I am looking for a generic estimate to the following problem coming from biology:

I am solving the ODE

$$y'(t)=Ay(t)+zf(t), y(0)=0.$$

where $f$ is an external force determined by us and $z$ a constant vector.

Now $A$ and $z$ come from some measurement, so in general they will be perturbed $\widetilde{A}$ and $\widetilde{z}.$

That is, I am actually solving

$$x'(t)=\widetilde{A}x(t)+\widetilde{z}f(t), x(0)=0$$

on my computer. Obviously, if any matrix $A$ or $\widetilde{A}$ had positive eigenvalues one could not say anything about the long-term dynamics, because there could be exponentially growing modes. Recall that by the Laplace transform the solution to these equations is then actually very simple:

$$\widehat{y}(t) = (t-A)^{-1}z\widehat{f}(t)$$ $$\widehat{x}(t) = (t-\widetilde{A})^{-1}\widetilde{z}\widehat{f}(t).$$

For small times, we can actually measure how close our model is to the true solution, that is by applying arbitrary forces in $L^1$ to the system we find for $t \in [0,T]$

$$\left\lVert y(t)-x(t) \right\rVert \le C \left\lVert f \right\rVert_{L^1[0,T]}$$

So we assume that both $A$ and $\widetilde{A}$ have only strictly negative eigenvalues.

Given that the error is known to satisfy a Lipschitz estimate for small times $t \in [0,T]$ and arbitrary controls in $L^1$.

Can we obtain any sharp ab-initio long-time estimates $t \in [0,\infty]$ of the form

$$\left\lVert y(t)-x(t) \right\rVert \le \widehat{C} \left\lVert f \right\rVert_{L^1[0,\infty]}$$ on this problem?

By ab-initio I mean estimates only depending on $C,\widetilde{A}$ and $\widetilde{z}$?

EDIT: If something similar would hold in any other $L^p$ space, I'd be interested as well. Or if there are any other global time estimates we can draw from this, please let me know.

  • $\begingroup$ You seem to assume that the eigenvalues of $A$ are real. Is that intentional? Do you know more about $A$? For instance, is it normal? $\endgroup$ – David Ketcheson Apr 29 '18 at 10:20

This is not an answer to your question, but a couple of remarks (much too long for a comment).

By the variation of constants formula, we have $$ y(t) = \int\limits_{0}^{t} f(s) e^{(t - s) A} z \, ds, \qquad x(t) = \int\limits_{0}^{t} f(s) e^{(t - s) \widetilde{A}} \tilde{z} \, ds, $$ consequently $$ \lVert y(t) - x(t) \rVert \le \int\limits_{0}^{t} \lvert f(s) \rvert \, \lVert e^{(t - s) A} (z - \tilde{z}) \rVert \, ds + \int\limits_{0}^{t} \lvert f(s) \rvert \, \lVert (e^{(t - s) A} - e^{(t - s) \widetilde{A}}) \tilde{z} \rVert \, ds $$ If the maximum of the real parts of the eigenvalues of $A$ is $- \mu$ with $\mu > 0$ then there exists $c_A \ge 1$ such that $$ \lVert e^{t A} \rVert \le c_A e^{- \mu t}, \quad t \ge 0. $$ Therefore we have $$ \lVert y(t) - x(t) \rVert \le \left(c_{A} \lVert z - \tilde{z} \rVert + (c_{A} + c_{\widetilde{A}}) \lVert \tilde{z} \rVert \right) \lVert f \rVert_{L^1(0, \infty)}. $$

The numbers $c_A$ are unbounded. So, if you ask whether the bound for the whole $[0, \infty)$ can be obtained from the bound on the (fixed) interval $[0, T]$, I would rather doubt that. But I have no counterexample.

  • $\begingroup$ If $\|\cdot\|=\|\cdot\|_2$ and $A, \tilde{A}$ are normal, then $c_A=c_\tilde{A}=1$, so this would be a complete answer (that was the motivation for my comment above). Even if $\tilde{A}$ is not guaranteed to be normal, but $A$ is and $\tilde{A}-A$ is small, then a reasonable bound could be obtained. $\endgroup$ – David Ketcheson Apr 30 '18 at 6:09
  • $\begingroup$ @DavidKetcheson I agree: certainly a reasonable bound can be obtained in terms of the distance between $A$ and $\widetilde{A}$. $\endgroup$ – user539887 Apr 30 '18 at 7:58

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.