NOTES ON CALCULUS OF VARIATIONS. September 13, 2012

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1 NOTES ON CALCULUS OF VARIATIONS JON JOHNSEN September 13, The basic problem In Calculus of Variations one is given a fixed C 2 -function F (t, x, u), where F is defined for t [, t 1 ] and x, u R, and the problem is to maximise (or minimise) the functional J(x) = that acts on functions x: [, t 1 ] R fulfilling F (t, x(t), ẋ(t)) dt (1) x( ) = x, x(t 1 ) = x 1. (2) Hereby x, x 1 are given numbers; and any C 1 -function 1 x(t) satisfying these two boundary conditions is said to be admissible. Example 1 (Ramsey 1928). In economics it is a task to determine the amount of capital K(t), say in a country, such that, when f(k(t)) denotes the gross domestic product created by having the capital K(t), one has max T U(f(K(t)) K(t))e ρt dt, K() = K, K(T ) = K T. (3) Here f(k) K is the consumption and U is a utility function, fixed so that U > > U on ], [. The factor e ρt, ρ >, gives a discount of future consumption in order to give priority to consumption in the near future. K and K T are given initial and terminal values of the available capital. By maximising the integral, the country is envisaged to benefit as much as possible from the change in capital from K to K T. The problem above can be seen as an optimisation problem in infinitely many variables (one for each t [, t 1 ]). But fortunately the possible minimising and maximising functions x(t) can be found among the solutions to a certain ordinary differential equation. This is the content of the following famous result: 1 Recall that a function f(t) is said to be a C k -function on an interval [, t 1 ] if all derivatives dm f dt with m m k exist and are continous on [, t 1 ]. The analogous definition applies to functions of several variables, such as F (t, x, u). 1

2 2 JON JOHNSEN Theorem 2 (Euler Lagrange equation). For an admissible C 1 -function x (t) to maximise or minimise J(x) = F (t, x(t), ẋ(t)) dt, such that x( ) = x, x(t 1 ) = x 1 (4) it is necessary that x (t) solves the ordinary differential equation d (F dt 3(t, x(t), ẋ(t))) = F 2(t, x(t), ẋ(t)). (5) To reveal the nature of the above differential equation, it is instructive to carry out the t-differentiation using the chain rule (this is allowed at least if the solution x(t) is a C 2 -function): F 33(t, x(t), ẋ(t))ẍ(t) + F 32(t, x(t), ẋ(t))ẋ(t) + F 31(t, x(t), ẋ(t)) F 2(t, x(t), ẋ(t)) =. (6) This is a homogeneous second order differential equation, which is said to be quasi-linear because the coefficients depend on the solution and its lower order derivatives. Being of second order (if F 33 ) the set of solutions usually involves two arbitrary (real or complex) constants. These are often determined by imposing the initial and terminal conditions in (2). To illustrate the usefulness of Theorem 2, one can take the simple Example 3. Consider J(x) = 1 (x(t) 2 + ẋ(t) 2 ) dt; x() =, x(1) = e 2 1. Here F (x, u) = x 2 + u 2 so the Euler Lagrange equation is = d (2ẋ) 2x = 2(ẍ x). dt The complete solution to this is given by the functions x(t) = Ae t + Be t. Invoking the boundary conditions it is seen that x(t) is admissible if and only if A + B = and Ae Be 1 = e 2 1, that is, precisely for A = e = B, so the only candidate for a minimiser or maximiser is x (t) = e 1+t e 1 t. However, x is not a maximising function (e.g. J(x) can be seen to take on arbitrarily large values), but it will follow later from sufficient conditions that x is a minimiser. The proof of Theorem 2 is given further below in case x (t) is a C 2 -function. The point of departure is to show the du Bois-Reymond lemma, which is also known as the fundamental lemma of calculus of variations: Lemma 4. Let f : [, t 1 ] R be a continuous function. When f has the property that, for every function µ in C 2 ([, t 1 ]) fulfilling µ( ) = = µ(t 1 ), f(t)µ(t) dt = (7) then f(t) = for every t.

