2
$\begingroup$

This question is cross-posted from Math StackExchange (link). I'm not sure it qualifies as research-level mathematics (although the application is to research) but it has been on MSE for several days, with a bounty for six of them, and failed to garner any sort of response.

It concerns a system of $n$ $n$-dimensional PDEs. \begin{equation*} -\frac{1}{2}\sum_{jk}a_{jk}x_{j}x_{k}u^{i}_{x_{j}x_{k}} +\sum_{jk} u^{j}\pi_{jk}u^{i}_{x_{k}} +c_{i}u^{i}=0. \end{equation*} Each $c_{i}$ is a positive real number. The matrix $A=\left(a_{jk}\right)$ is constant and positive definite while $\Pi=\left(\pi_{jk}\right)$ is constant with $1$s along the diagonal and nonpositive entries elsewhere such that each row sums to zero. Each component of $\mathbf{u}$ is given, with bounded nonnegative value, on the boundary of the domain $U\subset\mathbb{R}_{+}^{n}$.

These equations derive from a model of risk propagation on a network and a system of parabolic PDEs which might in theory be solved forwards in time. \begin{equation*} u^{i}_{t} -\frac{1}{2}\sum_{jk}a_{jk}x_{j}x_{k}u^{i}_{x_{j}x_{k}} +\sum_{jk} u^{j}\pi_{jk} u^{i}_{x_{k}}+c_{i}u^{i}=0. \end{equation*} But I don't have an initial condition and it's the long-time behaviour that I am interested in. That's why I am looking for a time-invariant solution. If there's a name for any of this I'd like to know. Maybe the time-dependent version looks a bit like Burgers'?

I have been reading Evans's book. But it seems there are about a thousand different ways to treat PDEs and I have no idea which is most appropriate to this system. I want to know about existence and uniqueness of solutions, and whether a "comparison" principle holds (i.e. whether $\mathbf{u}\preceq\mathbf{v}$ on $\partial U$ $\implies$ $\mathbf{u}\preceq\mathbf{v}$ in $U$). But any insight would be welcome, even if it depends on some additional assumption, e.g. $\Pi$ symmetric.

"If all you have is a hammer, everything looks like a nail." I am a poor mathematician in general but I do know of the Banach fixed point (contraction mapping) theorem. Could I build a sequence of solutions fixing $\sum_{j} u^{j}\pi_{jk}$ or $\sum_{jk} u^{j}\pi_{jk}u^{i}_{x_{k}}$ from the previous iteration? I guess in order to prove a contraction I would have to quantify the sensitivity of the solution to a change in that term.

A commenter has kindly suggested a change of variables to $\xi_{j}=\log x_{j}$. Then the differential equation becomes $$ -\frac{1}{2}\sum_{jk}a_{jk}u^{i}_{\xi_{j}\xi_{k}} +\sum_{k}\left( \frac{1}{2}a_{kk}+ e^{-\xi_{k}}\sum_{j} u^{j}\pi_{jk} \right)u^{i}_{\xi_{k}}+c_{i}u^{i} =0.$$

$\endgroup$
4
  • 2
    $\begingroup$ Looks like a non-linear (quasi-llinear in particular) elliptic PDE system. If you chance coordinates to $\xi_i = \log(x_i)$, the principal part (the one with second derivatives) will have constant coefficients. This observation might help to build an iteration scheme. There may be a "maximum principle" applicable to your system that could give something like the comparison inequality you are interested in. I'm no expert in these things, but these keywords might help you look things up in decent PDE books (Evans included). $\endgroup$ Jan 31 at 14:49
  • 2
    $\begingroup$ A maximum principle holds per component in your equation: for each component, suppose $\xi$ is a critical point, then $u^i_{\xi_k} = 0$, and since $a_{jk}$ is pos. def. and $c_i > 0$, you can conclude that each component $u^i$ cannot have a positive interior local max nor a negative interior local min. This doesn't immediately imply a comparison principle since your equation is non-linear. $\endgroup$ Jan 31 at 17:34
  • $\begingroup$ Thanks to both of you. Then with nonnegative boundary conditions on a bounded domain we have that a solution must be nonnegative. That makes sense. $\endgroup$
    – Ali
    Jan 31 at 21:05
  • $\begingroup$ :facepalm: I just noticed that I incorrectly identified this system as quasi-linear, it is in fact semi-linear, as already stated in the title of the question. $\endgroup$ Jan 31 at 22:34

0

Your Answer

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