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An observation on two methods of obtaining solutions to variational problems Dean Carlson The University of Toledo Dept. of Mathematics,Toledo, OH 43606–3390 October 9, 2001 The Third MidWest Optimization Seminar Western Michigan University Kalamazoo, MI

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1.

The basic problem

We consider a free problem in the calculus of variations described as minimizing an integral functional of the form Z x2 I(y) = (1) F (x, y(x), y 0 (x)) dx x1

over the class of all piecewise smooth functions (i.e., continuous functions with piecewise continuous derivatives) on the fixed interval [x1 , x2 ] satisfying the fixed endpoint conditions y(x1 ) = y1 and y(x2 ) = y2 . (2) We assume here that F : [x1 , x2 ] × R × R → R is a continuous function with continuous partial derivatives with respect to the third argument. We further assume that z → F (x, y, z) is a convex function for each fixed (x, y) ∈ [x1 , x2 ] × R.


2.

Carathéodory’s method of equivalent variational problems

In the 1935 book of Carathéodory [1] the following observation is made. If we let S : [x1 , x2 ] × R → R denote any fixed smooth function and define the new integrand ∂S ∂S (x, y) − (x, y)z F˜ (x, y, z) = F (x, y, z) − ∂x ∂y then the value of the integral functional Z x2 ˜ = I(y) F˜ (x, y(x), y 0 (x)) dx

(3)

(4)

x1

evaluated along a feasible trajectory (i.e. a piecewise smooth function satisfying the end conditions (2)) differs from the value of the original functional (1) by a fixed constant that depends only on S and the fixed end conditions (2). That is, Z x2 Z x2 0 ˜ I(y) − I(y) = F (x, y(x), y (x)) dx − F˜ (x, y(x), y 0 (x)) dx x1 Zx1x2 ∂S ∂S = (x, y(x)) + (x, y(x))y 0 (x) dx (5) ∂x ∂y Zx1x2 d = S(x, y(x)) dx x1 dx = S(x2 , y2 ) − S(x1 , y1 ).


An immediate consequence of this fact is that the minimizers of the original problem coincide with the minimizers of the new problem consisting of minimizing (4) over the set of piecewise smooth functions satisfying the fixed end conditions (2) Carathéodory referred to these two problems as equivalent. In particular he sought a function S so that F˜ (x, y, z) ≥ 0

(6)

for all (x, y, z) ∈ [x1 , x2 ] × R × R and for which one could obtain a continuous function p : [x1 , x2 ] × R → R for which F˜ (x, y, p(x, y)) = 0.

(7)

Lemma 2.1 Suppose that S : [x1 , x2 ] × R → R is a continuously differentiable function such that (6) holds and further assume that there exists a continuous function p : [x1 , x2 ] → R such that (7) holds. Then if y ∗ : [x1 , x2 ] → R is a solution of the two-point boundary value problem y 0 (x) = p(x, y(x)) y(x1 ) = y1 and y(x2 ) = y2 it is a minimizer of the problem of minimizing (1) over all piecewise smooth functions y : [x1 , x2 ] → R satisfying the fixed end conditions (2).


Carathéodory also gives conditions which permit him to obtain an appropriate function S and p. To see this we observe that since p(x, y) is a minimizer of z → F˜ (x, y, z) it follows that p and S are related by the necessary condition ∂ ∂S F (x, y, p(x, y)) − (x, y) = 0 ∂z ∂y

(8)

and that S is a solution of the Hamilton-Jacobi equation ∂S ∂S (x, y) + H(x, y, ) = 0, ∂x ∂y

(9)

in which H : [x1 , x2 ] × R × R → R is defined implicitly by the system of equations H(x, y, q) = qz − F (x, y, z) ∂F (x, y, z) = q. ∂x As a consequence of our convexity hypothesis this is equivalently expressed as

(10) (11)

H(x, y, q) = sup {qz − F (x, y, z)} . z∈R


Theorem 2.1 Suppose that S : [x1 , x2 ] × R → R is a solution of the Hamilton-Jacobi equation (9) and that p : [x1 , x2 ] → R is related to S through the equation (8). Then if y ∗ : [x1 , x2 ] → R is a solution of the two-point boundary value problem y 0 (x) = p(x, y(x)) y(x1 ) = y1 and y(x2 ) = y2 it is a minimizer of the problem of minimizing (1) over all piecewise continuous functions y : [x1 , x2 ] → R satisfying the end conditions (2).


