On the uniqueness of minimizers for a class of variational problems with Polyconvex integrand

TL;DR

This paper proves the existence and uniqueness of minimizers for a class of elastic energy functionals involving polyconvex integrands, including degenerate and incompressible cases, extending prior results in elasticity theory.

Contribution

It extends previous work by establishing uniqueness for a broader class of polyconvex integrands and analyzes minimizers in degenerate and incompressible scenarios.

Findings

Existence and uniqueness of minimizers for polyconvex energy functionals.

Minimizers in the degenerate case are measure-preserving maps.

Incompressible minimizers can be obtained as limits of general displacements.

Abstract

We prove existence and uniqueness of minimizers for a family of energy functionals that arises in Elasticity and involves polyconvex integrands over a certain subset of displacement maps. This work extends previous results by Awi and Gangbo to a larger class of integrands. First, we study these variational problems over displacements for which the determinant is positive. Second, we consider a limit case in which the functionals are degenerate. In that case, the set of admissible displacements reduces to that of incompressible displacements which are measure preserving maps. Finally, we establish that the minimizer over the set of incompressible maps may be obtained as a limit of minimizers corresponding to a sequence of minimization problems over general displacements provided we have enough regularity on the dual problems. We point out that these results defy the direct methods of the calculus of variations.

Equations

$$\{\begin{array}{ll}{\mathbf{u}}_t={\text{div}}_x{D}_{\xi }L(\nabla \mathbf{u})& \text{ on }\Omega \times (0,T),\\ \mathbf{u}(0,\cdot )={\mathbf{u}}_0& \text{ on }\Omega ,\end{array}$$

$$\underset{t\to 0^{+}}{\lim }H(t)=\underset{t\to \infty }{\lim }\frac{H(t)}{t}=+\infty .$$

$$\min \left\{{\int }_{\Omega }\left(f(\nabla u)+H(\beta )-F\cdot u\right)dx;(u,\beta )\in \mathcal{U}\right\}$$

$$\begin{split}\mathscr{U}=\Big{\{}(u,\beta):&u\in W^{1,p}(\Omega,\bar{\Lambda}),\;\beta:\Omega\to[0,\infty);\\ &\int_{\Omega}l(u)\beta dx=\int_{\Lambda}l(y)dy;\forall l\in C_{c}(\mathbb{R}^{d})\Big{\}}.\end{split}$$

$${\int }_{\Omega }f(G)dx$$

$${\mathcal{G}}_{\mathcal{S}}(u):=\left\{G\in L^p(\Omega ,{\mathbb{R}}^{d\times d}):{\int }_{\Omega }u\mathrm{div}\phi dx=-{\int }_{\Omega }⟨G,\phi ⟩dx\forall \phi \in \mathcal{S}\right\}.$$

$$\Big{|}\int_{\Omega}u\cdot\text{div}\,\varphi\;dx\Big{|}\leq c\|\varphi\|_{L^{q}({\Omega,\mathbb{R}^{d\times d}})},\;\quad\forall\varphi\in\mathcal{S}.$$

$$\inf \left\{{\int }_{\Omega }\left(f({\nabla }_{S}u)+H(\beta )-F\cdot u\right)dx;(u,\beta )\in \mathcal{U}\right\}.$$

$$\underset{n\to \infty }{\lim }{\int }_{\Omega }f({\nabla }_{{\mathcal{S}}_n}u)dx={\int }_{\Omega }f(\nabla u)dx.$$

$$W^{1,p}(\Omega ,{\mathbb{R}}^d)\ni u\mapsto V_{\mathcal{S}}^f[u]:=\underset{\phi \in \mathcal{S}}{\sup }{\int }_{\Omega }\left(-u\mathrm{div}\phi -f^{\ast }(\phi )\right)dx.$$

$$\underset{\tau \to \infty }{\lim }{V}_{{\mathcal{S}}_n}^f[u]=V_{W_0^{1,q}(\Omega ,{\mathbb{R}}^d)}^f[u]\left(={\int }_{\Omega }f(\nabla u)dx\right)$$

$$\underset{(u,\beta )\in {\mathcal{U}}_{\mathcal{S}}^{\ast }}{\inf }\left\{V_{\mathcal{S}}^f[u]+{\int }_{\Omega }H(\beta )-F\cdot udx\right\}$$

$$\begin{split}\mathscr{U}_{\mathcal{S}}^{*}=\Big{\{}(u,\beta):&u\in\mathscr{U}_{\mathcal{S}};\;\beta:\Omega\to[0,\infty);\\ &\int_{\Omega}l(u(x))\beta(x)\;dx=\int_{\Omega}l(y)\;dy\;\forall l\in C_{c}(\mathbb{R}^{d})\Big{\}}.\end{split}$$

$$k(v)+tl(u)+H(t)\ge u\cdot v\forall u,v\in {\mathbb{R}}^d,t>0.$$

$$J(k,l,\phi ):={\int }_{\Omega }k(F+\mathrm{div}\phi )dx+{\int }_{\Lambda }ldy+{\int }_{\Omega }{f}^{\ast }(\phi )dx.$$

$$F+\mathrm{div}\phi \text{ is non-degenerate.}$$

$$\{\begin{array}{ll}{u}_0=& \nabla k_0(F+\mathrm{div}{\phi }_0)\\ {\phi }_0\in & {\Phi }_{\mathcal{S}}(u_0).\end{array}$$

$$\underset{u\in {\mathcal{U}}_{\mathcal{S}}^1}{\inf }\left\{V_{\mathcal{S}}^f[u]-{\int }_{\Omega }F\cdot udx\right\}$$

$${\mathcal{U}}_{\mathcal{S}}^1=\left\{u\in {\mathcal{U}}_{\mathcal{S}}:{\int }_{\Omega }l(u(x))dx={\int }_{\Omega }l(y)dy\forall l\in C_c({\mathbb{R}}^d)\right\}$$

$$k(v)+l(u)\ge u\cdot v\forall u,v\in {\mathbb{R}}^d.$$

$$u_0=\nabla k_0(F+\mathrm{div}{\phi }_0).$$

$$I_n(u,\beta )=V_{\mathcal{S}}^f[u]+{\int }_{\Omega }{H}_n(\beta )-u\cdot Fdx$$

$$I_0(u)=V_{\mathcal{S}}^f[u]-{\int }_{\Omega }u\cdot Fdx.$$

$$\underset{(u,\beta )\in {\mathcal{U}}_{\mathcal{S}}^{\ast }}{\inf }{I}_n(u,\beta )$$

$${\chi }_A(x)=\{\begin{array}{l}0\text{ if }x\in A,\\ \infty \text{ otherwise.}\end{array}$$

$$f^{\ast }({\xi }^{\ast })=\underset{\xi \in {\mathbb{R}}^{d\times d}}{\sup }\left\{\xi \cdot {\xi }^{\ast }-f(\xi )\right\}.$$

$$|grad[h](x)|=\min \left\{|y|:y\in \partial h(x)\right\};x\in \text{Dom}(h).$$

$${\mathcal{S}}_f:=\left\{\phi \in \mathcal{S}:{\int }_{\Omega }{f}^{\ast }(\phi )\text{ is finite}\right\}.$$

$$c\frac{|\xi {|}^p}{p}+b\ge f(\xi )\ge a|\xi |-b$$

$$|{\xi }^{\ast }{|}^q\le c|\xi {|}^p+b.$$

Full text

Most of the work presented in this paper was carried out while R.A. was a postdoctoral fellow at the Institute for Mathematics and its Applications during the IMA’s annual program on Control Theory and its Applications.

M.S. gratefully acknowledges the support of the King Abdullah University of Science and Technology.

1. Introduction

We are interested in Euler-Lagrange equations, existence and uniqueness of minimizers for some problems in the vectorial calculus of variations emanating from elasticity theory. These variational problems are related to an open problem in Partial Differential Equations that we describe as follows : let $\displaystyle T>0$ and let $\displaystyle\Omega$ and $\displaystyle\Lambda$ be two open subsets of $\displaystyle\mathbb{R}^{d}$; suppose that $\displaystyle\mathbf{u}_{0}$ is a diffeomorphism between $\displaystyle\Omega$ and $\displaystyle\Lambda$; we seek $\displaystyle\mathbf{u}:\Omega\times(0,T)\longrightarrow\mathbb{R}^{d}$ such that $\displaystyle\mathbf{u}(\cdot,t)(\Omega)=\Lambda$ for each $\displaystyle t$ and

[TABLE]

in the sense of distributions. In (1.1), we assume that the map $\displaystyle\mathbb{R}^{d\times d}\ni\xi\mapsto L(\xi)$ is quasiconvex. We refer the reader to [2], [5], [6], [10], and [12] for further details on these gradient flows. Understanding variational problems associated to the time-discretization of (1.1) is arguably an important step toward the construction of a solution. In that regard, several partial results are available in the literature (See for instance [5] and [6]).

In [2], the authors have focused on a class of Lagrangians that arises in elastic materials. More precisely, they have considered polyconvex Lagrangians of the form $\displaystyle\xi\mapsto L(\xi)=f(\xi)+H(\det\xi)$. Here $\displaystyle f$ is a $\displaystyle C^{1}(\mathbb{R}^{d})$ strictly convex function with $\displaystyle p$-th order growth, and the map $\displaystyle H$ is a $\displaystyle C^{1}(0,\infty)$ convex function that satisfies

[TABLE]

As a result, a variational problem emerges from the time discretization and has a relaxation that takes the general form :

[TABLE]

where $\displaystyle F\in L^{1}(\Omega,\mathbb{R}^{d})$ and

[TABLE]

Although the existence of minimizers in (1.3) follows from the direct methods in the calculus of variations, the uniqueness is a rather challenging problem. Indeed, because of (1.2) and the non-convexity of the integrand, standard techniques in calculus of variations do not apply.

To bypass these difficulties, the authors of [2] have introduced a pseudo-projected gradient operator $\displaystyle\mathscr{U}_{\mathcal{S}}\ni u\mapsto\nabla_{\mathcal{S}}u$ defined as follows : for a given $\displaystyle u\in\mathscr{U}_{\mathcal{S}}$, the map $\displaystyle\nabla_{\mathcal{S}}u$ is the unique minimizer of

[TABLE]

over

[TABLE]

Here, $\displaystyle\mathcal{S}$ is a finite-dimensional subspace of $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$, $\displaystyle q$ is the conjugate of $\displaystyle p$, $\displaystyle\mathscr{U}_{\mathcal{S}}$ is the set of all $\displaystyle u:\Omega\to\bar{\Lambda}$ measurable such that there exists a $\displaystyle c=c(u,\Omega,\Lambda)>0$ satisfying :

[TABLE]

We point out that the pseudo-projected gradient operator depends also on $\displaystyle f$, though the dependence is not exhibited in its notation. As a first step to approaching (1.3), they have considered the following perturbed problem:

[TABLE]

The choice of problem (1.6) is justified by the construction of a family of finite dimensional subspaces $\displaystyle\left\{\mathcal{S}_{n}\right\}_{n}$ dense in $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ such that for $\displaystyle u\in W^{1,p}(\Omega,\mathbb{R}^{d})$, one has

[TABLE]

We note that a $\displaystyle L^{p}(\Omega,\;\mathbb{R}^{d})-$bounded subset of $\displaystyle\mathscr{U}_{\mathcal{S}}$ whose image by the operator $\displaystyle\nabla_{\mathcal{S}}$ is bounded in $\displaystyle L^{p}(\Omega,\;\mathbb{R}^{d\times d})$ is not in general strongly pre-compact with respect to the $\displaystyle L^{p}(\Omega,\;\mathbb{R}^{d})$ topology. As a result, compactness of level subsets of the functional in (1.6) can not be guaranteed. Nevertheless, the authors of [2] have successfully shown existence and, more importantly, uniqueness in (1.6) under the assumption that $\displaystyle F$ is non-degenerate (see definition below). This condition of non-degeneracy for uniqueness is crucial in a similar problem, the so-called Brenier polar factorization, and more generally, in optimal transport problems. Confer [1], [3], [11], [8], [9] and [14].

