Euler-Lagrange equation for a delay variational problem

Abstract We establish Euler-Lagrange equations for a problem of Calculus of Variations where the unknown variable contains a term of delay on a segment


Introduction
We consider the following problem of Calculus of Variations (P ) where r, T ∈ (0, +∞), r < T , F : [0, T ] × C 0 ([−r, 0], R n ) × R n → R is a functional, ψ ∈ C 0 ([−r, 0], R n ), ζ ∈ R n , and x t (θ) := x(t + θ) when θ ∈ [−r, 0] and t ∈ [0, T ].C 0 denotes the continuity and C 1 denotes the continuous differentiability.The aim of this paper is to establish a first-order necessary condition of optimality for problem (P ) which is analogous to the Euler-Lagrange equation of the variational problem without delay.Note that in other settings of delay variational problems, the question of the establishment of an Euler-lagrange equation was studied, for instance in [8] (see references therein), [9], [2].Now we describe the contents of the paper.In Section 2, we specify the notation of various functions spaces, we introduce an operator to represent the dual space of C 0 ([−r, 0], R n ) into a space of bounded variation functions (denoted by R n ) and we establish properties on this operator.In Section 3, we state the main theorem of the paper (Theorem 3.1) on the Euler-Lagrange equation.We provide comments on this theorem.In Section, we introduce function spaces and operators which are specific to the delayed functions and we establish several of their properties.In Section 5, we provide conditions to ensure the Fréchet differentiability of the criterion of (P ).Section 6 is devoted to the proof of the Theorem 3.1

Notation and recall
When X and Y are real normed vector spaces, L(X, Y ) is the space of the continuous linear mappings from X into Y .When Λ ∈ L(X, Y ), we use the writings Λ•x := Λ(x), Λ, x = Λ(x) when Y = R, and we write the norm of linear continuous operators as Λ L := sup{ Λ • x Y : x ∈ X, x X ≤ 1}.The topological dual space of X is denoted by X * := L(X, R).
When a < b are two real numbers, the space of the continuous functions from When E is a finite-dimensional normed vector space, and when a a (g) which defined as the supremum of the non negative numbers , and when β = b, γ(b+) := γ(b).The Lebesgue-Stieltjes integral build on γ is defined by . We denote by (e k ) 1≤k≤n the canonical basis of R n and by (e * k ) 1≤k≤n its dual basis.When The theorem of representation of F. Riesz of C 0 ([−r, 0], R) * permits to define the operator and it is an isometry: R 1 (ℓ where ϕ ∈ C 0 ([−r, 0], R).We set This formula defines an operator • φ(θ) and using (2.4) we obtain (2.5) Proof.R n is endowed by the norm n k=1 u k e k := max 1≤k≤n |u k |, and R n * is endowed by the norm We define the linear functional Hence we can build the linear operator and since R 1 is a linear isomorphism, we obtain g k = 0 for all k ∈ {1, ..., n}, hence g = 0. We have proven that L is injective.
)and that L = L(g), and so we have proven that L is surjective.Hence L is linear bijective and continuous.Using the Inverse mapping Theorem of Banach, we obtain that L −1 is continuous, and since R n = L −1 , we obtain the announced result.Now we consider a case with a dependence with respect to the time.
Proof.Assertion (i) is a straightforward consequence of Lemma 2.1.Assertions (ii) and (iii) are is proven in [6] (Theorem 4.1) in the case where is the space of the functions in BV ([−r, 0], M n (R)) (M n (R) being the space of the real n × n matrices) which are left-continuous on (0, T ] and equal to 0 at T .The modifications to do to adapt the proof to the case of the present paper are clear. We need the two following results to study the Nemytskii (or superposition) operators.
Lemma 2.3.Let E, F be two metric spaces, and Φ ∈ C 0 (E, F ). P c (E) denotes the set of the compacts subsets of E. Then we have: This result is established in [13], p. 355.It permits to compensate the lack for compact neighborhood of compact subset in non locally compact metric spaces, for instance in infinite-dimensional normed spaces.In [5] and in [4] we have called it "Lemma of Heine-Schwartz".Lemma 2.4.Let E, F be two metric spaces, A be a nonempty compact metric space, and Φ : A × E → F be a mapping.Then the two following assertions are equivalent.
We need to use the following classical Lemma of Dubois-Reymond.Lemma 2.5.Let α < β be two real numbers.Let p ∈ C 0 ([α, β], R n * ) and q ∈ C 0 ([α, β], R n * ).We assume that, for all This result is proven in [1] (p. 60) when n = 1.Working on coordinates, the extension to an arbiratry positive integer number is easy.

The main result
In this section we state the theorem on the Euler-Lagrange equation as a firstorder necessary condition of optimality for problem (P ).First we give assumptions which are useful to this theorem.
tial with respect to the second (function) variable, D 2 F (t, φ, v), exists and with respect to the third (vector) variable, D 3 F (t, φ, v), exists and Theorem 3.1.Under (A1, A2, A3) let x be a local solution of the problem (P).
Then the function is of class C 1 on [0, T ] , and we have The operator R n which is used in this theorem is defined in Section 2 (formulas (2.4), (2.5)).The Euler-Lagrange equation of this theorem can be written under the integral form as follows where c ∈ R n * is a constant which is independent of t.
Note the presence of an advance (the contrary of the delay) in this equation.In other settings, [2] and [9], the Euler-Lagrange also contains a term of advance.

