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I wish to solve the pde: $$-\frac{1}{1-t}\partial_x^2\phi+t^4(1-t)\partial_t^2\phi-t^4\partial_t\phi=\mu^2 \phi,$$ with initial conditions $\phi(x,0)=\cos(\mu x)$ and $\dot{\phi}(x,0)=0$ for some time period $t\in[0,0.9]$, for example. Lacking analytical methods, I thought I'd try numerically. However, numerically means I have to give a solver a spacial domain $[-a,a]$ on which to solve... and I have no idea what boundary conditions to impose on $x=\pm a$. In a sense, the value of $\phi$ on spacial slices like $x=a$ are exactly what I'm trying to find in the first place!

How does one get around this?

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  • $\begingroup$ Is your question about Mathematica (software) or mathematics? If it is about Mathematica could share the code you have written so far? $\endgroup$ – user21 Apr 30 at 4:03
  • $\begingroup$ Some progress can be made by Fourier decomposing the solution in x to obtain an ODE, which can be solved numerically as a function of the Fourier wavenumber. Note, however, that eventually boundary conditions in x are required. to obtain a unique solution. $\endgroup$ – bbgodfrey Apr 30 at 4:26
  • $\begingroup$ My question is, what does one do in principle to get around this difficulty in evolving a very wide class of problems numerically? I haven't included code because I know that Mathematica (or any solver) will not give me reasonable results as it won't know what to do on the artificial spatial boundaries I have to give to solve on a finite domain. $\endgroup$ – Rudyard Apr 30 at 10:21
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    $\begingroup$ Perhaps, since the initial condition is periodic, I could impose periodic boundary conditions to get away with it? $\endgroup$ – Rudyard Apr 30 at 10:22
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    $\begingroup$ “My question is, what does one do in principle to get around this difficulty in evolving a very wide class of problems numerically? ” As I've already mentioned under your question here, There's no general solution, AFAIK. The choosing of b.c. approximating infinity highly depends on the physical background of the PDE. $\endgroup$ – xzczd Apr 30 at 15:43

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