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# Coupled system involving eddy coefficients and a right hand side in L 1: Review chapter

Guillaume Leloup

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Guillaume Leloup. Coupled system involving eddy coefficients and a right hand side in L 1: Review chapter. 2024. hal-04405798v2

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# Coupled system involving eddy coefficients and a right hand side in $L^1$: review chapter

Guillaume LELOUP

October 18, 2024

## 1 Introduction

Let $\Omega$ be an open bounded subset of $\mathbb{R}^N$, this chapter focuses on the following problem. We search $u, k: \Omega \to \mathbb{R}$ satisfying

$$ (1) \qquad \left\{ \begin{array}{ll} -\operatorname{div}(A(k)\nabla u) = f & \text{in } \Omega, \\ -\operatorname{div}(B(k)\nabla k) = A(k)|\nabla u|^2 & \text{in } \Omega, \\ -A(k)\frac{\partial u}{\partial n} = \alpha u & \text{on } \partial\Omega, \\ k = 0 & \text{on } \partial\Omega. \end{array} \right. $$

In (1), $f \in L^2(\Omega)$ is a given source term, $\alpha > 0$ and there exists two constants $\nu, \mu > 0$ such that both viscosities $A$ and $B$ satisfy

$$ (2) \qquad \forall x \in \mathbb{R}, \quad \nu \le A(x), B(x) \le \mu. $$

The aim of this chapter is proving the existence of a couple $(u, k)$ of distributional solutions of (1). This is not the only kind of solutions commonly searched. Indeed, some articles deal with renormalized solutions, which are defined and studied in [3]. This notion is interesting for the study of a system having a source term in $L^1(\Omega)$, as shown in [6].

The source term of the equation in $k$ is in $L^1(\Omega)$. Without any better hypothesis, it remains possible to find distributional solutions of (1).

This figures in the works of L. Boccardo and T. Gallouët in [1], on which is based this chapter. The associated result, proved in section 4, is a consequence of the washer Lemma, stated and proved in section 2. This Lemma brings a bound for a solution in $W_0^{1,q}(\Omega)$, where $q < \frac{N}{N-1}$. Consequently, some particular function sets are needed.

Then, in section 4, we search weak solutions in $V = \bigcap_{1<q<N'} W_0^{1,q}(\Omega)$, where $N'$ refers to $\frac{N}{N-1}$; with test functions in $W = \bigcup_{q>N} W_0^{1,q}(\Omega)$. This method was also used by T. Gallouët and R. Herbin in [4].

Along this chapter, the viscosities $A$ and $B$ are bounded (assumption (2)). This does not really correspond to the physical situation, but is a right way to establish some interesting mathematical results, as a priori estimates, such as in section 3 and 4. The assumption of
bounded viscosities appears in [4], but also in the works of S. Clain and R. Touzani (see [2]), in which an existence result is proved in the two-dimensional case. In the present work, the existence results are proved when $N = 3$.

Of course it is possible to deal with unbounded viscosities, such as in [6], where they are only bounded from below. Otherwise, if they are not, other assumptions are required, concerning the decreasing. In addition, other function spaces need to be developed, like in [5].

In order to find the existence of a solution of (1), this work will be divided into two steps, that correspond to sections 3 and 4. Section 3 is devoted to the equation in $u$, which amounts to a linear problem. But when the problem is obviously non-linear, like the equation in $k$ considered in section 4, another method is required. This consists in regularizing the system, then solving it by a linearization and a fixed-point theorem. This method is already used in [2].

Given any borelian subset or $\mathbb{R}^N$ named $A$, the Lebesgue measure is simply written $|A|$.

# 2 Preliminary results

## 2.1 The washer lemma

This section aims at proving the following result (already proved by Boccardo-Gallouët in [1]).

**Lemma 1.** Let $q \in [1; N']$ where $N' = \frac{N}{N-1}$, and $u \in W_0^{1,q}(\Omega)$. We define

$$ (3) \qquad B_n = \{x \in \Omega / n \le |u(x)| < n + 1\}, $$

and assume that there is a constant $C > 0$ such that

$$ (4) \qquad \forall n \in \mathbb{N}, \quad \int_{B_n} |\nabla u|^2 \le C. $$

Then we have the inequality

$$ (5) \qquad \|u\|_{W^{1,q}(\Omega)} \le A, $$

where $A$ depends on $\Omega$, $q$ and $C$.

**Remark 2.** Both constants $C$ and $A$ do not depend on $u$. Actually, the result may be rephrased as: let $(u_j)_{j \in J}$ a family of functions in $W_0^{1,q}(\Omega)$ with $q \in [1; N']$ satisfying

$$ (6) \qquad \forall j \in J, \forall n \in \mathbb{N}, \quad \int_{B_n^j} |\nabla u_j(x)|^2 dx \le C, $$

where $B_n^j$ refers to $B_n^j = \{x \in \Omega / n \le |u_j(x)| < n + 1\}$.
Then there exists a constant $A > 0$ only depending on $\Omega$, $q$ and $C$ such that

$$ (7) \qquad \forall j \in J, \quad \|u_j\|_{W^{1,q}(\Omega)} \le A. $$

The proof of lemma 1 is based on a decomposition given by the following lemma.

**Lemma 3.** Let $u$ satisfy the assumptions of lemma 1. Let $q^*$ be the critical exponent in Sobolev embedding $W^{1,q}(\Omega) \hookrightarrow L^q^*(\Omega)$, and any index $n_0 \in \mathbb{N}$. There exist two constants $\lambda_1$ and $\lambda_2$ depending on $n_0$, $q$ and $\Omega$ satisfying

$$ (8) \qquad \|\nabla u\|_{L^q(\Omega)^N}^q \le C^{q/2} \left( \lambda_1 + \lambda_2 \|u\|_{L^{q^*}(\Omega)}^{\frac{q^*(2-q)}{2}} \right). $$

Moreover, we have: $\lim_{n_0 \to +\infty} \lambda_2(n_0) = 0$.

**Proof of lemma 3.** At first, we decompose $\int_{\Omega} |\nabla u|^q$ as a sum

$$ (9) \qquad \int_{\Omega} |\nabla u|^q = \sum_{n=0}^{+\infty} \int_{B_n} |\nabla u|^q. $$

Recalling that $B_n$ is bounded, Hölder inequality and (4) yield

$$ (10) \qquad \int_{B_n} |\nabla u|^q \le |B_n|^{\frac{2-q}{2}} \left( \int_{B_n} |\nabla u|^2 \right)^{\frac{q}{2}} \le C^{q/2} |B_n|^{\frac{2-q}{2}}. $$

We consider a fixed index $n_0 \in \mathbb{N}$.

If $n \le n_0$, (10) directly yields

$$ (11) \qquad \sum_{n=0}^{n_0} \int_{B_n} |\nabla u|^q \le C^{q/2} \sum_{n=0}^{n_0} |B_n|^{\frac{2-q}{2}} \le (n_0+1) C^{q/2} |\Omega|^{\frac{2-q}{2}}. $$

If $n > n_0$, the definition of $B_n$ yields

$$ (12) \qquad \int_{B_n} |u|^{q^*} \ge \int_{B_n} n^{q^*} = n^{q^*} |B_n|, \quad \text{hence } |B_n| \le \frac{1}{n^{q^*}} \int_{B_n} |u|^{q^*}. $$

Inequalities (10) and (12) yield

$$ (13) \qquad \forall n > n_0, \quad \int_{B_n} |\nabla u|^q \le C^{q/2} \left( \frac{1}{n^{q^*}} \right)^{\frac{2-q}{2}} \left( \int_{B_n} |u|^{q^*} \right)^{\frac{2-q}{2}}, $$

which yields by addition

$$ (14) \qquad \sum_{n=n_0+1}^{+\infty} \int_{B_n} |\nabla u|^q \le C^{q/2} \sum_{n=n_0+1}^{+\infty} n^{\frac{(q-2)q^*}{2}} \left( \int_{B_n} |u|^{q^*} \right)^{\frac{2-q}{2}}. $$

The discrete Hölder inequality applied on the right hand side of (14) is written

$$ (15) \qquad \sum_{n=n_0+1}^{+\infty} n^{\frac{(q-2)q^*}{2}} \left( \int_{B_n} |u|^{q^*} \right)^{\frac{2-q}{2}} \le \left[ \sum_{n=n_0+1}^{+\infty} \int_{B_n} |u|^{q^*} \right]^{\frac{2-q}{2}} \left[ \sum_{n=n_0+1}^{+\infty} n^{\frac{N(q-2)}{N-q}} \right]^{q/2}. $$
The inequality may be useful in the following if and only if the series $\sum_{n \ge n_0+1} n^{\frac{N(q-2)}{N-q}}$ converges, which is equivalent to $N(2-q) > N-q$. Some calculations show it is equivalent to $q < N'$.

Then we set $\lambda_2(n_0, q) = \left[ \sum_{n=n_0+1}^{+\infty} n^{\frac{N(q-2)}{N-q}} \right]^{q/2}$, so that we directly obtain $\lambda_2(n_0, q) \underset{n_0 \to +\infty}{\longrightarrow}$
0. With this notation, (14) and (15) become

$$ (16) \quad \sum_{n=n_0+1}^{+\infty} \int_{B_n} |\nabla u|^q \le C^{q/2} \lambda_2(n_0, q) \left[ \sum_{n=n_0+1}^{+\infty} \int_{B_n} |u|^{q^*} \right]^{\frac{2-q}{2}} \le C^{q/2} \lambda_2(n_0, q) \left( \int_{\Omega} |u|^{q^*} \right)^{\frac{2-q}{2}} \\ = C^{q/2} \lambda_2(n_0, q) \|u\|_{L^{q^*}(\Omega)}^{\frac{q^*(2-q)}{2}}. $$

By setting $\lambda_1(n_0, q) = (n_0 + 1)|\Omega|^{\frac{2-q}{2}}$, (11) and (16) yield

$$ (17) \quad \int_{\Omega} |\nabla u|^q \le C^{q/2} \left( \lambda_1(n_0, q) + \lambda_2(n_0, q) \|u\|_{L^{q^*}(\Omega)}^{\frac{q^*(2-q)}{2}} \right). $$

That corresponds to (8), which concludes the proof. $\square$

This result can now be used to prove Lemma 1.