In this paper, we investigate the respective roles played by the strict convexity of $\displaystyle f$, the convexity and smoothness of $\displaystyle H$, and the non-degeneracy of $\displaystyle F$ in problem (1.6). More precisely, we impose less stringent conditions so that either the map $\displaystyle F$ is allowed to be degenerate or $\displaystyle f$ is allowed to be merely convex or $\displaystyle H$ is neither smooth nor strictly-convex. These considerations are not just technicalities. Indeed we note that a prominent case of mere convexity, $\displaystyle f(\xi)=|\xi|$, is typical for the study of minimal surfaces as well as for the study of functionals involving the total variation (see for instance [4]) Furthermore, we observe that cases where $\displaystyle H$ is taken to be the characteristic function of a singleton of $\displaystyle\mathbb{R}$ arise in the study of incompressible deformations in Elasticity theory (see for instance [12] and [14]). Finally, the non degeneracy condition tests the extent to which one can hope for uniqueness in the variational problem we considered. To deal with these weaker assumptions, we introduce a family of operators $\displaystyle\left\{V_{\mathcal{S}}^{f}:\mathcal{S}\subset W_{0}^{1,q}(\Omega,\mathbb{R}^{d}),\;f\;\text{convex}\right\}$ defined by

[TABLE]

We note that the operator $\displaystyle V_{\mathcal{S}}^{f}$ is actually well defined on the set of measurable functions $\displaystyle u$ defined from $\displaystyle\Omega$ to $\displaystyle\bar{\Lambda}$ when the set $\displaystyle\mathcal{S}$ is a finite dimensional nonempty set and the function $\displaystyle f$ satisfies appropriate growth conditions. As a family, these operators extend the pseudo- projected gradient operators and the distributional gradient. Indeed, $\displaystyle V_{\mathcal{S}}^{f}[u]=\int_{\Omega}f(\nabla_{\mathcal{S}}u)$ if $\displaystyle\mathcal{S}$ is a finite dimensional subspace of $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ and $\displaystyle u\in\mathscr{U}_{\mathcal{S}}$ and furthermore $\displaystyle V_{\mathcal{S}}^{f}[u]=\int_{\Omega}f(\nabla u)$ if $\displaystyle\mathcal{S}=W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ and $\displaystyle u\in W^{1,p}(\Omega,\mathbb{R}^{d})$. These extensions are only valid under appropriate conditions on $\displaystyle f$. It is worth pointing out that if $\displaystyle f(\xi)=|\xi|$ and $\displaystyle\mathcal{S}=W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ then $\displaystyle V_{\mathcal{S}}^{f}(u)$ is nothing but the total variation of $\displaystyle u$ on the set $\displaystyle\Omega$. We show that for a collection of sets $\displaystyle\left\{\mathcal{S}_{n}\right\}_{n=1}^{\infty}$ of $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d})$ satisfying Hypothesis (H1) or Hypothesis (H2) (see section 2), we have a convergence result in the same spirit as (1.7):

[TABLE]

for any $\displaystyle u\in W^{1,p}(\Omega,\mathbb{R}^{d\times d})$ and appropriate conditions on $\displaystyle f$. We thus proceed to study a more general problem :

[TABLE]

where $\displaystyle\mathcal{S}$ is an element of a collection of sets satisfying Hypothesis (H1) or Hypothesis (H2), and

[TABLE]

Sublevel sets of the integrand in (1.10) are not compact. Nor is $\displaystyle f$ necessarily strictly convex. However, we show existence and uniqueness in Problem (1.10). In fact, this result holds for $\displaystyle F$ non-degenerate as well as for a class of degenerate $\displaystyle F$ provided that the set $\displaystyle\mathcal{S}$ is chosen accordingly ( see Corollaries 3.6 and 3.7). Unlike optimal transport theory, this analysis suggests that the non-degeneracy condition is not essential for a uniqueness result in (1.3).

Existence and uniqueness results for Problem (1.10) are established thanks to the discovery of suitable dual problems. Indeed, call $\displaystyle\mathcal{C}$ the set of all functions $\displaystyle(k,l)$ with $\displaystyle k,\;l:\mathbb{R}^{d}\rightarrow\mathbb{R}\cup\{\infty\}$ Borel measurable, finite at least at one point, and satisfying the relation $\displaystyle l\equiv\infty$ on $\displaystyle\mathbb{R}^{d}\setminus\bar{\Lambda}$ and such that

[TABLE]

Let $\displaystyle\mathscr{A}$ be the set of $\displaystyle(k,l,\varphi)$ such that $\displaystyle(k,l)\in\mathcal{C}$ and $\displaystyle\varphi\in\mathcal{S}$. Define the following functional over the set $\displaystyle\mathscr{A}$:

[TABLE]

Next, assume that the map $\displaystyle F$ and the set $\displaystyle\mathcal{S}$ are such that for all $\displaystyle\varphi\in\mathcal{S}$,

[TABLE]

Then $\displaystyle-J$ admits a maximizer $\displaystyle(k_{0},l_{0},\varphi_{0})$ with $\displaystyle k_{0}$ convex and $\displaystyle diam(\Lambda)$-Lipschitz. As a consequence, Problem (1.10) admits a unique minimizer $\displaystyle(u_{0},\beta_{0})$ and $\displaystyle u_{0}$ satisfies

[TABLE]

Here, we have denoted by $\displaystyle\Phi_{\mathcal{S}}(u_{0})$, the non-empty set of maximizers of problem (1.8) (see Proposition 2.8). In order to obtain condition (1.12), we consider two distinct situations.

First, we assume that $\displaystyle F$ has a countable range, thus degenerate. If $\displaystyle\mathcal{S}$ is an element of a collection of sets satisfying hypothesis (H2) then it holds that $\displaystyle F+\operatorname{div}\varphi$ is non degenerate.

Second, we assume $\displaystyle F$ non-degenerate and $\displaystyle\mathcal{S}$ is a finite dimensional vector space, as in [2]. It holds again that $\displaystyle F+\operatorname{div}\varphi$ is non degenerate. However, unlike the hypotheses in [2], we have allowed the map $\displaystyle f$ to be as singular as the map $\displaystyle\mathbb{R}^{d\times d}\ni\xi\mapsto|\xi|$.

We have also studied (1.10) when $\displaystyle H$ is replaced by $\displaystyle H_{0}:(0,\infty)\to\mathbb{R}\cup\{\infty\}$ defined by $\displaystyle H_{0}(1)=0$ and $\displaystyle H_{0}(t)=\infty$ if $\displaystyle t\neq 1$. This case corresponds to the case of measure preserving maps. Note that $\displaystyle H_{0}$ is not even continuous. However, it may be obtained as a limit of functions $\displaystyle H_{n}$ which are $\displaystyle C^{1}(0,\infty)$ convex functions and satisfy (1.2). We show that for such singular $\displaystyle H_{0}$, the corresponding problem

[TABLE]

with

[TABLE]

admits a unique minimizer. (See Theorem 4.3).

To obtain existence and uniqueness results in problem (1.14), we exploit a dual formulation and maximize $\displaystyle-J$ over the set that consists of $\displaystyle(k,l,\varphi)$ such that $\displaystyle\varphi\in\mathcal{S}$ and $\displaystyle k,\;l:\mathbb{R}^{d}\rightarrow\mathbb{R}\cup\{\infty\}$ are Borel measurable, finite at least at one point, and satisfy the relations $\displaystyle l\equiv\infty$ on $\displaystyle\mathbb{R}^{d}\setminus\bar{\Lambda}$ and

[TABLE]

One shows that $\displaystyle-J$ admits a maximizer $\displaystyle(k_{0},l_{0},\varphi_{0})$ with $\displaystyle k_{0}$ convex and Lipschitz and the unique minimizer of problem (1.14) is $\displaystyle u_{0}$ given by

[TABLE]

Finally, we show convergence of a sequence of problems of the form (1.10) to (1.14). More precisely, we show that the minimizer of problem (1.14) may be obtained as limit of minimizers of problems of the form (1.10) provided that the dual problems admit regular enough maximizers. In fact, suppose the map $\displaystyle F$ and the set $\displaystyle\mathcal{S}$ are such that for all $\displaystyle\varphi\in\mathcal{S}$, the map $\displaystyle F+\operatorname{div}\varphi$ is non-degenerate. For $\displaystyle(u,\beta)\in\mathscr{U}_{\mathcal{S}}$, define

[TABLE]

and set

[TABLE]

Thanks to Theorem 3.5, the problem

[TABLE]

admits a unique minimizer that we denote $\displaystyle(u_{n},\beta_{n})$ with $\displaystyle u_{n}=\nabla k_{n}(F+\operatorname{div}\varphi_{n})$ for some $\displaystyle k_{n}:\mathbb{R}^{d}\to\mathbb{R}$ convex and $\displaystyle\varphi_{n}\in\mathcal{S}$. Denote $\displaystyle u_{0}$ the unique minimizer of (1.14). If for all $\displaystyle n\in\mathbb{N}^{*}$ the map $\displaystyle k_{n}$ is differentiable then the sequence $\displaystyle\{u_{n}\}_{n\in\mathbb{N}^{*}}$ converges almost everywhere to $\displaystyle u_{0}$ and in addition, the minima $\displaystyle\{I_{n}(u_{n},\beta_{n})\}_{n\in\mathbb{N}^{*}}$ converge to $\displaystyle I_{0}(u_{0})$ (Cf. Theorem 4.7).

2. Preliminaries

2.1. Notation and definitions.

• $\displaystyle\bullet$

Throughout this manuscript, $\displaystyle\Omega$ and $\displaystyle\Lambda\subset\mathbb{R}^{d}$ are two bounded open convex sets; $\displaystyle r^{*}>1$ is such that $\displaystyle B(0,1/r^{*})\subset\Lambda\subset B(0,r^{*}/2)$; $\displaystyle p\in(1,\infty)$ and $\displaystyle q$ is its conjugate, that is, $\displaystyle p^{-1}+q^{-1}=1$.

• $\displaystyle\bullet$

Given $\displaystyle A\subset\mathbb{R}^{d}$, the indicator function of $\displaystyle A$ is defined as

[TABLE]

• $\displaystyle\bullet$

For any subset $\displaystyle\mathcal{S}$ of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$, we denote by $\displaystyle\text{span}(\mathcal{S})$ the linear subspace of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ generated by $\displaystyle\mathcal{S}$.

• $\displaystyle\bullet$

We denote by $\displaystyle f^{*}$ the Legendre transform of the map $\displaystyle f:\mathbb{R}^{d\times d}\longrightarrow\mathbb{R}$ so that

[TABLE]

• $\displaystyle\bullet$

If $\displaystyle h:\mathbb{R}^{d}\longrightarrow\mathbb{R}\cup\{\infty\}$ is convex then the subdifferential $\displaystyle\partial h(x)$ of $\displaystyle h$ at $\displaystyle x\in\text{Dom}(h)$ is closed and convex. If $\displaystyle\partial h(x)$ is non-empty we denote by $\displaystyle grad[h](x)$ the element of $\displaystyle\partial h(x)$ with minimum norm :

[TABLE]

• $\displaystyle\bullet$

Let $\displaystyle\mathcal{S}\subset W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$. We denote by $\displaystyle\mathscr{S}_{f}$ the set

[TABLE]

• $\displaystyle\bullet$

Let $\displaystyle F:\mathbb{R}^{d}\longrightarrow\mathbb{R}^{d}$ be measurable. We say that $\displaystyle F$ is non-degenerate if for any $\displaystyle N\subset\mathbb{R}^{d}$ such that $\displaystyle\mathcal{L}^{d}(N)=0$ we have $\displaystyle\mathcal{L}^{d}(F^{-1}(N))=0.$

2.2. Assumptions.

• (A0)

We additionally assume that there exists a strictly convex function that is $\displaystyle C^{1}(\bar{\Omega})$ and vanishes on the boundary of $\displaystyle\Omega$.

• (A1)

The set $\displaystyle\mathcal{S}$ is a subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$. In addition, the map $\displaystyle f:\mathbb{R}^{d\times d}\to\mathbb{R}$ is convex and satisfies the following three properties:

• (i)

There exist $\displaystyle a,b,c>0$ such that for all $\displaystyle\xi\in\mathbb{R}^{d\times d}$,

[TABLE]

and for all $\displaystyle\xi^{*}\in\partial f(\xi)$,

[TABLE]

• (ii)

The set $\displaystyle\mathscr{S}_{f}$ is non empty.

• (iii)

One of the following two conditions holds:

1. (a)

The map $\displaystyle f$ is such that $\displaystyle\partial f^{*}(x^{*})$ is non-empty and $\displaystyle grad[f^{*}](x^{*})=0$ for each $\displaystyle x^{*}\in\text{Dom}f^{*}$. 2. (b)

The map $\displaystyle f$ is strictly convex and there exist $\displaystyle\bar{a},\bar{b}>0$ such that for all $\displaystyle\xi^{*}\in\mathbb{R}^{d\times d}$, one has

[TABLE]

• (A2)

The map $\displaystyle H$ is $\displaystyle C^{1}(0,\infty)$, strictly convex, and such that

[TABLE]

• (A3)

The function $\displaystyle F$ is measurable and belongs to $\displaystyle L^{1}(\Omega)$.

Let $\displaystyle\mathcal{S}$ be a subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$. We say that $\displaystyle F$ satisfies the condition $\displaystyle\textbf{(ND)}_{\mathcal{S}}$ if

[TABLE]

for all $\displaystyle\varphi\in\mathcal{S}$.

Remark 2.1**.**

• (i)

As $\displaystyle f$ satisfies (2.2), we have

[TABLE]

for all $\displaystyle\xi^{*}\in\mathbb{R}^{d\times d}$.

• (ii)

If $\displaystyle f$ satisfies case (b) in (iii) of Assumption $\displaystyle\mathbf{(A1)}$, then $\displaystyle f^{*}$ is continuously differentiable. In that case, $\displaystyle grad[f^{*}]=\nabla f^{*}$.

• (iii)

If $\displaystyle f$ satisfies case (a) of Assumption $\displaystyle\mathbf{(A1)}$(iii) then $\displaystyle 0\in\partial f^{*}(x^{*})$ for every element $\displaystyle x^{*}\in\operatorname{Dom}(f^{*})$. Consequently, the map $\displaystyle f^{*}$ is constant on $\displaystyle\operatorname{Dom}(f^{*})$ and the following equalities are satisfied for all $\displaystyle x^{*}$ and $\displaystyle y^{*}$ in $\displaystyle\operatorname{Dom}(f^{*})$:

[TABLE]

• (iv)

The assumption $\displaystyle\mathbf{(A0)}$ is satisfied by $\displaystyle\Omega=B(0,1)\subset\mathbb{R}^{d}$ with the strictly convex function being the map $\displaystyle\mathbb{R}^{d}\ni x\mapsto|x|^{2}-1$.

• (v)

The map $\displaystyle f:\mathbb{R}^{d\times d}\to\mathbb{R}$ defined by $\displaystyle f(\xi)=|\xi|$ satisfies case (a) in (iii) of Assumption $\displaystyle\mathbf{(A1)}$. The map $\displaystyle f:\mathbb{R}^{d\times d}\to\mathbb{R}$ defined by $\displaystyle f(\xi)=|\xi|^{p}$ satisfies case (b) in (iii) of Assumption $\displaystyle\mathbf{(A1)}$.

