Boundary Value Problems

Lets say we have a spherical surface of radius \(A\) with a potential \(\Phi\) on the surface and \(\nabla^2\Phi=0\). Remember the general form for the potential is \(\Phi(\vec{x})=-\frac{1}{4\pi}\int\left[\Phi_S(\vec{x}')\frac{\partial G}{\partial n'}\right]\:da'\).

Remember, \(\nabla^2G(\vec{x}-\vec{x}')=-4\pi\delta(\vec{x}-\vec{x}')\) and \(G(\vec{x},\vec{x}')=\frac{1}{|\vec{x}-\vec{x}'|}+F(\vec{x},\vec{x}')\).

For Dirchlet boundary conditions if \(G(\vec{x},\vec{x}')=0\) either \(\vec{x}\) or \(\vec{x}\) on S. In our situation, that would mean \(r\) or \(r'=a\).

Let there be a point charge \(q\) at \(\vec{x}'\) then the Green’s function is the potential due to the charge at \(\vec{x}\). Remember that the mirror charge is \(q' = \frac{a}{r'}\) and placed at \(r''=\frac{a^2}{r'}\).

So, \(G(\vec{x},\vec{x}')=\frac{|\vec{x}-\vec{x}'|} = \frac{\frac{a}{r'}}{|\vec{x}-\frac{a^2}{r'}\hat{x}'}\). In spherical coordinates, \(G(\vec{x},\vec{x}')=\frac{1}{\sqrt{r^2+r'^2-2rr'\cos\gamma}}-\frac{1}{\sqrt{\frac{r^2r'^2}{a^2}+a^2-2rr'\cos\gamma}}\).

Plugging this in to our general form for the potential, gives us the general solution for the boundary value system.

\(\Phi(r,\theta,\varphi)= \frac{a(r^2-a^2)V}{4\pi}\int_0^{2\pi}\left[\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{r^2+a^2-2ar\cos\gamma}^3}-\int_{\frac{\pi}{2}}^\pi\frac{1}{\sqrt{r^2+a^2-2ar\cos\gamma}^3}\right]d\cos\theta'd\varphi'\frac{1}{\sqrt{r^2+a^2-2ar\cos\gamma}^3}\)

Solving Laplace’s Equation

\(\nabla^2\Phi=0\) in rectangular coordinates.


Lets say we have two infinite conducting plates that are grounded separated by a distance \(a\). Place a third plate at a potential \(V\gt 0\) that bridges the gap between them.

Then, \(\nabla^2\Phi=\frac{\partial^2\Phi}{\partial x^2}+\frac{\partial^2\Phi}{\partial y^2} = 0\) So, \(\Phi(x,y)=X(x)Y(y)\), then \(0 = \frac{1}{X}\frac{\partial^2 X}{\partial x^2} + \frac{1}{Y}\frac{\partial^2 Y}{\partial y^2} = \frac{1}{X}\frac{d^2 X}{dx^2} + \frac{1}{Y}\frac{d^2 Y}{dy^2}\). Since each partial will give a function of just \(x\) or just \(y\), then this must reduce to the problem where \(\frac{d^2 X}{dx^2}=-\alpha^2(x)\) and \(\frac{d^2 Y}{dy^2}=\alpha^2(y)\). This gives exponential solutions.

In our boundary conditions, we have a periodic boundary conditions along the x-axis. So, we have \(X(x) = \sum_i c_i\sin(\alpha_i x) + d_i\cos(\alpha_i x)\) and \(Y(y) = \sum_i e_i\exp(\alpha_i y) + f_i\exp(-\alpha_i y)\) These coefficients will be chosen so that our x-axis boundary is a square wave over a half period and the potential at the y-boundary is \(V\).

So, \(X(x) = \sum_i c_i\sin(\alpha_i x)\) Note, \(\alpha_n = \frac{n\pi}{a}\) for \(n\in\mathbb{Z}^+\).

Since, \(Y(\infty) = 0\), \(Y(y) = \sum_i f_i\exp(-\alpha_i y)\). If our $y$-boundary is placed at \(y=0\), then \(Y(0) = V\) implies that \(\sum_i f_i = V\).

Our partial solution is then, \(\Phi_n(x,y) = A_n\sin(\alpha_n x)\exp(-\alpha_n y)\). Then, \(\Phi(x,y) = \sum_n A_n\sin(\alpha_n x)\exp(-\alpha_n y)\).