*Proof of lemma 1.* We start with the case $N=2$, which is easier. In that case, we have $\frac{q^*(2-q)}{2} = \frac{Nq(2-q)}{2(N-q)} = q$. Then (8) becomes

$$ (18) \quad \|\nabla u\|_{L^q(\Omega)^N}^q \le C^{q/2} (\lambda_1 + \lambda_2 \|u\|_{L^{q^*}(\Omega)}^q). $$

Gagliardo-Nirenberg inequality and (18) yield

$$ (19) \quad \|u\|_{L^{q^*}(\Omega)^N}^q \le K \|\nabla u\|_{L^q(\Omega)^N}^q \le KC^{q/2} (\lambda_1 + \lambda_2 \|u\|_{L^{q^*}(\Omega)}^q), $$

where $K > 0$ depends on $\Omega$ and $q$. We recall that if $b < 1$, the following equivalence stands

$$ (20) \quad x \le a+bx \Leftrightarrow x \le \frac{a}{1-b}. $$

Since $\lambda_2(n_0, q) \underset{n_0 \to +\infty}{\longrightarrow} 0$, it is possible to choose $n_0 \in \mathbb{N}$ large enough to have $KC^{q/2}\lambda_2 < 1$. Then by (20), we obtain

$$ (21) \quad \|u\|_{L^{q^*}(\Omega)}^q \le \frac{KC^{q/2}\lambda_1}{1 - KC^{q/2}\lambda_2}. $$

Thus, from (18):

$$ (22) \quad \|\nabla u\|_{L^q(\Omega)^N}^q \le C^{q/2} \left( \lambda_1 + \frac{KC^{q/2}\lambda_1\lambda_2}{1 - KC^{q/2}\lambda_2} \right). $$
And Poincaré-Friedrichs inequality yields

$$ (23) \qquad \|u\|_{W^{1,q}(\Omega)}^q \le LC^{q/2} \left( \lambda_1 + \frac{KC^{q/2}\lambda_1\lambda_2}{1-KC^{q/2}\lambda_2} \right), $$

where L depends on q and Ω. This yields the result with

$$ (24) \qquad A = L^{1/q}C^{1/2} \left( \lambda_1 + \frac{KC^{q/2}\lambda_1\lambda_2}{1-KC^{q/2}\lambda_2} \right)^{1/q}. $$

We now focus on the case $N > 2$. Since $q < N' < N$, the Sobolev embedding used before yields

$$ (25) \qquad \|u\|_{L^{q^*}(\Omega)}^q \le K \|\nabla u\|_{L^q(\Omega)^N}^q \le KC^{q/2} \left( \lambda_1 + \lambda_2 \|u\|_{L^{q^*}(\Omega)}^{\frac{q^*(2-q)}{2}} \right). $$

Inequality (25) is of the form $x \le a + bx^\gamma$, with

$$ (26) \qquad \gamma = \frac{q^*(2-q)}{2q} = \frac{N(2-q)}{2(N-q)}. $$

We have already shown that if $q \in [1; N']$

$$ (27) \qquad 1 < \frac{N(2-q)}{N-q} < N' \quad \text{hence} \quad \frac{1}{2} < \gamma < \frac{N'}{2} < 1. $$

Thus, the inequality we need to solve is of the form: $x \le a + bx^\gamma$, with $\gamma \in ]0, 1[$ and $a, b > 0$.

The graph on the side shows the existence of $\tau \in ]0; +\infty[$ depending on a, b and $\gamma$ such that $x \le a + bx^\gamma \Leftrightarrow x \le \tau$.

<figure><img src="image_3.png" /></figure>

Let us show it rigorously by studying the variations of $F: x > 0 \mapsto a + bx^\gamma - x$. This function is $C^\infty$ on $]0; +\infty[$ and $|F'(x) = b\gamma x^{\gamma-1} - 1$ and $F''(x) = b\gamma(\gamma-1)x^{\gamma-2}$, therefore $F''(x) < 0$. We obtain the following variations
<figure><img src="image_4.png" /></figure>

<figure><img src="image_5.png" /></figure>

Inequality (25) finally ensures the existence of $\tau = \tau(\lambda_1, \lambda_2, C) > 0$ such that

$$ (28) \qquad \|u\|_{L^{q^*}(\Omega)}^q \le \tau. $$

Consequently, (8) and Poincaré-Friedrichs inequality yield

$$ (29) \qquad \|\nabla u\|_{L^q(\Omega)^N}^q \le C^{q/2} (\lambda_1 + \lambda_2 \tau) \quad \text{hence} \quad \|u\|_{W^{1,q}(\Omega)}^q \le LC^{q/2} (\lambda_1 + \lambda_2 \tau), $$

where $L$ only depends on $\Omega$ and $q$. With $A = L^{1/q}C^{1/q}(\lambda_1 + \lambda_2\tau)^{1/q}$, lemma 1 is proved in case $N > 2$. $\square$

The first hypothesis of lemma 1 is $q \in [1; N']$, where $N' = \frac{N}{N-1}$. It has been used many times, and we are able to show that is impossible to obtain similar results with higher values of $q$.

Indeed, considering the case $N > 2$, we recall that $\alpha$ is the only point where $F'$ vanishes, which means

$$ (30) \qquad b\gamma\alpha^{\gamma-1} = 1 \quad \text{thus} \quad \alpha = (b\gamma)^{\frac{1}{1-\gamma}}. $$

Replacing $b$ by the constants of lemma 3 yields $b(q) = M(q)\lambda_2(q)$, where $M(q)$ is bounded if $q \in [1; N']$. Consequently, $b(q) \underset{q \to N'}{\to} +\infty$ and then

$$ (31) \qquad \gamma(q) = \frac{N(2-q)}{2(N-q)} \underset{q \to N'}{\longrightarrow} 2 \quad \text{hence} \quad \frac{1}{1-\gamma} \underset{q \to N'}{\longrightarrow} \frac{1}{2}. $$

Therefore, (30) yields $\alpha \underset{q \to N'}{\longrightarrow} +\infty$. And since $\tau \ge \alpha$, we obtain $\tau \underset{q \to N'}{\longrightarrow} +\infty$, which means $A \underset{q \to N'}{\longrightarrow} +\infty$.

There is a similar result in the case $N = 2$. The proof of lemma 1 requires to choose an index $n_0$ great enough to have $b < 1$. But as $\lambda_2(n_0, q) \underset{q \to N'}{\longrightarrow} +\infty$, an increase of $q$ implies an increase of $n_0$.
Thus, as $\frac{2-q}{2} = 1 - \frac{q}{2} \xrightarrow{q \to N'} 1 - \frac{N'}{2} = 0$, it follows

$$ (32) \qquad \lambda_1(n_0, q) = (n_0 + 1)|\Omega|^{\frac{2-q}{2}} \xrightarrow[q \to N']{n_0 \to +\infty} +\infty \quad \text{hence} \quad A \xrightarrow{q \to N'} +\infty. $$

We have shown that $q$ cannot be larger than the assumptions of lemma 1.

The result shown in this first section will be useful in the following. Generally, solving a PDE system is divided into two steps:

* simplifying the problem (for example by moving to a finite-dimension space, or by using a truncation function),

* coming back to the initial problem with the help of compacity results.

## 2.2 A general result on norms

**Theorem 4** (Trace norm on $H^1$). Let the application $N_0$ be defined for any $u \in H^1(\Omega)$ by

$$ (33) \qquad N_0(u) = \left( \|\nabla u\|_{L^2(\Omega)^N}^2 + \|u\|_{L^2(\partial\Omega)}^2 \right)^{1/2}, $$

where "$u|_{\partial\Omega}$" refers to the trace of $u$ on $\partial\Omega$.

Then $N_0$ is a norm, which is equivalent to the classical norm on $H^1(\Omega)$.

*Proof.* $N_0$ is non negative, homogeneous and satisfies triangular inequality (direct consequence of the fact that the norms $L^2$ also do). Let us consider $u \in H^1(\Omega)$ such that $N_0(u) = 0$. It comes $\|\nabla u\|_{L^2(\Omega)^N} = 0$ and $\|u\|_{L^2(\partial\Omega)} = 0$.

First, this yields that $\nabla u = 0$ almost everywhere on $\Omega$. We write $u \stackrel{\text{ae.}}{=} \lambda$, where $\lambda$ is a real constant. We directly obtain that $\text{tr } u = \lambda$ a.e. on $\partial\Omega$, then $\lambda = 0$. We deduce that $N_0$ is a norm.

**Equivalence between both norms :**

At first, the trace operator $\text{tr}: H^1(\Omega) \to H^{1/2}(\partial\Omega)$ is surjective (see [7]), and Sobolev embeddings yield in particular $H^{1/2}(\partial\Omega) \hookrightarrow L^2(\partial\Omega)$. Hence

$$ (34) \qquad \exists C > 0, \forall u \in H^1(\Omega), \quad \|u\|_{L^2(\partial\Omega)}^2 \le C \|u\|_{H^1(\Omega)}^2. $$

We deduce

$$ (35) \qquad \forall u \in H^1(\Omega), \quad N_0(u)^2 \le (1+C) \|u\|_{H^1(\Omega)}^2. $$

Therefore $N_0$ is dominated by $\| \cdot \|_{H^1(\Omega)}$.

