Fourier transform, Fourier integral


### Eigenvalues of Laplacian in the disk and a ball

#### From wave equation to Helmholtz equation

Consider wave equation $$u_{tt}-c^2 \Delta u =0. \label{equ-LE.1}$$ Separating $t$ from spatial variables $(x,y,z)$ we get as usual $T(t)=e^{i\omega t}$ and $v= v(x,y,z)$ satisfying $$(\Delta +\lambda )v =0 \label{equ-LE.2}$$ with the corresponding boundary condition and with $\lambda =k^2=\omega^2 c^2$. Equation (\ref{equ-LE.2}) is called Helmholtz equation.

#### Separation of variables in polar coordinates

Using $\Delta v= v_{rr}+r^{-1}v_r + r^{-2}v_{\theta\theta}$ and separating $r$ and $\theta$: $v=R(r)\Theta (\theta)$ we arrive to \begin{equation*} (R'' + r^{-1}R'+k^2R) \Theta + r^{-2} R\Theta ''=0 \end{equation*} which can be rewritten as \begin{equation*} \frac{r^2R'' + rR'+r^2k^2R}{R} + \frac{\Theta ''}{\Theta}=0 \end{equation*} and then $\frac{\Theta ''}{\Theta}=-\mu$ and since we consider a disk we must assume that $\Theta$ is $(2\pi)$-periodic and therefore $\Theta =e^{\pm im\theta}$ and $\mu =-m^2$, $m=0,1,2,\ldots$. Then $$r^2R'' + rR'+(r^2k^2 - m^2) R=0. \label{equ-LE.3}$$ In this equation parameter $k$ is superficial and we can make it $1$. Indeed, scaling $x= k r$ (it is not an original Cartesian coordinate) we observe that equation becomes $$x^2R'' + xR'+(x^2 - m^2) R=0. \label{equ-LE.4}$$ This is Bessel equation and its solutions (bounded at $0$--as our domain is a disk $\mathcal{D}=\{r < a\}$) are Bessel functions $J_m (x)$ -- see Bessel function $J_m$.

Plugging $x=kr$ we get $R(r)=J_m (kr)$ and then

1. Under Dirichlet boundary condition $u|_{r=a}=0$ we have $R(a)=0$ which means $$J_m(ka)=0 \iff k_{mn}= a^{-1} s_{mn} \label{equ-LE.5}$$ where $s_{mn}$ is $n$-th $0$ of $J_m(x)$.

2. Under Neumann boundary condition $u|_{r=a}=0$ we have $R'(a)=0$ which means $$J'_m(ka)=0 \iff k_{mn}= a^{-1} s'_{mn} \label{equ-LE.6}$$ where $s'_{mn}$ is $n$-th $0$ of $J'_m(x)$.

#### Separation of variables in spherical coordinates

Using $\Delta v= v_{\rho\rho}+2\rho^{-1}v_r + \rho^{-2}\Lambda v$ where $\Lambda$ is a Laplace-Beltrami operator on the sphere and separating $\rho$ and $(\phi,\theta)$: $v=P(\rho)Y(\phi,\theta)$ we arrive to \begin{equation*} (P'' + 2\rho^{-1}P'+k^2P) Y + \rho{-2} R\Lambda Y=0 \end{equation*} which can be rewritten as \begin{equation*} \frac{P'' + 2\rho^{-1}P'+k^2P}{P} + \frac{\Lambda Y}{Y}=0 \end{equation*} and then $\frac{\Lambda Y}{Y}=-\mu$ and since we consider a ball we must assume that $Y$ is a spherical harmonic and $\mu =-l(l+1)$, $l=0,1,2,\ldots$. Then $$\rho ^2P'' + \rho P'+(\rho ^2k^2 - l(l+1)^2) P=0. \label{equ-LE.7}$$ In this equation parameter $k$ is superficial and we can make it $1$. Indeed, scaling $x= k \rho$ (it is not an original Cartesian coordinate) we observe that equation becomes $$x^2P'' + 2xP'+(x^2 - l(l+1)^2) P=0. \label{equ-LE.8}$$ One can rewrite this equation as (\ref{equ-LE.4}) with $R=x^{\frac{1}{2}} P$ and $m=(l+\frac{1}{2}$ and therefore we arrive to Bessel functions with half-integer parameter $m$: $R= J_{l+\frac{1}{2}}(x)$ (we are looking for its solutions bounded at $0$--as our domain is a ball $\mathcal{B}=\{\rho < a\}$) are Bessel functions $J_m (x)$.

However in contrast to Bessel functions with integer parameter $l$ solutions to (\ref{equ-LE.8}) known as spherical Bessel functions $j_l(x)$ are elementary functions--see spherical Bessel functions $j_l(x)$ .

Plugging $x=k\rho$ we get $P(\rho)=j_l (k\rho)$ and then

1. Under Dirichlet boundary condition $u|_{\rho=a}=0$ we have $P(a)=0$ which means $$j_l(ka)=0 \iff k_{ln}= a^{-1} s_{mn} \label{equ-LE.9}$$ where $s_{ln}$ is $n$-th $0$ of $j_l(x)$.

2. Under Neumann boundary condition $u|_{\rho=a}=0$ we have $P'(a)=0$ which means $$j'_l(ka)=0 \iff k_{ln}= a^{-1} s'_{ln} \label{equ-LE.10}$$ where $s'_{ln}$ is $n$-th $0$ of $j'_l(x)$.

#### Separation of variables in cylindrical coordinates

Consider cylinder $\{ r< a, 0< z < b\}$ and Laplace equation in it with Dirichlet boundary conditions on both lids: $u|_{z=0}=u_{z=b}=0$. Separating variable $z$ from other spatial variables $u(x,y,z)=v(x,y)Z(x)$ we arrive to $Z''+\mu Z=0$, $(\Delta _{x,y} +\mu )v=0$ and therefore $\mu = m^2\pi^2 b^{-2}$, $Z=\sin (m\pi z/b)$ and $(\Delta _{x,y}-k^2)v =0$ with $k= m\pi b^{-1}$. Continuing further we arrive to $$r^2R'' + rR'-(r^2k^2 + m^2) R=0. \label{equ-LE.11}$$ In this equation parameter $k$ is superficial and we can make it $1$. Indeed, scaling $x= k r$ (it is not an original Cartesian coordinate) we observe that equation becomes $$x^2R'' + xR'-(x^2 + m^2) R=0. \label{equ-LE.12}$$ This equation could be reduced to the original Bessel equation by replacing $x$ by $ix$ (thus going into complex domain).