The following Lemma summarizes some elementary properties of $\displaystyle H$. We refer the reader to Remark 2.1 in [2].

Lemma 2.2**.**

Assume $\displaystyle\mathbf{(A2)}$ holds. Then,

1. (i)

The map $\displaystyle H^{\prime}:(0,\infty)\to\mathbb{R}$ is a strictly increasing bijection. 2. (ii)

The Legendre transform $\displaystyle H^{*}$ of $\displaystyle H$ is a strictly increasing bijection from $\displaystyle\mathbb{R}$ to $\displaystyle\mathbb{R}$. 3. (iii)

Let $\displaystyle g:\mathbb{R}\to\bar{\mathbb{R}}$ be defined by $\displaystyle g(s)=\alpha s-\beta H^{*}(s)$, with $\displaystyle\alpha,\beta>0$. Then

[TABLE]

Define $\displaystyle H_{0}$ by

[TABLE]

and, for $\displaystyle n\geq 1,$

[TABLE]

The following Lemma is straightforward.

Lemma 2.3**.**

Assume $\displaystyle\mathbf{(A2)}$ holds. Then,

• (i)

There exists $\displaystyle\bar{H}\in\mathbb{R}$ such that

[TABLE]

• (ii)

The collection $\displaystyle\left\{H_{n}\right\}_{n=1}^{\infty}$ is a non decreasing sequence of functions that converges pointwise to $\displaystyle H_{0}$. In addition, for all $\displaystyle n\in\mathbb{N}^{*}$, the map $\displaystyle H_{n}$ is a $\displaystyle C^{1}(0,\infty)$ strictly convex function that satisfies

[TABLE]

• (iii)

Let $\displaystyle t>0$. If $\displaystyle\left\{H_{n}(t)\right\}_{n=1}^{\infty}$ is uniformly bounded above by a constant $\displaystyle c_{0}$ then

[TABLE]

and $\displaystyle t=1$.

2.3. **Hypothesis on the underlying sets of pseudo-gradients. **

We recall that in [2], the construction of $\displaystyle\nabla_{\mathcal{S}^{\tau}}u$ has relied on hypothesis on the underlying sets $\displaystyle\mathcal{S}^{\tau}$ that we summarize in Hypothesis (H1) below.

Hypothesis (H1).

A collection $\displaystyle\{\mathfrak{A}_{n}\}^{\infty}_{n=1}$ of subsets of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ satisfies Hypothesis (H1) if

• (i)

$\displaystyle\mathfrak{A}_{n}$ of a finite dimensional subspace of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ for each $\displaystyle n\in\mathbb{N^{*}}$.

• (ii)

The map $\displaystyle\nabla\varphi$ has a countable range whenever $\displaystyle\varphi\in\mathfrak{A}_{n}$, for any $\displaystyle n\in\mathbb{N}^{*}$.

• (iii)

The set $\displaystyle\cup_{n\in\mathbb{N}^{*}}\mathfrak{A}_{n}$ is dense in $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$.

• (iv)

For $\displaystyle i\leq j$, we have the inclusion $\displaystyle\mathfrak{A}_{i}\subset\mathfrak{A}_{j}.$

An explicit construction of sets satisfying Hypothesis (H1) is provided in [2]. Here, we build on the conditions of Hypothesis (H1) and we relax conditions on the underlying sets:

Hypothesis (H2).

A collection $\displaystyle\{\mathfrak{Q}_{n}\}^{\infty}_{n=1}$ of subsets of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ satisfies Hypothesis (H2) if

• (i)

$\displaystyle\text{Span}(\mathfrak{Q}_{n})$ is of finite dimension and $\displaystyle\mathfrak{Q}_{n}$ is a non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$.

• (ii)

The map $\displaystyle\operatorname{div}\varphi$ is non-degenerate whenever $\displaystyle\varphi\in\mathfrak{Q}_{n}$, for any $\displaystyle n\in\mathbb{N}^{*}$.

• (iii)

The set $\displaystyle\cup_{n\in\mathbb{N}^{*}}\mathfrak{Q}_{n}$ is dense in $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$.

• (iv)

For $\displaystyle i\leq j$, the inclusion $\displaystyle\mathfrak{Q}_{i}\subset\mathfrak{Q}_{j}$ holds.

The next lemma asserts that a collection of sets can be constructed to satisfy Hypothesis (H2).

Lemma 2.4**.**

*Assume $\displaystyle\mathbf{(A0)}$ holds. Then, there exists a collection of sets

$\displaystyle\{\mathfrak{Q}_{n}\}^{\infty}_{n=1}$ satisfying the requirements of Hypothesis (H2).*

Remark 2.5**.**

The condition $\displaystyle\mathbf{(A0)}$ in Lemma 2.4 is only needed for requirement (ii) of Hypothesis (H2).

Proof. Suppose that $\displaystyle\psi$ is a strictly convex function that is $\displaystyle C^{1}(\bar{\Omega})$ and vanishes on the boundary of $\displaystyle\Omega$ as given by Assumption (A0). Let $\displaystyle\varphi_{0}:\Omega\to\mathbb{R}^{d\times d}$ be defined by

[TABLE]

As $\displaystyle\psi$ is $\displaystyle C^{1}(\bar{\Omega})$, we have $\displaystyle\varphi_{0}\in W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ and it follows that $\displaystyle\operatorname{div}\varphi_{0}=\nabla\psi$. Thus, for almost every $\displaystyle x$ in $\displaystyle\Omega$, we have

[TABLE]

Thanks to Lemma 5.5.3 in [1], the map $\displaystyle\operatorname{div}\varphi_{0}$ is non-degenerate. Let $\displaystyle\{\mathfrak{A}_{n}\}^{\infty}_{n=1}$ be a collection of sets satisfying Hypothesis (H1). One readily checks that the family of sets defined by

[TABLE]

for $\displaystyle n\in\mathbb{N}^{*}$, satisfies hypothesis (H2).

$\displaystyle\square$

2.4. Special displacements.

To $\displaystyle\mathcal{S}\subset W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ we associate $\displaystyle\mathscr{U}_{\mathcal{S}}$, the set of all $\displaystyle u:\Omega\to\bar{\Lambda}$ measurable such that there exists $\displaystyle\bar{c}=\bar{c}(u,\Omega,\Lambda)>0$ satisfying :

[TABLE]

Remark that if $\displaystyle u\in\mathscr{U}_{\mathcal{S}}$, then $\displaystyle u$ belongs to $\displaystyle L^{\infty}(\Omega,\mathbb{R}^{d})$ since $\displaystyle u$ has values in $\displaystyle\bar{\Lambda}$ which is bounded. If $\displaystyle\text{span}(\mathcal{S})$ is of finite dimension then $\displaystyle\mathscr{U}_{\mathcal{S}}$ is the set of all measurable maps $\displaystyle u:\Omega\to\bar{\Lambda}$. In fact, the linear map $\displaystyle\text{span}(\mathcal{S})\ni\varphi\mapsto\int_{\Omega}u\operatorname{div}\varphi$ is continuous with respect to the $\displaystyle L^{q}$- norm as in finite dimension, all norms are equivalent. Therefore, we may find $\displaystyle c$ for which Equation (2.9) holds for all $\displaystyle\varphi\in\text{span}(\mathcal{S})$ and in particular for all $\displaystyle\varphi\in\mathcal{S}$.

At any rate, $\displaystyle\mathscr{U}_{\mathcal{S}}$ contains $\displaystyle W^{1,p}(\Omega,\mathbb{R}^{d})$. Indeed, notice that for a fixed $\displaystyle u\in W^{1,p}(\Omega,\mathbb{R}^{d})$, we have, for all $\displaystyle\varphi\in\mathcal{S}$:

[TABLE]

We introduce the following set

[TABLE]

and

[TABLE]

Notice that $\displaystyle\mathscr{U}_{\mathcal{S}}^{1}=\{u\in\mathscr{U}_{\mathcal{S}}:(u,1)\in\mathscr{U}_{\mathcal{S}}^{*}\}$. This corresponds to measure preserving displacements.

2.5. Extended pseudo-projected gradient

Let $\displaystyle\mathcal{S}\subset W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ and $\displaystyle u\in\mathscr{U}_{\mathcal{S}}$. Define

[TABLE]

Consider the operator

[TABLE]

We denote by $\displaystyle\Phi_{S}(u)$ the set of maximizers of Problem (2.10).

Lemma 2.6**.**

Let $\displaystyle\mathcal{S}\subset W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ and $\displaystyle u\in\mathscr{U}_{\mathcal{S}}$.

1. (1)

We have

[TABLE] 2. (2)

If $\displaystyle\operatorname{span}(\mathcal{S})$ is finite dimensional, then $\displaystyle\mathcal{G}_{\mathcal{S}}(u)$ is nonempty.

Proof. Set

[TABLE]

As $\displaystyle\mathcal{S}\subset\operatorname{span}(\mathcal{S})$, we have $\displaystyle\mathcal{\bar{G}}_{\mathcal{S}}(u)\subset\mathcal{G}_{\mathcal{S}}(u)$. Next, let $\displaystyle G\in\mathcal{G}_{\mathcal{S}}(u)$. Assume that $\displaystyle\varphi\in\operatorname{span}(\mathcal{S})$. We may find $\displaystyle n\in\mathbb{N}$, $\displaystyle\lambda_{1},\dots,\lambda_{n}\in\mathbb{R}$ and $\displaystyle\varphi_{1},\dots,\varphi_{n}\in\mathcal{S}$ such that $\displaystyle\varphi=\sum_{i=1}^{n}\lambda_{i}\varphi_{i}.$ Then

[TABLE]

and

[TABLE]

Thus $\displaystyle G\in\mathcal{\bar{G}}_{\mathcal{S}}(u)$. We deduce that $\displaystyle\mathcal{G}_{\mathcal{S}}(u)\subset\mathcal{\bar{G}}_{\mathcal{S}}(u)$. It follows that part 1.) holds. To obtain part 2.), we use part 1.) and the Riesz Representation Theorem.

$\displaystyle\square$

The following results are essentially found in Proposition 3.1 in [2].

Proposition 2.7**.**

*Suppose that the set $\displaystyle\mathcal{S}$ is a finite dimensional subspace of

$\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ and $\displaystyle f$ is $\displaystyle C^{1}$ and strictly convex. Suppose, in addition that there exist constants $\displaystyle c_{1},c_{2},c_{3}>0$ such that*

[TABLE]

for all $\displaystyle\xi\in\mathbb{R}^{d\times d}$. Then, there exists a unique map denoted $\displaystyle\nabla_{\mathcal{S}}u$ that minimizes

[TABLE]

Moreover, $\displaystyle\nabla_{\mathcal{S}}u$ uniquely satisfies $\displaystyle G\in\mathcal{G}_{\mathcal{S}}(u)$ and $\displaystyle Df(G)\in\mathcal{S}.$

In the next Proposition, we establish similar results as in Proposition 2.7 but under weaker assumptions on $\displaystyle\mathcal{S}$ and $\displaystyle f$ (except in part 4).

Proposition 2.8**.**

Assume $\displaystyle\mathbf{(A1)}$ holds. Assume $\displaystyle\mathcal{S}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ and let $\displaystyle u\in\mathscr{U}_{\mathcal{S}}$.

1. (1)

For all $\displaystyle G\in\mathcal{G}_{\mathcal{S}}(u)$, $\displaystyle\varphi\in\mathcal{S}$, we have

[TABLE] 2. (2)

The supremum in problem (2.10) is attained. 3. (3)

A map $\displaystyle\bar{\varphi}$ belongs to $\displaystyle\Phi_{\mathcal{S}}(u)$ if and only if $\displaystyle\bar{\varphi}$ belongs to $\displaystyle\mathscr{S}_{f}$ and

[TABLE]

for all $\displaystyle\varphi\in\mathscr{S}_{f}$. 4. (4)

Suppose that the hypotheses of Proposition 2.7 are satisfied. Then we have

[TABLE]

and $\displaystyle\Phi_{\mathcal{S}}(u)=\{Df(\nabla_{\mathcal{S}}u)\}$.

Proof. 1.) Let $\displaystyle\varphi\in\mathcal{S}$ and $\displaystyle G\in\mathcal{G}_{\mathcal{S}}(u)$, By using the Legendre transformation,

[TABLE]

2.) Let $\displaystyle\varphi\in\mathcal{S}$. We use (2.9) and (2.5) to get

[TABLE]

In light of (2.11), $\displaystyle q>1$ implies that the map

[TABLE]

is $\displaystyle L^{q}-$ coercive. Moreover, the convexity of $\displaystyle f^{*}$ guarantees that $\displaystyle T$ is lower semi-continuous. The direct methods of the calculus of variations thus yield the existence of a maximizer in problem (2.10).

