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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.





Table 1: Solutions to eigenvalue problem for various domains $\Omega$ and boundary conditions.
$\left\{\begin{array}{ll} DE & \lap w = -\lambda w\\ BC & \alpha w+\beta \grad w\cdot \unitnormal{n}=0 \end{array}\right\}$
$\Omega$ 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} } $




Table 2: Solutions for various domains and on nonhomogeneous boundary condition.
$\left\{\begin{array}{ll} DE & L(w) = 0\\ BC & \text{mostly homogeneous} \end{array}\right\}$
$\Omega$ 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.

\begin{align} &\langle u,-\lambda w_\lambda\rangle \\ &\quad= \langle w_\lambda,f\rangle+ \oint_{\partial\Omega} (u\grad w_\lambda-w_\lambda\grad u)\cdot \unitnormal{n}\upd s \end{align}

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

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.








Table 4: Various $n$-dimensional spacial Fourier transforms of function $u(\vec{x},t):\R^n\times [0,\infty)\to\R^n$
Fourier transform
$\mathscr{F}$
Cosine transform
$\mathscr{C}$
Sine transform
$\mathscr{S}$
$u(\vec{x},t)$ $\frac{1}{(2\pi)^n}\int_{\R^n} u(\vec{x},t)e^{i\vec\omega\cdot\vec{x}}\upd \vec{x}$ $\left(\frac{2}{\pi}\right)^n\int_{\R^n} u(\vec{x},t)\cos(\vec{\omega}\cdot\vec{x})\upd \vec{x}$ $\left(\frac{2}{\pi}\right)^n\int_{\R^n} u(\vec{x},t)\sin(\vec{\omega}\cdot\vec{x})\upd \vec{x}$
$\pd{u}{t}$ $\pd{\mathscr{F}[u]}{t}$ $\pd{\mathscr{C}[u]}{t}$ $\pd{\mathscr{S}[u]}{t}$
$\pd{^2 u}{t^2}$ $\pd{^2 \mathscr{F}[u]}{t^2}$ $\pd{^2 \mathscr{C}[u]}{t^2}$ $\pd{^2 \mathscr{S}[u]}{t^2}$
$\pd{u}{x_k}$ $-i\omega_k \mathscr{F}[u]$ $-\frac{2}{\pi}u(\vec{x},t)|_{x_k=0}+\omega_k \mathscr{S}[u]$ $-\omega_k \mathscr{C}[u]$
$\pd{^2 u}{x_k^2}$ $(-i\omega_k)^2 U$ $-\frac{2}{\pi}u(\vec{x},t)|_{x_k=0}+\omega_k \mathscr{C}[u]$ $-\omega_k \mathscr{S}[u]$
$\lap u=\sum_{k=1}^n \pd{^2 u}{x_k^2}$ \begin{align} &-\sum_{k=1}^n \omega_k^2 \mathscr{F}[u]\\ &\quad = (-i\sum_1^n \omega_k)^2 \mathscr{F}[u]\\ &\quad = -\vec{\omega}\cdot\vec{\omega} \mathscr{F}[u]\\ \end{align} \begin{align} \sum_{k=1}^n -\frac{2}{\pi}u(\vec{x},t)|_{x_k=0} + \omega_k \mathscr{C}[u] \end{align} \begin{align} -\sum_{k=1}^n -\omega_k \mathscr{S}[u]$ \end{align}
$e^{-\beta \vec{x}\cdot\vec{x}}$
(i.e. gaussian)
$\frac{1}{4\pi\beta}e^{-\frac{\vec\omega\cdot\vec\omega}{4\beta}}$ $\frac{2}{\sqrt{4\pi\beta}}e^{-\frac{\vec\omega\cdot\vec\omega}{4\beta}}$
convolution \begin{align} &\mathscr{F}\left[\frac{1}{2\pi}\int_\Omega f(\bar{\vec{x}})g(\vec{x}-\bar{\vec{x}}) \upd \bar{\vec{x}} \right]\\ &\quad=\mathscr{F}[f]\mathscr{F}[g] \end{align} \begin{align} &\mathscr{C}\left[ \int_\Omega f(\bar{\vec{x}})(g(\vec{x}-\bar{\vec{x}}) + g(\vec{x}+\bar{\vec{x}}) ) \upd \bar{\vec{x}} \right]\\ &\quad=\mathscr{C}[f]\mathscr{C}[g] \end{align} \begin{align} &\mathscr{S}\left[ \frac{1}{\pi} \int_\Omega f(\bar{\vec{x}})(g(\vec{x}-\bar{\vec{x}}) - g(\vec{x}+\bar{\vec{x}}) ) \upd \bar{\vec{x}} \right]\\ &\quad=\mathscr{S}[f]\mathscr{C}[g] \end{align}