3 NOTES ON CALCULUS OF VARIATIONS 3 Proof. Suppose the contrary, say f(s) > at some s, for simplicity. By continuity there is some subinterval I = ]a, b[ on which f(t) > and a < b t 1. On I one can then define µ(t) as µ(t) = (t a) 3 (b t) 3 >, and let µ(t) = outside I; the zeroes at a and b have so high order that µ (t) for t a + and for t b, so the Mean Value Theorem implies that µ is of class C 2 on [, t 1 ]. Denoting by P (t) a primitive function of f(t)µ(t), it follows by the Mean Value Theorem and the choice of I that = f(t)µ(t) dt = f(t)µ(t) dt = P (b) P (a) = f(τ)µ(τ)(b a) >. (8) I This is a contradiction; the case f(s) < is similar. Hence f. The lemma above is exploited by forming a so-called variation of the given solution, x(t, α) = x (t) + αµ(t). (9) Here α R is just a parameter, while µ is an arbitrary C 2 -function on [, t 1 ] such that µ( ) = = µ(t 1 ); clearly t x(t, α) is then admissible for every fixed α R. Such variations have of course given rise to the name of the whole theory Proof of the Euler Lagrange equation. To exploit the variation in (9), let I(α) = J(x(t, α)) = F (t, x(t, α), x(t,α) ) dt. (1) For simplicity the proof continues with the case of a maximum at x (the case of a minimum is similar). This means that I(α) I() for all α, whence I () =. (11) Of course this requires that I(α) can be differentiated. More precisely we shall proceed under the assumptions that α I(α) is differentiable for every α R, I (α) = α x F (t, x(t, α), (t, α)) dt. Here the second line means that one can simply differentiate under the integral sign. This non-trivial fact will be proved later in Lemma 5 below. However, granted this lemma, the easy consequences of it are given first: using an integration by parts in the last step below we get I () = = = (F 2(t, x, ẋ ) x (t, ) + F α 3(t, x, ẋ ) 2 x (t, )) dt α (F 2(t, x, ẋ )µ(t) + F 3(t, x, ẋ ) µ(t)) dt ( ) µ(t) F 2(t, x, ẋ ) d F dt 3(t, x, ẋ ) dt + [ µ(t)f 3(t, x, ẋ ) ] t=t 1 t=. (12)

4 4 JON JOHNSEN Since µ vanishes at the end points, the above yields = I () = µ(t) ( F 2(t, x, ẋ ) d dt F 3(t, x, ẋ ) ) dt. (13) Here µ is an arbitrary C 2 -function vanishing at the end points, and that F 2 d F dt 3 is continuous with respect to t can be seen from the terms in (6). So it follows from Lemma 4 that the function F 2 d F dt 3 equals for every t. This means that x (t) solves the Euler Lagrange equation, as desired. Now it only remains to account for the statements in the next lemma, which is lengthy to prove, but elementary in content: Lemma 5. The function I(α) is differentiable at every α R and I (α ) = x F (t, x(t, α), (t, α)) α α=α dt. (14) Proof. For simplicity the details are given in the main case, which is α =. We shall show that I(α) I() = Aα + o(α) with A equal to the above integral for α =. Now, it is clear that I(α) I() = (F (t, x(t, α), x(t,α) ) F (t, x, ẋ )) dt. (15) Here the integrand is a C 2 -function of α, as one can easily check. (The derivative of order 2 is given below.) So by using Taylor s formula (for fixed t), we find for some remainder R(α) that I(α) I() = α In fact, when we use the integral remainder α 2 1 x F (t, x(t, α), (t, α)) α α= dt + R(α). (16) (1 θ)f ( t, x(t, ), x(t, )) (θα) dθ, then the term R is by straightforward calculation seen to be given by R(α) = α 2 Therefore it fulfils R(α) α 2 1 (1 θ) ( µ(t) 2 F 22(t, x(t, θα), x (t, θα)) 1 + 2µ(t) µ(t)f 23(t, x(t, θα), x (t, θα)) + µ(t) 2 F 33(t, x(t, θα), x (t, θα))) dθdt. (17) ( µ 2 F 22(t, x(t, θα), x (t, θα)) + 2 µ µ F 23(t, x(t, θα), x (t, θα)) + µ 2 F 33(t, x(t, θα), x (t, θα)) ) dθdt, (18) To show that this double integral is a bounded function of α, we consider the map Φ(α, θ, t) = (t, x (t) + θαµ(t), ẋ (t) + θα µ(t)) (19)

