Prove that $\left \|x(t_0) \right \|\exp \left (-\int_{t_0}^{t} \left \|A(t_1) \right \|\mathrm{d}t_1 \right )\le \left \| x(t) \right \|$

Question

I have a problem:

For $\dfrac{dx}{dt}=A(t)x$, where $A(t)\in C\left [t_0,+\infty \right )$. Prove that: $$\left \|x(t_0) \right \|\exp \left (-\int_{t_0}^{t} \left \|A(t_1) \right \|\mathrm{d}t_1 \right )\le \left \| x(t) \right \|\le \left \|x(t_0) \right \|\exp \left (\int_{t_0}^{t} \left \|A(t_1) \right \|\mathrm{d}t_1 \right ), \forall t \ge t_0$$ I have thought about Bellman-Gronwall's inequality, but I still no solution :( . Can anyone have an idea or a solution? Any help will be appreciated! Thanks.

Answer 1

Hint: $\ln |x(t)|-\ln |x(t_0)| = \int_{t_0}^t A(t_1)dt_1.$

Answer 2

Ok! Here's my solution:

• Since $$\dfrac{dx}{dt}=A(t)x $$, $$\implies \int_{\tau}^{t}\dot{x}(t)\mathrm{d}t = \int_{\tau}^{t}A(t)x(t)\mathrm{d}t, \forall t \ge \tau \ge t_0$$ $$\implies x(t) =x(\tau)+\int_{\tau}^{t}A(t)x(t)\mathrm{d}t,\forall t \ge \tau \ge t_0\\$$

• Thus, $$\|x(t)\| =\|x(\tau)\|+\int_{\tau}^{t}\|A(t)\| \|x(t)\|\mathrm{d}t,\forall t \ge \tau \ge t_0\\$$

• Now, we apply general Bellman-Gronwall's inequality, we have: $$\left \|x(t_0) \right \|\exp \left (-\int_{t_0}^{t} \left \|A(t_1) \right \|\mathrm{d}t_1 \right )\le \left \| x(t) \right \|\le \left \|x(t_0) \right \|\exp \left (\int_{t_0}^{t} \left \|A(t_1) \right \|\mathrm{d}t_1 \right ), \forall t \ge t_0$$

QED