In Laplace's equation, why is it that there is only one solution for a particular boundary value? 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.
 A: 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. $\qquad \rule{0.7em}{0.7em}$
A: 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 $u_1(x) = u_2(x)$.
