Monotone discretization of elliptic problems with mixed derivatives on anisotropic meshes: a counterexample

TL;DR

This paper provides a counterexample demonstrating the challenges of monotone discretization for elliptic problems with mixed derivatives on anisotropic meshes, highlighting limitations in existing numerical methods.

Contribution

It introduces a specific counterexample that shows the failure of monotone discretization schemes in certain anisotropic mesh configurations for elliptic PDEs with mixed derivatives.

Findings

Counterexample shows monotone discretization failure on anisotropic meshes

Highlights limitations of current numerical schemes for mixed derivative problems

Provides insights for developing more robust discretization methods

Abstract

We present a counterexample concerning the monotone discretization of elliptic problems with mixed derivatives on anisotropic meshes.

Equations

$$C_1\le \frac{h_x}{h_y}\le C_2$$

$$Lu:=u_{xx}+u_{xy}+u_{yy}$$

$$L_{H,h}u$$

$$+{\beta }_{5}u(x_0-H,y_0-h)+{\beta }_{6}u(x_0+H,y_0-h)+{\beta }_{7}u(x_0-H,y_0+h)+{\beta }_{8}u(x_0+H,y_0+h).$$

$$\alpha +\sum {\beta }_i=0$$

$${\beta }_2-{\beta }_1+{\beta }_6+{\beta }_8-({\beta }_5+{\beta }_7)=0$$

$${\beta }_4-{\beta }_3+{\beta }_7+{\beta }_8-({\beta }_5+{\beta }_6)=0$$

$$\frac{H^2}{2}({\beta }_1+{\beta }_2+{\beta }_5+{\beta }_6+{\beta }_7+{\beta }_8)=1$$

$$\frac{h^2}{2}({\beta }_3+{\beta }_4+{\beta }_5+{\beta }_6+{\beta }_7+{\beta }_8)=1$$

$$hH({\beta }_5-{\beta }_7+{\beta }_8-{\beta }_6)=1.$$

$$H^2({\beta }_1+{\beta }_2)=h^2({\beta }_3+{\beta }_4),$$

$${\beta }_4-{\beta }_3=\mathcal{O}(\frac{1}{H^2}).$$

$$B+D={\beta }_4-{\beta }_3,-B+D=\frac{1}{hH}.$$

$${\beta }_6=-\frac{1}{2hH}+{\beta }_8+\frac{1}{2}({\beta }_4-{\beta }_3).$$

Full text

Monotone discretization of elliptic problems with mixed derivatives on anisotropic meshes:

a counterexample

Hans-G. Roos, TU Dresden

(January 2019)

In many papers of O’Riordan and Shishkin the authors underline the importance of numerical approximations for singularly perturbed elliptic problems which are free of spurious oscillations. Therefore they prefer (difference) methods where the associated system matrix is a monotone matrix, and they use, exclusively, the special class of M-matrices, which are monotone. But what about elliptic problems with mixed derivatives?

Not much is known concerning finite difference methods for singularly perturbed elliptic problems with mixed derivatives on layer-adapted meshes, in [4] and [5] is also nothing to find. It is well-known that on isotropic meshes one can generate an M-matrix (see, for instance, Theorem 10.1 in [3]). That means, that a condition of the type

[TABLE]

is sufficient to generate an M-matrix. But layer-adapted meshes are highly anisotropic. In [2] the authors state that a monotone scheme imposes a condition on the mesh ratio, but this is not proved so far. In [1] the authors avoid to discuss this question: they use simple an inconsistent approximation which approximates the mixed derivative only on the fine, isotropic part of a Shishkin mesh. The assumption $\varepsilon\leq N^{-1}$ then allows nevertheless to prove an error estimate.

In this paper we present an example which shows that a consistent first order scheme for an elliptic problem with mixed derivatives on a highly anisotropic mesh cannot generate an M-matrix.

Consider the elliptic operator

[TABLE]

on the rectangle $(x_{0}-H,x_{0}+H)\times(y_{0}-h,y_{0}+h)$ and the nine-point difference operator

[TABLE]

Taylor expansion yields the first order consistency conditions

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

Fixing $\beta_{2},\beta_{3},\beta_{4}$, the parameter $\beta_{1}$ is given by

[TABLE]

and it is easy to show that the remaining parameters are uniquely determined.

The question is: Can one choose the parameters in such a way that $\beta_{i}\geq 0$ for all $i$ ?

In the isotropic case this is possible, for instance, for $h=H$ one gets with the choice $\beta_{2}=\beta_{3}=\beta_{4}=\frac{1}{2h^{2}}$ the result $\beta_{1}=\beta_{5}=\beta_{7}=\frac{1}{2h^{2}}$, moreover $\beta_{6}=\beta_{8}=0$ and $\alpha=-\frac{3}{2h^{2}}$. The corresponding matrix is the negative of an M-matrix.

Now let us assume that (3)–(7) do have a solution with $\beta_{i}\geq 0$ for all $i$ in the case $h<<H$.

From (5) we obtain $0\leq\beta_{i}\leq\frac{2}{H^{2}}$ for $i=1,2,5,6,7,8$. Consequently, (4) implies

[TABLE]

The equations (4) and (7) yield a system for $B:=\beta_{6}-\beta_{8}$ and $D:=\beta_{5}-\beta_{7}$:

[TABLE]

Thus we get $B=-\frac{1}{2hH}+\frac{1}{2}(\beta_{4}-\beta_{3})$, respectively

[TABLE]

Because $\beta_{8}$ and $\beta_{4}-\beta_{3}$ are of order ${\cal O}(\frac{1}{H^{2}})$, they cannot compensate the large negative term $-\frac{1}{2hH}$ if $h$ is sufficiently small in comparison to a given $H$. Consequently, $\beta_{6}$ cannot be nonnegative for anisotropic meshes where $H/h$ is sufficiently large, and this is the case for Shishkin meshes and other types of layer-adapted meshes.