Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 31E (Dynamical Systems)

Let ${\varphi }_t(x_0)$ denote the flow map generated by $\dot{x}=f(x)$.

is nonpositive on $E$.

The First Lyapunov theorem says that the existence of such a function makes the origin Lyapunov stable. To prove it, choose a sufficiently small closed ball ${\overline{B}}_{\epsilon }\subset E$. By compactness and positive definiteness,By continuity, some $\delta >0$ satisfies $V(x)<\alpha$ whenever $\Vert x\Vert <\delta$. Along a trajectory beginning in $B_{\delta }$, $V$ cannot increase. Such a trajectory therefore cannot reach the sphere $\Vert x\Vert =\epsilon$, where $V\ge \alpha$. Hence it remains in $B_{\epsilon }$ for every $t\ge 0$, which is Lyapunov stability.

The LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set on which $\dot{V}\le 0$, then it approaches the largest invariant subset of $\{x:\dot{V}(x)=0\}$.

On the two vertical sides of the square, the $x$ component of the vector field satisfiesOn the horizontal sides,Thus the vector field never points outward through the boundary, so the closed square is a positively invariant set and trajectories cannot leave $S$.

At a fixed point, $\dot{y}=0$ gives $x=-y^3$. Substitution into $\dot{x}=0$ givesFor $0<y<1$, both $y^3(2-y^6)$ and $\sin (2y)$ are positive; for $-1<y<0$, both are negative. Hence $y=0$, and then $x=0$, is the unique equilibrium point in $S$.

This alone does not prove that $S$ lies in the domain of stability. A compact trapping region containing a unique equilibrium can also contain a periodic orbit or another nontrivial omega-limit set.

Forthe orbital derivative isChoosing $g^{\prime }(y)=\sin 2y$ cancels the mixed term. Taking $g(0)=0$ givessoThis function has positive definiteness near the origin, andFor $|x|<1$ and $|y|<1$, both terms are nonpositive and equality holds only at $(0,0)$. Thus $V$ is a strict Lyapunov function near the origin, and the Second Lyapunov theorem proves that the origin is asymptotically stable.

The factor $x^2(2-x^2)$ is positive for $0<|x|<\sqrt{2}$. The product $y^3\sin 2y$ is positive for $0<|y|<\pi /2$. Therefore $\dot{V}<0$ away from the origin throughoutNeither side length can be enlarged while retaining a strict Lyapunov function on the whole rectangle: at $(x,0)$ with $|x|>\sqrt{2}$, the first term makes $\dot{V}>0$, while at $(0,y)$ immediately beyond $|y|=\pi /2$, the second term makes $\dot{V}>0$.

Strict decrease on this rectangle does not by itself put the whole rectangle in the domain of stability, because the rectangle has not been shown to be positively invariant; a trajectory may leave it before convergence.

To obtain a basin estimate from a Lyapunov sublevel set, take the largest sublevel component around the origin that remains inside the strict-decrease rectangle. On its boundary $|x|=\sqrt{2}$ or $|y|=\pi /2$, the minimum value of $V$ is $1$. Hence the maximal open estimate isEvery smaller closed sublevel set $V\le c<1$ is compact and positively invariant, and $\dot{V}<0$ there except at the origin. The LaSalle invariance principle therefore implies convergence to the origin. Taking the union over $c<1$ shows that ${\mathcal{D}}_V$ lies in the domain of stability. No larger sublevel component can be certified from this $V$ alone, because every level $c>1$ contains points $(0,y)$ just beyond $|y|=\pi /2$ where $\dot{V}>0$.