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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!

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

1 Answer 1

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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 $t \ge 0$.

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  • $\begingroup$ Thanks! I was looking for something like that but never thought to treat B as the inhomogeneous part... $\endgroup$
    – ChanaG
    Feb 25, 2019 at 7:35

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