Second-Order Evolution Problems with Time-Dependent Maximal Monotone Operator and Applications

Abstract

We consider at first the existence and uniqueness of solution for a general second-order evolution inclusion in a separable Hilbert space of the form

$$\displaystyle 0\in \ddot u(t) + A(t) \dot u(t) + f(t, u(t)), \hskip 2pt t\in [0, T] $$

where A(t) is a time dependent with Lipschitz variation maximal monotone operator and the perturbation f(t, .) is boundedly Lipschitz. Several new results are presented in the sense that these second-order evolution inclusions deal with time-dependent maximal monotone operators by contrast with the classical case dealing with some special fixed operators. In particular, the existence and uniqueness of solution to

$$\displaystyle 0= \ddot u(t) + A(t) \dot u(t) + \nabla \varphi (u(t)), \hskip 2pt t\in [0, T] $$

where A(t) is a time dependent with Lipschitz variation single-valued maximal monotone operator and ∇φ is the gradient of a smooth Lipschitz function φ are stated. Some more general inclusion of the form

$$\displaystyle 0\in \ddot u(t) + A(t) \dot u(t) + \partial \Phi (u(t)), \hskip 2pt t\in [0, T] $$

where ∂ Φ(u(t)) denotes the subdifferential of a proper lower semicontinuous convex function Φ at the point u(t) is provided via a variational approach. Further results in second-order problems involving both absolutely continuous in variation maximal monotone operator and bounded in variation maximal monotone operator, A(t), with perturbation f : [0, T] × H × H are stated. Second- order evolution inclusion with perturbation f and Young measure control ν _t

$$\displaystyle \left \{ \begin {array}{lll} 0\in \ddot u_{x, y, \nu }(t) + A(t) \dot u_{x, y, \nu }(t) + f(t, u_{x, y, \nu }(t))+ \operatorname {{\mathrm {bar}}}(\nu _t), \hskip 2pt t \in [0, T] \\ u_{x, y, \nu }(0) = x, \dot u_{x, y, \nu } (0) =y \in D(A(0)) \end {array} \right . $$

where $\operatorname {{\mathrm {bar}}}(\nu _t)$ denotes the barycenter of the Young measure ν _t is considered, and applications to optimal control are presented. Some variational limit theorems related to convex sweeping process are provided.

Work partially supported by Fundação para a Ciência e a Tecnologia, UID/MAT/04561/2013.

JEL Classification: C61, C73

Mathematics Subject Classification (2010): 34A60, 34B15

Article type: Research Article

Received: March 15, 2018

Revised: March 30, 2018

1 Introduction

Let H be a separable Hilbert space. In this paper, we are mainly interested in the study of the perturbed evolution problem

$$\displaystyle \begin{aligned}0\in \ddot u(t) + A(t) \dot u(t) + \partial \Phi(u(t)), \hskip 2pt t\in [0, T]\end{aligned}$$

where ∂ Φ(u(t)) denotes the subdifferential of a proper lower semicontinuous convex function Φ at the point u(t), A(t) : D(A(t)) → 2^H is a maximal monotone operator in the Hilbert space H for every t ∈ [0, T], and the dependence t↦A(t) has Lipschitz variation, in the sense that there exists α ≥ 0 such that

$$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}(A(t), A(s)) \leq \alpha (t -s), \hskip 3pt \forall s, t \in [0, T]\;(s\leq t)\end{aligned}$$

$\operatorname {{\mathrm {dis}}}(., .)$ being the pseudo-distance between maximal monotone operators (m.m.o.) defined by A. A. Vladimirov [53] as

$$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}(A, B) = \sup \bigg\{ \frac{ \langle y-\hat y, \hat x-x \rangle} {1+||y||+ ||\hat y|| } : x \in D(A), y \in Ax, \hat x \in D(B), \hat y \in B\hat x \bigg\} \end{aligned}$$

for m.m.o. A and B with domains D(A) and D(B), respectively; the dependence t↦A(t) has absolutely continuous variation, in the sense that there exists β ∈ W ^{1, 1}([0, T]) such that

$$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}(A(t), A(s)) \leq |\beta(t) -\beta(s)|, \hskip 3pt \forall t, s \in [0, T],\end{aligned}$$

the dependence t↦A(t) has bounded variation in the sense that there exists a function r : [0, T] → [0, +∞[ which is continuous on [0, T[ and nondecreasing with r(T) < +∞ such that

$$\displaystyle \begin{aligned} \operatorname{{\mathrm{dis}}}(A(t), A(s))\leq dr(]s, t])=r(t)-r(s)\;\; {for}\;\;0\leq s\leq t\leq T \end{aligned}$$

The paper is organized as follows. Section 2 contains some definitions, notation and preliminary results. In Sect. 3, we recall and summarize (Theorem 3.2) the existence and uniqueness of solution for a general second-order evolution inclusion in a separable Hilbert space of the form

$$\displaystyle \begin{aligned} 0\in \ddot u(t) + A(t) \dot u(t) + f(t, u(t)), \hskip 2pt t\in [0, T] \end{aligned}$$

where A(t) is a time dependent with Lipschitz variation maximal monotone operator and the perturbation f(t, .) is dt-boundedly Lipschitz (short for dt-integrably Lipschitz on bounded sets). At this point, Theorem 3.2 and its corollaries are new results in the sense that these second-order evolution inclusions deal with time-dependent maximal monotone operators by contrast with the classical case dealing with some special fixed operators; cf. Attouch et al. [4], Paoli [43], and Schatzman [48]. In particular, the existence and uniqueness of solution, based on Corollary 3.2, to

$$\displaystyle \begin{aligned} 0= \ddot u(t) + A(t) \dot u(t) + \nabla \varphi(u(t)), \hskip 2pt t\in [0, T] \end{aligned}$$

where A(t) is a time dependent with Lipschitz variation single-valued maximal monotone operator and ∇φ is the gradient of a smooth Lipschitz function φ, have some importance in mechanics [40], which may require a more general evolution inclusion of the form

$$\displaystyle \begin{aligned} 0\in \ddot u(t) + A(t) \dot u(t) + \partial \Phi(u(t)), \hskip 2pt t\in [0, T] \end{aligned}$$

where ∂ Φ(u(t)) denotes the subdifferential of a proper lower semicontinuous convex function Φ at the point u(t).

We provide (Proposition 3.1) the existence of a generalized $W^{1, 1}_{BV}([0, T], H)$ solution to the second-order inclusion $0\in \ddot u(t) + A(t) \dot u(t) + \partial \Phi (u(t))$ which enjoys several regularity properties. The result is similar to that of Attouch et al. [4], Paoli [43], and Schatzman [48] with different hypotheses and a different method that is essentially based on Corollary 3.2 and the tools given in [22, 23, 27] involving the Young measures and biting convergence [9, 22, 32]. By $W^{1, 1}_{BV}([0, T], H)$, we denote the space of all absolutely continuous mappings y : [0, T] → H such that $\dot y$ are BV. Further results on second-order problems involving both the absolutely continuous in variation maximal monotone operators and the bounded in variation maximal monotone operator A(t) with perturbation f : [0, T] × H × H are stated.

Finally, in Sect. 4, we present several applications in optimal control in a new setting such as Bolza relaxation problem, dynamic programming principle, viscosity in evolution inclusion driven by a Lipschitz variation maximal monotone operator A(t) with Lipschitz perturbation f, and Young measure control ν _t

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \ddot u_{x, y, \nu}(t) + A(t) \dot u_{x, y, \nu}(t) + f(t, u_{x, y, \nu }(t))+ \operatorname{{\mathrm{bar}}}(\nu_t), \hskip 2pt t \in [0, T] \\ u_{x, y, \nu}(0) = x, \dot u_{x, y, \nu} (0) =y \in D(A(0)) \end{array} \right. \end{aligned}$$

where $\operatorname {{\mathrm {bar}}}(\nu _t)$ denotes the barycenter of the Young measure ν _t in the same vein as in Castaing-Marques-Raynaud de Fitte [25] dealing with the sweeping process. At this point, the above second-order evolution inclusion contains the evolution problem associated with the sweeping process by a closed convex Lipschitzian mapping $C : [0, T] \rightarrow \operatorname {{\mathrm {cc}}}(H)$

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll}0\in \ddot u(t) + N_{C(t)} (\dot u(t)) + f(t, u(t))+ \operatorname{{\mathrm{bar}}}(\nu_t), \hskip 2pt t \in [0, T]\\ u(0) = u_0, \dot u (0) = \dot u_0\in C(0) \end{array} \right. \end{aligned}$$

(where $\operatorname {{\mathrm {cc}}}(H)$ denotes the set of closed convex subsets of H) by taking A(t) = ∂ Ψ_C(t) and noting that if C(t) is a closed convex moving set in H, then the subdifferential of its indicator function is A(t) = ∂ Ψ_C(t) = N _C(t), the outward normal cone operator. Since for all s, t ∈ [0, T]

$$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}\big(A(t), A(s)\big) = \operatorname{\mathrm{\mathcal{H}}}\big(C(t), C(s)\big),\end{aligned}$$

where ${\mathcal {H}}$ denotes the Hausdorff distance; it follows that our study of these time-dependent maximal monotone operators includes as special cases some related results for evolution problems governed by sweeping process of the form

$$\displaystyle \begin{aligned}0\in \ddot u(t) +N_{C(t)} (\dot u(t)) + f(t, u(t)), \hskip 2pt t\in [0, T].\end{aligned}$$

Since now sweeping process has found applications in several fields in particular to economics [29, 31, 35], we present also some variational limit theorems related to convex sweeping process; see [1, 3, 34] and the references therein.

There is a vast literature on evolution inclusions driven by the sweeping process and the subdifferential operators. See [2, 5, 6, 10, 17, 18, 20, 21, 25, 26, 28, 30, 37, 39,40,41, 45, 47, 49,50,51,52] and the references therein. We refer to [9, 12, 13, 54] for the study of maximal monotone operators.

2 Notation and Preliminaries

In the whole paper, I := [0, T] (T > 0) is an interval of $\mathbb {R}$, and H is a real Hilbert space whose scalar product will be denoted by 〈⋅, ⋅〉 and the associated norm by ∥⋅∥. $\mathcal {L}([0, T])$ is the Lebesgue σ-algebra on [0, T], and $\mathcal {B}(H)$ is the σ-algebra of Borel subsets of H. We will denote by $\mathbf {\overline {B}}_H(x_0, r)$ the closed ball of H of center x ₀ and radius r > 0 and by $\mathbf {\overline {B}}_H$ its closed unit ball. C(I, H) denotes the Banach space of all continuous mappings u : I → H equipped with the norm $\|u\|{ }_C=\max \limits _{t\in I} \|u(t)\|$. For q ∈ [1, +∞[, $L^q_H([0, T], dt)$ is the space of (classes of) measurable u : [0, T] → H, with the norm $\|u(\cdot )\|{ }_q=(\int _0^T \|u(t)\|{ }^q dt)^{\frac {1}{q}}$, and $L^\infty _H([0, T], dt)$ is the space of (classes of) measurable essentially bounded u : [0, T] → H equipped with ∥.∥_∞.

If E is a Banach space and E ^∗ its topological dual, we denote by σ(E, E ^∗) the weak topology on E and by σ(E ^∗, E) the weak star topology on E ^∗. For any C ⊂ E, we denote by δ ^∗(., C) the support function of C, i.e.

$$\displaystyle \begin{aligned}\delta^*(x^*, C)=\sup_{x\in C}\langle x^*, x\rangle \hskip 2pt, \forall x^*\in E ^*.\end{aligned}$$

A set-valued map A : D(A) ⊂ H → 2^H is monotone if 〈y ₁ − y ₂, x ₁ − x ₂〉≥ 0 whenever x _i ∈ D(A) and y _i ∈ A(x _i), i = 1, 2. A monotone operator A is maximal if A is not contained properly in any other monotone operator, that is, for all λ > 0, R(I _H + λA) = H, with $R(A)=\bigcup \{A x,\;x\in D(A)\}$ the range of A and I _H the identity mapping of H. In the whole paper, I := [0, T] (T > 0) is an interval of $\mathbb {R}$, and H is a real Hilbert space whose scalar product will be denoted by 〈⋅, ⋅〉 and the associated norm by ∥⋅∥. Let A : D(A) ⊂ H → 2^H be a set-valued map. We say that A is monotone, if 〈y ₁ − y ₂, x ₁ − x ₂〉≥ 0 whenever $x_i \in \mathcal {D}(A)$ and y _i ∈ A(x _i), i = 1, 2. If 〈y ₁ − y ₂, x ₁ − x ₂〉 = 0 implies that x ₁ = x ₂, we say that A is strictly monotone. A monotone operator A is said to be maximal if A could not be contained properly in any other monotone operator.

If A is a maximal monotone operator, then, for every x ∈ D(A), A(x) is nonempty closed and convex. So the set A(x) contains an element of minimum norm (the projection of the origin on the set A(x)). This unique element is denoted by A ⁰(x). Therefore A ⁰(x) ∈ A(x) and ∥A ⁰(x)∥ =inf_{y ∈ A(x)}∥y∥. Moreover the set $\overline {D(A)}$ is convex.

For λ > 0, we define the following well-known operators:

$$\displaystyle \begin{aligned} \begin{array}{rcl} J_{\lambda}^A &\displaystyle =&\displaystyle (I + \lambda A)^{-1} \text{ (the resolvent of }A\text{),}\\ A_{\lambda} &\displaystyle =&\displaystyle \frac{1}{\lambda} (I- J_{\lambda}^A) \text{(the Yosida approximation of }A\text{).} \end{array} \end{aligned} $$

The operators $J_{\lambda }^A$ and A _λ are defined on all of H. For the terminology of maximal monotone operators and more details, we refer the reader to [9, 13], and [54].

Let A : D(A) ⊂ H → 2^H and B : D(B) ⊂ H → 2^H be two maximal monotone operators, and then we denote by $\operatorname {{\mathrm {dis}}}(A, B)$ the pseudo-distance between A and B defined by A. A. Vladimirov [53] as

$$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}(A,B)=\sup\bigg\{ \frac{\langle y-y', x'-x\rangle}{1+\| y\|+\| y'\|}:\;x\in D(A),\;y\in Ax,\;x'\in D(B),\;y'\in Bx'\bigg\}.\end{aligned} $$

Our main results are established under the following hypotheses on the operator A:

• (H1) The mapping t↦A(t) has Lipschitz variation, in the sense that there exists α ≥ 0 such that

  $$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}(A(t), A(s)) \leq \alpha (t -s), \hskip 3pt \forall s, t \in [0, T]\;(s\leq t).\end{aligned} $$

• (H2) There exists a nonnegative real number c such that

  $$\displaystyle \begin{aligned}\|A^0(t,x)\|\leq c(1+\| x\|)\;\; {for}\;\;t\in [0, T],\;x\in D(A(t)).\end{aligned}$$

We recall some elementary lemmas, and we refer to [38] for the proofs.

Lemma 2.1

Let A and B be maximal monotone operators. Then

• (1) $\operatorname {{\mathrm {dis}}}(A, B)\in [0,+\infty ]$ , $\operatorname {{\mathrm {dis}}}(A, B)= \operatorname {{\mathrm {dis}}}(B,A)$ and $\operatorname {{\mathrm {dis}}}(A,B)=0$ iff A = B.

• (2) $\|x-Proj(x, \overline {D(B)}\|\leq \operatorname {{\mathrm {dis}}}(A, B)$ for $x\in \overline {D(A)}$.

• (3) ${\mathcal {H}}(D(A), D(B))\leq \operatorname {{\mathrm {dis}}}(A, B)$.

Lemma 2.2

Let A be a maximal monotone operator. If x, y ∈ H are such that

$$\displaystyle \begin{aligned}\langle A^0(z)-y, z-x\rangle\geq 0\;\;\forall z\in D(A),\end{aligned}$$

then x ∈ D(A) and y ∈ A(x).

Lemma 2.3

Let $A_n\ (n\in \mathbb {N})$ and A be maximal monotone operators such that $\operatorname {{\mathrm {dis}}}(A_n, A)\to 0$ . Suppose also that x _n ∈ D(A _n) with x _n → x and y _n ∈ A _n(x _n) with y _n → y weakly for some x, y ∈ H. Then x ∈ D(A) and y ∈ A(x).

Lemma 2.4

Let A and B be maximal monotone operators. Then

• (1) for λ > 0 and x ∈ D(A)

  $$\displaystyle \begin{aligned}\|x-J_{\lambda}^{B}(x)\| \leq \lambda\|A^0(x)\|+\operatorname{{\mathrm{dis}}}(A,B)+\sqrt{\lambda\big(1+\|A^0(x)\|\big)\operatorname{{\mathrm{dis}}}(A, B)}.\end{aligned}$$

• (2) For λ > 0 and x, x′∈ H

  $$\displaystyle \begin{aligned}\|J_{\lambda}^{A}(x)-J_{\lambda}^{B}(x')\|{}^2\leq \|x-x'\|{}^2+2\lambda\big(1+\|A_{\lambda}(x)\|+\|B_{\lambda}(x')\|\big)\operatorname{{\mathrm{dis}}}(A,B).\end{aligned}$$

• (3) For λ > 0 and x, x′∈ H

  $$\displaystyle \begin{aligned}\|A_{\lambda}(x)-B_{\lambda}(x')\|{}^2\leq \frac{1}{\lambda^2}\|x-x'\|{}^2+\frac{2}{\lambda}\big(1+\|A_{\lambda}(x)\|+\|B_{\lambda}(x')\|\big)\operatorname{{\mathrm{dis}}}(A, B).\end{aligned}$$

3 Second-Order Evolution Problems Involving Time-Dependent Maximal Monotone Operators

In the sequel, H is a separable Hilbert space. For the sake of completeness, we summarize and state the following result. We say that a function f = f(t, x) is dt-boundedly Lipschitz (short for dt-integrably Lipschitz on bounded sets) if, for every R > 0, there is a nonnegative dt-integrable function $\lambda _R \in L^1([0, T], \mathbb {R}; dt)$ such that, for all t ∈ [0, T]

$$\displaystyle \begin{aligned}\|f(t,x)-f(t,y)\|\leq \lambda_R (t) ||x-y||, \hskip 3pt \forall x, y\in \mathbf{\overline{B}}(0,R).\end{aligned}$$

Theorem 3.1

Let for every t ∈ [0, T], A(t) : D(A(t)) ⊂ H → 2^H be a maximal monotone operator satisfying

• (H1) there exists a real constant α ≥ 0 such that

  $$\displaystyle \begin{aligned} \operatorname{{\mathrm{dis}}}(A(t), A(s))\leq \alpha(t-s)\;\; {for}\;\;0\leq s\leq t\leq T. \end{aligned}$$

• (H2) there exists a nonnegative real number c such that

  $$\displaystyle \begin{aligned}\|A^0(t,x)\|\leq c(1+\| x\|), t\in [0, T], x \in D(A(t))\end{aligned}$$

  Let f : [0, T] × H → H satisfying the linear growth condition

• (H3) there exists a nonnegative real number M such that

  $$\displaystyle \begin{aligned}\|f(t,x)\|\leq M(1+\| x\|)\;\; {for}\;\;t\in [0, T],\;x\in H.\end{aligned}$$

  and assume that f(., x) is dt-integrable for every x ∈ H. Assume also that f is dt-boundedly Lipschitz, as above.

Then for all u ₀ ∈ D(A(0)), the problem

$$\displaystyle \begin{aligned}- \frac {du} {dt} (t) \in A(t) u(t) + f(t, u(t)) \hskip 3pt dt-\mathrm{a.e.}\, t \in [0, T],\;\;u(0)=u_0\end{aligned}$$

has a unique Lipschitz solution with the property: $||u(t)-u(\tau )|| \leq K \max \{1, \alpha \}|t-\tau |$ for all t, τ ∈ [0, T] for some constant K ∈ ]0, ∞[.

Proof

See [7, Theorem 3.1 and Theorem 3.3].

Theorem 3.2

Let for every t ∈ [0, T], A(t) : D(A(t)) ⊂ H → 2^H be a maximal monotone operator satisfying

• (H1) there exists a real constant α ≥ 0 such that

  $$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}(A(t), A(s))\leq \alpha(t-s)\;\; {for}\;\;0\leq s\leq t\leq T.\end{aligned}$$

• (H2) there exists a nonnegative real number c such that

  $$\displaystyle \begin{aligned}\|A^0(t,x)\|\leq c(1+\| x\|), t\in [0, T], x \in D(A(t))\end{aligned}$$

  Let f : [0, T] × H → H satisfying the linear growth condition:

• (H3) there exists a nonnegative real number M such that

  $$\displaystyle \begin{aligned}\|f(t,x)\|\leq M(1+\| x\|)\;\; {for}\;\;t\in [0, T],\;x\in H.\end{aligned}$$

  and assume that f(., x) is dt-integrable for every x ∈ H. Assume also that f is dt-boundedly Lipschitz.

Then the second-order evolution inclusion

$$\displaystyle \begin{aligned}({\mathcal S}_1) \left\{ \begin{array}{lll}0\in \ddot u(t) + A(t) \dot u(t) + f(t, u(t)), \hskip 2pt t \in [0, T]\\ u(0) = u_0, \dot u (0) = \dot u_0\in D(A(0)) \end{array} \right. \end{aligned}$$

admits a unique solution $u \in W^{2, \infty }_H ([0, T], dt)$.

