Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 32A (Dynamical Systems)

Let $x_{\ast }$ be an equilibrium point of a dynamical system. A Lyapunov function on a neighbourhood of $x_{\ast }$ is a continuously differentiable function $V$ such thatand whose derivative along trajectories satisfies

The First Lyapunov theorem says that the existence of such a positive-definite $V$ proves Lyapunov stability of $x_{\ast }$. If $\dot{V}<0$ away from $x_{\ast }$, the Second Lyapunov theorem strengthens this to asymptotic stability.

LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set $\Omega$ on which $\dot{V}\le 0$, then it approaches the largest invariant subset ofIn particular, if that largest invariant subset consists only of $x_{\ast }$, every trajectory in $\Omega$ approaches $x_{\ast }$.

SetThe stated identity givesHence, along the system,Near the origin, $E$ is positive definite because $U(x)>0$ for $x\ne 0$. The First Lyapunov theorem therefore gives stability.

Choose a sufficiently small compact invariant sublevel setOn $\{\dot{E}=0\}$ we have $y=0$. A trajectory can remain there only ifwhich occurs at $x=0,\pm 1$. The choice $c<1/4=U(\pm 1)$ excludes the two latter points, so the largest invariant subset of $\{\dot{E}=0\}\cap {\Omega }_c$ is the origin. LaSalle invariance principle now gives convergence to the origin. Thus, for every $k>0$,This is an instance of damped mechanical energy as a Lyapunov function.

When $k=0$, the system is conservative andis constant. The fixed points areThe potential $U$ has a strict minimum $U(0)=0$, strict maximaand tends to zero as $|x|\to \infty$. The conservative planar phase portrait therefore has:

The portrait is symmetric under $(x,y)\mapsto (-x,-y)$ and under time reversal $(x,y,t)\mapsto (x,-y,-t)$.

For $0<k\ll 1$, the same three fixed points remain. At the origin the Jacobian isIts eigenvalues have negative real part; for small $k$ they are complex, so the origin is a stable spiral. At $x=\pm 1$, $U^{\prime \prime }(\pm 1)<0$, making the Jacobian determinant negative, so both points remain saddles.

Every nonstationary trajectory moves across the former energy contours in the direction of decreasingThe unstable branches of the two saddles directed into the central region spiral into the origin, while their outward branches escape toward $x=\pm \infty$. The stable manifolds of $(\pm 1,0)$ are the separatrices that form the boundary of the basin of attraction of the origin. Initial points between these stable separatrices lose enough energy to remain between the barriers and spiral to the origin; points outside escape to one of the two infinities.

This gives the weakly damped damped rational double-barrier phase portrait and identifies the requested domain of stability as the open region bounded by those stable manifolds.

At $(1,y_0)$ the trajectory initially moves into $x>1$. Suppose $y$ had a first zero at time $t_{\ast }>0$. Before that time $y>0$, hence $x(t_{\ast })>1$. Since $U^{\prime }(x)<0$ for $x>1$,which is incompatible with crossing from positive values to zero. Thusfor all $t>0$.

The energy from part (b) decreases strictly becauseSince $U\ge 0$,Therefore

Finally, fix $\epsilon >0$. If the trajectory never entered $0<y<\epsilon$, positivity would force $y(t)\ge \epsilon$ for all $t>0$. Thensowhich becomes negative for large $t$, contradicting $E\ge 0$. Hence the trajectory must enter every such strip. This proves outward escape from the rational potential barrier.