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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