Proof

The proof is a careful application of Theorem 3.1. In the new variables $X= (x, \dot x)$, let us set for all t ∈ I

$$\displaystyle \begin{aligned}B(t) X =\{0\}\times A(t) \dot x,\; g(t, X) = (- \dot x, f(t, x)).\end{aligned}$$

For any u ∈ W ^{2, ∞}(I, H;dt), define $X(t)=(u(t), \frac {du}{dt}(t))$ and $\dot X(t)=\frac {dX}{dt}(t)$. Then the evolution inclusion $({\mathcal S}_1)$ can be written as a first-order evolution inclusion associated with the Lipschitz maximal monotone operator B(t) and the locally Lipschitz perturbation g:

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0 \in \dot X(t) + B(t) X(t) + g(t, X(t)), \hskip 2pt t \in [0, T]\\ X (0)= (u_0, \dot u_0) \in H\times D(A(0)). \end{array} \right. \end{aligned}$$

So the existence and uniqueness solution to the second-order evolution inclusion under consideration follows from Theorem 3.1.

There are some useful corollaries to Theorem 3.2.

Corollary 3.1

Assume that for every t ∈ [0, T], A(t) : H → H is a single-valued maximal monotone operator satisfying (H1) and (H2). Let f : [0, T] × H → H be as in Theorem 3.2 . Then the second-order evolution equation

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll}0= \ddot u(t) + A(t) \dot u(t) + f(t, u(t)), \hskip 2pt t \in [0, T]\\ u(0) = u_0, \dot u (0) =\dot u_0 \end{array} \right. \end{aligned}$$

admits a unique solution $u \in W^{2, \infty }_H ([0, T])$.

Corollary 3.2

Assume that for every t ∈ [0, T], A(t) : H → H is a single-valued maximal monotone operator satisfying (H1) and (H2). Assume further that A(t) satisfies

• (i) (t, x)↦A(t)x is a Caratheodory mapping, that is, t↦A(t)x is Lebesgue measurable on [0, T] for each fixed x ∈ H, and x↦A(t)x is continuous on H for each fixed t ∈ [0, T],

• (ii) 〈A(t)x, x〉≥ γ||x||² , for all (t, x) ∈ [0, T] × H, for some γ > 0.

Let $\varphi \in C ^1(H,\mathbb {R})$ be Lipschitz and such that ∇φ is locally Lipschitz. Then the evolution equation

$$\displaystyle \begin{aligned} ({\mathcal S}_2) \left\{ \begin{array}{lll} 0= \ddot u(t) + A(t) \dot u(t) +\nabla \varphi (u(t)) , \hskip 2pt t \in [0, T]\\ u(0) = u_0, \dot u (0) = \dot u_0 \end{array} \right. \end{aligned}$$

admits a unique solution u ∈ W ^{2, ∞}([0, T], H;dt); moreover, u satisfies the energy estimate

$$\displaystyle \begin{aligned} \varphi(u(t)) -\frac{ 1}{2} ||\dot u(t)||{}^2 \leq \varphi(u(0)) -\frac{ 1}{2} ||\dot u(t)||{}^2 -\gamma \int_0 ^t ||\dot u(s)||{}^2 ds, \hskip 2pt t \in [0, T]. \end{aligned}$$

Proof

Existence and uniqueness of solution follows from Theorem 3.2 or Corollary 3.1. The energy estimate is quite standard. Multiplying the equation by $\dot u(t)$ and applying the usual chain rule formula gives for all t ∈ [0, T]

$$\displaystyle \begin{aligned}\frac{d} {dt} \bigg(\varphi (u(t)) + \frac{1} {2} ||\dot u(t)||{}^2\bigg)= - \big\langle A(t) \dot u(t) , \dot u (t) \big\rangle.\end{aligned}$$

By (i) and (ii) and by integrating on [0, t], we get the required inequality

$$\displaystyle \begin{aligned} \varphi (u(t)) + \frac{1} {2} ||\dot u(t)||{}^2 &= \varphi (u(0)) + \frac{1} {2} ||\dot u(0)||{}^2- \int_0 ^t \big\langle A(s) \dot u(s) , \dot u (s) \big\rangle ds\\ &\leq \varphi (u(0)) + \frac{1} {2} ||\dot u(0)||{}^2-\gamma \int_0 ^t ||\dot u(s)||{}^2 ds, \hskip 2pt t \in [0, T], \end{aligned} $$

which completes the proof.

It is worth mentioning that the uniqueness of the solution to the equation $(\mathcal {S}_1)$ is quite important in applications, such as models in mechanics, since it contains the classical inclusion of the form

$$\displaystyle \begin{aligned}0\in \ddot u(t) + \partial \Phi(\dot u(t)) + \nabla g(u(t))\end{aligned}$$

where ∂ Φ is the subdifferential of the proper lower semicontinuous convex function Φ and g is of class C ¹ and ∇g is Lipschitz continuous on bounded sets. We also note that the uniqueness of the solution to the equation $(\mathcal {S}_2)$ and its energy estimate allow to recover a classical result in the literature dealing with finite dimensional space H and A(t) = γI _H, t ∈ [0, T], where I _H is the identity mapping in H. See Attouch et al. [4]. The energy estimate for the solution of

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0= \ddot u(t) + \gamma \dot u(t) +\nabla \varphi (u(t)) , \hskip 2pt t \in I\\ u(0) = u_0 , \dot u (0) = \dot u_0 \end{array} \right. \end{aligned}$$

is then

$$\displaystyle \begin{aligned} \varphi (u(t)) + \frac{1} {2} ||\dot u(t)||{}^2= \varphi (u_0) + \frac{1} {2} ||\dot u_0||{}^2-\gamma \int_0 ^t ||\dot u(s)||{}^2 ds. \end{aligned}$$

Actually the dynamical system $({\mathcal S}_1)$ given in Theorem 3.2 has been intensively studied by many authors in particular cases. See Attouch et al. [4] dealing with the inclusion

$$\displaystyle \begin{aligned} 0\in \ddot u(t) + \gamma \dot u(t) + \partial \varphi (u(t)) \end{aligned}$$

and Paoli [43] and Schatzman [48] dealing with the second-order dynamical systems of the form

$$\displaystyle \begin{aligned} 0\in \ddot u(t) + \partial \varphi (u(t)) \end{aligned}$$

and

$$\displaystyle \begin{aligned} 0\in \ddot u(t) + A\dot u(t) + \partial \varphi (u(t)) \end{aligned}$$

where A is a positive autoadjoint operator. The existence and uniqueness of solutions in $({\mathcal S}_2)$ are of some importance since they allow to obtain the existence of at least a $W^{1, 1}_{BV}([0, T], H)$ solution with conservation of energy (see Proposition 3.1 below) for a second-order evolution inclusion of the form

$$\displaystyle \begin{aligned}({\mathcal S}_3) \left\{ \begin{array}{lll}0\in \ddot u(t) + A(t) \dot u(t) + \partial \Phi(u(t), \hskip 2pt t \in I\\ u(0) = u_0 \in \operatorname{{\mathrm{dom}}} \Phi , \dot u (0) =\dot u_0\in D(A(0)) \end{array} \right. \end{aligned}$$

where ∂ Φ is the subdifferential of a proper convex lower semicontinuous function; the energy estimate is given by

$$\displaystyle \begin{aligned} \Phi (u(t)) + \frac{1} {2} ||\dot u(t)||{}^2= \Phi (u(0)) + \frac{1} {2} ||\dot u(0)||{}^2- \int_0 ^t \big\langle A(s) \dot u(s), \dot u (s) \big\rangle ds. \end{aligned}$$

Taking into account these considerations, we will provide the existence of a generalized solution to the second-order inclusion of the form

$$\displaystyle \begin{aligned} 0\in \ddot u(t) + A(t) \dot u(t) + \partial \phi(u(t)) \end{aligned}$$

which enjoy several regular properties. The result is similar to that of Attouch et al. [4], Paoli [43], and Schatzman [48] with different hypotheses and a different method that is essentially based on Corollary 3.2 and the tools given in [22, 23, 27] involving the Young measures [9, 32] and biting convergence.

Let us recall a useful Gronwall-type lemma [21].

Lemma 3.5 (A Gronwall-like inequality.)

Let p, q, r : [0, T] → [0, ∞[ be three nonnegative Lebesgue integrable functions such that for almost all t ∈ [0, T]

$$\displaystyle \begin{aligned} r(t) \leq p(t) + q(t)\int_0 ^t r(s)\,ds. \end{aligned}$$

Then

$$\displaystyle \begin{aligned} r(t) \leq p(t) + q(t) \int_0 ^t \left[p(s)\,\mathrm{exp}\left(\int_s^t q(\tau) \,d\tau\right)\right]\,ds \end{aligned}$$

for all t ∈ [0, T].

Proposition 3.1

Assume that $H= \mathbb {R} ^d$ and that, for every t ∈ [0, T], A(t) : H → H is single-valued maximal monotone satisfying

• (H1) there exists α > 0 such that

  $$\displaystyle \begin{aligned} \operatorname{{\mathrm{dis}}}(A(t), A(s))\leq \alpha(t-s)\;\; {for}\;\;0\leq s\leq t\leq T, \end{aligned}$$

• (H2) there exists a nonnegative real number c such that

  $$\displaystyle \begin{aligned}\|A(t,x)\|\leq c(1+\| x\|)\;\; {for}\;\;t\in [0, T],\;x\in H.\end{aligned}$$

Assume further that A(t) satisfies

• A-1. (t, x) → A(t)x is a Caratheodory mapping, that is, t↦A(t)x is Lebesgue-measurable on [0, T] for each fixed x ∈ H, and x↦A(t)x is continuous on H for each fixed t ∈ [0, T],

• A-2. 〈A(t)x, x〉≥ γ||x||² , for all (t, x) ∈ [0, T] × H, for some γ > 0.

Let $n\in \mathbb {N}$ and $\varphi _n : H\rightarrow \mathbb {R}^+$ be a C ¹ , convex, Lipschitz function and such that ∇φ _n is locally Lipschitz, and let φ _∞ be a nonnegative l.s.c proper function defined on H with φ _n(x) ≤ φ _∞(x), ∀x ∈ H. For each $n\in \mathbb {N}$ , let u ⁿ be the unique $W^{2,\infty }_{ H}([0, T])$ solution to the problem

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0= \ddot u^n (t)+A(t) \dot u^n (t) + \nabla \varphi_n(u^n(t)), t \in [0, T] \\ u ^n(0) = u^n_0, \hskip 2pt \dot u^n(0) = \dot u ^n_0 \end{array} \right. \end{aligned}$$

Assume that

• (i) φ _n epiconverges to φ _∞ ,

• (ii) $u^n(0) \rightarrow u^\infty _0 \in \operatorname {{\mathrm {dom}}} \varphi _\infty$ and $\lim _n \varphi _n (u^n (0)) = \varphi _\infty (u^\infty _0)$ ,

• (iii) $\sup _{v \in {\overline B}_{L^\infty _{H}([0, T])}}\int _0^T\varphi _\infty (v(t)) dt < +\infty$ , where ${\overline B}_{L^\infty _{H}([0, T])}$ is the closed unit ball in $L^\infty _{H}([0, T])$.

• (a) Then up to extracted subsequences, (u ⁿ) converges uniformly to a $W^{1, 1}_{BV} ([0, T], \mathbb {R}^d)$ -function u ^∞ with $u^\infty (0) \in \operatorname {{\mathrm {dom}}} \varphi _\infty$ , and $(\dot u ^n)$ pointwisely converges to a BV function v ^∞ with $v^\infty = \dot u ^\infty$ , and $(\ddot u ^n)$ biting converges to a function $\zeta ^\infty \in L^1_{\mathbb {R}^d}([0, T])$ so that the limit function $u ^\infty , \dot u^\infty$ and the biting limit ζ ^∞ satisfy the variational inclusion

  $$\displaystyle \begin{aligned}-A(.) \dot u^ \infty - \zeta^\infty \in \partial I_{\varphi_\infty}(u^\infty)\end{aligned}$$

  where $\partial I_{\varphi _\infty }$ denotes the subdifferential of the convex lower semicontinuous integral functional $I_{\varphi _\infty }$ defined on $L^\infty _{\mathbb {R}^d}([0, T])$

  $$\displaystyle \begin{aligned}I_{\varphi_\infty}(u):= \int_0^T \varphi_\infty( u(t)) \,dt, \hskip 4pt \forall u \in L^\infty_{\mathbb{R}^d}([0, T]).\end{aligned}$$

• (b) $(\ddot u^n)$ weakly converges to a vector measure $m \in {\mathcal M}^b_H([0, T])$ so that the limit functions u ^∞(.) and the limit measure m satisfy the following variational inequality:

  $$\displaystyle \begin{aligned} \int_0^T \varphi_\infty( v(t)) \,dt \geq & \int_0^T \varphi_\infty( u ^\infty(t)) \,dt + \int_0 ^T \langle -A(t) \dot u^ \infty(t) ,v(t)- u ^\infty(t)\rangle \,dt \\ &+ \langle -m, v-u ^\infty \rangle_{ ({\mathcal M}^b_{\mathbb{R} ^d} ([0, T]), {\mathcal C}_E([0, T])) }. \end{aligned} $$

• (c) Furthermore $\lim _n \int _0^T \varphi _n( u^n(t)) dt = \int _0^T \varphi _\infty ( u ^\infty (t)) dt$ . Subsequently the energy estimate

  $$\displaystyle \begin{aligned}\varphi_\infty(u^\infty(t)) +\frac{1}{2} ||\dot u^\infty(t)||{}^2 = \varphi_\infty(u^\infty_0) +\frac{1}{2} ||\dot u^\infty_0||{}^2 +\int_0^t \langle -A(s) \dot u^ \infty(s) , u ^\infty (s)\rangle ds \end{aligned}$$

  holds a.e.

• (d) There is a filter ${\mathcal U}$ finer than the Fréchet filter $l \in L^\infty _{\mathbb {R}^d}([0, T])'$ such that

  $$\displaystyle \begin{aligned}{\mathcal U}-\lim_n [-A(.) \dot u^n-\ddot u ^n] = l \in L^\infty_{\mathbb{R}^d}([0, T])'_{\mathrm{weak}}\end{aligned}$$

where $L^\infty _{\mathbb {R}^d}([0, T])'_{\mathrm {weak}}$ is the second dual of $L^1_{\mathbb {R}^d} ([0, T])$ endowed with the topology $\sigma (L^\infty _{\mathbb {R}^d}([0, T])', L^\infty _{\mathbb {R}^d}([0, T]))$ , and $\mathbf {n} \in {\mathcal C}_{\mathbb {R}^d} ([0, T])'_{\mathrm {weak}}$ such that

$$\displaystyle \begin{aligned}\lim_n [-A(.) \dot u^n-\ddot u ^n] = \mathbf{n} \in {\mathcal C}_{\mathbb{R}^d} ([0, T])'_{\mathrm{weak}}\end{aligned}$$

where ${\mathcal C}_{\mathbb {R}^d} ([0, T])'_{\mathrm {weak}}$ denotes the space ${\mathcal C}_{\mathbb {R}^d} ([0, T])'$ endowed with the weak topology $\sigma ({\mathcal C}_{\mathbb {R}^d} ([0, T])', {\mathcal C}_{\mathbb {R}^d} ([0, T]))$ . Let l _a be the density of the absolutely continuous part l _a of l in the decomposition l = l _a + l _s in absolutely continuous part l _a and singular part l _s . Then

$$\displaystyle \begin{aligned}l_a(f) = \int_0 ^T \langle f(t), -A(t) \dot u^\infty(t) -\zeta^\infty(t)\rangle dt\end{aligned}$$

for all $f \in L^\infty _{\mathbb {R}^d}([0, T])$ so that

$$\displaystyle \begin{aligned}I_{\varphi_\infty}^*(l) = I_{\varphi_\infty^* }(- A(.) \dot u^\infty-\zeta^\infty) +\delta^* (l_s ,\operatorname{{\mathrm{dom}}} I_{\varphi_\infty} )\end{aligned}$$

where $\varphi _\infty ^*$ is the conjugate of φ _∞ , $I_{\varphi _\infty ^* }$ the integral functional defined on $L^1_{\mathbb {R}^d} ([0, T])$ associated with $\varphi _\infty ^*$ , $I_{\varphi _\infty }^*$ the conjugate of the integral functional $I_{\varphi _\infty }$ , $\operatorname {{\mathrm {dom}}} I_{\varphi _\infty } := \{ u \in L^\infty _{\mathbb {R}^d}([0, 1]) : I_{\varphi _\infty }(u) < \infty \}$ , and

$$\displaystyle \begin{aligned}\langle \mathbf{n}, f \rangle = \int_0 ^T \langle - A(t) \dot u^\infty(t) -\zeta^\infty(t), f(t)\rangle dt + \langle {\mathbf{n}}_s, f \rangle, \quad \forall f \in {\mathcal C}_{\mathbb{R}^d}([0, T]). \end{aligned}$$

with 〈n _s, f〉 = l _s(f), $\forall f \in {\mathcal C}_{\mathbb {R}^d}([0, T])$ . Further n belongs to the subdifferential $\partial J_{\varphi _\infty }(u^\infty )$ of the convex lower semicontinuous integral functional $J_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbb {R}^d}([0, T])$

$$\displaystyle \begin{aligned}J_{\varphi_\infty}(u):= \int_0^T \varphi_\infty( u(t)) \,dt, \hskip 4pt \forall u \in {\mathcal C}_{\mathbb{R}^d}([0, T]).\end{aligned}$$

Consequently the density $- A(.) \dot u^\infty -\zeta ^\infty$ of the absolutely continuous part n _a

$$\displaystyle \begin{aligned}{\mathbf{n}}_a (f):= \int_0 ^T \langle - A(t) \dot u^\infty(t) -\zeta^\infty(t), f(t)\rangle dt, \quad \forall f\in {\mathcal C}_{\mathbb{R}^d}([0, T])\end{aligned}$$

satisfies the inclusion

$$\displaystyle \begin{aligned}- A(t) \dot u ^\infty(t) - \zeta^\infty(t) \in \partial \varphi_\infty(u^\infty(t)), \quad \mathrm{a.e.}\end{aligned}$$

and for any nonnegative measure θ on [0, T] with respect to which n _s is absolutely continuous

$$\displaystyle \begin{aligned}\int_0^T r_{\varphi_\infty^*}\left(\frac{d{\mathbf{n}}_s}{d\theta}(t)\right) d\theta(t) = \int_0^T \left\langle u^\infty(t), \frac{d{\mathbf{n}}_s}{d \theta}(t)\right\rangle d\theta(t)\end{aligned}$$

where $r_{\varphi _\infty ^*}$ denotes the recession function of $\varphi _\infty ^*$.

Proof

The proof is long and based on the existence and uniqueness of $W_{H}^{2 ,\infty }([0, T])$ solution to the approximating equation (cf. Corollary 3.2)

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0= \ddot u^n (t)+A(t) \dot u^n (t) + \nabla \varphi_n(u^n(t)), t \in [0, T] \\ u ^n(0) = u^n_0, \hskip 2pt \dot u^n(0) = \dot u ^n_0 \end{array} \right. \end{aligned}$$

and the techniques developed in [22, 23, 27]. Nevertheless we will produce the proof with full details, since the techniques employed can be applied to further related results.

