~ a ~ 2 , and we have Mt[(< - a;)-] . (1 - * 2 )-«- 2 = j t [{t - X )'{1 - fi) — 1 ] The path of integration must be chosen so that the quantity within the last square bracket has the same value at the two ends of the path. When a = - n - 2 and n + 1 is positive, the conditions may be satisfied by taking t = - 1 and t = + 1 as the limits of inte- SOLUTION BY DEFINITE INTEGRALS 267 gration ; hence, if x does not belong to the interval ( - 1 < x < 1) we may conclude that the integral U{X > "J. a (t-z)-+i satisfies Legendre's equation. This integral may be identified with the function 2"+ 1 Q„(x). The conditions may also be satisfied by taking the path of inte- gration to be a closed contour in the plane of the complex variable t. If this contour encloses the points t = 1 and t = x, but not the point t — — 1, the expression (1 - {8)n+l (t - x) n + 2 will return to its initial value after describing the contour. To see this we notice that the arguments of t - 1, t - x increase by 2tc after one description of the closed contour, while the argument of t + 1 returns to its initial value. The argument of the expression l\ _ fi)n+l -, i— -- thus changes by (t - z)"+ 2 5 y (n + 1)2tc - (» + 2)2tc = - 2tt and its modulus is unaltered. Since e~ 2iri = 1, we see that the expression returns to its initial value. The definite integral u = ^— . I -^ r— rr is thus a solution of Legendre's equation whether n is integral or not ; it can be identified with the function 2 n P„(#). Linear differential equations have been solved recently with great success by means of definite integrals of type lafM>(s) ds in which w(a) is the ratio of two products of T functions of type T(a ± s). An account of this method, which is due to Pincherle, is given in a paper by E. W. Barnes, Proc. London Math. Soc, Ser. 2, Vol. VI. p. 141. EXAMPLES. 1. Prove that the differential equation d 2 y 2 dy [" n(n + 1)" da; 2 x dx |_ x 2 is satisfied by a definite integral of type 2/= [«<-p„(0«a y=o 268 DIFFERENTIAL EQUATIONS where P„(i) is a solution of Legendre's equation. Hence deduce Ray- leigh's formula . — , f+i . „_ , , , 6 V2TO"a;-4J n+i (a:) = eT„(n) dy. 2. Prove that the differential equation dx 2 x dx \ a; 2 / is satisfied by y = I e-*< XBh 'i> cosh nip dtp 3. Prove that the integral J-i *- (^ satisfies Legendre's equation. 4. Prove that the equation *± + 3 ^ + r* - 4w2 ~ i N >^ + *« = o da; 3 a: da: 2 \ x 2 J dx x y /•CO is satisfied by y = e-*P m (l + £« 2 ) * when n = m + \ and m is a positive integer. 5. Prove that I e"* J^ +i (aj) da: = - Q„(l + W). (H. M. Macdonald.) Jo t 6. If 3- denotes the operator x — dx |>F(3-) + G(»)>«w(s) = [F(s)a: s+1 + G(s)ai s ]w(s) Prove that the quantity on the right-hand side of this equation is a perfect difference $(s + 1) - $(s) if w(s) satisfies the difference equation G(s)w(s) + F(s - l)w(s - 1) = 7. Prove that equation (10) of § 81 is satisfied by an integral of type J T(c + s) when | arg( — x) | < 7t and the contour of integration is parallel to the imaginary axis in the complex s-plane, with loops, if necessary, to ensure that the points, 0, 1, 2, ... are to the right, and the points - a, - a - 1, - a - 2, ... , - b, -6-1, - 6 - 2, ... are to the left of the contour. § 83. The solution of partial differential equations by means of definite integrals. There are two types of definite integrals which are used to represent solutions of linear partial differential equations. In the first type the integrand is itself a solution of the partial differential equation for all values of the parameters with respect to which the integration takes place. In the second type of definite integral the integrand is not a solution of the partial differential equation. SOLUTION BY DEFINITE INTEGRALS 269 To obtain a solution of a given linear partial differential equation in the form of an integral of the first type we must first of all find a particular solution involving one or a number of arbitrary para- meters. If the differential equation is of the form *&£)'-• <» a particular solution may be found by writing V = e ix+r ®. The differential equation is then satisfied if F(£, 7)) = 0. Let us suppose that the coordinates of points on this curve F = can be expressed as functions of a parameter s ; then V = e*^ s)+!ft,w is a particular solution involving a parameter s, and this solution may be generalised by writing V = [%*««)+»!