Solve the wave equation for waves travelling in the surface of a sphere $f(\theta,\phi ,t)$ where $f$ is a small amplitude vibration
we take $r=a$
Start with the general form of $\nabla^2$ in spherical polar coordinate
$$ \nabla^2 f=\frac{1}{r^2}\frac{\partial}{\partial r}\left ( r^2 \frac{\partial f}{\partial r} \right )+\frac{1}{r^2 \sin \theta}\frac{\partial }{\partial \theta}\left ( \sin \theta \frac{\partial f}{\partial \theta} \right ) +\frac{1}{r^2 \sin^2 \theta} \frac{\partial ^2 f}{\partial \phi^2}=\frac{1}{c^2}\frac{\partial^2 f}{\partial t^2} $$
where $f=f(\theta,\phi,t)$ is the amplitude of the disturbance at $(a,\theta,\phi)$
Hence we have
$$ \cancel{\frac{1}{r^2}\frac{\partial}{\partial r}\left ( r^2 \frac{\partial f}{\partial r} \right )}+\frac{1}{a^2 \sin \theta}\frac{\partial }{\partial \theta}\left ( \sin \theta \frac{\partial f}{\partial \theta} \right ) +\frac{1}{a^2 \sin^2 \theta} \frac{\partial ^2 f}{\partial \phi^2}=\frac{1}{c^2}\frac{\partial^2 f}{\partial t^2} $$
we need to solve an equation in the form
$$ \frac{1}{a^2 \sin\theta}\frac{\partial}{\partial \theta}\left ( \sin\theta\frac{\partial f}{\partial \theta} \right ) +\frac{1}{a^2 \sin^2 \theta} \frac{\partial ^2 f}{\partial \phi^2}=\frac{1}{c^2}\frac{\partial^2 f}{\partial t^2} $$
We will use separation of variables twice 🗒️ Note: we have a boundary condition that the function is regular everywhere and periodic in $\phi$
$$ f(\theta,\phi,t)=Y(\theta,\phi)T(t) $$
noting we will separate $Y(\theta,\phi)$ in a moment. If we substitute in and divide by $YT$ we obtain
$$ \frac{1}{Ya^2 \sin\theta}\frac{\partial}{\partial \theta}\left ( \sin\theta\frac{\partial Y}{\partial \theta} \right ) +\frac{1}{Ya^2 \sin^2 \theta} \frac{\partial ^2 Y}{\partial \phi^2}=\frac{1}{c^2T}\frac{\text d^2 T}{\text d t^2}=\underbrace{-\frac{\ell(\ell+1)}{a^2}}_{=-k^2} $$
🗒️ Note: the constant is like this for a reason which will become apparent later
solving for $T$ we get
$$ T(t)=A\cos\omega t+B\sin\omega t $$
where
$$ \omega ^2=\frac{c^2}{a^2}l(l+1) $$
The angular equation for $Y$ is
$$ \frac{1}{ \sin\theta}\frac{\partial}{\partial \theta}\left ( \sin\theta\frac{\partial Y}{\partial \theta} \right ) +\frac{1}{ \sin^2 \theta} \frac{\partial ^2 Y}{\partial \phi^2}=-\ell(\ell+1)Y $$
🗒️ Note: the eigenfunctions of this equation $Y(\theta,\phi)$ are known as spherical harmonics.
We use separation of variables again $Y(\theta,\phi)=\Theta(\theta)\Phi(\phi)$
$$ \underbrace{\frac{\sin\theta}{ \Theta}\frac{\text d}{\text d \theta}\left ( \sin\theta\frac{\text d \Theta}{\text d \theta} \right ) +\ell(\ell+1)\sin^2\theta}{=+m^2}+\underbrace{\frac{1}{\Phi} \frac{\text d ^2 \Phi}{\text d \phi^2}}{=-m^2}=0 $$
We choose to pick out the solutions which do not depend on $\phi$ ($m=0$)
so for the $m=0$ we are looking for $f(\theta,\phi,t)$ and
$$ f(\theta,t)=\Theta(\theta)T(t) $$
So our angular equation is of the form
$$ \frac{\sin\theta}{ \Theta}\frac{\text d}{\text d \theta}\left ( \sin\theta\frac{\text d \Theta}{\text d \theta} \right ) +\ell(\ell+1)\sin^2\theta=0 $$
which we can write in the form of a Legendre’s equation but substituting $x=\cos\theta$ to obtain
$$ \frac{\text d}{\text dx}\left ( (1-x^2)\frac{\text d\Theta}{\text d x}\right )+\ell(\ell+1)\Theta=0 $$
we need $\ell\in\Z$ to obtain regular solution and we get
$$ \Theta=P_\ell(x) $$
Now the general solution is
$$ f(\theta,t)=\sum^\infin_{l=0}P_\ell(\cos\theta)(a_\ell\cos\omega_\ell t+B_\ell \sin\omega _\ell t) $$
where
$$ \omega_\ell=\frac{c}{a}\sqrt{\ell(\ell+1)} $$
To fix our constants we can use the initial conditions. They are
$$ \begin{aligned} f(\theta,0)&=g(x) \\ \dot f(\theta,0)&=0 \end{aligned} $$
remembering $x=\cos\theta$.
