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7
votes
1answer
123 views

How to solve this 1st-order linear ODE system with a few discrete eigenvalues?

I am trying to solve the eigensystem of a 1st-order linear ODE system in the region $(-\infty,\infty)$ and with Dirichlet boundary condition at the infinities $$ -i\partial_xu(x)+f^*(x)v(x)=\lambda u(...
3
votes
1answer
314 views

1st-order linear ODE system gives inaccurate/biased solutions

Consider an ODE eigensystem $$ t(y+\frac{1}{s})a(y)+[(q+\frac{1}{2}+\frac{s}{2}y)+s(y\partial_y+\frac{1}{2})]b(y)=\lambda a(y)\\ t(y+\frac{1}{s})b(y)+[(q+\frac{1}{2}+\frac{s}{2}y)-s(y\partial_y+\frac{...
2
votes
1answer
118 views

How to solve this 2nd-order ODE with quadratic coefficients?

Consider an ODE eigensystem $$ \begin{bmatrix} 0 & d_1-\mathrm id_2 \\ d_1+\mathrm id_2 & 0 \end{bmatrix} \begin{bmatrix} a(y) \\ b(y) \end{bmatrix} = \lambda \begin{bmatrix} a(y) \\ b(...
9
votes
1answer
323 views

How to solve this 2nd-order ODE with singularity?

I tried solving the eigenvalue problem of a 2nd-order ODE $$[b^2(k-2)^2y^2-2b(k-2)(1+2ky)+4k^2+b^2(k-2)3y]f(y) \\- 3b(3by-2)f'(y)\\-(3by-2)^2f''(y)=\lambda f(y)$$ with ...
2
votes
0answers
130 views

Boundary value problem, multiple dimensional shooting, coupled eigenvalue problem

Following the one dimensional boundary value problem here, I would like to understand the easiest way to solve a BVP for a coupled system. In the 1D case, BVP can be converted to an initial value ...
13
votes
3answers
827 views

An ODE system easily polluted with spurious eigenvalues

I tried solving the eigenvalue problem of a 1st-order ODE system (see the code below) with NDEigenvalue. (One option I found in it seems to be ...
6
votes
1answer
425 views

NDSolve eigenvalue problem of bound state

I am trying to solve this eigenvalue problem: \begin{align} \mu \Psi(r) & = -\frac{1}{2}\left ( \Psi^{\prime \prime}(r) + \frac{2}{r} \Psi' (r)\right ) -4\pi \Psi(r) \int _0^\infty dr' r'^2 \frac{...
3
votes
2answers
305 views

Why ODE's naive finite difference matrix works well for different boundary conditions

We know finite difference method (FDM) can replace $y''(x)$ as $\frac{1}{h^2}[y(x+h)+y(x-h)-2y(x)]$ or so. The naive way to write down the matrix of the differential operator is like the following, ...