Propagation error for ODEs 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.
 A: 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.
