Here, $p'$ is the Hölder conjugate of $p$. When $\psi$ run through a Schauder basis of $W_0^{1,p'}(V_i)$ we obtain that for a.e. $X_1 \in U_i$

$$ \int_{V_i} A_{22}(X_1, \cdot) \nabla_{X_2} u(X_1, \cdot) \cdot \nabla_{X_2} \psi dX_2 = \int_{V_i} f(X_1, \cdot) \psi dX_2, \forall \psi \in W_0^{1,p'}(V_i). \quad (4.13) $$

Since $\Pi_1\Omega$ could be covered by a countable set of $U_i$ then form (4.13) we get

$$ \text{For a.e. } X_1 \in \Pi_1\Omega: \int_{V_i} A_{22}(X_1, \cdot) \nabla_{X_2} u(X_1, \cdot) \cdot \nabla_{X_2} \psi dX_2 = \int_{V_i} f(X_1, \cdot) \psi dX_2, \forall \psi \in W_0^{1,p'}(V_i). \quad (4.14) $$

Now, let $\psi \in D(\Omega_{X_1})$ then the Supp$\psi$ could be covered by a finite set of $V_i$, by using a partition of unity subordinate to such a cover, and by using (4.14) we get

$$ \int_{\Omega_{X_1}} A_{22}(X_1, \cdot) \nabla_{X_2} u(X_1, \cdot) \cdot \nabla_{X_2} \psi dX_2 = \int_{\Omega_{X_1}} f(X_1, \cdot) \varphi dX_2, \psi \in D(\Omega_{X_1}), $$

for a.e $X_1 \in \Pi_1\Omega$. Finally, since $u(X_1, \cdot) \in W_0^{1,p}(\Omega_{X_1})$ (as we proved above), then $u(X_1, \cdot)$ is a weak solution of (4.3)-(4.4). $\square$

## 1.2 Strong convergence of the entropy solutions

Theorem 0.16 will be proved in three steps. It is well known that the entropy solution are the solution constructed by approximation [31, 38]. The proof is based on the use of the approximation problem (4.5). Throughout this subsection C denotes a positive generic constant independent of $n$ and $\epsilon$.

### Step 1

In this step we construct $u$, the entropy solution of the limit problem (4.3). Let $u_\epsilon^n \in H_0^1(\Omega)$ be the unique solution to (4.5), we have the following proposition.

**Proposition 4.2.** Under assumptions of Theorem 0.16 there exists a unique sequence $(u^n)_{n \in \mathbb{N}} \in H_0^1(\Omega; \Pi_2\Omega)^N$ such that $\epsilon \nabla_{X_1} u_\epsilon^n \to 0$ in $L^2(\Omega)$, $u_\epsilon^n \to u^n$ in $H_0^1(\Omega; \Pi_2\Omega)$ as $\epsilon \to 0$, for every $n \in \mathbb{N}$, in particular the two convergences hold in $L^p(\Omega)$ and $W_0^{1,p}(\Omega; \Pi_2\Omega)$ respectively, where $u^n$ is the unique solution in $H_0^1(\Omega; \Pi_2\Omega)$ to the problem

$$ \text{For a.e. } X_1 \in \Pi_1\Omega, \quad \int_{\Omega_{X_1}} \beta(u^n)\varphi dX_2 + \int_{\Omega_{X_1}} A_{22} \nabla_{X_2} u^n \cdot \nabla_{X_2} \varphi dX_2 = \int_{\Omega_{X_1}} f_n \varphi dX_2, \forall \varphi \in H_0^1(\Omega_{X_1}). \quad (4.15) $$
*Proof.* This result follows from the $L^2$-theory Proposition 2.7. The convergences in $W_0^{1,p}(\Omega; \Pi_2\Omega)$ and $L^p(\Omega)$ follow from the continuous embeddings $H_0^1(\Omega; \Pi_2\Omega) \hookrightarrow W_0^{1,p}(\Omega; \Pi_2\Omega)$, $L^2(\Omega) \hookrightarrow L^p(\Omega)$, $1 < p < 2$. $\square$

Now, we construct $u$. Testing with $\varphi = \theta(u^n(X_1, \cdot))$ in (4.15) and estimating like in proof of Proposition 4.1 where we use assumptions (5), (10), we obtain as in (4.6). For a.e. $X_1 \in \Pi_1\Omega$

$$ \forall n \in \mathbb{N} : \|\nabla_{X_2} u^n(X_1, \cdot)\|_{L^p(\Omega_{X_1})} \le \left( \frac{\|f_n(X_1, \cdot)\|_{L^p(\Omega_{X_1})}}{\lambda(p-1)} \right)^{\frac{1}{2}} \times \left( \int_{\Omega_{X_1}} (1 + |u^n(X_1, \cdot)|)^p dX_2 \right)^{\frac{1}{2p}} \quad (4.16) $$

Integrating (4.16) over $\Pi_1\Omega$ and using Cauchy-Schwaz's inequality we deduce

$$ \forall n \in \mathbb{N} : \|\nabla_{X_2} u^n\|_{L^p(\Omega)}^p \le C \|f_n\|_{L^p(\Omega)}^{\frac{p}{2}} \left( \int_{\Omega} (1 + |u^n|)^p dx \right)^{\frac{1}{2}}, $$

and therefore

$$ \forall n \in \mathbb{N} : \|\nabla_{X_2} u^n\|_{L^p(\Omega)}^2 \le C(1 + \|u^n\|_{L^p(\Omega)}). $$

Using Poincaré's inequality (57) (which holds since $u^n \in W_0^{1,p}(\Omega; \Pi_2\Omega)$), we obtain that

$$ \forall n \in \mathbb{N} : \|u^n\|_{L^p(\Omega)} \le C \quad (4.17) $$

Now, using the monotonicity of the problem (assumption (10)) and testing in (4.15) with the test function $\theta(u^n(X_1, \cdot) - u^m(X_1, \cdot))$, $m, n \in \mathbb{N}$, one can obtain like in (4.16)

$$ \|\nabla_{X_2} (u^n(X_1, \cdot) - u^m(X_1, \cdot))\|_{L^p(\Omega_{X_1})} \le \left( \frac{\|f_n(X_1, \cdot) - f_m(X_1, \cdot)\|_{L^p(\Omega_{X_1})}}{\lambda(p-1)} \right)^{\frac{1}{2}} \times \left( \int_{\Omega_{X_1}} (1 + |u^n(X_1, \cdot) - u^m(X_1, \cdot)|)^p dX_2 \right)^{\frac{1}{2p}}. $$

Integrating over $\Pi_1\Omega$ and using the Cauchy-Schwarz inequality and (4.17) we get

$$ \forall n, m \in \mathbb{N} : \|\nabla_{X_2} (u^n - u^m)\|_{L^p(\Omega)} \le C \|f_n - f_m\|_{L^p(\Omega)}^{\frac{1}{2}}, $$

where $C$ is independent of $m$ and $n$. Since $(f_n)_{n \in \mathbb{N}}$ is a converging sequence in $L^p(\Omega)$ then this last inequality shows that $(u^n)_{n \in \mathbb{N}}$ is a Cauchy sequence in the Banach space $W_0^{1,p}(\Omega; \Pi_2\Omega)$, consequently
there exists $u \in W_0^{1,p}(\Omega; \Pi_2\Omega)$ such that $u^n \to u$ in $W_0^{1,p}(\Omega; \Pi_2\Omega)$. Therefore, there exists a subsequence still labelled $(u^n)$ such that for a.e. $X_1 \in \Pi_1$ we have $u^n(X_1, \cdot) \to u(X_1, \cdot)$ in $W_0^{1,p}(\Omega_{X_1})$, therefore for a.e. $X_1 \in \Pi_1$, $u(X_1, \cdot)$ is the entropy solution of (4.3) (It is well known that, up to a subsequence, we have for a.e. $X_1 \in \Pi_1\Omega$, $u^n(X_1, \cdot) \to u(X_1, \cdot)$ in $L^r(\Omega_{X_1})$ for every $1 \le r < \frac{N}{N-1}$). Whence we have proved the following

**Proposition 4.3.** Under assumptions of Proposition 4.2 there exists $u \in W_0^{1,p}(\Omega; \Pi_2\Omega)$ unique such that for a.e. $X_1 \in \Pi_1\Omega$, $u(X_1, \cdot)$ is the entropy solution to (4.3), and $u^n \to u$ in $W_0^{1,p}(\Omega; \Pi_2\Omega)$ where $(u^n)_{n \in \mathbb{N}}$ is the sequence given in Proposition 4.2.

### 1.2.1 Step 2

We construct the sequence $(u_\epsilon)_{0<\epsilon\le 1}$ of the entropy solutions to (4.1), one can prove the following

**Proposition 4.4.** Let $(u_\epsilon)_{0<\epsilon\le 1}$ be a sequence of the entropy solutions to (4.1), then we have $u_\epsilon^n \to u_\epsilon$ in $W^{1,p}(\Omega)$ as $n \to \infty$, for every $\epsilon$ fixed. Moreover, $u_\epsilon^n \to u_\epsilon$ in $W_0^{1,p}(\Omega; \Pi_2\Omega)$ and $\epsilon\nabla_{X_1}u_\epsilon^n \to \nabla_{X_1}u_\epsilon$ in $L^p(\Omega)$, uniformly in $\epsilon$.

*Proof.* Testing with $\theta(u_\epsilon^n - u_\epsilon^m)$, $m, n \in \mathbb{N}$ in (4.5), and using assumptions (5) and (10), we obtain as in (4.6)

$$ \|\nabla_{X_2}(u_{\epsilon}^n - u_{\epsilon}^m)\|_{L^p(\Omega)} \le \left( \frac{\|f_n - f_m\|_{L^p}}{\lambda(p-1)} \right)^{\frac{1}{2}} \left( \int_{\Omega} (1 + |u_{\epsilon}^n - u_{\epsilon}^m|^p) \right)^{\frac{1}{2p}}. $$

Therefore (4.7) gives

$$ \forall m, n : \|\nabla_{X_2}(u_{\epsilon}^n - u_{\epsilon}^m)\|_{L^p(\Omega)} \le C \|f_n - f_m\|_{L^p}^{\frac{1}{2}}, \quad (4.18) $$

Similarly, we obtain

$$ \forall m, n : \|\epsilon \nabla_{X_1}(u_{\epsilon}^n - u_{\epsilon}^m)\|_{L^p(\Omega)} \le C \|f_n - f_m\|_{L^p}^{\frac{1}{2}}. \quad (4.19) $$

It follows that

$$ \forall m, n : \|u_{\epsilon}^n - u_{\epsilon}^m\|_{W^{1,p}(\Omega)} \le \frac{C}{\epsilon} \|f_n - f_m\|_{L^p}^{\frac{1}{2}}. $$

The last inequality implies that for every $\epsilon$ fixed $(u_\epsilon^n)_{n \in \mathbb{N}}$ is a Cauchy sequence in $W_0^{1,p}(\Omega)$, then there exists $u_\epsilon \in W_0^{1,p}(\Omega)$ such that $u_\epsilon^n \to u_\epsilon$ in $W^{1,p}(\Omega)$, notice that $u_\epsilon$ is the entropy solution to (4.1) (it
is well known that, up to a subsequence in $n$, we have $u_{\epsilon}^n \to u_{\epsilon}$ in $L^r(\Omega)$ for every $1 \le r < \frac{N}{N-1}$). Finally (4.18) and (4.19) show that $u_{\epsilon}^n \to u_{\epsilon}$ (resp $\epsilon \nabla_{X_1} u_{\epsilon}^n \to \epsilon \nabla_{X_1} u_{\epsilon}$) in $W_0^{1,p}(\Omega; \Pi_2\Omega)$ (resp in $L^p(\Omega)$), uniformly in $\epsilon$. $\square$

### 1.2.2 Step 3

Now, we are ready to conclude. Propositions 4.2, 4.3, 4.4 combined with the triangle inequality show that $u_{\epsilon} \to u$ in $W_0^{1,p}(\Omega; \Pi_2\Omega)$ and $\epsilon \nabla_{X_1} u_{\epsilon} \to 0$ in $L^p(\Omega)$, and the proof of Theorem 0.16 is completed.

