Finite Interior Boundary Value Problem: Cavity Vibrations

Lecture 44

Let us extend our study of wave amplitudes from the two-dimensional Euclidean plane to three-dimensional Euclidean space plus temporal dimension as determined by the wave equation.

$$\nabla^2\psi -\frac{1}{c^2}~\frac{\partial^2\psi}{\partial t^2} = 0\,.$$

The spatial domain we consider is the interior of a finite cylinder of length $L$ and radius $a$

Figure 5.14: Cylindrical Cavity.

Its geometry demands that the wave equation, which governs the wave amplitude inside that cylinder for all times, be expressed relative to cylindrical coordinates,

$$\frac{1}{r}~\frac{\partial}{\partial r} r\frac{\partial\psi}{\par... ...partial^2\psi} {\partial z^2}-\frac{1}{c^2}~\frac{\partial^2\psi}{\partial t^2} = 0\,.$$ (525)

The wave field $\psi$ is finite and single valued (obviously!) inside the cylinder, and vanishes on its boundary. These observations are expressed by the fact that $\psi$ satisfies the following three pairs of boundary conditions for all times.

$$\psi = 0~~\textrm{at}~~z = 0$$ (526)

$$\psi = 0~~\textrm{at}~~z = L\,;$$

$$~$$

$$\psi = \textrm{finite~at}~~r=0$$ (527)

$$\psi = 0~~\textrm{at}~~r=a\,;$$

$$~$$

$$\psi (\theta =0) = \psi (\theta =2\pi )$$ (528)

$$\psi '(\theta =0) = \psi '(\theta =2\pi )\,.$$

We shall see that these homogeneous boundary conditions characterize three Sturm-Liouville eigenvalue problems, with their three sets of eigenvalues.

Suppose we know in addition the initial amplitude and velocity profiles

$$\psi (r,\theta ,z,t=0) =f(r,\theta ,z)$$
and

$$\frac{\partial \psi}{\partial t}(r,\theta,z,t=0) =g(r,\theta,z)$$

at $t=0$ . The functions $f$ and $g$ are called the initial value data for the wave equation. The problem before us is to determine, from this initial value data, $\psi (r,\theta ,z,t)$ , namely, the amplitude inside the cylinder for all times.

The first step is to solve the wave equation, Eq. 5.25, by the method of ``separation of variables''. It consists of finding those solutions which have the product form

$$\psi =R(r)\Theta (\theta )Z(z)T(t)\,.$$

Introducing it into the wave equation, dividing by the product of these four factors, one obtains

$$\frac{1}{R}~\frac{1}{r}~\frac{d}{dr}r\frac{dR}{dr} +\frac{1}{r^2}... ...{1}{Z}~\frac{d^2Z}{dz^2} - \frac{1}{c^2}~\frac{1}{T}~\frac{d^2T}{dt^2} = 0\,.$$

Bring the $z$ -term to the right hand side. The resulting equality holds for all $r,\theta ,z$ , and $t$ . Thus the right hand must be independent of $r$ , $\theta$ and $t$ , while the left hand side must be independent of $z$ . But the two sides are equal. Thus the common quantity must be independent of $r,\theta ,t$ , and $z$ , i.e., it must be some constant. Call it $k^2_z$ . Consequently, $Z(z)$ satisfies

$$\frac{1}{Z}~\frac{d^2Z}{dz^2} = -k^2_z\,.$$ (529)

Next isolate the $\Theta$ term, and by the analogous argument obtain

$$\frac{1}{\Theta}~\frac{d^2\Theta}{d\theta^2} = -\nu^2\,.$$ (530)

Similarly obtain

$$\frac{1}{R}\left[\frac{1}{r}~\frac{d}{dr} r \frac{dR}{dr} - \frac{\nu^2} {r^2} R\right] = -k^2\,.$$ (531)

Here $k^2_z$ , $\nu^2$ and $k^2$ are three arbitrary constants. For obvious reasons they are called separation constants. Finally, the wave equation, together with these three equations, implies

$$\frac{d^2T}{dt^2} +\omega^2T=0$$

where

$$\omega^2 = (k^2_z+k^2)c^2\,.$$

The initial value data $f$ and $g$ is nonzero at $t=0$ . Consequently, none of the four factors, whose products constitutes the solution to the wave equation, is allowed to be identically zero. Thus the boundary conditions Equations 5.26, 5.27, and 5.28, are conditions on the solutions to the differential equations, Equations 5.29, 5.30, and 5.31. There are three of each. They give rise to three Sturm-Louiville systems

1. $\frac{d^2Z}{dz^2} +k^2_zZ=0$             $Z(0)=0$             $Z(L)=0$
2. $\frac{d^2\Theta}{dz^2} +\nu^2\Theta = 0$              $\Theta (0)= \Theta (2\pi)$              $\Theta '(0)=\Theta '(2\pi )$
3. $\frac{1}{r}~\frac{d}{dr} r\frac{dR}{dr} +\left( k^2-\frac{\nu^2}{r^2} \right) R=0$             $R(0)=$ finite            R(a)=0.

