Equilibrium Solutions — Question 3

Question 3 – Solution

Strategy. When the linear term vanishes, retain the nonlinear term. Separation gives both a stability proof and the actual time scales.

Step 1: Classify from signs. For A, $-y^3$ is positive below $0$ and negative above, so trajectories move toward $0$. For B the signs reverse, so trajectories move away. Thus $0$ is asymptotically stable for A and unstable for B. The shared value $f'(0)=0$ is an inconclusive test, not evidence of neutral stability.

Step 2: Solve with the initial sign preserved. For $a\ne 0$, differentiating $y^{-2}$ gives $2$ for A and $-2$ for B. Applying the initial value yields $$\boxed{y_A(t)=\frac{a}{\sqrt{1+2a^2t}},\qquad t\ge 0},$$ $$\boxed{y_B(t)=\frac{a}{\sqrt{1-2a^2t}},\qquad 0\le t<\frac 1{2a^2}}.$$ The full maximal intervals containing $0$ are $(-1/(2a^2),\infty)$ and $(-\infty,1/(2a^2))$, respectively. Direct differentiation verifies the equations. If $a=0$, the unique solution of either smooth equation is zero for all real time.

Step 3: Prove stability and attraction for A. For $t\ge 0$, $|y_A(t)|\le|a|$. Given $\varepsilon>0$, choosing $\delta=\varepsilon$ ensures $|a|<\delta$ implies $|y_A(t)|<\varepsilon$ for all future time. Also $y_A(t)\to 0$ for every $a$, so the attraction is global. More precisely, $$\sqrt{2t}\,|y_A(t)|\longrightarrow 1\quad(a\ne 0).$$ The decay is algebraic rather than exponential.

Step 4: Prove instability for B. Choose any fixed $\varepsilon>0$ and any nonzero $|a|<\varepsilon$. The magnitude first reaches $\varepsilon$ at $$t_\varepsilon=\frac 1{2a^2}-\frac 1{2\varepsilon^2}>0,$$ then exceeds it. Arbitrarily small nonzero perturbations therefore leave that neighborhood. Blow-up occurs later, at $1/(2a^2)$, with the sign of $a$. Instability only requires departure from a prescribed neighborhood; it does not require an infinite solution value.