5 NOTES ON CALCULUS OF VARIATIONS 5 with the domain D = [ 1, 1] [, 1] [, t 1 ]. Since D is a compact set, and since Φ is continuous, its range B = Φ(D) is necessarily compact. Consequently the functions F 22, F 23 and F 33 are all bounded on B (they are continuous because F C 2 ), and it follows that the double integral is less than or equal to the constant C = (max F 22 + max B B F 23 + max B F 33 ) ( µ(t) + µ(t) ) 2 dt. (2) Therefore R(α) C α α for α, (21) which means that R(α) is o(α) for α. Altogether (16), (21) account for both the differentiability of I(α) at α = and that the derivatives can be calculated under the integral sign. For general α R the main changes are that one should replace θα everywhere by α + θ(α α ) and x by x(, α ). This completes the proof of Theorem 2 on the necessity of the Euler Lagrange equation Further necessary conditions. In analogy with the second order conditions for functions of real variables one has the next result, which will only be given here without proof. Theorem 6 (Legendre). If F is a C 2 -function, it is a necessary condition for the functional J(x) = t 1 F (t, x(t), ẋ(t)) dt to have an extreme value at an admissible function x (t) that F 33(t, x (t), ẋ (t)) in case x maximises, (22) F 33(t, x (t), ẋ (t)) in case x minimises. (23) (These inequalities are required to hold for all t [, t 1 ].) The Legendre condition is often useful, when one wants to show that a solution candidate is not a maximiser (or a minimiser). This is elucidated by the next example. Example 7. In the above Example 3 where J(x) = 1 (x2 + ẋ 2 ) dt one finds at once that F 33 = 2 >, and this rules out that the admissible function x = e 1+t e 1 t can be a maximiser. (It is still open whether x is a minimiser; this cannot be concluded from the fact that F 33 > 2.) One of the possible complications in practise is that, for good reasons, there are further constraints on the admissible functions. This can e.g. leave us with the problems of finding a C 1 -function x: [, t 1 ] R such that max F (t, x, ẋ) dt; x( ) = x, x(t 1 ) = x 1 ; (24) h(t, x(t), ẋ(t)) > for all t [, t 1 ]. (25) Hereby h is a suitable C 1 -function defining the constraint. However, one can show that the Euler Lagrange and Legendre conditions are necessary also for such problems.

6 6 JON JOHNSEN In fact the preimage O = h 1 ( ], [ ) (26) is an open set in [, t 1 ] R n R n, and the image K = { (t, x (t), ẋ (t)) t [, t 1 ] } is a compact subset of O. Therefore K has distance d > to the boundary of O, so by restricting α (as we may) to be so close to that α d/2 max t [t,t 1 ] 1 + µ(t) 2 + µ(t), (27) 2 the curve t (t, x(t, α), x (t, α)) will for each α consequently have distance < d/2 to K, hence have its range completely inside O: that is, < h(t, x(t, α), x (t, α)) for all t, α. This suffices e.g. for the proof of the Euler equation s necessity also under the constraint given by h. The next example shows how such constraints can appear. Example 8. In the above Example 1 from macro economics, one has the integrand F (t, K(t), K(t)) = U(f(K(t)) K(t))e ρt. (28) But the utility function U is typically only a C 2 -function on ], [ ; for example U( ) = ( ) meets the requirements that U > > U. So here it is natural to impose the constraint that h(t, K(t), K(t)) := f(k(t)) K(t) >. This is not just a kind of mathematical obstruction (as U a, already a C 1 - extension of U to the whole axis R is impossible), for negative values of the consumption f(k(t)) K(t) does not make sense at all in the model. 2. General Terminal Conditions A more radical change is met, if one considers the problem of having max together with one of the following terminal conditions: (i) x(t 1 ) free (t 1 given); (ii) x(t 1 ) x 1 (t 1 and x 1 given); (iii) x(t 1 ) = g(t 1 ) (t 1 free, but g given in C 1 )). F (t, x(t), ẋ(t)) dt, x() = x, (29) Note that in order for case (iii) to make sense, the functions F and g must be defined on [, T ] R R, respectively [, T ], for some T in the range < T. However, this point will not be stressed in the following, where T = is used for simplicity s sake. Correspondingly the admissible functions are now required to be C 1 -functions that fulfill the stated initial and terminal conditions. It is clear that a maximising function x still fulfills the Euler Lagrange equations, for x also maximises among the admissible functions that satisfy x(t 1 ) = x (t 1 ). But one has to add some transversality conditions:

