VARIATIONAL PRINCIPLES

Transcription

1 CHAPTER - II VARIATIONAL PRINCIPLES Unit : Euler-Lagranges s Differential Equations: Introduction: We have seen that co-ordinates are the tools in the hands of a mathematician. With the help of these co-ordinates the motion of a particle and also the path followed by the particle can be discussed. The piece wise information of the path y = f, whether it is minimum or maimum at a point can be obtained from differential calculus by putting y =. The function is either maimum or minimum at the point depends upon the value of second derivative of the function at that point. The function is maimum at a point if its second derivative is negative at the point, and is minimum at the point if its second derivative is positive at that point. However, if we want to know the information about the whole path, we use integral calculus. i.e., the techniques of calculus of variation and are called variational principles. Thus the calculus of variation has its origin in the generalization of the elementary theory of maima and minima of function of a single variable or more variables. The history of calculus of variations can be traced back to the year 696, when John Bernoulli advanced the problem of the brachistochrone. In this problem one has to find the curve connecting two given points A and B that do not lie on a vertical line, such that a particle sliding down this curve under gravity from A reaches point B in the shortest time. Apart from the problem of brachistochrone, there are three other problems eerted great influence on the development of the subject and are:. the problem of geodesic,. the problem of minimum surface of revolution and Classical Mechanics Page No. 95

2 3. the isoperimetric problem. Thus in calculus of variation we consider the motion of a particle or system of particles along a curve y = f() joining two points P (, y ) and (, ) infinitesimal distance between two points on the curve is given by Q y. The Q(, y ) ds P(, y ) ds = d + dy. Hence the total distance between two point P and Q along the curve is given by ( ) ( ) dy I y = + y d, y = d In general the integrand is a function of the independent variable, the dependent variable y and its derivative y. Thus the most general form of the integral is given by I y = f, y, y d.... () ( ) ( ) This integral may represents the total path between two given points, the surface area of revolution of a curve, the time for quickest decent etc. depending upon the situation of the problem. The functional I in general depends upon the starting point (, y ), the end point (, ) y and the curve between two points. The question is what function y is of so that the functional I ( y ( )) has stationary value. Thus in this chapter we first find the condition to be satisfied by y() such that the functional I ( y ( )) defined in () must have etremum value. The fascinating principle in calculus of variation paves the way to find the curve of etreme distance between two points. Its object is to etremize the values of the functional. This is one of the most fundamental and beautiful principles in applied mathematics. Because from this principle one can determine the Classical Mechanics Page No. 96

3 (a) (b) (c) (d) (e) (f) Newton s equations of motion, Lagrange s equations of motion, Hamilton s equations of motion, Schrödinger s equations of motion, Einstein s field equations for gravitation, Hoyle-Narlikar s equations for gravitation and so on and so forth by slightly modifying the integrand. Note : A functional means a quantity whose values are determined by one or several functions. i.e., domain of a functional is a set of all admissible functions. e.g. The length of the path l between two points is a function of curves y(), which it self is a function of. Such functions are called functional. Basic Lemma : If and ( > ) are fied constants and function for and if G G is a particular continuous η d = for every choice of continuous differentiable function η such that η ( ) = = η ( ), then in. G = identically Proof : Let the lemma be not true. Let us assume that there is a particular value of in the interval such that G ( ). Let us assume that G ( ) >. Since [ ] G is continuous function in and in particular it is continuous at =. Hence there must eist an interval surrounding say in which G > everywhere. Classical Mechanics Page No. 97

4 Let us now see whether the integral G of η. We choose η such that η = for η d = permissible choice = for = for For this choice of η which also satisfies η the integral G η ( ) η ( ) = =, d becomes η = η + η + η G d G d G d G d... () Since G η d ( ) ( ) G d... () = G > in, R. H. S. of equation () is definitely positive. η G d >, This is contradiction to the hypothesis G η d =. If G ( ) <, Classical Mechanics Page No. 98

5 we obtain the similar contradiction. This contradiction arises because of our assumption that This implies that G for in. G = identically in. This completes the proof. Theorem : Find the Euler- Lagrange differential equation satisfied by twice differentiable function y() which etremizes the functional where y is prescribed at the end points. Proof: Let P (, y ) and (, ) = ( ) (,, ) I y f y y d Q y be two fied points in y plane. The points P and Q can be joined by infinitely many curves. Accordingly the value of the integral I will be different for different paths. We shall look for a curve along which the functional I has an etremum value. Let c be a curve between P and Q whose equation is given by y = y (,). O y c : y (, α) = y (, ) + α η() P(, y ) c : y = y (, ) Q(, y ) Let also the value of the functional along the curve c be etremum and is given by I y = f y y d... () ( ) (,, ) We can label all possible paths starting from P and ending at Q by the family of equations (, α ) (, ) αη y = y +,... () where α is a parameter and η is any differentiable function of. Classical Mechanics Page No. 99

6 For different values of α we get different curves. Accordingly the value of the integral I will be different for different paths. Since y is prescribed at the end points, this implies that there is no variation in y at the end points. i.e., all the curves of the family must be identical at fied points P and Q. ( ) η ( ) η = =... (3) Conversely, the condition (3) ensures us that the curves of the family that all pass through the points P and Q. Let the value of the functional along the neighboring curve be given by I y = f y y d... (4) ( (, α )), (, α ), (, α ) From differential calculus, we know the integral I is etremum if I α α = =, since for α = the neighboring curve coincides with the curve which gives etremum values of I. Thus I α α = = Integrating the second integration by parts, we get f f + d = y y., η η f f d f η d + η η d = y y d y... (5) As y is prescribed at the end points, hence on using equations (3) we obtain Classical Mechanics Page No. η d = y d y. By using the basic lemma of calculus of variation we get =.... (6) y d y This is required Euler- Lagrange differential equation to be satisfied by y() for which the functional I has etremum value.

