I am reading Partial differential equations written by Jurgen Just (Third edition) chapter 4, and I want to do some exercises. I think that I need some hints to solve the problem 4.2:

Let $x_0 \in \Omega_h$ have neighbours $x_1,\ldots ,x_{2d}$.We consider a difference operator Lu for u:$\bar\Omega_h\rightarrow\Bbb R$,


satisfying following assumptions:

$\hspace{4cm}b_{\alpha}\geq0 \hspace{0.2cm}for \hspace{0.2cm}{\alpha=1,\ldots,2d}, \hspace{0.2cm}\sum_{\alpha=1}^{2d}b_{\alpha}>0,\hspace{0.2cm}\sum_{\alpha=0}^{2d}b_{\alpha}\leq0.$

Prove the weak maximum principle: $Lu\geq0 \hspace{0.2cm}in\hspace{0.2cm} \Omega_h$ implies,

$$\max_{\Omega_h} u\leq \max_{\Gamma_h} u$$.

Note that in the above ${\Omega_h}$ is discretization of ${\Omega}$ and ${\Gamma_h}$ is the boundary of ${\bar\Omega_h}$. (c.f. pages 59, 60 of Partial differential equations written by Jurgen Just, Third edition)

Thanks in advance...


Let $x_0$ be an interior point, i.e. $x_0\in\Omega_h$. By hypothesis

$$Lu(x_0)=\sum_{\alpha=0}^{2d} b_{\alpha}u(x_{\alpha})\geq0,$$ hence \begin{equation} -b_0u(x_0)\leq\sum_{\alpha=1}^{2d} b_{\alpha}{u(x_{\alpha})}.\quad{(1)} \end{equation}

On the other hand by hypothesis $$\sum_{\alpha=0}^{2d} b_{\alpha}\leq0,\, \sum_{\alpha=1}^{2d} b_{\alpha}>0.$$

hence $$\left\{0<\sum_{\alpha=1}^{2d} b_{\alpha}\leq-b_0\atop {1\over-b_0}\leq{1\over \sum_{\alpha=1}^{2d} b_{\alpha}}\quad (2)\right.$$

From (1) and (2) we conclude $$u(x_0)\leq {1\over -b_0}{\sum_{\alpha=1}^{2d} b_{\alpha}u(x_{\alpha})}\leq {\sum_{\alpha=1}^{2d} b_{\alpha}u(x_{\alpha})\over \sum_{\alpha=1}^{2d} b_{\alpha}}$$therefore $$u(x_0)\leq max_{\alpha=1,\ldots,2d} u(x_{\alpha})$$

with equality only if $$u(x_0)=u(x_\alpha)\quad for\, all\quad \alpha\in\{1,\ldots,2d\}.$$

Thus, if u assumes an interior maximum at a vertex $x_0$, it does so at all neighbors of $x_0$ as well, and repeating this reasoning, then also at all neighbors of neighbors, etc. Since $\Omega_{h}$ is discretely connected by assumption, $u_h$ has to be constant in $\bar\Omega_{h}$. Otherwise the problem is trivial.


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