4.2 Extension to finite regions

If $x$ is restricted by finite boundaries, the D'Alembert solution does not really apply. To use it anyway, we must somehow extend the problem to a doubly infinite $x$-range without boundaries. But our solution without boundaries should still satisfy the boundary conditions for the finite range. That is often possible by clever use of symmetry. An example can clarify that.

4.2.1 The physical problem

The problem is to find the pressure for sound wave propagation in a tube with one end closed and one end open:

$$\hbox{\epsffile{dalex1.eps}}$$

4.2.2 The mathematical problem

$$\hbox{\epsffile{dalex2.eps}}$$

• There is a finite domain $\bar \Omega$ given by $0\le x\le\ell$
• There is an unknown pressure $u=u(x,t)$ to be found from the wave equation
  
  $$u_{tt} = a^2 u_{xx}$$
  
  

• The constant $a$ is the speed of sound $a$. The equation is only valid for normal acoustics in which the gas velocities are much less than the speed of sound.
• This equation is second order in time so it needs two initial conditions. They are:
  
  $$u(x,0) = f(x) \quad u_t(x,0) = g(x) \qquad \mbox{for} 0 \le x \le \ell$$
  
  

• This equation is second order in space so it needs two boundary conditions. Given are one homogeneous Neumann boundary condition at $x=0$ and one homogeneous Dirichlet condition at $x=\ell$.

4.2.3 Dealing with the boundary conditions

The D'Alembert solution applies to an infinite domain $-\infty<x<\infty$. So to use the D'Alembert solution, the given initial conditions, that are valid for $0<x<\ell$ must be extended to all $x$. In other words, functions $f(x)$ and $g(x)$ must be converted into functions $\bar{f}(x)$ and $\bar{g}(x)$ that have values for all $x$. Of course, in the interval $0<x<\ell$, they must stay the same as $f(x)$ and $g(x)$. Assume now for example that $f(x)$ looks as sketched below:

$$\hbox{\epsffile{dalex3.eps}}$$

You might think that you could now simply take $\bar{f}(x)$ to be zero for all $x$ outside the range of the pipe. The corresponding D'Alembert solution will satisfy the wave equation everywhere, including inside the pipe $0<x<\ell$. That is good, because the wave equation must indeed be satisfied. Unfortunately, the solution you get that way will not satisfy the boundary conditions at $x=0$ and $x=\ell$. So it will still be wrong.

You must select the extension $\bar{f}(x)$ of $f(x)$ to all $x$ so that the correct boundary conditions become automatic.

The way to do it is as follows:

• To make the boundary condition $u_x=0$ at $x=0$ automatic, create symmetry around $x=0$. Symmetric functions have zero derivative at the symmetry point.
• To make the boundary condition $u=0$ at $x=\ell$ automatic, create antisymmetry (odd symmetry) around $x=\ell$. Anti-symmetric functions have zero derivative at the symmetry point.

The process is shown for $f(x)$ below:

$$\hbox{\epsffile{dalex4.eps}}$$

$$\hbox{\epsffile{dalex5.eps}}$$

$$\hbox{\epsffile{dalex6.eps}}$$

$$\hbox{\epsffile{dalex7.eps}}$$

$$\hbox{\epsffile{dalex8.eps}}$$

Create the extended function $\bar g(x)$ or $G$ the same way.

It is OK if you get kinks or discontinuities in your functions $\bar f$ and $\bar g$ while creating (anti)symmetry. This happens when $f$ and/or $g$ does not satisfy the given boundary conditions. While then $u$ or $u_x$ may not have a unique value at the initial time, that problem will disappear when the time becomes greater than zero.

4.2.4 The final solution

$$\hbox{\epsffile{dalex8.eps}}$$

$$u(x,t) = \frac{\bar f(x-at) + \bar f(x+at)}{2} + \frac{1}{2a} \int_{x-at}^{x+at} \bar g(\xi) { \rm d}\xi$$

This is pretty easy to evaluate for simple functions $f$ and $g$. You will have fun doing it.

$$\hbox{\epsffile{dalex9.eps}}$$

In the range $0\le x\le\ell$, the found solution is exactly the same as for the finite pipe! The solution outside that range can simply be ignored.