## 1.3 A regularity result for the entropy solution of the limit problem

In this subsection, we assume that $\Omega = \omega_1 \times \omega_2$ where $\omega_1, \omega_2$ are two bounded open sets of $\mathbb{R}^q, \mathbb{R}^{N-q}$ respectively. We introduce the space

$$W_p = \{v \in L^p(\Omega) \mid \nabla_{X_1} v \in L^p(\Omega)^q\}$$

We suppose the following

$$f \in W_p. \qquad (4.20)$$

**Theorem 4.5.** Assume (5), (6), (22), (4.20). Suppose that $\beta = 0$, then $u \in W^{1,p}(\Omega)$, where $u$ is the entropy solution of (4.3) – (4.4).

*Proof.* The proof of this theorem is based on the difference quotient method (see for instance [20] page 168). Let $(u^n)$ be the sequence of Proposition 4.3, then we have $u^n \to u$ in $W_0^{1,p}(\Omega; \omega_2)$, where $u$ is the entropy solution of (4.3).

Let $1 \le i \le q$ be fixed. Let $\omega'_1 \subset \subset \omega_1$ be an open subset, for $0 < h < d(\partial\omega_1, \omega'_1)$ and for a.e. $X_1 \in \omega'_1$ we set $\tau_h^i u^n = u^n(X_1 + he_i, X_2)$ where $e_i = (0, \dots, 1, \dots, 0)$ then we have by (4.15) (with $\beta = 0$)

$$\int_{\omega_2} A_{22} \nabla_{X_2} (\tau_h^i u^n - u^n) \cdot \nabla_{X_2} \varphi dX_2 = \int_{\omega_2} (\tau_h^i f_n - f_n) \varphi dX_2, \forall \varphi \in H_0^1(\omega_2),$$

where we have used the fact that $A_{22}(x) = A_{22}(X_2)$. We introduce the function $\theta_{\delta}(t) = \int_{0}^{t} (\delta + |s|)^{p-2} ds$, $\delta > 0$, $t \in \mathbb{R}$ we have $0 < \theta'_{\delta}(t) = (\delta + |t|)^{p-2} \le \delta^{p-2}$ and $|\theta_{\delta}(t)| \le \frac{2(\delta+|t|)^{p-1}}{p-1}$. Testing with $\varphi = \frac{1}{h}\theta_{\delta}(\frac{\tau_h^i u^n - u^n}{h}) \in H_0^1(\omega_2)$. To make the notations less heavy we set

$$U = \frac{\tau_h^i u^n - u^n}{h}, \quad \frac{(\tau_h^i f_n - f_n)}{h} = F$$
Then we get

$$ \int_{\omega_2} \theta'_{\delta}(U) A_{22} \nabla_{X_2} U \cdot \nabla_{X_2} U dX_2 = \int_{\omega_2} F \theta_{\delta}(U) dX_2 . $$

Using the ellipticity assumption (5) for the left hand side, and Hölder's inequality for the right hand side of the above integral equality we deduce

$$ \lambda \int_{\omega_2} \theta'_{\delta}(U) |\nabla_{X_2} U|^2 dX_2 \le \frac{2}{p-1} \|F\|_{L^p(\omega_2)} \left( \int_{\omega_2} (\delta + |U|)^p dX_2 \right)^{\frac{p-1}{p}} . $$

Using Hölder's inequality we derive

$$ \begin{aligned} \|\nabla_{X_2} U\|_{L^p(\omega_2)}^p &\le \left( \int_{\omega_2} \theta'_{\delta}(U) |\nabla_{X_2} U|^2 dX_2 \right)^{\frac{p}{2}} \left( \int_{\omega_2} \theta'_{\delta}(U)^{\frac{p}{p-2}} dX_2 \right)^{\frac{2-p}{p}} \\ &\le \left( \frac{2}{\lambda(p-1)} \|F\|_{L^p(\omega_2)} \left( \int_{\omega_2} (\delta + |U|)^p dX_2 \right)^{\frac{p-1}{p}} \right)^{\frac{p}{2}} \times \\ &\qquad \left( \int_{\omega_2} \theta'_{\delta}(U)^{\frac{p}{p-2}} dX_2 \right)^{\frac{2-p}{p}}. \end{aligned} $$

Then we deduce

$$ \|\nabla_{X_2} U\|_{L^p(\omega_2)}^2 \le \frac{2}{\lambda(p-1)} \|F\|_{L^p(\omega_2)} \left( \int_{\omega_2} (\delta + |U|)^p dX_2 \right)^{\frac{1}{p}}. $$

Now passing to the limit as $\delta \to 0$ using the Lebesgue theorem we deduce

$$ \|\nabla_{X_2} U\|_{L^p(\omega_2)}^2 \le \frac{2}{\lambda(p-1)} \|F\|_{L^p(\omega_2)} \left( \int_{\omega_2} |U|^p dX_2 \right)^{\frac{1}{p}}, $$

and Poincaré's inequality (57) gives

$$ \|\nabla_{X_2} U\|_{L^p(\omega_2)} \le \frac{2C_{\omega_2}}{\lambda(p-1)} \|F\|_{L^p(\omega_2)}. $$

Now, integrating over $\omega'_1$ yields

$$ \left\| \frac{\tau_h^i u^n - u^n}{h} \right\|_{L^p(\omega'_1 \times \omega_2)} \le \frac{2C_{\omega_2}}{\lambda(p-1)} \left\| \frac{(\tau_h^i f_n - f_n)}{h} \right\|_{L^p(\omega'_1 \times \omega_2)} . $$

Passing to the limit as $n \to \infty$ using the invariance of the Lebesgue measure under translations we get

$$ \left\| \frac{\tau_h^i u - u}{h} \right\|_{L^p(\omega'_1 \times \omega_2)} \le \frac{2C_{\omega_2}}{\lambda(p-1)} \left\| \frac{(\tau_h^i f - f)}{h} \right\|_{L^p(\omega'_1 \times \omega_2)} . $$
Whence, since $f \in W_p$ then

$$ \left\| \frac{\tau_h^i u - u}{h} \right\|_{L^p(\omega_1' \times \omega_2)} \le C, $$

where $C$ is independent of $h$, therefore we have $\nabla_{X_1} u \in L^p(\Omega)^q$. Combining this with $u \in W_0^{1,p}(\Omega; \omega_2)$ we get the desired result. □

## 2 Estimation of the rate of convergence in cylindrical domains

In this section we prove Theorem 0.17 and Theorem 0.18. We suppose that $\Omega = \omega_1 \times \omega_2$ where $\omega_1, \omega_2$ are two bounded open sets of $\mathbb{R}^q$ and $\mathbb{R}^{N-q}$ respectively. Throughout this section, we suppose that $\beta = 0$ and that $A_{22}, A_{12}$, and $f$ satisfy assumptions (24), (58), and (59) respectively i.e. $\forall x \in \Omega$, $A_{12}(x) = A_{12}(X_2)$, $A_{22}(x) = A_{22}(X_2)$, and $f(x) = f(X_2) \in L^p(\omega_2)$ ($1 < p < 2$).

### 2.1 A polynomial rate

Let $u_\epsilon$, $u$ be the entropy solutions of (4.1), (4.3) respectively (recall that $\beta = 0$), we use the approximated sequence $(u_\epsilon^n)_{\epsilon,n}$, $(u^n)_n$ introduced in Section 1. Subtracting (4.15) from (4.5) we obtain

$$ \int_{\Omega} A_{\epsilon} \nabla (u_{\epsilon}^{n} - u^{n}) \cdot \nabla \varphi dx = 0, $$

where we have used that $u^n$ is independent of $X_1$ (since $f$ and $A_{22}$ are independent of $X_1$) and that $A_{12}$ is independent of $X_1$. Let $\omega_1' \subset \subset \omega_1$ then there exists $\omega_1' \subset \subset \omega_1'' \subset \subset \omega_1$. We introduce the function $\rho \in \mathcal{D}(\omega_1)$ such that $\text{Supp}(\rho) \subset \omega_1''$ and $\rho = 1$ on $\omega_1'$ (we can choose $0 \le \rho \le 1$) Testing with $\varphi = \rho^2 \theta_\delta(u_\epsilon^n - u^n) \in H_0^1(\Omega)$ (we can check that this function belongs to $H_0^1(\Omega)$ by using a density argument) in the above integral equality we get

$$ \begin{aligned} & \int_{\Omega} \rho^2 \theta_{\delta}^{\prime}(u_{\epsilon}^{n} - u^{n}) A_{\epsilon} \nabla (u_{\epsilon}^{n} - u^{n}) \cdot \nabla (u_{\epsilon}^{n} - u^{n}) dx \\ & = - \int_{\Omega} \rho \theta_{\delta}(u_{\epsilon}^{n} - u^{n}) A_{\epsilon} \nabla (u_{\epsilon}^{n} - u^{n}) \cdot \nabla \rho dx \\ & = - \epsilon^2 \int_{\Omega} \rho \theta_{\delta}(u_{\epsilon}^{n} - u^{n}) A_{11} \nabla_{X_1} (u_{\epsilon}^{n} - u^{n}) \cdot \nabla_{X_1} \rho dx \\ & \quad - \epsilon \int_{\Omega} \rho \theta_{\delta}(u_{\epsilon}^{n} - u^{n}) A_{12} \nabla_{X_2} (u_{\epsilon}^{n} - u^{n}) \cdot \nabla_{X_1} \rho dx. \end{aligned} $$
where we have used that $\rho$ is independent of $X_2$. Using the ellipticity assumption (5) for the left hand side and assumption (6) for the right hand side of the previous integral equality we deduce

$$ \epsilon^2 \lambda \int_{\Omega} \theta'_{\delta}(u_{\epsilon}^n - u^n) |\rho \nabla_{X_1}(u_{\epsilon}^n - u^n)|^2 dx + \lambda \int_{\Omega} \theta'_{\delta}(u_{\epsilon}^n - u^n) |\rho \nabla_{X_2}(u_{\epsilon}^n - u^n)|^2 dx \\ \le \epsilon^2 C \int_{\Omega} \rho |\theta_{\delta}(u_{\epsilon}^n - u^n)| |\nabla_{X_1}(u_{\epsilon}^n - u^n)| dx \\ + \epsilon C \int_{\Omega} \rho |\theta_{\delta}(u_{\epsilon}^n - u^n)| |\nabla_{X_2}(u_{\epsilon}^n - u^n)| dx. $$

Using Young's inequality for the two terms in the right hand side of the previous inequality, we obtain

$$ \epsilon^2 \frac{\lambda}{2} \int_{\Omega} \theta'_{\delta}(u_{\epsilon}^n - u^n) |\rho \nabla_{X_1}(u_{\epsilon}^n - u^n)|^2 dx + \frac{\lambda}{2} \int_{\Omega} \theta'_{\delta}(u_{\epsilon}^n - u^n) |\rho \nabla_{X_2}(u_{\epsilon}^n - u^n)|^2 dx \\ \le \epsilon^2 C \int_{\omega_1'' \times \omega_2} |\theta_{\delta}(u_{\epsilon}^n - u^n)|^2 \theta'_{\delta}(u_{\epsilon}^n - u^n)^{-1} dx. $$

Whence

$$ \epsilon^2 \frac{\lambda}{2} \int_{\Omega} \theta'_{\delta}(u_{\epsilon}^n - u^n) |\rho \nabla_{X_1}(u_{\epsilon}^n - u^n)|^2 dx + \frac{\lambda}{2} \int_{\Omega} \theta'_{\delta}(u_{\epsilon}^n - u^n) |\rho \nabla_{X_2}(u_{\epsilon}^n - u^n)|^2 dx \\ \le \epsilon^2 C \int_{\omega_1'' \times \omega_2} (\delta + |u_{\epsilon}^n - u^n|^p) dx, $$

Now, using Hölder's inequality and the above inequality we deduce

$$ \epsilon^2 \frac{\lambda}{2} \|\rho \nabla_{X_1}(u_{\epsilon}^n - u^n)\|_{L^p(\Omega)}^2 + \frac{\lambda}{2} \|\rho \nabla_{X_2}(u_{\epsilon}^n - u^n)\|_{L^p(\Omega)}^2 \\ \le \left[ \epsilon^2 \frac{\lambda}{2} \left( \int_{\Omega} \theta'_{\delta}(u_{\epsilon}^n - u^n) |\rho \nabla_{X_1}(u_{\epsilon}^n - u^n)|^2 dx \right) \right. \\ \left. + \frac{\lambda}{2} \left( \int_{\Omega} \theta'_{\delta}(u_{\epsilon}^n - u^n) |\rho \nabla_{X_2}(u_{\epsilon}^n - u^n)|^2 dx \right) \right] \times \\ \left( \int_{\omega_1'' \times \omega_2} (\delta + |u_{\epsilon}^n - u^n|^p) dx \right)^{\frac{2-p}{p}} \\ \le C \epsilon^2 \left( \int_{\omega_1'' \times \omega_2} (\delta + |u_{\epsilon}^n - u^n|^p) dx \right)^{\frac{2}{p}}. $$

