Examples of solutions to wave equation

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LAST TIME

Derived the wave equation

$\begin{displaymath}{{\partial ^2\xi }\over{\partial t^2}}=c^2 {{\partial ^2\xi }\over{\partial x^2}} \end{displaymath}$

using propagation of sound as an example.
Argued that one could express the general solution of the wave equation

$\begin{displaymath}\xi (x,t)=f_1(x-ct)+f_2(x+ct) \end{displaymath}$

in terms of traveling waves.

TODAY
We will present an example of a solution to the wave equation and indicate some important generalizations.

EXAMPLE:
An organ pipe is open in one end (x=0) and closed at the other end (x=a) At the closed end the amplitude $\xi$ has to be zero while at the open end the excess pressure is zero. Since the excess pressure is proportional to the derivative of the amplitude we have

$\begin{displaymath}\left.\frac{\partial\xi}{\partial x}\right\vert _{x=0}=0\end{displaymath}$

For the eigenvalue problem we can use the method of separation of variables as we did for the heat problem. Let us try to find solutions to the wave equation on the form

$\begin{displaymath}\xi(x,t)=\phi(x)\tau(t)\end{displaymath}$

Substitution gives

$\begin{displaymath}c^2\tau(t)\frac{d^2\phi(x)}{dx^2}=\phi(x)\frac{d^2\tau}{dt^2}\end{displaymath}$

Again using $\prime$ to indicate differentiation

$\begin{displaymath}\frac{\phi^{\prime\prime}(x)}{\phi(x)}=\frac{\tau^{\prime\prime}(t)} {c^2\tau(t)}=const=c_1\end{displaymath}$

The boundary conditions on $\phi(x)$ are

$\begin{displaymath}\phi^\prime(0)=0;\;\; \phi(a)=0\end{displaymath}$

If c₁>0 the solution to

$\begin{displaymath}\phi^{\prime\prime}(x)-c_1\phi(x)=0\end{displaymath}$

will be on the form

$\begin{displaymath}\phi(x)=A\sinh(x\sqrt{c_1})+B\cosh(x\sqrt{c_1})\end{displaymath}$

The boundary condition at x=0 requires

A=0

The condition at x=a is then impossible to satisfy.
Hence

$\begin{displaymath}0>c_1=-\lambda^2\end{displaymath}$

The solution to the differential equation for $\phi(x)$ is now

$\begin{displaymath}\phi(x)=A\sin(\lambda x)+B\cos(\lambda x)\end{displaymath}$

The boundary condition at x=0 gives A=0. The boundary condition at x=a gives

$\begin{displaymath}\lambda_n=\frac{(2n+1)\pi}{2a},\;\;n=integer\end{displaymath}$

The differential equation for the time dependent part is

$\begin{displaymath}\tau^{\prime\prime}(t)+c^2\lambda_n^2\tau(t)=0\end{displaymath}$

with solution

$\begin{displaymath}\tau(t)=\alpha_n\cos(c\lambda_n t)+\beta_n\sin(c\lambda_nt)]\end{displaymath}$

The solution to the problem is thus

$\begin{displaymath}\xi(x,t)=\sum_{n=1}^{\infty}\cos(\lambda_nx)[\alpha_n\cos(c\lambda_n t)+\beta_n\sin(c\lambda_nt)]\end{displaymath}$

In practice, one is often only interested in knowing the frequencies

$\begin{displaymath}\omega_n=\frac{c(2n+1)\pi}{2a}\end{displaymath}$

If we want a detailed solution we can determine $\alpha_n$ and $\beta_n$ from initial conditions. If

$\begin{displaymath}\xi(x,0)=f(x)\end{displaymath}$

is known, we can determine the coefficients $\alpha_n$ from

$\begin{displaymath}\alpha_n=\frac{2}{a}\int_0^adx\;f(x)\cos(\lambda_nx)\end{displaymath}$

To determine the coefficients $\beta_n$ we need to know the velocity of the displacements. Suppose

$\begin{displaymath}\frac{\partial\xi(x,t)}{\partial t}\vert _{t=0}=g(x)\end{displaymath}$

Then,

$\begin{displaymath}\beta_n=\frac{2}{a}\int_0^adx\;g(x)\cos(\lambda_nx)\end{displaymath}$

TRAVELING vs. STANDING WAVES
Our solution using the method of separation of variables involves standing waves, while the general solution which we found last time involved traveling waves. To see that the two approaches are equivalent we use the trigonometric identities

$\begin{displaymath}\cos A \cos B=\frac{1}{2}[\cos(A+B)+\cos(A-B)]\end{displaymath}$

$\begin{displaymath}\cos A \sin B=\frac{1}{2}[\sin(A+B)-\sin(A-B)]\end{displaymath}$

The term

$\begin{displaymath}\alpha_n\cos(\lambda_nx)\cos(c\lambda_nt)\end{displaymath}$

can thus be written

$\begin{displaymath}\frac{1}{2}\alpha_n[\cos(\lambda_n(x+ct))+\cos(\lambda_n(x-ct))\end{displaymath}$

Similarly

$\begin{displaymath}\beta_n\cos(\lambda_nx)\sin(c\lambda_nt)= \frac{1}{2}\beta_n[\sin(\lambda_n(x+ct))-\sin(\lambda_n(x-ct))\end{displaymath}$

REFLECTION AT A BOUNDARY
It is instructive to interprete the boundary conditions at closed or open ends in terms of traveling waves. To do this we write the solution to the wave equation approaching a boundary at

x=0

f₁(ct+x)+f₂(ct-x)

to indicate a wave f₁ which moves to the left towards x=0 and then is reflected into a wave f₂ which moves towards positive x.

