Mathematica Asked on June 9, 2021
I encountered this when trying to solve the PDE mentioned here. I’ve transformed the equation to the following:
With[{u = u[x, t]},
neweq = D[u, t] == Inactive[Div][{{u^2}}.Inactive[Grad][u, {x}], {x}] +
{1}.Inactive[Grad][Sign[x] u, {x}]]
This is the formal PDE for FiniteElement
method as far as I can tell, but it doesn’t lead to the correct solution. After checking with GetInactivePDE
, I found the Inactive[Grad][Sign[x] u, {x}]]
term is simply lost in parsing stage:
(* Definition of GetInactivePDE isn't included in this post,
please find it in the link above. *)
showFormalPDE[a__] :=
Module[{state}, {state} = NDSolve`ProcessEquations[a];
GetInactivePDE[state["FiniteElementData"]@"PDECoefficientData",
state@"VariableData"] == 0 // Thread];
{bc, ic} = {{u[-7, t] == 0, u[7, t] == 0}, u[x, 0] == Exp[-x^2]};
showFormalPDE[{neweq, ic, bc}, u, {t, 0, 2}, {x, -7, 7}]
(* {Inactive[Div][-{{u[x]^2}} . Inactive[Grad][u[t, x], {x}], {x}] +
Derivative[1, 0][u][t, x] == 0} *)
Is this a bug? Or separate Inactive[Grad][……]
term in PDE is not allowed at the moment?
Tested on v12.1.1.
Just a simpler sample reflecting the underlying issue:
eq = D[u[x, t], t] == Inactive[Grad][aaaa[x], {x}];
ic = u[x, 0] == 0;
showFormalPDE[{eq, ic}, u, {x, 0, 1}, {t, 0, 2}]
(* {Derivative[1, 0][u][t, x] == 0} *)
No, this is not a bug. You can not have Inactive[Grad][Sign[x] u[x], {x}]
you can only have Inactive[Grad][u[x], {x}]
the rest needs to go into some coefficient.
Correct answer by user21 on June 9, 2021
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