Method of Green's Function

Method of Green's functions can be applied to the general problem $$ \left\{ \begin{array}{ll} PDE & L(u)=f(\vec{x},t) \\ BC & \alpha u+\beta \grad u\cdot \unitnormal{n}=\gamma(\vec{x},t) \\ & \text{i.e. Dirichlet, Neuman, or Robin}\\ IC & u(\vec{x},0)=\zeta(\vec{x}) \\ & u_t(\vec{x},0)=\eta(\vec{x}) \end{array} \right\} $$ 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} \\ &\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.

Method of Green's functions offers structure by mapping IC, BC, and forcing term $f$ to separate terms of the solution. The steps are

  1. Use Green's formula (note: Green's formula $\neq$ Green's function) to obtain general solution $u$ in terms of some function $G$, called Green's function.
  2. Find a Green's function $G$. There are many methods, resulting in different representations of $G$.
  3. Finally, plug $G$ into general solution.
Methods to find Green's function $G$ include Intuitively, $G(\vec{x},t;\vec{x_0},t_0)$ represents the effect of a concentrated source at $(\vec{x}_0,t_0)$ on the position $(\vec{x},t)$.

Wave Equation

Structure of solution $u$

To write solution $u$ in terms of $G$, forcing, BC, and IC, use Green's formula for the wave equation \begin{align} &\int_{t_i}^{t_f} \int_\Omega (uL(G)- GL(u) \upd \omega \upd t \\ &=\quad \int_\omega (uG_t-Gu_t)|_{t_i}^{t_f} \upd \omega\\ &\quad- c^2\int_{t_i}^{t_f}\oint_{\partial\Omega} (u\grad G - G\grad u)\cdot \unitnormal{n}\upd s \upd t \end{align} Plug $\lap u=f$ and $\lap G=\delta$, note that $\langle u,\delta\rangle=u$, and rearrange. \begin{align} &u(\vec{x},t) \\ &\quad =\int_{t_i}^{t_f} \int_\Omega G f(\vec{x}_0,t_0) \upd \omega_0 \upd t_0 \\ &\quad + \int_\Omega (u_{t_0}(\vec{x}_0,0)G|_{t_0=0}-u(\vec{x}_0,0)G_{t_0}|_{t_0=0} \upd \omega_0\\ &\quad - c^2\int_{t_i}^{t_f}\oint_{\partial\Omega} (u\grad G - G\grad u)\cdot \unitnormal{n}\upd s_0 \upd t_0 \end{align} The surface integral can be simplified with the appropriate BC, e.g. zero for homogeneous BC.

Method 1: Solving mostly homogeneous BVP, bringing space integral outside

Talk talk talk.

Method 2: Directly solving Green's BVP

Solving the related semi-homogeneous problem $$ \left\{ \begin{array}{ll} PDE & G_{tt}-c^2\lap G = 0 \\ BC & G=0 \\ IC & G=0 \\ & G_t=\delta{\vec{x}-\vec{x}_0} \end{array} \right\} $$ is sufficient since $G$ must work for $$ u=\int_\Omega -u(\vec{x}_0,0)G_{t_0}|_{t_0=0} \upd \omega_0 $$

Heat Equation

Structure of solution $u$

To write solution $u$ in terms of $G$, forcing, BC, and IC, use Green's formula for the heat equation \begin{align} &\int_{t_i}^{t_f} \int_\Omega (uL^*(G)- GL(u) \upd \omega \upd t \\ &\quad= -\int_\omega uG|_{t_i}^{t_f} \upd \omega\\ &\quad + k\int_{t_i}^{t_f}\oint_{\partial\Omega} (u\grad G - G\grad u)\cdot \unitnormal{n}\upd s \upd t \end{align} Where $L^*$ is the adjoint heat operator. Plug $\lap u=f$ and $\lap G=\delta$, note that $\langle u,\delta\rangle=u$, and rearrange. \begin{align} u(\vec{x},t) &= \int_{t_i}^{t_f} \int_\Omega G f(\vec{x}_0,t_0) \upd \omega_0 \upd t_0 \\ &\quad + \int_\Omega (u(\vec{x}_0,0)G|_{t_0=0} \upd \omega_0\\ &\quad + k\int_{t_i}^{t_f}\oint_{\partial\Omega} ( G\grad u-u\grad G)\cdot \unitnormal{n}\upd s_0 \upd t_0 \end{align} The surface integral can be simplified with the appropriate BC, e.g. zero for homogeneous BC.

