I need to solve the PDE for a complex function $A(x,t)=A_r(x,t)+iA_i(x,t)$

eq = D[A[x, t], t] + 1/4*Conjugate[A[x, t]]*A[x, t]^2 - D[A[x, t], {x, 2}] - 2*A[x, t] == 0;

over $[-L,L]$ and $[0,t_\text{max}]$. The equation is subject to a random initial condition and the boundary conditions as follows: $A_r(-L,t)=A_r(L,t)$ and $A_i(-L,t)=-A_i(L,t)$

L = 30; tmax = 30;
ini[x_] = 1/10*BSplineFunction[RandomReal[{-1, 1}, 20], SplineClosed -> True, SplineDegree -> 5][x/(2*L)];
ibcs = {Re[A[-L, t]] == Re[A[L, t]], Im[A[-L, t]] == -Im[A[L, t]], A[x, 0] == ini[x]};

Then, I solve it with NDSolve

sol = NDSolve[{eq, ibcs}, A, {x, -L, L}, {t, 0, tmax}, 
             Method -> {"MethodOfLines",
             "SpatialDiscretization" -> {"TensorProductGrid", 
             "MinPoints" -> 201, "MaxPoints" -> 201, 
             "DifferenceOrder" -> "Pseudospectral"}}, AccuracyGoal -> 20]

But I received the error

NDSolve::bcedge: Boundary condition Im[A[-30,t]]==-Im[A[30,t]] is not specified on a single edge of the boundary of the computational domain.>>

I didn't understand the error. Why the boundary conditions (bcs) must be specified on a single edge. Should not we set the bcs at both sides? Any suggestion is highly appreciated.

Thank for @xzczd's comment:

I just knew that NDSolve could not handle anti-periodic bc. Yes, the equation can be solved with a periodic bc:

periodbcs = {A[-L, t] == A[L, t], A[x, 0] == ini[x]}

But the solution should be incorrect because the solution is a real function by observing its imaginary part.

 ContourPlot[Evaluate[Im[A[x, t] /. sol]], {x, -L, L}, {t, 0, tmax}, 
 Contours -> 10, PlotRange -> All, PlotLegends -> Automatic, 
 ColorFunction -> Hue, FrameLabel -> {"x", "t"}, PlotLabel -> "Ai", ImageSize -> 200]

enter image description here

Response to @user64494's comment:

Yes, I can split the real and imaginary parts by writing the 2nd term as

$(A^\ast A)A=\vert A\vert^2A=(A_r^2+A_i^2)(A_r+i A_i)=A_r^3+A_i^2A_r+i(A_r^2A_i+A_i^3)$

Then the equation can be split into

eqs = {D[Ar[x, t], t] + 1/4*(Ar[x, t]^3+Ai[x, t]^2*Ar[x, t]) - D[Ar[x, t], {x, 2}] - 2*Ar[x, t] == 0,
D[Ai[x, t], t] + 1/4*(Ai[x, t]^3+Ar[x, t]^2*Ai[x, t]) - D[Ai[x, t], {x, 2}] - 2*Ai[x, t] == 0};

But I don't know how to make an anti-periodic initial condition (Ai[x, 0] = inianti[x]) to be consistent with the boundary condition.

ibcs = {Ar[-L, t] == Ar[L, t], Ai[-L, t] == -Ai[L, t], Ar[x, 0] == ini[x], Ai[x, 0] = inianti[x]};
  • $\begingroup$ @xzczd many thanks. Please see my update, where I can explain more clearly. Any workaround will be highly appreciated! $\endgroup$
    – Nobody
    Jul 17 '19 at 15:59
  • $\begingroup$ Is it possible to split the real and imaginary parts in eq and ibcs? $\endgroup$
    – user64494
    Jul 17 '19 at 16:27
  • $\begingroup$ @user64494 great idea. Please see my update. Any suggestion :) $\endgroup$
    – Nobody
    Jul 17 '19 at 16:48
  • $\begingroup$ Did you try to solve the split problem? $\endgroup$
    – user64494
    Jul 17 '19 at 16:56
  • $\begingroup$ I haven’t because I don’t know how to get a random anti-periodic initial condition. Please help. $\endgroup$
    – Nobody
    Jul 17 '19 at 16:59

Pre-v12 Solution

The approach here is fully applicable to your problem. Anyway, the corresponding coding isn't trivial, so let me give an answer.

We start from the splitted equation system because Re, Im, Conjugate isn't that convenient for subsequent coding. The form of b.c.s are slightly modified, because both periodic b.c. and anti-periodic b.c. are set with one-sided difference formula in this method (which is different from using PeriodicInterpolation of NDSolve`FiniteDifferenceDerivative) and we need 4 constraints in x direction in total:

Clear[ini, inianti, Ai]
eqs = {D[Ar[x, t], t] + 1/4 (Ar[x, t]^3 + Ai[x, t]^2 Ar[x, t]) - D[Ar[x, t], {x, 2}] - 
     2 Ar[x, t] == 0, 
       D[Ai[x, t], t] + 1/4 (Ai[x, t]^3 + Ar[x, t]^2 Ai[x, t]) - D[Ai[x, t], {x, 2}] - 
     2 Ai[x, t] == 0};
ic = {Ar[x, 0] == ini[x], Ai[x, 0] == inianti[x]};
bc = {Ar[-L, t] == Ar[L, t], Ai[-L, t] == -Ai[L, t], 
   Derivative[1, 0][Ar][-L, t] == Derivative[1, 0][Ar][L, t], 
   Derivative[1, 0][Ai][-L, t] == -Derivative[1, 0][Ai][L, t]};


Derivative[1, 0][Ar][-L, t] == Derivative[1, 0][Ar][L, t] is added because periodic b.c. implies the solution is smooth enough across the boundary, but frankly speaking, I'm not familiar with anti-periodic b.c. and not sure if Derivative[1, 0][Ai][-L, t] == -Derivative[1, 0][Ai][L, t] is correct, but do remember a supplement for derivative of x of Ai at the boundary is necessary, or a particular solution won't be determined.

