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# Method of Separation of Variables

The scope of this method is to find $u(\vec{x},t)$ satisfying: $$\left\{ \begin{array}{ll} PDE & L(u)=0 \\ BC & \alpha u+\beta \grad u\cdot \unitnormal{n}=0 \\ & \text{i.e. Dirichlet, Neuman, or Robin}\\ IC & u(\vec{x},0)=\zeta(\vec{x}) \\ & u_t(\vec{x},0)=\eta(\vec{x}) \end{array} \right\}$$ where \begin{align} &\Omega\subset\field{R}^N \\ &u(\vec{x},t):\Omega\times [0,\infty) \rightarrow\field{R} \\ &L = \color{green}{\lap}, \color{red}{\partial_t-k\lap} \text{ or } \color{blue}{\partial_{tt}-c^2\lap} \\ &\alpha,\beta:\partial \Omega\to\R \text{ or simply constant}\\ &\zeta,\eta:\Omega\to\R \\ &\unitnormal{n}= \text{ unit outward normal} \end{align} Note that the heat equation requires only one IC, and Laplace's equation is independent of time.

The steps are:

1. Assume the solution $u$ can be separated into some algebraic combination of functions with fewer argument, e.g. $u(x,t)=w(x)v(t)$.
2. Plug $u$ into BVP to obtain separated BVP e.g. for $w(x)$ and $v(t)$.
3. Solve the separated BVPs. Some of these BVP can be eigenvalue problems with many solutions. e.g. solutions are $w_\lambda(x)$ and $v(t)$.
4. Unseparate and superpose solutions with arbitrary constants e.g. $u=\sum_\lambda a_\lambda w_\lambda(x)v(t)$.
5. Solve for arbitrary constants using orthogonality and IC or BC.

First homogeneous heat and wave equations will be considered, and then Laplace's equation. More general PDE will not be discussed here.

## Heat and Wave Equations

To allow unified analysis of heat and wave equation, rewrite the PDE. $$T(u) = k \lap u$$ where $T=\color{red}{\partial_t}$ or $\color{blue}{\partial_{tt}}$ and $k=\color{red}{k}$ or $\color{blue}{c^2}$

### Separation

Assume the solution looks like $u(\vec{x},t)=v(t)w(\vec{x})$ and plug into the BVP, then divide by $kvw$. \begin{align} T(v)w &= kv\lap w \\ \frac{T(v)}{kv} &= \frac{\lap w}{w} \end{align} Both sides are independent, so they must equal a constant, say $-\lambda$. So we have two separated BVP.

### Solve Separated BVP

#### Space-part (eigenvalue problem)

$$\left\{ \begin{array}{ll} DE & \lap w = -\lambda w\\ BC & \alpha w+\beta \grad w\cdot \unitnormal{n}=0 \end{array} \right\}$$ where the BC is from plugging $u=v(t)w(\vec{x})$ to the original BC to get $v(\alpha w+\beta \grad w\cdot \vec{n})=0$, and $v\neq 0$ for nontrivial solution. If the DE has a singularity, impose bounded BC at that point. One might need to separate variables multiple times until left with an ODE for each spacial variable.

Solutions to this problem are eigenvalues $\lambda$ and corresponding eigenfunctions $w_\lambda(\vec{x})$. The properties of these eigenvalues and eigenfunctions follow.

Theorem (Sturm-Liouville): The above eigenvalue problem has

1. all real eigenvalues $\lambda$,
2. an infinite number of eigenvalues, with a smallest one, but no largest one,
3. each eigenvalue $\lambda$ corresponds to a eigenfunction $w_\lambda$,
4. eigenfunctions form a complete set (i.e. can represent any piecewise smooth function in eigenfunction series), and
5. eigenfunctions corresponding to different eigenvalues are orthogonal i.e. $\langle w_{\lambda}, w_{\lambda'} \rangle = \delta(\lambda,\lambda')$ where the brackets represent the integral over $\Omega$ of their product.
Table 1 shows solutions $w_\lambda$ for various domains and boundary conditions.

#### Time-part

$$\left\{ \begin{array}{ll} ODE & T(v) = -\lambda kv\\ IC & \text{can ignore} \end{array} \right\}.$$ Case $T=\color{red}{\partial_t}$: plug $v=e^{rkt}$ to get characteristic equation $rk=-k\lambda$. End up with $$v(t)=ce^{-\lambda kt}.$$

Case $T=\color{blue}{\partial_{tt}}$: plug $v=e^{rc^2t}$ to get characteristic eqn $r^2c^4=-c^2\lambda$. End up with $$v(t)=c_1\cos(\sqrt{\lambda}ct)+c_2\sin(\sqrt{\lambda}ct).$$

### Superposition

The most general solution is a superposition of eigenfunctions, $$u(\vec{x},t) = \sum_\lambda c_\lambda v(t) w_\lambda(\vec{x})$$ where $\lambda$ can be a multi-index with a sum over each index. For a continuous spectrum of $\lambda$, the sum becomes an integral.

### Resolve $c_\lambda$ with IC and orthogonality

Impose the initial condition and orthogonality of eigenfunctions. \begin{align} u(\vec{x},0)=\alpha(\vec{x})&= \sum_\lambda c_\lambda v(0) w_\lambda(\vec{x})\\ c_\lambda &= \frac{1}{v(0)} \frac{\langle \alpha, w_\lambda \rangle}{\langle w_\lambda,w_\lambda \rangle} \end{align} And similarly for $u_t(\vec{x},0)=\beta(\vec{x})$, if necessary.

## Laplace's Equation

Laplace's BVP is to find $u(\vec{x}):\Omega \rightarrow \field{R}$ which solves $$\left\{ \begin{array}{ll} PDE & \lap u=0 \\ BC & \alpha u+\beta \grad u\cdot \unitnormal{n}=\gamma(\vec{x}) \end{array} \right\}$$ Laplace's equation for $\Omega\subseteq\R$ degenerates to an ODE.

