# Solving matrix PDE with a range of free parameters

I've a matrix PDE which looks like, $$\partial_r M(r,k) = M(r,k) \, A(k) \, M(r,k) \, + B(k) \quad, \quad \lim_{r \rightarrow 0} \, M(r,k) = C(k)$$ Here $M,A,B,C$ are all $4\times4$ matrices and functions of $r$ and/or $k$. One more thing is this equation has to be solved for a range of $k$ and in the end I want a plot of $\text{Tr} \,(\text{Im}\, M)$ vs $k$. I'm not sure how to solve a matrix equation and that to with a range of parameters.

Let me put this simple $2 \times 2$ example as reference (say for $k \in [-2,2]$ ): $$A(k) = \begin{pmatrix} k & -1 \\ -1 & -k \end{pmatrix} \, , \, B(k) = \begin{pmatrix} k & 1 \\ 1 & -k \end{pmatrix} \, , \, C(k) = \begin{pmatrix} \sqrt{k} & 0 \\ 0 & -\sqrt{k} \end{pmatrix} \, .$$ (Please feel free to change the $A$, $B$, and $C$ matrices if needed, as I'm not sure of the existence of the solution for these choices.)

Thanks!

• Use ParametricNDSolve and treat k as the parameter. – bbgodfrey Nov 6 '15 at 19:48
• please post Mathematica code, not just TeX – Dr. belisarius Nov 6 '15 at 20:27

NB. I am not an expert in PDE's, but I would like to put an answer here in hopes that it would lead to better answers.

Since $k$ is a parameter, I have removed it from the arguments to $M$. Then, I specified the matrices directly in the differential equation and the inital value.

soln = ParametricNDSolve[
{
M'[r] == M[r]. {{k, -1}, {-1, k}} + {{k, 1}, {1, -k}},
M == {{Sqrt[k], 0}, {0, Sqrt[k]}}
}
, M, {r, 0, 1}, {k}] The solution to $M$ at any $r$ (here, between 0 and 1) and parameter value $k$ can be determined with:

M[k][r] /. soln


Let's define $G(r,k)=\textrm{Tr}[\textrm{Im} M(r,k)]$. (I think the solutions to $M$ are real matrices for $k>0$.)

G[r_, k_] := Tr[Im[Evaluate[M[k][r] /. soln]]


Finally, the $G(r,k)$ can be plotted with a function like:

ContourPlot[
G[r, k],
{k, -5, 0}, {r, 0, 1},  FrameLabel -> {"k", "Tr[Im M]"}] • Thanks for your effort, this is precisely what I was looking for. I've made an edit to the question, $G$ is actually just $M$ (sorry for the typo). You may want to make an edit to your answer accordingly. – Skylar15 Nov 9 '15 at 7:08
• You're welcome, this was fun. I've made a few changes to my answer since I had the arguments to the ParametricFunction reversed. – Eric Brown Nov 9 '15 at 12:16