Conversely, we intend to show

$$ (36) \qquad \exists C > 0, \forall u \in H^1(\Omega), \quad \|u\|_{H^1(\Omega)}^2 \le C N_0(u)^2. $$

By contradiction, let us suppose the negation of (36), which means in particular

$$ (37) \qquad \forall n \in \mathbb{N}, \exists u_n \in H^1(\Omega), \quad \|u_n\|_{H^1(\Omega)}^2 > n N_0(u_n)^2. $$
Inequality (37) yields that $\forall n \in \mathbb{N}^*$, $\|u_n\|_{L^2(\Omega)} > 0$. Then, both sides of (37) can be divided by $\|u_n\|_{L^2(\Omega)}$. Which means, keeping the same notation

$$ (38) \qquad \forall n \in \mathbb{N}, \exists u_n \in S, \quad \|u_n\|_{H^1(\Omega)}^2 > nN_0(u_n)^2, $$

where $S$ refers to the "$L^2$-sphere" defined by $S = \{v \in H^1(\Omega) / \|v\|_{L^2(\Omega)} = 1\}$. Then (38) can be rewritten as

$$ (39) \qquad \forall n \ge 2, \quad \frac{1}{n} + \frac{1}{n} \|\nabla u_n\|_{L^2(\partial\Omega)^N}^2 > \|u_n\|_{L^2(\partial\Omega)}^2 + \|\nabla u_n\|_{L^2(\Omega)^N}^2. $$

Since $\|u_n\|_{L^2(\partial\Omega)}^2 \ge 0$, (39) yields in particular

$$ (40) \qquad \left(1 - \frac{1}{n}\right) \|\nabla u_n\|_{L^2(\Omega)^N}^2 \le \frac{1}{n}, \quad \text{then} \quad \|\nabla u_n\|_{L^2(\Omega)^N}^2 \le \frac{1}{n-1}. $$

Therefore, the sequence $(\nabla u_n)_{n \in \mathbb{N}^*}$ is bounded in $L^2(\Omega)$; and as $(u_n)_{n \in \mathbb{N}^*}$ is a sequence of points of $S$, $(u_n)_{n \in \mathbb{N}^*}$ is also bounded in $L^2(\Omega)$. We deduce $(u_n)_{n \in \mathbb{N}^*}$ is bounded in $H^1(\Omega)$, and there exists a subsequence also denoted $(u_n)_{n \in \mathbb{N}^*}$ satisfying $u_n \xrightarrow{n \to +\infty} u$ in $H^1(\Omega)$.

However, (39) yields

$$ (41) \qquad \forall n \ge 2, \quad \frac{1}{n} > \|u_n\|_{L^2(\partial\Omega)}^2 + \left(1 - \frac{1}{n}\right) \|\nabla u_n\|_{L^2(\Omega)^N}^2 \ge \frac{1}{2} N_0(u_n)^2, $$

hence $N_0(u_n) \xrightarrow{n \to +\infty} 0$, then $u = 0$.

Rellich theorem yields that $u_n \xrightarrow{n \to +\infty} 0$ in $L^2(\Omega)$, meaning $\|u_n\|_{L^2(\Omega)} \xrightarrow{n \to +\infty} 0$. But for all $n \in \mathbb{N}^*$, $\|u_n\|_{L^2(\Omega)} = 1$, which is a contradiction. □

# 3 Analysis of the first equation

In this section, given $p \in ]1; +\infty[$ and $k \in W_0^{1,p}(\Omega)$, we consider the system

$$ (42) \qquad \left\{ \begin{array}{ll} -\mathrm{div}(A(k)\nabla u) = f & \text{in } \Omega, \\ -A(k)\frac{\partial u}{\partial n} = \alpha u & \text{on } \partial\Omega, \end{array} \right. $$

with $f \in L^2(\Omega)$, $\alpha > 0$ and $A : \mathbb{R} \to \mathbb{R}_+$ satisfying

$$ (43) \qquad \forall k \in \mathbb{R}, \quad 0 < \nu \le A(k) \le \mu < +\infty. $$

## 3.1 Variational formulation and a priori estimates

We remind the Green formula, which is likely to be used all along this thesis.

**Theorem 5 (Green formula).** Let $f \in H^1(\Omega)$ and $g \in H^1(\Omega)^N$. Assume that $n$ is the normal vector of $\Omega$, then the following formula stands

$$ (44) \qquad \int_{\Omega} \mathrm{div}(g) f = - \int_{\Omega} g \cdot \nabla f + \int_{\partial\Omega} (g \cdot n) f \, dS. $$
Multiplying the first equation of (42) by $w \in H^1(\Omega)$ and then integrating yields

$$ (45) \qquad - \int_{\Omega} \mathrm{div}(A(k)\nabla u)w = \int_{\Omega} fw. $$

Then (44) and the second equation of (42) yield

$$ (46) \qquad - \int_{\Omega} \mathrm{div}(A(k)\nabla u)w = \int_{\Omega} A(k)\nabla u \cdot \nabla w + \alpha \int_{\partial\Omega} uw \, dS. $$

We obtain the variational formulation of (42)

$$ (47) \qquad \begin{cases} u \in H^1(\Omega) \\ \forall w \in H^1(\Omega), \quad \int_{\Omega} A(k)\nabla u \cdot \nabla w + \alpha \int_{\partial\Omega} uw \, dS = \int_{\Omega} fw. \end{cases} $$

All the terms of (47) are well defined. Moreover, the trace theorem (see [7]) yields that $u, w \in H^{1/2}(\partial\Omega) \hookrightarrow L^2(\partial\Omega)$.

The main result of this section is the following.

**Theorem 6.** There exists a unique function $u \in H^1(\Omega)$ satisfying (47).

*Proof.* This proof uses the Lax-Milgram theorem on the Hilbert space $H^1(\Omega)$. Let $a_k$ be the bilinear form given by

$$ (48) \qquad \forall u, w \in H^1(\Omega), \quad a_k(u, w) = \int_{\Omega} A(k)\nabla u \cdot \nabla w + \alpha \int_{\partial\Omega} uw \, dS. $$

Let $u, w \in H^1(\Omega)$. First we have

$$ (49) \qquad \begin{aligned} |a_k(u, w)| &\le \int_{\Omega} A(k) |\nabla u \cdot \nabla w| + \alpha \int_{\partial\Omega} |uw| \, dS \\ &\le \mu \|\nabla u\|_{L^2(\Omega)^N} \|\nabla w\|_{L^2(\partial\Omega)^N} + \alpha \|u\|_{L^2(\partial\Omega)} \|w\|_{L^2(\partial\Omega)} \\ &\le (\mu + \alpha M) \|u\|_{H^1(\Omega)} \|w\|_{H^1(\Omega)}, \end{aligned} $$

where $M$ does not depend on $u$. Therefore, $a_k$ is continuous on $H^1(\Omega) \times H^1(\Omega)$.

Then we show the coercivity. Let $w \in H^1(\Omega)$, we have

$$ (50) \qquad \begin{aligned} a_k(w, w) &= \int_{\Omega} A(k) |\nabla w|^2 + \alpha \int_{\partial\Omega} |w|^2 \, dS \ge \nu \|\nabla w\|_{L^2(\Omega)^N}^2 + \alpha \|w\|_{L^2(\partial\Omega)}^2 \\ &\ge \inf(\nu, \alpha) N_0(w)^2, \end{aligned} $$

hence Theorem 4 yields the coercivity of $a_k$.

Finally, it is clear that the linear form $[w \mapsto \int_{\Omega} fw]$ is continuous. Problem (47) then admits a unique solution $u \in H^1(\Omega)$, by Lax-Milgram theorem. □
Now the existence and uniqueness of $u$ is proved, it is relevant to state some a priori estimates on it.

**Proposition 7.** Let $k \in W_0^{1,p}(\Omega)$ with $p \in [1; +\infty[$. The solution $u = u(k)$ of (47) satisfies the inequalities

$$ (51) \qquad \|u\|_{H^1(\Omega)} \le C_1(\Omega, \alpha, \nu, f), $$

and

$$ (52) \qquad \|A(k)^{1/2} \nabla u(k)\|_{L^2(\Omega)^N} \le C_2(\Omega, \alpha, \mu, \nu, f). $$

*Proof.* Taking $w = u$ as a test in (47) leads to

$$ (53) \qquad \int_{\Omega} A(k) |\nabla u|^2 + \alpha \int_{\partial \Omega} |u|^2 dS = \int_{\Omega} f u. $$

Cauchy-Schwarz inequality then yields

$$ (54) \qquad \nu \|\nabla u\|_{L^2(\Omega)^N}^2 + \alpha \|u\|_{L^2(\partial \Omega)}^2 \le \|f\|_{L^2(\Omega)} \|u\|_{L^2(\Omega)}, $$

hence

$$ (55) \qquad \min(\nu, \alpha) N_0(u)^2 \le \|f\|_{L^2(\Omega)} \|u\|_{H^1(\Omega)}. $$

Theorem 4 directly yields

$$ (56) \qquad \frac{\min(\nu, \alpha)}{C} \|u\|_{H^1(\Omega)}^2 \le \|f\|_{L^2(\Omega)} \|u\|_{H^1(\Omega)}, $$

hence (51).