Step 1. Multiplying scalarly the equation

$$\displaystyle \begin{aligned} -A(t) \dot u^n (t)-\ddot u^n(t) = \nabla \varphi_n(u^n(t)) \end{aligned}$$

by $\dot u^n(t)$ and applying the chain rule theorem [42, Theorem 2] yields

$$\displaystyle \begin{aligned}-\langle \dot u^n(t), A(t) \dot u^n (t)\rangle -\langle \dot u^n(t), \ddot u^n(t) \rangle = \frac{d}{dt} [\varphi_n(u_n(t))],\end{aligned}$$

that is,

$$\displaystyle \begin{aligned}-\langle \dot u^n(t), A(t) \dot u^n (t) \rangle = \frac{d}{dt}\left[ \varphi_n(u^n(t)) +\frac{1}{2} ||\dot u^n(t)||{}^2\right].\end{aligned}$$

By integrating on [0, t] this equality and using the condition (ii), we get

$$\displaystyle \begin{aligned} \varphi_n(u^n(t)) +\frac{1}{2} ||\dot u^n(t)||{}^2 &= \varphi_n(u^n(0)) +\frac{1}{2} ||\dot u^n(0)||{}^2 - \int_0^t \langle \dot u^n(s), A(s) \dot u^n (s)\rangle ds\\ &\leq \varphi_n(u^n(0)) +\frac{1}{2} ||\dot u^n(0)||{}^2+\gamma \int_0^t || \dot u^n(s||{}^2 ds. \end{aligned} $$

Then, from our assumption, φ _n(u ⁿ(0)) ≤ positive constant < +∞ and $\frac {1}{2} ||\dot u^n(0)||{ }^2 \leq \text{ positive constant }< +\infty$ so that

$$\displaystyle \begin{aligned}\varphi_n(u^n(t)) +\frac{1}{2} ||\dot u^n(t)||{}^2 \leq p+ \gamma \int_0^t || \dot u^n(s||{}^2 ds, \hskip 2pt t \in [0, T]\end{aligned}$$

where p is a generic positive constant. So by the preceding estimate and the Gronwall inequality [21, Lemma 3.1] , it is immediate that

$$\displaystyle \begin{aligned} \sup_{n \geq 1} \sup_{t \in [0, T] } ||\dot u^n(t)|| < +\infty \quad \mathrm{and} \quad \sup_{n \geq 1} \sup_{t \in [0, T] }\varphi_n(u^n(t)) < +\infty. \end{aligned} $$

(1)

Step 2. Estimation of $||\ddot u^n(.)||$. For simplicity, let us set $z^n(t)=-A(t) \dot u^n (t) - \ddot u^n (t), \forall t \in [0, T]$. As

$$\displaystyle \begin{aligned}z^n(t) :=-A(t) \dot u^n (t) -\ddot u^n (t) =\nabla \varphi_n(u^n(t))\end{aligned}$$

by the subdifferential inequality for convex lower semicontinuous functions, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \varphi_n (x) \geq \varphi_n (u^n(t)) + \langle x-u^n(t), z^n(t) \rangle\vspace{-4pt} \end{array} \end{aligned} $$

for all $x \in \mathbb {R}^d$. Now let $v\in {\overline B}_{L^\infty _{\mathbb {R}^d}([0, T])}$, the closed unit ball of $L^\infty _{\mathbb {R}^d}[0, T])$. By taking x = v(t) in the preceding inequality, we get

$$\displaystyle \begin{aligned} \begin{array}{rcl} \varphi_n (v(t)) \geq \varphi_n (u^n(t)) + \langle v(t)-u^n(t), z^n(t) \rangle.\vspace{-4pt} \end{array} \end{aligned} $$

Integrating the preceding inequality gives

$$\displaystyle \begin{aligned} \int_0^T \langle v(t)-u^n(t), z^n(t) \rangle dt \leq \int_0^T \varphi_n (v(t)) dt -\int_0^T \varphi_n (u^n(t)) dt.\end{aligned} $$

Whence follows

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} &\displaystyle &\displaystyle \int_0^T \langle v(t) , z^n(t) \rangle dt\\ &\displaystyle &\displaystyle \qquad \qquad \quad \leq \int_0^T \varphi_n (v(t)) dt -\int_0^T \varphi_n (u^n(t)) dt+\int_0^1 \langle u^n(t), z^n(t) \rangle dt.\vspace{-4pt} \end{array} \end{aligned} $$

(2)

We compute the last integral in the preceding inequality. By integration and taking account of (1), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} &\displaystyle &\displaystyle \int_0^T \langle u^n(t), z^n(t) \rangle dt\\ &\displaystyle &\displaystyle \quad = \int_0^T \langle u^n(t), -A(t) \dot u^n (t)- \ddot u^n(t) \rangle dt\\ &\displaystyle &\displaystyle \quad =-[\langle u^n(t),\dot u^n(t)]_0^T + \int_0^T \langle \dot u^n(t), \dot u^n(t) \rangle dt - \int_0^T \langle u^n(t), A(t) \dot u^n (t) \rangle dt \\ &\displaystyle &\displaystyle \quad = -\langle u^n(T), \dot u^n(T)\rangle+ \langle u^n(0), \dot u^n(0) \rangle\\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad + \int_0^T ||\dot u^n(t)||{}^2 dt - \int_0^T \langle u^n(t), A(t) \dot u^n (t) \rangle dt.\vspace{-4pt} \end{array} \end{aligned} $$

(3)

As $||A(t) \dot u^n (t) || \leq c( 1+||\dot u_n(t) ||)$ by (H ₂), so that by (1) it is immediate that $\int _0^T \langle u^n(t), A(t) \dot u^n (t) \rangle dt$ is uniformly bounded so that by (1), (2), and (3), we get

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} &\displaystyle &\displaystyle \int_0^T \langle v(t) , z^n(t) \rangle dt \leq \int_0^T \varphi_n ( v(t)) dt +L\\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \quad \leq \sup_{v \in {\overline B}_{L^\infty_{ \mathbb{R}^d}([0, T])}} \int_0^T \varphi_\infty (v(t)) dt+L < \infty \end{array} \end{aligned} $$

(4)

for all $v \in {\overline B}_{L^\infty _{ \mathbb {R}^d}([0, T])}$. Here L is a generic positive constant independent of $n\in \mathbb {N}$. By (4), we conclude that $(z ^n = -A(.) \dot u^n- \ddot u^n)$ is bounded in $L^1_{\mathbb {R} ^d}([0, T])$, and then so is $(\ddot u^n)$. It turns out that the sequence $(\dot u^n)$ of absolutely continuous functions is uniformly bounded by (1) and bounded in variation and by Helly’s theorem; we may assume that $(\dot u^n)$ pointwisely converges to a BV function $v^ \infty : [0, T] \rightarrow \mathbb {R} ^d$ and the sequence (u ⁿ) converges uniformly to an absolutely continuous function u ^∞ with $\dot u^\infty = v ^\infty$ a.e. At this point, it is clear that $A(t) \dot u^n(t) \rightarrow A(t) v ^\infty (t)$ so that $A(t) \dot u^n(t) \rightarrow A(t) \dot u ^\infty (t)$ a.e. and $A(.) \dot u^n(.)$ converges in $L^1_{\mathbb {R} ^d}([0, T])$ to $A(.) \dot u ^\infty (.)$, using (1) and the dominated convergence theorem.

Step 3. Young measure limit and biting limit of $\ddot u_n$. As $(\ddot u_n)$ is bounded in $L^1_{\mathbb {R} ^d}([0, T])$, we may assume that $(\ddot u^n)$ stably converges to a Young measure $\nu \in \mathcal Y([0, T]); \mathbb {R}^d)$ with $\operatorname {{\mathrm {bar}}}( \nu ): t \mapsto \operatorname {{\mathrm {bar}}}( \nu _t) \in L ^1_{\mathbb {R}^d}([0, T])$ (here $\operatorname {{\mathrm {bar}}}( \nu _t)$ denotes the barycenter of ν _t). Further, we may assume that $(\ddot u ^n)$ biting converges to a function $\zeta ^\infty : t \mapsto \operatorname {{\mathrm {bar}}}( \nu _t)$, that is, there exists a decreasing sequence of Lebesgue-measurable sets (B _p) with lim_p λ(B _p) = 0 such that the restriction of $(\ddot u_n)$ on each $B_p^c$ converges weakly in $L^1_{\mathbb {R} ^d}([0, T])$ to ζ ^∞. Noting that $(A(.) \dot u^n)$ converges in $L^1_{\mathbb {R} ^d}([0, T])$ to $A(.) \dot u^\infty$. It follows that the restriction of $z^n = -A(.) \dot u^n- \ddot u^n$ to each $B_p^c$ weakly converges in $L^1_{\mathbb {R} ^d}([0, T])$ to $z^\infty := -A(.) \dot u^\infty -\zeta ^\infty$, because $(-A(.) \dot u^n)$ converges in $L^1_{\mathbb {R} ^d}([0, T])$ to $A(.) \dot u^\infty$ and $(\ddot u^n)$ biting converges to $\zeta ^\infty \in L^1_{\mathbb {R} ^d}([0, T])$. It follows that

$$\displaystyle \begin{aligned} \lim_n \int_B \langle -A(.) \dot u^n -\ddot u^n, w(t) -u^n(t) \rangle = \int_B \langle -A(.) \dot u^\infty -\operatorname{{\mathrm{bar}}}( \nu_t) , w(t)-u(t) \rangle dt \end{aligned} $$

(5)

for every $B \in B_p^c \cap \mathcal L ([0, T])$ and for every $w \in L^\infty _{\mathbb {R}^d} ([0, T])$. Indeed, we note that (w(t) − u ⁿ(t)) is a bounded sequence in $L ^\infty _{\mathbb {R} ^d}([0, T])$ which pointwisely converges to w(t) − u ^∞(t), so it converges uniformly on every uniformly integrable subset of $L ^1_{\mathbb {R}^d}([0, T])$ by virtue of a Grothendieck Lemma [33], recalling here that the restriction of $-A(.) \dot u^n-\ddot u^n$ on each $B_p^c$ is uniformly integrable. Now, since φ _n lower epiconverges to φ _∞, for every Lebesgue-measurable set A in [0, T], by virtue of [23, Corollary 4.7], we have

$$\displaystyle \begin{aligned} + \infty >\liminf_n \int_A \varphi_n (u^n(t)) dt \geq \int_A \varphi_\infty (u^\infty(t)) dt. \end{aligned} $$

(6)

Combining (1), (2), (3), (4), (5), and (6) and using the subdifferential inequality

$$\displaystyle \begin{aligned}\varphi_n(w(t)) \geq \varphi_n(u^n(t))+ \langle -A(.) \dot u^n- \ddot u^n(t), w(t) - u^n(t) \rangle,\end{aligned}$$

we get

$$\displaystyle \begin{aligned}\int_B \varphi_\infty( w(t)) \,dt \geq \int_B \varphi_\infty( u ^\infty(t))\, dt +\int_B \langle -A(.) \dot u^\infty -\operatorname{{\mathrm{bar}}}( \nu_t), w(t)-u^\infty(t) \rangle \,dt.\end{aligned}$$

This shows that $t \mapsto -A(.) \dot u^\infty - \operatorname {{\mathrm {bar}}}( \nu _t)$ is a subgradient at the point u ^∞ of the convex integral functional $I_{\varphi _\infty }$ restricted to $L^\infty _{\mathbb {R}^d} (B_p^c)$, consequently,

$$\displaystyle \begin{aligned}-A(.) \dot u^\infty- \operatorname{{\mathrm{bar}}}(\nu_t) \in \partial \varphi_\infty( u ^\infty(t)), \, \mathrm{a.e.} \hskip 2pt \text{ on } B_p^c.\end{aligned} $$

As this inclusion is true on each $B_p^c$ and $B_p^c \uparrow [0, T]$, we conclude that

$$\displaystyle \begin{aligned}-A(.) \dot u^\infty - \operatorname{{\mathrm{bar}}}( \nu_t) \in \partial \varphi_\infty( u ^\infty(t)), \, \mathrm{a.e.} \hskip 2pt \text{ on } [0, T].\end{aligned} $$

Step 4. Measure limit in $\mathcal M^b_{\mathbb {R}^d}([0, T])$ of $\ddot u ^n$. As $(\ddot u_n)$ is bounded in $L^1_{\mathbb {R} ^d}([0, T])$, we may assume that $(\ddot u^n)$ weakly converges to a vector measure $m \in {\mathcal M}^b_{\mathbb {R}^d}([0, T])$ so that the limit functions u ^∞(.) and the limit measure m satisfy the following variational inequality:

$$\displaystyle \begin{aligned} \int_0^T \varphi_\infty( v(t)) \,dt \geq& \int_0^T \varphi_\infty( u ^\infty(t)) \,dt + \int_0 ^T \langle -A(t) \dot u^\infty(t) ,v(t)- u ^\infty(t)\rangle \,dt \\ &+ \langle -m, v-u ^\infty \rangle_{ ({\mathcal M}^b_E([0, T]), {\mathcal C}_{\mathbb{R}^d}([0, T])) }. \end{aligned} $$

In other words, the vector measure $-m -A(t) \dot u^\infty (t) dt$ belongs to the subdifferential $\partial J_{\varphi _\infty }(u ^\infty )$ of the convex functional integral $J_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbb {R}^d}([0, T])$ by $J_{\varphi _\infty }(v)=\int _0 ^T \varphi _\infty (v(t))\ dt$, $\forall v\in {\mathcal C}_{\mathbb {R}^d}([0, T])$. Indeed, let $w \in \mathcal C_{\mathbb {R}^d}([0, T])$. Integrating the subdifferential inequality

$$\displaystyle \begin{aligned}\varphi_n(w(t)) \geq \varphi_n(u^n(t))+ \langle -A(t) \dot u^n(t) -\ddot u^n(t) , w(t) - u^n(t) \rangle\end{aligned}$$

and noting that φ _∞(w(t)) ≥ φ _n(w(t)) gives immediately

$$\displaystyle \begin{aligned} \int_0 ^T \varphi_\infty (w(t)) dt \geq& \int_0 ^T \varphi_n(w(t)) dt \\ \geq& \int_0 ^T \varphi_n(u^n(t)) dt + \langle -A(t) \dot u^n(t) -\ddot u^n(t), w(t) - u^n(t) \rangle dt. \end{aligned} $$

We note that

$$\displaystyle \begin{aligned}\lim_n \int_0 ^T \langle -A(t) \dot u^n(t), w(t) - u^n(t) \rangle dt = \int_0 ^T \langle A(t) \dot u^\infty(t), w(t) - u^\infty(t) \rangle dt\end{aligned}$$

because $(-A(.) \dot u^n)$ is uniformly integrable and converges in $L^1_H([0, T])$ to $A(.) \dot u^\infty$ and the sequence in (w − u ⁿ) converges uniformly to w − u ^∞. Whence follows

$$\displaystyle \begin{aligned} \int_0 ^T \varphi_\infty (w(t)) dt \geq& \int_0 ^T \varphi_\infty (u^\infty(t)) dt + \int_0 ^T \langle -A(t) \dot u^\infty(t), w(t) -u^\infty (t) \rangle dt\\ &+\langle -m, w-u^\infty \rangle_{({\mathcal M}^b_{\mathbb{R}^d}([0, T]), {\mathcal C}_{\mathbb{R}^d}([0, T])) }, \end{aligned} $$

which shows that the vector measure $-m -A(.) \dot u^\infty dt$ is a subgradient at the point u ^∞ of the of the convex integral functional $J_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbb {R}^d}([0, T]))$ by $J_{\varphi _\infty }(v) := \int _0^T \varphi _\infty (v(t))dt, \forall v \in {\mathcal C}_{\mathbb {R}^d}([0, T])$.

Step 5. Claim lim_n φ _n(u ⁿ(t)) = φ _∞(u ^∞(t)) < ∞ a.e. and $\lim _n \int _0^T \varphi _n(u^n(t)) dt = \int _0^T \varphi _\infty (u^\infty (t)) dt < \infty$, and subsequently, the energy estimate holds for a.e. t ∈ [0, T]:

$$\displaystyle \begin{aligned}\varphi_\infty(u^\infty(t)) +\frac{1}{2} ||\dot u^\infty(t)||{}^2 = \varphi_\infty(u^\infty_0) +\frac{1}{2} ||\dot u^\infty_0||{}^2 -\int_0^t \langle A(s)(\dot u^\infty(s), \dot u^\infty (s)\rangle ds.\end{aligned} $$

With the above stated results and notations, applying the subdifferential inequality

$$\displaystyle \begin{aligned}\varphi_n(w(t)) \geq \varphi_n(u^n(t))+ \langle -A(t) \dot u^n(t) -\ddot u^n(t) , w(t) - u^n(t) \rangle\end{aligned} $$

with w = u ^∞, integrating on $B\in B_p^c\cap {\mathcal L}([0, T])$, and passing to the limit when n goes to ∞, gives the inequality

$$\displaystyle \begin{aligned} \int_B \varphi_\infty (u^\infty (t)) dt \geq & \liminf_n \int_B \varphi_n(u^n(t)) dt \\ \geq & \int_B \varphi_\infty (u^\infty (t)) dt \geq \limsup_n \int_B \varphi_n(u^n(t)) dt \end{aligned} $$

so that

$$\displaystyle \begin{aligned} \lim_n \int_B \varphi_n(u^n(t)) dt = \int_B \varphi_\infty (u^\infty (t)) dt \end{aligned} $$

(7)

on $B\in B_p^c\cap {\mathcal L}([0, T])$. Now, from the chain rule theorem given in Step 1, recall that

$$\displaystyle \begin{aligned}-\langle \dot u^n(t), A(t) \dot u^n(t) \rangle -\langle \dot u^n(t), \ddot u^n(t) \rangle = \frac{d}{dt} [\varphi_n(u_n(t))],\end{aligned}$$

that is,

$$\displaystyle \begin{aligned}\langle \dot u^n(t), z^n(t) \rangle = \frac{d}{dt} [\varphi_n(u_n(t))].\end{aligned}$$

By the estimate (1) and the boundedness in $L ^1_{\mathbb {R}^d}([0, T])$ of (z ⁿ), it is immediate that $(\frac {d}{dt} [\varphi _n(u_n(t))])$ is bounded in $L ^1_{\mathbb {R}}([0, T])$ so that (φ _n(u _n(.)) is bounded in variation. By Helly’s theorem, we may assume that (φ _n(u _n(.)) pointwisely converges to a BV function ψ. By (1), (φ _n(u _n(.)) converges in $L ^1_{\mathbb {R}}([0, T])$ to ψ. In particular, for every $k \in L^\infty _{\mathbb {R}^+}([0, T])$, we have

$$\displaystyle \begin{aligned} \lim_{n \rightarrow \infty} \int_0^T k(t) \varphi_n(u_n(t)) dt = \int_0^T k(t) \psi(t) dt. \end{aligned} $$

(8)

Combining with (7) and (8) yields

$$\displaystyle \begin{aligned}\int_B \psi(t) \,dt =\lim_{n \rightarrow \infty} \int_B \varphi_n( u^n(t)) \,dt = \int_B \varphi_\infty( u^\infty(t)) \,dt\end{aligned}$$

for all $\in B_p^c\cap {\mathcal L}([0, T])$. As this inclusion is true on each $B_p^c$ and $B_p^c \uparrow [0, T]$, we conclude that

$$\displaystyle \begin{aligned}\psi(t)= \lim_n \varphi_n(u_n(t)) = \varphi_\infty(u^ \infty(t)) \, \mathrm{a.e.}\end{aligned}$$

Subsequently, using (iii), the passage to the limit when n goes to ∞ in the equation

$$\displaystyle \begin{aligned}\varphi_n(u^n(t)) +\frac{1}{2} ||\dot u^n(t)||{}^2 = \varphi_n(u^n(0)) +\frac{1}{2} ||\dot u^n(0)||{}^2 - \int_0^t \langle A(s) \dot u^n(s), \dot u^n(s)\rangle ds\end{aligned}$$

yields for a.e. t ∈ [0, T]

$$\displaystyle \begin{aligned}\varphi_\infty(u^\infty(t)) +\frac{1}{2} ||\dot u^\infty(t)||{}^2 = \varphi_\infty(u ^\infty_0) + \frac{1}{2} ||\dot u^\infty_0)||{}^2 -\int_0^t \langle A(s) \dot u^\infty(s), \dot u^\infty(s)\rangle ds . \end{aligned}$$

Step 6. Localization of further limits and final step.

As $(z^n = -A(.) \dot u^n -\ddot u ^n)$ is bounded in $L ^1_{\mathbb {R} ^d}([0, T])$ in view of Step 3, it is relatively compact in the second dual $L^\infty _{\mathbb {R}^d}([0, T])'$ of $L ^1_{\mathbb {R} ^d}([0, T])$ endowed with the weak topology $\sigma (L^\infty _{\mathbb {R}^d}([0, T])', L^\infty _{\mathbb {R}^d}([0, T]))$. Furthermore, (z ⁿ) can be viewed as a bounded sequence in ${\mathcal C}_{\mathbb {R}^d} ([0, T])'$. Hence there is a filter ${\mathcal U}$ finer than the Fréchet filter $l \in L^\infty _{\mathbb {R}^d}([0, T])'$ and $\mathbf {n} \in {\mathcal C}_{\mathbb {R}^d} ([0,T])'$ such that

$$\displaystyle \begin{aligned} {\mathcal U}-\lim_n z^n = l \in L^\infty_{\mathbb{R}^d}([0, T])'_{\mathrm{weak}} \end{aligned} $$

(9)

and

$$\displaystyle \begin{aligned} \lim_n z^n = \mathbf{n} \in {\mathcal C}_{\mathbb{R}^d} ([0, T])'_{\mathrm{weak}} \end{aligned} $$

(10)

where $L^\infty _{\mathbb {R}^d}([0, T])'_{\mathrm {weak}}$ is the second dual of $L^1_{\mathbb {R}^d} ([0, T])$ endowed with the topology $\sigma (L^\infty _{\mathbb {R}^d}([0, T])', L^\infty _{\mathbb {R}^d}([0, T]))$ and ${\mathcal C}_{\mathbb {R}^d} ([0, T])'_{\mathrm {weak}}$ denotes the space ${\mathcal C}_{\mathbb {R}^d} ([0, T])'$ endowed with the weak topology $\sigma ({\mathcal C}_{\mathbb {R}^d} ([0, T])', {\mathcal C}_{\mathbb {R}^d} ([0, T]))$, because ${\mathcal C}_{\mathbb {R}^d}([0, T])$ is a separable Banach space for the norm sup, so that we may assume by extracting subsequences that (z ⁿ) weakly converges to $\mathbf {n}\in {\mathcal C}_{\mathbb {R}^d}([0, T])'$. Let l _a be the density of the absolutely continuous part l _a of l in the decomposition l = l _a + l _s in absolutely continuous part l _a and singular part l _s, in the sense there is a decreasing sequence (A _n) of Lebesgue-measurable sets in [0, T] with A _n ↓∅ such that $l_s(f) = l_s(1_{A_n} f)$ for all $h\in L^\infty _{\mathbb {R}^d} ([0, T])$ and for all n ≥ 1. As $(z ^n =-A(.) \dot u^n -\ddot u ^n)$ biting converges to $z ^\infty =-A(.) \dot u^\infty -\zeta ^\infty$ in Step 4, it is already known [22] that

$$\displaystyle \begin{aligned}l_a(f) = \int_0 ^T \langle f(t),-A(t) \dot u^\infty(t)-\zeta^\infty(t) \rangle dt\end{aligned}$$

for all $f \in L^\infty _{\mathbb {R}^d}([0, T])$, shortly $z^\infty =-A(t) \dot u^\infty (t) -\zeta ^\infty (t)$ coincides a.e. with the density of the absolutely continuous part l _a. By [19, 46], we have

$$\displaystyle \begin{aligned}I_{\varphi_\infty}^*(l) = I_{\varphi_\infty^* }(-A(.) \dot u^\infty-\zeta^\infty) +\delta^* (l_s ,\operatorname{{\mathrm{dom}}} I_{\varphi_\infty} )\end{aligned}$$

where $\varphi _\infty ^*$ is the conjugate of φ _∞, $I_{\varphi _\infty ^* }$ is the integral functional defined on $L^1_{\mathbb {R}^d} ([0, T])$ associated with $\varphi _\infty ^*$, $I_{\varphi _\infty }^*$ is the conjugate of the integral functional $I_{\varphi _\infty }$, and

$$\displaystyle \begin{aligned}\operatorname{{\mathrm{dom}}} I_{\varphi_\infty} := \{ u \in L^\infty_{\mathbb{R}^d}([0, T]) : I_{\varphi_\infty}(u) < \infty\}.\end{aligned}$$