3.) Let $\displaystyle\bar{\varphi}\in\Phi_{\mathcal{S}}(u)$ so that $\displaystyle\bar{\varphi}\in\mathscr{S}_{f}$. Let $\displaystyle\varphi\in\mathscr{S}_{f}$ and $\displaystyle\epsilon\in(0,1)$. The convexity of $\displaystyle f^{*}$ ensures that $\displaystyle\bar{\varphi}+\epsilon(\varphi-\bar{\varphi})\in\mathscr{S}_{f}$ and the maximality property of $\displaystyle\bar{\varphi}$ implies that

[TABLE]

We rewrite (2.12), in turn, as

[TABLE]

Note that $\displaystyle grad[f^{*}](\bar{\varphi}+\epsilon(\varphi-\bar{\varphi}))$ belongs to the set $\displaystyle\partial f^{*}((\bar{\varphi}+\epsilon(\varphi-\bar{\varphi})))$ whenever $\displaystyle(\bar{\varphi}+\epsilon(\varphi-\bar{\varphi}))$ is in the domain of $\displaystyle f^{*}$. It follows that

[TABLE]

that is,

[TABLE]

We combine (2.13) and (2.14) to get

[TABLE]

First, we assume that $\displaystyle\mathbf{(A1)}$(iii)(a) holds. In light of (2.6), we have $\displaystyle grad[f^{*}](\bar{\varphi}+\epsilon(\varphi-\bar{\varphi})=grad[f^{*}](\bar{\varphi})$. Equation (2.15) becomes

[TABLE]

Second, we assume that $\displaystyle\mathbf{(A1)}$(iii)(b) holds. In light of Remark 2.1 (ii), we use the growth condition on $\displaystyle\nabla f^{*}$ in (2.4), the Lebesgue dominated convergence theorem and let $\displaystyle\epsilon$ go to $\displaystyle 0$ in (2.15) to obtain that:

[TABLE]

We next show the converse implication. Let $\displaystyle\varphi\in\mathscr{S}_{f}$ such that

[TABLE]

for all $\displaystyle\varphi\in\mathscr{S}_{f}$. We notice that, as $\displaystyle f^{*}$ is convex, the range of the map $\displaystyle grad[f^{*}](\bar{\varphi})$ lies in the sub-differential of $\displaystyle f^{*}$ so that $\displaystyle f^{*}(\varphi)-f^{*}(\bar{\varphi})\geq grad[f^{*}](\bar{\varphi})\left(\varphi-\bar{\varphi}\right)$ for all $\displaystyle\varphi\in\mathscr{S}_{f}$. Then, the inequality (2.16) implies that

[TABLE]

for all $\displaystyle\varphi\in\mathscr{S}_{f}$, that is,

[TABLE]

for all $\displaystyle\varphi\in\mathscr{S}_{f}$. We conclude that $\displaystyle\bar{\varphi}\in\Phi_{\mathcal{S}}(u)$.

4.) Thanks to Proposition 2.7, $\displaystyle Df(\nabla_{\mathcal{S}}u)\in\mathcal{S}$. Next, we set $\displaystyle\varphi_{0}:=Df(\nabla_{\mathcal{S}}u)$. By definition of $\displaystyle f^{*}$,

[TABLE]

for all $\displaystyle\varphi\in\mathcal{S}$. As $\displaystyle f$ is convex and $\displaystyle\varphi_{0}=Df(\nabla_{\mathcal{S}}u)$, we have

[TABLE]

Thus,

[TABLE]

and

[TABLE]

We deduce that $\displaystyle\varphi_{0}\in\Phi_{\mathcal{S}}(u)$. Since $\displaystyle f^{*}$ is strictly convex, we conclude that $\displaystyle\Phi_{\mathcal{S}}(u)=\{Df(\nabla_{\mathcal{S}}u)\}$ and moreover, $\displaystyle\int_{\Omega}f(\nabla_{\mathcal{S}}u)=V^{f}_{\mathcal{S}}(u)$.

$\displaystyle\square$

In the next Proposition, we establish a convergence result in the spirit of (1.7). We also connect the operator $\displaystyle V_{\mathcal{S}}^{f}$ with the usual notions of gradient and total variation.

Proposition 2.9**.**

Assume $\displaystyle\mathbf{(A1)}$ holds. Assume that $\displaystyle\mathcal{S}_{n}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ for each $\displaystyle n\geq 1$. The following holds.

1. (1)

If $\displaystyle\left\{\mathcal{S}_{n}\right\}_{n=1}^{\infty}$ is a monotonically increasing family of subsets of some set $\displaystyle\mathcal{S}_{0}$ and $\displaystyle\cup_{n\in\mathbb{N}^{*}}\mathcal{S}_{n}$ is dense in $\displaystyle\mathcal{S}_{0}$ with respect to the $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ norm then

[TABLE]

for any $\displaystyle u\in\mathscr{U}_{\mathcal{S}_{0}}.$ 2. (2)

If $\displaystyle\mathcal{S}=W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ and $\displaystyle u\in W^{1,p}(\Omega,\mathbb{R}^{d})$ then $\displaystyle V_{\mathcal{S}}^{f}[u]=\int_{\Omega}f(\nabla u)dx$. 3. (3)

Assume $\displaystyle u\in BV(\Omega,\mathbb{R}^{d\times d})$ and $\displaystyle f(\xi)=|\xi|$ for all $\displaystyle\xi\in\mathbb{R}^{d\times d}$. If

$\displaystyle\mathcal{S}=W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$* then $\displaystyle V_{\mathcal{S}}^{f}[u]$ is the total variation of $\displaystyle u$.*

Remark 2.10**.**

A consequence of Proposition 2.9 is the following : If the sequence of sets $\displaystyle\{\mathcal{S}_{n}\}_{n\in\mathbb{N}^{*}}$ is monotonically increasing to $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ and $\displaystyle u\in W^{1,p}(\Omega,\mathbb{R}^{d})$ we have

[TABLE]

Proof.

1.) Recall that

[TABLE]

As $\displaystyle\left\{\mathcal{S}_{n}\right\}_{n=1}^{\infty}$ is a monotonically increasing, $\displaystyle\lim_{n\rightarrow\infty}V_{S_{n}}^{f}[u]$ exists. Moreover, since $\displaystyle\mathcal{S}_{n}\subset\mathcal{S}_{0}$ for all $\displaystyle n\geq 1$,

[TABLE]

Let $\displaystyle\epsilon>0$ and choose $\displaystyle\varphi^{\epsilon}\in\mathcal{S}_{0}$ such that

[TABLE]

Let $\displaystyle\{\varphi^{\epsilon}_{n}\}_{n\in\mathbb{N}^{*}}$ be a sequence converging to $\displaystyle\varphi^{\epsilon}$ in $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ and such that $\displaystyle\varphi_{n}^{\epsilon}\in\mathcal{S}_{n}$ for all $\displaystyle n\in\mathbb{N}^{*}$. Then, we use the growth conditions on $\displaystyle f^{*}$ in (2.4) and (2.5), the continuity of $\displaystyle f^{*}$ on its domain and the Lebesgue dominated convergence theorem to obtain that

[TABLE]

It follows that

[TABLE]

As $\displaystyle\epsilon$ is arbitrary, we have

[TABLE]

From (2.17) and (2.18), we conclude that $\displaystyle\lim_{n\rightarrow\infty}V_{\mathcal{S}_{n}}^{f}[u]=V_{\mathcal{S}_{0}}^{f}[u].$

2.) One has

[TABLE]

The inequality above is obtained by using the definition of the Legendre transform $\displaystyle f^{*}$ of $\displaystyle f$. Let $\displaystyle\bar{\varphi}\in\partial f(\nabla u)$. Then $\displaystyle f^{*}(\bar{\varphi})+f(\nabla u)=\nabla u\cdot\bar{\varphi}$. Thanks to the growth conditions (2.2) and (2.3) on $\displaystyle f$, it holds that $\displaystyle\bar{\varphi}\in L^{q}(\Omega,\mathbb{R}^{d\times d})$. Since $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ is dense in $\displaystyle L^{q}(\Omega,\mathbb{R}^{d\times d})$ for the $\displaystyle L^{q}(\Omega,\mathbb{R}^{d\times d})$ norm, we get

[TABLE]

We conclude that $\displaystyle V^{f}_{S}[u]=\int_{\Omega}f(\nabla u)\;dx$.

3.) The total variation of $\displaystyle u\in BV(\Omega,\mathbb{R}^{d\times d})$ is

[TABLE]

while, using the Legendre transform of $\displaystyle f(\xi)=|\xi|$, we obtain

[TABLE]

It follows directly from (2.19) and (2.20) that $\displaystyle\|Du\|(\Omega)\leq V^{f}_{\mathcal{S}}[u]$. The converse inequality $\displaystyle\|Du\|(\Omega)\geq V^{f}_{\mathcal{S}}[u]$ follows from the density of $\displaystyle C_{c}^{1}({\Omega,\mathbb{R}^{d\times d}})$ in

$\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ and an argument similar to the one made in the proof of (2) in the proposition.

$\displaystyle\square$

3. Minimization with general displacements.

We consider the following :

[TABLE]

This problem will be studied via a dual problem that we will formulate next. We assume in this section that Assumption (A2) holds.

3.1. An auxiliary problem

For $\displaystyle l,k:\mathbb{R}^{d}\rightarrow(-\infty,\infty]$, define for $\displaystyle u,v\in\mathbb{R}^{d}$

[TABLE]

and

[TABLE]

Under Assumption (A2), it is known that $\displaystyle((l^{\#})_{\#})^{\#}=l^{\#}$ and $\displaystyle((k_{\#})^{\#})_{\#}=k_{\#}$ (see for instance Lemma A1 of [10] ). Call $\displaystyle\mathcal{C}$ the set of all functions $\displaystyle(k,l)$ with $\displaystyle k,\;l:\mathbb{R}^{d}\rightarrow\mathbb{R}\cup\{\infty\}$ Borel measurable, finite at least at one point, and satisfying $\displaystyle l\equiv\infty$ on $\displaystyle\mathbb{R}^{d}\setminus\bar{\Lambda}$ and such that

[TABLE]

Call $\displaystyle\mathcal{C}^{\prime}$ the set of all functions $\displaystyle(k,l)\in\mathcal{C}$ such that $\displaystyle l=k_{\#}$ and $\displaystyle k=l^{\#}$. The set $\displaystyle\mathcal{C}^{\prime}$ is nonempty. Indeed, $\displaystyle(\chi_{\bar{\Lambda}}^{\#},(\chi_{\bar{\Lambda}}^{\#})_{\#})\in\mathcal{C}^{\prime}$ as $\displaystyle((\chi_{\bar{\Lambda}}^{\#})_{\#}))^{\#}=\chi_{\bar{\Lambda}}^{\#}$.

Let $\displaystyle\mathscr{A}$ be the set of $\displaystyle(k,l,\varphi)$ such that $\displaystyle(k,l)\in\mathcal{C}$ and $\displaystyle\varphi\in\mathcal{S}$. Consider the following functional defined on $\displaystyle\mathscr{A}$:

[TABLE]

The following problem will play an important role in this section:

[TABLE]

The value in Equation (3.5) is the opposite of the value in the following equation:

[TABLE]

Let $\displaystyle\mathscr{A}^{\prime}$ denote the subset of $\displaystyle\mathscr{A}$ consisting of all $\displaystyle(k,l,\varphi)\in\mathscr{A}$ that satisfy $\displaystyle(k,l)\in\mathcal{C}^{\prime}$. It holds that

[TABLE]

Indeed, the key observation to this end is that for $\displaystyle(k,l,\varphi)\in\mathscr{A}$, one has $\displaystyle l\geq k_{\#}$ and $\displaystyle k\geq(k_{\#})^{\#}$ so that

[TABLE]

For $\displaystyle R>0$, we set

[TABLE]

Lemma 3.1**.**

Assume $\displaystyle\mathbf{(A1)}$, $\displaystyle\mathbf{(A2)}$ and $\displaystyle\mathbf{(A3)}$ hold. Let $\displaystyle(k,l,\varphi)\in\mathscr{A}_{R}$. Set $\displaystyle s_{l}:=-\inf_{u\in\bar{\Lambda}}l(u)$. Then,

[TABLE]

Moreover, there exists $\displaystyle M:=M(R,F,f,\Omega,\Lambda)>0$ such that

[TABLE]

Proof. As $\displaystyle\Lambda$ is bounded and $\displaystyle l$ is convex, we choose $\displaystyle u_{l}\in\overline{\Lambda}$ such that $\displaystyle-l(u_{l})=s_{l}$. Since $\displaystyle k:=l^{\#}$, in view of (3.2), we have

[TABLE]

Using the last inequality in (3.9), one gets

[TABLE]

We have used the fact that $\displaystyle u_{l}$ is a constant vector and $\displaystyle\varphi\in W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$ to obtain the equality in (3.11). Hence,

[TABLE]

Thus,

[TABLE]

Thanks to Lemma 2.2 (iii), $\displaystyle s_{l}$ is bounded uniformly in $\displaystyle l$.

$\displaystyle\square$

Lemma 3.2**.**

Assume $\displaystyle\mathbf{(A1)}$, $\displaystyle\mathbf{(A2)}$ and $\displaystyle\mathbf{(A3)}$ hold.

1. (1)

There exists $\displaystyle M>0$ such that for all $\displaystyle(k,l,\varphi)\in\mathscr{A}_{R}$ one has

[TABLE] 2. (2)

There exist $\displaystyle a_{0},b_{0},c_{0}>0$ such that for all $\displaystyle(k,l,\varphi)\in\mathscr{A}_{R}$, the map $\displaystyle k$ is $\displaystyle r^{*}$-Lipschitz, and one has for all $\displaystyle v\in\mathbb{R}^{d}$

[TABLE]

Proof. 1.) Recall that for $\displaystyle(k,l,\varphi)\in\mathcal{A}_{R}$, one has

[TABLE]

By Lemma 3.1, for all $\displaystyle(k,l,\varphi)\in\mathcal{A}_{R}$, if we define $\displaystyle s_{l}:=-\inf_{u\in\bar{\Lambda}}l(u)$, we get

[TABLE]

Rearranging the terms, we get:

[TABLE]

By definition of $\displaystyle s_{l}$ we also have $\displaystyle-s_{l}\mathcal{L}^{d}{(\Omega)}\leq\int_{\Lambda}l(y)dy$ and thus

[TABLE]

Lemma 3.1 also ensures that. We consider the negative part of $\displaystyle l$ defined by $\displaystyle l^{-}:=\max\{-l,0\}$ and note that

[TABLE]

Observe that, by the definition of $\displaystyle s_{l}$, we have $\displaystyle l^{-}\leq|s_{l}|$. This, combined with (3.14), (3.15) and (3.8) yields (3.12).