<•>$)(«) ds (2) where s v s a are arbitrary constants and 0, we have the solution V = f" e-3" a+iM q>(s) ds (3) J —CD In particular, when 0) . . (4) J — oo 270 DIFFERENTIAL EQUATIONS Transforming this integral by the substitution a = x + 2%y/y, we obtain a solution in the form of a definite integral of the second type, viz. e<° V = 2 \ x(x + 2u^y)e- v? du J — 00 If the function j[x) is such that the definite integral represents a continuous function of y as y -> 0, the solution V satisfies the condition <•» V = 2i(x) e~ u 'du = 2Ax(x) for # = I J -CO It is shown in books on definite integrals that A = \/7r ; hence a solution which satisfies the condition V = x(x) when y = is given by Laplace's formula V = -i-[ 'y{x + 2uWy)e~ u 'du . ... (5) The same problem may be solved in another way with the aid of formula (3). If we suppose that the function cp(s) is of such a nature that the definite integral represents a continuous function of y for y = 0, the function 0 SOLUTION BY DEFINITE INTEGRALS 271 In the case of Laplace's equation d 2 u d 2 u dhi ,,„> a? + 355 + 5?"° (10) a definite integral solution may be built up from the particular solution u _ j( x cos a + y s in a + iz) where a is an arbitrary constant and / is an arbitrary function. A more general solution is J02 f(x cos a + y sin a + iz, a) da where a 1( « 2 are constants. If we take a x = 0, a 2 = 2tc, we obtain Whittaker's solution J27T /(a; cos a + y sin a + iz, a) da o where /(£, vj) is an arbitrary function of \ and ■»]. If in the preceding integral we integrate round a closed contour and make a special choice of the function /, we obtain u = J_ f 9(") rfoc 2ro Jc a; cos a + ?/ sin a + iz + this solution u generally reduces to 27c/(tz). 272 DIFFERENTIAL EQUATIONS Another type of solution of Laplace's equation may be obtained as follows : We have seen in § 66 that if F m (s) is a solution of Bessel's equation and k and m are arbitrary constants, then u = e± iz+im *~F m {kp) is a solution of Laplace's equation. When m = |, we may take F»(*p) = (*?)-»*"* hence u = e^^p-M** is a solution for all values of k, and this indicates that u = /(z ± ip)p-*e^ (14) is a solution when / is an arbitrary function. In particular, we obtain the solution f_J I Vie** \z - ip z + ip/ _ 2ipie* i * _ 2a Va + zy ~ z 2 + p 2 ~ x % + «/ 2 + z 2 Generalising this by a transformation of rectangular axes, we obtain the solution _ [*(» - I) + m(y - 7]) + to(z - £)f . (15) (x _ Q. + (y _ ^a + ( Z _ Q 2 where £, tj, £ are arbitrary constants and I, m, n are constants such that ¥ + m 2 + n 2 = 0. Now, let 5. f\, X» l> m > n be regarded as functions of a parameter 0, and let us integrate the expression V = -L J«(x, 2/, z, 0)F(0) e*0 .... (16) round a closed contour containing only one root of the equation [x - urn* + \y - >](9)] 2 + \t- m? ■ (") The integral may be evaluated by means of Cauchy's theorem, and we obtain the solution v = _ xt? ia\ [*(s - I) + My - n) + n i z - QY where is a root of equation (17) and £', tj', £' denote the derivatives of 5, ■*), £ with regard to 0. In the particular case when I, m, n also satisfy the relation 11' + mr{ + «£' = SOLUTION BY DEFINITE INTEGRALS 273 the preceding solution may be expressed in a simpler form, for it is possible to find quantities X, y., v depending only on and such that [l{x - I) + m{y - 7)) + n(z - Z,)][\(x - £) + y{y - i\) + v(z - £)] = K'(* -I) + -n'iy - ri) + l'(z - K)f It is sufficient, in fact, to write and similar equations. It is easy to verify that X£' + (AT)' + v£' = X 2 + y? + v 2 = Hence, if X, [i, v are functions of satisfying these equations, the function v = m ,. . . (18) [\(x - I) + y.{y - 7]) + v(a - Qf is a solution of Laplace's equation. This is the result to which we have referred in § 66. The general method of solving linear partial differential equations by integrals of the second type may be described briefly as follows. Let ~L Xi y {u) =0 (19) be the given linear partial differential equation in which we shall suppose for simplicity that there are only two independent variables. The method is, however, not limited to this case. Let us suppose that a solution of the generalised harmonic equation I*, r (») =Me(«) (20) is known, and that this particular solution v is not simply a multiple of a function of x, y and a function of t. Now consider the definite integral u(x, y) = v(x, y, t)cp(t) dt .... (21) Assuming that the definite integral can be differentiated under the integral sign by the rule of Leibnitz, we find that Lr,»(tt) = L Xiy (v)(p{t)dt = f 2 Mj(t;)(t) dt is an exact differential dW ; and if the limits t lt t 2 are chosen so that W vanishes at both limits, the definite integral (21) will be a solution of the partial differential 274 DIFFERENTIAL EQUATIONS equation (19). To illustrate this method, let us consider the partial differential equation d 2 u n du d 2 u . dx 2 x dx dy 2 It is easy to verify that the function v = f(x cos t ± y) satisfies the partial differentia] equation d 2 v n dv~ .dy 2 dx 2 x dx_ Now sin" -1 J is an integrating factor of the expression on the left-hand side. Multiplying by this factor, the expression becomes equal to 3 r dv\ The quantity within square brackets is zero for t = and t = n, provided n > 0. Hence we obtain the solution La d 2 v . , . dv — + { n-l)cott w I f(x cos I ± y) sin™ -1 1 dt Jo This result is due to Poisson. EXAMPLES. 1. Prove that Laplace's equation is satisfied by a definite integral of type /•» m = I e~ az J (ap)

