MATHEMATICAL TRACTS FOR PHYSICISTS INTRODUCTION TO THE CALCULUS OF VARIATIONS BY WILLIAM ELWOOD BYERLY PERKINS PROFESSOR OF MATHEMATICS EMERITUS HARVARD UNIVERSITY CAMBRIDGE HARVARD UNIVERSITY PRESS 1917 COPYRIGHT, 1917 HARVARD UNIVERSITY PRESS CALCULUS OF VARIATIONS CALCULUS OF VARIATIONS CHAPTER I INTRODUCTION 1. The Calculus of Variations owed its origin to the attempt to solve a very interesting and rather narrow class of problems in Maxima and Minima, in which it is required to find the form of a function such that the definite integral of an expression involving that function and its derivative shall be a maximum or a minimum. Let us consider three simple examples: The Shortest Line, The Curve of Quickest Descent, and The Minimum Surface of Revolution. (a) The Shortest Line. Let it be required to find the equation of the shortest plane curve joining two given points. We shall use rectangular coordinates in the plane in question taking one of the points as the origin. Call the coordinates of the second point xl, yl. If y = f(x) is a curve through (0, 0) and (xl, yl) and I is the length of the arc between the points, obviously XI I = f dx2 + dy2 0 X! or I = 1 + y'2dx, (1) 0 and we Vish to determine the form of the function f so that this integral shall be a minimum. (b) The Curve of Quickest Descent. Let it be required to find the form of a smooth curve lying in a vertical plane and joining two I 2 CALCULUS OF VARIATIONS given points, down which a particle starting from rest will slide under gravity from the first point to the second in the least possible time. We shall use rectangular axes in the vertical plane taking the higher point as the origin and taking the axis of X downward. Call the coordinates of the second point xi, yi. Y xV X Y' Let y = f(x) be a curve through (0, 0) and (xl, yl) and use the ds well-known fact that d- the velocity of the moving particle at any time, is / 2gx. ds We have = 2gx, d = ds _1 + whence dt = dx - 2gx +1 2gx X1 and t=f Wi + y'2dx. Let I= f + y'dx, (2) 0 and the form of the function f is to be determined so that this integral shall be a minimum. (c) The Minimum Surface of Revolution. Given two points and a line which are co-planar, let it be required to find the form of a curve terminated by the two points and lying in the plane which, by its revolution about the given line, shall generate a surface of the least possible area. INTRODUCTION 3 Take the line as the axis of X and use an axis of Y through one of the points. Call the coordinates of the points 0, yo, and xl, yl. Let y = f(x) be a curve through (0, Yo) and (xi, yi). If S is the area of the surface of revolution generated by the curve, X1 XI S = 27r fyds = 2r fy V 1 + y'2dx. 0 0 XI Let I= f y + y'2dx, (3) 0 and we wish to determine the form of the function f so that I shall be a minimum. 2. The three problems just considered are special cases of what we shall call our fundamental problem which is, to determine the X1 form of the fu te functionso that if y f(x), (x, y, ') dx shall be a maximum or a minimum; being a given function and and x maximum or a minimum; j being a given function and x0 and xi being given constants, as are yo and yi, the corresponding values of y. In ordinary problems in maxima and minima y = f(x) is a given function and we wish to find a value, xo, of x for which y is greater, if we seek a maximum, less, if we seek a minimum, than for neighboring values of x; that is, for values of x differing from xo by a sufficiently small amount whether that amount is positive or negative. In our new problems, to speak in geometrical language, we have to find the form of a curve for which our integral, I, is greater or less than for any neighboring curve having the same end-points. 3. Let us now attack our first problem, that of the shortest line. We have to find the form of the function f so that if XI I = SV 1 + y' dx I shall be a minimum when y = f(x v. Art. 1 (a). I shall be a minimum when y = f(x). v. Art. 1 (a). 4 CALCULUS OF VARIATIONS Let y = F(x) be any other continuous curve joining the given points, and let fl(x) = F(x) - f(x). Then y = f(x) + 7(x) is our curve y = F(x). Consider the curve y = f(x) + a17(x) where a is a parameter independent of x. y=F(x) Y, 0 X For any particular value of a the curve y = f(x) + al(x) is one of a family of curves including y = f(x) for a = 0 and y = F(x) for a = 1. By taking a sufficiently small value for a we can make aqr(x) less in absolute value for that and all less values of a, and for all values of x between 0 and xl, than any previously chosen quantity 4; and for such values of a the curve y = f(x) + aq(x) is said to be a curve in the neighborhood of y = f(x). If y = f(x) and y = F(x) are given, I(a), the I for any one of our X1 curves y = f(x) + ar)(x), islf 1 + (y' + a (x))2dx, and I(a) is a function of a only. o A necessary condition that I(a) should be a minimum when d a = 0 is well known to be that - I(a) should be zero when a = 0. da This condition we shall express as I'(0) = 0. Since the limits 0 and xl are constants XI I'(a) = fd 1 ~+ [y' + a',(x)]2dx o 0 =a) y' + av'(x) 7' (x) dx, 1 + [y' + a'(x)]2 and '() =f i'(x)dx. and I'W 77 > r'(x)dx. ( -1 + y'2 0 INTRODUCTION 5 Integrating by parts I Xi I'(0) = [xdx- +' y -( dx L ^ 1+ P/'2 J J dx 1,1 + y '2 0 0 f= - Y 'f 7 (x)dx, J/1x yV 2 0 since ((x) vanishes when x = 0 and when x = xl. A necessary condition that I(0) shall be less than I(a) for some value of a and for all less values of a no matter what F(x) may be, in which case the length of the curve y = f(x) is less than that of any neighboring curve, is that I'(0) = 0 independently of r7(x), Xl r d v' i.e. thatf i 1r 7r(x)dx = 0 no matter what the form of the J dx -I+ y,2 0 arbitrary function r(x). d y' This condition will be satisfied if and only if d ' =0; dx /1 4- y'2 and we thus are led to a differential equation of the second order between y and x. Its solution will express y as a function of x involving two arbitrary constants. ~/1 + 2+y2 whence y' = K, a constant; y =Kx+L. The required curve is to pass through (0, 0) and (xl, yi) and thus we are able to determine K and L. L=0,K =y Xi Hence y = Y1 x; and our curve is a straight line through the given points. If certain other conditions, depending on the fact that when I(a) is a minimum I"(a) must be positive, are satisfied, y = Yx must be the required shortest line. X As our necessary conditions gave us but a single solution it is clear that if there is any shortest line it must be our line y = x. xi 6 CALCULUS OF VARIATIONS We may note in passing that in simplifying I'(o) by integration by parts we tacitly assumed that f(x) and F(x) were continuous and had continuous first derivatives over the range of integration. 