If 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 -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.
The problem is to find the pressure for sound wave propagation in a
tube with one end closed and one end open:
The D'Alembert solution applies to an infinite domain
. So to use the D'Alembert solution, the given
initial conditions, that are valid for must be extended to
all . In other words, functions and must be
converted into functions and that have
values for all . Of course, in the interval , they must
stay the same as and . Assume now for example that
looks as sketched below:
You might think that you could now simply take to be zero for all outside the range of the pipe. The corresponding D'Alembert solution will satisfy the wave equation everywhere, including inside the pipe . 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 and . So it will still be wrong.
You must select the extension of to all so that the correct boundary conditions become automatic.
The way to do it is as follows:
The process is shown for below:
It is OK if you get kinks or discontinuities in your functions and while creating (anti)symmetry. This happens when and/or does not satisfy the given boundary conditions. While then or may not have a unique value at the initial time, that problem will disappear when the time becomes greater than zero.
In the range , the found solution is exactly the same as for the finite pipe! The solution outside that range can simply be ignored.