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I'm new to the PDE toolbox in Matlab. From the PDE specification window of the toolbox, it looks like one can only solve PDE with constant coefficients.

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How can I use Matlab's PDE toolbox to solve PDE like

$$ \partial_t P(x,y,t)=x\partial_xP(x,y,t)+(y-1)\partial_yP(x,y,t)+2P(x,y,t)? $$

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  • $\begingroup$ This question would best be asked on the Mathworks forums. $\endgroup$
    – Bill Barth
    Commented Jul 24, 2014 at 9:54

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In general, PDE Toolbox is able to solve 2D PDE in the form shown on this page:

PDE Form

Any of those coefficients (e.g. c, a, f, d) can be functions of x, y, t, u, $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}$.

If the coefficients are relatively simple expressions, the simplest way to define them is by typing a string expression in text boxes above (e.g. 1+x.^2+4*y). That approach is documented on this page:

Coefficients defined with string expressions

If the coefficients are more complicated, it is usually more straightforward to define them as MATLAB functions. That approach is documented on this page:

Coefficients defined with a function

For your equation, f could be defined as the string "x.^ux+(y-1).*uy" (without the "), a=-2, d=1, and c=0.

However, a strong caveat is in order. The algorithms in PDE Toolbox are designed for the case where the second-order c-term is significant relative to the first-order derivative terms. So the success for this equation will depend very much on the boundary and initial conditions. It might also be necessary to set c to some small number; this is sometimes referred to as adding "artificial diffusion."

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  • $\begingroup$ Correct me if I'm wrong, from the PDE Form documentation, it looks like all equations types that can be solved by PDE toolbox involve only 2nd-order spatial derivatives but no 1st-order spatial derivatives. $\endgroup$
    – wdg
    Commented Jul 23, 2014 at 16:53
  • $\begingroup$ Oh, sorry, I read your second sentence and didn't look as closely at your PDE as I should have. In PDE Toolbox, the coefficients, themselves, can be functions of first-order spatial derivatives of the dependent variables so you could include them in the f-coefficient. But you are correct that it is designed to handle PDE that have a significant diffusion term. $\endgroup$
    – Bill Greene
    Commented Jul 23, 2014 at 18:49
  • $\begingroup$ How do I express, for example, f=$x\partial_xu$ in Matlab? $\endgroup$
    – wdg
    Commented Jul 24, 2014 at 1:52
  • $\begingroup$ I updated my original response to cover this question and say a bit more about using PDE Toolbox for first-order equations. $\endgroup$
    – Bill Greene
    Commented Jul 24, 2014 at 12:44
  • $\begingroup$ Indeed, PDE tool cannot solve the PDE when $c$=0. $\endgroup$
    – wdg
    Commented Jul 25, 2014 at 1:05
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I'm not familiar with this PDE toolbox in Matlab, but the software COMSOL Multiphysics is developed from this toolbox. You can dig a bit to find out their close relation. Also if you check out COMSOL you will find how these two look alike.

At least in COMSOL, I know your PDE is in principle possible to solve.

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  • $\begingroup$ I'm happy to know such package but it looks like an overkill. $\endgroup$
    – wdg
    Commented Jul 23, 2014 at 16:48

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