Since the boundary condition is nonhomogeneous, the trick is to isolate the nonhomogeneity to only one variable. This can be done by choosing an appropriate coordinate system. Also, for a polyhedron, the problem is split into $u=u_1+u_2+\cdots+u_k$, where each $u_i$ solves a related BVP with only one nonhomogeneous BC. Each $u_i$ can be solved similarly to the heat and wave eqn above, except the nonhomogeneous BC is ignored until it is used to resolve constants.

 $\left\{\begin{array}{ll} DE & \lap w = -\lambda w\\ BC & \alpha w+\beta \grad w\cdot \unitnormal{n}=0 \end{array}\right\}$ $\Omega$ BC Separated Eigenvalues $\lambda$ Eigenfunctions $w_\lambda$ $[0,L]$ $w(0)=0$ $w(L)=0$ - $\left(\frac{n\pi}{L}\right)^2$ $n=1,2,3,...$ $\sin(\sqrt{\lambda}x)$ $w_x(0)=0$ $w_x(L)=0$ - $\left(\frac{n\pi}{L}\right)^2$ $n=0,1,2,...$ $\cos(\sqrt{\lambda}x)$ $w(-L)=w(L)$ $w_x(-L)=w_x(L)$ - $\left(\frac{n\pi}{L}\right)^2$ $n=0,1,2,...$ $\left\{\begin{array}{ll} \cos(\sqrt{\lambda}x) \\ \sin(\sqrt{\lambda}x) \end{array}\right\}$ $(-\infty,\infty)$ $\lim\limits_{x\to\pm\infty}\abs{w}<\infty$ - $\lambda \geq 0$ let $\lambda=\omega^2$, $\omega\geq 0$ note: $\omega$ will be Fourier transform variable $\left\{\begin{array}{ll} \cos(\omega x) \\ \sin(\omega x) \end{array}\right\}$ or $\left\{\begin{array}{ll} e^{-i\omega x} \\ e^{i\omega x} \end{array}\right\}$ or $e^{-i\omega x}$, $\omega\in(-\infty,\infty)$ $[0,\infty)$ $w(0)=0$ $\lim\limits_{x\to\infty}\abs{w}<\infty$ - $\lambda > 0$ let $\lambda=\omega^2$, $\omega > 0$ $\sin(\omega x)$ plane: $x\in(-\infty,\infty)$ $y\in(-\infty,\infty)$ $\lim\limits_{x\to\pm\infty}\abs{w(x,y)}<\infty$ $\lim\limits_{y\to\pm\infty}\abs{w(x,y)}<\infty$ $w(x,y)=X(x)Y(y)$ $\left\{\begin{array}{ll} \text{ODE} &X_{xx}=-\lambda X \\ \text{BC} &\lim\limits_{x\to\pm\infty}X(x)=0 \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &Y_{yy}=-(\mu-\lambda) Y \\ \text{BC} &\lim\limits_{y\to\pm\infty}Y(y)=0\end{array}\right\}$ $\lambda=\omega_1^2+\omega_2^2$, $\omega_1,\omega_2\geq 0$ note: $\omega_1, \omega_2$ will be Fourier transform variable $\left\{\begin{array}{ll} \cos(\omega_1 x) \\ \sin(\omega_1 x) \end{array}\right\}\left\{\begin{array}{ll} \cos(\omega_2 y) \\ \sin(\omega_2 y) \end{array}\right\}$ or $\left\{\begin{array}{ll} e^{-i\omega_1 x} \\ e^{i\omega_1 x} \end{array}\right\}\left\{\begin{array}{ll} e^{-i\omega_2 y} \\ e^{i\omega_2 y} \end{array}\right\}$ or $e^{-i\omega_1 x}e^{-i\omega_2 y}$, $\omega_1,\omega_2\in(-\infty,\infty)$ 3d space: $x\in(-\infty,\infty)$ $y\in(-\infty,\infty)$ $z\in(-\infty,\infty)$ $\lim\limits_{x\to\pm\infty}\abs{w(x,y,z)}<\infty$ $\lim\limits_{y\to\pm\infty}\abs{w(x,y,z)}<\infty$ $\lim\limits_{z\to\pm\infty}\abs{w(x,y,z)}<\infty$ $w(x,y,z)=X(x)Y(y)Z(z)$ $\left\{\begin{array}{ll} \text{ODE} &X_{xx}=-\lambda X \\ \text{BC} &\lim\limits_{x\to\pm\infty}X(x)=0 \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &Y_{yy}=-(\mu-\lambda) Y \\ \text{BC} &\lim\limits_{y\to\pm\infty}Y(y)=0\end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &Z_{zz}=-(\nu-\mu-\lambda) Y \\ \text{BC} &\lim\limits_{z\to\pm\infty}Z(z)=0\end{array}\right\}$ $\lambda=\omega_1^2+\omega_2^2+\omega_3^2$, $\omega_1,\omega_2,\omega_3\geq 0$ $\left\{\begin{array}{ll} \cos(\omega_1 x) \\ \sin(\omega_1 x) \end{array}\right\}\left\{\begin{array}{ll} \cos(\omega_2 y) \\ \sin(\omega_2 y) \end{array}\right\}\left\{\begin{array}{ll} \cos(\omega_3 z) \\ \sin(\omega_3 z) \end{array}\right\}$ or $\left\{\begin{array}{ll} e^{-i\omega_1 x} \\ e^{i\omega_1 x} \end{array}\right\}\left\{\begin{array}{ll} e^{-i\omega_2 y} \\ e^{i\omega_2 y} \end{array}\right\} \left\{\begin{array}{ll} e^{i\omega_3 z} \\ e^{-i\omega_3 z} \end{array}\right\}$ or $e^{-i\omega_1 x}e^{-i\omega_2 y}e^{-i\omega_3 z}$ where $\omega_1,\omega_2,\omega_3\in(-\infty,\infty)$ rectangle: $x\in[0,L]$ $y\in[0,H]$ $w((x,0))=0$ $w((x,H))=0$ $w((0,y))=0$ $w((L,y))=0$ $w(x,y)=X(x)Y(y)$ $\left\{\begin{array}{ll} \text{ODE} &X_{xx}=-\mu X \\ \text{BC} &X(0)=0 \\ &X(L)=0 \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &Y_{yy}=-(\lambda-\mu) Y \\ \text{BC} &Y(0)=0 \\ &Y(H)=0 \end{array}\right\}$ $\left(\frac{n\pi}{H}\right)^2+\left(\frac{m\pi}{L}\right)^2$ $n,m=1,2,3,...