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Mathematics LibreTexts

4.5.2: Oscillation of a Membrane

Let \(\Omega\subset\mathbb{R}^2\) be a bounded domain. We consider the initial-boundary value problem

\begin{eqnarray}
\label{mem1}\tag{4.5.2.1}
u_{tt}(x,t)&=&\triangle_xu\ \ \mbox{in}\ \Omega\times\mathbb{R}^1,\\
\label{mem2} \tag{4.5.2.2}
u(x,0)&=&f(x),\ \ x\in\overline{\Omega},\\
\label{mem3} \tag{4.5.2.3}
u_t(x,0)&=&g(x),\ \ x\in\overline{\Omega},\\
\label{mem4} \tag{4.5.2.4}
u(x,t)&=&0\ \ \mbox{on}\ \partial\Omega\times\mathbb{R}^1.
\end{eqnarray}

As in the previous subsection for the string, we make the ansatz (separation of variables)

$$
u(x,t)=w(t)v(x)
$$

which leads to the eigenvalue problem

\begin{eqnarray}
\label{evpmem1} \tag{4.5.2.5}
-\triangle v&=&\lambda v\ \ \mbox{in}\ \Omega,\\
\label{evpmem2} \tag{4.5.2.6}
v&=&0\ \ \mbox{on}\ \partial\Omega.
\end{eqnarray}

Let \(\lambda_n\) are the eigenvalues of \((\ref{evpmem1})\), \((\ref{evpmem2})\) and \(v_n\) a complete associated orthonormal system of eigenfunctions. We assume \(\Omega\) is sufficiently regular such that the eigenvalues are countable, which is satisfied in the following examples. Then the formal solution of the above initial-boundary value problem is

$$
u(x,t)=\sum_{n=1}^\infty\left(\alpha_n\cos(\sqrt{\lambda_n}t)+\beta_n\sin(\sqrt{\lambda_n}t)\right)v_n(x),
$$

where

\begin{eqnarray*}
\alpha_n&=&\int_\Omega\ f(x)v_n(x)\ dx\\
\beta_n&=&\frac{1}{\sqrt{\lambda_n}}\int_\Omega\ g(x)v_n(x)\ dx.
\end{eqnarray*}

Note

In general, eigenvalues of (\ref{evpmem1}), (\ref{evpmem1}) are not known explicitly. There are numerical methods to calculate these values. In some special cases, see next examples, these values are known.

Examples

Example 4.5.2.1: Rectangle membrane

Let
$$
\Omega=(0,a)\times (0,b).
$$
Using the method of separation of variables, we find all eigenvalues of (\ref{evpmem1}), (\ref{evpmem2}) which are given by
$$
\lambda_{kl}=\sqrt{\frac{k^2}{a^2}+\frac{l^2}{b^2}},\ \ k,l=1,2,\ldots
$$ and associated eigenfunctions, not normalized, are
$$
u_{kl}(x)=\sin\left(\frac{\pi k}{a}x_1\right)\sin\left(\frac{\pi l}{b}x_2\right).
$$