7 NOTES ON CALCULUS OF VARIATIONS 7 Theorem 9. If x (t) is an admissible function solving the above maximisation problem, then x satisfies the Euler Lagrange equation and the corresponding transversality condition, (i) F 3(t 1, x (t 1 ), ẋ (t 1 )) = ; (ii) F 3(t 1, x (t 1 ), ẋ (t 1 )) (and = holds if x (t 1 ) > x 1 ); (iii) F (t 1, x (t 1 ), ẋ (t 1 )) (ġ(t 1 ) ẋ (t 1 ))F 3(t 1, x (t 1 ), ẋ (t 1 )) =. The conclusion in (iii) is valid with the proviso that ẋ (t 1 ) ġ(t 1 ). In case (ii) the inequality is reversed if x solves the minimisation problem. Proof of case (i),(ii). Set y 1 = x (t 1 ). Since x maximises J, the inequality J(x ) J(y) holds in particular for all admissible functions y(t) that satisfy y(t 1 ) = y 1. Therefore x is also a solution of the basic problem on the fixed time interval [, t 1 ] and with data x, y 1. Consequently x satisfies the Euler Lagrange equation. The transversality condition (i) can be proved as a corollary to the proof of Theorem 2: since the terminal value x(t 1 ) is not fixed in this context, there is the freedom to take the variation µ(t) such that µ(t 1 ) (but µ( ) = is still required); again it is seen that I () =. Since it is already known that x solves the Euler Lagrange equation, formula (12) therefore yields µ(t 1 )F 3(t 1, x (t 1 ), ẋ (t 1 )) =. (3) Taking µ such that µ(t 1 ) = 1 the conclusion in (i) follows. With a little more effort also (ii) can be obtained along these lines. Indeed, in case x (t 1 ) > x 1 the only modification needed is a restriction on α, which should be taken in an interval so narrow around α = that the variation t x(t, α) is admissible, i.e. such that x(t 1, α) x 1. Setting ε = x (t 1 ) x 1 >, this is obtained by requiring that α µ(t 1 ) < ε. (31) In the remaining cases one has x (t 1 ) = x 1, whence αµ(t 1 ) is necessary for admissibility. To proceed here it is useful to specialise to variations with µ(t 1 ) > ; then I(α) is only defined for α, so the maximality gives I(α) I() for all α. As I(α) is differentiable on [, α[ according to Lemma 5, this implies I (). So this time (12) gives µ(t 1 )F 3(t 1, x (t 1 ), ẋ (t 1 )), (32) which implies (ii). As a preparation for the somewhat longer proof of (iii), one may now deduce Lemma 1. Let f(t, α) be a continuous function of (t, α) R 2 and set I(α, β) = β f(t, α) dt. (33) If f exists and is continuous on α R2 and, moreover, I exists, equals β f α (t, α) dt and is α continuous with respect to α for each β, then I(α, β) is differentiable at every (α, β) and it holds that I = f(α, β). β