7 Important Note : If however, y is not prescribed at the end points then there is a difference in y even at the end points and hence η. As the value of the functional I is taken only on the etremal between two points and hence we must have the Euler- Lagrange equation is true. Consequently, in this case we must have from equation (5) that f f f and = = =.... (7) y y y We will prove this result a little latter in Theorem No.. Aliter : (Proof of the above Theorem ()): Let P (, y ) and (, ) Q y be two fied points in y plane. Let c be the curve between P and Q whose equation is given by y = y().let the etremum value of the functional along the curve c be given by I y = f y y d.... () ( ) (,, ) To find the condition to be satisfied by y(), let the curve c be slightly deformed from the original y position such that any point y on the curve c is displaced to y + δ y, where δ y is the variation in the path for an arbitrary choice of α, at any point ecept at the end points, as y prescribed there. Mathematically this means that O P(, y ) y + δy δy c : y = y(, ) Q(, y ), y = y y = y ( δ y) =.... () Classical Mechanics Page No.

8 Thus the value of the functional along the varied path is given by (, δ, δ ) I = f y + y y + y d.... (3) Hence the change in the value of the functional due to change in the path is given by (, δ, δ ) (,, ) I I = f y + y y + y f y y d, I I I f y y y y f y y d.... (4) Let = δ = (, + δ, + δ ) (,, ) We recall the Taylor s series epansion for the function of two variables f f f (, y δ y, y δ y ) f (, y, y ) y y... y δ y δ + + = Since δ y is very small, therefore by neglecting the higher order terms in δ y and δ y we have f f f (, y δ y, y δ y ) f (, y, y ) y y y δ y δ + + = +. Substituting this in the equation (4) we get δ f f I y y d y δ y δ = + We know hence we have dy d δ = δ y, d d δ f f d I y ( y) d y δ y d δ = +. Integrating the second integral by parts we get f f d f δ I δ y d δ y = + δ y d y y d y. Classical Mechanics Page No.

9 On using equation () we get f d f δ I = δ y d y d y.... (5) δ =. If I is etremum along the curve y = y() then change in I is zero. ( i. e., I ) This resembles very closely with a similar condition of etremum of a function in differential calculus. δ y d = y d y. Since δ y is arbitrary, we have =. y d y This is the Euler-Lagrange s differential equation to be satisfied by y() for the etremum of the functional between two points. Generalization of Theorem () : Euler-Lagrange s equations for several dependent variables. Theorem a : Derive the Euler-Lagrange s equations that are to be satisfied by twice differential functions y, y,..., y n that etremize the integral = (,,,...,,,,..., ) I f y y y y y y d n with respect to those functions y, y,..., y n which achieve prescribed values at the fied points,. Proof: The functional which is to be etremized can be written as ( ) I = f, yi, yi d, i =,,..., n. Choose the family of neighboring curves as (, α ) = (,) + αη y y i i i Classical Mechanics Page No. 3 n

10 and repeating the procedure delineated either in the Theorem () or in the alternate proof we arrive the following set of Euler-Lagrange s equations =, i =,,..., n. yi d y i Geodesic : Geodesic is defined as the curve of stationary (etremum) length between two points. Worked Eamples Eample : Show that the geodesic (shortest distance between two points) in a Euclidian plane is a straight line. Solution: Take P (, y ) and (, ) Q y be two fied points in a Euclidean plane. Let y = f be the curve between P and Q. Then the element of distance between two neighboring points on the curve y = f joining P and Q is given by ds = d + dy Hence the total distance between the point P and Q along the curve is given by I Q = P ds ( ) dy I = + y d, y =... () d Here the functional I is etremum if the integrand f = + y... () must satisfy the Euler-Lagrange s differential equation = y d y Now from equation () we find that... (3) Classical Mechanics Page No. 4

11 Integrating we get Squaring we get Integrating we get f f y = and = y y + y d y =. d + y y c y = +. y = c, where c = + c Classical Mechanics Page No. 5 c y = c + c.... (4) This is the required straight line. Thus the shortest distance between two points in a Euclidean plane is a straight line. Eample : Show that the shortest distance between two polar points in a plane is a straight line. Solution: Define a curve in a plane. If A (, y ) and B( d, y dy). + + are infinitesimal points on the curve, then an element of distance between A and B is given by ds = d + dy.... () Let θ = θ ( r) be the polar equation of the curve and P ( r, ) and Q( r, ) polar points on it. Recall the relations Hence equation () becomes = r cos θ, y = r sin θ. θ θ be two ds = dr + r dθ.... () Thus the total distance between the points P and Q becomes

12 r ( ) dθ I = + r θ dr, θ =.... (3) dr r The functional I is shortest if the integrand f = + r θ... (4) must satisfy the Euler-Lagrange s differential equation =, (5) θ dr θ d r θ =, dr + r θ θ θ r = h + r. Squaring and solving for θ we get On integrating we get dθ = ± dr r r θ h h h r = ± cos +. θ, where θ is a constant of integration. We write this as h = r cos( θ θ ).... (6) This is the polar form of the equation of straight line. Hence the shortest distance between two polar points is a straight line. Note : If r r ( θ ) given by = is the polar equation of the curve, then the length of the curve is θ dr I = r + dθ. dθ θ Since the integrand f = r + r does not containθ, we therefore have Classical Mechanics Page No. 6

13 f f r = h. r Solving this equation we readily obtain the same polar equation of straight line as the geodesic. Eample 3 : Show that the geodesic φ φ ( θ ) the great circle. = on the surface of a sphere is an arc of Solution : Consider a sphere of radius r described by the equations = r sinθ cos φ, y = r sinθ sin φ, z = r cos θ. If A (, y, z ) and B( d, y dy, z dz)... () be two neighboring points on the curve joining the points P and Q. Then the infinitesimal distance between A and B along the curve is given by where from equation () we find ds = d + dy + dz,... () d = r cosθ cosφdθ r sinθ sin φdφ, dy = r cosθ sinφdθ + r sinθ cos φdφ, dz = r sin θdθ. Squaring and adding these equations we readily obtain... (3) ds = r dθ + r sin θdφ.... (4) Hence the total distance between the points P and Q along the curve φ φ ( θ ) given by where θ = is ( sin ) dφ I = r + θ φ dθ, φ =... (5) dθ θ sin f = r + θ φ.... (6) Classical Mechanics Page No. 7