Consider the class of problems in which the integrand is given by F (x, y, z) = a(x)z 2 + b(x, y)z + c(x, y), in which a : R → R is continuous satisfying a(x) > 0 and b : R2 → R and c : R2 → R are both continuous. In this case the Hamiltonian is given by (11) which becomes H(x, y, q) = qz − a(x)z 2 + b(x, y)z + c(x, y) 2a(x)z + b(x, y) = q. Eliminating z in this system gives H(x, y, q) =

(q − b(x, y))2 4a(x)

− c(x, y).

Consequently in this case we obtain the following Hamilton-Jacobi equation 2 ∂S 1 ∂S + − b(x, y) − c(x, y) = 0, ∂x 4a(x) ∂y and the function p(x, y) is given explicitly by 1 ∂S p(x, y) = (x, y) − b(x, y) . 2a(x) ∂y

(12)

Substituting this fact into the Hamilton-Jacobi equation we also obtain the relationship a(x)p(x, y)2 = c(x, y) −

∂S . ∂x


To illustrate the above theorem we observe, for a solution S of the Hamilton-Jacobi equation and p as given above, F (x, y, z) −

∂S ∂S − z = a(x) (z − p(x, y))2 ≥ 0, ∂x ∂y

from which we see that the solutions to the differential equation y 0 (x) = p(x, y(x)) provide us with minimizers to the variational problem. Example 1We suppose now that F (x, y, z) = z 2 − 2yz + 2y 2 . Following the above we see that the Hamilton-Jacobi equation is 2 ∂S 1 ∂S + + 2y − 2y 2 = 0. ∂x 4 ∂y Assuming that a solution has the form S(x, y) = A(x) + B(y) in which A(·) and B(·) are differentiable functions we see that 1 2 A0 (x) = 2y 2 − (B 0 (y) + 2y) 4 so that the variables separate. Consequently we must have that A0 (x) = λ

and 2y 2 −

1 0 2 (B (y) + 2y) = λ 4


for some constant λ. Thus we obtain A(x) = λx + µ1 and p B 0 (y) = −2y ± 2 2y 2 − λ. Integrating this last expression we obtain B(y) = (−1 ±

√

2)y 2 + µ2

when λ = 0 and

√ 2−λ p √ 2y 2 B(y) = −y 2 ± ∓ λ ln | 2y + 2y 2 − λ| + µ2 λ 2 when λ 6= 0. Therefore solutions to the Hamilton-Jacobi equation can be obtained as y

p

S(x, y) = A(x) + B(y) for different values of the separation constant λ. To obtain solutions to the variational equation however it is not necessary to find S(x, y). All that is required is ∂S ∂y , since according equation (12) in the above development we must have p p 1 2 p(x, y) = −2y ± 2 2y − λ + 2y = ± 2y 2 − λ. 2 Therefore we can obtain the minimizers by solving the ordinary differential equation p y 0 (x) = ± 2y(x)2 − λ. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

It is an easy matter to see that the solutions to this equation must have the form (for any value of λ). √ √ y(x) = c1 e 2x + c2 e− 2x for arbitrary constants c1 and c2 . Consequently when the two fixed end conditions (2) are specified we obtain the unique solution. We conclude this section by observing that the Euler-Lagrange equation for this variational problem is given by the second order differential equation y 00 (x) − 2y(x) = 0, from which it is verified (as a necessary condition only) that the only candidates for the solution would be given by the solution obtained above.


3.

Leitmann’s coordinate transformation method

In 1967 Leitmann [2] established the following lemma. Lemma 3.1 Let y = z(x, y˜) be a transformation having a unique inverse y˜ = z˜(x, y) for x ∈ [x1 , x2 ] such that there is a one-to-one correspondence y(x) ⇔ y˜(x)

(13)

for all piecewise smooth functions y : [x1 , x2 ] → R satisfying the end conditions (2) and all piecewise smooth functions y˜ : [x1 , x2 ] → R satisfying the end conditions y˜(x1 ) = z˜(x1 , y1 ) and

y˜(x2 ) = z˜(x2 , y2 ).

(14)

If the transformation y = z(x, y˜) has the additional property that there exists a function G : [x1 , x2 ] × R → R such that F (x, y(x), y 0 (x)) − F (x, y˜(x), y˜0 (x)) =

d G(x, y˜(x)) dx

(15)

holds for all piecewise smooth functions y(·) satisfying (2) (here y˜(·) is related to y(·) through (13)), then there is a one-to-one correspondence between the minimizers of the original problem and those of minimizing (1) over the piecewise smooth functions satisfying the fixed end conditions (14). Moreover the transformation y = z(x, y˜) and its inverse y˜ = z˜(x, y) provide the desired correspondence.


