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?

To make it possible to talk about volumesdoesn'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