Some Problems in Second Order Evolution Inclusions with Boundary Condition: A Variational Approach

Abstract

We prove, under appropriate assumptions, the existence of solutions for a second order evolution inclusion with boundary conditions via a variational approach.

JEL Classification: C61, C73.

Mathematics Subject Classifications (2010): 34A60, 34B15.

• Article type: Research Article

• Received: September 12, 2016

• Revised: January 6, 2017

1 Introduction

In the present paper, we prove, under appropriate assumptions, the existence of solutions for a second order evolution inclusion with boundary conditions governed by subdifferential operators of the form

$$\begin{aligned} f(t) \in \ddot{u} (t)+M \dot{u}(t) +\partial \varphi (u(t)), t \in [0, T].\qquad (\mathrm{I}) \end{aligned}$$

Here, M is positive, $\varphi$ is a lower semicontinuous convex proper function defined on $\mathbf {R}^d$ and $\partial \varphi (u(t))$ is the subdifferential of the function $\varphi$ at the point u(t) and the perturbation f belongs to $L^2_{\mathbf {R}^d}([0, T])$. It is well known that this problem is difficult and needs a specific treatment via the Moreau-Yosida approximation or epiconvergence approach. See Attouch–Cabot–Redon [4] and Schatzmann [24] for a deep study of these problems, Castaing–Raynaud de Fitte–Salvadori [11], Castaing–Le Xuan Truong [8] dealing with second order evolution with m-point boundary conditions via the epiconvergence approach. These considerations lead us to consider the variational limits of a fairly general approximating problem

$$\begin{aligned} f ^n(t) \in \ddot{u}^n (t)+M \dot{u} ^n(t) +\partial \varphi _n (u ^n(t)), t \in [0, T]\qquad (\mathrm{II}) \end{aligned}$$

where $u^n$ is a $W ^{2, 1}_{\mathbf {R}^d}([0, T])$-solution, $f ^n$ weakly converging in $L^2_{\mathbf {R}^d}([0, T])$ to $f ^\infty$, $\varphi _n$ is a convex Lipschitz function which epiconverges to a lower semicontinuous convex proper function $\varphi _\infty$. This approximating problem covers various type of problems of practical interest in several dynamic systems, evolution inclusion, control theory etc. Here we focus on several variational limits of solutions via the Biting Lemma and Young measures and other tools occurring in this approach by showing under suitable limit assumption on the boundary conditions that $(\ddot{u}^n)$ is $L^1_{\mathbf {R}^d}([0, T])$-bounded. This main fact allows to study the variational limit of solutions in this problem, in particular, the traditional estimated energy for the variational limit solutions is conserved almost everywhere. The applicability of our abstract framework given therein (Proposition 3.3) will be exemplified in considering the existence of solution for second order differential inclusions

$$ f(t) \in \ddot{u} (t)+M \dot{u}(t) +\partial \varphi (u(t)), t \in [0, T] $$

under m-point boundary condition or anti-periodic conditions and further related second order evolution inclusions in the literature. This will be done by applying our abstract result to the single valued approximating problem

$$\begin{aligned} f ^n(t) = \ddot{u}^n (t)+M \dot{u} ^n(t) +\nabla \varphi _n (u ^n(t)), t \in [0, T]\qquad (\mathrm{III}) \end{aligned}$$

where $\nabla \varphi _n$ is the gradient of the $C^1$, Lipschitz, convex function $\varphi _n$ that epi-converges to a proper convex lower semicontinuous function $\varphi _\infty$ and $f ^n$ weakly converges in $L^2_{\mathbf {R}^d}([0, T])$ to $f ^\infty$ so that the variational limit solutions $u^\infty$ to (III) are generalized solutions to the inclusion

$$f ^\infty (t) \in \ddot{u}^\infty (t)+M \dot{u} ^\infty (t) +\partial \varphi _\infty (u ^\infty (t)), t \in [0, T] $$

with appropriate properties, namely, the solution limit $u^\infty$ is $W ^{1, 1}_{BV} ([0, T])$, that is, $u^\infty$ is continuous and its derivative $\dot{u} ^\infty$ is bounded variation (BV for short) and the estimated energy holds almost everywhere

$$\begin{aligned} \varphi _\infty ( u ^\infty (t) ) +\frac{1}{2} ||\dot{u} ^\infty (t) ||^2&= \varphi _\infty (u_0)+ \frac{1}{2} ||\dot{u}_0) ||^2\\&\quad - M\int _0^t ||\dot{u} ^\infty (s)||^2 ds+ \int _0^t \langle f ^\infty (s), \dot{u} ^\infty (s) \rangle ds \end{aligned}$$

with further related variational inclusion, in particular,

$$ f ^\infty (t) \in \zeta ^\infty (t)+M u ^\infty (t) +\partial \varphi _\infty (u ^\infty (t)), t \in [0, T] $$

almost everywhere, $\zeta ^\infty$ being the biting limit of the $L^1_{\mathbf {R}^d}([0, T])$-bounded sequence $(\ddot{u}^n)$. Section 3 is devoted to second order evolution inclusion with boundary conditions. We present the variational limits of the general approximating problem (II) and the applications of variational limits of the approximating problem (III) to the existence problem of second order evolution inclusion (I) involving variational techniques, the Biting Lemma, the characterization of the second dual of $L^1_{\mathbf {R}^d}$ and Young measures. It is worth to mention that the approximation (III) occurs in practical cases of second order evolution inclusion governed by subdifferential operators. For instance, Attouch–Cabot–Redon [4] considered the approximating problem

$$ 0= \ddot{u}^n (t)+\gamma \dot{u} ^n(t) +\nabla \varphi _n (u ^n(t)), t \in [0, T] $$

$$ u^n(0) = u^n_0, \dot{u}^n (0) = \dot{u}^n_1 $$

where $\gamma$ is positive, $\nabla \varphi _n$ is the gradient of a $C ^1$, smooth function. Schatzmann [24] considered the approximating problem

$$ f(t) = \ddot{u}_\lambda (t) +\partial \varphi _\lambda (u_\lambda (t)), t \in [0, T] $$

$$ u_\lambda (0)= u_0, \dot{u}_\lambda (0) = u_1 $$

where $f \in L ^2_{\mathbf {R}^d}([0, T])$ and $\partial \varphi _\lambda$ is the Moreau-Yosida approximation to the lower semicontinuous convex proper function $\varphi$. M. Mabrouk [19] continued the work of M. Schatzmann [24] by considering the approximating problem

$$\begin{aligned} f_\lambda (t) =\;&\ddot{u}_\lambda (t)+\nabla \varphi _\lambda (u_\lambda (t)), t \in [0, T]\\&\;u_\lambda (0)= u_0, \dot{u}_\lambda (0) = u_1, \end{aligned}$$

with $f_\lambda \in L^1_{\mathbf {R}^d}([0, T])$. In Sect. 4, we apply our techniques to the study of both first order and second order evolution equations with anti-periodic boundary condition using the approximating problem

$$\begin{aligned} f ^n(t) = \ddot{u}^n (t)\;+\;&M \dot{u} ^n(t) +\nabla \varphi _n (u ^n(t)), t \in [0, T]\\&u^n(0) = - u^n(T), \end{aligned}$$

where $u^n \in W^{2, 2}_{\mathbf {R}^d}([0, T])$ and $f^n \in L^2_{\mathbf {R}^d}([0,T])$, see H. Okochi [22], A. Haraux [17], Aftabizadeh, Aizicovici and Pavel [1, 2], Aizicovici and Pavel [3] and the references therein.

A general analysis of some related problems in Hilbert space is available, c.f K. Maruo [19] and M. Schatzmann [24].

2 Some Existence Theorems in Second Order Evolution Inclusions with m-Point Boundary Condition

We will use the following definitions and notations and summarize some basic results.

• Let E be a separable Banach space, ${\overline{B}}_E(0, 1)$ is the closed unit ball of E.

• c(E) (resp. cc(E)) (resp. ck(E))(resp. cwk(E)) is the collection of nonempty closed (resp. closed convex) (resp. compact convex) (resp. weakly compact convex) subsets of E.

• If A is a subset of E, $\delta ^{*}(.,A)$ is the support function of A.

• ${\mathcal L}([0,T])$ is the $\sigma$-algebra of Lebesgue measurable subsets of [0, T].

• If X is a topological space, ${\mathcal B}(X)$ is the Borel tribe of X.

• $L_E^1([0,T],dt)$ (shortly $L_E^1([0,T])$) is the Banach space of Lebesgue–Bochner integrable functions $f : [0,T]\rightarrow E$.

• A mapping $u :[0,T] \rightarrow E$ is absolutely continuous if there is a function $\dot{u} \in L^1_E([0,T])$ such that $u(t) = u(0) +\int _0^t \dot{u}(s)\,ds, \forall t\in [0,T]$.

• If X is a topological space, ${\mathcal C}_E(X)$ is the space of continuous mappings $u :X\rightarrow E$ equipped with the norm of uniform convergence.

• A set-valued mapping $F: [0,T]\rightrightarrows E$ is measurable if its graph belongs to ${\mathcal L}([0,T])\otimes {\mathcal B}(E)$.

• A convex weakly compact valued mapping $F : X \rightarrow ck(E)$ defined on a topological space X is scalarly upper semicontinuous if for every $x ^* \in E ^*$, the scalar function $\delta ^*(x ^*, F(.))$ is upper semicontinuous on X.

We refer to [13] for measurable multifunctions and Convex Analysis.

For the sake of completeness, we recall and summarize some results developed in [9]. By $W^{2, 1}_{E} ([0, T])$ we denote the set of all continuous functions in ${\mathcal C}_{E} ([0, T])$ such that their first derivatives are continuous and their second derivatives belong to $L ^1_{E} ([0, T])$.

Lemma 2.1

Assume that E is a separable Banach space. Let $0< \eta _1<\eta _{2}<\cdots<\eta _{m-2}<1$, $\gamma > 0$, $m > 3$ be an integer number, and $\alpha _{i}\in \mathbf {R}$ $\left( i=1,\ldots ,m-2\right)$ satisfying the condition

$$ \sum _{i=1}^{m-2}\alpha _{i}-1+\exp \left( -\gamma \right) -\sum _{i=1}^{m-2}\alpha _{i} \exp \left( -\gamma \eta _{i} )\right) \ne 0. $$

Let $G : [0, 1]\times [0, 1] \rightarrow \mathbf {R}$ be the function defined by

$$\begin{aligned} G(t,s) = \left\{ \begin{array}{lll} \frac{1}{\gamma } \left( 1- \exp (-\gamma (t-s))\right) , &{} 0\le s \le t \le 1 \\ 0, &{} t < s \le 1 \end{array} \right. + \frac{A}{\gamma } \left( 1-\exp (-\gamma t)\right) \phi (s), \end{aligned}$$

(2.1)

where

$$\begin{aligned} \phi (s) = \left\{ \begin{array}{ll} 1- \exp (-\gamma (1-s))-\sum \nolimits _{i = 1} ^{m-2} \alpha _i \left( 1-\exp (-\gamma (\eta _i-s))\right) , &{} 0 \le s < \eta _1, \\ \\ 1- \exp (-\gamma (1-s))-\sum \nolimits _{i = 2} ^{m-2} \alpha _i \left( 1- \exp (-\gamma (\eta _i-s))\right) , &{} \eta _1\le s\le \eta _2, \\ \\ ......\\ \\ 1- \exp (-\gamma (1-s)), &{} \eta _{m-2} \le s\le 1, \end{array} \right. \end{aligned}$$

(2.2)

and

$$\begin{aligned} A = \left( \sum _{i=1}^{m-2}\alpha _{i}-1+ \exp (-\gamma ) -\sum _{i=1}^{m-2}\alpha _{i}\exp (-\gamma \eta _{i} )\right) ^{-1}. \end{aligned}$$

(2.3)

Then the following assertions hold

• (i) For every fixed $s\in [0, 1]$, the function G(., s) is right derivable on [0, 1[ and left derivable on ]0, 1]. Its derivative is given by

  $$\begin{aligned} \left( \frac{\partial G }{\partial t}\right) _{+}(t, s) = \left\{ \begin{array}{lll} \exp (-\gamma (t-s)), &{} 0 \le s \le t< 1 \\ 0, &{} 0 \le t< s < 1 \end{array} \right. +A \exp (-\gamma t)\phi (s), \end{aligned}$$

  (2.4)
  $$\begin{aligned} \left( \frac{\partial G_{\tau }}{\partial t}\right) _{-}(t, s) = \left\{ \begin{array}{lll} \exp (-\gamma (t-s)), &{} 0\le s< t \le 1 \\ 0, &{} 0 < t \le s \le 1 \end{array} \right. +A \exp (-\gamma t)\phi (s). \end{aligned}$$

  (2.5)

• (ii) $G(\cdot ,\cdot )$ and $\frac{\partial G }{\partial t}(\cdot , \cdot )$ satisfies

  $$ \left| G(t, s)\right| \le M_{G } \quad \mathrm{and } \quad \left| \frac{\partial G}{\partial t}(t, s)\right| \le M_{G } \quad \forall (t, s) \in [0,1]\times [0, 1], $$

  where

  $$ M_{G } = \max \{{\gamma ^{-1}}, 1 \} \left[ 1+|A|\left( 1+\sum _{i = 1} ^{m-2} |\alpha _i|\right) \right] . $$

• (iii) If $u \in W^{2, 1}_{E }([0, 1])$ with $u(0) = x$ and $u(1) = \sum _{i = 1} ^{m-2} \alpha _i u(\eta _i)$, then

  $$ u(t) = e_{x }(t) + \int _0 ^1 G(t, s) (\ddot{u} (s) +\gamma \dot{u}(s) ) ds, \quad \forall t \in [0, 1], $$

  where

  $$ e_{x}(t) = x + A\left( 1-\sum _{i=1}^{m-2}\alpha _{i}\right) (1-\exp (-\gamma t ))x . $$

• (iv) Let $f\in L^1_E([0, 1])$ and let $u_f : [0, 1] \rightarrow E$ be the function defined by

  $$ u_f(t) = e_{x }(t) + \int _0 ^1 G(t, s) f(s) ds \quad \forall t \in [0, 1]. $$

  Then we have

  $$ u_f(0) = x \quad u_f(1) = \sum _{i = 1} ^{m-2} \alpha _i u_f( \eta _i). $$

  Further the function $u_f$ is weakly derivable on [0, 1] and its weak derivative $\dot{u}_f$ is defined by

  $$ \dot{u}_f(t) = \lim _{h \rightarrow 0}\frac{ u_f(t+h) -u_f(t) }{h} = \dot{e}_{ x }(t) + \int _\tau ^1 \frac{\partial G}{\partial t}(t, s) f(s) ds, $$

  with

  $$ \dot{e}_{ x}(t) = \gamma A\left( 1-\sum _{i=1}^{m-2}\alpha _{i} \right) \exp (-\gamma t) x . $$

• (v) If $f\in L^1_{E}([0, 1])$, the function $\dot{u}_f$ is weakly derivable, and its weak derivative $\ddot{u}_f$ satisfies

  $$ \ddot{u}_f(t) +\gamma \dot{u}_f(t) = f(t) \quad \mathrm{a.e.}\ t \in [0, 1]. $$

The following is a direct consequence of Lemma 2.1.

Proposition 2.1

Let $f\in L ^1_E([0, 1])$. The m-point boundary problem

$$ \left\{ \begin{array}{lll} \ddot{u}(t) +\gamma \dot{u} (t) = f(t), \, t\in [0, 1]\\ u (0) = x, u (1) = \sum \nolimits _{i =1}^{m-2}\alpha _i u (\eta _i) \end{array} \right. $$

has a unique $W^ {2, 1}_{ E}([0, 1])$-solution $u_{ f}$, with integral representation formulas

$$ \left\{ \begin{array}{l} u_{ f} (t) = e_{x} (t)+ \int _{0}^{1} G(t, s) f(s) ds , \, t \in [0, 1]\\ \dot{u}_{ f} (t) =\dot{e}_{x} (t)+ \int _{0}^{1}\frac{\partial G}{\partial t}(t,s) f(s) ds, \, t \in [0, 1].\\ \end{array} \right. $$

where

$$ \left\{ \begin{array}{ll} e_{x } (t) &{}= x + A (1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}) (1-\exp (-\gamma t) )x \\ \dot{e}_{x }(t) &{}= \gamma A \left( 1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}\right) \exp {\left( -\gamma t \right) }x\\ A &{}= \left( \sum \nolimits _{i=1}^{m-2}\alpha _{i}-1+ \exp (-\gamma ) -\sum \nolimits _{i=1}^{m-2}\alpha _{i}\exp (-\gamma (\eta _{i}))\right) ^{-1}. \end{array} \right. $$

The following result and its notation will be used in the next section.

Proposition 2.2

With the hypotheses and notations of Proposition 2.1, let E be a separable Banach space and let $X : [0, 1] \rightrightarrows E$ be a measurable convex weakly compact valued and integrably bounded mapping. Then the solution set of $W ^{2, 1}_{ E}([0, 1])$-solutions to

$$ \left\{ \begin{array}{l} \ddot{u}_f (t)+\gamma \dot{u}_f(t) = f(t), \, f \in S^1_X \\ u_f(0) =x, \quad u_f(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u_f(\eta _i) \end{array} \right. $$

is bounded, convex, equicontinuous and sequentially weakly compact in $\mathcal C_{E}([0, 1])$.

