The Vibrating Circular Membrane Problem
The wave equation can be easily generalized to a two-dimensional space. Particularly, we are often interested in the so-called vibrating circular membrane problem, e.g. a drumhead, so it is natural to expand the Laplacian operator in polar coordinates. The Laplace equation in this case is then expressed as
\begin{align}
\frac{\partial^2 z}{\partial t^2} -b^2\nabla^2 z = \frac{\partial^2 z}{\partial t^2} -b^2 (\frac{\partial^2 z}{\partial r^2} + \frac{1}{r}\frac{\partial z}{\partial r} + \frac{1}{r^2} \frac{\partial^2 z}{\partial \theta^2}) = 0 \tag{1}
\end{align}
where \( z \) is the vertical displacement of the membrane and \( b^2 = T/m \) is the ratio of the tension to mass per unit area. Again, we will need the method of Separation of Variables, and suppose that
\begin{align}
z(r, \theta, t) = R(r)\Theta(\theta)T(t) \tag{2}
\end{align}
Substituting (2) into (1), we have
\begin{align}
\begin{aligned} R(r)\Theta(\theta)T^{\prime\prime}(t) -b^2 (R^{\prime\prime}(r)\Theta(\theta)T(t) \\
+ \frac{1}{r}R'(r)\Theta(\theta)T(t) + \frac{1}{r^2} R(r)\Theta^{\prime\prime}(\theta)T(t)) \end{aligned} &= 0 \\
\frac{1}{b^2}\frac{T^{\prime\prime}(t)}{T(t)} -(\frac{R^{\prime\prime}(r)}{R(r)} + \frac{1}{r}\frac{R'(r)}{R(r)} + \frac{1}{r^2} \frac{\Theta^{\prime\prime}(\theta)}{\Theta(\theta)}) &= 0 \tag{3}
\end{align}
divided by \(R\Theta T\) as before. We now first separate the first term that depends on \(t\) only and the other terms that depend on \(r, \theta\):
\begin{align}
\frac{T^{\prime\prime}(t)}{b^2T(t)} = -k^2 = \frac{R^{\prime\prime}(r)}{R(r)} + \frac{1}{r}\frac{R'(r)}{R(r)} + \frac{1}{r^2} \frac{\Theta^{\prime\prime}(\theta)}{\Theta(\theta)} \tag{4}
\end{align}
The time part is simply solved via
\begin{align}
T^{\prime\prime} + k^2b^2 T = 0 \tag{5}
\end{align}
hence
\begin{align}
T = E\cos(kbt) + F\sin(kbt) \tag{6}
\end{align}
The second part in (4) can be further separated into \(r\) and \(\theta\) components respectively:
\begin{align}
r^2\frac{R^{\prime\prime}(r)}{R(r)} + r\frac{R'(r)}{R(r)} + k^2r^2 + \frac{\Theta^{\prime\prime}(\theta)}{\Theta(\theta)} &= 0 \tag{7}
\end{align}
\begin{align}
r^2\frac{R^{\prime\prime}(r)}{R(r)} + r\frac{R'(r)}{R(r)} + k^2r^2 = m^2 = -\frac{\Theta^{\prime\prime}(\theta)}{\Theta(\theta)} \tag{8}
\end{align}
The azimuthal part is also easily solved as
\begin{align}
\Theta^{\prime\prime} + m^2\Theta = 0 \tag{9}
\end{align}
\begin{align}
\Theta = C \cos(m\theta) + D \sin(m\theta) \tag{10}
\end{align}
where \( m = 0,1,2,\ldots \) by single-valuedness. Finally, the radial part is processed in a way very similar to what we have seen in this tutorial ((8)-(13)). We have
