[NOTES/EM-04006] Uniqueness Theorems

Uniqueness Theorem-I
Let $V$ be a volume bounded by a surface $S$. The solution of Laplace equation
\begin{align}
\nabla^2\Phi =0
\end{align}
in region $V$ subject to specified value of $\Phi$ on the boundary $S$ is unique.


In another words if $\Phi_1$ and $\Phi_2$ are two solutions of Laplace equation ($\otimes$)
\begin{align}
\nabla^2\Phi=0,
\end{align}
and of $\Phi_1=\Phi_2$ on the boundary $S$ then $\Phi_1=\Phi_2$ every where in volume $V$.

Thus if charge density is known in volume $V$ and the potential is specified on boundary $S$, then the potential is uniquely determined.

Proof of Uniqueness

Let \(\Phi_1, \Phi_2\) be two solutions satisfying the Laplace equation
\begin{equation}
 \nabla^2\Phi_1=0\qquad \nabla^2\Phi_2=0,
\end{equation}
in a volume \(V\) enclosed by a surface \(S\). Let \(\Phi_1=\Phi_2\) on the surface \(S\). Then \(\Phi=\Phi_1-\Phi_2\) satisfies the following restrictions.
\begin{eqnarray}
 \nabla^2 \Phi &=& 0 \qquad \text{in volume } V,\\
 \Phi &=& 0 \qquad \text{ on surface } S.
\end{eqnarray}
Now consider
\begin{align*}
\int_V\Phi\big(\nabla^2\Phi\big) d^3r = & \int_V
\nabla(\Phi\nabla\phi)d^3x-\int_V(\nabla\Phi)(\nabla\phi) dV\\
=&\int \hat{n}\cdot(\Phi\nabla\Phi)dS - \int (\nabla\Phi)^2dV\\
0 =& \int(\nabla\Phi)^2dV
\end{align*}
$\nabla\Phi=0$ every where inside the volume and  hence\(\Phi=\) constant. The constant here must vanish due to continuity of potential and vanishing of  \(\Phi\) on the boundary.

Uniqueness Theorem-II
In a region containing conductors and filled with a specified charge density $\rho$, the electric field is uniquely determined if the total charge on each conductor
is given.  We skip the proof of this theorem.