# Uniform estimations for conforming Galerkin method for anisotropic
singularly perturbed elliptic problems

David Maltesea and Chokri Ogabib b Corresponding author. Email:
<EMAIL_ADDRESS>a,bLAMA, Univ. Gustave Eiffel, Univ. Paris Est
Créteil, CNRS, F-77454 Marne-la-Vallée, France.

###### Abstract

In this article, we study some anisotropic singular perturbations for a class
of linear elliptic problems. A uniform estimations for conforming $Q_{1}$
finite element method are derived, and some other results of convergence and
regularity for the continuous problem are proved.

###### keywords:

Galerkin method, finite element method, singular perturbations, elliptic
problem, uniform estimations, global rate of convergence.

††articletype: Research Article

## Introduction

The numerical study of singular perturbations problems keeps an important
place numerical analysis. Consider an elliptic problem
$P_{\epsilon}(u_{\epsilon})=0$ where $\epsilon\in(0,1]$ is a small parameter.
Let $P_{\epsilon,h}(u_{\epsilon,h})=0$ be the numerical approximation of the
continuous problem $P_{\epsilon}$. The estimations obtained by a classical
analysis, for instance, by the Céa’s lemma, is of the form

$\|u_{\epsilon,h}-u_{\epsilon}\|_{\Omega}\leq C\frac{h}{\epsilon^{\alpha}},$
(1)

where $\|\cdot\|_{\Omega}$ is some norm on a suitable space on $\Omega$. To
ensure a good numerical approximation of the exact solution $u_{\epsilon}$
when $\epsilon$ is very small one must take $h$ very small than $\epsilon$
which is impractical from the numerical point of view. For some kind of
numerical scheme which called asymptotically preserved, that is when we have
$\lim_{\epsilon}\lim_{h}u_{\epsilon,h}=\lim_{h}\lim_{\epsilon}u_{\epsilon,h}$,
one can obtain $\epsilon-$uniform estimations. In [1] J. Sin gave a simple
method to obtain such estimations. The idea is combining (1) with an other
estimation of the form

$\|u_{\epsilon,h}-u_{\epsilon}\|_{\Omega}\leq C(\epsilon^{\beta}+h^{\gamma}),$
(2)

to obtain the $\epsilon-$uniform estimation

$\|u_{\epsilon,h}-u_{\epsilon}\|_{\Omega}\leq
Ch^{\min(\frac{\beta}{\alpha+\beta},\gamma)}.$

For isotropic singular perturbations, which model diffusion phenomena in
isotropic medium, many authors studied conforming and non conforming Galerkin
methods for the following problem

$-\epsilon^{2}\Delta u_{\epsilon}+au_{\epsilon}=f\text{, with }u=0\text{ on
}\partial\Omega,$

where $\Omega$ is square or a cube and $f$ is sufficiently regular (of class
$C^{2}$) with some compatibility condition, that is when $f$ is zero on the
edge of the vertices of $\Omega$. For instance, in [2] a uniform estimation of
the form $|||u_{\epsilon,h}-u_{\epsilon}|||_{\Omega}=O(h^{1/2})$ is proved,
where $|||\cdot|||_{\Omega}$ is a variant of the energy norm. Some other
quasi-uniform logarithmic estimations have been proved, see for instance the
references [4], [5], [6] and those cited therein.

In this article, we deal with anisotropic singular perturbations, which model
diffusion phenomena in anisotropic medium. We will prove some uniform
estimations for the conforming Galerkin method in 2D and 3D for an elliptic
problem with a general diffusion matrix, and a source term with low
regularity. A prototype of such problems is given by the following bi-
dimensional equation

$-\epsilon^{2}\partial_{x_{1}}^{2}u_{\epsilon}-\partial_{x_{2}}^{2}u_{\epsilon}=f.$
(3)

For the several results proved in this paper, we will consider different
assumptions for the regularity of $f$, we give here the principal ones

$f\in L^{2}(\Omega),$ (4) $f\in H^{2}(\Omega).$ (5)

We will deal with the following general set-up of (3) [7]

$-\text{div}(A_{\epsilon}\nabla u_{\epsilon})=f\leavevmode\nobreak\
\text{in}\leavevmode\nobreak\ \Omega,$ (6)

supplemented with the boundary condition

$u_{\epsilon}=0\leavevmode\nobreak\ \text{in}\leavevmode\nobreak\
\partial\Omega.$ (7)

Here, $\Omega=\omega_{1}\times\omega_{2}$ where $\omega_{1}$ and $\omega_{2}$
are two bounded open sets of $\mathbb{R}^{q}$ and $\mathbb{R}^{N-q},$ with
$N>q\geq 1$. We denote by
$x=(x_{1},...,x_{N})=(X_{1},X_{2})\in\mathbb{R}^{q}\times\mathbb{R}^{N-q}$
i.e. we split the coordinates into two parts. With this notation we set

$\nabla=(\partial_{x_{1}},...,\partial_{x_{N}})^{T}=\binom{\nabla_{X_{1}}}{\nabla_{X_{2}}},\text{
}$

where

$\nabla_{X_{1}}=(\partial_{x_{1}},...,\partial_{x_{q}})^{T}\text{ and
}\nabla_{X_{2}}=(\partial_{x_{q+1}},...,\partial_{x_{N}})^{T}$

The matrix-valued function $A=(a_{ij})_{1\leq i,j\leq
N}:\Omega\rightarrow\mathcal{M}_{N}(\mathbb{R})$ satisfies the classical
ellipticity assumptions

  * •

There exists $\lambda>0$ such that for a.e. $x\in\Omega$

$A(x)\xi\cdot\xi\geq\lambda\left|\xi\right|^{2}\leavevmode\nobreak\ \text{for
any}\leavevmode\nobreak\ \xi\in\mathbb{R}^{N}.$ (8)

  * •

The elements of $A$ are bounded that is

$a_{ij}\in L^{\infty}(\Omega)\leavevmode\nobreak\ \text{for
any}\leavevmode\nobreak\ (i,j)\in\\{1,2,....,N\\}^{2}.$ (9)

For the $H^{2}$ estimations, we suppose that $A$ satisfies the regularity
assumption

$a_{ij}\in W^{1,\infty}(\bar{\Omega})\leavevmode\nobreak\ \text{for
any}\leavevmode\nobreak\ (i,j)\in\\{1,2,....,N\\}^{2},$ (10)

and the boundary condition

$\text{ For every }i\neq j:a_{ij}=0\text{ on }\partial\Omega.$ (11)

We have decomposed $A$ into four blocks

$A=\begin{pmatrix}A_{11}&A_{12}\\\ A_{21}&A_{22}\end{pmatrix}$

where $A_{11}$, $A_{22}$ are $q\times q$ and $(N-q)\times(N-q)$ matrices
respectively. Notice that (8) implies that $A_{22}$ satisfies the ellipticity
assumption

$\text{ For a.e.
}x\in\Omega:A_{22}(x)\xi_{2}\cdot\xi_{2}\geq\lambda\left|\xi_{2}\right|^{2}\leavevmode\nobreak\
\text{for any}\leavevmode\nobreak\ \xi_{2}\in\mathbb{R}^{N-q}.$ (12)

For $\epsilon\in(0,1]$, we have set

$A_{\epsilon}=\begin{pmatrix}\epsilon^{2}A_{11}&\epsilon A_{12}\\\ \epsilon
A_{21}&A_{22}\end{pmatrix}$

For the estimation of the global rate of convergence for the continuous
problem, we suppose the following additional assumption

$\text{ The block }A_{22}\text{ depends only on }X_{2}$ (13)

The weak formulation of the problem (6)-(7) is

$\left\\{\begin{array}[]{l}\int_{\Omega}A_{\epsilon}\nabla
u_{\epsilon}\cdot\nabla\varphi{\rm d}x=\int_{\Omega}f\text{ }\varphi{\rm
d}x,\quad\forall\varphi\in H_{0}^{1}(\Omega)\text{\ \ \ \ }\\\ u_{\epsilon}\in
H_{0}^{1}(\Omega),\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\end{array}\right.$ (14)

where the existence and uniqueness is a consequence of the assumptions
$(\ref{hypA1})$, $(\ref{hypA2})$. The limit problem is given by

$-\text{div}_{X_{2}}(A_{22}\nabla u)=f\leavevmode\nobreak\
\text{in}\leavevmode\nobreak\ \Omega,$ (15)

supplemented with the boundary condition

$u(X_{1},\cdot)=0\leavevmode\nobreak\ \text{in}\leavevmode\nobreak\
\partial\omega_{2},\leavevmode\nobreak\ \text{for}\leavevmode\nobreak\
X_{1}\in\omega_{1}.$ (16)

We recall the Hilbert space [8]

$H_{0}^{1}(\Omega;\omega_{2})=\left\\{\begin{array}[]{cc}v\in
L^{2}(\Omega)\leavevmode\nobreak\ \text{such that}\leavevmode\nobreak\
\nabla_{X_{2}}v\in L^{2}(\Omega)^{N-q}\\\ \text{ and for a.e.
}\leavevmode\nobreak\ X_{1}\in\omega_{1},\leavevmode\nobreak\
v(X_{1},\cdot)\in H_{0}^{1}(\omega_{2})\end{array}\right\\},$

equipped with the norm
$\left\|\nabla_{X_{2}}(\cdot)\right\|_{L^{2}(\Omega)^{N-q}}$. Notice that this
norm is equivalent to

$\left(\left\|(\cdot)\right\|_{L^{2}(\Omega)}^{2}+\left\|\nabla_{X_{2}}(\cdot)\right\|_{L^{2}(\Omega)^{N-q}}^{2}\right)^{1/2},$

thanks to Poincaré’s inequality

$\left\|v\right\|_{L^{2}(\Omega)}\leq
C_{\omega_{2}}\left\|\nabla_{X_{2}}v\right\|_{L^{2}(\Omega)^{N-q}},\leavevmode\nobreak\
\text{for any}\leavevmode\nobreak\ v\in H_{0}^{1}(\Omega;\omega_{2}).$ (17)

The space $H_{0}^{1}(\Omega)$ will be normed by
$\left\|\nabla(\cdot)\right\|_{L^{2}(\Omega)^{N}}$. One can check immediately
that the embedding $H_{0}^{1}(\Omega)\hookrightarrow
H_{0}^{1}(\Omega,\omega_{2})$ is continuous. The weak formulation of the limit
problem (15)-(16) is given by

$\left\\{\begin{array}[]{ll}\begin{array}[]{c}\int_{\omega_{2}}A_{22}\mathbf{(}X_{1},\cdot)\nabla_{X_{2}}u\mathbf{(}X_{1},\cdot)\cdot\nabla_{X_{2}}\psi
dX_{2}=\int_{\omega_{2}}f\mathbf{(}X_{1},\cdot)\text{ }\psi dX_{2}\text{,
}\forall\psi\in H_{0}^{1}(\omega_{2})\text{\ \ \ \ }\end{array}&\\\
u\mathbf{(}X_{1},\cdot)\in H_{0}^{1}(\omega_{2})\text{,\ for a.e.
}X_{1}\in\omega_{1}\text{, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }&\text{
}\end{array}\right.$ (18)

where the existence and uniqueness is a consequence of the assumptions
(9),(12). Recall that we have $u\in H_{0}^{1}(\Omega;\omega_{2})$ and
$u_{\epsilon}\rightarrow u\text{ in }H_{0}^{1}(\Omega;\omega_{2})$ as
$\epsilon\rightarrow 0$ [8]. Notice that for $\varphi\in
H_{0}^{1}(\Omega;\omega_{2})$, then for a.e $X_{1}$ in $\omega_{1}$ we have
$\varphi(X_{1},\cdot)\in H_{0}^{1}(\omega_{2})$, testing with it in (18) and
integrating over $\omega_{1}$ yields

$\int_{\Omega}A_{22}\nabla_{X_{2}}u\cdot\nabla_{X_{2}}\varphi
dx=\int_{\Omega}f\text{ }\varphi dx,\text{\ }\forall\varphi\in
H_{0}^{1}(\Omega;\omega_{2}).$ (19)

Notice that (19) could be seen as the weak formulation of the limit problem
(15)-(16) in the Hilbert space $H_{0}^{1}(\Omega;\omega_{2})$ (thanks to
Poincaré’s inequality (17)).
Finally, for the estimation of the global rate of convergence, let us recall
this result proved in our paper [8]

$\|\nabla_{X_{2}}(u_{\epsilon}-u)\|_{L^{2}(\Omega)^{N-q}}\leq C\epsilon,\text{
for }f\in H^{1}_{0}(\Omega;\omega_{1}),$

here $H^{1}_{0}(\Omega;\omega_{1})$ is similarly defined as
$H^{1}_{0}(\Omega;\omega_{2})$. The main results of the article are given in
two sections:

  * •

In the first section, we show a version of the previous estimation for more
general $f$, and we show some high order regularity estimations for the
solutions $u_{\epsilon}$ and $u$, for general domains in arbitrary dimension
$N\geq 2$.

  * •

In the second section, we use the results of the first section to analyse a
$Q_{1}$ finite element method for problem (14) when $\Omega$, is a square or a
cube . In the case of regular perturbation (i.e. no boundary layers), that is
when $f\in H^{1}_{0}\cap H^{2}(\Omega)$, we derive a uniform estimation of the
form

$\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
Ch^{\frac{1}{3}}.$

In the case of singular perturbation (i.e. with boundary layer formation),
that is when $f\in H^{2}(\Omega)$, we derive a uniform estimation of the form

$\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
Ch^{\frac{1}{5}}.$

Notice that throughout this article $C_{{f,A\text{ ... etc.}}}$ denotes a
generic positive constant depending only on the objects $f,A\text{ ... etc.}$.

##  1 Some results for the continuous problem

###  1.1 Rate of convergence for general data

In this subsection, we suppose that the block $A_{12}$ satisfies the
assumption

$\partial_{i}a_{ij}\in L^{\infty}(\Omega),\partial_{j}a_{ij}\in
L^{\infty}(\Omega)\text{ for }i=1,...,q\text{ and }j=q+1,...,N.$ (20)

In [8] (Theorem 2.3), we have proved the following estimation.

