While solving one research paper published in Physical Review Letters, I came across the following equation and I am unable to solve it.

$$\frac{\partial f}{\partial t}−(\mathcal{H}(f)\left(\frac{\partial f}{\partial x}\right)=0 $$

where $ \displaystyle [\mathcal{H}(f)] \stackrel{\text{def}}{=} \text{p.v.} \frac{1}{\pi} \int_{- \infty}^{\infty} \frac{f(x')}{x - x'} ~ d{x'} $.

and $f=f(x,t)$ and initial condition is $f(x,0)=\cos(x)$.

In the paper it is given that the solution of the above mentioned equation is obtained with periodic conditions using pseudospectral method given below, $$F_k\{H_x\{f(x')\}\}=i \cdot\text{sgn}(k) F_k\{f(x)\}$$ where $$F_k\{f(x)\}=\frac{1}{\sqrt{2 \pi}}\int_{- \infty}^\infty e^{-ikx}f(x)dx $$ x∈[0,2Pi], t∈[0,1.275]

So I am thinking of application of Fourier transforms on both sides of the equation but I am unable proceed forward. Please solve the equation and can give the code for the same in mathematica.


1 Answer 1


I used the method of solving integro-differential equations proposed by Michael E2 on Solving an integro-differential equation with Mathematica I added new options to his code to solve this problem. The right figure in Figure 1 corresponds to Figure 1 of the article Viscous Flow at Infinite Marangoni Number by A. Thess, D. Spirn, and B. Juttner - see journals.aps.org/prl/pdf/10.1103/PhysRevLett.75.4614

L = Pi; tmax = 1.;
sys = {D[u[x, t], t] + 1/(Pi)*int[u[x, t], x, t]*D[u[x, t], x] == 0, 
   u[-L, t] == u[L, t], u[x, 0] == -Cos[x]};
periodize[data_] := 
 Append[data, {N@L, data[[1, 2]]}];(*for periodic interpolation*)
Block[{int},(*the integral*)
  int[u_, x_?NumericQ, t_ /; t == 0] := (cnt++;
    NIntegrate[-Cos[xp]/ (x - xp), {xp, x - L, x, x + L}, 
      Method -> {"InterpolationPointsSubdivision", 
        Method -> "PrincipalValue"}, PrecisionGoal -> 8, 
      MaxRecursion -> 20, AccuracyGoal -> 20] // Quiet);
  int[uppp_?VectorQ, xv_?VectorQ, t_] := Function[x, cnt++;
       Interpolation[periodize@Transpose@{xv, uppp}, xp, 
         PeriodicInterpolation -> True]/ (x - xp), {xp, x - L, x, 
        x + L}, Method -> {"InterpolationPointsSubdivision", 
         Method -> "PrincipalValue"}, PrecisionGoal -> 8, 
       MaxRecursion -> 20] (*adjust to suit*)] /@ xv // Quiet;
  (*monitor while integrating pde*)Clear[foo];
  cnt = 0;
  PrintTemporary@Dynamic@{foo, cnt, Clock[Infinity]};
  (*broken down NDSolve call*)
  Internal`InheritedBlock[{MapThread}, {state} = 
    NDSolve`ProcessEquations[sys, u, {x, -L, L}, {t, 0, tmax}, 
     StepMonitor :> (foo = t), 
     Method -> {"MethodOfLines", 
       "SpatialDiscretization" -> {"TensorProductGrid", 
         "MinPoints" -> 41, "MaxPoints" -> 81, 
         "DifferenceOrder" -> 2}}];
   MapThread[f_, data_, 1] /; ! FreeQ[f, int] := f @@ data;
   NDSolve`Iterate[state, {0, tmax}];
   sol = NDSolve`ProcessSolutions[state]]] // AbsoluteTiming
{Plot3D[u[x, t] /. sol, {x, -Pi, Pi}, {t, 0., 1.}, Mesh -> None, 
   ColorFunction -> Hue, AxesLabel -> Automatic] // Quiet, 
 Plot[Evaluate[Table[u[x, t] /. sol, {t, 0., 1., .2}]], {x, -Pi, 
    Pi}] // Quiet}

Figure 1 For this equation, we can apply another solution method by decomposing the desired function in a Fourier series:

u= Sum[f[m][t] Exp[I m x], {m, -Infinity, Infinity}]

Then the integral is exactly calculated for each mode. As a result, we find the system of equations and the numerical model

nn = 137; tm = 1.2; eq = 
 Table[f[m]'[t] - 
    Sum[ If[Abs[m - k] <= nn, (k - m) f[m - k][t], 0] Sign[k] f[k][
       t], {k, -nn, nn}] == 0, {m, -nn, nn}];
ic = Table[
   f[m][0] == (KroneckerDelta[m, 1] + KroneckerDelta[m, -1])/
     2, {m, -nn, nn}];
var = Table[f[i], {i, -nn, nn}];

sol1 = NDSolveValue[{eq, ic}, var, {t, 0, tm}];

     Sum[sol1[[m + 1]][t] Exp[I (-nn + m) x], {m, 0, 2*nn}]], {t, 0, 
     tm, .2}]], {x, 0, 2*Pi}, Mesh -> None, ColorFunction -> Blue, 
  AxesLabel -> Automatic, PlotLegends -> Automatic], 
   Sum[sol1[[m + 1]][t] Exp[I (-nn + m) x], {m, 0, 2*nn}]], {t, 0., 
   tm}, {x, 0, 2*Pi}, Mesh -> None, ColorFunction -> Hue, 
  AxesLabel -> Automatic]}

Figure 2

The results of calculations for the two models are the same, but the second model takes less time. So, for example, 341 seconds were spent on the test example for the first model, and only 0.49 seconds for the second model (on my laptop).

  • $\begingroup$ Sir, Can I know how you have mentioned the periodic boundary conditions in the above code? $\endgroup$ Commented Jun 17, 2019 at 6:56
  • $\begingroup$ @MohanAdityaSabbineni Then read carefully on mathematica.stackexchange.com/questions/192123/… $\endgroup$ Commented Jun 17, 2019 at 10:11
  • $\begingroup$ I got to know about the code and problems at present I am facing are 1. The domain of x here is [0,2Pi] but in the code it is taken as [-Pi,Pi] and proceeded which gave the false plot and is different from that of in research paper, I tried to formulate it but mathematica is giving many warnings while evaluating. 2. Here NIntegrate evaluates at the singularity when the singularity is already mentioned in the code. In code it was written {xp,x-L,x,x+L} and when I am trying to change it as {xp,0,x,x+2*L} the evaluation is not smooth and resulting in false plot. $\endgroup$ Commented Jun 24, 2019 at 10:20
  • $\begingroup$ @MohanAdityaSabbineni First, the data in Figure 1 correspond exactly to the data in Figure 1 of the article. I just shifted the area by $\pi $ using the periodicity of the solution. It seems that the curves are different in my Fig 1 and in the article. But it is an illusion. In the article under Figure 1 there is an explanation: successive curves are shifted. In other words, they shifted each curve upwards. I did not move the curves up, and left as it is. $\endgroup$ Commented Jun 24, 2019 at 12:31
  • 1
    $\begingroup$ I know that the curves were shifted, but my doubt is why we are getting different solution if we change the domain of x as 0 to 2*Pi? and also the curve may be different in range -Pi to Pi than that of in 0 to 2*Pi as we don't know the nature of solution, so I think we cannot shift the area on basis of periodicity of initial condition. $\endgroup$ Commented Jun 24, 2019 at 12:48

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