To obtain (52), the higher bound on $A$ and Poincaré inequality yield

$$ (57) \qquad \|A(k)^{1/2} \nabla u\|_{L^2(\Omega)^N} \le \sqrt{\mu} \|\nabla u\|_{L^2(\Omega)^N} \le c \sqrt{\mu} \|u\|_{H^1(\Omega)}. $$

Finally, (51) yields (52). □

## 3.2 An energy method

This subsection is devoted to show the $H^1$ convergence of solutions of (42). More precisely, if $(k_n)_{n \in \mathbb{N}}$ is a sequence converging to a certain $k$ in a sense precised after, a subsequence of $(u(k_n))_{n \in \mathbb{N}}$ converges in $H^1(\Omega)$ to $u(k)$. An intermediary result is first needed.

**Theorem 8 (L$^2$ convergence).** Let $p \in [1; +\infty[$ and $(k_n)_{n \in \mathbb{N}}$ a sequence of functions converging weakly to $k$ in $W_0^{1,p}(\Omega)$. Let $u_n = u(k_n)$ and $u = u(k)$ the associated solutions of problem (42). There exists a subsequence still denoted $(u_n)_{n \in \mathbb{N}}$ satisfying

$$ (58) \qquad A(k_n)^{1/2} \nabla u_n \underset{n \to +\infty}{\longrightarrow} A(k)^{1/2} \nabla u \quad \text{in } L^2(\Omega). $$
Proof. The weak convergence of $(k_n)_{n \in \mathbb{N}}$ and compact Sobolev embeddings yield that $k_n \xrightarrow{n \to +\infty} k$ in $L^1(\Omega)$. By Lebesgue inverse theorem, a subsequence named the same way converges a.e. in $\Omega$ to $k$. The continuity of $A$ yields

$$ (59) \qquad A(k_n) \xrightarrow{n \to +\infty} A(k) \quad \text{a.e. in } \Omega. $$

We recall that $u_n$ and $u$ satisfy

$$ (60) \qquad \forall v \in H^1(\Omega), \forall n \in \mathbb{N}, \quad \int_{\Omega} A(k_n) \nabla u_n \cdot \nabla v + \alpha \int_{\partial \Omega} u_n v \, dS = \int_{\Omega} fv. $$

$$ (61) \qquad \forall v \in H^1(\Omega), \quad \int_{\Omega} A(k) \nabla u \cdot \nabla v + \alpha \int_{\partial \Omega} uv \, dS = \int_{\Omega} fv. $$

Inequality (51) yields that $(u_n)_{n \in \mathbb{N}}$ is bounded in $H^1(\Omega)$, which ensures the existence of a same-named subsequence converging weakly to a certain $u^* \in H^1(\Omega)$. The continuity of the trace operator yields in particular

$$ (62) \qquad \forall v \in H^1(\Omega), \quad \int_{\partial \Omega} u_n v \, dS \xrightarrow{n \to +\infty} \int_{\partial \Omega} u^* v \, dS. $$

Then, in order to show the convergence of the first term in (60), an integration lemma is needed.

**Lemma 9 (Convergence of integrals).** Let $(f_n)_{n \in \mathbb{N}}$ and $(g_n)_{n \in \mathbb{N}}$ be two sequences of $L^2(\Omega)$ functions, and $f, g \in L^2(\Omega)$ such that

$$ (63) \qquad f_n \xrightarrow{n \to +\infty} f \quad \text{and} \quad g_n \xrightarrow{n \to +\infty} g \quad \text{in } L^2(\Omega). $$

Then we have

$$ (64) \qquad \int_{\Omega} f_n g_n \xrightarrow{n \to +\infty} \int_{\Omega} fg. $$

This result will be proved after finishing the proof of Theorem 8.

The assumptions of Lemma 9 have to be checked.

* Since the sequence $(A(k_n))_{n \in \mathbb{N}}$ converges almost everywhere to $A(k)$, the domination $A(k_n)|\nabla v| \le \mu|\nabla v|$ and the Lebesgue theorem yield $A(k_n)\nabla v \xrightarrow{n \to +\infty} A(k)\nabla v$ in $L^2(\Omega)$.

* As a consequence of Rellich theorem, $\nabla u_n \xrightarrow{n \to +\infty} \nabla u^*$ in $L^2(\Omega)$.

Therefore, Lemma 9 yields

$$ (65) \qquad \forall v \in H^1(\Omega), \quad \int_{\Omega} A(k_n) \nabla u_n \cdot \nabla v \xrightarrow{n \to +\infty} \int_{\Omega} A(k) \nabla u^* \cdot \nabla v. $$

Both convergences (62) and (65) ensure that the limit $u^*$ satisfies (61). And since the solution of (61) is unique (by Theorem 6), $u^* = u = u(k)$.
Hence, $\nabla u_n \underset{n \to +\infty}{\rightharpoonup} \nabla u$ in $L^2(\Omega)$, and (52) ensures that there exists a subsequence of $(A(k_n)^{1/2}\nabla u_n)_{n \in \mathbb{N}}$ converging weakly in $L^2$. Therefore, Lemma 9 yields

$$ (66) \qquad A(k_n)^{1/2} \nabla u_n \underset{n \to +\infty}{\rightharpoonup} A(k)^{1/2} \nabla u \text{ in } L^2(\Omega). $$

In order to obtain the strong convergence of this sequence, the convergence of the norms remains to be proved. Let us write (60) with $v = u_n$ and (61) with $v = u$.

$$ (67) \qquad \forall n \in \mathbb{N}, \quad \int_{\Omega} A(k_n) |\nabla u_n|^2 + \alpha \int_{\partial \Omega} |u_n|^2 dS = \int_{\Omega} f u_n. $$

$$ (68) \qquad \int_{\Omega} A(k) |\nabla u|^2 + \alpha \int_{\partial \Omega} |u|^2 dS = \int_{\Omega} f u. $$

The strong convergence of $(u_n)_{n \in \mathbb{N}}$ in $L^2$ directly yields

$$ (69) \qquad \int_{\Omega} f u_n \underset{n \to +\infty}{\rightharpoonup} \int_{\Omega} f u. $$

The compact Sobolev embedding $H^{1/2}(\partial \Omega) \hookrightarrow L^2(\partial \Omega)$ yields in particular

$$ (70) \qquad \int_{\partial \Omega} |u_n|^2 dS \underset{n \to +\infty}{\rightharpoonup} \int_{\partial \Omega} |u|^2 dS. $$

Convergences (69) and (70) yield

$$ (71) \qquad \int_{\Omega} A(k_n) |\nabla u_n|^2 \underset{n \to +\infty}{\rightharpoonup} \int_{\Omega} A(k) |\nabla u|^2. $$

The result of Theorem 8 is proved.

<figure><img src="image_6.png" /></figure>

It remains to prove Lemma 9, who that the proof is complete.

*Proof of Lemma 9.* The difference between the two integrals is written

$$ (72) \qquad \int_{\Omega} (f_n g_n - fg) = \int_{\Omega} f(g_n - g) + \int_{\Omega} (f_n - f)g_n. $$

The weak convergence of $(g_n)_{n \in \mathbb{N}}$ yields the convergence of the first integral of (72), and the fact that $(g_n)_{n \in \mathbb{N}}$ is bounded in $L^2(\Omega)$ yields

$$ (73) \qquad \left| \int_{\Omega} (f_n - f)g_n \right| \le \|f_n - f\|_{L^2(\Omega)} \|g_n\|_{L^2(\Omega)} \le M \|f_n - f\|_{L^2(\Omega)} \underset{n \to +\infty}{\rightharpoonup} 0. $$

<figure><img src="image_7.png" /></figure>

**Theorem 10 (H¹ convergence).** With the same assumptions as in Theorem 8, there is a subsequence of $(u_n)_{n \in \mathbb{N}}$ still named the same such that

$$ (74) \qquad u_n \underset{n \to +\infty}{\rightharpoonup} u \text{ in } H^1(\Omega). $$
Proof of Theorem 10. The weak convergence and Rellich Theorem ensure that $u_n \xrightarrow{n \to +\infty} u$ in $L^2(\Omega)$. It remains to prove that $\nabla u_n \xrightarrow{n \to +\infty} \nabla u$ in $L^2(\Omega)$.

$$ (75) \qquad \begin{aligned} \sqrt{\nu} \|\nabla u_n - \nabla u\|_{L^2(\Omega)^N} &\le \|A(k_n)^{1/2} \nabla u_n - A(k_n)^{1/2} \nabla u\|_{L^2(\Omega)^N} \\ &\le \|A(k_n)^{1/2} \nabla u_n - A(k)^{1/2} \nabla u\|_{L^2(\Omega)^N} + \| (A(k_n)^{1/2} - A(k)^{1/2}) \nabla u \|_{L^2(\Omega)^N} \xrightarrow{n \to +\infty} 0, \end{aligned} \qquad \square $$
by Theorem 8 and Lebesgue Theorem.

# 4 Analysis of the second equation

The system considered in this section is the following

$$ (76) \qquad \left\{ \begin{array}{ll} -\operatorname{div}(B(k)\nabla k) = A(k)|\nabla u(k)|^2 & \text{in } \Omega, \\ k = 0 & \text{on } \partial\Omega. \end{array} \right. $$

The notations are lightened by defining

$$ (77) \qquad D(k) = A(k)|\nabla u(k)|^2. $$

## 4.1 Variational formulation and description of the method

In order to write the variational formulation of (76), we start by defining two function sets, and stating a regularity result.

