# On the proof of the maximum principle for elliptic equations

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.

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.;)
• 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? – EpsilonDelta Feb 1 at 14:41
• @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. :) – Zixiao_Liu Feb 2 at 0:10