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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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You got the idea, now to write it more formally, you can say that the solution of the equation which satisfies $x(0)=x_0$ is $x(t,x_0)=x_0e^{Dt}$, then use the formal definition of asymptotic stability. –  Davide Giraudo Feb 11 '12 at 11:57
    
Thanks. Would the following work: Let $D_1 < 0$ and $D_2 < 0$ be two separate values. Then there exists a $\eta > 0$ such that $||e^{D_1 t_0} - e^{D_2 t_0}|| < \eta$. Since $D_1 < 0$ and $D_2 < 0$ it then follows that $\lim_{t \to \infty}||e^{D_1 t} - e^{D_2 t}|| = 0$, and thus the criterion for asymptotic stability is fulfilled. –  Kristian Feb 11 '12 at 12:07
    
Can you give the definition of asymptotic stability you have? –  Davide Giraudo Feb 11 '12 at 12:55
    
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). –  Kristian Feb 11 '12 at 13:46

1 Answer 1

up vote 1 down vote accepted

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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Cheers! Thanks a lot for the help! –  Kristian Feb 11 '12 at 16:43
    
Hi, just one more brief question. So in order to perturbate the initial condition, we need to set up a different initial condition for $x(0)$? Such as $x(0) = 2$? We should not alter $D$? (I guess this makes sense, since regardless of what value we choose for $D$, the $e$-term will become equal to 1 when t = 0, so the only way to obtain a different initial value would be to alter the $C$-term. Is this correct? (Sorry for asking so much, but I just want to make sure I understand this completely :)) –  Kristian Feb 11 '12 at 16:52
    
I worried about this when I was answering. Since there are two parameters that can be adjusted for the solution to $x\ddot{x}-\dot{x}^2=0$, for each initial condition, $x(0)$, we have a family of solutions. Should we alter $D$ change when we change initial conditions? I figured that since $C=x(0)$ no matter what $D$ was, the two were orthogonal, and $D$ should be fixed. However, that is a bit of an ambiguity to the question. In any case, I reread the question and it specifies that $x>1$ (which I assume means for $t>0$ since $x(0)=1$). –  robjohn Feb 11 '12 at 18:45
    
To satisfy that condition, we must have $D>0$, and the solution is definitely not asymptotically stable whether we allow D to vary or not. –  robjohn Feb 11 '12 at 18:45
    
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! –  Kristian Feb 11 '12 at 19:23

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