Passing to the limit as $\delta \to 0$ using the Lebesgue theorem. Passing to the limit as $n \to \infty$ we get

$$ \epsilon^2 \|\nabla_{X_1}(u_{\epsilon} - u)\|_{L^p(\omega_1' \times \omega_2)}^2 + \|\nabla_{X_2}(u_{\epsilon} - u)\|_{L^p(\omega_1' \times \omega_2)}^2 \le C \epsilon^2 \|(u_{\epsilon} - u)\|_{L^p(\omega_1'' \times \omega_2)}^2. \quad (4.21) $$

Using Poincaré's inequality (57) we obtain

$$ \epsilon^2 \|\nabla_{X_1}(u_{\epsilon} - u)\|_{L^p(\omega_1' \times \omega_2)}^2 + \|\nabla_{X_2}(u_{\epsilon} - u)\|_{L^p(\omega_1' \times \omega_2)}^2 \le C \epsilon^2 \|\nabla_{X_2}(u_{\epsilon} - u)\|_{L^p(\omega_1'' \times \omega_2)}^2. \quad (4.22) $$
Let $d \in \mathbb{N}^*$ then there exist $\omega'_1 \subset \subset \omega''_1 \subset \subset \dots \omega_1^{(d+1)} \subset \subset \omega_1$. Iterating (4.22) $d$-times we deduce

$$ \epsilon^2 \|\nabla_{X_1}(u_\epsilon - u)\|^2_{L^p(\omega'_1 \times \omega_2)} + \|\nabla_{X_2}(u_\epsilon - u)\|^2_{L^p(\omega'_1 \times \omega_2)} \le C \epsilon^{2d} \|\nabla_{X_2}(u_\epsilon - u)\|^2_{L^p(\omega_1^{(d)} \times \omega_2)}. $$

Now, from (4.21) (with $\omega'_1$ and $\omega''_1$ replaced by $\omega_1^{(d)}$ and $\omega_1^{(d+1)}$ respectively) we deduce

$$ \epsilon^2 \|\nabla_{X_1}(u_\epsilon - u)\|^2_{L^p(\omega'_1 \times \omega_2)} + \|\nabla_{X_2}(u_\epsilon - u)\|^2_{L^p(\omega'_1 \times \omega_2)} \le C \epsilon^{2(d+1)} \|u_\epsilon - u\|^2_{L^p(\omega_1^{(d+1)} \times \omega_2)}. $$

Since $u_\epsilon \to u$ in $L^p(\Omega)$ then $\|u_\epsilon - u\|_{L^p(\Omega)}$ is bounded and therefore we obtain

$$ \|u_\epsilon - u\|_{W^{1,p}(\omega'_1 \times \omega_2)} \le C\epsilon^d, $$

where the constant $C$ only depends on $d, A, \lambda, f, \Omega, \omega'_1$.

**Remark 4.1.** It is not easy to estimate the convergence rate for general data. However, under assumptions of Theorem 0.17 and when $f = f_1 + f_2$ with $f_1 \in L^p(\omega_2)$ and $f_2 \in W_2$ we only have the estimation

$$ \epsilon \|\nabla_{X_1}(u_\epsilon - u)\|_{L^p(\omega'_1 \times \omega_2)} + \|\nabla_{X_2}(u_\epsilon - u)\|_{L^p(\omega'_1 \times \omega_2)} + \|u_\epsilon - u\|_{L^p(\omega'_1 \times \omega_2)} \le C\epsilon. $$

This follows from the linearity of the equation, Theorem 0.17 and the estimation (20).

## 2.2 Elliptic problems in a cylinder becoming unbounded

Anisotropic singular perturbations problems could be seen as problems in a cylinder becoming unbounded [4], [5]. Indeed the two category of problems could be connected to each other via a scaling $\epsilon = 1/\ell$. Let us consider the problem:

$$ \left\{ \begin{array}{l} -\nabla \cdot (\tilde{A}\nabla u_\ell) = f \\ u_\ell = 0 \quad \text{on } \partial\Omega_\ell \end{array} \right. \qquad (4.23) $$

where $\tilde{A} = (\tilde{a}_{ij})$ is a $N \times N$ matrix valued function which has the same block structure as in (7), and such that

$$ \tilde{a}_{ij} \in L^\infty(\mathbb{R}^q \times \omega_2), \qquad (4.24) $$

$$ \exists \lambda > 0 : \tilde{A}(x)\xi \cdot \xi \ge \lambda |\xi|^2 \quad \forall \xi \in \mathbb{R}^N \text{ for a.e } x \in \mathbb{R}^q \times \omega_2. \qquad (4.25) $$
For $\ell > 0$ we set $\Omega_{\ell} = \ell\omega_{1} \times \omega_{2}$ where $\omega_{1}, \omega_{2}$ are two bounded open sets with $\omega_{1}$ convex and containing 0. We consider the limit problem:

$$ \begin{cases} -\nabla_{X_2} \cdot (\tilde{A}_{22} \nabla_{X_2} u_{\infty}) = f \\ u_{\infty} = 0 \quad \text{on } \partial\omega_2 \end{cases} \qquad (4.26) $$

We have the following result which gives Theorem 0.18 as a corollary.

**Theorem 4.2.** Assume that the blocks $\tilde{A}_{22}, \tilde{A}_{12}$ satisfy (24), (58) respectively. Suppose (59), (4.24), (4.25). Let $u_{\ell}, u_{\infty}$ be the unique entropy solutions to (4.23) and (4.26) then for every $\alpha \in (0, 1)$ there exist $C > 0, c > 0$ independent of $\ell$ such that

$$ \forall \ell > 0 : \|\nabla(u_{\ell} - u_{\infty})\|_{W^{1,p}(\Omega_{\alpha\ell})} \le C e^{-c\ell} $$

*Proof.* Let $u_{\ell}, u_{\infty}$ be the unique entropy solutions to (4.23) and (4.26) respectively, and let $(u_{\ell}^{n})$ and $(u_{\infty}^{n})$ the approximation sequences (as in Section 1, Chapter 4). we have $u_{\ell}^{n} \to u_{\ell}$ in $W_{0}^{1,p}(\Omega_{\ell})$ and $u_{\infty}^{n} \to u_{\infty}$ in $W_{0}^{1,p}(\omega_{2})$. Subtracting the associated approximated problems to (4.23) and (4.26), and taking the weak formulations. We get

$$ \int_{\Omega_{\ell}} \tilde{A} \nabla (u_{\ell}^{n} - u_{\infty}^{n}) \nabla \varphi dx = 0, \forall \varphi \in H_{0}^{1}(\Omega_{\ell}). \qquad (4.27) $$

Where we have used that $\tilde{A}_{22}, \tilde{A}_{12}, u_{\infty}^{n}$ are independent of $X_{1}$. Now we will use the iteration technique introduced in [35]. Let $0 < \ell_{0} \le \ell - 1$ fixed, and let $\rho \in \mathcal{D}(\mathbb{R}^{q})$ a bump function such that

$$ 0 \le \rho \le 1, \rho = 1 \text{ on } \ell_{0}\omega_{1} \text{ and } \rho = 0 \text{ on } \mathbb{R}^{q} \setminus (\ell_{0} + 1)\omega_{1}, |\nabla_{X_{1}}\rho| \le c_{0}. $$

where $c_{0}$ is a positive constant which only depends in $\omega_{1}$ ([4], Chapter 6). Testing with $\rho^{2}\theta_{\delta}(u_{\ell}^{n} - u_{\infty}^{n}) \in H_{0}^{1}(\Omega_{\ell})$ in (4.27) we get

$$ \int_{\Omega_{\ell}} \rho^{2} \theta'_{\delta}(u_{\ell}^{n} - u_{\infty}^{n}) \tilde{A} \nabla (u_{\ell}^{n} - u_{\infty}^{n}) \cdot \nabla (u_{\ell}^{n} - u_{\infty}^{n}) dx + \int_{\Omega_{\ell}} \rho \theta_{\delta}(u_{\ell}^{n} - u_{\infty}^{n}) \tilde{A} \nabla (u_{\ell}^{n} - u_{\infty}^{n}) \cdot \nabla \rho dx = 0. $$

Using the ellipticity assumption (4.25). We obtain that

$$ \int_{\Omega_{\ell}} \rho^{2} \theta'_{\delta}(u_{\ell}^{n} - u_{\infty}^{n}) |\nabla(u_{\ell}^{n} - u_{\infty}^{n})|^2 dx \le 2 \int_{\Omega_{\ell}} \rho |\theta_{\delta}(u_{\ell}^{n} - u_{\infty}^{n})| |\tilde{A} \nabla(u_{\ell}^{n} - u_{\infty}^{n})| |\nabla \rho| dx. $$
Notice that $\nabla \rho = 0$ on $\Omega_{\ell_0}$, and $\Omega_{\ell_0} \subset \Omega_{\ell_0+1}$ (since $\omega_1$ is convex and containing 0). Then, by the Cauchy-Schwarz inequality we get

$$ 
\begin{aligned}
\int_{\Omega_{\ell}} \rho^2 \theta'_{\delta}(u_{\ell}^n - u_{\infty}^n) |\nabla(u_{\ell}^n - u_{\infty}^n)|^2 dx &\le 2c_0 \|\tilde{A}\|_{\infty} \int_{\Omega_{\ell_0+1} \setminus \Omega_{\ell_0}} \rho |\theta_{\delta}(u_{\ell}^n - u_{\infty}^n)| |\nabla(u_{\ell}^n - u_{\infty}^n)| dx \\
&\le 2c_0 \|\tilde{A}\|_{\infty} \left( \int_{\Omega_{\ell}} \rho^2 \theta'_{\delta}(u_{\ell}^n - u_{\infty}^n) |\nabla(u_{\ell}^n - u_{\infty}^n)|^2 dx \right)^{\frac{1}{2}} \times \\
&\qquad \left( \int_{\Omega_{\ell_0+1} \setminus \Omega_{\ell_0}} |\theta_{\delta}(u_{\ell}^n - u_{\infty}^n)|^2 \theta'_{\delta}(u_{\ell}^n - u_{\infty}^n)^{-1} dx \right)^{\frac{1}{2}}.
\end{aligned}
$$

where we have used (4.24). Whence, we get (since $\rho = 1$ on $\Omega_{\ell_0}$)

$$ 
\begin{aligned}
\int_{\Omega_{\ell_0}} \theta'_{\delta}(u_{\ell}^n - u_{\infty}^n) |\nabla(u_{\ell}^n - u_{\infty}^n)|^2 dx &\le \int_{\Omega_{\ell}} \rho^2 \theta'_{\delta}(u_{\ell}^n - u_{\infty}^n) |\nabla(u_{\ell}^n - u_{\infty}^n)|^2 dx \\
&\le \left( \frac{4c_0 \|\tilde{A}\|_{\infty}}{p-1} \right)^2 \int_{\Omega_{\ell_0+1} \setminus \Omega_{\ell_0}} (\delta + |u_{\ell}^n - u_{\infty}^n|)^p dx.
\end{aligned}
$$

From Hölder's inequality it holds that

$$ 
\begin{aligned}
\|\nabla(u_{\ell}^n - u_{\infty}^n)\|_{L^p(\Omega_{\ell_0})}^2 &\le \left( \int_{\Omega_{\ell_0}} \theta'_{\delta}(u_{\ell}^n - u_{\infty}^n) |\nabla(u_{\ell}^n - u_{\infty}^n)|^2 dx \right) \left( \int_{\Omega_{\ell_0}} (\delta + |u_{\ell}^n - u_{\infty}^n|)^p dx \right)^{\frac{2-p}{p}} \\
&\le \left( \frac{4c_0 \|\tilde{A}\|_{\infty}}{p-1} \right)^2 \left( \int_{\Omega_{\ell_0+1} \setminus \Omega_{\ell_0}} (\delta + |u_{\ell}^n - u_{\infty}^n|)^p dx \right) \left( \int_{\Omega_{\ell_0}} (\delta + |u_{\ell}^n - u_{\infty}^n|)^p dx \right)^{\frac{2-p}{p}}.
\end{aligned}
$$

