Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, JANUARY 30 Population Dynamics
Logistic Equations. We discuss an application of autonomous equations by considering population growth.
Say that we have a population which has a size P = P (t) at time t. Assume that the rate of change of
the size is proportional to both the size of the population and the diﬀerence of the size from some maximal
sustainable size K . That is
dP
P
∝ P (K − P ) = K · P 1 −
.
dt
K
Hence there exists a positive constant r such that
dP
P
= rP 1−
.
dt
K
This is known as the logistic equation. The equilibrium solutions P = PL can be found by computing the
critical points of the function
P
f (P ) = r P 1 −
.
K
Clearly these are PL = 0 and PL = K . Sometimes we refer to K as the saturation level or the environmental
carrying capacity. Today we will study the solution to this problem.
Solution to Initial Value Problem. We solve this problem explicitly. To this end, it will be useful to
consider the “normalized” function and the “normalized” initial condition
P (t)
P0
y (t) =
and
y0 =
.
K
K
We substitute P = K y into the diﬀerential equation above:
dP
P
= rP 1−
dt
K
d
Ky
[K y ] = r K y 1 −
dt
K
dy
dy
K
= r K y (1 − y )
=⇒
= r y (1 − y ).
dt
dt
Hence we ﬁnd the equivalent initial value problem
dy
= r y (1 − y ) ,
y (0) = y0 .
dt
Fortunately this diﬀerential equation is a separable equation:
dy
= r y (1 − y )
dt =⇒ 1
dy
= r.
y (1 − y ) dt We wish to compute the antiderivative of the function on the lefthand side. To this end, we assume that it
can be expressed in a partial fraction expansion :
1
A
B
=+
y (1 − y )
y
1−y for some constants A and B yet to be determined. The righthand side can be simpliﬁed by ﬁnding a common
denominator:
A
B
A (1 − y )
By
A (1 − y ) + B y
A + ( B − A) y
+
=
+
=
=
.
y
1−y
y (1 − y )
y (1 − y )
y (1 − y )
y (1 − y )
1 ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.
 Spring '09
 EdrayGoins
 Equations

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