Show that all the roots of $\frac{dx}{dt}=A(t)x$ are bounded in $[t_0, \infty)$. For real system of equations$$\frac{dx}{dt}=A(t)x,(1)$$ where $A(t) \in C[t_0, +\infty)$.
Prove that if $\int_{t_0}^{\infty} \|A(t_1)+A^T(t_1)\|< +\infty$ then all the roots of (1) are bounded in $[t_0, \infty)$.
Can anyone help me? Thanks.
 A: How 'bout this:
Starting with the equation (1),
$\dot x = \frac{dx}{dt} = A(t)x \tag{1}$
we take the inner product of each side with $x$:
$\langle x,\dot x \rangle = \langle x, A(t)x \rangle, \tag{2}$
and then use the fact that
$\frac{d}{dt}\langle x, x \rangle = \langle \dot x, x \rangle +  \langle x,  \dot x \rangle = \langle x,\dot x \rangle +  \langle x,  \dot x \rangle = 2 \langle x,  \dot x \rangle \tag{3}$
to re-write (2) as
$\frac{d}{dt}\langle x, x \rangle = 2\langle x, A(t)x \rangle, \tag{4}$
and then go to work re-arranging (4):
$\frac{d}{dt}\langle x, x \rangle = 2\langle x, A(t)x \rangle = \langle A(t)x, x \rangle + \langle x, A(t)x \rangle$
$= \langle x,  A^T(t)x \rangle + \langle x, A(t)x \rangle = \langle x, (A^T(t) + A(t))x \rangle, \tag{5}$
and then keep on working at it:
$\frac{d}{dt}\langle x, x \rangle = \langle x, (A^T(t) + A(t))x \rangle$
$\le \Vert x \Vert \Vert (A^T(t) + A(t))x \Vert \le \Vert (A^T(t) + A(t)) \Vert \Vert x \Vert^2, \tag{6}$
and then finally use
$\frac{d}{dt}\langle x, x \rangle = \frac{d}{dt}\Vert x \Vert^2 \tag{7}$
to convert (6) to
$\frac{d}{dt}\Vert x \Vert^2 \le \Vert (A^T(t + A(t)) \Vert \Vert x \Vert^2, \tag{8}$
and then use the following fact, which follows from the uniqueness of the solutions of (1), which follows from the Lipschitz continuity of its right-hand side with respect to $x$, and its continuity with respect to $x$ and $t$, which I'll leave y'all to work out for yourselves, or else ask yet another question:  any $x(t)$ satisfying (1) is either never zero, or it is always zero, zero identically, since the only solution (by uniqueness!) with $x(t) = 0$ somewhere is the one with $x(t) = 0$ everywhere.  Well, the solution with $x(t) = 0$ everywhere looks pretty bounded to me!  So that means we can take $x(t) \ne 0$ everywhere, which means we can divide (8) through by $\Vert x \Vert^2$ to obtain
$\frac{d}{dt} \ln (\Vert x \Vert^2) \le  \Vert (A^T(t)+ A(t)) \Vert, \tag{9}$
which means that we can integrate (9) to obtain
$\ln (\Vert x(t) \Vert^2) - \ln (\Vert x(t_0) \Vert^2) \le \int_{t_0}^t \Vert (A^T(s)+ A(s)) \Vert ds, \tag{10}$
and then do even more re-arranging, this time of the logarithmo-algebraic sort, to get
$2\ln (\frac{\Vert x(t) \Vert}{\Vert x(t_0) \Vert}) \le \int_{t_0}^t \Vert (A^T(s)+ A(s)) \Vert ds, \tag{10}$
whence
$\Vert x(t) \Vert \le \Vert x(t_0) \Vert \exp(\frac{1}{2}\int_{t_0}^t \Vert (A^T(s)+ A(s)) \Vert ds), \tag{11}$
and since by hypothesis the integral occurring in the exponential on the right is bounded as $t \to \infty$, it follows that $x(t)$ is as well.  QED.
Hope this helps.  Cheerio, and as always
Fiat Lux!!!
A: Thank you for your answer!
It seems that you are very busy, and don't have much time to ANSWER? 
(asked Sep 14 at 8:37 $\to$ answered  Sep 22)
Here's my solution:


*

*Since $\dot{x(t)}=A(t)x(t)\implies \mathrm{\dot{x}^T(t)=x^T(t)A^T(t)}$ $\implies \left\{\begin{matrix}
 & \mathrm{ x^T\dot{x}=x^TAx} \\ 
 & \mathrm{\dot{x}^Tx=x^TA^Tx}
\end{matrix}\right.
$


Thus $\dfrac{d}{dt} \mathrm{\left ( x^Tx \right)}=\mathrm{ x^T \dot{x}}+\mathrm{\dot{x}^Tx}=\mathrm{x^T(A+A^T)x} \implies \dfrac{d}{dt}\left ( \|x(t)\|^2 \right )=x^T(t)\left [ A(t)+A^T(t) \right ]x(t)$


*

*By integration on $\left [ t_0 , t \right ]$, one gets:$$\|x(t)\|^2 =\|x(t_0)\|^2 +\int_{t_0}^{t}x^T(s)\left [ A(s)+A^T(s) \right ]x(s)\mathrm{d}s$$

*Whence $$\|x(t)\|^2 \le \|x(t_0)\|^2 +\int_{t_0}^{t}\|x^T(s)\|\left  \|A(s)+A^T(s) \right \| \|x(s)\| \mathrm{d}s  $$  $$= \|x(t_0)\|^2 +\int_{t_0}^{t}\left  \|A(s)+A^T(s) \right \| \|x(s)\|^2\mathrm{d}s$$

*Applying Gronwall - Bellman's inequality, we have:
$$\|x(t)\|^2 \le \|x(t_0)\|^2 \cdot\exp \int_{t_0}^{t}\left  \|A(s)+A^T(s) \right \| \mathrm{d}s  \\\\ \le \|x(t_0)\|^2 \cdot\exp \int_{t_0}^{+\infty}\left  \|A(s)+A^T(s) \right \| \mathrm{d}s< +\infty, \forall t \ge t_0 \blacksquare $$ QED.