4. We can deal with the general problem formulated in Art. 2 precisely as we have dealt with the shortest line problem. Let it be required to determine the form of the function f so that X1 y = f(x) shall make f (x, y, y') dx a maximum or a minimum; Xo given that y = yo when x = Xo and y = yl when x = xl. Y=f(x) As in the last article let 7r(x) = F(x) - f(x), and consider the family of curves y = f(x) + an(x). X1 Let J(a) = jf(x, y + afr(x), y' + a7'(x))dx. Xo X1 I'(a) = d (x, y + al(x), y' + a'(x))dx. do Let u = y + anr(x), v = y' + ait'(x). du dv.q(x), i (x). da = dax, Xl I'(a) =fd f (x, u, v) dx da4 Xo = [-a x) + -^ n\Jdx. xo INTRODUCTION 7 Xl I'(0) =f[L (x) + a'(x)J dx (1) x0 and I'(O) must be made equal to zero. Integrating by parts, I/ () = Jy 7j(x)dx + [, ()]Jdx y(x), xl x r(o)= ([_ _ d 1 I'(O) Jag_ d 0' (x)dx lay dz ay '" x0 since 7(x) vanishes when x = Xo and when x = x1. I'(O) must be zero independently of the form of q(x) therefore ao_ d a_, dx ay - ', (2) ay dx ay' and as before we are led to a differential equation of the second order which if solved gives y as a function of x involving two arbitrary constants which must be determined from the facts that y = Yo when x = xo and y = y when x = xl. This differential equation is known as Lagrange's Equation and it is a necessary condition that I should be a maximum or a minimum. Any particular solution of Lagrange's Equation is called an extremal, and if the given problem has a solution it is that extremal which passes through the given end-points. If 0 is a function of y' only Lagrange's Equation becomes a e n(y) = C. ay' Whence y' = k, a constant, and the extremals are straight lines; and therefore the required solution is the straight line through the given end-points as in Art. 3. The problems of Art. 1 (b) and (c) can be solved by substituting in Lagrange's Equation the appropriate value of 0 and then solving the resulting equation. 8 CALCULUS OF VARIATIONS. For the curve of quickest descent l ---y'2~ v.Art.1(b)(2). 00 01 ay ay T I + y12 Lagrange's Equation becomes d 1 -. dx<~ ~' For the minimum surface y=y -1-y'2 v. Art. 1(c) (3). ay' Lagrange's Equation becomes d YYI d - = 0. (4) dx -r v -o 2 We shall reserve the solving of (3) and (4) for a later article. CHAPTER II VARIATIONS. NOTATION AND NOMENCLATURE. ILLUSTRATIVE PROBLEMS 5. If y = f(x) and y = F(x) are the equations of two curves and r(x) = F(x) - f(x), v(x) can be regarded as the increment produced in y by changing f(x) into F(x) without changing x. We shall call it the variation of y and represent it by by. It is a function of x and usually a wholly arbitrary function of x. F(X) r(x) '. xo x, (x) == - by and is easily shown to be the increment produced in y' by changing f(x) into F(x), x being held fast, since when y becomes y + v(x) y' becomes y' + -'(x) and is changed by v'(x). We shall call this increment the variation of y' and represent it by by'. Like by it is a function of x. As we have seen y' dy. (1) If y is increased by b1y d dy(Y) is an approximation to the increment produced in q(y), the closeness of the approximation depending on the magnitude of by. If by is infinitesimal this approximation differs from the actual increen*_ byr an infinitesimal of higher order than by. 9 10 CALCULUS OF VARIATIONS We shall call this approximate increment the variation of ~ (y) and represent it by 65(y); so that 6(y) = ((y)ay. (2) 1v\ Ike trvajer 0(y, y') by + (Y, y') ay ay' is -la r an approximation to the increment produced in 4(y, y') by giving y the increment by and y' the increment 6y'; the closeness of the approximation depending on the magnitudes of by and by'. We shall call it the variation of k(y, y') and represent it by 5((y, y'), so that 0(y, y') _ (a, y') y + 4V(Y, y') ay/. (3) aJy ay ' As all our variations are supposed to be caused by changing the form of the function f in y = f(x) without changing x, a0(x, y, y') ~ y (x, y, y') ^(x, ') y a- ( y + ay, y'. (4) a1y ay1' If in (2) and (3) we replace the symbol a by the symbol d we obtain the familiar formulas for d+. Hence variations are calculated by the use of the familiar formulas and rocesses of the Differential Calculus. Indeed if y is a function, of x, df and 60 are approximate increments of precisely the same type but differently caused, do, by changing x without changing the form of the function and 60, by changing the form of the function without changing x. With our new notation our necessary condition that I, the Xi, f )(x y, y) d, shall be a maximum or a minimum can be written xo X1 f (x, y, y')dx = 0, (5) v. Art. 4 (1). If we define the variation of a definite integral as the integral of the variation of the integrand, so that ILLUSTRATIVE PROBLEMS 11 XXI fb0.dx = 6f&.dx, our condition (5) * xo xo may be abridged into 61 = 0. (6) 6. To illustrate the ordinary technique of the Calculus of Variations we shall now work out at length the problem of Art. 1 (b); the brachistochrone or curve of quickest descent. We wish to make I a minimum where (v. Art. 1 (b)) I S l+y2dx. We must make I1 = 0. xI x! frl y' d 'J Y - 10 + y'2 [ d _; IX + y,2 ydx. dl xl Integrating by parts, ri = f4 -= J -r —d1 LY V I -+ yi12 J dx Lx,1+ y'2 0 0 Since our end points are fixed by = 0 when x = 0 and when x = x1. Therefore Cd l y' 61 - -L; f1 ' ydx. As 