Skip to main content

numerics - Fourier Collocation For Heat Equation



In fact My problem is this $$\frac{\partial u}{\partial t}+\ sin(y)\frac{\partial u}{\partial x}=\nu(\frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2})$$ But I wanted to test the method first to the heat equation and check if the L^2 norm of the solution behaves like this $$|u|_{L^2} =(\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} u^2 dx dy)^{1/2} \leq e^{-\nu t}$$ Given that $$\frac{\partial u}{\partial t}=\nu\Bigl(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\Bigr)$$ With the following periodic boundary conditions: $$u(-\pi,y,t)=u(\pi,y,t) \\ u(x,-\pi,t)=u(x,\pi,t) \\u_x(-\pi,y,t)=u_x(\pi,y,t)\\ u_y(x,-\pi,t)=u_y(x,\pi,t)\\ u(x,y,0)=\sin(x)$$



I have tried to solve this using fourier collocation method in mathematica And then using NDSolve to solve the system of ODe.


n = 11;
ν = 1;
T = 100;
u[x_, y_, t_] := \!\(
\*UnderoverscriptBox[\(∑\), \(k = 0\), \(n - 1\)]\(
\*UnderoverscriptBox[\(∑\), \(l = 0\), \(n - 1\)]\(a[k, l]\)[t]*
     EXP[I*k*x]*EXP[I*l*y]\)\);
R[x_, y_, t] = 
  D[u[x, y, t], t] - ν*(D[u[x, y, t], x, x] + D[u[x, y, t], y, y]);

{S1} = Table[
   R[(2 πk)/n, (2 πl)/n, mT/n] == 0, {k, 1, n - 2}, {l, 1, 
    n - 2}, {m, 1, n - 1}];
S2 = Table[
   u[(2 πk)/n, -π, t] == u[(2 πk)/n, π, t], {k, 1, 
    n - 2}];

S3 = Table[
     D[u[(2 πk)/n, -π, t], y] == 
      D[u[(2 πk)/n, π, t], y], {k, 1, n - 1}];[] ( {

   {\[Placeholder], \[Placeholder]}
  } )
S4 = Table[
   u[-π, (2 πl)/n, t] == u[π, (2 πl)/n, t], {l, 1, 
    n - 2}];

S5 = Table[
   D[u[-π, (2 πl)/n, t], x] == 
    D[u[π, (2 πl)/n, t], x], {l, 1, n - 1}];
S6 = Table[u[(2 πk)/n, y, 0] == Sin[(2 πk)/n], {k, 1, n - 2}];

Sys = Join[S1, S2, S3, S4, S5, S6];
Dimensions[Sys];

I have a problem plotting the solution using NDSovle . And How to plot the L^2 norm of the solution ?


Edited


n = 11;
ν = 1;
T = 100;
u[x_, y_, t_] := \!\(
\*UnderoverscriptBox[\(∑\), \(k = 0\), \(n - 1\)]\(

\*UnderoverscriptBox[\(∑\), \(l = 0\), \(n - 1\)]\(a[k, l]\)[t]*
     Exp[I*k*x]*Exp[I*l*y]\)\);
R[x_, y_, t] = 
  D[u[x, y, t], t] + 
   Sin[y]*D[u[x, , y, t], 
     x] - ν*(D[u[x, y, t], x, x] + D[u[x, y, t], y, y]);
S1 = Table[
   R[(2 πk)/n, (2 πl)/n, t] == 0, {k, 1, n - 2}, {l, 1, 
    n - 2}];
S2 = Table[

   u[(2 πk)/n, -π, t] == u[(2 πk)/n, π, t], {k, 1, 
    n - 2}];

S3 = Table[
   D[u[(2 πk)/n, -π, t], y] == 
    D[u[(2 πk)/n, π, t], y], {k, 1, n - 1}];
S4 = Table[
   u[-π, (2 πl)/n, t] == u[π, (2 πl)/n, t], {l, 1, 
    n - 2}];


S5 = Table[
   D[u[-π, (2 πl)/n, t], x] == 
    D[u[π, (2 πl)/n, t], x], {l, 1, n - 1}];
S6 = Table[u[(2 πk)/n, y, 0] == Sin[(2 πk)/n], {k, 1, n - 2}];
Sys = Join[S1, S2, S3, S4, S5, S6];
Dimensions[Sys]

Answer



First, we do not need periodic boundary conditions when implementing the Fourier method, since the functions used are periodic by definition. Secondly, we cannot use two sets of boundary conditions for the heat equation. Thus, the implementation of the Fourier method is such


n = 11;
\[Nu] = 1;

T = 100;
u[x_, y_, t_] := \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = \(-n\)\), \(n\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(l = \(-n\)\), \(n\)]\(a[k, l]\)[t]*
     Exp[I*k*x]*Exp[I*l*y]\)\);
eq = Flatten[
   Table[a[k, l]'[t] + \[Nu] a[k, l][t] (k^2 + l^2) == 0, {k, -n, 
     n}, {l, -n, n}]];
ic = Flatten[
   Table[a[k, l][0] == 