######  Theorem 1.1.

[8] Let $\Omega=\omega_{1}\times\omega_{2}$ where $\omega_{1}$ and
$\omega_{2}$ are two bounded open sets of $\mathbb{R}^{q}$ and
$\mathbb{R}^{N-q}$ respectively, with $N>q\geq 1.$ Suppose that $A$ satisfies
$(\ref{hypA1})$, $(\ref{hypA2})$, $(\ref{A22})$ and $(\ref{hypAd1})$. Let
$f\in H^{1}_{0}(\Omega,\omega_{1})$, then there exists
$C_{\lambda,\Omega,A}>0$ such that:

$\forall\epsilon\in(0,1]:\left\|\nabla_{X_{2}}(u_{\epsilon}-u)\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{\lambda,\Omega,A}(\left\|\nabla_{X_{1}}f\right\|_{L^{2}(\Omega)^{q}}+\left\|f\right\|_{L^{2}(\Omega)})\times\epsilon,$
(21)

where $u_{\epsilon}$ is the unique solution of $(\ref{u-perturbed-weak})$ in
$H_{0}^{1}(\Omega)$ and $u$ is the unique solution to $(\ref{u-limit-weak})$
in $H_{0}^{1}(\Omega;\omega_{2}),$ moreover we have $u\in H^{1}_{0}(\Omega).$

When $f$ does not have the $H^{1}_{0}$ regularity in the $X_{1}$ direction
i.e. $f\notin H^{1}_{0}(\Omega,\omega_{1})$, we will show a rate of
convergence of order $O(\epsilon^{s})$, $0<s<1$. The argument is based on the
interpolation trick (see for instance [10]). In the above reference, Lions
shows that every $H^{1}$ function could be splitted into a
$H^{1}_{0}$-function with a big $H^{1}$ norm, and a $H^{1}$-function with a
small $L^{2}$ norm, when the domain $\Omega$ is regular (of class $C^{1}$, or
of class $C^{2}$ for the decomposition of $H^{2}$ functions). Here, we will
prove some decomposition lemmas for functions with a partial $H^{1}$
regularity and for domains with a very low regularity.

######  Definition 1.2.

Let $\mathcal{O}$ be a bounded open set of $\mathbb{R}^{d}$. we say that
$\mathcal{O}$ satisfies the decomposition hypothesis if there exist positive
constants $c_{1}$ and $c_{2}$ such that for any $\delta\in(0,1)$ there exists
$\mathcal{O}_{\delta}\subset\subset\mathcal{O}$ open such that

$\text{{meas}}(\mathcal{O}\textbackslash\mathcal{O}_{\delta})\leq
c_{1}\delta\text{ and
}\text{dist}(\mathcal{O}_{\delta},\partial\mathcal{O})\geq c_{2}\delta$ (D)

One can show that polygonal domains, or more generally Lipschitz domains,
satisfy the decomposition hypothesis (D). That is still true for some non-
Lipschitz domains, for example, for the following open set of
$\mathbb{R}^{2}$, $\mathcal{O}=\\{|y|\leq x^{2},\text{ }0<x<1\\}$. Notice that
there exist bounded open sets which do not satisfy the decomposition
propriety, for instance in dimension $d\geq 1$, the open set
$\mathbb{B}_{d}(0,1)\backslash\\{\cup_{k=1}^{\infty}\mathbb{S}_{d}(0,\frac{1}{k})\\}$,
where $\mathbb{B}_{d}(0,1)$ is the unite euclidean ball of $\mathbb{R}^{d}$
and $\mathbb{S}_{d}(0,\frac{1}{k})$ is the euclidean sphere of
$\mathbb{R}^{d}$ of center $0$ and ray $\frac{1}{k}$, gives such an example.
Now, we have to prove the following

######  Lemma 1.3.

Let $\Omega=\omega_{1}\times\omega_{2}$ be a bounded open set of
$\mathbb{R}^{q}\times\mathbb{R}^{N-q}$. Let $f\in L^{r}(\Omega)$ for some
$r>2$ such that $\nabla_{X_{1}}f\in L^{2}(\Omega)^{q}$. Suppose that
$\omega_{1}$ satisfies (D), then there exist
$C_{{}_{f,c_{1},c_{2},\omega_{2}}},C_{{}_{f,c_{1},\omega_{2}}}>0$ such that
for any $\delta\in(0,1)$ there exist $f^{1}_{\delta}\in
H^{1}_{0}(\Omega,\omega_{1})$ and $f^{2}_{\delta}\in L^{2}(\Omega)$ with
$\nabla_{X_{1}}f^{2}_{\delta}\in L^{2}(\Omega)^{q}$ such that
$f=f^{1}_{\delta}+f^{2}_{\delta}$ with:

$\|\nabla_{X_{1}}f^{1}_{\delta}\|_{L^{2}(\Omega)^{q}}\leq
C_{{}_{f,c_{1},c_{2},\omega_{2}}}\delta^{-\frac{1}{2}-\frac{1}{r}}\text{, and
}\|f^{2}_{\delta}\|_{L^{2}(\Omega)}\leq
C_{{}_{f,c_{1},\omega_{2}}}\delta^{\frac{1}{2}-\frac{1}{r}}.$

###### Proof.

Let us recall the space

$H_{0}^{1}(\Omega;\omega_{1})=\left\\{\begin{array}[]{cc}v\in
L^{2}(\Omega)\leavevmode\nobreak\ \text{such that}\leavevmode\nobreak\
\nabla_{X_{1}}v\in L^{2}(\Omega)^{q}\\\ \text{ and for a.e.
}\leavevmode\nobreak\ X_{2}\in\omega_{2},v(\cdot,X_{2})\in
H_{0}^{1}(\omega_{1})\end{array}\right\\},$

and the anisotropic Poincaré’s inequality

$\left\|v\right\|_{L^{2}(\Omega)}\leq
C_{\omega_{1}}\left\|\nabla_{X_{1}}v\right\|_{L^{2}(\Omega)^{q}},\leavevmode\nobreak\
\text{for any}\leavevmode\nobreak\ v\in H_{0}^{1}(\Omega;\omega_{1}).$ (22)

Let $\delta\in(0,1)$, since $\omega_{1}$ satisfies (D) then there exists
$\omega_{1}^{\delta}\subset\subset\omega_{1}$ such that

$\text{{meas}}(\omega_{1}\textbackslash\omega_{1}^{\delta})\leq
c_{1}\delta\text{ and }\text{dist}(\omega_{1}^{\delta},\partial\omega_{1})\geq
c_{2}\delta$

Let $K_{\delta}=\\{X_{1}\in\omega_{1}\text{,
dist}(X_{1},\omega_{1}^{\delta})\leq\frac{c_{2}\delta}{3}\\}$. We define the
bump function $\rho_{\delta}\in\mathcal{D}(\omega_{1})$ by

$\rho_{\delta}(X_{1})=\int_{K_{\delta}}\theta_{\frac{c_{2}\delta}{3}}(X_{1}-y)dy,$

where
$\theta_{\mu}(\cdot)=\Big{(}\frac{1}{\mu}\Big{)}^{q}\theta(\frac{\cdot}{\mu})$,
with $\theta\in\mathcal{D}(\mathbb{R}^{q})$ such that
$\int_{\mathbb{R}^{q}}\theta(X_{1})dX_{1}=1$ and $\text{Supp}(\theta)\subset
B_{q}(0,1)$. We can check that

$\rho_{\delta}=1\text{ on }\omega_{1}^{\delta},\text{
supp}(\rho_{\delta})\subset\omega_{1}\text{, and
}\|\rho_{\delta}\|_{\infty}\leq 1\text{,
}\|\nabla_{X_{1}}\rho_{\delta}\|_{\infty}\leq\frac{3\|\nabla_{X_{1}}\theta\|_{\infty}}{c_{2}\delta}.$

We define

$f_{\delta}^{1}:=\rho_{\delta}f\text{ and }f_{\delta}^{2}:=f-\rho_{\delta}f.$

It is clear that $f_{\delta}^{1}\in H^{1}_{0}(\Omega;\omega_{1})$ and
$f_{\delta}^{2}\in L^{2}(\Omega)$ with $\nabla_{X_{1}}f_{\delta}^{2}\in
L^{2}(\Omega)^{q}$.
Now, let $1\leq i\leq q$, we have

$\|\partial_{i}f^{1}_{\delta}\|_{L^{2}(\Omega)}\leq\|\partial_{i}f\|_{L^{2}(\Omega)}+\frac{3\|\partial_{i}\theta\|_{\infty}}{c_{2}\delta}\|f\|_{L^{2}((\omega_{1}\textbackslash\omega_{1}^{\delta})\times\omega_{2})},$

whence by Hölder inequality we get

$\|\partial_{i}f^{1}_{\delta}\|_{L^{2}(\Omega)}\leq\|\partial_{i}f\|_{L^{2}(\Omega)}+\frac{3\|\partial_{i}\theta\|_{\infty}}{c_{2}\delta}\|f\|_{L^{r}(\Omega)}\times\Big{(}c_{1}|\omega_{2}|\delta\Big{)}^{\frac{1}{2}-\frac{1}{r}},$

and the first inequality of Lemma 1.3 follows. Similarly, we obtain the second
inequality of Lemma 1.3. ∎

Now, we are able to prove the following theorem

######  Theorem 1.4.

Assume that $\Omega$ and $A$ are given as in Theorem 1.1, assume in addition
that $\omega_{1}$ satisfies (D). Suppose that $f\in L^{r}(\Omega)$ for some
$\infty>r>2$ such that $\nabla_{X_{1}}f\in L^{2}(\Omega)$, then there exists
$C_{\lambda,f,A,\Omega}>0$ such that

$\forall\epsilon\in(0,1],\leavevmode\nobreak\
\left\|\nabla_{X_{2}}(u_{\epsilon}-u)\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\times\epsilon^{\frac{1}{2}-\frac{1}{r}}.$

In particular, when $r=\infty$ we have

$\forall\epsilon\in(0,1],\leavevmode\nobreak\
\left\|\nabla_{X_{2}}(u_{\epsilon}-u)\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\times\epsilon^{\frac{1}{2}}.$

###### Proof.

Let $f=f^{1}_{\delta}+f^{2}_{\delta}$ be the decomposition of Lemma 1.3. Let
$u_{\epsilon}^{1}$ (resp. $u_{\epsilon}^{2}$) be the solution of (14) with $f$
replaced by $f_{\delta}^{1}$ ( resp. $f_{\delta}^{2}$). The linearity of the
problem shows that

$u_{\epsilon}=u_{\epsilon}^{1}+u_{\epsilon}^{2}.$ (23)

Similarly, Let $u^{1}$ (resp. $u^{2}$) be the solution of (19) with $f$
replaced by $f_{\delta}^{1}$ ( resp. $f_{\delta}^{2}$), we also have

$u=u^{1}+u^{2}.$ (24)

Now, according to Theorem 1.1 we have

$\left\|\nabla_{X_{2}}(u_{\epsilon}^{1}-u^{1})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,\Omega,A}}(\left\|\nabla_{X_{1}}f_{\delta}^{1}\right\|_{L^{2}(\Omega)^{q}}+\left\|f_{\delta}^{1}\right\|_{L^{2}(\Omega)})\times\epsilon,$
(25)

By using the anisotropic Poincaré’s inequality (22), which holds because
$f^{1}_{\delta}\in H^{1}_{0}(\Omega,\omega_{1})$), we obtain from (25)

$\left\|\nabla_{X_{2}}(u_{\epsilon}^{1}-u^{1})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,\Omega,A}}\times\left\|\nabla_{X_{1}}f_{\delta}^{1}\right\|_{L^{2}(\Omega)^{q}}\times\epsilon.$
(26)

By applying Lemma 1.3 on the right hand side of the previous inequality we get

$\left\|\nabla_{X_{2}}(u_{\epsilon}^{1}-u^{1})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,\Omega,A}}\times\delta^{-\frac{1}{2}-\frac{1}{r}}\times\epsilon.$
(27)

Now, let us estimate
$\left\|\nabla_{X_{2}}(u_{\epsilon}^{2}-u^{2})\right\|_{L^{2}(\Omega)^{N-q}}$.
We test by $u_{\epsilon}^{2}$ (resp. $u^{2}$) in the corresponding weak
formulation i.e. (14) with $f$ replaced by $f_{\delta}^{2}$ ( resp. (19) with
$f$ replaced by $f_{\delta}^{2}$), we get

$\left\|\nabla_{X_{2}}u_{\epsilon}^{2}\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\Omega,\lambda}}\|f_{\delta}^{2}\|_{L^{2}(\Omega)}\text{ and
}\left\|\nabla_{X_{2}}u^{2}\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\Omega,\lambda}}\|f_{\delta}^{2}\|_{L^{2}(\Omega)}.$

Therefore, by the triangle inequality and Lemma 1.3 we get

$\left\|\nabla_{X_{2}}(u_{\epsilon}^{2}-u^{2})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{f,\Omega,\lambda}}\times\delta^{\frac{1}{2}-\frac{1}{r}}$ (28)

Finally, by using the decompositions (23), (24), and inequalities (27), (28)
and the triangle inequality we get

$\left\|\nabla_{X_{2}}(u_{\epsilon}-u)\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{f,\Omega,A,\lambda}}\Big{(}\delta^{-\frac{1}{2}-\frac{1}{r}}\epsilon+\delta^{\frac{1}{2}-\frac{1}{r}}\Big{)}$

Notice that this last inequality holds for every $\delta\in(0,1)$ and every
$\epsilon\in(0,1]$. Whence, we can choose $\delta=\epsilon$ and we get the
expected estimation. The case $r=\infty$ follows immediately by letting
$r\rightarrow\infty$ in the first estimation. ∎

Let us finish by this particular case which will be used in the analysis of
the numerical scheme. In fact, we have the following analogous of Lemma 1.3

######  Lemma 1.5.

Let $\Omega$ be a bounded open set of $\mathbb{R}^{N}$. Let $f\in
H^{2}(\Omega)\cap L^{\infty}(\Omega)$. Suppose that $\Omega$ satisfies
$(\ref{D})$, then the exist
$C_{{}_{f,c_{1},\Omega}},C_{{}_{f,c_{1},c_{2},\Omega}}>0$ such that for any
$\delta\in(0,1)$ there exist $f^{1}_{\delta}\in H^{1}_{0}\cap H^{2}(\Omega)$
and $f^{2}_{\delta}\in H^{2}(\Omega)$ such that
$f=f^{1}_{\delta}+f^{2}_{\delta}$ with:

$\|f^{1}_{\delta}\|_{H^{1}(\Omega)}\leq
C_{{}_{f,c_{1},c_{2},\Omega}}\delta^{-1/2},\text{
}\|f_{\delta}^{1}\|_{L^{2}(\Omega)}\leq\|f\|_{L^{2}(\Omega)},$

and

$\|f_{\delta}^{1}\|_{H^{2}(\Omega)}\leq
C_{{}_{f,c_{1},c_{2},\Omega}}\delta^{-\frac{3}{2}}\text{ and
}\|f^{2}_{\delta}\|_{L^{2}(\Omega)}\leq C_{{}_{f,c_{1},\Omega}}\delta^{1/2}.$

The proof is similar to that of Lemma 1.3. Here, we can notice that if
$\Omega$ is Lipschitz then the assumption $f\in L^{\infty}(\Omega)$, is
automatically satisfied in dimension 2 and 3 thanks to Sobolev embeddings.