Using the inclusion

$$\displaystyle \begin{aligned}z^\infty = -A(.) \dot u^\infty-\zeta^\infty \in \partial I_{\varphi_\infty}(u ^\infty),\end{aligned}$$

that is,

$$\displaystyle \begin{aligned}I_{\varphi_\infty^* }(-A(.) \dot u^\infty-\zeta^\infty) = \langle -A(.) \dot u^\infty -\zeta^\infty, u^\infty \rangle -I_{\varphi_\infty} (u^\infty),\end{aligned}$$

we see that

$$\displaystyle \begin{aligned} I_{\varphi_\infty}^*(l) = \langle -A(.) \dot u^\infty -\zeta^\infty, u^\infty \rangle -I_{\varphi_\infty} (u^\infty) +\delta^* (l_s, \operatorname{{\mathrm{dom}}} I_{\varphi_\infty}). \end{aligned}$$

Coming back to z ⁿ(t) = ∇φ _n(u ⁿ(t)), we have

$$\displaystyle \begin{aligned} \varphi_n (x) \geq \varphi_n (u^n(t)) + \langle x-u^n(t), z^n(t) \rangle \end{aligned}$$

for all $x \in \mathbb {R}^d$. Substituting x by h(t) in this inequality, where $h\in {\mathcal C}_{\mathbb {R}^d}([0, T])$, and integrating, we get

$$\displaystyle \begin{aligned} \int_0^T \varphi_n ( h(t))\, dt \geq \int_0^T \varphi_n ( u^n(t)) \,dt + \int_0^T \langle h(t)-u^n(t), z^n(t) \rangle \,dt. \end{aligned}$$

Arguing as in Step 4 by passing to the limit in the preceding inequality, involving the epiliminf property for integral functionals (cf. (6)), it is easy to see that

$$\displaystyle \begin{aligned} \int_0^T \varphi_\infty(h(t)) \,dt \geq \int_0^T \varphi_\infty( u^\infty(t)) \,dt + \langle h-u^\infty, \mathbf{n} \rangle. \end{aligned}$$

Whence n belongs to the subdifferential $\partial J_{\varphi _\infty }(u^\infty )$ of the convex lower semicontinuous integral functional $J_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbb {R}^d}([0, T])$ by

$$\displaystyle \begin{aligned}J_{\varphi_\infty}(u):= \int_0^T \varphi_\infty( u(t)) \,dt, \hskip 4pt \forall u \in {\mathcal C}_{\mathbb{R}^d}([0, T]).\end{aligned} $$

Now let $B : {\mathcal C}_{\mathbb {R}^d}([0, T]) \rightarrow L^\infty _{\mathbb {R}^d}([0, T])$ be the continuous injection, and let $B^* : L^\infty _{\mathbb {R}^d}([0, T])' \rightarrow {\mathcal C}_{\mathbb {R}^d}([0, T])'$ be the adjoint of B given by

$$\displaystyle \begin{aligned}\langle B^*l , f \rangle = \langle l, Bf \rangle = \langle l, f \rangle, \quad \forall l \in L^\infty_{\mathbb{R}^d}([0, T])', \quad \forall f \in {\mathcal C}_{\mathbb{R}^d}([0, T]).\end{aligned} $$

Then we have B ^∗ l = B ^∗ l _a + B ^∗ l _s, $l\in L^\infty _{\mathbb {R}^d}([0, T])'$ being the limit of z _n under the filter ${\mathcal U}$ given in Sect. 4 and l = l _a + l _s being the decomposition of l in absolutely continuous part l _a and singular part l _s. It follows that

$$\displaystyle \begin{aligned} \langle B^*l, f \rangle = \langle B^*l_a, f\rangle + \langle B^*l_s, f\rangle =\langle l_a, f \rangle +\langle l_s, f \rangle \end{aligned}$$

for all $f\in {\mathcal C}_{\mathbb {R}^d}([0, T])$. But it is already seen that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \langle l_a, f \rangle = \langle -A(.) \dot u^\infty -\zeta^\infty, f \rangle \\ &\displaystyle &\displaystyle \qquad \qquad \quad =\int_0 ^T \langle -A(.) \dot u^\infty(t) - \zeta^\infty(t) , f(t)\rangle dt, \quad \forall f \in L^\infty_{\mathbb{R}^d}([0, T]) \end{array} \end{aligned} $$

so that the measure B ^∗ l _a is absolutely continuous

$$\displaystyle \begin{aligned}\langle B^*l_a, h \rangle = \int_0 ^T \langle -A(.) \dot u^\infty(t) - \zeta^\infty(t), f(t)\rangle dt, \quad \forall f \in {\mathcal C}_{\mathbb{R}^d}([0, T])\end{aligned}$$

and its density $-A(.) \dot u^\infty - \zeta ^\infty$ satisfies the inclusion

$$\displaystyle \begin{aligned}-A(t) \dot u^\infty(t)- \zeta^\infty(t) \in \partial \varphi_\infty(u^\infty(t)), \quad \mathrm{a.e.}\end{aligned}$$

and the singular part B ^∗ l _s satisfies the equation

$$\displaystyle \begin{aligned}\langle B^*l_s, f \rangle = \langle l_s, h \rangle , \quad \forall f \in {\mathcal C}_{\mathbb{R}^d}([0, T]).\end{aligned}$$

As B ^∗ l = n, using (9) and (10), it turns out that n is the sum of the absolutely continuous measure n _a with

$$\displaystyle \begin{aligned}\langle {\mathbf{n}}_a, f \rangle = \int_0 ^T \langle -A(t) \dot u^\infty(t) - \zeta^\infty(t) , f(t)\rangle dt, \quad \forall f \in {\mathcal C}_{\mathbb{R}^d}([0, T])\end{aligned}$$

and the singular part n _s given by

$$\displaystyle \begin{aligned}\langle {\mathbf{n}}_s, f \rangle = \langle l_s, f \rangle, \quad \forall f \in {\mathcal C}_{\mathbb{R}^d}([0, T]).\end{aligned}$$

which satisfies the property: for any nonnegative measure θ on [0, T] with respect to which n _s is absolutely continuous

$$\displaystyle \begin{aligned}\int_0^T r_{\varphi_\infty^*}\left( \frac{d{\mathbf{n}}_s}{d\theta}(t)\right) d\theta(t)= \int_0^T \left\langle u^\infty(t), \frac{d{\mathbf{n}}_s}{d \theta}(t)\right\rangle d\theta(t)\end{aligned} $$

where $r_{\varphi _\infty ^*}$ denotes the recession function of $\varphi _\infty ^*$. Indeed, as n belongs to $\partial J_{\varphi _\infty }(u^\infty )$ by applying [46, Theorem 5], we have

$$\displaystyle \begin{aligned} J^*_{\varphi_\infty}(n) = I_{\varphi_\infty^*}\left(\frac{d{\mathbf{n}}_a}{dt}\right) +\int_0^T r_{\varphi_\infty^*}\left( \frac{d{\mathbf{n}}_s}{d \theta}(t)\right) d\theta(t),\end{aligned} $$

(11)

with

$$\displaystyle \begin{aligned}I_{\varphi_\infty^*} (v):=\int_0^T \varphi_\infty^*(v(t)) dt , \forall v \in L^1_{\mathbb{R}^d}([0, T]).\end{aligned}$$

Recall that

$$\displaystyle \begin{aligned}\frac{d{\mathbf{n}}_a}{dt}=-A(.) \dot u^\infty - \zeta^\infty\in \partial I_{\varphi_\infty}(u^\infty),\end{aligned}$$

that is,

$$\displaystyle \begin{aligned} I_{\varphi_\infty^*} \left(\frac{d{\mathbf{n}}_a}{dt}\right) = \langle -A(.) \dot u^\infty - \zeta^\infty, u^\infty\rangle_{\langle L^1_{\mathbb{R} ^d}([0, T]), L^\infty_{\mathbb{R} ^d}([0, T])\rangle} - I_{\varphi_\infty}(u^\infty). \end{aligned} $$

(12)

From (12), we deduce

$$\displaystyle \begin{aligned} J^*_{\varphi_\infty}(n) =& \langle u ^\infty, \mathbf{n}\rangle_{\langle {\mathcal C}_{\mathbb{R} ^d}([0, T]),{\mathcal C}_{\mathbb{R} ^d}([0, T])'\rangle}- J_{\varphi_\infty}(u ^\infty)\\ =& \langle u ^\infty, \mathbf{n}\rangle_{\langle {\mathcal C}_{\mathbb{R} ^d}([0, T]),{\mathcal C}_{\mathbb{R} ^d}([0, T])'\rangle}- I_{\varphi_\infty}(u ^\infty)\\ =&\int_0 ^T \langle u ^\infty(t), -A(.) \dot u^\infty -\zeta^\infty(t) \rangle dt \\ &+ \int_0 ^T \left\langle u ^\infty(t), \frac{d{\mathbf{n}}_s}{d \theta}(t)\right\rangle {d \theta}(t)-I_{\varphi_\infty}(u ^\infty)\\ =& I_{\varphi_\infty ^*}\left(\frac{d{\mathbf{n}}_a}{dt}\right) +\int_0 ^T \left\langle u ^\infty(t), \frac{d{\mathbf{n}}_s}{d \theta}(t) \right\rangle {d \theta}(t)). \end{aligned} $$

Coming back to (11), we get the equality

$$\displaystyle \begin{aligned}\int_0^T r_{\varphi_\infty^*}\left( \frac{d{\mathbf{n}}_s}{d \theta}(t)\right) d\theta(t) = \int_0^T \left\langle u ^\infty(t), \frac{d{\mathbf{n}}_s}{d \theta}(t)\right\rangle{d \theta}(t)).\end{aligned}$$

The proof is complete.

Comments

Some comments are in order. In Proposition 3.1, using the existence and uniqueness of $W^{2, \infty }_H(]0, T])$ of the approximating second-order equation

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0= \ddot u^n (t)+A(t) \dot u^n (t) + \nabla \varphi_n(u^n(t)), t \in [0, T] \\ u ^n(0) = u^n_0, \hskip 2pt \dot u^n(0) = \dot u ^n_0, \end{array} \right. \end{aligned}$$

we state the existence of a generalized solution u ^∞ to the second-order evolution inclusion

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \ddot u (t)+A(t) \dot u(t) + \partial \varphi _\infty(u(t)), t \in [0, T] \\ u(0)= u_0 \in \operatorname{{\mathrm{dom}}} \, \varphi_\infty , \hskip 2pt \dot u(0) = \dot u_0 \end{array} \right. \end{aligned}$$

via an epiconvergence approach involving the structure of bounded sequences in $L ^1_H([0, T]$ space [22] and describe various properties of such a generalized solution. In particular, we show that such a generalized solution u ^∞ is $W^{1, 1}_{BV} ([0, T])$ and satisfies the energy conservation and there exists a Young measure ν _t with barycenter $\operatorname {{\mathrm {bar}}}(\nu _t ) \in L^1_{H} ([0, T])$ such that $-A(t) \dot u^\infty (t) - \operatorname {{\mathrm {bar}}}(\nu _t ) \in \partial \varphi _\infty (u \infty (t))$ a.e. In this vein, compare with Attouch et al. [4, 27], Paoli [43], and Schatzman [48].

Now we deal at first with $W^{1, 1}_{BV}([0, T], H)$ solution for a second-order evolution problem.

Theorem 3.3

Let for every t ∈ [0, T], A(t) : D(A(t)) ⊂ H → 2^H be a maximal monotone operator with D(A(t)) ball compact for every t ∈ [0, T] satisfying

• (H1) there exists a function r : [0, T] → [0, +∞[ which is continuous on [0, T[ and nondecreasing with r(T) < +∞ such that

  $$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}(A(t), A(s))\leq dr(]s, t])=r(t)-r(s)\;\; {for}\;\;0\leq s\leq t\leq T\end{aligned}$$

• (H2) there exists a nonnegative real number c such that

  $$\displaystyle \begin{aligned}\|A^0(t,x)\|\leq c(1+\| x\|)\;\; {for}\;\;t\in [0, T] ,\;x\in D(A(t))\end{aligned}$$

Let f : [0, T] × H × H → H be such that for every x, y ∈ H × H the mapping f(., x, y) is Borel-measurable on [0, T] and for every t ∈ [0, T], f(t, ., .) is continuous on H × H and satisfying

• (i) ||f(t, x, y)||≤ M(1 + ||x||), ∀t, x, y ∈ [0, T] × H × H.

• (ii) ||f(t, x, z) − f(t, y, z)||≤ M||x − y||, ∀t, x, y, z ∈ [0, T] × H × H × H.

Then for u ₀ ∈ D(A(0))andy ₀ ∈ H, there are a BVC mapping u : [0, T] → H and a $W^{1, 1}_{BV}([0, T], H)$ mapping y : [0, T] → H satisfying

$$\displaystyle \begin{gathered} y(t) = y_0+ \int_0^t u(s) ds, \hskip 3pt \; t \in [0, T],\\ -\frac {du} {dr}(t) \in A(t) u(t) +f(t, u(t), y(t) ) \hskip 3pt dr\mathit{\text{-a.e.}} \hskip 3pt t \in [0, T],\\ u(0)=u_0 \end{gathered} $$

with the property: |u(t) − u(τ)|≤ K|r(t) − r(τ)| for all t, τ ∈ [0, T] for some constant K ∈ ]0, ∞[.

Proof

By [8, Theorem 3.1] and the assumptions on f, for any continuous mapping h : [0, T] → H, there is a unique BVC solution v _h to the inclusion

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &v_{h} (0) = u_0 \in D(A(0))\\ &-\frac {dv_h} {dr}(t) \in A(t) v_{h} (t) + f(t, v_h(t), h(t)) \hskip 3pt dr \text{-a.e.} \end{aligned} \right . \end{aligned}$$

with ||v _h(t)||≤ K, t ∈ [0, T] and ||v _h(t) − v _h(τ)||≤ K(r(t) − r(τ)), t, τ ∈ [0, T] so that

$$\displaystyle \begin{aligned}dv_h = \frac{ dv_h}{dr} dr\end{aligned}$$

with $\frac { dv_h}{dr} \in K {\overline B}_H$, consequently $\frac { dv_h}{dr} \in L ^\infty _H([0, T], dr)$. Let consider the closed convex subset ${\mathcal X}$ in the Banach space ${\mathcal C} _H([0, T])$ defined by

$$\displaystyle \begin{aligned}{\mathcal X} : = \{ u : [0, T] \rightarrow H : u (t)= u_0+ \int_0^t \dot u(s) ds, \hskip 3pt \dot u \in S ^1_{K{\overline B}_H}, \hskip 3pt t \in [0, T] \}\end{aligned}$$

where $S ^1_{K{\overline B}_H}$ denotes the set of all integrable selections of the convex weakly compact valued constant multifunction $K{\overline B}_H$. Now for each $h \in {\mathcal X}$, let us consider the mapping

$$\displaystyle \begin{aligned}\Phi(h) (t) := u_0 + \int_0^t v_h(s) ds, \hskip 3pt t \in [0, T].\end{aligned}$$

Then it is clear that $\Phi (h) \in {\mathcal X}$. Our aim is to prove the existence theorem by applying some ideas developed in Castaing et al. [24] via a generalized fixed point theorem [36, 44]. Nevertheless this needs a careful look using the estimation of the BVC solution given above. For this purpose, we first claim that $\Phi : \mathcal X \rightarrow \mathcal X$ is continuous and for any $h \in \mathcal X$ and for any t ∈ [0, T] the inclusion holds

$$\displaystyle \begin{aligned}\Phi(h) (t) \in u_0 + \int_0^t \overline{co} [D(A(s)) \cap K{\overline B}_H] ds.\end{aligned}$$

Since $s \mapsto \overline {co} [D(A(s)) \cap K{\overline B}_H]$ is a convex compact valued and integrably bounded multifunction using the ball-compactness assumption, the second member is convex compact valued [14] so that $\Phi (\mathcal X )$ is equicontinuous and relatively compact in the Banach space ${\mathcal C} _H([0, T])$. Now we check that Φ is continuous. It is sufficient to show that, if (h _n) converges uniformly to h in $\mathcal X$, then BVC solution $v_{h_n}$ associated with h _n

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &v_{h_n} (0) = u_0 \in D(A(0))\\ &- \frac{ dv_{h_n }} {dr} (t) \in A(t) v_{h_n } (t) + f(t, v_{h_n} (t), h_n(t)) \hskip 3pt dr\text{-a.e.} \end{aligned} \right .\end{aligned} $$

pointwisely converges to the BVC solution v _h associated with h

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &v_{h} (0) = u_0 \in D(A(0))\\ &- \frac{ dv_h} {dr} (t) \in A(t) v_{h} (t) + f(t, v_h(t), h(t)) \hskip 3pt dr \text{-a.e.} \end{aligned} \right .\end{aligned} $$

As D(A(t)) is ball compact, $(v_{h_n})$ is uniformly bounded, and bounded in variation since $|| v_{h_n}(t) -v_{h_n}(\tau )|| \leq K (r(t) -r(\tau )), \hskip 3pt t, \tau \in [0, T]$, we may assume that $(v_{h_n})$ pointwisely converges to a BVC mapping v. As $v_{h_n } = v_0+ \int _{]0, t]} \frac { dv_{h_n }} {dr} dr, \hskip 3pt t \in [0, T]$ and $\frac { dv_{h_n }} {dr} (s) \in K {\overline B}_H, \hskip 2pt s \in [0, T]$, we may assume that $( \frac { dv_{h_n }} {dr})$ converges weakly in $L^1_H([0, T], dr)$ to $w \in L^1_H([0, T], dr)$ with $w(t)\in K {\overline B}_H, \hskip 3pt t \in [0, T]$ so that

$$\displaystyle \begin{aligned} \mathrm{weak-}\lim_n v_{h_n }= u_0 + \int_{]0, t]} w dr:= z(t), \hskip 3pt t \in [0, T].\end{aligned} $$

By identifying the limits, we get

$$\displaystyle \begin{aligned} v(t) = z(t) = u_0 + \int_{]0, t]} w dr \end{aligned}$$

with $\frac {dv} {dr} = w$ so that $\lim _n f(t, v_{h_n}(t), h_n (t)) = f(t, v (t), h(t)), \hskip 3pt t \in [0, T]$. Consequently we may assume that $( \frac { dv_{h_n }} {dr} +f(., v_{h_n}( .), h_n (.)) )$ Komlos converges to $\frac {dv} {dr} -f(., v(.), h (.))$. For simplicity, set $g_n(t) = f(t, v_{h_n}( t), h_n (t))$ and g(t) = f(t, v(t), h(t)). There is a dr-negligible set N such that for t ∈ I ∖ N and

$$\displaystyle \begin{aligned} \lim_{n\to\infty} \frac{1} {n} \sum_{j= 1}^n \left(\frac{ dv_{h_j }} {dr}(t) + g_j(t)\right) = \frac{dv} {dr} (t) +g(t). \end{aligned}$$

Let η ∈ D(A(t)). From

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \left\langle \frac{ dv_{h_n }} {dr} (t) +g_n(t) , v(t) - \eta \right\rangle\\ &\displaystyle &\displaystyle \qquad \ \ = \left\langle \frac{ dv_{h_n }} {dr} (t)+g_n(t) , v_{h_n}(t) -\eta\right\rangle + \left\langle \frac{ dv_{h_n }} {dr} (t)+g_n(t) , v(t) -v_{h_n}(t) \right\rangle, \end{array} \end{aligned} $$

let us write

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \ \ \frac{1}{n} \sum_{j= 1}^n \left\langle \frac{ dv_{h_j }} {dr} (t) +g_j (t), v(t) - \eta \right\rangle\\ &\displaystyle &\displaystyle = \frac{1}{n} \sum_{j= 1}^n \left\langle \frac{ dv_{h_j }} {dr} (t) +g_j (t) , v_{h_j} (t)-\eta \right\rangle + \frac{1}{n} \sum_{j= 1}^n \left\langle \frac{ dv_{h_j }} {dr} (t) +g_j(t) , v(t) -v_{h_j} (t) \right\rangle, \end{array} \end{aligned} $$

so that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \frac{1}{n} \sum_{j= 1}^n \left\langle \frac{ dv_{h_j }} {dr} (t)+g_j(t), v(t) - \eta \right\rangle\\ &\displaystyle &\displaystyle \qquad \qquad \leq \frac{1}{n} \sum_{j= 1}^n \left\langle A^0( t , \eta) , \eta - v_{h_j}(t) \big\rangle +\right.\big(\mathrm{Constant} \big) \frac{1}{n} \sum_{j= 1}^n \|v(t) -v_{h_j}(t))\|. \end{array} \end{aligned} $$

Passing to the limit when n →∞, this last inequality gives immediately

$$\displaystyle \begin{aligned} \left\langle \frac{dv} {dr} (t)+g(t) , v(t) - \eta \right\rangle \leq \big\langle A^0(t, \eta) , \eta-v(t) \big\rangle \text{ a.e.} \end{aligned}$$

As a consequence, by Lemma 2.2, $-\frac {dv} {dr} (t) \in A(t) v(t)+g(t) = A(t) v(t) + f(t, v( t), h(t))$ a.e. with v(0) = u ₀ ∈ D(A(0)) so that by uniqueness v = v _h. Now let us check that $\Phi : \mathcal X \rightarrow \mathcal X$ is continuous. Let h _n → h. We have

$$\displaystyle \begin{aligned} \Phi (h_n) (t) - \Phi (h) (t) = \int_0 ^t v _{h_n}(s) ds - \int_0 ^t v _{h}(s) ds = \int_0 ^t [v _{h_n}(s)- v _{h}(s)] ds \end{aligned}$$

As $||v _{h_n}(.)- v _{h}(.)|| \rightarrow 0$ pointwisely and is uniformly bounded : $||v _{h_n}(.)- v _{h}(.)|| \leq 2K$, by we conclude that

$$\displaystyle \begin{aligned} \sup_{t \in [0, T]} ||\Phi (h_n) (t) - \Phi (h) (t) || \leq \sup_{t \in [0, T]} \int_0^t ||v _{h_n}(.)- v _{h}(.)|| ds \rightarrow 0\end{aligned}$$

so that Φ(h _n) − Φ(h) → 0 in ${\mathcal C}_H([0, T])$. Here one may invoke a general fact that on bounded subsets of L ^∞, the topology of convergence in measure coincides with the topology of uniform convergence on uniformly integrable sets, i.e., on relatively weakly compact subsets, alias the Mackey topology. This is a lemma due to Grothendieck [33, Ch.5 §4 no 1 Prop. 1 and exercice] (see also [15] for a more general result concerning the Mackey topology for bounded sequences in $L^\infty _{E^*}$). Since $\Phi : \mathcal X \rightarrow \mathcal X$ is continuous and $\Phi (\mathcal X)$ is relatively compact in $\mathcal C_H([0, T])$, by [36, 44] Φ has a fixed point, say $h = \Phi (h) \in \mathcal X$, that means

$$\displaystyle \begin{gathered} h(t) = \Phi(h) (t) = u_0+\int_0^t v_h (s) ds, \hskip 3pt t \in [0, T],\\ \left\{ \begin{aligned} &v_{h} (0) = u_0 \in D(A(0))\\ &-\frac{dv_h} {dr} (t) \in A(t) v_{h} (t) + f(t, v_h(t), h(t)) \hskip 3pt dr \text{-a.e.} \end{aligned} \right . \end{gathered} $$

The proof is complete.

