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

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


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

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

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


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

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).

If one is more interested in the electrostatic field (vs. potential), we use {\bf E} = - \nabla \Phi({\bf x}) and we can find

    \[ E_x({\bf x}) =\frac{3}{2 \pi} \int_S \Phi_S(x',y') \frac{(x-x')z}{\left( (x-x')^2 + (y-y')^2 + z^2 \right)^{5/2}} dx'dy'\]

    \[ E_y({\bf x}) =\frac{3}{2 \pi} \int_S \Phi_S(x',y')  \frac{(y-y')z}{\left( (x-x')^2 + (y-y')^2 + z^2 \right)^{5/2}} dx'dy'\]

    \[ E_z({\bf x}) = \frac{-1}{2 \pi} \int_S \Phi_S(x',y') \frac{(x-x')^2+(y-y')^2 - 2 z^2}{\left( (x-x')^2 + (y-y')^2 + z^2 \right)^{5/2}} dx'dy'\]

where again we can view these as 2D convolutions. We also note that \nabla \cdot {\bf E} = 0 can be shown for z>0, as required.

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