Phase Plane — Question 9

Question 9 – Solution

Strategy. Keep equilibria separate from nearby nonconstant orbits, and retain the open domain when eliminating time.

Step 1: Identify the connecting trajectory. The equilibria are $\boxed{(-1,0),(1,0)}$. The function $(\tanh t,0)$ solves both equations, has $x'>0$, and tends to $(-1,0)$ backward in time and $(1,0)$ forward in time. Its orbit is the open horizontal segment $-1<x<1$, directed rightward.

Step 2: Separate the orbit from its closure. For every finite $t$, $-1<\tanh t<1$, so neither endpoint is reached. Each endpoint is its own constant trajectory. Uniqueness for the smooth field also prevents a solution from reaching an equilibrium and then leaving it. The closed segment is the union of three trajectories, not one.

Step 3: Recover the whole one-parameter family. For initial $(0,b)$, $$\boxed{x=\tanh t,\quad y=be^{-t},\qquad y=b\sqrt{\frac{1-x}{1+x}},\quad -1<x<1.}$$ The inverse hyperbolic tangent formula proves the eliminated equation. Every member tends to $(1,0)$ as $t\to\infty$. If $b=0$, the backward limit is $(-1,0)$. If $b\ne 0$, then $x\to-1$ while $y\to+\infty$ for $b>0$ or $-\infty$ for $b<0$; there is no finite backward limiting state.

Step 4: Check the invariant boundary lines. On either $x=\pm 1$, the first coordinate stays fixed and $y=y_0e^{-(t-t_0)}$. For $y_0\ne 0$, each vertical half-line is an orbit directed toward $(\pm 1,0)$, reached only as $t\to\infty$. At $y_0=0$ the state is already the equilibrium. The figure marks equilibria as separate filled points; connecting arrows and nearby trajectories approach them without asserting finite-time arrival.