Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 6C (Differential Equations)

Use the complete system in the PDF: the TeX omits $\dot{y}=y(4x-1)/8$. Both coordinate axes are invariant, and the equilibrium points are $(0,0)$, $(1,0)$ and $(1/4,3/4)$. The Jacobian matrix isAt $(0,0)$ its eigenvalues are $1,-1/8$, and at $(1,0)$ they are $-1,3/8$: both points are saddle equilibria. At the coexistence point,Thus the interior point is a stable spiral. The nontrivial nullclines are $x+y=1$ and $x=1/4$: $x$ increases below the first and decreases above it, while $y$ increases to the right of the second and decreases to the left. The rotation near coexistence is counterclockwise.

A local linearization by itself does not prove that every interior solution has the same limit. For this logistic predator-prey model, let $x_{\ast }=1/4,y_{\ast }=3/4$ and use the Lyapunov functionThe inequality $u-1-\log u\ge 0$ makes $V$ nonnegative, with equality only at coexistence. Its sublevel sets are compact inside the positive quadrant, because $V$ diverges both at the axes and at infinity. Direct differentiation cancels the predator-prey cross terms and givesOn $\{\dot{V}=0\}$ one has $x=x_{\ast }$, and remaining there requires $\dot{x}=-x_{\ast }(y-y_{\ast })=0$. The largest invariant subset is therefore the coexistence point. The LaSalle invariance principle provesOn $x=0$, $y$ decays as $e^{-t/8}$ toward the origin. On $y=0$, the logistic differential equation takes every $x(0)>0$ to one. The origin itself stays fixed. These boundary trajectories are the exceptions to the interior convergence statement.

For $r>0$, differentiate $r^2=x^2+y^2$ and use the angular derivative $(x\dot{y}-y\dot{x})/r^2$. The polar coordinates satisfyHence the rotation is clockwise. The origin is an unstable equilibrium, while $r=1$ is a limit cycle of period $2\pi$. Writing $u=r^2$ gives the logistic differential equation $\dot{u}=2u(1-u)$. For $r(t_0)=r_0>0$,All nonzero solutions tend to the unit-circle attracting limit cycle as $t\to \infty$, spiralling outward for $r_0<1$ and inward for $r_0>1$. For $0<r_0<1$ the solution exists for every real time and approaches the origin as $t\to -\infty$; for $r_0=1$ it remains periodic in both time directions. For $r_0>1$, the denominator first vanishes atThis finite-time blow-up of an ordinary differential equation occurs backward in time. Such exterior solutions have no $t\to -\infty$ behavior: their maximal time interval is $(t_{\ast },\infty )$. The zero solution exists for all time and has no defined polar angle.