1
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Here, my problem is that

$$ \left(\int_{-L_0}^{L_0} \left(\int_{-L_0}^{L_0}\mathrm e^{-(x-x_1)^2-(y-y_1)^2} ({\bf u_{\lambda}}(x_1, y_1) + {\bf v_{\lambda}}(x_1, y_1)) \, \mathrm dx_1\right) \, \mathrm dy_1\right)\\-\beta ^2 {\bf u_{\lambda}}(x,y) + \sin x \cos y \ {\bf u_{\lambda}}(x,y) + (\partial_{xx} + \partial_{yy}) {\bf u_{\lambda}}(x,y) = \lambda {\bf u_{\lambda}}(x,y) $$ and $$ -\left(\int_{-L_0}^{L_0} \left(\int_{-L_0}^{L_0}\mathrm e^{-(x-x_1)^2-(y-y_1)^2} ({\bf u_{\lambda}}(x_1, y_1) + {\bf v_{\lambda}}(x_1, y_1)) \, \mathrm dx_1\right) \, \mathrm dy_1\right)\\-\beta ^2 {\bf v_{\lambda}}(x,y) + \sin x \cos y \ {\bf v_{\lambda}}(x,y) + (\partial_{xx} + \partial_{yy}) {\bf v_{\lambda}}(x,y) = \lambda {\bf v_{\lambda}}(x,y) $$ where $\beta$ is constant. $u(x,y)$ and $v(x,y)$ is what we need. I want to solve the eigenvalues, and I have tried to do with as follows:

enter image description here

But it does not work, can you give me some advice about the partial-integral equation's eigenvalue? Thanks.

Clear["Global'*"] 
laplace1 = Laplacian[u[x, y], {x, y}];
laplace2 = Laplacian[v[x, y], {x, y}];
A = 1;
w2 = 1;
a = 0.1;
Subscript[\[Beta], 1] = 0.1;
Subscript[\[Beta], 2] = 0.1;
\[Beta] = Sqrt[Subscript[\[Beta], 1]^2 + Subscript[\[Beta], 2]^2];
L0 = 10;
Iter0 = 10^3;
N0 = 1000;(*本征值数目*)
L1u = Subscript[\[Beta], 1] D[u[x, y], x] + 
  Subscript[\[Beta], 2] D[u[x, y], y];
L2u = Sin[y] Cos[x] D[u[x, y], x] + Sin[x] Cos[y] D[u[x, y], y];
L3u = Subscript[\[Beta], 1] Sin[y] Cos[x] u[x, y] + 
   Subscript[\[Beta], 2] Sin[x] Cos[y] u[x, y];
L11u = 2 I L1u + 2 I a L2u - 2 a L3u;
(**********************)
L1v = Subscript[\[Beta], 1] D[v[x, y], x] + 
   Subscript[\[Beta], 2] D[v[x, y], y];
L2v = Sin[y] Cos[x] D[v[x, y], x] + Sin[x] Cos[y] D[v[x, y], y];
L3v = Subscript[\[Beta], 1] Sin[y] Cos[x]*v[x, y] + 
   Subscript[\[Beta], 2] Sin[x] Cos[y]*v[x, y];
L22v = -2 I L1v + 2 I a L2v - 2 a L3v;
eq1 = laplace1 - \[Beta]^2 u[x, y] + L11u + A^2 w2 \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(-L0\), \(+L0\)]\(\((
\*SubsuperscriptBox[\(\[Integral]\), \(-L0\), \(+L0\)]\((
\*SuperscriptBox[\(E\), \(\(-
\*SuperscriptBox[\((x - x1)\), \(2\)]\) - 
\*SuperscriptBox[\((y - y1)\), \(2\)]\)] \((v[x1, y1] + 
             u[x1, y1])\))\) \[DifferentialD]x1)\) \[DifferentialD]y1\
\)\);
eq2 = -laplace2 + \[Beta]^2 v[x, y] + L22v - A^2 w2 \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(-L0\), \(+L0\)]\(\((
\*SubsuperscriptBox[\(\[Integral]\), \(-L0\), \(+L0\)]\((
\*SuperscriptBox[\(E\), \(\(-
\*SuperscriptBox[\((x - x1)\), \(2\)]\) - 
\*SuperscriptBox[\((y - y1)\), \(2\)]\)] \((v[x1, y1] + 
             u[x1, y1])\))\) \[DifferentialD]x1)\) \[DifferentialD]y1\
\)\);
{vals, funs} = 
  NDEigensystem[{eq1, eq2}, {u[x, y], v[x, y]}, {x, -L0, L0}, {y, -L0,
     L0}, N0];
ListPlot[Table[{Re[vals[[i]]], Im[vals[[i]]]}, {i, 1, N0}]]