**Definition 11.** Let the sets $V$ and $W$ be defined by

$$ (78) \qquad V = \bigcap_{1 \le r < N'} W_0^{1,r}(\Omega) \quad \text{and} \quad W = \bigcup_{N < r \le +\infty} W_0^{1,r}(\Omega). $$

Note that $V$ is a vectorial space, but not $W$.

We aim at writing a variational formulation searching solutions in $V$ with test functions in $W$.

**Proposition 12.** Let $k \in V$ and $\psi \in W$. Then $B(k)\nabla k \cdot \nabla \psi \in L^1(\Omega)$ and $D(k)\psi \in L^1(\Omega)$.

*Proof.* Let $k \in V$ and $\psi \in W$. By applying Hölder inequality with $r > N$ such that $\psi \in W_0^{1,r}(\Omega)$, we obtain

$$ (79) \qquad \int_{\Omega} B(k) |\nabla k \cdot \nabla \psi| \le \mu \int_{\Omega} |\nabla k \cdot \nabla \psi| \le \mu \|\nabla k\|_{L^{r'}(\Omega)} \|\nabla \psi\|_{L^r(\Omega)}. $$

Moreover, (52) yields $\|D(k)\|_{L^1(\Omega)} \le C$, then $D(k) \in L^1(\Omega)$ and $\psi \in L^\infty(\Omega)$. Hence $D(k)\psi \in L^1(\Omega)$. $\square$
**Remark 13.** The compact Sobolev embedding

$$ (80) \qquad V \hookrightarrow L^2(\Omega) $$

stands if $N \in \{2; 3\}$.

**Definition 14** (Variational formulation). A function $k$ is a weak solution of (76) if

$$ (81) \qquad \left\{ \begin{array}{l} k \in V \\ \forall \psi \in W, \quad \int_{\Omega} B(k) \nabla k \cdot \nabla \psi = \int_{\Omega} D(k) \psi. \end{array} \right. $$

As proved before, $D(k) \in L^1(\Omega)$, but it seems difficult to obtain a better regularity. In addition, (81) is not a linear problem, and could not be solved using only Lax-Milgram theorem, even if $D(k)$ was in $L^2(\Omega)$. These problems lead to use another method, which is divided into three steps:

* Regularizing the equation, by changing the source term.

* Linearizing it to solve the regularized problem with a fixed-point theorem.

* Taking the limit when the regularizing parameter goes to zero.

## 4.2 Two first steps: solving the regularized problem

Before linearizing Problem (81), the source term needs to be approached by a sequence of regular functions.

**Definition 15.** If $w \in L^r(\Omega)$ with $r \in [1, +\infty]$, $\tilde{w} \in L^r(\mathbb{R}^N)$ refers to the extension by zero

$$ \tilde{w} : x \mapsto \begin{cases} w(x) & \text{if } x \in \Omega \\ 0 & \text{if } x \notin \Omega. \end{cases} $$

**Definition 16** (Mollifier). Let $\rho: \mathbb{R}^N \to \mathbb{R}$ be a $C^\infty$ function satisfying

* $\text{Supp}(\rho) = \bar{B}(0, 1)$,

* $\int_{\mathbb{R}^N} \rho(y) dy = 1$.

* $\rho \ge 0$,

Let the family $(\rho_\varepsilon)_{\varepsilon>0}$ be defined by

$$ (82) \qquad \forall x \in \mathbb{R}^N, \quad \rho_\varepsilon(x) = \frac{1}{\varepsilon^N} \rho\left(\frac{x}{\varepsilon}\right). $$

Then $(\rho_\varepsilon)_{\varepsilon>0}$ is called a mollifier.

**Remark 17.** From now on, the term "mollifier" always refers to Definition 16.
**Definition 18 (Regularized problem).** Let the approximated source term $D_{\varepsilon}(k)$ be defined for any $\varepsilon > 0$ by

$$ (83) \qquad D_{\varepsilon}(k) = |\widetilde{U(k)} * \rho_{\varepsilon}|^2, $$

where $U(k) = A(k)^{1/2} \nabla u(k)$.

A function $k$ is called a weak solution of the regularized problem if it satisfies

$$ (84) \qquad \begin{cases} k \in H_0^1(\Omega) \\ \forall \psi \in H_0^1(\Omega), \quad \int_{\Omega} B(k) \nabla k \cdot \nabla \psi = \int_{\Omega} D_{\varepsilon}(k) \psi. \end{cases} $$

Linearizing Problem (84) is useful if and only if Lax-Milgram theorem can be applied in the Hilbert space $H_0^1(\Omega)$. To do so, the linear form $[\psi \mapsto \int_{\Omega} D_{\varepsilon}(k) \psi]$ must be continuous on $H_0^1(\Omega)$.

**Theorem 19.** Let $\varepsilon > 0$. The function $D_{\varepsilon}(k)$ defined just above is in $L^2(\Omega)$ and there exists a constant $C > 0$ independent of $k$ and $\varepsilon$ such that

$$ (85) \qquad \|D_{\varepsilon}(k)\|_{L^2(\Omega)} \le \frac{C}{\varepsilon^2}. $$

The proof of Theorem 19 is based on a general result.

**Lemma 20.** Let $v \in H^m(\mathbb{R}^N)$, $\ell \in \mathbb{N}$ and $(\rho_{\varepsilon})_{\varepsilon>0}$ a mollifier. For $\varepsilon > 0$ sufficiently close to 0, we have

$$ (86) \qquad \|v * \rho_{\varepsilon}\|_{H^{m+\ell}(\mathbb{R}^N)} \le \frac{C(m, \ell)}{\varepsilon^{\ell}} \|v\|_{H^m(\mathbb{R}^N)}. $$

*Proof of Lemma 20.* The $H^{m+\ell}$ norm can be defined by

$$ (87) \qquad \|v * \rho_{\varepsilon}\|_{H^{m+\ell}(\mathbb{R}^N)}^2 = \sum_{\substack{|\alpha| \le m \\ |\beta| \le \ell}} \|\partial^{\alpha+\beta} (v * \rho_{\varepsilon})\|_{L^2(\mathbb{R}^N)}^2. $$

Let $\alpha, \beta \in \mathbb{N}^d$ such that $|\alpha| \le m$ and $|\beta| \le \ell$. Plancherel theorem yields

$$ (88) \qquad \|\partial^{\alpha+\beta} (v * \rho_{\varepsilon})\|_{L^2(\mathbb{R}^N)}^2 = \frac{1}{(2\pi)^N} \|\mathcal{F}(\partial^{\alpha+\beta} (v * \rho_{\varepsilon}))\|_{L^2(\mathbb{R}^N)}^2, $$

where $\mathcal{F}$ refers to the Fourier transform operator.

Since $v \in H^m(\mathbb{R}^N)$, the derivation formula $\partial^{\alpha+\beta}(v * \rho_{\varepsilon}) = (\partial^{\alpha}v) * (\partial^{\beta}\rho_{\varepsilon})$ and the Fourier transform yields

$$ (89) \qquad \mathcal{F}(\partial^{\alpha+\beta}(v * \rho_{\varepsilon})) = \mathcal{F}(\partial^{\alpha}v)\mathcal{F}(\partial^{\beta}\rho_{\varepsilon}). $$

Then with norms

$$ (90) \qquad \|\mathcal{F}(\partial^{\alpha+\beta}(v * \rho_{\varepsilon}))\|_{L^2(\mathbb{R}^N)}^2 = \int_{\mathbb{R}^N} |\mathcal{F}(\partial^{\alpha}v)(\xi)|^2 \cdot |\mathcal{F}(\partial^{\beta}\rho_{\varepsilon})(\xi)|^2 d\xi. $$
A linear change of variables yields

$$ (91) \qquad \mathcal{F}(\rho_{\varepsilon})(\xi) = \frac{1}{\varepsilon^N} \int_{\mathbb{R}^N} e^{-ix\xi} \rho\left(\frac{x}{\varepsilon}\right) dx = \int_{y=x/\varepsilon} \int_{\mathbb{R}^N} e^{-i\varepsilon y\xi} \rho(y) dy = \mathcal{F}(\rho)(\varepsilon\xi). $$

Then we obtain

$$ (92) \qquad \begin{aligned} |\mathcal{F}(\partial^\beta \rho_\varepsilon)(\xi)|^2 &= |\xi^\beta \mathcal{F}(\rho_\varepsilon)(\xi)|^2 = |\xi^\beta \mathcal{F}(\rho)(\varepsilon\xi)|^2 \\ &= \frac{1}{\varepsilon^{2|\beta|}} |(\varepsilon\xi)^\beta \mathcal{F}(\rho)(\varepsilon\xi)|^2 = \frac{1}{\varepsilon^{2|\beta|}} |\mathcal{F}(\partial^\beta \rho)(\varepsilon\xi)|^2. \end{aligned} $$

Yet, $\partial^\beta \rho \in \mathcal{D}(\mathbb{R}^N) \subset L^1(\mathbb{R}^N)$, which yields $\forall \xi \in \mathbb{R}^N$, $|\mathcal{F}(\partial^\beta \rho)(\xi)| \le \|\partial^\beta \rho\|_{L^1(\mathbb{R}^N)}$. Then (90) ensures the existence of a constant $c_0 > 0$ independent of $v$ such that

$$ (93) \qquad \| \mathcal{F}(\partial^{\alpha+\beta} (v * \rho_{\varepsilon})) \|_{L^2(\mathbb{R}^N)}^2 \le \frac{c_0}{\varepsilon^{2|\beta|}} \| \mathcal{F}(\partial^{\alpha} v) \|_{L^2(\mathbb{R}^N)}^2. $$

Plancherel Theorem then yields

$$ (94) \qquad \| \partial^{\alpha+\beta} (v * \rho_{\varepsilon}) \|_{L^2(\mathbb{R}^N)}^2 \le \frac{c_0}{\varepsilon^{2|\beta|}} \| \partial^{\alpha} v \|_{L^2(\mathbb{R}^N)}^2. $$

Then, by addition

$$ (95) \qquad \sum_{|\alpha| \le m} \| \partial^{\alpha+\beta} (v * \rho_{\varepsilon}) \|_{L^2(\mathbb{R}^N)}^2 \le \frac{c_0}{\varepsilon^{2|\beta|}} \| v \|_{H^m(\mathbb{R}^N)}^2. $$

Taking $\varepsilon \in ]0, 1[$ yields $\frac{1}{\varepsilon^{2|\beta|}} \le \frac{1}{\varepsilon^{2\ell}}$. Thus, adding for $|\beta| \le \ell$ yields

$$ (96) \qquad \| v * \rho_{\varepsilon} \|_{H^{m+\ell}(\mathbb{R}^N)}^2 \le \frac{c(\ell)}{\varepsilon^{2\ell}} \| v \|_{H^m(\mathbb{R}^N)}^2, $$

hence (86).

