Question

Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable. Consider the equation dy/dt = y^2(4 − y 2 ) = f(y), where y(0) = y0 and −∞ < y0 < ∞.

(i) Sketch the graph of f(y) versus y.

(ii) Determine the critical points. (iii) Classify each one as asymptotically stable, unstable, or semistable. (iv) Illustrate several solutions in the ty-plane that illustrate how the different solutions depend upon y0.

Answer 1

(a) If y is an equilibrium solution, then dy/dt = 0 identically, so 0 = dy dt = k(1 − y) 2 hence y(t) = 1 identically. (b) 1 2 k y f(y) = dy dt Since dy/dt = k(1 − y) 2 > 0 for y 6= 1, then y is increasing for y < 1 and y > 1. (c) Separating the variables, if y 6= 1 we have kt + C = Z k dt = Z dy (1 − y) 2 = − Z du u 2 = 1 u = 1 1 − y 1 where we have made the substitution u = 1 − y, du = −dy. Then 1 − y = 1 kt + C =⇒ y = 1 − 1 kt + C and the initial condition gives y0 = y(0) = 1 − 1 C =⇒ 1 C = 1 − y0 =⇒ C = 1 1 − y0 . Thus y = 1 − 1 kt + 1 1−y0 = 1 − 1 − y0 k(1 − y0)t + 1 or y = 1 since we previously excluded this case. Since 0 = k(1 − y0)t + 1 =⇒ t = − 1 k(1 − y0) then y(t) has a vertical asymptote at t = − 1 k(1 − y0) and a horizontal asymptote at y = 1. If y > 1, then 1 < y = 1 − 1 kt + 1 1−y0 =⇒ 0 < − 1 kt + 1 1−y0 =⇒ 1 kt + 1 1−y0 < 0 . Then the numerator and denominator have different signs, so the denominator is negative: kt + 1 1 − y0 < 0 =⇒ kt < − 1 1 − y0 =⇒ t < − 1 k(1 − y0) since k > 0. Then we are to the left of the vertical asymptote, so y(t) diverges to ±∞ as t → − 1 k(1 − y0) . On the other hand, if y > 1, then we are to the right of the vertical asymptote and y(t) → 1 as t → ∞ since y(t) has a horizontal asymptote at y = 1. (See below for a plot of y(t) for the values k = 1, y0 = 2.)