I'm currently looking at the PDE \begin{align*} u_t + \left[x(1-y) - (1-x)\right]u_x - (1-y) u_y + (z-xy) u_z = (z-xy) u_{xy} - (1-x)u& \\ \end{align*} with \begin{align*} u(x,y,z,0) = 1& \\ (x,y,z) \in [0,1]\times [0,1] \times [0,1]&, \end{align*} which I found via the conversion of a chemical master equation. The solution $u$ of this PDE is the probability generating function of the master equation. Is there any way to solve this numerically? It resembles advection-diffusion with a source, but it's not parabolic due to the mixed derivative term and variable coefficients. I don't believe the lack of boundary conditions is an issue, since if we just look at the advective terms, characteristics flow out of the boundary. Any help/suggestions would be greatly appreciated!

  • $\begingroup$ The presence of a mixed derivative does not mean the equation cannot be parabolic; that boils down to what the equation looks like in the canonical form. However the coefficients of the equation change sign in the domain, so the type of the equation would change from one location to another. The lack of boundary conditions is a big issue actually, whether it is parabolic or not. $\endgroup$ Commented Dec 16, 2020 at 23:01
  • $\begingroup$ If you let $x' = x+y$, $y' = x - y$, the mixed derivative becomes $u_{x'x'} - u_{y'y'}$, so not parabolic right? Would imposing numerical boundary conditions lead to instability? $\endgroup$
    – Mike D
    Commented Dec 16, 2020 at 23:05
  • $\begingroup$ It should be possible to set up boundary conditions so that a numerical instability would be caused; but you'd need to know the details of the time-integration algorithm to design such unstable boundary conditions. For time integration, one should note that the problem after all is linear, so it can be expressed as d/dt f = A*f, where f is the state vector and A is a matrix. So the solution can be formally written as a matrix exponent, and easily found numerically. However, you'd still need to make some assumptions on what happens on the boundary so you'd need those boundary conditions. $\endgroup$ Commented Dec 17, 2020 at 2:54
  • $\begingroup$ I guess it is fair to say that initial conditions and boundary conditions is the same kind of thing, we don't know (and we don't care) if an independent variable describes time or space. So we need initial and boundary conditions for a PDE (and for an ODE which is a particular case of PDE in 1D). $\endgroup$ Commented Dec 17, 2020 at 3:00


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