14 The curve is geodesic if the functional I is stationary. This is true if the function f must satisfy the Euler-Lagrange s equations. = φ dθ φ... (7) d r sin θ φ =. dθ + sin θ φ Integrating we get Solving for φ we get sin θ φ + sin θ φ = c, φ = On simplifying we get c cos ec θ ( c cos ec θ ). dφ k cos ec θ = dθ ( k cot θ ).... (8) Put Therefore we have dφ = Integrating we get φ α dt t = sin t, or φ α sin ( k cotθ ) =, cotθ = cos θ θ =, k t ec d dt. k cotθ = sin α φ, k cosθ = sinα sinθ cosφ cosα sinθ sin φ, kz = sinα y cosα.... (9) Classical Mechanics Page No. 8

15 This is the first-degree equation in, y, z, which represents a plane. This plane passes through the origin, hence cutting the sphere in a great circle. Hence the geodesic on the surface of a sphere is an arc of a great circle. Eample 4 : Show that the curve is a catenary for which the area of surface of revolution is minimum when revolved about y-ais. Solution: Consider a curve between two points (, y ) and (, y ) in the y plane y whose equation is y y =. We form a surface z A ds O area of the strip is equal to π ds. by revolving the curve about y-ais. Our claim is to find the nature of the curve for which the surface area is minimum. Consider a small strip at a point A formed by revolving the arc length ds about y ais. If the distance of the point A on the curve from y-ais is, then the surface But we know the element of arc ds is given by ds = + y d. Thus the surface area of the strip ds is equal to π + y d. Hence the total area of the surface of revolution of the curve y y is given by I = + y d This surface area will be minimum if the integrand = about y- ais π.... () f π y must satisfy Euler-Lagrange s equation = +... () Classical Mechanics Page No. 9

16 =,... (3) y d y d π y =, d + y Integrating we get d y =. d + y y a y = +. Solving for y we get dy a = d a Integrating we get y = a cosh + b. a Or a cosh y b = a.... (4) This shows that the curve is the catenary. The Brachistochrone Problem : The Brachistochrone is the curve joining two points not lie on the vertical line, such that the particle falling from rest under the influence of gravity from higher point to the lower point in minimum time. The curve is called the cycloid. Eample 5: Find the curve of quickest decent. Or A particle slides down a curve in the vertical plane under gravity. Find the curve such that it reaches the lowest point in shortest time.. Classical Mechanics Page No.

17 Solution: Let A and B be two points on the curve not lie on the vertical line. Let ds v = be the speed of the particle along the curve. Then the time required to fall an dt arc length ds is given by A v B ds dt = v dt = + y v d. Therefore the total time required for the particle to go from A to B is given by t AB B + y = d... () v A Since the particle falls freely under gravity, therefore its potential energy goes on decreasing and is given by and the kinetic energy is given by V T = mg, = mv. Now from the principle of conservation of energy we have T + V = constant. Initially at point A, we have = and v =. Hence the constant is zero. Hence equation () becomes y mv = mg, v = g.... () t AB + y = d.... (3) g Thus t AB is minimum if the integrand Classical Mechanics Page No.

18 f + y =,... (4) g must satisfy Euler-Lagrange s equation = y d y... (5) Integrating we get d y = d g( y ) + d y = d ( + y ) y c y = ( + ). Solving it for y we get Integrating we get dy = d a Put Hence y = d + b... (6) a = asin ( θ / ).... (7) d = a sin( θ / ) cos( θ / ) dθ. y = a sin ( θ / ) dθ + b a y = ( θ sinθ ) + b, If y =, θ = b =, hence Classical Mechanics Page No.

19 a y = ( θ sinθ ).... (8) Thus from equations (7) and (8) we have ( θ ) = b cos, a y = b( θ sin θ ), for b = This is a cycloid. Thus the curve is a cycloid for which the time of decent is minimum. Eample 6 : Find the etremal of the functional π ( + ) y y y d subject to the conditions that y =, y π =. Solution: Let the functional be denoted by π ( ) I = y y + y d.... () The functional is etremum if the integrand = +... () f y y y must satisfy the Euler-Lagrange s equation =,... (3) y d y d ( y) ( y ) =, d y + y =.... (4) Classical Mechanics Page No. 3

20 This is second order differential equation, whose complementary function (C.F.) is given by y = c cos + c sin... (5) where c and c are arbitrary constants. The particular integral (P.I.) is y =. ( ) y = + D Hence the general solution is given by y = c cos + c sin (6) This shows that the etremals of the functional are the two-parameter family of curves. On using the boundary conditions we obtain c y = =, π π y = c =. Hence the required etremal is π y = sin.... (7) Eample 7 : Find the etremal of the functional 3 d y subject to the conditions that y y =, = 3. Solution: Let the functional be denoted by I = 3 d... () y The functional is etremum if the integrand Classical Mechanics Page No. 4

21 must satisfy the Euler-Lagrange s equation Integrating we get 3 f =... () y =. (3) y d y 3 3 d = d y = cy 3 3 or y = a. Integrating we get y b Now using the boundary conditions we get. a = (4) a y = + b =, y = 3 a + b = 3. Solving these two equations we obtain a =, b =. Hence the required functional becomes y =.... (5) Eample 8 : Show that the time taken by a particle moving along a curve y = y ds with velocity =, from the point (,) to the point (,) is minimum if the curve dt is a circle having its center on the -ais. Solution: Let a particle be moving along a curve y = y() from the point (, ) to the point (, ) with velocity Classical Mechanics Page No. 5

22 ds = dt ds dt =. Therefore the total time required for the particle to move from the point (, ) to the point (, ) is given by ds t =... () where the infinitesimal distance between two points on the path is given by Hence the equation () becomes dy = + =. d ds y d, y + y t = d... () Time t is minimum if the integrand + y f =... (3) must satisfy the Euler-Lagrange s equation =.... (4) y d y d y =, d + y Solving for y we get y = c + y. Integrating we get y =, for a =. a c Classical Mechanics Page No. 6