TV as a -> -, F'(£) denoting the derivative of F(£). 5. Prove that the partial differential equation dx* + dy* is satisfied by the definite integrals rzir V= e kix °°' a+! " ina) + (2/ - a;)*]<-e(l - t)P~ l dt + P<|/[>+ (1/ - x)t]t-P(l - rtP-Uog^tl - t)(y- x)]dt Jo § 84. The Green's function of a linear differential equation with assigned boundary conditions. Let Jj x {u) denote the self-adjoint linear differential expression T , d I du\ where q(x), p(x) and p'(x) are continuous functions of x in a finite 276 DIFFERENTIAL EQUATIONS real interval a< *< b. If u and v are two functions whose first two derivatives are continuous in the interval (a, b), we have the so-called Green's formula * JW«) - uh x {v)}dx =[p(v^ - u^) (1) Now let y(x, £) be a function of a; and £, which has a continuous second derivative with regard to x at all points of the interval (a, b) with the exception of the point x = £. At this point the derivative J- is discontinuous in such a way that e -+o \-\dx/ x =£ +e \dx/ x= g- J If y is a solution of the differential equation L^w) = for all points of the interval (a, b) other than x = \, it is called a ground- solution t (Grundlosung) for the interval (a, 6). A ground-solution may be expressed in terms of two linearly independent solutions u L (x), u 2 (x) of the equation ~L x {u) = by means of the formula 2 a; - I u 2 {l)u^{l) - u^Qu^Q where A and B are arbitrary constants. These constants may be chosen so that y(x, £) satisfies certain supplementary conditions. If these involve the values of y and its derivatives at the end points of the interval in a manner which will be specified later, the function a( x I) = iM) will be called a Green's function for the differential equation and the given supplementary conditions. Now let f(x) be a function which is continuous in the interval a < x < b, and let u(x) denote a solution of the differential equation I'xiu) + f{x) = 0, which satisfies supplementary conditions of the type just mentioned and has a continuous second derivative for the interval (a 0, we find on making use of the equations Ls(w) = - /(«) L» = v{x) = g{x, I) M9=(>%M)^[,(4;-«|)];. . (2 ) Thus the value of w(£) is expressed linearly in terms of the values of g(x, $;) and its derivative with regard to x. In some cases «(£) can be expressed by means of the definite integral alone. For instance, if the supplementary conditions state that u = when x = a and when x = b, then we have also g(a, £) = g(b, £) = 0, and the second term on the right-hand side of (2) vanishes. The same thing happens when the supplementary conditions state that h^r- + hi = for x = a H-=- + K.u = for x = b ax ax where h, k, H and K are constants. If p(x) is zero when x = a, a condition which states that u and t- must be finite for x = a will cause p ( q -j- - u -£- ) to be zero ax \ ax dxl when x =a. A similar condition for x = b or a condition such as die H j- + Km = for x = b will then make the right-hand side of (2) zero. If in any of these cases a Green's function exists, we can conclude that a solution of l> x {u) + /(^) = 0, which satisfies the supple- mentary conditions and possesses a continuous second derivative, is given uniquely by the formula «(5) = (V) 9fr. Z)d* (3) J a for {a < ^ < b). It should be remarked, however, that a Green's function does not always exist. This may be seen by considering the simple case when the supplementary conditions take one of the forms just specified. Let ol{x) be a solution of li x {u) = 0, which satisfies the condition associated with the end point x = a, and is such that a and 3- are continuous functions of x throughout the interval a < x < b. Also, let p(ar) be a solution of L(«) = 0, which satisfies the condition associated with the end point x = b, and is such that (3 and j- are continuous functions of x throughout the interval a < x < b. If (5) 278 DIFFERENTIAL EQUATIONS a.{x) is not a constant multiple of $(x), the arbitrary constant which occurs in the general expression for a (a;) may be chosen so that a'(Q m - a® m) = ^y • • • (4) for we have ^[p(Q {a'(5)P(Q - a(£)P'($)}] = It can now be shown that the function g(x,%=a(x)p(® (s9 " is a Green's function of li x (u) for the given supplementary con- ditions.* The function evidently satisfies the supplementary con- ditions, and it is clear from (4) that the product jp(£) g(x, %) is a ground-solution of the differential equation lijv) = 0. Hence the function satisfies our definition of a Green's function. This method of constructing a Green's function breaks down when a (as) is a constant multiple of (5{sr), for then it is no longer possible to satisfy the relation (4). It should be noticed that in this case the equation L(w) = possesses a. solution a.