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happy holidays, I am solving a coupled ODE - PDE system -for simplicity all coefficients are taken as 1-. I checked other questions on the issue but couldn't find any suitable answer or comparison. The model

$$ \frac{\partial \text{x1}(t,x)}{\partial t}=\frac{\partial ^2\text{x1}(t,x)}{\partial x\, \partial x}+\text{x1}(t,x) \text{x2}(t,x) \text{x3}(t,x)-\text{x1}(t,x)+1,\\ \frac{\partial \text{x2}(t,x)}{\partial t}=\frac{\partial ^2\text{x2}(t,x)}{\partial x\, \partial x}-\text{x2}(t,x),\\ \frac{\partial \text{x3}(t,x)}{\partial t}=1-\text{x1}(t,x)^2-\text{x3}(t,x) $$ The code

Clear["Global`*"]
solveme = NDSolve[{
    D[x1[t, x], t] == 
     D[x1[t, x], x, x] + x1[t, x] x2[t, x] x3[t, x] - x1[t, x] + 1,
    D[x2[t, x], t] == D[x2[t, x], x, x] - x2[t, x],
    D[x3[t, x], t] == (1 - x1[t, x]^2) - x3[t, x],
    x1[0, x] == If[-20 < x < 20, 30, 0],
    x1[t, -200] == 0,
    x1[t, 200] == 0,
    x2[0, x] == If[-20 < x < 20, 2, 0],
    x2[t, -200] == 0,
    x2[t, 200] == 0,
    x3[0, x] == 1
    },
   {x1, x2, x3}, {t, 0, 200}, {x, -200, 200}];



DensityPlot[
 Evaluate[{x1[t, x]} /. solveme], {t, 0, 200}, {x, -200, 200}, 
 ColorFunction -> "Rainbow", PlotLegends -> Automatic, 
 Method -> {"MethodOfLines", 
   "SpatialDiscretization" -> {"TensorProductGrid", 
     "MaxPoints" -> 101}}, PlotPoints -> 100]

Question Can NDSolve be used for this problem, or do you know\suggest any other approach for coupled PDE - ODE systems?

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    $\begingroup$ Your NDSolve runs without complaint. What's the problem? (BTW, your DensityPlot code has got the wrong options, but it seems irrelevant since you're not asking only about NDSolve.) $\endgroup$
    – Michael E2
    Dec 27, 2020 at 17:41
  • $\begingroup$ @MichaelE2 thanks for the comment. I was wondering whether having an ODE-PDE coupled system NDSolve is the correct way to solve it or not. Would you mind telling me which options are wrong? $\endgroup$
    – confused
    Dec 28, 2020 at 8:20
  • $\begingroup$ The Method option is appropriate for NDSolve but not for DensityPlot. Are you thinking the last equation is an ODE? The function $x1$ and therefore $x3$ depend on both $t$ and $x$, so it's not an ODE of the form $du/dt = F(t,u)$. The other equations are clearly PDEs. $\endgroup$
    – Michael E2
    Dec 28, 2020 at 14:21
  • $\begingroup$ @MichaelE2 Yes you are right, let's call it "pseudo-ODE approach" for $x3(x,t)$. If I wish to keep it as an ODE (lets call this "ODE approach") then $dx1(x,t)/dt$ needs $x3(t)$ as $f(x)*x3(t)$ and $dx3(t)/dt$ and $x1(x,t)$ as $\int_{domain} x1(x,t) dx$. How one can choose between these approaches? $\endgroup$
    – confused
    Dec 28, 2020 at 19:02
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    $\begingroup$ (1) I'm pretty sure NDSolve won't handle dependent variables with different arguments. So if you have x1[t, x], then it must be x3[t, x] and not just x3[t]. (This came up before on this site, and I couldn't figure out a way around it.) (2) If, for example, you have D[x3[t, x], t] == 1 - t^2 - x3[t, x], then NDSolve will construct x3 by integrating the ODE $du/dt = 1 - t^2 - u^2$ for every value of $x$ in the spatial grid. Everything should be ok, despite the extra work. (I'm not sure I understood your last comment, so I'm taking two shots at it, in hopes that they help.) $\endgroup$
    – Michael E2
    Dec 28, 2020 at 19:15

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