The following results are sharp variants of Theorem 3.3.

Theorem 3.4

Let for every t ∈ [0, T], A(t) : D(A(t)) ⊂ H → 2^H be a maximal monotone operator with D(A(t) ball compact for every t ∈ [0, T] satisfying (H2) and

• (H1)′ there exists a function $\beta \in W^{1,1}([0, T], \mathbb {R}; dt)$ which is nonnegative on [0, T] and non-decreasing with β(T) < ∞ such that

  $$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}(A(t), A(s)) \leq | \beta(t) -\beta(s)| , \hskip 2pt \forall s, t \in [0, T].\end{aligned} $$

• (H1)∗ For any t ∈ [0, T] and for any x ∈ D(A(t)), A(t)x is cone-valued.

Let f : [0, T] × H × H → H be such that for every x, y ∈ H × H the mapping f(., x, y) is Lebesgue-measurable on [0, T] and for every t ∈ [0, T], f(t, ., .) is continuous on H × H and satisfying

• (i) ||f(t, x, y)||≤ M(1 + ||x||), ∀t, x, y ∈ [0, T] × H × H.

• (ii) ||f(t, x, z) − f(t, y, z)||≤ M||x − y||, ∀t, x, y, z ∈ [0, T] × H × H × H.

Then, for all u ₀ ∈ D(A(0)), y ₀ ∈ H, there are an absolutely continuous mapping u : [0, T] → H and an absolutely continuous mapping y : [0, T] → H satisfying

$$\displaystyle \begin{gathered} y(t) = y_0+ \int_0^t u(s) ds, \hskip 3pt \; t \in [0, T],\\ -\frac {du} {dt}(t) \in A(t) u(t) +f(t, u(t), y(t) ) \hskip 3pt dt-\mathrm{a.e.}\; t \in [0, T], u(0)=u_0,\vspace{-4pt} \end{gathered} $$

with

$$\displaystyle \begin{aligned} \|\dot u(t)\| \leq \big(K+ M(1+K)\big) (\dot \beta (t)+1) + M(1+K) \end{aligned}$$

for a.e. t ∈ [0, T], for some positive constant K.

Proof

By [7, Theorem 3.4] and the assumptions on f, for any continuous mapping h : [0, T] → H, there is a unique AC solution v _h to the inclusion

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &v_{h} (0) = u_0 \in D(A(0))\\ &- \dot v_h(t) \in A(t) v_{h} (t) + f(t, v_h(t), h(t)) \hskip 3pt dt \text{-a.e.} \end{aligned} \right . \end{aligned}$$

with $||\dot v_h(t) || \leq \gamma (t):=(K+M(1+K))(\dot \beta (t)+1)+M(1+K)$ a.e. t ∈ [0, T] so that $\gamma \in L^1_{\mathbb {R}}([0, T])$ and ||v _h(t)||≤ L = Constant, t ∈ [0, T]. Let us consider the closed convex subset ${\mathcal X}$ in the Banach space ${\mathcal C} _H([0, T])$ defined by

$$\displaystyle \begin{aligned} {\mathcal X} : = \{ u : [0, T] \rightarrow H : u (t)= u_0+ \int_0^t \dot u(s) ds, \hskip 3pt \dot u \in S ^1_{L{\overline B}_H}, \hskip 3pt t \in [0, T] \} \end{aligned}$$

where $S ^1_{L{\overline B}_H}$ denotes the set of all integrable selections of the convex weakly compact valued constant multifunction $L{\overline B}_H$. Now for each $h \in {\mathcal X}$, let us consider the mapping

$$\displaystyle \begin{aligned} \Phi(h) (t) := u_0 + \int_0^t v_h(s) ds, \hskip 3pt t \in [0, T]. \end{aligned}$$

Then it is clear that $\Phi (h) \in {\mathcal X}$. Our aim is to prove the existence theorem by applying some ideas developed in Castaing et al. [24] via a generalized fixed point theorem [36, 44]. Nevertheless this needs a careful look using the estimation of the AC solution given above. For this purpose, we first claim that $\Phi : \mathcal X \rightarrow \mathcal X$ is continuous for any $h \in \mathcal X$ and for any t ∈ [0, T], the inclusion holds

$$\displaystyle \begin{aligned} \Phi(h) (t) \in u_0 + \int_0^t \overline{co} [D(A(s)) \cap L{\overline B}_H] ds. \end{aligned}$$

Since $s \mapsto \overline {co} [D(A(s)) \cap L{\overline B}_H]$ is a convex compact valued and integrably bounded multifunction, the second member is convex compact valued [14] so that $\Phi (\mathcal X )$ is equicontinuous and relatively compact in the Banach space ${\mathcal C} _H([0, T])$. Now we check that Φ is continuous. It is sufficient to show that, if h _n converges uniformly to h in $\mathcal X$, then the AC solution $v_{h_n}$ associated with h _n

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &v_{h_n} (0) = u_0 \in D(A(0))\\ &- \dot v_{h_n }(t) \in A(t) v_{h_n } (t) + f(t, v_{h_n} (t), h_n(t)) \hskip 3pt dt \text{-a.e.} \end{aligned} \right . \end{aligned}$$

converges uniformly to the AC solution v _h associated with h

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &v_{h} (0) = u_0 \in D(A(0))\\ &- \dot v_h (t) \in A(t) v_{h} (t) + f(t, v_h(t), h(t)) \hskip 3pt dt \text{-a.e.} \end{aligned} \right . \end{aligned}$$

We have

$$\displaystyle \begin{aligned} -\dot v_{h_n}(t) \in A(t) v_{h_n}( t) + f(t, v_{h_n}( t), h_n (t)), \text{ a.e. }t \in [0, T], \end{aligned}$$

with the estimation $||\dot v_{h_n}(t)|| \leq \gamma (t)$ and $\gamma \in L^1_{\mathbb {R}} ([0, T])$ for all $n\in \mathbb {N}$. As D(A(t)) is ball compact and $(\dot v_{h_n})$ is relatively weakly compact in $L^1_H ([0, T])$, we may assume that $(v_{h_n})$ converges uniformly to an absolutely continuous mapping v such that $v(t) = u_0+ \int _0^t \dot v(s) ds, \hskip 3pt t \in [0, T]$, $||\dot v(t) || \leq \gamma (t), \hskip 3pt t \in [0, T]$, and $(\dot v_{h_n}) \hskip 2pt \sigma ( L^1_H, L^\infty _H)$ converges to $\dot v$ so that $\lim _n f(t, v_{h_n}(t), h_n (t)) = f(t, v (t), h(t)), \hskip 3pt t \in [0, T]$. Consequently we may assume that $(\dot v_{h_n} +f(., v_{h_n}( .), h_n (.)) )$ Komlos converges to $\dot v -f(., v(.), h (.))$. Let us set $g_n(t) = f(t, v_{h_n}( t), h_n (t))$ and g(t) = f(t, v(t), h(t)). There is a negligible set N such that for t ∈ [0, T] ∖ N and

$$\displaystyle \begin{aligned} \lim_{n\to\infty} \frac{1} {n} \sum_{j= 1}^n \big(\dot v_{h_j}(t) + g_j(t)\big) = \dot v (t) +g(t). \end{aligned}$$

Let η ∈ D(A(t)). From

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \big\langle \dot v_{h_n}(t) +g_n(t) , v(t) - \eta \big\rangle\\ &\displaystyle &\displaystyle \qquad \qquad \qquad = \big\langle \dot v_{h_n}(t) +g_n(t) , v_{h_n}(t) -\eta\big\rangle + \big\langle \dot v_{h_n}(t)+g_n(t) , v(t) -v_{h_n}(t) \big\rangle \end{array} \end{aligned} $$

let us write

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \frac{1}{n} \sum_{j= 1}^n \big\langle \dot v_{h_j}(t) +g_j (t), v(t) - \eta \big\rangle\\ &\displaystyle &\displaystyle \ = \frac{1}{n} \sum_{j= 1}^n \big\langle \dot v_{h_j}(t)+g_j (t) , v_{h_j} (t)-\eta \big\rangle + \frac{1}{n} \sum_{j= 1}^n \big\langle \dot v_{h_j}(t)+g_j(t) , v(t) -v_{h_j} (t) \big\rangle, \end{array} \end{aligned} $$

so that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \frac{1}{n} \sum_{j= 1}^n \big\langle \dot v_{h_j}(t) +g_j(t), v(t) - \eta \big\rangle \\ &\displaystyle &\displaystyle \quad \leq \frac{1}{n} \sum_{j= 1}^n \big\langle A^0( t , \eta) , \eta - v_{h_j}(t) \big\rangle +\big(\gamma(t) +\mathrm{Constant} \big) \frac{1}{n} \sum_{j= 1}^n \|v(t) -v_{h_j}(t))\|. \end{array} \end{aligned} $$

Passing to the limit when n →∞, this last inequality gives immediately

$$\displaystyle \begin{aligned}\big\langle \dot v(t)+g(t) , v(t) - \eta \big\rangle \leq \big\langle A^0(t, \eta) , \eta-v(t) \big\rangle\text{ a.e.}\end{aligned}$$

As a consequence, $-\dot v(t) \in A(t) v(t)+g(t) = A(t) v(t) + f(t, v( t), h(t))$ a.e. with v(0) = u ₀ ∈ D(A(0)) so that by uniqueness v = v _h. Since $\Phi : \mathcal X \rightarrow \mathcal X$ is continuous and $\Phi (\mathcal X)$ is relatively compact in ${\mathcal C}_H([0, T])$, by [36, 44] Φ has a fixed point, say $h = \Phi (h) \in \mathcal X$, that means

$$\displaystyle \begin{gathered} h(t) = \Phi(h) (t) = u_0+\int_0^t v_h (s) ds, \hskip 3pt t \in [0, T],\\ \left\{ \begin{aligned} &v_{h} (0) = u_0 \in D(A(0))\\ &-\dot v_h(t) \in A(t) v_{h} (t) + f(t, v_h(t), h(t)) \hskip 3pt dt \text{-a.e.} \end{aligned} \right . \end{gathered} $$

The proof is complete.

Comments

The use of a generalized fixed point theorem is initiated in [24] dealing with some second-order sweeping process associated with a closed moving set C(t, u). Actually it is possible to obtain a variant of Theorem 3.4 by assuming that A(t) : D(A(t)) ⊂ H → 2^H is a maximal monotone operator with D(A(t) ball compact for every t ∈ [0, T] satisfying (H2) and

(H1)′ there exists a function $\beta \in W^{1,2}([0, T], \mathbb {R}; dt)$ which is nonnegative on I and non-decreasing with β(T) < ∞ such that

$$\displaystyle \begin{aligned} \operatorname{{\mathrm{dis}}}(A(t), A(s)) \leq | \beta(t) -\beta(s)| , \hskip 2pt \forall s, t \in [0, T].\end{aligned} $$

Here using fixed point theorem provides a short proof with new approach involving the continuous dependance of the trajectory v _h associated with the control $h \in \mathcal X$ and also the compactness of the integral of convex compact integrably bounded multifunctions [14].

4 Evolution Problems with Lipschitz Variation Maximal Monotone Operator and Application to Viscosity and Control

Now, based on the existence and uniqueness of $W^{2, \infty }_H([0, T])$ solution to evolution inclusion

$$\displaystyle \begin{aligned}({\mathcal S}_1) \left\{ \begin{array}{lll}0\in \ddot u(t) + A(t) \dot u(t) + f(t, u(t)), \hskip 2pt t \in [0, T]\\ u(0) = u_0, \dot u (0) = \dot u_0\in D(A(0)) \end{array} \right.\end{aligned} $$

we will present some problems in optimal control in a second-order evolution inclusion driven by a Lipschitz variation maximal monotone operator A(t) in the same vein as in Castaing-Marques-Raynaud de Fitte [25] dealing with the sweeping process. Before going further, we note that $({\mathcal S}_1)$ contains the evolution problem associated with the sweeping process by a closed convex Lipschitzian mapping $C : [0, T] \rightarrow \operatorname {{\mathrm {cc}}}(H)$

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll}0\in \ddot u(t) + N_{C(t)} (\dot u(t)) + f(t, u(t)), \hskip 2pt t \in [0, T]\\ u(0) = u_0, \dot u (0) = \dot u_0\in C(0) \end{array} \right. \end{aligned}$$

by taking A(t) = ∂ Ψ_C(t) in $({\mathcal S}_1)$.

We need some notations and background on Young measures in this special context. For the sake of completeness, we summarize some useful facts concerning Young measures. Let $(\Omega , {\mathcal F}, P)$ be a complete probability space. Let X be a Polish space, and let ${\mathcal C}^b(X)$ be the space of all bounded continuous functions defined on X. Let ${\mathcal M} ^1_+(X)$ be the set of all Borel probability measures on X equipped with the narrow topology. A Young measure $\lambda :\Omega \rightarrow {\mathcal M}^1_+({X})$ is, by definition, a scalarly measurable mapping from Ω into ${\mathcal M}^1_+({X})$, that is, for every $f \in {\mathcal C}^b({X})$, the mapping ω↦〈f, λ _ω〉 :=∫_X f(x) dλ _ω(x) is ${\mathcal F}$-measurable. A sequence (λ ⁿ) in the space of Young measures ${\mathcal Y}(\Omega ,{\mathcal F}, P; {\mathcal M} ^1_+(X) )$ stably converges to a Young measure $\lambda \in {\mathcal Y}(\Omega ,{\mathcal F}, P; {\mathcal M} ^1_+(X) )$ if the following holds:

$$\displaystyle \begin{aligned}\lim_{n \rightarrow \infty} \int_A \left[\int_X f(x) \,d\lambda^n_\omega(x)\right]\,dP(\omega) = \int_A \left[\int_X f(x) \,d\lambda_\omega(x)\right]\,dP(\omega)\end{aligned} $$

for every $A\in {\mathcal F}$ and for every $f\in {\mathcal C} ^b(X)$. We recall and summarize some results for Young measures.

Theorem 4.5 ( [22, Theorem 3.3.1])

Assume that S and T are Polish spaces. Let (μ ⁿ) be a sequence in ${\mathcal Y}(\Omega ,{\mathcal F}, P; {\mathcal M}^1_+ (S))$ , and let (ν ⁿ) be a sequence in ${\mathcal Y}(\Omega ,{\mathcal F}, P; {\mathcal M}^1_+ (T))$ . Assume that

• (i) (μ ⁿ) converges in probability to $\mu ^\infty \in {\mathcal Y}(\Omega ,{\mathcal F}, P; {\mathcal M}^1_+ ( S))$ ,

• (ii) (ν ⁿ) stably converges to $\nu ^\infty \in {\mathcal Y}(\Omega ,{\mathcal F}, P; {\mathcal M}^1_+ ( T))$.

Then (μ ⁿ ⊗ ν ⁿ) stably converges to μ ^∞⊗ ν ^∞ in ${\mathcal Y}(\Omega ,{\mathcal F}, P; {\mathcal M}^1_+ (S\times T))$.

Theorem 4.6 ( [22, Theorem 6.3.5])

Assume that X and Z are Polish spaces. Let (u ⁿ) be sequence of ${\mathcal F}$ -measurable mappings from Ω into X such that (u ⁿ) converges in probability to a ${\mathcal F}$ -measurable mapping u ^∞ from Ω into X, and let (v ⁿ) be a sequence of ${\mathcal F}$ -measurable mappings from Ω into Z such that (v ⁿ) stably converges to $\nu ^\infty \in {\mathcal Y}(\Omega ,{\mathcal F}, P; {\mathcal M}^1_+ (Z))$ . Let $h :\Omega \times X \times Z \rightarrow \mathbb {R}$ be a Carathéodory integrand such that the sequence (h(., u _n(.), v _n(.)) is uniformly integrable. Then the following holds:

$$\displaystyle \begin{aligned} \lim_{n\rightarrow \infty} \int_\Omega h(\omega, u^n(\omega), v^n(\omega))\,dP(\omega) = \int_\Omega \left[\int_Z h(\omega, u^\infty(\omega), z) \,d\nu ^\infty_\omega(z)\right]\,dP(\omega).\end{aligned} $$

In the remainder, Z is a compact metric space, and ${\mathcal M} ^1_+(Z)$ is the space of all probability Radon measures on Z. We will endow ${\mathcal M} ^1_+(Z)$ with the narrow topology so that ${\mathcal M} ^1_+(Z)$ is a compact metrizable space. Let us denote by ${\mathcal Y} ([0, T]; {\mathcal M} ^1_+(Z))$ the space of all Young measures (alias relaxed controls) defined on [0, T] endowed with the stable topology so that ${\mathcal Y} ([0, T]; {\mathcal M} ^1_+(Z))$ is a compact metrizable space with respect to this topology. By the Portmanteau Theorem for Young measures [22, Theorem 2.1.3], a sequence (ν ⁿ) in ${\mathcal Y} ([0, T]; {\mathcal M} ^1_+(Z))$ stably converges to $\nu \in {\mathcal Y} ([0, T]; {\mathcal M} ^1_+(Z))$ if

$$\displaystyle \begin{aligned}\lim_{n \rightarrow \infty} \int_0 ^T \left[\int_Z h_t(z) d\nu ^n_t(z)\right]\,dt = \int_0 ^T \left[\int_Z h_t(z) d\nu_t(z)\right]\,dt\end{aligned}$$

for all $h \in L^1_{{\mathcal C }(Z)}([0, T])$, where ${\mathcal C }(Z)$ denotes the space of all continuous real-valued functions defined on Z endowed with the norm of uniform convergence. Finally let us denote by ${\mathcal Z}$ the set of all Lebesgue-measurable mappings (alias original controls) z : [0, T] → Z and ${\mathcal R}:= {\mathcal Y} ([0, T]; {\mathcal M} ^1_+(Z))$ the set of all relaxed controls (alias Young measures) associated with Z. In the remainder, we assume that $H = \mathbb {R} ^d$ and Z is a compact subset in H.

For simplicity, let us consider a mapping f : [0, T] × H → H satisfying

• (i) for every x ∈ H × Z, f(., x) is Lebesgue-measurable on [0, T],

• (ii) there is M > 0 such that

  $$\displaystyle \begin{aligned}||f(t, x)|| \leq M(1+||x||)\end{aligned}$$

  for all (t, x) in [0, T] × H, and

  $$\displaystyle \begin{aligned}||f(t, x) -f(t, y)|| \leq M ||x-y||\end{aligned}$$

  for all (t, x, y) ∈ [0, T] × H × H.

We consider the $W^{2, \infty }_H([0, T])$ solution set of the two following control problems

$$\displaystyle \begin{aligned}({\mathcal S}_{\mathcal O}) \left\{ \begin{array}{lll}0\in \ddot u_{ x, y,\zeta}(t) +A(t) \dot u_{x, y, \zeta}(t)) + f(t, u_{x, y, \zeta}(t))+ \zeta(t), \hskip 2pt t \in [0, T]\\ u_{ x,y, \zeta}(0) =x\in H, \dot u_{x, y, \zeta}(0) = y\in D(A(0)) \end{array} \right. \end{aligned}$$

and

$$\displaystyle \begin{aligned}({\mathcal S}_{\mathcal R}) \left\{ \begin{array}{lll}0\in \ddot u_{x, y, \lambda}(t) +A(t) \dot u_{x, y, \lambda}(t)) + f(t, u_{x, y, \lambda}(t))+ \operatorname{{\mathrm{bar}}}(\lambda_t) , \hskip 2pt t \in [0, T]\\ u_{x,y, \lambda}(0) =x\in H, \dot u_{x, y, \lambda}(0) = y\in D(A(0)) \end{array} \right. \end{aligned}$$

where ζ belongs to the set $\mathcal Z$ of all Lebesgue-measurable mappings (alias original controls) ζ : [0, T] → Z original and λ belongs to the set $\mathcal R$ of all relaxed controls. Taking $({\mathcal S}_1)$ into account, for each $(x, y, \zeta ) \in H\times D(A(0))\times {\mathcal Z}$ (resp. $(x, y, \lambda ) \in H\times D(A(0))\times {\mathcal R}$, there exists a unique $W^{2, \infty }_H(]0, T])$ solutions, solution u _x,y,ζ (resp. u _x,y,λ), to $({\mathcal S}_{\mathcal O})$ (resp. $({\mathcal S}_{\mathcal R}))$. We aim to present some problems in the framework of optimal control theory for the above inclusions. In particular, we state a viscosity property of the value function associated with these evolution inclusions. Similar problems driven by evolution inclusion with perturbation containing Young measures are initiated by [22, 23]. However, the present study deals with a new setting in the sense that it concerns a second-order evolution inclusion involving time-dependent maximal monotone operator.

