Equilibrium Solutions — Question 7

Question 7 – Solution

Strategy. Verify explicit joined solutions instead of invoking a uniqueness theorem whose hypothesis fails.

Step 1: Find the zero and signs. The only equilibrium is $y=0$. The right-hand side is positive for every $y\ne 0$, so motion is upward on either side. Below zero the direction is toward the equilibrium level, but arrival need not force the solution to stay there.

Step 2: Verify delayed departures. For any $c\ge 0$, define $$\boxed{y_c(t)=\begin{cases}0,&0\le t\le c,\\(t-c)^2,&t\ge c.\end{cases}}$$ Before $c$ both sides of the equation vanish; afterward $y_c'=2(t-c)=2\sqrt{|y_c|}$. At $c$, both values and both one-sided derivatives are zero. The difference quotient also gives derivative zero, so the joined solution is continuously differentiable. The identically zero function is another solution.

Step 3: Cross from below. A crossing solution is $$\boxed{y(t)=\begin{cases}-(1-t)^2,&0\le t\le 1,\\(t-1)^2,&t\ge 1.\end{cases}}$$ Before $1$, $y'=2(1-t)=2\sqrt{|y|}$; afterward the same check as above applies. Both derivatives at the join are zero, and $y(0)=-1$.

Step 4: Diagnose instability and failed uniqueness. For any $\delta>0$, initial value $0$ is within $\delta$ of the equilibrium, yet the solution $y_0(t)=t^2$ leaves the fixed neighborhood $(-1,1)$ after $t=1$. This violates the stated requirement for every forward solution, so the equilibrium is unstable.

Local Lipschitz regularity fails since $|f(y)-f(0)|/|y|=2/\sqrt{|y|}$ is unbounded near zero. The crossing and constant solutions agree at $(1,0)$ but have different continuations. The usual noncrossing proof requires local uniqueness; continuity alone does not provide it.