$ $\sin\frac{n\pi x}{L}\sin\frac{m\pi y}{H}$ box: $x\in[0,L]$ $y\in[0,H]$ $z\in[0,D]$ $w(\partial \omega)=0$ $w(x,y)=X(x)Y(y)Z(z)$ $\left\{\begin{array}{ll} \text{ODE} &X_{xx}=-\mu X \\ \text{BC} &X(0)=0 \\ &X(L)=0 \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &Y_{yy}=-(\eta-\mu) Y \\ \text{BC} &Y(0)=0 \\ &Y(H)=0 \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &Z_{zz}=-(\lambda-\eta-\mu) Z \\ \text{BC} &Z(0)=0 \\ &Z(D)=0 \end{array}\right\}$ $\left(\frac{n\pi}{H}\right)^2+\left(\frac{m\pi}{L}\right)^2+\left(\frac{l\pi}{D}\right)^2$ $l,n,m=1,2,3,...$ $\sin\frac{n\pi x}{L}\sin\frac{m\pi y}{H}\sin\frac{l\pi z}{D}$ disk: $r\in[0,a]$ $\theta\in[-\pi,\pi]$ $w(a,\theta)=0$ $w(r,\theta)=R(r)\Theta(\theta)Z(z)$ $\left\{\begin{array}{ll} \text{ODE} &\Theta_{\theta\theta}=-\mu \Theta \\ \text{BC} &\Theta(-\pi)=\Theta(\pi) \end{array}\right\}$ Bessel problem: $\left\{\begin{array}{ll} \text{ODE} &r(rR_r)_r+(\lambda r^2-\mu)R=0 \\ \text{BC} &R(a)=0 \\ &\abs{R(0)}<\infty \end{array}\right\}$ $\mu=m^2$ $m=0,1,2,\cdots$ $\lambda_{mn}=\left(\frac{z_{mn}}{a}\right)^2$ where $z_{mn}$ is the $m$th zero of $J_{m}$ $n=1,2,3,\cdots$ $J_m(\frac{\lambda_{mn}r}{a})\left\{\begin{array}{ll} \cos m\theta \\ \sin m\theta \end{array}\right\}$ cylinder: $r\in[0,a]$ $\theta\in[-\pi,\pi]$ $z\in[0,D]$ $w(r,\theta,D)=0$ $w(r,\theta,0)=0$ $w(a,\theta,z)=0$ periodic in $\theta$ bounded at $r=0$ $w(r,\theta)=R(r)\Theta(\theta)Z(z)$ $\left\{\begin{array}{ll} \text{ODE} &Z_{zz}=(\mu-\lambda) Z \\ \text{BC} &Z(0)=0 \\ &Z(D)=0 \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &\Theta_{\theta\theta}=-\nu \Theta \\ \text{BC} &\Theta(-\pi)=\Theta(\pi) \\ &\pd{\Theta}{\theta}(-\pi)=\pd{\Theta}{\theta}(\pi) \end{array}\right\}$ Bessel problem $\left\{\begin{array}{ll} \text{ODE} &r(rR_r)_r+(\mu r^2-\underbrace{\nu}_{m^2})R=0 \\ \text{BC} &\abs{R(0)}<\infty \\ &R(a)=0 \end{array}\right\}$ $\nu=m^2$ $m=0,1,2,\cdots$ $\mu_{mn}=\left(\frac{z_{mn}}{a}\right)^2$ where $z_{mn}$ is the $m$th zero of $J_{m}$ $n=1,2,3,\cdots$ $\mu-\lambda=-\left( \frac{l\pi}{D} \right)^2$ $l=1,2,3,\cdots$ $J_m(\frac{\mu_{mn}r}{a})\left\{\begin{array}{ll} \cos m\theta \\ \sin m\theta \end{array}\right\}\sin\frac{l\pi z}{d}$ sphere: $r\in[0,a]$ $\theta\in[-\pi,\pi]$ $\phi\in[0,\pi]$ (from North pole) $w(a,\theta,\phi)=0$ $w(r,\theta,\phi)=R(r)Y(\theta,\phi)$ Spherical harmonic BVP $\left\{\begin{array}{ll} \text{ODE} &\frac{1}{\sin^2\phi}Y_{\theta\theta}+\frac{1}{\sin\phi}(\sin\phi Y_\phi)_\phi+\gamma Y=0 \\ \text{BC} &\text{Periodic, bdd at poles} \end{array}\right\}$ $Y(\theta,\phi) = \Theta(\theta)\Phi(\phi)$ $\left\{\begin{array}{ll} \text{ODE} &\Theta_{\theta\theta}=-m^2 \Theta \\ \text{BC} &\text{periodic} \end{array}\right\}$ Associated Legendre BVP $\left\{\begin{array}{ll} \text{ODE} &[(1-x^2)\Phi_x]_x+(\mu-\frac{m^2}{1-x^2})\Phi=0\\ &\quad\text{ where }x=\sin\phi, x\in{-1,1}\\ \text{BC} &\abs{\phi(x=\pm1)}<\infty \end{array}\right\}$ Nearly Bessel BVP $\left\{\begin{array}{ll} \text{ODE} &(r^2R_r)_r+(\lambda r-\underbrace{\mu}_{n(n+1)})R=0 \\ \text{BC} &R(a)=0 \\ &\abs{R(0)}<\infty \end{array}\right\}$ $\lambda_{nml}=\left(\frac{z_{nl}}{a}\right)^2$ where $z_{nl}$ is the $l$th zero of $J_{n+\frac{1}{2}}$ $m=0,1,2,\cdots$ $n=m,m+1,m+2,\cdots$ $l=1,2,3,\cdots$ $r^{-\frac{1}{2}}J_{n+\frac{1}{2}}(\lambda r) \underbrace{ \left\{\begin{array}{ll} \cos m \theta \\ \sin m\theta \end{array}\right\} P_n^m(\cos\phi) }_{ Y_n^m=\text{spherical harmonics} }$

