4
$\begingroup$

I tried searching for a solution to this type of problem online but was unsuccessful. I almost found a solution here, but it requires the coefficient matrix (the square one in the equation below) to have distinct eigenvalues which doesnt hold for this problem. Anyway here's the problem:

$$ \begin{bmatrix} \partial \rho/\partial t \\[0.3em] \partial V/\partial t \end{bmatrix} + \begin{bmatrix} V & \rho \\[0.3em] 0 & V\end{bmatrix} \begin{bmatrix} \partial \rho/\partial r \\[0.3em] \partial V/\partial r \end{bmatrix}=\begin{bmatrix} -2\rho V/r \\[0.3em] A\rho r \end{bmatrix} $$

Here $A$ is a constant. Any help or trick to modify the mehod in the link to suit this problem would be appreciated.

$\endgroup$
5
  • $\begingroup$ Is $V_r \equiv \partial V / \partial r$? $\endgroup$ May 7, 2017 at 13:49
  • $\begingroup$ @Rumplestillskin no it isn't.Now that you pointed out i guess its confusing, i'll change the name. $\endgroup$
    – alex
    May 7, 2017 at 13:50
  • $\begingroup$ Also, I'm not sure if it helps but once you multiply the matrix out for the first equation you can write it as the derivative of a product i.e $\partial / \partial r ( V \rho) $. $\endgroup$ May 7, 2017 at 13:55
  • $\begingroup$ I don't think it helps much. I actually expanded that term while working it out to reach this form since this seemed like the standard form used in many places $\endgroup$
    – alex
    May 7, 2017 at 13:59
  • $\begingroup$ These equations write $$ \partial_t {\bf U} + \partial_r {\bf F}({\bf U}) = {\bf R}(r,{\bf U}) $$ where ${\bf U} = (\rho, V)^\top$, ${\bf F}({\bf U}) = (\rho V, \frac{1}{2}V^2)^\top$, and ${\bf R}(r,{\bf U}) = (-2\rho V/r, A\rho r)^\top$. Both the physical flux and the r.h.s. are nonlinear. Thus, it seems hopeless to look for analytical solutions. $\endgroup$
    – EditPiAf
    Aug 8, 2017 at 13:31

0

You must log in to answer this question.