Solving Laplace’s equation given a surface potential

Interesting problem appeared recently. It’s equivalent to solving the electrostatic potential above a plane given the surface potential on that plane.

Say the plane is at $z=0$ and we wish to find the electrostatic potential $\Phi(x,y,z)$ for $z>0$ given $\Phi(x,y,0) = \Phi_S(x,y)$ . We start with the divergence theorem

$\int_V {\bf \nabla} \cdot {\bf A} \,d^3x = \int_S {\bf A} \cdot {\bf n} \,da$

where ${\bf A}$ is a well behaved vector field defined in the volume $V$ bounded by the closed surface $S$ . Let ${\bf A} = \phi \nabla \psi$ where $\phi$ and $\psi$ are arbitrary scalar fields. Then

${\bf \nabla} \cdot \left(\phi {\bf \nabla} \psi \right) = \phi {\bf \nabla}^2 \psi + {\bf \nabla}\phi \cdot {\bf \nabla} \psi$

and

$\phi {\bf \nabla} \psi \cdot {\bf n} = \phi \frac{\partial \psi}{\partial n}$

where $\partial/\partial n$ is the normal derivative at the surface $S$ (directed outwards from inside the volume $V$ ). Substituting the above equations back into the divergence equation, we find Green’s first identity:

$\int_V \left( \phi \nabla^2 \psi+\nabla \phi \cdot \nabla \psi \right) \, d^3x = \int_S \phi \frac{\partial \psi}{\partial n} \, da$

Interchanging $\phi$ and $\psi$ in the above and subtracting the two equations (the $\nabla \phi \cdot \nabla \psi$ term cancels) we obtain Green’s theorem (or Green’s second identity):

$\[ \int_V \left( \phi \nabla^2 \psi-\psi \nabla^2 \phi \right) \, d^3x = \int_S \left[ \phi \frac{\partial \psi}{\partial n} - \psi \frac{\partial \phi}{\partial n}\right] \, da$

We now choose $\psi = 1/R = 1/|{\bf x} - {\bf x}'|$ where ${\bf x}$ is the observation point and ${\bf x}'$ is the integration variable and let $\phi = \Phi$ , the electrostatic potential. Now use the facts that $\nabla^2 \Phi = - 4 \pi \rho$ and $\nabla^2(1/R) = - 4 \pi \delta({\bf x} - {\bf x}')$ and we find

$\int_V \left[ -4\pi \Phi({\bf x}')\delta({\bf x} - {\bf x}') + \frac{4\pi}{R}\rho({\bf x}')\right] \, d^3x' = \int_S \left[ \Phi\frac{\partial}{\partial n'} \left(\frac{1}{R}\right) - \frac{1}{R}\frac{\partial \Phi}{\partial n'}\right] \, da'$

If ${\bf x}$ lies within the volume $V$ , we obtain:

$\Phi({\bf x}) = \int_V \frac{\rho({\bf x}')}{R}\, d^3x' + \frac{1}{4 \pi} \int_S \left[ \frac{1}{R}\frac{\partial \Phi}{\partial n'} - \Phi \frac{\partial}{\partial n'}\left(\frac{1}{R}\right) \right] \, da'$

If ${\bf x}$ lies outside of $V$ then the left hand side of the above equation is 0.

This equation is not a solution of the boundary value problem but is simply an integral statement since the right hand side is an overspecification of the problem. We can define a Green’s function

$\nabla'\,^2 G({\bf x},{\bf x}') = - 4 \pi \delta({\bf x} - {\bf x}')$

where

$G({\bf x},{\bf x}') = \frac{1}{|{\bf x} - {\bf x}'|} + F({\bf x},{\bf x}')$

and the function $F$ satisfies Laplace equation inside the volume $V$ :

$\nabla'\,^2 F({\bf x},{\bf x}') = 0$

We see if we use $G$ and the freedom that $F$ allows us, we can now use Green’s theorem to eliminate one or the other surface terms. If we take $\phi=\Phi, \psi = G({\bf x},{\bf x}')$ and plug this back into Green’s theorem we find

$\Phi({\bf x}) = \int_V \rho({\bf x}')G({\bf x},{\bf x}') \, d^3x' + \frac{1}{4 \pi} \int_S \left[ G({\bf x},{\bf x}')\frac{\partial \Phi}{\partial n'} - \Phi \frac{\partial G({\bf x},{\bf x}')}{\partial n'} \right] \, da'$

Since we are interested in the Dirichlet boundary condition (that is we are given the potential on $S$ ), we demand

$G_D({\bf x}, {\bf x}') = 0$

for ${\bf x}'$ on $S$ . Then we have

$\Phi({\bf x}) = \int_V \rho({\bf x}')G_D({\bf x},{\bf x}')\, d^3x' - \frac{1}{4 \pi} \int_S \Phi({\bf x}') \frac{\partial G_D({\bf x},{\bf x}')}{\partial n'} \, da'$

If one is interested in the Neuman boundary condition (that is we have the normal component of the field on $S$ ), then we can define the corresponding Green’s function:

$\frac{\partial G_N({\bf x}, {\bf x}')}{\partial n'} = - \frac{4 \pi}{S}$

for ${\bf x}'$ on $S$ . Similarly as before, we can use Green’s theorem to find

$\Phi({\bf x}) = \langle\Phi\rangle_S + \int_V \rho({\bf x}')G_N({\bf x},{\bf x}')\, d^3x' + \frac{1}{4 \pi} \int_S \frac{\partial \Phi({\bf x}') }{\partial n'}G_N \, da'$

where $\langle\Phi\rangle_S$ is the average value of the potential over the surface- typically an irrelevant constant.

For the Dirichlet case, it can be shown via Green’s theorem, that $G_D({\bf x},{\bf x}') = G_D({\bf x}',{\bf x})$ while for the Neuman case, $G_N$ can be forced to have the same property.

In our problem, $V$ consists of the upper half infinite volume $z>0$ . If we assume that $\Phi_s$ falls to zero at large $\sqrt{x'^2 + y'^2}$ the surface integral only survives for $z=0$ . Since we are specifying the potential on the $z=0$ plane we need to find $G_D$ . By the method of images, we can immediately write it down

$G_D({\bf x},{\bf x}') = \frac{1}{|{\bf x} - {\bf x}'|} - \frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z+z')^2}}$

We see that $G_D({\bf x}, {\bf x}') = 0$ when $z' = 0$ as required. We also note that the second term on the right hand side does satisfy Laplace’s equation in $V$ . Therefore by uniqueness we have the correct Green’s function. We next need to compute

$-\nabla' G_D \cdot {\bf n'} = \frac{\partial G_D}{\partial z'} = \frac{z-z'}{|{\bf x} - {\bf x}'|^3} + \frac{z+z'}{\left( (x-x')^2 + (y-y')^2 + (z+z')^2\right)^{3/2}}$

Plugging this back into the equation for $\Phi$ and assuming there are no charges in the volume ( $\rho=0$ ) we find

$\Phi({\bf x}) = \frac{1}{2 \pi} \int_S \Phi_S(x',y') \frac{z}{\left((x-x')^2+(y-y')^2 + z^2\right)^{3/2}} \, dx'dy'$

This is the solution to our problem. We see that given $\Phi_S$ we have an integral equation to find $\Phi$ for $z>0$ . Furthermore, we see the Green’s function only depends on the differences of $x-x'$ and $y-y'$ , or equivalently, $\Phi$ is simply a 2D convolution of $\Phi_S$ and the kernel ( $z$ being viewed as simply a parameter for the surface integral to determine where in $z$ the potential should be evaluated at).