23 y = a d + b Put = asinθ d = a cosθdθ... (5) Therefore, y = asinθdθ + b, y = a cosθ + b.... (6) Squaring and adding equations (5) and (6) we get + y b = a, which is the circle having center on y ais. Eample 9 : Show that the geodesic on a right circular cylinder is a heli. Solution: We know the right circular cylinder is characterized by the equations + y = a, z = z... () The parametric equations of the right circular cylinder are = a cos θ, y = asin θ, z = z. where a is a constant. The element of the distance (metric) ds = d + dy + dz on the surface of the cylinder becomes ds = a dθ + dz, Hence the total length of the curve on the surface of the cylinder is given by θ s a z dθ, for z dz = + =.... () dθ For s to be etremum, the integrand θ f = a + z.... (3) Classical Mechanics Page No. 7

24 must satisfy the Euler-Lagrange s equation =.... (4) z dθ z d z =. dθ f Integrating the equation and solving for z we get z = a (constant). Integrating we get z = aθ + b, a,... (5) where a, b are constants. Equation (5) gives the required equation of heli. Thus the geodesic on the surface of a cylinder is a heli. Eample : Find the differential equation of the geodesic on the surface of an inverted cone with semi-vertical angle θ. Solution: The surface of the cone is characterized by the equation y z θ θ const + = tan, =.... () The parametric equations of the cone are given by = ar cos φ, y = ar sin φ, z = br. where for a = sin θ, b = cosθ are constant. Thus the metric ds = d + dy + dz on the surface of the cone becomes... () ds = dr + a r dφ... (3) Hence the total length of the curve φ φ ( r) dφ s = + a r dr = dr = on the surface of the cone is given by φ, φ... (4) Classical Mechanics Page No. 8

25 The length s is stationary if the integrand f a r φ = +... (5) must satisfy the Euler-Lagrange s equation = φ dr φ.... (6) d a r φ =, dr f Solving for φ we get dφ c = dr ar a r c,... (7) where c = constant. This is the required differential equation of geodesic, and the geodesic on the surface of the cone is obtained by integrating equation (7). This gives φ φ c dr α. = + ar a r c sec ar = + a c α, c r = sec a( φ α ). a Eample : Find the curve for which the functional π 4 = ( ) I y y y d can have etrema, given that y()=, while the right hand end point can vary along the line π =. 4 Classical Mechanics Page No. 9

26 Solution: To find the etremal curve of the functional π 4 = ( ) I y y y d,... () we must solve Euler s equation where =,... () y d y f = y y... (3) y + y =.... (4) This is the second order differential equation, whose solution is given by y = a cos + bsin.... (5) The boundary condition y = gives a =. y = bsin.... (6) The second boundary point moves along the line f = y π = 4 π = 4 ( y ) =, where from equation (6) we have y = bcos. Thus π =. 4 b=. This implies the etremal is attained on the line y =. Eample : If f satisfies Euler-Lagrange s equation π y at = gives 4 =. y d y Then show that f is the total derivative dg d conversely. of some function of and y and Classical Mechanics Page No.

27 Solution: Given that f satisfies Euler-Lagrange s equation =.... () y d y dg We claim that f =, d where g g (, y) =. As f f (, y, y ) we write equation () eplicitly as y y y y y =, f f f f y y =.... () We see from equation () that the first three terms on the l. h. s. of () contain at the highest the first derivative of y. Therefore equation () is satisfied identically if the coefficient of y vanishes identically. f =. y Integrating w. r. t. y we get f = q(, y). y Integrating once again we get where p(, y ) and (, ) (, ) (, ) f = q y y + p y,... (3) q y are constants of integration and may be function of and y only. Then the function f so determined must satisfy the Euler Lagrange s equation (). From equation (3) we find f q p = y + y y y, f =. y and q(, y) Classical Mechanics Page No.

28 Therefore equation () becomes q p d y + ( q(, y) ) =. y y d q p q q y + y =. y y y p q = y..(4) This is the condition that the equation pd + qdy is an eact differential equation dg. Therefore, dg = pd + qdy, dg = p + qy = f by d This proves the necessary part. Conversely, assume that equation Since Therefore, we find (3) dg f =.... (5) d =. y d y dg f =. We prove that f satisfies the Euler-Lagrange s d dg g g f = = + y, d y y y y f g g = +, y g g =. y y Classical Mechanics Page No.

29 Consider now g g d g = + y y d y y y d y. g g g g = + y y. y d y y y y y = y d y dg f = d satisfies Euler-Lagrange s equation. Eample 3 : Show that the Euler-Lagrange s equation of the functional = ( ) (,, ) I y f y y d f has the first integral f y = const, if the integrand does not depend on. y Solution: The Euler-Lagrange s equation of the functional = ( ) (,, ) I y f y y d to be etremum is given by =.... () y d y = y y y y y f f f f y y f If f does not involve eplicitly, then =. Therefore, we have Classical Mechanics Page No. 3

30 y y y y f f f y y = Multiply equation () by y we get.... () But we know that f f f y y y y = y y y y d y y y y y y y d f f f f f f f y = y + y y y y y,... (3) d y y y y y d f f f f f y = y y y y.... (4) From equations (3) and (4) we see that d f f y =, d y f f y = const.... (5) y This is the first integral of Euler-Lagrange s equation, when the functional (, ) f = f y y. Worked Eamples Eample 4 : Find the minimum of the functional I y = y + yy + y + y d ( ) if the values at the end points are not given. Solution: For the minimum of the functional ( ) the integrand Classical Mechanics Page No. 4 I y = y + yy + y + y d... ()

31 = () f y yy y y must satisfy the Euler-Lagrange s equation Integrating we get Further integrating we get =.... (3) y d y d y + ( y + y + ) =, d y =.... (4) y = + c,... (5) y = + c + c,... (6) where c, c are constants of integration and are to be determined. However, note that the values of y at the end points are not prescribed. In this case the constants are determined from the conditions. f f =, and =.... (7) y y = = These two conditions will determine the values of the constants. ( y y ) and ( y y ) + + =, + + =,... (8) = = where from equation (5) and (6) we have y = c and y = c, y = + c and y = + c + c. similarly, Thus the equations (8) become 5 c + c + =, c + c + =. Classical Mechanics Page No. 5

