As long as algebra is taught in school, there will be prayer in school, Cokie Roberts

Don’t panic, The Hitchhiker’s Guide to the Galaxy

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

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, $\frac{dy}{dx}$

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 y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, 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}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:

- y is the dependent variable (a function of the independent variable t),
- y′ and y′′ are the first and second derivatives of y with respect to t,
- t is the independent variable,
- A and B are constants.

This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.

In ecological modeling, predator-prey interactions are crucial for understanding population dynamics within ecosystems. One classic model that describes these interactions is the Lotka-Volterra equations. This model captures the cyclical nature of predator and prey populations, such as sharks (predators) and fish (prey).

Let:

- a, b, c, d: positive constant representing interaction rates.
- x(t): population of predators (e.g., sharks) at time t.
- y(t): Population of prey (e.g., fish) at time t.

**Predator and Prey dynamics**

In the absence of prey, predators (sharks) die out due to lack of food: x’ = -ax, where a > 0 is the natural rate of predators. Growth due to consuming prey: x’ -ax + bxy, where b > 0 represents the rate at which predators consume prey.

Without predators, prey reproduce and their population grows exponentially: y′ = cy where c > 0 is the natural growth rate of prey. Prey population decreases as they are consumed by predators: y′= cy − dxy where d > 0 represents the rate at which prey are consumed.

The system of equations modeling the interaction is: $\begin{cases} x’ = -ax + bxy \\ y’ = cy -dxy \end{cases}$

Critical points occur where the population sizes do not change over time (x′= 0 and y′= 0).

Setting the Equations to Zero: $\begin{cases} x(-a + by) = 0 ~(i)\\ y(c-dx) = 0 ~(ii)\end{cases}$

If x = 0 (no predators) ⇒[(ii)] yc = 0 ⇒[c is a positive constant] y = 0 (no fish, both populations are extinct).

Otherwise, by (i) -a + by = 0 ⇒ y = ^{a}⁄_{b} > 0 (a and b are positive constants). By (ii) c -dx = 0 ⇒ x = ^{c}⁄_{d}. It only produces two critical points: (0, 0) and (^{c}⁄_{d}, ^{a}⁄_{b}).

Critical point: (^{c}⁄_{d}, ^{a}⁄_{b}), both populations are stable and non-zero.

To understand the system’s behavior near the origin, we linearize it by ignoring the non-linear terms (bxy and −dxy). At (0, 0), the terms bxy and dxy as negligible since x and y are near zero.

**Linearized System**
$\begin{cases} x’ = -ax \\ y’ = cy \end{cases}$

**Matrix Representation**
$(\begin{smallmatrix}x’\\ y’\end{smallmatrix})=(\begin{smallmatrix}-a & 0\\ 0 & c\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix})$

If we have a diagonal matrix, the eigenvalues are the diagonal elements, λ_{1} = -a < 0, λ_{2} = c > 0.

For complementary sake, det(A -λI) = 0 ↭ $(\begin{smallmatrix}-a-λ & 0\\ 0 & c-λ\end{smallmatrix}) = (-a -λ)(c-λ) = 0$. Hence, λ = -a, c. Given that a > 0 and c > 0, λ_{1} = -a is a negative eigenvalue, and λ_{2} = c is a positive eigenvalue.
Detailed Analysis of the Matrix with Two distinct real Eigenvalues
, general solution: x(t) = x_{0}e^{-at}, y(t) = y_{0}e^{ct}

- Negative Eigenvalue (λ
_{1}= -a < 0). Corresponds to a**stable**direction. Solutions along this direction (x-axis, predator population x) decay exponentially toward zero as time t increases. In other words,**the predator population decreases without prey**. - Positive Eigenvalue (h
_{2}= c > 0). Corresponds to an**unstable**direction (y-axis, prey population). Solutions along this direction grow exponentially as time t increases.**The prey population increases in the absence of predators**.

Since one eigenvalue is negative and the other is positive, the critical point (0,0) is a saddle point. **Saddle points are unstable because trajectories move toward the critical point along the stable direction (x-axis, predator population) and away from it along the unstable direction (y-axis, prey population)**. (Refer to Figure ii for a visual representation and aid in understanding it).

For simplicity, let’s assume that a = b = c = d = 1 (set all parameters to 1). The system becomes:

$\begin{cases} x’ = -x + xy \\ y’ = y -xy \end{cases}$

**Critical points with simplified parameters**: (0, 0) and (1, 1).

**Linearizing at the Non-Zero Critical Point (1, 1)**

In other to linearize the system at the critical point (^{c}⁄_{d}, ^{a}⁄_{b}) = (1, 1), we are going to use the Jacobian matrix:

$J = (\begin{smallmatrix}\frac{∂x’}{∂x} & \frac{∂x’}{∂y}\\ \frac{∂y’}{∂x} & \frac{∂y’}{∂y}\end{smallmatrix}) = (\begin{smallmatrix}-1+y & x\\ -y & 1-x\end{smallmatrix})$

**Evaluate at (1, 1)**
$(\begin{smallmatrix}-1+y & x\\ -y & 1-x\end{smallmatrix})\bigg|_{(1, 1)} = (\begin{smallmatrix}0 & 1\\ -1 & 0\end{smallmatrix})$. The Linearized system is:

$\begin{cases} x’ = y \\ y’ = -x \end{cases}$.

