Let $\rho(\mathbf x)$ be a function on a volume $V$ and $f(\mathbf x)$ a function on its boundary $S=\partial V$. Show that a solution $\phi (\mathbf x)$ to the following problem is unique: $\nabla^2 \phi -\phi=\rho$ on $V$ , $\frac{\partial \phi}{\partial n}=f$ on $S$.
I would usually put my thought so far, but I have none.
Of course, anything is appreciated, but hints may be wasted on me, because at this point I think there's no hope of me getting anywhere here, so I pose this as a question for anyone to tackle for their own sake, rather than assisting me. Having said that a full solution to look at would be useful for me.
Thanks
Suppose $\phi_1$ and $\phi_2$ are both solutions to the Neumann problem \begin{align} \begin{cases} \nabla^2 \phi - \phi = \rho & \text{ in } V\\ \frac{\partial \phi}{\partial n} = f & \text{ on } \partial V \end{cases} \ \ (*) \end{align} then $\psi:=\phi_1-\phi_2$ is a solution to the problem \begin{align} \begin{cases} \nabla^2 \psi - \psi = 0 & \text{ in } V\\ \frac{\partial \psi}{\partial n} = 0 & \text{ on } \partial V \end{cases}. \ \ (**) \end{align}
Now, multiply the PDE in $(**)$ by $\psi$ and perform integration by parts yields \begin{align} \int_V \psi\nabla^2 \psi - \psi^2\ dx = \int_{\partial V} \psi\frac{\partial \psi}{\partial n}\ dS-\int_V |\nabla\psi|^2+\psi^2\ dx = 0. \end{align} Since $\frac{\partial \psi}{\partial n}=0$, then it follows \begin{align} \int_V |\nabla \psi|^2+\psi^2\ dx = 0 \end{align} which is only possible provided $\psi=\phi_1-\phi_2 \equiv 0$ on $V$. Hence $\psi_1\equiv \psi_2$ on $V$. Thus we have our uniqueness result.