Limit of a system of nonhomogeneous ODEs

Question

I have a system of ordinary differential equations with the following structure, $$ y'(t) = A(\lambda) y(t) + \lambda b(t), \qquad y(t)\geq 0, \> b(t)\geq 0 $$ Here the $4 \times 4$ matrix $A$ depends on some parameter $\lambda > 0$. I want to estimate $\lambda_{c}$ s.t. if $\lambda < \lambda_{c}$, $\lim_{t \to \infty} y(t) = 0 $. I know $b_{1}=b_{2}=0$ and $b_{3}$, $b_{4}$ can be bounded in terms of $y_{1}$ or $y_{2}$. Would transforming the system into $$ y'(t) = \tilde A(\lambda) y(t) + \lambda \tilde b(t) \> \text{where} \> y(t)\geq 0, \> \tilde b(t)\leq 0 $$ Be sufficient to conclude that if $\tilde A(\lambda)$ has all negative eigenvalues, $\lim_{t \to \infty} y(t) = 0 $? If this is the case, would making $|\tilde b(t)|$ smaller make my estimate for $\lambda_{c}$ better?

Answer 1

Hint.

Assuming that $b(t)$ is Laplace transformable, first, be sure that the eigenvalues for $A(\lambda)$ are non-positive and after that use the Laplace final value theorem. With null initial conditions we have

$$ \lim_{s\to 0}s\lambda \left(I s-A(\lambda)\right)^{-1}B(s) = y_{\infty} $$