8 8 JON JOHNSEN Proof. For each fixed α the fundamental theorem of calculus states that the integral is a differentiable function of the upper limit (since the integrand is continuous) and moreover that I = f(α, β). By assumption on f this function is continuous on β R2, and the first partial derivative I is continuous with respect to α. By a standard result for functions α of several variables this implies the differentiability. (Here a small well-known sharpening of the usual C 1 -condition is used: in the proof, that uses the Mean Value Theorem on increments along the coordinate axes, continuity of the first partial derivative is only needed along the first axis.) It is left for the reader to formulate and prove a local version of the lemma, in which both α, β belong to proper open subintervals of the real line. Proof of (iii). To obtain (iii), it is convenient to invoke the Implicit Function Theorem, which is applied to the mapping Φ(t, α) = x (t) + αµ(t) g(t), t [, [, α R. (34) (Hereby the function x is artificially continued for t t 1 as the quadratic polynomial x (t 1 ) + ẋ (t 1 )(t t 1 ) + 1 2ẍ (t 1 )(t t 1 ) 2 in order that it be C 2 on the interval [, [.) The function µ is in C 2 ([, [) and need only satisfy µ( ) =. Now, since the pair (x, t 1 ) solves the maximum problem under condition (iii), the point (t, α) = (t 1, ) is obviously on the level surface Φ(t, α) =. (35) Since Φ (t 1, ) = ẋ (t 1 ) ġ(t 1 ) by assumption, there exists according to the Implicit Function Theorem some α >, τ > and a C 1 -function ψ(α), mapping [ α, α ] to [t 1 τ, t 1 + τ ], with ψ() = t 1 and the crucial property that its graph { (ψ(α), α) R 2 α α α } (36) constitutes all the solutions of (35) in the rectangle [t 1 τ, t 1 + τ ] [ α, α ]. By its construction the function ψ fulfills Φ(ψ(α), α) for α α, that is, x (ψ(α)) + αµ(ψ(α)) g(ψ(α)) =. (37) This relation is important because it shows that t x(t, α), defined for t [, ψ(α)], actually fulfills the terminal condition (iii), hence is admissible. Moreover, differentiation of (37) gives so in particular for α =, ẋ (ψ(α)) ψ(α) + µ(ψ(α)) + (α µ(ψ(α)) ġ(ψ(α))) ψ(α) =, (38) (ẋ (t 1 ) ġ(t 1 )) ψ() = µ(t 1 ). (39) Choosing µ from now on so that µ(t 1 ), it follows that ψ(). To exploit this knowledge, it is first noted that for the above variation, I(α) = ψ(α) F (t, x (t) + αµ(t), ẋ (t) + α µ(t)) dt. (4)

9 NOTES ON CALCULUS OF VARIATIONS 9 In view of Lemma 5 and Lemma 1, the chain rule gives the differentiability of α (ψ(α), α) I(α) and that I (α) = F (t, x(t, α), x (t, α)) t=ψ(α) ψ(α) + ψ(α) (µf 2 + µf 3) dt. (41) With the usual partial integration this gives, since x solves the Euler equation, I () = F (t 1, x (t 1 ), ẋ (t 1 )) ψ() + [ µf 3 ] t1 t=t1 t= + µ(f 2 d dt F 3) dt = F (t 1, x (t 1 ), ẋ (t 1 )) ψ() + µ(t 1 )F 3(t 1, x (t 1 ), ẋ (t 1 )). Since I(α) I() also here implies I () =, an insertion of the above formula for µ(t 1 ) now gives that = ( F t=t1 + (ġ(t 1 ) ẋ (t 1 ) f t=t1 ) ) ψ(). (43) Because ψ() was seen earlier, the transversality condition (iii) now follows at once. Remark 11. The additional assumption made in Theorem 9 for the transversality condition (iii) that ẋ (t 1 ) ġ(t 1 ) is hardly severe. Indeed, it is most likely always fulfilled as a consequence of the fact that the pair (x, t 1 ) maximises the functional: if ẋ (t 1 ) = ġ(t 1 ) it is conceivable that x could be continued along the graph of g for t > t 1 so that a larger integral would be obtained on the interval [, t 1 + ε] for some ε >, for one can arrange that the integrand is positive at t 1 by adding a suitable constant to F. (The continuation along the tangent of course only gives a C 1 -function, so that framework for admissibility would be the natural one for a final discussion of this point.) Sufficient conditions for solutions Finally, a result on sufficient conditions for a solution is given. However, it holds only in case the basic function F (t, x, u) is concave with respect to (x, u). This means that the Hessian matrix is negative semi-definite at all points, i.e. ( 2 F ) ; that is, for all t, x, x u u, this matrix has eigenvalues λ 1, λ 2 in ], ]. Theorem 12. Suppose F (t, x, u) is concave with respect to (x, u), and that x (t) fulfils the Euler Lagrange equation and one of the terminal conditions a) x(t 1 ) = x 1, b) x(t 1 ) x 1, or c) x(t 1 ) free, together with the corresponding transversality condition b) F 3(t 1, x (t 1 ), ẋ (t 1 )) (and = if x (t 1 ) > x 1 ), respectively c) F 3(t 1, x (t 1 ), ẋ (t 1 )) =. Then x is a global maximiser, i.e. for every other admissible function x(t) it holds true that F (t, x(t), ẋ(t)) dt ẋ (42) F (t, x (t), ẋ (t)) dt. (44) Since minimisation of F (t, x, ẋ) dt is achieved by maximising for F, one has the analogous minimisation result for convex functions.

