The Wave Equation
Wave Equation on a Disk
Consider a thin circular membrane free to vibrate except at the edge (like a drum). We will solve the wave equation
$$ \ddot{u} = c^2\nabla^2 u $$where \( u(r, \theta, t) \) describes the displacement of the membrane, for \( r < R \), with boundary conditions
$$ \frac{\partial u}{\partial r}(R,\theta,t) = -\mu u(R,\theta,t) \quad \: \text{(boundary has some give)} $$and initial conditions
$$ u(r,\theta,0) = u_0(r,\theta) \quad \: \text{(a general deformation)} $$ $$ \frac{\partial u}{\partial t}(r,\theta,0) = 0 \quad \: \text{(motionless)} $$We know that the eigenvectors of the Laplacian, \( \nabla^2 \), on a disk are of the form
$$ u_n(r,\theta) = J_n(k r) e^{in\theta} $$with eigenvalues \( \lambda = - k^2 \).
These are the natural vibration modes of the disk (see why).
The boundary condition gives
$$ \frac{\partial}{\partial r} J_n(k r) \Bigg|_{r=R} = -\mu J_n(k R) $$Since
$$ \frac{\partial}{\partial r} J_n(k r) = k J_n'(k r) $$we get
$$ k J_n'(k R) = -\mu J_n(k R) $$The allowed values \( k \) are found by solving
$$ k J_n'(k R) + \mu J_n(k R) = 0 $$For each angular mode \( n \), this equation has a discrete sequence of solutions. We denote these solutions by \( k_{n,m} \) where \( m = 1,2,3,... \) labels the successive roots.
$$ u_{n,m}(r,\theta) = J_n (k_{n,m} r) e^{in\theta} $$The boundary condition therefore selects a discrete set of allowed vibration frequencies.
Since the Laplacian is Hermitian under the following inner product
$$ \langle u_{n,m},u_{a,b}\rangle = \int_0^{2\pi} \int_0^R u_{n,m}^*(r,\theta) u_{a,b}(r,\theta) r\,dr\,d\theta = \delta_{na}\delta_{mb}\|u_{n,m}\|^2 $$the eigenfunctions are orthogonal. The solution is obtained by decomposing the initial displacement into Laplacian eigenmodes on the disk. We expand the solution as
$$ u(r,\theta,t) = \sum_{n=-\infty}^{\infty} \sum_{m=1}^{\infty} u_{n,m}(r,\theta) \psi_{n,m}(t) $$That is,
$$ u(r,\theta,t) = \sum_{n=-\infty}^{\infty} \sum_{m=1}^{\infty} J_n(k_{n,m} r) e^{in\theta} \psi_{n,m}(t) $$Each spatial eigenmode evolves independently in time.
Substitute this into the wave equation
$$ \ddot{u} = c^2\nabla^2 u $$ $$ \sum_{n,m} u_{n,m}(r,\theta) \psi_{n,m}''(t) = c^2 \sum_{n,m} \nabla^2 u_{n,m}(r,\theta) \psi_{n,m}(t) $$Since \( \nabla^2 u_{n,m} = \lambda_{n,m} u_{n,m}, \) this becomes
$$ \sum_{n,m} u_{n,m}(r,\theta) \psi_{n,m}''(t) = c^2 \sum_{n,m} \lambda_{n,m} u_{n,m}(r,\theta) \psi_{n,m}(t) $$Therefore each mode satisfies
$$ \psi_{n,m}''(t) = c^2\lambda_{n,m} \psi_{n,m}(t) $$Each mode therefore behaves like an independent harmonic oscillator. And so we have
$$ \psi_{n,m}(t) = a_{n,m} \cos(c k_{n,m} t) + b_{n,m} \sin(c k_{n,m} t) $$ $$ \psi'(0) = 0 \; \Rightarrow \; b_{n,m}=0 $$ $$ \psi_{n,m}(t) = a_{n,m} \cos(c k_{n,m} t) $$Therefore
$$ u(r,\theta,t) = \sum_{n=-\infty}^{\infty} \sum_{m=1}^{\infty} a_{n,m} u_{n,m}(r,\theta) \cos(c k_{n,m} t) $$Apply the initial condition:
$$ u(r,\theta,0) = \sum_{n=-\infty}^{\infty} \sum_{m=1}^{\infty} a_{n,m} u_{n,m}(r,\theta) = u_0(r,\theta) $$By orthogonality,
$$ a_{n,m} = \frac{\langle u_{n,m},u_0\rangle}{\langle u_{n,m},u_{n,m}\rangle} $$Hence
$$ a_{n,m} = \frac{ \int_0^{2\pi} \int_0^R u_{n,m}^*(r,\theta) u_0(r,\theta) \,r\,dr\,d\theta }{ \int_0^{2\pi} \int_0^R |u_{n,m}(r,\theta)|^2 \,r\,dr\,d\theta } $$The final solution is
$$ u(r,\theta,t) = \sum_{n=-\infty}^{\infty} \sum_{m=1}^{\infty} \frac{ \int_0^{2\pi} \int_0^R u_{n,m}^*(r,\theta) u_0(r,\theta) \,r\,dr\,d\theta }{ \int_0^{2\pi} \int_0^R |u_{n,m}(r,\theta)|^2 \,r\,dr\,d\theta } u_{n,m}(r,\theta) \cos(c k_{n,m} t) $$The membrane vibration is therefore a superposition of independent normal modes, each oscillating with its own characteristic frequency.