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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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