###  1.2 Some Regularity estimations for the solution of the perturbed problem

In this subsection, we suppose that $\Omega$ is of the form

$\Omega=\prod\limits_{i=1}^{N}(0,l_{i}),$ (29)

where $l_{1},...,l_{N}$ are positive real numbers, and $N\geq 2$. We will
prove the following

######  Theorem 1.6.

Assume $(\ref{hypA1})$, $(\ref{A-class-C1})$, and $(\ref{Anullbord})$. Let
$\Omega$ such that $(\ref{Omega-cuboid})$. Let $f$ such that $(\ref{fL2})$ is
satisfied. Let $u_{\epsilon}$ be the solution of $(\ref{u-perturbed-weak})$
then there exists a constant $C_{{}_{\lambda,f,A,\Omega}}>0$ such that, for
any $\epsilon\in(0,1]$

$\epsilon^{2}\|D^{2}_{X_{1}}u_{\epsilon}\|_{L^{2}(\Omega)^{q^{2}}}+\epsilon\|D^{2}_{X_{1}X_{2}}u_{\epsilon}\|_{L^{2}(\Omega)^{q(N-q)}}+\|D^{2}_{X_{2}}u_{\epsilon}\|_{L^{2}(\Omega)^{(N-q)^{2}}}\leq
C_{{}_{\lambda,f,A,\Omega}}$

In addition, if $A\in C^{1}(\bar{\Omega})$ then the strong convergences

$\epsilon^{2}D^{2}_{X_{1}}u_{\epsilon}\rightarrow 0,\text{
}D^{2}_{X_{1}X_{2}}u_{\epsilon}\rightarrow 0\text{ and
}D^{2}_{X_{2}}u_{\epsilon}\rightarrow D^{2}_{X_{2}}u,$

hold in $L^{2}(\Omega)^{q^{2}}$, $L^{2}(\Omega)^{q(N-q)}$ and
$L^{2}(\Omega)^{(N-q)^{2}}$ respectively.

Here, we used the notation: $D^{2}_{X_{1}}:=(\partial_{ij}^{2})_{1\leq i,j\leq
q}$, $D^{2}_{X_{2}}:=(\partial_{ij}^{2})_{1+q\leq i,j\leq N}$, and
$D^{2}_{X_{1}X_{2}}:=(\partial_{ij}^{2})_{1\leq i\leq q,\text{ }q+1\leq j\leq
N}$
In [9], we have a local version of this result. The proof of Theorem 1.6 is
based on a symmetrization trick and the application of the local result.

We introduce $\widetilde{\Omega}=\prod\limits_{i=1}^{N}(-l_{i},2l_{i})$, and
we denote $sub(\widetilde{\Omega})$ the set of all disjoint subsets of
$\widetilde{\Omega}$ of the form $I=\prod\limits_{i=1}^{N}I_{i}$ where each
$I_{i}$ has one of the forms $(-l_{i},0),$ $(0,l_{i}),$ $(l_{i},2l_{i}).$ For
$0\leq j\leq N$ fixed we denote $sub(\widetilde{\Omega})_{-1,j}$,
$sub(\widetilde{\Omega})_{1,j},$ $sub(\tilde{\Omega})_{2,j}$ the subsets of
$sub(\widetilde{\Omega})$ such that $I_{j}$ is of the form $(-l_{j},0),$
$(0,l_{j}),$ $(l_{j},2l_{j})$ respectively. It is clear that these three
subsets define a partition of $sub(\widetilde{\Omega}).$ For $x=(x_{i})_{1\leq
i\leq N}\in I$, we denote

$y=(s_{i}(I)+r_{i}(I)x_{i})_{1\leq i\leq N}\in\Omega$

where $r_{i}$ and $s_{i}$, ${1\leq i\leq N}$, are defined on
$sub(\widetilde{\Omega})$ by :

$r_{i}(I)=\left\\{\begin{array}[]{c}{1\text{ if }I\in
sub(\widetilde{\Omega})_{1,i}}\\\ {-1\text{ else\ \ \ \ \ \ \ \ \ \ \
}}\end{array}\right.$ $s_{i}(I)=\left\\{\begin{array}[]{c}0\text{ if }I\in
sub(\widetilde{\Omega})_{-1,i}\cup sub(\widetilde{\Omega})_{1,i}\\\
2l_{i}\text{ if }I\in sub(\widetilde{\Omega})_{2,i}\text{ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ }\end{array}\right.$

Now, we define the function $\widetilde{f}\in L^{2}(\widetilde{\Omega})$ by

$\text{ For }x\in I\in
sub(\widetilde{\Omega}):\widetilde{f}(x)=\left(\prod\limits_{i=1}^{N}r_{i}(I)\right)\times
f(y)_{1\leq i\leq N}),\\\ \text{ and }\widetilde{f}=0\text{ else}.$

Similarly, we define the extension $\widetilde{u}_{\epsilon}$ of
$u_{\epsilon}$ by:

$\text{ For }x\in I\in
sub(\widetilde{\Omega}):\widetilde{u}_{\epsilon}(x)=\left(\prod\limits_{i=1}^{N}r_{i}(I)\right)\times
u_{\epsilon}(y),\\\ \text{ and }\widetilde{u}_{\epsilon}=0\text{ else}.$

We define $\widetilde{A}=(\widetilde{a_{ij}})$ the extension of $A$ as
follows: For $x\in\overline{\widetilde{\Omega}}$, there exists $I\in
sub(\widetilde{\Omega})$ such that $x\in\overline{I},$ in this case we set

$\widetilde{a_{ij}}(x)=r_{i}(I)r_{j}(I)a_{ij}(y),\text{ }i,j=1,...,N.$

Notice that assumption (11) implies that the value of each $\widetilde{A}(x)$
does not depend in the choice of $I$ so $\widetilde{A}$ is well-defined.
Notice also that (10) implies that $\widetilde{A}$ is Lipschitz on
$\overline{\widetilde{\Omega}}$. Moreover, we can check immediately that
$\widetilde{A}$ satisfies the ellipticity assumption (8) with the same
constant. Finally, we define $\widetilde{A}_{\epsilon}$ as we have defined
$A_{\epsilon}$ (see the introduction).
Under the above notations we have the following Lemma:

######  Lemma 1.7.

Suppose that assumptions of Theorem 1.6 hold. Let $\widetilde{u}_{\epsilon}$
constructed as above, then $\widetilde{u}_{\epsilon}$ is the unique weak
solution in $H_{0}^{1}(\widetilde{\Omega})$ to the elliptic equation

$-\text{div}(\widetilde{A}_{\epsilon}\nabla\widetilde{u}_{\epsilon})=\widetilde{f}\text{.
}$

Moreover, we have $\widetilde{u}_{\epsilon}\in
H^{2}_{loc}(\widetilde{\Omega})$.

###### Proof.

At first, one can check immediately that the restriction of
$\widetilde{u}_{\epsilon}$ to each $I\in Sub(\widetilde{\Omega})$ belongs to
$H_{0}^{1}(I)$ and hence $\widetilde{u}_{\epsilon}\in
H_{0}^{1}(\widetilde{\Omega}),$ and moreover for $1\leq j\leq N$ and $I\in
sub(\widetilde{\Omega}),$ we have :

$\text{For a.e. }x\in
I:\partial_{j}\widetilde{u}_{\epsilon}(x)=r_{j}(I)\left(\prod\limits_{i=1}^{N}r_{i}(I)\right)\partial_{j}u_{\epsilon}(y).$

Now, let $\varphi\in\mathcal{D}(\widetilde{\Omega})$ we have

$\int_{\widetilde{\Omega}}\widetilde{A}_{\epsilon}\nabla\widetilde{u}_{\epsilon}\cdot\nabla\varphi
dx=\sum_{I\in
Sub(\widetilde{\Omega})}\int_{I}\widetilde{A}_{\epsilon}\nabla\widetilde{u}_{\epsilon}\cdot\nabla\varphi
dx,$

by a change of variables we get

$\int_{\widetilde{\Omega}}\widetilde{A}_{\epsilon}\nabla\widetilde{u}_{\epsilon}\cdot\nabla\varphi
dx=\sum_{I\in Sub(\widetilde{\Omega})}\int_{\Omega}A_{\epsilon}\nabla
u_{\epsilon}\cdot\nabla\widetilde{\varphi}_{I}dx,$ (30)

where $\widetilde{\varphi}_{I}$ is defined on $\overline{\Omega}$ by

$\widetilde{\varphi}_{I}(x)=\left(\prod\limits_{i=1}^{N}r_{i}(I)\right)\varphi_{I}((s_{i}(I)+r_{i}(I)x_{i})_{1\leq
i\leq N}),$

and $\varphi_{I}$ is the restriction of $\varphi$ on $I$. Let us show that

$\sum_{I\in Sub(\widetilde{\Omega})}\widetilde{\varphi}_{I}\in
H_{0}^{1}(\Omega).$ (31)

It is clear that $\sum_{I\in
sub(\widetilde{\Omega})}\widetilde{\varphi}_{I}\in H^{1}(\Omega)\cap
C(\bar{\Omega}),$ it is enough to show that it vanishes on $\partial\Omega.$
So, let $x^{0}=(x_{i}^{0})_{1\leq i\leq N}$ be an element of $\partial\Omega,$
then there exists at least $1\leq j\leq N$ such that $x^{0}_{j}=0$ or
$x^{0}_{j}=l_{j}.$

1) If $x_{j}^{0}=0:$ For any $I\in sub(\widetilde{\Omega})_{2,j}$, we have
$y_{j}^{0}=s_{j}(I)+r_{j}(I)x_{j}^{0}=2l_{j}$, then
$y^{0}\in\partial\widetilde{\Omega}$ therefore $\widetilde{\varphi}_{I}(x)=0.$
Now, for any $I\in sub(\widetilde{\Omega})_{-1,j}$, we have
$y_{j}^{0}=s_{j}(I)+r_{j}(I)x_{j}^{0}=-x_{j}^{0}=0$, and any $I\in
sub(\widetilde{\Omega})_{1,j}$, we have :
$y_{j}^{0}=s_{j}(I)+r_{j}(I)x_{j}^{0}=x_{j}^{0}=0$, notice that there is a
bijection from $sub(\widetilde{\Omega})_{1,j}$ onto
$sub(\widetilde{\Omega})_{-1,j}$ defined by : $I\mapsto I^{\prime}$ such that
$I$ and $I^{\prime}$ have the same intervals except for the $j^{th}$ one we
have $I_{j}=(0,l_{j})$ and $I_{j}^{\prime}=(-l_{j},0).$ For such $I$ and
$I^{\prime}$ we have $r_{j}(I)=1$ and $r_{j}(I^{\prime})=-1$, then
$\widetilde{\varphi}_{I}(x^{0})+\widetilde{\varphi}_{I^{\prime}}(x^{0})=0.$
Finally, we get $\sum_{I\in
Sub(\widetilde{\Omega})}\widetilde{\varphi}_{I}(x^{0})=0.$

2) If $x_{j}^{0}=l_{j}:$ For any $I\in sub(\widetilde{\Omega})_{-1,j}$, $\
y_{j}^{0}=s_{j}(I)+r_{j}(I)x_{j}^{0}=-l_{j}$, then
$y^{0}\in\partial\widetilde{\Omega}$ therefore $\widetilde{\varphi}_{I}(x)=0$.
Now, for any $I\in sub(\widetilde{\Omega})_{1,j}$, we have
$y_{j}^{0}=s_{j}(I)+r_{j}(I)x_{j}^{0}=l_{j}$, and any $I\in
sub(\widetilde{\Omega})_{2,j}$, we have : $y_{j}^{0}=2l_{j}-l_{j}=l_{j}$,
notice that there is a bijection from $sub(\widetilde{\Omega})_{1,j}$ onto
$sub(\widetilde{\Omega})_{2,j}$ defined by : $I\longmapsto I^{\prime}$ such
that $I$ and $I^{\prime}$ have the same intervals except for the $j^{th}$ one
we have $I_{j}=(0,l_{j})$ and $I_{j}^{\prime}=(l_{j},2l_{j}).$ For such $I$
and $I^{\prime}$ we have $r_{j}(I)=1$ and $r_{j}(I^{\prime})=-1$, then
$\widetilde{\varphi}_{I}(x^{0})+\widetilde{\varphi}_{I^{\prime}}(x^{0})=0$.
Finally, we get $\sum_{I\in
Sub(\widetilde{\Omega})}\widetilde{\varphi}_{I}(x^{0})=0$.
At the end, (31) follows from the two points above.
Now, since $u_{\epsilon}$ is the solution of $(\ref{u-perturbed-weak})$ then
(30) and (31) give

$\int_{\widetilde{\Omega}}\widetilde{A}_{\epsilon}\nabla\widetilde{u}_{\epsilon}\cdot\nabla\varphi
dx=\int_{\Omega}f\sum_{I\in
Sub(\widetilde{\Omega})}\widetilde{\varphi}_{I}dx.$

By using an other variables change in the second member of the above equality
we get the first affirmation of the Lemma. Finally, as we have mentioned
above, the function $\widetilde{A_{\epsilon}}$ is Lipschitz on
$\widetilde{\Omega}$ (thanks to (10)), then the $H^{2}$ interior elliptic
regularity gives the second affirmation of the Lemma. ∎

Now, we can prove Theorem 1.6. Let $\omega\subset\subset\Omega$ an open set.
According to Corollary 2.3 of [9] we have, for any $\epsilon\in(0,1]$

$\epsilon^{2}\|D^{2}_{X_{1}}u_{\epsilon}\|_{L^{2}(\omega)^{q^{2}}}+\epsilon\|D^{2}_{X_{1}X_{2}}u_{\epsilon}\|_{L^{2}(\omega)^{q(N-q)}}+\|D^{2}_{X_{2}}u_{\epsilon}\|_{L^{2}(\omega)^{(N-q)^{2}}}\leq
C_{{}_{\lambda,f,A,\omega}}.$ (32)

Now, let us show the same estimation near the boundary of $\Omega$. Let
$\omega^{\prime}\subset\subset\widetilde{\Omega}$, then by using Corollary 2.3
of [9] ,thanks to Lemma 1.7, we get

$\epsilon^{2}\|D^{2}_{X_{1}}\widetilde{u}_{\epsilon}\|_{L^{2}(\omega^{\prime})^{q^{2}}}+\epsilon\|D^{2}_{X_{1}X_{2}}\widetilde{u}_{\epsilon}\|_{L^{2}(\omega^{\prime})^{q(N-q)}}+\|D^{2}_{X_{2}}\widetilde{u}_{\epsilon}\|_{L^{2}(\omega^{\prime})^{(N-q)^{2}}}\leq
C_{{}_{\lambda,f,A,\omega^{\prime}}}.$

Therefore,

$\epsilon^{2}\|D^{2}_{X_{1}}u_{\epsilon}\|_{L^{2}(\omega^{\prime}\cap\Omega)}+\epsilon\|D^{2}_{X_{1}X_{2}}u_{\epsilon}\|_{L^{2}(\omega^{\prime}\cap\Omega)}+\|D^{2}_{X_{2}}u_{\epsilon}\|_{L^{2}(\omega^{\prime}\cap\Omega)}\leq
C_{{}_{\lambda,f,A,\omega^{\prime}}},$ (33)

By compacity, we can cover $\bar{\Omega}$ by a finite cover of open subsets of
$\omega$-type and $\omega^{\prime}\cap\Omega$-type, then we use (32)-(33), and
the estimations of the Theorem 1.6 follows.
For the convergences of Theorem 1.6, we use the same trick, in fact when $A\in
C^{1}(\bar{\Omega})$, then $\widetilde{A}$ satisfies eq. (13) in [9].

