Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 8B (Differential Equations)

Write the vector field asThe equilibrium points of a dynamical system in the closed first quadrant areThe Jacobian matrix is

At $(0,0)$ its eigenvalues are $8,-1$, so this is a saddle equilibrium. Its unstable direction is the positive $x$-axis and its stable direction is the positive $y$-axis; to first order, nearby trajectories satisfy $\dot{x}=8x$, $\dot{y}=-y$.

At $(4,0)$ the eigenvalues are $-8,3$, so it is also a saddle. The $x$-axis is the stable direction, while trajectories entering the quadrant in the $y$ direction move away.

At $(1,\sqrt{3})$,whose eigenvalues areIt is therefore a stable spiral. A point immediately to its right moves upward, so nearby trajectories spiral counterclockwise into the equilibrium. These eigendirections and the inward spiral give the requested local phase-portrait sketches.

The only positive equilibrium solves $1-y=0$ and $x-1=0$, hence isAway from a nullcline,This is separable:Integration gives the first integralThus one convenient conserved quantity is

Its gradient isso $(1,1)$ is its only stationary point in the positive quadrant. Its Hessian matrix iswhich is positive definite everywhere. Hence $(1,1)$ is a strict, indeed global, minimum.

The nearby level curves of $E$ are closed curves surrounding this minimum. Since $E$ is constant along every trajectory, a solution starting on a sufficiently small nearby level set cannot leave the region bounded by a slightly larger level set. This proves Lyapunov stability of the equilibrium: solutions initially close to $(1,1)$ remain close for all time.