There are three conditions when we want to get eigenfunctions in Cartesian coordinates, similar to eigenfunctions in cylindrical coordinates. The first is the correspondence of boundaries. The second is the azimuth number match, eg $l_1=l_2=0$.Third, the radius of the outer circle must meet the boundary condition. All three conditions are violated in the author’s code. I show how to find the eigenfunction with $\beta = 5.336$.
<< NDSolve`FEM`
r = 0.8; ne = 10; om = 0.0; kap = 1000;
reg = ImplicitRegion[x^2 + y^2 <= 2.0928^2, {x, y}]; f =
Function[{vertices, area},
Block[{x, y}, {x, y} = Mean[vertices];
If[x^2 + y^2 <= r^2, area > 0.001, area > 0.01]]];
mesh = ToElementMesh[reg, MeshRefinementFunction -> f];
glass = 1.45; air = 1.; k0 = (2 \[Pi])/1.55; b = 5;
n[R_] := ( .5*(1 - Tanh[kap*(R - r)])*(glass^2 - air^2) + air^2)*k0^2
helm = -Laplacian[u[x, y], {x, y}] - (b^2 + n[Sqrt[x^2 + y^2]])*
u[x, y] + I*om*(x*D[u[x, y], y] - y*D[u[x, y], x]);
boundary = DirichletCondition[u[x, y] == 0, True];
{vals, funs} =
NDEigensystem[{helm, boundary}, u[x, y], {x, y} \[Element] mesh,
ne];
Sqrt[Re[vals] + b^2]
(* {5.01271, 5.01285, 5.03804, 5.03825, 4.92974, 4.92969, \
5.28019, 5.28066, 5.33615, 5.60683}*)
Here we can see that the 9-th eigenvalue is equal to 5.33615, which corresponds to the desired $\beta = 5.336$. Figure 1 shows the mesh and eigenfunction along with the cylinder bounding the fiberglass.
{Show[ mesh["Wireframe"],
ContourPlot[x^2 + y^2 == r^2, {x, -1, 1}, {y, -1, 1},
ColorFunction -> Hue]],
Show[Plot3D[Re[funs[[9]]], {x, y} \[Element] mesh, PlotRange -> All,
PlotLabel -> Sqrt[vals[[9]] + b^2], Mesh -> None,
ColorFunction -> Hue],
Graphics3D[{Gray, Opacity[.4],
Cylinder[{{0, 0, -1}, {0, 0, 1.}}, r]}]]}
Figure 2 shows the remaining functions with $l\ne 0$ and desired eigenfunction with $l=0$
To isolate monotone solutions in the clad with l = 1, we add to the Helmholtz operator (b^2 + l^2/(x^2 + y^2))*u[x, y] and choose eigenfunctions that fade out in the outer region what is achieved when b = I*Sqrt[glass]*k0. Figure 3 shows one of the eigenfunctions. In this case, the desired value $\beta = 5.336$ is achieved with increasing size of the clad. In fig. 4 shows the same eigenfunction with a 2-fold increase in the size of the region of integration.
<< NDSolve`FEM`
r = 0.8; ne = 10; om = 0.0; kap = 1000; l = 1;
reg = ImplicitRegion[x^2 + y^2 <= 2^2, {x, y}]; f =
Function[{vertices, area},
Block[{x, y}, {x, y} = Mean[vertices];
If[x^2 + y^2 <= r^2, area > 0.001, area > 0.01]]];
mesh = ToElementMesh[reg, MeshRefinementFunction -> f];
glass = 1.45; air = 1.; k0 = (2 \[Pi])/1.55; b = I*Sqrt[glass]*k0;
n[R_] := ( .5*(1 - Tanh[kap*(R - r)])*(glass^2 - air^2) + air^2)*k0^2
helm = -Laplacian[
u[x, y], {x, y}] - (b^2 + n[Sqrt[x^2 + y^2]] + l^2/(x^2 + y^2))*
u[x, y] + I*om*(x*D[u[x, y], y] - y*D[u[x, y], x]);
boundary = DirichletCondition[u[x, y] == 0, True];
{vals, funs} =
NDEigensystem[{helm, boundary}, u[x, y], {x, y} \[Element] mesh,
ne];
Sqrt[vals + b^2]
(*{0. + 4.93777 I, 0. + 5.29335 I, 0. + 5.29463 I,
0. + 3.9743 I, 0. + 3.97351 I, 0. + 3.51044 I, 0. + 3.50924 I,
0. + 3.23389 I, 0. + 2.86891 I, 0. + 2.86774 I}*)
{Show[ mesh["Wireframe"],
ContourPlot[x^2 + y^2 == r^2, {x, -1, 1}, {y, -1, 1},
ColorFunction -> Hue]],
Show[Plot3D[Im[funs[[3]]], {x, y} \[Element] mesh, PlotRange -> All,
PlotLabel -> Row[{"\[Beta] = ", Im[Sqrt[vals[[3]] + b^2]]}],
Mesh -> None, ColorFunction -> Hue],
Graphics3D[{Gray, Opacity[.4],
Cylinder[{{0, 0, -1}, {0, 0, 1.}}, r]}]]}
Table[Plot3D[Im[funs[[i]]], {x, y} \[Element] mesh, PlotRange -> All,
PlotLabel -> Sqrt[vals[[i]] + b^2], Mesh -> None,
ColorFunction -> Hue], {i, Length[vals]}]
