Skip to main content

Finding eigenvalues of a differential operator



I am trying to get the eigenvalues of the following differential operator


$$L\psi(r) = -f \partial_r (f \partial_r \psi(r)) + V \psi(r)$$


which must satisfy (obviously)


$$L \psi(r) = \omega^2 \psi(r)$$


where I want to acquire both the real and imaginary part of $\omega$. To my problem, we have


$$ f = 1 - \frac{2M}{r}$$ and $$ V = f \left( \frac{l (l-1)}{r^2} + \frac{2 (1-S^2) M}{r^3} \right) $$ with $ M = 1, l = 2, S=2$. The boundary conditions are $\psi(2M) = 0, \psi(\inf) = 0$. (To whomever may care, I am getting the real and imaginary oscillations of a Schwarzschild black hole. It is well studied in the literature, but I need to recover the result).


I tried three different ways to do it:


1) Using NDEigensystem:


f = 1 - 2*(M/r); 
V = f*(l*((l - 1)/r^2) + (2*(1 - S^2)*M)/r^3);

M = 1; l = 2; S = 2;
\[ScriptCapitalL] = f*D[f*D[\[Chi][r], r], r] + V*\[Chi][r];
\[ScriptCapitalB] = DirichletCo - ndition[\[Chi][r] == 0, True];
boundarydistance = 10;
{ev, ef} =
NDEigensystem[{\[ScriptCapitalL], \[ScriptCapitalB]}, \[Chi][
r], {r, 2*M, boundarydistance}, 3];
Print["The eigenvalues are = ", ev]
Print["The real part is = ", Re[Sqrt[ev]]]
Print["The imaginary part is = ", Im[Sqrt[ev]]]

Plot[Evaluate[ef], {r, 0, boundarydistance}]

THE PROBLEM:It works perfectly, for a condition which is not at infinity. When I try to set boundarydistance = :inf:, it gives me error.


2) NDSolve with "Shooting Method":


f = 1 - 2*(M/r); 
V = f*(l*((l - 1)/r^2) + (2*(1 - S^2)*M)/r^3);
M = 1; l = 2; S = 2;
\[ScriptCapitalL] = (-f)*D[f*D[\[Chi][r], r], r] + V*\[Chi][r];
NDSolve[{\[ScriptCapitalL] == \[Lambda]*\[Chi][r], \[Chi][2*M] == 0, \[Chi][Infinity] == 0}, \[Chi][r], {r, 0, 10},
Method -> {"Shooting"}]

Plot[%, {r, 0, 10}]

THE PROBLEM: It gives the following error: NDSolve::ndsv: Cannot find starting value for the variable [Chi]^[Prime]. Besides this, how to I recover the eigenvalue? It seems like, to use NDSolve, I need to put a value for it.


3) Using the magical package in one of the answers in here:


Needs["PacletManager`"]
PacletInstall["CompoundMatrixMethod",
"Site" -> "http://raw.githubusercontent.com/paclets/Repository/master"];
Needs["CompoundMatrixMethod`"]
f = 1 - 2*(M/r);
V = f*(l*((l - 1)/r^2) + (2*(1 - S^2)*M)/r^3);

M = 1; l = 2; S = 2;
sys = ToMatrixSystem[f*D[f*D[\[Chi][r], r], r] - V*\[Chi][r] == \[Omega]*\[Chi][r],
{\[Chi][3*M] == 0, \[Chi][Infinity] == 0}, \[Chi], {r, 3, 10}, \[Omega]];
Plot[Evans[\[Omega], sys], {\[Omega], 0, 15}]

THE PROBLEM: Well, it actually runs. But I need to start from somewhere far from zero (due to the singularity) and I cannot really understand the answer which is:


{{{0, 1}, {(12 - 10 r + 2 r^2 + \[FormalLambda] r^4)/((-2 + r)^2 r^2), (4 r - 2 r^2)/((-2 + r)^2 r^2)}}, {}, {}, {r, 2.5, 10}}

and a plot, which does not resembles me anything that I would like to get.


EDIT:



Since I am not converging to an answer, it may help to add some more details. The real equation that I am trying to solve is $$\frac{d^2 \psi(r)}{dr_*^2} + (\omega^2 - V(r)) \psi(r) =0$$ where the (tortoise) coordinate is defined via $$\frac{dr}{dr_*} = f(r)$$ with the definitions of both $V(r)$ and $f(r)$ are given above. If you use these two equations together, you will get in the first equation of this page.


I thought that working with a single differential equation, although a bit more complicate, might be easier. However, I have no idea, even so, how I would work with the two differential equations separately, since the problem with this tortoise coordinate is that it cannot give me a function $r(r_*)$, because it is a transcendental equation.




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....