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Let $U$ be a bounded, open subset of $\mathbb{R}^n$. Prove that there exists a constant $C$, depending on only $U$, such that $$\max_{\bar{U}} |u|\le C(\max_{\partial U} |g|+\max_{\bar{U}} |f|)$$ wherever $u$ is a smooth solution of \begin{cases}-\Delta u=f & \text{in }U\\ \quad \, \, \, u=g & \text{on } \partial U \end{cases}

This is from PDE Evans (2nd edition), Chapter 2 Exercise 6. (Note that $\bar{U}$ is the closure of $U$.)

First, I tried proving for the boundary case, that is, suppose $x \in \partial U$. Then $u=g$. We have $$|u|=|g|\le \max_{\partial U} |g| \le \max_{\partial U}|g|+\max_{\bar{U}}|f|.$$ Hence, $|u|$ is bounded by $\max_{\partial U}|g|+\max_{\bar{U}}|f|$. So there exists a $C > 0$ such that $|u| \le C(\max_{\partial U}|g|+\max_{\bar{U}}|f|)$. Thus, $$\max_{\partial U} |u| \le C(\max_{\partial U}|g|+\max_{\bar{U}}|f|),$$ as required.

Now, how can I prove for the interior case, that is, for $x \in U$? I wanted to use strong maximum principle, but that applies only to harmonic solutions, so I can't use that here.

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Here is a hint. You already noticed that we want to use Maximum principle and Maximum principle only happens on harmonic functions.

Let's divide your problem into too cases. Namely,

\begin{cases}-\Delta u_1=0 & \text{in }U\\ \quad \, \, \, u_1=g & \text{on } \partial U \end{cases}

\begin{cases}-\Delta u_2=f & \text{in }U\\ \quad \, \, \, u_2=0 & \text{on } \partial U \end{cases}

You already have the estimation of $u_1$ by M-P. Now we work on $u_2$.

Since $U$ is bounded, let's assume that $U\subset \{x\in R^N,\,\, |x\cdot \xi|\leq d\}$ where $|\xi|=1$ and $d>0$ is an constant.

Next we define $$ k(x):= e^{Lx\cdot \xi} $$ and we notice that $$ -\Delta k(x)=-L^2\xi_i^2e^{Lx\cdot\xi}=-L^2 e^{Lx\cdot\xi}\leq 1 $$ for large enough $L>0$.

Now we send $$ v(x):= (e^{Ld}-e^{Lx\cdot \xi})\sup_{U}f^+ $$

Now we compute and find out $$ -\Delta v(x)\geq \sup_{U}f^+ $$ and hence we have $$-\Delta (u_2(x)-v(x))\leq f-\sup_U f^+\leq 0 $$ hence we have $u-v$ is subharmonic and by M-P we have $u(x)\leq v(x)$ for all $x\in U$ and hence we have $$ \sup u_2\leq C\sup f^+$$

Now write $u=u_1+u_2$, I am sure you could take it from here.

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  • $\begingroup$ So we can use the maximum principle, even if a solution like $v(x)$ is only subharmonic, rather than fully harmonic? $\endgroup$ – Cookie Dec 22 '14 at 4:16
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    $\begingroup$ This is right. I suggest you to work out an proof yourself, although it is almost equal to the proof of harmonic case $\endgroup$ – spatially Dec 22 '14 at 4:20
  • $\begingroup$ I worked out the proof of the maximum principle as you suggested (and this is Chapter 2, Exercise 5B), and instead of $$M=v(x_0) = \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{B(x_0,r)} u \, dy \le M$$ in the harmonic case with $u(x_0)=M:=\max_{\bar{U}} u$ (see page 27 in 2nd edition), we have instead $$M=v(x_0) \color{red}{\le} \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{B(x_0,r)} u \, dy \le M$$ in the subharmonic case. Both the statements force $u \equiv M$ to remain true. This is how the maximum principle still holds in both the harmonic and subharmonic cases? $\endgroup$ – Cookie Dec 23 '14 at 6:08
  • $\begingroup$ Yes. This is a nice proof $\endgroup$ – spatially Dec 23 '14 at 6:11

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