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On the validity of the Euler–Lagrange equation
J. Math. Anal. Appl. 304 (2005) 356–369 www.elsevier.com/locate/jmaa
On the validity of the Euler–Lagrange equation A. Ferriero ∗ , E.M. Marchini Dipartimento di Matematica e Applicazioni, Università degli Studi di Milano-Bicocca, Via R. Cozzi 53, 20125 Milano, Italia Received 9 September 2003 Available online 14 October 2004 Submitted by B.S. Mordukhovich
Abstract The purpose of the present paper is to establish the validity of the Euler–Lagrange equation for the solution xˆ to the classical problem of the calculus of variations. 2004 Elsevier Inc. All rights reserved. Keywords: Euler–Lagrange equation
1. Introduction
The purpose of the present paper is to establish (under Carathéodory’s conditions) the validity of the Euler–Lagrange equation (E–L) for the solution xˆ to the classical problem of the calculus of variations consisting in minimizing the functional J (x) =
L t, x(t), x (t) dt,
I
* Corresponding author.
E-mail addresses: [email protected] (A. Ferriero), [email protected] (E.M. Marchini). 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.09.029
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357
where I = (a, b) ⊂ R, on the set of those absolutely continuous functions x : I → R s satisfying the boundary conditions x(a) = A, x(b) = B. Establishing the validity of the Euler–Lagrange equation amounts to proving that ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt = 0 ∇ξ L t, x(t), I
for every variation η in some suitable class. A large number of papers has been devoted to this classical problem, e.g., [4–6,8–10]. The example obtained by Ball and Mizel [1], modifying an earlier example of Maniá [7], provides a variational problem where the inteˆ xˆ (·)) does not hold and, as a consequence, (E–L) is not true along grability of ∇x L(·, x(·), ˆ xˆ (·)) has to be imposed in the solution. Hence, some condition on the term ∇x L(·, x(·), order to ensure the validity of (E–L). A result of Clarke [5] implies that the following ˆ xˆ (·)): assumption on the term ∇x L(·, x(·), there exists a function S(t) integrable on I such that, for y in a neighborhood of the solution, ∇x L t, y, xˆ (t) S(t) is sufficient to establish the validity of (E–L). This condition implies that, locally along the solution, x → L(t, x, x ) is Lipschitzian of Lipschitz constant S(t). However, there are simple and meaningful examples of variational problems where this Lipschitzianity condition is not verified. √ Consider the Lagrangian defined by L(x, ξ ) = (ξ |x| − 2/3)2 , and the problem (P) of minimizing 1
L x(t), x (t) dt
0
over the absolutely continuous functions x with x(0) = 0, x(1) = 1. ˆ xˆ (t)) = 0 One can easily verify that x(t) ˆ = t 2/3 is a minimizer for (P) (indeed, L(x(t), on [0, 1], and L is non-negative everywhere). In this case, although L is not differentiable ˆ xˆ (t)) exists a.e. (it is a.e. zero) and it is integrable. The purpose everywhere, Lx (x(t), of the present paper is to provide a result on the validity of (E–L) that is satisfied by Lagrangians that are Lipschitzian in x, but that applies as well to the non-Lipschitzian cases as the example before. In the proof we first show that the fact that xˆ is a solution implies the integrability ˆ xˆ (·)). Then, using this result, we establish the validity of (E–L) under of ∇ξ L(·, x(·), Carathéodory’s condition. Note that we do not assume any convexity hypothesis on the Lagrangian. Moreover, no growth condition whatsoever is assumed so that, as far as we know, relaxation theorems cannot be applied.
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2. Integrability of ∇ ξ L(·, x(·), ˆ xˆ (·)) Consider the problem of minimizing the functional J (x) = L t, x(t), x (t) dt I
on the set of those absolutely continuous functions x : I → R s satisfying the boundary conditions x(a) = A, x(b) = B. Let xˆ be a (weak local) minimizer yielding a finite value ˆ for the functional J , and set µ = supt∈[a,b] x(t). Our results will depend on the following assumption. Assumption A. (i) L is differentiable in x along x, ˆ for a.e. t, and the map ∇x L(·, x(·), ˆ xˆ (·)) is integrable on I ; (ii) there exists a function S(t) integrable on I such that, for any y ∈ B(0, µ + 1), ˆ . ˆ xˆ (t) + S(t)y − x(t) L t, y, xˆ (t) L t, x(t), Consider problem (P) as presented in the Introduction. L and xˆ satisfy Assumption A: Lx (x(t), ˆ xˆ (t)) exists a.e. (identically zero, hence integrable), S(t) = t −2/3 verifies the inequality ˆ , L y, xˆ (t) S(t) y − x(t) since
2 −1/3 2 t |y| − 3 3
2
√ 4( |y| − t 1/3 ) 4 |y| − t 2/3 2/3 |y − t 2/3 |. = 2/3 √ 1/3 9t 9t ( |y| + t )
This is our first result on the term ∇ξ L(·, x(·), ˆ xˆ (·)). In what follows, R¯ denotes R ∪ {+∞}. Theorem 2.1. Suppose that L : I × R s × R s → R¯ is an extended valued function, finite on its effective domain of the form dom L = I × R s × G, where G ⊂ R s is an open set, and that it satisfies Carathéodory’s conditions, i.e., L(·, x, ξ ) is measurable for fixed (x, ξ ) and L(t, ·, ·) is continuous for almost every t. Moreover assume that L is differentiable in ξ on dom L and that ∇ξ L satisfies Carathéodory’s conditions on dom L. Suppose that Assumption A holds. Then, ∇ξ L t, x(t), ˆ xˆ (t) dt < +∞. I
Proof. (1) By assumption, L(·, x(·), ˆ xˆ (·)) ∈ L1 (I ), hence setting S0 = {t ∈ I : xˆ (t) ∈ / G}, we have m(S0 ) = 0. Given > 0, we can cover S0 by an open set O1 of measure m(O1 ) < /2. We have also that ∇ξ L is a Carathéodory’s function and that xˆ is measurable in I . Hence, by the theorems of Scorza Dragoni and of Lusin, for the given > 0 there exists an open set O2 such that m(O2 ) < /2 and at once xˆ is continuous in I \ O2 , and ∇ξ L is continuous in (I \ O2 ) × R s × G. By taking K = I \ (O1 ∪ O2 ), we have that K
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is a closed set such that on it xˆ is continuous with values in G, ∇ξ L is continuous on s n+1 and K = K ; set also K × R n n
× G and m(I \ K ) < . For n 1 set n = (b − a)/2 Cn = nj=1 Kj . Then Cn are closed sets, Cn ⊂ Cn+1 , xˆ is continuous on Cn with values in G, ∇ξ L is continuous in Cn × R s × G and limn→+∞ m(I \ Cn ) = 0. From these properties it follows that there exists kn > 0 such that, for all t ∈ Cn , ∇ξ L t, x(t), ˆ xˆ (t) < kn . There is no loss of generality in assuming kn kn−1 . Moreover, we have that m(C1 ) (b − a)/2 and ∞ n=2 m(Cn \ Cn−1 ) (b − a)/2.