 $\left\{\begin{array}{ll} DE & L(w) = 0\\ BC & \text{mostly homogeneous} \end{array}\right\}$ $\Omega$ BC Separated Eigenvalues Eigenfunctions $[0,L]$ $w_x(0)=c_1$ $w_x(L)=c_2$ - - $c_1+\frac{c_1-c_2}{L}x$ $w_x(0)=0$ $w_x(L)=0$ - - const rectangle: $x\in[0,L]$ $y\in[0,H]$ $w((0,y))=g(y)$ $w((L,y))=0$ $w((x,H))=0$ $w((x,0))=0$ $w(x,y)=X(x)Y(y)$ $\left\{\begin{array}{ll} \text{ODE} &X_{xx}=\lambda X \\\text{BC} & X(L)=0 \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &Y_{yy}=-\lambda Y \\ \text{BC} &Y(0)=0 \\ &Y(H)=0 \end{array}\right\}$ $\left(\frac{n\pi}{H}\right)^2$ $n=1,2,3,...$ $\sin\frac{n\pi y}{H}\sinh\frac{n\pi (x-L)}{H}$ box: $x\in[0,L]$ $y\in[0,H]$ $z\in[0,D]$ \begin{align} w((0,y,z))&=0\\ w((L,y,z))&=0\\ w((x,0,z))&=0\\ w((x,H,z))&=0\\ w((x,y,0))&=\gamma(x,y)\\ w((x,y,D))&=0 \end{align} $w(x,y)=X(x)Y(y)Z(z)$ $\left\{\begin{array}{ll} \text{ODE} &X_{xx}=-\eta X \\ \text{BC} &X(0)=0 \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &Y_{yy}=-\mu Y \\ \text{BC} &Y(0)=0 \\ &Y(H)=0 \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &Z_{zz}=(\eta+\mu) Z \\ \text{BC} &Z(0)=0 \\ &Z(D)=0 \end{array}\right\}$ $\mu=\left(\frac{n\pi}{H}\right)^2$ $\eta=\left(\frac{m\pi}{L}\right)^2$ $n,m=1,2,3,...$ $\sin\frac{n\pi x}{L}\sin\frac{m\pi y}{H}\sinh\left( \left[ \left(\frac{n\pi}{H}\right)^2 + \left(\frac{m\pi}{L}\right)^2\right] z \right)$ disk: $r\in[0,a]$ $\theta\in[-\pi,\pi]$ $w(a,\theta)=f(\theta)$ $\abs{w(0,\theta)}<\infty$ $w(r,-\pi)=w(r,pi)$ $\pd{w}{\theta}(r,-\pi)=\pd{w}{\theta}(r,pi)$ $w(r,\theta)=R(r)\Theta(\theta)$ $\left\{\begin{array}{ll} \text{ODE} &\Theta_{\theta\theta}=-\lambda \Theta \\ \text{BC} &\Theta(-\pi)=\Theta(\pi) \\ &\pd{\Theta}{\theta}(-\pi)=\pd{\Theta}{\theta}(\pi) \end{array}\right\}$ Equidimensional (Cauchy-Euler) BVP $\left\{\begin{array}{ll} \text{ODE} &r(rR_r)_r=\lambda\Theta \\ \text{BC} &\abs{R(0)}<\infty \end{array}\right\}$ $\lambda_{n}=n^2$ $n=1,2,3,\cdots$ $r^n \left\{\begin{array}{ll} \cos n\theta \\ \sin n\theta \end{array}\right\}$ cylinder: $r\in[0,a]$ $\theta\in[-\pi,\pi]$ $z\in[0,D]$ $w(r,\theta,D)=0$ $w(r,\theta,0)=0$ $w(a,\theta,z)=\gamma(\theta,z)$ bounded at $r=0$ periodic in $\theta$ $w(r,\theta)=R(r)\Theta(\theta)Z(z)$ $\left\{\begin{array}{ll} \text{ODE} &Z_{zz}=\lambda Z \\ \text{BC} &Z(0)=0 \\ &Z(D)=0 \text{ or } \eta \end{array}\right\}$ $\left\{\begin{array}{ll} \text{ODE} &\Theta_{\theta\theta}=-\mu \Theta \\ \text{BC} &\Theta(-\pi)=\Theta(\pi) \\ &\pd{\Theta}{\theta}(-\pi)=\pd{\Theta}{\theta}(\pi) \end{array}\right\}$ Bessel or modified Bessel BVP $\left\{\begin{array}{ll} \text{ODE} &r(rR_r)_r+(\lambda r^2-\underbrace{\mu}_{m^2})R=0 \\ \text{BC} &\abs{R(0)}<\infty \\ &R(a)=\gamma \text{ or } 0 \end{array}\right\}$ $\mu=m^2$ $m=0,1,2,\cdots$ $\lambda_{n}=\left(\frac{n\pi}{D}\right)^2$ $n=1,2,3,\cdots$ $I_m\left(\frac{n\pi r}{D}\right) \sin\frac{n\pi z}{D} \left\{\begin{array}{ll} \cos m\theta \\ \sin m\theta \end{array}\right\}$ $w(r,\theta,D)=\eta(r,\theta)$ $w(r,\theta,0)=0$ $w(a,\theta,z)=0$ bounded at $r=0$ periodic in $\theta$ $\mu=m^2$ $m=0,1,2,\cdots$ $\lambda_{nm}=$ $n^\text{th}$ zero of $J_m$ $n=1,2,3,\cdots$ $J_m(\sqrt{\lambda}r) \sinh(\sqrt{\lambda}z) \left\{\begin{array}{ll} \cos m\theta \\ \sin m\theta \end{array}\right\}$ sphere: $r\in[0,a]$ $\theta\in[-\pi,\pi]$ $\phi\in[0,\pi]$ (from North pole) $w(a,\theta,\phi)=\gamma(\theta,\phi)$ periodic $w(r,\theta,\phi)=R(r)Y(\theta,\phi)$ Spherical harmonic BVP $\left\{\begin{array}{ll} \text{ODE} &\frac{1}{\sin^2\phi}Y_{\theta\theta}+\frac{1}{\sin\phi}(\sin\phi Y_\phi)_\phi+\gamma Y=0 \\ \text{BC} &\text{Periodic, bdd at poles} \end{array}\right\}$ $Y(\theta,\phi) = \Theta(\theta)\Phi(\phi)$ $\left\{\begin{array}{ll} \text{ODE} &\Theta_{\theta\theta}=-m^2 \Theta \\ \text{BC} &\text{periodic} \end{array}\right\}$ Associated Legendre BVP $\left\{\begin{array}{ll} \text{ODE} &[(1-x^2)\Phi_x]_x+(\mu-\frac{m^2}{1-x^2})\Phi=0\\ &\quad\text{ where }x=\sin\phi, x\in{-1,1}\\ \text{BC} &\abs{\phi(x=\pm1)}<\infty \end{array}\right\}$ Equidimensional (Cauchy Euler) BVP $\left\{\begin{array}{ll} \text{ODE} &(r^2R_r)_r+\underbrace{\mu}_{n(n+1)}R=0 \\ \text{BC} &R(a)=0 \\ &\abs{R(0)}<\infty \end{array}\right\}$ $\lambda_{nml}=\left(\frac{z_{nl}}{a}\right)^2$ where $z_{nl}$ is the $l$th zero of $J_{n+\frac{1}{2}}$ $m=0,1,2,\cdots$ $n=m,m+1,m+2,\cdots$ $l=1,2,3,\cdots$ $r^n \underbrace{ \left\{\begin{array}{ll} \cos m \theta \\ \sin m\theta \end{array}\right\} P_n^m(\cos\phi) }_{ Y_n^m=\text{spherical harmonics} }$