Proof

Let us set

$$ \mathcal X :=\left\{ u_f \in \mathcal C_E([0, 1] : u_f(t) = e_{ x } (t)+ \int _{0}^{1}G(t, s) f(s) ds, t \in [0, 1], f \in S ^1_X \right\} $$

with

$$ \left\{ \begin{array}{ll} e_{x } (t) &{}= x + A (1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}) (1-\exp (-\gamma t) )x, \, t \in [0, 1]\ \\ \dot{e}_{x}(t) &{}= \gamma A\left( 1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}\right) \exp {\left( -\gamma t \right) }x, \, \, t \in [0, 1]\\ A&{}= \left( \sum \nolimits _{i=1}^{m-2}\alpha _{i}-1+ \exp (-\gamma ) -\sum \nolimits _{i=1}^{m-2}\alpha _{i}\exp (-\gamma (\eta _{i}))\right) ^{-1}. \end{array} \right. $$

Taking account of the properties of G in Lemma 2.1, it is not difficult to show that $\mathcal X$ is bounded, convex, equicontinuous and relatively weakly compact in $\mathcal C_E([0, 1])$ because for each $t \in [0, T]$, $\int _{0}^{1}G(t, s) X(s) ds$ is convex and weakly compact, see e.g. [11]. We only need to check the compactness property since other properties are obvious. Indeed, let $u_{f_n } \in \mathcal X$. As $S ^1_X$ is $\sigma (L ^1_{E}, L^\infty _{E ^*_s})$ sequentially compact, we may assume that $(f_n)$ $\sigma (L ^1_{E}, L^\infty _{E ^*_s})$ converges to $f_\infty \in S ^1_X$. Then we have for each $t \in [0, 1]$,

$$\begin{aligned} { \mathrm w-}\lim _n u_{f_n} (t)&= e_{ x } (t) + { \mathrm w}-\lim _n \int _{0}^{1}G(t, s)f_n(s) ds \\&= e_{ x } (t)+\int _{0}^{1} G(t, s)f_\infty (s) ds := u_{f_\infty }(t). \end{aligned}$$

This means that $u_{f_n}(t)$ converges to $u_{f_\infty }(t)$ in $E_\sigma$ for every $t \in [0, 1]$. Hence $u_{f_n}$ converges weakly in $\mathcal C_{E}([0, 1])$ to $u_{f_\infty } \in \mathcal X$. Similarly using the properties of $\frac{\partial G}{\partial t}$ in Lemma 2.1,

$$ \mathcal Y := \left\{ \dot{u}_f \in \mathcal C_E([0, 1] : \dot{u}_f(t) = \dot{e}_{x} (t)+ \int _{0}^{1}\frac{\partial G}{\partial t}(t, s)f(s) ds, t \in [0, 1], f \in S ^1_X \right\} $$

is bounded, convex, equicontinuous and sequentially weakly compact in $\mathcal C_{E}([0, 1])$ with

$$ \left\{ \begin{array}{l} \dot{e}_{x}(t) = \gamma A\left( 1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}\right) \exp {\left( -\gamma t \right) }x, \, \, t \in [0, 1]\\ A = \left( \sum \nolimits _{i=1}^{m-2}\alpha _{i}-1+ \exp (-\gamma ) -\sum \nolimits _{i=1}^{m-2}\alpha _{i}\exp (-\gamma (\eta _{i}))\right) ^{-1}, \end{array} \right. $$

and we have

$$\begin{aligned} { \mathrm w}-\lim _n \dot{u}_{f_n} (t)&= \dot{e}_{ x } (t) + { \mathrm w}-\lim _n \int _{0}^{1}\frac{\partial G}{\partial t}(t, s)f_n(s) ds\\&= \dot{e}_{ x } (t)+ \int _{0}^{1}\frac{\partial G}{\partial t}(t, s)f_\infty (s) ds := u_{f_\infty }(t). \end{aligned}$$

This means that $\dot{u}_{f_n}(t)$ converges to $\dot{u}_{f_\infty }(t)$ in $E_\sigma$ for every $t \in [0, 1]$.$\blacksquare$

Remark

In the context of Control Theory, we have stated in the proof of Proposition 2.2, the dependence of the solution with respect to the control $f \in S ^1_X$. Namely, if $u_{ f_n}$ is the $W ^{2, 1}_{ E}([0, 1])$-solution to

$$ \left\{ \begin{array}{lll} \ddot{u}_{ f_n} (t)+\gamma \dot{u}_{f_n}(t) = f_n(t), \quad t \in [0, 1] \\ u_{ f_n}(0) =x, \quad u_{ f_n}(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u_{f_n}(\eta _i) \end{array} \right. $$

and if $(f_n)$ converges $\sigma (L ^1_E, L ^\infty _{E ^*_s})$ to $f_\infty \in S^{1}_X$, then $(u_{ f_n}(t))$ converges to $u_{ f_\infty }(t)$ and $(\dot{u}_{ f_n}(t))$ converges to $\dot{u}_{ f_\infty }(t)$, in $E_\sigma$ for every $t \in [0, 1]$ where $u_{ f_\infty }$ is the $W ^{2, 1}_{ E}([0, 1])$-solution to

$$ \left\{ \begin{array}{lll} \ddot{u}_{ f_\infty } (t)+\gamma \dot{u}_{ f_\infty } (t) = f_\infty (t), \quad t \in [0, 1] \\ u_{ f_\infty }(0) =x, \quad u_{ f_\infty }(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u_{ f_\infty }(\eta _i). \end{array} \right. $$

The above remark is of importance since it allows to prove further results. Here is an application to the existence of $W^{2,1}_{E}([0, 1])$-solution to a second order differential inclusion with m-point boundary condition.

Proposition 2.3

Let $X : [0, 1] \rightrightarrows E$ be a convex weakly compact valued measurable and integrably bounded mapping, $F: [0, 1]\times E\times E \rightrightarrows E$ be a convex weakly compact valued mapping satisfying

• (1) For each $x ^*\in E ^*$, the scalar function $\delta ^*( x^*, F(., ., .))$ is ${\mathcal L}_\lambda ([0, 1])\otimes {\mathcal B}(E_\sigma ) \otimes {\mathcal B}(E_\sigma )$-measurable,¹

• (2) For each $x ^*\in E ^*$ and for each $t \in [0, 1]$, the scalar function $\delta ^*( x^*, F(t, ., .) )$ is sequentially weakly upper semicontinuous, i.e., for any sequence $(x_n)$ in E weakly converging to $x \in E$, for any sequence $(y_n)$ in E weakly converging to $y \in E$, $\limsup _n \delta ^*( x^*, F(t, x_n, y_n) ) \le \delta ^*( x^*, F(t, x, y))$,

• (3) $F(t, x, y) \in X(t)$ for all $(t, x, y ) \in [0, 1]\times E\times E$.

  Then the $W ^{2, 1}_{E}([0, 1])$-solutions set to

  $$ \left\{ \begin{array}{lll} \ddot{u} (t)+\gamma \dot{u} (t) \in F(t, u(t), \dot{u}(t))), t \in [0, 1] \\ u(0) =x, \quad u(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u (\eta _i) \end{array} \right. $$

  is non empty and weakly compact in the space $C_E([0, 1])$.

Proof

The sets

$$\begin{aligned} \mathcal X&:=\left\{ u_f \in \mathcal C_E([0, 1] : u_f(t) =e_x(t) + \int _0 ^1 G(t, s) f(s) ds, f \in S ^1_X, t \in [0, 1]\right\} \end{aligned}$$

(2.3.1)

and

$$\begin{aligned} \mathcal Y&:=\left\{ \dot{u}_f \in \mathcal C_E([0, 1] : \dot{u}_f(t) = \dot{e}_{x} (t)+ \int _{0}^{1}\frac{\partial G}{\partial t}(t, s)f(s) ds, t \in [0, 1], f \in S ^1_X \right\} \end{aligned}$$

(2.3.2)

are bounded, convex, equicontinuous and weakly compact in $C_E([0, 1])$. By condition (3), it is clear that

$$\begin{aligned} F(t, u_f(t), \dot{u}_f(t)) \subset X(t) \end{aligned}$$

(2.3.4)

for all $t \in [0, 1]$ and for all $f \in S ^1_X$. Further, recall that $S ^1_X$ is $\sigma ( L^1_E, L^\infty _{E^*})$-compact (see e.g. [10]). Using (1)–(3), for each $f \in S ^1_X$, let us consider the convex $\sigma ( L^1_E, L^\infty _{E^*})$-compact valued mapping $\Psi : S ^1_X \rightrightarrows S ^1_X$ defined by

$$ \Psi (f):= \{g \in S ^1_X: g(t) \in F(t, u_f(t), \dot{u}_f(t)), \mathrm{a.e.}\ t \in [0, 1]\}. $$

Now we are going to show that $\Psi$ is upper semi continuous on the convex $\sigma ( L^1_E, L^\infty _{E^*})$-compact set $S ^1_X$. We need to check that the graph of $\Psi$ is $\sigma ( L^1_E, L^\infty _{E^*})$-closed in $S ^1_X\times S ^1_X$. Let $g_n\in \Psi (f_n)$ such that $f_n$, $\sigma ( L^1_E, L^\infty _{E^*})$-converges to $f \in S ^1_X$ and $g_n$ $\sigma ( L^1_E, L^\infty _{E^*})$-converges to $g \in S ^1_X$. By compactness of $\mathcal X$ and $\mathcal Y$, it follows that $u_{f_n} (t) \rightarrow u_{f} (t)$ in $E_\sigma$ and $\dot{u}_{f_n} (t) \rightarrow \dot{u}_{f} (t)$ in $E_\sigma$ for every $t\in [0, 1]$. From the inclusion $g_n\in \Psi (f_n)$, we have, for each $x ^*\in E ^*$ and for each $A\in {\mathcal L}_\lambda ([0, 1])$

$$ \langle 1_A(t) x^*, g_n(t) \rangle \le 1_A(t) \delta ^*(x^*, F(t, u_{f_n}(t), \dot{u}_{f_n}(t))), $$

so that, by integration,

$$ \int _A \langle x^*, g_n(t) \rangle dt \le \int _A \langle x^*, F(t, u_{f_n}(t), \dot{u}_{f_n}(t)) \rangle dt. $$

We thus have

$$\begin{aligned} \int _A \langle x^*, g(t) \rangle dt&= \lim _n \int _A \langle x^*, g_n(t) \rangle dt\\&\le \limsup _n \int _A \delta ^*(x^*, F(t, u_{f_n}(t), \dot{u}_{f_n}(t)) dt\\&\le \int _A \delta ^*(x^*, F(t, u_{f}(t), \dot{u}_{f}(t)) \rangle dt. \end{aligned}$$

Whence we get

$$ \int _A \langle x^*, g(t) \rangle dt \le \int _A \delta ^*(x^*, F(t, u_{f}(t), \dot{u}_{f}(t)) dt $$

for every $A \in {\mathcal L}_\lambda ([0, 1])$. Consequently

$$ \langle x^*, g (t)\rangle \le \delta ^*(x^*, F(t, u_{f}(t), \dot{u}_{f}(t))\mathrm{a.e.}$$

Taking a dense sequence $(e ^*_k)$ in $E^*$ with respect to the Mackey topology $\tau ( E ^*, E)$, we get

$$ \langle e^*_k, g(t) \rangle \le \delta ^*(e ^*_k, F(t, u_{f}(t), \dot{u}_{f}(t))\mathrm{a.e.}$$

for all $k\in \mathbf N$. By [13, Proposition III.35], we get finally

$$ g(t) \in F(t, u_{f}(t), \dot{u}_{f}(t)))\mathrm{a.e.}$$

which proves that $g\ in \Psi (f)$. Whence by Kakutani-Ky Fan fixed point theorem $\Psi$ admits a fixed point $f \in S ^1_X$. This is a solution to the second order differential inclusion under consideration. Using Lemma 2.1, such a fixed point f verifies

$$ \left\{ \begin{array}{lll} \ddot{u}_f(t)+\gamma \dot{u}_f(t) \in F(t, u_{f}(t), \dot{u}_{f}(t)), \mathrm{a.e.}\ t \in [0, 1] \\ u_f(0) =x, \quad u_f(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u (\eta _i). \end{array} \right. $$

The compactness of the solution set follows from the compactness of $\mathcal X$. $\blacksquare$

Second Order Evolution Inclusions Governed by Subdifferential Operators

We need to recall and summarize some notions on the subdifferential mapping of local Lipchitz functions developed by L. Thibault [25]. Let E be a separable Banach space. Let $f : E \rightarrow \mathbf {R}$ be a locally Lipschitz function. By Christensen [14, Theorem 7.5], there is a set $D_f$ such that its complementary is Haar-nul (hence $D_f$ is dense in E) such that for all $x \in D_f$ and for all $v \in E$

$$ r_f(x, v) = \lim _{ \delta \rightarrow 0} \frac{ f(x+\delta v) -f(x)}{\delta } $$

exists and $v \mapsto r_f(x, v)$ is linear and continuous. Let us set $\nabla f(x) = r_f(x, .) \in E ^*$. Then $r_f(x, v)= \langle \nabla f(x), v\rangle$, $\nabla f(x)$ is the gradient of f at the point x. Let us set

$$ { \mathcal L}_f(x) = \{ \lim _{j \rightarrow \infty } \nabla f(x_j) | x_j \in D_f, x_j \rightarrow x \}. $$

By definition, the subdifferential $\partial f(x)$ in the sense of Clarke [15] at the point $x \in E$ is defined by

$$ \partial f (x) = {\overline{co} } \, { \mathcal L}_f(x). $$

The generalized directional derivative of f at a point $x \in E$ in the direction $v \in E$ is denoted by

$$ f^.(x, v) = \limsup _{h \rightarrow 0, \delta \rightarrow 0 }\frac{ f(x+h+\delta v)-f(x+h)}{\delta }. $$

Proposition 2.4

Let $f : E \rightarrow \mathbf {R}$ be a locally Lipchitz function. Then the subdifferential $\partial f (x)$ at the point $x \in E$ is convex weak star compact and

$$ f^.(x, v) = \sup \{ \langle \zeta ^*, v\rangle | \zeta ^* \in \partial f(x) \} \quad \forall v \in E $$

that is, $f^.(x, .)$ is the support function of $\partial f(x)$.

Proof

See Thibault [25, Proposition I.12].$\blacksquare$

Here are some useful properties of the subdifferential mapping.

Proposition 2.5

Let $f : E \rightarrow \mathbf {R}$ be a locally Lipchitz function. Then the convex weak star compact valued subdifferential mapping $\partial f$ is upper semicontinuous with respect to the weak star topology.

Proof

See [25, Proposition I.17]. Indeed we have

$$ \delta ^*(v, \partial f (x)) = f^.(x; v) =\limsup _{h \rightarrow 0, \delta \rightarrow 0 }\frac{ [f (x+h+\delta v)-f (x+h)]}{\delta }. $$

As $f^.(.;v)$ is upper semicontinuous and $\partial f$ is convex compact valued in $E ^*_s$, by [13], $\partial f$ is upper semicontinuous in $E ^*_s$. $\blacksquare$

Proposition 2.6

Let $(T, {\mathcal T})$ a measurable space, and let $f : T\times E \rightarrow \mathbf {R}$ such that

$f(., \zeta )$ is ${\mathcal T}$-measurable, for every $\zeta \in E$.

f(t, .) is locally Lipschitz for every $t\in T$.

Let $f^._t(x; v)$ be the directional derivative of $f(t, .) := f_t$ in the direction v for every fixed $t \in T$. Let x and v be two ${\mathcal T}$-measurable mappings from T to E. Then the following hold:

• (a) the mapping $t \mapsto f^._t (x(t); v(t))$ is ${\mathcal T}$-measurable.

• (b) the mapping $t \mapsto \partial f_t(x(t))$ is graph measurable, that is, its graph belongs to ${\mathcal T}\otimes {\mathcal B}(E ^*_s)$.

Proof

See Thibault [25, Proposition I.20 and Corollary I.21]. Note that the convex weak star compact valued mapping $t \mapsto \partial f_t(x(t))$ is scalarly ${\mathcal T}$-measurable, and so enjoys good measurability properties because $E ^*_s$ is a locally convex Lusin space. $\blacksquare$

We begin with a second order differential inclusion involving the subdifferential operator.

Proposition 2.7

Assume that $E= \mathbf {R}^d$, and $h : [0, 1]\times \mathbf {R}^d \times \mathbf {R}^d \rightarrow \mathbf {R}^d$ be a bounded Carathéodory mapping, that is, h is separately Lebesque-measurable on [0, 1], separately continuous on $\mathbf {R}^d\times \mathbf {R}^d$, $||h(t, x, y)|| \le \alpha (t), \forall (t, x, y ) \in [0, T]\times \mathbf {R}^d \times \mathbf {R}^d$ where $\alpha$ is positive Lebesque-integrable. Let $f : [0, 1]\times E \rightarrow \mathbf {R}$ be a mapping such that

• (1) $\forall x \in E, f(., x)$ is Lebesgue-measurable,

• (2) There exists $\beta \in L^1_{\mathbf {R}^+}([0, 1])$ such that, for all $t \in [0, 1]$, for all $x, y \in E$,

  $$ ||f(t, x)-f(t, y)|| \le \beta (t) ||x-y||. $$

Then the following hold

• (a) $\partial f_t(x) \subset \beta (t) {\overline{B}}_E$, for all $(t, x) \in [0, 1]\times E$,

• (b) The $W^{2, 1}_{E}([0, 1])$-solution set to

  $$ \left\{ \begin{array}{l} \ddot{u}(t)+\gamma \dot{u}(t) \in \partial f_t(u(t))+h(t, u(t), \dot{u}(t)) , \mathrm{a.e.}\ t \in [0, 1] \\ u(0) =x, \quad u(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u (\eta _i) \end{array} \right. $$

  is compact in the space $\mathcal C_E([0, T])$.

Proof

The proof is immediate by applying Proposition 2.3 to the convex compact valued mapping $(t, x, y) \mapsto \partial f_t(x)+h(t, x, y)$, taking account of the properties of the subdifferential mapping and its measurable properties given in Proposition 2.6. $\blacksquare$

We finish this section with a variant which has some importance in the study of epiconvergence problem for the approximating system

$$ \ddot{u}(t)+\gamma \dot{u}(t)= h(t, u(t), \dot{u}(t))- \nabla \varphi (u(t)) $$

where $\varphi$ is $C ^1$ and Lipschitz.

Proposition 2.8

Assume that $E= \mathbf {R}^d$, $\varphi : E \rightarrow \mathbf {R}$ is $C ^1$, Lipschitz, and that $h : [0, 1]\times \mathbf {R}^d \times \mathbf {R}^d \rightarrow \mathbf {R}^d$ is a bounded Carathéodory mapping, that is, h is separately Lebesque-measurable on [0, 1], separately continuous on $\mathbf {R}^d\times \mathbf {R}^d$, $||h(t, x, y)|| \le \alpha (t), \forall (t, x, y ) \in [0, T]\times \mathbf {R}^d \times \mathbf {R}^d$ where $\alpha$ is positive Lebesque-integrable.Then the $W^{2, 1}_{E}([0, 1])$-solution set to

$$ \left\{ \begin{array}{lll} \ddot{u}(t)+\gamma \dot{u}(t) = h(t,u(t), \dot{u}(t) ) - \nabla \varphi (u(t))\, \mathrm{a.e.}\ t \in [0, 1] \\ u(0) =x, \quad u(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u (\eta _i) \end{array} \right. $$

is compact in the space $\mathcal C_E([0, T])$.

