Logistic Growth

## **Problem**

The logistic growth model is a population model that incorporates Malthusian-like growth in the early stage and in-species competition for resources at the later stage. This model includes a growth rate

r

{\displaystyle r}

 and carrying capacity

K

{\displaystyle K}

. The differential equation of this growth model is

d
N

d
t

=
r
N

(

1
−

N
K

)

{\displaystyle \frac{dN}{dt} = rN\left(1 - \frac{N}{K} \right) }

with the initial condition

N
(
0
# )

N
0

{\displaystyle N(0) = N\_0 }

.

**Part 1:** What are the steady-state solutions?

**Part 2:** When is the population growing the fastest?

**Part 3:** Solve the differential equation with the aforementioned initial condition.

**Part 4:** Show that if the carrying capacity approaches infinity

K
→
∞

{\displaystyle K \to \infty}

, then the solution in Part 3 is equivalent to Malthusian growth

N
(
t
# )

N
0

e

r
t

{\displaystyle N(t) = N\_0 e^{rt}}

.

## **Solution**

**Part 1**

Steady-state solutions occur when

d
N

d
t

=
0

{\displaystyle \frac{dN}{dt} = 0 }

 for all time

t

{\displaystyle t}

. For the logistic growth model,

# 0
r
N

(

1
−

N
K

)

.

{\displaystyle 0 = rN\left(1 - \frac{N}{K} \right). }

Solving for

N

{\displaystyle N}

 shows that the steady-state solutions occur at

# N
0

{\displaystyle N =0 }

 and

# N
K

{\displaystyle N = K }

.

**Part 2**

To figure out when the population is growing rate the fastest, we must find the maximum of the differential euqation

d
N

d
t

=
r
N

(

1
−

N
K

)

{\displaystyle \frac{dN}{dt} = rN\left(1 - \frac{N}{K} \right) }

. Since the differential equation is a parabola, the maximum must occur on the axis of symmetry. The axis of symmetry is located at the arithmetic mean of the two steady-state solutions. This means the population is growing the fastest when it is at half the carrying capacity

# N

K
2

.

{\displaystyle N = \frac{K}{2}. }

**Part 3**

Rewrite the differential equation

d
N

d
t

=
r
N

(

1
−

N
K

)

{\displaystyle \frac{dN}{dt} = rN\left(1 - \frac{N}{K} \right) }

 as

d
N

d
t

=
r
N

(

K
−
N

K

)

{\displaystyle \frac{dN}{dt} = rN\left(\frac{K - N}{K} \right) }

. Integrate both sides

∫

K

N
(
K
−
N
)

d
# N
∫
r

d
t

{\displaystyle \int \frac{K}{N(K - N)} \,dN = \int r \,dt }

by partial fractions using the fact

K

N
(
K
−
N
)

=

1
N

1

K
−
N

.

{\displaystyle \frac{K}{N(K - N)} = \frac{1}{N} + \frac{1}{K-N}. }

Consequently,

∫

(

1
N

1

K
−
N

)

d
# N
∫
r

d
t

{\displaystyle \int \left(\frac{1}{N} + \frac{1}{K-N} \right) \,dN = \int r \,dt }

∫

(

−

1
N

−

1

K
−
N

)

d
# N
−
∫
r

d
t

{\displaystyle \int \left(-\frac{1}{N} - \frac{1}{K-N} \right) dN = -\int r \,dt }

−
ln
⁡

N

ln
⁡

K
−
N

=
−
r
t
C

ln
⁡

K
−
N

N

=
−
r
t
C

{\displaystyle \ln\left|\frac{K-N}{N}\right| = -rt + C }

K
−
N

N

=

e

−
r
t
C

{\displaystyle \frac{K-N}{N} = {e}^{-rt + C} }

K
−
N

N

=

e

−
r
t

e

C

.

{\displaystyle \frac{K-N}{N} = {e}^{-rt}{e}^{C}. }

Let

# A

e

C

{\displaystyle A = {e}^{C} }

, then

K
N

−
# 1
A

e

−
r
t

{\displaystyle \frac{K}{N} - 1 = A{e}^{-rt} }

K
N

=
1
A

e

−
r
t

{\displaystyle \frac{K}{N} = 1 + A{e}^{-rt} }

# N

K

1
A

e

−
r
t

.

{\displaystyle N = \frac{K}{1 + A{e}^{-rt}}. }

Solve the initial-value problem

N
(
0
# )

N
0

{\displaystyle N(0) = N\_0 }

:

N
(
0
# )

K

1
A

e

−
r
(
0
)

{\displaystyle N(0) = \frac{K}{1 + A{e}^{-r(0)}} }

N
0

=

K

1
A

{\displaystyle N\_0 = \frac{K}{1 + A} }

# A

K

N
0

−
1

{\displaystyle A = \frac{K}{N\_0} - 1 }

Consequently, the solution of the logistic growth model is

N
(
t
# )

K

1

(

K

N
0

−
1

)

e

−
r
t

.

{\displaystyle N(t) = \frac{K}{1 + \left(\frac{K}{N\_0} - 1 \right){e}^{-rt}}. }

**Part 4**

N
(
t
)

=

lim

N
→
∞

K

1

(

K

N
0

−
1

)

e

−
r
t

=

lim

N
→
∞

K

K

(

1
K

(

1

N
0

−

1
K

)

e

−
r
t

)

=

lim

N
→
∞

1

1
K

(

1

N
0

−

1
K

)

e

−
r
t

=

lim

N
→
∞

1

1

N
0

e

−
r
t

=

N
0

e

r
t

{\displaystyle \begin{align}
N(t) &= \lim\_{N \to \infty} \frac{K}{1 + \left(\frac{K}{N\_0} - 1 \right){e}^{-rt}} \\[5 pt]
&= \lim\_{N \to \infty} \frac{K}{K \left(\frac{1}{K} + \left(\frac{1}{N\_0} - \frac{1}{K} \right){e}^{-rt} \right)} \\[5 pt]
&= \lim\_{N \to \infty} \frac{1}{\frac{1}{K} + \left(\frac{1}{N\_0} - \frac{1}{K} \right){e}^{-rt}} \\[5 pt]
&= \lim\_{N \to \infty} \frac{1}{\frac{1}{N\_0} {e}^{-rt}} \\[5 pt]
&= N\_0 {e}^{rt}
\end{align} }