To find the electric potential $\phi: \mathbb{R^3} \to \mathbb{R}$ of a uniformly charged sphere with total charge $Q$ $S=\partial B_0(r)$ of radius $r > 0$ one needs to solve the Poisson differential equation $$ -\Delta \phi = \frac{Q}{\sigma(S)} \delta_S$$ on $\mathbb{R^3}$, where $\sigma$ is the surface measure. The $\delta_S$ denotes the surface dirac distribution of $S$, such that in integral form the equation becomes $$ \int_{\Omega} - \Delta \phi(x) d\lambda^3(x) = \delta_S(\Omega)Q, $$ for any sufficiently regular $\Omega$.

Now in physics classes I have seen the following argument (although with different notation):

(i) We write $-\Delta \phi = -\operatorname{div}(\nabla \phi)$, and first find $-\nabla \phi =: E$ (the electric field). With this it follows for any $\rho > r$ that $$ Q = \int_{B_0(\rho)} \operatorname{div}(E) d\lambda^3 = \int_{\partial B_{0}(\rho)} \langle E, \nu \rangle d\sigma,$$ where $\nu$ is the surface normal on $\partial B_{0}(\rho)$. So far so good.

(ii) Then a "symmetry argument" is invoked stating that since the sphere is rotationally symmetric, it should follow that its electric field $E$ is as well and so $\langle E, \nu \rangle = \| E \|$. Furthermore, by the same argument $\| E \|$ is constant on $\partial B_{0}(\rho)$, so the surface integral evaluates to $\sigma(\partial B_{0}(\rho))\| E \|,$ so in total we have $$ E(x) = \frac{Q}{\sigma(\partial B_{0}(\|x\|))} \frac{\partial}{\partial r}(x)$$ for any $x$ outside of $B_0(r).$ The potential then can easily be found from this (up to an additive constant). This is not entirely clear to me:

What we have shown is that there exists a radially symmetric solution to the Poisson equation. Is this solution unique? Or in other words, does the symmetry of the solution follow from the equation itself or do we need to maybe specify them as additional boundary conditions? What results are there for uniqueness of the Poisson equation on unbounded domains? In my PDE class we only covered the bounded case.


1 Answer 1


$$\epsilon >0$$

$$\text{if the boundary condition }\lim_{r \to \infty}|\nabla\phi(r,\theta,\varphi)| \le \frac{1}{r^{2+\epsilon}}=0 $$ $$\text{ is required, then the electric field $E=\nabla \phi(r,\theta,\varphi)$ is unique}$$

let the two solutions be $\phi_1$ and $\phi_2$ and let $\phi=\phi_1-\phi_2$

Apply Divergence theorem between a sphere of radial vector $r'$ centered on point $r$ with surface $S_1$ and another with radial vector $R$ also centered on point $r$ with surface $S_2$,$ V$ is the space between $S_1$ and $S_2$, $|R-r|>|r'-r|$

$$\int_V \nabla^2\phi dV=0=-\int_{S_1} \nabla\phi dS+\int_{S_2} \nabla\phi dS$$

$$\int_{S_1} \nabla\phi dS=\lim_{|R-r| \to \infty}\int_{S_2} \nabla\phi dS=0$$

$$\lim_{|r' - r|}\int_{S_1} \nabla\phi dS=0$$

$$\text{using continuity of }\nabla\phi \text{ it follows $\phi_1=\phi_2$}$$


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