The i.c.s are simply generated randomly, they don't satisfy the b.c.s of course, but this should not be a big deal because the i.c.s will be slightly modified at the boundary to satisfy the b.c.s in the upcoming disretization step. (For more information about handling inconsistency between i.c. and b.c., you may want to check this post. )

L = 30; tmax = 30;
ini = ListInterpolation[RandomReal[{-1, 1}, 20], {{-L, L}}];
inianti = ListInterpolation[RandomReal[{-1, 1}, 20], {{-L, L}}];

Finally, discretize the PDE system to an ODE system and solve, with the help of pdetoode:

points = 200; domain = {-L, L}; difforder = 4;
grid = Array[# &, points, domain];
(* Definition of pdetoode isn't included in this code piece,
   please find it in the link above. *)
ptoofunc = pdetoode[{Ar, Ai}[x, t], t, grid, difforder];
odebc = Map[ptoofunc, bc, {2}]
del = #[[2 ;; -2]] &;
odeic = del /@ ptoofunc@ic;
ode = del /@ ptoofunc@eqs;
sollst = NDSolveValue[{ode, odeic, odebc}, 
   Table[v[x], {v, {Ar, Ai}}, {x, grid}], {t, 0, tmax}];
{solAr, solAi} = rebuild[#, grid, -1] & /@ sollst;


Plot[{solAr[-L, t], solAr[L, t], solAi[-L, t], solAi[L, t]}, {t, 0, tmax}, 
 PlotStyle -> {Automatic, {Thick, Red, Dashed}, Dotted, Dotted}]

enter image description here

With[{d = Derivative[1, 0]}, 
 Plot[{d[solAr][-L, t], d[solAr][L, t], d[solAi][-L, t], d[solAi][L, t]}, {t, 0, 2}, 
  PlotStyle -> {Automatic, {Thick, Red, Dashed}, Dotted, Dotted}, PlotRange -> All]]

enter image description here

Suspicious v12 Solution

Since v12, "FiniteElement" method can handle nonlinear PDE, so it's possible to solve the problem with PeriodicBoundaryCondition in principle. Nevertheless, the v12 solution is suspicious:

test = NDSolveValue[{eqs, ic, 
     PeriodicBoundaryCondition[Ar[x, t], x == L, Function[x, x - 2 L]], 
     PeriodicBoundaryCondition[-Ai[x, t], x == L, Function[x, x - 2 L]]}, {Ar, Ai}, {t, 
     0, tmax}, {x, -L, L}, 
    Method -> {"MethodOfLines", 
      "SpatialDiscretization" -> {"FiniteElement", 
        "MeshOptions" -> "MaxCellMeasure" -> 0.01}}]; // AbsoluteTiming

With[{d = Derivative[1, 0]}, 
 Plot[{d[test[[1]]][-L, t], d[test[[1]]][L, t], d[test[[2]]][-L, t], 
   d[test[[2]]][L, t]}, {t, 0, 2}, 
  PlotStyle -> {Automatic, {Thick, Red, Dashed}, Dotted, Dotted}, PlotRange -> All]]

enter image description here

It's clear Derivative[1, 0][Ar][-L, t] == Derivative[1, 0][Ar][L, t] isn't satisfied. (Zero NeumannValue is set at $x=-L$? ) I guess the underlying issue may be related to that in this post.

  • $\begingroup$ Thank you very much @xzczd. I noticed that you didn't write a semicolon at the end of odebc = Map[ptoofunc, bc, {2}] on purpose to show me why you Map at level 2 :) Thanks! $\endgroup$
    – Nobody
    Jul 18 '19 at 11:03
  • $\begingroup$ a question: in eqs you defined the sequence of independent vars as [x,t], why does in the solution, say, solAr[t, -L], the sequence become [t,x]? $\endgroup$
    – Nobody
    Jul 18 '19 at 11:16
  • $\begingroup$ @Nobody By default, the first variable of the function generated by rebuild is the variable amounts to time, if you want the original order, use e.g. {solAr, solAi} = rebuild[#, grid, 2] & /@ sollst. $\endgroup$
    – xzczd
    Jul 18 '19 at 11:50
  • $\begingroup$ @Nobody Oops, I've made a mistake in my original check, and the original points turns out to be too small. Corrected. $\endgroup$
    – xzczd
    Jul 18 '19 at 12:10
  • $\begingroup$ why did you say that the original result was incorrect? How did you identify that the problem results from the low mesh resolution? Thank you! $\endgroup$
    – Nobody
    Jul 18 '19 at 14:50

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