Passing to the limit as $\delta \to 0$ (using the Lebesgue theorem) we get

$$ \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^p(\Omega_{\ell_0})}^2 \le C \left( \int_{\Omega_{\ell_0+1} \setminus \Omega_{\ell_0}} |u_{\ell}^{n} - u_{\infty}^{n}|^p dx \right) \times \left( \int_{\Omega_{\ell_0}} |u_{\ell}^{n} - u_{\infty}^{n}|^p dx \right)^{\frac{2-p}{p}}, $$

where we have used $0 \le \rho \le 1$. Notice that in the rest of this proof $C$ is a generic positive constant independent of $n$ and $\ell$. Using the following Poincaré's inequality

$$ \|u_{\ell}^{n} - u_{\infty}^{n}\|_{L^{p}(\Omega_{\ell_{0}})} \le C_{\omega_{2}} \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\ell_{0}})}, $$

we get

$$ \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^p(\Omega_{\ell_0})}^p \le C \|u_{\ell}^{n} - u_{\infty}^{n}\|_{L^p(\Omega_{\ell_0+1} \setminus \Omega_{\ell_0})}^p. $$

Using the following Poincaré's inequality

$$ \|u_{\ell}^{n} - u_{\infty}^{n}\|_{L^{p}(\Omega_{\ell_{0}+1} \setminus \Omega_{\ell_{0}})} \le C_{\omega_{2}} \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\ell_{0}+1} \setminus \Omega_{\ell_{0}})}, $$
we get

$$ \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\ell_{0}})}^{p} \leq C \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\ell_{0}+1} \setminus \Omega_{\ell_{0}})}^{p}. $$

Whence

$$ \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\ell_{0}})}^{p} \leq \frac{C}{C+1} \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\ell_{0}+1})}^{p}. $$

Let $\alpha \in (0, 1)$ and $\ell \ge \frac{1}{1-\alpha}$. Iterating the previous formula starting from $\alpha\ell$ we get

$$ \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\alpha\ell})}^{p} \leq \left(\frac{C}{C+1}\right)^{[(1-\alpha)\ell]} \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\alpha\ell+[(1-\alpha)\ell]})}^{p}, $$

where $[(1 - \alpha)\ell]$ is the integer part of $(1 - \alpha)\ell$. Whence

$$ \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\alpha\ell})} \leq C e^{-c(1-\alpha)\ell} \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\ell})}, \quad (4.28) $$

where $c > 0$ is independent of $\ell$ and $n$. Now we have to estimate the right hand side of (4.28). Testing with $\theta(u_{\ell}^{n})$ in the approximated problem associated to (4.23) one can obtain as in Section 1

$$ \|\nabla u_{\ell}^{n}\|_{L^{p}(\Omega_{\ell})} \leq C \ell^{\frac{q}{2}}. \quad (4.29) $$

Similarly, testing with $\theta(u_{\infty}^{n})$ in the approximated problem associated to (4.26), we get

$$ \|\nabla u_{\infty}^{n}\|_{L^{p}(\Omega_{\ell})} \leq C \ell^{\frac{q}{2}}. \quad (4.30) $$

Apply the triangle inequality to the right hand side of (4.28), and use (4.29), (4.30). Passing to the limit as $n \to \infty$ we obtain that

$$ \forall \ell \ge \frac{1}{1-\alpha} : \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\alpha\ell})} \le C e^{-c(1-\alpha)\ell} \quad (4.31) $$

Since the left hand side of (4.31) is bounded when $\ell$ goes through the interval $(0, \frac{1}{1-\alpha})$, then we get

$$ \forall \ell > 0 : \|\nabla(u_{\ell}^{n} - u_{\infty}^{n})\|_{L^{p}(\Omega_{\alpha\ell})} \le C e^{-c\ell}, $$

for some positives $C$ and $c$ independent of $\ell$. $\square$
### 3 Non standard neutron transport equation with L$^p$ data

Suppose that $\Omega = \omega_1 \times \omega_2$ where $\omega_1, \omega_2$ are two bounded open sets of $\mathbb{R}^q, \mathbb{R}^{N-q}$ respectively. Consider the following nonlinear problem:

$$ \begin{cases} -\nabla \cdot (A_{\epsilon} \nabla u_{\epsilon}) = f + B(u_{\epsilon}) \\ u_{\epsilon} = 0 \quad \text{on } \partial\Omega \end{cases} \qquad (4.32) $$

Where $f \in L^p(\Omega)$, $1 < p < 2$ and $B: L^p(\Omega) \to L^p(\Omega)$ is a continuous nonlinear operator. We suppose that

$$ \exists M \ge 0, \forall u \in L^p(\Omega) : \|B(u)\|_{L^p} \le M \qquad (4.33) $$

**Proposition 4.1.** Assume (5), (6). Suppose that $f \in L^p(\Omega)$. Suppose that $B: L^p(\Omega) \to L^p(\Omega)$ is continuous such that (4.33). Then:

1) There exists a sequence $(u_{\epsilon})_{0<\epsilon \le 1} \in W_0^{1,p}(\Omega)^{\mathbb{N}}$ of entropy solutions to (4.32) which are also weak solutions such that

$$ \epsilon \|\nabla_{X_1} u_{\epsilon}\|_{L^p(\Omega)}, \|\nabla_{X_2} u_{\epsilon}\|_{L^p(\Omega)}, \|u_{\epsilon}\|_{L^p(\Omega)} \le C_0, $$

where $C_0 \ge 0$ is independent of $\epsilon$ (the constant $C_0$ depends only on $\Omega, \lambda, f$ and $M$).

2) If $(u_{\epsilon})_{0<\epsilon \le 1}$ is a sequence of entropy solutions to (4.32), then we have the above estimates.

*Proof.* 1) The existence of $u_{\epsilon}$ is based on the Schauder fixed point theorem, we define the mapping $\Gamma: L^p(\Omega) \to L^p(\Omega)$ by

$$ v \in L^p(\Omega) \to \Gamma(v) = v_{\epsilon} \in W_0^{1,p}(\Omega) $$

where $v_{\epsilon}$ is the entropy solution (which is also a weak solution) of the linearised problem:

$$ \begin{cases} -\nabla \cdot (A_{\epsilon} \nabla v_{\epsilon}) = f + B(v) \\ v_{\epsilon} = 0 \quad \text{on } \partial\Omega \end{cases} \qquad (4.34) $$

Since the entropy solution is unique then $\Gamma$ is well defined, we can prove that $\Gamma$ is continuous, from $L^p(\Omega)$ to $L^p(\Omega)$, as a composition of continuous mappings. As in Section 1, Chapter 4 we can obtain the estimates

$$ \epsilon \|\nabla_{X_1} v_{\epsilon}\|_{L^p(\Omega)}, \|\nabla_{X_2} v_{\epsilon}\|_{L^p(\Omega)}, \|v_{\epsilon}\|_{L^p(\Omega)} \le C_0 $$
where $C_0$ is independent of $\epsilon$ and $v$ (thanks to (4.33)) Now, for $\epsilon$ fixed we define the subset

$$K_{\epsilon} = \{ v \in W_{0}^{1,p}(\Omega) : \epsilon \|\nabla_{X_1} v\|_{L^p(\Omega)}, \|\nabla_{X_2} v\|_{L^p(\Omega)}, \|v\|_{L^p(\Omega)} \le C_0 \}$$

The subset $K_\epsilon$ is convex and compact in $L^p(\Omega)$ thanks to the Sobolev compact embedding $W_0^{1,p}(\Omega) \hookrightarrow L^p(\Omega)$. Since $C_0$ is independent of $v$ then the subset $K_\epsilon$ is stable under $\Gamma$. Whence $\Gamma$ admits at least a fixed point $u_\epsilon \in K_\epsilon$, in other words $u_\epsilon$ is an entropy solution to (4.32) which is also a weak solution.

2) Let $(u_\epsilon)_{0<\epsilon\le 1}$ be a sequence of entropy solutions to (4.32), $u_\epsilon$ is the unique entropy solution (which is also a weak solution) to (4.34) with $v$ replaced by $u_\epsilon$ and therefore we obtain the desired estimates as proved in 1). $\square$

**Remark 4.2.** In the general case, the uniqueness theory of [31] could not ensure that the entropy solution $u_\epsilon$ of (4.32) is unique.

Now, we assume that for every $E \subset W_p$ bounded in $L^p(\Omega)$ we have

$$\overline{conv}\{B(E)\} \subset W_2, \qquad (4.35)$$

where $\overline{conv}\{B(E)\}$ is the closed convex-hull of $B(E)$ in $L^p(\Omega)$. Assumption (4.35) has been introduced in Chapter 3 (see assumption (40)). We have the following

**Theorem 4.3.** Assume that $\Omega$, $A$, $f$ satisfy assumptions of Theorem 0.17. Suppose that $B: L^p(\Omega) \to L^p(\Omega)$ is continuous such that (4.33) and (4.35). Let $(u_\epsilon)_{0<\epsilon\le 1} \subset W_0^{1,p}(\Omega)$ be a sequence of entropy solutions to (4.32), then for every $\Omega' \subset \Omega$ there exists $C_{\Omega'} \ge 0$ independent of $\epsilon$ such that

$$\forall \epsilon \in (0, 1] : \|u_{\epsilon}\|_{W^{1,p}(\Omega')} \le C_{\Omega'}.$$

*Proof.* The proof is similar to the one of Lemma 3.5. Let $(\Omega_i)_{i \in \mathbb{N}}$ be an open cover of $\Omega$ such that $\overline{\Omega_j} \subset \Omega_{j+1}$. We equip the space $W_{\text{loc}}^{1,p}(\Omega)$ with the topology generated by the family of seminorms $(p_j)_{j \in \mathbb{N}}$ defined by

$$p_j(u) = \|u_\epsilon\|_{W^{1,p}(\Omega_j)}$$

We define the family of linear continuous mappings

$$\Lambda_{\epsilon}: L^{p}(\Omega) \to W_{\text{loc}}^{1,p}(\Omega)$$
by: g ∈ L<sup>p</sup>(Ω), Λ<sub>ε</sub>(g) = v<sub>ε</sub> where v<sub>ε</sub> is the unique entropy solution to

$$
\begin{cases}
-\nabla \cdot (A_{\epsilon} \nabla v_{\epsilon}) = g \\
v_{\epsilon} = 0 \quad \text{on } \partial\Omega
\end{cases}
$$

The continuity of Λ_ε follows immediately if we observe Λ_ε as a composition of the continuous mapping
Λ_ε : L^p(Ω) → W^1,p(Ω) and the canonical injection W^1,p(Ω) → W^1,p<sub>loc</sub>(Ω) Now, we denote L^p(Ω)<sup>weak</sup>,
W^1,p<sub>loc</sub>(Ω)<sup>weak</sup> the spaces L^p(Ω), W^1,p<sub>loc</sub>(Ω) equipped with their weak topology respectively. Therefore,
Λ_ε : L^p(Ω)<sup>weak</sup> → W^1,p<sub>loc</sub>(Ω)<sup>weak</sup> is also continuous. Consider the bounded (in L^p(Ω)) subset

$$
E_0 = \{u \in W_p \mid \|u\|_{L^p(\Omega)} \le C_0\},
$$

where $C_0$ is the constant of Proposition 4.1. Consider the subset $G = f + \overline{\text{conv}}\{B(E_0)\}$ where the closure is taken in the $L^p$-topology. $G$ is convex closed in $L^p(\Omega)$ and it is bounded thanks to (4.33). According to (4.35) we have $G \subset L^p(\omega_2) + W_2$, it follows from Remark 4.1 that for every $g \in G$ the orbit $\{\Lambda_\epsilon g\}_\epsilon$ is bounded in $W_{loc}^{1,p}(\Omega)$, and therefore $\{\Lambda_\epsilon g\}_\epsilon$ is bounded in $W_{loc}^{1,p}(\Omega)^{\text{weak}}$. It is clear that the set $G$ is compact in $L^p(\Omega)^{\text{weak}}$. Then, it follows by Theorem 3.4 that there exists a bounded subset $F$ in $W_{loc}^{1,p}(\Omega)^{\text{weak}}$ such that

$$
\forall \epsilon \in (0, 1] : \Lambda_{\epsilon}(G) \subset F.
$$

The boundedness of F in $W_{loc}^{1,p}(\Omega)^{weak}$ implies its boundedness in $W_{loc}^{1,p}(\Omega)$.i.e For every $j \in \mathbb{N}$ there
exists $C_j \ge 0$ independent of $\epsilon$ such that

$$
\forall \epsilon \in (0, 1] : p_j(\Lambda_{\epsilon}(G)) \le C_j. \tag{4.36}
$$

