Equilibrium Solutions — Question 5

Question 5 – Solution

Strategy. Compare a horizontal parameter with $y^2$, and treat the double zero separately from the two simple roots.

Step 1: Classify the parameter regimes. If $\mu<0$, there are no equilibria and $y'<0$ everywhere. If $\mu=0$, the only equilibrium is $0$: $-y^2<0$ on both sides, so it attracts from above and repels from below. If $\mu>0$, the equilibria are $$\boxed{y=-\sqrt\mu\text{ (unstable)},\qquad y=\sqrt\mu\text{ (asymptotically stable)}}.$$ Indeed $\mu-y^2$ is positive between the roots and negative outside them. The derivative $f_y=-2y$ confirms the classifications at the simple roots.

Step 2: Solve the selected IVP. Separation and $y(0)=0$ give the following cases, with $a>0$ as indicated: $$\boxed{ \begin{array}{c|c|c} \text{parameter}&y(t)&\text{forward time range}\\\hline \mu>0,\ a=\sqrt\mu&a(1-e^{-2at})/(1+e^{-2at})&0\le t<\infty\\ \mu=0&0&0\le t<\infty\\ \mu<0,\ a=\sqrt{-\mu}&-a\tan(at)&0\le t<\pi/(2a) \end{array}}$$ For $\mu>0$, integration on $-a<y<a$ gives $\ln((a+y)/(a-y))=2at$. For the last case, integration gives $\arctan(y/a)=-at$ on the initial branch. Differentiating the displayed expressions verifies $y'=\mu-y^2$ and their zero initial values.

Step 3: Describe the future. For $\mu>0$, the selected solution increases to $\sqrt\mu$; at $\mu=0$ it stays at zero; for $\mu<0$ it decreases to $-\infty$ as $t\uparrow\pi/(2a)$, so no finite continuous extension exists.

Step 4: Interpret the branch merger. As $\mu$ decreases to zero, the attracting and repelling equilibria merge into a semistable double zero; for negative $\mu$ neither remains. At the merger $f_y(0)=0$, so the derivative test is inconclusive. The sign of $-y^2$ supplies the one-sided classification.