(x), which satisfies the supplementary conditions and is such that -=- is a continuous function of x throughout the interval a < a; < 6. Furthermore, if in this case the equation L„(w) + f[x) = possesses a solution satisfying the supplementary conditions and having a continuous derivative for all points of the interval a < x < b, this solution is not unique ; for if k is an arbitrary constant, u(x) + ka.(x) is a second solution with similar properties. It should be remarked that a solution with the properties just mentioned only exists when f{x) satisfies a certain condition. The above method of constructing Green's functions cannot be d 2 u used in every case. For instance, if we take the equation t-j = and the supplementary conditions x w(0) = - «(1) m'(0) = - ti'(l) a solution a (a;) which satisfies the first is given by oc(ai) = x - \, and a solution $(x) which satisfies the second is given by (3(a;) = 1. The function ' g{x>l) = x _ ^ (jeasQ does not, however, satisfy the supplementary conditions, and so is not the correct Green's function. In the case when a Green's function can be constructed by Kneser's method, it appears that the function is a symmetric junction of x 'A. Kneser, Math. Ann., Bd. 63. SOLUTION BY DEFINITE INTEGRALS 279 and \ (i.e. it is unaltered when x and \ are interchanged), and it is the only Green's function corresponding to the supplementary con- ditions. Indeed, if there were two different Green's functions g(x, £) and G(x, £), the difference g(x, 5) - G(x, £) would have a continuous derivative throughout the range a < x < b ; it would also be a solution of the differential equation l> x {u) = 0, and would satisfy both boundary conditions. Such a function would have to be a constant multiple of both a.(x) and $(x) ; consequently, a.(x) would be a constant multiple of (3 (a;), which is contrary to hypothesis. It should be noticed that the function g(x, £) as defined by equa- tions (5) can have a perfectly definite value when \ lies outside the interval a < !; < b, but now the integral '(5) = f/0%(*, I) dx Ja v= no longer satisfies the equation L^(m) + /(£) = 0. Instead of this it satisfies the equation Lj(m) = 0, for it is a constant multiple of either a(^) or p(^). This theorem is analogous to the following well-known property of a potential function, fff pg.-g, 0<£ like the function -, where r is the distance from {x, y, z) to (£, vj, Q. The Green's function also satisfies a linear boundary condition such as g = at points of a closed surface. These developments are given in books on potential theory for the case of Laplace's equation. A brief account of the general theory is given in H. B. Heywood and M. Frechet's L'equation de Fredholm et ses applications a la Physique Mathematique, Paris (1912), pp. 121-140. 280 DIFFERENTIAL EQUATIONS EXAMPLES. 1. Prove that the Green's function for the expression and the supplementary conditions u = when x = and when x = 1 is _. sin Xx . sin X(l - Z) , _ s . '<*'*> = X.sinA " ( *~ 5) siaXg.sinX(l-») ( „ {Hilbert) X.sinX v ' d 2 u 2. A Green's function for the expression — - + Iht and the condition that the function should have the period 1 is g[x, 5) = - gx sin X l ^ " 5 1 + cot - cos X(a; - Q d 2 u 3. The Green's function for the expression ~L x {u) = -r—^ - X 2 u and the conditions that u(x) is finite for x = + oo and x = - oo is g[x, 5) = L e -\\x-t\ (Hilbert.) § 85. Riemann's method.* Let L I; y {u) be used to denote the differential expression , d 2 u du ,du ... + 9; h j- + CM (1) da; 9«/ ' dx dy where a, b, c are functions of x and y with first derivatives which are continuous in a region R of the {x, y) plane. The adjoint expression li x> y {v) is defined by the property that T / N T I \ 9M SN /ON vL Xt „(«) - wL*, y {v) = — + — . . (2) where M and N are certain quantities which can be expressed in terms of u, v and their first derivatives. The expression for L x>y (v) is suggested at once by the form of the expression adjoint to an ordinary linear differential expression, §18, and it is easy to verify that if * Gottinger Abhandlungen, Bd. 8 (1860) ; Gesammelte Werke, p. 145. See also Darboux, Thiorie des Surfaces, t. 2. SOLUTION BY DEFINITE INTEGRALS then equation (2) is satisfied with 281 M = auv + -!(• du dy N = buv + -(dt- - u ) dy) dy Bv\ + Sx 3a;/ 9a; where ^ is an arbitrary function. We shall find it convenient to put x = 0. Now, if C is a closed curve in the (x, y) plane which lies entirely within the region R, and if u and v, together with their first deriva- tives, are supposed to be continuous in R, we have by Green's theorem * !.<»*-**> -!!