Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2011 / 8 / Solution

Both components vanish atThe Jacobian matrix of the vector field isIts eigenvalues classify these equilibria by linearization of a dynamical system:All real parts are nonzero, so the linear classifications determine the local stability of the nonlinear autonomous system.

At the origin, the stable direction is the $y$-axis and the unstable direction is the $x$-axis. At $(1,0)$ the eigendirections are $(1,0)$ for $-1$ and $(1,-2)$ for $-3$. At $(0,1)$,has only one eigendirection, the $y$-axis; its linearization of a dynamical system has solutions $\delta x=A{e}^{2t}$, $\delta y=(B-At)e^{2t}$, explaining the improper node. At $(4,3)$, stable and unstable tangent vectors are respectively $(4,5-\sqrt{13})$ and $(4,5+\sqrt{13})$.

For the phase portrait, the axes are invariant, and the remaining nullclines are $y=x-1$ and $y=x/2+1$. Off these lines, the complete directional information isIn the positive quadrant, arrows point right above $y=x-1$ and left below it, and point up above $y=x/2+1$ and down below it. On the positive $x$-axis, trajectories tend to $(1,0)$. On the $y$-axis, $0<y<1$ flows toward the origin and $y>1$ flows upward. These signs and the saddle directions give the following sketch; the plotted curves illustrate the phase portrait, with the marked saddle curves tangent to the calculated eigenvectors.