I would like to solve the PDE $\frac{\partial u}{\partial t}=(1-x)\frac{\partial u}{\partial x}$, in the interval $[0,1]$, with initial condition $u(0,x)=e^{-218(\frac{-2}{3}-ln(1-x))^2}$, by manually implementing a pseudospectral method with Chebyshev nodes for spatial discretization and a Taylor method for time integration. The first ListPlot appears in a matter of seconds, but after that the code keeps running forever, to the point that I do not even know if it produces the correct results at the end. Is the code correct, and if so, can it become quicker? Please let me know for any clarifications.


M = (II+dt*L+((dt^2)/2)*L2+((dt^3)/6)*L3);
M= Developer`ToPackedArray[M, Real];
U=ConstantArray[0.,{Nt, Nx}];
U[[1, All]]=f[X];
U= Developer`ToPackedArray[U, Real];
ListPlot[Transpose[{X, U[[1, All]]}], PlotRange->All, Joined->True]
Do[U[[it+1,All]]=M.U [[it,All]],{it,1,Nt-1}]
ListPlot[Transpose[{X,U[[Nt, All]]}], PlotRange->All, Joined->True]
ListPlot[Transpose[{X,U[[Nt, All]]-f[X-T[[Nt]]]}], PlotRange->All, Joined->True]
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    $\begingroup$ The time consuming step is: Do[U[[it+1,All]]=M.U [[it,All]],{it,1,Nt-1}]You could try Compilâtion or ParallelEvaluate. $\endgroup$ Jan 14, 2022 at 15:35
  • $\begingroup$ @DanielHuber I see, thank you! Is the code running correctly for you? It is returning the two latter ListPlots empty for me $\endgroup$
    – JBuck
    Jan 14, 2022 at 16:09
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    $\begingroup$ Sorry, but I do not have the time (and nerve ): ) to wait for the loop to end. $\endgroup$ Jan 14, 2022 at 16:14
  • $\begingroup$ @DanielHuber Thanks anyway! $\endgroup$
    – JBuck
    Jan 14, 2022 at 16:17
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    $\begingroup$ @JBuck Sorry I didn't check your code in detail, I only tried to find an alternative solution using "MethodOfLines" $\endgroup$ Jan 14, 2022 at 16:47

1 Answer 1


There are typos in the code that x in L = (1 - x)*D1 is not defined and also in the initial data last point is Indeterminate. General structure of the code should be revisited as follows



b = -2/3; a = 218;
x0 = 0;
x1 = 1;
u[x_, t_, a1_, b1_] := Exp[-a1*(b1 + t - Log[1 - x])^2];
f[x_] := u[x, 0, a, b];
D[u, t] - (1 - x)*D[u, x]

der1[x0_] := 
  NDSolve`FiniteDifferenceDerivative[Derivative[1], x0, 
    "DifferenceOrder" -> "Pseudospectral"]@"DifferentiationMatrix";
Nx = 2^8;
II = IdentityMatrix[Nx];
n = Nx - 1;
theta = N[Table[i*Pi/n, {i, 0, n}]];
X = N[Chop[((x0 + x1)/2) + ((x0 - x1)/2)*Cos[theta]]];
dt = N[0.01*(x1 - x0)/n];
D1 = der1[X];
L = (II - DiagonalMatrix[X]) . D1;
L2 = L . L;
L3 = L2 . L;
M = (II + dt*L + ((dt^2)/2)*L2 + ((dt^3)/6)*L3);
M = Developer`ToPackedArray[M, Real];
Nt = Round[1/dt];
T = Table[it*dt, {it, 1, Nt}];
U[0] = Chop[f[X] /. Indeterminate -> 0] // Quiet;

Iteration on time takes about 0.25 s

Do[U[it] = M . U[it - 1], {it, 1, Nt}]

Visualization of function and logarithm of absolute error on several steps

    Table[{ListPlot[Transpose[{X, U[i]}], PlotRange -> All, 
   Joined -> True, PlotLabel -> i dt],
  ListPlot[Transpose[{X, Abs[U[i] - u[X, T[[i]], a, b] // Quiet]}], 
   PlotRange -> All, Joined -> True, ScalingFunctions -> "Log"]}, {i, 
  1, Nt, Round[Nt/10]}]

Figure 1

  • $\begingroup$ Thank you so much! I next tried to plot the energy error $E(t)-E(0)$ as a function of t, where the energy is $\Int__{0}^{1} (1-x) (\partial_x u)^2 \,dx \$. I used the Clenshaw-Curtis integration rule h[x_]:=(1/n)*(1-Sum[2*Cos[2*i*x]/(4*i^2 - 1), {i, (n-1)/2}]) weights=h[X]; weights[[1]]=1/(2*n^2); weights[[n]]=1/(2*n^2); weights.U[0] NIntegrate[f[x],{x,x0,x1}] energy=ConstantArray[0.,{Nt}]; Do[energy[[it]]=weights.(D1.U[[it,All]])^2,{it,1,Nt}] ListPlot[energy-energy[[1]]], to compute the energy,but weights.U[0] is nowhere near the actual integral value, any clue why? $\endgroup$
    – JBuck
    Jan 19, 2022 at 18:53
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    $\begingroup$ @JBuck Please, check that weights . ConstantArray[1, Length[weights]]=0.704224, and therefore weights is not defined as the weights of the Clenshaw-Curtis rule. We can use weights1 = NIntegrateClenshawCurtisRuleData[Nx/2, 20][[2]], but this is also not so precise since weights1 . (Drop[U[0], -1] + Drop[U[0], 1])/2=0.0619452 , and we expecting about 0.0617043. We should increase Nx up to 2^12 to get about 0.06172. $\endgroup$ Jan 19, 2022 at 23:39
  • $\begingroup$ Nvm I found the mistake about the weights, they should be defined as h[theta] instead of h[X]. On the other hand, when I try to compute the energy energy=ConstantArray[0.,{Nt}]; Do[energy[[it]]=weights.(D1.U[[it,All]])^2,{it,1,Nt}] ListPlot[energy-energy[[1]]], I get errors like Part::partd: Part specification U[[1,All]] is longer than depth of object, do you have any idea why this is happening? $\endgroup$
    – JBuck
    Jan 20, 2022 at 14:57
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    $\begingroup$ @JBuck Ok! It looks much better {weights . U[0], NIntegrate[f[x], {x, x0, x1}]} is about {0.0617043, 0.0617043}. To calculate energy we can use Do[energy[it] = weights . (D1 . U[it])^2, {it, 0, Nt}]; ListPlot[Table[energy[i] - energy[0], {i, 0, Nx}]] $\endgroup$ Jan 20, 2022 at 16:59
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    $\begingroup$ @JBuck Don't use energy=ConstantArray[0.,{Nt}];! Run code first to compute U and weights. Also in the last line use Do[energy[it] = weights . (D1 . U[it])^2, {it, 0, Nt}]; ListPlot[[Table[energy[i] - energy[0], {i, 0, Nt}]]. $\endgroup$ Jan 23, 2022 at 2:47

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