(£♦£)** ti dy. \vL Xi y{u) - u!j Xi y(v)]dxdy where the double integral extends over the area bounded by the curve C. Let u and v be chosen so that the equations ljx, y {v) = L*,y{«) = are satisfied ; then I (M dy - N dx) = Fig. We shall now apply this theorem to the case in which the curve C consists of two lines AX, AY parallel to the axes of x and y respec- tively and a portion YX of a curve I\ * See Goursat-Hedrick, Mathematical Analysis, Vol. 1, p. 262. 282 DIFFERENTIAL EQUATIONS We easily find that [ Y Mdy + f (Mdy - Nda;) - f N dx = Ja Jy Jx Now \Mdy = J A [^M - »{ Ty - a»JJ^ = |[(w)y - Ma] - J A M (g^ - au )^ Similarly, I Ndx = f[(m>) x - (mu)aJ _ I M (a 6m J da; Hence {uv) A = $[(uv) x + (im')t] ~ I (Md# - Ncfe;) Jx -fr(£- 6 *) & -C*(§r° B )*' Now let us suppose that the function v can be chosen in such a way that dv dv bv = and -=- - av = oa; 9y on the lines AX, AY respectively ; then the value of u at the point A is given by the formula Mi = |[(w) x + My] + I (Mdy - ~N dx) It is easy to see from this equation that if the value of u and one of its derivatives is known for all points of the curve V, then the Fig. 9. value of u at certain outside points can be found. Notice that if u and =- are given, the value of -=- can be found from the equation du du dx du dy ds dx ds dy ds SOLUTION BY DEFINITE INTEGRALS 283 where s denotes the length of an arc of the curve, or, in fact, any parameter on which the coordinates of a point of the curve F depend. Everything depends of course on the possibility of finding a suitable function v. If (x a , y ) denote the coordinates of A, we may write v = g(x, y ; x , y ) and call it a Green's function of the differential expression L^ ,,(«). Now let us consider the case when the curve T consists of a line BY parallel to the axis of x and a line BX parallel to the axis of y ; we then have (Mdy - Ndx) = M.dy - N dx Jy Jb Jy But \ B mx - £[](«£ - «g) + buv]dx = \][-\h uv)+v ^Tx + bu )\ dx = HMi _ Mb] + J v \fi- + bujdx Similarly, rx j*x /g M \ J Mdy = l[(uv) B - {uv) x ] + I «( j- + awjcfo/ Hence (wv) A = (mv) b - I vl-^ + bujdx - I v(~ — I- aujdy Now let a function u = h(x, y ; x lt y t ) be supposed to exist having the properties that — +&w=0 — - + «m = on the lines BY, BX respectively, and L^^m) = elsewhere. If (x v y x ) are the coordinates of B, the arbitrary constant which occurs as a multiplier in the expression of the general function of the required type may be chosen so that u B = 1, i.e. h(x 1 , y 1 ; x x , y x ) = 1 We shall also suppose that the function g is chosen so that 9{z . Vo : x o- Vo) = l Our formula then gives %o> Vo ; x i> ¥i) = 9( x i> Vi > x o- Vol Hence the Green's function g(x, y ; x , y ) is a solution of the equation ~L x>y (v) = when considered as a function of its first two arguments, and a solution of li Xoi!/o (u) = when considered 284 DIFFERENTIAL EQUATIONS as a function of its last two arguments ; it is, in fact, a Green's function for each of the two adjoint partial differential equations. To illustrate the method, let us consider the case when a = 0, 6=0; we then require a solution of the equation d 2 v ___ + cv = ox ay under the supplementary conditions v = 1 when x = x , y = y , ^- = when y = y -5- = when x = x dx a '" ay Let us try v = ¥[(x - x )(y - y )] = F(z) then z- rY +-j-+cF = The supplementary conditions may be satisfied by assuming F = 1 + Kl z + a 2 g-j + ... a„— + ... We easily find that wa K + cxt. n -. y = ; hence * = 1 - % + i^ - - " J.<*^> where J (£) is the Bessel function of zero order. 9 2 M A solution of the equation -=— = — h cm = is now given in terms of the values, which it and one of its derivatives assume on a curve r, by means of the equation 2u(x ,y ) = u x+ u Y+ | y [(F| - u^)dy - (pg - «g)i* Certain equations which occur in mathematical physics may be 9 2 W reduced to the form _ . + cu = by simple substitutions, the telegraphic equation * x & dhi 1 d 2 u 9m 9a 2 = c 2 " W + G dt d 2 w d 2 w and the equation -=-=- - c 2 ■=-* + a 2 c 2 w = at 2 az 2 which occurs in the theory of the vertical propagation of waves in the atmosphere f may be mentioned as examples. *For this equation see G. Kirchhoff, Pogg. Ann., Bd. 100 (1857) ; O. Heaviside, Phil. Mag., Aug. (1876); Electrical Papers, Vol. 1, p. 53; H. Poincare, Comptes Rendus, t. 117 (1893), p. 1027 ; E. Picard, Bull, de la Soctttt math, de France, t. 22 (1894), p. 2; M. I. Pupin, Trans. Amer. Math. Soc, Vol. 1 (1900), p. 259. tH. Lamb, Proc. London Math. Soc, Ser. 2, Vol. 7 (1908) ; S. Sano, Bull, of the Central Meteorological Observatory of Japan, Vol. II. (1913), No. 2. SOLUTION BY DEFINITE INTEGRALS 285 Riemann's method has recently been extended in such a way that it can be applied to linear partial differential equations of parabolic type.* Some of the results are given in the examples. EXAMPLES. 