Proof

The proof is immediate by applying Proposition 2.3 with $F(t, x, y )= h(t, x, y)- \nabla \varphi (x), \forall (t, x, y) \in [0, 1]\times E\times E$ and by observing that the subdifferential $x \mapsto \partial \varphi (x) = \nabla \varphi (x)$ is bounded and continuous. $\blacksquare$

3 Applications. Towards the Variational Convergence in Second Order Evolution Inclusions

Let us recall a useful Gronwall type lemma [12].

Lemma 3.1

(A Gronwall-like inequality) Let $p, q, r : [0, T] \rightarrow [0, \infty [$ be three nonnegative Lebesgue integrable functions such that for almost all $t\in [0, T]$

$$ r(t) \le p(t) + q(t)\int _0 ^t r(s)\,ds. $$

Then

$$ r(t) \le p(t) + q(t) \int _0 ^t [p(s)\,\mathrm{exp}(\int _s^t q(\tau ) \,d\tau )]\,ds $$

for all $t \in [0, T].$

We recall below some notations and summarize some results which describe the limiting behavior of a bounded sequence in $L^1_H([0, T])$. See [10, Proposition 6.5.17].

Proposition 3.1

Let H be a separable Hilbert space. Let $(\zeta _n)$ be a bounded sequence in $L^1_H([0, T])$. Then the following hold:

• (1) $(\zeta _n)$ (up to an extracted subsequence) stably converges to a Young measure $\nu$ that is, there exist a subsequence $(\zeta '_n)$ of $(\zeta _n)$ and a Young measure $\nu$ belonging to the space of Young measure ${\mathcal Y}([0, T] ; H_\sigma )$ with $t\mapsto \mathop {\text {bar}}(\nu _t) \in L^1_H ([0, T])$ (here $\mathop {\text {bar}}(\nu _t)$ denotes the barycenter of $\nu _t$) such that

  $$ \lim _{n\rightarrow \infty } \int _0^T h(t, \zeta '_n(t)))\,dt) = \int _0^T\left[ \int _H h(t,x)\,\nu _t(dx)\right] \,dt $$

  for all bounded Carathéodory integrands $h :[0, T] \times H_\sigma \rightarrow \mathbf {R}$,

• (2) $(\zeta _n)$ (up to an extracted subsequence) weakly biting converges to an integrable function $f\in L^1_H([0, T])$, which means that there is a subsequence $(\zeta '_m)$ of $(\zeta _n)$ and an increasing sequence of Lebesgue-measurable sets $(A_p)$ with $\lim _p \lambda (A_p) = 1$ and $f\in L^1_H([0, T])$ such that, for each p,

  $$ \lim _{m\rightarrow \infty } \int _{A_p} \langle h(t), \zeta '_m(t) \rangle \,dt = \int _{A_p} \langle h(t), f(t) \rangle \,dt $$

  for all $h\in L^\infty _H([0, T])$,

• (3) $(\zeta _n)$ (up to an extracted subsequence) Komlós converges to an integrable function $g\in L^1_H([0, T])$, which means that there is a subsequence $(\zeta _{\beta (m)})$ and an integrable function $g\in L^1_H([0, T])$, such that

  $$ \lim _{n\rightarrow \infty } \frac{1}{n} \Sigma _{j=1}^n \zeta _{\gamma (j)}(t) = g(t), \, \mathrm{a.e.}\, \in [0, T], $$

  for every subsequence $(f_{\gamma (n)})$ of $(f_{\beta (n)}).$

• (4) There is a filter ${\mathcal U}$ finer than the Fréchet filter such that ${\mathcal U}-\lim _n \zeta _n = l \in (L^\infty _H)'_{weak}$ where $(L^\infty _H)'_{weak}$ is the second dual of $L^1_H ([0, T])$.

  Let $w_{l_a}\in L^1_H([0, T])$ 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 $l_s$.

  If we have considered the same extracted subsequence in (1)–(4), then one has

  $$ f(t) = g(t) = \mathop {\text {bar}}(\nu _t) = w_{l_a}(t) \mathrm{a.e.}\ t \in [0, T]. $$

By $W^{2, 1}_{\mathbf {R}^d} ([0, T])$ (resp. $W^{2, 2}_{\mathbf {R}^d} ([0, T]$) we denote the set of all continuous functions in ${\mathcal C}_{\mathbf {R}^d} ([0, T])$ such that their first derivatives are continuous and their second derivatives belong to $L ^1_{\mathbf {R}^d} ([0, T])$ (resp. $L ^2_{\mathbf {R}^d} ([0, T])$) and by $W^{1, 1}_{BV} ([0, T])$ we denote the set of all continuous functions in ${\mathcal C}_{\mathbf {R}^d} ([0, T])$ such that their first derivatives are of bounded variation (BV for short).

We begin with a preliminary result which shows the limiting properties of $W^{2, 1}_{\mathbf {R}^d}([0, 1])$-solutions for a second order ordinary differential equation with m-point boundary conditions.

Proposition 3.2

Let $E= \mathbf {R}^d$. Let $(f_n)_{n\in \mathbf N}$ be a bounded sequence in $L^1_E([0, 1])$. For each $n\in \mathbf N$, let us consider the $W^{2, 1}_E([0, 1])$-solution $u_n : [0, 1]\rightarrow E$ of the equation

$$ \ddot{u}_n(t)+\gamma \dot{u}_n(t) =f_n(t), t\in [0, 1];\quad u_n(0) =x, \quad u_n(1)= \sum _{i = 1} ^{m-2} \alpha _i u_n (\eta _i). $$

Then there exist a subsequence of $(u_n)$ still denoted by $(u_n)$, a $W^{1, 1}_{BV}([0, 1])$-function $u : [0, 1]\rightarrow E$ and a Young measure $\nu \in {\mathcal Y}([0, 1]; E)$ such that $t\mapsto \mathop {\text {bar}}(\nu _t)$ belongs to $L^1_E([0, 1])$ which satisfy the following conditions:

• (a) $(u_n(.))$ converges in ${\mathcal C}_E([0, 1])$ to u(.) with $u(0) =x, u(1)= \sum _{i = 1} ^{m-2} \alpha _i u (\eta _i).$

• (b) $(\dot{u}_n(.))$ converges in $L^1_E([0, 1])$ to $\dot{u}(.)$.

• (c) $(\delta _{\ddot{u}_n})$ stably converges in ${\mathcal Y}([0, 1], E)$ to $\nu$.

• (d) Assume further that the negative parts $\langle u_n, \ddot{u}_n \rangle ^-$ of the functions $\langle u_n, \ddot{u}_n \rangle$ are uniformly integrable in $L^1_{\mathbf {R}}([0, 1])$.

  Then

  $$\begin{aligned} \liminf _{n\rightarrow \infty } \int _0 ^1 \langle u_n(t), \ddot{u}_n(t)\rangle \,dt \ge \int _0 ^1 \langle u(t), \mathop {\text {bar}}(\nu _t) \rangle \,dt = \int _0 ^1\left[ \int _E \langle u(t), x \rangle \,\nu _t(dx)\right] \,dt. \end{aligned}$$

Proof

Existence and uniqueness of a $W^{2, 1}_E([0, 1])$-solution for the equation

$$ \ddot{u}_n(t)+\gamma \dot{u}_n(t) =f_n(t), t\in [0, 1];u(0) =x, \quad u(1)= \sum _{i = 1} ^{m-2} \alpha _i u (\eta _i). $$

are ensured by Proposition 2.1 with integral representation formulas

$$ \left\{ \begin{array}{lll} u_n (t) = e_{x} (t)+ \int _{0}^{1} G(t, s) f_n(s) ds , \, t \in [0, 1]\\ \dot{u}_n (t) =\dot{e}_{x} (t)+ \int _{0}^{1}\frac{\partial G}{\partial t}(t,s) f_n(s) ds, \, t \in [0, 1]\\ \end{array} \right. $$

where

$$ \left\{ \begin{array}{ll} e_{x } (t) &{}= x + A (1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}) (1-\exp (-\gamma t) )x \\ \dot{e}_{x }(t) &{}= \gamma A \left( 1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}\right) \exp {\left( -\gamma t \right) }x\\ A &{}= \left( \sum \nolimits _{i=1}^{m-2}\alpha _{i}-1+ \exp (-\gamma ) -\sum \nolimits _{i=1}^{m-2}\alpha _{i}\exp (-\gamma (\eta _{i}))\right) ^{-1}. \end{array} \right. $$

Since $(f_n(.))$ is bounded in $L^1_E([0, 1])$ by assumption, $(\dot{u}_n(.))$ is uniformly bounded by using the integral formula for $\dot{u}_n$ and the boundedness of the Green function G given in Lemma 3.1. So $(\dot{u}_n(.))$ is uniformly bounded and bounded in variation. In view of the Helly–Banach theorem (see e.g. [20, p. 11]), we may, by extracting a subsequence, assume that $(\dot{u}_n(.))$ pointwise converges to a BV function v(.). Let us set $u(t) =\int _0 ^t v(s)\,ds$ for all $t\in [0, 1]$. Then $u \in W^{1, 1}_{BV}([0, 1])$ with $\dot{u}(t) =v(t)$ for almost every $t \in [0, 1]$. Then $(\dot{u}_n(.))$ is uniformly bounded and pointwise converges to v(.). By Lebesgue’s theorem, we conclude that $(\dot{u}_n(.))$ converges in $L^1_E([0, 1])$ to $\dot{u}(.)$. Hence $(u_n(.))$ converges uniformly to u(.) with $u(0) =x, u(1)= \sum _{i = 1} ^{m-2} \alpha _i u (\eta _i)$. It remains to check (c) and (d). Since $(\ddot{u}_n(.))$ is bounded, in view of Proposition 3.1, we may assume that the sequence $(\delta _{\ddot{u}_n})$ of associated Young measures stably converges in ${\mathcal Y}([0, 1], E)$ to a Young measure $\nu$ such that $t \mapsto \mathop {\text {bar}}(\nu _t)$ belongs to $L^1_E([0, 1])$. Let us prove the last Fatou property (d). We may suppose that

$$ a:=\lim _{n\rightarrow \infty } \int _0^1\, \langle u_n(t), \ddot{u}_n(t)\rangle \,dt\in \mathbf {R}. $$

Furthermore, since $(\ddot{u}_n(.))$ is bounded in $L^1_E([0, 1])$, in view of Proposition 3.1 we may suppose that $(\ddot{u}_n(.))$ weakly biting converges to a function $f\in L^1_E([0, 1])$, that is, there exist a subsequence (still denoted by $(\ddot{u}_n(.))$) of $(\ddot{u}_n(.))$ and an increasing sequence of measurable sets $(A_p)$ in [0, 1] such that $\lim _{p\rightarrow \infty } \lambda (A_p) =1,$ and such that, for each p and for each $g\in L^{\infty }_E(A_p, A_p\cap {\mathcal L} ([0, 1]), \lambda |_{A_p}),$ the following holds:

$$ \lim _{n\rightarrow \infty } \int _{A_p}\,\langle \ddot{u}_n(t), g(t)\rangle \,dt =\int _{A_p}\,\langle f(t), g(t)\rangle \,dt. $$

Let $\varepsilon >0$ be given. Pick $N\in \mathbf N$ such that

$$ \int _{A_N}\, \langle u(t), f(t)\rangle \,dt \ge \int _{[0, 1]} \langle u(t), f(t)\rangle \,dt-\varepsilon , $$

and that

$$ \limsup _{n\rightarrow \infty }\int _{[0,1]\setminus A_N}\, \langle u_n(t), \ddot{u}_n(t)\rangle ^-\,dt \le \varepsilon $$

(this is possible because $(\langle u_n, \ddot{u}_n\rangle ^-)_n$ is uniformly integrable by hypothesis). As $||u_n(.)-u(.)|| \rightarrow 0$ uniformly, it is easy to see that

$$ \lim _{n\rightarrow \infty } \int _{A_N}\,||u_n(t)-u(t)||||\ddot{u}_n(t)||\,dt =0. $$

See [6, 16] for a more general case. Whence

$$ \lim _{n\rightarrow \infty } \big [ \int _{A_N}\,\langle u_n(t), \ddot{u}_n(t)\rangle \,dt -\int _{A_N}\,\langle u(t), \ddot{u}_n(t) \rangle \,dt\big ]=0. $$

An easy computation gives

$$\begin{aligned} a\ge & {} \lim _{n\rightarrow \infty }\int _{A_N}\,\langle u_n(t), \ddot{u}_n(t)\rangle -\limsup _{n\rightarrow \infty }\int _{[0, 1]\setminus A_N}\,\langle u_n(t), \ddot{u}_n(t)\rangle ^- \,dt \\\ge & {} \lim _{n\rightarrow \infty } \int _{A_N}\,\langle u_n(t), \ddot{u}_n(t)\rangle \,dt -\varepsilon . \end{aligned}$$

Finally we get

$$\begin{aligned} a\ge & {} \lim _{n\rightarrow \infty } \int _{A_N}\,\langle u_n(t), \ddot{u}_n(t)\rangle \,dt -\varepsilon \\= & {} \lim _{n\rightarrow \infty } \int _{A_N}\,\langle u(t), \ddot{u}_n(t) \rangle \,dt -\varepsilon \\= & {} \int _{A_N}\,\langle u(t) , f(t)\rangle \,dt -\varepsilon \\\ge & {} \int _{[0, 1]}\,\langle u(t), f(t)\rangle \,dt -2\varepsilon . \end{aligned}$$

By virtue of Proposition 3.1 $f(t) = \mathop {\text {bar}}(\nu _t)$ a.e. The proof is therefore complete because

$$ \int _0 ^1 \langle u(t), \mathop {\text {bar}}(\nu _t) \rangle \,dt = \int _0 ^1\left[ \int _E \langle u(t), x \rangle \,\nu _t(dx)\right] \,dt. $$

$\blacksquare$

The above techniques can be used to prove the existence of a solution for second order evolution inclusion with boundary conditions governed by subdifferential operators of the form

$$\begin{aligned} f(t) \in \ddot{u} (t)+M u(t) +\partial \varphi (u(t)), t \in [0, T]\qquad (\mathrm{I}) \end{aligned}$$

where M is positive, $\varphi$ is a proper convex proper lower semicontinuous function defined on $\mathbf {R}^d$, and $\partial \varphi (u(t))$ is the subdifferential of the function $\varphi$ at the point u(t) and the perturbation f belongs to $L^2_{\mathbf {R}^d}([0, T])$. Similar results in this direction are obtained by [1,2,3,4, 11].

Now we present a fairly general result for the approximating problem via the epiconvergence approach in a second order evolution problem. The applicability of our abstract results will be exemplified below.

Proposition 3.3

Assume that $M>0, \beta \in L^2_{\mathbf {R}^+} ([0, T])$. For each $n\in \mathbf N$, let $\varphi _n : \mathbf {R}^d \rightarrow \mathbf {R}^+$ be a convex, Lipschitz function and let $\varphi _\infty$ be a nonnegative l.s.c proper function defined on $\mathbf {R}^d$ such that $\varphi _n(x) \le \varphi _\infty (x)$ for all $n\in \mathbf N$ and for all $x\in \mathbf {R}^d$. Let $f ^n \in L^2_{\mathbf {R}^d}([0, T])$ such that $||f_n(t)|| \le \beta (t), \forall n \in \mathbf N, \forall t \in [0, T]$. For each $n\in \mathbf N$, let $u^n$ be a $W^{2,1}_{ \mathbf {R}^d}([0, T])$-solution to the problem

$$ \left\{ \begin{array}{lll} f^n(t)\in \ddot{u}^n (t)+M\dot{u}^n(t)+\partial \varphi _n(u^n(t)), t \in [0, T] \\ u^n(0) = u^n_0;\dot{u}^n(0)= \dot{u}^n_0. \end{array} \right. $$

Assume that

• (i) $f^n \, \sigma (L^2_{\mathbf {R}^d}, L^2_{\mathbf {R}^d})$-converges to $f ^\infty \in L^2_{\mathbf {R}^d}([0, T])$,

• (ii) $\varphi _n$ epi-converges to $\varphi _\infty$,

• (iii) $\lim _n u^n (0)=u ^ \infty _0 \in \mathop {\text {dom}}\, \varphi _\infty , \,\lim _n \varphi _n(u^n (0)) = \varphi _\infty (u^ \infty _0 )$, and $\lim _n \dot{u}^n (0)=\dot{u}^ \infty _0$,

• (iv) There exist $r_0 >0$ and $x_0 \in \mathbf {R}^d$ such that

  $$ \sup _{n\in \mathbf N} \sup _{v \in {\overline{B}}_{L^\infty _{ \mathbf {R}^d}([0, T])}}\int _0^T\varphi _\infty (x_0+r_0v(t)) < +\infty $$

  where ${\overline{B}}_{L^\infty _{ \mathbf {R}^d}([0, T])}$ is the closed unit ball in $L^\infty _{ \mathbf {R}^d}([0, T])$.