###  1.3 $H^{2}$ Regularity of the solution of the limit problem

In this subsection, we consider a general bounded domain
$\Omega=\omega_{1}\times\omega_{2}$.

######  Theorem 1.8.

Let $\Omega=\omega_{1}\times\omega_{2}$ be an open bounded subset of
$\mathbb{R}^{N}$, where $\omega_{1}$ and $\omega_{2}$ are two bounded open
subsets of $\mathbb{R}^{q}$ and $\mathbb{R}^{N-q},$ with $N>q\geq 1.$ Let us
assume that $\omega_{2}$ is convex. Let $f$ such that $(\ref{fL2})$ is
satisfied, and such that $\nabla_{X_{1}}f\in L^{2}(\Omega)^{q},$
$D_{X_{1}}^{2}f\in L^{2}(\Omega)^{q^{2}}$. Assume that $A_{22}$ satisfies
$(\ref{hypA_221})$ and $(\ref{A22})$. Assume in addition $(\ref{A-class-C1})$
and let $u$ be the unique solution in $H_{0}^{1}(\Omega;\omega_{2})$ of
$(\ref{u-limit-weak})$, then $u\in H^{2}(\Omega)$ and

$\|u\|_{H^{2}(\Omega)}\leq
C_{{}_{\lambda,A,\Omega}}\Big{(}\|f\|_{L^{2}(\Omega)}+\|\nabla_{X_{1}}\|_{L^{2}(\Omega)^{q}}+\|D^{2}_{X_{1}}f\|_{L^{2}(\Omega)^{q^{2}}}\Big{)}.$

We will proceed in several steps to prove Theorem 1.8. In the following lemma
we prove that $D^{2}_{X_{2}}u$ is a function of $L^{2}(\Omega)^{N-q}$.

######  Lemma 1.9.

Under assumptions of Theorem 1.8 we have

$D_{X_{2}}^{2}u\in L^{2}(\Omega)^{(N-q)^{2}}\leavevmode\nobreak\ \text{ and
}\leavevmode\nobreak\ \|D_{X_{2}}^{2}u\|_{L^{2}(\Omega)^{(N-q)^{2}}}\leq
C_{{}_{\lambda,A_{22},\omega_{2}}}\|f\|_{L^{2}(\Omega)}.$

###### Proof.

We proceed in several steps.

Step 1. Let us assume that $f(x)=f_{1}(X_{1})f_{2}(X_{2})$ where $f_{1}\in
L^{2}(\omega_{1})$ and $f_{2}\in L^{2}(\omega_{2})$. Let $u_{f_{2}}$ be the
unique solution of

$\left\\{\begin{array}[]{l}\int_{\omega_{2}}A_{22}\nabla_{X_{2}}u_{f_{2}}\cdot\nabla_{X_{2}}\varphi_{2}dX_{2}=\int_{\omega_{2}}f_{2}\varphi_{2}dX_{2}\text{,
\ }\forall\varphi_{2}\in H_{0}^{1}(\omega_{2}),\\\ u_{f_{2}}\in
H_{0}^{1}(\omega_{2}).\end{array}\right.,$

According to assumption (10) it follows that $A_{22}\in
W^{1,\infty}(\bar{\omega_{2}})$, and since $\omega_{2}$ is convex we obtain by
the elliptic regularity in $\omega_{2}$ that the function $u_{f_{2}}$ belongs
to $H^{2}(\omega_{2})$. We multiply the previous identity by
$f_{1}(X_{1})\varphi_{1}(X_{1})$ where $\varphi_{1}\in H^{1}_{0}(\omega_{1})$
and we integrate over $\omega_{1}$, then we use assumption (13) to obtain

$\int_{\Omega}A_{22}\nabla_{X_{2}}(f_{1}u_{f_{2}})\cdot\nabla_{X_{2}}(\varphi_{1}\varphi_{2})dx=\int_{\Omega}f\varphi_{1}\varphi_{2}dx\text{,
\ }\forall(\varphi_{1},\varphi_{2})\in H_{0}^{1}(\omega_{1})\times
H_{0}^{1}(\omega_{2}),$

which gives by linearity

$\int_{\Omega}A_{22}\nabla_{X_{2}}(f_{1}u_{f_{2}})\cdot\nabla_{X_{2}}\varphi
dx=\int_{\Omega}f\varphi dx\text{, \ }\forall\varphi\in
H_{0}^{1}(\omega_{1})\otimes H_{0}^{1}(\omega_{2}).$

Using the fact that $H_{0}^{1}(\omega_{1})\otimes H_{0}^{1}(\omega_{2})$ is
dense in $H_{0}^{1}(\Omega;\omega_{2})$ [8], we obtain

$\left\\{\begin{array}[]{l}\int_{\Omega}A_{22}\nabla_{X_{2}}(f_{1}u_{f_{2}})\cdot\nabla_{X_{2}}\varphi
dx=\int_{\Omega}f\varphi dx\text{, \ }\forall\varphi\in
H_{0}^{1}(\Omega;\omega_{2})\\\ f_{1}u_{f_{2}}\in
H_{0}^{1}(\Omega;\omega_{2}).\end{array}\right.$

Consequently we obtain that

$u=f_{1}u_{f_{2}}\leavevmode\nobreak\ \text{a.e. in}\leavevmode\nobreak\
\Omega.$

Step 2. Let us assume that $f=\sum_{i=1}^{m}f_{1,i}f_{2,i}\in
L^{2}(\omega_{1})\otimes L^{2}(\omega_{2})$ where $(f_{1,i},f_{2,i})\in
L^{2}(\omega_{1})\times L^{2}(\omega_{2})$ for any $i\in\\{1,...,m\\}$. By
linearity we obtain that

$u=\sum_{i=1}^{m}f_{1,i}u_{f_{2,i}}\leavevmode\nobreak\ \text{a.e.
in}\leavevmode\nobreak\ \Omega.$

Using this identity and the above step we obtain that $D^{2}_{X_{2}}u\in
L^{2}(\Omega)^{(N-q)^{2}}$, in particular one has, for a.e.
$X_{1}\in\omega_{1}$

$u(X_{1},\cdot)\in H^{2}(\omega_{2})\text{,}$

Now, from (18) the elliptic regularity on $\omega_{2}$ shows that there exists
$C_{{}_{\lambda,A_{22},\omega_{2}}}>0$ independent of $X_{1}$, such that for
a.e. $X_{1}\in\omega_{1}$ :

$\|D^{2}_{X_{2}}u(X_{1},\cdot)\|_{L^{2}(\omega_{2})^{(N-q)^{2}}}\leq
C_{{}_{\lambda,A_{22},\omega_{2}}}\|f(X_{1},\cdot)\|_{L^{2}(\omega_{2})}.$

We integrate this identity over $\omega_{1}$ and we obtain

$\|D^{2}_{X_{2}}u\|_{L^{2}(\Omega)^{(N-q)^{2}}}\leq
C_{{}_{\lambda,A_{22},\omega_{2}}}\|f\|_{L^{2}(\Omega)}.$

Step 3. Let $f\in L^{2}(\Omega)$. There exists a sequence $(f_{n})_{n\geq 0}$
of functions of $L^{2}(\omega_{1})\otimes L^{2}(\omega_{2})$ converging to $f$
in $L^{2}(\Omega)$ as $n$ goes to infinity. Let $u_{n}$ be the unique solution
in $H_{0}^{1}(\Omega;\omega_{2})$ of

$\int_{\Omega}A_{22}\nabla_{x_{2}}u_{n}\cdot\nabla_{x_{2}}\varphi\mathop{}\\!\mathrm{d}x=\int_{\Omega}f_{n}\varphi\mathop{}\\!\mathrm{d}x\text{,
\ }\forall\varphi\in H_{0}^{1}(\Omega;\omega_{2}).$ (34)

Subtracting (34) from (19) and taking $\varphi=u_{n}-u$, then by using (12)
and (17) we obtain that

$u_{n}\longrightarrow u\text{ in }L^{2}(\Omega)\text{ as
}n\rightarrow\infty\text{ strongly}.$ (35)

From the previous step we obtain that $D_{X_{2}}^{2}u_{n}\in
L^{2}(\Omega)^{(N-q)^{2}}$ for any $n\geq 0$ and

$\|D^{2}_{X_{2}}u_{n}\|_{L^{2}(\Omega)^{(N-q)^{2}}}\leq
C_{\lambda,A_{22},\omega_{2}}\|f_{n}\|_{L^{2}(\Omega)},\leavevmode\nobreak\
\text{for any}\leavevmode\nobreak\ n\geq 0.$ (36)

Using the fact that the sequence $(f_{n})_{n\geq 0}$ is bounded in
$L^{2}(\Omega)$ we obtain that the sequence $(D^{2}_{X_{2}}u_{n})_{n\geq 0}$
is bounded in $L^{2}(\Omega)^{(N-q)^{2}}$. Therefore, there exists a
subsequence still labelled $(D_{X_{2}}^{2}u_{n})$ such that for any
$i,j\in\llbracket q+1,N\rrbracket$ there exists $u_{ij}^{\infty}\in
L^{2}(\Omega)$ such that

$\partial^{2}_{ij}u_{n}\rightharpoonup u_{ij}^{\infty}\text{ as
}n\rightarrow\infty\text{, weakly in }L^{2}(\Omega).$ (37)

Now, for $\phi\in\mathcal{D}(\Omega)$ and $i,j\in\llbracket q+1,N\rrbracket$
we have

$\int_{\Omega}\partial^{2}_{ij}u_{n}\phi
dx=\int_{\Omega}u_{n}\partial^{2}_{ij}\phi dx$

Passing to the limit in the above identity by using (35) and (37) to obtain

$\int_{\Omega}u_{ij}^{\infty}\phi dx=\int_{\Omega}u\partial^{2}_{ij}\phi
dx\text{, for any }i,j\in\llbracket q+1,N\rrbracket$

Therefore, we obtain that $\partial^{2}_{ij}u=u^{\infty}_{ij}\in
L^{2}(\Omega)$ for any $i,j\in\llbracket q+1,N\rrbracket$. Finally, passing to
the limit in (36) we get

$\|D^{2}_{X_{2}}u\|_{L^{2}(\Omega)^{(N-q)^{2}}}\leq\liminf\|D^{2}_{X_{2}}u_{n}\|_{L^{2}(\Omega)^{(N-q)^{2}}}\leq
C_{\lambda,A_{22},\omega_{2}}\lim\|f_{n}\|_{L^{2}(\Omega)}=C_{\lambda,A_{22},\omega_{2}}\|f\|_{L^{2}(\Omega)},$

and the Lemma 1.9 follows. ∎

At the next step we prove in the following lemma that $D^{2}_{X_{1}X_{2}}u\in
L^{2}(\Omega)^{q(N-q)}$

######  Lemma 1.10.

Let $\Omega=\omega_{1}\times\omega_{2}$ be an open bounded subset of
$\mathbb{R}^{N}$, where $\omega_{1}$ and $\omega_{2}$ are two bounded open
subsets of $\mathbb{R}^{q}$ and $\mathbb{R}^{N-q},$ with $N>q\geq 1$. Assume
that $(\ref{hypA2})$, and $(\ref{hypA_221})$ are satisfied. Suppose that $f\in
L^{2}(\Omega)$ such that $\nabla_{X_{1}}f\in L^{2}(\Omega)^{q}$. Then:

$\nabla_{X_{1}}u\in L^{2}(\Omega)^{q}\leavevmode\nobreak\
\text{and}\leavevmode\nobreak\ D^{2}_{X_{1}X_{2}}u\in L^{2}(\Omega)^{q(N-q)},$

with:

$\|\nabla_{X_{1}}u\|_{L^{2}(\Omega)^{q}}\leq
C_{{}_{\lambda,\omega_{2}}}\|\nabla_{X_{1}}f\|_{L^{2}(\Omega)^{q}}\text{ and
}\|D_{X_{1}X_{2}}^{2}u\|_{L^{2}(\Omega)^{q(N-q)}}\leq
C_{{\lambda,\omega_{2}}}\|\nabla_{X_{1}}f\|_{L^{2}(\Omega)^{q}},$

and for a.e. $X_{1}\in\omega_{1}:\nabla_{X_{1}}u(X_{1},\cdot)\in
H_{0}^{1}(\omega_{2})$.

