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The solutions to exercise set 3 of the math 308 differential equations course offered in the fall 2003 semester. It covers various problems related to equilibrium points, stability, and the behavior of solutions to differential equations.
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Math 308 Differential Equations, Fall 2003 Exercise Set 3 Solutions
โ0.
โ0.
โ0.
0
f(y)
0.5 1 1.5 2 y
The equilibrium points are y = 0, y = 1 and y = 2. y = 0 is unstable; y = 1 is asymptotically stable; and y = 2 is unstable.
0
1
2
3
f(y)
โ3 โ2 โ1 1 y
The only equilibrium point is y = 0, and it is unstable.
(a) We have f (y) = k(1 โ y)^2. The critical points are the solutions to f (y) = 0, and the only solution to k(1 โ y)^2 = 0 (when k > 0) is y = 1. Thus y = 1 is a critical point; the corresponding equilibrium solution is the constant function ฯ(t) = 1. (Or simply y(t) = 1; the book often uses ฯ to refer to specific solutions.)
(b) For any k > 0, the graph of f (y) is a parabola the opens upwards, with its minimum at the point (1, 0). For example, here is the plot with k = 1:
0
1
2
f(y)
โ0.5 0.5 1 1.5 2 2. y Clearly f (y) > 0 for y < 0 and for y > 0. Thus, unless y = 0, the differential equation dydt = f (y) tells us that y(t) must be an increasing function of t.
(c) We already know that y(t) = 1 is an equilibrium solution. Now assume y 6 = 1. By separating, we find
dy (1 โ y)^2
= k dt
Integrate to obtain 1 1 โ y
= kt + C.
To satisfy the initial condition y(0) = y 0 , we must have
1 1 โ y 0
Now solve for y:
y = 1 โ
kt + C
kt + 1/(1 โ y0)
1 โ y 0 (1 โ y 0 )kt + 1
Consider the term (^) (1โ^1 yโ 0 )ykt^0 +1. If y 0 < 1, then the denominator increases monotonically (from the value
1 when t = 0), and so the quotient decreases monotonically and approaches 0 asymptotically. Thus y(t) increases monotonically and approaches 1 asymptotically. If y 0 > 1, then the denominator of (^) (1โ^1 yโ 0 )ykt^0 +1 decreases from 1 (when t = 0) and becomes 0 when
t = (^) k(y 01 โ1). Since the denominator goes to zero, the quotient must โblow upโ; and since 1 โ y 0 < 0, it
approaches negative infinity. Therefore y(t) is increasing, and has a vertical asymptote at t = (^) k(y^10 โ1).
(b) We could compute f โฒ(y) and evaluate at y 1 and y 2 , but in this case we can simply point out that the graph of f is a parabola that opens downward, so the slope at y 1 = 0 (the left equilibrium) must be postive and the slope at y 2 (the right equilibrium) must be negative. Therefore, by the result of problem 14, y 1 is unstable and y 2 is asymptotically stable.
(c) The sustainable yield is
Y = Ey 2 = KE(1 โ E/r) = โ
r
The graph of Y (E) is a parabola opening downwards, with zeros are E = 0 and E = r.
(d) The maximum of Y (E) occurs when E = r/2, and the yield at this value is Ym = Y (r/2) = Kr/4. This is the maximum sustainable yield.
2.5/ 21. The differential equation is
dy dt
= r
y K
y โ h.
Let f (y) = r
1 โ (^) Ky
y โ h = โ (^) Kr y^2 + ry โ h.
(a) By solving f (y) = 0 we find y = โK
โr ยฑ
r^2 โ 4 rh/K
/(2r) = K
1 โ 4 h/(rk)
/2. There
are two real distinct solutions if 1 โ 4 h/rK > 0, and this holds if h < rK/4. Thus, if h < rK/4, the
equilibrium solutions are y 1 = K
1 โ 4 h/(rk)
/2 and y 2 = K
1 โ 4 h/(rk)
A representative graph of f (y) when h < rK/4 is shown here. In this example, r = 1, K = 2 and h = 0.1. Since rK/4 = 0.5 and h < 0 .5, there are two equilibria, as expected.
โ0.
โ0.
โ0.
0
f(y)
โ0.5 0.5 1 1.5 2 2. y
(b) We make a simple argument. The graph of f (y) is a parabola opening downward, so the slope of the graph of f must be positive at y 1 and negative at y 2. Therefore, by problem 14, y 1 is unstable and y 2 is asymptotically stable.
(c) If the initial population y 0 is between y 1 and y 2 , then f (y) is always positive, so y(t) will increase and approach y 2 asymptotically. If the y 0 > y 2 , then f (y) < 0, and y(t) will decrease, but it also ap- proaches y 2 asymptotically. Thus for any initial population y 0 larger than y 1 , the population will approach y 2 asymptotically. If y 0 < y 1 , then y(t) will monotonically decrease, and it will eventually reach zero. (Mathematically, the solution would go through zero and become negative, because zero is not an equilibrium solution in this case. However, a negative population is not meaningful. Once the population reaches zero, there are no more fish, so it is pointless to continue from there.)
(d) If h > rK/4, then f (y) = 0 has no solutions; there are no equilibria. In fact, in this case f (y) < 0 for all y. This means that all solutions to the differential equation decrease monotonically, and all solutions will eventually reach zero. The following plot shows a representative graph of f (y) when h > rK/4. In this example, r = 1, K = 2, and h = 0. 6 > rK/4. Note that there are no equilibria, and f (y) < 0 for all y.
โ0.
โ0.
โ0.
โ0.
f(y)
โ0.5 0.5 1 1.5 2 2. y
(e) If h = rK/4, then f (y) = โ (^) Kr
y โ K 2
. This is that situation discussed in problem 7; the only equilibrium is y = K/2 and it is semi-stable. The following plot shows a representative graph of f (y) when h = rK/4. In this example, r = 1, K = 2, and h = 0.5 = rK/4. There is just one equilibrium at y = 1, and for all other y, f (y) < 0.
โ0.
โ0.
โ0.
โ0.
f(y)
โ0.5 0.5 1 1.5 2 2. y
