# Proving uniqueness of the Dirichlet problem

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?

• 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. Commented Sep 29, 2018 at 22:40
• @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? Commented Sep 29, 2018 at 22:45
• if u is harmonic, so is -u. So the maximum principle also implies that the minimum is on the boundary... Commented Sep 29, 2018 at 22:52

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 $$\overline{\Omega}$$.
• 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? Commented Sep 29, 2018 at 23:31
• @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?
• @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.