# Eigenvalue-like problem with coupled ODEs

I am looking at the following system of ODEs:

$$\begin{array}{r}{\left[c_{2}(k)-\partial_{\tau}^{2}\right] \varphi_{2}\left(\tau \right)=f_{21}(\tau) \varphi_{1}\left(\tau \right)} \\ {\left[c_{1}(k)-\partial_{\tau}^{2}\right] \varphi_{1}\left(\tau \right)=f_{11}(\tau) \varphi_{1}\left(\tau \right)+f_{12}(\tau) \varphi_{2}\left(\tau \right)} \end{array}$$

where $$c_1(k)$$ and $$c_2(k)$$ are known scalar functions of an unknown number $$k$$ and $$f_{21}, f_{12}, f_{11}$$ are known functions. The unknowns are $$k, \varphi_{1,2}$$ and I'd like to find $$k$$ numerically.

When there's only one ODE, $$c(k)$$ can be obtained by discretising time and solving a matrix eigenvalue problem:

$$\begin{array}{r}{\left[c(k)-\partial_{\tau}^{2}\right] \varphi\left(\tau \right)=f(\tau) \varphi\left(\tau \right)} \end{array}$$

I am curious if I can recast the coupled problem as a (generalized)eigenvalue problem as well? Otherwise, another possibility is to try a bisection-like method to find the correct $$k$$ so that the matrix is singular, but this seems clunky.

• Where did you get this problem from? – nicoguaro Sep 6 at 23:37
• It comes up in the context of studying growth rate of perturbations in certain dynamical systems. But, I have a feeling it appears more generally! – KartMan Sep 7 at 3:14
• I think that depending on the form of $c(k)$ it can be written as a polynomial or nonlinear eigenvalue problem. – nicoguaro Sep 7 at 22:05
• @KartMan, I have a Mathematica package that can solve this kind of eigenvalue problem, if it is of interest to you – KraZug Nov 12 at 8:20

Assume the equations are discretized on the $$\tau$$ grid, and introduce several column-vectors, using the notation where the superscript stands for the grid point index i $$\in$$ [1,...,n].

$$\vec{\tau}=\left[ \tau^1,\tau^2,...,\tau^{n} \right]$$

$$\vec{\phi_1}=\left[ \phi_1^1,\phi_1^2,...,\phi_1^{n} \right]$$

$$\vec{\phi_2}=\left[ \phi_2^1,\phi_2^2,...,\phi_2^{n} \right]$$

$$\vec{f_{11}}=\left[ f_{11}^1,f_{11}^2,...,f_{11}^{n} \right]$$

$$\vec{f_{12}}=\left[ f_{12}^1,f_{12}^2,...,f_{12}^{n} \right]$$

$$\vec{f_{21}}=\left[ f_{21}^1,f_{21}^2,...,f_{21}^{n} \right]$$

Next, the second derivative operator on the grid is a matrix $$\hat{\Delta}$$ of size $$n \times n$$.

Also, use $$\alpha (k) = c_2(k) / c_1(k)$$, and denote $$c_1(k) = \lambda$$ and $$c_2(k) = \alpha \lambda$$

Now, let's write both equations as a single linear system of size $$2n \times 2n$$ for the compound state column-vector $$\vec{\phi}$$ which is

$$\begin{equation} \vec{\phi} = [\phi_1^1,\phi_1^2,...,\phi_1^{n},\phi_2^1,\phi_2^2,...,\phi_2^{n}] \end{equation}$$

We'll use here several matrices of size $$2n \times 2n$$:

The left hand side matrix $$L$$ $$\begin{bmatrix} \hat{I} & \hat{0}\\ \hat{0} & \alpha \hat{I} \end{bmatrix}$$

the differential operator matrix $$D$$ $$\begin{bmatrix} \hat{\Delta} & \hat{0} \\ \hat{0} & \hat{\Delta} \end{bmatrix}$$

the cross-coupling matrix $$C$$ $$\begin{bmatrix} \hat{0} & \hat{I} \vec{f_{12}} \\ \hat{I} \vec{f_{21}} & \hat{0} \end{bmatrix}$$

and the forcing term matrix $$F$$ $$\begin{bmatrix} \hat{I} \vec{f_{11}} & \hat{0} \\ \hat{0} & \hat{0} \end{bmatrix}$$

Here $$\hat{I}$$ is the unit matrix of size $$n \times n$$, $$\hat{0}$$ is the null matrix of size $$n \times n$$.

Now our problem can be written as

$$\begin{equation} \lambda L \vec{\phi} = \left( D + C + F \right) \vec{\phi} \end{equation}$$

or

$$\begin{equation} \lambda \vec{\phi} = L^{-1} \left( D + C + F \right) \vec{\phi} \tag{*}\label{*} \end{equation}$$

This a linear eigenvalue problem, where the right hand side matrix contains the parameter $$\alpha$$, so the eigenvalues are functions of $$\alpha$$, which in turn is a function of $$k$$.

For a given value of $$k$$ there is a spectrum of eigenvalues, $$\lambda_0, \lambda_1$$ etc. Let's say we are interested in the smallest eigenvalue $$\lambda_0$$ which we'd just call $$\lambda$$. Solution of the eigenvalue problem $$\eqref{*}$$ by standard methods of linear algebra defines a function $$\lambda(k)$$, on the other hand by construction $$\lambda$$ must be equal to $$c_1(k)$$. The root of the equation $$\lambda(k) = c_1(k)$$ (if it exists) defines the solution of the problem. Since $$\lambda(k)$$ and $$c_1(k)$$ are two nonlinear functions the root of the equation $$\lambda(k) = c_1(k)$$ has to be sought by some iterative numerical technique.

• How is this a linear eigevalue problem when both $\lambda$ and $L$ are functions of the unknown $k$ which needs to be determined? – KartMan Sep 9 at 19:10
• It is a linear eigenvalue problem for determining $\lambda$, as long as $\alpha$ is given. But if you need to find $k$ then you have to solve the nonlinear problem $c_1(k) = \lambda(k)$ using the above machinery. – Maxim Umansky Sep 9 at 21:01
• well, as I had described in the orginal post, $k$ and thus $c(k)$ are unknown and need to be determined. I don't see how this formulation casts the problem of finding $k$ into an eigenvalue/generalised eigenvalue problem. – KartMan Sep 10 at 1:03
• I added some explanations on how that could be done. To determine $k$ you have to solve a nonlinear equation; but for your example with a single ODE you'd have to do it as well, to solve $c(k)=\lambda$. – Maxim Umansky Sep 10 at 2:57
• if you can find $\lambda$ getting $k$ is quite easy in my case. There's no numerics required. Also, I had indicated in my earlier post that you can do some iterative (e.g. bisection method) procedure to get $k$ but this is not the eigenvalue formulation. I don't think what you wrote describes this as any eigenvalue problem. – KartMan Sep 10 at 11:53