0
$\begingroup$

For a linear system of ODEs in $\mathbb{R}^{n}$ (with the usual inner product), say $\dot{V}(t) = A(t) V(t)$, we know that if $\xi_{1},\ldots,\xi_{k} \in \mathbb{R}^{n}$ and $V_{j}(t)=V_{j}(t,\xi_{j})$ are the corresponding solutions such that $V_{j}(0)=\xi_{j}$, the evolution of the $k$-dimensional volume $|V_{1}(t) \wedge\ldots \wedge V_{k}(t)|_{\bigwedge^{k}\mathbb{R}^{n}}$ satisfies the Liouville trace formula $$|V_{1}(t) \wedge\ldots \wedge V_{k}(t)|_{\bigwedge^{k}\mathbb{R}^{n}} = |\xi_{1} \wedge \ldots \wedge \xi_{k}|_{\bigwedge^{k}\mathbb{R}^{n}} \operatorname{exp}\left( \int\limits_{0}^{t} \operatorname{Tr} (A(s) \circ \Pi(s)) ds \right),$$ where $\Pi(s)=\Pi(s,\xi_{1},\ldots,\xi_{k})$ is the orthogonal projector onto the space $L(s)$ spanned by $V_{1}(s),\ldots,V_{k}(s)$. To provide effective estimates for the volume decay one usually considers the orthogonal basis $e_{1}(s),\ldots,e_{n}(s)$ of $\mathbb{R}^{n}$ such that $e_{1}(s),\ldots,e_{k}(s)$ span $L(s)$ and proceeds as $$\operatorname{Tr}(A(s) \circ \Pi(s)) = \sum\limits_{j=1}^{k}(A(s)e_{j},e_{j}) = \sum\limits_{j=1}^{k}\left(\frac{1}{2}(A(s)+A(s)^{*})e_{j},e_{j}\right) \leq \alpha_{1}(s) + \ldots + \alpha_{k}(s),$$ where $\alpha_{i}(s)$ is the $i$-th eigenvalue of $(A(s)+A(s)^{*})/2$ such that $\alpha_{1}(s) \geq \alpha_{2}(s) \geq \ldots \geq \alpha_{n}(s)$. If the linear system comes from the linearization along a trajectory of some nonlinear system, the matrix $(A(s)+A(s)^{*})/2$ is called the symmetrized Jacobi matrix of $A(s)$.

I am interested in such an analog and corresponding estimates for delay equations. Below I explain my attempts and several problems arising on this way.

