The scientists Verhulst (1828)
where is the natural growth rate of the population when resources are limited and is a positive constant called carrying capacity of the environment.
(a) Using sign analysis, find all constant solution and discuss the general behavior of solution.
(b) Solve the equation when Compute the limit as of your solution.
(c) Solve the equation when Compute the limit as of your solution.
(d) Discuss why is called carrying capacity
Your question is incomplete
(a) To find the constant solutions, we set the equation equal to zero:
= 0
This implies that either is equal to zero or is equal to the carrying capacity .
If = 0, the population growth rate is zero. This means that the population size remains constant and does not increase or decrease.
If = , the population growth rate is also zero. In this case, the population size has reached its maximum capacity, and the growth rate is balanced with the available resources.
Now let's discuss the general behavior of the solution. When the population size is below the carrying capacity , the growth rate is positive. As the population size approaches the carrying capacity , the growth rate decreases. At the carrying capacity , the growth rate becomes zero, indicating a steady state where the population size remains constant. When the population size is above the carrying capacity , the growth rate becomes negative, indicating a decline in the population size.
(b) To solve the equation when , we can rearrange it as follows:
=
Integrating both sides of the equation:
= ln| | +
Simplifying further:
= ln| | +
Applying initial conditions when :
= ln| | +
Solving for :
=
(c) To solve the equation when , we can rearrange it as follows:
=
Integrating both sides of the equation:
= ln| | +
Simplifying further:
= ln| | +
Applying initial conditions when :
= ln| | +
Solving for :
=
(d) The carrying capacity is the maximum population size that can be sustained by the available resources in the environment. When the population size is equal to the carrying capacity , the growth rate is zero, indicating that the population has reached an equilibrium where the available resources can support the existing population size without any further growth. This is why the carrying capacity is called carrying capacity, as it represents the capacity or limit of the environment to carry or sustain a certain population size.
(a) To find the constant solutions of the equation, we need to set the derivative of the population growth equation equal to zero.
The population growth equation is modeled by the equation:
P' = kP(1 - P/C)
Where P' is the derivative of the population with respect to time (dP/dt), k is the natural growth rate of the population, P is the population size, and C is the carrying capacity of the environment.
Setting the derivative equal to zero, we get:
0 = kP(1 - P/C)
This equation will hold true if either k = 0 or (1 - P/C) = 0.
If k = 0, then the population growth is zero and remains constant.
If (1 - P/C) = 0, then P = C. This is the constant solution, where the population reaches its carrying capacity, and there is no net population growth.
To analyze the general behavior of the solutions, we can consider the signs of the derivative for different values of P.
For P < C, the factor (1 - P/C) will be positive, and the derivative will be positive if k > 0 and negative if k < 0. This indicates that the population will increase if k is positive and decrease if k is negative.
For P > C, the factor (1 - P/C) will be negative, and the derivative will be negative if k > 0 and positive if k < 0. This implies that the population will decrease if k is positive and increase if k is negative.
(b) To solve the equation P' = kP(1 - P/C), we can separate the variables and integrate both sides:
∫(1/(P(1 - P/C))) dP = ∫k dt
The left-hand side can be rewritten using partial fraction decomposition:
∫((1/C)/(P(1 - P/C)) - (1/C)/(P)) dP = ∫k dt
This simplifies to:
(1/C)ln|P(1 - P/C)| - (1/C)ln|P| = kt + C1
Combining the logarithms, we get:
ln|P(1 - P/C)/P| = C2 - (k/C)t
Taking the exponential of both sides:
|P(1 - P/C)/P| = e^(C2 - (k/C)t)
Since the population size P cannot be negative, we can remove the absolute value:
P(1 - P/C)/P = e^(C2 - (k/C)t)
Simplifying the equation:
1 - P/C = e^(C2 - (k/C)t)
Multiplying both sides by C to isolate P:
C - P = Ce^(C2 - (k/C)t)
P = C(1 - e^(C2 - (k/C)t))
This equation represents the solution to the population growth equation when k ≠ 0. The constant C represents the carrying capacity of the environment.
We can compute the limit as t approaches infinity (lim(t→∞)) of this solution:
lim(t→∞) P = lim(t→∞) C(1 - e^(C2 - (k/C)t))
For large values of t, the term e^(C2 - (k/C)t) goes to zero, because the exponent becomes very negative. Therefore, the limit of the solution as t approaches infinity is the carrying capacity of the environment: lim(t→∞) P = C.
(c) To solve the equation when P' = 0, we can set the derivative equal to zero:
0 = kP(1 - P/C)
This equation holds true if either k = 0 or (1 - P/C) = 0.
If k = 0, the population growth is zero, and there is no change in population size.
If (1 - P/C) = 0, then P = C. This is the constant solution where the population reaches its carrying capacity.
To compute the limit as t approaches infinity (lim(t→∞)) of this solution:
lim(t→∞) P = lim(t→∞) C
For both cases (k = 0 and P = C), the limit of the solution as t approaches infinity is the carrying capacity of the environment: lim(t→∞) P = C.
(d) The carrying capacity, denoted by C, is the maximum population size that an environment can sustain indefinitely given its available resources.
In the population growth equation, (1 - P/C) represents the fraction of resources left for growth when the population size is P. When the population size is less than the carrying capacity, this fraction is positive, indicating that there are resources available for population growth.
However, when the population size reaches the carrying capacity, the fraction (1 - P/C) becomes zero. This means that there are no more available resources for further growth, and the population size stabilizes, reaching a constant value.
Thus, the carrying capacity represents the maximum sustainable population size, beyond which the population cannot grow due to resource limitations.