Skip to main content

differential equations - Numerically Solving Helmholtz over the Rectangle - Why does this code only give eigenfunctions of the form $u_{m1}$



I have been following the method for numerically solving the Helmholtz equation in this example (the answer by User21) and have come across two problems. I have been implementing the method for a 2x1 rectangle domain. The code I have used is as follows:


 Needs["NDSolve`FEM`"]


boundaryMesh =
ToBoundaryMesh[Polygon[{{1, 1/2}, {1, -1/2}, {-1, -1/2}, {-1, 1/2}}]]
boundaryMesh["Wireframe"]


mesh = ToElementMesh[boundaryMesh, "MeshOrder" -> 1,
"MaxCellMeasure" -> 0.001];
mesh["Wireframe"]


k = 1/10;
pde = D[u[t, x, y], t] - Laplacian[u[t, x, y], {x, y}] +
k^2 u[t, x, y] == 0;
\[CapitalGamma] = DirichletCondition[u[t, x, y] == 0, True];


nr = ToNumericalRegion[mesh];
{state} =
NDSolve`ProcessEquations[{pde, \[CapitalGamma], u[0, x, y] == 0},
u, {t, 0, 1}, {x, y} \[Element] nr];

femdata = state["FiniteElementData"]

initBCs = femdata["BoundaryConditionData"];
methodData = femdata["FEMMethodData"];

initCoeffs = femdata["PDECoefficientData"];


vd = methodData["VariableData"];
sd = NDSolve`SolutionData[{"Space" -> nr, "Time" -> 0.}];


discretePDE = DiscretizePDE[initCoeffs, methodData, sd];
discreteBCs = DiscretizeBoundaryConditions[initBCs, methodData, sd];



load = discretePDE["LoadVector"];
stiffness = discretePDE["StiffnessMatrix"];
damping = discretePDE["DampingMatrix"];


DeployBoundaryConditions[{load, stiffness, damping}, discreteBCs]


nDiri = First[Dimensions[discreteBCs["DirichletMatrix"]]];


numEigenToCompute = 5;
numEigen = numEigenToCompute + nDiri;


res = Eigensystem[{stiffness, damping}, -numEigen];
res = Reverse /@ res;
eigenValues = res[[1, nDiri + 1 ;; Abs[numEigen]]];
eigenVectors = res[[2, nDiri + 1 ;; Abs[numEigen]]];
(*res=Null;*)



evIF = ElementMeshInterpolation[{mesh}, #] & /@ eigenVectors;

I then plot the 2nd eigenfunction (for example) using:


Plot3D[Evaluate[evIF[[2]][x, y]], {x, y} \[Element] mesh, 
PlotRange -> All, Axes -> None, ViewPoint -> {0, -2, 1.5},
Boxed -> False, BoxRatios -> {1, 1, 1/4}, ImageSize -> 612,
Mesh -> All]


and I have (attempted) to plot the nodal lines of the eigenfunction using:


ContourPlot[Evaluate[evIF[[2]][x, y]]==0, {x, y} ∈ mesh, 
PlotRange -> All, Axes -> None, ImageSize -> 612,
PlotPoints -> 40]

The first problem is this:


It is well known that the nodal lines of the rectangle have the form $u_{mn}$, but the code seems to only give the eigenfunctions for $u_{m1}$ ($n$ is fixed at $1$) [see the below image]. I would be really grateful if somebody could tell me why this is happening. I am just starting to learn the basics of mathematica, so my knowledge is quite limited. I have been reading over the code but cannot seem to understand why it produces only eigenvalues of this form. In the image below, you can see that the code fails to produce eigenfunctions for $u_{12}$, $u_{21}$ and $u_{13}$.


Eigenfunctions


The second issue is not so much of a problem, but more myself seeking some advice from the experts. Basically, I would like to know whether my contour plot for the nodal lines is the best way to approach this task. I have naively asked Mathematica to plot when the eigenfunction vanishes. Is there a better way of doing this?


I am very new to mathematica and I am not the most experienced of individuals when it comes to mathematics. Hence, if there are obvious flaws in my understanding of mathematica, or the mathematical material, then I would be more than happy for you to point them out.



I look forward to reading your responses.




Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....