For all n > 1, we set An = Cn \ Cn−1 . Hence we obtain that Cm = C1 m n=2 An and that
I = E ∪ C1 n>1 An , where m(E) = 0. (2) Consider the function 0 if ∇ξ L(t, x(t), ˆ xˆ (t)) = 0, θ (t) = ∇ξ L(t,x(t), ˆ xˆ (t)) otherwise, ∇ L(t,x(t), ˆ xˆ (t)) ξ
and
vn =
θ (t) dt, An
so that vn m(An ). There exists a closed set Bn ⊆ C1 such that m(Bn ) = vn . Set vn χB (t). θn (t) = −θ (t)χAn (t) + vn n We have that θn (t) dt = − θ (t) dt + vn = 0. I
An
t Hence, setting θn (t) = a θn (τ ) dτ , we see that the functions θn (t) are admissible variations. Moreover we obtain t θn ∞ sup θ (τ ) dτ θ (τ ) dτ 2m(An ). t∈I
n
a
n
I
ˆ xˆ (t)) < kn ; for t in Bn , ∇ξ L(t, x(t), ˆ xˆ (t)) < (3) For t in An , we have ∇ξ L(t, x(t), ¯ ¯ k1 kn . Recalling that An ⊂ Cn , we infer that, for all t ∈ An ∪ Bn , ∇ξ L t, x(t), ˆ xˆ (t) kn . We wish to obtain an uniform bound for ∇ξ L computed in a suitable neighborhood of the solution (x(·), ˆ xˆ (·)). Consider the set (A¯ n ∪ Bn ) × R s × G as a metric space Mn with distance d((t, x, ξ ), (t , x , ξ )) = sup(|t − t |, |x − x |, |ξ − ξ |). On Mn , ∇ξ L is continuous. Moreover, its subset ˆ xˆ (t) : t ∈ A¯ n ∪ Bn Gn = t, x(t), is compact and, on Gn , ∇ξ L is bounded by kn . Hence there exists δn > 0 such that, for ˆ xˆ (t))) < δn , we have ∇ξ L(t, x, ξ ) < kn + 1. (t, x, ξ ) ∈ Mn with d((t, x, ξ ), (t, x(t),
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(4) For |λ| < min{1/2m(An ), δn /2m(An ), δn }, consider the integrals 1 L t, x(t) ˆ + λθn (t), xˆ (t) + λθn (t) − L t, x(t), ˆ xˆ (t) dt λ I 1 = L t, x(t) ˆ + λθn (t), xˆ (t) + λθn (t) − L t, x(t) ˆ + λθn (t), xˆ (t) dt λ An ∪Bn 1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), + ˆ xˆ (t) dt. λ I
For every t ∈ An ∪ Bn there exists ζλ (t) ∈ (0, λ) such that 1 L t, x(t) ˆ + λθn (t), xˆ (t) + λθn (t) − L t, x(t) ˆ + λθn (t), xˆ (t) λ = ∇ξ L t, x(t) ˆ + λθn (t), xˆ (t) + ζλ (t)θn (t) , θn (t) ˆ + λθn (t), xˆ (t) + ζλ (t)θ (t) ∇ξ L t, x(t) n
and from the choice of λ, ∇ξ L(t, x(t) ˆ + λθn (t), xˆ (t) + ζλ (t)θn (t)) < kn + 1. Hence, we can apply the dominated convergence theorem to obtain that 1 L t, x(t) ˆ + λθn (t), xˆ (t) + λθn (t) − L t, x(t) ˆ + λθn (t), xˆ (t) dt lim λ→0 λ An ∪Bn = ˆ xˆ (t) , θn (t) dt. ∇ξ L t, x(t), An ∪Bn
(5) Set f + (s) = max{0, f (s)}, f − (s) = max{0, −f (s)}. Since + 1 ˆ + λθn (t), xˆ (t) − L t, x(t), 0 L t, x(t) ˆ xˆ (t) S(t)θn (t), λ by the dominated convergence theorem, + 1 lim L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ→0 λ I + 1 ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt. = lim L t, x(t) λ→0 λ I
By the Fatou’s lemma, − 1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt lim inf λ→0 λ I − 1 ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt. lim inf L t, x(t) λ→0 λ I
We have obtained that
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1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ λ→0 I + 1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), lim ˆ xˆ (t) λ→0 λ I − 1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), − lim inf ˆ xˆ (t) dt λ→0 λ I 1 ˆ + λθn (t), xˆ (t) − L t, x(t), lim sup L t, x(t) ˆ xˆ (t) dt λ→0 λ I ˆ xˆ (t) , θn (t) dt. = ∇x L t, x(t),
lim sup
I
(6) Since xˆ is a minimizer, we have ∇ξ L t, x(t), ˆ xˆ (t) , θn (t) dt 0 An ∪Bn
1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ λ→0 I ∇ξ L t, x(t), ˆ xˆ (t) , θn (t) dt + ∇x L t, x(t), ˆ xˆ (t) , θn (t) dt. + lim sup
An ∪Bn
(∗)
I
θn (t)
Since = −∇ξ L(t, x(t), ˆ xˆ (t))/∇ξ L(t, x(t), ˆ xˆ (t)), for any t in An , it follows − ∇ξ L(t, x(t), ˆ xˆ (t)), θn (t) χAn (t) = ∇ξ L(t, x(t), ˆ xˆ (t))χAn (t). Hence, we obtain
that that
(∗) can be written as ∇ξ L t, x(t), ˆ xˆ (t) dt An
Bn
∇ξ L t, x(t), ˆ xˆ (t) , θn (t) dt +
∇x L t, x(t), ˆ xˆ (t) , θn (t) dt.
I
On Bn , ∇ξ L(t, x(t), ˆ xˆ (t)) is bounded by k1 ; from Hölder’s inequality and the estimate on θn ∞ obtained in (2) we have that there exists a constant C (independent of n) such that ∇ξ L t, x(t), ˆ xˆ (t) dt Cm(An ). An
(7) As m → +∞, the sequence of functions (∇ξ L(t, x(t), ˆ xˆ (t))χ{ mn=2 An } (t))m con verges monotonically to the function ∇ξ L(t, x(t), ˆ xˆ (t))χ{ n>1 An } (t). From the estimate above and monotone convergence, we obtain
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∇ξ L t, x(t), ˆ xˆ (t) dt =
I \C1
∇ξ L t, x(t), ˆ xˆ (t) χ{
n>1 An }
dt
I
Cm An . n>1
L(t, x(t), ˆ xˆ (t)) < k
On C1 , ∇ξ 1 . Hence ∇ξ L t, x(t), ˆ xˆ (t) dt < +∞.
