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.

Example

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

$\nabla^2\Phi(r,\theta,\phi)=0$

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

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

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