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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:

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

Davide Giraudo
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rmh52
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2 Answers2

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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 $|x| \leq 1$.

Ray Yang
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    Could you please explain your steps in the first few lines? – rmh52 Mar 31 '13 at 23:53
  • (+1) your answer is more clear and direct than mine. – Tomás Apr 01 '13 at 00:18
  • @rmh52: which step is confusing? -R – Ray Yang Apr 01 '13 at 01:19
  • The expression for u+. Im not sure how you decided to do that – rmh52 Apr 01 '13 at 01:21
  • Oh. The idea is, I picked $B$ for the boundary term, and for the $A$ term I wanted a function that had 0 on the boundary, and a very negative laplacian so I could invoke the maximum principle on the difference with $u$. Since $\Delta (|x|^2) = 2n$, I basically took the standard quadratic of the right sign as a function with the right properties. – Ray Yang Apr 01 '13 at 01:26
  • I see... I'm not familiar with the notation you've used but I assume the infinity norm is the maximum of abs value? How do I take the laplacian of u+ to verify that the difference is non-positive? – rmh52 Apr 01 '13 at 01:53
  • Yeah, the infinity norm is the essential supremum of a function (in this case, for $f$, on $|x| \leq 1$, and for $g$, on $|x|=1$). The reason I picked these functions is that you can explicitly calculate the laplacian: $\Delta u_+ = \Delta B + \Delta A(1-|x|^2) = -2An$ since $A$ and $B$ are just constants. – Ray Yang Apr 01 '13 at 02:02
  • Got it. Thank you for your help. – rmh52 Apr 01 '13 at 02:20
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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.

Tomás
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