The proof of this result proceeds as follows. Let y(·) be piecewise smooth satisfying the end conditions (2) and define y˜(·) through the formula, 3.0.0.1. Proof:

y˜(x) = z˜(x, y(x)),

x1 ≤ x ≤ x 2 .

Then it is easy to see that y˜(·) is piecewise smooth and satisfies the endpoint conditions (14). Moreover, as a result of (15) it follows that Z x2 Z x2 0 I(y) − I(˜ y) = F (x, y(x), y (x)) dx − F (x, y(x), y 0 (x)) dx x1 Zx1x2 d = G(x, y˜(x)) dx dx x1 = G(x2 , z˜(x2 , y2 )) − G(x1 , z˜(x1 , y1 )) (16) from which the desired conclusion follows immediately since the right hand side of the above equality is a constant depending only on the fixed endpoint conditions (2). 2


As a consequence of this lemma we observe that the problem of minimizing I(˜ y) over the endpoint conditions (14) is equivalent to our original problem, since there is a one-to-one correspondence between the extremals of each problem. We further observe that there is a similarity between Carathéodory’s observation in (5) and Leitmann’s observation (16) although they are indeed different. In (5) the two equivalent variational problems have the same endpoint conditions but different integrands while in (16) the problems have the same integral functional. Corollary 3.1 In order for a differentiable function G to have the properties given in the above lemma it is necessary the integrand F (·, ·, ·) together with the transformation y = z(x, y˜) be such that the left-hand side of (15) be linear in y˜0 (x).


To illustrate the utility of Leitmann’s observations we consider the same examples as in the previous section beginning with the F (x, y, z) = a(x)z 2 + b(x, y)z + c(x, y), with a(·) 6= 0, b(·, ·), and c(·, ·) all continuous on [x1 , x2 ] and [x1 , x2 ] × R respectively. For this integrand we observe that the class of admissible transformations required to satisfy Corollary (3.1) have the form z(x, y˜) = ±˜ y + f (x) for some smooth function f (·). 2 ∂z ∂z ∂z(x, y˜) ∂z(x, y˜) + p˜) − F (x, y˜, p˜) = a(x) (x, y˜) + (x, y˜)˜ p F (x, z(x, y˜), ∂x ∂ y˜ ∂x ∂ y˜ ∂z ∂z (x, y˜) + (x, y˜)˜ p + b(x, z(x, y˜)) ∂x ∂ y˜ +c(x, z(x, y˜)) − a(x)˜ p2 − b(x, y˜)˜ p − c(x, y˜) ! 2 ∂z ∂z ∂z = a(x) (x, y˜) + 2 (x, y˜) (x, y˜)˜ p ∂x ∂x ∂ y˜ ! ! 2 ∂z + (x, y˜) − 1 p˜2 ∂ y˜ ∂z ∂z +b(x, z(x, y˜)) (x, y˜) + (x, y˜)˜ p ∂x ∂ y˜ •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

−b(x, y˜)˜ p + c(x, z(x, y˜)) − c(x, y˜) which is linear in p˜ if and only if

2 ∂z (x, y˜) = 1. ∂ y˜

From this we immediately obtain z(x, y˜) = f (x) ± y˜ in which f (·) is a differentiable function of x. Substituting this form into the above gives us F (x, f (x) ± y˜, f 0 (x) ± p˜) − F (x, y˜, p˜) =

a(x)(f 0 (x))2 + b(x, f (x) ± y˜)f 0 (x) + c(x, f (x) ± y˜) − c(x, y˜)) ± (2a(x)f 0 (x) + b(x, f (x) ± y˜) − b(x, y˜)) p˜

from which we extract (assuming G(·, ·) is smooth enough) that ∂G(x, y˜) = a(x)(f 0 (x))2 + b(x, f (x) ± y˜)f 0 (x) + c(x, f (x) ± y˜) − c(x, y˜) ∂x ∂G(x, y˜) = ± (2a(x)f 0 (x) + b(x, f (x) ± y˜) − b(x, y˜)) ∂ y˜ Assuming additional smoothness (i.e., G(·, ·) is twice continuously differentiable) we •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

have that

∂ 2 G(x, y˜) ∂ 2 G(x, y˜) = ∂ y˜∂x ∂x∂ y˜

or ∂b(x, f (x) ± y˜) ∂c(x, f (x) ± y˜) ∂c(x, y˜) = ± ± − ∂y ∂y ∂y ∂b(x, f (x) ± y˜) = ± 2a(x)f 00 (x) + 2a0 (x)f 0 (x) + ∂x ∂b(x, f (x) ± y˜) 0 ∂b(x, y˜) + f (x) − . ∂y ∂x

