Irony is wasted on the stupid, Oscar Wilde

To err is human, to blame it on someone else is even more human, Jacob’s Law

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:

- y = y(x) is the unknown function of the independent variable x.
- y′ = $\frac{dy}{dx}$ is the derivative of y with respect to x.
- a(x), b(x), and c(x) are known functions of x.

These equations are termed **“linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together** or composed in any nonlinear way.

If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

- The function f(x, y) (the right-hand side of the ODE) in the equation y’ = f(x, y) is continuous in some neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, denoted as $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}), meaning that it satisfies the initial condition y(x_{0}) = y_{0}.

This theorem ensures that under these conditions, the solution exists and is unique near x = x_{0}.

Definition. A first order autonomous ordinary differential equation (ODE) is a differential equation in which the rate of change of a dependent variable depends solely on y itself, and not explicitly on the independent variable, typically denoted by t. In mathematical terms, it has the general form: $\frac{dy}{dt}=f(y)$.

Here, f(y) is a function that depends only on y, and there is no explicit dependence on the independent variable t. This absence of explicit dependence on t is what characterizes the equation as autonomous.

The simplest model of population growth is described by the differential equation: $\frac{dy}{dt} = ky$ where:

- y(t) represents the population at time k,
- k is the intrinsic growth rate (a positive constant).

This equation states that **the rate of change of a given population is proportional to the population size itself. As the population increases, the rate of growth also increases exponentially**. The solution to this equation is y(t) = y_{0}e^{kt} where y_{0} is the initial population at time t = 0.

While this model captures the essence of unchecked population growth, it is overly simplistic and unrealistic because it does not account for environmental limitations and Earth’s limited resources, which are crucial in real-world scenarios. In reality, resources such as food, space, and other environmental factors limit population growth.

Two things are infinite: the universe and human stupidity; and I’m not sure about the universe, Albert Einstein.

This quote humorously highlights the limitations of assuming infinite resources, which is an underlying assumption of the exponential growth model.

To address the limitations of the exponential model, the logistic growth model introduces the concept of a carrying capacity. Logistic growth introduces the idea that as the population increases, the net growth rate decreases due to limited resources. The logistic equation accounts for limited resources by incorporating a term that decreases the growth rate as the population size increases.

The logistic equation models population growth with a carrying capacity, accounting for limited resources. The logistic differential equation is given by: $\frac{dy}{dt} = ay -by^2$ where:

- y(t) represents the population at time t.
- a is the intrinsic growth rate (the rate at which the population would grow without any limitations),
- b is a positive constant representing how the population growth is limited by limited resources.

**Interpretation of terms**:

- The term ay represents the natural growth of the population, proportional to its current size.
- The term -by
^{2}represents the limitation imposed by finite resources (such as food, space, competition, environmental issues, etc.). As the population grows larger, this term -by^{2}grows, slowing down the rate of growth until the population reaches a stable size.

The logistic model is also used in modeling the spread of diseases, adoption of technologies, and other phenomena exhibiting saturation behavior.

1. **Autonomous Nature**. The logistic equation is an autonomous differential equation because the rate of change $\frac{dy}{dt}$ depends only on the dependent variable y and not explicitly on the independent variable t. Autonomous ODEs like this allow for the use of qualitative techniques to analyze the behavior of solutions.

Solutions are invariant under time translation, meaning the behavior depends only on the state y, not on time t.

2. **Finding Critical Points (Equilibrium Solutions)**. Critical points occur where the rate of change of the population is zero: $\frac{dy}{dt} = ay -by^2 = 0↭[\text{Factor out y}] y(a -by) = 0 ↭ y = 0, y = \frac{a}{b}$.

In this critical points, y = 0 and y = a/b, the population remains constant (Refer to Figure 3 for a visual representation and aid in understanding it) because the rate of change of the population is zero.

y = 0 represents an unstable critical point, meaning if the population is very small, it is likely to grow. However, **if the population reaches zero, it stays at zero (extinction of the population)**.

y = a/b is a stable critical point, meaning the population stabilizes at this value. This is called the carrying capacity of the population —essentially the maximum population that the environment can sustain given its limited resources.

If the parabola $ay -by^2 = 0$ crosses the x-axis in a single point, there’s only a single point (Refer to Figure 4 for a visual representation and aid in understanding it) which is semi-stable.

3. **Behaviour of the solution. Stability of Critical Points**

The graph of $\frac{dy}{dt} = ay-by^2$ is a downward-opening parabola with roots at y = 0 and
y = ^{a}⁄_{b}. The maximum growth rate occurs at y = ^{a}⁄_{2b}, which is the vertex of the parabola.