2.) Let $\displaystyle(k,l,\varphi)\in\mathscr{A}_{R}$. Since $\displaystyle k=l^{\#}$, by equation (3.2), $\displaystyle k$ is a $\displaystyle r^{*}$-Lipschitz as $\displaystyle\Lambda$ has diameter less or equal to $\displaystyle r^{*}$. Next

[TABLE]

As $\displaystyle s_{l}$ is uniformly bounded, the growth condition on $\displaystyle H$ ensures that $\displaystyle|k(0)|$ is uniformly bounded say by some $\displaystyle b_{0}>0$. We get then the inequality $\displaystyle k(v)\leq b_{0}+r^{*}|v|$ for all $\displaystyle v\in\mathbb{R}^{d}$.

Because of the hypothesis on the domain $\displaystyle\Lambda$, we take $\displaystyle a_{0}>0$ such that $\displaystyle B(0,a_{0})\subset\Lambda$. As $\displaystyle(k,l,\varphi)\in\mathcal{A}_{R}$, we use relation (3.4) to obtain for $\displaystyle v\neq 0$

[TABLE]

Thanks to equation (3.12), $\displaystyle\int_{\Lambda}|l|dy$ is uniformly bounded in $\displaystyle l$. We use in addition the fact that $\displaystyle l$ is bounded to deduce that $\displaystyle\sup_{y\in\bar{B}(0,a_{0})}|l|(y)$ is bounded by a constant independent of $\displaystyle l$ (see for instance Theorem 1, p. 236 in [7]). Thus equation (3.16) implies that there exists $\displaystyle c_{0}>0$ such that $\displaystyle k(v)\geq a_{0}|v|-c_{0}$ for all $\displaystyle v\in\mathbb{R}^{d}$.

$\displaystyle\square$

Proposition 3.3**.**

Assume $\displaystyle\mathbf{(A1)}$, $\displaystyle\mathbf{(A2)}$, and $\displaystyle\mathbf{(A3)}$ hold. Assume $\displaystyle\mathcal{S}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$. Then, the functional $\displaystyle J$ admits a minimizer $\displaystyle(k_{0},l_{0},\varphi_{0})$ in $\displaystyle\mathscr{A}^{\prime}$.

Proof. Let $\displaystyle(\bar{k},\bar{l},\bar{\varphi})\in\mathscr{A}$. Set $\displaystyle R=J(\bar{k},\bar{l},\bar{\varphi})$. Take a minimizing sequence

$\displaystyle\{(k_{n},l_{n},\varphi_{n})\}_{n\in\mathbb{N}^{*}}$ of Problem (3.5) that is in $\displaystyle\mathscr{A}_{R}$. By Lemma 3.1 and the growth condition on $\displaystyle f^{*}$ we may assume without loss of generality that $\displaystyle\{\varphi_{n}\}^{\infty}_{n=1}$ converges to some $\displaystyle\varphi_{0}\in\mathcal{S}$ weakly in $\displaystyle L^{q}(\Omega,\mathbb{R}^{d\times d})$. Since $\displaystyle\text{Span}(\mathcal{S})$ is finite dimensional, $\displaystyle\{\varphi_{n}\}^{\infty}_{n=1}$ converges to some $\displaystyle\varphi_{0}\in\mathcal{S}$ strongly in the $\displaystyle L^{q}(\Omega,\mathbb{R}^{d\times d})$ norm. We deduce

[TABLE]

From Lemma 3.2, as $\displaystyle l_{n}$ is convex, we use Ascoli-Arzéla Theorem together with Theorem 1, p. 236 in [7] to deduce that up to a subsequence, we may assume that $\displaystyle(k_{n},l_{n})$ converges locally uniformly $\displaystyle\mathbb{R}^{d}\times\Lambda$ to $\displaystyle(k_{0},l_{0})\in\mathcal{C}^{\prime}$ on . The Lebesgue dominated convergence together with inequality (3.13) yield

[TABLE]

Since $\displaystyle\{l_{n}\}^{\infty}_{n=1}$ is uniformly bounded below (thanks to Lemma 3.1 ), by Fatou’s Lemma we get

[TABLE]

By inequalities (3.17), (3.18) and (3.19), we get

[TABLE]

and $\displaystyle(k_{0},l_{0},\varphi_{0})$ is a minimizer of $\displaystyle J$ over $\displaystyle\mathscr{A}^{\prime}$.

$\displaystyle\square$

3.2. A uniqueness result

Here, we prove the main result of this section. We will need the following lemma which is in the spirit of Lemma 4.3 and Lemma 4.4 in [2]. A proof of Lemma 3.4 is given in subsection 5.1.

Lemma 3.4**.**

Assume assumption (A2) holds. Consider a lower semicontinuous function $\displaystyle l_{0}:\mathbb{R}^{d}\to\bar{\mathbb{R}}$ such that $\displaystyle\inf_{\bar{\Lambda}}l_{0}>-\infty$; $\displaystyle l_{0}$ is finite on $\displaystyle\Lambda$ and $\displaystyle l_{0}\equiv+\infty$ on $\displaystyle\mathbb{R}^{d}\setminus\bar{\Lambda}$. Set $\displaystyle k_{0}=({l_{0}})^{\#}$. Let $\displaystyle v\in\mathbb{R}^{d}$ be such that $\displaystyle k_{0}$ is differentiable at $\displaystyle v$.

1. (1)

There exist unique $\displaystyle u_{0}\in\bar{\Lambda}$ and $\displaystyle t_{0}>0$ such that $\displaystyle k_{0}(v)=-t_{0}l_{0}(u_{0})-H(t_{0})-u_{0}\cdot v.$ In addition, $\displaystyle u_{0}$ and $\displaystyle t_{0}$ are characterized by $\displaystyle u_{0}=\nabla k_{0}(v)$ and $\displaystyle H^{\prime}(t_{0})+l(u_{0})=0$. 2. (2)

Let $\displaystyle\hat{l}\in C_{b}(\mathbb{R}^{d})$ and let $\displaystyle 1\geq\epsilon>0$. Define $\displaystyle l_{\epsilon}=l_{0}+\epsilon\hat{l}$ and $\displaystyle k_{\epsilon}={\left(l_{\epsilon}\right)}^{\#}$.

1. (a)

There exists a constant $\displaystyle M$ independent of $\displaystyle v$ and $\displaystyle\epsilon$ such that

[TABLE] 2. (b)

We have

[TABLE]

Next, we give the main result of this section.

Theorem 3.5**.**

Assume $\displaystyle\mathbf{(A1)}$, $\displaystyle\mathbf{(A2)}$, and $\displaystyle\mathbf{(A3)}$ hold. Assume $\displaystyle\mathcal{S}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$. Assume $\displaystyle F$ satisfies the condition $\displaystyle\textbf{(ND)}_{\mathcal{S}}$. Then, Problems (3.1) and (3.6) are dual. Problem (3.6) admits a maximizer $\displaystyle(k_{0},l_{0},\varphi_{0})$ with $\displaystyle k_{0}=l_{0}^{\#}$ and $\displaystyle l_{0}=(k_{0})_{\#}$. Problem (3.1) admits a unique minimizer $\displaystyle(u_{0},\beta_{0})$. Moreover $\displaystyle u_{0}$ satisfies

[TABLE]

Proof. Step 1. For $\displaystyle(u,\beta)\in\mathscr{U}^{*}_{\mathcal{S}}$ and $\displaystyle(k,l,\varphi)\in\mathscr{A}$, one has

[TABLE]

Thus $\displaystyle I(u,\beta)\geq-J(k,l,\varphi)$ with equality if and only if $\displaystyle\varphi\in\Phi_{\mathcal{S}}(u)$ and

[TABLE]

Note that if $\displaystyle k$ is convex, the map $\displaystyle\nabla k(F+\operatorname{div}\varphi)$ is well defined as the map $\displaystyle F+\operatorname{div}\varphi$ is non-degenerate. Using Lemma 3.4(i), it follows that if $\displaystyle k$ is convex, then $\displaystyle I(u,\beta)=-J(k,l,\varphi)$ if and only if

[TABLE]

Step 2. Thanks to equation (3.7), we may find a maximizer $\displaystyle(k_{0},l_{0},\varphi_{0})$ of Problem (3.5) satisfying $\displaystyle k_{0}=l_{0}^{\#}$ and $\displaystyle l_{0}=(k_{0})_{\#}$. The function $\displaystyle u_{0}=\nabla k_{0}(F+\operatorname{div}\varphi_{0})$ is well defined as $\displaystyle k_{0}$ is convex and we set $\displaystyle\beta_{0}=(H^{\prime})^{-1}(-l(u_{0}))$. We are to show that $\displaystyle(u_{0},\beta_{0})\ \in\mathscr{U}^{*}_{\mathcal{S}}$ and $\displaystyle\varphi_{0}\in\Phi_{\mathcal{S}}(u_{0})$.

Step 3. Let $\displaystyle\bar{l}\in C_{c}(\mathbb{R}^{d})$. For $\displaystyle\epsilon\in(0,1)$, define $\displaystyle l_{\epsilon}=l_{0}+\epsilon\bar{l}$ and $\displaystyle k_{\epsilon}=(l_{\epsilon})^{\#}$. Using Lemma 3.4, one has

[TABLE]

Since $\displaystyle J(k_{0},l_{0},\varphi_{0})\leq J(k_{\epsilon},l_{\epsilon},\varphi_{0})$, we deduce that $\displaystyle-\int_{\Lambda}\bar{l}\;dy+\int_{\Omega}\beta_{0}\bar{l}(u_{0})\;dx\leq 0$. As we can replace $\displaystyle\bar{l}$ by $\displaystyle-\bar{l}$, one deduces that $\displaystyle\int_{\Lambda}\bar{l}\;dy=\int_{\Omega}\beta_{0}\bar{l}(u_{0})\;dx$. Therefore $\displaystyle(u_{0},\beta_{0})\in\mathscr{U}_{S}^{*}$.

Step 4. Let $\displaystyle\varphi\in\mathcal{S}$. For $\displaystyle\epsilon\in(0,1)$, set $\displaystyle\varphi_{\epsilon}=\epsilon\varphi+(1-\epsilon)\varphi_{0}$. By the convexity of $\displaystyle\mathcal{S}$, the map $\displaystyle\varphi_{\epsilon}$ belongs to $\displaystyle\mathcal{S}$. As $\displaystyle J(k_{0},l_{0},\varphi_{0})\leq J(k_{0},l_{0},\varphi_{\epsilon})$, we have

[TABLE]

Thanks to Lemma 3.4, Inequality (3.22) implies

[TABLE]

It follows from Proposition 2.8 that $\displaystyle\varphi_{0}\in\Phi_{\mathcal{S}}(u_{0})$.

Step 5. Since $\displaystyle(u_{0},\beta_{0})\ \in\mathscr{U}^{*}_{\mathcal{S}}$, $\displaystyle\varphi_{0}\in\Phi_{\mathcal{S}}(u_{0})$, $\displaystyle u_{0}=\nabla k_{0}(F+\operatorname{div}\varphi_{0})$, and $\displaystyle\beta_{0}=(H^{\prime})^{-1}(-l(u_{0}))$, we deduce that $\displaystyle I(u_{0},\beta_{0})=J(k_{0},l_{0},\varphi_{0})$ and $\displaystyle u_{0}$ is a minimizer of Problem (3.1) thanks to relation (3.20). Suppose $\displaystyle(u_{1},\beta_{1})\ \in\mathscr{U}^{*}_{S}$ is another minimizer of Problem (3.1). Then we have $\displaystyle I(u_{1},\beta_{1})=J(k_{0},l_{0},\varphi_{0})$ and by relation (3.20), we get $\displaystyle u_{1}=\nabla k_{0}(F+\operatorname{div}\varphi_{0})$ which implies $\displaystyle u_{1}=u_{0}$. Next the strict convexity of $\displaystyle H$ yields that $\displaystyle\beta_{0}=\beta_{1}$. We conclude that $\displaystyle(u_{0},\beta_{0})$ is the unique minimizer of Problem (3.1) and $\displaystyle u_{0}$ is characterized by

[TABLE]

$\displaystyle\square$

Corollary 3.6**.**

Assume $\displaystyle\mathbf{(A0)}$, $\displaystyle\mathbf{(A1)}$, $\displaystyle\mathbf{(A2)}$, and $\displaystyle\mathbf{(A3)}$ hold. Assume $\displaystyle\mathcal{S}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ and $\displaystyle\nabla\varphi$ is non-degenerate whenever $\displaystyle\varphi\in\mathcal{S}$. Suppose $\displaystyle F$ has a countable range (thus degenerate) Then, $\displaystyle F$ satisfies the condition $\displaystyle\textbf{(ND)}_{\mathcal{S}}$ and problem (3.1) admits a unique solution.

Corollary 3.7**.**

Assume $\displaystyle\mathbf{(A1)}$, $\displaystyle\mathbf{(A2)}$, and $\displaystyle\mathbf{(A3)}$ hold. Assume $\displaystyle\mathcal{S}$ is a finite dimensional subspace of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ and $\displaystyle\nabla\varphi$ has a countable range whenever $\displaystyle\varphi\in\mathcal{S}$. Suppose $\displaystyle F$ is non-degenerate. Then, $\displaystyle F$ satisfies the condition $\displaystyle\textbf{(ND)}_{\mathcal{S}}$ and problem (3.1) admits a unique solution.

4. The incompressible case

Throughout this section, we assume that $\displaystyle\mathcal{S}$ is a subset of $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$. We consider the following problem:

[TABLE]

and we recall that the set $\displaystyle\mathcal{U}_{\mathcal{S}}^{1}$ is defined as

[TABLE]

We assume $\displaystyle\mathcal{L}^{d}(\Omega)=\mathcal{L}^{d}(\Lambda)$ so that $\displaystyle\mathscr{U}_{\mathcal{S}}^{1}$ is non-empty.

