Generalized Eigenvector in Dynamical System in Infinite Dimensions

Question

Consider a system of linear delay differential equations:

$$ \dot{z_1}(t) = z_1(t) + z_2(t-1) $$ $$ \dot{z_2}(t) = z_2(t) + z_3(t-1) $$ $$ \dot{z_3}(t) = z_3(t) - z_1(t-1) $$

The characteristic matrix is: $\Delta(\sigma) = \sigma \cdot Id - (Id + J e^{-\sigma})$, where $Id$ is the $3\times 3$ identity matrix, and $$ J = \left(\begin{array}{rrr} 0 & 1 & 0; \\ 0 & 0 & 1; \\ -1 & 0 & 0 \end{array}\right) $$ Clearly, the characteristic equation is: $p(\sigma) = \det(\Delta(\sigma)) = 0$, i.e. $p(\sigma) = (\sigma - 1)^3 + e^{-3\sigma} = 0$. It is easy to see that $\sigma=0$ is a characteristic root of algebraic multiplicity 2, as $$ p(0) = (-1)^3 + e^0 = -1 + 1 = 0, $$ $$ p'(\sigma) = 3(\sigma - 1)^2 - 3 e^{-3\sigma}, p'(0) = 3 - 3 = 0, $$ and $$ p"(\sigma) = 6(\sigma - 1) + 9 e^{-3\sigma}, p"(0) = -6 + 9 = 3 \not= 0 $$

However, when I tried to find the two generalized eigenvectors by solving $\Delta(0) \phi_2 = \phi_1$, where $\phi_1 = (1 -1 1)^T$, and $\phi_1$ is derived by solving $\Delta(0)\phi_1 = 0$, I found that the equation $\Delta(0) \phi_2 = \phi_1$ is inconsistent, i.e., there is no solution!

I did realize that $\Delta(0)$ is a matrix of rank 2, that is, the null space of $\Delta(0)$ is only one dimensional. But unfortunately, the null space of $(\Delta(0))^2$ is one dimensional too! This makes me unable to find $\phi_2$. I believe I must have missed something, or have misunderstood something. Any comment or suggestion would be highly appreciated!

Answer 1

Ordinary generalized eigenvector has nothing to do with the problem. Ordinary characteristic polynomial of the matrix $\Delta(0)$ is $\det(\lambda I-\Delta(0))=(\lambda+1)^3-1$ has simple root at $0$, and no generalized eigenvector.

To solve your equation, you proceed as follows.

1. Look for a solution of the form $e^{\sigma t}c$, where $c$ is a vector. You obtain $((\sigma-1)I-e^{-\sigma}J)c=0$. So $\sigma$ is a root of the "characteristic" equation, same as yours. This root $\sigma=0$ gives you a constant solution $(1,-1,1)^T$. But this root is of multiplicity $2$, so you want a second linearly independent solution.

2. This second linearly independent solution must in general be of the form $Y(t)=e^{\sigma t}(c_0+c_1t)$, where $c_0,c_1$ are vectors. In your case, $\sigma=0$, but I will keep it and make computation in the general case, for the equation $$Y'(t)=Y(t)+JY(t-1),$$ where $J$ can be arbitrary matrix. Substitute our form of the solution to the equation, divide by $e^{\sigma t}$ and group the terms with $t$ and without $t$. You obtain $$(\sigma I-I-e^{-\sigma}J)c_1=0,$$ and $$(\sigma I-I-e^{-\sigma}J)c_0=-(I+e^{-\sigma}J)c_1.$$ First equation means that $c_1$ is an "eigenvector" which we found on step 1. Using the first equation, we can transform the right hand side of the second equation: $$(\sigma I-I-e^{-\sigma}J)c_0=-\sigma c_1.$$ THIS is the equation of the "generalized eigenvector", adapted to our differential-difference equation. In our case $\sigma=0$, so we obtain the second solution in the form $$Y(t)=t(1,-1,1)^T.$$ You can verify that this is a solution.