Phase Plane — Question 10

Question 10 – Solution

Strategy. A proper conserved function confines the positive orbit, and nonzero tangential velocity turns its level curve into a periodic trajectory.

Step 1: Read the positive nullclines. In the positive quadrant, $x'=0$ exactly at $y=1$, and $y'=0$ exactly at $x=1$. The only positive equilibrium is $(1,1)$. At $(2,1)$ the velocity is $(0,1)$, so the orbit initially moves upward. The positive axes are not included in the stated domain.

Step 2: Derive the conserved level. Differentiating yields $$H'=(1-1/x)x(1-y)+(1-1/y)y(x-1)=0.$$ For the stated data, $\boxed{H=3-\ln 2}$. The Hessian is diagonal with entries $1/x^2,1/y^2>0$, so $H$ is strictly convex with unique minimum $H(1,1)=2$.

Step 3: Justify a closed periodic orbit. The function $z-\ln z$ tends to infinity as $z\downarrow 0$ or $z\to\infty$. Thus this level is compact within the positive quadrant. Its gradient vanishes only at $(1,1)$, below the stated level; it is regular. The given convex-level fact makes it a simple closed curve. The vector field is tangent by conservation and never zero on this curve. Its continuous speed therefore has a positive minimum, so traversal of the finite-length smooth closed curve takes finite time. Uniqueness then gives a periodic orbit. It is counterclockwise, as its velocity at the rightmost point $(2,1)$ is upward.

Step 4: Locate and compare the extrema. Extrema of $x$ occur at $y=1$, hence $x-\ln x=2-\ln 2$. This has roots $2$ and a unique $a\in(0,1)$, since $z-\ln z$ decreases on $(0,1)$ and increases on $(1,\infty)$. By symmetry, $\boxed{x_{\min}=y_{\min}=a,\quad x_{\max}=y_{\max}=2}$. The four extreme points are $(a,1),(2,1),(1,a),(1,2)$. In particular the two maxima occur at different states, so they are not simultaneous. The plotted level uses the exact conserved equation.