Let $(u_\epsilon)_\epsilon$ be a sequence of entropy solutions to (4.32), then we have $(u_\epsilon)_\epsilon \subset E_0$ as proved in Proposition 4.1, then $f + B(u_\epsilon) \in G$ and $\Lambda_\epsilon(f + B(u_\epsilon)) = u_\epsilon \in F$ for every $\epsilon \in (0, 1]$, therefore by (4.36) we get

$$
\forall \epsilon \in (0, 1] : \|u_{\epsilon}\|_{W^{1,p}(\Omega_j)} \le C_j.
$$

Whence, for every $\Omega' \subset \subset \Omega$ there exists $C_{\Omega'} \ge 0$ independent of $\epsilon$ such that

$$
\forall \epsilon \in (0, 1] : \|u_{\epsilon}\|_{W^{1,p}(\Omega')} \le C_{\Omega'}.
$$

□
Now, we are ready to prove the convergence theorem. Assume that

$$
\exists r \in [1, p) \text{ such that } B : L^r(\Omega) \to L^p(\Omega) \text{ is continuous.} \tag{4.37}
$$

Then, we have the following

**Theorem 4.4.** Under assumptions of Theorem **4.3**, assume in addition (4.37), then there exist $u \in W_0^{1,p}(\Omega; \omega_2)$ and a subsequence $(u_{\epsilon_k})_{k \in \mathbb{N}}$ of entropy solutions to (4.32) such that

$$
\epsilon_k \nabla_{X_1} u_{\epsilon_k} \to 0, \nabla_{X_2} u_{\epsilon_k} \to \nabla_{X_2} u_0 \text{ in } L^p(\Omega) - \text{weak}
$$

$$
\text{and } u_{\epsilon_k} \to u_0 \text{ in } L^r(\Omega) - \text{strong}
$$

Moreover $u$ satisfies in $D'(\omega_2)$ the equation

$$
-\nabla_{X_2} \cdot (A_{22} \nabla_{X_2} u(X_1, \cdot)) = f + B(u)(X_1, \cdot),
$$

for a.e $X_1 \in \omega_1$.

*Proof.* The estimations given in Proposition **4.1** show that there exist $u_0 \in W_0^{1,p}(\Omega; \omega_2)$ and a sequence $(u_{\epsilon_k})_{k \in \mathbb{N}}$ solutions to (4.32) such that

$$
\epsilon_k \nabla_{X_1} u_{\epsilon_k} \to 0, \nabla_{X_2} u_{\epsilon_k} \to \nabla_{X_2} u_0 \text{ and } u_{\epsilon_k} \to u_0 \text{ in } L^p(\Omega) - \text{weak} \quad (4.38)
$$

Now, we pass will to the limit in the nonlinear term. As in proof of the convergence (3.12) we show
by using Theorem 4.3 and the regularity of the Lebesgue measure that there exists a subsequence
still labelled $(u_{\epsilon_k})_k$ such that $u_{\epsilon_k} \to u$ in $L^r(\Omega)$, therefore by using assumption (4.37) we get

$$
B(u_{\epsilon_k}) \to B(u) \text{ in } L^p(\Omega) - \text{strong.} \tag{4.39}
$$

Passing to the limit in the weak formulation of (4.32) by using (4.38), (4.39). We deduce

$$
-\nabla_{X_2} \cdot (A_{22} \nabla_{X_2} u_0(X_1, \cdot)) = f + B(u_0)(X_1, \cdot), \text{ in } \mathcal{D}'(\omega_2),
$$

and the proof of the theorem is completed.

We can check as in Section 2 of Chapter 3 that the operator $B_2$ of Theorem 0.19 satisfies the
same assumptions of the operator $B$ of Theorem 4.4. Therefore, Theorem 0.19 is a direct corollary
of Theorem 4.4.
# Conclusion

Nous terminons ce mémoire par une courte description de quelques travaux en cours, et par la donnée de quelques questions ouvertes.

Les résultats des Chapitres 1 et 2 constituent la base théorique d'une étude numérique (en cours) du problème linéaire (4). Nous construirons un schéma numérique asymptotiquement préservant au sens de J. Sin [19] et nous dériverons des estimations uniformes de l'erreur de la forme

$$ \forall \epsilon \in (0, 1] : \| \nabla_{X_2} (u_{\epsilon,h} - u_{\epsilon}) \|_{L^2(\Omega)^{N-q}} = O(h^{\alpha}), $$

des questions de régularité elliptique sont également étudiée.

Le Théorème 0.3 donne un développement asymptotique d'ordre 0 de la solution $u_\epsilon$ dans l'espace $H_0^1(\Omega; \omega_2)$, avec plus de régularité sur $f$ dans la direction $X_1$ à savoir $f(\cdot, X_2) \in H_0^d(\omega_1)$ nous pouvons obtenir un développement asymptotique (dans le cas du laplacien perturbé) d'ordre $d-1$ dans l'espace $H_0^1(\Omega; \omega_2)$, les arguments utilisent la technique du produit tensoriel introduite dans le Chapitre 1. Nous pensons qu'avec plus de régularité sur les coefficients de $A$, nous pouvons obtenir un développement asymptotique d'ordre $d-1$ pour le problème général. L'existence d'un développement asymptotique local (i.e. loin du bord) a été étudiée dans [14].

Concernant les questions ouvertes, commençons par le cas des problèmes à données $L^p$. Nous avons démontré des convergences fortes dans le cas $1 < p < 2$ pour le problème (8). Dans le cas $p > 2$, nous avons une convergence forte triviale dans l'espace $H_0^1(\Omega; \omega_2)$. Par régularité elliptique de degré $p$ la solution du problème perturbé vit dans l'espace $W^{1,p}(\Omega)$. Peut-on utiliser les techniques de la théorie de Caldéron-Zygmund pour démontrer une convergence dans $W^{1,p}(\Omega, \omega_2)$?

Le cas des problèmes à données $L^1$ est plus délicat. En effet, d'une manière générale la solution entropique $u_\epsilon$ ne peut pas converger dans $W_0^{1,r}(\Omega, \omega_2)$ avec $1 \le r < \frac{N}{N-1}$. Observons l'exemple
suivant :

On choisit un $f_1 \in L^1(0,1) \setminus (\cup_{1<r<2} L^r(0,1))$. Soit $f_2 \in L^1(0,1)$. On note $f = f_1 \otimes f_2$. et soit $u_2$ l'unique solution faible de

$$ - \partial_{x_2}^2 u_2 = f_2, \text{ avec } u_2(0) = u_2(1) = 0 \qquad (4.40) $$

Soit $u_\epsilon$ l'unique solution faible de

$$ -\epsilon^2 \partial_{x_1}^2 u_\epsilon - \partial_{x_2}^2 u_\epsilon = f, \text{ avec } u_\epsilon = 0 \text{ sur } \partial(0,1)^2. $$

Pour tout $r \in [1,2)$, au moins l'une des deux suites $(\epsilon \partial_{x_1} u_\epsilon)$, $(\partial_{x_2} u_\epsilon)$ n'est pas bornée dans $L^r(\Omega)$. En effet, supposons le contraire, alors on peut extraire une sous suite $(u_{\epsilon_k})$ faiblement convergente dans $W_0^{1,r}((0,1)^2; (0,1))$, notons $u$ sa limite. Nous montrons immédiatement que $u(x_1, \cdot)$ est solution faible de (4.40) pour p.p. $x_1 \in (0,1)$. Ainsi, Par unicité de la solution faible de (4.40) on obtient $u = f_1 \otimes u_2$ ce qui est absurde car $f_1 \otimes u_2 \notin L^r(\Omega)$. Nous pouvons conclure que pour un problème elliptique avec une donnée $L^1$ générale, il est impossible de prouver que les suites $(\epsilon \nabla_{X_1} u_\epsilon)$, $(\nabla_{X_2} u_\epsilon)$ sont bornées dans $L^r(\Omega)$ simultanément, et donc prouver la convergence dans $L^r(\Omega)$, vers la solution du problème limite, en d'autres termes il est inutile d'utiliser les techniques d'estimation de la théorie $L^1$ pour chercher des estimations dans $L^r(\Omega)$. Cependant, il reste une seule issue possible, celle de démontrer la convergence (même faible) directement dans $L^1(\Omega)$ et celle-ci reste une question ouverte.

Maintenant, considérons l'écoulement d'un fluide dans un cylindre dont la longueur tend vers l'infinie. Soit $\Omega = (0, \ell) \times \omega$ un Ouvert borné de $\mathbb{R}^3$. L'écoulement est modélisée par l'équation de Navier-Stokes: $\partial_t \vec{u} - \nu \Delta \vec{u} + (\vec{u} \cdot \nabla)\vec{u} + \nabla p = \vec{f}$, avec $\nabla \cdot \vec{u} = 0$, $\vec{u} = \vec{u}_0$, $\vec{u} = 0$ sur $\partial \Omega$. Supposons que $f$ ne dépend pas de $x_1$. On souhaite étudier le comportement asymptotique de $\vec{u}$ quand $\ell$ tend vers $+\infty$. Avec le rescaling $\epsilon = \frac{1}{\ell}$. Nous pouvons considérer le problème de perturbations singulières suivant:

$$ \partial_t \vec{u}_\epsilon - \nu \Delta_\epsilon \vec{u}_\epsilon + (\vec{u}_\epsilon \cdot \nabla_\epsilon) \vec{u}_\epsilon + \nabla_\epsilon p_\epsilon = \vec{f}, $$

avec $\nabla_\epsilon \cdot \vec{u}_\epsilon = 0$, $\vec{u}_\epsilon = \vec{u}_{0,\epsilon}$, $\vec{u}_\epsilon = 0$ sur $\partial((0,1) \times \omega)$. Ici, $\nabla_\epsilon := \begin{pmatrix} \epsilon \partial_{x_1} \\ \partial_{x_2} \\ \partial_{x_3} \end{pmatrix}$, et $\Delta_\epsilon := \epsilon^2 \partial_{x_1}^2 + \partial_{x_2}^2 + \partial_{x_3}^2$. Passons à la limite $\epsilon \to 0$ formellement, nous pouvons conjecturer que le champs de vecteurs ayant comme composantes la deuxième et la troisième composante de $\vec{u}_\epsilon$, tend vers la solution de l'équation
de Navier-Stokes bidimensionnelle sur $\omega$. Afin d'étudier ce problème de convergence rigoureusement,
nous pouvons étudier, dans un premier temps, le problème stationnaire. Nous pouvons étudier aussi,
à la lumière du Théorème 0.4, le comportement asymptotique du semi-groupe généré par l'opérateur
de Stokes perturbé dont le domaine est donné par $\{\vec{u} \in H_0^1((0,1) \times \omega))^3 : \nabla_\epsilon \cdot \vec{u} = 0\} \cap H^2((0,1) \times \omega)^3$.
# Appendix A

## Density lemmas

Let $\omega_1$ and $\omega_2$ be two open bounded subsets of $\mathbb{R}^q$ and $\mathbb{R}^{N-q}$ respectively. Recall that

$$H_0^1(\Omega; \omega_2) = \{u \in L^2(\Omega) \mid \nabla_{X_2} u \in L^2(\Omega)^{N-q}, \text{ and for a.e. } X_1 \in \omega_1, u(X_1, \cdot) \in H_0^1(\omega_2)\},$$
normed by $\|\nabla_{X_2}(\cdot)\|_{L^2(\Omega)}$. We have the following

**Lemma A.1.** The space $H_0^1(\Omega)$ is dense in $H_0^1(\Omega; \omega_2)$.

*Proof.* Let $u \in H_0^1(\Omega; \omega_2)$ fixed. Let $l$ be the linear form defined on $H_0^1(\Omega)$ by

$$\forall \varphi \in H_0^1(\Omega) : l(\varphi) = \int_{\Omega} \nabla_{X_2} u \cdot \nabla_{X_2} \varphi dx.$$

