Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question

1 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.