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Consider an advection-diffusion problem

$$u_t+au_{xx}+bu_{x}=0,\quad x>0.$$

Now I want to remove the drift and rewrite the problem for the new domain. I use the change of variables $y=x-bt$, and the equation can be rewritten as

$$v_t+av_{yy}=0$$

but what is the new domain in terms of variables $t$ and $y$?

Assume $b>0$, then $t\in [0,T]$ as before. But for $y$ the domain is not rectangular anymore, because it is like a triangle on one side. That is, for every fixed $t$, my $y$ is between $-bt$ and $\infty$? Did I compute the new domain correctly? If so, how do I solve this numerically? This is not a square box anymore, so do I have to place the grid for finite differences on such domain shape?

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Assuming your old domain is $x\in[0,L]$, then $y(t)\in[-bt,L-bt]$. In your case, you could either solve it by hand since it is just the heat equation or you directly solve the original equation numerically and avoid the change of variables altogether (it doesn't change the computational cost anyway).

I would expect that rather than your domain, the boundary conditions are of importance.

If you stay far enough from the boundary until your final time, you could just assume a fixed mesh for $y\in[0,L]$ and shift the solution which would be the case for an open boundary, where both domains are equal $D(x)=D(y)=(-\infty,\infty)$

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  • $\begingroup$ thanks, I was interested solving numerically and if there is a benefit of simplifying the equation this way. On one hand, I remove the drift, so I don't need to care about how to discretize that advection term(in case it is large and then I have to be careful), but looks like than I end up with non square domain and it gets harder to implement it for the new equation numerically. However, could not I create a romb grid in this case? I have never seen this in the books, in that way the transformed equation can be solved by the usual FDM as if I had a square grid...? $\endgroup$ – Kamil Aug 11 '16 at 13:16

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