# Method of Eigenfunction Expansion

The method of eigenfunction expansion generalizes separation of variables to allow some nonhomogeneities. The problem is to find $u(\vec{x},t)$ satisfying: $$\left\{ \begin{array}{ll} PDE & L(u)=f(\vec{x},t) \\ BC & \alpha u+\beta \grad u\cdot \unitnormal{n}=\gamma(\vec{x},t) \\ & \text{i.e. Dirichlet, Neuman, or Robin}\\ IC & u(\vec{x},0)=\zeta(\vec{x}) \\ & u_t(\vec{x},0)=\eta(\vec{x}) \end{array} \right\}$$ where \begin{align} &\Omega\subset\field{R}^N \\ &u(\vec{x},t):\Omega\times [0,\infty) \rightarrow\field{R} \\ &L = \color{green}{\lap}, \color{red}{\partial_t-k\lap} \text{ or } \color{blue}{\partial_{tt}-c^2\lap} \\ &f(\vec{x},t):\Omega\times [0,\infty) \rightarrow\field{R}\\ &\alpha,\beta:\partial \Omega\to\R \text{ or simply constant}\\ &\zeta,\eta:\Omega\to\R \\ &\unitnormal{n}= \text{ unit outward normal} \end{align} Note that the heat equation requires only one IC, and Laplace's equation is independent of time.

The steps are:

1. Solve related homogeneous spacial eigenvalue problem $\lap w=-\lambda w$ for eigenvalues $\lambda$ and eigenfunctions $w_\lambda$.
2. Assume the solution $u$ can be expanded in eigenfunction series where coefficients are not necessarily constant.
3. Create a problem to solve for the coefficents. This can be done in several ways.
4. Solve the resulting problem for coefficients.
First heat and wave equations will be considered, and then Poisson's equation.

## Heat and Wave Equations

To allow unified analysis of heat and wave equation, rewrite the PDE. $$T(u) = k \lap u + f$$ where $T=\color{red}{\partial_t}$ or $\color{blue}{\partial_{tt}}$ and $k=\color{red}{k}$ or $\color{blue}{c^2}$

### Related spacial eigenvalue problem

Obtain eigenvalues $\lambda$ and corresponding eighenfunctions $w_\lambda(\vec{x})$ from the related problem $$\left\{\begin{array}{ll} DE & \lap w = -\lambda w\\ BC & \alpha w+\beta \grad w\cdot \unitnormal{n}=0 \end{array}\right\}.$$ The eigenvalues and eigenvectors have all the properties listed in the section on separation of variables. See table 1 for example solutions.

### Eigenfunction Expansion

Assume that the solution can be expanded like $$u(\vec{x},t) = \sum_\lambda a_\lambda(t)w_\lambda(\vec{x}).$$ where $\lambda$ can be a multi-index with a sum over each index. For a continuous spectrum of $\lambda$, the sum becomes an integral.

### Create a problem to solve for the coefficents

The two methods are (1) using Green's formula and (2) homogenizing the BC then plugging.