61 must be equal to zero no matter what the value of our arbitrary function by we must have d 1 Y (1) dx+ V 2 (1) 12 CALCULUS OF VARIATIONS and (1) is our differential equation for y, i. e. the differential equation of our required curve. v. Art. 4 (3). (1) is easily integrated. We have 1 y' C; - Cl4x x whence Y= - X 1 - C2x C2X-X x/1x - x2 If we let 2a = y = I) xdx = - / 2ax - x2 + a vers — C. Y1 2ax -x2 a Since y = 0 when x = 0, C = 0; and y = - 2ax - x2 + a vers-' - (2) a where a is to be determined so that the curve shall pass through the point (xl, y1), is our required curve of quickest descent. (2) is the equation of a cycloid having its base in the axis of Y, its cusp at the origin, and lying on the positive side of the axis of Y. o Y X Our solution, then, is an inverted cycloid with a horizontal base and it has a cusp at the higher of the given points. Our treatment has assumed that in the required curve y is a single-valued function of x. Our solution is, therefore, apparently valid only if a cycloid suitably situated can be drawn so that (xi, yi) shall lie between the origin and the vertex of the curve. As a matter of fact this limitation is unnecessary as may be shown by interchanging the axes of X and Y and working the problem anew. ILLUSTRATIVE PROBLEMS 13 If this is done I = = 1 + Y- dx and it is not difficult to make 0 AI = 0. The resulting differential equation 2yy" 1 + y'2 = 0 can be solved by the method employed in the following section and we find as the equation of our required curve x = - / 2ay - y2 a vers-ly- (3) a Throughout the whole arch of the cycloid (3) y is a single valued function of x, and (xi, yl) is, therefore, not restricted to lying in the first half of the arch. In any concrete problem the numerical determination of a is a matter of some difficulty. This, however, may be avoided by Newton's famous solution for the curve of quickest descent from A toB. "Through A on the horizontal line AC as a base draw any cycloid cutting the straight line AB in D. Through A on the line AC draw A C a second cycloid whose base and altitude are to the base and altitude of the first as AB is to AD; it will pass through B and will be the curve required." 7. As another example we shall take the problem of Art. 1 (c), the minimum surface of revolution. Here (v. Art. 1 (c), (3)), X1 I = y iv1 + y'2dx. 0 1 =j f[y 1 + y'2]dx 0 14 CALCULUS OF VARIATIONS Xl = l 1 + y2y + yy' A dx 0 Xl = Cl —'+- 5 + / b 1yy'2 dx. Integrating by parts, XX fI = [ i y y /I1 + yt'2 1 + ' ] - 1ydx. 0 To make 61 = 0 we must make _d yy' =0 ( I -k- y, dx,2 () and this is the differential equation of our required curve. v. Art. 4 (4). Performing the differentiation indicated in (1) and reducing, the equation becomes yy - I - y'2 = 0. (2) Equation (2) can be integrated by a regulation device as follows: Let = y'. dz dz ddy,dz dz dx dy dx Y dy dy dz (2) becomes yz - = 1+ z2 dy dz dy 1 + 2 y. log (1 + Z2) - log y = C. lI+ Z2 1 -K -— K y a where a is an arbitrary constant. ILLUSTRATIVE PROBLEMS 15 y = y/y2 - 1 dy dx a V[y2 - a2 a cosh-l _ + b, a a y= a cosh +b). (3) (3) is the equation of a catenary having the axis of x as its directrix. The constants a and b must be determined so that (3) shall pass through the two given points (0, yo) and (xl, yi). To calculate their numerical values in a concrete case is not easy but the curve required may be obtained mechanically by the following process. Insert smooth pegs, at the two given points, in the vertical plane through these points. Throw over the pegs a uniform flexible string so long that one end will hang below the given axis; and then pull down on the other end until the free end is drawn up to the axis. Then the portion of the string between the pegs will be the required catenary. For it is well known that if one end of a uniform flexible string is fastened and the other end is thrown over a smooth horizontal peg, the curved portion of the string when in equilibrium will be a catenary whose directrix is a horizontal line through the hanging end of the string. 8. Since any function of a single variable can be geometrically represented by a graph by treating the independent variable as an abscissa and the corresponding value of the function as an ordinate, our theory covers the case where we are to make the definite integral of a given function of the independent variable, and of the dependent variable and its derivative a maximum or a minimum, no matter what the ordinary interpretation of the variables in question. For example let us work the shortest line problem (Art. 1 (a)) using polar coordinates. 16 CALCULUS OF VARIATIONS Let r = f(0) be the required shortest plane curve connecting the -points (ro, 40) and (rl, q1). Since ds = V/ dr2 + r2d02 01 I =f Vr2 + r'2d4, SI = 0. r1 81 = 8 r2 + r'd 2 00 rr +1 r 'rr 00 r~r d, -rd~ 2+ r2 f2 ' J r 2 d '2 Integrating by parts, +~~~41 4 ' 1 ~1 f r d = f r' - d r r2 + r,'2 df L r2 + r'2 d r2 + rt2 _r/ r jr + r — 00o,00 00 8r vanishes when < = qo and when 5 = 01. r d r' Therefore 81= \ r r I 8rd. I r2 + r'2 d % /r2+r, J 00 To make I1 = 0 we must make r d r' r+ - d r 2 =0; (1) r/2r2 + r do / 1 r 2 and this is the differential equation of our required line. If we perform the indicated differentiation and reduce, (1) assumes the comparatively simple form rr" -2r'2 _ (2) (r2 + r(2) (r2 + r'2) ILLUSTRATIVE PROBLEMS 17 We can clear of fractions and solve without difficulty, but if we happen to remember the Calculus formula for the curvature of a curve in polar coordinates [r2r (dr)] 2 r2-r +2W - we see that our differential equation (2) represents a curve of zero curvature, i. e. a straight line. 9. As another example let us find a geodesic line joining two given points on the surface of a sphere. The element of arc of a curve on the spherical surface is ds = V a2d62 + a2 sin2 Odp2 = a V1 + sin2 0'2dO; 01 and we may take I = f /1 + sin2 60'2 dO. 00 01 =r sin2 0'/6 dO J V 1 + sin2 o0'2 0o -= f sin2 0 d.d -V1 + sin2 0"'2 dO Integrating by parts and remembering