     1/(2 I) (KroneckerDelta[k, 1] - 
        KroneckerDelta[k, -1]) KroneckerDelta[l, 0], {k, -n, 
     n}, {l, -n, n}]];
var = Flatten[Table[a[k, l], {k, -n, n}, {l, -n, n}]];

sol = NDSolve[{eq, ic}, var, {t, 0, 100}];

The solution for t = 0, 5, 10 has the form of a sinusoid damping in amplitude


Table[Plot3D[
  Evaluate[Re[u[x, y, t] /. sol]], {x, -Pi, Pi}, {y, -Pi, Pi}, 

  Mesh -> None, ColorFunction -> "Rainbow"], {t, 0, 10, 5}]

Figure 1


The solution of the same problem obtained by the automatic method NDSolve


sol1 = NDSolveValue[{D[u1[x, y, t], 
      t] - \[Nu] Laplacian[u1[x, y, t], {x, y}] == 0, 
   u1[-Pi, y, t] == u1[Pi, y, t], u1[x, -Pi, t] == u1[x, Pi, t], 
   u1[x, y, 0] == Sin[x]}, u1, {x, -Pi, Pi}, {y, -Pi, Pi}, {t, 0, 100}]

Table[Plot3D[sol1[x, y, t], {x, -Pi, Pi}, {y, -Pi, Pi}, Mesh -> None, 

  ColorFunction -> "Rainbow"], {t, 0, 10, 5}]

Figure 2


Compare two solutions at one point x=Pi/2, y=0. We see that the solutions diverge at t> 5. Increasing the number of modes to n=22 does not change this picture.


LogLogPlot[{Evaluate[Abs[u[Pi/2, 0, t] /. sol]], 
  Abs[sol1[Pi/2, 0, t]]}, {t, 0, 100}, AxesLabel -> Automatic, 
 PlotLegends -> {"Fourier", "Automatic"}]

Figure 3


Consider the solution of the modified equation taking into account the term $\sin (y) u_x$. Fourier method



n = 22;
\[Nu] = 1;
T = 100;
u[x_, y_, t_] := \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = \(-n\)\), \(n\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(l = \(-n\)\), \(n\)]\(a[k, l]\)[t]*
     Exp[I*k*x]*Exp[I*l*y]\)\);
Table[{a[k, n + 1][t_] := 0, a[k, -n - 1][t_] := 0}, {k, -n, n}];
eq = Flatten[
   Table[a[k, l]'[t] + 

      k (a[k, l + 1][t] - a[k, l - 1][t])/2 + \[Nu] a[k, l][
        t] (k^2 + l^2) == 0, {k, -n, n}, {l, -n, n}]];
ic = Flatten[
   Table[a[k, l][0] == 
     1/(2 I) (KroneckerDelta[k, 1] - 
        KroneckerDelta[k, -1]) KroneckerDelta[l, 0], {k, -n, 
     n}, {l, -n, n}]];
var = Flatten[Table[a[k, l], {k, -n, n}, {l, -n, n}]];

sol = NDSolve[{eq, ic}, var, {t, 0, 10}];


Table[Plot3D[
  Evaluate[Re[u[x, y, t] /. sol]], {x, -Pi, Pi}, {y, -Pi, Pi}, 
  Mesh -> None, ColorFunction -> "Rainbow"], {t, 0, 10, 5}]

Figure 4


The automatic method NDSolve


sol1 = NDSolveValue[{D[u1[x, y, t], t] + 
      Sin[y] D[u1[x, y, t], x] - \[Nu] Laplacian[
        u1[x, y, t], {x, y}] == 0, u1[-Pi, y, t] == u1[Pi, y, t], 

    u1[x, -Pi, t] == u1[x, Pi, t], u1[x, y, 0] == Sin[x]}, 
   u1, {x, -Pi, Pi}, {y, -Pi, Pi}, {t, 0, 10}];

Table[Plot3D[sol1[x, y, t], {x, -Pi, Pi}, {y, -Pi, Pi}, Mesh -> None, 
  ColorFunction -> "Rainbow"], {t, 0, 3, 1}]

Figure 5


The solutions are quite different in appearance due to the uncertainty of periodic boundary conditions (solutions differ in phase). Although at a point x=Pi/2, y=0 the difference appears only when t>5 Figure 6


The calculation of the L2 norm and comparison with c Exp[-t]


f = Re[u[x, y, t] /. sol];

L2norm = Table[{t, 
   First[Sqrt[NIntegrate[f^2, {x, -Pi, Pi}, {y, -Pi, Pi}]]]}, {t, 0, 
   5, .2}];

c = Sqrt[NIntegrate[Sin[x]^2, {x, -Pi, Pi}, {y, -Pi, Pi}]];
Show[Plot[c Exp[-t], {t, 0, 5}, Frame -> True, PlotRange -> {-1, 4.5},
   Axes -> False], ListPlot[L2norm, PlotStyle -> Red, Axes -> False]]

Figure 7


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more