$l$ is continuous on $H_0^1(\Omega)$, indeed we have

$$\forall \varphi \in H_0^1(\Omega) : |l(\varphi)| \le \|\nabla_{X_2} u\|_{L^2(\Omega)} \|\nabla_{X_2} \varphi\|_{L^2(\Omega)},$$

and then,

$$\forall \varphi \in H_0^1(\Omega) : |l(\varphi)| \le \|\nabla_{X_2} u\|_{L^2(\Omega)} \|\nabla \varphi\|_{L^2(\Omega)}.$$

For every $n \in \mathbb{N}^*$, we denote $u_n$ the unique solution to

$$\left\{ \begin{array}{l} \frac{1}{n^2} \int_{\Omega} \nabla_{X_1} u_n \cdot \nabla_{X_1} \varphi dx + \int_{\Omega} \nabla_{X_2} u_n \cdot \nabla_{X_2} \varphi dx = l(\varphi), \forall \varphi \in H_0^1(\Omega) \\ u_n \in H_0^1(\Omega), \end{array} \right. \qquad (A.1)$$

where the existence and the uniqueness follow from the Lax-Milgram theorem. Testing with $u_n$ in (A.1) we get, for every $n \in \mathbb{N}^*$

$$\frac{1}{n^2} \int_{\Omega} |\nabla_{X_1} u_n|^2 dx + \int_{\Omega} |\nabla_{X_2} u_n|^2 dx \le \|\nabla_{X_2} u\|_{L^2(\Omega)} \|\nabla_{X_2} u_n\|_{L^2(\Omega)},$$
then, we deduce that

$$ \forall n \in \mathbb{N}^* : \|\nabla_{X_2} u_n\|_{L^2(\Omega)} \le \|\nabla_{X_2} u\|_{L^2(\Omega)}, \qquad (A.2) $$

and

$$ \forall n \in \mathbb{N}^* : \frac{1}{n} \|\nabla_{X_1} u_n\|_{L^2(\Omega)} \le \|\nabla_{X_2} u\|_{L^2(\Omega)}. \qquad (A.3) $$

Using (A.2) and Poincaré's inequality we obtain:

$$ \forall n \in \mathbb{N}^* : \|u_n\|_{L^2(\Omega)} \le C_{\omega_2} \|\nabla_{X_2} u\|_{L^2(\Omega)}. \qquad (A.4) $$

Reflexivity of $L^2(\Omega)$ shows that there exists, $u_\infty, u'_\infty, u''_\infty \in L^2(\Omega)$ and a subsequence still labelled $(u_n)$ such that

$$ u_n \to u_\infty, \nabla_{X_2} u_n \to u'_\infty \text{ and } \frac{1}{n} \nabla_{X_1} u_n \to u''_\infty \text{ in } L^2(\Omega), \text{ weakly.} $$

Using the continuity of derivation on $D'(\Omega)$ we get

$$ u_n \to u_\infty, \nabla_{X_2} u_n \to \nabla_{X_2} u_\infty \text{ and } \frac{1}{n} \nabla_{X_1} u_n \to 0 \text{ in } L^2(\Omega), \text{ weakly.} \qquad (A.5) $$

1) we have $u_\infty \in H_0^1(\Omega; \omega_2)$ : By the Mazur Lemma, there exists a sequence $(U_n)$ of convex combinations of $\{u_n\}$ such that

$$ \nabla_{X_2} U_n \to \nabla_{X_2} u_\infty \text{ in } L^2(\Omega) \text{ strongly,} \qquad (A.6) $$

then by the Lebesgue theorem there exists a subsequence $(U_{n_k})$ such that:

$$ \text{For a.e. } X_1 \in \omega_1 : \nabla_{X_2} U_{n_k}(X_1, \cdot) \to \nabla_{X_2} u_\infty(X_1, \cdot) \text{ in } L^2(\omega_2) \text{ strongly.} \qquad (A.7) $$

Now, since $(U_{n_k}) \in H_0^1(\Omega)^N$ then

$$ \text{For a.e. } X_1 \in \omega_1 : (U_{n_k}(X_1, \cdot)) \in H_0^1(\omega_2)^N. \qquad (A.8) $$

Combining (A.7) and (A.8) we deduce:

$$ \text{For a.e. } X_1 \in \omega_1, u_\infty(X_1, \cdot) \in H_0^1(\omega_2), $$

and the proof of $u_\infty \in H_0^1(\Omega; \omega_2)$ is finished.
2) we have $u_{\infty} = u$: Passing to the limit in (A.1) by using (A.5) we obtain

$$ \int_{\Omega} \nabla_{X_2} u_{\infty} \cdot \nabla_{X_2} \varphi dx = \int_{\Omega} \nabla_{X_2} u \cdot \nabla_{X_2} \varphi dx, \forall \varphi \in H_0^1(\Omega). \quad (A.9) $$

For every $\varphi_1 \in H_0^1(\omega_1)$ and $\varphi_2 \in H_0^1(\omega_2)$ take $\varphi = \varphi_1 \otimes \varphi_2$ in (A.9) we obtain, for a.e. $X_1 \in \omega_1$

$$ \int_{\omega_2} \nabla_{X_2} u_{\infty}(X_1, \cdot) \cdot \nabla_{X_2} \varphi_2 dX_2 = \int_{\omega_2} \nabla_{X_2} u(X_1, \cdot) \cdot \nabla_{X_2} \varphi_2 dX_2, \forall \varphi_2 \in H_0^1(\omega_2). $$

For a.e. $X_1 \in \omega_1$, take $\varphi_2 = u_{\infty}(X_1, \cdot) - u(X_1, \cdot)$ (which belongs to $H_0^1(\omega_2)$) in the above equality, we get:

$$ \int_{\omega_2} |\nabla_{X_2}(u_{\infty}(X_1, \cdot) - u(X_1, \cdot))|^2 dX_2 = 0. $$

Integrating over $\omega_1$ we deduce

$$ \int_{\Omega} |\nabla_{X_2}(u_{\infty} - u)|^2 dx = 0. $$

Finally, since $\|\nabla_{X_2}(\cdot)\|_{L^2(\Omega)}$ is a norm on $H_0^1(\Omega; \omega_2)$ we get,

$$ u_{\infty} = u. \quad (A.10) $$

Combining (A.6) and (A.10) we get the desired result. □

**Remark A.2.** By symmetry, $H_0^1(\Omega)$ is dense in the space

$$ H_0^1(\Omega; \omega_1) = \{u \in L^2(\Omega) \mid \nabla_{X_1} u \in L^2(\Omega), \text{ and for a.e. } X_2 \in \omega_2, u(\cdot, X_2) \in H_0^1(\omega_1)\}, \\ \text{normed by } \|\nabla_{X_1}(\cdot)\|_{L^2(\Omega)}. $$

**Lemma A.3.** The space $H_0^1(\omega_1) \otimes H_0^1(\omega_2)$ is dense in $H_0^1(\Omega)$.

*Proof.* It is well known that $D(\omega_1) \otimes D(\omega_2)$ is dense in $D(\omega_1 \times \omega_2)$. Here, $D(\omega_1 \times \omega_2)$ is equipped with its natural topology (the inductive limit topology). It is clear that the injection of $D(\omega_1 \times \omega_2)$ in $H_0^1(\omega_1 \times \omega_2)$ is continuous, thanks to the inequality

$$ \forall u \in D(\Omega) : \left( \int_{\Omega} |\nabla u|^2 dx \right)^{\frac{1}{2}} \le \sqrt{N \times \text{mes}(\Omega)} \times \left( \max_{1 \le i \le N} \sup_{\text{Support}(u)} |\partial_{x_i} u| \right). $$

Hence, by the density rule we obtain the density of $D(\omega_1) \otimes D(\omega_2)$ in $H_0^1(\Omega)$, and the lemma follows. □
**Lemma A.4.** Let $(V_n^{(1)})$ and $(V_n^{(2)})$ be two sequences of subspaces (not necessarily of finite dimension) of $H_0^1(\omega_1)$ and $H_0^1(\omega_2)$ respectively. If $\cup V_n^{(1)}$ and $\cup V_n^{(2)}$ are dense in $H_0^1(\omega_1)$ and $H_0^1(\omega_2)$ respectively, then $\text{vec}(\cup_n (V_n^{(1)} \otimes V_n^{(2)}))$ is dense in $H_0^1(\omega_1) \otimes H_0^1(\omega_2)$ for the induced topology of $H_0^1(\Omega)$. In particular, if $(V_n^{(1)})$ and $(V_n^{(2)})$ are nondecreasing then $\cup_n (V_n^{(1)} \otimes V_n^{(2)})$ is dense in $H_0^1(\omega_1) \otimes H_0^1(\omega_2)$.

*Proof.* Let us start by a useful inequality. For $u \otimes v$ in $H_0^1(\omega_1) \otimes H_0^1(\omega_2)$ we have :

$$ \begin{aligned} \|u \otimes v\|_{H_0^1(\Omega)}^2 &= \int_{\Omega} |\nabla_{X_1}(u \otimes v)|^2 dx + \int_{\Omega} |\nabla_{X_2}(u \otimes v)|^2 dx \\ &= \left( \int_{\omega_2} v^2 dX_2 \right) \times \left( \int_{\omega_1} |\nabla_{X_1} u|^2 dX_1 \right) \\ &\quad + \left( \int_{\omega_1} u^2 dX_1 \right) \times \left( \int_{\omega_2} |\nabla_{X_2} v|^2 dX_2 \right) \\ &\le C \|u\|_{H_0^1(\omega_1)}^2 \times \|v\|_{H_0^1(\omega_2)}^2, \end{aligned} \quad (A.11) $$

where we have used Fubini's theorem and Poincaré's inequality. Here, $C = C_{\omega_1}^2 + C_{\omega_2}^2 > 0$. Now, fix $\eta > 0$ and let $\varphi \otimes \psi \in H_0^1(\omega_1) \otimes H_0^1(\omega_2)$, by density of $\cup V_n^{(1)}$ in $H_0^1(\omega_1)$ there exists $n \in \mathbb{N}$ and $\varphi_n \in V_n^{(1)}$ such that:

$$ \|\psi\|_{H_0^1(\omega_2)} \times \|\varphi_n - \varphi\|_{H_0^1(\omega_1)} \le \frac{\eta}{2\sqrt{C}}. $$

Similarly by density of $\cup V_n^{(2)}$ in $H_0^1(\omega_2)$, there exists $m \in \mathbb{N}$ (which depends on $n$ and $\psi$) and $\psi_m \in V_m^{(2)}$ such that

$$ \|\varphi_n\|_{H_0^1(\omega_1)} \times \|\psi_m - \psi\|_{H_0^1(\omega_2)} \le \frac{\eta}{2\sqrt{C}}. $$

Whence, by using the triangle inequality and (A.11) we obtain

$$ \|\varphi \otimes \psi - \varphi_n \otimes \psi_m\|_{H_0^1(\Omega)} \le \eta. \quad (A.12) $$

Now, since every element of $H_0^1(\omega_1) \otimes H_0^1(\omega_2)$ could be written as $\sum_{i=1}^l \varphi_i \otimes \psi_i$, then by using the triangle inequality and using (A.12) with $\eta$ replaced by $\eta_l$, one gets the desired result. $\square$

**Corollary A.5.** $\text{vec}(\cup_n (V_n^{(1)} \otimes V_n^{(2)}))$ is dense in $H_0^1(\Omega)$. in particular, if $(V_n^{(1)})$ and $(V_n^{(2)})$ are nondecreasing, then $\cup_n (V_n^{(1)} \otimes V_n^{(2)})$ is dense in $H_0^1(\Omega)$.
# Appendix B

## Semigroups

**Lemma B.1.** Assume (5), (6), (24) and let $f_1 \in L^2(\omega_1)$, $f_2 \in L^2(\omega_2)$, then for every $\mu > 0$ we have:

$$R_{0,\mu}(f_1 \otimes f_2) = f_1 \otimes (R_{0,\mu}f_2).$$

Notice that $R_{0,\mu}f_2 \in H_0^1(\omega_2)$. Moreover, we have

$$A_{0,\mu}(f_1 \otimes f_2) = f_1 \otimes (A_{0,\mu}f_2).$$

Notice also that $A_{0,\mu}f_2 \in L^2(\omega_2)$. Here, $A_{0,\mu}$ is the Yosida approximation of $A_0$, recall that $A_{0,\mu} = \mu A_0 R_{0,\mu}$.