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

  $$ f ^\infty \in \zeta ^\infty + M\dot{u}^\infty + \partial I_{\varphi _\infty }(u^\infty ) $$

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

  $$ I_{\varphi _\infty }(u):= \int _0^T \varphi _\infty ( u(t)) \,dt, \forall u \in L^\infty _{\mathbf {R}^d}([0, T]). $$

  Furthermore $\lim _n \varphi _n( u^n(t)) = \varphi _\infty ( u^\infty (t)) < \infty$ a.e. and $\lim _n \int _0^T \varphi _n( u^n(t)) dt = \int _0^T \varphi _\infty ( u ^\infty (t)) dt$. Subsequently, the energy estimate holds true almost everywhere $t \in [0, T]$,

  $$\begin{aligned} \varphi _\infty (u^\infty (t)) \,+\,&\frac{1}{2} ||\dot{u}^\infty (t)||^2 = \varphi _\infty (u^\infty _0)) +\frac{1}{2} ||\dot{u}^\infty _0||^2\\&\qquad \qquad -\int _0^t \langle M \dot{u}^\infty (s), \dot{u}^\infty (s) \rangle ds+ \int _0^t \langle \dot{u}^\infty (s), f ^\infty (s)\rangle ds. \end{aligned}$$

  Further $(\ddot{u}^n)$ weakly converges to the vector measure $m\in {\mathcal M}^b_{\mathbf {R}^d} ([0, T])$ so that the limit functions $u ^\infty (.)$ and the limit measure m satisfy the following variational inequality:

  $$\begin{aligned} \int _0^T \varphi _\infty ( v(t)) \,dt \ge&\int _0^1 \varphi _\infty ( u ^\infty (t)) \,dt + \int _0 ^1 \langle -M \dot{u} ^\infty (t)+f^\infty (t) ,v(t)- u ^\infty (t)\rangle \,dt \\&+ \langle -m, v-u ^\infty \rangle _{ ({\mathcal M}^b_{\mathbf {R}^d}([0, T]), {\mathcal C}_{\mathbf {R}^d}([0, T]))}. \end{aligned}$$

  In other words, the vector measure $-m+[-M \dot{u}^\infty +f ^\infty ] \,dt$ belongs to the subdifferential $\partial J_{\varphi _\infty }(u ^\infty )$ of the convex functional integral $J_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbf {R}^d}([0, T])$ by $J_{\varphi _\infty }(v)=\int _0 ^1 \varphi _\infty (t, v(t))\ dt$, $\forall v\in {\mathcal C}_{\mathbf {R}^d}([0, T])$.

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

  $$ {\mathcal U}-\lim _n [f^n-\ddot{u} ^n-M\dot{u} ^n] = l \in L^\infty _{\mathbf {R}^d}([0, T])'_{weak} $$

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

  $$ \lim _n [f ^n-\ddot{u} ^n-M\dot{u} ^n] = \mathbf n\in {\mathcal C}_{\mathbf {R}^d} ([0, T])'_{weak} $$

  where ${\mathcal C}_{\mathbf {R}^d} ([0, T])'_{weak}$ denotes the space ${\mathcal C}_{\mathbf {R}^d} ([0, T])'$ endowed with the weak topology $\sigma ({\mathcal C}_{\mathbf {R}^d} ([0, T])', {\mathcal C}_{\mathbf {R}^d} ([0, T]))$. 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

  $$ l_a(h) = \int _0 ^T \langle h(t),f ^\infty (t) -\zeta ^\infty (t) -M\dot{u}^\infty (t) \rangle dt $$

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

  $$ I_{\varphi _\infty }^*(l) = I_{\varphi _\infty ^* }(f^\infty -\zeta ^\infty -M\dot{u}^\infty ) +\delta ^* (l_s ,\mathop {\text {dom}}I_{\varphi _\infty } ) $$

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

  $$ \langle \mathbf n, h \rangle = \int _0 ^T \langle f ^\infty (t) -\zeta ^\infty (t)-M\dot{u}^\infty (t) , h(t)\rangle dt + \langle \mathbf n_s, h \rangle , \quad \forall h \in {\mathcal C}_{\mathbf {R}^d}([0, T]). $$

  with $\langle \mathbf n_s, h \rangle = l_s(h)$, $\forall h \in {\mathcal C}_{\mathbf {R}^d}([0, T])$. Further $\mathbf n$ belongs to the subdifferential $\partial J_{\varphi _\infty }(u^\infty )$ of the convex lower semicontinuous integral functional $J_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbf {R}^d}([0, T])$

  $$ J_{\varphi _\infty }(u):= \int _0^T \varphi _\infty ( u(t)) \,dt, \forall u \in {\mathcal C}_{\mathbf {R}^d}([0, T]). $$

• (c) Consequently the density $f ^\infty -\zeta ^\infty -M\dot{u}^\infty$ of the absolutely continuous part $n_a$

  $$ \mathbf n_a (h):= \int _0 ^T \langle f^\infty (t) -\zeta ^\infty (t)-M\dot{u}^\infty (t) , h(t)\rangle dt, \quad \forall h\in {\mathcal C}_{\mathbf {R}^d}([0, T]) $$

  satisfies the inclusion

  $$ f ^\infty (t)- \zeta ^\infty (t) - M\dot{u}^\infty (t) \in \partial \varphi _\infty (u^\infty (t)), \quad \mathrm{a.e.} $$

  and for any nonnegative measure $\theta$ on [0, T] with respect to which $\mathbf n_s$ is absolutely continuous

  $$ \int _0^T h_{\varphi _\infty ^*}( \frac{d\mathbf n_s}{d\theta }(t)) d\theta (t) = \int _0^T \langle u^\infty (t), \frac{d\mathbf n_s}{d \theta }(t)\rangle d\theta (t) $$

  here $h_{\varphi _\infty ^*}$ denotes the recession function of $\varphi _\infty ^*$.

Proof

Step 1 $||\dot{u}^n(.)||$ and $\varphi _n(u_n(.))$ are uniformly bounded.

Multiplying scalarly the inclusion

$$ f^n(t)-\ddot{u}^n(t)- M \dot{u}^n(t) \in \partial \varphi _n(u^n(t)) $$

by $\dot{u}^n(t)$ and applying the chain rule theorem [21, Theorem 2] yields

$$ \langle \dot{u}^n(t), f ^n(t)\rangle -\langle \dot{u}^n(t), \ddot{u}^n(t) \rangle - \langle \dot{u}^n(t), M \dot{u}_n(t) \rangle = \frac{d}{dt} [\varphi _n(u_n(t))] $$

that is,

$$\begin{aligned} -\langle M \dot{u}^n(t), \dot{u}^n(t) \rangle + \langle \dot{u}^n(t), f ^n(t)\rangle = \frac{d}{dt}\left[ \varphi _n(u_n(t)) +\frac{1}{2} ||\dot{u}^n(t)||^2\right] . \end{aligned}$$

(3.3.1)

Integrating this equality on [0, t], we get

$$\begin{aligned} \begin{aligned} \varphi _n(u^n(t))\; +\;&\frac{1}{2} ||\dot{u}^n(t)||^2\\ =\;&\varphi _n(u^n(0)) +\frac{1}{2} ||\dot{u}^n(0)||^2\\&-\int _0^t \langle M \dot{u}^n(s), \dot{u}^n(s) \rangle ds+ \int _0^t \langle \dot{u}^n(s), f ^n(s)\rangle ds \\ \le \;&\varphi _n(u^n(0)) +\frac{1}{2} ||\dot{u}^n(0)||^2 \\&+ M\int _0^t ||\dot{u}^n(s)||^2 ds+ ||f^n||_{L^2_{\mathbf {R}^d}([0, T])} \left( \int _0^t ||\dot{u}^n(s)|| ^2 ds \right) ^{\frac{1}{2}} \\ \le \;&\varphi _n(u^n(0)) +\frac{1}{2} ||\dot{u}^n(0)||^2 \\&+M \int _0^t ||\dot{u}^n(s)||^2 ds+ \frac{1}{2} ||f^n||_{L^2_{\mathbf {R}^d}([0, T])}\left( 1+\int _0^t ||\dot{u}^n(s)|| ^2 ds\right) \\ \le \;&\varphi _n(u^n(0)) +\frac{1}{2} ||\dot{u}^n(0)||^2 \\&+M\int _0^t ||\dot{u}^n(s)||^2 ds+ \frac{1}{2} ||\beta ||_{L^2_{\mathbf {R}}([0, T])}\left( 1+\int _0^t ||\dot{u}^n(s)|| ^2 ds\right) . \end{aligned} \end{aligned}$$

Then, from (iii), the preceding estimate and the Gronwall like inequality (Lemma 3.1), it is immediate that

$$\begin{aligned} \sup _{n \ge 1} \sup _{t \in [0, T] } ||\dot{u}^n(t)||< +\infty \quad \mathrm{and} \quad \sup _{n \ge 1} \sup _{t \in [0, T] }\varphi _n(u^n(t)) < +\infty . \end{aligned}$$

(3.3.2)

Step 2 Estimation of $||\ddot{u}^n(.)||$. As

$$ z^n(t) :=f^n(t) -\ddot{u}^n (t) -M \dot{u}^n(t) \in \partial \varphi _n(u^n(t)) $$

by the subdifferential inequality for convex lower semi continuous functions we have

$$\begin{aligned} \varphi _n (x) \ge \varphi _n (u^n(t)) + \langle x-u^n(t), z^n(t) \rangle \end{aligned}$$

for all $x \in \mathbf {R}^d$. Now let $v\in {\overline{B}}_{L^\infty _{\mathbf {R}^d}([0, T])}$, the closed unit ball of $L^\infty _{\mathbf {R}^d}[0, T])$. Taking $x= w(t):= x_0+ r_0v(t)$ in the preceding inequality we get

$$\begin{aligned} \varphi _n (w(t)) \ge \varphi _n ( u^n(t)) + \langle w(t)-u^n(t), z^n(t) \rangle . \end{aligned}$$

Integrating the preceding inequality gives

$$\begin{aligned} \begin{aligned} \int _0^T \langle x_0+ r_0v(t)-u^n(t)&, z^n(t) \rangle dt \\&= \int _0^T \langle x_0-u^n(t), z^n(t) \rangle dt + r_0 \int _0^T \langle v(t) , z^n(t) \rangle dt \\&\le \int _0^T \varphi _n (x_0+ r_0v(t)) dt -\int _0^T \varphi _n (u^n(t)) dt. \end{aligned} \end{aligned}$$

Whence follows

$$\begin{aligned} r_0\int _0^T&\langle v(t) , z^n(t) \rangle dt \le \int _0^T \varphi _n (x_0+ r_0v(t)) dt\\&-\int _0^T \varphi _n (u^n(t)) dt-\int _0^T \langle x_0-u^n(t), z^n(t) \rangle dt.\nonumber \end{aligned}$$

(3.3.3)

We compute the last integral in the preceding inequality. For simplicity, let us set $v^n(t) = u^n(t)-x_0$ for all $t\in [0, T]$. By integration by parts and taking into account (3.3.2), we have

$$\begin{aligned}&-\int _0^T \langle x_0-u^n(t), z^n(t) \rangle dt = -\int _0^T \langle v^n(t), \ddot{v}^n(t)+M \dot{v}^n(t) \rangle -f^n(t) \rangle dt\\ =&-[\langle v^n(t),\dot{v}^n(t)+ Mv^n(t)]_0^T + \int _0^T \langle \dot{v}^n(t), \dot{v}^n(t)+M v^n(t) \rangle dt + \int _0^T \langle v^n(t), f^n(t) \rangle dt \nonumber \\ \le&-\langle v^n(T), \dot{v}^n(T)\rangle + \langle v^n(0), \dot{v}^n(0) \rangle -\langle Mv^n(T), v^n(T) \rangle \nonumber \\&+ \langle Mv ^n(0), v^n(0) \rangle + \int _0^T ||\dot{v}^n(t)||^2 dt + \int _0^T \langle \dot{v}^n(t), M v^n(t) \rangle dt+ \int _0^T \langle v^n(t), f^n(t) \rangle dt.\nonumber \end{aligned}$$

(3.3.4)

By (3.3.2)–(3.3.4), we get

$$\begin{aligned} \quad r_0\int _0^T \langle v(t) , z^n(t) \rangle dt \le \int _0^T \varphi _\infty (x_0+ r_0v(t)) dt +L \end{aligned}$$

(3.3.5)

for all $v \in {\overline{B}}_{L^\infty _{ \mathbf {R}^d}([0, T])}$, where L is a generic positive constant independent of $n\in \mathbf N$. By (iv) and (3.3.5) we conclude that $(z ^n = f^n- \ddot{u}^n-M\dot{u}^n)$ is bounded in $L^1_{\mathbf {R}^d}([0, T])$, then so is $(\ddot{u}^n)$. It turns out that the sequence $(\dot{u}^n)$ is uniformly bounded by using (3.3.2) and is bounded in variation. By Helly theorem, we may assume that $(\dot{u}^n)$ pointwisely converges to a BV function $v^ \infty : [0, T] \rightarrow \mathbf {R}^d$ and the sequence $(u^ n)$ converges uniformly to an absolutely continuous function $u ^\infty$ with $\dot{u}^\infty = v ^\infty$ a.e. At this point, it is clear that $(\dot{u}_n)$ converges in $L^1_{\mathbf {R}^d}([0, T])$ to $v^ \infty$, using (3.3.2) and the dominated convergence theorem. Hence $(M \dot{u}^n(.))$ converges in $L^1_{\mathbf {R}^d}([0, T])$ to $Mv^ \infty (.)$.

Step 3 Young measure limit and biting limit of $\ddot{u}_n$. As $(\ddot{u}_n)$ is bounded in $L^1_{\mathbf {R}^d}([0, T])$, we may assume that $(\ddot{u}^n)$ stably converges to a Young measure $\nu \in \mathcal Y([0, T]); \mathbf {R}^d)$ with $\mathop {\text {bar}}(\nu ): t \mapsto \mathop {\text {bar}}(\nu _t) \in L ^1_{\mathbf {R}^d}([0, T])$ (here $\mathop {\text {bar}}(\nu _t)$ denotes the barycenter of $\nu _t$). Further by Proposition 3.1, we may assume that $(\ddot{u} ^n)$ biting converges to a function $\zeta ^\infty : t \mapsto \mathop {\text {bar}}(\nu _t)$ that is, there exists a decreasing sequence of Lebesgue-measurable sets $(B_p)$ with $\lim _p\lambda ( B_p) =0$ such that the restriction of $(\ddot{u}_n)$ on each $B_p^c$ converges weakly in $L^1_{\mathbf {R}^d}([0, T])$ to $\zeta ^\infty$. Note that $(M\dot{u}^n)$ converges in $L^1_{\mathbf {R}^d}([0, T])$ to $Mv^\infty$. It follows that the restriction of $(z^n = f ^n- \ddot{u}^n- M\dot{u}^n)$ to each $B_p^c$ weakly converges in $L^1_{\mathbf {R}^d}([0, T])$ to $z^\infty := f^\infty -\zeta ^\infty -Mv^\infty$, because $(f^n)$ weakly converges in $L^1_{\mathbf {R}^d}([0, T])$ to $f ^\infty$, $(M\dot{u}^n)$ converges in $L^1_{\mathbf {R}^d}([0, T])$ to $Mv^\infty$ and $(\ddot{u}^n)$ biting converges to $\zeta ^\infty \in L^1_{\mathbf {R}^d}([0, T])$. It follows that

$$\begin{aligned} \lim _n \int _B \langle -\ddot{u}^n-W^n(t), w(t) -u^n(t) \rangle = \int _B \langle -\mathop {\text {bar}}(\nu _t)-W(t) , w(t)-u(t) \rangle dt \end{aligned}$$

(3.3.6)

for every $B \in B_p^c \cap \mathcal L ([0, T])$, and for every $w \in L^\infty _{\mathbf {R}^d} ([0, T])$, where $W^n(t)= M\dot{u}^n(t) -f^n(t)$ and $W(t) = M \dot{u}^\infty (t) -f^\infty (t)$. Indeed, we note that $(w(t) -u^n(t))$ is a bounded sequence in $L ^\infty _{\mathbf {R}^d}([0, 1])$ which pointwisely converges to $w(t) -u^\infty (t)$, it converges uniformly on every uniformly integrable subset of $L ^1_{\mathbf {R}^d}([0, T])$ by virtue of a Grothendieck Lemma [16], recalling here that the restriction of $-\ddot{u}^n-W^n$ on each $B_p^c$ is uniformly integrable. Now, since $\varphi _n$ lower epiconverges to $\varphi _\infty$, for every Lebesgue-measurable set A in [0, T], by virtue of Corollary 4.7 in [11], we have

$$\begin{aligned} + \infty >\liminf _n \int _A \varphi _n (u^n(t)) dt \ge \int _A \varphi _\infty (u^\infty (t)) dt. \end{aligned}$$

(3.3.7)

Combining (3.3.2)–(3.3.5)–(3.3.6)–(3.3.7) and using the subdifferential inequality

$$ \varphi _n(w(t)) \ge \varphi _n(u^n(t))+ \langle -\ddot{u}^n(t) -W^n(t) , w(t) - u^n(t) \rangle $$

gives

$$ \int _B \varphi _\infty ( w(t)) \,dt \ge \int _B \varphi _\infty ( u ^\infty (t))\, dt +\int _B \langle -\mathop {\text {bar}}(\nu _t) -W(t), w(t)-u^\infty (t) \rangle \,dt. $$

This shows that $t \mapsto - \mathop {\text {bar}}( \nu _t) -W(t)$ is a subgradient at the point $u ^\infty$ of the convex integral functional $I_{\varphi _\infty }$ restricted to $L^\infty _{\mathbf {R}^d} (B_p^c)$, consequently,

$$ - \mathop {\text {bar}}(\nu _t) -W(t) \in \partial \varphi _\infty ( u ^\infty (t)), \, \mathrm{a.e.} \mathrm{on } B_p^c. $$

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

$$ - \mathop {\text {bar}}( \nu _t)-W(t) \in \partial \varphi _\infty ( u ^\infty (t)), \, \mathrm{a.e.} \mathrm{on } [0, T]. $$

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

$$\begin{aligned} \int _0^T \varphi _\infty ( v(t)) \,dt \ge&\int _0^1 \varphi _\infty ( u ^\infty (t)) \,dt + \int _0 ^1 \langle -M \dot{u} ^\infty (t)+f^\infty (t) ,v(t)- u ^\infty (t)\rangle \,dt \\&+ \langle -m, v-u ^\infty \rangle _{ ({\mathcal M}^b_E([0, T]), {\mathcal C}_{\mathbf {R}^d}([0, T])) }. \end{aligned}$$