Now we present a lemma which is useful for our purpose.

Lemma 4.6

Let for all t ∈ [0, T], A(t) : D(A(t)) ⊂ H → 2^H be a maximal monotone operator satisfying (H1) and (H2). Let f : [0, T] × H → H be a mapping satisfying

• (i) for every x ∈ H × Z, f(., x) is Lebesgue-measurable on [0, T],

• (ii) there is M > 0 such that

  $$\displaystyle \begin{aligned}||f(t, x)|| \leq M(1+||x||)\end{aligned} $$

  for all (t, x) in [0, T] × H, and

  $$\displaystyle \begin{aligned}||f(t, x) -f(t, y)|| \leq M ||x-y||\end{aligned} $$

  for all (t, x, y) ∈ [0, T] × H × H.

Let $h_n , h \in L^\infty _H([0, T], dt)$ with ||h _n(t)||≤ 1 for all t ∈ [0, T], for all $n \in \mathbb {N}$ and ||h(t)||≤ 1 for all t ∈ [0, T]. Let us consider the two following second-order evolution inclusions:

$$\displaystyle \begin{gathered} {\mathcal S}(A, f, h_n, x, y) \left\{ \begin{array}{lll}0\in \ddot u_{x, y, h_n}(t) + A(t) \dot u_{x, y, h_n}(t) + f(t, u_{x, y, h_n}(t))+ h_n(t) , \hskip 2pt t \in [0, T] \\ u_{x, y, h_n}(0) = x, \dot u_{x, y, h_n} (0) =y \in D(A(0)) \end{array} \right. \\ \rule{0cm}{.8cm} {\mathcal S}(A, f, h, x, y) \left\{ \begin{array}{lll}0\in \ddot u_{x, y, h}(t) + A(t) \dot u_{x, y, h}(t) + f(t, u_{x, y, h}(t))+ h(t) , \hskip 2pt t \in [0, T] \\ u_{x, y, h}(0) = x, \dot u_{x, y, h} (0) =y \in D(A(0)) \end{array} \right.\vspace{-4pt} \end{gathered} $$

where $u_{x, y, h_n}$ (resp. u _x,y,h ) is the unique $W ^{2, \infty }_H([0, T])$ solution to $({\mathcal S}(A, f, h_n, x, y))$ (resp. $({\mathcal S}(A, f, h_n, x, y))$ ). Assume that (h _n) σ(L ¹, L ^∞) converges to h. Then $(u_{x, y, h_n})$ converges pointwisely to u _x,y,h.

Proof

We note that $\ddot u_{x, y, h_n}$ is uniformly bounded, so there is $u \in W ^{2, \infty }_H([0, T])$ such that

• $u_{x, y, h_n}\ \rightarrow u$ pointwisely with u(0) = x,

• $\dot u_{x, y, h_n} \rightarrow \dot u$ pointwisely with $\dot u(0) = y$,

• $\ddot u_{x, y, h_n} \rightarrow \ddot u$ with respect to σ(L ¹, L ^∞).

Using Lemma 2.3, it is not difficult to see that $\dot u(t) \in D(A(t))$ for every t ∈ [0, T]. As $f(t, u_{x, y, h_n}(t)) \rightarrow f(t, u(t) )$ pointwisely so that $f( . , u_{x, y, h_n}(.)) \rightarrow f(.., u(.) )$ with respect to σ(L ¹, L ^∞). Since (h _n) σ(L ¹, L ^∞) converges to h, so that $f( . ,u_{x, y, h_n}(.)) + h _n \rightarrow f(t., u(.) )+h$ with respect to σ(L ¹, L ^∞). And so $\ddot u_{x, y, h_n}(.) + f(., u_{x, y, h_n}(.))+ h_n(.)$ σ(L ¹, L ^∞) converges to $\dot u + f(.., u(.) )+h$. As a consequence, we may also assume that $\ddot u_{x, y, h_n}(.) +f(., u_{x, y, h_n}(.))+ h_n(.)$ Komlos converges to $\dot u + f(.., u(.) )+h$. Coming back to the inclusion $-\ddot u_{x, y, h_n}(t) - f(t, u_{x, y, h_n}(t))- h_n(t) \in A(t) \dot u_{x, y, h_n}(t)$, we have by the monotonicity of A(t)

$$\displaystyle \begin{aligned}\langle \ddot u_{x, y, h_n}(t) + f(t, u_{x, y, h_n}(t))+ h_n(t), \dot u_{x, y, h_n}(t) -\eta \rangle \leq \langle A^0(t, \eta), \eta- \dot u_{x, y, h_n}(t) \rangle\end{aligned} $$

for any η ∈ D(A(t)). For notational convenience, set

$$\displaystyle \begin{aligned} v_n(t) =& \ddot u_{x, y, h_n}(t) + f(t, u_{x, y, h_n}(t))+ h_n(t), \forall t \in [0, T],\\ v(t) = & \ddot u(t) + f(t, u(t) )+h(t), \forall t \in [0, T]. \end{aligned} $$

There is a negligible set N such that

$$\displaystyle \begin{aligned}\lim_n \frac{1}{n} \sum_{i= 1} ^n v_i (t) = v(t)\end{aligned} $$

for t∉N. Let us write

$$\displaystyle \begin{aligned}\langle v_n (t), \dot u(t) -\eta \rangle = \langle v_n (t), \dot u_{x, y, h_n}(t) -\eta \rangle + \langle v_n (t), \dot u(t) - \dot u_{x, y, h_n}(t) \rangle\end{aligned} $$

so that

$$\displaystyle \begin{aligned} \frac{1}{n} \sum_{i{=} 1} ^n \langle v_i (t), \dot u(t) {-}\eta \rangle {=}& \frac{1}{n} \sum_{i= 1} ^n \langle v_i (t), \dot u_{x, y, h_i}(t) {-}\eta \rangle {+} \frac{1}{n} \sum_{i= 1} ^n \langle v_i (t), \dot u(t) {-} \dot u_{x, y, h_i}(t) \rangle\\ \leq& \frac{1}{n} \sum_{i= 1} ^n \langle A^0(t, \eta), \eta{-} \dot u_{x, y, h_i}(t) \rangle {+}L \frac{1}{n} \sum_{i= 1} ^n || \dot u(t) {-} \dot u_{x, y, h_i}(t) ||, \end{aligned} $$

where L is a positive generic constant. Passing to the limit when n goes to ∞ in this inequality gives immediately

$$\displaystyle \begin{aligned}\langle v(t), \dot u(t) -\eta \rangle \leq \langle A^0(t, \eta), \eta- \dot u(t) \rangle\end{aligned}$$

so that by Lemma 2.2 we get

$$\displaystyle \begin{aligned}-\ddot u(t) - f(t, u_{x, y, h}(t))- h(t) \in A(t) \dot u(t) \text{ a.e.}\end{aligned}$$

with u(0) = x and $\dot u(0) = y$. Due to the uniqueness of solution, we get u(t) = u _x,y,h(t) for all t ∈ [0, T]. The proof is complete.

The following shows the continuous dependence of the solution with respect to the control.

Theorem 4.7

Let for all t ∈ [0, T], A(t) : D(A(t)) ⊂ H → 2^H be a maximal monotone operator satisfying (H1) and (H2). Let f : [0, T] × H → H be a mapping satisfying

• (i) for every x ∈ H × Z, f(., x) is Lebesgue-measurable on [0, T],

• (ii) there is M > 0 such that

  $$\displaystyle \begin{aligned}||f(t, x)|| \leq M(1+||x||) \end{aligned}$$

  for all (t, x) in [0, T] × H, and

  $$\displaystyle \begin{aligned} ||f(t, x_1) -f(t, x_2)|| \leq M ||x_1-x_2|| \end{aligned}$$

  for all (t, x ₁, (t, x ₂, ) ∈ [0, T] × H × H.

Let Z be a compact subset of H. Let us consider the control problem

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \ddot u_{x, y, \nu}(t) + A(t) \dot u_{x, y, \nu}(t) + f(t, u_{x, y, \nu }(t))+ \operatorname{{\mathrm{bar}}}(\nu_t), \hskip 2pt t \in [0, T] \\ u_{x, y, \nu}(0) = x, \dot u_{x, y, \nu} (0) =y \in D(A(0)) \end{array} \right.\end{aligned} $$

where $\operatorname {{\mathrm {bar}}}(\nu _t)$ denotes the barycenter of the measure $\nu _t \in \mathcal M ^1_+(Z)$ and u _x,y,ν is the unique $W ^{2, \infty }_H([0, T])$ solution associated with to $\operatorname {{\mathrm {bar}}}(\nu _t)$ . Then, for each t ∈ [0, T], the mapping ν↦u _x,y,ν is continuous from $\mathcal R$ to C _H([0, T], where $\mathcal R$ is endowed with the stable topology and C _H([0, T] is endowed with the topology of pointwise convergence.

Proof

(a) Let $\nu \in \mathcal R$ and let $\operatorname {{\mathrm {bar}}}(\nu ) : t \mapsto \operatorname {{\mathrm {bar}}}(\nu _t), \; t \in [0, T]$. It is easy to check that $\nu \mapsto \operatorname {{\mathrm {bar}}}(\nu )$ from $\mathcal R$ to $L^\infty _H([0, T])$ is continuous with respect to the stable topology and the $\sigma (L^1 _H, L^\infty _H)$, respectively. Note that $\mathcal R$ is compact metrizable for the stable topology. Now let (ν ⁿ) be a sequence in $\mathcal R$ which stably converges to $\nu \in \mathcal R$. Then $\operatorname {{\mathrm {bar}}}(\nu ^n)$ $\sigma ( L^1_H, L ^\infty _H)$ converges to $\operatorname {{\mathrm {bar}}}(\nu )$. By Lemma 4.6, we see that $u_{x,y, \nu ^n}$ pointwisely converges to u _x,y,ν. The proof is complete.

We are now able to relate the Bolza type problems associated with the maximal monotone operator A(t) as follows:

Theorem 4.8

With the hypotheses and notations of Theorem 4.7 , assume that $J : [0, T]\times H \times Z\rightarrow \mathbb {R}$ is a Carathéodory integrand, that is, J(t, ., .) is continuous on H × Z for every t ∈ [0, T] and J(., x, z) is Lebesgue-measurable on [0, T] for every (x, z) ∈ H × Z, which satisfies the condition $({\mathcal C})$ : for every sequence (ζ _n) in $\mathcal Z$ , the sequence $(J(., u_{x, y, \zeta ^n}(.), \zeta ^n(.))$ is uniformly integrable in $L^1_{\mathbb {R}} ([0, T],\,dt)$ , where $u_{x , y, \zeta ^n}$ denotes the unique $W ^{2, \infty }_H([0, T])$ solution associated with ζ ⁿ to the evolution inclusion

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \ddot u_{x, y, \zeta ^n}(t) + A(t) \dot u_{x, y, \zeta ^n}(t) + f(t, u_{x, y, \zeta ^n }(t))+ \zeta ^n(t), \hskip 2pt t \in [0, T] \\ u_{x, y, \zeta ^n}(0) = x, \dot u_{x, y, \zeta ^n} (0) =y \in D(A(0)) \end{array} \right.\end{aligned} $$

Let us consider the control problems

$$\displaystyle \begin{aligned}\inf (P_{\mathcal Z}):=\hskip 3pt \inf_{\zeta \in \mathcal Z } \int_0^T J(t, u_{x,y, \zeta}(t),\zeta(t))\,dt\end{aligned}$$

and

$$\displaystyle \begin{aligned}\inf(P_{\mathcal R} ):=\hskip 3pt \inf_{\lambda \in \mathcal R } \int_0 ^T \left[\int_Z J(t, u_{x, y, \lambda}(t), z) \,\lambda_t(dz)\right]\,dt\end{aligned}$$

where u _{x, y, ζ} (resp. u _{x, y, λ} ) is the unique $W ^{2, \infty }_H([0, T])$ solution associated with ζ ( resp. λ) to

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \ddot u_{x, y, \zeta }(t) + A(t) \dot u_{x, y, \zeta }(t) + f(t, u_{x, y, \zeta }(t))+ \zeta (t), \hskip 2pt t \in [0, T] \\ u_{x, y, \zeta} (0) = x, \dot u_{x, y, \zeta } (0) =y \in D(A(0)) \end{array} \right. \end{aligned}$$

and

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \ddot u_{x, y, \lambda}(t) + A(t) \dot u_{x, y, \nu}(t) + f(t, u_{x, y, \nu }(t))+ \operatorname{{\mathrm{bar}}}(\lambda_t), \hskip 2pt t \in [0, T] \\ u_{x, y, \lambda}(0) = x, \dot u_{x, y, \lambda} (0) =y \in D(A(0)) \end{array} \right. \end{aligned}$$

respectively. Then one has

$$\displaystyle \begin{aligned}\inf (P_{ \mathcal Z}) = \inf (P_{ \mathcal R} ).\end{aligned}$$

Proof

Take a control $\lambda \in {\mathcal R}$. By virtue of the denseness with respect to the stable topology of $\mathcal Z$ in $\mathcal R$, there is a sequence $(\zeta ^n)_{n \in \mathbb {N}}$ in $\mathcal Z$ such that the sequence $(\delta _{\zeta ^n})_{n \in \mathbb {N}}$ of Young measures associated with $(\zeta ^n)_{n \in \mathbb {N}}$ stably converges to λ. By Theorem 4.7, the sequence $(u_{x ,y, \zeta ^n})$ of $W ^{2, \infty }_H([0, T])$ solutions associated with ζ ⁿ pointwisely converges to the unique $W ^{2, \infty }_H([0, T])$ solution u _x,y,λ. As $(J(t, u_{x, y, \zeta ^n}(t),\zeta ^n(t)))$ is uniformly integrable by assumption $({\mathcal C})$, using Theorem 4.6 (or [22, Theorem 6.3.5]), we get

$$\displaystyle \begin{aligned}\lim_{n \rightarrow \infty} \int_0 ^T J(t, u_{x, y, \zeta ^n}(t), \zeta^n(t))\,dt = \int_0 ^T \left[\int_Z J(t, u_{x, y, \lambda}, z) d\lambda_t(z)\right]\,dt.\end{aligned}$$

As

$$\displaystyle \begin{aligned}\int_0 ^T J(t, u_{x, y, \zeta ^n}(t), \zeta^n(t))\,dt \geq \inf (P_{{\mathcal Z}})\end{aligned}$$

for all $n\in \mathbb {N}$, so is

$$\displaystyle \begin{aligned}\int_0 ^T \left[\int_ Z J(t, u_{x, y, \lambda}, z) d\lambda_t(z)\right]\,dt \geq \inf (P_{{\mathcal Z}}) ;\end{aligned}$$

by taking the infimum on $\mathcal R$ in this inequality, we get

$$\displaystyle \begin{aligned}\inf (P_{{\mathcal R}}) \geq \inf (P_{{\mathcal O}})\end{aligned}$$

As $\inf (P_{ {\mathcal O}}) \geq \inf (P_{{\mathcal R}})$, the proof is complete.

In the framework of optimal control, the above considerations lead to the study of the value function associated with the evolution inclusion

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \ddot u_{\tau, x, y, \nu}(t) + A(t) \dot u_{\tau, x, y, \nu}(t) + f(t, u_{\tau, x, y, \nu}(t))+ \operatorname{{\mathrm{bar}}}(\nu_t), \\ u_{\tau, x, y, \nu} (\tau) = x, \dot u_{\tau, x, y, \nu}(\tau) = y \in D(A(\tau)). \end{array} \right. \end{aligned}$$

The following shows that the value function satisfies the dynamic programming principle (DPP).

Theorem 4.9

(of dynamic programming principle). Assume the hypothesis and notations of Theorem 4.7 , and let x ∈ E, τ < T and σ > 0 such that τ + σ < T. Assume that $J : [0, T]\times H \times Z\rightarrow \mathbb {R}$ is bounded and continuous. Let us consider the value function

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle V_J(\tau, x, y) = \sup_{\nu \in \mathcal R} \int_\tau^T \left[\int_Z J(t, u_{\tau, x, y, \nu}(t), z) \nu_t(dz)\right] \,dt,\\ &\displaystyle \qquad (\tau, x, y ) \in [0, T]\times H\times D(A(\tau))\vspace{-4pt}\end{array} \end{aligned} $$

where u _τ,x,y,ν is the $W ^{2, \infty }_H([0, T])$ solution to the evolution inclusion defined on [τ, T] associated with the control $\nu \in \mathcal R$ starting from x, y at time τ

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \ddot u_{\tau, x, y, \nu}(t) + A(t) \dot u_{\tau, x, y, \nu}(t) + f(t, u_{\tau, x, y, \nu}(t))+ \operatorname{{\mathrm{bar}}}(\nu_t), \\ u_{\tau, x, y, \nu} (\tau) = x, \dot u_{\tau, x, y, \nu}(\tau) = y \in D(A(\tau)) \end{array} \right.\end{aligned} $$

Then the following holds:

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle V_J(\tau, x, y ) = \sup _{\nu \in \mathcal R} \Bigg\{ \int_{\tau}^{\tau + \sigma}\left[\int_Z J (t, u_{\tau, x, y, \nu}(t), z )\nu_t(dz) \right]dt\\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \quad +V_J(\tau + \sigma, u_{\tau, x, y, \nu}(\tau + \sigma), \dot u_{\tau, x, y, \nu}(\tau + \sigma)\Bigg\} \end{array} \end{aligned} $$

with

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle V_J(\tau + \sigma, u_{\tau,x, \nu }(\tau + \sigma), \dot u_{\tau,x, \nu }(\tau + \sigma))\\ &\displaystyle &\displaystyle \quad \qquad =\sup_{ \mu \in \mathcal R} \int_{\tau+\sigma}^{T}\left[ \int_Z J (t, v_{\tau+\sigma, u_{\tau, x,y, \nu}(\tau+\sigma), \dot u_{\tau, x, y, \nu}(\tau+\sigma), \mu}(t), z) \mu_t(dz)\right] \,dt \end{array} \end{aligned} $$

where $v_{\tau +\sigma , u_{\tau , x, y, \nu }(\tau +\sigma ), \dot u_{\tau , x, y, \nu }(\tau +\sigma ), \mu }$ ¹ is the $W^{2, \infty }_H(\tau +\sigma , T)$ solution defined on [τ + σ, T] associated with the control $\mu \in \mathcal R$ starting from $u_{\tau , x, \nu }(\tau +\sigma ), \dot u_{\tau , x, \nu }(\tau +\sigma )$ at time τ + σ

(13)

Proof

Let

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle &\displaystyle W_J(\tau, x, y) := \sup _{\nu \in \mathcal R } \Bigg\{ \int_{\tau}^{\tau + \sigma}\left[\int_Z J (t, u_{\tau, x, y, \nu}(t), z )\nu_t(dz) \right]dt \\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad +V_J(\tau + \sigma, u_{\tau, x, y, \nu }(\tau + \sigma))\Bigg\}.\vspace{-4pt}\end{array} \end{aligned} $$

For any $\nu \in \mathcal R$, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \int_\tau^T \left[\int_Z J (t, u_{\tau, x, y, \nu}(t), z )\nu_t(dz) \right]dt {=} \int_\tau^{\tau+\sigma} \left[\int_Z J (t, u_{\tau, x, y, \nu}(t), z )\nu_t(dz) \right]dt\\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad + \int_{\tau+\sigma}^T \left[\int_Z J (t, u_{\tau, x, y, \nu}(t), z )\nu_t(dz) \right]dt.\vspace{-4pt} \end{array} \end{aligned} $$

By the definition of $V_J(\tau + \sigma , u_{\tau ,x, y, \nu }(\tau + \sigma ),\dot u_{\tau ,x, y, \nu }(\tau + \sigma )$, we have

$$\displaystyle \begin{aligned}V_J(\tau + \sigma, u_{\tau,x, y, \nu }(\tau +\sigma), \dot u_{\tau,x, y, \nu }(\tau + \sigma) \geq \int_{\tau+\sigma}^T \left[\int_Z J (t, u_{\tau, x, y, \nu}(t), z )\nu_t(dz) \right]dt. \end{aligned}$$

It follows that

By taking the supremum on $\nu \in \mathcal R$ in this inequality, we get

$$\displaystyle \begin{aligned} V_J(\tau, x, y ) \leq& \sup_{\nu \in \mathcal R }\bigg\{ \int_\tau^{\tau+\sigma} \left[\int_Z J (t, u_{\tau, x, y, \nu}(t), z )\nu_t(dz) \right]dt\\ &+ V_J(\tau + \sigma, u_{\tau,x, y, \nu }(\tau + \sigma), \dot u_{\tau,x, y, \nu }(\tau + \sigma) )\bigg\}\\ =& W_J(\tau, x, y). \end{aligned} $$

Let us prove the converse inequality.

Main fact: $\nu \mapsto V_J(\tau + \sigma , u_{\tau ,x, \nu }(\tau + \sigma ), \dot u_{\tau ,x, \nu }(\tau + \sigma ) )$ is continuous on $\mathcal R$.

