How can one solve a Parabolic PDE (like the wave or diffusion equations) if the boundary conditions were given over ranges? Here is an example: How to solve the equation $u_{xx}+u_{yy}-\alpha^{2}u_{t}=0$ over the following boundary conditions, assuming zero initial condition, variables $x\in [0,a], y\in [0,a]$, and $b<a$:

$u=0$ for $x=0$;

$u=0$ for $y=0$;

$u=0$ for $x=a\ \ \text{and}\ y\in[0,b]$;

$u=0$ for $y=a\ \ \text{and}\ x\in[0,b]$;

$u=0$ for $x=[(a+b)-y]\ \ \text{for}\ x\in[b,a]$ and $y\in[b,a]$

Note that these boundary ranges look geometrically like a square with a cut at its far corner. Also note that $b$ should not be assumed to be very small (as in $b<<a$) as to blindly grant the usage of perturbation techniques (though this would also be of interest, if generalisations can be shown).

I tried asking this question first on stackexchange, but no answer was given (since March 2014): https://math.stackexchange.com/questions/713298/solving-a-parabolic-pde-with-boundary-conditions-given-over-ranges

  • $\begingroup$ You have a standard 'abstract evolution equation' $u_t = \alpha^{-2} Lu$, where $L$ is a self-adjoint elliptic operator, under your described boundary conditions. Each eigenmode of $L$ evolves in time with the multiplicative factor of $e^{-\lambda t/alpha^2}$, where $\lambda$ is the eigenvalue. Don't expect any explicit solutions for the eigenmodes, unless you can make some kind of approximation. $\endgroup$ – Igor Khavkine Jul 12 '14 at 10:41
  • $\begingroup$ @Igor Many thanks. I will look it up (the abstract evolution equation) for more details. Do you see a place for variational principles here? Any advice? $\endgroup$ – user135626 Jul 12 '14 at 22:11
  • $\begingroup$ @Igor Also, when thinking about your previous comment about the compatibility between the boundary conditions and coordinates, what if I replaced the cut at the far corner (in the question above) with some elliptic cut (or other similar curve), then took the limit of some of its parameter (focus radius, etc) to approach the orginal line cut at the corner? Could this approach make it treatable, to find a solution, then we take the limit? $\endgroup$ – user135626 Jul 12 '14 at 22:28
  • $\begingroup$ Unlikely. If you want explicit solutions/approximations rather than just knowing they exist, numerics might be your best bet. $\endgroup$ – Igor Khavkine Jul 13 '14 at 17:52

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