A function space and operators
We define the following function space On X we consider the following norm Proof.We can also write ∞,[0,T ] = 0. Using the uniqueness of the limit we obtain v | [0,T ] = u.Therefore we have v ∈ X and from the inequality We define the set A := {x ∈ X : A is a non empty closed affine subset of X and the unique vector subspace which is parallel to A is V := {h ∈ X : h 0 = 0, h(T ) = 0}.
Proof.Setting y(t) := ψ(t) when t ∈ [−r, 0] and y(t) := t T (ζ − ψ(0)) + ψ(0), we see that y ∈ A which proves that A is nonempty.From the inequalities x(T ) ≤ x X and x | [−r,0] ∞,[−r,0] ≤ x X , we obtain that A is closed in X.It is easy to verify that A is affine.The unique vector subspace of X is V = A − u where u ∈ A. and we can easily verify the announced formula for V.
Proof.Using a Heine's theorem, since [−r, T ] is compact and x is continuous, x is uniformly continuous on [−r, T ], i.e. ∀ǫ > 0, After Lemma 4.3 we can define the operator The continuity of S 1 results from the inequality Now we consider the following operator Proof.The linearity of D is clear.When x ∈ X, we have which implies the continuity of D.
When V and W are normed vector spaces we consider the operator B is bilinear continuous, and when I is a compact interval of R, we consider the Nemytskii operator defined on B where we have assimilated C 0 (I, L(V, W ))×C 0 (I, V ) and C 0 (I, L(V, W ))×V ).N B is bilinear and the following inequality holds This inequality shows that N B is continuous and consequently it is of class C 1 .

The differentiability of the criterion
First we establish a general result on the differentiability of the Nemytskii operators.
Lemma 5.1.Let I be a compact interval of R, V , W be two normed vector spaces, and Φ : I × V → W be a mapping.We assume that the following conditions are fulfilled.
(a) Φ ∈ C 0 (I × V, W ). (b) For all t ∈ I, the partial Fréchet differential of Φ with respect to the second variable, D 2 Φ(t, x), exists for all x ∈ V , and Then the operator N Φ defined by Proof.Under our assumptions, from Lemma 2.4 the following assertions hold.
We arbitrarily fix v ∈ C 0 (I, V ).The set K := {(t, v(t)) : t ∈ I} is compact as the image of a compact by a continuous mapping.Let ǫ > 0; using Lemma 2.4 we have which implies which implies, taking the supremum on the t ∈ I, And so we have proven that N Φ is Fréchet differentiable at v and When v, v 1 , δv ∈ C 0 (I, V ), using (4.8) we have and taking the supremum on the δv ∈ C 0 (I, V ) such that δv ∞,I ≤ 1 we obtain and (5.2) implies the continuity of DN Φ .
In different frameworks, similar results of differentiability of Nemytskii operators were proven in [4] (for almost periodic functions) , in [5] (for bounded sequences), in [3](for continuous functions which converge to zero at infinite).

Proof of the main result
To abridge the writing, we write D , and in the proofs we write g(t, θ) Proof.Using Proposition 3.2 in [6] and g(t, −r) = 0, we have, for all t ∈ [0, T ], For each ξ, we consider A .,ξ := {t ∈ [0, T ] : (t, ξ) ∈ A}.We have Using the Fubini theorem, we obtain Using (6.6) in (6.1) we obtain the announced formula.
< b are two real numbers, BV ([a, b], E) denotes the space of the bounded variation functions from [a, b] into E. N BV ([a, b], E) denotes the space of the g ∈ BV ([a, b], E) which are left-continuous on [a, b) and which satisfy g(a) = 0.When g ∈ BV ([a, b], E), the total variation of g is V b ) by setting in(x) := x, and the functional I : C 0 ([0, T ], R) → R by setting I(f ) := T 0 f (t)dt the Riemann integral of f on [0, T ].The operator in is clearly linear and from the inequality • ∞,[−r,T ] ≤ • X , it is continuous.I is linear and by using the mean value theorem, it is continuous.Note that J = I • N F • (S • in, D).Since in, S, D and I are linear continuous, they are of class C 1 , and so (S • in, D) is of class C 1 .Using Lemma 5.2, N F is of class C 1 , and so J is of class C 1 as a composition of C 1 mappings.The calculation of DJ is a simple application of the Chain Rule : , ξ − t) • h ′ (ξ)dξdt.(6.1)We set A := {(t, ξ) : 0 ≤ t ≤ T, t − r ≤ ξ ≤ t} and from the Fubini-Tonelli theorem we have , ξ − t) • h ′ (ξ)dξdt = A g(t, ξ − t)dξdt (6.2)
Lemma 5.3.Under (A1, A2, A3), J ∈ C 1 (X, R) and for all x ∈ A and for all h ∈ V, we have DJ