Maxwell’s Equations

Say we have Dirchlet (the potential on the surface, \(\Phi\) on \(S\)) or the Neumann condition (the derivative on the surface, \(\frac{\partial\Phi}{\partial n}=\nabla\Phi\cdot\hat{n}\)) then you can uniquely determine the potential inside the region.

Let \(\Phi,\psi\) be scalar field and let \(\vec{A}=\phi\nabla\psi\) be a vector field. Then, \(\int\nabla\cdot(\phi\nabla\psi)dV=\oint_S\phi(\nabla\psi)\hat{n}da\). By the chain rule, we have \(\int(\nabla\phi\nabla\psi+\phi\nabla^2\psi)dV = \oint_S\phi\frac{\partial\psi}{\partial n}da\).

Let \(\vec{B}=\psi\nabla\phi\) be another vector field. Then, \(\int(\nabla\psi\nabla\phi+\psi\nabla^2\phi)dV = \oint_S\psi\frac{\partial\phi}{\partial n}da\).

Subtracting these two expressions, we have \(\int_V(\psi\nabla^2\phi-\phi\nabla^2\psi)dV = \oint_S\left(\phi\frac{\partial\Psi}{\partial n}-\psi\frac{\partial\phi}{\partial n}\right)da\).

Let \(\psi=\frac{1}{|\vec{x}-\vec{x}'|} = \frac{1}{R}\) and \(\phi=\Phi\).

\(\nabla^2\frac{1}{r} = -4\pi\delta^3(\vec{x}-\vec{x}')\)

Integrating over \(x'\), \(\int_V\left(\frac{1}{R}\frac{\rho}{\varepsilon_0}-\Phi(\vec{x}')(-4\pi\delta^3(R))\right)\:dV = \int_V\left(\frac{1}{R}\frac{\rho}{\varepsilon_0}\right)\:dV +4\pi\Phi(\vec{x}) = \int_S\left(\Phi(\vec{x}')\frac{\partial}{\partial n'}\frac{1}{R}-\frac{1}{R}\frac{\partial\Phi}{\partial n'}\right)\:da'\). Thus, \(\Phi(\vec{x}) = \frac{1}{4\pi\varepsilon_0}\int_V\frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}\:d^3x' + \int_S\cdots da'\) Note: We integrate over the enclosed volume and we must know both BC in this formulation.

Note: \(\frac{\partial\psi}{\partial n} = \nabla\psi\cdot\hat{n}\).

1. \(-\nabla^2\Phi=-\frac{\rho}{\varepsilon_0}\) and either of the Dirchlet or Neumann boundary conditions are satisfied, then \(\Phi\) is unique.
2. …

Proof of 1. Assume we have \(\Phi_1,\Phi_2\) which satisfy the equation and the BC. Then, let \(U=\Phi_2-\Phi_1\). So, \(\nabla^2U=0\), satisfying Laplace’s equation and at the surface since both boundaries are the same, then either \(U=0\) on the surface or the derivative is zero on the surface. So, \(\int_V(U\nabla^2+\nabla U\cdot\nabla U)dV = \int_S U\left(\frac{\partial U}{\partial n}\right)d a = 0\). So, \(\int|\nabla U|^2 dV=0\). Then, \(\nabla U=0\) implies that \(U(x)=C\). If we have Dirchlet, \(C=0\) since the boundaries must be equal otherwise it could be some arbitrary \(C\). QED.

G(\vec{x},\vec{x}’) = G(\vec{x}’,\vec{x})

1. \(\nabla^2G(\vec{x},\vec{x}')=-4\pi\delta(\vec{x}-\vec{x}')\)
2. \(G(\vec{x},\vec{x}) = \frac{1}{|\vec{x}-\vec{x}'|} + F(\vec{x},\vec{x}')\) such that \(\nabla'^2F=0\) inside \(V\). I.e. homogeneous in \(V\).

Types:

1. Dirchlet: \(G_D=0\) on surface
2. Neumann: \(G_N(\vec{x},\vec{x}')\) \(\frac{\partial G_N}{\partial n}=-\frac{4\pi}{A_S}\).

\(\nabla'^2G_N=-4\pi\delta(\vec{x}-\vec{x}'), \int_V\nabla'(\nabla G_N)dV = -4\pi = \int_S(\nabla' G_N)\hat{n}dS\). So, \(\frac{\partial G_N}{\partial n}=C\) such that \(C\) times the area of the surface is equal to \(-4\pi\).

\(\Phi(\vec{x})=\frac{1}{4\pi\varepsilon_0}\int\rho(\vec{x}')G(\vec{x},\vec{x}')d^3x' + \frac{1}{4\pi}\int_S\left[G(\vec{x},\vec{x}')\frac{\partial\Phi}{\partial n}-\Phi(\vec{x}')\frac{\partial G}{\partial n'}\right]\)

If we have Dirchlet BC. Then note that \(G_D=0\) implies that the term with the Neumann boundary condition vanishes and so we can solve for the Potential exactly without knowledge of the Neumann BC. I.e. \(\Phi(\vec{x})=\frac{1}{4\pi\varepsilon_0}\int_V\rho(\vec{x}')G_D\:dV-\frac{1}{4\pi}\int_S\Phi(\vec{x}')\frac{\partial G_D}{\partial n'}\:da'\).

Similarly, if we have Neumann BC, then \(\partial_n G_N=\frac{-4\pi}{A_S}\) implies …