The coefficients can be found by ensuring that it is a square wave of width \(a\) and height \(V\) at \(y=0\). Ansatz: \(A_n = V\frac{4}{n\pi}\)

\(\int_0^a\sin\alpha_m x(V)dx=\int_0^a \Phi(x,0)\sin\alpha_m x dx = \frac{a}{2}A_m = V\frac{-a}{m\pi}((-1)^m-1) = \frac{4V}{m\pi}\) for \(m\) odd and zero for \(m\) even.

Thus, \(\Phi(x,y) = \sum_{n odd}\frac{4V}{n\pi}\exp(-\alpha_n y)\sin(\alpha_n x)\).

For \(y\gg a\), \(\Phi(x,y)\approx\frac{4V}{\pi}\exp(-\pi y/a)\sin\frac{\pi}{a}x\).

3D Example

\(\nabla^2\Phi(x,y,z) = \frac{1}{X}\frac{d^2 X}{d X^2} + \frac{1}{Y}\frac{d^2 Y}{d Y^2} + \frac{1}{Z}\frac{d^2 Z}{d Z^2} = f(x) + g(y) + h(z) = 0\). Thus, each term has only x, y, or z dependence.

Assume \(g(y) + h(z) = \alpha^2\). Then, \(f(x)=-\alpha^2\Rightarrow X(x) = A\exp(i\alpha x) + B\exp(-i\alpha x)\). If \(\alpha^2 \lt 0\), then note that these become real exponentials. Let \(f(x) = -\alpha^2\), \(g(y) = -\beta^2\), \(h(z) = \gamma^2\) such that \(\gamma^2 = \alpha^2+\beta^2\).

  • Box

    Let the side lengths a, b, c for the x, y, z directions. Assume the top \(\Phi(x,y,z=c)=V(x,y)\) and assume the rest are grounded. Thus, \(X(x)\) and \(Y(y)\) ought to be sinusoidal (specifically cosines) and so the \(Z(z)\) should be real exponentials (sinh due to the boundary condition).

    \(X(x) = \sum_{n=0}^\infty x_n\sin\left(\frac{n\pi}{a}x\right)\)

    \(Y(y) = \sum_{m=0}^\infty y_m\sin\left(\frac{m\pi}{b}y\right)\)

    \(Z_{nm}(z) = \sinh\gamma_{nm}z\)

    \(\Phi_{nm}(x,y,z) = \sin\left(\frac{n\pi}{a}x\right)\sin\left(\frac{m\pi}{b}y\right)\sinh\gamma_{nm}z\)

    \(\Phi(x,y,z) = \sum_{n,m}a_{nm}\Phi_{nm}(x,y,z)\).

    Using our last boundary condition and the ortogonality, \(\Phi(x,y,c) = V(x,y)\), \(\iint_{0,0}^{a,b}\Phi(x,y,c)\sin(\alpha_p x)\sin(\beta_q y)dxdy = \iint_{0,0}^{a,b}V(x,y)\sin(\alpha_p x)\sin(\beta_q y)dxdy\)

    Recall: \(\gamma_{nm}^2 = \alpha_n^2 + \beta_n^2\)

    \(\frac{ab}{4}A_{pq}\sinh\gamma_{pq} = \iint_{0,0}^{a,b}V(x,y)\sin(\alpha_p x)\sin(\beta_q y)dxdy\)

    \(\int_0^a\sin\alpha_n x\sin\alpha_m x dx = \frac{a}{2}\delta_{nm}\) \(\int_0^b\sin\beta_n x\sin\beta_m x dx = \frac{b}{2}\delta_{nm}\)

Spherical Coordinates


\(\left[\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right]\Phi = 0\)

\(\Phi(r,\theta,\varphi) = R(r)Y(\Omega)\)

Then we get a radial equation, \(\frac{1}{R(r)}\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right) = \lambda\)

\(\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)=r^2\frac{d^2R}{dr^2}+2r\frac{dR}{dr}=\lambda R\). This has power series solutions since each derivative may lose a power but it regains it from the coefficient.

So, let \(R(r)=r^\alpha\). Then, \(\alpha(\alpha-1)+2\alpha-\lambda=\alpha^2+\alpha-\lambda=(\alpha-a)(\alpha-b)=\alpha^2-(a+b)\alpha+ab=0\) gives a quadratic in \(\alpha\) gives \(a+b=-1\) and \(ab=-\lambda\). Thus, there are two solutions. \(\lambda=\ell(\ell+1)\). So, \(\alpha=0\) or \(\alpha=-(\ell+1)\). Thus, \(R_\ell(r) = A_\ell r^\ell + B_\ell\frac{1}{r^{\ell+1}}\).

