# Consider the discrete-time dynamical system with the following rule: N(t+1) = (rN(t)) / (N(t)-3)?

Consider the discrete-time dynamical system with the following rule:

N(t+1) = (rN(t)) / (N(t)-3)

where r is the parameter, r cannot =0, and N(t) may take any real value except N(t) cannot =3. Find the value(s) (if they exist) of the parameter r for which

a) the system has no equilibrium

b) the system has exactly one equilibrium, and

c) the system has nonnegative equailibrium

### 2 Answers

- MaryLv 77 years agoFavorite Answer
Given

N(t 1) = (rN(t))/(N(t) - 3)

r ≠ 0, N(t) ≠ 3

To find equilibrium points, set N(t 1) = N(t)

(rN(t))/(N(t) - 3) = N(t)

rN(t) = N(t)(N(t) - 3) ... transposing N(t) - 3

= N^2(t) - 3N(t)

0 = N^2(t) - 3N(t) - rN(t) ... transposing rN(t)

0 = N^2(t) - (3 r)N(t)

= N(t)[N(t) - (3 r)]... factoring N(t)

N(t) = 0 or N(t) = 3 r

Since N(t) cannot be 3, r is either less than 3 or greater than 3.

Let r = -10 and suppose N(1) = 2

N(1) = 2

N(2) = 20

N(3) = -11.76

N(4) = -7.968

N(5) = -7.265

N(6) = -7.077

N(7) = -7.023

N(8) = -7.0069

N(9) = -7.0021

The values are slowly approaching -7

Graph http://i42.tinypic.com/2jf0oly.png

Let r = -1 and suppose N(1) = 2.5

N(1) = 2.5

N(2) = 5

N(3) = -2.5

N(4) = -0.454

N(5) = -0.132

N(6) = -0.0420

The values are slowly approaching 0

Graph http://i43.tinypic.com/2mxeu78.png

Let r = -1 and suppose N(1) = -1

N(1) = -1

N(2) = -0.25

N(3) = -0.0769

N(4) = -0.0250

N(5) = -0.00826

N(6) = -0.00275

The values are slowly approaching 0 again

Let r = 1 and suppose N(1) = 2

N(1) = 2

N(2) = -2

N(3) = 0.4

N(4) = -0.154

N(5) = 0.0488

N(6) = -0.0165

The values are slowly approaching 0

Graph http://i40.tinypic.com/24g367d.png

Let r = 5 and suppose N(1) = 2

N(1) = 2

N(2) = -10

N(3) = 3.85

N(4) = 22.7

N(5) = 5.76

N(6) = 10.4

N(7) = 7.02

N(8) = 8.73

N(9) = 7.62

The values are slowly approaching 8

Graph http://i39.tinypic.com/2hol4e0.png

a) the system has no equilibrium - none

b) the system has exactly one equilibrium - r = -10

c) the system has nonnegative equilibrium - r = -1, 1, 5

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- mcbengtLv 77 years ago
An equilibrium is a number k for which, if N(t) attains that number, it stays that number past that point (ie, once N(t) = k, you also have N(t + 1) = k). You can find equilibria by putting the same symbol in for both N(t) and N(t+1) and solving the resulting equation in that symbol, e.g. if N(t) = N(t+1) = k, then

k = r k/(k - 3),

which implies that k(k - 3) = rk, which implies that k^2 - 3k = rk, which implies that k^2 - 3k - rk = 0, which implies that

k^2 + (-3 - r) k = 0,

where I am writing it in this form so it is clear that it is a quadratic equation in k. We can easily solve this for k because the left hand side factors as k (k - 3 - r), so we see the equilibria are the values k = 0 and k = 3 + r.

This shows that the system always has an equilibrium, so the answer to (a) is that no value of r exists for which the system has no equilibrium.

The system will have exactly one equilibrium when 0 = 3 + r, or when r = -3. (In this case, the one equilibrium value is 0.)

Since the system always has 0 as an equilibrium, the system always has a nonnegative equilibrium. If you wanted there to be a positive equilibrium, this will happen only when 3 + r > 0, or when r > -3.

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