#### Method 1: Green's Formula

Plug the eigenfunction expansion into time part of the PDE using term-by-term differentiation, $$\sum_\lambda T(a_\lambda)w_\lambda = k\lap u+f.$$ Employ orthogonality of eigenfunctions, $$T(a_\lambda) = \frac{k\langle \lap u,w_\lambda\rangle}{\langle w_\lambda,w_\lambda \rangle} + \frac{\langle f,w_\lambda \rangle }{\langle w_\lambda,w_\lambda \rangle}$$ Note that the brackets represent inner product, i.e. the integral over $\Omega$ of their product. To simplify this equation, recall Green's formula, \begin{align} \langle u,\lap w_\lambda\rangle - &\langle w_\lambda,\lap u\rangle \\ &= \oint_{\partial\Omega} (u\grad w_\lambda - w_\lambda\grad u)\cdot \unitnormal{n}\upd s \end{align} Use $\lap w_\lambda=-\lambda w_\lambda$ and $w_\lambda(\partial\Omega)=0$, and rearrange, \begin{align} \langle w_\lambda,\lap u\rangle = &-\lambda\langle u,w_\lambda\rangle \\ &\quad- \oint_{\partial\Omega} (u\grad w_\lambda - w_\lambda\grad u)\cdot \unitnormal{n}\upd s. \end{align} Note that for Diriclet or Neuman BC, either $w_\lambda(\partial\Omega)=0$ or $\grad w_\lambda(\partial\Omega)=0$, allowing part of the surface integral to disappear.

Plug into the equation to be simplified, employ $a_\lambda=\frac{\langle u,w_\lambda\rangle}{\langle w_\lambda,w_\lambda\rangle}$, and rearrange, \begin{align} T(a_\lambda) &+k\lambda a_\lambda \\ & = \frac{-k\oint_{\partial\Omega} (u\grad w_\lambda - w_\lambda\grad u)\cdot \unitnormal{n}\upd s}{\langle w_\lambda,w_\lambda\rangle} \\ &\quad+ \frac{\langle f,w_\lambda\rangle}{\langle w_\lambda,w_\lambda\rangle} \end{align} This is an ODE for $a_\lambda(t)$. Notice on the the right hand side, the first term is from the nonhomogeneous boundary condition and the second is from the forcing term in the PDE.

#### Method 2: Homogenize BC then plug eigenfunction expansion

Let $v(\vec{x},t)=u(\vec{x},t)-r(\vec{x},t)$ where $r$ kills the boundary values of $u$, choose the simplest $r$ possible. The resulting is a BVP with homogeneous BC. $$\left\{ \begin{array}{ll} PDE & T(v)=k\lap v+f-T(r)-k\lap r \\ BC & \alpha v+\beta \grad v\cdot \unitnormal{n}=0 \\ IC & v(\vec{x},0)=\alpha(\vec{x})-r(\vec{x},0) \\ & v_t(\vec{x},0)=\beta(\vec{x}) \end{array} \right\}$$ Plug the eigenfunction expansion $v(\vec{x},t)=\sum_\lambda a_\lambda(t) w_\lambda(\vec{x})$ into this PDE using term-by-term differentiation, $$\sum_\lambda [T(a_\lambda)+k\lambda a_\lambda]w_\lambda = f-T(r)-k\lap r.$$ Employ orthogonality of eigenfunctions, $$T(a_\lambda)+k\lambda a_\lambda = \frac{\langle f-T(r)-k\lap r,w_\lambda\rangle}{\langle w_\lambda,w_\lambda\rangle}.$$ This is an ODE for $a_\lambda(t)$. Notice that the right hand side is from the nonhomogeneity in the PDE for $v$. The nonhomogeneous boundary condition will be accounted for when the solution is recovered, $u=v+r$.

Note that coefficients $a_\lambda(t)$ from method 2 provide fast converging eigenfunction expansion since $v$ and $w_\lambda$ satisfy the same boundary conditions.

### Solve the resulting problem for coefficients

The above methods resulted in a linear homogeneous ODE with constant coefficent $k\lambda$, $$T(a_\lambda)+k\lambda a_\lambda = g(\vec{x},t),$$ where $g$ is $\frac{-k\oint_{\partial\Omega} (u\grad w_\lambda-w_\lambda\grad u)\cdot \unitnormal{n}\upd s}{\langle w_\lambda,w_\lambda\rangle} + \frac{\langle f,w_\lambda\rangle}{\langle w_\lambda,w_\lambda\rangle}$ when using Green's formulas and $\frac{\langle f-T(r)-k\lap r,w_\lambda\rangle}{\langle w_\lambda,w_\lambda\rangle}$ when using homogenized BC.

Case $T=\color{red}{\partial_t}$: use integrating factor $e^{-\lambda kt}$ to end up with $$a_\lambda(t) = a_\lambda(0)e^{-\lambda kt} + e^{-\lambda kt} \int_0^t g(\vec{x},\tau)e^{\lambda k\tau}k\tau$$ where each $a_\lambda(0)$ is found from IC and orthogonality.

Case $T=\color{blue}{\partial_{tt}}$: use variation of parameters or method of moments \begin{align} a_\lambda(t)&=a_\lambda(0)\cos(\sqrt{\lambda}ct)+a_\lambda '(0)\sqrt{\lambda}c\sin(\sqrt{\lambda}ct)\\ &\quad+\int_0^t \frac{\langle g(\vec{x},\tau),w_\lambda\rangle}{\langle w_\lambda,w_\lambda \rangle}\frac{sin(\sqrt{\lambda}c(t-\tau))}{\sqrt{\lambda}c}\upd\tau. \end{align}

In both cases, notice that the particular solution corresponds to the PDE nonhomogeneity, and $g=0$ results in homogeneous solutions.

Finally, plug $a_n$, impose IC, and employ orthogonality to obtain coefficients of homogeneous solutions. \begin{align} u(\vec{x},0)=\alpha(\vec{x})&= \sum_\lambda a_\lambda(0) w_\lambda(\vec{x})\\ a_\lambda(0) &= \frac{\langle \alpha(\vec{x}), w_\lambda(\vec{x}) \rangle}{\langle w_\lambda,w_\lambda \rangle} \end{align} And similarly for $u_t(\vec{x},0)=\beta(\vec{x})$, if necessary.

## Poisson's Equation

Poisson's equation is $$\left\{ \begin{array}{ll} PDE & \lap u = f(\vec{x}) \\ BC & \alpha u(\vec{x})+\beta \grad u(\vec{x})\cdot \unitnormal{n}=\gamma(\vec{x}) \\ \end{array} \right\}.$$ Three strategies will be presented.