###### Proof.

We use the difference quotient method of Nirenberg (see for instance [11]).
Let $\omega_{1}^{\prime}\subset\subset\omega_{1},$ and $0<\eta<\text{dist}($
$\omega_{1}^{\prime},\partial\omega_{1}).$ Let $i\in\llbracket 1,q\rrbracket$.
For a.e $X_{1}\in$ $\omega_{1}^{\prime},$ we obtain from (18) the following
identity

$\displaystyle\int_{\omega_{2}}A_{22}\left(\frac{\nabla_{X_{2}}u(X_{1}+\eta
e_{i},\cdot)-\nabla_{X_{2}}u(X_{1},\cdot)}{eta}\right)\cdot\nabla_{X_{2}}\varphi_{2}dX_{2}=\leavevmode\nobreak\
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\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ }$
$\displaystyle\int_{\omega_{2}}\left(\frac{f(X_{1}+\eta
e_{i},\cdot)-f(X_{1},\cdot)}{\eta}\right)\varphi_{2}dX_{2}\text{, \ }$
$\displaystyle\forall\varphi_{2}\in H_{0}^{1}(\omega_{2}),$

where $e_{i}$ is the $i$-th element of the canonical basis of
$\mathbb{R}^{N}$. Testing with $u(X_{1}+\eta e_{i},\cdot)-u(X_{1},\cdot)\in
H_{0}^{1}(\omega_{2})$ in the above equality and using (12), (17) we obtain

$\int_{\omega_{2}}\left|\frac{\nabla_{X_{2}}u(X_{1}+\eta
e_{i},\cdot)-\nabla_{X_{2}}u(X_{1},\cdot)}{\eta}\right|^{2}dX_{2}\leq\frac{C_{\omega_{2}}^{2}}{\lambda^{2}}\int_{\omega_{2}}\left|\frac{f(X_{1}+\eta
e_{i},\cdot)-f(X_{1},\cdot)}{\eta}\right|^{2}dX_{2},$

where we have used Poincaré’s inequality (17). Integrating over
$\omega_{1}^{\prime}$ yields

$\displaystyle\int_{\omega_{1}^{\prime}\times\omega_{2}}\left|\frac{\nabla_{X_{2}}u(X_{1}+\eta
e_{i},X_{2})-\nabla_{X_{2}}u(X_{1},X_{2})}{\eta}\right|^{2}dx\leq\text{\leavevmode\nobreak\
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$\displaystyle\frac{C_{\omega_{2}}^{2}}{\lambda^{2}}\int_{\omega_{1}^{\prime}\times\omega_{2}}\left|\frac{f(X_{1}+\eta
e_{i},X_{2})-f(X_{1},X_{2})}{\eta}\right|^{2}dx$
$\displaystyle\leq\frac{C_{\omega_{2}}^{2}}{\lambda^{2}}\left\|\partial_{i}f\right\|_{L^{2}(\Omega)}^{2},$
(38)

and

$\int_{\omega_{1}^{\prime}\times\omega_{2}}\left|\frac{u(X_{1}+\eta
e_{i},X_{2})-u(X_{1},X_{2})}{\eta}\right|^{2}dx\leq\frac{C_{\omega_{2}}^{4}}{\lambda^{2}}\left\|\partial_{i}f\right\|_{L^{2}(\Omega)}^{2}$
(39)

The inequalities (38) and (39) imply that $D^{2}_{X_{1}X_{2}}u\in
L^{2}(\Omega)$ and $\nabla_{X_{1}}u\in L^{2}(\Omega)^{q}$, with

$\left\|D^{2}_{X_{1}X_{2}}u\right\|_{L^{2}(\Omega)^{q(N-q)}}\leq\frac{C_{\omega_{2}}}{\lambda}\left\|\nabla_{X_{1}}f\right\|_{L^{2}(\Omega)^{q}}\text{,
and
}\|\nabla_{X_{1}}u\|_{L^{2}(\Omega)^{q}}\leq\frac{C_{\omega_{2}}^{2}}{\lambda}\left\|\nabla_{X_{1}}f\right\|_{L^{2}(\Omega)^{q}}.$

Finally, let us show that $\nabla_{X_{1}}u(X_{1},\cdot)\in
H_{0}^{1}(\omega_{2})$ for a.e. $X_{1}\in\omega_{1}.$ Let $i\in\llbracket
1,q\rrbracket$, and let $\omega_{1}^{\prime}\subset\subset\omega_{1},$ be
open, and set $\tau_{\eta}u(X_{1},X_{2})=u(X_{1}+\eta e_{i},X_{2}),$ for
$x\in\omega_{1}^{\prime}\times\omega_{2}$, and for
$0<\eta<\text{dist}(\omega_{1}^{\prime},\partial\omega_{1})$. Let
$j\in\llbracket q+1,N\rrbracket$, then one can check that :

$\frac{\partial_{j}\tau_{\eta}u-\partial_{j}u}{\eta}\rightarrow\partial_{ij}^{2}u\text{,
and }\frac{\tau_{\eta}u-u}{\eta}\rightarrow\partial_{i}u\text{ as
}\eta\rightarrow 0\text{ in
}\mathcal{D}^{\prime}(\omega_{1}^{\prime}\times\omega_{2})\text{ }.$ (40)

Let $(\eta_{n})$ be a sequence such that, for every $n\in\mathbb{N},$
$0<\eta_{n}<dist($ $\omega_{1}^{\prime},\partial\omega_{1})$ and
$\eta_{n}\rightarrow 0.$ The sequences
$\left(\frac{\tau_{\eta_{n}}u-u}{\eta_{n}}\right)$,
$\left(\frac{\partial_{j}(\tau_{\eta_{n}}u-u)}{\eta_{n}}\right)$ are bounded
in $L^{2}(\omega_{1}^{\prime}\times\omega_{2})$ thanks to (39) and
$(\ref{est:gradX1X2u})$, therefore it follows from (40) that

$\frac{\partial_{j}(\tau_{\eta_{n}}u-u)}{\eta_{n}}\rightharpoonup\partial_{ij}^{2}u\text{
and }\frac{\tau_{\eta_{n}}u-u}{\eta_{n}}\rightharpoonup\partial_{i}u\text{ in
}L^{2}(\omega_{1}^{\prime}\times\omega_{2})\text{ weakly.}$

Finally, Mazur’s Lemma shows that there exists a sequence $(U_{n})$ of convex
combination of $\\{\frac{\tau_{\eta_{n}}u-u}{\eta_{n}}\\}_{n\in\mathbb{N}}$
such that
$\partial_{j}U_{n}\longrightarrow\partial_{j}(\partial_{i}u)=\partial_{ij}^{2}u$
and $U_{n}\longrightarrow\partial_{i}u$ as $n\rightarrow\infty$ strongly in
$L^{2}(\omega_{1}^{\prime}\times\omega_{2})$. Now, since $(U_{n})\in
H_{0}^{1}(\omega_{1}^{\prime}\times\omega_{2};\omega_{2})^{\mathbb{N}}$ and
the space $H_{0}^{1}(\omega_{1}^{\prime}\times\omega_{2};\omega_{2})$ is
complete with the norm
$\left\|\nabla_{X_{2}}(\cdot)\right\|_{L^{2}(\omega_{1}^{\prime}\times\omega_{2})}$
then we deduce that $\partial_{i}u\in
H_{0}^{1}(\omega_{1}^{\prime}\times\omega_{2};\omega_{2})$ i.e. for a.e.
$X_{1}\in\omega_{1}^{\prime}$, $\partial_{i}u(X_{1},\cdot)\in
H_{0}^{1}(\omega_{2})$. Notice that $\omega_{1}$ could be covered by a
countable family of $\omega_{1}^{\prime}\subset\subset\omega_{1}$, whence for
a.e. $X_{1}\in\omega_{1}$, $\partial_{i}u(X_{1},\cdot)\in
H_{0}^{1}(\omega_{2}).$ and finally we obtain that
$\nabla_{X_{1}}u(X_{1},\cdot)\in H_{0}^{1}(\omega_{2})$ for a.e.
$X_{1}\in\omega_{1}.$ ∎

We finish by the following lemma

######  Lemma 1.11.

Let $\Omega=\omega_{1}\times\omega_{2}$ be an open bounded subset of
$\mathbb{R}^{N}$, where $\omega_{1}$ and $\omega_{2}$ are two bounded open
subsets of $\mathbb{R}^{q}$ and $\mathbb{R}^{N-q},$ with $N>q\geq 1$. Assume
that $(\ref{hypA2})$, and $(\ref{hypA_221})$ are satisfied. Let $f\in
L^{2}(\Omega)$ such that $\nabla_{X_{1}}f\in L^{2}(\Omega)^{q},$ and
$D_{X_{1}}^{2}f\in L^{2}(\Omega)^{q^{2}}$, then:

$D_{X_{1}}^{2}u\in L^{2}(\Omega)^{q^{2}}\text{ and
}\left\|D_{X_{1}}^{2}u\right\|_{L^{2}(\Omega)^{q^{2}}}\leq\frac{C_{\omega_{2}}^{2}}{\lambda}\left\|D_{X_{1}}^{2}f\right\|_{L^{2}(\Omega){q^{2}}}.$

###### Proof.

Let $\varphi_{1}\otimes\varphi_{2}\in H_{0}^{1}(\omega_{1})\otimes
H_{0}^{1}(\omega_{2})$. Let $i\in\\{1,...,q\\}$, testing with
$\partial_{x_{i}}\varphi_{1}\varphi_{2}$ in (19) we obtain

$\int_{\Omega}A_{22}\nabla_{X_{2}}u\cdot\nabla_{X_{2}}\varphi_{2}\partial_{x_{i}}\varphi_{1}dx=\int_{\Omega}f\partial_{x_{i}}\varphi_{1}\varphi_{2}dx.$

According to Lemma 1.10 we have $\partial_{ij}^{2}u\in L^{2}(\Omega)$ for any
$j\in\llbracket 1+q,N\rrbracket$ then, by integration by part we get

$\int_{\Omega}A_{22}(X_{2})\nabla_{X2}\partial_{i}u(X_{1},X_{2})\cdot\varphi_{1}(X_{1})\nabla_{X_{2}}\varphi(X_{2})dx=\int_{\Omega}\partial_{i}f(X_{1},X_{2})\varphi_{1}(X_{1})\varphi(X_{2})dx.$

and hence, for a.e. $X_{1}\in\omega_{1}$ we obtain

$\int_{\omega_{2}}A_{22}(x_{2})\nabla_{X_{2}}\partial_{i}u(x_{1},x_{2})\cdot\nabla_{X_{2}}\varphi_{2}dX_{2}=\int_{\omega_{2}}\partial_{i}f(X_{1},X_{2})\varphi_{2}dX_{2}\text{,
\ }\forall\varphi_{2}\in H_{0}^{1}(\omega_{2}).$

Repeating the same method as in proof of Lemma 1.10. Then, for
$i^{\prime}\in\\{1,...,q\\}$ and
$\omega_{1}^{\prime}\subset\subset\omega_{1},$ and for
a.e.$X_{1}\in\omega_{1}^{\prime}$ and for $0<\eta<\text{dist}($
$\omega_{1}^{\prime},\partial\omega_{1})$ we obtain that

$\displaystyle\int_{\omega_{2}}\left|\nabla_{X_{2}}\left(\frac{\partial_{i}u(X_{1}+\eta
e_{i^{\prime}},X_{2})-\partial_{i}u(X_{1},X_{2})}{\eta}\right)\right|^{2}dX_{2}\leq\text{\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
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\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ }$
$\displaystyle\frac{C_{\omega_{2}}^{2}}{\lambda^{2}}\int_{\omega_{2}}\left|\frac{\partial_{i}f(X_{1}+\eta
e_{i^{\prime}},X_{2})-\partial_{i}f(X_{1},X_{2})}{\eta}\right|^{2}dX_{2}.$

By the above lemma we have for
a.e.$X_{1}\in\omega_{1}^{\prime}:\partial_{i}u(X_{1}+\eta
e_{i^{\prime}},\cdot)-\partial_{i}u(X_{1},\cdot)\in H_{0}^{1}(\omega_{2}).$ We
integrate over $\omega_{1}^{\prime}$ and we apply (17) to obtain

$\displaystyle\int_{\omega_{1}^{\prime}\times\omega_{2}}\left|\frac{\partial_{i}u(X_{1}+\eta
e_{i^{\prime}},X_{2})-\partial_{i}u(X_{1},X_{2})}{\eta}\right|^{2}dx\leq\text{\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
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\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ }$
$\displaystyle\frac{C_{\omega_{2}}^{4}}{\lambda^{2}}\int_{\omega_{1}^{\prime}\times\omega_{2}}\left|\frac{\partial_{i}f(X_{1}+\eta
e_{i^{\prime}},X_{2})-\partial_{i}f(x_{1},x_{2})}{\eta}\right|^{2}dx.$

Whence, $\partial_{ii^{\prime}}^{2}u\in L^{2}(\Omega)$ and
$\left\|\partial_{ii^{\prime}}^{2}u\right\|_{L^{2}(\Omega)}\leq\frac{C_{\omega_{2}}^{2}}{\lambda}\left\|\partial_{ii^{\prime}}^{2}f\right\|_{L^{2}(\Omega)}.$
∎

In conclusion, Theorem 1.8 follows from Lemmas 1.9, 1.10, and 1.11.

##  2 The Analysis of the numerical scheme

###  2.1 Numerical scheme and the main result

In this section, we assume that $N\in\\{2,3\\}$ and that the computational
domain is $\Omega=(0,1)^{N}$.
Let $M_{i}\in\mathbb{N}$, $M_{i}\geq 2$, $i\in\llbracket 1,N\rrbracket$ and
let $\\{h_{k_{i}}^{x_{i}}>0,k_{i}=1,...,M_{i}\\}$, for $i\in\llbracket
1,N\rrbracket$. such that

$\sum_{k_{i}=1}^{M_{i}}h_{k_{i}}^{x_{i}}=1,\text{ for }i\in\llbracket
1,N\rrbracket,$

and let us define the step size $h$ of the discretization by:

$h=\max_{i\in\llbracket 1,N\rrbracket}(h_{1}^{x_{i}},...,h_{M_{i}}^{x_{i}}).$

Let $(x^{i}_{k_{i}})_{0\leq k_{i}\leq M_{i}}$, $i\in\llbracket 1,N\rrbracket$
be the families of real numbers defined by

$x^{i}_{k_{i}}-x^{i}_{k_{i}-1}=h_{k_{i}}^{x_{i}},\text{ for }i\in\llbracket
1,N\rrbracket,\text{ and }k_{i}=1,..,M_{i},$

with $x_{0}^{i}=0$ for $i\in\llbracket 1,N\rrbracket$. We define a rectangular
mesh $\mathcal{R}_{h}=(R_{k_{1},...,k_{N}})_{\begin{subarray}{c}1\leq
k_{i}\leq M_{i}\\\ i\in\llbracket 1,N\rrbracket\end{subarray}}$ on $\Omega$ by
letting

$R_{k_{1},...,k_{N}}=\prod_{{i}=1}^{N}(x^{i}_{k_{i}-1},x^{i}_{k_{i}})$

We denote by $\mathcal{S}$ the set of the nodes of the mesh that is

$\mathcal{S}=\\{(x^{i}_{k_{i}})_{1\leq i\leq N},k_{i}\in\\{0,...,M_{i}\\}\\}.$

We denote $\mathbb{Q}_{1}(K)$ the space of real polynomials in two variables
of partial degree less or equal to $1$ over $K\subset\mathbb{R}^{N}$. We
define the finite dimensional spaces $W_{h}\subset H^{1}(\Omega)$ by

$W_{h}=\left\\{v\in C(\bar{\Omega})\text{,
}v_{|R}\in\mathbb{Q}_{1}(R)\leavevmode\nobreak\ \text{for
any}\leavevmode\nobreak\ R\in\mathcal{R}_{h}\right\\}.$

As usual at the continuous level or for variational discrete formulation (as
in the finite element context), the Dirichlet boundary conditions are
incorporated in the definition of the discrete space $V_{h}\subset
H_{0}^{1}(\Omega)$ defined by

$V_{h}=\left\\{v\in W_{h},\text{ and }v=0\text{ on }\partial\Omega\right\\}.$

Mention that the discrete space $V_{h}$ can be written as a tensor product
that is

$V_{h}=\otimes_{i=1}^{N}V_{h}^{i},$ (41)

where, for $i\in\llbracket 1,N\rrbracket$:

$V_{h}^{i}=\left\\{v\in C([0,1]),\leavevmode\nobreak\
v_{|(x_{k_{i}-1},x_{k_{i}})}\in\mathbb{P}_{1}((x_{k_{i}-1},x_{k_{i}})),\leavevmode\nobreak\
k_{i}\in\\{1,...,M_{i}\\},\leavevmode\nobreak\ v(0)=v(1)=0\right\\},$

where $\mathbb{P}_{1}(I)$ is the space of real polynomials in one variable of
degree less or equal to $1$ over $I\subset\mathbb{R}$. Recall the Sobolev
embedding

