Past exam of the differential equations 1 course of the University of Oxford / 2023 / 2 / a / ii / Solution

At an equilibrium point, $x=-y^2$ and $x^2=a^{2}y$, so $y(y^3-a^2)=0$. With $d=|a{|}^{2/3}>0$, the two equilibrium points and their Jacobian matrix areAt $(-d^2,d)$, $\det J=-3a^2<0$. The trace-determinant stability criterion therefore gives an unstable saddle equilibrium. At the origin, the eigenvalues are $\lambda =\pm i|a|$: its linearization is a center equilibrium. The linearization stability theorem does not classify this nonhyperbolic equilibrium of the nonlinear system.

Nonlinear classification. In fact the answer depends on $|a|$:Here is an explicit derivation, rather than an inference from the imaginary eigenvalues. When $a^2=1$, set $L=1-x+y$. Near the origin $L>0$, and the first integralhasIts Hessian matrix at the origin is the identity. Small level curves are closed ovals on which the vector field is nonzero, proving the center equilibrium classification and Lyapunov stability.

For $a^2\ne 1$, a higher-order Lyapunov function calculation settles the issue. Put $k=|a|$, $U=x$, $V=ky$, $\tau =kt$, $p=k^{-1}$ and $q=k^{-2}$. ThenWrite $R^2=U^2+V^2$ and introduce the differential operatorsSuccessively cancel the terms of degrees three, four and five in the derivative of $R^2+H_3+H_4+H_5$, using these homogeneous polynomials:Direct differentiation verifies $\mathcal{L}{H}_3=-\mathcal{N}{R}^2$, $\mathcal{L}{H}_4=-\mathcal{N}{H}_3$ and $\mathcal{L}{H}_5=-\mathcal{N}{H}_4$. The remaining degree-six term has circular averageFor example, the only nonzero averages of degree-six monomials needed here are $⟨U^6⟩=⟨V^6⟩=5/16$ and $⟨U^4{V}^2⟩=⟨U^2{V}^4⟩=1/16$. Since $\mathcal{L}={\partial }_{\theta }$ in polar coordinates, the zero-average part can be cancelled by a degree-six homogeneous polynomial $H_6$ satisfyingIndeed its restriction to the unit circle is obtained by integrating a zero-average trigonometric polynomial; its Fourier modes have orders $2,4,6$, so multiplication by $R^6$ gives a polynomial. Consequently the positive Lyapunov function $W=R^2+H_3+H_4+H_5+H_6$ satisfiesFor $k>1$ this is strictly negative near the origin, giving asymptotic stability. For $k<1$ it is strictly positive, and $W_{\tau }\ge C{W}^3$ in a sufficiently small neighbourhood; every nonzero starting point must eventually leave that neighbourhood. Finally ${\theta }_{\tau }=1+O(R)>0$, so the nearby trajectories rotate: these are weak foci, with the stability stated above. Merely calling the nonlinear origin a center for every $a\ne 0$ would miss this distinction.