that 6/ vanishes at both limits 0o and 01, we have 01 / 1d sin2' 04' 0J 1+ si 00 18 CALCULUS OF VARIATIONS Since 8I = 0 whatever the form of 50 d sjn26q$' = 0. (1) do V 1 + sin2 q5'2 Integrating, sin2 00' = C 1 + sin2 Oc'/2, C csc 0 C csc2 0 C csc2 0 sin20 - C2 i/ 1 - C2 csc2 0 ~ - C2 - C2 ctn2 0 If we let z = ctn and K2= - C 1 -,C= = -I d = - sin-1 Kz + a = - sin-l (K ctn 0) + a VK2 z where a is an arbitrary constant. sin (a - 0) = K ctn, sin a sin 0 cos 0 - cos a sin 0 sin 0 = K cos 0. (2) If we multiply by a and change to rectangular coordinates we get y sin a - z cos o. = Kx a plane through the centre of the sphere. The constants must be determined so that this plane shall pass through the given points and then the great circle in which it cuts the sphere is the required geodesic. EXAMPLES (1) Work the brachistochrone problem taking the axis of Y vertically downward. v. Art. 6, page 12. (2) Integrate rr" - 2r'2 - r2 = 0, the polar differential equation of the shortest line. v. Art. 8 (2). Ans. r cos (O - a) = p; a straight line. (3) Find the shortest line that can be drawn on the surface of a cylinder of revolution and joining two given points on the surface; a geodesic. Suggestion. Use cylindrical coordinates a, 0, z, taking z as the independent variable. ILLUSTRATIVE PROBLEMS 19 Z! Then I= f 1 + a2~'2dz. Z0 Ans. The geodesic is the line of intersection of the cylinder with the helicoid o = bz + c; and is a helix. (4) Find a geodesic on a cone of revolution. Suggestion. Use spherical coordinates r, a, 0, taking r as the independent variable. Then I = fi 1 + r2 sin2 a2dr. ro0 Ans. The geodesic is the intersection of the cone with the surface r cos [(0 - I) sin a] = c. (5)' A mountain is in the shape of a hemisphere and the velocity of a man walking upon it varies as his height above its base; show that his path of least time between two given points upon its surface lies in the vertical plane through the points. 10. Isoperimetrical Problems. There is an important class of problems in which it is required to make I a maximum or a minimum while keeping constant the integral of a second given function of x, y, and y' taken between the same limits. (a) For example take the problem to find the closed plane curve of given perimeter and maximum area. We shall use polar coordinates and take the origin within the required curve. If r = f(4) is our curve, its area, A, and its perimeter, S, are equal 27r 27r respectively to r2d and fVr + r'2d4. 0 0 20 CALCULUS OF VARIATIONS We are to make A a maximum while S is kept equal to a given constant 1. If X is a constant multiplier at present undetermined A + XS will be a maximum if our problem is correctly solved. 27r Let, then, I = [r2 + X / r2 + r'2]do, 0 and proceed as usual. 2 r =I f[rr +; rr +r d r 0 J +[ r~t^p2 2 + r2 i r'2 ] 27r J -] Vr2+r'2) d -Vr2- +rd2 Since 1I = 0, I d r/ r + V 0. ra(j + r1+r _ d r2 -r = 0o \X VA r2 + ry2/ d2 d -r2 -+ ry'2 Performing the indicated differentiation and reducing we get rr" - 2r'2 - r2 1 (r2 + r'2)2 ~ X as the differential equation of our required curve. The first member is the curvature of the curve, and as the curvature is i, a constant, the required curve is a circle. v. Art. 8, page 17. As X is the radius of the circle its given perimeter 1 is 2rX and our X proves to be1 2' ILLUSTRATIVE PROBLEMS 21 (b) The ends of a uniform string are fastened at given points; find the equation of the curve in which it must hang that its centre of gravity may be as low as possible. Let (xo, yo) and (xi, yi) be the end points, and let 1 be the length of the string. fi 1 + y'2dx = 1. xo If y is the ordinate of the centre of gravity x1 Y = { y l - y'2dx. Xo We must make xI y + Xl, i. e. I(y + Xl) 1 + y'2dx, a minimum. xo X1 Let K = X1 and let I = (y + K) V1 + y'2dx. Xo Turning to Art. 7 we see that we have only to replace y by y + K in the solution there given to get the solution of the present problem, and that our required curve is y + K = a cosh ( b), a catenary. Unlike the catenary in Art. 7 it does not generally have the axis of X for its directrix. In a numerical problem a and b would have to be determined so that the catenary would pass through (xo,^yo) and (xl, yi) and K would have to be found by the condition that X1 fI 1 + y'2dx = 1. Xo Problems like (a) and (b) are known as isoperimetrical problems and can all be handled by the device we have illustrated. 22 CALCULUS OF VARIATIONS EXAMPLES (1) Find the form of a curve if the area bounded by an arc of given length, the ordinates of its given end points, and the axis of X is to be a maximum. Ans. A circle. (2) Find the form of a curve if the sectorial area bounded by an arc of given length and the radii vectores of its end points is to be a maximum. Ans. A circle. CHAPTER III PROBLEMS INVOLVING SEVERAL DEPENDENT VARIABLES 11. It is easy to extend our theory to problems where instead of a single dependent variable y and its derivative there are several variables to be determined as functions of the same independent varia-t XI ble. That is, when I = 0f(x, y, z,., y *, z',.. )dx and I is to be xo made a maximum or a minimum. Let 6- = ay + a z+... + 8y' - +.. and by the method of Art. 4 it can be shown that we must make 61 = 0. d d Note that y' = y, ' = z, z~ ~ As an example let it be required to find the shortest curve not necessarily plane joining (0, 0, 0) and (xi, yi, z1). ds = V dx2 + dy2 + dz2 = 1 + y2 y + 2dx. X1 I = J1 + yy2 + z,2dx. 0 i = z' + z dx. -J 1 + y'2 + z'2 By integration by parts 61 =- - __-_ y -dx _ _y' _-.'2 d. ) dx 4 1 + y'2 + z2 d 1 + y2 + 2 3 24 CALCULUS OF VARIATIONS To make 6I = 0 we must make d y' d z =y +- 3z dx 1 + y'2 + z22 dx 1 + yt 2 + z2 vanish when by and 8z are independent arbitrary functions of x. Hence d Y' = dx V 1 + y'2 + z2 d z/ and = 0. dx V i +- y'2 + z'2 These are a pair of simultaneous differential equations of the second order connecting y and z with x; and can be solved without difficulty. We get -= =, a straight line. 12. Hamilton's Principle, called by Professor