$H^{2}(\Omega)\hookrightarrow C(\bar{\Omega})$

which holds in dimension 2 and 3 for Lipschitz domains. We define the
classical interpolation operator

$I_{h}:H^{2}(\Omega)\longrightarrow W_{h}$

by

$I_{h}(v)(x)=\sum\limits_{s\in\mathcal{S}}v(x)N_{s}(x)$

where $(N_{s})_{s\in\mathcal{S}}$ is the nodal basis. There exists
$C_{{}_{N}}>0$ such that for any $v\in H^{2}(\Omega)$ [12]:

$\|\nabla(v-I_{h}(v))\|_{L^{2}(\Omega)^{N}}\leq
C_{{}_{N}}h\Big{(}\sum_{i=1}^{N}\|\partial_{i}^{2}v\|_{L^{2}(\Omega)}^{2}\Big{)}^{1/2},$
(42)

and

$\|v-I_{h}(v)\|_{L^{2}(\Omega)^{N}}\leq
C_{{}_{N}}h\Big{(}\sum_{i=1}^{N}\|\partial_{i}v\|_{L^{2}(\Omega)}^{2}\Big{)}^{1/2},$
(43)

The numerical scheme to approximate problem (14) is

$\left\\{\begin{array}[]{l}\int_{\Omega}A_{\epsilon}\nabla
u_{\epsilon,h}\cdot\nabla
v\mathop{}\\!\mathrm{d}x=\int_{\Omega}I_{h}(f)v\mathop{}\\!\mathrm{d}x,\quad\quad\forall
v\in V_{h},\text{\ \ \ \ \ \ }\\\ u_{\epsilon,h}\in V_{h}.\text{\ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
}\end{array}\right.,$ (44)

Now, we are ready to give the main theorem of this section

######  Theorem 2.1.

Let $\Omega=(0,1)^{N}$, with $N\in\\{2,3\\}$. Assume that $A$ satisfies (8),
(10), (11), and (13). Let $f$ such that (5) is satisfied, then there exists a
positive constant $C_{{}_{\lambda,f,\Omega,A}}$ independent of $h$ and
$\epsilon$ such that

$\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,\Omega,A}}h^{\frac{1}{5}},$ (45)

where $u_{\epsilon,h}$ and $u_{\epsilon})$ are the solutions of (44) and (14)
respectively. Moreover, if we assume, in addition, that $f\in
H^{1}_{0}(\Omega)$ then we have

$\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,\Omega,A}}h^{\frac{1}{3}}.$ (46)

This theorem follows immediately from estimations of type (1) and (2). Each
one of these two estimations will be the subject of the next subsections.

###  2.2 The first estimation of type (1)

In this subsection we prove the following

######  Proposition 2.2.

Suppose that $N\in\\{2,3\\}$. Assume that $A$ satisfies assumptions (8), (10),
and (11). Assume that $f$ satisfies assumption (5), then there exists
$C_{{}_{\lambda,f,A,\Omega}}>0$ independent of $\epsilon$ and $h$ such that

$\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\frac{h}{\epsilon^{2}}.$

###### Proof.

Let $w_{\epsilon,h}$ be the solution of the following

$\left\\{\begin{array}[]{l}\int_{\Omega}A_{\epsilon}\nabla
w_{\epsilon,h}\cdot\nabla vdx=\int_{\Omega}fvdx,\quad\quad\forall v\in
V_{h},\text{\ \ \ \ \ \ }\\\ w_{\epsilon,h}\in V_{h}.\text{\ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\end{array}\right.,$
(47)

By subtracting (47) from (44), and by testing by
$(u_{\epsilon,h}-w_{\epsilon,h})$ to get

$\|\nabla_{X_{2}}(u_{\epsilon,h}-w_{\epsilon,h})\|_{L^{2}(\Omega)^{N-q}}\leq\frac{C_{\omega_{2}}}{\lambda}\|I_{h}(f)-f\|_{L^{2}(\Omega)}.$

Since $f\in H^{2}(\Omega)$ (thanks to assumption (5)), then by applying (43)
to the second member of the above inequality, we obtain

$\|\nabla_{X_{2}}(u_{\epsilon,h}-w_{\epsilon,h})\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,\omega_{2},N}}h,$ (48)

notice that $h^{2}\leq h$ since $h\in(0,1]$.
Now, subtracting (14) from (47) and using the Galerkin orthogonality and (8),
we obtain for any $v\in V_{h}$,

$\displaystyle\lambda\epsilon^{2}\int_{\Omega}|\nabla_{X_{1}}(w_{\epsilon,h}-u_{\epsilon})|^{2}dx+\lambda\int_{\Omega}|\nabla_{X_{2}}(w_{\epsilon,h}-u_{\epsilon})|^{2}dx\leq\int_{\Omega}A_{\epsilon}\nabla(w_{\epsilon,h}-u_{\epsilon})\cdot\nabla(u_{\epsilon}-v)dx.$

Remark that by a direct application of classical Céa’s lemma we obtain an
estimation of order $O(\frac{h}{\epsilon^{4}})$. We will improve that by using
the anisotropic nature of the perturbation, so let us develop the right hand
side of the above inequality to obtain for any $v\in V_{h}$ the following

$\displaystyle\lambda\epsilon^{2}\int_{\Omega}|\nabla_{X_{1}}(w_{\epsilon,h}-u_{\epsilon})|^{2}dx+\lambda\int_{\Omega}|\nabla_{X_{2}}(w_{\epsilon,h}-u_{\epsilon})|^{2}dx\leq\text{
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\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
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\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ }$
$\displaystyle\epsilon^{2}\int_{\Omega}A_{11}\nabla_{X_{1}}(w_{\epsilon,h}-u_{\epsilon})\cdot\nabla_{X_{1}}(u_{\epsilon}-v)dx+\epsilon\int_{\Omega}A_{12}\nabla_{X_{2}}(w_{\epsilon,h}-u_{\epsilon})\cdot\nabla_{X_{1}}(u_{\epsilon}-v)dx+$
$\displaystyle\epsilon\int_{\Omega}A_{21}\nabla_{X_{1}}(w_{\epsilon,h}-u_{\epsilon})\cdot\nabla_{X_{2}}(u_{\epsilon}-v)dx+\int_{\Omega}A_{22}\nabla_{X_{2}}(w_{\epsilon,h}-u_{\epsilon})\cdot\nabla_{X_{2}}(u_{\epsilon}-v)dx.$

By using boundedness of $A$ (thanks to (9) or (10)), and Young’s inequality to
each term in the right hand side of the previous inequality, we obtain for any
$v\in V_{h}$

$\displaystyle\frac{\lambda\epsilon^{2}}{2}\int_{\Omega}|\nabla_{X_{1}}(w_{\epsilon,h}-u_{\epsilon})|^{2}dx+\frac{\lambda}{2}\int_{\Omega}|\nabla_{X_{2}}(w_{\epsilon,h}-u_{\epsilon})|^{2}dx\leq
C_{{}_{A,\lambda}}\int_{\Omega}|\nabla(u_{\epsilon}-v)|^{2}dx.$

Now, we take $v=I_{h}(u_{\epsilon})$ (which belongs to $H^{1}_{0}\cap
H^{2}(\Omega)$) in the previous inequality, then we obtain

$\|\nabla_{X_{2}}(w_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{A,\lambda}}\|\nabla(I_{h}(u_{\epsilon})-u_{\epsilon})\|_{L^{2}(\Omega)^{N}}.$

Applying (42) to right hand side of the above inequality to obtain

$\|\nabla_{X_{2}}(w_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{A,\lambda}}h\Big{(}\sum_{i=1}^{N}\|\partial_{i}^{2}u_{\epsilon}\|_{L^{2}(\Omega)}^{2}\Big{)}^{1/2}.$

Therefore, by applying Theorem 1.6 to the right hand side of the previous
inequality we obtain

$\|\nabla_{X_{2}}(w_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\frac{h}{\epsilon^{2}}.$

Finally, we combine the above inequality with (48) and we use the triangle
equality to obtain the expected result. ∎

###  2.3 The second estimation of type (2)

The proof of the estimation is based on the following theoretical result
proved in our previous work (see proof of Lemma 3.9 in [8]), that is the
analogous discrete version of the continuous version given in Theorem 1.1.

######  Lemma 2.3.

[8] Suppose that assumptions of Theorem 1.1 hold. Let $V=V_{1}\otimes V_{2}$
where $V_{1}\subset H^{1}_{0}(\omega_{1})$ and $V_{2}\subset
H^{1}_{0}(\omega_{2})$ are finite dimension spaces. Let $g\in V$. Let
$u_{\epsilon,V,g}$ and $u_{V,g}$ be the solutions of

$\left\\{\begin{array}[]{l}\int_{\Omega}A_{\epsilon}\nabla
u_{\epsilon,V,g}\cdot\nabla\varphi dx=\int_{\Omega}g\text{ }\varphi dx\text{,
}\forall\varphi\in V\text{\ \ \ \ \ }\\\ u_{\epsilon,V,g}\in
V\text{.}\end{array}\right.$

and

$u_{V,g}\in V\text{,
}\int_{\Omega}A_{22}\nabla_{X_{2}}u_{V,g}\cdot\nabla_{X_{2}}\varphi
dx=\int_{\Omega}g\text{ }\varphi dx,\text{\ }\forall\varphi\in V.$

Then,

$\displaystyle\|\nabla_{X_{2}}u_{\epsilon,V,g}-\nabla_{X_{2}}u_{V,g}\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,A,\Omega}}\Big{(}\|\nabla_{X_{1}}g\|_{L^{2}(\Omega)^{q}}+\|g\|_{L^{2}(\Omega}\Big{)}\times\epsilon.$

######  Remark 2.4.

The space $V_{h}$ has the same tensor structure of the space $V$ of Lemma 2.3.
In fact:
\- When $N=2$, identity (41) is $V_{h}=V_{h}^{1}\otimes V_{h}^{2}$.
\- When $N=3$, identity (41) is $V_{h}=V_{h}^{1}\otimes(V_{h}^{2}\otimes
V_{h}^{3})=(V_{h}^{1}\otimes V_{h}^{2})\otimes V_{h}^{3}$, the first equality
corresponds to the case $q=1$, and the second equality corresponds to the case
$q=2$.

Now, we are ready to prove the following:

######  Proposition 2.5.

Let $\Omega=(0,1)^{N}$, with $N\in\\{2,3\\}$. Assume that $A$ satisfies (8),
(10), and (13). Let $f$ such that (5) is satisfied, then there exist a
positive constant $C_{{}_{\lambda,f,A,\Omega}}>0$ such that

$\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}(h^{1/4}+\epsilon^{1/2}).$

In particular, if we assume in addition that $f\in H^{1}_{0}(\Omega)$ then we
have

$\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}(h+\epsilon).$

We process by several steps.
Step 1. Let $u_{h}$ be the solution of the following problem

$\left\\{\begin{array}[]{l}\int_{\Omega}A_{22}\nabla_{X_{2}}u_{h}\cdot\nabla_{X_{2}}v\mathop{}\\!\mathrm{d}x=\int_{\Omega}I_{h}(f)\text{
}vdx,\quad\forall v\in V_{h}\text{\ \ \ \ \ \ }\\\ u_{h}\in V_{h}.\text{\ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
}\end{array}\right.$ (49)

We have the following

######  Lemma 2.6.

Suppose that assumptions of Proposition 2.5 hold then:
\- For $f\in H^{1}_{0}\cap H^{2}(\Omega)$ we have

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\times\epsilon.$ (50)

\- For $f\in H^{2}(\Omega)$ we have

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\,f,A,\Omega}}(\epsilon+h^{1/3}).$ (51)

###### Proof.

1) Suppose that $f\in H_{0}^{1}(\Omega)\cap H^{2}(\Omega)$, then $I_{h}(f)\in
V_{h}$. Therefore, according to Lemma 2.3 and Remark 2.4 with $g$ and $V$
replaced by $I_{h}(f)$ and $V_{h}$ respectively, we have

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,A,\Omega}}\left(\left\|\nabla_{X_{1}}I_{h}(f)\right\|_{L^{2}(\Omega)^{q}}+\left\|I_{h}(f)\right\|_{L^{2}(\Omega)}\right)\times\epsilon.$

Since $I_{h}(f)\in H^{1}_{0}(\Omega)$, then by using Poincaré’s inequality
(22) we obtain

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,A,\Omega}}\times\left\|\nabla_{X_{1}}I_{h}(f)\right\|_{L^{2}(\Omega)^{q}}\times\epsilon.$
(52)

In the other hand (42) gives

$\displaystyle\left\|\nabla_{X_{1}}I_{h}(f)\right\|_{L^{2}(\Omega)^{q}}\leq\left\|f\right\|_{L^{2}(\Omega)}+C_{N}h\left\|f\right\|_{H^{2}(\Omega)}\leq
C_{N}\|f\|_{H^{2}(\Omega)},$ (53)

Therefore, from (52) and (53) yields

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,\Omega,A}}\epsilon.$ (54)

2) Now, we suppose that $f\in H^{2}(\Omega)$, from Lemma 1.5 we use the
decomposition $f=f_{\delta}^{1}+f_{\delta}^{2}$, and let $u_{\epsilon,h}^{i}$,
$u_{h}^{i}\text{, }i=1,2$ be the solutions of (44) and (49) respectively with
$f$ replaced by $f_{\delta}^{i}$. The linearity of the equations and the
operator $I_{h}$ gives

$u_{\epsilon,h}=u_{\epsilon,h}^{1}+u_{\epsilon,h}^{2}\text{ and
}u_{h}=u_{h}^{1}+u_{h}^{2}$ (55)

As in (52) we have

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}^{1}-u_{h}^{1})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,\Omega,A}}\left\|\nabla_{X_{1}}I_{h}(f_{\delta}^{1})\right\|_{L^{2}(\Omega)^{q}}\times\epsilon.$
(56)

According to (42) we obtain

$\displaystyle\left\|\nabla_{X_{1}}I_{h}(f_{\delta}^{1})\right\|_{L^{2}(\Omega)^{q}}\leq\left\|f_{\delta}^{1}\right\|_{L^{2}(\Omega)}+C_{N}h\left\|f_{\delta}^{1}\right\|_{H^{2}(\Omega)},$

and therefore, by applying Lemma 1.5 we get

$\displaystyle\left\|\nabla_{X_{1}}I_{h}(f_{\delta}^{1})\right\|_{L^{2}(\Omega)^{q}}\leq
C_{{}_{f,\Omega}}\Big{(}1+h\delta^{-3/2}\Big{)}.$

Combining this with (56) to obtain

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}^{1}-u_{h}^{1})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\Big{(}1+h\delta^{-3/2}\Big{)}\times\epsilon.$ (57)

In the other hand, testing by $u_{\epsilon,h}^{2}$ and $u_{h}^{2}$ is the
corresponding formulations (44) and (49), with $f$ replaced by
$f_{\delta}^{2}$, we obtain this basic estimation