2
I
3. Additional regularity and the validity of the Euler–Lagrange equation Corollary 3.1. Under the same assumptions as in Theorem 2.1, for every variation η, η(a) = 0, η(b) = 0 and η ∈ L∞ (I ), we have ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt = 0. I
Proof. We shall prove that, for every η in AC(I ) with bounded derivative, such that η(a) = η(b) = 0, we have ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt 0. I
Fix η, let η (t) K for almost every t in I . (1) Define Cn and kn as in point (1) of the proof of Theorem 2.1. Set vn = η (t) dt. I \Cn
We have that limn→+∞ vn = 0. In particular, for n ν, there exists Bn ⊆ C1 such that m(Bn ) = vn . Set for t ∈ I \ Cn , 0 (t) for t ∈ Cn \ Bn , η (ηn ) (t) = vn + η (t) for t ∈ B . n vn We obtain ηn (t) dt = I
η (t) dt +
Cn \Bn
= I
Bn
vn + η (t) dt = η (t) dt + vn vn Cn
η (t) dt = 0.
t Hence, setting ηn (t) = a ηn (τ ) dτ , we have that the functions ηn (t) are variations and that, for almost every t in I , ηn (t) (1 + K), so that ηn ∞ (1 + K)(b − a).
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(2) As in point (3) of the proof of Theorem 2.1, there exists δn > 0 such that for ˆ xˆ (t))) < δn , we have ∇ξ L(t, x, ξ ) < (t, x, ξ ) ∈ Cn × R s × G, with d((t, x, ξ ), (t, x(t), kn + 1. (3) For |λ| < min{1/(1 + K)(b − a), δn /(1 + K)(b − a), δn /(1 + K)}, consider the integrals 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t), ˆ xˆ (t) dt λ I 1 = L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) dt λ Cn 1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), + ˆ xˆ (t) dt. λ I
For almost every t ∈ Cn , there exists ζλ (t) ∈ (0, λ) such that 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) λ = ∇ξ L t, x(t) ˆ + ληn (t), xˆ (t) + ζλ (t)ηn (t) , ηn (t) ˆ + ληn (t), xˆ (t) + ζλ (t)η (t) (1 + K) < (kn + 1)(1 + K). ∇ξ L t, x(t) n
Hence, we can apply the dominated convergence theorem to obtain that 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) lim ˆ + ληn (t), xˆ (t) dt λ→0 λ I ˆ xˆ (t) , ηn (t) dt = ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) dt. = ∇ξ L t, x(t), Cn
I
(4) Following the point (5) of Theorem 2.1, we obtain that 1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt lim sup λ λ→0 I ˆ xˆ (t) , ηn (t) dt. ∇x L t, x(t), I
(5) Since xˆ is a minimizer, we have ˆ xˆ (t) , ηn (t) dt 0 ∇ξ L t, x(t), I
+ lim sup λ→0
I
I
1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ
∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) + ∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt.
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(6) Since lim η (t) = η (t) n→+∞ n and
∇ξ L t, x(t), ˆ xˆ (t) η (t) ∇ξ L t, x(t), ˆ xˆ (t) (1 + K), n
and lim ηn (t) = η(t)
n→+∞
and
∇x L t, x(t), ˆ xˆ (t) ηn (t) ∇x L t, x(t), ˆ xˆ (t) (1 + K)(b − a),
by the dominated convergence we obtain lim ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) dt = ∇ξ L t, x(t), ˆ xˆ (t) , η (t) dt n→+∞
I
and
lim
n→+∞
I
∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt =
I
∇x L t, x(t), ˆ xˆ (t) , η(t) dt.
I
It follows that ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt 0.
2
I
Even if the validity of the Euler–Lagrange equations already follows by the previous Corollary 3.1 and the DuBois–Reymond’s lemma [3], we give an alternative proof in Corollary 3.3. In the following theorem we prove an additional regularity result for the Lagrangian evaluated along the minimizer. ˆ xˆ (·)) Theorem 3.2. Under the same assumptions as in Theorem 2.1, the map ∇ξ L(·, x(·), ∞ is in L (I ). ˆ xˆ (·)) Proof. Using an iteration process, we shall prove that for every p in N ,∇ξ L(·, x(·), p p ˆ xˆ (·)) ∈ p1 Lp (I ). is in L (I ). Since the L are nested, this proves that ∇ξ L(·, x(·), At the same time, we shall prove that there exists a constant K > 0 such that, for every ˆ xˆ (·)) is in L∞ (I ). 1 p < +∞, ∇ξ Lp K, thus proving that ∇ξ L(·, x(·), ˆ xˆ (·)) is in L1 (I ). StartSuppose that: (1) From Theorem 2.1, we know that ∇ξ L(·, x(·), ˆ xˆ (·)) ∈ Lp (I ), to prove ing the iteration process, fix p ∈ N and suppose that ∇ξ L(·, x(·), p+1 ˆ xˆ (·)) ∈ L (I ). that ∇ξ L(·, x(·), (2) We can assume ∇ξ Lp = 0. Define Cn , An , kn as in point (1) of the proof of Theorem 2.1. For all n > 1, set p−1 (b − a) p vn = ∇ξ L t, x(t), ˆ xˆ (t) ∇ξ L t, x(t), ˆ xˆ (t) dt. p 2∇ξ Lp An
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365
Since m(C1 ) (b − a)/2 and
p vn (b − a) p 2∇ξ Lp
p ∇ξ L t, x(t), ˆ xˆ (t) (b − a)/2,
An
p Bn
p
p
⊂ C1 such that m(Bn ) = vn , so that p p p−1 2∇ξ Lp vn ∇ξ L t, x(t), ˆ xˆ (t) ∇ξ L t, x(t), ˆ xˆ (t) . dt = p (b − a) vn
there exists a set
p
An
Bn
Set
p p−1 θn (t) = −∇ξ L t, x(t), ˆ xˆ (t) ∇ξ L t, x(t), ˆ xˆ (t) χAn (t) p
2∇ξ Lp vnp χ p (t) + (b − a) vnp Bn
t p p p and θn (t) = a (θn ) (τ ) dτ . We obtain that θn ∞ 2 An ∇ξ L(t, x(t), ˆ xˆ (t))p dt. The p variations θn have bounded derivatives so we can apply Corollary 3.1 to obtain that p p ˆ xˆ (t) , θn (t) + ∇x L t, x(t), ˆ xˆ (t) , θn (t) dt = 0. ∇ξ L t, x(t), I
It follows that p+1 ∇ξ L t, x(t), ˆ xˆ (t) dt An p
2∇ξ Lp = (b − a) + I
An
p
Bn
p ∇x L t, x(t), ˆ xˆ (t) , θn (t) dt
+2 C˜
p p ∇ξ L t, x(t), ˆ xˆ (t) , vn vn dt
p ∇ξ L t, x(t), ˆ xˆ (t) dt
k1
An
p ∇ξ L t, x(t), ˆ xˆ (t) dt
∇x L t, x(t), ˆ xˆ (t) dt
I
p ∇ξ L t, x(t), ˆ xˆ (t) dt,
An
where C˜ is independent of n and p (suppose C˜ 1). The sequence of maps p+1 ∇ξ L t, x(t), ˆ xˆ (t) χ mn=2 An (t) m
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converges monotonically to ∇ξ L(t, x(t), ˆ xˆ (t))p+1 χ{ n>1 An } (t), and each integral p+1
ˆ xˆ (t)) χ{ mn=2 An } (t) dt is bounded by the same constant I ∇ξ L(t, x(t), C˜
