Show that all the roots of $\frac{dx}{dt}=A(t)x$ are bounded in $[t_0, \infty)$.

Question

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.

Answer 1

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!!!

Answer 2

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.