The characteristic equation is obtained from det(J -λI) = $(\begin{smallmatrix}-λ & 1\\ -1 & -λ\end{smallmatrix}) = λ^2+1 = 0$. Eigenvalues are λ = ±i (purely imaginary).

Detailed Analysis of the Matrix with Complex Eigenvalues
, The general solution to this system (λ = α±βi) is $x(t) = e^{αt}[C_1(\begin{smallmatrix}cos(βt)\\ -sin(βt)\end{smallmatrix})+ C_2(\begin{smallmatrix}sin(βt)\\ cos(βt)\end{smallmatrix})] = [λ = ±i, α = 0, β = 1] C_1(\begin{smallmatrix}cos(t)\\ -sin(t)\end{smallmatrix})+ C_2(\begin{smallmatrix}sin(t)\\ cos(t)\end{smallmatrix})$ where c_{1} and c_{2} are constant determined by initial conditions.

Both x(t) and y(t) involve sine and cosine functions. There is no exponential growth or decay factor. The solutions are periodic with a period of 2π. The amplitude of oscillations remains constant over time. Plotting x(t) versus y(t) yields closed curves (concentric circles or ellipses centered at the origin). This is characteristic of center behavior in dynamical systems.

Since there is no exponential term (no real part in eigenvalues), trajectories neither spiral inward nor outward. The system remains in perpetual oscillation around the critical point. At (1, 1), the trajectories represent oscillations of predator and prey populations around their equilibrium values. Populations return to their starting values after each cycle.

Starting from the linearized system:

$\begin{cases} x’ = y \\ y’ = -x \end{cases}$.

Differentiate equation (i) with respect to t: x’’ = y’. Substitute y′ from equation (ii): x’’ = -x. We get a second-Order ODE: x’’ + x = 0.

Characteristic Equation: r^{2} + 1 = 0 ⇒λ = ±i. General Solution: x(t) = Acos(t) + Bsin(t). Since y = x’ (i), y(t) = -Asin(t) + Bcos(t).

Characteristic Equation: λ^{2}-Tλ + D = 0 where T is the trace Trace (sum of diagonal elements) and D is the determinant of the Jacobian matrix.

At (1, 1): Jacobian Matrix: $(\begin{smallmatrix}0 & 1\\ -1 & 0\end{smallmatrix})$, T = 0, D = 1, Δ = T^{2}-4D = 0^{2}-4·1 = -4.

**Interpretation**

- Negative discriminant indicates
**complex eigenvalues**. **Centers Lie Along T = 0**. Small changes in parameters can shift the discriminant, altering the eigenvalues’ real parts (Refer to Figure iii for a visual representation and aid in understanding it).- Purely Imaginary eigenvalues (T = 0): Center.
- Eigenvalues have positive Real Part (T > 0): Spiral source (unstable).
- Eigenvalues have negative Real Part (T < 0): Spiral sink (stable).

Centers are across the line T = 0. A small change of the parameters, I might get a sink or a spiral source. In other words, you cannot predict the behaviour when you change a little its parameters (The behavior near (1, 1) is sensitive to changes in the parameters a, b, c, d). The non-linear system at (1, 1) might exhibit center behaviour, a spiral sink, or a spiral source depending on parameter values.

$\begin{cases} x’ = -ax + bxy ~(i)\\ y’ = cy -dxy ~(ii)\end{cases}$

Given

$\begin{cases} x’ = -x + xy ~(i)\\ y’ = y -xy ~(ii)\end{cases}$

$\begin{cases} x’ = x(-1 + y) ~(i)\\ y’ = y(1 -x) ~(ii)\end{cases}$

To find the trajectories (relationship between x and y), eliminate time t. Divide (ii) and (i): $\frac{dy}{dx} = \frac{y’}{x’} = \frac{y(1-x)}{x(-1+y)} = \frac{y(1-x)}{x(y-1)}$

Separate variables: $\frac{y-1}{y}dy = \frac{1-x}{x}dx ↭[\text{Simplify}] (1-\frac{1}{y})dy = (\frac{1}{x}-1)dx$

Integrating Both Sides: $\int (1-\frac{1}{y})dy = \int (\frac{1}{x}-1)dx ⇒ y -ln(y) = ln(x) -x + c_1$