Angular equation, \(\frac{1}{Y(\Omega)}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right]Y(\Omega)=-\lambda Y(\Omega) = -\ell(\ell+1)Y(\Omega)\)

Separating \(\phi\) dependence is as simple as multiplying by \(\sin^2\theta\). Then, \(Y(\Omega)=P(\theta)Q(\phi)\)

\(\frac{1}{Q}\frac{d^2Q}{d\phi^2} = \alpha=2m\). Due to periodicity, \(\alpha\) must be such that we get \(Q_m(\phi)=A\sin(2m\phi) + B\cos\left(2m\phi\right)\) with \(m\) an integer. Then \(Q_m(\phi) = A\sin2m\phi + B\cos2m\phi\)

For the polar angle, we have the associated legendre polynomials, \(P_\ell^m(\theta)\) to give the total angular solution \(Y_\ell(\Omega)=P_\ell^m(\theta)Q_m(\phi)\)

We get the Associated Legendre polynomials by changing variables, \(x=\cos\theta\) with \(dx=-\sin\theta\:d\theta\). So, \(\frac{1}{\sin\theta}\frac{d}{d\theta}\left(\frac{\sin^2\theta}{\sin\theta}\frac{dP}{d\theta}\right) = \frac{d}{dx}\left((1-x^2)\frac{dP}{dx}\right) + \left[\ell(\ell+1)-\frac{m^2}{1-x^2}\right]P(x)= 0\)

These give the solution, \(P_\ell^m(x)\) for \(|x|\leq 1\). The solution is finite and continuous for \(\ell\geq 0\) and integer. And \(|m|\leq \ell\).

\(\int_{-1}^1P_\ell^m(x)P_{\ell'}^{m}(x)dx = \frac{2}{2\ell+1}\frac{(\ell+|m|)!}{(\ell-|m|)!}\delta_{\ell,\ell'}\)

\(P_\ell^m(x)=(1-x^2)^\frac{|m|}{2}\sum_{n=0}^{\ell-|m|}a_nx^n = \frac{(-1)^m}{2^\ell\ell!}(1-x^2)^\frac{|m|}{2}\frac{d^{m+\ell}}{dx^{m-\ell}}(x^2-1)^\ell\), for \(x\gt 0\)

Note \(P_{\ell}^m(-x) = (-1)^{\ell-m}P_{\ell}^m(x)\)

For \(m=0\), then there is no \(\phi\) dependence. We then get the legendre polynomials.

\(P_\ell^m(x) = \exp(-im\phi)P_{\ell}(x)\), \(P_\ell(1) = 1\).

\(\int P_\ell P_{\ell'}dx = \frac{2}{2\ell+1}\delta_{\ell\ell'}\)

The total Spherical Harmonic solution is then,

\(Y_\ell^m(\Omega) = \sqrt{\frac{(2\ell+1)}{4\pi}\frac{(\ell-m)!}{(\ell+m)!}}P_\ell^m(\cos\theta)\exp(im\phi)\), m positive. \(Y_\ell^{-m}(\Omega) = (-1)^mY_{\ell}^{m*}\). They have the inner product, \(\int Y_{\ell'm'}^*Y_{\ell m}d\Omega = \delta_{\ell\ell'}\delta_{mm'}\)

\(Y(-\Omega) = (-1)^\ell Y(\Omega)\). \(-\Omega\) implies \(\theta\to\pi-\theta\) and \(\phi\to\pi+\phi\)

Some examples, \(Y_0^0(\Omega) = \frac{1}{\sqrt{4\pi}}\)

\(Y_1^{\pm1}(\Omega) = \mp\sqrt{\frac{3}{8\pi}}\sin\theta\exp(\pm i\phi)\)

\(Y_1^0(\Omega) = \frac{3}{4\pi}\cos\theta\).

General Spherical Solutions

\(\nabla^2\Phi = 0\)

\(\Phi(r,\theta,\phi) = R(r)Y(\Omega)\), \(R_\ell(r) = Ar^\ell+\frac{B}{r^{\ell+1}}\), \(Y_\ell^m(\Omega) = N_{\ell,m}P_\ell^m(\cos\theta)\exp(im\phi)\).

So, \(\Phi(r,\theta,\phi) = \sum_{\ell=0}^\infty\sum_{m=-\ell}^\ell (A_{\ell m}r^\ell+\frac{B_{\ell m}}{r^{\ell+1}})Y_\ell^m(\theta,\phi)\).

Azimuthal Symmetry

Polar Laplace Equation

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:27