Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 5A (Differential Equations)

Combine the real variables into $z=x+iy$. The linear system of ordinary differential equations becomes $z^{\prime }=(1-i\mu )z$, hence $z=(A+iB)e^{(1-i\mu )t}$ for arbitrary real $A,B$. Taking real and imaginary parts givesThe two independent constants cover every real initial condition. In the phase plane, the distance from the origin grows as $e^t$ and the polar angle changes at rate $-\mu$. For $\mu \ne 0$ the trajectories are outward logarithmic spirals; for $\mu =0$ they are outward straight rays.

At a fixed point, the first equation gives $y=-x$. Substitution into the second gives $2x(x^2-1)=0$, so the equilibrium points are $(0,0),(1,-1),(-1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1\pm i$, so the origin is an unstable spiral. The local rotation is clockwise: on the positive $x$-axis the velocity points downward. At either nonzero equilibrium point,Thus both nonzero equilibria are saddles, with eigenvalues $-2,2$. The stable manifold is tangent to $(1,-3)$, and the unstable manifold to $(1,1)$, at each saddle equilibrium. Since all these equilibrium points are hyperbolic, linear stability of a planar equilibrium gives their actual local types.

For the phase portrait, the nullclines are $y=-x$ and $y=x/(1-2x^2)$, the latter defined away from $x=\pm 1/\sqrt{2}$. The horizontal direction has the sign of $x+y$; the vertical direction has the sign of $(1-2x^2)y-x$. The system is invariant under $(x,y)\mapsto (-x,-y)$, so trajectories occur in centrally symmetric pairs. The arrows below show these directions; the highlighted stable manifolds enter the saddle equilibria, whereas the highlighted unstable manifolds leave them.