Let us focus on the expression of $V_J(\tau + \sigma , u_{\tau ,x, \nu }(\tau + \sigma ), \dot u_{\tau ,x, \nu }(\tau + \sigma ))$:

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle V_J(\tau + \sigma, u_{\tau,x, \nu }(\tau + \sigma), \dot u_{\tau,x, \nu }(\tau + \sigma) )\\ &\displaystyle &\displaystyle \qquad \qquad \qquad =\sup_{ \mu \in \mathcal R} \int_{\tau+\sigma}^{T}\bigg[ \int_Z J (t, v_{\tau+\sigma, u_{\tau, x, \nu}(\tau+\sigma), \dot u_{\tau,x, \nu }(\tau + \sigma), \mu}(t), z) \mu_t(dz)\bigg] \,dt \end{array} \end{aligned} $$

where $v_{\tau +\sigma , u_{\tau , x, \nu }(\tau +\sigma ), \dot u_{\tau ,x, \nu }(\tau + \sigma ), \mu }$ denotes the trajectory solution on [τ + σ, T] associated with the control $\mu \in \mathcal R$ starting from $u_{\tau , x, \nu }(\tau +\sigma ), \dot u_{\tau , x, \nu }(\tau +\sigma ),$ at time τ + σ in (13). Using the continuous dependence of the solution with respect to the state and the control, it is readily seen that the mapping $(\nu , \mu ) \mapsto v_{\tau +\sigma , u_{\tau , x, \nu }(\tau +\sigma ), \dot u_{\tau , x, \nu }(\tau +\sigma ), \mu }(t)$ is continuous on $\mathcal R \times \mathcal R$ for each t ∈ [τ, T], namely, if ν ⁿ stably converges to $\nu \in \mathcal R$ and μ ⁿ stably converges to $\mu \in \mathcal R$, then $v_{\tau +\sigma , u_{\tau , x, \nu ^n}(\tau +\sigma ), \dot u_{\tau , x, \nu }(\tau +\sigma ), \mu ^n}$ pointwisely converges to $v_{\tau +\sigma , u_{\tau , x, \nu }(\tau +\sigma ), \dot u_{\tau , x, \nu }(\tau +\sigma ), \mu }$. By using the fiber product of Young measure (see Theorem 4.5 or [22, Theorem 3.3.1]), we deduce that

$$\displaystyle \begin{aligned}(\nu, \mu) \mapsto \int_{\tau+\sigma}^{T}\bigg[ \int_Z J (t, v_{\tau+\sigma, u_{\tau, x, \nu}(\tau+\sigma), \dot u_{\tau, x, \nu}(\tau+\sigma) \mu}(t), z) \mu_t(dz)\bigg] \,dt\end{aligned}$$

is continuous on $\mathcal R \times \mathcal R$. Consequently $\nu \mapsto V_J(\tau + \sigma , u_{\tau ,x, \nu }(\tau + \sigma ), \dot u_{\tau ,x, \nu }(\tau + \sigma ))$ is continuous on $\mathcal R$. Hence the mapping $\nu \mapsto \int _\tau ^{\tau + \sigma } [\int _Z J (t, u_{\tau , x, \nu }(t), z )\nu _t(dz) ]dt + V_J(\tau + \sigma , u_{\tau ,x, \nu }(\tau + \sigma ), \dot u_{\tau ,x, \nu }(\tau + \sigma ))$ is continuous on $\mathcal R$. By compactness of $\mathcal R$, there is a maximum point $\nu ^1 \in \mathcal R$ such that

Similarly there is $\mu ^2 \in \mathcal R$ such that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle V_J(\tau + \sigma, u_{\tau, x, \nu^1 }(\tau + \sigma), \dot u_{\tau, x, \nu^1 }(\tau + \sigma))\\ &\displaystyle &\displaystyle \qquad \qquad \quad = \int_{\tau+\sigma}^{T}\bigg[\int_Z J(t, v_{\tau+\sigma, u_{\tau, x, \nu^1} (\tau+\sigma), \dot u_{\tau, x, \nu^1} (\tau+\sigma), \mu ^2}(t), z)\mu^2_t(dz)\bigg]dt \end{array} \end{aligned} $$

where

$$\displaystyle \begin{aligned}v_{\tau+\sigma, u_{\tau, x, \nu^1} (\tau+\sigma), \dot u_{\tau, x, \nu^1} (\tau+\sigma), \mu^2}(t)\end{aligned}$$

denotes the trajectory solution associated with the control $\mu ^2 \in \mathcal R$ starting from $u_{\tau , x, \nu ^1} (\tau +\sigma ), \dot u_{\tau , x, \nu ^1} (\tau +\sigma )$ at time τ + σ defined on [τ + σ, T]

(14)

Let us set

$$\displaystyle \begin{aligned}\overline \nu := 1_{[\tau, \tau+\sigma]} \nu ^1+ 1_{[ \tau+\sigma, T]} \mu^2.\end{aligned}$$

Then $\overline \nu \in \mathcal R$. Let $w_{\tau , x, y, \overline \nu }$ be the trajectory solution on [τ, T] associated with $\overline \nu \in \mathcal R$, that is,

$$\displaystyle \begin{aligned}\left\{\begin{aligned} &0\in \ddot w_{\tau, x, y, \overline \nu } (t) {+} A(t)\dot w_{\tau, x, y, \overline \nu}(t) {+} f(t, w_{\tau, x, y, \overline \nu }(t) ) {+} \operatorname{{\mathrm{bar}}}(\overline \nu_t )\ &w_{\tau, x, y, \overline \nu} (\tau) = x \\ &\dot w_{\tau, x, y, \overline \nu} (\tau) = y \in D(A(\tau) ) \end{aligned}\right.\end{aligned}$$

By uniqueness of the solution, we have

$$\displaystyle \begin{aligned} w_{\tau, x, y, \overline \nu}(t) &= u_{\tau, x, y, \nu^1}(t), \, \forall t \in [\tau, \tau+\sigma],\\ w_{\tau, x, y, \overline \nu}(t) &= v_{\tau+\sigma, u_{\tau, x, y, \nu^1} (\tau+\sigma), \dot u_{\tau, x, y, \nu^1} (\tau+\sigma), \mu^2}(t), \, \forall t \in [\tau+\sigma , T]. \end{aligned} $$

Coming back to the expression of V _J and W _J, we have

$$\displaystyle \begin{aligned} W_J(\tau, x, y) =& \int_{\tau} ^{\tau+\sigma} \bigg[\int_Z J( t, u_{\tau, x, y, \nu^1}(t) , z)\nu ^1_t(dz)\bigg]\,dt\\ &+ \int_{\tau+\sigma } ^T \bigg[\int_Z J(t, v_{\tau+\sigma, u_{\tau, x, \nu^1} (\tau+\sigma), \dot u_{\tau, x, \nu^1} (\tau+\sigma), \mu^2}(t), z)\mu ^2_t(dz)\bigg]\,dt\\ = &\int_{\tau} ^{T } \bigg[\int_Z J (t, w_{\tau, x, y, \overline \nu}(t), z) \overline \nu_t(dz)\bigg]\,dt\\ \leq& \sup_{\nu \in \mathcal R} \bigg\{ \int_{\tau}^T \bigg[\int_Z J (t, u_{\tau, x, y, \nu }(t), z)\nu_t(dz)\bigg]\,dt\bigg\} = V_J(\tau, x, y). \end{aligned} $$

The proof is complete.

In the above evolution problem, we deal with second-order inclusion of the form

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \ddot u_{x, y, \lambda}(t) + A(t) \dot u_{x, y, \lambda}(t) + f(t, u_{x, y, \lambda }(t))+ \operatorname{{\mathrm{bar}}}(\lambda_t), \hskip 2pt t \in [0, T] \\ u_{x, y, \lambda}(0) = x, \dot u_{x, \lambda} (0) =y \in D(A(0)) \end{array} \right. \end{aligned}$$

with perturbed term f and $\operatorname {{\mathrm {bar}}}(\lambda _t)$. Now we focus to the evolution inclusion of the form

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \dot u_{x, \lambda}(t) + A(t) u_{x, \lambda}(t) + f(t, u_{x, \lambda }(t))+ \operatorname{{\mathrm{bar}}}(\lambda_t), \hskip 2pt t \in [0, T] \\ u_{x, \lambda}(0) = x \in D(A(0)) \end{array} \right. \end{aligned}$$

By Theorem 3.1, there is a unique Lipschitz solution u _x,λ to this inclusion. Using the above techniques and Theorem 3.1, we have a result of dynamic principle that is similar to Theorem 4.9.

Theorem 4.10 (of dynamic programming principle)

Assume the hypothesis and notations of Theorem 3.1 , and let x ∈ E, τ < T and σ > 0 such that τ + σ < T. Assume that $J : [0, T]\times H \times Z\rightarrow \mathbb {R}$ is bounded and continuous. Let us consider the value function

where u _τ,ν is the Lipschitz solution to the evolution inclusion defined on [τ, T] associated with the control $\nu \in \mathcal R$ starting from x, at time τ

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \dot u_{\tau, x, \nu}(t) + A(t) u_{\tau, x, \nu}(t) + f(t, u_{\tau, x, \nu}(t))+ \operatorname{{\mathrm{bar}}}(\nu_t), \\ u_{\tau, x, \nu} (\tau) = x \in D(A(\tau)). \end{array} \right. \end{aligned}$$

Then the following holds:

$$\displaystyle \begin{aligned}V_J(\tau, x ) = \sup _{\nu \in \mathcal R} \bigg\{ \int_{\tau}^{\tau + \sigma}\left[\int_Z J (t, u_{\tau, x, \nu}(t), z )\nu_t(dz) \right]dt +V_J(\tau + \sigma, u_{\tau, x, \nu}(\tau + \sigma))\bigg\} \end{aligned}$$

with

$$\displaystyle \begin{aligned}V_J(\tau + \sigma, u_{\tau,x, \nu }(\tau + \sigma)) =\sup_{ \mu \in \mathcal R} \int_{\tau+\sigma}^{T}\bigg[ \int_Z J (t, v_{\tau+\sigma, u_{\tau, x, \nu}(\tau+\sigma), \mu}(t), z) \mu_t(dz)\bigg] \,dt\end{aligned}$$

where $v_{\tau +\sigma , u_{\tau , x, \nu }(\tau +\sigma ), \mu }$ ² is the Lipschitz solution defined on [τ + σ, T] associated with the control $\mu \in \mathcal R$ starting from u _τ,x,ν(τ + σ) at time τ + σ

Let us mention a useful lemma. See also [16, 22, 23] for related results.

Lemma 4.7

Assume the hypothesis and notations of Theorem 3.1 . Let Z be a compact subset in H, and ${\mathcal M} ^1_+(Z)$ is endowed with the narrow topology and $\mathcal R$ the space of relaxed controls associated with Z. Let $\Lambda : [0, T]\times H \times {\mathcal M} ^1_+(Z) \rightarrow \mathbb {R}$ be an upper semicontinuous function such that the restriction of Λ to $[0, T]\times B \times {\mathcal M} ^1_+(Z)$ is bounded on any bounded subset B of H. Let (t ₀, x ₀) ∈ [0, T] × E. If $\max _{\mu \in {\mathcal M} ^1_+(Z) } \Lambda (t_0, x_0, \mu ) < -\eta < 0$ for some η > 0, then there exist σ > 0 such that

$$\displaystyle \begin{aligned}\sup_{\nu \in \mathcal R} \int_{t_0} ^{t_0+\sigma} \Lambda (t, u_{t_0, x_0, \nu}(t), \nu_t)\,dt < -\frac{\sigma \eta}{2}\end{aligned}$$

where $u_{t_0, x_0, \nu }$ is the trajectory solution associated with the control $\nu \in \mathcal R$ and starting from x ₀ at time t ₀

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &0\in \dot u_{t_0, x_0, \nu} (t)+ A(t) u_{t_0, x_0, \nu} (t) + f(t, u_{t_0, x_0, \nu} (t)+ \operatorname{{\mathrm{bar}}}(\nu_t), \hskip 2pt t\in [t_0, T], \\ & u_{t_0, x_0, \nu} (t_0) = x_0 \in D(A(t_0)). \end{aligned} \right. \end{aligned}$$

Proof

By our assumption $\max _{\mu \in {\mathcal M} ^1_+(Z) } \Lambda (t_0, x_0, \mu ) < -\eta < 0$ for some η > 0. As the function (t, x, μ)↦ Λ(t, x, μ) is upper semicontinuous, so is the function

$$\displaystyle \begin{aligned}(t, x) \mapsto \max_{\mu \in {\mathcal M} ^1_+(Z)} \Lambda(t, x, \mu).\end{aligned}$$

Hence there exists ζ > 0 such that

$$\displaystyle \begin{aligned}\max_{\mu \in {\mathcal M} ^1_+(Z)} \Lambda(t, x, \mu) < -\frac{\eta} {2}\end{aligned}$$

for 0 < t − t ₀ ≤ ζ and ||x − x ₀||≤ ζ. Thus, for small values of σ, we have

$$\displaystyle \begin{aligned}||u_{t_0,x_0, \nu} (t)- u_{t_0, x_0, \nu} (t_0) || \leq \zeta\end{aligned}$$

for all t ∈ [t ₀, t ₀ + σ] and for all $\nu \in \mathcal R$ because $||\dot u_{t_0, x_0, \nu } (t)|| \leq K= \mathrm {Constant}$ for all $\nu \in \mathcal R$ and for all t ∈ [0, T] so that $|| u_{t_0, x_0, \nu } (t)|| \leq L= \mathrm {Constant}$ for all $\nu \in \mathcal R$ and for all t ∈ [0, T] Hence $t\mapsto \Lambda (t, u_{t_0, x_0, \nu } (t), \nu _t)$ is bounded and Lebesgue-measurable on [t ₀, t ₀ + σ]. Then by integrating

$$\displaystyle \begin{aligned}\int_{t_0}^{t_0+\sigma} \Lambda(t, u_{t_0, x_0, \nu} (t), \nu_t)\,dt \leq \int_{t_0}^{t_0+\sigma} \bigg[ \max_{ \mu \in {\mathcal M} ^1_+(Z)} \Lambda(t, u_{t_0, x_0, \nu} (t), \mu)\bigg]\,dt < -\frac{\sigma \eta}{2}.\end{aligned}$$

The proof is complete.

Now to finish the paper, we provide a direct application to the viscosity solution to the evolution inclusion of the form

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lll} 0\in \dot u_{x, \lambda}(t) + A(t) u_{x, \lambda}(t) + f(t, u_{x, \lambda }(t))+ \operatorname{{\mathrm{bar}}}(\lambda_t), \hskip 2pt t \in [0, T] \\ u_{x, \lambda}(0) = x \in D(A(0)) \end{array} \right. \end{aligned}$$

where A(t) is a convex weakly compact valued H → cwk(H) maximal monotone operator.

Theorem 4.11

Let for every t ∈ [0, T], A(t) : H → cwk(H) be a convex weakly compact valued maximal monotone operator satisfying

• (H1) there exists a real constant α ≥ 0 such that

  $$\displaystyle \begin{aligned}\operatorname{{\mathrm{dis}}}(A(t), A(s))\leq \alpha(t-s)\;\; {for}\;\;0\leq s\leq t\leq T.\end{aligned}$$

• (H2) there exists a nonnegative real number c such that

  $$\displaystyle \begin{aligned}\|A^0(t,x)\|\leq c(1+\| x\|), t\in [0, T], x \in H\end{aligned}$$

• (H3) (t, x)↦A(t)x is scalarly upper semicontinuous on [0, T] × H.

Let Z be a compact subset in H, and let $\mathcal R$ be the space of relaxed controls associated with Z. Let f : [0, T] × H → H be a continuous mapping satisfying

• (i) there is M > 0 such that ||f(t, x)||≤ M(1 + ||x||) for all (t, x) in [0, T] × H,

• (ii) ||f(t, x) − f(t, y)||≤ M||x − y|| for all (t, x, y) ∈ [0, T] × H × H.

Assume that $J : [0, T]\times H \times Z\rightarrow \mathbb {R}$ is bounded and continuous. Let us consider the value function

$$\displaystyle \begin{aligned}V_J(\tau, x) = \sup_{\nu \in \mathcal R} \int_{\tau}^T \bigg[\int_Z J(t, u_{{\tau}, x, \nu}(t), z) \nu_t(dz)\bigg] \,dt, \, (\tau, x) \in [0, T]\times H\end{aligned}$$

where u _τ,x,ν is the trajectory solution on [τ, T] of the evolution inclusion associated with A(t) and the control $\nu \in \mathcal R$ and starting from x ∈ H at time τ

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &0\in \dot u_{\tau, x, \nu} (t) + A(t , u_{\tau, x, \nu} (t))+ f(t, u_{\tau, x, \nu}(t))+ \operatorname{{\mathrm{bar}}}(\nu_t), \, t\in [\tau, T] \\ &u_{\tau, x, \nu} (\tau) = x\in H \end{aligned} \right. \end{aligned}$$

and the Hamiltonian

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \ \ H(t, x, \rho) \\ &\displaystyle &\displaystyle = \sup_{ \mu \in {\mathcal M}^1_+(Z)} \big\lgroup -\langle \rho, \operatorname{{\mathrm{bar}}}(\mu)\rangle +\int_Z J(t, x, z)\mu(dz) \big\rgroup + \delta^*( \rho,-f(t, x) -A(t, x)) \end{array} \end{aligned} $$

where (t, x, ρ) ∈ [0, T] × H × H. Then, V _J is a viscosity subsolution of the HJB equation

$$\displaystyle \begin{aligned} \frac{\partial U}{\partial t} (t, x) + H(t, x, \nabla U(t, x))= 0,^{3} \end{aligned}$$

^{3 3}

that is, for any φ ∈ C ¹([0, T]) × H) for which V _J − φ reaches a local maximum at (t ₀, x ₀) ∈ [0, T] × H, we have

$$\displaystyle \begin{aligned}H(t_0, x_0, \nabla \varphi (t_0, x_0)) +\frac{\partial \varphi}{\partial t}(t_0, x_0) \geq 0.\end{aligned} $$

Proof

Assume by contradiction that there exists a φ ∈ C ¹([0, T] × H) and a point (t ₀, x ₀) ∈ [0, T] × H for which

$$\displaystyle \begin{aligned}\frac{\partial \varphi}{\partial t}(t_0, x_0)+ H( t_0, x_0, \nabla \varphi(t_0, x_0)) \leq -\eta < 0 \quad \mathrm{for} \quad \eta >0.\end{aligned} $$

Applying Lemma 3.5 by taking

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \Lambda(t, x, \mu) = - \langle \nabla \varphi (t, x), \operatorname{{\mathrm{bar}}}(\mu) \rangle+ \int_Z J(t, x, z)\mu(dz) \\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad +\delta^*(\nabla \varphi(t, x), -f(t, x) -A(t, x)) + \frac{\partial \varphi} {\partial t}(t, x)\vspace{-4pt} \end{array} \end{aligned} $$

yields some σ > 0 such that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \sup_{\nu \in \mathcal R} \bigg\lgroup \int_{t_0} ^{t_0+\sigma} \bigg[\int_Z J(t, u_{t_0, x_0, \nu} (t),z)\nu_t(dz)\bigg]\,dt -\int_{t_0} ^{t_0+\sigma} \langle {\nabla \varphi} (t, u_{t_0, x_0, \nu} (t), \operatorname{{\mathrm{bar}}}(\nu_t) \rangle \,dt\\ &\displaystyle &\displaystyle \qquad \ +\int_{t_0}^{t_0+\sigma} \delta ^*( \nabla \varphi (t, u_{t_0, x_0, \nu} (t)),-f(t, u_{t_0, x_0, \nu^n} (t))- A(t, u_{t_0, x_0, \nu}(t) ) )\,dt\\ &\displaystyle &\displaystyle \qquad \ + \int_{t_0} ^{t_0+\sigma} \frac{\partial \varphi} {\partial t}(t, u_{t_0, x_0, \nu} (t))\,dt \bigg\rgroup\\ &\displaystyle &\displaystyle \ \ \leq -\frac{\sigma \eta}{2} {}\vspace{-4pt} \end{array} \end{aligned} $$

(15)

where $u_{t_0, x_0, \nu }$ is the trajectory solution associated with the control $\nu \in \mathcal R$ starting from x ₀ at time t ₀

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &0\in \dot u_{t_0, x_0, \nu} (t) +A(t, u_{t_0, x_0, \nu} (t)) + f(t, u_{t_0, x_0, \nu}(t)) + \operatorname{{\mathrm{bar}}}(\nu_t), \, \, t\in [t_0, T] \\ & u_{t_0, x_0, \nu} (t_0) = x_0. \end{aligned} \right.\end{aligned} $$

Applying the dynamic programming principle (Theorem 4.10) gives

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} &\displaystyle &\displaystyle V_J(t_0, x_0) = \sup_{\nu \in \mathcal R} \big\lgroup \int_{t_0}^{t_0 + \sigma}\bigg[\int_ZJ (t, u_{t_0, x_0, \nu}(t), z) \nu_t(dz)\bigg]dt +V_J(t_0 \\ &\displaystyle &\displaystyle \qquad \qquad \qquad + \sigma, u_{t_0, x_0, \nu }(t_0 + \sigma)) \big\rgroup. \\ \end{array} \end{aligned} $$

(16)

Since V _J − φ has a local maximum at (t ₀, x ₀), for small enough σ

$$\displaystyle \begin{aligned} V_J(t_0, x_0) -\varphi (t_0, x_0) \geq V_J(t_0+\sigma, u_{t_0, x_0, \nu} (t_0+\sigma))-\varphi (t_0+\sigma, u_{t_0, x_0, \nu} (t_0+\sigma)) \end{aligned} $$

(17)

for all $\nu \in \mathcal R$. By (16), for each $n\in \mathbb {N}$, there exists $\nu ^n \in \mathcal R$ such that

