# Modeling a Vibrating Drumhead III

In the previous discussion, we finally obtained the solution of the vibrating drumhead problem:
$$u(r,\theta,t)=\sum_{n=0}^\infty\sum_{m=1}^\infty J_n(\lambda_{nm}r)\cos(n\theta)[A_{nm}\cos(\lambda_{nm} ct)+B_{nm}\sin(\lambda_{nm}ct)].$$
In this lecture, we determine the Fourier coefficients $A_{nm}$ and $B_{nm}$ using the initial conditions $u(r,\theta,0)$ and $u_t(r,\theta,0)$. Before we go on, we need to mention two types of orthogonalities: the orthogonality of cosine functions and the orthogonality of Bessel functions. First note that
$$\int_0^{2\pi}\cos(n\theta)\cos(k\theta)d\theta=\left\{\begin{array}{ccc}0 & \mbox{if} & n\ne m,\\\pi & \mbox{if} & n=m.\end{array}\right.$$
The reason this property is called an orthogonality is that if $V$ is the set of all (Riemann) integrable real-valued functions on the interval $[a,b]$, then $V$ forms a vector space over $\mathbb R$. This vector space is indeed an inner product space with the inner product $$\langle f,g\rangle=\int_a^bf(x)g(x)dx\ \mbox{for}\ f,g\in V.$$
Bessel functions are orthogonal as well in the following sense:
$$\int_0^1J_n(\lambda_{nm}r)J_n(\lambda_{nl}r)rdr=\left\{\begin{array}{ccc}0 & \mbox{if} & m\ne l,\\\frac{1}{2}[J_{n+1}(\lambda_{nm})]^2 & \mbox{if} & m=l.\end{array}\right.$$

From the solution $u(r,\theta,t)$, we obtain the initial position of the drumhead:
$$u(r,\theta,0)=\sum_n\sum_mJ_n(\lambda_{nm}r)\cos(n\theta)A_{nm}.$$
On the other hand, $u(r,\theta,0)=f(r,\theta)$. Multiply
$$\sum_n\sum_mJ_n(\lambda_{nm}r)\cos(n\theta)A_{nm}=f(r,\theta)$$
by $\cos(k\theta)$ and integrate with respect to $\theta$ from $0$ to $2\pi$:
$$\sum_n\sum_mJ_n(\lambda_{nm}r)A_{nm}\int_0^{2\pi}\cos(n\theta)\cos(k\theta)d\theta=\int_0^{2\pi}f(r,\theta)\cos(k\theta)d\theta.$$ The only nonvanishing term of the above series is when $n=k$, so we obtain
$$\pi\sum_mJ_k(\lambda_{km}r)A_{km}=\int_0^{2\pi}f(r,\theta)\cos(k\theta)d\theta.$$ Multiply this equation by $J_k(\lambda_{kl}r)$ and integrate with respect to $r$ from $0$ to $1$:
$$\pi\sum_mA_{km}\int_0^1J_k(\lambda_{km}r)J_k(\lambda_{kl}r)rdr=\int_0^{2\pi}\int_0^1f(r,\theta)\cos(k\theta)J_k(\lambda_{kl}r)rdrd\theta.$$ The only nonvanishing term of this series is when $m=l$. As a result we obtain:
$$A_{kl}=\frac{1}{\pi L_{kl}}\int_0^{2\pi}\int_0^1f(r,\theta)\cos(k\theta)J_k(\lambda_{kl}r)rdrd\theta$$
or
$$A_{nm}=\frac{1}{\pi L_{nm}}\int_0^{2\pi}\int_0^1f(r,\theta)\cos(n\theta)J_n(\lambda_{nm}r)rdrd\theta,\ n,m=1,2,\cdots$$
where
$$L_{nm}=\int_0^1J_n(\lambda_{nm}r)^2rdr=\frac{1}{2}[J_{n+1}(\lambda_{nm})]^2, n=0,1,2,\cdots.$$
For $n=0$ we obtain
$$A_{0m}\frac{1}{2\pi L_{0m}}\int_0^1f(r,\theta)J_0(\lambda_{0m}r)rdrd\theta,\ m=1,2,\cdots.$$
Using
$$u_t(r,\theta,0)=\sum_n\sum_mJ_n(\lambda_{nm}r)\cos(n\theta)B_{nm}\lambda_{nm}c=g(r,\theta),$$
we obtain
\begin{align*}
B_{nm}&=\frac{1}{c\pi L_{nm}\lambda_{nm}}\int_0^{2\pi}\int_0^1g(r,\theta)\cos(n\theta)J_n(\lambda_{nm}r)rdrd\theta,\ n,m=1,2,\cdots,\\
B_{0m}&=\frac{1}{2c\pi L_{nm}\lambda_{nm}}\int_0^{2\pi}\int_0^1g(r,\theta)J_0(\lambda_{0m}r)rdrd\theta,\ m=1,2,\cdots.
\end{align*}
Unfortunately at this moment I do not know if I can make an animation of the solution using an open source math software package such as Maxima or Sage. I will let you know if I find a way. In the meantime, if any of you have an access to Maple, you can download a Maple worksheet I made here and run it for yourself. In the particular example in the Maple worksheet, I used $f(r,\theta)=J_0(2.4r)+0.10J_0(5.52r)$ and $g(r,\theta)=0$. For an animation of the solution, click here.