4.1. Existence and uniqueness via duality

We study Problem (4.1) via duality. Let $\displaystyle u\in\mathscr{U}^{1}_{\mathcal{S}}$, $\displaystyle\varphi\in\mathcal{S}$, $\displaystyle l\in C(\Lambda)$ and $\displaystyle k:\mathbb{R}^{d}\to\mathbb{R}$ satisfy $\displaystyle k(v)+l(u)\geq u\cdot v$ for all $\displaystyle u\in\Lambda$ and all $\displaystyle v\in\mathbb{R}^{d}$. One has

[TABLE]

This suggests that we consider the dual problem

[TABLE]

with $\displaystyle A_{0}$ being the set of all $\displaystyle(k,l,\varphi)$ such that $\displaystyle\varphi\in\mathcal{S}$, $\displaystyle l\in C(\Lambda)$, $\displaystyle\inf_{\Lambda}l=0$ and $\displaystyle k:\mathbb{R}^{d}\to\mathbb{R}$ satisfies $\displaystyle k(v)+l(u)\geq u\cdot v$ for all $\displaystyle u\in\bar{\Lambda},$ and all $\displaystyle v\in\mathbb{R}^{d}$. Remark that we have

[TABLE]

4.1.1. Existence and regularity of minimizers of Problem (4.5)

Denote by $\displaystyle\mathcal{C}$ the set of all $\displaystyle(k,l)$ such that $\displaystyle k:\mathbb{R}^{d}\to\mathbb{R}$ and $\displaystyle l:\mathbb{R}^{d}\to\mathbb{R}\cup\{\infty\}$ satisfy

[TABLE]

and $\displaystyle l\equiv\infty$ on $\displaystyle\mathbb{R}^{d}\setminus\bar{\Lambda}$. Consider the subset $\displaystyle\mathcal{C}_{0}$ of $\displaystyle\mathcal{C}$ consisting of $\displaystyle(k,l)\in\mathcal{C}$ such that $\displaystyle l\in C(\Lambda)$ and $\displaystyle\inf_{\Lambda}l=0$. The following lemma is standard:

Lemma 4.1**.**

Let $\displaystyle(k,l)\in\mathcal{C}$. It holds that $\displaystyle(l^{*},l^{**})\in\mathcal{C}$, $\displaystyle l^{*}\leq k$, $\displaystyle 0\leq l^{**}\leq l$ and $\displaystyle l^{***}=l^{*}$. If $\displaystyle(k,l)\in\mathcal{C}_{0}$ then $\displaystyle l^{*}(0)=0$.

Let us denote by $\displaystyle C_{0}^{\prime}$ the set of all $\displaystyle(k,l)\in C_{0}$ such that $\displaystyle l^{*}=k$, $\displaystyle k^{*}=l$, $\displaystyle k(0)=0$, and $\displaystyle l\geq 0$, and by $\displaystyle A^{\prime}_{0}$ the set of all $\displaystyle(k,l,\varphi)$ with $\displaystyle(k,l)\in C_{0}^{\prime}$ and $\displaystyle\varphi\in\mathcal{S}$. Remark that an element in $\displaystyle C_{0}^{\prime}$ is the couple $\displaystyle(\chi_{\bar{\Lambda}},(\chi_{\bar{\Lambda}})^{*})$. Hence $\displaystyle A^{\prime}_{0}$ is nonempty when $\displaystyle\mathcal{S}$ is nonempty. One readily checks that, in light of Lemma 4.1, Problem (4.5) has the same infimum value as

[TABLE]

We recall that $\displaystyle r^{*}$ is such that $\displaystyle B(0,1/r^{*})\subset\Lambda\subset B(0,r^{*}/2)$;

Lemma 4.2**.**

Assume $\displaystyle\mathbf{(A1)}$ and $\displaystyle\mathbf{(A3)}$ hold. Assume that the set $\displaystyle\mathcal{S}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$. Then, problem (4.8) admits a minimizer $\displaystyle(k_{0},l_{0},\varphi_{0})\in A_{0}^{\prime}$ with $\displaystyle k_{0}$ convex and $\displaystyle r^{*}$-Lipschitz and $\displaystyle k_{0}(0)=0$.

Proof. Consider a minimizing sequence $\displaystyle\left\{(k_{n},l_{n},\varphi_{n})\right\}_{n=1}^{\infty}$ of Problem (4.8). Since $\displaystyle k_{n}=l_{n}^{*}$ and $\displaystyle l_{n}=(k_{n})^{*}$, $\displaystyle k_{n}$ is $\displaystyle r^{*}$-Lipschitz. As $\displaystyle k_{n}(0)=0$, we use Ascoli-Arzéla theorem to deduce that a subsequence of $\displaystyle\left\{k_{n}\right\}_{n=1}^{\infty}$ converges locally uniformly to some $\displaystyle k_{0}$. Next, using the growth condition (2.5) on $\displaystyle f^{*}$ as well as the facts that $\displaystyle k_{n}$ is $\displaystyle r^{*}$-Lipschitz, $\displaystyle k_{n}(0)=0$, we establish the following estimate :

[TABLE]

As the left hand side of (4.9) is bounded, $\displaystyle l_{n}\geq 0$ and $\displaystyle\mathcal{S}$ is finite dimensional, we deduce from (4.9) that a subsequence of $\displaystyle\left\{\varphi_{n}\right\}_{n=1}^{\infty}$ converges strongly to some $\displaystyle\varphi_{0}$ in $\displaystyle W_{0}^{1,q}(\Omega,\mathbb{R}^{d\times d})$. Invoking (4.9) again, we show that $\displaystyle\left\{\int_{\Lambda}l_{n}(y)dy\right\}_{n=1}^{\infty}$ is bounded. This, combined with the fact that $\displaystyle l_{n}$ is non negative and convex, yields the existence of a subsequence of $\displaystyle\left\{l_{n}\right\}_{n=1}^{\infty}$ that converges locally uniformly to some $\displaystyle l_{0}$ (see for instance Theorem 1, p. 236 in [7]). One readily checks that $\displaystyle(k_{0},l_{0},\varphi_{0})\in A_{0}^{\prime}$. We next exploit lower semi-continuity properties of the functional $\displaystyle J$ to conclude that $\displaystyle(k_{0},l_{0},\varphi_{0})$ is a minimizer of $\displaystyle J$ over $\displaystyle A_{0}^{\prime}$.

$\displaystyle\square$

4.1.2. A duality result

We have the following theorem.

Theorem 4.3**.**

Assume $\displaystyle\mathbf{(A1)}$ and $\displaystyle\mathbf{(A3)}$ hold. Assume $\displaystyle\mathcal{S}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$. Suppose that the map $\displaystyle F$ satisfies the condition $\displaystyle\textbf{(ND)}_{\mathcal{S}}$. Then Problems (4.1) and (4.6) are dual. Problem (4.6) admits a maximizer $\displaystyle(k_{0},l_{0},\varphi_{0})$ with $\displaystyle k_{0}=l_{0}^{*}$ and $\displaystyle l_{0}=(k_{0})^{*}$. Problem (4.1) admits a unique minimizer $\displaystyle u_{0}$. Moreover $\displaystyle u_{0}$ satisfies

[TABLE]

Proof. Suppose $\displaystyle u\in\mathscr{U}_{S}^{1}$ and $\displaystyle(k,l,\varphi)\in A_{0}$. Using (4.3) and (4.4), we see that $\displaystyle I_{0}(u)\geq-J(k,l,\varphi)$ with equality if and only if $\displaystyle\varphi\in\Phi_{S}(u)$ and $\displaystyle l(u)+k(F+\operatorname{div}\varphi)=u\cdot(F+\operatorname{div}\varphi)$ for almost every $\displaystyle x\in\Omega$. The latter condition reduces to $\displaystyle u(x)=\nabla k(F(x)+\operatorname{div}\varphi(x))$ if $\displaystyle k$ is convex, under the assumption $\displaystyle F+\operatorname{div}\varphi$ is non-degenerate. Now, let $\displaystyle(k_{0},l_{0},\varphi_{0})\in A_{0}^{\prime}$ be a minimizer of $\displaystyle J$ over $\displaystyle A_{0}$. Since $\displaystyle F+\operatorname{div}\varphi_{0}$ is non-degenerate and $\displaystyle k_{0}$ is convex, the map $\displaystyle u_{0}=\nabla k_{0}(F+\operatorname{div}\varphi_{0})$ is well defined.

Variation around $\displaystyle l_{0}$. Let $\displaystyle\bar{l}\in C_{c}(\mathbb{R}^{d})$. For $\displaystyle\epsilon\in(0,1)$, set $\displaystyle l_{\epsilon}=l_{0}+\epsilon\bar{l}$ and $\displaystyle k_{\epsilon}=(l_{\epsilon})^{*}.$ Let $\displaystyle v\in\mathbb{R}^{d}$ be a point where $\displaystyle k_{0}$ is differentiable. Using the measurable selection theorem, one deduces that there exists $\displaystyle T_{\epsilon}:\mathbb{R}^{d}\to\mathbb{R}^{d}$ measurable such that for all $\displaystyle\epsilon\in[0,1)$

[TABLE]

Then, for $\displaystyle\epsilon\in(0,1),$ we have

[TABLE]

and

[TABLE]

Moreover,

[TABLE]

We refer the reader to Lemma 5.3 for (4.10)- (4.12). Hence, as

[TABLE]

using again (4.12), one has

[TABLE]

Since $\displaystyle J(k_{0},l_{0},\varphi_{0})\leq J(k_{\epsilon},l_{\epsilon},\varphi_{0})$, we deduce from (LABEL:eq:lim_intom_k_epsilon_bis_bis) that

$\displaystyle-\int_{\Lambda}\bar{l}+\int_{\Omega}\bar{l}(u_{0})\leq 0$. By replacing $\displaystyle l$ by $\displaystyle-l$ in the above argument, one deduces that $\displaystyle\int_{\Lambda}\bar{l}=\int_{\Omega}\bar{l}(u_{0})$. As a result, $\displaystyle u_{0}\in\mathscr{U}_{S}^{1}$.

Variation around $\displaystyle\varphi_{0}$. Let $\displaystyle\varphi\in\mathcal{S}$. For $\displaystyle\epsilon\in(0,1)$, by convexity of $\displaystyle\mathcal{S}$, we have $\displaystyle\varphi_{\epsilon}:=\epsilon\varphi+(1-\epsilon)\varphi_{0}\in\mathcal{S}$. Then $\displaystyle J(k_{0},l_{0},\varphi_{0})\leq J(k_{0},l_{0},\varphi_{\epsilon})$. This implies that

[TABLE]

As $\displaystyle\epsilon$ tends to $\displaystyle 0^{+}$, the above equation yields

[TABLE]

It follows from Proposition 2.8 that $\displaystyle\varphi_{0}\in\Phi_{\mathcal{S}}(u_{0})$.

$\displaystyle\square$

Corollary 4.4**.**

Assume $\displaystyle\mathbf{(A0)}$, $\displaystyle\mathbf{(A1)}$, and $\displaystyle\mathbf{(A3)}$ hold. Assume $\displaystyle\mathcal{S}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ and $\displaystyle\nabla\varphi$ is non-degenerate whenever $\displaystyle\varphi\in\mathcal{S}$. Suppose $\displaystyle F$ has a countable range (thus degenerate). Then, $\displaystyle F$ satisfies the condition $\displaystyle\textbf{(ND)}_{\mathcal{S}}$ and problem (4.1) admits a unique solution.

Corollary 4.5**.**

Assume $\displaystyle\mathbf{(A1)}$ and $\displaystyle\mathbf{(A3)}$ hold. Assume $\displaystyle\mathcal{S}$ is a finite dimensional subspace of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$ and $\displaystyle\nabla\varphi$ has a countable range whenever $\displaystyle\varphi\in\mathcal{S}$. Suppose $\displaystyle F$ is non-degenerate. Then, $\displaystyle F$ satisfies the condition $\displaystyle\textbf{(ND)}_{\mathcal{S}}$ and problem (4.1) admits a unique solution.