Method 1: Solving mostly homogeneous BVP, bringing space integral outside

Talk talk talk.

Method 2: Directly solving Green's BVP

Solving the related semi-homogeneous problem $$ \left\{ \begin{array}{ll} PDE & G_t-k\lap G = 0 \\ BC & G=0 \\ IC & G=\delta{\vec{x}-\vec{x}_0} \end{array} \right\} $$ is sufficient since $G$ must work for $$ u=\int_\Omega (u(\vec{x}_0,0)G|_{t_0=0} \upd \omega_0 $$

Poisson's Equation

Consider the general Poisson BVP $$ \left\{ \begin{array}{ll} PDE & \lap u = f(\vec{x}) \\ BC & \alpha u+\beta \grad u\cdot \unitnormal{n}=\gamma(\vec{x}) \\ \end{array} \right\}. $$ First the structure of solution $u$, then four methods to solve for various representations of $G$.

Structure of solution $u$

To write solution $u$ in terms of $G$, $f$, and $\gamma$, use Green's formula (note: Green's formula $\neq$ Green's function) \begin{align} &\langle u,\lap G\rangle-\langle G,\lap u\rangle \\ &\quad= \oint_{\partial\Omega} (u\grad G - G\grad u)\cdot \unitnormal{n}\upd s \end{align} Plug $\lap u=f$ and $\lap G=\delta$, note that $\langle u,\delta\rangle=u$, and rearrange. $$ u = \langle f,G \rangle + \oint_{\partial\Omega} (u\grad G - G\grad u)\cdot \unitnormal{n}\upd s $$ The surface integral can be simplified with the appropriate BC, e.g. zero for homogeneous BC. This formula must be why Poisson's problem can be split up $u=u_1+u_2$ into mostly nonhomogeneous problems.

Method 1: Solving homogeneous BVP, interchanging sum and integral

Green's function can be ellicited from the homogeneous BC problem. So WLOG, consider $$ \left\{ \begin{array}{ll} PDE & \lap u = f(\vec{x}) \\ BC & \alpha u(\partial \Omega)+\beta \grad u(\partial \Omega)\cdot \unitnormal{n}=0 \\ \end{array} \right\}. $$ This will be solved similarly as before. First solve the related eigenvalue problem $$ \left\{ \begin{array}{ll} PDE & \lap w = -\lambda w \\ BC & \alpha w+\beta \grad w\cdot \unitnormal{n}=0 \\ \end{array} \right\} $$ for evals $\lambda$ and corresponding eigenfunctions $w_\lambda$ (see table 1). Expand $u$ in these efuncs. [justified since $G$ and $w_\lambda$ satisfy same BC]. $$ u = \sum_\lambda a_\lambda w_\lambda $$ Plug into DE, term-by-term differentiation, plug $\lap w_\lambda = -\lambda w_\lambda$ $$ \sum_\lambda -\lambda a_\lambda \lap w_\lambda = f $$ Impose orthogonality of eigenfunctions to isolate $a_\lambda $ $$ a_\lambda = -\frac{1}{\lambda} \frac{\langle f, w_\lambda \rangle}{\langle w_\lambda, w_\lambda \rangle} $$ Plug $a_\lambda $ into series for $u$, interchange $\sum$ and $\int$ $$ u = \int_\Omega f(\vec{x}) \left( \sum_\lambda -\frac{1}{\lambda} \frac{ w_\lambda(\vec{x}) w_\lambda(\vec{x}_0)}{\langle w_\lambda, w_\lambda \rangle} \right) \upd \omega_0 $$ The term in parenthesis is Green's function. Note: the singular case when $\lambda=0$ is discussed above using Fredholm alternative theorem.

This sum over all $\lambda$ may be difficult to work with. Often cant simplify the sum. Next methods will try to do better.

Method 2: Directly solving Green's BVP

Green's function $G(\vec{x},\vec{x}_0)$ is defined such that $$ \left\{ \begin{array}{ll} PDE & \lap G = \delta(\vec{x}-\vec{x}_0) \\ BC & \alpha G+\beta \grad G\cdot \unitnormal{n}=0 \\ \end{array} \right\}. $$ Properties of $G$ are continuity at $x_0$ and $\grad G$ jumps at $x_0$.