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}^{2}-u_{h}^{2})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,\Omega}}\|f_{\delta}^{2}\|_{L^{2}(\Omega)}.$ (58)

Now, applying Lemma 1.5 to the right hand side of (58) we obtain

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}^{2}-u_{h}^{2})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,\Omega}}\times\delta^{1/2}$ (59)

The combination of (55), (57) and (59) gives, by the triangle inequality, the
following

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\Big{(}\epsilon+h\delta^{-3/2}\epsilon+\delta^{1/2}\Big{)}.$
(60)

Finally, since $\delta$ is arbitrary in $(0,1)$, then by setting
$\delta=h^{2/3}$ in the previous inequality we get

$\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\Big{(}\epsilon+h^{1/3}\Big{)}.$ (61)

∎

Step 2. We denote $w_{h}$ the solution to following the problem

$\left\\{\begin{array}[]{l}\int_{\Omega}A_{22}\nabla_{X_{2}}w_{h}\cdot\nabla_{X_{2}}vdx=\int_{\Omega}f\text{
}vdx,\quad\forall v\in V_{h}\text{\ \ \ \ \ \ }\\\ w_{h}\in V_{h}.\text{\ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
}\end{array}\right.$ (62)

We have the following

######  Lemma 2.7.

Assume that assumptions of Proposition 2.5 hold, then:
\- If $f\in H^{1}_{0}\cap H^{2}(\Omega)$, we have :

$\|\nabla_{X_{2}}(w_{h}-u)\|_{L^{2}(\Omega)}\leq C_{{}_{\lambda,f,A,\Omega}}h$

\- If $f\in H^{2}(\Omega)$, we have:

$\|\nabla_{X_{2}}(w_{h}-u)\|_{L^{2}(\Omega)}\leq
C_{{}_{\lambda,f,A,\Omega}}h^{1/4}$

###### Proof.

1) Suppose that $f\in H^{1}_{0}\cap H^{2}(\Omega)$:
By using the classical Céa’s Lemma we obtain from (19) and (62) the following

$\left\|\nabla_{X_{2}}(w_{h}-u)\right\|_{L^{2}(\Omega)^{N-q}}\leq\frac{\left\|A_{22}\right\|_{L^{\infty}}}{\lambda}\inf_{v\in
V_{h}}\left\|\nabla_{X_{2}}(v-u)\right\|_{L^{2}(\Omega)}.$

Now, according to Theorem 1.8 and Theorem 1.1 we have $u\in
H_{0}^{1}(\Omega)\cap H^{2}(\Omega)$, then $I_{h}(u)\in V_{h}$. Therefore,
from the above inequality we obtain

$\left\|\nabla_{X_{2}}(w_{h}-u)\right\|_{L^{2}(\Omega)^{N-q}}\leq\frac{\left\|A_{22}\right\|_{L^{\infty}}}{\lambda}\left\|\nabla_{X_{2}}(I_{h}(u)-u)\right\|_{L^{2}(\Omega)}$

Finally, by applying (42) and Theorem 1.8 to the right hand side of the above
inequality, we get

$\left\|\nabla_{X_{2}}(w_{h}-u)\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}h.$

2)Suppose that $f\in H^{2}(\Omega)$:
We use the decomposition trick of Lemma 1.5, and let $w_{h}^{i}$, $u^{i}$,
$i=1,2$ be the solutions of (62) and (19) with $f$ replaced by
$f_{\delta}^{i}$ respectively. As in the previous case we have

$\left\|\nabla_{X_{2}}(w_{h}^{1}-u^{1})\right\|_{L^{2}(\Omega)^{N-q}}\leq\frac{\left\|A_{22}\right\|_{L^{\infty}}}{\lambda}\left\|\nabla_{X_{2}}(I_{h}(u^{1})-u^{1})\right\|_{L^{2}(\Omega)},$

and the application of (42) to the right hid side of the previous inequality
gives

$\left\|\nabla_{X_{2}}(w_{h}^{1}-u^{1})\right\|_{L^{2}(\Omega)}\leq\frac{\left\|A_{22}\right\|_{L^{\infty}}}{\lambda}\left\|u^{1}\right\|_{H^{2}(\Omega)}h,$

and by Theorem 1.8 we get

$\left\|\nabla_{X_{2}}(w_{h}^{1}-u^{1})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,A,\Omega}}\left\|f_{\delta}^{1}\right\|_{H^{2}(\Omega)}h,$

whence by applying Lemma 1.5 to the right hand side of the previous inequality
we obtain

$\left\|\nabla_{X_{2}}(w_{h}^{1}-u^{1})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\delta^{-3/2}h,$

Testing with $w_{h}^{2}$, $u^{2}$ in (62) and (19) (with $f$ replaced by by
$f_{\delta}^{2}$) we get

$\left\|\nabla_{X_{2}}(w_{h}^{2}-u^{2})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{f,\Omega}}\delta^{1/2}.$

Combining these two last inequalities and using the triangle inequality to
obtain

$\left\|\nabla_{X_{2}}(w_{h}-u)\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}\Big{(}\delta^{-3/2}h+\delta^{1/2}\Big{)}.$

Finally, we choose $\delta=h^{1/2}$ we obtain

$\left\|\nabla_{X_{2}}(w_{h}-u)\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}h^{1/4}.$

∎

Step 3. We have the following

######  Lemma 2.8.

Assume that assumptions of Proposition 2.5 hold, then:

$\left\|\nabla_{X_{2}}(w_{h}-u_{h})\right\|_{L^{2}(\Omega)}\leq
C_{{}_{f,\Omega}}h.$

###### Proof.

Subtracting (62) from (49) and testing with $w_{h}-u_{h}$, we get

$\left\|\nabla_{X_{2}}(w_{h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\Omega}}\|f-I_{h}(f)\|_{L^{2}(\Omega)}.$

By using (43) in the right hand side of the previous inequality we get

$\left\|\nabla_{X_{2}}(w_{h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\Omega}}\|f\|_{H^{1}(\Omega)}h.$

∎

Step 4.  Now, we are ready to conclude. By using the triangle inequality we
get

$\displaystyle\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\right\|_{L^{2}(\Omega)^{N-q}}\leq\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{h})\right\|_{L^{2}(\Omega)^{N-q}}+\left\|\nabla_{X_{2}}(u_{h}-w_{h})\right\|_{L^{2}(\Omega)^{N-q}}$
$\displaystyle+\left\|\nabla_{X_{2}}(w_{h}-u)\right\|_{L^{2}(\Omega)^{N-q}}+\left\|\nabla_{X_{2}}(u-u_{\epsilon})\right\|_{L^{2}(\Omega)^{N-q}}$

Finally, from Lemmas 2.6, 2.7, 2.8 and Theorem 1.4 we get

$\displaystyle\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}(\epsilon^{1/2}+h^{1/4})\text{, when }f\in
H^{2}(\Omega),$

and

$\displaystyle\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\right\|_{L^{2}(\Omega)^{N-q}}\leq
C_{{}_{\lambda,f,A,\Omega}}(\epsilon+h)\text{, when }f\in H^{1}_{0}\cap
H^{2}(\Omega),$

and the proof of Proposition 2.5 is achieved.

In conclusion, we combine the estimations given in Proposition 2.2 and
Proposition 2.5 to obtain

$\displaystyle\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\right\|_{L^{2}(\Omega)}\leq
C_{{}_{\lambda,f,A,\Omega}}h^{1/5},\text{ when }f\in H^{2}(\Omega),$

and

$\displaystyle\left\|\nabla_{X_{2}}(u_{\epsilon,h}-u_{\epsilon})\right\|_{L^{2}(\Omega)}\leq
C_{{}_{\lambda,f,A,\Omega}}h^{1/3},\text{ when }f\in H^{1}_{0}\cap
H^{2}(\Omega),$

and the proof of Theorem 2.1 is finished.

## References

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Explicit Expressions for Moments of the Duration of a 3-Player Gambler’s Ruin

Shalosh B. EKHAD and Doron ZEILBERGER

Dedicated to Jean-Paul Delahaye, the French analog of Martin Gardener

Abstract: Using experimental mathematics and symbolic computation, we derive
many moments for the duration of a $3$-player (fair) gambler’s ruin.

How It All Started: Jean-Paul Delahaye’s Wonderful Monthly Column

We are fortunate to be on the emailing list of Jean-Paul Delahaye, who sends
us each month his fascinating Logique et Calcul column for the French analog
of Scientific American, called Pour la Science. The August 2023 piece [De] was
dedicated to the Gambler’s Ruin problem, that is one of our favorite topics.
The classical two-player case goes back to 1657, when Christiaan Huygens
solved it. It is covered extensively in Feller’s probability bible [F].

Thanks to Delahaye’s column, we learned about exciting new research on the
three-player case, by Persi Diaconis and Stewart Ethier [DiE], and, via the
latter paper, about earlier work [E] [BLT] [S] [R].

Warm-Up: On the Duration of The 2-player Gambler’s Ruin

Alice and Bob start out with $A$ dollars and $B$ dollars respectively. A fair
coin is tossed, and if it lands on Heads, Alice gives Bob a dollar, so now
their capitals are $A-1$ and $B+1$ respectively. On the other hand, if it
lands on Tails, Bob gives Alice a dollar, and now their respective capitals
are $A+1$ and $B-1$. Sooner or later, with probability one, one of them goes
broke, and the other one goes home with $A+B$ dollars. The duration of the
game is a certain random variable, let’s call is $D$. Huygnes proved (see [F])
the famous result

$E[D]\,=\,AB\quad.$

Less well known is an explicit expression for the variance (see [F])

$Var(D)={AB\left(A^{2}+B^{2}-2\right)\over 3}\quad.$

In [Z] we computed the first ten moments (about the mean), but now we have
forty(!) of them, and can easily compute more. Let’s state a few of them here.
Let $m_{r}$ be the $r$-th moment about the mean. We have

$m_{3}(D)={AB\left(3A^{4}+10A^{2}B^{2}+3B^{4}-10A^{2}-10B^{2}+4\right)\over
15}\quad,$

$m_{4}(D)={AB\over 105}\cdot$
$(17A^{6}+35A^{5}B+119A^{4}B^{2}+70A^{3}B^{3}+119A^{2}B^{4}+35A\,B^{5}+17B^{6}-84A^{4}-140A^{3}B-280A^{2}B^{2}$
$-140A\,B^{3}-84B^{4}+84A^{2}+140AB+84B^{2}+8)\quad,$

$m_{5}(D)={AB\over 945}\cdot$
$(155A^{8}+630A^{7}B+1860A^{6}B^{2}+2730A^{5}B^{3}+3906A^{4}B^{4}+2730A^{3}B^{5}+1860A^{2}B^{6}+630A\,B^{7}+155B^{8}$
$-1020A^{6}-3360A^{5}B-7140A^{4}B^{2}-8400A^{3}B^{3}-7140A^{2}B^{4}-3360A\,B^{5}-1020B^{6}+1764A^{4}$
$+5040A^{3}B+5880A^{2}B^{2}+5040A\,B^{3}+1764B^{4}-440A^{2}-1680AB-440B^{2}-144)\quad,$

$m_{6}(D)={AB\over 10395}\cdot$
$(2073A^{10}+12573A^{9}B+43780A^{8}B^{2}+95040A^{7}B^{3}+154143A^{6}B^{4}+172326A^{5}B^{5}+154143A^{4}B^{6}+95040A^{3}B^{7}$
$+43780A^{2}B^{8}+12573A\,B^{9}+2073B^{10}-17050A^{8}-86130A^{7}B-239250A^{6}B^{2}-418110A^{5}B^{3}-498960A^{4}B^{4}-$
$418110A^{3}B^{5}-239250A^{2}B^{6}-86130A\,B^{7}-17050B^{8}+42636A^{6}+182028A^{5}B$
$+367752A^{4}B^{2}+489720A^{3}B^{3}+367752A^{2}B^{4}+182028A\,B^{5}+42636B^{6}-29040A^{4}-116160A^{3}B$
$-143000A^{2}B^{2}-116160A\,B^{3}-29040B^{4}-3344A^{2}-528AB-3344B^{2}-1440)\quad.$

To see all the moments up to the $40$-th, look at the output file

https://sites.math.rutgers.edu/~zeilberg/tokhniot/oGR3moms3a.txt .

Arthur Engel’s Beautiful Expression for the Expected Duration of a (fair)
$3$-player Gambler’s Ruin

In a gambler’s ruin game with three players, the players have initial capitals
$A,B,C$. At each round, one of the players is chosen uniformly at random (i.e.
with probability ${1\over 3}$) to give one dollar to one of the other two
players (each of whom is equally likely to be the recipient). Sooner or later,
one of the players goes broke. How soon?

Equivalently (and that is how Engel thought about it) you have three towers
with $A$, $B$, and $C$ pegs, and at each round, uniformly at random, you
remove a peg from one of the towers and put it (each with probability ${1\over
2}$), on one of the other towers (It is like the Tower of Hanoi, but all pegs
have the same diameter). Sooner or later, one of the towers is empty. How
soon?

Using simulation, and guessing, Arthur Engel [E] conjectured that the expected
duration, $E[D]$, is given by the lovely, explicit formula

$E[D]\,=\,{3ABC\over A+B+C}\quad.$

Once discovered, he had no trouble proving it, since calling this quantity
$E(A,B,C)$, it obviously satisfies the equation

$E(A,B,C)\,=\,1\,+\,{1\over 6}\,(E(A+1,B-1,C)\,+\,E(A-1,B+1,C)$
$+\,E(A+1,B,C-1)\,+\,E(A-1,B,C+1)$ $+\,E(A,B+1,C-1)\,+\,E(A,B-1,C+1))\quad,$

subject to the boundary conditions

$E(0,B,C)=0\quad,\quad E(A,0,C)=0\quad,\quad E(A,B,0)=0\quad,$

and this is a routine verification. Proving uniqueness is fairly easy, using a
maximum principle (see [A], p. 40).

But, in hindsight, there was no need for simulation. Because of the boundary
conditions, it is natural to postulate the ansatz

$E(A,B,C)\,=\,{kABC\over A+B+C}\quad,$

(note that the denominator $A+B+C$ is the total number of pegs (dollars) and
is an invariant). Here $k$ is an undetermined coefficient, that is to be
determined. Plugging it in the equation, we get that indeed $k=3$.

Bruss et. al. [BLT] worked pretty hard to find the variance.

$Var(D)={3ABC\left(A^{2}B+A^{2}C+A\,B^{2}-3ABC+A\,C^{2}+B^{2}C+B\,C^{2}-A-B-C\right)\over
2\left(A+B+C\right)^{2}}\quad.$

David Stirzaker [S] later gave a simpler proof, using martingales. We will
soon describe an even simpler (and much more elementary!) proof, using
computer algebra. Using our method, we will also discover, and prove, explicit
expressions for many higher moments (and one can keep going indefinitely).