<figure><img src="image_8.png" /></figure>

We can now apply this result to prove Theorem 19, recalling that

$$ (97) \qquad \| U(k) \|_{L^2(\Omega)^N} \le C, $$

where $C > 0$ depends on $f$, $\Omega$, $\alpha$, $\mu$ and $\nu$. This is a rewriting of (52).

*Proof of Theorem 19.* The definition of $D_\varepsilon$ yields

$$ (98) \qquad \| D_\varepsilon(k) \|_{L^2(\Omega)} = \left( \int_{\Omega} |\widetilde{U(k)} * \rho_\varepsilon|^4 \right)^{1/2} = \| \widetilde{U(k)} * \rho_\varepsilon \|_{L^4(\Omega)^N}^2. $$

Sobolev embedding $H^1(\Omega) \hookrightarrow L^4(\Omega)$ and Lemma 20 then yield

$$ (99) \qquad \| \widetilde{U(k)} * \rho_{\varepsilon} \|_{L^4(\Omega)^N} \le c_1 \| \widetilde{U(k)} * \rho_{\varepsilon} \|_{H^1(\Omega)^N} \le \frac{c_2}{\varepsilon} \| U(k) \|_{L^2(\Omega)^N}. $$

Finally, (97) yields the result.

<figure><img src="image_9.png" /></figure>
In order to solve the regularized problem (84), we first linearize it, as said before.

**Definition 21.** A function $g = g(k)$ is said to be a weak solution of the linearized problem associated to (76) if

$$ (100) \qquad \begin{cases} g = g(k) \in H_0^1(\Omega) \\ \forall \psi \in H_0^1(\Omega), \quad \int_{\Omega} B(k) \nabla g \cdot \nabla \psi = \int_{\Omega} D_{\varepsilon}(k) \psi. \end{cases} $$

**Lemma 22.** Let $k \in L^1(\Omega)$, there exists a unique solution to Problem (100).

*Proof.* This is a consequence of Lax-Milgram theorem. Theorem 19 ensures that $[\psi \mapsto \int_{\Omega} D_{\varepsilon}(k)\psi]$ is a continuous linear form on $H_0^1(\Omega)$. Let the bilinear form $b$ be defined by

$$ (101) \qquad b : (g, \psi) \mapsto \int_{\Omega} B(k) \nabla g \cdot \nabla \psi. $$

First, let $g, \psi \in H_0^1(\Omega)$

$$ (102) \qquad |b(g, \psi)| \le \mu \|\nabla g\|_{L^2(\Omega)^N} \|\nabla \psi\|_{L^2(\Omega)^N} \le \|g\|_{H^1(\Omega)} \|\psi\|_{H^1(\Omega)}. $$

The continuity of $b$ is proved.

Then, let $\psi \in H_0^1(\Omega)$, we have

$$ (103) \qquad b(\psi, \psi) = \int_{\Omega} B(k) |\nabla \psi|^2 \ge \nu \|\nabla \psi\|_{L^2(\Omega)^N}^2 \ge c(\Omega) \nu \|\psi\|_{H^1(\Omega)}^2, $$

as a consequence of Poincaré inequality. The coercivity of $b$ is proved.

Lax-Milgram theorem yields the existence and unicity of the solution $g$. $\square$

Now, we aim at showing the existence of a solution to problem (84). To do so, we recall the Leray-Schauder fixed-point theorem (see [9]).

**Theorem 23 (Leray-Schauder).** Let $X$ a Banach space, and $B$ a convex closed bounded nonempty subset of $X$. Let $f: B \to B$ a continuous function with $f(B)$ compact.

Then $f$ admits at least one fixed point in $B$.

This result is the main ingredient of the proof of the following statement.

**Theorem 24.** The regularized problem (84) admits a solution.

*Proof.* Let the operator $T$ be defined by

$$ (104) \qquad T : k \mapsto g(k), $$

where $g(k)$ refers to the solution of Problem (100). In other words, $Tk \in H_0^1(\Omega)$ satisfies

$$ (105) \qquad \forall \psi \in H_0^1(\Omega), \quad \int_{\Omega} B(k) \nabla Tk \cdot \nabla \psi = \int_{\Omega} D_{\varepsilon}(k) \psi. $$

The method consists in applying Theorem 23 to the operator $T$. First, taking $\psi = Tk$ in (105) yields

$$ (106) \qquad \int_{\Omega} B(k) |\nabla Tk|^2 = \int_{\Omega} D_{\varepsilon}(k) Tk, \quad \text{hence } \nu \|\nabla Tk\|_{L^2(\Omega)^N} \le c_1 \|D_{\varepsilon}(k)\|_{L^2(\Omega)}^2. $$
Then Theorem 19 yields

$$ (107) \qquad \|\nabla Tk\|_{L^2(\Omega)^N} \le c_2 \|D_\varepsilon(k)\|_{L^2(\Omega)} \le M, $$

where $M > 0$ does not depend on $k$.

Considering the notations of Theorem 23, in this problem we take $X = H_0^1(\Omega)$, equipped with the scalar product

$$ (108) \qquad (f,g)_{H_0^1(\Omega)} = \int_{\Omega} \nabla f \cdot \nabla g, $$

and its associated topology.

Moreover, we take $B = \bar{B}(0, M)$ which is a convex closed bounded nonempty subset of $H_0^1(\Omega)$. It has been shown that $T(B) \subset B$. Now it remains to prove that $T$ is continuous and $T(B)$ is a compact set.

### Compactness of $T(B)$:

Let $(k_n)_{n \in \mathbb{N}}$ be any sequence of $B$. We aim at proving the existence of a subsequence (named the same way) converging in $H_0^1(\Omega)$ to a certain $Tk$, with $k \in B$. Given that $B$ is bounded, there is a subsequence satisfying

$$ (109) \qquad k_n \underset{n \to +\infty}{\to} k \quad \text{in } H_0^1(\Omega), $$

where $k \in B$.

The assumptions on $B$ and (107) yield that $(B(k_n)^{1/2} \nabla Tk_n)_{n \in \mathbb{N}}$ is bounded in $L^2(\Omega)^N$, so there exists a subsequence converging weakly to a certain function. In the following, we state this limit is $B(k)^{1/2} \nabla Tk$.

Let $(k_n)_{n \in \mathbb{N}}$ be a sequence of elements of $B$. Since it is bounded, there is a subsequence named the same satisfying $k_n \underset{n \to +\infty}{\to} k$ in $H_0^1(\Omega)$.

We first aim at showing that for any $\varepsilon > 0$ and $\psi \in H_0^1(\Omega)$

$$ (110) \qquad \int_{\Omega} D_{\varepsilon}(k_n) \psi \underset{n \to +\infty}{\longrightarrow} \int_{\Omega} D_{\varepsilon}(k) \psi. $$

The difference is rewritten as follows

$$ (111) \qquad \begin{aligned} D_{\varepsilon}(k_n) - D_{\varepsilon}(k) &= |\widetilde{U(k_n)} * \rho_{\varepsilon}|^2 - |\widetilde{U(k)} * \rho_{\varepsilon}|^2 \\ &= [(\widetilde{U(k_n)} - \widetilde{U(k)}) * \rho_{\varepsilon}] \cdot [(\widetilde{U(k_n)} + \widetilde{U(k)}) * \rho_{\varepsilon}]. \end{aligned} $$

Hölder and then Gagliardo-Nirenberg inequality yield

$$ (112) \qquad \begin{aligned} & \left| \int_{\Omega} (D_{\varepsilon}(k_n) - D_{\varepsilon}(k)) \psi \right| \\ & \le \left\| (\widetilde{U(k_n)} - \widetilde{U(k)}) * \rho_{\varepsilon} \right\|_{L^3(\mathbb{R}^N)^N} \left\| (\widetilde{U(k_n)} + \widetilde{U(k)}) * \rho_{\varepsilon} \right\|_{L^3(\mathbb{R}^N)^N} \|\psi\|_{L^3(\Omega)} \\ & \le c_1 \left\| (\widetilde{U(k_n)} - \widetilde{U(k)}) * \rho_{\varepsilon} \right\|_{H^1(\mathbb{R}^N)^N} \left\| (\widetilde{U(k_n)} + \widetilde{U(k)}) * \rho_{\varepsilon} \right\|_{H^1(\mathbb{R}^N)^N} \|\psi\|_{L^3(\Omega)}. \end{aligned} $$
Lemma 20 and Young convolution inequality yield

$$
(113) \quad \left| \int_{\Omega} (D_{\varepsilon}(k_n) - D_{\varepsilon}(k)) \psi \right| \le \frac{c_2}{\varepsilon^2} \|U(k_n) - U(k)\|_{L^2(\Omega)^N} \|U(k_n) + U(k)\|_{L^2(\Omega)^N} \|\psi\|_{L^3(\Omega)}.
$$

According to Theorem 8, the right hand side of (113) converges to 0 when $n$ goes to $+\infty$.
This yields (110).