4.2. A Link between Problem (3.1) and Problem (4.1)

Here, we explore the relationships between problem (3.1) and problem (4.1). For this purpose, we make a further assumption of the domains $\displaystyle\Omega$ and $\displaystyle\Lambda$ by requiring that $\displaystyle\Omega=\Lambda$. Assume $\displaystyle\mathbf{(A1)}$ holds and recall $\displaystyle\left\{H_{n}\right\}_{n=0}^{\infty}$ as defined in (2.7) and (2.8). Then, Lemma 2.3 ensures that $\displaystyle\mathbf{(A2)}$holds for $\displaystyle H_{n}$ for all $\displaystyle n\geq 1$. Define

[TABLE]

and

[TABLE]

Recall that $\displaystyle C_{0}$ is the set of all $\displaystyle(k,l)$ such that $\displaystyle l\in C(\bar{\Lambda})$, $\displaystyle\inf l=0$ and $\displaystyle k:\mathbb{R}^{d}\to\mathbb{R}$ satisfies for all $\displaystyle u\in\Lambda$ and all $\displaystyle v\in\mathbb{R}^{d}$:

[TABLE]

Let $\displaystyle C_{n}$ be the set of all $\displaystyle(k,l)$ such that $\displaystyle l\in C(\bar{\Lambda})$ and $\displaystyle k:\mathbb{R}^{d}\to\mathbb{R}$ satisfy:

[TABLE]

We denote by $\displaystyle\mathcal{A}_{0}$ the set of all $\displaystyle(k,l,\varphi)$ satisfying $\displaystyle(k,l)\in C_{0}$ and $\displaystyle\varphi\in S$. Similarly $\displaystyle\mathcal{A}_{n}$ denotes the set of all $\displaystyle(k,l,\varphi)$ satisfying $\displaystyle(k,l)\in C_{n}$ and $\displaystyle\varphi\in S$. If $\displaystyle(k,l,\varphi)\in\mathcal{A}_{0}\cup\mathcal{A}_{n}$, we still set

[TABLE]

Lemma 4.6**.**

Assume $\displaystyle\mathbf{(A1)}$, $\displaystyle\mathbf{(A2)}$, and $\displaystyle\mathbf{(A3)}$ hold. Assume $\displaystyle\mathcal{S}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$. For each $\displaystyle n\in\mathbb{N}$, let $\displaystyle(u_{n},\beta_{n})$ be the unique minimizer of $\displaystyle I_{n}$ over $\displaystyle\mathscr{U}_{\mathcal{S}}^{*}$ as given by Theorem 3.5 and let $\displaystyle(k_{n},l_{n},\varphi_{n})$ be a minimizer of $\displaystyle J$ over $\displaystyle\mathcal{A}_{n}$ with $\displaystyle k_{n}$ convex and $\displaystyle r^{*}$-Lipschitz as ensured by Proposition 3.3 and Lemma 4.2. Then,

1. (1)

The sequence $\displaystyle\{I_{n}(u_{n},\beta_{n})\}_{n\in\mathbb{N}^{*}}$ is bounded. 2. (2)

The sequence $\displaystyle\{\beta_{n}\}_{n\in\mathbb{N}^{*}}$ converges to $\displaystyle 1$ in $\displaystyle L^{2}(\Omega)$. 3. (3)

The sequence $\displaystyle\{\varphi_{n}\}_{n\in\mathbb{N}^{*}}$ admits a subsequence that converges to some $\displaystyle\bar{\varphi}$ in $\displaystyle S$ with respect to the $\displaystyle W_{0}^{1,q}\left(\Omega,\mathbb{R}^{d\times d}\right)-$ norm.

Proof. Step 1. Let $\displaystyle\bar{u}\in\mathscr{U}_{\mathcal{S}}^{1}$. We have $\displaystyle(\bar{u},1)\in\mathscr{U}_{\mathcal{S}}^{*}$ and thus $\displaystyle I_{n}(u_{n},\beta_{n})\leq I_{n}(\bar{u},1)$ for all $\displaystyle n\geq 1$. As $\displaystyle H_{n}(1)=0$, it holds that $\displaystyle I_{n}(\bar{u},1)=V_{\mathcal{S}}^{f}(\bar{u})-\int_{\Omega}\bar{u}\cdot F\;dx$ which is finite. Hence

[TABLE]

On the other hand, we use growth condition (2.5) to get

[TABLE]

Finally, we use (4.16) and (4.17) to prove (1).

Step 2. Let $\displaystyle\varphi_{0}\in\mathcal{S}$. As $\displaystyle u_{n}$ has values in $\displaystyle\Lambda$, it holds that

[TABLE]

and

[TABLE]

We combine (4.16), (4.17), (4.18), (4.19) to get

[TABLE]

Setting $\displaystyle c_{0}\mathcal{L}^{d}(\Omega):=R_{0}-R_{2}+r^{*}\|F\|_{L^{1}(\Omega,\mathbb{R}^{d})},$ we use lemma 2.3 and (4.20) to obtain

[TABLE]

This establishes (2).

Step 3. As $\displaystyle\left\{H_{n}\right\}_{n=1}^{\infty}$ is a non decreasing sequence that converges to $\displaystyle H_{0}$, it holds that $\displaystyle C_{n+1}\subset C_{n}\subset C_{0}$ for all $\displaystyle n\in\mathbb{N}$. Thus, as $\displaystyle(k_{n},l_{n})\in C_{n}$, we have $\displaystyle(k_{n},l_{n})\in C_{0}$ so that

[TABLE]

Since $\displaystyle-J(k_{n},l_{n},\varphi_{n})=I_{n}(u_{n},\beta_{n})$, we have $\displaystyle J(k_{n},l_{n},\varphi_{n})\leq R_{1}$ for all $\displaystyle n\in\mathbb{N}^{*}$. This, combined with $\displaystyle\Omega=\Lambda$, and (4.21) yields

[TABLE]

In view of the growth condition (2.5) and boundedness of $\displaystyle\Omega$, (4.22) implies

[TABLE]

As, $\displaystyle\mathcal{S}$ is of finite dimension and the $\displaystyle\operatorname{div}$ operator is continuous on $\displaystyle\mathcal{S}$, we conclude that $\displaystyle\left\{\varphi_{n}\right\}_{n=1}^{\infty}$ is convergent up to a subsequence in $\displaystyle W_{0}^{1,q}\left(\Omega,\mathbb{R}^{d\times d}\right)$ which allows us to conclude (3).

$\displaystyle\square$

Theorem 4.7**.**

Assume $\displaystyle\mathbf{(A1)}$, $\displaystyle\mathbf{(A2)}$, and $\displaystyle\mathbf{(A3)}$ hold. Assume $\displaystyle\mathcal{S}$ is a finite dimensional non-empty closed and convex subset of $\displaystyle W_{0}^{1,q}({\Omega,\mathbb{R}^{d\times d}})$. Assume $\displaystyle F$ satisfies the condition $\displaystyle\mathbf{(ND)_{\mathcal{S}}}$. For each $\displaystyle n\in\mathbb{N}$, let $\displaystyle(u_{n},\beta_{n})$ be the unique minimizer of $\displaystyle I_{n}$ over $\displaystyle\mathscr{U}_{\mathcal{S}}^{*}$ as given by Theorem 3.5 and let $\displaystyle(k_{n},l_{n},\varphi_{n})$ be a minimizer of $\displaystyle J$ over $\displaystyle\mathcal{A}_{n}$with $\displaystyle k_{n}$ is convex and $\displaystyle r^{*}$-Lipschitz as ensured by Proposition 3.3 and Lemma 4.2. Suppose that $\displaystyle k_{n}$ is differentiable for all $\displaystyle n\in\mathbb{N}^{*}$. Then, the sequence $\displaystyle\{u_{n}\}_{n\in\mathbb{N}^{*}}$ converges almost everywhere to the unique minimizer $\displaystyle u_{0}$ of $\displaystyle I_{0}$ over $\displaystyle\mathscr{U}_{S}^{1}$. In addition, the minima $\displaystyle\left\{I_{n}(u_{n},\beta_{n})\right\}_{n=1}^{\infty}$ converge to $\displaystyle I_{0}(u_{0})$.

Proof. Step 1. For $\displaystyle n\in\mathbb{N}^{*}$, set $\displaystyle\bar{k}_{n}=k_{n}-k_{n}(0)$. Note that we have $\displaystyle\bar{k}_{n}(0)=0$. Since the functions $\displaystyle k_{n}$ are $\displaystyle r^{*}$-Lipschitz, so are the functions $\displaystyle\bar{k}_{n}$ and we obtain that, up to a subsequence, the sequence $\displaystyle\left\{\bar{k}_{n}\right\}_{n=1}^{\infty}$ converges locally uniformly to a certain function $\displaystyle\bar{k}$. Since $\displaystyle F+\operatorname{div}\varphi_{n}$ is non-degenerate, we have that $\displaystyle\nabla\bar{k}_{n}(F+\operatorname{div}\varphi_{n})$ is well-defined. Furthermore, Lemma 4.6 ensures that $\displaystyle\left\{\varphi_{n}\right\}_{n=1}^{\infty}$ converges up to a subsequence to some $\displaystyle\bar{\varphi}\in S$ with respect to the $\displaystyle W^{1,q}\left(\Omega,\mathbb{R}^{d}\right)-$ norm. As a result, $\displaystyle\left\{\operatorname{div}\varphi_{n}\right\}_{n=1}^{\infty}$ converges to $\displaystyle\operatorname{div}\bar{\varphi}$ in $\displaystyle L^{q}(\Omega,\mathbb{R}^{d})$. Since $\displaystyle\mathcal{S}$ is of finite dimension, the $\displaystyle L^{q}$ convergence of $\displaystyle\left\{\operatorname{div}\varphi_{n}\right\}_{n=1}^{\infty}$ reduces to a pointwise convergence. Next, using the convexity of the $\displaystyle\bar{k}_{n}$ and the pointwise convergence of $\displaystyle\left\{\operatorname{div}\varphi_{n}\right\}_{n=1}^{\infty}$ to $\displaystyle\operatorname{div}\varphi$, we deduce that up to a subsequence $\displaystyle\left\{\nabla\bar{k}_{n}(F+\operatorname{div}\varphi_{n})\right\}_{n=1}^{\infty}$ converges a.e to $\displaystyle\nabla\bar{k}(F+\operatorname{div}\bar{\varphi})$ ( cf. [13] Theorem 25.7).

As a duality result, Theorem 3.5 ensures that $\displaystyle\nabla\bar{k}_{n}(F+\operatorname{div}\varphi_{n})=u_{n}$. If we denote $\displaystyle\bar{u}:=\nabla\bar{k}(F+\operatorname{div}\bar{\varphi})$, then, up to a subsequence, the sequence $\displaystyle\{u_{n}\}_{n\in\mathbb{N}}$ converges a.e to $\displaystyle\bar{u}$.

Step 2. Let $\displaystyle l\in C_{b}(\mathbb{R}^{d})$. The strong convergence in $\displaystyle L^{2}(\Omega)$ of $\displaystyle\left\{\beta_{n}\right\}_{n=1}^{\infty}$ to 1 established in Lemma 4.6 and the almost everywhere convergence of $\displaystyle\{u_{n}\}_{n\in\mathbb{N}}$ to $\displaystyle\bar{u}$ obtained in Step 1 ensure that $\displaystyle\lim_{n\rightarrow\infty}\int_{\Omega}\beta_{n}l(u_{n})dx=\int_{\Omega}l(\bar{u}(x))dx$. As $\displaystyle(u_{n},\beta_{n})\in\mathscr{U}_{S}^{*}$, $\displaystyle\int_{\Omega}\beta_{n}(x)l(u_{n})dx=\int_{\Omega}l(y)dy$ for all $\displaystyle l\in C_{b}(\mathbb{R}^{d})$. It follows that in the limit $\displaystyle\int_{\Omega}l(\bar{u})dx=\int_{\Omega}l(y)dy$ for all $\displaystyle l\in C_{b}(\mathbb{R}^{d})$ and thus $\displaystyle\bar{u}\in\mathscr{U}_{S}^{1}$.

Step 3. We recall that

[TABLE]

Since $\displaystyle u\mapsto V_{S}^{f}(u)$ is lower-semicontinuous as a supremun of affine functions, by applying the Fatou’s Lemma, we have

[TABLE]

Let $\displaystyle u_{0}$ be the unique minimizer of $\displaystyle I_{0}$ over $\displaystyle\mathscr{U}_{S}^{1}$ as given by Theorem 4.3. Then,

[TABLE]

Meanwhile, as $\displaystyle C_{n}\subset C_{0}$ and $\displaystyle(k_{0},l_{0},\varphi_{0})$ is a minimizer of $\displaystyle J$ over $\displaystyle C_{0}$, we have

[TABLE]

This, along with the duality established in Theorem 3.5 imply that

[TABLE]

We combine (4.25) and (4.26) to obtain $\displaystyle I_{0}(\bar{u})=I_{0}(u_{0})$. As $\displaystyle u_{0}$ is the unique minimizer of $\displaystyle I_{0}$ over $\displaystyle\mathscr{U}_{S}^{1}$ we have $\displaystyle u_{0}=\bar{u}$. We note that the limit $\displaystyle\bar{u}$ does not depend on the subsequence of $\displaystyle\{u_{n}\}_{n}$ chosen. Thus, the whole sequence $\displaystyle\{u_{n}\}_{n}$ converges a.e. to $\displaystyle u_{0}$. In addition, $\displaystyle\{I_{n}(u_{n},\beta_{n})\}_{n}$ converges to $\displaystyle I_{0}(u_{0})$.

$\displaystyle\square$

5. Appendix

5.1. Proof of Lemma 3.4

We will prove Lemma 3.4 through two lemmas. The results of the first lemma can be found in Lemma 4.3 of [2]. We give here a sketch of the proof for the convenience of the reader.

Lemma 5.1**.**

Assume $\displaystyle\mathbf{(A2)}$ holds. Consider a lower semicontinuous function $\displaystyle l:\mathbb{R}^{d}\to\bar{\mathbb{R}}$ such that $\displaystyle\inf_{\bar{\Lambda}}l>-\infty$; $\displaystyle l$ is finite on $\displaystyle\Lambda$ and $\displaystyle l\equiv+\infty$ on $\displaystyle\mathbb{R}^{d}\setminus\bar{\Lambda}$. Set $\displaystyle k=l^{\#}$ and let $\displaystyle w\in\mathbb{R}^{d}$. Then:

1. (1)

There exist $\displaystyle\bar{u}\in\bar{\Lambda}$ and $\displaystyle\bar{t}>0$ such that

[TABLE]

Moreover, $\displaystyle\bar{u}$ and $\displaystyle\bar{t}$ satisfy $\displaystyle\bar{u}\in\partial k(w)$ and $\displaystyle H^{\prime}(\bar{t})+l(\bar{u})=0$. 2. (2)

If $\displaystyle k$ is differentiable at $\displaystyle w$ then $\displaystyle\bar{u}$ and $\displaystyle\bar{t}$ are uniquely determined by $\displaystyle\bar{u}=\nabla k(w)$ and $\displaystyle\bar{t}=(H^{\prime})^{-1}(-l(\bar{u}))$.

