The Vibration of a Drumhead can be modeled with the weave equation in two dimensions. In this article we solve specific boundary value problems of the two-dimensional wave equation for which we encountered the solutions.

\begin{align} T(t)=a_1 \, cos \left( \lambda c t \right) + a_2 \, sin \left( \lambda c t \right) \end{align}

\begin{align} \Theta(\theta)=c_{\mu} \, cos \left( {\mu} \theta \right)+ d_{\mu} \, sin \left( {\mu} \theta \right) \end{align}

\begin{align} R_1(r)=c_1 J_{\mu}\left( \lambda r \right)+c_2 J_{- \mu}\left( \lambda r \right) \end{align} or \begin{align} R_2(r)=c_1 J_{\mu}\left( \lambda r \right) + c_2 Y_{\mu}\left( \lambda r \right) \end{align}

• Say we have a circular drum of radius $b$ and the drum holds a membrane that is free to vibrate. The first thing to notice is that $\Theta(\theta)$ is $2\pi$ periodic, so we replace $\mu$ by an integer we call $n$ (now only $R_2$ is allowed, why?).
• Moreover we assume that $R(r)$ is finite at $r=0$, thus we only have $R(r)=c_1 J_{n}\left( \lambda r \right)$ and $c_2$ must be zero $c_2=0$ otherwise the solution would blow up at $r=0$.
• In order for $J_n$ to remain positive $n$ is not allowed to take negative values, $n$ must be a positive integer, this requires $c_2$ to be zero otherwise $Y$ would tend to infinity for $r \rightarrow 0$.

For reasons we will clear later, we substitute $\lambda=\gamma /b$. Now we have the equations \begin{align} T(t)=a_1 \, cos \left( \frac{ \gamma ct}{b} \right) + a_2 \, sin \left( \frac{ \gamma ct}{b} \right) \end{align}

\begin{align} \Theta(\theta)=c_n \, cos \left( n\theta \right)+ d_n \, sin \left( n \theta \right) \end{align}

\begin{align} R(r)=c \, J_{n}\left( \frac{ \gamma r}{b} \right)= \sum_{k=0}^{\infty} \frac{(-1)^k }{k! \, \Gamma(k+n +1)} \left( \frac{ \gamma r }{2b} \right)^{2k+n} \end{align} For $u(t,r,\theta)=u(t,b,\theta)=0$ we need $R(b)$ to be zero $R(b)=0$. After the transformation $\lambda=\gamma /b$, \begin{align} R(b)=J_n\left( \frac{ \gamma b}{b} \right)=J_n(\gamma)=0, \tag{1} \end{align} so that the boundary condition $R(b)=0$ is satisfied, if the Bessel function $J_n(\gamma_m)=0$ is zero. For each $n$ of the Bessel function there will be $m$ roots denoted $\gamma_{n,m}$. Suppose for the moment that $b=1$ so we would need no transformation. Then the roots of the bessel function are.

Mathematica:

BesselJZero[0, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}] // N

$\gamma_{n,m}=\gamma_{0,1-10}=${2.40483, 5.52008, 8.65373, 11.7915, 14.9309, 18.0711, 21.2116, 24.3525, 27.4935, 30.6346}


Now, suppose that $b > 1$ what is the difference? The intuitive way to think about $b$ is, that it scales the Bessel functions that the roots are zero at the radius of the drum. Compare the Bessel function for $b=1$ and $b=2$