The second condition fixed $B_\ell=0$. Then we can fix $A_\ell$ as
$$ g(x,0)=\sum^\infin_{\ell=0}P_\ell(x)A_\ell $$
and hence
$$ \int^{+1}{-1}g(x)P\ell (x)\,\text dx=\int^{+1}{-1} P^2\ell (x) \,\text dx A_\ell=\frac{2 A_\ell}{2l+1} $$
The second integral is equal to $2/(2l+1)$ from the properties of Legendre polynomials we also used the Legendre polynomial orthogonality integral
$$ \int^{+1}{-1}P\ell(x)P_n(x)\,\text dx=0 \quad \forall l\ne n $$
This gives an integral expression for $A_n$ in exactly the same way as the expression we are used to for Fourier series
$$ A_\ell=\frac{2l+1}{2}\int^{+1}{-1}g(x)P\ell(x)\,\text dx $$
An actual problem requires specifying $g(x)$ and then computing the coefficients $A_\ell$
💼 Case: consider an asteroid impact on earth, We assume initial conditions are a Dirac delta function and write the impact as
$$ g(x)=I\delta(x-1) $$
which models a huge disturbance in a very localised area and in which $x=\cos\theta$. Our system is axially symmetric and the location fo the impact is $x=1$ meaning the $z$-axis passes through the impact sight.
As we have a non-zero $g(x)=f(x,t=0)$ we get $B_\ell$ coefficients are zero (for $\dot f(x,0)=0$). We now try and find $A_\ell$
$$ A_\ell=\frac{2l+1}{2}\int^{+1}{-1}g(x)P\ell(x)\,\text dx $$
giving
$$ A_\ell=\frac{2l+1}{2}I\int^{+1}{-1}\delta(x-1)P\ell(x)\,\text dx $$
Here we used the fact that
$$ \int^{+1}{-1}P^2\ell (x)\,\text dx=\frac{2}{2l+1} $$
This gives for coefficients
$$ A_\ell=\frac{2l+1}{2}I $$
as $P_\ell(1)=1$
The resulting surface waves ( earthquake) is
$$ f(\theta,t)=\frac{I}{2}\sum^\infin_{l=0}(2l+1)P_\ell(x)\cos\omega_\ell t $$
with
$$ \omega_\ell =\frac ca \sqrt{l(l+1)} $$
🗒️ Note:
$l/a$ plays the role of $k$ at large $l$. This gives the wave as a function of $\theta$ and $t$.
On the opposite side of the globe at $x=-1$ here we have $P_\ell(-1)=(-1)^l$ so terms generally oscillate in sign and interfere destructively except when
$$ \cos(\omega _\ell t)=(-1)^l $$
when they add constructively. This will occur at $\omega_\ell t=l\pi$
For large $l$ we have
$$ \omega_\ell\sim cl/a $$
This means $t=\pi a/c$ and is the time taken to travel halfway around the globe at speed $c$.
We obtain the following separated equation for the angular part
$$ \frac{\sin\theta}{\Theta}\frac{\text d}{\text d\theta}\left ( \sin(\theta) \frac{\text d \Theta}{\text d \theta} \right ) +l(l+1)\sin^2\theta+\frac{1}{\Phi}\frac{\text d^2 \Phi}{\text d\phi^2}=0 $$
We solve for $\Phi(\phi)$
$$ A\cos m\phi+B\sin m\phi $$
with $m=0,1,2,3$ due to periodic boundary conditions
🗒️ Note: this could be written in exponential form as
$$ \Phi(\phi)=A\exp(im\phi) $$
with $m=-2,-1,0,1,2$
The equation for $\Phi$ can be rewritten
$$ \frac{1}{\sin\theta}\frac{\text d}{\text dx} \left ( \sin\theta \frac{\text d \Theta}{\text d \theta} \right ) +\left [ l(l+1)-\frac{m^2}{\sin^2\theta} \right ]\Theta=0 $$
To solve we write $x=\cos\theta$
$$ \frac{\text d}{\text dx} \left ( (1-x)^2\frac{\text d\Theta}{\text d x} \right )+\left ( l(l+1)-\frac{m^2}{1-x^2} \right )\Theta=0 $$
if we set $m=0$ e get Legendre’s equation
$$ \frac{\text d}{\text dx}\left ( (1-x)^2\frac{\text d\Theta}{\text dx} \right )+l(l+1)\Theta=0 $$
This give Legendre polynomials $P_\ell(x)$ however we can have $m\ne0$ and these solutions will be called associated Legendre polynomials obeying
$$ \frac{\text d}{\text dx} \left ( (1-x)^2\frac{\text d\Theta}{\text d x} \right )+\left ( l(l+1)-\frac{m^2}{1-x^2} \right )\Theta=0 $$
These solutions are called associated Legendre functions ( $P^m_\ell (x))$ and carry two indices. We need the condition $|m|\le l$
The first four spherical harmonics $Y(\theta,\phi)$ are given
$$ \begin{aligned} Y^0_0(\theta,\phi)&=\frac 12 \frac{1}{\sqrt{\pi}} \\ Y^0_{-1}(\theta,\phi)&=\frac 12 \sqrt{\frac{3}{2\pi}} \sin(\theta) \, e^{-i\phi} \\ Y^1_{0}(\theta,\phi)&=\frac 12 \sqrt{\frac{3}{\pi}} \cos(\theta) \\ Y^1_{1}(\theta,\phi)&=-\frac 12 \sqrt{\frac{3}{2\pi}} \sin(\theta) \, e^{i\phi}
\end{aligned} $$
💼 Case: wish to solve for waves in a 3D spherical system, (example a sound waves in a spherical cavern) where we solve for pressure. This pressure is a function of $r, \theta,\phi,t$