*Proof.* Let $v_2 \in H_0^1(\omega_2)$ be the unique solution in $H_0^1(\omega_2)$ to

$$\mu \int_{\omega_2} v_2 \varphi_2 dX_2 + \int_{\omega_2} A_{22}(X_2) \nabla_{X_2} v_2 \cdot \nabla_{X_2} \varphi_2 dX_2 = \int_{\omega_2} f_2 \varphi_2 dX_2, \forall \varphi_2 \in H_0^1(\omega_2), \quad (B.1)$$

Let $\varphi \in H_0^1(\Omega; \omega_2)$, then $\varphi(X_1, \cdot) \in H_0^1(\omega_2)$ for a.e. $X_1 \in \omega_1$. Let $f_1 \in L^2(\omega_1)$, multiplying (B.1) by $f_1$, testing in (B.1) with $\varphi(X_1, \cdot)$ and integrating over $\omega_1$ yields

$$\mu \int_{\Omega} f_1 v_2 \varphi dx + \int_{\Omega} A_{22}(X_2) \nabla_{X_2} (f_1 v_2) \cdot \nabla_{X_2} \varphi dx = \int_{\Omega} f_1 f_2 \varphi dx.$$

It is clear that $f_1 v_2 \in H_0^1(\Omega; \omega_2)$ whence, $R_{0,\mu}(f_1 \otimes f_2) = f_1 \otimes v_2$, in particular when $f_1 = 1$ we have $R_{0,\mu}(f_2) = v_2$, and therefore $R_{0,\mu}(f_1 \otimes f_2) = f_1 \otimes R_{0,\mu}(f_2)$. The second assertion follows immediately from the first one, in fact

$$A_{0,\mu}(f_1 \otimes f_2) = \mu A_0 R_{0,\mu}(f_1 \otimes f_2) = \mu A_0(f_1 \otimes R_{0,\mu}f_2).$$
We have $R_{0,\mu}f_2 \in D(\mathcal{A}_0) \cap H_0^1(\omega_2)$ then by using (24) we get

$$ \mathcal{A}_0(f_1 \otimes R_{0,\mu}f_2) = f_1 \otimes \mathcal{A}_0(R_{0,\mu}f_2), $$

Notice that the operator $\mathcal{A}_0$ is independent of the $X_1$ direction and that $\mathcal{A}_0(R_{0,\mu}f_2) \in L^2(\omega_2)$. Finally we get

$$ \mathcal{A}_{0,\mu}(f_1 \otimes f_2) = \mu f_1 \otimes \mathcal{A}_0(R_{0,\mu}f_2) = f_1 \otimes \mathcal{A}_{0,\mu}(f_2). $$

□

Now, let $s \ge 0$, $\mu > 0$ and $g \in L^2(\Omega)$. To simplify the notations, we denote $f_g := e^{s\mathcal{A}_{0,\mu}}g$ instead of $f_{g,s,\mu}$.

**Lemma B.2.** Assume (5), (6), (24). Let $g = g_1 \otimes g_2 \in L^2(\omega_1) \otimes L^2(\omega_2)$, then for $s \ge 0$, $\mu > 0$ we have:

$$ f_g = g_1 \otimes e^{s\mathcal{A}_{0,\mu}}g_2. $$

Notice that $e^{s\mathcal{A}_{0,\mu}}g_2 \in L^2(\omega_2)$.

*Proof.* we have

$$ f_g = e^{s\mathcal{A}_{0,\mu}}g = \sum_{k=0}^{\infty} \frac{s^k}{k!} \mathcal{A}_{0,\mu}^k g, $$

where the series converges in $L^2(\Omega)$. By an immediate induction we get by using Lemma B.1

$$ \forall k \in \mathbb{N}: \mathcal{A}_{0,\mu}^k g = g_1 \otimes \mathcal{A}_{0,\mu}^k g_2, $$

with $\mathcal{A}_{0,\mu}^k g_2 \in L^2(\omega_2)$ for every $k \in \mathbb{N}$, and the Lemma follows.

□

**Lemma B.3.** Assume (5), (6), (24). Let $g \in H^2(\omega_1) \otimes L^2(\omega_2)$, then for $s \ge 0$, $\mu > 0$, $i, j = 1, ..., q$ we have $D_{x_ix_j}^2 f_g$, $D_{x_i} f_g \in L^2(\Omega)$, with:

$$ D_{x_ix_j}^2 f_g = e^{s\mathcal{A}_{0,\mu}}(D_{x_ix_j}^2 g), \quad D_{x_i} f_g = e^{s\mathcal{A}_{0,\mu}}(D_{x_i} g). \qquad (B.2) $$

$$ \|D_{x_ix_j}^2 f_g\|_{L^2(\Omega)} \le \|D_{x_ix_j}^2 g\|_{L^2(\Omega)}, \quad \|D_{x_i} f_g\|_{L^2(\Omega)} \le \|D_{x_i} g\|_{L^2(\Omega)}. \qquad (B.3) $$
Proof. 1) Suppose the simple case when $g = g_1 \otimes g_2$. So, let $g = g_1 \otimes g_2 \in H^2(\omega_1) \otimes L^2(\omega_2)$ and let us prove assertions (B.2). By Lemma B.2 we get

$$f_g = g_1 \otimes e^{sA_{0,\mu}}(g_2),$$

with $e^{sA_{0,\mu}}g_2 \in L^2(\omega_2)$. Hence, for $i, j = 1, ..., q$ we get $D_{x_ix_j}^2 f_g \in L^2(\Omega)$ and $D_{x_ix_j}^2 f_g = (D_{x_ix_j}^2 g_1) \otimes e^{sA_{0,\mu}}g_2$, and applying Lemma B.2 we get

$$D_{x_ix_j}^2 f_g = e^{sA_{0,\mu}}(D_{x_ix_j}^2 g).$$

Similarly we get $D_{x_i}f_g = e^{sA_{0,\mu}}(D_{x_i}g)$, and assertion (B.2) follows when $g = g_1 \otimes g_2$.

2) Now, let $g \in H^2(\omega_1) \otimes L^2(\omega_2)$, since $g$ is a finite sum of elements of the form $g_1 \otimes g_2$, then by linearity we get $D_{x_ix_j}^2 f_g$, $D_{x_i}f_g \in L^2(\Omega)$ and

$$D_{x_ix_j}^2 f_g = e^{sA_{0,\mu}}(D_{x_ix_j}^2 g), \quad D_{x_i}f_g = e^{sA_{0,\mu}}(D_{x_i}g), \quad \text{for } i, j = 1, ..., q,$$

therefore

$$\|D_{x_ix_j}^2 f_g\|_{L^2(\Omega)} \le \|e^{sA_{0,\mu}}\| \|D_{x_ix_j}^2 g\|_{L^2(\Omega)} \le \|D_{x_ix_j}^2 g\|_{L^2(\Omega)}, \quad \text{for } i, j = 1, ..., q,$$

and similarly we obtain the second inequality of (B.3). □

**Lemma B.4.** Assume (5), (6), (24) and (26). Let $g \in (H_0^1 \cap H^2(\omega_1)) \otimes (H_0^1 \cap H^2(\omega_2))$, then for $s \ge 0$, $\mu > 0$ we have:

$$f_g \in D(\mathcal{A}_0), \mathcal{A}_0(f_g) \in H_0^1(\Omega; \omega_1), \text{ and } D_{x_i}(\mathcal{A}_0 f_g) = e^{s\mathcal{A}_{0,\mu}}(D_{x_i}\mathcal{A}_0 g), \quad i = 1, ..., q, \quad (B.4)$$

$$\|(\mathcal{A}_0 f_g)\|_{L^2(\Omega)} \le \|\mathcal{A}_0 g\|_{L^2(\Omega)} \text{ and } \|D_{x_i}(\mathcal{A}_0 f_g)\|_{L^2(\Omega)} \le \|D_{x_i}\mathcal{A}_0 g\|_{L^2(\Omega)}, \quad i = 1, ..., q. \quad (B.5)$$

Proof. 1) Suppose $g = g_1 \otimes g_2 \in (H_0^1 \cap H^2(\omega_1)) \otimes (H_0^1 \cap H^2(\omega_2))$ and let us prove (B.4). Since $g \in D(\mathcal{A}_0)$, thanks to (26), then $f_g = e^{s\mathcal{A}_{0,\mu}}g \in D(\mathcal{A}_0)$ and $\mathcal{A}_0 f_g = e^{s\mathcal{A}_{0,\mu}}\mathcal{A}_0 g$ (thanks to (1.38)). Now, we have

$$\mathcal{A}_0 f_g = \mathcal{A}_0(e^{s\mathcal{A}_{0,\mu}}g) = \mathcal{A}_0(g_1 \otimes e^{s\mathcal{A}_{0,\mu}}g_2).$$

Notice that, $g_2 \in D(\mathcal{A}_0)$, thanks to (26), then $e^{s\mathcal{A}_{0,\mu}}g_2 \in D(\mathcal{A}_0)$ (thanks to (1.38)), hence

$$\mathcal{A}_0 f_g = g_1 \mathcal{A}_0 e^{s\mathcal{A}_{0,\mu}} g_2,$$
where we have used the fact that $\mathcal{A}_0$ is independent of the $X_1$ - direction. Since $e^{s\mathcal{A}_{0,\mu}}$ and $\mathcal{A}_0$ commute on $D(\mathcal{A}_0)$, then

$$ \mathcal{A}_0 f_g = g_1 e^{s\mathcal{A}_{0,\mu}} \mathcal{A}_0 g_2. $$

Now, we have $\mathcal{A}_0 g_2 \in L^2(\omega_2)$ then $e^{s\mathcal{A}_{0,\mu}} \mathcal{A}_0 g_2 \in L^2(\omega_2)$ (thanks to Lemma B.2), however $g_1 \in H_0^1(\omega_1)$, then $\mathcal{A}_0 f_g \in H_0^1(\Omega; \omega_1)$. Whence, for $i = 1, ..., q$ we have

$$ D_{x_i}(\mathcal{A}_0 f_g) = D_{x_i} g_1 \otimes e^{s\mathcal{A}_{0,\mu}} \mathcal{A}_0 g_2, $$

and hence by, Lemma B.2 we get

$$ \begin{aligned} D_{x_i}(\mathcal{A}_0 f_g) &= e^{s\mathcal{A}_{0,\mu}} (D_{x_i} g_1 \otimes \mathcal{A}_0 g_2) \\ &= e^{s\mathcal{A}_{0,\mu}} (D_{x_i} \mathcal{A}_0 g). \end{aligned} $$

(Remark that $D_{x_i} \mathcal{A}_0 g \in L^2(\Omega)$ since $g_1 \in H_0^1(\omega_1)$ and $\mathcal{A}_0 g_2 \in L^2(\omega_2)$).