\begin{align}
r^2R^{\prime\prime} + rR’ + (k^2r^2-m^2)R = 0 \tag{11}
\end{align}
that is the same as the Bessel equation ((1) of this tutorial), again with \( r \rightarrow kr \) and \( \nu = m\). Therefore
\begin{align}
R = AJ_m(kr) + BY_m(kr) \tag{12}
\end{align}
The physical condition requires that the solution is bounded at \(r = 0\), so \(B=0\) in (12). Meanwhile, the boundary condition of a membrane is clearly a Dirichlet one, fixed to the rim: \( z(a,\theta,t) = 0, R(a) = 0 \) where \(a\) is the radius of the membrane. This means that
\begin{align}
R = J_m(\frac{k_{mn}}{a}r) \tag{13}
\end{align}
where \( k_{mn} = ka \) will be the \(n\)-th zero of the Bessel function of the first kind of order \(m\), \(J_m\), just as (17) of this tutorial. The full solution is then a superposition of
\begin{align}
z(r,\theta,t) = \sum_{m=0}^{\infty}\sum_{n=1}^{\infty} J_m(\frac{k_{mn}}{a}r) (C_{mn} \cos(m\theta) + D_{mn} \sin(m\theta))(E_{mn}\cos(\frac{k_{mn}b}{a}t) + F_{mn}\sin(\frac{k_{mn}b}{a}t)) \tag{14}
\end{align}
Now, if we are given the I.C.s that correspond to releasing the membrane from some position to rest, i.e.
\begin{align}
\left\{\begin{aligned}
z(r,\theta,0) &= f(r,\theta) \\
z_t(r,\theta,0) &= 0
\end{aligned}\right. \tag{15}
\end{align}
where \( f(r,\theta) \) can be a function of both \(r\) and \(\theta\), then we can match them by using the technique in the same tutorial, (19)-(22). First, notice that the second I.C. means that \( F = 0 \). Then, we can assume that the first I.C. can be expressed as Fourier modes with certain radial structures:
\begin{align}
z(r,\theta,0) = f(r,\theta) = \sum_{m=0}^{\infty} (C_m(r) \cos(m\theta) + D_m(r)\sin(m\theta)) \tag{16}
\end{align}
Comparing the Fourier modes in (16) with (14) at \( t = 0 \) shows that
\begin{align}
C_m(r) = \sum_{n=1}^{\infty}C_{mn}J_m(\frac{k_{mn}}{a}r) \tag{17}
\end{align}
where \( E_{mn} \) is absorbed. It is similar for \(D_m\) and \( D_{mn} \). \( C_{mn} \) is then given as a Bessel series following the orthonormality of the Bessel function ((30) of this tutorial):
\begin{align}
C_{mn} &= \frac{\int_{0}^{a} \frac{r}{a} C_m(r) J_m(k_{mn}\frac{r}{a}) d\frac{r}{a}}{(\int_{0}^{a} \frac{r}{a} [J_m(k_{mn}\frac{r}{a})]^2 d\frac{r}{a})} \\
&= \frac{2\int_{0}^{a} r C_m(r) J_m(k_{mn}\frac{r}{a}) dr}{a^2[J_{m+1}(k_{mn})]^2} \tag{18}
\end{align}
And \(D_{mn}\) will be computed in the exact same way. The full solution will therefore be
\begin{align}
z(r,\theta,t) &= \sum_{m=0}^{\infty}\sum_{n=1}^{\infty} J_m(\frac{k_{mn}}{a}r) ( \frac{2\int_{0}^{a} r C_m(r) J_m(k_{mn}\frac{r}{a}) dr}{a^2[J_{m+1}(k_{mn})]^2} \cos(m\theta) \\