E. B. Wilson " the most fundamental and important single theorem in mathematical physics," and closely allied to the so-called principle of Least Action is that in the actual motion of any conservative system the time-integral of the sum of the Energy and the force function is a minimum. to Formulating, [T + U] dt = 0, if U is the force function and T to the kinetic energy. We give a proof of Hamilton's Principle for a single moving particle. It can be extended without difficulty to a moving system. Let a particle of mass m moving under forces X, Y, Z, actually move from its initial to its final position in the time ti. Suppose that by the introduction of additional forces it had been made to move from its initial to its final position in the same time tj, but bya slightly different path. Let (x, y, z) be its position at any time t in the actual motion and (x + 8x, y + 8y, z + 3z) its position in the hypothetical motion after the lapse of the same time. If t is our independent variable we have from Mechanics mx" = X, my" = Y, mz" = Z; and if the forces are conservative and U is theforce function Xax + Y6y + Z z = bU. SEVERAL DEPENDENT VARIABLES 25 Xax = mx"8x = m -(x'8) - x'']= m x2d(X) _ Lati'b) *'bI dt 2 bU = md[xbx + y'by + z'az]- -T, where T = r [x'2 + y'2 + '2] and is the Kinetic Energy. t ti t=tt t = O Jb(T + U)dt = m[x'xa + y'y + z'z] = 0, 0 since the initial and final positions of the particle are the same in the actual motion and the hypothetical motion. As an illustration of the use of Hamilton's Principle we shall obtain the differential equations for planetary motion by its aid. If we use polar coordinates and take the sun as origin T, the energy of the planet, is - (r'2 + r2'2). The force, F, exerted by the sun is an attractive force directed toward the sun, proportional to the product of the masses and inversely proportional to the square of the distance. uM F =- 7m The work done in any motion of the planet isfFdr and JFdr = - rMmf = -r + C. The simplest force function, U,:is equal to -Mm. r By Hamilton's Principle (r2 + r2p'2) + dt = 0. r br'6' + r2&S' + rk'26r - brM rdt = 0, or [r ' + r2' dt 6 + (r2 - ) r]dt = 0. 0 26 CALCULUS OF VARIATIONS Integrating by parts, tl d d J d[-t dt (r')8- + r'2 - _ rdt = 0. o Whence r" - r'2 = - M and r2' = k. EXAMPLE A heavy cube containing a smooth spherical cavity rests on a smooth horizontal plane, and a heavy particle lies at the bottom of the cavity. The cube is given a horizontal velocity. Find the equations for the subsequent motion of the system. Suggestion. Let M be the mass of the cube and m the mass of the particle, and use as coordinates x, the distance the cube has moved, and 0, the angle a radius of the cavity drawn through the particle makes with the vertical. The horizontal, velocity of the particle is easily seen to be x'- a cos 00', its vertical velocity is a sin 00'. T = ~(M + m)x'2 + ma?20'2 - 2a cos Ox'O'. U = mga cos 0. The equations required are d [(M + m)x' - ma cos 00'] = 0 d (aO' - cos Ox') - sin 0 (x'0' - g) = 0, or (M + m)x' - ma cos 00' = C a0" - cos Ox" + g sin 0 = 0. 13. If in the proof of Hamilton's Principle in the last article we do not replace Xax + Y6y + Z5z by its equal SU we get tl f[aT + XSx + Yhy + Z8z]dt = 0, 0 and this equation is valid even if the forces are not conservative. It is to be noted that X~x + Y6y + Z5z is the work that would be done SEVERAL DEPENDENT VARIABLES 27 by the actual forces if the particle were displaced from its actual position at any time to its corresponding position in the hypothetical path. The differential equations in polar coordinates for the motion of a particle in a plane are easily obtained by the aid of our new equation. Let the component of the force along the radius vector be R and perpendicular to the radius vector be 1. If (r, 0) is the actual position of the particle at the time t the corresponding position in the hypothetical path is (r + Sr, + 5- &). Xax + Y5y + Z~z becomes RMr + 4rbo. T = - [r'2 + r252]. tl rf(r12 + r20'2) + Rar + bPrWj dt = 0,O tl F[m (r'r' + r20'6$' + r'2 Sr) + Rar + dJ rSk]dt = 0, 0 tl ffEt-'r" + r~'2 + R)6r - (d t(r2,) _- r)S]dt = O. - r2= Rd r d-(r2') = i, EXAMPLES (1) Find the differential equations for the motion of a spherical pendulum. Ans. 0" - sin cos ' + sin0 = 0 a sin2 a0' = C. (2) Obtain the differential equations for the motion of a particle in space, using cylindrical coordinates. m d Ans. m(r" - r0'2) = R, m d (r2') =, mz" = Z. r dt 28 CALCULUS OF VARIATIONS (3) Obtain the differential equations for the motion of a particle in space, using spherical coordinates. Ans. m[r - r (0'2 + sin2 06'2)] = R m[d (r20') - r2 sin 0 cos 0/2] = 0 rm d r i 0dt (r2 sin2 60') = -. r sin 6 di CHAPTER IV MULTIPLE INTEGRALS 14. If I is a multiple integral no serious additional difficulty is presented. We have still to make 5I = 0. If I = jf (x, y, z, p, q)dxdy, Oz where P = ax Oz and q = -y and the integration is over a given area in the XY plane, let it be required to make I a maximum or a minimum by suitably determining z as a function, f, of x and y, z being given for all values of x and y corresponding to points in the boundary of the area. Here 6z is the increment produced in z by changing the form of the function f, x and y being held fast, and is a function of x and y. a a If we let p = - 3z, and bq = z then ap and aq are the increments produced in p and q. z = f(x, y) + acz is any one of a family of surfaces of which = f(x, y) is one. I(a) = )(x, y, z + a5z, p + ap, q + a8q)dxdy. I'(O) = [ z + ap p + aq qj d xdy:- CSdxdy = ffcdxdy = I, 30 CALCULUS OF VARIATIONS and, as before, the necessary condition that I shall be a maximum or a minimum is 81 = 0. 15. Minimal Surface. As an example let us find the form of the surface of least area that can be bounded by a given closed curve not necessarily plane. I = ff 1 + p2 + q2 dxdy, the integral being taken over the area bounded by the projection of the given curve on the XY plane. 