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My teacher made the following to prove that the solution for the Dirichlet problem:

Let $\Omega$ be a bounded open set

Given $f\in C(\partial\Omega)$, find $u\in C^2(\Omega)\cap C(\Omega)$ such that

$$\Delta u = 0 \mbox{ in }\Omega\\ u = f\mbox{ in } \partial\Omega$$

is unique.

Suppose that there exists $2$ solutions $u_1,u_2$. Then $\Delta (u_1-u_2) = 0$ in $\Omega$, and $u_1-u_2 = f-f = 0$ in $\partial\Omega\implies u_1=u_2$

Well, it just proves that $u_1=u_2$ in the boundary, not inside. I know that both $u_1$ and $u_2$ have $Delta=0$ inside, but they could be different functions just being equal in the boundary.

Is this proof wrong?

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    $\begingroup$ You can use the maximum principle for harmonic functions. Namely, since the max and min of $u_1 - u_2$is always achieved on the boundary, because that function is harmonic, it is constantly zero. $\endgroup$
    – Elle Najt
    Commented Sep 29, 2018 at 22:40
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    $\begingroup$ @Lorenzo sorry, I think I missed some step. The principle of the maximum says that the maximum value of the function $u_1-u_2$ is achieved at the border, so the maximum of $u_1-u_2$ is $0$. Now what? $\endgroup$
    – Paprika
    Commented Sep 29, 2018 at 22:45
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    $\begingroup$ if u is harmonic, so is -u. So the maximum principle also implies that the minimum is on the boundary... $\endgroup$
    – Elle Najt
    Commented Sep 29, 2018 at 22:52

1 Answer 1

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Let $\Omega$ be bounded and let $u \in {C^2({\Omega})} \cup C(\overline{\Omega})$ and $f \in C(\partial \Omega)$ such that \begin{align} (*)\left\{ \begin{array}{ll} \Delta u = 0 & \mbox{in } \Omega \\ \ \ \ u = f & \mbox{on } \partial \Omega \end{array} \right. \end{align} Now suppose that $\widetilde{u}$ is another solution to the Dirichlet problem $(*)$. Define $w := \widetilde{u} - u$ and due to the linearity of the Laplace operator we obtain \begin{align} \left\{ \begin{array}{ll} \Delta w = 0 & \mbox{in } \Omega \\ \ \ \ w = 0 & \mbox{on } \partial \Omega \end{array} \right. \implies w \equiv 0 \implies u \equiv \widetilde{u} \end{align} You can see that this is directly implied from the Strong Maximum Principle. In fact, if for any point $x \in \Omega$ we had $w > 0$ or $w < 0$ then $w$ would attain a maximum/minimun in $\Omega$ and thus from s.m.p. $w$ must be constant. Since the boundary value is $0$, it must be $w \equiv 0$. It is not that $\widetilde{u} = u$ only at the boundary $\partial \Omega$ but that $u$ is identical to $\widetilde{u}$ in $\overline{\Omega}$ because $w$ is identical to zero in $\overline{\Omega}$.

You can see that with the same approach we obtain the uniqueness of the Dirichlet problem for the Poisson equation.

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  • $\begingroup$ Could you elaborate more on why the maximum principle implies $w=0$ in the interior without using that $u$ must be constant if harmonic? We know that the maximum is attained in the boundary, so $w$ is at max $0$. The minimum is also attained in the boundary, so $w$ is at min $0$. So $w$ is $0$. Is this it? $\endgroup$
    – Paprika
    Commented Sep 29, 2018 at 23:31
  • $\begingroup$ @Paprika Well, you see that $u$ satisfies $(*)$ and that $\widetilde{u}$ also satisfies $(*)$. Then $w = \widetilde{u} - u$ is a linear combination of harmonic functions and thus harmonic. Now you apply S.M.P (See Evans Chap. 2.2 Thm. 4) to $w$. You deduce that $w$ is identically zero, and by its definition $\widetilde{u} \equiv u$. Is that more clear? $\endgroup$
    – ares
    Commented Sep 29, 2018 at 23:38
  • $\begingroup$ I didn't want to use strong maximum principle, only the weak one, because this proof follows right after the weak maximum principle in my teacher's notes. Does the min and max argument work? $\endgroup$
    – Paprika
    Commented Sep 29, 2018 at 23:41
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    $\begingroup$ @Paprika The Maximum Principle says that the maximum/minimum must be attained on the boundary for harmonic functions. If the value on the boundary is $0$ then if you have any nonzero value for $w$ in $\Omega$ then you have a contradiction. Thus $w$ must be identically $0$. Yes, it works. $\endgroup$
    – ares
    Commented Sep 29, 2018 at 23:44

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