Expand $G$ in efuncs of related eigenvalue problem $\lap w=-\lambda w$ with the same BC [justified since satisfy same BC]. But leave out one of the variables and let $a_\lambda$ be a function of this remaining variable. $$ G(\vec{x}) = \sum_\lambda a_\lambda(x^*) w_\lambda(\vec{x}^*) $$ Plug into DE, term-by-term differentiation $$ \sum_\lambda \lap( a_\lambda w_\lambda) = \delta $$ Impose orthogonality of eigenfunctions to isolate $a_\lambda $ part, use $\lap w_\lambda = -\lambda w_\lambda$ \begin{align} a_\lambda '' -\lambda a_\lambda &= \delta(x^*-x_0^*)\delta(\vec{x}^*-\vec{x}_0^*) \\ a_\lambda '' -\lambda a_\lambda &= \frac{\langle \delta\delta, w_\lambda \rangle}{\langle w_\lambda, w_\lambda \rangle}\\ &=\frac{\delta(x^*-x_0^*) w_\lambda(\vec{x}_0^*)}{\langle w_\lambda, w_\lambda \rangle} \end{align} Solve ODE for $a_\lambda$ by splitting into homogeneous problems away from $\vec{x}_0$. Use BC from Green's problem. Find consts of ODE soln by using continuity of $G$ and jump of $\grad G|_{\vec{x}_0}$. Plug $a_\lambda $ into series for $G$. This $G$ should be piecewise, and have one less summation than method 1.

Method 3: Using infinite domain Green's functions

Consider two problems, the ``infinite domain'' Green's BVP $$ \left\{ \begin{array}{ll} PDE & \lap G_\infty = \delta(\vec{x}-\vec{x}_0) \\ BC & \text{negligible} \\ \end{array} \right\}, $$ and the related Laplace's problem, $$ \left\{ \begin{array}{ll} PDE & \lap v = 0 \\ BC & \alpha v+\beta \grad v\cdot \unitnormal{n}=0 \\ \end{array} \right\}. $$ Green's function is $$ G(\vec{x},\vec{x}_0) = G_\infty(\vec{x},\vec{x}_0) + v(\vec{x},\vec{x}_0) $$ $v$ can be solved for with standard separation of variables above. To solve for $G_\infty$, note that the problem is spherically symmetric, so the non-radial derivatives of the $\lap$ operator disappear. Consider the problem away from $\vec{x}_0$, $\lap G_\infty=0$. Solve the ODE in the radial variable. Resolve arbitrary consts in $\lim\limits_{r\to\infty}$ and using Green's formula. Extra restrictions as $r\to\infty$

Method 4: Using Method of Images

Modify Green's BVP to look like $$ \left\{ \begin{array}{ll} PDE & \lap G = \delta(\vec{x}-\vec{x}_0) \pm \delta(\vec{x}-\vec{x}_0^*) \\ BC & \alpha G+\beta \grad G\cdot \unitnormal{n}=0 \\ \end{array} \right\}, $$ where the $\pm$ is chosen based on the problem, and $\vec{x}_0^*$ cancels out $\vec{x}_0$. The solution will be $G(\vec{x}-\vec{x}_0) = G_\infty(\vec{x}-\vec{x}_0)\pm G_\infty(\vec{x}-\vec{x}_0^*)$. May be difficult to prove there exists such a $\vec{x}_0^*$ and provide formula for it.








Table 4: Green's functions for various wave BVP.
$\Omega$ BC $G(\vec{x},t;\vec{x}_0,t_0)$ Notes
General $\R$ $g(\vec{x}-\vec{x}_0-c(t-t_0))+h(\vec{x}-\vec{x}_0+c(t-t_0))$
  • used method of characteristics
  • $g,h$ are functions moving left and right with velocity $c$
  • $g,h$ depend on the problem
$\R$ negligible $\frac{1}{2c} ( H(x-x_0 + c(t-t_0)) - H(x-x_0 - c(t-t_0) )$
where $H$ is heavyside finction
This is a pulse expanding in both directions
$\R^2$ negligible $$ \left\{ \begin{array}{ll} 0 & \text{if } r>c(t-t_0) \\ \frac{1}{2\pi c\sqrt{c^2(t-t_0)-r^2}} & \text{if } r \lt c(t-t_0)\\ \end{array} \right. $$ where $r=\abs{\vec{x}-\vec{x}_0}$
  • Used method of descent, 3d solution with a 2d source
  • Largest at radius of expanding circle, then decays inside
$\R^3$ negligible $\frac{1}{4\pi c\rho}\delta(\rho-c(t-t_0))$
where $\rho=\abs{\vec{x}-\vec{x}_0}$
  • Used transformation to convert to 1d wave equation
  • only nonzero at radius (Huygens' principle)








