# Liapunov and asymptotically stable

I have three maps $F:\mathbb R\rightarrow\mathbb R$

$F(x)=-x, F(x)=x+x^2, F(x)=x-x^3$

I want to check whether the non-hyperbolic fixed point at the origin is Liapunov stable, asymptotically stable or not.

Well in all three cases the non-hyperbolic fixed point is $0$, because we know that for being non-hyperbolic the eigenvalues of the linearized system need to have a non-zero real part.

The definition for Liapunov stable I use is: $x$ is Liapunov stable iff for all $\epsilon>0$ there exists $\delta >0$ s.t if $|x-y|<\delta$ then $|\phi(x,t)-\phi(y,t)|<\epsilon$

Quasi-asyptotically stable iff there $\exists \delta>0$ such that if $|x-y|<\delta$ then $|\phi(x,t)-\phi(y,t)|\rightarrow 0$ as $t\rightarrow \infty$

Asymptotically stable if both defintions above hold.

Question: How can find out now explicitly in my three cases whether there is some kind of stability? Simply by solving the diff equations?

• So $F$ is the right-hand side of an ODE?
– fgp
Commented Jan 13, 2014 at 16:03
• Yes $F$ is the RHS of an ODE Commented Jan 13, 2014 at 16:08
• In this case, yes, I think explicitely solving the ODEs is the simplest way to go. Non-hyperbolic situations are hard in general... Commented Jan 13, 2014 at 16:13
• The solutions are $e^{-t}, -\frac{e^t}{e^t-1}, +-\frac{e^t}{\sqrt{e^{2t}}}$. I set the constants equal to zero. How can I proceed? Commented Jan 13, 2014 at 16:28

If you look at $F=x+x^2$, i.e. the ODE $$x' = x+x^2$$ you can see that if you start out with $x(0)=-1$, you get a constant solution, since then $x'(0)=0$. The same holds for $x(0)=0$.

Now watch what happens if $x(0)$ is slightly smaller than $-1$. You then get $x'(0) > 0$, i.e. $x$ will get larger as time progresses. Due to the uniqueness theorem, the solution can't get larger than $-1$, though, so it'll converge to something between $x(0)$ and $-1$.

Now let $x(0)$ be slightly larger than $-1$ (but less than $0$). You then get $x'(0) < 0$, so the solution will fall, and again due to uniqueness converge to something between $-1$ and $x(0)$.

Finally, look at $x(0) > 0$. Then, $x'(0) > 0$, so the solution will grow, which will increase $x'$ even further, leading to faster growth, and so on.

Can the system thus be lyapunov-stable? No - an arbitrary small change in the initial condition, i.e. $x(0)=0$ vs. $x(0)=0+\delta$ leads to arbitrary large differences after enough time has passed.

A more formal method of showing (but not disproving) lyapunov stability around the equilibrium point $0$ is to find a lyapunov function, i.e. a function $V(x) \geq 0$ with $V(x)=0$ only for $x=0$ and whose time derivative $\frac{df}{dt}V(x) \leq 0$, again taking zero only for $x=0$. For $F(x)=-x$ that's particularly easy - just take $$V(x)=x \text{.}$$ Since $\frac{df}{dt}V(x) = x' = -x$, that function is indeed a lyapunov function, and thus the system is lyapuniv-stable around $0$. It's also asymptotically stable, because the lyapunov function's property don't only hold locally around $0$ (as would be sufficient for lyapunov stability), but globally, and because whenever $|x|\to\infty$ then $V(x) \to \infty$.

• Thanks for your analysis, so in the case $F(x)=-x$ we obviously have both, Liapnuov and asymptotical stability since every little change in the initial condition wont make any differences in the behaviour of the solution $x(t)=e^{-t}+c$, correct? F(x)=x-x^3 is also both I think? Commented Jan 13, 2014 at 16:42
• For $F(x)=-x$ at least, yes. For an $x(0)>0$, the solution will fall, and for $x(0)<0$ it will grow, so on both cases the solution's absolute value is bounded by your $\epsilon$. Your explicit solution shows that too, of course - $e^{-t}$ always goes to zero. Didn't look at $F(x)=x-x^3$ yet.
– fgp
Commented Jan 13, 2014 at 17:10
• @TIJones You need to be a bit carefull with asymptotic stability, though - my rather simplistic arguments don't really show that for $F(x)=-x$.. You have to argue that if $x(0) < 0$, the solution reall converges to zero, and isn't just bounded by $0$ and $x(0)$. It's easy to see that convergence can only be monontonic, because uniqueness prevents it from crossing $0$, and $F$ is positive for all $x < 0$. So can it converge to something other than 0? No, because the derivative needs to go to zero for $x(t)$ to converge...
– fgp
Commented Jan 13, 2014 at 17:24
• Thanks for your edited answer, would you also have a suggestion for a Liapunov function for $x-x^3$ ? Then we do not have to consider a lot of cases. Commented Jan 13, 2014 at 19:49
• You have in idea for $x-x^3$? Commented Jan 16, 2014 at 16:41