32 3 Solving these equations for c and c we obtain c = and c = Hence the required curve for which the functional given in () becomes minimum is y = ( 3 + ).... (9) Theorem 3 : Find the Euler- Lagrange differential equation satisfied by four times differentiable function y() which etremizes the functional = ( ) (,,, ) I y f y y y d under the conditions that both y and y are prescribed at the end points. Proof : Let P (, y ) and (, ) Q y be two fied points in y plane. The points P and Q can be joined by infinitely many curves. Accordingly the value of the integral I will be different for different paths. We shall look for a curve along which the functional I has an etremum value. Let c be a curve between P and Q whose equation is given by y = y(, ). Let also the value of the functional along the curve c be etremum and is given by I y = f y y y d... () ( ) (,,, ) We can label all possible paths starting from P and ending at Q by the family of equations (, α ) (, ) αη y = y +,... () where α is a parameter and η is any differentiable function of. For different values of α we get different curves. Accordingly the value of the integral I will be different for different paths. Since y and y are prescribed at the end points, this implies that there is no variation in y and y at the end points. Classical Mechanics Page No. 6

33 i.e., all the curves of the family and their derivative must be identical at fied points P and Q. This implies that and also ( α ) ( α ) y, = y, = y, y, = y, = y, ( α ) (, α ) (,) y, = y, = y, y = y = y ( ) η ( ) η = =,... (3a) and η ( ) η ( ) = =.... (3b) Conversely, the conditions (3) ensure us that the curves of the family that all pass through the points P and Q. Let the value of the functional along the neighboring curve be given by I y = f y y y d.... (4) ( (, α )), (, α ), (, α ), (, α ) From differential calculus, we know the integral I is etremum if I α α = =, because for α = the neighboring curve coincides with the curve which gives etremum values of I. Thus I α α = =, f f f + + d = y y y. η η η Integrating the second and the third integrations by parts, we get Classical Mechanics Page No. 7

34 f f d f f η d + η η d + η y y d y y d f d f η + η d =, d y d y f d f η d + η η d + y y d y d y f d f + η + η d =. y d y... (5)... (6) As y and y are both prescribed at the end points, hence on using equations (3) we obtain d f + η d = y d y d y.... (7) By using the basic lemma of calculus of variation we get y d y d y f d f d f + =.... (8) This is required Euler- Lagrange differential equation to be satisfied by y() for which the functional I has etremum value. Note : If the functions y and y are not prescribed at the end points then we must have unlike the fied end point problem, η and η need no longer vanish at the points and. In order that the curve y =y() to be a solution of the variable end point problem, y must be an etremal. i.e., y must be a solution of Euler s equation (8). Thus for the etremal we have from equation (6) = b = b f,. y = = y d y a = = a... (9) Classical Mechanics Page No. 8

35 Thus to solve the variable end point problem, we must first solve Euler s equation (8) and then use the conditions (9) to determine the values of the arbitrary constants. The above result can be summarized in the following theorem (3a). Theorem (3a) : Derive the differential equation satisfied by four times differentiable function y ( ), which etremizes the integral = (,,, ) I f y y y d under the condition that both y, y are prescribed at both the ends. Show that if neither y nor y is prescribed at either end points then f f = = y y = = = = y d y = Remark: (General case of Theorem (3)) If the integrand in equation () of the n Theorem (3) is of the form f f (, y, y, y,..., y ) = with the boundary conditions n n y = y, y = y,..., y = y, n y = y, y = y,..., y = y, n then the Euler-Lagrange s differential equation is n d f n d f ( ) n n =. y d y d y d y Worked Eamples Eample 5 : Find the curve, which etremizes the functional π 4 ( ) = ( + ) I y y y d Classical Mechanics Page No. 9

36 under the conditions that y y =, =, π π y = y = 4 4 Solution : For the etremization of the functional the integrand π 4 ( ) = ( + ) I y y y d... () f y y = +... () must satisfy the Euler-Lagrange s equation i.e., y d y d y f d f d f + = d y + ( y ) =, d 4 d y y (3) d =.... (4) The solution of equation (4) is given by y = ae + be + c cos + d sin... (5) where a, b, c, d are constants of integration and are to be determined. Thus y = a + b + c =, π y = ae + be + c + d = 4 y = a b + d =, π π 4 4, π = + = 4 π π 4 4 y ae be c d Classical Mechanics Page No. 3

37 Solving these equations we get a = b = c = and d =. Hence the required curve is y = sin. Eample 6 : Minimize the functional satisfies I = dt, Solution : To minimize the functional the integral =, =, =, =. I = dt,... () f =... () must satisfy the Euler-Lagrange s equation This implies dt dt f d f d f + = 4 d 4 dt Integrating we get.... (3) =.... (4) 3 t t = c + c + c3t + c4.... (5) 6 where given in (5) must satisfy the conditions = c =, = c =, 4 = 4c + 6c = 3, 3 = c + c =. Classical Mechanics Page No. 3

38 Solving for c and c we get the required functional is t = + t +. 4 Eample 7 : Find the function on which the functional can be etremized such that and y is not prescribed. ( ) = ( ) I y y y d, y = y =, y = Solution : For the etremization of the functional the integrand ( ) = ( ) I y y y d... () =... () f y y must satisfy the Euler-Lagrange s equation 4 d y y d y d y f d f d f + = d + ( y ) =, d =. 4 d Integrating we obtain y = + a, 3 y = + a + b, (3) Classical Mechanics Page No. 3

39 4 a y b c = + + +, 4 5 a 3 b y = c + d,... (4) 6 where a, b, c, d are constants to be determined. Given that y = d =, y = c =, y = a + 3b =. Since y is not prescribed. i.e., y at = is not given, then we have the condition that f y = = y =. This gives from above equation that 6a + 6b =. Solving the equations for a and b we get a =, and b =. 4 Substituting these values in equation (4) we get This is the required curve. 5 3 y = + ( ).... (5) 4 When integrand is a function more than two dependent variables : Eample 8: Prove that the shortest distance between two points in a Euclidean 3- space is a straight line. Solution: Define the curve y y, z z space. Let P (, y, z ) and Q ( d, y dy, z dz) = = in the 3-dimentional Euclidean be two neighboring points on the Classical Mechanics Page No. 33