In other words, the vector measure $-m+[-M \dot{u}^\infty +f ^\infty ] \,dt=-m-W. dt$ belongs to the subdifferential $\partial J_{\varphi _\infty }(u ^\infty )$ of the convex functional integral $J_{f_\infty }$ defined on ${\mathcal C}_{\mathbf {R}^d}([0, T])$ by $J_{\varphi _\infty }(v)=\int _0 ^1 \varphi _\infty (v(t))\ dt$, $\forall v\in {\mathcal C}_{\mathbf {R}^d}([0, T])$. Indeed, let $w \in \mathcal C_{\mathbf {R}^d}([0, T])$. Integrating the subdifferential inequality

$$ \varphi _n(w(t)) \ge \varphi _n(u^n(t))+ \langle -\ddot{u}^n(t) -W^n(t) , w(t) - u^n(t) \rangle $$

and noting that $\varphi _\infty (w(t))\ge \varphi _n(w(t))$ gives immediately

$$\begin{aligned} \int _0 ^T \varphi _\infty (w(t)) dt&\ge \int _0 ^T \varphi _n(w(t)) dt \\&\ge \int _0 ^T \varphi _n(u^n(t)) dt + \langle -\ddot{u}^n(t) -W^n(t) , w(t) - u^n(t) \rangle dt . \end{aligned}$$

We note that

$$ \lim _n \int _0 ^T \langle -W^n(t) , w(t) - u^n(t) \rangle dt = \int _0 ^T \langle -W(t) , w(t) - u^\infty (t) \rangle dt $$

because $(W ^n := M\dot{u}^n -f^n)$ is uniformly integrable, and weakly converges to $W := M\dot{u}^\infty -f^\infty$ and the bounded sequence in $w(t) - u^n(t)$ pointwise converges to $w - u^\infty$ so that it converges uniformly on uniformly integrable subsets by virtue of Grothendieck lemma. Whence follows

$$\begin{aligned} \int _0 ^T \varphi _\infty (w(t)) dt \ge \int _0 ^T \varphi _\infty (u^\infty (t)) dt + \int _0 ^T \langle -W(t), w(t) -u^\infty (t) \rangle dt\\ +\,\langle -m, w-u^\infty \rangle _{({\mathcal M}^b_{\mathbf {R}^d}([0, T]), {\mathcal C}_{\mathbf {R}^d}([0, T])) }, \end{aligned}$$

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

Step 5 Claim $\lim _n \varphi _n(u^n(t)) = \varphi _\infty (u^\infty (t)) < \infty$ 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. $t \in [0, T]$:

$$\begin{aligned} \varphi _\infty (u^\infty (t))&+\frac{1}{2} ||\dot{u}^\infty (t)||^2 = \varphi _\infty (u^\infty (0)) +\frac{1}{2} ||\dot{u}^\infty (0)||^2\\&\qquad \qquad \qquad -\int _0^t \langle M \dot{u}^\infty (s), \dot{u}^\infty (s) \rangle ds+\int _0^t \langle \dot{u}^\infty (s), f ^\infty (s)\rangle ds. \end{aligned}$$

With the above results and notations, applying the subdifferential inequality

$$ \varphi _n(w(t)) \ge \varphi _n(u^n(t))+ \langle -\ddot{u}^n(t) -W^n(t) , w(t) - u^n(t) \rangle $$

with $w = u ^\infty$, integrating on [0, T], and passing to the limit when n goes to $\infty$, gives the inequalities

$$\begin{aligned} \int _B \varphi _\infty (u^\infty (t)) dt \ge \liminf _n&\int _B \varphi _n(u^n(t)) dt \\&\ge \int _B \varphi _\infty (u^\infty (t)) dt \ge \limsup _n \int _B \varphi _n(u^n(t)) dt \end{aligned}$$

on $B\in B_p^c\cap {\mathcal L}([0, T])$ so that

$$\begin{aligned} \lim _n \int _B \varphi _n(u^n(t)) dt = \int _B \varphi _\infty (u^\infty (t)) dt \end{aligned}$$

(3.3.8)

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

$$ \langle \dot{u}^n(t), f^n(t) \rangle -\langle \dot{u}^n(t), \ddot{u}^n(t) - M \dot{u}_n(t) \rangle = \frac{d}{dt} [\varphi _n(u_n(t))], $$

that is,

$$ \langle \dot{u}^n(t), z^n(t) \rangle = \frac{d}{dt} [\varphi _n(u_n(t))]. $$

By the estimate (3.3.2) and the boundedness in $L ^1_{\mathbf {R}^d}([0, T])$ of $(z^n)$, it is immediate that $(\frac{d}{dt} [\varphi _n(u_n(t))])$ is bounded in $L ^1_{\mathbf {R}}([0, T])$ so that $(\varphi _n(u_n(.))$ is bounded in variation. By Helly theorem, we may assume that $(\varphi _n(u_n(.))$ pointwisely converges to a BV function $\psi$. By (3.3.2), $(\varphi _n(u_n(.))$ converges in $L ^1_{\mathbf {R}}([0, T])$ to $\psi$. In particular, for every $k \in L^\infty _{\mathbf {R}^+}([0, T])$ we have

$$\begin{aligned} \lim _{n \rightarrow \infty } \int _0^T k(t) \varphi _n(u_n(t)) dt = \int _0^T k(t) \psi (t) dt. \end{aligned}$$

(3.3.9)

Combining (3.3.8) and (3.3.9) yields

$$ \int _B \psi (t) \,dt =\lim _{n \rightarrow \infty } \int _B \varphi _n( u^n(t)) \,dt = \int _B \varphi _\infty ( u^\infty (t)) \,dt $$

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

$$ \psi (t)= \lim _n \varphi _n(u_n(t)) = \varphi _\infty (u^ \infty (t)) \, \mathrm{a.e.} $$

Hence we get $\lim _n \varphi _n(u_n(t)) = \varphi _\infty (u^ \infty (t))$ a.e. Subsequently, using (iii) the passage to the limit when n goes to $\infty$ in the equation

$$\begin{aligned} \varphi _n(u^n(t)) +\frac{1}{2} ||\dot{u}^n(t)||^2&= \varphi _n(u^n(0)) +\frac{1}{2} ||\dot{u}^n(0)||^2\\&\quad -\int _0^t \langle M \dot{u}^n(s), \dot{u}^n(s) \rangle ds+ \int _0^t \langle \dot{u}^n(s), f ^n(s)\rangle ds \end{aligned}$$

yields for a.e. $t \in [0, T]$

$$\begin{aligned} \varphi _\infty (u^\infty (t)) +&\frac{1}{2} ||\dot{u}^\infty (t)||^2 = \varphi _\infty (u^\infty _0) +\frac{1}{2} ||\dot{u}^\infty _0||^2\\&\qquad \qquad -\int _0^t \langle M \dot{u}^\infty (s), \dot{u}^\infty (s) \rangle ds+ \int _0^t \langle \dot{u}^\infty (s), f ^\infty (s)\rangle ds. \end{aligned}$$

Noting that $(f ^n)$ is uniformly integrable and $\dot{u}^n$ is uniformly bounded and pointwise converges to $\dot{u}^\infty$, by virtue of Grothendieck lemma [16], it converges uniformly on uniformly integrable ($=$relatively weakly compact) subsets of $L^1_{\mathbf {R}^d}([0, T])$, so that

$$ \lim _n \int _0^t \langle \dot{u}^n(s), f ^n(s)\rangle ds =\int _0^t \langle \dot{u}^\infty (s), f ^\infty (s)\rangle ds. $$

Step 6 Localization of further limits and final step.

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

$$\begin{aligned} {\mathcal U}-\lim _n z^n = l \in L^\infty _{\mathbf {R}^d}([0, T])'_{weak} \end{aligned}$$

(3.3.10)

and

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

(3.3.11)

where $L^\infty _{\mathbf {R}^d}([0, T])'_{weak}$ is the second dual of $L^1_{\mathbf {R}^d} ([0, T])$ endowed with the topology $\sigma (L^\infty _{\mathbf {R}^d}([0, T])', L^\infty _{\mathbf {R}^d}([0, T]))$ and ${\mathcal C}_{\mathbf {R}^d} ([0, T])'_{weak}$ denotes the space ${\mathcal C}_{\mathbf {R}^d} ([0, T])'$ endowed with the weak topology $\sigma ({\mathcal C}_{\mathbf {R}^d} ([0, T])', {\mathcal C}_{\mathbf {R}^d} ([0, T]))$, because ${\mathcal C}_{\mathbf {R}^d}([0, T])$ is a separable Banach space for the norm sup, so that we may assume by extracting subsequences that $(z^n)$ weakly converges to $\mathbf n\in {\mathcal C}_{\mathbf {R}^d}([0, T])'$. Using Step 4, we note that $\mathbf n= -m-W.dt=-m-(M\dot{u} ^\infty -f ^\infty ). dt$. 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 an decreasing sequence $(A_n)$ of Lebesgue measurable sets in [0, T] with $A_n \downarrow \emptyset$ such that $l_s(h) = l_s(1_{A_n} h)$ for all $h\in L^\infty _{\mathbf {R}^d} ([0, T])$ and for all $n \ge 1$. As $(z ^n =f ^n -\ddot{u} ^n -M \dot{u} ^n)$ biting converges to $z ^\infty =f ^\infty -\zeta ^\infty -M\dot{u}^\infty$ in Step 4, it is already seen (cf. Proposition 3.1) that

$$ l_a(h) = \int _0 ^T \langle h(t), f ^\infty (t)-\zeta ^\infty (t) -M\dot{u}^\infty (t) \rangle dt $$

for all $h \in L^\infty _{\mathbf {R}^d}([0, T])$, shortly $z^\infty =f ^\infty -\zeta ^\infty -M\dot{u}^\infty$ coincides a.e. with the density of the absolutely continuous part $l_a$. By [13, 23], we have

$$ I_{\varphi _\infty }^*(l) = I_{\varphi _\infty ^* }(f ^\infty -\zeta ^\infty -M\dot{u}^\infty ) +\delta ^* (l_s ,\mathop {\text {dom}}I_{\varphi _\infty } ), $$

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

$$ \mathop {\text {dom}}I_{\varphi _\infty } := \{ u \in L^\infty _{\mathbf {R}^d}([0, T]) : I_{\varphi _\infty }(u) < \infty \}. $$

Using the inclusion

$$ z^\infty = f ^\infty -\zeta ^\infty - M \dot{u}^\infty \in \partial I_{\varphi _\infty }(u ^\infty ), $$

that is,

$$ I_{\varphi _\infty ^* }(f ^\infty -\zeta ^\infty -M\dot{u}^\infty ) = \langle f ^\infty -\zeta ^\infty -M\dot{u}^\infty , u^\infty \rangle -I_{\varphi _\infty } (u^\infty ), $$

we see that

$$\begin{aligned} I_{\varphi _\infty }^*(l) = \langle f ^\infty -\zeta ^\infty -M\dot{u}^\infty , u^\infty \rangle -I_{\varphi _\infty } (u^\infty ) +\delta ^* (l_s, \mathop {\text {dom}}I_{\varphi _\infty }). \end{aligned}$$

Coming back to the inclusion $z^n(t) \in \partial \varphi _n( u^n(t))$, we have

$$\begin{aligned} \varphi _n (x) \ge \varphi _n (u^n(t)) + \langle x-u^n(t), z^n(t) \rangle \end{aligned}$$

for all $x \in \mathbf {R}^d$. By substituting x by h(t) in this inequality, where $h\in L^\infty _{\mathbf {R}^d}([0, T])$, and by integrating

$$\begin{aligned} \int _0^T \varphi _n ( h(t))\, dt \ge \int _0^T \varphi _n ( u^n(t)) \,dt + \int _0^T \langle h(t)-u^n(t), z^n(t) \rangle \,dt. \end{aligned}$$

Arguing as in Step 4 by passing to the limit in the preceding inequality, involving the epiliminf property for integral functionals $\int _0^T \varphi _n(h(t)) dt$ defined on $L^\infty _{\mathbf {R}^d}([0, T])$, it is easy to see that

$$\begin{aligned} \int _0^T \varphi _\infty ( h(t)) \,dt \ge \int _0^T \varphi _\infty ( u^\infty (t)) \,dt + \langle h-u^\infty , \mathbf n\rangle . \end{aligned}$$

Since this holds, in particular, when $h \in {\mathcal C}_{\mathbf {R}^d}([0, T])$, we conclude that $\mathbf n$ belongs to the subdifferential $\partial J_{\varphi _\infty }(u^\infty )$ of the convex lower semicontinuous integral functional $J_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbf {R}^d}([0, T])$

$$ J_{\varphi _\infty }(u):= \int _0^T \varphi _\infty ( u(t)) \,dt, \forall u \in {\mathcal C}_{\mathbf {R}^d}([0, T]). $$

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

$$ \langle B^*l , h \rangle = \langle l, Bh \rangle = \langle l, h \rangle , \quad \forall l \in L^\infty _{\mathbf {R}^d}([0, T])', \quad \forall h \in {\mathcal C}_{\mathbf {R}^d}([0, T]). $$

Then we have $B^*l = B^*l_a+ B^*l_s$, $l\in L^\infty _{\mathbf {R}^d}([0, T])'$ being the limit of $(z_n =f ^n -\ddot{u}^n -M \dot{u} ^n)$ 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

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

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

$$\begin{aligned} \langle l_a, h \rangle&= \langle f ^\infty -\zeta ^\infty -M\dot{u}^\infty , h \rangle \\&\quad =\int _0 ^T \langle f^\infty (t) - \zeta ^\infty (t)-M\dot{u}^\infty (t) , h(t)\rangle dt, \quad \forall h \in L^\infty _{\mathbf {R}^d}([0, T]) \end{aligned}$$

so that the measure $B^*l_a$ is absolutely continuous

$$ \langle B^*l_a, h \rangle = \int _0 ^T \langle f ^\infty (t)- \zeta ^\infty (t)-M\dot{u}^\infty (t) , h(t)\rangle dt, \quad \forall h \in {\mathcal C}_{\mathbf {R}^d}([0, T]) $$

and its density $f ^\infty - \zeta ^\infty -M\dot{u}^\infty$ satisfies the inclusion

$$ f ^\infty (t)- \zeta ^\infty (t) - M\dot{u}^\infty (t) \in \partial \varphi _\infty (u^\infty (t)), \quad \mathrm{a.e.} $$

and the singular part $B^*l_s$ satisfies the equation

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

As $B^*l =\mathbf n$, using (3.3.10) and (3.3.11), it turns out that $\mathbf n$ is the sum of the absolutely continuous measure $\mathbf n_a$ with

$$ \langle \mathbf n_a, h \rangle = \int _0 ^T \langle f ^\infty (t) - \zeta ^\infty (t)-M\dot{u}^\infty (t) , h(t)\rangle dt, \quad \forall h \in {\mathcal C}_{\mathbf {R}^d}([0, T]) $$

and the singular part $\mathbf n_s$ given by

$$ \langle \mathbf n_s, h \rangle = \langle l_s, h \rangle , \quad \forall h \in {\mathcal C}_{\mathbf {R}^d}([0, T]), $$

which satisfies the property: for any nonnegative measure $\theta$ on [0, T] with respect to which $\mathbf n_s$ is absolutely continuous,

$$ \int _0^T h_{\varphi _\infty ^*}\left( \frac{d\mathbf n_s}{d\theta }(t)\right) d\theta (t)= \int _0^T \langle u^\infty (t), \frac{d\mathbf n_s}{d \theta }(t)\rangle d\theta (t), $$

where $h_{\varphi _\infty ^*}$ denotes the recession function of $\varphi _\infty ^*$. Indeed, as $\mathbf n$ belongs to $\partial J_{\varphi _\infty }(u^\infty )$ by applying Theorem 5 in [23] we have

$$\begin{aligned} J^*_{\varphi _\infty }(n) = I_{\varphi _\infty ^*}\left( \frac{d\mathbf n_a}{dt}\right) +\int _0^T h_{\varphi _\infty ^*}\left( \frac{d\mathbf n_s}{d \theta }(t)\right) d\theta (t) \end{aligned}$$

(3.3.12)

with

$$ I_{\varphi _\infty ^*} (v):=\int _0^T \varphi _\infty ^*(v(t)) dt , \forall v \in L^1_{\mathbf {R}^d}([0, T]). $$

Recall that

$$ \frac{d\mathbf n_a}{dt}=f ^\infty - \zeta ^\infty - M\dot{u}^\infty \in \partial I_{\varphi _\infty }(u^\infty ), $$

that is,

$$\begin{aligned} I_{\varphi _\infty ^*} \left( \frac{d\mathbf n_a}{dt}\right) = \langle f ^\infty - \zeta ^\infty - M\dot{u}^\infty , u^\infty \rangle _{\langle L^1_{\mathbf {R}^d}([0, T]), L^\infty _{\mathbf {R}^d}([0, T])\rangle } - I_{\varphi _\infty }(u^\infty ). \end{aligned}$$

(3.3.13)

From (3.3.13), we deduce

$$\begin{aligned} J^*_{\varphi _\infty }(n) =\;&\langle u ^\infty , \mathbf n\rangle _{\langle {\mathcal C}_{\mathbf {R}^d}([0, T]),{\mathcal C}_{\mathbf {R}^d}([0, T])'\rangle }- J_{\varphi _\infty }(u ^\infty )\\ =\;&\langle u ^\infty , \mathbf n\rangle _{\langle {\mathcal C}_{\mathbf {R}^d}([0, T]),{\mathcal C}_{\mathbf {R}^d}([0, T])'\rangle }- I_{\varphi _\infty }(u ^\infty )\\ =\;&\int _0 ^T \langle u ^\infty (t), f ^\infty (t) -\zeta ^\infty (t) -M\dot{u}^\infty (t) \rangle dt \\&+ \int _0 ^T \langle u ^\infty (t), \frac{d\mathbf n_s}{d \theta }(t)\rangle {d \theta }(t)-I_{\varphi _\infty }(u ^\infty )\\ =\;&I_{\varphi _\infty ^*}\left( \frac{d\mathbf n_a}{dt}\right) +\int _0 ^T \langle u ^\infty (t), \frac{d\mathbf n_s}{d \theta }(t) \rangle {d \theta }(t)). \end{aligned}$$

Coming back to (3.3.12) we get the equality

$$ \int _0^T h_{\varphi _\infty ^*}\left( \frac{d\mathbf n_s}{d \theta }(t)\right) d\theta (t) = \int _0^T \langle u ^\infty (t), \frac{d\mathbf n_s}{d \theta }(t)\rangle {d \theta }(t)). $$

$\blacksquare$

Actually, Proposition 3.3 completes Proposition 4.6 in [7], which is a precursor of some results we present here.

