Does there exist a library (in any programming language) for solving (numerically) systems of multidimensional first-order linear PDEs in the form

$$\mathbf{u}_{t}+\hat{A}(\mathbf{x})\mathbf{u}_{\mathbf{x}}=\mathbf{f}(\mathbf{u}),$$ $$\mathbf{u}(\mathbf{x},0)=\mathbf{g}(\mathbf{x}),$$

where $\mathbf{u}=(u_1(\mathbf{x}),\ldots,u_m(\mathbf{x})),\mathbf{x}=(x_1,\ldots,x_n),\mathbf{f}(\mathbf{u})=(f_{1}(\mathbf{u}),\ldots,f_{m}(\mathbf{u})),$ and $\hat{A}(\mathbf{x})$ is a matrix of variable coefficients?

  • 2
    $\begingroup$ I believe CLAWPACK can do this, but it has been some time since I worked with it. Randy LeVeque worked on it and he is a leader in numerical PDEs $\endgroup$ – whpowell96 Dec 25 '18 at 22:24
  • 1
    $\begingroup$ Since most current methods for solving hyperbolic pdes are based only on one-dimensional concepts, they are pretty much independent of dimensionality. If you are happy wita cartesian grid, Im sure you could write your own $\endgroup$ – Philip Roe Dec 27 '18 at 18:48
  • 1
    $\begingroup$ It seems a bit strange that you specify the number of equations and the number of spatial dimensions to be the same. That's very rarely the case for physically meaningful systems. And what do you mean by $u_x$ when $x$ is a vector? Clawpack works with any number of equations. Clawpack does not have code for more than 3 spatial dimensions. You could certainly implement the same algorithms in higher dimensions, though. $\endgroup$ – David Ketcheson Dec 29 '18 at 18:30
  • $\begingroup$ @DavidKetcheson Sorry, my question indeed was a little misleading, I am not specifically studying the case where the number of spatial dimensions equals the number of equations; I've edited the question. By writing $\mathbf{u}_{\mathbf{x}}$, where $\mathbf{x}$ is a vector, I mean that there are derivatives with respect to all spacial variables in every equation. Since Clawpack does not support more than 3 dimensions, I actually ended up writing my own solver and for my first model system where n = 4 it performs quite well. Nevertheless, your comment is much appreciated! $\endgroup$ – Yashman Dec 30 '18 at 23:44

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.