Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 7

A critical point of a dynamical system or equilibrium is a point $(a,b)$ with $f(a,b)=g(a,b)=0$, so the constant function $(x,y)=(a,b)$ is a solution. Put $u=x-a$, $v=y-b$. For a continuously differentiable vector field, its linearization iswith remainder $o(\sqrt{u^2+v^2})$ in the original nonlinear system. For distinct eigenvalues, $M$ is diagonalizable over $\mathbb{C}$, and the linear solutions are combinations of $e^{{\lambda }_{j}t}{w}_j$, where $w_j$ are corresponding eigenvectors. Real parts determine growth or decay; nonzero imaginary parts give rotation. For hyperbolic equilibria, meaning no eigenvalue has zero real part, the Hartman-Grobman theorem transfers the local stability and qualitative orbit pattern to the nonlinear system. A purely imaginary pair is not hyperbolic: it determines a linear centre, while higher-order terms decide whether the nonlinear system actually has a centre.

A stable node hasBoth exponential modes decay as $t\to \infty$, so the equilibrium is asymptotically stable. Generic linear orbits approach tangent to the eigenvector for the less negative eigenvalue; the eigenvector directions themselves are invariant straight lines. There is no rotation because both eigenvalues are real.

An unstable node hasBoth linear modes grow forward in time. Nearby nonconstant orbits move away from the equilibrium, and approach it under reversed time; their local directions come from the two real eigenvectors.

A saddle equilibrium hasThe negative eigenvalue gives a stable manifold direction and the positive eigenvalue an unstable manifold direction. Perturbations along the latter grow, so the equilibrium is unstable even though there are special trajectories approaching it.

A centre equilibrium of the linearized real planar system hasAfter a real invertible linear change of coordinates, the system is a constant angular rotation. Its nonzero orbits are ellipses of period $2\pi /|\omega |$, so it is stable but not asymptotically stable. This is a classification of the linearization: a centre in the linearization need not be a nonlinear centre. In particular, imaginary eigenvalues alone do not prove a nonlinear centre equilibrium, which requires a punctured neighbourhood filled with periodic orbits. Part (c) illustrates exactly this limitation.

A stable spiral has the complex conjugate eigenvaluesIn suitable real coordinates the radius decays like $e^{\alpha t}$ while the angle changes at rate $\beta$. The orbit therefore spirals towards the equilibrium, which is asymptotically stable. This is also called a stable focus.

An unstable spiral hasRotation is accompanied by exponentially increasing radius in suitable real coordinates. Thus nearby nonconstant orbits spiral away, and the equilibrium is unstable. This is also called an unstable focus.

The first component of the vector field vanishes only when $y=0$ or $x=\pm 1$. The second vanishes only when $y=1$ or $x+y=3$. Intersecting these alternatives gives exactly five equilibria. Their Jacobian matrix isAt $(3,0)$,The real parts are negative and the imaginary parts are nonzero, so $(3,0)$ is a stable spiral.

At $(1,1)$ the Jacobian matrix is $\mathrm{diag}(2,-1)$. Its real eigenvalues have opposite signs, so $(1,1)$ is a saddle equilibrium.

At $(1,2)$,Both eigenvalues are positive, so $(1,2)$ is an unstable node.

At $(-1,1)$ the Jacobian matrix is $\mathrm{diag}(-2,-3)$, giving a stable node.

At $(-1,4)$,so $(-1,4)$ is a saddle equilibrium. All five equilibria are hyperbolic, so these linear classifications determine their nonlinear local stability as well.

The Bendixson-Dulac theorem states that if $(P,Q)$ is a continuously differentiable planar vector field on a simply connected open region $D$, and a continuously differentiable Dulac function $B$ makeshave one weak sign throughout $D$, without vanishing identically on any nonempty open subset, then there is no nonconstant periodic orbit contained in $D$. To see why, a hypothetical simple periodic orbit bounds a region in $D$. The divergence theorem gives the integral of this expression as the outward flux of $B(P,Q)$ across the orbit. The flux is zero because the vector field is tangent to its orbit, whereas the one-sign area integral is nonzero, a contradiction. The usual stricter version assumes a strictly positive or strictly negative expression everywhere.

For the printed polynomial system, the origin is an equilibrium andThus its linearization is a centre equilibrium. For the full system take $B=1$. The divergence isOn the open disk $x^2+y^2<3$ this is nonpositive, and is negative wherever $x\ne 0$. Its zero set inside the disk is a line, so it cannot vanish on any open subset. The Bendixson-Dulac theorem therefore provesThis also exposes a terminology problem in the source: a nonlinear centre equilibrium has surrounding closed orbits, so the printed claim of a centre cannot hold in that sense. The imaginary eigenvalues establish only a centre of the linearized system. In fact the full equilibrium is an asymptotically stable weak focus, as the following direct check shows.

A quartic Lyapunov correction for a weak focus makes the nonlinear damping visible. Put $r^2=x^2+y^2$ and consider the Lyapunov functionIt is positive definite sufficiently close to zero. Differentiating along the polynomial vector field givesFor clarity, the quartic terms come from $-x^4$ in the derivative of $r^2/2$, and from applying $y{\partial }_x-x{\partial }_y$ to the quartic correction. Their sum is $-3x^4/8-3x^2{y}^2/4-3y^4/8$; all other terms have degree at least six. Therefore $\dot{V}<0$ in a sufficiently small punctured neighbourhood, proving asymptotic stability by Second Lyapunov theorem. Finallyso nonconstant nearby orbits turn clockwise as they converge. The equilibrium is a weak focus despite the centre in its linearization. This interpretation resolves both the local dynamics and the absence of closed orbits without changing the printed coefficients.