- If $y > \frac{a}{b}$ or y < 0, the function y(a -by) = 0 is negative, meaning $\frac{dy}{dt} < 0$. The solution is decreasing, which means the population decreases over time. This suggests that if the population exceeds the carrying capacity a/b, it will start to decrease due to resource constraints. Similarly, populations that start with negative (or maybe very small) values (negative population sizes are not biologically meaningful but can be considered mathematically) will move further away from y = 0.
- If $0 < y < \frac{a}{b}$, the function y(a -by) = 0 is positive, meaning $\frac{dy}{dt} > 0$. The solution is increasing, which means the population increases over time, growing toward the stable carrying capacity a/b.

Basically, we can observe **arrows pointing away from y = 0 (unstable equilibrium) and toward y = L (stable equilibrium)**. Curves starting above L decrease toward L. Curves starting below L and above 0 increase toward L.

The general behavior is that populations starting below the carrying capacity grow and populations starting above the carrying capacity shrink, both moving towards the stable equilibrium y = a/b.

1. **The logistic equation is separable**: $\frac{dy}{dt} = ay -by^2 ↭[\text{Factor out y}] \frac{dy}{dt} = y(a-by) ↭ [\text{Separable variables}] \frac{dy}{y(a-by)} = dt$.

2. **Partial Fraction Decomposition**: $\frac{1}{y(a-by)} = \frac{A}{y} + \frac{B}{a-by}$

Multiply both sides by y(a−by): $1 = A(a-by)+By$. Set up equations to solve for A and B:

$\begin{cases} -Ab + B = 0~\text{Coefficient of y (i)} \\ Aa = 1 ~\text{Constant term (ii)}\end{cases}$

(ii) ⇒ $A = \frac{1}{a}$. (i)⇒ $B = Ab = \frac{b}{a}$, hence $\frac{1}{y(a-by)} = \frac{1}{a}\frac{1}{y} + \frac{b}{a}\frac{1}{a-by}$

3. **Integrate Both Sides**: $\int (\frac{1}{a}\frac{1}{y} + \frac{b}{a}\frac{1}{a-by})dy = \int dt$

$\int \frac{1}{a-by}dy =$[Let t = a-by, dt = -bdy] $\frac{-1}{b}\int \frac{1}{t}dt = \frac{-1}{b}ln|t| = \frac{-1}{b}ln|a-by|$

4. **Combine both integral and simplify the expression**: $\frac{1}{a}ln|y| + \frac{b}{a}·\frac{-1}{b}ln|a-by| = t + C ↭ \frac{1}{a}ln|y| + \frac{-1}{a}ln|a-by| = t + C ↭[\text{Multiply both sides by a to eliminate the fractions:}] ln|y|-ln|a-by| = a(t+C)↭ ln(y)-ln(a-by) = a(t+C)$, this is only valid between the equilibrium points y = 0 and y = L = $\frac{a}{b}$ (both y and a-by are positive, $a-by > 0 ↭ a > by ↭ y < \frac{a}{b} = L$).

Combine Logarithms: $ln(\frac{y}{a-by})=a(t+C)↭[\text{Exponentiate both sides}] \frac{y}{a-by} = Ae^{at}$, where A = e^{aC} (absorbing the constant).

5. **Solve for y**: $\frac{y}{a-by} = Ae^{at} ↭ y = Ae^{at}(a-by)↭[\text{Distribute to the right}] y = aAe^{at} -bAe^{at}y ↭[\text{Bring terms involving y to one side}] y + bAe^{at}y = aAe^{at} ↭[\text{Factor out y}] y(1+bAe^{at}) = aAe^{at} ↭ [\text{Solve for y}] y = \frac{aAe^{at}}{1+bAe^{at}} = \frac{aA}{e^{-at}+bA}$

6. **Apply Initial Conditions** y(0) = y_{0}: $y_0=\frac{aA}{e^{-a·0}+bA} = \frac{aA}{1+bA}⇒y_0(1+bA) = aA ⇒ y_0+y_0bA = aA ⇒y_0 = A(a-by_0)⇒ A = \frac{y_0}{a-by_0}$

7. **Final Explicit Solution**. Substitute A back into the expression for y(t): $y(t)=\frac{a(\frac{y_0}{a-by_0})}{e^{-at}+b(\frac{y_0}{a-by_0})} = \frac{(\frac{ay_0}{a-by_0})}{(\frac{e^{-at}(a-by_0)+by_0}{a-by_0})} = \frac{ay_0}{e^{-at}(a-by_0)+by_0} = \frac{ay_0e^{at}}{a-by_0 +by_0e^{at}} = \frac{ay_0e^{at}}{a+by_0(e^{at}-1)}$

8. **Interpretation of the Solution**. $\lim_{t \to \infty} \frac{ay_0e^{at}}{a+by_0(e^{at}-1)} = \frac{a}{b}$ (the leading terms in numerator and denominator are ay_{0}e^{at} and by_{0}e^{at}) respectively. The population y(t) approaching the carrying capacity y = ^{a}⁄_{b} as t → ∞, regardless of the initial population y_{0}, provided y_{0}> 0 (no extinction).