∞ p p ∇ξ L t, x(t), ∇ξ L t, x(t), ˆ xˆ (t) dt = C˜ ˆ xˆ (t) dt. n=2 A
I \C1
n
Hence, by the monotone convergence theorem, p+1 p ∇ξ L t, x(t), ∇ξ L t, x(t), ˆ xˆ (t) dt C˜ ˆ xˆ (t) dt < +∞. I \C1
I \C1
ˆ xˆ (t)) < k1 , proving that ∇ξ L(·, x(·), ˆ xˆ (·)) ∈ Lp+1 (I ). Moreover, On C1 , ∇ξ L(t, x(t), we have also obtained that p+1 ∇ξ L t, x(t), ∇ξ L t, x(t), ˆ xˆ (t) dt C˜ p ˆ xˆ (t) dt, I \C1
so that
I \C1
p+1 ∇ξ L t, x(t), ˆ xˆ (t) dt
I \C1
where S = max{1, that, for all p ∈ N ,
I \C1
˜ CS,
∇ξ L(t, x(t), ˆ xˆ (t)) dt}. Setting T = max{1, m(C1 )} we have
∇ξ L(p+1)
1/(p+1)
(p+1) k1 m(C1 ) +
p+1 ∇ξ L t, x(t), ˆ xˆ (t) dt
1/(p+1)
I \C1
k1 m(C1 )1/(p+1) +
p+1 ∇ξ L t, x(t), ˆ xˆ (t) dt
1/(p+1)
I \C1
˜ = K. k1 T + CS
2
Corollary 3.3. Under the same conditions as in Theorem 2.1, for every variation η, η(a) = 0, η(b) = 0 and η ∈ L1 (I ), we have ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt = 0. I
As a consequence, t → ∇ξ L(t, x(t), ˆ xˆ (t)) is absolutely continuous. Proof. We shall prove that, for every η in AC(I ), such that η(a) = η(b) = 0, we have ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt 0. I
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(1) Fix η. Through the same steps as in point (1) of the proof of Theorem 2.1, for every n ∈ N we can define a closed set Cn such that on it η is continuous, xˆ is continuous with values in G, ∇ξ L is continuous in Cn × R s × G and limn→+∞ m(I \ Cn ) = 0. In particular, it follows that there are constants kn and cn > 0 such that, for all t ∈ Cn , ∇ξ L t, x(t), ˆ xˆ (t) < kn and η (t) < cn . Define vn , Bn , ηn and ηn as in the proof of Corollary 3.1. Since, for all t ∈ I , ηn (t) 1 + η (t), it follows that ηn 1 (b − a) + η 1 . Moreover, ηn ∞ ηn 1 (b − a) + η 1 . (2) As in point (3) of the proof of Theorem 2.1, there exists δn > 0 such that for ˆ xˆ (t))) < δn , we have ∇ξ L(t, x, ξ ) < (t, x, ξ ) ∈ Cn × R s × G, with d((t, x, ξ ), (t, x(t), kn + 1. (3) For |λ| < min{1/((b − a) + η 1 ), δn /((b − a) + η 1 ), δn /(cn + 1)}, consider the integrals 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t), ˆ xˆ (t) dt λ I 1 = L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) dt λ I 1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt + λ I 1 = L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) dt λ Cn
+
1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt. λ
I
For every t ∈ Cn , there exists ζλ (t) ∈ (0, λ) such that 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) λ = ∇ξ L t, x(t) ˆ + ληn (t), xˆ (t) + ζλ (t)ηn (t) , ηn (t) ˆ + ληn (t), xˆ (t) + ζλ (t)η (t) cn < (kn + 1)cn . ∇ξ L t, x(t) n
Hence, we can apply the dominated convergence theorem to obtain that 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) lim ˆ + ληn (t), xˆ (t) dt λ→0 λ I ˆ xˆ (t) , ηn (t) dt = ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) dt. = ∇ξ L t, x(t), Cn
I
(4) Following the point (5) of Theorem 2.1, we obtain that
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1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ
lim sup λ→0
I
∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt
I
and, since xˆ is a minimizer, we have 0 ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) I
+ lim sup λ→0
1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ
I
∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) + ∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt.
I
(5) Finally we have lim η (t) = η (t) n→+∞ n and
∇ξ L t, x(t), ˆ xˆ (t) ηn (t) ∇ξ L ·, x(·), ˆ xˆ (·) ∞ 1 + η (t) ,
and lim ηn (t) = η(t)
n→+∞
and
∇x L t, x(t), ˆ xˆ (t) ηn (t) ∇x L t, x(t), ˆ xˆ (t) (b − a) + η 1 ,
so that, by dominated convergence, we obtain lim ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) dt = ∇ξ L t, x(t), ˆ xˆ (t) , η (t) dt n→+∞
I
and
lim
n→+∞ I
I
∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt =
∇x L t, x(t), ˆ xˆ (t) , η(t) dt.
I
Hence, it follows that ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt 0. I
ˆ xˆ (t)) is classical (e.g., [2]). The absolute continuity of t → ∇ξ L(t, x(t),
2
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References [1] J.M. Ball, V.J. Mizel, One-dimensional variational problems whose minimizers do not satisfy the Euler– Lagrange equation, Arch. Rational Mech. Anal. 90 (1985) 325–388. [2] H. Brezis, Analyse fonctionnelle, théorie et applications, Masson, Paris, 1983. [3] G. Buttazzo, M. Giaquinta, S. Hildebrant, One-Dimensional Variational Problems. An Introduction, Oxford Lecture Series in Mathematics and Its Applications, vol. 15, Oxford Univ. Press, New York, 1998. [4] L. Cesari, Optimization. Theory and Applications, Springer, Berlin, 1983. [5] L. Clarke, Optimization and Nonsmooth Analysis, Wiley–Interscience, New York, 1983, reprinted in: Classics in Applied Mathematics, vol. 5, SIAM, Philadelphia, 1990. [6] A.D. Ioffe, R.T. Rockafellar, The Euler and Weierstrass conditions for nonsmooth variational problems, Calc. Var. Partial Differential Equations 4 (1996) 59–87. [7] B. Maniá, Sopra un esempio di Lavrentieff, Boll. Unione Mat. Ital. 13 (1934) 147–153. [8] C. Marcelli, E. Outkine, M. Sytchev, Remarks on necessary conditions for minimizers of one-dimensional variational problems, Nonlinear Anal. 48 (2002) 979–993. [9] B.S. Mordukhovich, Discrete approximations and refined Euler–Lagrange conditions for nonconvex differential inclusions, SIAM J. Control Optim. 33 (1995) 882–915. [10] R. Vinter, H. Zheng, The extended Euler–Lagrange condition for nonconvex variational problems, SIAM J. Control Optim. 35 (1997) 56–77.
