In a square, heat-conducting sheet, insulated from above and below
(5.29) |
If we are looking for a steady state solution, i.e. we take the time derivative does not contribute, and we get Laplace’s equation
(5.30) |
an example of an elliptic equation. Let us once again look at a square plate of size , and impose the boundary conditions
(This choice is made so as to be able to evaluate Fourier series easily. It is not very realistic!) We once again separate variables,
(5.32) |
and define
(5.33) |
Or explicitly
(5.34) |
With boundary conditions , . The 3rd boundary conditions remains to be implemented.
Once again distinguish three cases:
,
,
. We find
Since
we find
(
).
No solutions
So we have
(5.36) |
The one remaining boundary condition gives
(5.37) |
This leads to the Fourier series of ,
So, in short, we have
(5.39) |
Question: The dependence on enters through a trigonometric function, and that on through a hyperbolic function. Yet the differential equation is symmetric under interchange of and . What happens?
Answer: The symmetry is broken by the boundary conditions.