Maximum Principle for Poisson Equation

Question

For a smooth $u(x)$, $x \in \mathbb{R}^n$, satisfying:

$\Delta u = -f$ for $||x||<1$ , $u=g$ on $||x||=1$

I want to show that there exists a constant $C$ such that:

$$\max\{|u|:||x||\leq 1\} \leq C(\max\{|g|:||x||=1\}+\max\{|f|:||x||<1\}).$$

Answer 1

Well, implicitly, $A = \|f\|_\infty$ and $B = \|g\|_\infty$ must exist.

Thus, let us consider $$u_+(x) = B + A (1-|x|^2)$$ It is a simple matter to check that $\Delta(u_+ - u) \leq 0$ where $|x| < 1$ and that $u_+ - u \geq 0$ on $|x|=1$. The maximum principle tells you the rest.

You can analogously construct $$u_-(x) = -B - A (1-|x|^2)$$

Using these as super and subsolutions for your problem, you can see that $$|u(x)| \leq A + B$$ where $|x| \leq 1$.

Answer 2

If $u$ is a smooth function, then also $f$ and $g$ are smooth. Defined the Newtonian Potential $v$ of $f$ by $$v(x)=\int_B \Gamma(x-y)f(y)dy$$

where $B$ is the unit open ball about the origin and $\Gamma$ is the fundamental sollution of the Laplace equation. Consider the solution of the auxiliary problem $$ \left\{ \begin{array}{rl} -\Delta w=0 &\mbox{ in $B$} \\ w=v-g &\mbox{ in $\partial B$} \end{array} \right. $$

By unicity we conclude that $u=v-w$. Also, from the maximum principle, we have that $$\|w\|_{L^\infty(\overline{B})}\leq \|v\|_{L^\infty(B)}+\|g\|_{L^\infty(\partial B)}$$

From here, you have two things to do:

1 - Show that there exist a constan $C_1$, such that $\|v\|_{L^\infty(B)}\leq C_1\|f\|_{L^\infty(B)}$

2 - Use the representation of $u$ in terms of $v$ and $w$ to conclude.