1. The solution of 3— - c 2 5— + a'c'w = ot 3 oz 2 which satisfies the initial conditions w=f(z) -_- = 0, and the value of w at the at plane z = is maintained equal to F(t), a solution of the partial dif- ferential equation considered in the last example is given by : ^[ CJ ° atj)F(r)dT t> - ° z> *<- c where a 2 = c 2 (« - t) 2 - z 2 . (S. Sano.) 3. If, on the other hand, the val F(t) at z = 0, the corresponding value of w is 3. If, on the other hand, the value of ^- is maintained equal to oz -V J (ao)F(T) d,T t>~ t< Z - c 4. Prove that the Green's function for the equation + — — - =- = dxdy x - y dx x - y By is g(x, y ; *„ y t ) = (y„ - x) -?(y - x)W{y - x,) -flP(p, p', 1 ; o) *Wera Lebedeff, Dissertation Gbttingen (1906); E. E. Levi, Annali di Mate- matica (3), t. 14 (1908), p. 187 ; E. Holmgren, Arkiv for Matematik, Bd. Ill and IV (1907-8); E. Goursat, Traiti d' Analyse, t. Ill; M. Gevrey, Liouville's Journal (6), t. 9 (1913), pp. 305-475, t. 10 (1914), pp. 105-148 ; G. C. Evans, American Journal, Vol. 37 (1915), p. 431. 286 DIFFERENTIAL EQUATIONS where q = (» - *.) fr - y.) (* - 2/o) (2/ - *o) and F denotes the hypergeometric function. (Darboux.) 5. Prove that equation (2) of § 85 is satisfied if x , . d 2 u du - , d 2 v . dv UU) = W ~ By UV) = dx* + dy HI ^ M ^ -NT ox ox 6. Let C denote the curve (Fig. 10) formed of a segment AjA a of a line y = constant and two curves C v C 2 through the points A v A 2 . Let Fia. 10. S„ denote the region bounded by C and a straight line through (x, y) parallel to the axis of x, also let C y denote the boundary of this region. Let zbea solution of the equation d 2 z dz ,, . W~dy =Ax ' V) such that the derivatives entering into this equation are continuous in S u ; then, if h(x, y; 5, tj) = (y - 7)) - *exp{(a: - 5) 2 /4(v) - y)} we have J h{z(*, v,)dg + |J? - g 2( y~^] *]} - jj h.M, ^dZdr) = te(a, V) where X is 2a/tt, Vtc, or according as the point (x, y) lies inside, on the boundary, or outside the region bounded by C. If x = X^y), x = X t (y) are the equations of the curves C 1 and C 2 the functions Xj and X 2 are assumed to be continuous with their first derivatives and to have only a finite number of maxima and minima for the considered range of values of y. (This theorem has been established under less stringent conditions. See the'memoirs cited above.) CHAPTER XI THE MECHANICAL INTEGRATION OF DIFFERENTIAL EQUATIONS § 86. The reduction of a differential equation to a convenient form. Simple mechanisms for solving differential equations are generally suitable only for differential equations of certain special types, and so it is important to know in the first place whether the equa- tions which can be reduced to these special types are of frequent occurrence. One type of equation of the first order considered by E. Pascal* is P = *{Q(aO - y} (i) where Q and «3? are arbitrary functions ; it is of special importance because C. Ajellof has shown that a general Riccatian equation p = Ay 2 + By + C can be reduced to the above form. The appropriate transformation is £ = - \A$dx v) = \Cydx - yy = Q© - yy where log [3 = B dx log y = - I B dx The reduced equation is The first type of equation considered by Lord Kelvin \ is s{r2} + '" < 2 > where U is an arbitrary function of x. The general linear differential equation of the second order 8^2 + *=° *Rend. R. Ace. delle Sc. fis. e mat. di Napoli (3), v. 17 (1911) ; Ibid. (3), v. 18 (1912). flbid. (3), v. 18 (1912). %Proc. Roy. Soc, Vol. 24 (1876). 288 DIFFERENTIAL EQUATIONS may be reduced to this form by writing dz = Qe\ Fdx dx U(z).Qe 2 f F ' ta = l Equation (2) may be solved by a method of successive approxima- tions by writing y % = \ U(C - f yjdx) dx Jo Jo y 3 = f U(C - f y 2 dx) dx Jo Jo where y 1 is any function of x, to begin with, as, for example, % = x ; then y 2 , y 3 , etc., are successive approximations converging to that one of the solutions of (2) which vanishes when x = 0. To carry out the approximations mechanically an instrument is needed for the calculation of an integral of type f(x) = 1, and so the path of H is the integral curve of the function tp(a;)^(a;). The integrating machine invented by James Thomson is based on an entirely different principle. Motion is transmitted from a disk to a cylinder by the intervention of a loose ball, which presses by its gravity on the disk and cylinder, the pressure being sufficient to give the necessary frictional coherence at each point of rolling contact ; and the axis of the disk and that of the cylinder being both held fixed in position by bearings in stationary framework, and the arrangement of these axes being such that when the disk and the cylinder are kept steady, or, in other words, without rotation on their axes, the ball can roll along them in contact with both, so that the point of rolling contact between the ball and the cylinder shall describe a straight line on the cylindric surface parallel neces- sarily to the axis of the cylinder — and so