But let’s first state some of these beautiful formulas.

$m_{3}(D)={3ABC\over 40(A+B+C)^{3}}\cdot$
$(9A^{5}B+9A^{5}C+54A^{4}B^{2}+108A^{4}BC+54A^{4}C^{2}+90A^{3}B^{3}-225A^{3}B^{2}C-225A^{3}B\,C^{2}+90A^{3}C^{3}+54A^{2}B^{4}$
$-225A^{2}B^{3}C+1080A^{2}B^{2}C^{2}-225A^{2}B\,C^{3}+54A^{2}C^{4}+9A\,B^{5}+108A\,B^{4}C-225A\,B^{3}C^{2}-225A\,B^{2}C^{3}+108AB\,C^{4}$
$+9A\,C^{5}+9B^{5}C+54B^{4}C^{2}+90B^{3}C^{3}+54B^{2}C^{4}+9B\,C^{5}-90A^{3}B-90A^{3}C-180A^{2}B^{2}+90A^{2}BC-180A^{2}C^{2}$
$-90A\,B^{3}+90A\,B^{2}C+90AB\,C^{2}-90A\,C^{3}-90B^{3}C-180B^{2}C^{2}-90B\,C^{3}+12A^{2}$
$+24AB+24AC+12B^{2}+24BC+12C^{2})\quad,$

$m_{4}(D)={3ABC\over 280(A+B+C)^{4}}\cdot$
$(6A^{8}B+6A^{8}C+54A^{7}B^{2}+108A^{7}BC+54A^{7}C^{2}+195A^{6}B^{3}+354A^{6}B^{2}C+354A^{6}B\,C^{2}+195A^{6}C^{3}$
$+357A^{5}B^{4}+63A^{5}B^{3}C-588A^{5}B^{2}C^{2}+63A^{5}B\,C^{3}+357A^{5}C^{4}+357A^{4}B^{5}-378A^{4}B^{4}C$
$+3423A^{4}B^{3}C^{2}+3423A^{4}B^{2}C^{3}-378A^{4}B\,C^{4}+357A^{4}C^{5}+195A^{3}B^{6}+63A^{3}B^{5}C+3423A^{3}B^{4}C^{2}$
$-8505A^{3}B^{3}C^{3}+3423A^{3}B^{2}C^{4}+63A^{3}B\,C^{5}+195A^{3}C^{6}+54A^{2}B^{7}+354A^{2}B^{6}C-588A^{2}B^{5}C^{2}+3423A^{2}B^{4}C^{3}$
$+3423A^{2}B^{3}C^{4}-588A^{2}B^{2}C^{5}+354A^{2}B\,C^{6}+54A^{2}C^{7}+6A\,B^{8}+108A\,B^{7}C+354A\,B^{6}C^{2}+63A\,B^{5}C^{3}$
$-378A\,B^{4}C^{4}+63A\,B^{3}C^{5}+54B^{2}C^{7}+6B\,C^{8}+354A\,B^{2}C^{6}+108AB\,C^{7}+6A\,C^{8}+6B^{8}C$
$+54B^{7}C^{2}+195B^{6}C^{3}+357B^{5}C^{4}+357B^{4}C^{5}+195B^{3}C^{6}-63A^{6}B-63A^{6}C-441A^{5}B^{2}-882A^{5}BC$
$-441A^{5}C^{2}-1008A^{4}B^{3}-819A^{4}B^{2}C-819A^{4}B\,C^{2}-1008A^{4}C^{3}-1008A^{3}B^{4}$
$-3150A^{3}B^{2}C^{2}-1008A^{3}C^{4}-441A^{2}B^{5}-819A^{2}B^{4}C-3150A^{2}B^{3}C^{2}-3150A^{2}B^{2}C^{3}-819A^{2}B\,C^{4}$
$-441A^{2}C^{5}-63A\,B^{6}-882A\,B^{5}C-819A\,B^{4}C^{2}-819A\,B^{2}C^{4}-882AB\,C^{5}-63A\,C^{6}-63B^{6}C-441B^{5}C^{2}$
$-1008B^{4}C^{3}-1008B^{3}C^{4}-441B^{2}C^{5}-63B\,C^{6}+161A^{4}B+161A^{4}C+483A^{3}B^{2}+791A^{3}BC$
$+483A^{3}C^{2}+483A^{2}B^{3}+1260A^{2}B^{2}C+1260A^{2}B\,C^{2}+483A^{2}C^{3}+161A\,B^{4}+791A\,B^{3}C$
$+1260A\,B^{2}C^{2}+791AB\,C^{3}+161A\,C^{4}+161B^{4}C+483B^{3}C^{2}+483B^{2}C^{3}$
$+161B\,C^{4}+16A^{3}+48A^{2}B+48A^{2}C+48A\,B^{2}+96ABC+48A\,C^{2}+16B^{3}+48B^{2}C+48B\,C^{2}+16C^{3})\quad.$

To see the first $20$ moments (about the mean), enjoy the following output
file:

https://sites.math.rutgers.edu/~zeilberg/tokhniot/oGR3moms4a.txt .

How we used Symbol-Crunching to Derive these Nice Formulas for the Moments?

The Two-Player Case

Let’s first consider the classical case. Let $D(A,B)$ be the random variable

“duration until one of the players goes broke if the initial capitals were $A$
and $B$”,

and let $F(A,B)(y)$ be its probability generating function.

Obviously, $F(A,B)(y)$ satisfies the following equation:

$F(A,B)(y)\,=\,{y\over 2}(F(A+1,B-1)(y)+F(A-1,B+1)(y))\quad,$ $None$

subject to the boundary conditions

$F(A,0)(y)=1\quad,\quad F(0,B)(y)=1\quad.$

Let’s expand $F(A,B)(y)$ as a Taylor series around $y=1$,

$F(A,B)(y)=\sum_{i=0}^{\infty}f_{i}(A,B)(y-1)^{i}\quad.$ $None$

$f_{i}(A,B)$ is equal to the binomial moment

$E\left[{{D(A,B)}\choose{i}}\right]\quad.$

Once you have the binomial moments, you can easily compute the moments

$E[D(A,B)^{i}]=\sum_{j=0}^{i}j!\,S(i,j)\,E\left[{{D(A,B)}\choose{j}}\right]\quad,$

where $S(i,j)$ are the Stirling numbers of the second kind.

Once you have the moments, including the most important one, $\mu=E[D(A,B)]$,
you can get the moments about the mean by

$E[(D(A,B)-\mu)^{i}]=\sum_{j=0}^{i}(-\mu)^{i-j}\,{{i}\choose{j}}E[D(A,B)]^{j}\quad.$

Plugging-in $(2)$ into $(1)$ and writing, for convenience $y-1=z$, we have

$\sum_{i=0}^{\infty}f_{i}(A,B)z^{i}\,=\,{1+z\over
2}\left(\sum_{i=0}^{\infty}f_{i}(A+1,B-1)z^{i}+\sum_{i=0}^{\infty}f_{i}(A-1,B+1)z^{i}\right)\quad.$
$None$

Comparing the coefficient of $z^{i}$, we have the equation

$f_{i}(A,B)={1\over 2}(f_{i}(A+1,B-1)+f_{i}(A-1,B+1))+{1\over
2}(f_{i-1}(A+1,B-1)+f_{i-1}(A-1,B+1))\quad,$

implying that

$f_{i}(A,B)-{1\over 2}(f_{i}(A+1,B-1)+f_{i}(A-1,B+1))={1\over
2}(f_{i-1}(A+1,B-1)+f_{i-1}(A-1,B+1))\quad.$

Assuming that we already know $f_{i-1}(A,B)$ (of course $f_{0}(A,B)=1$), we
postulate an ansatz for $f_{i}(A,B)$ as a polynomial in $A,B$ of degree $2i$,
using undetermined coefficients, letting Maple do the tedious high-school
algebra, getting a system of linear equations satisfied by these, so far
undetermined, coefficients. Then we ask Maple to kindly solve that system that
it generated for us. To our delight, there was always a unique solution. The
uniqueness (in general) is easy by a “maximum principle”.

The Three-Player (fair) Gambler’s Ruin Until One Player Goes Broke

Everything is analogous. Letting $f_{i}(A,B,C)$ be the binomial moment of the
random variable ‘duration (until one of the players goes broke)’, with initial
capitals $A,B,C$, we have

$f_{i}(A,B,C)-{1\over 6}(f_{i}(A+1,B-1,C)+f_{i}(A-1,B+1,C)$
$+f_{i}(A+1,B,C-1)+f_{i}(A-1,B,C+1)$ $+f_{i}(A,B+1,C-1)+f_{i}(A,B-1,C+1))$
$={1\over 6}(f_{i-1}(A+1,B-1,C)+f_{i-1}(A-1,B+1,C)$
$+f_{i-1}(A+1,B,C-1)+f_{i-1}(A-1,B,C+1)$
$+f_{i-1}(A,B+1,C-1)+f_{i-1}(A,B-1,C+1))\quad.$

Inspired by Engel’s formula for the first moment, we took the ansatz to be a
polynomial in $A,B,C$ of degree $2i+1$ divided by $A+B+C$. Since the
denominator is invariant, it is retained, and we can look at the numerator and
do the linear algebra with undetermined coefficients, like we did for the two-
player case.

Duration for a $k$-Player (fair) Gambler’s Ruin until Only One Player survives

The Expectation is easy, as already pointed out by Sheldon Ross [R] (of the
popular probability textbooks fame). If $k$ players have initial capitals
$A_{1},\dots,A_{k}$, the expected duration until $k-1$ of them drop out, is

$\sum_{1\leq i<j\leq k}A_{i}\,A_{j}\quad.$

Alas, the variance, and higher moments, do not seem to be nice, let alone
polynomials (or even rational functions).

Probability of Winning

For the classical, two-player, gambler’s ruin, famously (and trivially) the
probability of winning the whole pot for the player whose initial capital was
$A$ dollars (and the other had $B$ dollars) is ${A\over A+B}$. For the three-
player case, there does not seem to be a nice formula, but the fascinating
article [DiE] has some very good approximations and asymptotics. In
particular, they proved that with initial capitals $1,1,N$, the probability is
$O({1\over N^{3}})$.

Recall that the scaled moments (about the mean) are $m_{i}/m_{2}^{i/2}$.

The limit of the scaled moments with equal capitals for two players

If both players start out with the same capital $a$, then, of course

$m_{1}=E[D]=a^{2}\quad,\quad\sigma^{2}=m_{2}={2\over 3}a^{4}+O(a^{3})\quad,$

and the limits of the scaled moments (about the mean), from the third to the
$20^{th}$, as $a$ goes to infinity, expressed in decimals, are as follows.

$1.9595917942265424786,8.8285714285714285714,42.737763893131259770,255.69610389610389610,$
$1775.6935988640577894,14103.678121878121878,126010.34642879135344,1250957.5292367075649,$
$13660640.862506747726,162737708.10778675241,2100232711.7489454497,29189813843.019299016,$
$434668787296.09441079,6904215035796.8421121,116519631313642.16450,$
$2082128256599445.4176,39273259761289810.763,779763346464591374.25\quad.$

In particular, the limiting skewness (scaled third-moment about the mean) is
$1.9595917942265424786$, and the limiting kurtosis (scaled fourth-moment about
the mean) is $8.8285714285714285714$, as already computed in [Z].

For the exact values, see the output file

https://sites.math.rutgers.edu/~zeilberg/tokhniot/oGR3moms5.txt .

The Limit of the Scaled moments with equal capitals for three players

If all three players start out with the same capital $a$, then, of course

$m_{1}=E[D]=a^{2}\quad,\quad\sigma^{2}=m_{2}={1\over 2}a^{4}+O(a^{3})\quad,$

and the limits of the scaled moments (about the mean), from the third to the
$15^{th}$, as $a$ goes to infinity, in decimals are

$1.8384776310850235634,8.2714285714285714286,38.719147032686238015,226.13701298701298701,$
$1529.4150671558283928,11835.953251748251748,103028.09771971592326,$
$996496.58001633815101,10601990.334510850051,123051719.84505166456,$
$1547212853.2328220329,20950631863.915761840,303953682939.86889196\quad.$

In particular, the limiting skewness is $1.8384776310850235634$, a little less
than the two-player case, and the limiting kurtosis is
$8.2714285714285714286$, also a little less than its counterpart. For the
exact values, see the output file

https://sites.math.rutgers.edu/~zeilberg/tokhniot/oGR3moms6.txt .

More than Three-Players?

With the exception of Ross’s formula (mentioned above), that only handles the
expectation, for the $k$-player game until only one player survives, our
approach does not seem to work with more than three players. Of course,
perhaps there is yet-another ansatz, and perhaps one can get, by symbol-
crunching, at least asymptotic results, but this remains to be seen. Again we
strongly recommend the very interesting Diaconis-Ethier article [DiE], of
whose existence we learned thanks to Jean-Paul Delahaye’s engaging column.

References

[A] Jiri Andrei, “Mathematics and Chance”, Wiley, 2001.

[BLT] F.T. Bruss, G. Louchard, and J.W. Turner, On the $N$-tower problem and
related problems. Adv. Appl. Probab. 35 (2003), 278-294.

[De] Jean-Paul Delahaye, Ruiner le casino ou se ruiner, Pour la Science 550
(August 2023), 80-85.
https://medias.pourlascience.fr/api/v1/files/64b5166b903f8e26c07a14f5?alt=file

[DiE] Persi Diaconis and Stewart N. Ethier, Gambler’s Ruin and the ICM,
Statist. Sci. 3, 289-305, August 2022. https://arxiv.org/abs/2011.07610 .

[E] Arthur Engel, The computer solves the three tower problem, Amer. Math.
Monthly 100 (1993), 62-64.

[F] William Feller, “An Introduction to Probability Theory and Its
Application”, volume 1, three editions. John Wiley and sons. First edition:
1950. Second edition:1957. Third edition: 1968. Volume 2: two editions.

[R] Sheldon M. Ross, A simple solution to a multiple player gambler’s ruin
problem, Amer. Math. Monthly 116 (2009), 77-81.

[S] David Stirzaker, Three-handed gambler’s ruin, Adv. Appl. Probab. 38
(2006), 284-286.

[Z] Doron Zeilberger, Symbol-crunching with the gambler’s ruin problem, in:
“Tapas in Experimental Mathematics”, Contemporary Mathematics 457 (2008),
285-292, (Tewodros Amdeberhan and Victor Moll, eds.)
https://sites.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/ruin.html .

Shalosh B. Ekhad, c/o D. Zeilberger, Department of Mathematics, Rutgers
University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen Rd.,
Piscataway, NJ 08854-8019, USA. Email: ShaloshBEkhad at gmail dot com .

Doron Zeilberger, Department of Mathematics, Rutgers University (New
Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway, NJ
08854-8019, USA. Email: DoronZeil at gmail dot com .

Sept. 15, 2023.