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} &\displaystyle &\displaystyle V_J(t_0, x_0) \leq \int_{t_0}^{t_0+\sigma} \bigg[\int_Z J(t, u_{t_0, x_0, \nu^n} (t)) , z)\nu ^n_t(dz)\bigg]\,dt\\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \quad +V_J(t_0+\sigma, u_{t_0, x_0, \nu^n} (t_0+\sigma))+ \frac{1}{n}. \end{array} \end{aligned} $$

(18)

From (17) and (18), we deduce that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle V_J(t_0+\sigma, u_{t_0, x_0, \nu^n} (t_0+\sigma)) -\varphi (t_0+\sigma, u_{t_0, x_0, \nu ^n} (t_0+\sigma))\\ &\displaystyle &\displaystyle \qquad \quad \qquad \leq \int_{t_0}^{t_0+\sigma} \bigg[\int_Z J(t, u_{t_0, x_0, \nu^n} (t)) , z)\nu ^n_t(dz)\bigg]dt + \frac{1}{n}\\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \quad - \varphi (t_0, x_0) + V_J(t_0+\sigma, u_{t_0, x_0, \nu^n} (t_0+\sigma)). \end{array} \end{aligned} $$

Therefore we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle 0\leq \int_{t_0}^{t_0+\sigma} \bigg[\int_Z J(t, u_{t_0, x_0, \nu^n} (t)),z)\nu ^n_t(dz)\bigg]\,dt\\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \quad \ \ + \varphi (t_0+\sigma, u_{t_0, x_0, \nu ^n} (t_0+\sigma)) -\varphi(t_0, x_0) + \frac{1}{n}. {} \end{array} \end{aligned} $$

(19)

As φ ∈ C ¹([0, T] × H), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} &\displaystyle &\displaystyle \varphi(t_0+\sigma, u_{t_0, x_0, \nu^n}(t_0+\sigma))- \varphi (t_0, x_0) \\ &\displaystyle &\displaystyle \ =\int_{t_0}^{t_0+\sigma} \langle \nabla \varphi (t, u_{t_0, x_0, \nu^n} (t)), \dot u_{t_0, x_0, \nu^n} (t) \rangle\,dt +\int_{t_0}^{t_0+\sigma} \frac{\partial \varphi}{\partial t} (t, u_{t_0, x_0, \nu ^n} (t) )\,dt. \end{array} \end{aligned} $$

(20)

Since $u_{t_0, x_0, \nu ^n}$ is the trajectory solution starting from x ₀ at time t ₀

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &0\in \dot u_{t_0, x_0, \nu ^n} (t) +A (t, u_{t_0, x_0, \nu ^n} (t)) + f(t, u_{t_0, x_0, \nu ^n}(t)) + \operatorname{{\mathrm{bar}}}(\nu ^n_t), \, t\in [t_0, T] \\ & u_{t_0, x_0, \nu^n} (t_0) = x_0 \end{aligned} \right. \end{aligned}$$

so that (20) yields the estimate

$$\displaystyle \begin{aligned} \begin{array}{rcl} {} &\displaystyle &\displaystyle \varphi(t_0+\sigma, u_{t_0, x_0, \nu^n}(t_0+\sigma))- \varphi (t_0, x_0) \\ &\displaystyle &\displaystyle =\int_{t_0}^{t_0+\sigma} \langle \nabla \varphi (t, u_{t_0, x_0, \nu^n} (t)), \dot u_{t_0, x_0, \nu^n} (t) \rangle\,dt +\int_{t_0}^{t_0+\sigma} \frac{\partial \varphi}{\partial t} (t, u_{t_0, x_0, \nu ^n} (t) )\,dt\\ &\displaystyle &\displaystyle \leq -\int_{t_0}^{t_0+\sigma} \langle \nabla \varphi (t, u_{t_0, x_0, \nu^n} (t)), \operatorname{{\mathrm{bar}}}(\nu ^n_t) + f(t, u_{t_0, x_0, \nu ^n}(t)) \rangle \,dt\\ &\displaystyle &\displaystyle \qquad \qquad \quad +\int_{t_0}^{t_0+\sigma} \delta^* ( \nabla \varphi (t, u_{t_0, x_0, \nu^n} (t)), -A (t,u_{t_0, x_0, \nu^n} (t)))\,dt\\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \quad +\int_{t_0}^{t_0+\sigma} \frac{\partial \varphi}{\partial t} (t, u_{t_0, x_0, \nu^n} (t) )\,dt. \end{array} \end{aligned} $$

(21)

Inserting the estimate (21) into (19), we get

$$\displaystyle \begin{aligned} 0\leq & \int_{t_0}^{t_0+\sigma} \bigg[\int_Z J(t, u_{t_0, x_0, \nu^n} (t)) , z)\nu ^n_t(dz)\bigg]dt \\ &- \int_{t_0}^{t_0+\sigma} \langle \nabla \varphi (t, u_{t_0, x_0, \nu^n} (t)), \operatorname{{\mathrm{bar}}}(\nu ^n_t) + f(t, u_{t_0, x_0, \nu^n} (t)) \rangle\,dt \\ &+\int_{t_0}^{t_0+\sigma} \delta ^*( \nabla \varphi (t, u_{t_0, x_0, \nu ^n} (t)), -A(t, u_ {t_0, x_0, \nu^n} (t) )\,dt \\ & +\int_{t_0}^{t_0+\sigma} \frac{\partial \varphi}{\partial t} (t, u_{t_0, x_0, \nu ^n} (t) )\,dt+\frac{1}{n}.\end{aligned} $$

(22)

Then (15) and (22) yield $0\leq - \frac {\sigma \eta }{2} +\frac {1}{n}$ for all $n \in \mathbb {N}$. By passing to the limit when n goes to ∞ in this inequality, we get a contradiction: $0\leq - \frac {\sigma \eta }{2}$. The proof is therefore complete.

Existence results for evolution inclusion driven by time-dependent maximal monotone operators A(t) with single-valued perturbation f or convex weakly compact valued perturbation F of the form

$$\displaystyle \begin{aligned}-\dot u(t) \in A(t) u(t) + f(t, u(t))\end{aligned}$$

or

$$\displaystyle \begin{aligned}-\dot u(t) \in A(t) u(t) + F(t, u(t))\end{aligned}$$

are developed in [7, 8], while existence results for convex or nonconvex sweeping process in the form

$$\displaystyle \begin{aligned}-\dot u(t) \in N_{C(t)} (u(t)) + f(t, u(t))\end{aligned}$$

or

$$\displaystyle \begin{aligned}-\dot u(t) \in N_{C(t)} (u(t)) + F(t, u(t))\end{aligned} $$

where C(t) is a closed convex (or nonconvex) moving set and N _C(t)(u(t)) is the normal cone of C(t) at the point u(t) is much studied so that our tools developed above allow to treat some further variants on the viscosity solution dealing with some specific maximal monotone operators A(t) or convex or nonconvex sweeping process such as

$$\displaystyle \begin{aligned} \left\{ \begin{aligned} &0\in \dot u_{t_0, x_0, \nu} (t) +N_{C(t)} (u_{t_0, x_0, \nu} (t)) + f(t, u_{\tau, x, \nu}(t)) + \operatorname{{\mathrm{bar}}}(\nu_t), \, t\in [t_0, T] \\ & u_{t_0, x_0, \nu} (t_0) = x_0 \end{aligned} \right.\end{aligned} $$

using the subdifferential of the distance function d _C(t) x.

We end the paper with some variational limit results which can be applied to further convergence problems in state-dependent convex sweeping process or second-order state-dependent convex sweeping process. See [1, 3, 34] and the references therein.

Theorem 4.12

Let C _n : [0, T] → H and C : [0, T] → H be a sequence of convex weakly compact valued scalarly measurable bounded mappings satisfying

• (i) $\sup _n \sup _{t \in [0, T]} {\mathcal {H}}\big (C_n(t), C(t)\big ) \leq M < \infty$ ,

• (ii) $\lim _n {\mathcal {H}}\big (C_n(t), C(t)\big )= 0$ , for each t ∈ [0, T].

Let (v _n) be a uniformly integrable sequence in $L ^1_H([0, T])$ such that v _n converges for $\sigma (L ^1_H([0, T]), L ^\infty _H([0, T])$ to $v \in L ^1_H([0, T])$ , and let (u _n) be a uniformly bounded sequence $L ^\infty _H([0, T])$ which pointwisely converges to $u \in L ^\infty _H([0, T])$ . Assume that $-v_n(t) \in N_{C_n(t)} (u_n(t))$ a.e., then

$$\displaystyle \begin{aligned} u(t) \in C(t) \mathit{\text{ a.e. and }} -v(t) \in N_{C(t)} (u(t)) \mathit{\text{ a.e.}} \end{aligned}$$

Proof

For simplicity, let $\rho _n (t) = {\mathcal {H}}\big (C_n(t), C(t)\big )$ for each t ∈ [0, T]. Firstly it is clear that the mappings ρ _n, t↦δ ^∗(−v _n(t), C _n(t)), t↦δ ^∗(−v _n(t), C(t)), and t↦δ ^∗(−v(t), C(t)) are measurable on [0, T] and integrable by boundedness. By the Hormander formula for convex weakly compact set (see [19]), we have

$$\displaystyle \begin{aligned}|\delta^*( -v_n(t) , C_n(t))- \delta^*( -v_n(t) , C(t))| \leq ||v_n(t)|| \rho_n(t)\end{aligned}$$

so that

$$\displaystyle \begin{aligned}\delta^*( -v_n(t) , C_n(t))- \delta^*( -v_n(t) , C(t)) \geq - ||v_n(t)|| \rho_n(t).\end{aligned}$$

By $-v_n(t) \in N_{C_n(t)} (u_n(t))$, we have δ ^∗(−v _n(t), C _n(t)) + 〈v _n(t), u _n(t)〉≤ 0 so we get the estimation

$$\displaystyle \begin{aligned}- ||v_n(t)|| \rho_n(t) + \delta^*( -v_n(t) , C(t))+\langle v_n(t, u_n(t) \rangle \leq 0\end{aligned}$$

or

$$\displaystyle \begin{aligned}\delta^*( -v_n(t) , C(t))+ \langle v_n(t) , u_n(t) \rangle \leq ||v_n(t)|| \rho_n(t).\end{aligned} $$

Note that the mappings t↦δ ^∗(−v _n(t), C(t)) + 〈v _n(t), u _n(t)〉, and t↦||v _n(t)||ρ _n(t) are integrable on [0, T]. Let B a measurable set in [0, T] and then by integrating

$$\displaystyle \begin{aligned}\int_B \delta^*( -v_n(t) , C(t)) dt+ \int_B \langle v_n(t), u_n(t) \rangle dt \leq \int_B ||v_n(t)|| \rho_n(t) dt.\end{aligned} $$

We note that the convex integrand H(t, e) = δ ^∗(e, C(t)) defined on [0, T] × H is normal because t↦H(t, e) is continuous on [0, T] and e↦H(t, e) is convex continuous on H, with H(t, e) ≥〈e, u(t)〉 for all (t, e) ∈ [0, T] × H. Consequently H(t, −v _n(t)) = δ ^∗(−v _n(t), C(t)) ≥〈−v _n(t), u(t)〉. But (〈−v _n, u〉) is uniformly integrable in $L^1_{\mathbb {R}}([0, T], dt)$, so that by virtue of the lower semicontinuity of the integral convex functional [22, Theorem 8.1.16], we have

$$\displaystyle \begin{aligned} \liminf_{n\to\infty}\int_B \delta^*(-v_n(t) , C(t) ) dt \geq \int_B \delta^*( -v(t), C(t)) dt.\end{aligned} $$

(23)

Note that the sequence (u _n(.) − u(.)) is uniformly bounded and pointwisely converges to 0, so that it converges to 0 uniformly on any uniformly integrable subset of $L^1_H([0, T], dt)$; in other terms, it converges to 0 with respect to the Mackey topology $\tau (L^\infty _H([0, T], dt), L^1_H([0, T], dt))$ (see [15]),⁴ so that

$$\displaystyle \begin{aligned}\lim_{n\to\infty} \int_B \langle v_n(t) , u_n(t)-u(t) \rangle dt = 0\end{aligned}$$

because (v _n) is uniformly integrable. Consequently

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \lim_{n\to\infty} \int_B\langle v_n(t) , u_n(t)\rangle dt &\displaystyle =&\displaystyle \lim_{n\to\infty} \int_B \langle v_n(t), u_n(t)-u(t) \rangle dt + \lim_{n\to\infty} \int_B \langle v_n(t), u(t) \rangle dt\\ &\displaystyle =&\displaystyle \lim_{n\to\infty} \int_B \langle v_n(t), u(t) \rangle dt = \int_B \langle \dot v(t), u(t) \rangle dt. \end{array} \end{aligned} $$

(24)

By our assumptions, ρ _n(t) is bounded measurable and pointwisely converges to 0 and ||v _n(t)|| is uniformly integrable; then similarly we have

$$\displaystyle \begin{aligned} \lim_n \int_B ||v_n(t)|| \rho_n(t) dt = 0. \end{aligned} $$

(25)

Finally by passing to the limit when n goes to ∞ in

$$\displaystyle \begin{aligned}\int_B \delta^*( -v_n(t) , C(t)) dt+ \int_B \langle v_n(t), u_n(t) \rangle dt \leq \int_B ||v_n(t)|| \rho_n(t) dt\end{aligned} $$

and taking into account the above convergence limits (23), (24), and (25), we get

$$\displaystyle \begin{aligned}\int_B \delta^*( -v(t) , C(t)) dt+ \int_B \langle v(t), u(t)\rangle dt \leq 0.\end{aligned} $$

As the function t↦δ ^∗(−v(t), C(t)) + 〈v(t), u(t) is integrable on [0, T] and this holds for every B measurable set in [0, T], we get

$$\displaystyle \begin{aligned}\delta^*( -v(t) , C(t)) ) + \langle v(t), u(t) \rangle \leq 0 \text{ a.e.}\end{aligned}$$

Furthermore, it is not difficult to check that u(t) ∈ C(t) a.e. using (ii) and the fact that u _n(t) ∈ C _n(t) for all $n \in \mathbb {N}$ and a.e. t ∈ [0, T]; therefore, we conclude that − v(t) ∈ N _C(t)(u(t)) a.e. The proof is complete.

Our tools allow to treat the variational limits for further evolution variational inequalities such as

Proposition 4.2

Let C _n : [0, T] → H and $C : [0, T] \rightrightarrows H$ be a sequence of convex weakly valued scalarly measurable bounded mappings satisfying

• (i) $\sup _n \sup _{t \in [0, T]} {\mathcal {H}}\big (C_n(t), C(t)\big ) \leq M < \infty$ ,

• (ii) $\lim _n {\mathcal {H}}\big (C_n(t), C(t)\big )= 0$ , for each t ∈ [0, T].

Let B : H → H be a linear continuous operator such that 〈Bx, x〉 > 0 for all x ∈ H ∖{0}. Let (v _n) be a uniformly bounded sequence in $L ^\infty _H([0, T])$ such that $v_n \hskip 3pt \sigma (L ^\infty _H([0, T]), L ^1_H([0, T])$ converges to $v \in L ^\infty _H([0, T])$ , and let (u _n) be a uniformly bounded sequence $L ^\infty _H([0, T])$ which pointwisely converges to $u \in L ^\infty _H([0, T])$ . Assume that $-v_n(t) \in N_{C_n(t)} (u_n(t) +Bv_n(t))$ for all $n\in \mathbb {N}$ and for a.e. t ∈ [0, T]. Then

$$\displaystyle \begin{aligned} u(t) +Bv(t) \in C(t) \mathit{\text{ a.e. and }}-v(t) \in N_{C(t)} (u(t) +Bv(t)) \mathit{\text{ a.e.}} \end{aligned}$$

Proof

Apply the notations of the proof of Theorem 4.12. Let $\rho _n (t) = {\mathcal {H}}\big (C_n(t), C(t)\big )$ for each t ∈ [0, T]. It is clear that the mappings ρ _n, t↦δ ^∗(−v _n(t), C _n(t)), t↦δ ^∗(−v _n(t), C(t)), and t↦δ ^∗(−v(t), C(t)) are measurable and integrable on [0, T]. By the Hormander formula for convex weakly compact sets (see [19]), we have

$$\displaystyle \begin{aligned}|\delta^*( -v_n(t) , C_n(t))- \delta^*( -v_n(t) , C(t))| \leq ||v_n(t)|| \rho_n(t)\end{aligned}$$

so that

$$\displaystyle \begin{aligned}\delta^*( -v_n(t) , C_n(t))- \delta^*( -v_n(t) , C(t)) \geq - ||v_n(t)|| \rho_n(t).\end{aligned}$$

By $-v_n(t) \in N_{C_n(t)} (u_n(t) +Bv_n(t))$, we have

$$\displaystyle \begin{aligned}\delta ^*(-v_n(t), C_n(t)) + \langle v_n(t) , u_n(t) + Bv_n(t) \rangle \leq 0.\end{aligned} $$

Whence

$$\displaystyle \begin{aligned}\delta ^*(-v_n(t), C(t)) + \langle v_n(t) , u_n(t) + Bv_n(t) \rangle \leq ||v_n(t)|| \rho_n(t)\end{aligned} $$

Note that the mappings t↦δ ^∗(−v _n(t), C(t)) + 〈v _n(t), u _n(t) + Bv _n(t)〉, and t↦||v _n(t)||ρ _n(t) are integrable on [0, T] so that by integrating on any measurable set L ⊂ [0, T]

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \int_L \delta^*(-v_n(t), C(t)) dt + \int_L \langle v_n(t) , u_n(t) \rangle dt + \int_L \langle v_n(t), Bv_n(t) \rangle dt\\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \leq \int_L ||v_n(t)|| \rho_n(t) dt.\vspace{-4pt} \end{array} \end{aligned} $$

Since $(v_n )\ \sigma (L ^\infty _H([0, T]), L ^1_H([0, T])$ converges to $v \in L ^\infty _H([0, T])$, it is not difficult to check that (Bv _n) converges for $\sigma (L ^\infty _H([0, T]), L ^1_H([0, T])$ to $Bv \in L ^1_H([0, T])$, arguing as in [11, Theorem 4.1]. As a consequence, the sequence (u _n + Bv _n) converges for $\sigma (L ^\infty _H([0, T]), L ^1_H([0, T])$ to $u+Bv \in L ^\infty _H([0, T])$. From u _n(t) + Bv _n(t) ∈ C _n(t), we deduce

$$\displaystyle \begin{aligned}\int_L \langle e, u_n(t) +Bv_n(t) \rangle dt \leq \int_L \delta^*(e, C_n(t)) dt\end{aligned}$$

for every e ∈ H and for every measurable set L ⊂ [0, T]. By passing to the limit in this inequality, we get

$$\displaystyle \begin{aligned}\int_L \langle e, u(t) +Bv(t) \rangle dt \leq \limsup_n \int_L \delta^*(e, C_n(t)) dt \leq \int_L \delta^*(e, C(t)) dt.\end{aligned}$$

It follows that

$$\displaystyle \begin{aligned}\langle e, u(t) +Bv(t) \rangle \leq \delta^*(e, C(t))\text{ a.e.}\end{aligned}$$

By [19, Proposition III.35], we deduce that u(t) + Bv(t) ∈ C(t) a.e. As in Theorem 3.1, we have already stated that for every measurable set L ⊂ [0, T],

$$\displaystyle \begin{gathered} \lim_n \int_L \langle u_n(t) , v_n(t) \rangle dt = \int_L \langle u(t) , v(t) \rangle dt,{} \end{gathered} $$

(26)

$$\displaystyle \begin{gathered} \lim_n \int_L |v_n(t)|| \rho_n(t) dt = 0,{} \end{gathered} $$

(27)

$$\displaystyle \begin{gathered} \liminf_n \int_B \delta ^*(-v(t) , C_n(t)) dt \geq \int_B \delta ^*(-v(t) , C_n(t)) dt. {} \end{gathered} $$

(28)

Now set φ(x) = 〈x, Bx〉 for all x ∈ H. Then φ(x) is a nonnegative lower semicontinuous and convex function defined on H. So we have

$$\displaystyle \begin{aligned}\int _L \langle v_n(t) , Bv_n(t) \rangle dt = \int_L \varphi ( v_n(t)) dt.\end{aligned}$$

By lower semicontinuity of convex integral functional [19, 22, 23], we get

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \liminf_n \int _L \langle v_n(t) , Bv_n(t) \rangle dt\\ &\displaystyle &\displaystyle \qquad \qquad \qquad = \liminf_n \int_L \varphi ( v_n(t)) dt \geq \int_L \varphi ( v(t)) dt = \int _L \langle v(t) , Bv(t) \rangle dt. \end{array} \end{aligned} $$

Taking into consideration the above stated limits (26), (27), (28) and passing to the limit when n goes to ∞ in the inequality

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \int_L \delta^*(-v_n(t), C(t)) dt + \int_L \langle v_n(t) , u_n(t) \rangle dt + \int_L \langle v_n(t), Bv_n(t) \rangle dt \\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \leq \int_L ||v_n(t)|| \rho_n(t) dt, \end{array} \end{aligned} $$

we get

$$\displaystyle \begin{aligned}\int_L \delta ^*( -v(t), C(t))dt+ \int_L \langle v(t) , u(t) +Bv(t) \rangle dt \leq 0\end{aligned}$$

for every measurable set L ⊂ [0, T]. Since the mapping t↦δ ^∗(−v(t), C(t)) + 〈v(t), u(t) + Bv(t)〉 is integrable on [0, T], we have

$$\displaystyle \begin{aligned}\delta ^*( -v(t), C(t))+ \langle v(t) , u(t) +Bv(t) \rangle \leq 0 \text{ a.e.} \end{aligned}$$

As u(t) + Bv(t) ∈ C(t) a.e., this yields − v(t) ∈ N _C(t)(u(t) + Bv(t)) a.e. The proof is complete.