# Show that System $(I)$ is stable iff $X(t)$ is bounded.

I have a theorem:

For a linear homogeneous system: $$\dfrac{dx}{dt}=A(t)x \tag{I}$$

Where $A(t)=(a_{ij}(t))_{n \times n} \in C(\mathbb{R}^+,\mathbb{R}^{n \times n})$

Suppose that $X(t)$ be the fundamental matrix solution of the following reference system $(I)$.

Let $K(t,s)=X(t)X^{-1}(s)$ be the Cauchy matrix of the following reference system $(I)$.

Prove that:

• a/ System $(I)$ is stable iff $X(t)$ is bounded, it means $\exists M>0$ such that $$\|X(t)\| \le M, \forall t \ge 0$$

• b/ System $(I)$ is asymptotically stable iff $$\lim_{t \to +\infty}X(t)=0$$

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I have stuck when I try to show this theorem.

I have tried using the definition of stable, asymptotically stable. But I have no solution.

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Ps: (Or) if somebody knows/reads this theorem (book/pdf/djvu...) then you can post it.

Can anyone help me!

Any help will be appreciated! Thanks/

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Maybe a hint $X(t)$ is formed by linearly independent solutions. Then just follow the definitions. I can/will elaborate an answer in case you still need it. –  PepeToro Sep 25 '13 at 17:08
Of course, user58533. I need it. If you have an ANS, it's great. You're welcome! –  kimtahe6 Sep 25 '13 at 17:18

## 2 Answers

Here's my solution for Question a:

a/ A linear homogeneous system $\dot{x}=A(t)x$ is stable iff $X(t)$ is bounded.

Without loss of generality we can assume that $X(t_0)=E$, then $x(t)=X(t)x_0$.

• If $X(t)$ is bounded. Whence $$\left \| x(t) \right \|=\left \| X(t)x_0 \right \| \le \left \| X(t) \right \| \left \| x_0 \right \| \le M\left \| x_0 \right \|<\epsilon, \text{for}\left \| x_0 \right \|<\delta:=\frac{\epsilon}{M}$$ Hence, $x \equiv 0$ is stable.

• If $x \equiv 0$ is stable. Whence

$$\forall \varepsilon >0, \exists \delta >0 \ \text{such that}\ \left \| x_0 \right \| \le \delta \implies \left \| x(t) \right \| = \left \| X(t)x_0 \right \| < \epsilon, \forall t \ge t_0$$ Thus $$\left \| X(t)\alpha \right \|=\left \| X(t)\alpha \cdot \dfrac{\delta}{\left \| \alpha \right \|}\right \|\cdot \dfrac{\left \| \alpha \right \|}{\delta}<\dfrac{\epsilon}{\delta}\left \| \alpha \right \|, \forall \alpha \in \mathbb{R}^n$$ Hence, $\left \| X(t) \right \| \le M:=\dfrac{\epsilon}{\delta} \blacksquare$.

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My solution for Question b:

• If $\lim_{t \to \infty}X(t)=0$ then since $x(t)=X(t)X^{-1}(t_0)x_0$ we have: $$\left \| x(t) \right \| \le \left \| X(t) \right \|\cdot \left \| X^{-1}(t_0)\right \|\cdot \left \| x_0 \right \|$$ Thus, $\lim_{t \to \infty}\left \| x(t;t_0,x_0) \right \| =0$, forall $x(t)$ of $(I)$.

Hence, $(I)$ is asymptotically stable.

• If $(I)$ is asymptotically stable. How we can show that $\lim_{t \to \infty}X(t)=0$?
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The key step is to note that $x(t)=K(t,s)x_0$ solves the initial value problem

$$\dot x(t)=A(t)x(t)\\ x(s)=x_0.$$

Now just take the definitions of stability and use $x(t)=K(t,s)x_0=X(t)X^{-1}(s)x_0$. Furthermore, note that $X^{-1}(s)x_0$ is a constant vector so it can be absorbed in some other constant when you take the norm.

Assume the equilibrium point is at the origin (you can always do it).

Stable $\iff$ $||x(t)||< M$ $\iff$ $|| X(t)X^{-1}(s)x_0 ||=||X(t)||\cdot||X^{-1}(s)x_0||<M$ $\iff$ $|| X(t)||<\frac{M}{||X^{-1}(s)x_0||}=P$.

The division is well defined, so $P$ is another constant.

Exponential Stability $\iff$ $||x(t)||< Me^{-ct}, \, c>0, \, t>s$ $\iff$ $|| X(t)X^{-1}(s)x_0 ||=||X(t)||\cdot||X^{-1}(s)x_0||<Me^{-ct}$ $\iff$ $|| X(t)||<\frac{Me^{-ct}}{||X^{-1}(s)x_0||}=Pe^{-ct}$. Next, $\displaystyle\lim_{t\to\infty}Pe^{-ct}=0$.

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I agree with you: $x(t)=K(t,t_0)x_0=X(t)X^{-1}(t_0)x_0$, ok! But Why do you have: $x(t)=K(t,s)x_0=\boxed{X(t)X^{-1}(s)x_0 \cdot X^{-1}(s)x_0}$? –  kimtahe6 Sep 25 '13 at 17:27
Sorry, the dot is not multiplication, it is the end of the sentence. I will edit. –  PepeToro Sep 25 '13 at 17:30
You mean $$x(t)=X(t) \cdot C \implies X^{-1}(s)x_0=C=const$$. So $$\|x(t)\| \le \|X(t)\| \|X^{-1}(s)x_0\| =C \cdot \|X(t)\| <\epsilon$$, $\forall t \le t_0, \exists \delta >0, \|x_0\|< \delta$. Hence $M:=\epsilon /C$ –  kimtahe6 Sep 25 '13 at 17:40
The first $\leq$ is equality, everything else seems correct. –  PepeToro Sep 25 '13 at 17:46
I mean you can show that asymptotically stable iff $\|x(t)\|< Me^{-ct}, \, c>0, \, t>s$, when you get it in your proof. –  kimtahe6 Sep 25 '13 at 18:24