81r= ff PSP + q8q a a 61 = If t x- ^-dxdy V1 + p2 + q2dxdy ++ p + q JJ~ YV 1 + lp2 + q2 Integrating by parts, r p z = p~z r' p _ _ J p2 + q2 1 + p 2 + qa2 X 1 + p2 + q2 Since in our x-integration y is held fast our limits are the abscissas of the points on the projection of the given boundary which correspond to the value of y in question, and for these points 8z = 0. Therefore the term outside of the sign of integration vanishes and r a a p I -d = _- q adx P - 8zdx. f -1j 2+q2 dx- -J axl+p2+q2 In like manner it may be shown that - az I-f q-y^ - dy Szdy. J/ l + p2 + q2 d = - - 1 + V p2 + q2 MULTIPLE INTEGRALS 31 Hence I= - fL V 1 + p2 + q2 ~ a W + p2 + q2] bzdxdy 9P /q IqP\ Iq (1 ) a - Pq-aq + y (1 + p2) a y. - JJ ~ (1 + p2 + q2)2 AI = 0 when and only when [1 (y) aX2 ax ay axay + x ay2 andx Y L xy -0, (1)if and (1) is the differential equation of the required minimal surface. It is shown in works on Differential Geometry that the first member of (1) is the sum of the curvatures of any two mutually perpendicular normal sections of the surface, and its half is called the mean curvature of the surface at any point. Therefore a minimal surface is a surface of zero mean curvature. EXAMPLES (1) Show that the mean curvature of a helicoid z = k tan-1 y is x zero. (2) -A cylindrical cup with a plane bottom of any shape and a given upper rim of any shape is to be capped so that the closed cup shall have a given volume and a minimum upper surface. Show that the cap must be a surface of constant mean curvature. 16. The differential equation for small transverse oscillations of a stretched elastic string can now be obtained from Hamilton's Principle. Let 1 be the length and p the tension of the string in its position of equilibrium, and let the motion be so small that higher powers than the first of the slope of the string at any time may be neglected. Let us assume also that longitudinal motions and the change in tension due to the stretching of the string during its oscillation are negligible. 32 CALCULUS OF VARIATIONS If m is the mass of a unit length of the string the kinetic energy T is equal to - ) dx. The force function U is the negative of 0 the potential energy, V, of the string, and V is equal to the work that would be done by the tension, p, in restoring the string to its length when in equilibrium: i. e. V = p (s - 1), 1 l or V=-p 1 \ ) dx- ] =P (1+ ( ) +... dx-l] 0 0 so that approximately V = f~Y) dx. 0 By Hamilton's Principle 6I = 0, t where I = (T - V) dt. 0 t I,:ssR~ 2 ff[; 2\axdt. -Jffr (aYt 2 /- ( 0 0 t 1 JI = [m t at p x dxd 0 0 t I ay a by - P ay a by dxdt 0 0 I t t I =- fm eaY a by dtdx - p aY a ay dxdt. at at O 00 00 t t may a ydt - fm a ydt at at Ot2 o o since Sy = O when t = 0 and when t = t. v. Art. 12. Z I P ay a ydx = - 8ydx. 0 0 MULTIPLE INTEGRALS 33 t I 61 -J -[mj a-t2-P 2 ydxdt = 0. 0 0 92y p a2y Hence at2 m ox2 EXAMPLE Obtain the differential equation for small transverse oscillations of a stretched elastic membrane. Suggestion. V = pjj\il + (-) + () dxd - A where A = jfdxdy is the area of the membrane at rest. A n 2' p [a2z -2z1 Ans. -= - a + y W~ m [8x2 y y2 CHAPTER V VARIATION OF THE LIMITS. PRINCIPLE OF LEAST ACTION 17. In our fundamental problem it is required to determine y as X1 a function of x so that f(x, y, y')dx shall be a maximum or a Xo minimum, xo, xl, and the corresponding values'yo and yi being given, so that 5y = 0 when x = x0 and when x = x1. If while keeping the range of integration fixed we remove the restriction that yo and yl are given the old reasoning by which we established the necessary condition X1 PI =fgLraf by + go, by'] dx = O0 Xo still holds good; but when we integrate by parts and get XI X1 b1 = I \ - T b,ydx + by 0 1 -ay dx ] y'd a] o Xo XO the last term no longer disappears, but becomes c0] -y - [ra1ay x = X1 X = ao where 3yo and byi are entirely arbitrary; so that 81 = 0 when and only when ~a d y [10, 0= and o, =0, ay dx 9y' 'Ly- J - Ly- aJ X = X0 = 21 and we must determine the two arbitrary constants in the solution of the first equation by the aid of the second and third equations. 34 VARIATION OF THE LIMITS 35 The problem of sailing to windward is an interesting illustration. The speed of a sailing vessel is a function, more or less complicated, of the angle her course makes with the direction of the wind. It is required to find how she must sail to get a given distance, a, to windward in the least possible time. Let the starting point be taken as origin and an axis of X be drawn directly to windward. If y = f(x) is the required path y' is 0 a the tangent of the angle the course makes at any time with the wind's direction, and the velocity can be expressed in terms of y'. ds 1 _ — y'2 dt = - ) +.d d F(y') F(y') a."~/1 -% y'2d. and I = WF(y' dx. 0 a a1 - _-I + y_2yd JI y= f F(y') 'd 0 _d ___ /1- yI ~ /l1+y'21 - dx a9y' F(y') aydx / La' F(y') 0 x- =a Since we must make 5I = 0 a y,2 = C (1) ay' F(y') and [w' j ](2) ~and [by'l aF(y') 36 CALCULUS OF VARIATIONS (1) gives us y' = c, and the required path is a straight line. (v. Art. 3.) (2) reduces to y'F(y') - (1 + y'2)F'(y') = 0, or since y' = c to cF(c) - (1 + c2)F'(c) = 0, which determines c, the constant slope of the required path. As a rough approximation of the speed law for a rather sluggish boat let v = F(y') = k (tan- y' - a). F'(y') = 1 + and ck (tan-c - a) = k. ( - a) tan0 = 1, then, gives us the angle, 0, the course (known to sailors as full and by) should make with the direction of the wind; and as 0 does not depend upon k this course is independent of the strength of the wind. If the point aimed at lies within the angle formed by lines through the starting point making angles 0 and - 0 with the wind's direction it is reached by tacking, both tacks being steered full and by. If it lies outside of this angle it should be steered for directly. EXAMPLES (1) If a 1= 3 - 1 = =620~. 