I would like to show that the Poisson's equation, i.e.,
$\nabla^2 \Phi = \rho$,
has a unique solution for given boundary conditions, namely, Dirichlet and Neumann boundary conditions.
To this end, we assume $\Phi_1$ and $\Phi_2$ are two different solutions of the Poisson's equation which satisfy the boundary conditions. Then we define $\phi = \Phi_1 – \Phi_2$ which satisfies $\nabla^2 \phi = 0$.
Now, we have:
$\nabla . (\phi \nabla \phi) = \nabla \phi . \nabla \phi + \phi \nabla^2 \phi = | \nabla \phi |^2$.
We integrate the two sides of the above over the Volume $V$ and exploit the divergence Theorem, to arrive at
$\iint_S da \, (\phi \nabla \phi) . \hat{e}_n = \iint_S da \, (\Phi_1 – \Phi_2) (\hat{e}_n . \nabla \Phi_1 – \hat{e}_n . \nabla \Phi_2) = \iiint_V dV \, |\nabla \phi|^2$.
In the case of the Dirichlet boundary condition where $\Phi_1$ and $\Phi_2$ take the same value on the bounding surface, the middle equation in the above vanishes, so we have:
$| \nabla \phi |^2 = 0$.
Thus, $\Phi_1 – \Phi_2 = \text{Constant}$, for all points in the volume $V$. Now, because $\phi$ is zero on the bounding surface (the given boundary condition), this constant must be zero. So, $\Phi_1 = \Phi_2$ for all points in the volume $V$.
We can do the same for the Neumann boundary condition, that is, the normal derivatives take the same value on the bounding surface, so again, the middle equation in the above vanishes, and thus, $\Phi_1 – \Phi_2 = \text{Constant}$. However, in this case, how can one show that the constant is zero, thus the solution is unique?
Best Answer
For Newmann BV problem, Solution is unique up to constant. So we can not show what you are saying until unless we have more data.