Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 2 / 6C (Differential Equations)

$\nabla U=(3x^2-3+y^2,2y(x+1))$, so the critical points are $(\pm 1,0)$. The Hessian matrix at $(1,0)$ is $\mathrm{diag}(6,4)$, making it the local minimum; $(-1,0)$ is degenerate and not a minimum.

$dU/dt=\nabla U\cdot (-\nabla U)=-|\nabla U{|}^2\le 0$. The descending gradient flow trajectory from $(2,1)$ stays in its bounded contour basin and tends to the only critical point there, $(1,0)$.

Writing $\xi =x-1$, $\eta =y$, the linearization of a dynamical system is $\dot{\xi }=-6\xi$, $\dot{\eta }=-4\eta$. Eliminating $t$ gives $\eta =C{\xi }^{2/3}$.

Now $\dot{U}=-|\nabla U{|}^2+U_y\cos \omega t$, whose second term has either sign. Near the minimum, $\xi =C{e}^{-6t}$ and $\dot{\eta }+4\eta =\cos \omega t$, so