Would you agree that no population can increase w/o bound???

We must take into account environmental limitations to growth

such as: * Food Scarcity

* Competition Between Species

* Climate Change

* The most widely used population model is:

The Logistic Differential Eq uation

1dy y

kydt A

populationy

Growth ConstantCarrying Capacity

, where

1dy y

ky

y

dt

t

A

Next Slide

The Logistics (Integral) Curve

The slope field shows clearly that there are two families of solutions, depending on the initial value y0 = y (0).

• If y0 > A, then y (t) is decreasing and approaches A as t → ∞.

• If 0 < y0 < A, then y (t) is increasing and approaches A as t → ∞.

• If y0 < 0, then y (t) is decreasing and lim .t b

y t

2 0 2 4 6 8 The population diverges from...

Slope field for 1dy y

kydt A

converges to a and

diverges from a

stable eqilibrium

unstable equilibrn .m iu

y t

Constant Solution

The population converges to...

1dy y

kydt A

Graph

has constant solu 0tions and .y t y y A

0

EquilibriumStable Equilibrium

Unstable Equilibrium

1 /

1 1

dykdt

y y A

dyy y A

1 2

1 2

1 2

2 1

1

1 / 1 /

1 /1 /

1 / 1 /

1 1 /

1; 0 1

B B

y y A y y A

y y A B By y A

y y A y y A

B y A B y

y A B y BA

1 1ln ln

ln kt

dy kdt y y A kt Cy y A

y ykt C Ce

y A y A

1/ 1 1 1

1 / 1 /

A

y A A y A A y

Let’s now find the nonequilibrium solutions explicitly, using separation of variables. Assuming that y 0 and y A, we have ≠ ≠

1dy y

kydt A

1 /

dykdt

y y A

Let a # which

drops a parameter

y

Partial Fractions

kdt

This is for Jewel

With the differential equation

for the logistics growth function,

we can come up w/ the general

population function and

an expression for our constant of

integration .

y t

C

1 1 1 /

kt

kt kt

kt kt

kt kt

kt kt kt

y y A Ce

y yCe ACe

y yCe ACe

ACe ACe Ay

Ce Ce e C

0

0

kt yyCe C

y A y A

For t = 0, we have a useful relation between C and the initial value y0 = y (0):

So is easy to find!CSolve for ...y

As C 0, we may divide by Cekt to obtain the general nonequilibrium solution:

≠

11 /kt

dy y Aky y t

dt A e C

1 kt

kt

e

Ce C

We can clean this up!

Solve 0.3 4 with initial condition 0 1.y y y y

We'll rewrite 0.3 4 as 1.2 14

dy yy y y y y y

dt

0

0

11 /kt

ydy y Aky y t C

dt A e C y A

1.2

thus, 1.2 & 4

4so,

1 /t

k A

ye C

0

0

1

3

yC

y AC

1.2

4

1 3 ty

e

0

0

1 & 1 /kt

ydy y Aky y C

dt A e C y A

0.4

0.4 & 1000

0.4 11000

1000

1 /t

k A

dP PP

dt

P te C

0.40.4

1000500 1000 500 4500

1 9t

te

e

0.4 ln 1/ 91

9 0.45.493 yearste t

0

00.4

1000

1 9

1

9 ty

yC

y A eC

Deer Population A deer population grows logistically with growth constant k = 0.4 year−1 in a forest with a carrying capacity of 1000 deer.(a) Find the deer population P(t) if the initial population is P0 = 100.(b) How long does it take for the deer population to reach 500?

0

0

2 22

0.8959 0.89590.8959

1

1 /0.1 1

0 1/10 and 0.9 9

1

1 92 2 1 5 1

2 1 95 5 1 9 2

' (1 )

0.89

61

2 ln6

3 3 1 4 11 9

4 4 1 9 3 27

59

3.679 days

kt

kt

k kk

t tt

y te C

yy C C

y A

y te

y e ee

k

y t

y t ky

e e

y

k

te