Both (110) and (105) imply

$$
(114) \quad \forall \psi \in H_0^1(\Omega), \quad \int_{\Omega} B(k_n) \nabla T k_n \cdot \nabla \psi \underset{n \to +\infty}{\longrightarrow} \int_{\Omega} B(k) \nabla T k \cdot \nabla \psi.
$$

The uniqueness of the limit yields that

$$
(115) \qquad B(k_n)^{1/2} \nabla T k_n \underset{n \to +\infty}{\longrightarrow} B(k)^{1/2} \nabla T k \quad \text{in } L^2(\Omega)^N.
$$

In addition, Lebesgue inverse theorem and the continuity of 1/B yield $\frac{1}{B(k_n)} \xrightarrow{n \to +\infty} \frac{1}{B(k)}$ a.e. in $\Omega$, and in addition $\frac{1}{B(k_n)}|\nabla\psi| \le \frac{1}{\nu}|\nabla\psi|$. Then we obtain

$$
(116) \quad \frac{1}{B(k_n)} \nabla \psi \xrightarrow{n \to +\infty} \frac{1}{B(k)} \nabla \psi \quad \text{in } L^2(\Omega)^N.
$$

Convergences (115) and (116), and Lemma 9 then yield for any $\psi \in H_0^1(\Omega)$

$$
(117) \qquad \int_{\Omega} \nabla T k_n \cdot \nabla \psi \underset{n \to +\infty}{\longrightarrow} \int_{\Omega} \nabla T k \cdot \nabla \psi.
$$

In other words, $Tk_n \xrightarrow{n \to +\infty} Tk$ in $H_0^1(\Omega)$.

The convergence of the norms has to be established. To do so, we have to prove

$$
(118) \qquad \int_{\Omega} D_{\varepsilon}(k_n) T k_n \underset{n \to +\infty}{\longrightarrow} \int_{\Omega} D_{\varepsilon}(k) T k.
$$

We start by writing the inequality

$$
(119) \quad \begin{aligned}
\| D_{\varepsilon}(k_n) T k_n - D_{\varepsilon}(k) T k \|_{L^1(\Omega)} &\le \|(D_{\varepsilon}(k_n) - D_{\varepsilon}(k)) T k\|_{L^1(\Omega)} \\
&\quad + \|D_{\varepsilon}(k_n) (T k_n - T k)\|_{L^1(\Omega)}.
\end{aligned}
$$

On one hand, (110) yields $\|(D_{\varepsilon}(k_n) - D_{\varepsilon}(k))Tk\|_{L^1(\Omega)} \xrightarrow{n \to +\infty} 0$.

On the other hand, by compact Sobolev embedding, we have $Tk_n \xrightarrow{n \to +\infty} Tk$ in $L^2(\Omega)$.
Hölder inequality then yields

$$
(120) \quad \| D_{\varepsilon}(k_n) (Tk_n - Tk) \|_{L^1(\Omega)} \le \| D_{\varepsilon}(k_n) \|_{L^2(\Omega)} \| Tk_n - Tk \|_{L^2(\Omega)} \xrightarrow{n \to +\infty} 0.
$$

Hence, (118) is proved. Reporting in (105) yields

$$
(121) \qquad \int_{\Omega} B(k_n) |\nabla T k_n|^2 \underset{n \to +\infty}{\longrightarrow} \int_{\Omega} B(k) |\nabla T k|^2.
$$
Both (115) and (121) yield

$$ (122) \qquad B(k_n)^{1/2} \nabla T k_n \underset{n \to +\infty}{\longrightarrow} B(k)^{1/2} \nabla T k \quad \text{in } L^2(\Omega)^N. $$

Lebesgue inverse theorem ensures that another subsequence of $(\nabla T k_n)_{n \in \mathbb{N}}$ is dominated by a $L^1(\Omega)^N$ function. Lebesgue theorem then yields

$$ (123) \qquad \int_{\Omega} (B(k_n) - B(k)) |\nabla T k_n|^2 \underset{n \to +\infty}{\longrightarrow} 0. $$

Consequently, the following decomposition

$$ (124) \qquad \int_{\Omega} (B(k_n)|\nabla T k_n|^2 - B(k)|\nabla T k|^2) = \int_{\Omega} B(k) (|\nabla T k_n|^2 - |\nabla T k|^2) \\ \qquad + \int_{\Omega} (B(k_n) - B(k)) |\nabla T k_n|^2 $$

implies that

$$ (125) \qquad \int_{\Omega} B(k) (|\nabla T k_n|^2 - |\nabla T k|^2) \underset{n \to +\infty}{\longrightarrow} 0. $$

Let $N_B$ be defined by

$$ (126) \qquad N_B : f \in H_0^1(\Omega) \mapsto \|B(k)\nabla f\|_{L^2(\Omega)^N}. $$

The norms $\| \cdot \|_{H_0^1(\Omega)}$ and $N_B$ are equivalent, and (125) ensures that $(Tk_n)_{n \in \mathbb{N}}$ converges strongly for the norm $N_B$, so as for the $H_0^1(\Omega)$ norm.

We now have proved that $T(B)$ is compact.

### Continuity of $T$:

Let $(k_n)_{n \in \mathbb{N}}$ be a sequence of $B$ such that $k_n \underset{n \to +\infty}{\longrightarrow} k$ in $H_0^1(\Omega)$. The computations made before ensure that a subsequence of $(Tk_n)_{n \in \mathbb{N}}$ converges to $Tk$ in $H_0^1(\Omega)$. Moreover, $Tk$ is the only accumulation point of this bounded sequence. As a result, the whole sequence satisfies

$$ (127) \qquad Tk_n \underset{n \to +\infty}{\longrightarrow} Tk \quad \text{in } H_0^1(\Omega). $$

All the assumptions of Theorem 23 are satisfied. Consequently, $T$ admits a fixed point in $B$, which corresponds to a solution to Problem (84). □

## 4.3 Last step: Taking the limit when $\varepsilon$ goes to zero

This subsection aims at proving the existence of a solution to the initial variational TKE problem, by analyzing the convergence of the approximated solutions found just before.

**Theorem 25.** There exists a function $k \in V$ satisfying (81).
Proof. Let $(k_\epsilon)_{\epsilon>0}$ be the sequence of approximation functions given in the previous subsection. Then for any $\epsilon > 0$, $k_\epsilon$ satisfies

$$ (128) \qquad \forall \psi \in H_0^1(\Omega), \quad \int_{\Omega} B(k_{\epsilon}) \nabla k_{\epsilon} \cdot \nabla \psi = \int_{\Omega} D_{\epsilon}(k_{\epsilon}) \psi. $$

We intend to show that $(k_\epsilon)_{\epsilon>0}$ is bounded for a certain topology. To do so, we use Lemma 1. A general theorem from Stampacchia (see [8]) is needed before.

**Theorem 26 (Regularity and truncation function).** Let $H \in W^{1,\infty}(\mathbb{R})$ with a finite number of discontinuities, and such that $H(0) = 0$. Let $q \in [1; +\infty[$ and $v \in W_0^{1,q}(\Omega)$.

Then $H(v) \in W_0^{1,q}(\Omega)$ and $\nabla H(v) = H'(v)\nabla v$.

Until the end of this chapter, for $j \in \mathbb{N}$, $H_j$ refers to the truncation function defined by

<figure><img src="image_10.png" /></figure>

This function satisfies the assumptions of Theorem 26, so that we can take $\psi = H_j(k_\epsilon)$ in (128) to obtain

$$ (129) \qquad \forall j \in \mathbb{N}, \quad \int_{\Omega} B(k_{\epsilon}) |\nabla k_{\epsilon}|^2 H'_{j}(k_{\epsilon}) = \int_{\Omega} D_{\epsilon}(k_{\epsilon}) H_{j}(k_{\epsilon}). $$

On one hand,

$$ (130) \qquad \int_{\Omega} B(k_{\epsilon}) |\nabla k_{\epsilon}|^2 H'_{j}(k_{\epsilon}) = \int_{B_{j}^{\epsilon}} B(k_{\epsilon}) |\nabla k_{\epsilon}|^2 \geq \nu \int_{B_{j}^{\epsilon}} |\nabla k_{\epsilon}|^2, $$

where $B_j^\epsilon$ refers to the subset $\{x \in \Omega / j \leq |k_\epsilon(x)| < j + 1\}$.

One the other hand, Young convolution inequality yields

$$ (131) \qquad \int_{\Omega} D_{\epsilon}(k_{\epsilon}) \underbrace{H_j(k_{\epsilon})}_{\leq 1} \leq \|\widetilde{U(k_{\epsilon})} * \rho_{\epsilon}\|_{L^2(\Omega)^N}^2 \leq \|U(k_{\epsilon})\|_{L^2(\Omega)^N}^2. $$

Inequalities (130), (131) and (97) reported in (129) yield the existence of a constant $C > 0$ not depending on $k$ and $\epsilon$ such that

$$ (132) \qquad \forall j \in \mathbb{N}, \quad \int_{B_j^\epsilon} |\nabla k_\epsilon|^2 \leq C. $$

Lemma 1 then yields $\|k_\epsilon\|_{W^{1,q}(\Omega)} \leq A$, for $q \in [1; N']$. And given that $C$ does not depend on $\epsilon$, it is the same for $A$.

