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.

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