# Bounds for solutions of $x'=(A+B)x$

Let $$A(t),B(t)$$ be two matrix-valued functions, continuous in $$[0,\infty)$$, such that:

1) The ODE $$x'(t)=A(t)x(t)$$ has a fundamental matrix $$\Phi(t)$$ satisfying $$|\Phi(t)\Phi^{-1}(s)|\leq M$$ for all $$0\leq s\leq t$$, and

2) $$B(t)$$ satisfies $$\int_{0}^{\infty}|B(t)|dt<\infty$$.

Show that every solution of $$x'(t)=(A(t)+B(t))x(t)$$ is bounded on $$[0,\infty)$$.

Any ideas? Thanks!

• @ChanG: Just want to confirm that you intend $\mathbf{X}'(t)=[\mathbf{A}(t)+\mathbf{B}(t)]\mathbf{X}(t)$ and not $\mathbf{X}'(t)=\mathbf{A}(t)\mathbf{X}(t) +\mathbf{B}(t)$. Feb 24, 2019 at 15:38

The idea is to treat the original equation as if it were a nonhomogeneous linear equation $$\begin{equation*} x'(t) = A(t) x(t) + f(t) \end{equation*}$$ with $$f(t) = B(t) x(t)$$. Applying variation of constants formula we obtain $$\begin{equation*} x(t) = \Phi(t) \Phi^{-1}(0) x(0) + \int\limits_{0}^{t} \Phi(t) \Phi^{-1}(s) B(s) x(s) \, ds, \quad t \ge 0, \end{equation*}$$ consequently $$\begin{equation*} \lvert x(t) \rvert \le M \,\lvert x(0) \rvert + \int\limits_{0}^{t} M \, \lvert B(s) \rvert \, \lvert x(s) \rvert \, ds, \quad t \ge 0. \end{equation*}$$ Putting $$u(t) = \lvert x(t) \rvert$$, $$\alpha(t) \equiv M \,\lvert x(0) \rvert$$ and $$\beta(t) = M \, \lvert B(s) \rvert$$ in an integral form of Grönwall's inequality we obtain $$\begin{equation*} \lvert x(t) \rvert \le M \,\lvert x(0) \rvert \,\exp\!{\Bigl(M \int\limits_{0}^{t} \lvert B(s) \rvert \, ds\Bigr)} \le M \,\lvert x(0) \rvert \,\exp\!{\Bigl(M \int\limits_{0}^{\infty} \lvert B(s) \rvert \, ds\Bigr)} \end{equation*}$$ for all $$t \ge 0$$.