Let’s exponentiate: $e^{y -ln(y)} = \frac{e^y}{e^{ln(y)}} = \frac{e^{ln(x)}}{e^x}·e^{c_1} ⇒\frac{e^y}{y} = \frac{x}{e^x}c_2 ⇒ \frac{x}{e^x}·\frac{y}{e^y} = c$ The integral curves (trajectories of the system) are graphs of this equation. These are contour curves of $xe^{-x}·ye^{-y} = h(x, y)$ (Refer to Figure i for a visual representation and aid in understanding it)

$f(x) = xe^{-x}$ has a maximum at 1 (f(1) = ^{1}⁄_{e}), $ye^{-y}$ has a maximum at 1, too. h(x, y) is the product of both functions, h(x, y) = f(x)·f(y), hence h(x, y) has a maximum at (1, 1) [$h(1, 1) = (\frac{1}{e})^2$], that is, as a figure of speech, the peak of the mountain.

As x or y approach 0 or infinity, h(x,y) approaches 0. The contours h(x, y) = C are closed curves around (1, 1). These are concentric closed curves that move clockwise around (1, 1). These represent the closed orbits in the phase plane, confirming the oscillatory behavior of the system (populations oscillate around the equilibrium point (1, 1)).

**High Predator, Low Prey**: Predators consume prey rapidly. Prey population declines sharply. Predator population eventually declines due to lack of food.**Low Predator, High Prey**: Reduced predation allows prey population to grow. Predator population begins to recover due to increased food availability.- This leads to an
**endless cycle of population rises and falls**. The system reflects natural oscillations observed in predator-prey relationships.

Modified System: $\begin{cases} x’ = -ax + bxy -kx ~(i)\\ y’ = cy -dxy -ky ~(ii)\end{cases}$ where the terms −kx and -ky represents additional deaths due to fishing.

Combining terms:

$\begin{cases} x’ = -(a+k)x + bxy ~(i)\\ y’ = (c-k)y -dxy ~(ii)\end{cases}$

**New Critical Point:**

- From x′ = 0: -(a+k)x + bxy = 0 ⇒x[−(a+k)+by]=0 ⇒ x = 0 or y = $\frac{a+k}{b}$
- From y’ = 0: cy -dxy -ky = 0 ⇒ y[(c−k)−dx]=0 ⇒ y = 0 or x = $\frac{c-k}{d}$

The old critical point was $(\frac{c}{d}, \frac{a}{b})$. **Critical point shifts to: $(\frac{c-k}{d}, \frac{a+k}{b})$** (Refer to Figure ii for a visual representation and aid in understanding it).

**Effect on Populations:**

**Predator Population (x) Decreases**: Since c -k < c (because k > 0) ↭ $x = \frac{c-k}{d} < \frac{c}{d}$, so x decreases. Predator population at equilibrium decreases.**Prey Population (y) Increases**: Since a + k > a ↭ $\frac{a+k}{b} > \frac{a}{b}$, so y increases. This shift reflects the decrease in predators and increase in prey.

The effect of fishing with nets is that reduces the predator (shark) population and increases the prey (fish) population.

Fishing removes prey and predators directly (-ky, -kx). Reduced predator numbers lead to fewer prey being eaten. The indirect positive effect outweighs the direct negative effect, resulting in a higher prey and lower predator population at equilibrium.

Statement of Volterra’s Principle: **If both predator and prey populations are subjected to a proportional external mortality (e.g., fishing, pollution) that affects them equally, the prey population will increase on average, while the predator population will decrease on average**.

The Lotka-Volterra equations provide valuable insights into predator-prey dynamics, demonstrating how interactions between species can lead to oscillatory behaviors. By analyzing the system’s critical points, linearizing near these points, and understanding the effects of external factors like fishing, we gain a deeper appreciation for the complex balance within ecosystems.

Suppose a fish pond experiences a mosquito infestation. What should we do? Let’s assume that the decision is to spray pesticides (e.g., DDT) to kill those pesky mosquitoes.

**Unintended Consequences**. The result is quite the same as before because:

**Pesticides harm both mosquitoes and fish.**When pesticides are applied, they do not discriminate between target and non-target species. Both mosquitoes and fish in the pond are affected, leading to unintended mortality in fish populations.**Fish are natural predators of mosquitoes.**Fish help control mosquito populations by preying on their larvae. Reducing the fish population diminishes this natural control mechanism.**Reducing the fish population can lead to a mosquito population increase due to reduced predation.**With fewer fish to consume mosquito larvae, mosquito numbers can rebound or even increase, exacerbating the original infestation problem.

In our analysis, the equilibrium (critical point) moves in the same direction, so the population of mosquitoes (prey) increases as the population of fish (predator) decreases, hence spraying pesticides may worsen the mosquito problem. This analysis highlights the complexity of ecological interventions.

Interventions like fishing or pesticide use can have profound and sometimes unintended effects. Careful analysis and consideration of ecological interactions are essential for sustainable management.

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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