We begin with a second order evolution equation with m-point boundary condition

Proposition 3.4

Assume that $E =\mathbf {R}^d$, $M>0, \beta \in L^2_{\mathbf {R}^+} ([0, T])$. For each $n\in \mathbf N$, let $\varphi _n : \mathbf {R}^d \rightarrow \mathbf {R}^+$ be a $C ^1$, convex, Lipschitz function and let $\varphi _\infty$ be a nonnegative l.s.c proper function defined on $\mathbf {R}^d$ such that $\varphi _n(x) \le \varphi _\infty (x)$ for all $n\in \mathbf N$ and for all $x\in \mathbf {R}^d$. Let $f : [0, T]\times E\times E \rightarrow E$ satisfying

• (1) For each $(x, y) \in E\times E$ the scalar function $t \mapsto f(t, x, y) \rangle$ is Lebesgue measurable,

• (2) For each $t \in [0, 1]$, function f(t, ., .) is continuous on $E\times E$,

• (3) $||f(t, x, y)|| \le \beta (t)$ for all $(t, x, y ) \in [0, 1]\times E\times E$.

  For each $n\in \mathbf N$, let $u^n$ be a $W^{2, 1}_{ \mathbf {R}^d}([0, 1])$-solution to the approximating problem

  $$ ({\mathcal P_n}) \left\{ \begin{array}{lll} f(t, u^n(t), \dot{u}^n(t) )= \ddot{u}^n (t)+M\dot{u}^n(t)+\nabla \varphi _n(u^n(t)), t \in [0, 1]\\ u^n(0) = x \in \mathop {\text {dom}}\, \varphi _\infty , \quad u_n(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u_n (\eta _i) \end{array} \right. $$

Assume that

• (i) $\varphi _n$ epi-converges to $\varphi _\infty$,

• (ii) $\lim _n \dot{u}^n (0)=\dot{u}^ \infty _0$,

• (iii) There exist $r_0 >0$ and $x_0 \in \mathbf {R}^d$ such that

  $$ \sup _{v \in {\overline{B}}_{L^\infty _{ \mathbf {R}^d}([0, 1])}}\int _0^T\varphi _\infty (x_0+r_0v(t)) < +\infty $$

  where ${\overline{B}}_{L^\infty _{ \mathbf {R}^d}([0, 1])}$ is the closed unit ball in $L^\infty _{ \mathbf {R}^d}([0, 1])$.

  • (a) Then, up to extracted subsequences, $(u ^n)$ converges uniformly to a $W^{1, 1}_{BV} ([0, 1])$-function $u ^\infty$ with $u^\infty (0) = x \in \mathop {\text {dom}}\, \varphi _\infty , \quad u^\infty (1)= \sum _{i = 1} ^{m-2} \alpha _i u ^\infty (\eta _i)$ and $(\dot{u} ^n)$ pointwisely converges to a BV function $v^\infty$ with $v^\infty = \dot{u} ^\infty$, and $(\ddot{u} ^n)$ biting converges to a function $\zeta ^\infty \in L^1_{\mathbf {R}^d}([0, 1])$ so that the limit function $u ^\infty , \dot{u}^\infty$ and the biting limit $\zeta ^\infty$ satisfy the variational inclusion

    $$ ({\mathcal P_\infty })\qquad f ^\infty \in \zeta ^\infty + M\dot{u}^\infty + \partial I_{\varphi _\infty }(u^\infty ) $$

    where $f ^\infty (t):= f(t, u ^\infty (t), \dot{u}^\infty (t), \forall t \in [0, 1]$, $\partial I_{\varphi _\infty }$ denotes the subdifferential of the convex lower semicontinuous integral functional $I_{\varphi _\infty }$ defined on $L^\infty _{\mathbf {R}^d}([0, 1])$ by

    $$ I_{\varphi _\infty }(u):= \int _0^1 \varphi _\infty ( u(t)) \,dt, \forall u \in L^\infty _{\mathbf {R}^d}([0, 1]). $$

  • (b) $(\ddot{u}^n)$ weakly converges to the vector measure $m \in {\mathcal M}^b_E([0, 1])$ so that the limit functions $u ^\infty (.)$ and the limit measure m satisfy the following variational inequality:

    $$\begin{aligned} \int _0^1 \varphi _\infty ( v(t)) \,dt \ge&\int _0^1 \varphi _\infty ( u ^\infty (t)) \,dt + \int _0 ^1 \langle -M \dot{u} ^\infty (t)+f^\infty (t) ,v(t)- u ^\infty (t)\rangle \,dt \\&+ \langle -m, v-u ^\infty \rangle _{ ({\mathcal M}^b_{\mathbf {R}^d} ([0, 1]), {\mathcal C}_E([0, 1])) }. \end{aligned}$$

  • (c) Furthermore $\displaystyle { \lim _n \int _0^1 \varphi _n( u^n(t)) dt = \int _0^T \varphi _\infty ( u ^\infty (t)) dt}$. Subsequently the energy estimate

    $$\begin{aligned} \varphi _\infty \left( u^\infty (t)) +\frac{1}{2} ||\dot{u}^\infty (t)||^2 \right.&\left. \le \varphi _\infty (x) +\frac{1}{2} ||\dot{u}^\infty _0\right) ||^2\\&-\int _0^t \langle M \dot{u}^\infty (s), \dot{u}^\infty (s) \rangle ds+\int _0^t \langle \dot{u}^\infty (s), f ^\infty (s)\rangle ds \end{aligned}$$

    holds a.e.

  • (d) There are a filter ${\mathcal U}$ finer than the Fréchet filter, $l \in L^\infty _{\mathbf {R}^d}([0, 1])'$ such that

    $$ {\mathcal U}-\lim _n [f^n-\ddot{u} ^n-M\dot{u} ^n] = l \in L^\infty _{\mathbf {R}^d}([0, 1])'_{weak} $$

    where $L^\infty _{\mathbf {R}^d}([0, 1])'_{weak}$ is the second dual of $L^1_{\mathbf {R}^d} ([0, 1])$ endowed with the topology $\sigma (L^\infty _{\mathbf {R}^d}([0, 1])', L^\infty _{\mathbf {R}^d}([0, 1]))$ and $\mathbf n\in {\mathcal C}_{\mathbf {R}^d} ([0, 1])'_{weak}$ such that

    $$\lim _n [f ^n-\ddot{u} ^n-M\dot{u} ^n] = \mathbf n\in {\mathcal C}_{\mathbf {R}^d} ([0, 1])'_{weak} $$

    where ${\mathcal C}_{\mathbf {R}^d} ([0, 1])'_{weak}$ denotes the space ${\mathcal C}_{\mathbf {R}^d} ([0, 1])'$ endowed with the weak topology $\sigma ({\mathcal C}_{\mathbf {R}^d} ([0, 1])', {\mathcal C}_{\mathbf {R}^d} ([0, 1]))$ so that $\mathbf n=-m -(M\dot{u}^\infty -f ^\infty )dt$. 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

    $$ l_a(h) = \int _0 ^T \langle h(t),f ^\infty (t) -\zeta ^\infty (t) -M\dot{u}^\infty (t) \rangle dt $$

    for all $h \in L^\infty _{\mathbf {R}^d}([0, 1])$ so that

    $$ I_{\varphi _\infty }^*(l) = I_{\varphi _\infty ^* }(f^\infty -\zeta ^\infty -M\dot{u}^\infty ) +\delta ^* (l_s ,\mathop {\text {dom}}I_{\varphi _\infty } ) $$

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

    $$ \langle \mathbf n, h \rangle = \int _0 ^1 \langle f ^\infty (t) -\zeta ^\infty (t)-M\dot{u}^\infty (t) , h(t)\rangle dt + \langle \mathbf n_s, h \rangle , \quad \forall h \in {\mathcal C}_{\mathbf {R}^d}([0, 1]) $$

    with $\langle \mathbf n_s, h \rangle = l_s(h)$, $\forall h \in {\mathcal C}_{\mathbf {R}^d}([0, 1])$. Further $\mathbf n$ belongs to the subdifferential $\partial J_{\varphi _\infty }(u^\infty )$ of the convex lower semicontinuous integral functional $J_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbf {R}^d}([0, 1])$

    $$ J_{\varphi _\infty }(u):= \int _0^1 \varphi _\infty ( u(t)) \,dt, \forall u \in {\mathcal C}_{\mathbf {R}^d}([0, 1]). $$

    (c) Consequently the density $f ^\infty -\zeta ^\infty -M\dot{u}^\infty$ of the absolutely continuous part $\mathbf n_a$

    $$ \mathbf n_a (h):= \int _0 ^1 \langle f^\infty (t) -\zeta ^\infty (t)-M\dot{u}^\infty (t) , h(t)\rangle dt, \quad \forall h\in {\mathcal C}_{\mathbf {R}^d}([0, 1]) $$

    satisfies the inclusion

    $$ f ^\infty (t)- \zeta ^\infty (t) - M\dot{u}^\infty (t) \in \partial \varphi _\infty (u^\infty (t)), \quad \mathrm{a.e.} $$

    and for any nonnegative measure $\theta$ on [0, T] with respect to which $\mathbf n_s$ is absolutely continuous

    $$ \int _0^1 h_{\varphi _\infty ^*}( \frac{d\mathbf n_s}{d\theta }(t)) d\theta (t) = \int _0^T \langle u^\infty (t), \frac{d\mathbf n_s}{d \theta }(t)\rangle d\theta (t) $$

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

Proof

Existence of a $W^{2, 1}_{\mathbf {R}^d}([0, 1])$-solution for the approximating equation

$$ \left\{ \begin{array}{lll} \ddot{u}_n(t)+M \dot{u}_n(t) +\nabla \varphi _n(u^n(t) = f(t, u^n(t), \dot{u}^n(t) ), \mathrm{a.e.}\ t \in [0, 1] \\ u_n(0) =x, \quad u_n(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u_n (\eta _i) \end{array} \right. $$

is ensured by Proposition 2.8 with integral representation formulas

$$ \left\{ \begin{array}{ll} u_{ n} (t) &{}= e_{x} (t)+ \int _{0}^{1} G(t, s) [\ddot{u}_n(t)+M \dot{u}_n(s)] ds , \, t \in [0, 1]\\ \dot{u}_{ n} (t) &{}=\dot{e}_{x} (t)+ \int _{0}^{1}\frac{\partial G}{\partial t}(t,s)[\ddot{u}_n(t)+M \dot{u}_n(s)] ds, \, t \in [0, 1]\\ \end{array} \right. $$

$$ \left\{ \begin{array}{ll} e_{x } (t) &{}= x + A (1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}) (1-\exp (-\gamma t) )x \\ \dot{e}_{x }(t) &{}= \gamma A\left( 1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}\right) \exp {\left( -\gamma t \right) }x\\ A &{}= \left( \sum \nolimits _{i=1}^{m-2}\alpha _{i}-1+ \exp (-\gamma ) -\sum \nolimits _{i=1}^{m-2}\alpha _{i}\exp (-\gamma (\eta _{i}))\right) ^{-1} \end{array} \right. $$

where G is the Green function given by Lemma 2.1. Then $u^n(0) = x$ and $u_n(1)= \sum _{i = 1} ^{m-2} \alpha _i u_n (\eta _i)$.

The rest of the proof follows the same lines as that of Proposition 3.3. $\blacksquare$

The following is a new variant on the existence of solutions for the second order evolution inclusion with m-point boundary condition.

Proposition 3.5

Let $(\partial \varphi _n)$ $(n\in \mathbf N\cup \{\infty \})$ be a sequence of subdifferential operators associated with a sequence of nonnegative normal convex integrands $( \varphi _n)$ $(n\in \mathbf N\cup \{\infty \})$. Assume that the following conditions are satisfied:

• (1) For each $n \in \mathbf N$, $|\varphi _n(t,x)-\varphi _n(t,y)| \le \beta _n(t) ||x-y||$ for all $t\in [0, 1]$ and for all $x, y \in \mathbf {R}^d$, where $\beta _n$ is a nonnegative integrable functions.

• (2) For each Lebesgue-measurable set $A \in [0,1]$, for each $w\in L^\infty _{\mathbf {R}^d}([0, 1])$,

  $$ \limsup _n \int _A \varphi _n(t, w(t))\, dt \le \int _A \varphi _\infty (t, w(t))\, dt. $$

• (3) For each $t \in [0, 1]$, $\varphi _n (t, .)$ lower epiconverges to $\varphi _\infty (t, .)$, that is, for each fixed $t \in [0, 1]$, for each $(x_n)$ in $\mathbf {R}^d$, converging to $x\in \mathbf {R}^d$, $\liminf \varphi _n(t, x_n) \ge \varphi _\infty (t, x)$.

  For each $n\in \mathbf N$, let $u ^n : [0, 1] \rightarrow \mathbf {R}^d$ be a $W^{2, 1}_ {\mathbf {R}^d}([0, 1])$-solution to

  $$ \left\{ \begin{array}{lll} \ddot{u}^n(t)+\gamma \dot{u}^n(t) \in \partial \varphi _n(t, u^n(t)), \mathrm{a.e.}\ t \in [0, 1] \\ u ^n(0) =x, \quad u ^n(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u^n (\eta _i). \end{array} \right. $$

• (4) Assume further that

  $$ \sup _{n \in \mathbf N} \int _0 ^1 \varphi _n(t, u_n(t)) dt < + \infty $$

  and

  $$ \sup _{n \in \mathbf N} \int _0^1 |\partial \varphi _n(t, u^n(t))| dt <+\infty . $$

Then the following hold:

• (a) Up to extracted subsequences, $(u ^n)$ converges uniformly to a $W^{1, 1}_{BV}([0, 1])$ function $u ^\infty$ with $u^\infty (0) = x, u ^\infty (1) = \sum _{i = 1} ^{m-2} \alpha _i u ^\infty (\eta _i)$ and $(\dot{u} ^n)$ pointwisely converges to the BV function $\dot{u} ^\infty$, and $(\ddot{u} ^n)$ stably converges to a Young measure $\nu ^\infty \in \mathcal Y([0, 1]; \mathbf {R}^d)$ with $t \mapsto \mathop {\text {bar}}(\nu ^\infty _t )\in L^1_ {\mathbf {R}^d}([0, 1])$ (here $\mathop {\text {bar}}(\nu ^\infty _t )$ denotes the barycenter of $\nu ^\infty _t$) such that the limit functions $u^\infty (.)$, $\dot{u} ^\infty (.)$ and the Young limit measure $\nu ^\infty$ satisfy

  $$ \int _0^1 \varphi _\infty (t, u ^\infty (t)) dt \le \liminf _n \int _0^1 \varphi _n (t, u ^n(t)) dt $$

  consequently

  $$ \lim _n \int _0^1 \varphi _n (t, u ^n(t)) dt = \int _0^1 \varphi _\infty (t, u ^\infty (t)) dt < \infty $$

  and

  $$ \mathop {\text {bar}}(\nu ^\infty _t ) + \gamma \dot{u}^\infty (t) \in \partial \varphi _\infty (t, u^\infty (t)), \mathrm{a.e.}$$

  equivalently the function $t \mapsto \mathop {\text {bar}}(\nu ^\infty _t ) + \gamma \dot{u}^\infty (t)$ belongs to the subdifferential $\partial I_{\varphi _\infty }(u^\infty )$ of the convex lower semicontinuous integral functional $I_{\varphi _\infty }$ defined on $L^\infty _{\mathbf {R}^d}([0, T])$

  $$ I_{\varphi _\infty }(u):= \int _0^T \varphi _\infty (t, u(t)) \,dt, \forall u \in L^\infty _{\mathbf {R}^d}([0, T]). $$

• (b) Up to extracted subsequences, $(u ^n)$ converges uniformly to a $W^{1, 1}_{BV}([0, 1])$ function $u ^\infty$ with $u^\infty (0) = x, u ^\infty (1) = \sum _{i = 1} ^{m-2} \alpha _i u ^\infty (\eta _i)$ and $(\dot{u} ^n)$ pointwisely converges to the BV function $\dot{u} ^\infty$, $(\ddot{u}^n )$ weakly converges to $m ^\infty \in {\mathcal M}^b_{\mathbf {R}^d} ([0, 1])$ so that the limit functions $u ^\infty (.)$ and the limit measure $m ^\infty$ satisfy the variational inequality: for every $v \in {\mathcal C}_{ \mathbf {R}^d} ([0, 1])$,

  $$\begin{aligned} \int _0^1 \varphi _\infty (t, v(t)) \,dt \ge&\int _0^1 \varphi _\infty (t, u ^\infty (t)) \,dt + \int _0 ^1 \langle \gamma \dot{u} ^\infty (t)),v(t)- u ^\infty (t)\rangle \,dt \\&+ \langle m^\infty , v-u ^\infty \rangle _{ ({\mathcal M}^b_ {\mathbf {R}^d}([0, 1]), {\mathcal C}_ {\mathbf {R}^d} ([0, 1])) }. \end{aligned}$$

  In other words, the vector measure $m ^\infty + \gamma \dot{u}^\infty \,dt$ belongs to the subdifferential $\partial I_{\varphi _\infty }(u)$ of the convex functional integral $I_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbf {R}^d} ([0, 1])$ by $I_{\varphi _\infty }(v)=\int _0 ^1 \varphi _\infty (t, v(t))\ dt$, $\forall v\in {\mathcal C}_ {\mathbf {R}^d} ([0, 1])$.

Proof

Existence of a $W^{2, 1}_{\mathbf {R}^d} ([0, 1])$-solution $u ^n$ to

$$ \left\{ \begin{array}{lll} \ddot{u}^n(t)+\gamma \dot{u}^n(t) \in \partial \varphi _n(t, u^n(t)), \mathrm{a.e.}\ t \in [0, 1] \\ u ^n(0) =x, \quad u ^n(1)= \sum \nolimits _{i = 1} ^{m-2} \alpha _i u^n (\eta _i) \end{array} \right. $$

is ensured by Proposition 2.7 with integral representation formulas

$$ \left\{ \begin{array}{ll} u^n (t)&{} = e_{x} (t)+ \int _{0}^{1} G(t, s) [\ddot{u}^n(s)+\gamma \dot{u} ^n(s)] ds , \, t \in [0, 1]\\ \dot{u}^{ n} (t) &{}=\dot{e}_{x} (t)+ \int _{0}^{1}\frac{\partial G}{\partial t}(t,s) [\ddot{u}^n(s)+\gamma \dot{u}^n(s)] ds, \, t \in [0, 1]\\ \end{array} \right. $$

where

$$ \left\{ \begin{array}{ll} e_{x } (t) &{}= x + A (1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}) (1-\exp (-\gamma t) )x \\ \dot{e}_{x }(t) &{}= \gamma A\left( 1-\sum \nolimits _{i=1}^{m-2}\alpha _{i}\right) \exp {\left( -\gamma t \right) }x\\ A &{}= \left( \sum \nolimits _{i=1}^{m-2}\alpha _{i}-1+ \exp (-\gamma ) -\sum \nolimits _{i=1}^{m-2}\alpha _{i}\exp (-\gamma (\eta _{i}))\right) ^{-1} \end{array} \right. $$

where G is the Green function given by Lemma 2.1.