- If y
_{0}<^{a}⁄_{b}: The population increases over time, approaching the carrying capacity from below. - If y
_{0}>^{a}⁄_{b}: The population decreases over time, approaching the carrying capacity from above. - If y
_{0}= 0: The population remains at zero (extinction).

In real-world scenarios, populations may be subject to harvesting, where a portion of the population is removed at a constant rate. This situation commonly arises in fisheries, wildlife management, and forestry, where humans harvest animals or plants for resources.

The logistic equation with harvesting modifies the original equation to account for this constant removal rate. The equation becomes: $\frac{dy}{dt} = ay -by^2 -h$ where:

- y(t) represents the population at time t.
- a is the intrinsic growth rate (the rate at which the population would grow without any limitations),
- b is a positive constant representing how the population growth is limited by limited resources.
- h is the harvest rate, representing the number of individuals removed from the population per unit time (assumed to be a positive constant, refer to Figure i, ii for a visual representation and aid in understanding it).

To understand how harvesting affects the population dynamics, we analyze the equation by finding its critical points (equilibrium solutions) and examining the stability of these points.

- Case when h = 0. When there is no harvesting, the equation reduces to the standard logistic equation: $\frac{dy}{dt} = ay-by^2$. This case has been previously analyzed, showing that the population will grow and stabilize at the carrying capacity y =
^{a}⁄_{b}. - Case when h > 0. When harvesting occurs, the critical points and behavior of the population change. To find the critical points, we solve the quadratic equation in y: ay -by
^{2}-h = 0 ↭ by^{2}-ay + h = 0 ↭ $y = \frac{a ± \sqrt{a^2-4bh}}{2b}$.

Depending on the value of D(discriminant) -h-, we get different behaviors.

**Analysis Based on the Discriminant D**:

- Case 1. $D = a^2-4bh > 0 \text{ i.e., } h = h_1 < h_m =\frac{a^2}{4b}$, there are two equilibrium points $y_1 = \frac{a - \sqrt{a^2-4bh}}{2b}, y_2 = \frac{a + \sqrt{a^2-4bh}}{2b}$ where y
_{1}is the lower equilibrium point and y_{2}is the higher equilibrium point. Both equilibrium points are positive and meaningful in a biological context.

If h is small, i.e., h = h_{1} < h_{m}, where h_{m} is the maximum sustainable harvest rate, the solutions behaves as previously described but the critical points (y_{1} and y_{2} represented in red and h is labelled as h_{1}) are between [0, a/b]. The population will stabilize between these two points.

If y < y_{1} (unstable equilibrium point), $\frac{dy}{dt} < 0$, the population decreases and declines toward extinction. If y_{1} < y < y_{2}, $\frac{dy}{dt} > 0$, the population increases and grows toward the stable equilibrium at y_{2}. If y > y_{2} (stable equilibrium point), $\frac{dy}{dt} < 0$, the population decreases toward y_{2} ↭ The population can sustain itself and stabilize at y_{2} if the initial population is above y_{1}.

Conclusion: Harvesting is sustainable as long as the rate is below h_{m}

- Case 2. When the discriminant is zero $D = a^2-4bh = 0 ↭ h = h_m = \frac{a^2}{4b}$, there is one real repeated root y = $\frac{a}{2b}$. It is a single equilibrium critical point, which is semi-stable. It result in a precarious equilibrium.

h = **h _{m} represents the maximum sustainable harvest rate**. There is a single equilibrium point $y = \frac{a}{2b}$. This point is a semi-stable equilibrium; small perturbations lead the population to decline.

f’’(y)=-2b< 0, there is a maximum at y = $\frac{a}{2b}$. Any small decrease in y, $\frac{dy}{dt}< 0$ population decreases, leads to decline towards extinction. Any small increase in y, $\frac{dy}{dt}< 0$ population decreases, leads to decline towards the equilibrium point h_{m}.

- Case 3. When the discriminant is negative $D = a^2-4bh < 0 ↭ h = h_2 > h_m = \frac{a^2}{4b}$, no real solutions exist. The quadratic equation yields complex roots, which are not biologically meaningful.

If h > h_{m}, $\frac{dy}{dt}<0$ for all y > 0. No critical points exist (there are no population sizes where growth and harvesting balance out), meaning that the population decreases continuously over time and it will eventually collapse to zero (extinction is inevitable).

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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