On the validity of the Euler–Lagrange equation A. Ferriero ∗ , E.M. Marchini Dipartimento di Matematica e Applicazioni, Università degli Studi di Milano-Bicocca, Via R. Cozzi 53, 20125 Milano, Italia Received 9 September 2003 Available online 14 October 2004 Submitted by B.S. Mordukhovich
Abstract The purpose of the present paper is to establish the validity of the Euler–Lagrange equation for the solution xˆ to the classical problem of the calculus of variations. 2004 Elsevier Inc. All rights reserved. Keywords: Euler–Lagrange equation
1. Introduction
The purpose of the present paper is to establish (under Carathéodory’s conditions) the validity of the Euler–Lagrange equation (E–L) for the solution xˆ to the classical problem of the calculus of variations consisting in minimizing the functional J (x) =
L t, x(t), x (t) dt,
I
* Corresponding author.
E-mail addresses: [email protected] (A. Ferriero), [email protected] (E.M. Marchini). 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.09.029
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where I = (a, b) ⊂ R, on the set of those absolutely continuous functions x : I → R s satisfying the boundary conditions x(a) = A, x(b) = B. Establishing the validity of the Euler–Lagrange equation amounts to proving that ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt = 0 ∇ξ L t, x(t), I
for every variation η in some suitable class. A large number of papers has been devoted to this classical problem, e.g., [4–6,8–10]. The example obtained by Ball and Mizel [1], modifying an earlier example of Maniá [7], provides a variational problem where the inteˆ xˆ (·)) does not hold and, as a consequence, (E–L) is not true along grability of ∇x L(·, x(·), ˆ xˆ (·)) has to be imposed in the solution. Hence, some condition on the term ∇x L(·, x(·), order to ensure the validity of (E–L). A result of Clarke [5] implies that the following ˆ xˆ (·)): assumption on the term ∇x L(·, x(·), there exists a function S(t) integrable on I such that, for y in a neighborhood of the solution, ∇x L t, y, xˆ (t) S(t) is sufficient to establish the validity of (E–L). This condition implies that, locally along the solution, x → L(t, x, x ) is Lipschitzian of Lipschitz constant S(t). However, there are simple and meaningful examples of variational problems where this Lipschitzianity condition is not verified. √ Consider the Lagrangian defined by L(x, ξ ) = (ξ |x| − 2/3)2 , and the problem (P) of minimizing 1
L x(t), x (t) dt
0
over the absolutely continuous functions x with x(0) = 0, x(1) = 1. ˆ xˆ (t)) = 0 One can easily verify that x(t) ˆ = t 2/3 is a minimizer for (P) (indeed, L(x(t), on [0, 1], and L is non-negative everywhere). In this case, although L is not differentiable ˆ xˆ (t)) exists a.e. (it is a.e. zero) and it is integrable. The purpose everywhere, Lx (x(t), of the present paper is to provide a result on the validity of (E–L) that is satisfied by Lagrangians that are Lipschitzian in x, but that applies as well to the non-Lipschitzian cases as the example before. In the proof we first show that the fact that xˆ is a solution implies the integrability ˆ xˆ (·)). Then, using this result, we establish the validity of (E–L) under of ∇ξ L(·, x(·), Carathéodory’s condition. Note that we do not assume any convexity hypothesis on the Lagrangian. Moreover, no growth condition whatsoever is assumed so that, as far as we know, relaxation theorems cannot be applied.
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2. Integrability of ∇ ξ L(·, x(·), ˆ xˆ (·)) Consider the problem of minimizing the functional J (x) = L t, x(t), x (t) dt I
on the set of those absolutely continuous functions x : I → R s satisfying the boundary conditions x(a) = A, x(b) = B. Let xˆ be a (weak local) minimizer yielding a finite value ˆ for the functional J , and set µ = supt∈[a,b] x(t). Our results will depend on the following assumption. Assumption A. (i) L is differentiable in x along x, ˆ for a.e. t, and the map ∇x L(·, x(·), ˆ xˆ (·)) is integrable on I ; (ii) there exists a function S(t) integrable on I such that, for any y ∈ B(0, µ + 1), ˆ . ˆ xˆ (t) + S(t)y − x(t) L t, y, xˆ (t) L t, x(t), Consider problem (P) as presented in the Introduction. L and xˆ satisfy Assumption A: Lx (x(t), ˆ xˆ (t)) exists a.e. (identically zero, hence integrable), S(t) = t −2/3 verifies the inequality ˆ , L y, xˆ (t) S(t) y − x(t) since
2 −1/3 2 t |y| − 3 3
2
√ 4( |y| − t 1/3 ) 4 |y| − t 2/3 2/3 |y − t 2/3 |. = 2/3 √ 1/3 9t 9t ( |y| + t )
This is our first result on the term ∇ξ L(·, x(·), ˆ xˆ (·)). In what follows, R¯ denotes R ∪ {+∞}. Theorem 2.1. Suppose that L : I × R s × R s → R¯ is an extended valued function, finite on its effective domain of the form dom L = I × R s × G, where G ⊂ R s is an open set, and that it satisfies Carathéodory’s conditions, i.e., L(·, x, ξ ) is measurable for fixed (x, ξ ) and L(t, ·, ·) is continuous for almost every t. Moreover assume that L is differentiable in ξ on dom L and that ∇ξ L satisfies Carathéodory’s conditions on dom L. Suppose that Assumption A holds. Then, ∇ξ L t, x(t), ˆ xˆ (t) dt < +∞. I
Proof. (1) By assumption, L(·, x(·), ˆ xˆ (·)) ∈ L1 (I ), hence setting S0 = {t ∈ I : xˆ (t) ∈ / G}, we have m(S0 ) = 0. Given > 0, we can cover S0 by an open set O1 of measure m(O1 ) < /2. We have also that ∇ξ L is a Carathéodory’s function and that xˆ is measurable in I . Hence, by the theorems of Scorza Dragoni and of Lusin, for the given > 0 there exists an open set O2 such that m(O2 ) < /2 and at once xˆ is continuous in I \ O2 , and ∇ξ L is continuous in (I \ O2 ) × R s × G. By taking K = I \ (O1 ∪ O2 ), we have that K
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359
is a closed set such that on it xˆ is continuous with values in G, ∇ξ L is continuous on s n+1 and K = K ; set also K × R n n
× G and m(I \ K ) < . For n 1 set n = (b − a)/2 Cn = nj=1 Kj . Then Cn are closed sets, Cn ⊂ Cn+1 , xˆ is continuous on Cn with values in G, ∇ξ L is continuous in Cn × R s × G and limn→+∞ m(I \ Cn ) = 0. From these properties it follows that there exists kn > 0 such that, for all t ∈ Cn , ∇ξ L t, x(t), ˆ xˆ (t) < kn . There is no loss of generality in assuming kn kn−1 . Moreover, we have that m(C1 ) (b − a)/2 and ∞ n=2 m(Cn \ Cn−1 ) (b − a)/2.
