I've been trying to prove that this autonomous equation

$\ddot x + e^{cos(x)}\dot x + sin(x) = 0$

is asymptotically stable at $x_{k}=2k\pi$.

My main problem is that the only way I know for working this kind of problems is by lyapunov functions, but havent been able to find one. This seems really like the damped pendulum equation but Im not so sure of how to work this out.

And one extra (simple)question about asymptotic stability(in Lyapunov sense): If we prove asymptotic stability(Lyapunov sense), does that mean that we are proving that all trajectories are converges to 0?

Any help would be really appreciated :)

Thanks so much. :D


1 Answer 1


Rewrite the equation as a system: $$\tag{1} \left\{\begin{array}{ll} \dot x&=&y\\ \dot y &=& -e^{\cos x}y-\sin x. \end{array}\right. $$ Suppose that the Lyapunov function has the form $V(x,y)=\phi(x)+\psi(y)$. The derivative is $$ \dot V= \phi'(x)y-\psi'(y)e^{\cos x}y-\psi'(y)\sin x. $$ It can be noticed that the first and the last terms can be cancelled by choosing $\phi'(x)=\sin x$ and $\psi'(y)=y$. The correspondidng Lyapunov function is $$V(x,y)=1-\cos x+\frac{y^2}2.$$ It has zero value at $x_k=2\pi k$, $y_k=0$ and nonpositive derivative everywhere: $$ \dot V(x,y)= -e^{\cos x}y^2. $$ This shows Lyapunov stability. In order to prove asymptotic stability we need to show additionally that the set $$M=\{ (x,y):\; \dot V(x,y)=0 \}=\{ (x,y):\; y=0 \}$$ doesn't contain whole trajectories except for the equilibrium points $(\pi k,0)$. Substituting $y=0$ to the second equation of (1) gives $$ \dot x=0,\quad \dot y=-\sin x. $$ This means that at any point in $M$, except for $(\pi k,0)$, the tangent vector to the solution is $(0,-\sin x)$, so any solution, except for $(\pi k,0)$, does not stay in $M$.

Asymptotic stability implies that all trajectories starting in some neighborhood of the equilibrium point converges to this equilibrium point.

  • 1
    $\begingroup$ Amazing answer, thanks so much! $\endgroup$
    – asd123
    Commented Dec 6, 2019 at 20:58

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