40 curve joining the points A(, y, z ) and (,, ) distance between P and Q along the curve is given by ds = d + dy + dz. B y z.thus the infinitesimal Hence the total distance between the points A and B along the curve is given by ( ) dy I = + y + z d, y =... () d Let f = + y + z.... () We know the functional I is shortest if the function f must satisfy the Euler- Lagrange s equations. =,... (3) y d y and =.... (4) z d z and d y = o y = af, a = constant d f y a a z = a.... (5) ( ) Similarly, from equation (4) we obtain ( ) z b b y = b.... (6) Solving equations (5) and (6) for ẏ and ż we get a y = ±, and a b z = ± a a b, i.e., y = ± c, and z = ± c for c = a a b... (7) for c = b a b.... (8) Classical Mechanics Page No. 34

41 Integrating equations (7) and (8) we get where φ ( z) and ( y) y = ± c + φ z... (9), z = ± c + ψ y... (), ψ are constants of integration and may be functions of z and y respectively. Thus the required curve is given by equations (9) and (). But these equations represent a pair of planes. The common point of intersection of these planes is the straight line. Hence the shortest distance between two points in Euclidean 3-space is a straight line. Eample 9 : Show that the geodesic defined in the 3-dimentional Euclidean space by the equations ( t), y y ( t), z z ( t) Solution : Let = = = is a straight line.,, = t y = y t z = z t... () be a curve in 3-dimentional Euclidean space, where t is a parameter of the curve. The infinitesimal distance between two neighboring points on the curve () is given by ds = d + dy + dz, where from equation () we have Thus d = dt, dy = ydt, dz = zdt. ds = + y + z dt. Hence the total length of the curve between the points P ( t ) and t t P t is given by I = + y + z dt... () The curve is geodesic if the length of the curve I is etremum. This is true if the integrand Classical Mechanics Page No. 35

42 ... (3) f = + y + z must satisfy the Euler-Lagrange s equations. = i =,,3 i dt i where f f = i and = i =,,3. f i = af, y = bf, z = cf. Thus the Euler-Lagrange s equations become a + a y + a z =, b + b y + b z =, c + c y + c z =. These equations are consistent provided a a a b b b = c c c a + b + c =. i with (, y, z) Classical Mechanics Page No. 36 i =,... (4)... (5) Solving equations (5) we obtain = a a b z,... (6) y = b a b z, z.... (7) Integrating equations (6) and (7) we obtain = c z + φ y, c = y = c z + ψ, c = a a a b b b,... (8).... (9)

43 where ψ and ( y) φ are constants of integration and may be functions of and y respectively. Equations (8) and (9) represent planes. The locus of the common points of these planes is the straight line. Hence the geodesic in 3-dimentional Euclidean space is the straight line. Unit : Isoperimetric Problems: The problems in which the function which is eligible for the etremization of a given definite integral is required to confirm with certain restrictions that are given as the boundary conditions. Such problems are called isoperimetric problems. The method is eactly analogous to the method of finding stationary value of a function under certain conditions by Lagrange s multipliers method. Theorem 4 : Obtain the differential equation, which is satisfied by the functional f (,, ) y y which etremizes the integral = ( ) (,, ) I y f y y d subject to the conditions, y = y y = y, and the integral J = g y y d = constant. (,, ) Proof : Consider the functional between two points P (, y ) and (, ) by subject to the conditions ( ) (,, ) Classical Mechanics Page No. 37 Q y given I y = f y y d... (), y = y y = y,... ()

44 and J (,, ) O y where, such that = g y y d = constant.... (3) Y = y () + ε η () + P(, y ) The points P and Q can be joined by infinitely many curves. Accordingly the value of the integral I will be different for different paths. Let all possible paths starting from P and ending at Q be given by two parameters family of curves ε η ε η Y = y (4) ε ε are parameters and, η η ( ) = = η ( ), ( ) = = η ( ) η η are arbitrary differentiable functions of... (5) These conditions ensure us that the curves of the family that all pass through the points P and Q. ε η () y = y () Note that, we can not however, epress Y() as merely a one parameter family of curves, because any change in the value of the single parameter would in general alter the value of J, whose constancy must be maintained as prescribed. For this reason we introduce two parameter families of curves. We shall look for a curve along which the functional I has an etremum value under the condition (3). Let c be such a curve between P and Q whose equation is given by y = y () such that the functional () along the curve c has etremum value. The values of the integrals () and (3) along the neighboring curve (4) are obtained by replacing y by Y in both the equations () and (3). Thus we have I = f Y Y d... (6) ( ε, ε ) (,, ) Q(, y ) Classical Mechanics Page No. 38

45 and ( ε ε ) ( ) J, = g, Y, Y d = const.... (7) Equation (7) shows that ε and ε is not independent, but they are related by J ε, ε = const.... (8) Thus the changes in the value of the parameters are such that the constancy of (7) is maintained. Thus our new problem is to etremizes (6) under the restriction (7). To solve the problem we use the method of Lagrange s multipliers. Multiply equation (7) by λ and adding it to equation (6) we get * * ( ε, ε ) λ (,, ) I = I + J = f Y Y d,... (9) where λ is Lagrange s undetermined multiplier and = + λ... () * f f g Thus etremization of () subject to the condition (3) is equivalent to the etremization of (9). Thus from differential calculus, the integral * I ε j ε =, ε = =. Thus from equation (9) we have * I is etremum if * I ε j ε =, ε = = * * f f j + j d = y y. J=,... () η η Note here that by settingε = ε =, we replace Y and Y to y, y. Integrating the second integration by parts, we get * * * f f d f η d + η η d =.... () j j j y y d y As any curve is prescribed at the end points, hence on using conditions (5) we obtain Classical Mechanics Page No. 39