Let us consider the scalar equation with delay: $\dot{x} = x(t) - \alpha x(t-\tau)$. To make it possible to talk about volumes we consider this equation in the Hilbert space $\mathbb{H} = \mathbb{R} \times L_{2}(-\tau,0;\mathbb{R})$ with the usual inner product. Define the operator $A \colon \mathcal{D}(A) \subset \mathbb{H} \to \mathbb{H}$ as $$(x,\phi) \overset{A}{\mapsto} \left(\phi(0) - \alpha \phi(-\tau), \frac{\partial}{\partial\theta} \phi\right)$$ for $(x,\phi) \in \mathcal{D}(A) := \{ (x,\phi) \in \mathbb{H} \ | \ \phi \in W^{1,2}(-\tau,0;\mathbb{R}), \phi(0)=x \}$. The evolution equation in $\mathbb{H}$ $$\dot{V}(t)=AV(t)$$ is well-posed and for $\xi \in \mathcal{D}(A)$ we have classical solutions $V(t)=V(t,\xi)$ (see, for example, Bátkai A., and Piazzera S. Semigroups for Delay Equations). It is not hard to prove that for $\xi_{1},\ldots,\xi_{k} \in \mathcal{D}(A)$ we have an analog of the Liouville trace formula $$|V_{1}(t) \wedge\ldots \wedge V_{k}(t)|_{\bigwedge^{k}\mathbb{H}} = |\xi_{1} \wedge \ldots \wedge \xi_{k}|_{\bigwedge^{k}\mathbb{H}} \operatorname{exp}\left( \int\limits_{0}^{t} \operatorname{Tr} (A \circ \Pi(s)) ds \right)$$ The spectrum of $A$ is determined by the roots of $$1-\alpha e^{-\tau \lambda} - \lambda = 0$$ If $\alpha \in (0,1)$ there are two real roots $\lambda_{1} > 0$ and $\lambda_{2} < 0$ and the others are located to the left from $\lambda_{2}$. If $\lambda_{1} + \lambda_{2} < 0$ then using the dichotomy we have the exponential decay of $2$-volumes. How this fact can be obtained from the trace formula? It is not hard to see that the adjoint of $A$ is given by the formula $$ (y,\psi) \overset{A^{*}}{\mapsto} \left(y + \psi(0), -\frac{\partial }{\partial \theta} \psi\right),$$ where $(y,\psi) \in \mathcal{D}(A^{*}) = \{ (y,\psi) \in \mathbb{H} \ | \ \psi \in W^{1,2}(-\tau,0), \psi(-\tau) = \alpha y \}$. We see that for $(x,\phi) \in \mathcal{D}(A) \cap \mathcal{D}(A^{*})$ we have $$A+A^{*} = (\phi(0)-\alpha\phi(-\tau) + x + \phi(0),0) = (x-\alpha^{2}x + x + x,0).$$ Thus $A+A^{*}$ is a bounded self-adjoint operator in $\mathbb{H}$ with the kernel of codimension one. Clearly, its spectrum has nothing to do with the volumes decay. This is not surprising since to proceed in the second equality from $$\operatorname{Tr}(A \circ \Pi(s)) = \sum\limits_{j=1}^{k}(Ae_{j},e_{j}) = \sum\limits_{j=1}^{k}\left(\frac{1}{2}(A+A^{*})e_{j},e_{j}\right)$$ we must have $e_{j} \in \mathcal{D}(A) \cap \mathcal{D}(A^{*})$, but we usually have only $e_{j} \in \mathcal{D}(A)$. Moreover, if $e_{j}(s) = (x(s),\phi(s))$ then we have $$\sum\limits_{j=1}^{k}(Ae_{j},e_{j}) = \sum\limits_{j=1}^{k}\left(\frac{3}{2} |\phi_{j}(0)|^{2} - \alpha \phi_{j}(0) \phi_{j}(-\tau) - \frac{1}{2} |\phi_{j}(-\tau)|^{2}\right)$$ It is not obvious how one may obtain the exponential decay from this expression.

So what's the right approach for the symmetrization of $A$ and the corresponding trace estimates in the case of delay equations?

$\endgroup$
  • $\begingroup$ This isn't my area, so I do apologise if I'm speaking out of turn, but... the construction you introduce To make it possible to talk about volumes doesn't seem at all to me like a natural step from what you're talking about at the start of your question. Is your main interest in evolution of volumes in $\mathbb{R}^d$ or are you more interested in abstract analogs of the Liouville trace formula? $\endgroup$ – DCM May 23 at 10:08
  • $\begingroup$ @DCM, delay equations generate infinite-dimensional dynamical systems. There are two approaches for the well-posedness of delay equations and the construction of the corresponding dynamical systems. The first one treats such equations in the space of continuous functions $C([-\tau,0];\mathbb{R})$ and the second one considers the Hilbert space setting. The latter approach allows to talk about volumes and other things concerned with the Hilbert space structure. I'm interested in the evolution of $k$-dimensional volumes in the introduced Hilbert space $\mathbb{H}$. $\endgroup$ – demolishka May 23 at 10:39
  • $\begingroup$ @DCM, the problem is in that it is easy to obtain the trace formula, but it is not obvious how to derive effective estimates from it (as we could do in the case of ODEs using the eigenvalues of the symmetrized matrix). $\endgroup$ – demolishka May 23 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.