&\quad + \frac{2\int_{0}^{a} r D_m(r) J_m(k_{mn}\frac{r}{a}) dr}{a^2[J_{m+1}(k_{mn})]^2} \sin(m\theta))\cos(\frac{k_{mn}b}{a}t) \tag{19}
\end{align}
For example, if the initial displacement is axisymmetric and has the form \( z(r,\theta,0) = 1 -\frac{r}{a} \), then \( m = 0\) and simply \(\Theta = C_{0n}, C_0(r) = 1 -\frac{r}{a} \). The evaluated expression will be
\begin{align}
C_{0n} &= \frac{2\int_{0}^{a} r (1-\frac{r}{a}) J_0(k_{0n}\frac{r}{a}) dr}{a^2[J_{1}(k_{0n})]^2} \tag{20}
\end{align}
and the complete solution will be
\begin{align}
z(r,\theta,t) &= \sum_{n=1}^{\infty} J_0(\frac{k_{0n}}{a}r) \frac{2\int_{0}^{a} r (1-\frac{r}{a}) J_0(k_{0n}\frac{r}{a}) dr}{a^2[J_{1}(k_{0n})]^2} \cos(\frac{k_{0n}b}{a}t) \tag{21}
\end{align}
Exercise
If the I.C.s are now
\begin{align}
\left\{\begin{aligned}
z(r,\theta,0) &= 0\\
z_t(r,\theta,0) &= g(r,\theta)
\end{aligned}\right. \tag{22}
\end{align}
so that the instantaneous displacement of the membrane is zero but there is some non-zero velocity, find the new solution form. Hence, what will be the full solution if the I.C.s are in the general form of
\begin{align}
\left\{\begin{aligned}
z(r,\theta,0) &= f(r,\theta) \\
z_t(r,\theta,0) &= g(r,\theta)
\end{aligned}\right. \tag{23}
\end{align}
Answer
With the new I.C.s, this time \( E = 0 \), and the velocity is given by
\begin{align}
z_t(r,\theta,t) = \sum_{m=0}^{\infty}\sum_{n=1}^{\infty} J_m(\frac{k_{mn}}{a}r) (C_{mn} \cos(m\theta) + D_{mn} \sin(m\theta)) \frac{k_{mn}b}{a} \cos(\frac{k_{mn}b}{a}t)
\end{align}
where we have absorbed the \(F_{mn}\) coefficient. To match the second I.C., we again assume a decomposition in the form of (16)
\begin{align}
z_t(r,\theta,0) &= g(r,\theta) = \sum_{m=0}^{\infty} (C_m(r) \cos(m\theta) + D_m(r)\sin(m\theta))
\end{align}
Comparing these two expressions then yields
\begin{align}
C_m(r) = \sum_{n=1}^{\infty}C_{mn}\frac{k_{mn}b}{a}J_m(\frac{k_{mn}}{a}r)
\end{align}
and by the same argument as before
\begin{align}
C_{mn} &= \frac{a}{k_{mn}b}\frac{2\int_{0}^{a} r C_m(r) J_m(k_{mn}\frac{r}{a}) dr}{a^2[J_{m+1}(k_{mn})]^2} \\
&= \frac{1}{k_{mn}ab}\frac{2\int_{0}^{a} r C_m(r) J_m(k_{mn}\frac{r}{a}) dr}{[J_{m+1}(k_{mn})]^2}
\end{align}
It is similar for \(D_m\) and \(D_{mn}\). Therefore, the corresponding solution will be
\begin{align}
z(r,\theta,t) &= \sum_{m=0}^{\infty}\sum_{n=1}^{\infty} J_m(\frac{k_{mn}}{a}r) (\frac{2\int_{0}^{a} r C_m(r) J_m(k_{mn}\frac{r}{a}) dr}{[J_{m+1}(k_{mn})]^2} \cos(m\theta) \\
&\quad + \frac{2\int_{0}^{a} r D_m(r) J_m(k_{mn}\frac{r}{a}) dr}{[J_{m+1}(k_{mn})]^2} \sin(m\theta)) \frac{1}{k_{mn}ab}\sin(\frac{k_{mn}b}{a}t)
\end{align}
The full solution for the general I.C.s will then be the sum of this and (19) due to linearity.
Leave a Reply