Example 4.5.2.2: Disk membrane

Set
$$
\Omega=\{x\in\mathbb{R}^2:\ x_1^2+x_2^2<R^2\}.
$$
In polar coordinates, the eigenvalue problem (\ref{evpmem1}), (\ref{evpmem2}) is given by
\begin{eqnarray}
\label{evppol1} \tag{4.5.2.6}
-\frac{1}{r}\left((ru_r)_r+\frac{1}{r}u_{\theta\theta}\right)&=&\lambda u\\
\label{evppol2} \tag{4.5.2.7}
u(R,\theta)&=&0,
\end{eqnarray}
here is \(u=u(r,\theta):=v(r\cos\theta,r\sin\theta)\). We will find eigenvalues and eigenfunctions by separation of variables
$$
u(r,\theta)=v(r)q(\theta),
$$
where \(v(R)=0\) and \(q(\theta)\) is periodic with period \(2\pi\) since \(u(r,\theta)\) is single valued.
This leads to
$$
-\frac{1}{r}\left((rv')'q+\frac{1}{r}vq''\right)=\lambda v q.
$$
Dividing by \(vq\), provided \(vq\not=0\), we obtain
\begin{equation}
\label{disk1} \tag{4.5.2.8}
-\frac{1}{r}\left(\frac{(rv'(r))'}{v(r)}+\frac{1}{r}\frac{q''(\theta)}{q(\theta)}\right)=\lambda,
\end{equation}
which implies
$$
\frac{q''(\theta)}{q(\theta)}=const.=:-\mu.
$$
Thus, we arrive at the eigenvalue problem
\begin{eqnarray*}
-q''(\theta)&=&\mu q(\theta) \\
q(\theta)&=&q(\theta+2\pi).
\end{eqnarray*}
It follows that eigenvalues \(\mu\) are real and nonnegative. All solutions of the differential equation are given by
$$
q(\theta)=A\sin(\sqrt{\mu}\theta)+B\cos(\sqrt{\mu}\theta),
$$
where \(A\), \(B\) are arbitrary real constants. From the periodicity requirement
$$
A\sin(\sqrt{\mu}\theta)+B\cos(\sqrt{\mu}\theta)=A\sin(\sqrt{\mu}(\theta+2\pi))+B\cos(\sqrt{\mu}(\theta+2\pi))
$$
it follows\begin{eqnarray*}
\sin x-\sin y&=&2\cos\frac{x+y}{2}\sin\frac{x-y}{2}\\
\cos x-\cos y&=&-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}\end{eqnarray*}
$$
\sin(\sqrt{\mu}\pi)\left(A\cos(\sqrt{\mu}\theta+\sqrt{\mu}\pi)-B\sin(\sqrt{\mu}\theta+\sqrt{\mu}\pi)\right)=0,
$$
which implies, since \(A\), \(B\) are not zero simultaneously, because we are looking for \(q\) not identically zero,
$$
\sin(\sqrt{\mu}\pi)\sin(\sqrt{\mu}\theta+\delta)=0
$$
for all \(\theta\) and a \(\delta=\delta(A,B,\mu)\). Consequently the eigenvalues are
$$
\mu_n=n^2,\ \ n=0,1,\ldots\ .
$$
Inserting \(q''(\theta)/q(\theta)=-n^2\) into (\ref{disk1}), we obtain the boundary value problem
\begin{eqnarray}
\label{disk2} \tag{4.5.2.9}
r^2v''(r)+rv'(r)+(\lambda r^2-n^2)v&=&0\ \ \mbox{on}\ (0,R)\\
\label{disk3} \tag{4.5.2.10}
v(R)&=&0\\
\label{disk4} \tag{4.5.2.11}
\sup_{r\in(0,R)}|v(r)|&<&\infty.
\end{eqnarray}
Set \(z=\sqrt{\lambda}r\) and \(v(r)=v(z/\sqrt{\lambda})=:y(z)\), then, see (\ref{disk2}),
$$
z^2y''(z)+zy'(z)+(z^2-n^2)y(z)=0,
$$
where \(z>0\). Solutions of this differential equations which are bounded at zero are Bessel functions of first kind and \(n\)-th order \(J_n(z)\). The eigenvalues follows from boundary condition (\ref{disk3}), i. e., from \(J_n(\sqrt{\lambda}R)=0\). Denote by \(\tau_{nk}\) the zeros of \(J_n(z)\), then the eigenvalues of (\ref{evppol1})-(\ref{evppol1}) are
$$
\lambda_{nk}=\left(\frac{\tau_{nk}}{R}\right)^2
$$
and the associated eigenfunctions are
\begin{eqnarray*}
J_n(\sqrt{\lambda_{nk}}r)\sin(n\theta), &&\ n=1,2,\ldots\\
J_n(\sqrt{\lambda_{nk}}r)\cos(n\theta), &&\ n=0,1,2,\ldots.
\end{eqnarray*}
Thus the eigenvalues \(\lambda_{0k}\) are simple and \(\lambda_{nk},\ n\ge1\), are double eigenvalues.

Remark. For tables with zeros of \(J_n(x)\) and for much more properties of Bessel functions see \cite{Watson}. One has, in particular, the asymptotic formula
$$
J_n(x)=\left(\frac{2}{\pi x}\right)^{1/2}\left(\cos(x-n\pi/2-\pi/5)+O\left(\frac{1}{x}\right)\right)
$$
as \(x\to\infty\). It follows from this formula that there are infinitely many zeros of \(J_n(x)\).

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