This question is connected with reference to my previous question (Solution of Dimensionless Partial Differential Equation Over a Semi-Circular Domain), where I asked for a help to solve the following BVP.
I would like to thank Nasser (https://mathematica.stackexchange.com/users/70/nasser) for his kind help with the following code.
ClearAll["Global`*"];
k=5;
\[Alpha]=3*\[Pi]/4;
\[Eta]=0;
pde=Laplacian[u[r,\[Theta]],{r,\[Theta]},"Polar"]==k^2*u[r,\[Theta]];
bc=u[1,\[Theta]]==1-(1-\[Eta])*Piecewise[{{0,\[Theta]<\[Alpha]},{1,True}}]
sol=NDSolveValue[{pde,bc},u,{r,0,1},{\[Theta],0,Pi}]
ContourPlot[sol[Sqrt[x^2 + y^2], ArcTan[x, y]], {x, -1, 1}, {y, 0, 1},
Contours -> 30, ColorFunction -> "LakeColors", AspectRatio -> Automatic,
PlotLegends -> Automatic, PerformanceGoal -> "Quality",
PlotPoints -> 100,ContourStyle->None]
Now I have the following BVP which uses the solution of first one.
I am trying proceed using the following code.
ClearAll["Global`*"];
k=5;
\[Alpha]=3*\[Pi]/4;
\[Eta]=0;
L=1;
pde1=Laplacian[s[r,\[Theta]],{r,\[Theta]},"Polar"]==k^2*s[r,\[Theta]];
bc1=s[1,\[Theta]]==1-(1-\[Eta])*Piecewise[{{0,\[Theta]<\[Alpha]},{1,True}}];
sol1=NDSolveValue[{pde1,bc1},s,{r,0,1},{\[Theta],0,Pi}]
pde2=Laplacian[u[r,\[Theta]],{r,\[Theta]},"Polar"]==-(4.0*L)-(k^2*s[r,\[Theta]]/.sol1);
bc2=u[1,\[Theta]]==0;
sol2=NDSolveValue[{pde2,bc2},u,{r,0,1},{\[Theta],0,Pi}]
ContourPlot[sol2[Sqrt[x^2 + y^2], ArcTan[x, y]], {x, -1, 1}, {y, 0, 1},
Contours -> 20, ColorFunction -> "LakeColors", AspectRatio -> Automatic,
PlotLegends -> Automatic, PerformanceGoal -> "Quality",
PlotPoints -> 75,ContourStyle->None]
But it shows some errors. How should I proceed correctly?