Step 1 (a) As $\sup _n \int _0^1 |\partial \varphi _n(t, u^n(t))| dt <+\infty$, it follows that $(\ddot{u}^n+\gamma \dot{u}^n)$ is bounded in $L ^1_{\mathbf {R}^d}([0, 1])$, namely

$$ \sup _n \int _0^1 ||(\ddot{u}^n(t)+\gamma \dot{u}^n(t) || dt < +\infty , $$

so that, by the representation formulas given above, it is immediate that $(u^n)$ and $(\dot{u}^n)$ are uniformly bounded. Hence $(\ddot{u}^n)$ is bounded in $L ^1_{\mathbf {R}^d}([0, 1])$ and $(\dot{u}_n(.))$ is bounded in variation because $\sup _n \int _0^1 ||\ddot{u}_n(t)|| \,dt< +\infty$. In view of the Helly–Banach theorem, we may, by extracting a subsequence, assume that $(\dot{u}^n(.))$ converges pointwisely to a BV function $v ^\infty (.)$. Let us set $u ^\infty (t) =\int _0 ^t v ^\infty (s)\,ds$ for all $t\in [0, 1]$. Then $u ^\infty \in W^{1, 1}_{BV}([0, 1])$. As $(\dot{u}_n(.))$ is uniformly bounded and pointwise converges to $v ^\infty (.)$, by Lebesgue’s theorem, we conclude that $(\dot{u}^n(.))$ converges in $L^1_{\mathbf {R}^d}([0, 1])$ to $\dot{u}^\infty (.)$. Hence $u^n(.)$ converges uniformly to $u ^\infty (.)$ with $u ^\infty (0) =x, \quad u ^\infty (1)= \sum _{i = 1} ^{m-2} \alpha _i u ^\infty (\eta _i)$. So (a) is proved. From the general compactness result for Young measures, [5, 10] one may assume that $\ddot{u} ^n$ stably converge to an Young measure $\nu ^\infty$. Further, by virtue of Proposition 3.1 we may assume that $(\ddot{u}^n)$ biting converges to the integrable function $\mathop {\text {bar}}(\nu ^\infty ): t\mapsto \mathop {\text {bar}}(\nu ^\infty _t)$, that is, there exists a decreasing sequence $(B_p)$ of Lebesgue measurable sets with $\lambda (\cap B_p) =0$ such that the restriction of $(\ddot{u}^n)$ on each $B_p^c$ converges $\sigma (L^1,L^\infty )$ to $\mathop {\text {bar}}(\nu )$. It follows that

$$\begin{aligned} \lim _n\int _B \langle \ddot{u}^n +\gamma \dot{u}^n(t) , w(t)- u^n(t) \rangle \,dt = \int _B \langle \mathop {\text {bar}}(\nu _t) +\gamma \dot{u}^\infty (t), w(t)-u ^\infty (t) \rangle \,dt \end{aligned}$$

(3.5.1)

for every $B \in B_p^c\cap {\mathcal L}([0, 1])$, and for every $w\in L^\infty _E([0, 1])$ because the sequence $(w- u^n)$ in $L^\infty _{\mathbf {R}^d} ([0, 1])$ is bounded and pointwise converges to $w-u ^\infty$, so it converges uniformly on uniformly integrable subsets of $L^1_{\mathbf {R}^d}([0, 1])$. Since $(\varphi _n)$ lower epiconverges to $\varphi _\infty$, by Corollary 4.7 in [11], we have

$$\begin{aligned} \liminf _n \int _A \varphi _n(t, u^n(t)) \,dt \ge \int _A \varphi _\infty (t,u^\infty (t))\,dt \end{aligned}$$

(3.5.2)

for every Lebesgue-measurable set A in [0, 1]. Combining (3.5.1), (3.5.2) and Assumption (2), and integrating the subdifferential inequality

$$\begin{aligned} \varphi _n(t, w(t)) \ge \varphi _n(t, u^n(t))+ \langle \ddot{u}^n(t) +\gamma \dot{u}^n(t) , w(t) - u^n(t) \rangle \end{aligned}$$

(3.5.3)

on each $B\in B_p^c\cap {\mathcal L}([0, 1])$ and for every $w \in L^\infty _{\mathbf {R}^d}([0, 1])$, we get

$$ \int _B \varphi _\infty (t, w(t)) \,dt \ge \int _B \varphi _\infty (t, u ^\infty (t))\, dt +\int _B \langle \mathop {\text {bar}}(\nu ^\infty _t) +\gamma \dot{u} ^\infty (t), w(t)-u^\infty (t) \rangle \,dt. $$

This shows that $t \mapsto \mathop {\text {bar}}(\nu ^\infty _t) +\gamma \dot{u} ^\infty (t)$ is a subgradient at the point $u^\infty$ of the convex integral functional $I_{\varphi _\infty }$ restricted to $L^\infty _E(B_p^c)$, consequently,

$$ \mathop {\text {bar}}(\nu _t) +\gamma \dot{u} ^\infty (t) \in \partial \varphi _\infty (t, u ^\infty (t)), \, \text {a.e. on } B_p^c. $$

As this inclusion is true on each $B_p^c$ and $B_p^c \uparrow [0, 1]$, we conclude that

$$ \mathop {\text {bar}}(\nu ^\infty _t) +\gamma \dot{u} ^\infty (t) \in \partial \varphi _\infty (t, u ^\infty (t)), \, \text {a.e. on } [0, 1]. $$

Finally, applying the above subdifferential inequality, and putting $w= u^\infty$ in (3.5.3), we deduce

$$\begin{aligned}&\int _B \varphi _\infty (t, u^\infty (t) dt \\&\quad \qquad \qquad \ge \limsup _n\int _B \varphi _n(t, u^\infty (t)) dt\\&\quad \qquad \qquad \ge \limsup _n \int _B [\varphi _n(t, u^n(t))+ \langle \ddot{u}^n(t) +\gamma \dot{u}^n(t) , u^\infty (t) - u^n(t) \rangle ] dt \\&\quad \qquad \qquad =\limsup _n \int _B \varphi _n(t, u^n(t)) dt \ge \liminf _n \int _B \varphi _n(t, u^n(t)) dt \\&\quad \qquad \qquad \ge \int _B \varphi _\infty (t, u^\infty (t)) dt \end{aligned}$$

because

$$ \lim _n \int _B \langle \ddot{u}^n(t) +\gamma \dot{u}^n(t) , u^\infty (t) - u^n(t) \rangle ] dt = 0 $$

recalling that $1_B[ \ddot{u}^n +\gamma \dot{u}^n]$ is uniformly integrable. Whence follows

$$ \lim _n \int _B \varphi _n(t, u^n(t)) dt = \int _B \varphi _\infty (t, u^\infty (t)) dt . $$

As this inclusion is true on each B in $B_p^c$ and $B_p^c \uparrow [0, 1]$, we conclude that

$$ \lim _n \int _0^1 \varphi _n(t, u^n(t)) dt = \int _0^1 \varphi _\infty (t, u^\infty (t)) dt. $$

Step 2 (b) Repeating the results in Step 1, up to extracted subsequences, $(u ^n)$ converges uniformly to a $W^{1, 1}_{BV}([0, 1])$ function $u ^\infty$ with $u^\infty (0) = x, u ^\infty (1) = \sum _{i = 1} ^{m-2} \alpha _i u ^\infty (\eta _i)$ and $(\dot{u} ^n)$ pointwisely converges to the BV function $\dot{u} ^\infty$. As $(\ddot{u}_n )$ is $L^1$-bounded we may assume that $(\ddot{u}_n )$ weakly converges to a vector measure $m^\infty \in {\mathcal M}^b_{\mathbf {R}^d}([0, 1])$ since the Banach space ${\mathcal C}_{\mathbf {R}^d}([0, 1])$ is separable and its topological dual is ${\mathcal M}^b_{\mathbf {R}^d}([0, 1])$. Let $w\in {\mathcal C}_{\mathbf {R}^d}(([0, 1])$. Integrating the subdifferential inequality

$$ \varphi _n(t, w(t)) \ge \varphi _n(t, u^n(t)) + \langle \ddot{u}^n(t)+\gamma \dot{u}^n(.),w(t)- u^n(t) \rangle $$

and passing to the limit gives immediately

$$\begin{aligned} \int _0^1 \varphi _\infty (t, w(t)) \,dt \ge&\int _0^1 \varphi _\infty (t, u^\infty (t)) \,dt + \int _0 ^1 \langle \gamma \dot{u}^\infty (t), w(t)-u ^\infty (t)\rangle \,dt \\&+ \langle m ^\infty , w-u \rangle _{({\mathcal M}^b_{\mathbf {R}^d}([0, 1]), {\mathcal C}_{\mathbf {R}^d}([0, 1])) }, \end{aligned}$$

which shows that the vector measure $m ^\infty \ +\gamma \dot{u} ^\infty \,dt$ belongs to the subdifferential $\partial I_{\varphi _\infty }$ of the convex functional integral $I_{\varphi _\infty }$ defined on ${\mathcal C}_{\mathbf {R}^d}([0, 1])$ by $I_{\varphi _\infty }(v):= \int _0 ^1 \varphi _\infty (t, v(t))\ dt$, $\forall v\in {\mathcal C}_{\mathbf {R}^d}([0, 1])$. $\blacksquare$

4 Further Applications: Second Order Evolution Problems with Anti-periodic Boundary Condition

It is worth to focus on the main difference in discussing the various approximating problems.

$$\begin{aligned} f ^n(t)&= [\ddot{u}^n (t)+M \dot{u} ^n(t)] +\nabla \varphi _n (u ^n(t)), t \in [0, T] \end{aligned}$$

(4.1)

$$\begin{aligned} f ^n(t)&\in [\ddot{u}^n (t)+M \dot{u} ^n(t)] +\partial \varphi _n (u ^n(t)), t \in [0, T] \end{aligned}$$

(4.2)

$$\begin{aligned} f ^n(t)&= -[\ddot{u}^n (t)+M \dot{u} ^n(t)] +\nabla \varphi _n (u^n(t)), t \in [0, T]\end{aligned}$$

(4.3)

$$\begin{aligned} f ^n(t)&\in -[\ddot{u}^n (t)+M \dot{u} ^n(t)] +\partial \varphi _n (u ^n(t)), t \in [0, T]. \end{aligned}$$

(4.4)

Equations (4.1) and (4.2) are usual in second order dynamical systems. We refer to Attouch et al. [4] and Schatzmann [24] for a deep study of such models. See also the results developed in Propositions 3.2–3.5. Here, according to a traditional vein, we prove the existence of generalized solution with the conservation of energy in (3.3) and (3.4). Meanwhile (4.3) and (4.4) appear in the problem of anti-periodic solution developed in Aizicovici et al. [1,2,3]. Here in Proposition 4.3 we present a first result of the existence of generalized solution for the problem

$$ f(t) \in [\ddot{u} (t)+M \dot{u}(t)] +\partial \varphi (u(t) ) $$

using the approximating problem (4.2) with application (Proposition 3.4) to problem

$$ f(t,u(t), \dot{u}(t) )\in \ddot{u} (t)+M \dot{u}(t) +\partial \varphi (u(t)), t \in [0, T] $$

with m-point boundary condition using the approximating problem

$$ f(t, u ^n(t), \dot{u}^n(t))= \ddot{u}^n (t)+M \dot{u} ^n(t)] +\nabla \varphi _n (u(t)), t \in [0, T] $$

with m-point boundary condition. Here one can see that the techniques employed in (4.1) and (4.2) cannot be used to develop similar results to (4.3) and (4.4), in particular, we cannot obtain the conservation of energy for the variational limits in (4.3) and (4.4) by contrast with (4.1) and (4.2). So it is worth to mention that our tools allow to study the approximating problem of anti-periodic solution in the framework of Haraux–Okochi with anti-periodic solution

$$\begin{aligned} f ^n(t) = [\ddot{u}^n (t)\,+\,&M \dot{u} ^n(t)] +\nabla \varphi _n (u ^n(t)), t \in [0, T],\\&u_n(0) = -u_n(T). \end{aligned}$$

In our opinion, the general problem of the existence of energy conservation solution to second order evolution inclusion of the form

$$\begin{aligned} f(t) \in [\ddot{u} (t)+M \dot{u}(t)] +\partial \varphi (u(t)) \end{aligned}$$

(4.5)

where $\varphi$ is a lower semicontinuous convex proper function is a difficult problem when the perturbation $f \in L ^1_H([0, T])$ and H is a separable Hilbert space.

Now, to finish the paper, we show that our abstract result in Proposition 3.3 and the tool developed therein can be applied to the first order of evolution equation and also to the second order evolution equation with anti-periodic boundary conditions. H. Okochi initiated the study for anti-periodic solutions to evolution equations in Hilbert spaces. Following Okochi’s work, A. Haraux proved some existence and uniqueness theorems for anti-periodic solutions by using Brouwer’s or Schauder fixed point theorems. Aftabizadeh, Aizicovici and Pavel have studied the anti-periodic solutions to second order evolution equation in Hilbert spaces and Banach spaces by using monotone and accretive operator theory for equations of type (4.3) and (4.4). Here we show the applicability of our abstract result to the study of evolution equations of type (4.1) and (4.2) with anti-periodic boundary condition. For notational convenience let us denote by $\mathcal H$ the set of of functions $f \in L^2_{loc} (\mathbf {R}, H)$ such that f is anti-periodic, that is, $f(t+T) = -f(t)$ for all $t\in \mathbf {R}$ and

$$ \mathcal H_\beta ([0, T]):= \{ f\in \mathcal H : ||f(t)|| \le \beta (t), \beta \in L^2_\mathbf {R}([0, T]), t \in [0, T] \}. $$

We begin with some examples in the first order of evolution equation with anti-periodic condition.

Proposition 4.1

Let $H = \mathbf {R}^d$. Assume that $\varphi _n : \mathbf {R}^d \rightarrow [0, +\infty [$ are even, convex, Lipschitz and $\varphi _\infty : \mathbf {R}^d \rightarrow [0, +\infty ]$ is proper lower semicontinuous convex function such that $\varphi _n(x) \le \varphi _\infty (x)$ for all $n\in \mathbf N$ and for all $x\in \mathbf {R}^d$. Let $f^n$ be sequence in $\mathcal H_\beta ([0, T])$ and let $u^n$ be a $W^{1, 2}_{\mathbf {R}^d}([0, T])$-solution to the problem

$$ \left\{ \begin{array}{l} f^n (t)\in \dot{u}^n(t)+ \partial \varphi _n(u^n(t)) \quad t\in [0, T], \\ u_n(T) =-u_n(0) \end{array} \right. $$

Assume that the following conditions are satisfied:

• (i) $\varphi _n$ epiconverges to $\varphi _\infty$,

• (ii) $\lim _n u^n (0)=u ^ \infty _0 \in \mathop {\text {dom}}\, \varphi _\infty$ and $\lim _n \varphi _n (u^n (0)) = \varphi _\infty (u^ \infty _0 )$.

• (iii) $f^n$ $\sigma ( L^2_{\mathbf {R}^d}([0, T]), L^2_{\mathbf {R}^d}([0, T]))$-converges to $f ^\infty \in L^2_{\mathbf {R}^d}([0, T])$.

Then the following hold

• (a) Up to extracted subsequences, $(u ^n)$ converges pointwisely to an anti-periodic absolutely continuous mapping $u ^\infty$ with $u ^\infty (T)= - u ^\infty (0)$, $(\dot{u} ^n)$ $\sigma ( L^2_{\mathbf {R}^d}, L^2_{\mathbf {R}^d})$-converges to $\zeta ^\infty \in L^2_{\mathbf {R}^d}([0, T])$ with $\zeta ^\infty = \dot{u}^\infty$, $\lim _n \varphi _n( u^n(t)) = \varphi _\infty ( u^\infty (t)) < +\infty$ a.e. and $\lim _n \int _0^T \varphi _n( u^n(t)) dt = \int _0^T \varphi _\infty ( u ^\infty (t)) dt < +\infty$.

• (b) $f^\infty - \zeta ^\infty \in \partial I_{\varphi _\infty }(u^\infty )$ where $\partial I_{\varphi _\infty }$ denotes the subdifferential of the convex lower semicontinuous integral functional $I_{\varphi _\infty }$ defined on $L^\infty _{\mathbf {R}^d}([0, T])$

  $$ I_{\varphi _\infty }(u):= \int _0^T \varphi _\infty ( u(t)) \,dt, \forall u \in L^\infty _{\mathbf {R}^d}([0, T]). $$

Proof

Existence of $W^{1, 2}_{\mathbf {R}^d}([0, T])$-solution $u^n$ to the problem

$$ \left\{ \begin{array}{lll} f^n (t)\in \dot{u}^n(t)+ \partial \varphi _n(u^n(t)) \quad t\in [0, T], \\ u_n(T) =-u_n(0) \end{array} \right. $$

is ensured. See Haraux [17], Okochi [22].

