Equilibrium Solutions — Question 1

Question 1 – Solution

Strategy. Use sign intervals and uniqueness to trap solutions; bounded monotone trajectories reveal their limits.

Step 1: Find the zeros and arrows. The only zeros of $f(y)=y(1-y)(y-3)$ are $0,1,3$. Its signs are $$\begin{array}{c|cccc} y&(-\infty,0)&(0,1)&(1,3)&(3,\infty)\\\hline f(y)&+&-&+&- \end{array}$$

Step 2: Classify the equilibria. Arrows point toward $0$ and $3$ on both sides, so both are asymptotically stable. The arrows on either side of $1$ point away, so $1$ is unstable. Nearby solutions remain between their initial value and the attracting zero, which supplies stability as well as attraction.

Step 3: Determine every forward limit. The classification is $$\boxed{\lim_{t\to\infty}y(t)= \begin{cases}0,&a<1,\\1,&a=1,\\3,&a>1.\end{cases}}$$ In particular, the basins are $\boxed{(-\infty,1)}$ for $0$ and $\boxed{(1,\infty)}$ for $3$. These include the attracting equilibria themselves. The value $a=1$ stays exactly at the unstable equilibrium.

Step 4: Justify the global conclusions. Since $f$ is a polynomial, local solutions are unique. A nonconstant solution cannot meet an equilibrium at a finite time: local uniqueness there, in both time directions, would force the two solutions to agree.

Every trajectory is trapped in a bounded interval between its initial value and adjacent equilibria. A smooth right-hand side permits continuation while the solution remains bounded, so all forward solutions are global. Each nonconstant trajectory is monotone and bounded. If its limit $L$ had $f(L)\ne 0$, continuity would keep its speed bounded away from zero near $L$, contradicting convergence. The accessible zero in its direction of motion is therefore its limit.