in my way to understand ODE's I've found some problems that I have no idea how to tackle and how to relate to what I've learn so far. For example, this one:

Let $x(t)\in C^{1}([0,T])$ be a solution of the IVP $\dot x = A(t)x$, $x(0)=x_{0}$ with $(t,x) \in [0,T]\times \mathbb{R}^{n}$ and $x_{0} \in \mathbb{R}^{n}$. Suppose $A(t)v \cdot v \leq M|v|^{2}$ for all $v \in \mathbb{R}^{n}$ and $M>0$ some constant. Show that $|x(t)|\leq |x_{0}|e^{Mt}$.

Any help would be really appreciated. Thanks so much for all your help. :)

  • 1
    $\begingroup$ Is $A$ symmetric? $\endgroup$
    – copper.hat
    Commented Sep 5, 2019 at 5:37

3 Answers 3


This would be a fairly straightforward application of Gronwall Inequality:

$$x'\cdot x = Ax\cdot x \leq M|x|^2 \implies \left(\frac{1}{2}|x|^2\right)' \leq 2M\left(\frac{1}{2}|x|^2\right)$$

$$\frac{1}{2}|x|^2 \leq \frac{1}{2}|x_0|^2 e^{2Mt} \implies |x| \leq |x_0|e^{Mt}$$


We may assume

$\vert x(t) \vert \ne 0, \; \forall t \in [0, T], \tag 0$

for if $\vert x(t) \vert = 0$ for some $t$, then uniqueness of solutions implies $x(t) = 0$ for all $t \in [0, T]$, in which case the desired result trivially holds. We may then argue as follows:

$\dfrac{d}{dt} \vert x(t) \vert^2 = \dfrac{d}{dt} \langle x(t), x(t) \rangle = 2\langle x(t), \dot x(t) \rangle$ $= 2 \langle x(t), A(t)x(t) \rangle \le 2M \langle x(t), x(t) \rangle = 2M \vert x(t) \vert^2; \tag 1$


$\vert x(t) \vert \ne 0, \tag 2$

(1) may be written

$\dfrac{1}{\vert x(t) \vert^2}\dfrac{d}{dt} \vert x(t) \vert^2 \le 2M; \tag 3$

that is,

$\dfrac{d}{dt} \ln \vert x(t) \vert^2 \le 2M, \tag 4$


$2 \dfrac{d}{dt} \ln \vert x(t) \vert \le 2M,\tag 5$


$\dfrac{d}{dt} \ln \vert x(t) \vert \le M, \tag 6$

which we integrate 'twixt $0$ and $t$:

$\ln \left ( \dfrac{\vert x(t) \vert}{\vert x(0)\vert} \right ) = \ln \vert x(t) \vert - \ln \vert x(0) \vert = \displaystyle \int_0^t \dfrac{d}{ds} \ln \vert x(s) \vert \; ds \le \int_0^t M \; ds = Mt; \tag 7$


$ \dfrac{\vert x(t) \vert}{\vert x(0)\vert} \le e^{Mt}, \tag 8$


$\vert x(t) \vert \le \vert x(0) \vert e^{Mt}, \tag 9$



Let $\phi(t) = {1 \over 2} \|x(t)\|^2 $, $\eta(t) = e^{-2Mt} \phi(t)$.

$\dot{\phi}(t) = x^T(t) \dot{x}(t) = x^T A(t) x(t) \le M \|x(t)\|^2 = 2M \phi(t)$

$e^{-2Mt} (\dot{\phi}(t) - 2M \phi(t)) = \dot{\eta}(t) \le 0$, and so $\eta(t) \le \eta(0)$.

Then $\phi(t) \le e^{2Mt} \phi(0)$ and taking square roots gives the desired result.


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