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Given the problem:

Find the time solutions of

$x \ddot{x} - \dot{x}^2 = 0$, $x > 1$

which satisfy $x(0) = 1$. Decide which solutions are asymptotically stable.

OK. Finding the solution is quite simple, and it yields:

$x = Ce^{Dt}$

With the initial condition, we get:

$x = e^{Dt}$

What I am a little bit unsure about is the notion of asymptotic stability. The book presents the theory behding this, but has no proper examples. What I get is that if we start off with two solutions for $t_0$ close to one another, then we have asymptotic stability if the absolute value of the difference between the two solutions approach zero as $t$ tends to infinity. In the above example, my intution tells me that this occurs only when $D<0$. If $D>0$, then we have exponential growth, and two solutions will then move further away from each other. If $D=0$, then a slight disturbance which causes $D$ to become positive, will again ruin the stability. However, when $D<0$, $e^{Dt}$ will approach zero as $t$ tends to inifinity no matter what value we choose for $D$. Hence the absolute value of the difference between two solutions will also approach $0$.

I would truly appreciate it if anyone can confirm/disconfirm that my reasoning above is correct!

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    Sure. The definition is as follows: Let $\mathbf{x^*}$ be a stable (or uniformly stable) solution for $t \geq t_0$. If additionaly there exists \eta(t_0) > 0 such that ||\mathbf{x}(t_0) - \mathbf{x}^* (t_0)|| < \eta \to \lim_{t \to \infty}||\mathbf{x}(t) - \mathbf{x}^* (t)|| = 0, then the solution is said to be asymptotically stable (or uniformaly and asymptotically stable).2012-02-11

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Asymptotic stability has to do with perturbing the initial condition, not perturbing the solution in some way. So suppose we have a family of solutions $x(t)=x(0)e^{Dt}$.

When $D<0$, as you say, $\lim\limits_{t\to\infty}x(t)=0$, so the solution is asymptotically stable.

When $D=0$, $\lim\limits_{t\to\infty}x(t)=x(0)$; thus, the solution is not asymptotically stable.

When $D>0$, $\lim\limits_{t\to\infty}x(t)=\infty$; thus, the solution is not asymptotically stable.

Thus, you properly determined when the solutions were asymptotically stable, but I don't think that varying $x(0)$ should vary $D$.

Another Point:

The problem also specifies that $x>1$ (presumably for $t>0$ since $x(0)=1$). This means that $D>0$, and so the solution is not asymptotically stable.

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    Ah, you are absolutely correct! I simply forgot about the condition x>0. This shows that $D$, just like you say, has to be bigger than zero, and hence the given problem is not asymptotically stable. Thanks a lot!2012-02-11