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I am reading a passage from Perko's book about the Stable Manifold Theorem. Here is the problem:

Let $\dot x = f(x)$ be a system where $f: E \subset \mathbb{R}^{n} \rightarrow \mathbb{R}^n$ (with $E$ open and containing the origin). The linearization of this system at $0$ is $\dot x = Ax$ where $A = Df(0)$. Suppose $A$ has $k$ eigenvalues $\lambda_{j}$ with negative real part and $n - k$ eigenvalues $\lambda_{j}$ with positive real part. Then we can write the eigendecomposition as $A = C\Lambda C^{-1}$ with

$\Lambda = \begin{bmatrix} P & 0 \\ 0 & Q \end{bmatrix}$, where $P$ is the matrix for the first $k$ eigenvalues and $Q$ is the matrix for the remaining $n - k$ eigenvalues. For the eigenvalues in $P$, choose $\alpha > 0$ sufficiently small so that $Re(\lambda_{j}) < -\alpha < 0$.

If we define $U(t) = \begin{bmatrix} e^{Pt} & 0 \\ 0 & 0 \end{bmatrix}$ and $V(t) = \begin{bmatrix} 0 & 0 \\ 0 & e^{Qt} \end{bmatrix}$, then $\dot U = \Lambda U$ and $\dot V = \Lambda V$.

Now we get to something that Perko claims but does not prove: There exists $K > 0$ sufficiently large and $\sigma > 0$ sufficiently small that

$\|U(t)\| \leq Ke^{-(\alpha + \sigma)t}$ for $t \geq 0$ and $\|V(t)\| \leq Ke^{\sigma t}$ for $t \leq 0$.

There is a discrete version of the first part discussed here, but that problem only addresses a matrix $A$ with no unstable subspace. Can someone explain how to prove the full claim here? Thanks.

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    $\begingroup$ Why is $A$ necessarily disgonalisable? $\endgroup$ – copper.hat Apr 20 at 16:59
  • $\begingroup$ For sure, Perko does not assume diagonalizability. A general Jordan form could be used for $\Lambda$, $P$, and $Q$. I don't know how to prove the claim even for $A$ diagonalizable, though. $\endgroup$ – user231 Apr 20 at 17:01
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Note that if $\operatorname{re} \lambda < \beta$ for all $\lambda \in \sigma(B)$, then there is some $K$ such that $\|e^{Bt}\| \le K e^{\beta t}$ for all $t \ge 0$. (This is true even if $B$ is not diagonalisable.)

Let $q = \min_{\lambda \in \sigma(Q)} \lambda$, $p = \max_{\lambda \in \sigma(P)} \lambda$. Note that $p < -\alpha < 0 < q$.

Pick any $\sigma$ such that $p < -(\alpha+\sigma) < 0 < \sigma < q$.

Then there is a $K$ such that $\|e^{Pt} \| \le K e^{-(\alpha+\sigma)t}$ and $\|e^{-Qt}\| \le K e^{-\sigma t}$ for all $t \ge 0$.

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