Past exam of the differential equations 2 course of the University of Oxford / 2014 / 1 / a / Solution

At an equilibrium point, $x(1-x-y)=0$ and $2-2x-y=0$. The first equation gives $x=0$ or $y=1-x$. Substituting in the second gives, respectively,These are all the equilibrium points. The Jacobian matrix of the vector field isAt $(0,2)$,The eigenvalue $-1$ has algebraic multiplicity two and eigenspace $\mathrm{span}\{(0,1{)}^T\}$. For the linearized coordinates $u=x$, $v=y-2$,Both components tend to zero. The repeated eigenvalue with one independent eigenvector makes this a stable improper degenerate node, a sink equilibrium. The term $t{e}^{-t}$ is the signature of the nontrivial Jordan block. Since it is a hyperbolic equilibrium point, the linearization stability theorem gives local asymptotic stability of the nonlinear system. Approaching trajectories are tangent to the vertical eigendirection.

At $(1,0)$,One eigenvalue is positive and one negative, so this point is a saddle equilibrium. Its stable manifold is tangent to $y=\sqrt{2}(x-1)$, and its unstable manifold is tangent to $y=-\sqrt{2}(x-1)$.

For the phase portrait, the nullclines areAbove the second line the vertical flow is downward, and below it upward. For $x>0$, horizontal flow is to the right below $y=1-x$ and to the left above it; for $x<0$ the directions reverse. The line $x=0$ is invariant, with $\dot{y}=2-y$, so it flows toward $(0,2)$. Together with the saddle's two invariant directions, these signs determine the sketch.