Linear ODE When the Jacobian is Defective

112 Views Asked by At

Consider the linear ODE \begin{align} \dot x = J x, \: x\in \mathbb{R}^n \end{align} with the Jacobian $J$. Page 35 of Ordinary Differential Equations with Applications by Chicone asserts that the space $\mathbb{R}^n$ can always be decomposed as a direct sum of linear subspaces: the stable eigenspace (stable manifold) corresponding to the eigenvalues of $J$ with negative real parts, the unstable eigenspace (unstable manifold) corresponding similarly to the eigenvalues of $J$ with positive real parts, and the center eigenspace (center manifold) corresponding to the eigenvalues with zero real parts.

My question is: what are the implications of $J$ being defective? For example, suppose \begin{align} J = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \end{align} which has the eigenvalue of $-1$ with algebraic multiplicity of $2$, but the geometric multiplicity of $1$. The eigenspace is spanned by $[1,0]$. I was wondering what happens to the Hartman-Grobman Theorem in this case and how the space is partitioned.

Thank you, in advance, for your comments.

1

There are 1 best solutions below

0
On

What happens is that the existence of a topological conjugacy between linear equations whose matrices have only eigenvalues with nonzero real part depends solely on the number of eigenvalues with positive (or negative) real part. More precisely:

There exists a topological conjugacy between the flows of two linear equations $x'=Ax$ and $x'=Bx$ for some $n\times n$ matrices without eigenvalues on the vertical axis if and only if $A$ and $B$ have the same number of eigenvalues with positive (or negative) real part.

The proof becomes somewhat simple only after introducing specific norms related to $A$ and $B$.