Give the code that looks more directly

Clear["Global'*"]; 
Clear[A, w2, a, L0, Iter0, N0]; 
laplace1 = Laplacian[u[x, y], {x, y}]; 
laplace2 = Laplacian[v[x, y], {x, y}]; 
A = 1; 
w2 = 1; 
a = 0.1; 
Subscript[\[Beta], 1] = 0.1; 
Subscript[\[Beta], 2] = 0.1; 
Clear["Global'*"];
Clear[A,w2,a,L0,Iter0,N0];
(*u[x_,y_]:=u[x,y];v[x_,y_]:=v[x,y];*)
laplace1=Laplacian[u[x,y],{x,y}];
laplace2=Laplacian[v[x,y],{x,y}];
A=1;
w2=1;
a=0.1;
Subscript[\[Beta], 1]=0.1;
Subscript[\[Beta], 2]=0.1;
\[Beta]=Sqrt[Subscript[\[Beta], 1]^2+Subscript[\[Beta], 2]^2];
L0=10;
Iter0=10^3;
N0=100;
L1u=Subscript[\[Beta], 1]D[u[x,y],x]+Subscript[\[Beta], 2]D[u[x,y],y];
L2u=Sin[y]Cos[x]D[u[x,y],x]+Sin[x]Cos[y]D[u[x,y],y];
L3u=Subscript[\[Beta], 1]Sin[y]Cos[x]u[x,y]+Subscript[\[Beta], 2]Sin[x]Cos[y]u[x,y];
L11u=2I L1u+2I a L2u-2 a L3u;
(**********************)
L1v=Subscript[\[Beta], 1]D[v[x,y],x]+Subscript[\[Beta], 2]D[v[x,y],y];
L2v=Sin[y]Cos[x]D[v[x,y],x]+Sin[x]Cos[y]D[v[x,y],y];
L3v=Subscript[\[Beta], 1]Sin[y]Cos[x]*v[x,y]+Subscript[\[Beta], 2]Sin[x]Cos[y]*v[x,y];
L22v=-2I L1v+2I a L2v-2 a L3v;
eq1=laplace1-\[Beta]^2 u[x,y]+L11u+A^2 w2 NIntegrate[ E^(-(x-x1)^2-(y-y1)^2) (v[x1,y1]+u[x1,y1]),{x1,-L0,L0},{y1,-L0,L0}];
eq2=-laplace2+\[Beta]^2 v[x,y]+L22v-A^2 w2 NIntegrate[ E^(-(x-x1)^2-(y-y1)^2) (v[x1,y1]+u[x1,y1]),{x1,-L0,L0},{y1,-L0,L0}];
{vals,funs}=NDEigensystem[{eq1,eq2,DirichletCondition[{u[x,y]==0,v[x,y]==0},True]},{u[x,y],v[x,y]},{x,-L0,L0},{y,-L0,L0},N0];
ListPlot[Table[{Re[vals[[i]]],Im[vals[[i]]]},{i,1,N0}]]
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2
  • $\begingroup$ Can you add the full code? It might be a syntax problem based on that error messages. $\endgroup$
    – user59583
    Commented Dec 21, 2018 at 9:49
  • $\begingroup$ @yunshi Solution of what problem is used under the sign of the integral? $\endgroup$ Commented Dec 21, 2018 at 14:16