that the point of rolling contact of the ball with the disk shall traverse a straight line passing through the centre of the disk. It will thus readily be seen that, whether the cylinder and the disk be at rest or revolving on their axes, the two lines of rolling contact of the ball, one on the cylindric surface and the other on the disk, when both considered as lines traced out in space fixed relatively to the framing of the whole instrument, will be two parallel straight lines, and that the line of motion of the ball's centre will be straight and parallel to them. For facilitating explanations, the motion of the centre of the ball along its path parallel to the axis of the cylinder may be called the ball's longitudinal motion. Now for the integration of y dx : the distance of the point of contact of the ball with the disk from the centre of the disk in the ball's longitudinal motion is to represent y, while the angular space 292 DIFFERENTIAL EQUATIONS turned by the disk from any initial position represents x ; and then the angular space turned by the cylinder will, when multiplied by a suitable constant numerical coefficient, express the integral in terms of any required unit for its evaluation. The plane of the disk may suitably be placed inclined to the horizontal at some such angle as 45°. PLAN. FRONT ELEVATION. SIDE ELEVATION. Fig. 14. To calculate IcpCaOtpC*) dx, the rotating disk is to be displaced from J rx a zero or initial position through an angle equal to I cp(#) dx, while the rolling globe is moved so as always to be at a distance from its zero position equal to <\i(x). This being done, the cylinder turns through an angle equal to I (p(x)\\j(x) dx. One way of giving the required motions to the rotating disk and rolling globe is as follows : On two pieces of paper draw the curves y = I (p(x)dx and y = <\i(x) Jo Attach these pieces of paper to the circumference of two circular cylinders, or to different parts of the circumference of one cylinder, with the axis of x in each in the direction perpendicular to the axis of the cylinder. Let the two cylinders (if there are two) be geared MECHANICAL INTEGRATION 293 together so that their circumferences shall move with equal velocities. Attached to the framework let there be, close to the circumference of each cylinder, a slide or guide-bar to guide a movable point, moved by the hand of an operator, so as always to touch the curve on the surface of the cylinder, while the two cylinders are moved round. Two operators are required as one operator could not move the two points so as to fulfil this condition — at all events unless the motion were very slow. One of these points, by proper mechanism, gives an angular motion to the rotating disk equal to its own linear motion, the other gives a linear motion equal to its own to the centre of the rolling globe. Lord Kelvin has described a chain of integrators by means of which a differential equation of a very general type may be solved. His apparatus does not, however, seem to be quite so convenient to use as the mechanisms invented by Pascal. § 89. Pascal's mechanism for the solution of a linear equation of the first order. . In this case there are two articulated parallelograms by means of which the small wheel at H is made to move parallel to EG and HG is maintained parallel to KM. The bar EG and HG contains slots which enable them to slide over a pivot at G, which slides along a slot in the rod BC. The bar EF is slotted so that it slides over the pivot F which moves along a slot to BC. The bar EF is also constrained to touch a rectangular hyperbola, so that EN . FN = 1 If now as the rectangle moves the points G, M. F trace out the curves y =f{x), y = = F(a;)y + ^ {x) z > = ^ x)z + y , _ _ (1) where / + F = $ /' = F/ + -« .111"*" 2n+ 1 It. (2n+ 1)(2m+ 2) x* ~\ T(l - 2?i) 1 + n+ a E. j. (^ + a )(n + a- 1) a; 2 2t»-ll! (2w - l)(2n - 2) 21 + '"} Express the solution in terms of Bessel functions when a = - \, 0, \, 1 and - 1 respectively. (H. G. Savidge.) MISCELLANEOUS EXAMPLES 303 47. Prove that the roots of the equation u" - Su + 6z = satisfy the differential equation ("-!.£♦* J-. Find a power series which represents the solution which is zero when z =0. 48. The small transverse vibrations of a string of beads at equal distances apart are governed by an equation of type 2r = Wvm + 2/ s -i - «v.) where y s is the displacement of the sth bead from its equilibrium position. Determine the possible periods of vibration on the assumption that the ends of the string are fixed, i.e. y = 0, y n = 0, for all values of t. 49. The string of beads in the last example is supposed to extend to infinity in both directions and the bead for which s = is initially displaced through a distance s, while all the other beads are in their zero positions, i.e. 2/i=0 2/2=0 2/o= E 2/_i =0 2/-2 = The system then begins to move ; use the equation of Ex. 48 to prove that f~rl2r„ i %4- s =0 if r<3 and find a power series for y s . Hence show that y, = 3. zs {2kt) (W. R. Hamilton and T. H. Havelock.) 