Proof. 1) We have

[TABLE]

Consider a maximizing sequence $\displaystyle\{(u_{n},t_{n})\}_{n=1}^{\infty}$ in (5.2). As $\displaystyle 0\in\Lambda$, we may assume without loss of generality that

[TABLE]

for $\displaystyle n\geq 1$. It follows that

[TABLE]

for $\displaystyle n\geq 1$. In light of the growth condition on $\displaystyle H$ in (A2) there exists positive real numbers $\displaystyle\alpha$ such that $\displaystyle\{t_{n}\}_{n=1}^{\infty}\subset[\alpha,\alpha^{-1}]$. As $\displaystyle\Lambda$ is bounded, we may assume without loss of generality that the sequence $\displaystyle\{(u_{n},t_{n})\}_{n=1}^{\infty}$ converges to some $\displaystyle(\bar{u},\bar{t})\in\bar{\Lambda}\times[\alpha,\alpha^{-1}]$. We next use the lower semicontinuity of $\displaystyle H$ and $\displaystyle l$ to deduce that

[TABLE]

Note that $\displaystyle k(w)\geq\bar{u}\cdot w-l(\bar{u})t-H(t)\text{ for all }t>0$. In view of (5.3), it follows that $\displaystyle g:(0,\infty)\to\mathbb{R}$ defined by $\displaystyle g(t)=\bar{u}\cdot w-l(\bar{u})t-H(t)$ admits a maximum at $\displaystyle\bar{t}$. As $\displaystyle g$ is differentiable at $\displaystyle\bar{t}$, we have $\displaystyle g^{\prime}(\bar{t})=0$, that is, $\displaystyle l(\bar{u})+H^{\prime}(\bar{t})=0$. Next, observe that $\displaystyle k(z)\geq\bar{u}\cdot z-l(\bar{u})\bar{t}-H(\bar{t})\text{ for all }z\in\mathbb{R}^{d}$. In light of the convexity of $\displaystyle k$ we have that $\displaystyle\bar{u}\in\partial k(w)$.

2. Assume that $\displaystyle k$ is differentiable at $\displaystyle w$. Then, $\displaystyle\bar{u}$ is uniquely determined as $\displaystyle\bar{u}=\nabla k(w)$. As $\displaystyle H^{\prime}(\bar{t})=-l_{0}(\bar{u})$ and $\displaystyle H^{\prime}$ is a bijection, we obtain that $\displaystyle\bar{t}$ is also uniquely determined as $\displaystyle\bar{t}=(H^{\prime})^{-1}(-l(\bar{u}))$. ∎

The second lemma which is inspired by Lemma 4.4 in [2] is the following:

Lemma 5.2**.**

Assume assumption $\displaystyle\mathbf{(A2)}$ holds. Consider a lower semicontinuous function $\displaystyle l_{0}:\mathbb{R}^{d}\to\bar{\mathbb{R}}$ such that $\displaystyle\inf_{\bar{\Lambda}}l_{0}>-\infty$; $\displaystyle l_{0}$ is finite on $\displaystyle\Lambda$ and $\displaystyle l_{0}\equiv+\infty$ on $\displaystyle\mathbb{R}^{d}\setminus\bar{\Lambda}$. Set $\displaystyle k_{0}=({l_{0}})^{\#}$. Let $\displaystyle\hat{l}\in C_{b}(\mathbb{R}^{d})$ and let $\displaystyle 1\geq\epsilon>0$. Define $\displaystyle l_{\epsilon}=l_{0}+\epsilon\hat{l}$ and $\displaystyle k_{\epsilon}={\left(l_{\epsilon}\right)}^{\#}$. Let $\displaystyle v\in\mathbb{R}^{d}$ be such that $\displaystyle k_{0}$ is differentiable at $\displaystyle v$.

1. (1)

There exists a constant $\displaystyle M$ independent of $\displaystyle v$ and $\displaystyle\epsilon$ such that

[TABLE] 2. (2)

We have

[TABLE]

Proof. Note that the map $\displaystyle l_{\epsilon}=l_{0}+\epsilon\hat{l}$ is bounded below by $\displaystyle m-|\hat{l}|_{\infty}$. As $\displaystyle k_{\epsilon}={\left(l_{\epsilon}\right)}^{\#}$ and $\displaystyle k_{0}={\left(l_{0}\right)}^{\#}$, lemma 5.1 ensures that there exist $\displaystyle t_{0},t_{\epsilon}>0$ and $\displaystyle u_{0},u_{\epsilon}\in\bar{\Lambda}$ such that

[TABLE]

and

[TABLE]

We then have

[TABLE]

and

[TABLE]

We combine (5.6) and (5.7) to get

[TABLE]

Using again lemma 5.1 we have

[TABLE]

As $\displaystyle l_{\delta}$ is bounded below by $\displaystyle m-|l|_{\infty}$, we use the fact that $\displaystyle H^{\prime}$ is a continuous and strictly increasing bijection from $\displaystyle(0,\infty)$ to $\displaystyle\mathbb{R}$ to deduce that $\displaystyle t_{\delta}$ is bounded above by $\displaystyle M_{1}>0$ given by $\displaystyle M_{1}:=(H^{\prime})^{-1}(-m+|\hat{l}|_{\infty})$. This bound on $\displaystyle t_{\delta}$ combined with (5.8) yields a constant $\displaystyle M:=|\hat{l}|_{\infty}(H^{\prime})^{-1}(-m+|\hat{l}|_{\infty})$ such that (5.4) holds. As a result $\displaystyle\lim_{\epsilon\to 0^{+}}k_{\epsilon}(v)=k_{0}(v).$ Next, let $\displaystyle\{e_{n}\}_{n=1}^{\infty}\subset(0,1]$ converging to $\displaystyle 0$ such that $\displaystyle\limsup_{\epsilon\to 0}\hat{l}(u_{\epsilon})t_{\epsilon}=\lim_{n\to\infty}\hat{l}(u_{e_{n}})t_{e_{n}}$. Without loss of generality, we may assume that $\displaystyle\{u_{e_{n}}\}_{n=1}^{\infty}$ converges to some $\displaystyle\bar{u}\in\bar{\Lambda}$ and $\displaystyle\{t_{e_{n}}\}_{n=1}^{\infty}$ converges to $\displaystyle\bar{t}\in[0,M_{1}]$. Exploiting the lower semicontinuity of $\displaystyle l_{0}$, $\displaystyle\hat{l}$ and $\displaystyle H$, we get:

[TABLE]

It follows that $\displaystyle k_{0}(v)=\bar{u}v-l_{0}(\bar{u})\bar{t}-H(\bar{t})$. As $\displaystyle k_{0}$ is differentiable at $\displaystyle v$, we have $\displaystyle t_{0}=\bar{t}$ and $\displaystyle u_{0}=\bar{u}$. We use (5.8), the definition of $\displaystyle\{e_{n}\}_{n=1}^{\infty}$, the convergence of $\displaystyle\{u_{e_{n}}\}_{n=1}^{\infty}$ and $\displaystyle\{t_{e_{n}}\}_{n=1}^{\infty}$ to obtain

[TABLE]

As a result, $\displaystyle\lim_{\epsilon\to 0}-t_{\epsilon}\hat{l}(u_{\epsilon})=-t_{0}\hat{l}(u_{0})$. We invoke one more time equation (5.8) to obtain (5.5).

5.2. Some properties of the Legendre transform.

We have the following Lemma which is similar to Lemma 3.4 but uses the Legendre transform instead of the $\displaystyle(\cdot)^{\#}$ operator.

Lemma 5.3**.**

Consider a lower semicontinuous function $\displaystyle l_{0}:\mathbb{R}^{d}\to\bar{\mathbb{R}}$ such that $\displaystyle\inf_{\bar{\Lambda}}l_{0}>-\infty$; $\displaystyle l_{0}$ is finite on $\displaystyle\Lambda$ and $\displaystyle l_{0}\equiv+\infty$ on $\displaystyle\mathbb{R}^{d}\setminus\bar{\Lambda}$. Set $\displaystyle k_{0}=({l_{0}})^{*}$.

1. (1)

There exists a measurable map $\displaystyle T_{0}:\mathbb{R}^{d}\to\mathbb{R}^{d}$ such that $\displaystyle k_{0}(v)=v\cdot T_{0}(v)-l_{0}(T_{0}(v))$ for all $\displaystyle v\in\mathbb{R}^{d}$ and $\displaystyle T_{0}(v)=\nabla k_{0}(v)$ whenever $\displaystyle k_{0}$ is differentiable at $\displaystyle v\in\mathbb{R}^{d}$. 2. (2)

Let $\displaystyle\hat{l}\in C_{b}(\mathbb{R}^{d})$ and let $\displaystyle 1\geq\epsilon>0$. Define $\displaystyle l_{\epsilon}=l_{0}+\epsilon\hat{l}$ and $\displaystyle k_{\epsilon}={\left(l_{\epsilon}\right)^{*}}$.

1. (a)

For all $\displaystyle v\in\mathbb{R}^{d}$ we have :

[TABLE] 2. (b)

For $\displaystyle\epsilon\in(0,1)$, there exists a map $\displaystyle T_{\epsilon}:\mathbb{R}^{d}\to\mathbb{R}^{d}$ satisfying for all $\displaystyle v\in\mathbb{R}^{d}$: $\displaystyle k_{\epsilon}(v)=vT_{\epsilon}(v)-l_{\epsilon}(T_{\epsilon}(v))$. When $\displaystyle k_{0}$ is differentiable at $\displaystyle v\in\mathbb{R}^{d}$, we have $\displaystyle\lim_{\epsilon\to 0}T_{\epsilon}(v)=\nabla k_{0}(v)$ and

[TABLE]

Proof. 1.) Let $\displaystyle v\in\mathbb{R}^{d}$. We have

[TABLE]

We use the lower semicontinuity of $\displaystyle l_{0}$ and the compactness of $\displaystyle\bar{\Lambda}$ to deduce that there exists $\displaystyle\bar{u}\in\Lambda$ such that $\displaystyle k_{0}(v)=\bar{u}v-l_{0}(\bar{u}).$ We have $\displaystyle k_{0}(w)-(\bar{u}w-l_{0}(\bar{u}))\geq 0$ for all $\displaystyle w\in\mathbb{R}^{d}$ while $\displaystyle k_{0}(v)-(\bar{u}v-l_{0}(\bar{u}))=0$. Since $\displaystyle k_{0}$ is convex, we deduce that $\displaystyle\bar{u}\in\partial k_{0}(v)$.

Next, for $\displaystyle v\in\mathbb{R}^{d}$, define

[TABLE]

Assume $\displaystyle\{u_{n}\}_{n\in\mathbb{N}}\subset\mathbb{R}^{d}$ converges to $\displaystyle u$; $\displaystyle\{v_{n}\}_{n\in\mathbb{N}}\subset\mathbb{R}^{d}$ converges to $\displaystyle v$ and for all $\displaystyle n\in\mathbb{N}$, one has $\displaystyle u_{n}\in\Gamma(v_{n})$. Then $\displaystyle u\in\Gamma(v)$. Indeed, one has

[TABLE]

Therefore, $\displaystyle uv-l_{0}(u)=k_{0}(v)$ and $\displaystyle u\in\Gamma(v)$. As a result, the multifunction $\displaystyle\Gamma:\mathbb{R}^{d}\rightrightarrows\mathbb{R}^{d}$ is closed and nonempty valued. By the Measurable Selection Theorem [Corollary 14.6, [15]], there exists a measurable map $\displaystyle T_{0}:\mathbb{R}^{d}\to\mathbb{R}^{d}$ such that for all $\displaystyle v\in\mathbb{R}^{d}$, one has $\displaystyle T_{0}(v)\in\Gamma(v)$. That is $\displaystyle k_{0}(v)=vT_{0}(v)-l_{0}(T_{0}(v))$. As $\displaystyle T(v)\in\Gamma(v)\subset\partial k_{0}(v)$, we also have $\displaystyle T_{0}=\nabla k_{0}$ almost everywhere.

2.) For $\displaystyle\epsilon>0$, $\displaystyle l_{\epsilon}$ is bounded below and satisfies the hypothesis on $\displaystyle l_{0}$. Let $\displaystyle k_{\epsilon}=l_{\epsilon}^{*}$ and consider a map $\displaystyle T_{\epsilon}$ satisfying for all $\displaystyle v\in\mathbb{R}^{d}$: $\displaystyle k_{\epsilon}(v)=vT_{\epsilon}(v)-l_{\epsilon}(T_{\epsilon}(v))$ as given by part 1.). We have for $\displaystyle v\in\mathbb{R}^{d}$:

[TABLE]

Similarly, for $\displaystyle v\in\mathbb{R}^{d}$ we have

[TABLE]

We combine (5.10) and (5.11) to get

[TABLE]

which leads to

[TABLE]

Consider a sequence $\displaystyle\{\epsilon_{n}\}_{n}$ converging to $\displaystyle 0$. The sequence $\displaystyle\{T_{\epsilon_{n}}(v)\}_{n}$ is bounded so we may find a subsequence $\displaystyle\{\epsilon^{\prime}_{n}\}_{n}$ of $\displaystyle\{\epsilon_{n}\}_{n}$ such that the sequence $\displaystyle\{T_{\epsilon^{\prime}_{n}}(v)\}_{n}$ converges to $\displaystyle u\in\bar{\Lambda}$. We then have:

[TABLE]

We use (5.14) to obtain $\displaystyle k_{0}(v)=vu-l_{0}(u)$ and thus $\displaystyle u=\nabla k_{0}(v)$ as $\displaystyle k_{0}$ is differentiable at $\displaystyle v$. It follows that $\displaystyle\lim_{\epsilon\to 0}T_{\epsilon}(v)=\nabla k_{0}(v)$. We use Equation (5.12) and the continuity of $\displaystyle\hat{l}$ to obtain

[TABLE]

∎