For all n > 1, we set An = Cn \ Cn−1 . Hence we obtain that Cm = C1 m n=2 An and that
I = E ∪ C1 n>1 An , where m(E) = 0. (2) Consider the function 0 if ∇ξ L(t, x(t), ˆ xˆ (t)) = 0, θ (t) = ∇ξ L(t,x(t), ˆ xˆ (t)) otherwise, ∇ L(t,x(t), ˆ xˆ (t)) ξ
and
vn =
θ (t) dt, An
so that vn m(An ). There exists a closed set Bn ⊆ C1 such that m(Bn ) = vn . Set vn χB (t). θn (t) = −θ (t)χAn (t) + vn n We have that θn (t) dt = − θ (t) dt + vn = 0. I
An
t Hence, setting θn (t) = a θn (τ ) dτ , we see that the functions θn (t) are admissible variations. Moreover we obtain t θn ∞ sup θ (τ ) dτ θ (τ ) dτ 2m(An ). t∈I
n
a
n
I
ˆ xˆ (t)) < kn ; for t in Bn , ∇ξ L(t, x(t), ˆ xˆ (t)) < (3) For t in An , we have ∇ξ L(t, x(t), ¯ ¯ k1 kn . Recalling that An ⊂ Cn , we infer that, for all t ∈ An ∪ Bn , ∇ξ L t, x(t), ˆ xˆ (t) kn . We wish to obtain an uniform bound for ∇ξ L computed in a suitable neighborhood of the solution (x(·), ˆ xˆ (·)). Consider the set (A¯ n ∪ Bn ) × R s × G as a metric space Mn with distance d((t, x, ξ ), (t , x , ξ )) = sup(|t − t |, |x − x |, |ξ − ξ |). On Mn , ∇ξ L is continuous. Moreover, its subset ˆ xˆ (t) : t ∈ A¯ n ∪ Bn Gn = t, x(t), is compact and, on Gn , ∇ξ L is bounded by kn . Hence there exists δn > 0 such that, for ˆ xˆ (t))) < δn , we have ∇ξ L(t, x, ξ ) < kn + 1. (t, x, ξ ) ∈ Mn with d((t, x, ξ ), (t, x(t),
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(4) For |λ| < min{1/2m(An ), δn /2m(An ), δn }, consider the integrals 1 L t, x(t) ˆ + λθn (t), xˆ (t) + λθn (t) − L t, x(t), ˆ xˆ (t) dt λ I 1 = L t, x(t) ˆ + λθn (t), xˆ (t) + λθn (t) − L t, x(t) ˆ + λθn (t), xˆ (t) dt λ An ∪Bn 1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), + ˆ xˆ (t) dt. λ I
For every t ∈ An ∪ Bn there exists ζλ (t) ∈ (0, λ) such that 1 L t, x(t) ˆ + λθn (t), xˆ (t) + λθn (t) − L t, x(t) ˆ + λθn (t), xˆ (t) λ = ∇ξ L t, x(t) ˆ + λθn (t), xˆ (t) + ζλ (t)θn (t) , θn (t) ˆ + λθn (t), xˆ (t) + ζλ (t)θ (t) ∇ξ L t, x(t) n
and from the choice of λ, ∇ξ L(t, x(t) ˆ + λθn (t), xˆ (t) + ζλ (t)θn (t)) < kn + 1. Hence, we can apply the dominated convergence theorem to obtain that 1 L t, x(t) ˆ + λθn (t), xˆ (t) + λθn (t) − L t, x(t) ˆ + λθn (t), xˆ (t) dt lim λ→0 λ An ∪Bn = ˆ xˆ (t) , θn (t) dt. ∇ξ L t, x(t), An ∪Bn
(5) Set f + (s) = max{0, f (s)}, f − (s) = max{0, −f (s)}. Since + 1 ˆ + λθn (t), xˆ (t) − L t, x(t), 0 L t, x(t) ˆ xˆ (t) S(t)θn (t), λ by the dominated convergence theorem, + 1 lim L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ→0 λ I + 1 ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt. = lim L t, x(t) λ→0 λ I
By the Fatou’s lemma, − 1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt lim inf λ→0 λ I − 1 ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt. lim inf L t, x(t) λ→0 λ I
We have obtained that
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1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ λ→0 I + 1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), lim ˆ xˆ (t) λ→0 λ I − 1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), − lim inf ˆ xˆ (t) dt λ→0 λ I 1 ˆ + λθn (t), xˆ (t) − L t, x(t), lim sup L t, x(t) ˆ xˆ (t) dt λ→0 λ I ˆ xˆ (t) , θn (t) dt. = ∇x L t, x(t),
lim sup
I
(6) Since xˆ is a minimizer, we have ∇ξ L t, x(t), ˆ xˆ (t) , θn (t) dt 0 An ∪Bn
1 L t, x(t) ˆ + λθn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ λ→0 I ∇ξ L t, x(t), ˆ xˆ (t) , θn (t) dt + ∇x L t, x(t), ˆ xˆ (t) , θn (t) dt. + lim sup
An ∪Bn
(∗)
I
θn (t)
Since = −∇ξ L(t, x(t), ˆ xˆ (t))/∇ξ L(t, x(t), ˆ xˆ (t)), for any t in An , it follows − ∇ξ L(t, x(t), ˆ xˆ (t)), θn (t) χAn (t) = ∇ξ L(t, x(t), ˆ xˆ (t))χAn (t). Hence, we obtain
that that
(∗) can be written as ∇ξ L t, x(t), ˆ xˆ (t) dt An
Bn
∇ξ L t, x(t), ˆ xˆ (t) , θn (t) dt +
∇x L t, x(t), ˆ xˆ (t) , θn (t) dt.