Table 4: Green's functions for various heat BVP.
$\Omega$ BC $G(\vec{x},t;\vec{x}_0,t_0)$ Notes
$[0,L]$ $u(0)=0$
$u(L)=0$
$\frac{1}{\sqrt{4\pi k(t-t_0)}}\sum_{n=-\infty}^\infty \left( e^{-\frac{(x-x_0-2Ln)^2}{4k(t-t_0)} } - e^{-\frac{(x+x_0-2Ln)^2}{4k(t-t_0)} } \right)$ used method of images
$\sum_{n-1}^\infty \frac{2}{L}\sim\frac{n\pi x}{L}\sin\frac{n\pi x_0}{L} e^{-k\left(\frac{n\pi}{L}\right)^2(t-t_0) }$
  • used separation of variables then interhanging integral and sum
  • better for $t>>t_0$
$[0,\infty)$ $u(0,t)=0$
$\lim\limits_{x\to\infty}u(x,t)=0$
$\frac{1}{4\pi k(t-t_0)}\left( e^{-\frac{(x-x_0)^2}{4k(t-t_0)}} -e^{-\frac{(x+x_0)^2}{4k(t-t_0)}} \right)$
  • used method of images
$\R^N$ negligible $\left(\frac{1}{4\pi k(t-t_0)}\right)^{\frac{N}{2}} e^{- \frac{\abs{\vec{x}-\vec{x}_0}} {4k(t-t_0)} }$ used Fourier transform








Table 4: Solutions to Green's BVP related to Poisson's equation for various domains and boundary conditions.
$\left\{\begin{array}{ll} PDE & \lap G = \delta(\vec{x}-\vec{x}_0) \\ BC & \alpha G+\beta \grad G\cdot \unitnormal{n}=0 \\ \end{array}\right\}$
$\Omega$ BC $G(\vec{x}-\vec{x}_0)$ Notes
General bounded domain
e.g. $[0,H]$, rectangle, box, disk, etc.
- $\sum_\lambda -\frac{1}{\lambda} \frac{ w_\lambda(\vec{x}) w_\lambda(\vec{x}_0)}{\langle w_\lambda, w_\lambda \rangle}$ might be too much work to sum over each $\omega$
$[0,H]$ $G(0,x_0)=0$
$G(H,x_0)=0$
$\left\{\begin{array}{ll} -\frac{x}{H}(H-x_0) & x < x_0 \\ -\frac{x_0}{H}(H-x) & x > x_0 \end{array}\right\}$
  • Used method 2 above
  • Split into two ODE away from $x_0$
  • Get different linear functions for different BC
Box $G((0,y),(x_0,y_0))=0$
$G((H,y),(x_0,y_0))=0$
$\sum\frac{2\sin\frac{n\pi x_0}{H}\sin\frac{n\pi x}{L}}{n\pi\sinh\frac{n\pi L}{H}} \left\{\begin{array}{ll} \sinh\frac{n\pi (y_0-L)}{H} \sinh\frac{n\pi y}{H} & y < y_0 \\ \sinh\frac{n\pi (y-L)}{H} \sinh\frac{n\pi y_0}{H} & y > y_0 \\ \end{array}\right\} $
  • Used method 2
  • Can flip $x$ and $y$ and get similar result
$(-\infty,\infty)$ - - -
$\R^2$ negligible $\frac{1}{2}\ln\abs{\vec{x}-\vec{x}_0}+c$ used strategy in method 3 above
$\R^3$ negligible $\frac{1}{3\pi\abs{\vec{x}-\vec{x}_0}}$ used strategy in method 3 above
Disk $G(\vec{x},\vec{x}_0)=0$ on $\abs{\vec{x}-\vec{x}_0}=a$ $\frac{1}{3\pi}\left( a^2 \frac{r^2+r_0^2 -2rr_0\cos\theta}{r^2r_0^2+a^4-2rr_0a^2\cos\theta} \right)$
where $\phi=$ angle between $\vec{x},\vec{x}_0$
$r=\abs{\vec{x}}$
$r_0=\abs{\vec{x}_0}$
used method 4 above