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I would like to use EventLocator to stop NDSolve with the Event command

"Event" -> Abs[D[f[x,y,t],t]] < threshold

That is, I need to check the time derivative of function f[x,y,t] over a spatial domain a >= 0 && a <= L, b >= 0 && b <= L at each integration step. Beside, it is clear that at that moment the temporal boundary is t. When I run it, NDSolve gives:

"The function value \ Abs[InterpolatingFunction[...]+<<1>><<<1>> is not True or False \ when the arguments are {3.187144811900335`*^-8,<<4>>}"

I am not familiar with Event, but I understand that one should use a and b instead of x and y in the Abs[...], because x and y are NDSolve variables and it is better to avoid inserting them in the numerical functions.

Actually, what I need is something like this:

"Event" -> Abs[D[f[a,b,t],t], a >= 0 && a <= L, b >= 0 && b <= L] < threshold

Obviously, it is not correct in the syntax of Abs, how can I implement the 'Event' properly to this end. Thank you!

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1 Answer 1

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Here is an example for a 1D PDE modified from the documentation, where the integration will stop as long as Derivative[1, 0][u][t, x] > 1:

eqn = {
 D[u[t, x], t, t] == D[u[t, x], x, x] - Sin[u[t, x]], 
 u[0, x] == E^(-(x - 5)^2) + E^(-(x + 5)^2/2), Derivative[1, 0][u][0, x] == 0, 
 u[t, -50] == u[t, 50]
}

NDSolve[eqn, u, {t, 0, 100}, {x, -50, 50},
 Method -> {
   "MethodOfLines", "DiscretizedMonitorVariables" -> True,
   "SpatialDiscretization" -> {"TensorProductGrid", "DifferenceOrder" -> "Pseudospectral"},
   Method -> {
     "EventLocator",
     "Event" :> 
      If[And @@ (# < 1 & /@ Abs[Derivative[1, 0][u][t, x]]), 0, 1],
     "EventLocationMethod" -> "StepBegin"
   }
 }
]
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  • $\begingroup$ Perhaps you could add a link to the relevant documentation page as well. $\endgroup$
    – MarcoB
    Jun 1, 2016 at 17:15
  • $\begingroup$ @MarcoB Thanks for the editing. I have added the link. $\endgroup$
    – xslittlegrass
    Jun 1, 2016 at 17:51
  • $\begingroup$ @can Derivative[1, 0][u][t, x] represents a series of values corresponding to different x in the solving region at the current time t. For example, {u[t,x1],u[t,x2],...}.I'm trying to check whether there is at least one of those values breaks our condition, so we will stop the integration. $\endgroup$
    – xslittlegrass
    Jun 2, 2016 at 1:59
  • $\begingroup$ @ xslittlegrass, thanks for your answer. In the condition of the If I understand that you Map the pure function #<1 & onto Abs to give 0 or 1 when the condition is true or false, respectively. But I don't understand that why you Apply the And to (...)? What is the intention that And only Apply only one argument? $\endgroup$
    – Enter
    Jun 2, 2016 at 2:00
  • $\begingroup$ @xslittlegrass, in my real problem, when I run the NDSolve it gives >"The function value If[<<1>>,0,1] is not True or False when the arguments are {...}." After three warning like this, and this line: >Further output of NDSolve::nbnum1 will be suppressed during this calculation. >>. It runs normally so far, however I will accept it if the solution is numerical reasonably. Thank you! $\endgroup$
    – Enter
    Jun 4, 2016 at 3:13

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