I
On Bn , ∇ξ L(t, x(t), ˆ xˆ (t)) is bounded by k1 ; from Hölder’s inequality and the estimate on θn ∞ obtained in (2) we have that there exists a constant C (independent of n) such that ∇ξ L t, x(t), ˆ xˆ (t) dt Cm(An ). An
(7) As m → +∞, the sequence of functions (∇ξ L(t, x(t), ˆ xˆ (t))χ{ mn=2 An } (t))m con verges monotonically to the function ∇ξ L(t, x(t), ˆ xˆ (t))χ{ n>1 An } (t). From the estimate above and monotone convergence, we obtain
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∇ξ L t, x(t), ˆ xˆ (t) dt =
I \C1
∇ξ L t, x(t), ˆ xˆ (t) χ{
n>1 An }
dt
I
Cm An . n>1
L(t, x(t), ˆ xˆ (t)) < k
On C1 , ∇ξ 1 . Hence ∇ξ L t, x(t), ˆ xˆ (t) dt < +∞.
2
I
3. Additional regularity and the validity of the Euler–Lagrange equation Corollary 3.1. Under the same assumptions as in Theorem 2.1, for every variation η, η(a) = 0, η(b) = 0 and η ∈ L∞ (I ), we have ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt = 0. I
Proof. We shall prove that, for every η in AC(I ) with bounded derivative, such that η(a) = η(b) = 0, we have ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt 0. I
Fix η, let η (t) K for almost every t in I . (1) Define Cn and kn as in point (1) of the proof of Theorem 2.1. Set vn = η (t) dt. I \Cn
We have that limn→+∞ vn = 0. In particular, for n ν, there exists Bn ⊆ C1 such that m(Bn ) = vn . Set for t ∈ I \ Cn , 0 (t) for t ∈ Cn \ Bn , η (ηn ) (t) = vn + η (t) for t ∈ B . n vn We obtain ηn (t) dt = I
η (t) dt +
Cn \Bn
= I
Bn
vn + η (t) dt = η (t) dt + vn vn Cn
η (t) dt = 0.
t Hence, setting ηn (t) = a ηn (τ ) dτ , we have that the functions ηn (t) are variations and that, for almost every t in I , ηn (t) (1 + K), so that ηn ∞ (1 + K)(b − a).
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(2) As in point (3) of the proof of Theorem 2.1, there exists δn > 0 such that for ˆ xˆ (t))) < δn , we have ∇ξ L(t, x, ξ ) < (t, x, ξ ) ∈ Cn × R s × G, with d((t, x, ξ ), (t, x(t), kn + 1. (3) For |λ| < min{1/(1 + K)(b − a), δn /(1 + K)(b − a), δn /(1 + K)}, consider the integrals 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t), ˆ xˆ (t) dt λ I 1 = L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) dt λ Cn 1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), + ˆ xˆ (t) dt. λ I
For almost every t ∈ Cn , there exists ζλ (t) ∈ (0, λ) such that 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) λ = ∇ξ L t, x(t) ˆ + ληn (t), xˆ (t) + ζλ (t)ηn (t) , ηn (t) ˆ + ληn (t), xˆ (t) + ζλ (t)η (t) (1 + K) < (kn + 1)(1 + K). ∇ξ L t, x(t) n
Hence, we can apply the dominated convergence theorem to obtain that 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) lim ˆ + ληn (t), xˆ (t) dt λ→0 λ I ˆ xˆ (t) , ηn (t) dt = ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) dt. = ∇ξ L t, x(t), Cn
I
(4) Following the point (5) of Theorem 2.1, we obtain that 1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt lim sup λ λ→0 I ˆ xˆ (t) , ηn (t) dt. ∇x L t, x(t), I
(5) Since xˆ is a minimizer, we have ˆ xˆ (t) , ηn (t) dt 0 ∇ξ L t, x(t), I
+ lim sup λ→0
I
I
1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ
∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) + ∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt.
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(6) Since lim η (t) = η (t) n→+∞ n and
∇ξ L t, x(t), ˆ xˆ (t) η (t) ∇ξ L t, x(t), ˆ xˆ (t) (1 + K), n
and lim ηn (t) = η(t)
n→+∞
and
∇x L t, x(t), ˆ xˆ (t) ηn (t) ∇x L t, x(t), ˆ xˆ (t) (1 + K)(b − a),
by the dominated convergence we obtain lim ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) dt = ∇ξ L t, x(t), ˆ xˆ (t) , η (t) dt n→+∞
I
and
lim
n→+∞
I
∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt =
I
∇x L t, x(t), ˆ xˆ (t) , η(t) dt.
I
It follows that ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt 0.
2
I
Even if the validity of the Euler–Lagrange equations already follows by the previous Corollary 3.1 and the DuBois–Reymond’s lemma [3], we give an alternative proof in Corollary 3.3. In the following theorem we prove an additional regularity result for the Lagrangian evaluated along the minimizer. ˆ xˆ (·)) Theorem 3.2. Under the same assumptions as in Theorem 2.1, the map ∇ξ L(·, x(·), ∞ is in L (I ). ˆ xˆ (·)) Proof. Using an iteration process, we shall prove that for every p in N ,∇ξ L(·, x(·), p p ˆ xˆ (·)) ∈ p1 Lp (I ). is in L (I ). Since the L are nested, this proves that ∇ξ L(·, x(·), At the same time, we shall prove that there exists a constant K > 0 such that, for every ˆ xˆ (·)) is in L∞ (I ). 1 p < +∞, ∇ξ Lp K, thus proving that ∇ξ L(·, x(·), ˆ xˆ (·)) is in L1 (I ). StartSuppose that: (1) From Theorem 2.1, we know that ∇ξ L(·, x(·), ˆ xˆ (·)) ∈ Lp (I ), to prove ing the iteration process, fix p ∈ N and suppose that ∇ξ L(·, x(·), p+1 ˆ xˆ (·)) ∈ L (I ). that ∇ξ L(·, x(·), (2) We can assume ∇ξ Lp = 0. Define Cn , An , kn as in point (1) of the proof of Theorem 2.1. For all n > 1, set p−1 (b − a) p vn = ∇ξ L t, x(t), ˆ xˆ (t) ∇ξ L t, x(t), ˆ xˆ (t) dt. p 2∇ξ Lp An
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365
Since m(C1 ) (b − a)/2 and
p vn (b − a) p 2∇ξ Lp
p ∇ξ L t, x(t), ˆ xˆ (t) (b − a)/2,
An
p Bn
p
p
⊂ C1 such that m(Bn ) = vn , so that p p p−1 2∇ξ Lp vn ∇ξ L t, x(t), ˆ xˆ (t) ∇ξ L t, x(t), ˆ xˆ (t) . dt = p (b − a) vn