Observe that this is a second order ordinary differential equation for the unknown function f (·) (depending on the parameter y˜) from which one could obtain the desired function G. To illustrate this idea we consider the specific example discussed in the second section. 3.0.0.2. Example’

We consider the specific case when F (x, y, z) = z 2 − 2yz + 2y 2 .

Substituting into the above we have ∂G(x, y˜) = (f 0 (x))2 − 2f (x)(1 − f 0 (x)) ± (4f (x) − 2f 0 (x))˜ y ∂x •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

∂G(x, y˜) = ± (2f 0 (x) − 2(f (x)) ∂ y˜ which, when equating the mixed second partial derivatives, we obtain f 00 (x) − 2f (x) = 0. This gives us

√

f (x) = c1 e Substituting this value for f into

∂G ∂y

2x

√

+ c2 e−

2x

.

and integrating, is easy to see that

G(x, y) = ±2 (f 0 (x) − f (x)) y˜ + h(x) √ √ √ √ = ±2(c1 ( 2 − 1)e 2x − c2 ( 2 + 1)e− 2x )˜ y + h(x)

(17)

in which h(·) is still to be determined. Differentiating the above with respect to x and equating to ∂G ∂x given above we obtain √ √ √ √ √ √ 0 2 2 2x 2 −2 2x 2x − 2x h (x) = (2 2 + 1)c1 e − (2 2 − 1)c2 e − 2(c1 e + c2 e ) or, √ √ √ √ √ √ 1 √ 2 2 2x 2 −2 2x 2x − 2x + (2 2 − 1)c2 e − 2 c1 e − c2 e +C, h(x) = √ (2 2 + 1)c1 e 2 2 (18)


in which C is an arbitrary constant. Therefore we can obtain G(x, y˜) by combining (17) and (18). To arrive at the correct solution to the above we observe that we have Z x2 I(˜ y) = F (x, y˜(x), y˜0 (x)) dx Zx1x2 = (2˜ y 0 (x)2 − 2˜ y (x)˜ y 0 (x) + 2˜ y (x)2 ) dx Zx1x2 2 (˜ y 0 (x) − y˜(x)) + y˜(x)2 dx. = x1

It is now an easy matter to conclude that by choosing the boundary conditions y˜(x1 ) = 0 and

y˜(x2 ) = 0

we have the solution y˜∗ (x) ≡ 0. Therefore to obtain the solution to the original problem it suffices to choose c1 and c1 so that f (x1 ) = y1

and

f (x2 ) = y2

to arrive at the minimizer, ∗

∗

√

y (x) = y˜ (x) + f (x) = c1 e

2x

√ − 2x

+ c2 e

.


4.

A combination of the two methods

Theorem 4.1 Let F : [x1 , x2 ]×R×R → R and F˜ : [x1 , x2 ]×R×R → R be two continuous functions which are convex in their third arguments. Further let z : [x1 , x2 ] → R be a transformation having a unique inverse y˜ = z˜(x, y) for x ∈ [x1 , x2 ] such that there exists a one-to-one correspondence between the piecewise smooth functions y : [x1 , x2 ] → R satisfying the end conditions (2) and y˜[x1 , x2 ] → R satisfying the end conditions (14). If the transformation y = z(x, y˜) has the additional property that there exists a function G : [x1 , x2 ] × R → R such that d F˜ (x, y(x), y 0 (x)) − F (x, y˜(x), y˜0 (x)) = G(x, y˜(x)) dx

(19)

holds for all piecewise smooth functions y(·) satisfying (2) (here y˜(·) is related to y(·) through (13), then there is a one-to-one correspondence between the minimizers of the original problem and those of minimizing the (1) over the piecewise smooth functions satisfying the fixed end conditions (4). Moreover the transformation y = z(x, y˜) and its inverse y˜ = z˜(x, y) provide the desired correspondence.


