Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 4

An autonomous differential equation has no explicit dependence on the independent time variable: its vector field depends only on the current state. The two families of nullclines areOn the first family the horizontal velocity is zero; on the second the vertical velocity is zero. A critical point of a dynamical system is a state where both components vanish. It gives a constant solution, also called an equilibrium.

Translate the equilibrium to the origin by $z=(x-a,y-b{)}^T$. If the vector field is continuously differentiable, its first-order Taylor expansion isThe Jacobian matrix $M$ gives the linearization $\dot{z}=Mz$, whose general solution is $z(t)=e^{tM}z(0)$ using the matrix exponential. Its eigenvalues determine the exponential factors, while its eigenvectors and, when necessary, generalized eigenvectors determine their directions.

For distinct eigenvalues ${\lambda }_1,{\lambda }_2$ and corresponding eigenvectors $v_1,v_2$,For a real matrix with nonreal conjugate eigenvalues, choose conjugate coefficients to obtain real solutions. Equivalently, if $\lambda =a+ib$ and $v=p+iq$, a real pair of independent solutions isFor a repeated eigenvalue $\lambda$, there are two cases:

The factor $t$ in the second case is essential. These are exact solutions of the linearization; the remainder describes the error for the nonlinear differential equation. When no eigenvalue has zero real part, the hyperbolic equilibrium theory justifies the corresponding local qualitative classification. Zero real parts require additional nonlinear analysis.

Factor each component of the vector field. The horizontal-velocity nullclines and vertical-velocity nullclines are, respectively,The last curve is a hyperbola. In the nonnegative quadrant its branch is $y=2/x$ for $x>0$.

Intersect the two nullcline families. The axes yield $(0,0)$ and $(1,0)$, and $y=1$ with $xy=2$ yields $(2,1)$. The possible remaining intersection would have $x+y=1$ and $xy=2$, requiring $x^2-x+2=0$, whose discriminant is $-7$. Hence there are no further real equilibria.

The Jacobian matrix isEvaluate it at each equilibrium:

All three are hyperbolic equilibria, so their classifications persist for the nonlinear system. Immediately to the right of $(2,1)$ on $y=1$, the velocity is vertically downward, showing that the attracting spiral turns clockwise.

For $x,y>0$, the signs needed for the phase portrait follow directly from the factorizations:The coordinate axes are invariant sets. On $y=0$, the logistic differential equation $\dot{x}=x(1-x)$ directs positive states toward the boundary saddle point $(1,0)$. On $x=0$, $\dot{y}=2y$ directs positive states upward. At the three equilibria, use the local directions found in part (ii); interior sample trajectories near $(2,1)$ spiral clockwise inward.

The nullclines prescribe where arrows become vertical or horizontal; they are generally not themselves trajectories. The image samples the vector field and several solutions, rather than assuming every nullcline is an invariant set.