there exists a set
p
An
Bn
Set
p p−1 θn (t) = −∇ξ L t, x(t), ˆ xˆ (t) ∇ξ L t, x(t), ˆ xˆ (t) χAn (t) p
2∇ξ Lp vnp χ p (t) + (b − a) vnp Bn
t p p p and θn (t) = a (θn ) (τ ) dτ . We obtain that θn ∞ 2 An ∇ξ L(t, x(t), ˆ xˆ (t))p dt. The p variations θn have bounded derivatives so we can apply Corollary 3.1 to obtain that p p ˆ xˆ (t) , θn (t) + ∇x L t, x(t), ˆ xˆ (t) , θn (t) dt = 0. ∇ξ L t, x(t), I
It follows that p+1 ∇ξ L t, x(t), ˆ xˆ (t) dt An p
2∇ξ Lp = (b − a) + I
An
p
Bn
p ∇x L t, x(t), ˆ xˆ (t) , θn (t) dt
+2 C˜
p p ∇ξ L t, x(t), ˆ xˆ (t) , vn vn dt
p ∇ξ L t, x(t), ˆ xˆ (t) dt
k1
An
p ∇ξ L t, x(t), ˆ xˆ (t) dt
∇x L t, x(t), ˆ xˆ (t) dt
I
p ∇ξ L t, x(t), ˆ xˆ (t) dt,
An
where C˜ is independent of n and p (suppose C˜ 1). The sequence of maps p+1 ∇ξ L t, x(t), ˆ xˆ (t) χ mn=2 An (t) m
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converges monotonically to ∇ξ L(t, x(t), ˆ xˆ (t))p+1 χ{ n>1 An } (t), and each integral p+1
ˆ xˆ (t)) χ{ mn=2 An } (t) dt is bounded by the same constant I ∇ξ L(t, x(t), C˜
∞ p p ∇ξ L t, x(t), ∇ξ L t, x(t), ˆ xˆ (t) dt = C˜ ˆ xˆ (t) dt. n=2 A
I \C1
n
Hence, by the monotone convergence theorem, p+1 p ∇ξ L t, x(t), ∇ξ L t, x(t), ˆ xˆ (t) dt C˜ ˆ xˆ (t) dt < +∞. I \C1
I \C1
ˆ xˆ (t)) < k1 , proving that ∇ξ L(·, x(·), ˆ xˆ (·)) ∈ Lp+1 (I ). Moreover, On C1 , ∇ξ L(t, x(t), we have also obtained that p+1 ∇ξ L t, x(t), ∇ξ L t, x(t), ˆ xˆ (t) dt C˜ p ˆ xˆ (t) dt, I \C1
so that
I \C1
p+1 ∇ξ L t, x(t), ˆ xˆ (t) dt
I \C1
where S = max{1, that, for all p ∈ N ,
I \C1
˜ CS,
∇ξ L(t, x(t), ˆ xˆ (t)) dt}. Setting T = max{1, m(C1 )} we have
∇ξ L(p+1)
1/(p+1)
(p+1) k1 m(C1 ) +
p+1 ∇ξ L t, x(t), ˆ xˆ (t) dt
1/(p+1)
I \C1
k1 m(C1 )1/(p+1) +
p+1 ∇ξ L t, x(t), ˆ xˆ (t) dt
1/(p+1)
I \C1
˜ = K. k1 T + CS
2
Corollary 3.3. Under the same conditions as in Theorem 2.1, for every variation η, η(a) = 0, η(b) = 0 and η ∈ L1 (I ), we have ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt = 0. I
As a consequence, t → ∇ξ L(t, x(t), ˆ xˆ (t)) is absolutely continuous. Proof. We shall prove that, for every η in AC(I ), such that η(a) = η(b) = 0, we have ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt 0. I
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367
(1) Fix η. Through the same steps as in point (1) of the proof of Theorem 2.1, for every n ∈ N we can define a closed set Cn such that on it η is continuous, xˆ is continuous with values in G, ∇ξ L is continuous in Cn × R s × G and limn→+∞ m(I \ Cn ) = 0. In particular, it follows that there are constants kn and cn > 0 such that, for all t ∈ Cn , ∇ξ L t, x(t), ˆ xˆ (t) < kn and η (t) < cn . Define vn , Bn , ηn and ηn as in the proof of Corollary 3.1. Since, for all t ∈ I , ηn (t) 1 + η (t), it follows that ηn 1 (b − a) + η 1 . Moreover, ηn ∞ ηn 1 (b − a) + η 1 . (2) As in point (3) of the proof of Theorem 2.1, there exists δn > 0 such that for ˆ xˆ (t))) < δn , we have ∇ξ L(t, x, ξ ) < (t, x, ξ ) ∈ Cn × R s × G, with d((t, x, ξ ), (t, x(t), kn + 1. (3) For |λ| < min{1/((b − a) + η 1 ), δn /((b − a) + η 1 ), δn /(cn + 1)}, consider the integrals 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t), ˆ xˆ (t) dt λ I 1 = L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) dt λ I 1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt + λ I 1 = L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) dt λ Cn
+
1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt. λ
I
For every t ∈ Cn , there exists ζλ (t) ∈ (0, λ) such that 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) ˆ + ληn (t), xˆ (t) λ = ∇ξ L t, x(t) ˆ + ληn (t), xˆ (t) + ζλ (t)ηn (t) , ηn (t) ˆ + ληn (t), xˆ (t) + ζλ (t)η (t) cn < (kn + 1)cn . ∇ξ L t, x(t) n
Hence, we can apply the dominated convergence theorem to obtain that 1 L t, x(t) ˆ + ληn (t), xˆ (t) + ληn (t) − L t, x(t) lim ˆ + ληn (t), xˆ (t) dt λ→0 λ I ˆ xˆ (t) , ηn (t) dt = ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) dt. = ∇ξ L t, x(t), Cn
I
(4) Following the point (5) of Theorem 2.1, we obtain that
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1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ
lim sup λ→0
I
∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt
I
and, since xˆ is a minimizer, we have 0 ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) I
+ lim sup λ→0
1 L t, x(t) ˆ + ληn (t), xˆ (t) − L t, x(t), ˆ xˆ (t) dt λ
I
∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) + ∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt.
I
(5) Finally we have lim η (t) = η (t) n→+∞ n and
∇ξ L t, x(t), ˆ xˆ (t) ηn (t) ∇ξ L ·, x(·), ˆ xˆ (·) ∞ 1 + η (t) ,
and lim ηn (t) = η(t)
n→+∞
and
∇x L t, x(t), ˆ xˆ (t) ηn (t) ∇x L t, x(t), ˆ xˆ (t) (b − a) + η 1 ,
so that, by dominated convergence, we obtain lim ∇ξ L t, x(t), ˆ xˆ (t) , ηn (t) dt = ∇ξ L t, x(t), ˆ xˆ (t) , η (t) dt n→+∞
I
and
lim
n→+∞ I
I
∇x L t, x(t), ˆ xˆ (t) , ηn (t) dt =
∇x L t, x(t), ˆ xˆ (t) , η(t) dt.
I
Hence, it follows that ∇ξ L t, x(t), ˆ xˆ (t) , η (t) + ∇x L t, x(t), ˆ xˆ (t) , η(t) dt 0. I
ˆ xˆ (t)) is classical (e.g., [2]). The absolute continuity of t → ∇ξ L(t, x(t),
2
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369
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