1 Answer 1

2
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I do not see the correct formulation of the probes. Therefore, we solve the problem with zero boundary conditions.This is the zero iteration.

laplace1 = Laplacian[u[x, y], {x, y}];
laplace2 = Laplacian[v[x, y], {x, y}];
A = 0;
w2 = 1;
a = 0.1;
\[Beta]1 = 0.1;
\[Beta]2 = 0.1;
\[Beta] = Sqrt[\[Beta]1^2 + \[Beta]2^2];
L0 = 10;
Iter0 = 10^3;
N0 = 10;
L1u = \[Beta]1 D[u[x, y], x] + 
     \[Beta]2 D[u[x, y], y];
L2u = Sin[y] Cos[x] D[u[x, y], x] + Sin[x] Cos[y] D[u[x, y], y];
L3u = \[Beta]1 Sin[y] Cos[x] u[x, y] + 
      \[Beta]2 Sin[x] Cos[y] u[x, y];
L11u = 2 I L1u + 2 I a L2u - 2 a L3u;
(**********************)
L1v = \[Beta]1 D[v[x, y], x] + 
      \[Beta]2 D[v[x, y], y];
L2v = Sin[y] Cos[x] D[v[x, y], x] + Sin[x] Cos[y] D[v[x, y], y];
L3v = \[Beta]1 Sin[y] Cos[x]*v[x, y] + 
      \[Beta]2 Sin[x] Cos[y]*v[x, y];
L22v = -2 I L1v + 2 I a L2v - 2 a L3v;
int[0] = 0;
eq1 = laplace1 - \[Beta]^2 u[x, y] + L11u + int[0];
eq2 = -laplace2 + \[Beta]^2 v[x, y] + L22v - int[0];
{vals, funs} = 
    NDEigensystem[{eq1, eq2, 
    DirichletCondition[{u[x, y] == 0, v[x, y] == 0}, True]}, {u[x, 
     y], v[x, y]}, {x, -L0, L0}, {y, -L0,
         L0}, N0];


 ListPlot[Table[{Re[vals[[i]]], Im[vals[[i]]]}, {i, 1, N0}]]
Table[Plot3D[Re[First[funs[[i]]]], {x, -L0, L0}, {y, -L0,
        L0}, PlotRange -> All, PlotLabel -> vals[[i]], 
  ColorFunction -> "TemperatureMap"], {i, Length[vals]}]

Table[Plot3D[Re[Last[funs[[i]]]], {x, -L0, L0}, {y, -L0,
        L0}, PlotRange -> All, PlotLabel -> vals[[i]], 
  PlotPoints -> 50, ColorFunction -> Hue], {i, Length[vals]}]

fig1

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5
  • $\begingroup$ Actually, the perodic boundary can be considered, But the most challenged problem is how we can deal with the Convolution integral . $\endgroup$
    – yun shi
    Commented Dec 22, 2018 at 0:09
  • $\begingroup$ @yunshi Explain what function is used under the sign of the integral? This is one of funs[[i]] which is a solution to the equation with vals[[i]] or is it any solution? $\endgroup$ Commented Dec 22, 2018 at 4:34
  • $\begingroup$ Under the sign of the integral,it's a convolution integral with the kernel being Gaussian function, and another part is u which is what we should to determine. $\endgroup$
    – yun shi
    Commented Dec 22, 2018 at 4:50
  • $\begingroup$ @yunshi What is u[x1,y1]+v[x1,y1] under the sign of the integral? This is a solution to what problem? Can you add index like $u_\lambda (x,y)$ to distinguish the solution and eigenvalues? $\endgroup$ Commented Dec 22, 2018 at 13:29
  • $\begingroup$ under the sign of the integral, $u(x_1,y_1)=u_{\lambda}(x_1,y_1)$ which is unknown eigenvector corresponding to eigenvalue $\lambda$ $\endgroup$
    – yun shi
    Commented Dec 23, 2018 at 1:32

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