2 (2) Find the curve of quickest descent from a given point to a given vertical line. (v. Art. 9, Ex. 1.) Ans. An inverted cycloid with a horizontal base, a cusp at the given point, and the vertex in the given line. (3) Given two mutually perpendicular lines and a point in their plane; find the curve terminated by the point and one of the lines which by its revolution about the other line shall generate a surface of minimum area. (v. Art. 7.) Ans. A catenary having the axis of revolution as its directrix and having its vertex on the other given line. VARIATION OF THE LIMITITS 37 18. If I= (xy, y') dx and we vary xo and xi as well as yo and yi, so that the range of integration is no longer fixed, the problem is more complicated. For the sake of simplicity we shall suppose that xo and Yo are given but not xi or yi, and that we are to determine the curve y = f(x) so that I shall be less if we seek a minimum, greater if we seek a maximum than when y = f(x) is replaced by any neighboring curve and xi is replaced by any value, xi + ax1, differing from xi by a sufficiently small amount. With our ordinary notation y = f(x) + a~y is a neighboring curve if a is sufficiently small (v. Art. 3); and I(a), where xI + abxl 1(a) =f(x, y + a~y, y' + a~y') dx, XO must be a maximum or a minimum when a = 0. Let u = y + asy and v = y' + aby', X1 + axIx then I(a) =f(x, u, v)dx. xl xI + aux1 I'(a) f y + by' dx + [O(x, u, v)] ax1 q au eV x = x1 + axIx Xl and it(O) b(y+ by)dx + [4x, y,y')]6x1 I'(O) =J \ay ay' ~1~~~~~x =fc dx + [4(x, y, y')] xi [00 - d 00 bydx + x1 __ d l aydx+[f ya yl + [cP(x, Y, y') 6x1. Lay dx ay' [ay x= x XO x=xl 38 CALCULUS OF VARIATIONS To make I'(0) = 0 we have as usual Lagrange's Equation 2~ d 2~ -- 0 -; (1) 2y dx 2y' and we have the conditions y = yo when x = Xo, and by, ay] + [ (x, y, y')]x1 = 0, (2) x = x= to determine the two arbitrary constants in the solution of Lagrange's Equation. If (by),, the value by assumes when x = x1, and 3x1 are independent the solution is generally impossible, but in many interesting problems (by)1 depends upon 8x1. For instance suppose that the required curve y = f(x) which makes I a maximum or a minimum is to be terminated by a given point (x0, yo) and a given curve y = p(x), so that yl = {(xl). Let the second end point in the varied curve be (x + -6xl, yi + -by). Then by1 = f'(x1)6x1 + e and (6y)l = aY1 - y' 8x1 - r = ((x1) - y'D)ax + E -, where e and v depend upon 6x1, and - and can be made as small as we choose by taking a sufficiently small value of 8x1. The substitution of this value for (ay)l in (2) gives [ay/' ( ) - Y + E-n) + 4 (x, Y')],1 = 0, (3) X= X1 VARIATION OF THE LIMITS 39 and as (3) is to hold independently of the value of ax1, [, ('(x) - y') + +(x, y, ') = 0. (4) X = X1 Comparing (4) with (2) we see that it is (2) with 8y replaced by di(x) - dy, and Ax1 by dx. If xo is varied as well as x we have in addition to (2) the condition [~ ] tiona [by + (x,, y, Y') xo] = o x = XQ and if y = f(x) is to start at a point on the curve y = x(x) we must replace Sy and Ax0 by dx(x) - dy and dx respectively. 19. (a) Suppose that in our Minimum Surface problem of Art. 7 the required generating curve is to be terminated by the point (0, yo) and the given curve y =,/(x). We get as in Art. 7 d yy' y /1 + y,2 - y -= 0 dx j 1+ yf2 as the differential equation of the curve, and the condition [y1 + ay + y 1 + y2 axl = O, V = X1 which by Art. 18 gives us [iY Y+ ('(x) - y') + I + y'2]= 0 X = X1 or [y'p'(x)] + 1 = 0. X= X1 Therefore the required curve meets the given curve at right angles. It has been shown in Art. 7 that it is a catenary having the axis of revolution as its directrix. (b) As a second example let it be required to find the curve of given length and terminated by the axes, which with the axes encloses the maximum area. Here XI I = f[y + 1 + y'2] dx. 0 40 CALCULUS OF VARIATIONS SItf1 y 1 +yY + 'X d + y'2], 0 X = X1 Xl Z = X1 = f[l — — ^ — 5,+1~2l aydx + -ay ++X'V]ax, =0. j[ dx y' l+ (1 y [.kly"2/ [ [ y L X /26x= 0 x-=0 x=xI d Xy' 1 dx 1 (1) dxy-0 l '2 1 +- 0Y (2) x= xX f.^.(o - l) + y + x I T y = o. (3) 41 + y,2 (1) integrates into (x + C)2 + (y - K)2 = \2, a circle. (2) gives [y'] = 0, and the circle crosses the axis of Y at right angles. (3) gives[ ' =0 since y = 0 when x = x1, and the circle X = XI crosses the axis of X at right angles. Therefore the area in question is the quadrant of a circle. 20. THE PRINCIPLE OF LEAST ACTION. If T is the kinetic energy of a moving system the time-integral of 2T is called the action of the system. If V is the potential energy of the system and the forces are conservative it is well known that the so-called equation of energy T + V = h, where h is a constant, holds good during the motion. The Principle of Least Action is that in the actual motion of the system the action is less than in any hypothetical motion, between the same initial and final configurations, in which the equation of energy holds good. In Hamilton's Principle the actual motion is compared with a hypothetical motion in which initial and final configurations and. PRINCIPLE OF LEAST ACTION 41 time of transit are the same as in the actual motion, and to bring about the hypothetical motion additional forces would have to be imposed on the system, and these forces would do work. In the Principle of Least Action the actual motion is compared with a hypothetical motion in which initial and final configurations are the same as in the actual motion, but as the equation of energy holds good, the time of transit is different in the two motions; and the hypothetical motion could be brought about by imposing constraints that would do no work. The formulation of the Principle of Least Action is tl f2Tdt = O T + V = h. 0 As in the case of Hamilton's Principle we shall prove the Principle of Least Action for a single moving particle, and we shall use the notation and as far as possible the results of Art. 12. Keeping in mind that the potential energy is the negative of the force function, i. e. V = - U, and that ti, the time of transit, is varied. t l tj tj 5f2Tdt = af(T + h - V)dt =(T - V)dt + [T + h - V]tl. 0 0 0 t= tl - t=ti Jf(T - V)dt = m[x'ax + y y + z'6z], t=0 0 as in