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Consider a system with Hamiltonian:

$$H(p, q)=\frac{1}{2}p^2+1-\cos q$$

The corresponding system of equations is $\dot{q}=p, \dot{p}=-\sin q$. For this system, the critical point $(0,0)$ is a stable center but not asymptotically stable since the Hamiltonian is a constant. Now, we add energy dissipation so that the system becomes

$\dot{q}=p$,
$\dot{p}=-p-\sin q$.

We expect that $(0,0)$ becomes asymptotically stable. Which one of the following statements is correct for showing that $(0,0)$ is asymptotically stable?

A

Linearization of the system around $(0,0)$ does not work. Before adding $-p$, the eigenvalues are $\pm i$. After adding $-p$, the eigenvalues are still complex.

B

The energy function $V=\frac{1}{2}p^2+(1-\cos q)$ itself is a suitable Lyapunov function to show the asymptotic stability, because the trajectory derivative $\dot{V}$ is negative definite.

C

One can use the function $V=p^2+q^2$ as a Lyapunov function to show the asymptotic stability.

D

One can use the function $V=\frac{1}{2}(p+q)^2+q^2+\frac{1}{2}p^2$ to show the asymptotic stability.

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