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From Renardy - "An Introductionto Partial Differential Equations".

Let $$Lu=a_{ij}(x)\frac{\partial^2 u}{\partial x_i\partial x_j}+b_i(x)\frac{\partial u}{\partial x_i}+c(x)u,\ x\in\Omega\subset\mathbb{R}^n$$

The maximum principle states that for $Lu\ge0$ in a bounded domain $\Omega$ with $c(x)=0$ the maximum of $u$ is achieved on $\partial \Omega$.

The proof is done by contradiction. Suppose $Lu>0$ then $u$ cannot achieve its maximum anywhere in $\Omega$. Suppose it did at a point $x_0$. Then all first derivatives must vanish and one can show that $Lu(x_0)\le 0$, contradiction.

Now (for the case $Lu=0$) an approximation argument is used. Let $u_\epsilon=u+\epsilon\exp(\gamma x_1)$. We obtain $$Lu_\epsilon = Lu + \epsilon(\gamma^2a_11 + \gamma b_1)\exp(\gamma x_1)$$We can then choose $\gamma$ large enough s.t. $\gamma^2a_11 + \gamma b_1\ge0$

Then $Lu_\epsilon >0$ and we have that $$\max_{\overline \Omega}u_\epsilon = \max_{\partial \Omega}u_\epsilon$$ for every $\epsilon>0$. The theorem follows from $\epsilon \rightarrow 0$.

Now my question is: Why can I assume that this maximum property holds for the limes? It might be possible that when taking the limit, that the limit function does not satisfy this relation anymore.

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Due to $\Omega$ is a bounded domain, hence the added "error part" is uniformly (to $\varepsilon$) bounded (denoted as M). Hence we have the following dominate (note that $u_{\varepsilon}>u$):$$ \max_{\Omega}u\leq \max_{\Omega}u_{\varepsilon} =\max_{\partial_{\Omega}}u_{\varepsilon}\leq \max_{\partial\Omega}+\varepsilon\cdot M. $$ The last step comes from $\max(a+b)\leq\max(a)+\max(b)$. Now we take limit and the result follows immediately.

This is the reason why we demaned $\Omega$ is a bounded domain. This proof doesn't holds for unbounded $\Omega$. I hope I stated the reason clear.;)

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  • $\begingroup$ But since $\Omega$ is open, we need know if the error part does attain its supremum. But since the error part is $\exp(\gamma x_1)$ we can continue this function on a closed set, is this correct? $\endgroup$ – EpsilonDelta Feb 1 at 14:41
  • $\begingroup$ @EpsilonDelta Well, due to $\Omega$ is bounded, hence the supremum of the "error part" can be dominated by some constant (depending on $\overline{\Omega}$). And it is true that we can do the same process on $\overline{\Omega}$, there are no big differences here. :) $\endgroup$ – Zixiao_Liu Feb 2 at 0:10

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