If u is a solution of $\nabla^2 u =p(\boldsymbol{x})u$ show that u is unique

Question

Let D be a bounded region in $\mathbb{R}^3$, and suppose $p(\boldsymbol{x})>0 $ ON $D$.

If u is a solution of $\nabla^2 u =p(\boldsymbol{x})u$, $x \in D$ and $\nabla u \cdot n=g(\boldsymbol{x})$, $x \in \partial D$. Show that $u$ is unique

Answer 1

We assume that $p\ge 0$ in $\mathscr{D}$.

Suppose that $u$ and $v$ are distinct solutions to $\nabla^2u=p(\vec x)u$, for $\vec x\in \mathscr{D}$, and $\frac{\partial u}{\partial n}=g$ for $\vec x\in \partial \mathscr{D}$. Then, $w=u-v$ satisfies

$$\nabla ^2 w=pw\tag 1$$

for $x\in \mathscr{D}$ and

$$\frac{\partial w}{\partial n}=0 \tag 2$$

for $\vec x\in \partial \mathscr{D}$.

Multiplying $(1)$ by $w$ and integrating over $\mathscr{D}$ reveals

$$\begin{align} \int_{\mathscr{D}}w^2\,p\,dV&=\int_{\mathscr{D}}w\nabla ^2 w\,dV \tag 3\\\\ &=-\int_{\mathscr{D}}\nabla w\cdot \nabla w\,dV \tag 4 \end{align}$$

where we used integration by parts in going from $(3)$ to $(4)$ and made use of $(2)$. Note that the left-hand side of $(3)$ is non-negative while the right-hand side of $(4)$ is non-positive. The only way these terms can be equal is if both are zero. Therefore, we conclude that $w=0$ for $x\in \mathscr{D}$, which implies $u=v$. And we are done!