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I know that there is a unique solution to Laplace's equation that has particular boundary value. But I do not understand why this is the case. Thanks for your help in advance!

PS: this is a follow-up to this question.

GanTheMan
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2 Answers2

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Say we want to solve Laplace's equation $\nabla^2 T = 0$ in a region $R$, with $T = f$ on the boundary $\partial R$ (where $f$ is some function). To show that the solution is unique, I will assume that there are two solutions, say $T_1$ and $T_2$. Let $W = T_1 - T_2$. Then we have $\nabla^2 W = \nabla^2 T_1 - \nabla^2 T_2 = 0 - 0 = 0$ in $R$, and on $\partial R$, we have $W = T_1 - T_2 = f - f = 0$.

By Green's theorem, we have the following result: \begin{equation*}\iint\limits_{R}\nabla\cdot(W \nabla W) \mathop{}\!\mathrm{d}A = \int\limits_{\partial R}(W \nabla W)\cdot\boldsymbol{n} \mathop{}\!\mathrm{d}s\end{equation*} where I use $\nabla \cdot$ to denote the divergence operator, and $\boldsymbol{n}$ is the unit normal vector that points outwards on $\partial R$. But, by a well-known identity, it is the case that $\nabla\cdot(W\nabla W) = W\nabla^2 W + \nabla W \cdot \nabla W$. Substituting $\nabla^2 W = 0$, we get $\nabla\cdot(W\nabla W) = \nabla W \cdot \nabla W = \left\lVert \nabla W \right\rVert^2$. Moreover, since $W = 0$ on $\partial R$, we of course have $(W\nabla W) \cdot \boldsymbol{n} = 0$ on $\partial R$, so the right hand side is $0$.

Thus \begin{equation*}\iint\limits_{R}\left\lVert\nabla W\right\rVert^2 \mathop{}\!\mathrm{d}A = 0\end{equation*} Recall that if the integral of a continuous non-negative function is $0$, then the function itself must be $0$ everywhere on the region of integration. In this case, $T_1$ and $T_2$ must be at least twice differentiable in order to satisfy Laplace's equation, so we do have continuity. This means $\left\lVert\nabla W\right\rVert^2 = 0$, and thus $\nabla W = \boldsymbol{0}$ in $R$.

Another well-known result is that if a function has $0$ gradient on a path-connected open set, then it is a constant function. Accordingly, assuming that the region $R$ is path-connected (which it almost always will be in any practical example), $W$ is constant. Since $W$ is constant in $R$ and $0$ on the boundary $\partial R$, and $W$ is continuous, the constant must be $0$. That is, $W = 0$ on $R \cup \partial R$, so $T_1 = T_2$ on $R \cup \partial R$, showing that the solution is unique. $\qquad \rule{0.7em}{0.7em}$

Prasiortle
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  • understand that you've done a proof by Reductio Ad Absurdum. Can you please also provide an intuitive explanation as to why there is just one solution? Thanks a lot in advance. :) – GanTheMan May 08 '20 at 17:54
  • It's not reductio ad absurdum. To do that, you would assume that there are two non-equal solutions, and show that this leads to a contradiction. My proof was to just assume that there are two solutions (not assuming non-equal), and show that this implies they are equal. As for an intuitive explanation, I can't think of one in this case. One can intuitively see why other PDEs such as the wave equation or heat equation have unique solutions from the fact that the initial conditions are $0$. But I don't think such an explanation is possible for Laplace's equation. – Prasiortle May 08 '20 at 18:43
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Are you aware of the maximum principle? It basically states that a solution of Laplace’s equation takes its maximal and minimal values on the boundary.

Thus, let us suppose you have two solutions $u_1, u_2$ of Laplace’s equation on a domain $\Omega$ that fulfil $u_1(x) = u_2(x)$ on $\partial\Omega$. Then their difference $d(x):=u_1(x) - u_2(x)$ also solves Laplace’s equation. By the maximum principle, $d$ takes is maximal and minimal values on the boundary. But we have $d(x)=0$ on $\partial\Omega$. Thus $d\equiv 0$ and thus $u_1(x) = u_2(x)$.

numerics dude
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