I have a system of ordinary differential equations that comes up in control theory for $\mathbf{x}: [0, \infty) \to \mathbb{R}^n$ that looks like
$$\frac{d\mathbf{x}}{dt} = [A + B(t)]\mathbf{x}\\\mathbf{x}(0) = \mathbf {c} \neq 0$$
The matrix $A$ is real with distinct eigenvalues $\lambda_1, ...,\lambda_n$ where $\Re(\lambda_j) < 0$ for all $j$.
Without $B(t)$ I know the solution must behave as $\lim_{t \to \infty} \mathbf{x}(t) = 0.$ Now I am unable to solve the system in closed form with $B(t)$ present, but I would like to know if the solution has the same asymptotic behavior ($\to 0$) as $t \to \infty$.
Conditions on $B(t)$:
$\lim_{t \to \infty}b_{ij}(t) = 0 \\ b_{ij} \in L^1([0,\infty))$
I think this may be enough to ensure the asymptotic behavior $\mathbf{x}(t) \to 0$ but maybe more is needed.
Thank you for any help.
The claim is correct. In fact, convergence of the elements of $B(t)$ to zero suffices and the $L_1$ property of $b_{ij}$ is not needed.
Since the eigenvalues of $A$ have all negative real parts for every $Q=Q^T>0$ there exists some $P=P^T>0$ such that $$PA+A^TP=-Q$$ Consider now the Lyapunov function candidate $V:=(1/2)x^TPx$. Its derivative has the form $$\dot{V}=\frac{1}{2}x^T(PA+A^TP)x+x^TPB(t)x\\ \leq -\frac{1}{2}x^TQx+\|P\|\|B(t)\|\|x\|^2$$ Due to the limit property of $b_{ij}$ there exists time $T>0$ such that $$\|P\|\|B(t)\|\|x\|^2\leq \frac{1}{4}x^TQx \qquad \forall t\geq T$$ and therefore $$\dot{V}\leq -\frac{1}{4}x^TQx\leq -\frac{\lambda_{\min}(Q)}{2\lambda_{\max}(P)}V \qquad t\geq T$$ The above inequality proves the exponential convergence of $V$ (and therefore $x$) to zero.