Step 1 Estimation of $u^n$, $\dot{u}^n$, and $\varphi _n(u ^n(.)$ Multiplying scalarly the inclusion

$$f^n(t)-\dot{u}^n(t) \in \partial \varphi _n(u^n(t) $$

by $\dot{u}^n(t)$ and applying the chain rule formula [21] for the Lipschitz function $\varphi _n$ gives

$$\begin{aligned} \langle \dot{u}^n(t), f^n (t) \rangle -||\dot{u}^n(t)||^2 = \frac{d}{dt}[ \varphi _n(u^n(t))]. \end{aligned}$$

(4.1.1)

Hence by integration of (4.1.1) on [0, T] and anti-periodicity condition we get the estimate

$$\begin{aligned} ||\dot{u}^n||_{L^2_H([0, T])} \le ||f^n||_{L^2_H([0, T])}. \end{aligned}$$

(4.1.2)

From the Poincaré inequality

$$\begin{aligned} ||u ^n(t) || \le {\sqrt{T}}\, ||\dot{u}^n||_{L^2_H([0, T]) }, \forall t \in [0, T]. \end{aligned}$$

(4.1.3)

Integrating (4.1.1) on [0, t] we get

$$\begin{aligned} 0\le \varphi _n(u ^n(t)) = \varphi _n(u ^n(0)) -\int _0^t ||\dot{u} ^n(s) ||^2 ds +\int _0 ^t \langle \dot{u}^n(s), f ^n(s)\rangle ds\\ \le \varphi _n(u ^n(0))+\int _0 ^t \langle \dot{u}^n(s), f ^n(s)\rangle ds\nonumber \end{aligned}$$

(4.1.4)

so that by using the above estimates (4.1.2)–(4.1.3)–(4.1.4), the weak convergence of $f ^ n$ in $L^2_H([0, T])$ and (ii) we note that $\varphi _n(u ^n(t))$ is uniformly bounded.

Step 2 Using the results in Step 1, up to extracted subsequences $(u ^n)$ converges pointwisely to an anti-periodic absolutely continuous mapping $u ^\infty$ with $u ^\infty (T)= - u ^\infty (0)$, $(\dot{u} ^n)$ $\sigma ( L^2_{\mathbf {R}^d}, L^2_{\mathbf {R}^d})$-converges to $\zeta ^\infty \in L^2_{\mathbf {R}^d}([0, T])$ with $\zeta ^\infty = \dot{u}^\infty$. For simplicity set $z^n(t) := f^n(t) -\dot{u}^n(t)$. Since we have

$$ \langle \dot{u}^n (t), z^n(t) \rangle = \frac{d}{dt}[ \varphi _n(u^n(t))] $$

and $\langle \dot{u}^n (.), z^n(.) \rangle$ is bounded in $L^1_\mathbf {R}([0, T])$, $\varphi _n(u^n(t))$ is of bounded variation and uniformly bounded.

Claim $\lim _n \varphi _n(u_n(t)) = \varphi _\infty (u_\infty (t)) < \infty$ a.e and $\lim _n\int _0 ^T \varphi _n(u_n(t))dt = \int _0 ^T \varphi _\infty (u^\infty (t))dt < \infty$.

From the above estimates and Helly theorem, we may assume that $(\varphi _n(u_n(.))$ pointwisely converges to a BV function $\theta$ so that $(\varphi _n(u_n(.))$ converges in $L ^1_{\mathbf {R}}([0, T])$ to $\theta$. In particular, for every $k \in L^\infty _{\mathbf {R}^+}([0, T])$, we have

$$ \lim _{n \rightarrow \infty } \int _0^T k(t) \varphi _n(u_n(t)) dt = \int _0^T k(t) \theta (t) dt. $$

Coming back to the inclusion $z^n(t) \in \partial \varphi _n( u^n(t))$, and using the fact that $\varphi _n(x) \le \varphi _\infty (x), \forall n \in \mathbf N, \forall x \in \mathbf {R}^d$, we have

$$\begin{aligned} \varphi _\infty (x) \ge \varphi _n (x) \ge \varphi _n (u^n(t)) + \langle x-u^n(t), z^n(t) \rangle \end{aligned}$$

for all $x \in \mathbf {R}^d$. Let $h \in L ^\infty _{\mathbf {R}^d}([0, T])$. Substituting x by h(t) in this inequality and by integrating on each measurable set B gives

$$\begin{aligned} \int _B \varphi _\infty ( h(t))\, dt \ge \int _B \varphi _n ( h(t))\, dt \ge \int _B \varphi _n ( u^n(t)) \,dt + \int _B \langle h(t)-u^n(t), z^n(t) \rangle \,dt \end{aligned}$$

and passing to the limit in the preceding inequality when n goes to $+\infty$, we get

$$\begin{aligned} \int _B \varphi _\infty ( h(t)) \,dt \ge \int _B \theta (t) \,dt +\int _B \langle h(t)-u^\infty (t), z^\infty (t) \rangle \,dt \end{aligned}$$

(4.1.5)

with $z^\infty = f ^\infty - \dot{u}^\infty$. In particular, by taking $h = u^\infty$ we get the estimate

$$ \int _B \varphi _\infty ( u^\infty (t)) \,dt \ge \int _B \theta (t) \,dt $$

for all $B\in {\mathcal L}([0, T])$. By the epi-lower convergence result [11, Corollary 4.7], we have

$$ \int _B \theta (t) \,dt =\lim _{n \rightarrow \infty } \int _B \varphi _n( u^n(t)) \,dt \ge \liminf _{n \rightarrow \infty } \int _B \varphi _\infty ( u^n(t)) \,dt \ge \int _B \varphi _\infty ( u^\infty (t)) \,dt $$

for all $B\in {\mathcal L}([0, T])$. It turns out that $\varphi _\infty ( u^\infty (t))= \theta (t)$ a.e. and

$$\begin{aligned} \lim _{n \rightarrow \infty } \int _B \varphi _n( u^n(t)) \,dt = \int _B \varphi _\infty ( u^\infty (t)) \,dt < \infty . \end{aligned}$$

(4.1.6)

From (4.1.5) and (4.1.6) it follows that $f^\infty - \zeta ^\infty \in \partial I_{\varphi _\infty }(u^\infty )$ where $\partial I_{\varphi _\infty }$ denotes the subdifferential of the convex lower semicontinuous integral functional $I_{\varphi _\infty }$ defined on $L^\infty _{\mathbf {R}^d}([0, T])$

$$ I_{\varphi _\infty }(u):= \int _0^T \varphi _\infty ( u(t)) \,dt, \forall u \in L^\infty _{\mathbf {R}^d}([0, T]). $$

$\blacksquare$

Here is a variant of Proposition 4.1.

Proposition 4.2

Let $H = \mathbf {R}^d$. Assume that $\gamma > 0$, $\varphi _n : \mathbf {R}^d \rightarrow [0, +\infty ]$ is even, convex, Lipschitz, $\varphi _\infty : \mathbf {R}^d \rightarrow [0, +\infty ]$ is proper lower semicontinuous convex function such that $\varphi _n(x) \le \varphi _\infty (x)$ for all $n\in \mathbf N$ and for all $x\in \mathbf {R}^d$. Let $(f^n)$ be an anti-periodic sequence in $\mathcal H_\beta ([0, T])$. Let $u^n$ be a $W^{1, 2}_{\mathbf {R}^d}([0, T])$ anti-periodic solution to the problem

$$ \left\{ \begin{array}{lll} f^n (t)\in \dot{u}^n(t)+ \partial \varphi _n (u ^n(t))-\gamma u^n(t), \, t\in [0, T]\\ u_n(T) =-u_n(0). \end{array} \right. $$

Assume that the following conditions are satisfied:

• (i) $\varphi _n$ epiconverges to $\varphi _\infty$,

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

• (iii) $f^n$ $\sigma ( L^2_{\mathbf {R}^d}([0, T]), L^2_{\mathbf {R}^d}([0, T]))$-converges to $f ^\infty \in L^2_{\mathbf {R}^d}([0, T])$.

Then the following hold

• (a) Up to extracted subsequences, $(u ^n)$ converges pointwisely to an anti-periodic absolutely continuous mapping $u ^\infty$ with $u ^\infty (T)= - u ^\infty (0)$, $(\dot{u} ^n)$ $\sigma ( L^2_{\mathbf {R}^d}, L^2_{\mathbf {R}^d})$-converges to $\zeta ^\infty \in L^2_{\mathbf {R}^d}([0, T])$ with $\zeta ^\infty = \dot{u}^\infty$, $\lim _n \varphi _n( u^n(t)) = \varphi _\infty ( u^\infty (t)) < +\infty$ a.e. and $\lim _n \int _0^T \varphi _n( u^n(t)) dt = \int _0^T \varphi _\infty ( u ^\infty (t)) dt < +\infty$.

• (b) $f^\infty - \zeta ^\infty \in \partial I_{\varphi _\infty }(u^\infty )$ where $\partial I_{\varphi _\infty }$ denotes the subdifferential of the convex lower semicontinuous integral functional $I_{\varphi _\infty }$ defined on $L^\infty _{\mathbf {R}^d}([0, T])$

  $$ I_{\varphi _\infty }(u):= \int _0^T \varphi _\infty ( u(t)) \,dt, \forall u \in L^\infty _{\mathbf {R}^d}([0, T]). $$

Proof

Existence of $u ^n$ for the problem

$$ \left\{ \begin{array}{lll} f^n(t) -\dot{u}^n(t)+\gamma u^n(t) \in \partial \varphi _n(u^n(t)) \quad t\in [0, T], \\ u_n(T) =-u_n(0), \end{array} \right. $$

is ensured. See Haraux [17], Okochi [22].

Step 1 Estimation of $\dot{u}^n$ and $u^n$. Multiplying scalarly the inclusion

$$\begin{aligned} f^n(t) -\dot{u}^n(t)+\gamma u^n(t) \in \partial \varphi _n(u^n(t)) \end{aligned}$$

(4.2.1)

by $\dot{u}^n(t)$ and applying the chain rule formula [21] for the Lipschitz function $\varphi _n$ gives

$$\begin{aligned} \langle \dot{u}^n(t) , f^n(t) \rangle - || \dot{u}^n(t)|| ^2+ \gamma \langle \dot{u}^n(t), u^n(t ) \rangle = \frac{d}{dt}[ \varphi (u^n(t))]. \end{aligned}$$

(4.2.2)

Hence by integration in (4.2.1) and anti-periodicity conditions we get the estimate

$$\begin{aligned} ||\dot{u}^n||_{L^2_H([0, T])} \le ||f^n||_{L^2_H([0, T])}. \end{aligned}$$

(4.2.3)

From the Poincaré inequality,

$$\begin{aligned} ||u ^n(t) || \le {\sqrt{T}}\, ||\dot{u}^n||_{L^2_H([0, T]) } \le {\sqrt{T}}\, ||f^n||_{L^2_H([0, T])}. \end{aligned}$$

(4.2.4)

Integrating (4.2.2), we get

$$\begin{aligned} 0\le \varphi _n(u ^n(t)) = \varphi _n(u ^n(0))- \int _0^t ||\dot{u} ^n(s) ||^2 ds +&\int _0 ^t \langle \dot{u}^n(s), f ^n(s)\rangle ds\\&+ \gamma \int _0 ^t \langle \dot{u}^n(s), u^n(s ) \rangle ds \end{aligned}$$

We note that

$$\begin{aligned} \int _0 ^t \langle \dot{u}^n(s), f ^n(s)\rangle ds \le \frac{1}{2} ||f^n||_{L ^2_H([0, T])} (1+\int _ 0^t ||\dot{u}^n (s)|| ^2 ds) \le \mathrm{Const. }\\ \gamma \int _0 ^t \langle \dot{u}^n(s), u^n(s ) \rangle ds \le \mathrm{Const. } ||f^n||^2_{L^2_H([0, T])} \end{aligned}$$

so that by using the above estimate, the $\sigma ( L^2_{\mathbf {R}^d}([0, T]), L^2_{\mathbf {R}^d}([0, T]))$ convergence of $f ^ n$ and (ii), we conclude that $\varphi _n(u ^n(t))$ is uniformly bounded. Now the remainder of the proof is similar to that of Proposition 4.1.$\blacksquare$

We finish the paper with the approximating problem in second order evolution equation with anti-periodic condition

$$ \left\{ \begin{array}{ll} f^n(t) &{}=\ddot{u}^n (t)+M\dot{u}^n(t)+\nabla \varphi _n(u^n(t)), \\ u^n(T) &{}= -u^n(0). \end{array} \right. $$

where M is a positive constant, $\varphi _n$ are convex Lipschitz, $C ^1$, even, functions that epiconverges to a lower semicontinuous convex proper function $\varphi _\infty$, $(f_n)$ is a sequence in $L ^2_H([0, T])$ which weakly converges to a function $f_\infty \in L ^2_H([0, T])$. Existence of a $W^{2, 2}_{\mathbf {R}^d}([0, T])$ anti-periodic -solution to this approximating problem is well known. See Haraux [17], Okochi [22].

Proposition 4.3

Let $H = \mathbf {R}^d$, $M \in \mathbf {R}^+$. Assume that $\varphi _n : \mathbf {R}^d \rightarrow [0, +\infty [$ is ${\mathcal C}^1$, even, convex, Lipschitz and, $\varphi _\infty : \mathbf {R}^d \rightarrow [0, +\infty ]$ is proper convex lower semicontinuous with $\varphi _n(x) \le \varphi _\infty (x), \, \forall x \in \mathbf {R}^d$. Let $f ^n \in \mathcal H_\beta ([0, T])$ Let $u^n$ be a $W^{2, 2}_{\mathbf {R}^d}([0, T])$ anti-periodic solution to the approximated problem

$$ \left\{ \begin{array}{ll} f ^n(t)&{}= \ddot{u}^n(t)+M \dot{u}^n (t)+ \nabla \varphi _n (u^n(t)), t\in [0, T], \\ u_n(T) &{}=-u_n(0). \end{array} \right. $$

Assume that

• (i) $f^n \sigma (L^2_H, L^2_H)$ converges to $f^\infty \in L^2_H([0, T])$.

• (ii) $\lim _n u^n (0)=u ^ \infty _0 \in \mathop {\text {dom}}\, \varphi _\infty , \,\lim _n \varphi _n (u^n (0)) = \varphi _\infty (u^ \infty _0 )$, and $\lim _n \dot{u}^n (0)=\dot{u}^ \infty _0$,

• (iii) $\varphi _n$ epi-converges to $\varphi _\infty$,

• (iv) There exist $r_0 >0$ and $x_0 \in \mathbf {R}^d$ such that

  $$ \sup _{v \in {\overline{B}}_{L^\infty _{ \mathbf {R}^d}([0, T])}}\int _0^T\varphi _\infty (x_0+r_0v(t))) < +\infty $$

  where ${\overline{B}}_{L^\infty _{ \mathbf {R}^d}([0, 1])}$ is the closed unit ball in $L^\infty _{ \mathbf {R}^d}([0, T])$.

Then the following hold

• (a) Up to extracted subsequences, $(u ^n)$ converges uniformly to a $W^{1, 1}_{BV}([0, T])$ anti-periodic function $u ^\infty$ with $u ^\infty (T)= - u ^\infty (0)$, and $(\dot{u} ^n)$ pointwisely converges to the BV function $\dot{u} ^\infty$, and $(\ddot{u} ^n)$ biting converges to a function $\zeta ^\infty \in L^1_{\mathbf {R}^d}([0, T])$ which satisfy the variational inclusion

  $$ f^\infty -\zeta ^\infty - M\dot{u}^\infty \in \partial I_{\varphi _\infty }(u^\infty ) $$

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

  $$ I_{\varphi _\infty }(u):= \int _0^T \varphi _\infty ( u(t)) \,dt, \forall u \in L^\infty _{\mathbf {R}^d}([0, T]). $$

  Furthermore

  $$\begin{aligned} \lim _n \varphi _n( u^n(t))&= \varphi _\infty ( u ^\infty (t))< \infty \mathrm{a.e.} \\ \lim _n \int _0^T \varphi _n( u^n(t)) dt&=\int _0^T \varphi _\infty ( u ^\infty (t)) dt < \infty . \end{aligned}$$

  Subsequently, the estimated energy holds almost everywhere

  $$\begin{aligned} \varphi _\infty (u^\infty (t)) +\frac{1}{2} ||\dot{u}^\infty (t)||^2&= \varphi _\infty (u^\infty (0)) +\frac{1}{2} ||\dot{u}^\infty (0)||^2 \\&\quad -M\int _0^t ||\dot{u}^\infty (s)||^2\,ds+ \int _0^t \langle \dot{u}^\infty (s), f ^\infty (s)\rangle ds. \end{aligned}$$

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

  $$\begin{aligned} \int _0^T \varphi _\infty ( v(t)) \,dt \ge&\int _0^T \varphi _\infty ( u ^\infty (t)) \,dt + \int _0 ^T \langle -M \dot{u} ^\infty (t)+f^\infty (t) ,v(t)- u ^\infty (t)\rangle \,dt \\&+ \langle -m, v-u ^\infty \rangle _{ ({\mathcal M}^b_E([0, T]), {\mathcal C}_E([0, T])) }. \end{aligned}$$

  In other words, the vector measure $-m+[-M \dot{u}^\infty +f ^\infty ] \,dt$ belongs to the subdifferential $\partial I_{f_\infty }(u)$ of the convex functional integral $I_{f_\infty }$ defined on ${\mathcal C}_H([0, T])$ by $I_{\varphi _\infty }(v)=\int _0 ^T \varphi _\infty (t, v(t))\ dt$, $\forall v\in {\mathcal C}_H([0, T])$.

Proof

Existence of $W^{2, 2}_{\mathbf {R}^d}([0, T])$-solution $u ^n$ for the approximated problem

$$ \left\{ \begin{array}{ll} f ^n(t) &{}= \ddot{u}^n(t)+ M\dot{u}^n (t) +\nabla \varphi _n(u^n(t)) \quad t\in [0, T], \\ u_n(T) &{}=-u_n(0) \end{array} \right. $$

follows from Haraux [17]. Now we can finish the proof by repeating mutatis mutandis the machinery developed in Proposition 3.3. Therefore our $W^{1, 1}_{BV}([0, T])$ anti-periodic limit $u ^\infty$ of $(u^n)$ and biting limit $\zeta ^\infty$ of $(\ddot{u} ^n)$ satisfies the inclusion

$$ f ^\infty (t) - \zeta ^\infty (t) - M\dot{u}^\infty (t) \in \partial \varphi _\infty (u^\infty (t)) $$

and the energy estimate holds

$$\begin{aligned} \varphi _\infty (u^\infty (t)) +\frac{1}{2} ||\dot{u}^\infty (t)||^2 =\;&\varphi _\infty (u^\infty _0) +\frac{1}{2} ||\dot{u}^\infty _0||^2\\&-M\int _0^t ||\dot{u}^\infty (s)||^2\,ds+ \int _0^t \langle \dot{u}^\infty (s), f ^\infty (s)\rangle ds \end{aligned}$$

almost everywhere.$\blacksquare$