The root for $b=2$ are the roots of $b=1$ muliplied by the factor two. In other words $b>1$ stretches the Bessel function in a way that the boundary conditions are satisfied. We now know that $u$ must be of the form \begin{align} u(r,\theta,t)&=R(r)\Theta(\theta)T(t)=J_{n}\left( \frac{ \gamma_{n,m} r}{b} \right) \left[ c_{n,m} \, cos \left( n\theta \right)+ d_{n,m} \, sin \left( n \theta \right) \right] \left[a_{n,m} \, cos \left( \frac{ \gamma_{n,m} ct}{b} \right) + b_{n,m} \, sin \left( \frac{ \gamma_{n,m} ct}{b} \right) \right] \end{align} Superposition of this solution gives the equation \begin{align} u(r,\theta,t)&=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}J_{n}\left( \frac{ \gamma_{n,m} r}{b} \right) \left[ c_{n,m} \, cos \left( n\theta \right)+ d_{n,m} \, sin \left( n \theta \right) \right] \left[a_{n,m} \, cos \left( \frac{ \gamma_{n,m} ct}{b} \right) + b_{n,m} \, sin \left( \frac{ \gamma_{n,m} ct}{b} \right) \right] \end{align} that satisfies the 2D wave equation and (1). Now, we must determine the coefficients $a,b,c,d$ specific for our boundary value problem. At time $t=0$ the membrane needs to be deformed to see something happen, so $u(r,\theta,0)=f(r,\theta)$ where $f(r,\theta)$ is some function that gives the shape of the deformed membrane at $t=0$. Our equation for $u$ becomes at $t=0$ becomes \begin{align} u(r,\theta,0)&=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}J_{n}\left( \frac{ \gamma_{n,m} r}{b} \right) \left[ c_{n,m} \, cos \left( n\theta \right)+ d_{n,m} \, sin \left( n \theta \right) \right] \left[a_{n,m} \right]=f(r,\theta) \end{align} Furthermore $u_t(r,\theta,0)=0$, so the initial velocity of the membrane is zero. \begin{align} u_t(r,\theta,0)&=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}J_{n}\left( \frac{ \gamma_{n,m} r}{b} \right) \left[ c_{n,m} \, cos \left( n\theta \right)+ d_{n,m} \, sin \left( n \theta \right) \right] \left[ -a_{n,m} \frac{ \gamma_{n,m} c}{b} \, sin \left( \frac{ \gamma_{n,m} c0}{b} \right) + b_{n,m} \frac{ \gamma_{n,m} c}{b} \, cos \left( \frac{ \gamma_{n,m} c0}{b} \right)\right]=0 \end{align} Our expression for $u$ is now \begin{align} u(r,\theta,t)&=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}J_{n}\left( \frac{ \gamma_{n,m} r}{b} \right) \left[ c_{n,m} \, cos \left( n\theta \right)+ d_{n,m} \, sin \left( n \theta \right) \right] \, cos \left( \frac{ \gamma_{n,m} ct}{b} \right) \end{align} this means $b_{n,m}$ must be zero. We could define new constants $c'_{n,m}=c_{n,m}a_{n,m}$, $d'_{n,m}=d_{n,m}a_{n,m}$, instead we choose $a_{n,m}=1$ to keep the notation simple. Now we only have to solve for the coefficients. \begin{align} f(r,\theta)&=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}J_{n}\left( \frac{ \gamma_{n,m} r}{b} \right) \left[ c_{n,m} \, cos \left( n\theta \right)+ d_{n,m} \, sin \left( n \theta \right) \right] \\ f(r,\theta)&=\sum_{n=0}^{\infty} C_n \, cos \left( n\theta \right)+ D_n \, sin \left( n \theta \right) \end{align} with \begin{align} C_n=\sum_{m=0}^{\infty}c_{n,m} J_{n}\left( \frac{ \gamma_{n,m} r}{b} \right) \end{align} and \begin{align} D_n=\sum_{m=0}^{\infty}d_{n,m} J_{n}\left( \frac{ \gamma_{n,m} r}{b} \right) \end{align} $C_m$ and $D_m$ are now coefficients in a simple Fourier Series and are readily calculated \begin{align} C_0=\frac{1}{2\pi} \int_0^{2\pi} f(r,\theta) d\theta \end{align} \begin{align} C_n=\frac{1}{\pi} \int_0^{2\pi} f(r,\theta) cos(n\theta) d\theta \hspace{2cm} \text{for } n=1,2,3,\dots \end{align} \begin{align} D_n=\frac{1}{\pi} \int_0^{2\pi} f(r,\theta) sin(n\theta) d\theta \hspace{2cm} \text{for } n=0,1,2,\dots \end{align} Once we have the Fourier coefficients determined, we can tackle $c_{n,m}$ and $d_{n,m}$ by using the orthogonality of the Bessel functions \begin{align} c_{n,m}=\frac{2}{[J_{n+1}\left( \gamma_{n,m} \right)]^2} \int_0^b \rho \, J_n\left( \frac{ \gamma_{n,m} r}{b} \right) C_n dr \end{align} \begin{align} c_{0,m}= \frac{2}{[J_{1} ( \gamma_{0,m} ) ]^2} \int_0^b \int_0^{2\pi} \rho \, J_0 \left( \frac{ \gamma_{n,m} r}{b} \right) \frac{1}{2\pi} f(r,\theta) dr \, d\theta \end{align} \begin{align} c_{n,m}=\frac{2}{[J_{n+1} ( \gamma_{n,m} )]^2} \int_0^b \int_0^{2\pi} \rho \, J_n \left( \frac{ \gamma_{n,m} r}{b} \right) \frac{1}{\pi} f(r,\theta) cos(n\theta) dr \, d\theta \end{align} \begin{align} d_{n,m}=\frac{2}{[J_{n+1}\left( \gamma_{n,m} \right)]^2} \int_0^b \rho \, J_n\left( \frac{ \gamma_{n,m} r}{b} \right) D_n dr \end{align} \begin{align} d_{n,m}=\frac{2}{[J_{n+1} ( \gamma_{n,m} )]^2} \int_0^b \int_0^{2\pi} \rho \, J_n \left( \frac{ \gamma_{n,m} r}{b} \right) \frac{1}{\pi} f(r,\theta) sin(n\theta) dr \, d\theta \end{align} We could have obtained the expressions for $c_{n,m}$ and $d_{n,m}$ immediately as solution of the General Fourier Series. Notice, that $J_0$ the $0^{th}$ order Bessel functions have no $\theta$ dependence, thus they $J_0$ is the solution to boundary conditions with radial symmetry $f(r,\theta)=f(r)$. If $f$ depends on $\theta$, higher order bessel functions with the sum of respective cosine and sine terms solve the boundary value problem.

Solutons with $u(r,\theta)=J_{n}\left( \frac{ \gamma_{n,m} r}{b} \right)\, cos \left( n\theta \right)$