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Consider a vector space $E$ with finite dimension and linear map $A: E \rightarrow E$. The following statements are equivalent:

  • $x'(t)=A \circ x(t)$ defines an attractor.

  • All eigenvalues of $A$ have negative real part.

  • There are $\alpha, \beta, c, C>0$ and $k\geq0$ such that$$c|t^k|e^{-\alpha t}\parallel v\parallel \leq \parallel e^{tA}(v)\parallel\leq Ce^{-\beta t}\parallel v \parallel, \ \ \ v\in E$$

And, if $x'(t)=A\circ x(t)$ defines an attractor, then $$\displaystyle \lim_{t\rightarrow - \infty}\parallel e^{tA}(v) \parallel=+\infty , \ \ \ \ \ \forall v\in E$$

I have the proof for finite dimension, but I use the Jordan canonical form theorem. Clearly, I can't use this argument in infinite dimension.

Consider a linear map $A$ and $\phi :\mathbb{R}\times E \rightarrow E$ the main flow of differential equation $x'(t)=A\circ x(t)$. This equation defines an attractor when, $$\displaystyle\lim_{t\rightarrow\infty}\phi(t,v)=0$$ $\forall v \in E,$ fixed.

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  • $\begingroup$ I think by "application" you mean "linear application", that is (in English) "linear map". Also, could you define "attractor"? $\endgroup$ – Lior Silberman Apr 21 '14 at 0:43
  • $\begingroup$ Now, I define an attractor. Is it clear? @LiorSilberman $\endgroup$ – Henfe Apr 21 '14 at 1:13
  • $\begingroup$ My question is in disagreement with something? $\endgroup$ – Henfe Apr 21 '14 at 2:18
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As stated the Theorem is false, even for bounded self-adjoint operators on Hilbert space. Take an operator with eigenvalue $-1/k$ on the $k$th basis vector. All the spectrum has negative real parts, wich does imply $\lim_{t\to-\infty}\Vert e^{At} v \Vert = \infty$ for all (non-zero) vectors $v$, but the more precise bound above that doesn't hold -- there is no uniform exponential rate of decay.

A more precise answer will depend on the definition of "attractor".

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  • $\begingroup$ Can you explain more? Can you give me an example? I do not quite understand... Thank you @LiorSilberman $\endgroup$ – Henfe Apr 30 '14 at 20:42

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