If the boundary condition at x=0 is $\xi(0,t)=0$ we find

f₁(ct)=-f₂(ct)
that is the reflected wave has the opposite phase of the incoming wave.
If we have open end boundary conditions

$\begin{displaymath}\frac{\partial\xi(x,t)}{\partial x}\vert _{x=0}=0\end{displaymath}$

we find

$\begin{displaymath}f_1^\prime(ct)=f_2^\prime(ct)\end{displaymath}$

which we interprete by saying that the two waves are in phase.

DAMPING
There will always be some damping of the sound wave due to dissipative effects. (Some of the kinetic energy associated with the organized wave motion will be lost as heat).

$\begin{figure} \epsfysize=140pt \epsffile{airvo.eps} \end{figure}$

To take this into account we go back to the derivation of the wave equation. In the expression

$\begin{displaymath}Adx \rho _a{{\partial ^2\xi }\over{\partial t^2}} =Ac^2 dx{{\partial ^2\xi }\over{\partial x^2}}\rho _a \end{displaymath}$

(1)

the left hand side represented mass times acceleration, while the right hand side represented a restoring force.

It is natural to include the effect of damping by adding a term

$\begin{displaymath}-Adx\rho_ak\frac{\partial\xi}{\partial t}\end{displaymath}$

proportional to the velocity of the volume element opposing the motion, i.e. the constant k should be positive.

The modified wave equation thus becomes:

$\begin{displaymath}{{\partial ^2\xi }\over{\partial t^2}}+k\frac{\partial\xi}{\partial t}=c^2 {{\partial ^2\xi }\over{\partial x^2}} \end{displaymath}$

To get a slightly different situation from what we had before let us assume that the boundary conditions are

$\begin{displaymath}\xi(0,t)=\xi(a,t)=0\end{displaymath}$

Again we can solve this problem by the method of separation of variables. Put

$\begin{displaymath}\xi(x,t)=\phi(x)\xi(t)\end{displaymath}$

We find

$\begin{displaymath}\frac{1}{\phi(x)}\frac{d^2\phi}{dx^2}=\frac{1}{c^2\tau(t)}(\frac{d^2\tau}{dt^2}+k\frac{d\tau}{dt})=c_1=const\end{displaymath}$

As before the constant c₁ must be negative and we put $c_1=-\lambda^2$ and have

$\begin{displaymath}\phi^{\prime\prime}+\lambda^2\phi=0\end{displaymath}$

with general solution

$\begin{displaymath}\phi=A\sin(\lambda x)+B\cos(\lambda x)\end{displaymath}$

This time the boundary conditions require that B=0 and the eigenvalues are

$\begin{displaymath}\lambda_n=\frac{n\pi}{a}\end{displaymath}$

The differential equation for $\tau$ is now

$\begin{displaymath}\frac{d^2\tau}{dt^2}+k\frac{d\tau}{dt}+c^2\lambda^2\tau=0\end{displaymath}$

This is a differential equation with constant coefficients and it can be solved by trying solutions on the form

$\begin{displaymath}\tau(t)\propto e^{\gamma t}\end{displaymath}$

Substituting into the differential equation gives the characteristic equation

$\begin{displaymath}\gamma^2+k\gamma+\lambda^2_nc^2=0\end{displaymath}$

If the damping is not too large this equation has two complex conjugate solutions

$\begin{displaymath}\gamma=-\frac{k}{2}\pm i\sqrt{\lambda^2_nc^2-\frac{k^2}{4}}\end{displaymath}$

The general solution to our problem is then

$\begin{displaymath}\xi(x,t)=\sum_{n=1}^{\infty}\sin(\lambda_nx)e^{=kt/2}[\alpha_... ...ac{k^2}{4}})+\beta_n\cos(t\sqrt{\lambda^2_nc^2-\frac{k^2}{4}})]\end{displaymath}$

where the coefficients $\alpha_n$ and $\beta_n$ must be determined from the initial conditions.

SOLUTION WITH SOURCE TERM
As our final example let us consider the inhomogeneous wave equation

$\begin{displaymath}\frac{\partial^2\xi}{\partial t^2}-c^2\frac{\partial^2\xi}{\partial x^2}+k\frac{\partial\xi}{\partial t}=f(x)e^{i\omega t}\end{displaymath}$

i.e there is a source of sound with spatial extent f(x) generating waves at the frequency $\omega$ . As usual when we use the complex form it is understood that the actual solution is the real part of whatever solution that we find.

We have already solved the homogeneous problem f(x)=0, so all we need to do is to find a particular solution. We assume this term is on the form

$\begin{displaymath}\xi_p=\phi(x)e^{i\omega t}\end{displaymath}$

Substitution into the wave equation yields

$\begin{displaymath}-\omega^2\phi-c^2\phi^{\prime\prime}+i\omega\phi=f(x)\end{displaymath}$

We assume that f(x) has the complex Fourier series

$\begin{displaymath}f(x)=\alpha_n\sin(\frac{n\pi x}{a})\end{displaymath}$

and that

$\begin{displaymath}\phi(x)=\beta_n\sin(\frac{n\pi x}{a})\end{displaymath}$

Substituting into the differential equation for $\phi$ we find that the coefficients $\beta_n$ can be found from

$\begin{displaymath}(-\omega^2+\frac{c^2n^2\pi^2}{a^2}+ik\omega)\beta_n=\alpha_n\end{displaymath}$

and we can thus express $\phi(x)$ as a spatial Fourier series.

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