2) Now, for a general $g \in (H_0^1 \cap H^2(\omega_1)) \otimes (H_0^1 \cap H^2(\omega_2))$, assertion (B.4) follows by linearity. Finally, we show (B.5). We have

$$ \begin{aligned} \| (\mathcal{A}_0 f_g) \|_{L^2(\Omega)} &= \| e^{s\mathcal{A}_{0,\mu}} (\mathcal{A}_0 g) \|_{L^2(\Omega)} \le \| e^{s\mathcal{A}_{0,\mu}} \| \| \mathcal{A}_0 g \|_{L^2(\Omega)} \\ &\le \| \mathcal{A}_0 g \|_{L^2(\Omega)}. \end{aligned} $$

For $i = 1, ..., q$ we get

$$ \begin{aligned} \| D_{x_i}(\mathcal{A}_0 f_g) \|_{L^2(\Omega)} &= \| e^{s\mathcal{A}_{0,\mu}} (D_{x_i} \mathcal{A}_0 g) \|_{L^2(\Omega)} \le \| e^{s\mathcal{A}_{0,\mu}} \| \| D_{x_i} \mathcal{A}_0 g \|_{L^2(\Omega)} \\ &\le \| D_{x_i} \mathcal{A}_0 g \|_{L^2(\Omega)}. \end{aligned} $$

□

**Lemma B.5.** Assume (5), (6), (24) and (26). Let $g \in (H_0^1 \cap H^2(\omega_1)) \otimes (H_0^1 \cap H^2(\omega_2))$, then for $s \ge 0$, $\mu > 0$, $i = 1, ..., q$, $j = q + 1, ..., N$ we have $D_{x_j} f_g$, $D_{x_i x_j}^2 f_g \in L^2(\Omega)$ with:

$$ \| D_{x_j} f_g \|_{L^2(\Omega)}^2 \le \frac{1}{\lambda} \| \mathcal{A}_0 g \|_{L^2(\Omega)} \| g \|_{L^2(\Omega)}, \quad \| D_{x_i x_j}^2 f_g \|_{L^2(\Omega)}^2 \le \frac{1}{\lambda} \| D_{x_i} \mathcal{A}_0 g \|_{L^2(\Omega)} \| D_{x_i} g \|_{L^2(\Omega)}. \quad (B.6) $$
Proof. 1) Let us show the first inequality of(B.6). Suppose $g \in (H_0^1 \cap H^2(\omega_1)) \otimes (H_0^1 \cap H^2(\omega_2))$. Notice that $g \in D(\mathcal{A}_0)$, thanks to (26), then according to (1.38) we have $f_g \in D(\mathcal{A}_0) \subset H_0^1(\Omega; \omega_2)$, hence for $j \in \{q+1, ..., N\}$ the ellipticity assumption gives

$$ \begin{aligned} \|D_{x_j} f_g\|_{L^2(\Omega)}^2 &\le \frac{1}{\lambda} \langle -\mathcal{A}_0 f_g, f_g \rangle_{L^2(\Omega)} \\ &\le \frac{1}{\lambda} \|\mathcal{A}_0 f_g\|_{L^2(\Omega)} \|f_g\|_{L^2(\Omega)}. \end{aligned} $$

We have, $\|\mathcal{A}_0 f_g\|_{L^2(\Omega)} = \|\mathcal{A}_0 e^{s\mathcal{A}_{0,\mu}} g\|_{L^2(\Omega)} = \|e^{s\mathcal{A}_{0,\mu}} \mathcal{A}_0 g\|_{L^2(\Omega)} \le \|\mathcal{A}_0 g\|_{L^2(\Omega)}$, and $\|f_g\|_{L^2(\Omega)} \le \|g\|_{L^2(\Omega)}$, therefore

$$ \|D_{x_j} f_g\|_{L^2(\Omega)}^2 \le \frac{1}{\lambda} \|\mathcal{A}_0 g\|_{L^2(\Omega)} \|g\|_{L^2(\Omega)}. $$

2) Now, let $1 \le i \le q$ fixed, then according to Lemma B.3 we have $D_{x_i} f_g = e^{s\mathcal{A}_{0,\mu}}(D_{x_i} g)$, notice that $D_{x_i} g \in D(\mathcal{A}_0)$ and hence, $D_{x_i} f_g \in D(\mathcal{A}_0)$, in particular $D_{x_i} f_g \in H_0^1(\Omega; \omega_2)$, and for $q+1 \le j \le N$ we get

$$ \begin{aligned} \|D_{x_j x_i}^2 f_g\|_{L^2(\Omega)}^2 &\le \frac{1}{\lambda} \langle -\mathcal{A}_0 D_{x_i} f_g, D_{x_i} f_g \rangle_{L^2(\Omega)} \\ &\le \frac{1}{\lambda} \|\mathcal{A}_0 D_{x_i} f_g\|_{L^2(\Omega)} \|D_{x_i} f_g\|_{L^2(\Omega)}. \end{aligned} $$

We have,

$$ \begin{aligned} \|\mathcal{A}_0 D_{x_i} f_g\|_{L^2(\Omega)} &= \|\mathcal{A}_0 e^{s\mathcal{A}_{0,\mu}}(D_{x_i} g)\|_{L^2(\Omega)} = \|e^{s\mathcal{A}_{0,\mu}}(\mathcal{A}_0 D_{x_i} g)\|_{L^2(\Omega)} \\ &\le \|(D_{x_i} \mathcal{A}_0 g)\|_{L^2(\Omega)}. \end{aligned} $$

Finally, by using (B.3) and the above inequality we obtain

$$ \|D_{x_j x_i}^2 f_g\|_{L^2(\Omega)}^2 \le \frac{1}{\lambda} \|(D_{x_i} \mathcal{A}_0 g)\|_{L^2(\Omega)} \|D_{x_i} g\|_{L^2(\Omega)}. $$

**Lemma B.6.** Under assumptions of Lemma B.5, we have for $g \in (H_0^1 \cap H^2(\omega_1)) \otimes (H_0^1 \cap H^2(\omega_2))$:

$$ f_g \in H_0^1(\Omega) \cap D(\mathcal{A}_0), \qquad (B.7) $$

and

$$ \mathrm{div}_{X_1}(A_{11}\nabla_{X_1}f), \mathrm{div}_{X_1}(A_{12}\nabla_{X_2}f), \mathrm{div}_{X_2}(A_{21}\nabla_{X_1}f) \in L^2(\Omega). \qquad (B.8) $$
*Proof.* Let us prove (B.7). In Lemma B.4 we proved that $f_g \in D(\mathcal{A}_0)$. Let us show that $f_g \in H_0^1(\Omega)$. Suppose the simple case $g = g_1 \otimes g_2$, we have $f_g = g_1 \otimes e^{s\mathcal{A}_{0,\mu}}g_2$. Since $g_2 \in D(\mathcal{A}_0)$, then $e^{s\mathcal{A}_{0,\mu}}g_2 \in D(\mathcal{A}_0)$, in particular we have $e^{s\mathcal{A}_{0,\mu}}g_2 \in H_0^1(\Omega; \omega_2)$ however, according to Lemma B.2 $e^{s\mathcal{A}_{0,\mu}}g_2 \in L^2(\omega_2)$, hence $e^{s\mathcal{A}_{0,\mu}}g_2 \in H_0^1(\omega_2)$. Finally as $g_1 \in H_0^1(\omega_1)$ we get $f_g \in H_0^1(\Omega)$. For a general $g$ in the tensor product space, the proof follows by linearity.

Now, let us show (B.8). According to Lemmas B.3, B.5 all these derivatives $D_{x_i}f_g$, $D^2_{x_i x_j}f_g$ for $1 \le i, j \le q$, and $D_{x_j}f_g$, $D^2_{x_i x_j}f_g$ for $1 \le i \le q$, $q+1 \le j \le N$ belong to $L^2(\Omega)$. Whence, combining this with (26) we get (B.8). $\square$
# Appendix C

## Existence theorems

Let $V \subset H_0^1(\Omega)$ be a subspace. We consider the problem

$$ \left\{ \begin{array}{l} \int_{\Omega} \beta(u) \varphi dx + \int_{\Omega} A_{22} \nabla_{X_2} u \cdot \nabla_{X_2} \varphi dx = \int_{\Omega} f \varphi dx, \forall \varphi \in V \\ u \in V, \end{array} \right. \qquad (C.1) $$

with $A_{22}$ and $\beta$ as in the introduction.

**Proposition C.1.** If $V$ is closed in $H_0^1(\Omega; \omega_2)$, then there exists a solution to (C.1).

*Proof.* We consider the perturbed problem

$$ \left\{ \begin{array}{l} \int_{\Omega} \beta(u_{\epsilon}) \varphi dx + \int_{\Omega} \tilde{A}_{\epsilon} \nabla u_{\epsilon} \cdot \nabla \varphi dx = \int_{\Omega} f \varphi dx, \forall \varphi \in V \\ u_{\epsilon} \in V, \end{array} \right. \qquad (C.2) $$

with

$$ \tilde{A}_{\epsilon} = \begin{pmatrix} \epsilon^2 I_q & 0 \\ 0 & A_{22} \end{pmatrix}. $$

The space $V$ is closed in $H_0^1(\Omega)$, thanks to the continuous embedding $H_0^1(\Omega) \hookrightarrow H_0^1(\Omega; \omega_2)$. The function $\tilde{A}_{\epsilon}$ is bounded and coercive, then by using the Schauder fixed point theorem, one can show the existence of a solution $u_{\epsilon}$ to (C.2). This solution is unique in $V$ thanks to (10) and coercivity of $\tilde{A}_{\epsilon}$. Testing with $u_{\epsilon}$ in (C.2) we obtain

$$ \epsilon \| \nabla_{X_1} u_{\epsilon} \|_{L^2(\Omega)}, \quad \| \nabla_{X_2} u_{\epsilon} \|_{L^2(\Omega)}, \quad \| u_{\epsilon} \|_{L^2(\Omega)} \le C, $$

where $C$ is independent of $\epsilon$, we have used that $\int_{\Omega} \beta(u_{\epsilon})u_{\epsilon}dx \ge 0$ (thanks to (10)). By using (11), we get

$$ \| \beta(u_{\epsilon}) \|_{L^2(\Omega)} \le M (|\Omega|^{\frac{1}{2}} + C). $$
So, there exist $v \in L^2(\Omega)$, $u \in L^2(\Omega)$ with $\nabla_{X_2}u \in L^2(\Omega)$, and a subsequence $(u_{\epsilon_k})_{k \in \mathbb{N}}$ such that

$$ \beta(u_{\epsilon_k}) \to v, \epsilon_k \nabla_{X_1} u_{\epsilon_k} \to 0, \nabla_{X_2} u_{\epsilon_k} \to \nabla_{X_2} u, u_{\epsilon_k} \to u \text{ in } L^2(\Omega)\text{-weak.} \quad (C.3) $$

Passing to the limit in (C.2) we get

$$ \int_{\Omega} v \varphi dx + \int_{\Omega} A_{22} \nabla_{X_2} u \cdot \nabla_{X_2} \varphi dx = \int_{\Omega} f \varphi dx, \forall \varphi \in V. \quad (C.4) $$

Take $\varphi = u_{\epsilon_k}$ in (C.4) and passing to the limit we get

$$ \int_{\Omega} v u dx + \int_{\Omega} A_{22} \nabla_{X_2} u \cdot \nabla_{X_2} u dx = \int_{\Omega} f u dx \quad (C.5) $$

Let us consider the quantity

$$ \begin{aligned} 0 \le I_k &= \int_{\Omega} \epsilon^2 |\nabla_{X_1} u_{\epsilon_k}|^2 dx + \int_{\Omega} A_{22} \nabla_{X_2} (u_{\epsilon_k} - u) \cdot \nabla_{X_2} (u_{\epsilon_k} - u) \\ &\quad + \int_{\Omega} (\beta(u_{\epsilon_k}) - \beta(u))(u_{\epsilon_k} - u) dx \\ &= \int_{\Omega} f u_{\epsilon_k} dx - \int_{\Omega} A_{22} \nabla_{X_2} u_{\epsilon_k} \cdot \nabla_{X_2} u dx - \int_{\Omega} A_{22} \nabla_{X_2} u \cdot \nabla_{X_2} u_{\epsilon_k} dx \\ &\quad + \int_{\Omega} f u dx - \int_{\Omega} v u dx - \int_{\Omega} \beta(u) u_{\epsilon_k} dx \\ &\quad - \int_{\Omega} \beta(u_{\epsilon_k}) u dx + \int_{\Omega} \beta(u) u dx \end{aligned} $$

Remark that this quantity is nonnegative, thanks to the ellipticity and monotonicity assumptions.

Passing to the limit as $k \to \infty$ using (C.3), (C.5) we get

$$ \lim I_k = 0. $$

Therefore, the ellipticity assumption shows that

$$ \|\epsilon_k \nabla_{X_1} u_{\epsilon_k}\|_{L^2(\Omega)}, \|u_{\epsilon_k} - u\|_{L^2(\Omega)}, \|\nabla_{X_2} (u_{\epsilon_k} - u)\|_{L^2(\Omega)} \to 0, \quad (C.6) $$

and hence, by a contradiction argument one has

$$ \beta(u_{\epsilon_k}) \to \beta(u) \text{ strongly in } L^2(\Omega). $$

Whence (C.4) becomes

$$ \int_{\Omega} \beta(u) \varphi dx + \int_{\Omega} A_{22} \nabla_{X_2} u \cdot \nabla_{X_2} \varphi dx = \int_{\Omega} f \varphi dx, \forall \varphi \in V. $$

Finally, $\|\nabla_{X_2} (u_{\epsilon_k} - u)\|_{L^2(\Omega)} \to 0$ shows that $u \in H_0^1(\Omega; \omega_2)$, and therefore as $V$ is closed in $H_0^1(\Omega; \omega_2)$ we obtain that $u \in V$. $\square$
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