#### Method 1: Green's Formula

Expand the solution in related eigenfunctions using constant coefficients, $$u(\vec{x})=\sum_\lambda a_\lambda w_\lambda(\vec{x})$$ Employ orthogonality on the eigenfunction expansion to isolate each $a_\lambda$. $$a_\lambda = \frac{\langle u,w_\lambda\rangle}{\langle w_\lambda,w_\lambda\rangle}$$

$\langle u,w_\lambda\rangle$ can be simplified using Green's formula

\begin{align} &\langle u,\lap w_\lambda\rangle-\langle w_\lambda,\lap u\rangle\\ &\quad = \oint_{\partial\Omega} (u\grad w_\lambda - w_\lambda\grad u)\cdot \unitnormal{n}\upd s \end{align} Note that for Diriclet or Neuman BC, either $w_\lambda(\partial\Omega)=0$ or $\grad w_\lambda(\partial\Omega)=0$, allowing part of the surface integral to disappear.

Plug $\lap w_\lambda=-\lambda w_\lambda$ and $\lap u=f$ and rearrange.

Now can rewrite $a_\lambda$

\begin{align} a_\lambda = -\frac{1}{\lambda} \frac{\langle w_\lambda,f\rangle + \oint_{\partial\Omega} (u\grad w_\lambda-w_\lambda\grad u)\cdot \unitnormal{n}\upd s}{\langle w_\lambda,w_\lambda\rangle} \end{align} Notice that the right hand side has terms forcing and boundary conditions.

#### Method 2: Split the problem, full dimensional eigenfunctions

The general Poisson BVP can be split into $u(\vec{x})=u_1(\vec{x})+u_2(\vec{x})$ where $u_1$ solves $$\left\{ \begin{array}{ll} PDE & \lap u_1 = 0 \\ BC & \alpha u_1(\vec{x})+\beta \grad u_2(\vec{x})\cdot \unitnormal{n}=\gamma(\vec{x}) \\ \end{array} \right\}$$ and $u_2$ solves. $$\left\{ \begin{array}{ll} PDE & \lap u_2 = f(\vec{x}) \\ BC & \alpha u_2(\vec{x})+\beta \grad u_2(\vec{x})\cdot \unitnormal{n}=0 \\ \end{array} \right\}$$ $u_1$ solves Laplace's equation [see method of separation of variables]. Solving for $u_2$ requires related eigenvalues $\lambda$ and eigenfunctions $w_\lambda$. Expand the solution $u_2$ in eigenfunction series with $a_\lambda$ constant $$u_2(\vec{x})=\sum_\lambda a_\lambda w_\lambda(\vec{x})$$ Plug into the PDE using term-by-term differentiation. $$\sum_\lambda a_\lambda \lap w_\lambda = f$$ Note that $\lap w_\lambda=-\lambda w_\lambda$. Employ orthogonality to isolate each $a_\lambda$. $$a_\lambda=-\frac{1}{\lambda} \frac{\langle f,w_\lambda\rangle}{\langle w_\lambda,w_\lambda\rangle}$$ Recover $u$ by adding $u_1$ and $u_2$

#### Method 3: Split the problem, lower dimensional eigenfunctions

Alternatively, one can solve a lower dimensional eigenvalue problem, expand $u_2$ in eigenfunction series with coefficeints which are functions of the remaining variables, and solve for these coefficents using orthogonality or Green's formula to end up with a DE for the coefficents, which must be solved. This method is more complicated, but can result in faster converging series since there are less indices to sum over.

#### Singularity $\frac{1}{\lambda}$

If $\lambda=0$ is an eigenvalue, then there is a singlularity. The following theorem clarifies this dilemma.

Theorem (Fredholm alternative):
If $\lambda=0$, then either $\langle f,w_0\rangle \neq 0$ and there are no solutions or $\langle f,w_0\rangle = 0$ and there are infinitely many solutions.

# Methods of Transformation

Consider problems with infinite or semi-infinite domains and negligible BC at infinity. $$\left\{ \begin{array}{ll} PDE & L(u)=0 \\ BC & \lim\limits_{x_i\to\infty}u=0 \text{ for unbounded} \\ & u(\vec{x},t)=\gamma(\vec{x}) \text{ for bounded} \\ IC & u(\vec{x},0)=\zeta(\vec{x}) \\ & u_t(\vec{x},0)=\eta(\vec{x}) \end{array} \right\}$$ where \begin{align} &\Omega\subset\field{R}^N, \text{ unbdd in at least one direction} \\ &u(\vec{x},t):\Omega\times [0,\infty) \rightarrow\field{R} \\ &L = \color{green}{\lap}, \color{red}{\partial_t-k\lap} \text{ or } \color{blue}{\partial_{tt}-c^2\lap} \\ &\alpha:\partial \Omega\to\R \text{ or simply constant}\\ &\zeta,\eta:\Omega\to\R \\ &\unitnormal{n}= \text{ unit outward normal} \end{align} Note that the heat equation requires only one IC, and Laplace's equation is independent of time.

The idea is to transform the Problem to a domain where it is easily solved, solve it, then transform the solution back to the original domain. The chosen transform depends on the domain,

1. Fourier transform works well for infinite domains,
2. cosine transform is useful for semi-infinite domains where the finite BC is Neumann,
3. sine transform is useful on semi-infinite domains where the finite BC is Dirichlet,
4. Spherical harmonic transform is useful for a spherical domain,
5. Hankel transform is useful for circularly symmetric problems, and
6. Abel transform is useful for spherically symmetric problems.
The steps are:
1. transform the BVP,
2. solve the resulting problem in the transform domain, and
3. backward transform the solution to regular space.

Any of these steps could be very difficult, so these methods are only usable for convenient problems.

First heat and wave equations will be considered. Then Laplace's equation.