Each of these three S-L eigenvalue problems determines its own eigenvalue spectrum, namely

$$k_z = \frac{n\pi}{L} n=1,2,\dots$$

$$\nu = m m=0,\pm 1,\dots$$

$$k = k_{mj} J_m(ka)=0,~~ j\textrm{th~ root~of~ the~Bessel~function}~J_m(x)~~.$$

For each triplet of eigenvalues there is a corrsponding amplitude profile,

$$R_{mj} (r)\Theta_m (\theta )Z_n(z)~~.$$

The product of the first two factors,

$$R_{mj} (r)\Theta_m (\theta )~~,$$

is the amplitude profile in the transverse plane. The last factor, $Z_n(z)$ , is the amplitude profile along the longitudinal direction.

The eigenvalue spectra are an expression of the boundary condition. Change the boundary conditions, and the eigenvalue spectra and their amplitude profiles will change. However, the boundary conditions remain fixed for all times. Consequently, the eigenvalue spectra and the corresponding amplitude profiles remain the same for all times.

Each triplet of eigenvalues $\left( k_z = \frac{2\pi}{L}\,,~\nu =m\,,~k= k_{mj}\right)$ determines three corresponding eigenfunctions and hence a solution to the wave equation, whose consequent reduced form is

$$\frac{1}{c^2}\frac{\partial^2\psi_{mjn}}{\partial t^2} +c^2\left[ k^2_{mj} +\left(\frac{n\pi} {L}\right)^2\right] \psi_{mjn} = 0~~.$$

Here

$$\psi_{mjn} (r,\theta ,z,t) = R_{mj} (r)\Theta_m (\theta )Z_n(z)T(t)\,.$$

Such a product solution, $\psi_{mjn}$ , to the wave equation is called a normal mode. Normal modes have the same (oscillatory) time dependence at every point of its domain. The unique features of any particular normal mode are determined by the three integers $(m,j,n)$ of the three eigenvalues $(k_z, \nu ,k)$ . This is true not only for its spatial amplitude profile

$$R_{mj} (r)\Theta_m(\theta )Z_n(z)\,,$$

but also for its oscillatory frequency

$$\omega = c\left( k^2_{mj}+\left(\frac{n\pi}{L}\right)^2\right)^{1/2}\equiv \omega_{mjn}$$

which determines its oscillatory behavior as a function of time

$$\psi_{mjn}(r,\theta ,z,t)=R_{mn}(r)\Theta_m(\theta )Z_n(z)[A_{mjn}\cos \omega_{mjn}t+B_{mjn}\sin\omega_{mjn}t]\,.$$

In brief, the boundary conditions determine the spectrum of allowed oscillatory frequencies of its normal modes. Furthermore, a cylindrical cavity illustrates a universal feature which is shared by all linear systems governed by a wave equation: a finite system always has a discrete eigenvalue spectrum.

Any vibratory system governed by linear wave equation obeys the linear superposition principle. Consequently, the general solution to the wave equation is a linear combination of normal modes

$$\psi =\sum^\infty_{m=-\infty}~\sum^\infty_{j=1}~\sum^\infty_{n=1}... ... \Theta_m(\theta )Z_n(z)[A_{mjn}\cos\omega_{mjn}t+B_{mjn}\sin\omega_{mjn}t]\,.$$

This is a generalized triple Fourier series. The two sets of Fourier coefficients $\{ A_{mjn}\}$ and $\{ B_{mjn}\}$ are determined by the initial value data $f(r,\theta ,z)$ and $g(r,\theta ,z)$ :

$$A_{mjn} = \int^a_0\int^{2\pi}_0\int^L_0 R_{mj}(r)\Theta_m(\theta ) Z_n(z) f(r,\theta ,z)rdrd\theta dz$$

$$\omega_{mjn}\,B_{mjn} = \int^a_0\int^{2\pi}_0\int^L_0 R_{mj}(r)\Theta_m (\theta )Z_n(z)g(r,\theta ,z)rdrd\theta dz\,.$$