Equilibrium Solutions — Question 9

Question 9 – Solution

Strategy. Combine the energy identity with the phase line; energy decrease is informative but does not by itself choose a limiting equilibrium.

Step 1: Derive the flow and classify its zeros. Since $V'(y)=y(y^2-1)$, $$\boxed{y'=y-y^3=y(1-y^2)}.$$ Its zeros are $-1,0,1$. Its signs on the successive intervals are $+,-,+,-$, so $-1$ and $1$ are asymptotically stable and $0$ is unstable.

Step 2: Differentiate the energy. The chain rule gives $$\boxed{\frac{dV}{dt}=V'(y)y'=-[V'(y)]^2=-(y-y^3)^2\le 0}.$$ The derivative vanishes exactly at $y=-1,0,1$. On any nonconstant trajectory it is strictly negative, since uniqueness prevents meeting an equilibrium at a finite time.

Step 3: Interpret the energy landscape. The points $\pm 1$ are minima with $V=0$; $0$ is a local maximum with $V=1/4$. Since the flow is down the energy gradient, the minima attract and the maximum repels. Nevertheless, the initial value $y(0)=0$ stays at that maximum forever, with constant energy. Nonincreasing energy does not force every solution to reach a global minimum, nor distinguish which minimum will be selected.

Step 4: Determine the basins and future limits. The decreasing-energy bound confines a trajectory to $\{y:V(y)\le V(y(0))\}$, a bounded set because $V(y)\to\infty$ as $|y|\to\infty$. Smoothness then ensures global forward continuation. Phase-line trapping gives bounded monotonic motion for every nonconstant solution. Its limit must be a zero of $y-y^3$, and it cannot cross zero. Therefore $$\boxed{\lim_{t\to\infty}y(t)= \begin{cases}-1,&y(0)<0,\\0,&y(0)=0,\\1,&y(0)>0.\end{cases}}$$ The basins are $\boxed{(-\infty,0)}$ for $-1$ and $\boxed{(0,\infty)}$ for $1$.