## Heat and wave equation

To allow unified analysis of heat and wave equation, rewrite the PDE. $$T(u) = k \lap u$$ where $T=\color{red}{\partial_t}$ or $\color{blue}{\partial_{tt}}$ and $k=\color{red}{k}$ or $\color{blue}{c^2}$

This section will only consider $\Omega=\R^N$. A similar procedure is followed for other semi-infinite domains, but sine and cosine transforms are employed for those variables. The first step is to take the spacial Fourier transform, denoted with hat $\widehat{}$, of the PDE. $$\left\{ \begin{array}{ll} DE & \widehat{T(u)}(\vec{\omega},t) = k \widehat{\lap u}(\vec{\omega},t) \\ ICs & \widehat{u}(\vec{\omega},0)=\widehat{\zeta}(\vec{\omega}) \\ & \widehat{u_t}(\vec{\omega},0)=\widehat{\eta}(\vec{\omega}) \end{array} \right\}$$ The transform of the time derivative is trivial, since the spacial transform ignores the time part. The spacial transform takes its value from the table below. The PDE becomes $$T(\widehat{u}) = -k\vec{w}\cdot\vec{\omega}\widehat{u}$$ Next the ODE is solved for $T$.

Case $T=\color{red}{\partial_t}$: $$\widehat{u}=c(\omega)e^{-k\vec{\omega}\cdot\vec{\omega}t}$$ where imposing IC gives $c(\omega)=\widehat{\zeta}(\vec{\omega})$.

Case $T=\color{blue}{\partial_{tt}}$: $$\widehat{u}=a(\vec{\omega})\cos(c^2t\vec{\omega}\cdot\vec{\omega})+b(\vec{\omega})\sin(c^2t\vec{\omega}\cdot\vec{\omega})$$ where imposing IC gives $a(\vec{\omega}) = \widehat{\zeta}(\vec{\omega})$ and $b(\vec{\omega})=\frac{\widehat{\eta}(\vec{\omega})}{c^2\vec{\omega}\cdot\vec{\omega}}$

For both cases, the inverse transform of the solution is simplified using convolution theorem. \begin{align} \widehat{u}(\vec{\omega},t) &= \widehat{h}(\vec{\omega})\widehat{g}(\vec{\omega},t)\\ u(\vec{x},t) &= h(\vec{x}) \ast g(\vec{x},t) \end{align} Where $\widehat{h},\widehat{g}$ are specific to each case.

## Laplace's Problem

A similar precedure is followed for Laplace's problem $$\left\{ \begin{array}{ll} PDE & \lap u=0 \\ BC & \lim\limits_{x_i\to\infty}u=0 \text{ for unbounded} \\ & u(\vec{x},t)=\alpha(\vec{x}) \text{ for bounded} \\ \end{array} \right\}.$$ But specifics depend on the domain. For example
• half space should use one dimensional sine or cosine transform, depending on the IC,
• quarter spaces should split the problem up, and use one dimensional sine or cosine transforms, depending on each IC
• infinite strips should use a one-dimensional sine or cosine transform in the unbounded variable. This problem can also be split up.

## Expanding Functions in Some Basis

A strategy often used in math is to expand a function as a superposition of some basis functions. The choice of basis functions depends on the original function. For example, functions on $\R$ may be expanded as a superposition of basis functions $e^{i\omega x}$ over a spectrum of $\omega$. Other basis functions include $\sin(nx)$, $\cos(nx)$, $x^n$, spherical harmonics, etc. Below are four examples of expanding a function $u(x)$ in basis functions $e^{i\omega x}$ over a spectrum of $\omega$ with a weight $\widehat{u}(\omega)$ for each $\omega$.
1. For piecewise continuous function $u(x):\R\to\R$, use a continuous spectrum $\omega\in\R$, $$u(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \widehat{u}(\omega) e^{i\omega x}\upd \omega$$ where $$\widehat{u}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty u(x) e^{-i\omega x}\upd x$$ This is the Fourier transform and its inverse
2. If above $u(x)$ is also periodic with period $L$, then the above superposition degenerates to a sum over a discrete spectrum $n\in\N$ $$u(x) = a_0 +\sum_{n=1}^\infty a_n \cos \left( \frac{2\pi n}{L} x \right) + \sum_{n=1}^\infty b_n \sin \left( \frac{2\pi n}{L} x \right)$$ where \begin{align} a_0 &= \frac{1}{L}\int_{-\pi}^\pi u(x) \upd x\\ a_n &= \frac{2}{L}\int_{-\pi}^\pi u(x) \cos\left( \frac{2\pi n}{L}x \right) \upd x\\ b_n &= \frac{2}{L}\int_{-\pi}^\pi u(x) \sin\left( \frac{2\pi n}{L}x \right) \upd x \end{align} This is the Fourier series and its inverse. It looks cleaner when $L=2\pi$.
3. A discrete function $u(x_k)$, $k\in\Z$, can be expanded in a continuous spectrum $\omega\in\R$, $$u(x_k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \widehat{u}(\omega) e^{i\omega x_k}\upd \omega$$ where $$\widehat{u}(\omega) = \frac{1}{\sqrt{2\pi}}\sum_{k=-\infty}^\infty u(x_k) e^{-i\omega x_k}$$ This is the continuous Fourier transform, and its discrete inverse.
4. For a periodic discrete function $u_n=u(x_n)$ with period $L$ and sample points $x_n=\frac{nL}{N}$ for $n=0,...,N-1$, the spectral coefficients $\widehat{u}_\omega=\widehat{u}(\omega)$ are over spectum $\omega=0,...,N-1$ $$u_n = \frac{1}{\sqrt{N}}\sum_{\omega=0}^{N-1} \widehat{u}_\omega e^{i\frac{2\pi n}{N} \omega}$$ where $$\widehat{u}_\omega = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} u_k e^{-i\frac{2\pi n}{N}\omega}$$ This is the discrete Fourier transform, and its discrete inverse. Can put $x$ in the exponent using $x=\frac{nL}{N}$, and with $L=2\pi$, the exponent becomes the familiar $\pm i\omega x$.
Intuitively, the transform is finding the correlation between the original function and each basis function. These correlations are used as weights for expanding the original function in that basis.