Boundary Value Problems

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

Then, ${\mathrm{\nabla }}^2\mathrm{\Phi }=\frac{{\partial }^2\mathrm{\Phi }}{\partial x^2}+\frac{{\partial }^2\mathrm{\Phi }}{\partial y^2}=0$ So, $\mathrm{\Phi }(x,y)=X(x)Y(y)$, then $0=\frac{1}{X}\frac{{\partial }^{2}X}{\partial x^2}+\frac{1}{Y}\frac{{\partial }^{2}Y}{\partial y^2}=\frac{1}{X}\frac{d^{2}X}{d{x}^2}+\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}$. Since each partial will give a function of just $x$ or just $y$, then this must reduce to the problem where $\frac{d^{2}X}{d{x}^2}=-{\alpha }^2(x)$ and $\frac{d^{2}Y}{d{y}^2}={\alpha }^2(y)$. 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, $Y(\mathrm{\infty })=0$, $Y(y)=\sum _i{f}_i\exp (-{\alpha }_{i}y)$. 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, ${\mathrm{\Phi }}_n(x,y)=A_n\sin ({\alpha }_{n}x)\exp (-{\alpha }_{n}y)$. Then, $\mathrm{\Phi }(x,y)=\sum _n{A}_n\sin ({\alpha }_{n}x)\exp (-{\alpha }_{n}y)$.

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

${\int }_0^a\sin {\alpha }_{m}x(V)dx={\int }_0^a\mathrm{\Phi }(x,0)\sin {\alpha }_{m}xdx=\frac{a}{2}{A}_m=V\frac{-a}{m\pi }((-1{)}^m-1)=\frac{4V}{m\pi }$ for $m$ odd and zero for $m$ even.

Thus, $\mathrm{\Phi }(x,y)=\sum _{nodd}\frac{4V}{n\pi }\exp (-{\alpha }_{n}y)\sin ({\alpha }_{n}x)$.

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

${\mathrm{\nabla }}^2\mathrm{\Phi }(x,y,z)=\frac{1}{X}\frac{d^{2}X}{d{X}^2}+\frac{1}{Y}\frac{d^{2}Y}{d{Y}^2}+\frac{1}{Z}\frac{d^{2}Z}{d{Z}^2}=f(x)+g(y)+h(z)=0$. 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 ${\alpha }^2<0$, then note that these become real exponentials. Let $f(x)=-{\alpha }^2$, $g(y)=-{\beta }^2$, $h(z)={\gamma }^2$ such that ${\gamma }^2={\alpha }^2+{\beta }^2$.

${\mathrm{\nabla }}^2\mathrm{\Phi }(r,\theta ,\varphi )=0$

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

$\mathrm{\Phi }(r,\theta ,\phi )=R(r)Y(\mathrm{\Omega })$

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

$\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)=r^2\frac{d^{2}R}{d{r}^2}+2r\frac{dR}{dr}=\lambda R$. 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, $\frac{1}{Y(\mathrm{\Omega })}[\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }(\sin \theta \frac{\partial }{\partial \theta })+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\phi }^2}]Y(\mathrm{\Omega })=-\lambda Y(\mathrm{\Omega })=-\ell (\ell +1)Y(\mathrm{\Omega })$

Separating $\varphi$ dependence is as simple as multiplying by ${\sin }^2\theta$. Then, $Y(\mathrm{\Omega })=P(\theta )Q(\varphi )$

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

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

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

These give the solution, $P_{\ell }^m(x)$ for $|x|\le 1$. The solution is finite and continuous for $\ell \ge 0$ and integer. And $|m|\le \ell$.

${\int }_{-1}^1{P}_{\ell }^m(x)P_{{\ell }^{\prime }}^m(x)dx=\frac{2}{2\ell +1}\frac{(\ell +|m|)!}{(\ell -|m|)!}{\delta }_{\ell ,{\ell }^{\prime }}$

$P_{\ell }^m(x)=(1-x^2{)}^{\frac{|m|}{2}}\sum _{n=0}^{\ell -|m|}{a}_n{x}^n=\frac{(-1{)}^m}{2^{\ell }\ell !}(1-x^2{)}^{\frac{|m|}{2}}\frac{d^{m+\ell }}{d{x}^{m-\ell }}(x^2-1{)}^{\ell }$, for $x>0$

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

For $m=0$, then there is no $\varphi$ dependence. We then get the legendre polynomials.

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

$\int P_{\ell }{P}_{{\ell }^{\prime }}dx=\frac{2}{2\ell +1}{\delta }_{\ell {\ell }^{\prime }}$

The total Spherical Harmonic solution is then,

$Y_{\ell }^m(\mathrm{\Omega })=\sqrt{\frac{(2\ell +1)}{4\pi }\frac{(\ell -m)!}{(\ell +m)!}}{P}_{\ell }^m(\cos \theta )\exp (im\varphi )$, m positive. $Y_{\ell }^{-m}(\mathrm{\Omega })=(-1{)}^m{Y}_{\ell }^{m\ast }$. They have the inner product, $\int Y_{{\ell }^{\prime }{m}^{\prime }}^{\ast }{Y}_{\ell m}d\mathrm{\Omega }={\delta }_{\ell {\ell }^{\prime }}{\delta }_{m{m}^{\prime }}$

$Y(-\mathrm{\Omega })=(-1{)}^{\ell }Y(\mathrm{\Omega })$. $-\mathrm{\Omega }$ implies $\theta \to \pi -\theta$ and $\varphi \to \pi +\varphi$

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

$Y_1^{\pm 1}(\mathrm{\Omega })=\mp \sqrt{\frac{3}{8\pi }}\sin \theta \exp (\pm i\varphi )$

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