Art. 12. Since the initial positions are the same in the two motions Ax = by = 8z = 0 when t = 0. Since the final positions are the same but are reached in the time t1 in the actual motion and in the time t- + -tl in the hypothetical motion Ax = by = 8z = 0 when t = tl + itl, and are approximately - x' tl,, - y' btl, and - z' 3t, when t = tl. Therefore (v. Art. 18) tl fb(T - V)dt = - m[x'2 + y'2 + z2]tl =-[2T}]6t 0t = t t =t 0 42 CALCULUS OF VARIATIONS approximately, and tl af 2Tdt = - [T + V - h]tl approximately. t= t=tl 0 =0. In any concrete problem of course the Principle of Least Action leads to the same differential equations of motion as Hamilton's Principle. As an example we shall take the problem of planetary motion (v. Art. 12). m (r,2 _2012) V YMm T = (r'2 + r22), V - r as in Art. 12. By the Principle of Least Action tl I = r fnm(r'2 + r20'2)dt = 0 0 and (r,2 + r22) Mm = h. 2 r ti $I = $7 [2 (rg2 + r2~'2) + h + mdt = fm r'8r' + r 0'6(' + ro br 2 S rdt +M (r2 + rt22) + h + m] t = tl Integrating by parts tl 1 = M df- r'r dt- (- ') + (rS 2 - )6r dt dt d M 0 + m[r'6r + r2'6] + [ (r'2 + r2412) + h + m btl. 0 L t = t The terms outside the integral sign reduce to [ (r'2 + r24/2) + h + tMm] t t = t PRINCIPLE OF LEAST ACTION 43 and are equal to zero. Therefore tl = m S- dtd r —dt (r2') +(rp'2- r]dt = O o as in Art. 12. And we get r" - r12 = and r2~' = k, as in Art. 12. As Hamilton's Principle is more general than the older Principle of Least Action and is decidedly easier to use it is much more commonly employed in practice than the latter principle. 21. The whole subject of Variation of the Limits can be treated from a different point of view, that of parametric representation. Instead of trying to determine y as a function of x so that X1 I = ( (x, y, y')dx shall be a maximum or a minimum let us XO start to determine x and y as functions of a parameter r. dx dy We shall find it convenient to represent - by x and dY by,. in which case y' = -, and dx = Xdr. r of course will be our independent x variable, and we shall suppose that r0 and r1 are fixed while the corresponding values x0 and x1 may be varied. Xl T1 I = (x, y, y') dx = j(x, y, y')xdr. 0o To tl I = r[ + x +. a ay ay + /1dr to tl LXA -xx + I- + y +, dTto 44 44 ~~CALCULUS OF VARIATIONS Integrating by parts Ti ex d X y'Ja \ay dr ay/' To make 6I =0 we must make ay dry0(1 ~ao _ d / _ Ox dT ayf ~~~~~~~~(2) - f X + 6 = (3) when r To and (q ",ax, + ~aoy, 0 (4) when T r Dividing through by (1) and (2) become apo d _, _ (5) a0 d l __ sy)-0. (6) (3) and (4) reduce to (q f y' bXo + 8o Yo =0 (7).when x =xo, ( 4 - '~;ax, + by= 0 (8) when x = x1. (5) is Lagrange's Eqffation. (6) which can be shown to have the same solution as (5) we can disregard. (7) and (8), if we have a relation between bxo and 6yo, and between ax, and by,, serve to determine the two arbitrary constants- in the solution of (5). PRINCIPLE OF LEAST ACTION 45i For instance if the required curve is to be drawn from a given curve y = x(x) to a given curve y = 4/jx) as in Art. 18, 5Yo = X'(x) axo and by, = ik'(x) ax1, approximately, as in Art. 18; and (7) and (8) reduce to 0 + (XI W> - Y') = 0 when x = xo ayl and 0 + (IP'(x) - y') = 0 when x = x1 v. Art. 18. 22. To illustrate the practical working of this method we shall apply it to the second example in Art. 19, to the proof of the Principle of Least Action (v. Art. 20) and to the problem solved in Art. 20. XI (a) (v. Art. 19). I = [ + X- + y2] dX. 0 If x and y are functions of a parameter T, ax = 0 and by is arbitrary at the start, i. e. when r = ro; and ax is arbitrary and ay = 0 at the finish, i. e. when r = Ti. 7T T ai =Jh Lxay + yai1+ *2 + 1 2 J dr f + + - t -j -MT ~' 7 TO T Tj~~~~~~~~~~~~~~T ~ti + Y" lr + + ~3~z drjl Sincea=.1 _ d xY - TO= d d X.dT W 2+ d7 ~ +2 by + +ax0; dT 1/ 2 + Y2 46 CALCULUS OF VARIATIONS = 0 when 7 = -o J 2 + 1 2 'X. + + = 0 when r = ri. ~ 2~ + ~2 These equations reduce to dx V+ (1) Y' + d 0; (2) dx -VI + y'2 0 when x = 0 W 1 + y'2 0 when x = x1. ~11+ Y'2 (1) is (1) Art. 18 and integrates into (x + C)2 + (y - K)2 X2, as does (2). The curve is a circle; and as y' = 0 when x = 0 and y = oo when y = 0, the circle crosses both axes at right angles and the centre must be at the origin. (b) (v. Art. 20). We shall suppose that the time, t, and therefore the coordinates x and y, depends upon an independent variable r in such a way that when r = To, t = 0 in the actual and in the hypothetical motion, and when r = ri, t = t1 in the actual motion and t = t1 + St, in the hypothetical motion. Note that 8x and by are zero when r = ro and when r = -ri, and that 5t = 0 when r = To. Tdt (T + h - V)dt = [T + h - Vl]dr. o 0 To f2 Tdt =f[(6T - 6V)> + (T + h - V)S]dr 0 T. - [(ST - V)i + 2TM]dr. TO -V =M(X"8x + Y"by + z"5z). PRINCIPLE OF LEAST ACTION 47 x"6x = d (x'ax) - x' dx dt dt =t Ax y + ya+ zz 1 ^(T) m d xIx + Y6y + ix _ t m d ax7 y~.a a2 t drt t t m d &ax + yay + z 1 [T- V] + 2T6 = md x + y +- z tl a 2T6t = m[x'ax + y'ay + z'az] 0 t70~~~~ r=7~ TO 0 since ax = by = az = 0 when r = To and when r = r1. (c) Planetary motion (v. Art. 20). We are to make I =f[(r12 + r22) + h + jM] dt 0 a minimum, while keeping m (r'2 + r2q,2) _ Mm h. 2 r = mf r2 r22 + h KI dr TO TJ (m+ r - r + arjdrO M L _r_2 _ r* M + h2+r + =mJJ L i 2)r+ar+ t a -~ r - J To 48 CALCULUS OF VARIATIONS SI f [(r&2 MMt dr\ d r2O d (IM i2+r2 2~ ] 61 = m -- - d-r —_0 j2-' - _ - }t dr t r2 dr t dr i dr\ r 22 To + m ar + r0 + ( M A + r2-2) t] t?st.r V T = TO To make i6 = 0 we must make r02 lMit d d r20 dr ~ - 0, dr t d (M _ r2 + r22)=; dr\ r 2t2 h F tM r2 + r202 and w+ _-= o whenr = 1; m r 2 or r" - r'2 + M = 0 (1) r2' = k (2) dt[ rM - m(r'2 + r'2) = 0; (3) and h + Mm (r2 + r2~'2) = O (4) and gh+ Mm (4) r 2 when t = ti. (1) and (2) are the required equations of motion and (3) and (4) follow from the Equation of Energy. EXAMPLES (1) Work Art. 19 (a) by the method of Art. 21. (2) Work examples 2 and 3 of Art. 17 by the method of Art. 21. PRINTED AT THE HARVARD UNIVERSITY PRESS CAMBRIDGE, MASS., U.S.A.