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

## Polar Laplace Equation

Created: 2023-06-25 Sun 02:27

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