Elliptic PDE Uniqueness This is a problem which I am unable to solve. The problem is from an old competitive exam:
Let $G$ be a closed curve in xy-plane and let $S$ denote the region bounded by $G$. Let:
$ \frac{\partial ^2w}{\partial x^2} + \frac{\partial ^2w}{\partial y^2} = f(x, y) \text{ } \forall (x, y) \text{ in } S$.
If $f$ is prescribed at each point $(x, y)$ of $S$ and $w$ is prescribed on boundary $G$ of $S$, then prove that the solution $w = w(x, y)$ is unique.
This is an elliptic PDE but I couldn't come up with any proof. I tried using Monge's method but the roots were imaginary. Can someone provide references/solution to this problem?
Thanks
 A: Suppose that $w_1$ and $w_2$ both satisfy your PDE and boundary conditions and let $u :=w_1-w_2$. Then \begin{align*}\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}&= \frac{\partial^2w_1}{\partial x^2}+\frac{\partial^2w_1}{\partial y^2}-\frac{\partial^2w_2}{\partial x^2}-\frac{\partial^2w_2}{\partial y^2}\\&=f(x,y)-f(x,y)\\&=0.\end{align*} Since $w_1$ and $w_2$ are the same on the boundary, $u=0$ on $G$. Hence, uniqueness is equivalent to showing that the only solution to $$ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} =0  \text{ in } S, \quad u=0 \text{ on } G$$ is $u \equiv 0$. There are two standard ways to prove this (both can be found in practically any textbook on elliptic PDE, see for example Chapter 2 in Partial Differential Equations by Larence C. Evans). The first way is by the maximum principle. The second way is via integration by parts as follows. Multiply the above equation by $u$ then integrate by parts: \begin{align*}
0 &= \int_S u\bigg (\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} \bigg ) dxdy= - \int_S \bigg (\frac{\partial u}{\partial x} \bigg)^2+\bigg (\frac{\partial u}{\partial y} \bigg)^2 dxdy + \int_G u (\nabla u \cdot \nu) ds 
\end{align*} where $\nu$ is the outward pointing normal to $S$. Since $u=0$ on $G$ the boundary integral vanishes, so we have that$$0 = - \int_S \bigg (\frac{\partial u}{\partial x} \bigg)^2+\bigg (\frac{\partial u}{\partial y} \bigg)^2 dxdy \leqslant 0. $$ Hence, $$  \bigg (\frac{\partial u}{\partial x} \bigg)^2+\bigg (\frac{\partial u}{\partial y} \bigg)^2=0$$ so $u$ is a constant. Since $u=0$ on the boundary, we conclude that $u\equiv 0$.
