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.

Any comments/references appreciated!

  • $\begingroup$ Where did you get this problem from? $\endgroup$
    – nicoguaro
    Sep 6 '19 at 23:37
  • 2
    $\begingroup$ 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! $\endgroup$
    – KartMan
    Sep 7 '19 at 3:14
  • $\begingroup$ I think that depending on the form of $c(k)$ it can be written as a polynomial or nonlinear eigenvalue problem. $\endgroup$
    – nicoguaro
    Sep 7 '19 at 22:05
  • $\begingroup$ @KartMan, I have a Mathematica package that can solve this kind of eigenvalue problem, if it is of interest to you $\endgroup$
    – SPPearce
    Nov 12 '19 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}


\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.

  • $\begingroup$ How is this a linear eigevalue problem when both $\lambda$ and $L$ are functions of the unknown $k$ which needs to be determined? $\endgroup$
    – KartMan
    Sep 9 '19 at 19:10
  • $\begingroup$ 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. $\endgroup$ Sep 9 '19 at 21:01
  • $\begingroup$ 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. $\endgroup$
    – KartMan
    Sep 10 '19 at 1:03
  • $\begingroup$ 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 $c(k)=\lambda$. $\endgroup$ Sep 10 '19 at 2:57
  • $\begingroup$ 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. $\endgroup$
    – KartMan
    Sep 10 '19 at 11:53

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