10 1 JON JOHNSEN Example 13. Since F (x, u) = x 2 + u 2 is convex in Example 3, the solution x (t) = e 1+t e 1 t actually minimises the functional J(x) = 1 (x(t)2 + ẋ(t) 2 ) dt subject to the conditions x() = and x(1) = e 2 1. (As was claimed earlier.) Exercises Exercise 1 (i) Calculate the value of J(x) = 1 (x2 + ẋ 2 ) dt in the cases (a) x(t) = (e 2 1)t, (b) x(t) = e 2t 1, (c) x(t) = e 1+t e 1 t, (d) x(t) = at 2 + (e 2 1 a)t. (They all go through (, ) and (1, e 2 1), hence are admissible.) (ii) Let a and conclude that J(x) has no maximum on the curves joining (, ) to (1, e 2 1). Exercise 2 Let J(x) = 2 1 t 2 ẋ(t) 2 dt and consider the boundary conditions x(1) = 1, x(2) = 2. (i) Find the admissible solutions to the Euler Lagrange equation. (ii) Show that the maximisation problem for J has no solution. (Try x(t) = at 2 (1 3a)t + 2a.) (iii) Does the above imply that the solution in (i) minimises J(x)? Exercise 3 Consider the problem min 1 (t + x) 4 dt, x() =, x(1) = a. Find the solution of the Euler Lagrange equation and determine the value of a for which this solution is admissible. For this value of a, find the solution of the problem. Exercise 4 The length of the graph of a C 1 -function x(t) which connects (, x ) to (t 1, x 1 ) is given by L(x) = 1 + ẋ(t)2 dt. Prove that L(x) attains its minimum over the admissible functions exactly when x(t) has a straight line as its graph. Exercise 5 Let A(t) denote the assets (or wealth) of a person at time t, let w be the constant wage, and suppose money can be borrowed at the fixed interest rate r; thus the consumption at time t is modelled as C(t) = ra(t) + w A(t). Suppose the person wants to maximise consumption from now until the expected death date T, T U(C(t))e ρt dt. Hereby U is a certain utility function, U > > U, and ρ is a discount factor. While the present assests are A, the purpose is also to leave at least the amount A T to the heirs, i.e. to have A(T ) A T.

11 NOTES ON CALCULUS OF VARIATIONS 11 Apply the necessary conditions to this case. Show in particular that A(t) is only optimal if A(T ) = A T. (Is this understandable?) Solve Euler equation if U(C) = a e bc for constants a, b >. Exercise 6 Investigate what the Euler Lagrange equations give in the special cases when F = F (t, x), F = F (t, ẋ), F = F (x, ẋ). Prove that here d dt 3(x, ẋ) = F 2(x, ẋ) (45) = F (x, ẋ) ẋf 3(x, ẋ) is constant (46) = ẋ = or d dt F 3(x, ẋ) = F 2(x, ẋ). (47)