46 * * η j d =, j =, y d y. By using the basic lemma of calculus of variation we get * * =.... (3) y d y This is required Euler- Lagrange s differential equation to be satisfied by y () for which the functional I has etremum value under the condition (3). Remark : If y is not prescribed at either end point then from equation () we have * f y = at that end point. Generalization of Theorem 4 : Euler-Lagrange s equations for several dependent variables : Theorem 4a : Obtain the differential equations which must be satisfied by the function which etremize the integral = (,,,...,,,,..., ) I f y y y y y y d n with respect to the twice differentiable functions y, y,..., y n for which ( ) J = g, y, y,..., y, y, y,..., y d = const. n and with y, y i prescribed at points,. Proof : The functional which is to be etremized can be written as together with the conditions ( ) I = f, y, y d, i =,,..., n i i n n Classical Mechanics Page No. 4

47 ( ) J = g, y, y d = const. i =,,..., n i i and y, y i prescribed at points,. Repeating the procedure described in the Theorem (4) we arrive the following set of Euler-Lagrange s equations * * =, i =,,..., n yi d y i where = + λ. * f f g Theorem 5 : Obtain the differential equation, which is satisfied by four times differential function y () which etremizes the functional = ( ) (,,, ) I y f y y y d subject to the conditions that the integral J = g y y y d = constant. (,,, ) and both y and y are prescribed at the end points. Proof: Proof of the Theorem 5 runs eactly in the same manner as that of the proof of Theorem 3 and Theorem 4. The required Euler-Lagrange s differential equation in this case is given by d f y d y d y * * * + =, where = + λ. * f f g Classical Mechanics Page No. 4

48 Remarks :. If y is not prescribed at either end point, then we have the condition * * = y d y at that end point.. If y is not prescribed at either end point then we have 3. In general if (,,,,..., n ) (,,,,..., n ) f = f y y y y g = g y y y y * f = y at that point. with the boundary conditions that n y are prescribed at both the y, y, y,..., ends, then in this case the Euler-Lagrange s equation is * * * n * d f n d f ( ) n n =. y d y d y d y Worked Eamples Eample : Find the plane curve of fied perimeter that encloses maimum area. (The problem is supposed to have arisen from the gift of a king who was happy with a person and promised to give him all the land that he could enclose by running round in a day. The perimeter of his path was fied.) Solution: Let c : y = y () be a plane curve of fied perimeter l. l = ds,... () where the infinitesimal distance between two points on the curve is given by dy = + =. d ds y d, y Classical Mechanics Page No. 4

49 Hence the total length of the curve between two points P and Q becomes + y d = l.... () The area bounded by the curve c and the -ais is given by ( ) A y = yd.... (3) Thus we maimize (3) subject to the condition (). Hence the required Euler- Lagrange s equation to be satisfied is where * * =,... (4) y d y λ, * f = f + g * f = y + λ + y.... (5) Solving equation (4) we get d λ y =, d + y λ y = a. + y Solving for y we get or ( a) y = = λ λ y ( + y ) a ( a),.... (6) Integrating we get Classical Mechanics Page No. 43

50 ( a) y = d + b... (7) λ ( a) Put a = λ sin t,... (8) d = λ costdt y = λ sin tdt + b, y = λ cost + b... (9) Squaring and adding equations (8) and (9) we get a + y b = λ.... () This is a circle centered at (a, b) and of radius λ and is to be determined. To l determine λ, we know that the circumference of the circle is πλ = l λ =. π Eample : Find the shape of the plane curve of fied length l whose end points lie on the -ais and area enclosed by it and the -ais is maimum. Solution: Let c: y = y() be a plane curve of fied length l whose end points lie on y y = y() the -ais and the curve lies in the upper half plane. The area bounded by the curve c and the -ais is given by ( ) A y = yd... () O (, ) (, ) such that the length of the curve is fied and is given by J = ds = + y d = l... () The area given in equation () is maimum under the condition (), if Classical Mechanics Page No. 44

51 where * * =,... (3) y d y λ, * f = f + g f = y + λ + y *... (4) where λ is Lagrange s multiplier. Solving equation (3) we get d λ y =, d y + λ y + y = a Solving for y we get or ( a) y = = λ λ y ( + y ) a ( a),.... (5) Integrating we get ( a) y = d.... (6) λ ( a) Put a = λ sin t,... (7) d = λ costdt y = λ sin tdt + b, y = λ cost + b.... (8) Squaring and adding equations (7) and (8) we get a + y b = λ.... (9) Classical Mechanics Page No. 45

52 Thus the curve is a semi circle centered at (a, b) and of radius λ, and is to be determined. Since from the condition () the perimeter of the semi-circle is πλ = l This is the radius of the curve of fied length l which encloses maimum area. l λ =. π Eample : Find the etremals for an isoperimetric problem ( ) = ( + ) I y y d subject to the conditions that y d y y =, =, =. Solution : The functional, which is to be etremized, is given by such that ( ) = ( + ) I y y d... () and y d =... () y y =, =... (3) Thus for the etremizes of () under (), we know the condition to be satisfied is where * * =,... (4) y d y λ, * f = f + g f = y + + λ y *,... (5) where λ is Lagrange s multiplier. Classical Mechanics Page No. 46

53 Solving equation (4) we get This equation has roots Case : Let λ >. The solution of equation (6) is Conditions (3) give y λ y =.... (6) y ae be λ ± λ for λ > or ± i λ for λ <. λ = +,... (7) y = a + b =, λ λ y = ae + be =. Solving for a and b we have for b = λ, e λ + inπ e = λ + inπ =, λ = inπ. This is contradictory to λ is positive, hence b = and consequently a = proving that the equation has only trivial solution. Case : Let λ <. The solution of equation (6) is Boundary conditions (3) give y = a cos λ + bsin λ.... (8) y = a = y = bsin λ =, λ = nπ, for λ. Hence the required solution becomes we get y = bsin nπ.... (9) Classical Mechanics Page No. 47