We consider the problem of minimizing the integral functional Z x2 y(x)4 9y 0 (x)2 − 6y(x)y 0 (x) + 2y(x)2 J(y) = x1 − cos(x) cos(y(x)3 ) + 3y(x)2 sin(x) sin(y(x)3 )y 0 (x) dx

4.0.0.3. Example 2

over the class of all piecewise smooth trajectories satisfying the fixed end conditions y(x1 ) = y1

and

y(x2 ) = y2

To solve this problem we let F˜ : [x1 , x2 ] × R × R → R be given by F˜ (x, y, z) = y 4 9z 2 − 6yz + 2y 2 − cos(x) cos(y 3 ) + 3y 2 sin(x) sin(y 3 )z and consider the transformation y˜ = y 3 and observe that for each feasible trajectory, y : [x1 , x2 ] → R we have, F˜ (x, y(x), y 0 (x)) = y(x)4 9y 0 (x)2 − 6y(x)y 0 (x) + 2y(x)2 − cos(x) cos(y(x)3 ) + 3y(x)2 sin(x) sin(y(x)3 )y 0 (x) = (3y(x)2 y 0 (x))2 − 6y(x)5 y 0 (x) + 2y(x)6 − cos(x) cos(y(x)3 ) + 3y(x)2 sin(x) sin(y(x)3 )y 0 (x) = y˜0 (x)2 − 2˜ y (x)˜ y 0 (x) + 2˜ y (x)2 − cos(x) cos(˜ y (x)) + sin(x) sin(˜ y (x))˜ y 0 (x) •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

From this computation we identify F : [x1 , x2 ] × R × R → R as F (x, y˜, z˜) = z˜2 − 2˜ y z˜ + 2˜ y2 and G : [x1 , x2 ] × R → R by G(x, y˜) = − sin(x) cos(˜ y) to get d G(x, y˜(x)). F˜ (x, y(x), y 0 (x)) − F (x, y˜(x), y˜0 (x)) = dx Therefore, by appealing to Theorem 4.1 we see that the minimizers of the above problem are in one-to-one correspondence with the minimizers of the problem of minimizing Z x2 I(˜ y) = y˜0 (x)2 − 2˜ y (x)˜ y 0 (x) + 2˜ y (x)2 dx x1

over the class of piecewise smooth functions satisfying the fixed end conditions y˜(x1 ) = y13

and

y˜(x2 ) = y23 .

This of course is the specific example we discussed in the previous sections so that the solution is given explicitly as √

y˜(x) = c˜1 e

2x

√ − 2x

+ c˜2 e

in which c˜1 and c˜2 are chosen so that the end conditions are satisfied. The solution for the original problem is now given by, √ √ 1/3 2x − 2x + c˜2 e . y(x) = c˜1 e •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

Before concluding this example we note that the Euler-Lagrange equation for the original equation reduces to 0 = 6y(x)3 3y(x)y 00 (x) + 6(y 0 (x))2 − 2y(x)2 which is a nonlinear second order differential equation which is much more difficult to solve than the linear Euler-Lagrange equation associated with the equivalent problem.


5.

Additional Information

Recently G. Leitmann and others have produced a series of papers in which his coordinate transformation method has been refined and extended to include optimal control problems, open-loop Nash equilibria in dynamic games and to infinite horizon optimal control problems. In particular we mention Leitmann [3] and Dockner and Leitmann [4]. Similar extensions of Carathéodory’s method have been done by Carlson [5], [6] for optimal control problems and in [7] for feedback Nash equilibria in dyamic games


References [1] C. Carathéodory. Calculus of Variations and Partial Differential Equations. Chelsea, New York, New York, 1982. [2] G. Leitmann. A note on absolute extrema of certain integrals. International Journal of Non-Linear Mechanics, 2:55–59, 1967. [3] G. Leitmann. On a class of direct optimization problems. J. Optimization Theory and Applications, 108(3), 2001. [4] E. J. Dockner and G. Leitmann. Coordinate transformation and derivation of openloop nash equilibrium. Journal of Optimization Theory and Applications, 110(1):1– 16, July 2001. [5] Dean A. Carlson. Carathédory-hamilton-jacobi theory for nonsmooth problems of lagrange and optimal control. Journal of Mathematical Analysis and Applications, 162:259–273, 1985. [6] Dean A. Carlson. A carathéodory-hamilton-jacobi theory for infinite horizon optimal control problems. Journal of Optimization Theory and Applicaions, 48:265–287, 1986. [7] D. A. Carlson. Carathéodory’s method for a class of dynamic games dynamic games. Journal of Mathematical Analysis and Applications. too appear.
