# Show that a solution of the IVP is bounded

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. :)

• Is $A$ symmetric? Commented Sep 5, 2019 at 5:37

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:

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

with

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

(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,

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

or

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

or

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

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

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

thus

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

or

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

$$OE\Delta$$.

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.