54 Condition () gives b = ±. Therefore the required solution is y = ± sin nπ. Eample 3 : Find the etremals for an isoperimetric problem subject to the conditions that Solution : It is given that π ( ) = ( ) I y y y d π ( π ) yd =, y =, y =. π ( ) = ( ) I y y y d... () where the functional I is to be etremized under the conditions π yd =,... () and y y ( π ) =, =... (3) The corresponding Euler-Lagrange s equation is where * * =,... (4) y d y λ * f = f + g, λ * f = y y + y... (5) Hence the equation (4) becomes λ y + y =.... (6) The C.F. of equation (6) is given by y = a cos + bsin, Classical Mechanics Page No. 48

55 where as the P.I. is given by λ y =. Hence the general solution becomes λ y = a cos + bsin +... (7) To determine the arbitrary constants of integration we use the boundary conditions (3) λ y = a =, y ( π ) = λ =, a = To determine other constant of integration we use π yd =, π π π cos d b sin d d. + + = This gives the value of b as π b = 4. Hence the required curve is y = ( cos ) + ( π ) sin.... (8) 4 Eample 4 : Prove that the etremal of the isoperimetric problem subject to the condition 4 I y d y y =, = 3, 4 = 4, Classical Mechanics Page No. 49

56 4 yd = 36 is a parabola. Solution: Here f * = y + λ y. The corresponding Euler-Lagrange s differential equation is y λ =.... () Integrating two times we get λ = + +,... () 4 y a b where the constants of integration are to be determined. Now the boundary conditions Also the condition y = 3 λ + 4a + 4b =, y 4 = 4 4λ + 4a + b = (3) gives 4 yd = 36, 4 λ + a + b d = 36 4, λ + 3a + b = (4) Solving equations (3) and (4) we obtain a =, b =, λ = 4. Thus the required curve is obtain by putting these values in equation () and is y = +. We write this as ( ) + = y + Classical Mechanics Page No. 5

57 Or equivalently for X=(+), Y=(y+), we have Hence the curve is a parabola. X = Y. Eample 5 : Find the etremals for the isoperimetric problem t I = ( y y) dt... () t with the conditions that t J = + y dt = l,... () t t = t = y t = y t = y.,... (3) Solution : We wand to find the function for which equation () is etremum w. r. t. the functions (t), y(t) satisfying the conditions () and (3). We know the conditions that the integral () is etremum under () if the following Euler Lagrange s equations are satisfied. * * =,... (4) dt where * * =,... (5) y dt y λ * f = f + g, * f = ( y y ) + λ + y.... (6) Hence the equations (4) and (5) reduce to y d y λ + = dt y + Classical Mechanics Page No. 5

58 d λ y + =. dt + y Integrating these equations w. r. t. t we get y y λ + = a y + λ y a = + y and λ y + + = b, y + λ y b =. + y Squaring and adding above equations we get b + y a = λ.... (7) This is a circle of radius λ and centered at (b, a). Thus the closed curve for which the enclosed area is maimum is a circle. The length of the circle is πλ = l Eercise: λ = l π. This gives the radius of the curve.. Show that the shortest distance between two points along the curve, = t y = y t in a Euclidean plane is a straight line.. Show that the geodesic defined by r r ( t), θ θ ( t) line. = = in a plane is a straight 3. Show that the stationary (etremum) distance between two points along the curve θ θ ( t), φ φ ( t) = = on the sphere = r sinθ cos φ, y = r sinθ sin φ, z = r cosθ is an arc of the great circle. Classical Mechanics Page No. 5

59 4. Find the geodesic on the surface obtained by generating the parabola y = 4a about -ais. Ans.: The surface of revolution is representation is y z 4a = au, y = au sin v, z = au cos v The geodesic on this surface is obtain by solving the integral + u v = c du + c. u u c + =, whose parametric 5. Derive the Euler-Lagrange s equations that are to be satisfied by twice differential functions ( t), y ( t),..., z ( t ), that etremize the integral t d I = f (, y,..., z,, y,..., z, t) dt, = dt t which achieve prescribed values at the fied points t, t. Ans : =, =,..., =, dt y dt y z dt z 6. Find the curve which generates a surface of revolution of minimum area when it is revolved about ais. Ans : Area of revolution of a curve about -ais is I = y + y d. The curve is a catenary given by y c sec, or y a cosh ψ b = = a. 7. Find the function on which the functional can be etremized and y is not prescribed at both the ends. I y = ( y y) d, y =, y ( ) =, Classical Mechanics Page No. 53

60 Ans: 5 3 y = Find the stationary function of ( ) boundary conditions y y =, 4 = 3. 4 y y d, which is determined by the Ans : y = =, =, = 4. y 9. Find the etremum of I y d y y Ans: y =.. Find the etremal of y I y = d 3. Ans: 4 y = c + c. 4 π. Find the etremal of ( y y y ) d y y ( π ) Ans: y ( c ) = + sin.. Find the etremal of the functional Ans: + 4 cos, =, =. ( y y) d y y 3 y = Determine the curve z z, =, =. = for which the functional ( ) I = z z d, z =, z = is etremal. 3 Ans: z = Classical Mechanics Page No. 54

61 4. Determine the curve z z π = for which the functional π I = ( z z ) d, z =, z = is etremal. Ans: z = sin. 5. Find the etremal of the functional π π π I y = ( y y + ) d, y =, y =, y =, y =. Ans : y = cos.. 6. Obtain the differential equation in which the etremizing function makes the integral = ( ) (,, ) I y f y y d etremum subject to the conditions ( ) y = y, y = y, and J = g, y, y d = const. k =,,..., n k Ans : The problem would be etrmization of k * * =, (,, ) I f y y d where n * I I λk J k k = = + and n * f f λk gk k = = +. The required differential equation is * * =. y d y Classical Mechanics Page No. 55

62 7. Find the etremals of the isoperimetric problem Ans: y = λ + a + b. 4 I = y d, s. t. yd = c. 8. Find the etremal of the functional Ans: y = ( + ), =, =, =, I y y d y y y y. =. 9. Find the etremal of the functional = ( + ) I y, z yz y y z d. y = c + c cos + c + c sin, z = y + y. Ans: 3 4. Find the etremal of the functional = ( + + ) I y y y ye d. e Ans: y = + ce + ce. Classical Mechanics Page No. 56