50. If in Ex. 17 the rate at which the eggs go bad is a linear function of the time and proportional to the number of eggs present, prove that the solution of the problem depends on a linear equation of Laplace's type, and solve this equation by means of a definite integral. INDEX Abdank-Abakanowicz 288, 290 Abel 276 Abel's equation 288 Adjoint differential expressions 53, 280 Ajello 287 dAlembert's equation 176 Appell212, 215, 275 Approximate solution, method for 227, 248 Approximations, method of succes- sive 245 Baker 245 Ballif 100, 102 Barnes 235, 267 Bateman 159 Bendixon 252 Bernoulli's equation 21 Bertrand 164 Bessel's equation 76, 113, 198, 200, 221, 237, 272 Bilinear concomitant 53 Birkhoff 255 Bliss 245, 252 Block 175 Bocher 253 Bois Reymond 170 Bolza 75, 79 Boole 51, 62 Branch point, movable 69 Brill 212, 215 Burkhardt 15 Cailler 100 Canonical form of a harmonic equa- tion 218 Canonical form of a linear partial differential equation of the second order 170 Canonical form of a total differential expression 164 Carslaw 215, 256, 270 Cauchy 91, 93, 245, 271, 272 Cauchy's existence theorem 142, 223 Cayley 92 Characteristics 131, 172 differential equations of 133, 171 Chrystal 92, 94, 131 Clairaut 50 Clairaut's equation 63 Cohen 83, 93 Complementary function 54 Complete integral 122 Complete solution 122 Constants, number of, in general solution 3 Contact transformation 81, 148 Continuity, equation of 139 Convergence theorems 242 Cusp-locus 90, 95 Darboux 56, 90, 100, 101, 128, 174, 216, 219, 221, 280, 286 De Donder 159 Degree of an equation 2 Dependent variable 1 Dirichlet's conditions 257, 270 Dixon 159, 255 Donkin 210 Duality, principle of 81, 148 Dyck, 93, 295 Elliptic equation 170 Emden 229 Envelope 89, 92, 95 Euler 50, 179, 261 Euler's equation 103 Euler's transformation 265 Evans 285 Forsyth 37, 129, 142, 164, 175, 184, 204, 230, 235, 244 Fouret 88 Fourier 211 Fourier's inversion formula 270 Frechet 279 Frobenius 243 INDEX 305 Fuchs 243 Fundamental solution 269 Gaskin 115 Gegenbauer 115 General integral 122 General solution 3 Gevrey 285 Goursat 49, 117, 142, 143, 281, 285 Grade 80 Green's function 15, 276, 283 Guldberg 163 Haar 255 Hackett 8, 230 Hargreaves 159, 192 Harmonic equation 216, 270 Harmonic function 193 Harrison 229 Hayashi 295 Heaviside 284, 285 Heine 265 Hertz 186 Heun 228, 229 Heywood 279 Hilbert 255, 276, 280 Hill 92, 94, 96, 128 Hobson 255, 265, 274 Holmgren 285 Homogeneous equation 80 Hudson 92 Hypergeometric equation 111, 170, 232 Hypergeometric function 112, 241 Indicial equation 231 Integrability, condition of 153 Integral curve 88 Invariant 72, 76, 161 Inversion 199, 201 Jacobi 259 Jacobian 116 Jacobi's equation 83 Jordan 265 Kelvin 197, 199, 230, 287, 294 Kirchhoff 284 Kneser 255, 278 Kummer 112, 236, 237 Kutta 228, 229 Lagrange's auxiliary equation 135 linear equation 134 type of equation 67 Laguerre 110 Lamb 197 Lame 197 Laplace 170 Laplace's equation 193, 271 method 173 transformation 260 type of equation 75 Lebedeff 285 Lecornu 125 Legendre's equation 113, 198, 219, 240, 259 266, 268 Leibnitz 49, 91, 261, 273 Levi 285 Levi Civita 159, 204, 210 Levy 128 Lie 83 Lienard 186 Linear equation 17 Lines of force 190 Liouville 75 Lipschitz's condition 249 Lorentz 186 Lovett 163 Macdonald 268 Mason 255 Massau 94, 125, 248 Maxwell's equations 179 Melde 40 Mellor 11 Mercer 255 Monge 124 Morley 8 Multipliers 136, 138 Neumann 211 Node-locus 93, 95 d'Ocagne 248 Order 2, 17 Orthogonal trajectories 98 Painleve 249 Parabolic equation 170, 285 Particular integral 54 Pascal 287, 288, 293 Perron 98, 252 Petrovitch 93, 96, 294 Petzval 261 Pfaff's problem 164 type of equation 263 Picard 70, 245, 284, 295 Pochhammer 75, 237, 265 Poincare 284 Poisson 46, 210, 274 Potential function 193 Primitive 3 Pupin, 284 306 INDEX Rayleigh 38, 40, 230, 268 Riccati's equation 68, 101 Ricci 159 Richardson 230 Riemann 211, 280 Righi 186 Ritz 230 Rouquet 102 Runge 227 Rutherford 42 Saint Robert 125 Schafheitlin 73 Scherk-Lobatto equation 77 Schlomilch 75 Scribanti 295 Self -adjoint equation 54 Separation of variables 10, 146 Serret 86 Silberstein 185 Singular integral 122, 129 solution 88, 90 Special solution 140 Stackel 93 Standard forms 144 Stefan 100 Steimley 142 Stephenson 38 Stieltjes 201 Stokes 115 Symbolical differentiation 156 method 33 product 155 Tac-locus 89, 95, 97 Taylor 91, 228 Thomson, James, 288 Thomson, J. J. 179, 190 Turriere 140 Wave-equation 179 Weber 185 Wedderburn 94, 275 Weight 80 Weyr 70 Whittaker 271 Wilton 175 Wright 159 ■PRINTED IN GREAT BRITAIN BV ROBERT MACLEHOSE AND CO. LTD. AT THE UNIVERSITY PRESS, GLASGOW