Thus, there exists a sequence $(\epsilon_n)_{n \in \mathbb{N}}$ such that $\epsilon_n \xrightarrow{n \to +\infty} 0$ and $k_{\epsilon_n} \xrightarrow{n \to +\infty} k$ in $W_0^{1,q}(\Omega)$.
The function $k$ is in $V$.
We aim at showing that $k$ is a solution of (81), which means

$$ (133) \qquad \forall \psi \in H_0^1(\Omega), \quad \int_{\Omega} B(k) \nabla k \cdot \nabla \psi = \int_{\Omega} D(k) \psi. $$

We recall that the intermediary solutions $(k_{\varepsilon_n})$ satisfy

$$ (134) \qquad \forall \psi \in H_0^1(\Omega), \quad \int_{\Omega} B(k_{\varepsilon_n}) \nabla k_{\varepsilon_n} \cdot \nabla \psi = \int_{\Omega} D_{\varepsilon_n}(k_{\varepsilon_n}) \psi. $$

**Convergence of the left hand side:**

Given that $k_{\varepsilon_n} \underset{n \to +\infty}{\to} k$ in $V$, we have

$$ (135) \qquad \forall \psi \in W, \quad \int_{\Omega} \nabla k_{\varepsilon_n} \cdot \nabla \psi \underset{n \to +\infty}{\longrightarrow} \int_{\Omega} \nabla k \cdot \nabla \psi. $$

In order to add the term $B(k_n)$, it seems useful to state a variant of Lemma 9.

**Lemma 27** (Double convergence in $V$ and $W$). Let $\Omega$ be an open subset of $\mathbb{R}^N$, $f \in \bigcap_{1<r<N'} L^r(\Omega)$ and $g \in L^q(\Omega)$, where $q > N$. Let us consider two sequences $(f_n)_{n \in \mathbb{N}}$ and $(g_n)_{n \in \mathbb{N}}$ such that

$$ (136) \qquad \left\{ \begin{array}{ll} f_n \underset{n \to +\infty}{\longrightarrow} f & \text{in } L^r(\Omega) \text{ for } r \in [1; N'], \\ g_n \underset{n \to +\infty}{\longrightarrow} g & \text{in } L^q(\Omega). \end{array} \right. $$

Then we have

$$ (137) \qquad \int_{\Omega} f_n g_n \underset{n \to +\infty}{\longrightarrow} \int_{\Omega} fg. $$

*Proof of Lemma 27.* The proof of Lemma 27 is similar to the proof of Lemma 9. We write

$$ (138) \qquad \int_{\Omega} (f_n g_n - fg) = \int_{\Omega} (f_n - f)g_n + \int_{\Omega} f(g_n - g). $$

On one hand, the Hölder inequality yields

$$ (139) \qquad \left| \int_{\Omega} (f_n - f)g_n \right| \le \|f_n - f\|_{L^{q'}(\Omega)} \|g_n\|_{L^q(\Omega)} \le M \|f_n - f\|_{L^{q'}(\Omega)} \underset{n \to +\infty}{\longrightarrow} 0, $$

where $q'$ refers to the conjugate of $q$, which is in $[1; N']$.

On the other hand, the weak convergence of $(g_n)_{n \in \mathbb{N}}$ yields

$$ (140) \qquad \int_{\Omega} (g_n - g) f \underset{n \to +\infty}{\longrightarrow} 0. $$

Hence the result.

□
We now check if the assumptions of Lemma 27 are satisfied.

By compact Sobolev embedding, $k_{\varepsilon_n} \underset{n \to +\infty}{\longrightarrow} k$ in $L^1(\Omega)$. Thus, Lebesgue inverse theorem yields that a subsequence satisfies $k_{\varepsilon_n} \underset{n \to +\infty}{\longrightarrow} k$ a.e. in $\Omega$. The continuity of $B$ yields $B(k_{\varepsilon_n}) \underset{n \to +\infty}{\longrightarrow} B(k)$ a.e. in $\Omega$. And the domination $B(k_{\varepsilon_n})|\nabla\psi| \le \mu|\nabla\psi|$ stands. Consequently, Lebesgue theorem yields

$$ (141) \qquad \forall r \in [1; N'[, \forall \psi \in W, \quad B(k_{\varepsilon_n})\nabla\psi \underset{n \to +\infty}{\longrightarrow} B(k)\nabla\psi \quad \text{in } L^r(\Omega). $$

The weak convergence of $(k_{\varepsilon_n})_{n \in \mathbb{N}}$ being already established, Lemma 27 yields

$$ (142) \qquad \forall \psi \in W, \quad \int_{\Omega} B(k_{\varepsilon_n}) \nabla k_{\varepsilon_n} \cdot \nabla \psi \underset{n \to +\infty}{\longrightarrow} \int_{\Omega} B(k) \nabla k \cdot \nabla \psi. $$

**Convergence of the right hand side:**

We intend to show that

$$ (143) \qquad \forall \psi \in W, \quad \int_{\Omega} D_{\varepsilon_n}(k_{\varepsilon_n})\psi \underset{n \to +\infty}{\longrightarrow} \int_{\Omega} D(k)\psi, $$

at least for a subsequence.

Given that $\psi \in L^\infty(\Omega)$, it is sufficient to prove that

$$ (144) \qquad D_{\varepsilon_n}(k_{\varepsilon_n}) \underset{n \to +\infty}{\longrightarrow} D(k) \quad \text{in } L^1(\Omega). $$

The $L^1$ norm is decomposed as follows

$$ (145) \qquad \int_{\Omega} |D_{\varepsilon_n}(k_{\varepsilon_n}) - D(k)| \le \int_{\Omega} |D_{\varepsilon_n}(k_{\varepsilon_n}) - D_{\varepsilon_n}(k)| + \int_{\Omega} |D_{\varepsilon_n}(k) - D(k)|. $$

Cauchy-Schwarz inequality first yields

$$ (146) \qquad \int_{\Omega} |D_{\varepsilon_n}(k) - D(k)| = \int_{\Omega} |\widetilde{U(k)} * \rho_{\varepsilon_n}|^2 - |U(k)|^2 \\ \le \|\widetilde{U(k)} * \rho_{\varepsilon_n} - \widetilde{U(k)}\|_{L^2(\mathbb{R}^N)^N} \|\widetilde{U(k)} * \rho_{\varepsilon_n} + \widetilde{U(k)}\|_{L^2(\mathbb{R}^N)^N}. $$

We recall the following classical result.

**Theorem 28.** Let $p \in [1; +\infty[$, $f \in L^p(\mathbb{R}^N)$, and $(\rho_\varepsilon)_{\varepsilon>0}$ a mollifier. Then we have

$$ (147) \qquad \rho_{\varepsilon} * f \underset{\varepsilon \to 0}{\longrightarrow} f \quad \text{in } L^p(\mathbb{R}^N). $$

Theorem 28 and (146) yield

$$ (148) \qquad \int_{\Omega} |D_{\varepsilon_n}(k) - D(k)| \underset{n \to +\infty}{\longrightarrow} 0. $$

Similarly, we write

$$ (149) \qquad \int_{\Omega} |D_{\varepsilon_n}(k_{\varepsilon_n}) - D_{\varepsilon_n}(k)| \le \| (\widetilde{U(k_{\varepsilon_n})} - \widetilde{U(k)}) * \rho_{\varepsilon_n} \|_{L^2(\mathbb{R}^N)^N} \| (\widetilde{U(k_{\varepsilon_n})} + \widetilde{U(k)}) * \rho_{\varepsilon_n} \|_{L^2(\mathbb{R}^N)^N}. $$
On one hand, Young convolution inequality and (97) yield

$$ (150) \qquad \left\| \left( \widetilde{U(k_{\epsilon_n})} + \widetilde{U(k)} \right) * \rho_{\epsilon_n} \right\|_{L^2(\mathbb{R}^N)^N} \le \|U(k_{\epsilon_n})\|_{L^2(\Omega)^N} + \|U(k)\|_{L^2(\Omega)^N} \le C. $$

On the other hand, Young convolution inequality and Theorem 8 yield for a subsequence

$$ (151) \qquad \left\| \left( \widetilde{U(k_{\epsilon_n})} - \widetilde{U(k)} \right) * \rho_{\epsilon_n} \right\|_{L^2(\mathbb{R}^N)^N} \le \|U(k_{\epsilon_n}) - U(k)\|_{L^2(\Omega)^N} \underset{n \to +\infty}{\longrightarrow} 0. $$

We have obtained

$$ (152) \qquad \int_{\Omega} |D_{\epsilon_n}(k_{\epsilon_n}) - D_{\epsilon_n}(k)| \underset{n \to +\infty}{\longrightarrow} 0, $$

hence (144).

As a conclusion, $k$ satisfies (81).

<figure><img src="image_11.png" /></figure>

## 4.4 Conclusion: solution of the coupled problem

Let us consider the sequence $(\epsilon_n)$ mentioned previously, after all the extractions that were done. We have shown that $(k_{\epsilon_n})$ strongly converges in $L^2(\Omega)$ to a function $k$.

Moreover, $(k_{\epsilon_n})$ weakly converges to $k$ in every space $W_0^{1,q}(\Omega)$ with $q \in [1; N']$. Therefore, theorem 10 yields that $(u(k_{\epsilon_n}))$ strongly converges to $u(k)$ in $H^1(\Omega)$.

All of this ensures the existence of a couple $(u; k)$ of solutions of (1) at the distribution sense, with $u \in H^1(\Omega)$ and $k \in V$ (in particular $k \in L^2$).

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