It ain’t about how hard you can hit. Its about how hard you can get hit, and how much you can take, and keep moving forward, Rocky Balboa

If you hear a voice within you say ‘you cannot paint’, then by all means paint and that voice will be silenced, Vincent van Gogh

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 the study of differential equations, non-linear autonomous systems are of significant interest due to their complex and rich behaviors. Such systems can model a variety of real-world phenomena, including population dynamics, chemical reactions, and mechanical systems like pendulums.

An autonomous system is a system of differential equations where the independent variable (often time t) does not appear explicitly in the functions. The general form of a two-dimensional non-linear autonomous system is:

$\begin{cases} x’ = f(x, y) \\ y’ = g(x, y) \end{cases}$

where f(x, y) and g(x, y) are **non-linear functions governing the time evolution of the variables x and y**. Our goal is to analyze the behavior of the system by:
ur goal is to analyze the behavior of such systems by:

**Finding the critical points**(equilibrium points).**Linearizing the system near each critical point**to understand local dynamics.**Sketching trajectories**to visualize the system’s behavior

For clarity, we will apply the concepts to a specific example: the lightly damped pendulum.

Consider a lightly damped pendulum, which consists of a mass m attached to a rod of length l swinging back and force under the influence of gravity and a damping force. The angle θ between the rod and the vertical axis describes the pendulum’s position (Refer to Figure ii for a visual representation and aid in understanding it).

**Deriving the Equations of Motion**

The dynamics of the pendulum motion are governed by Newton’s Second Law, which states that the net force acting on the pendulum in the tangential direction is equal to the mass times the angular acceleration. The angular acceleration can be expressed as: Angular acceleration = θ’’ = $\frac{\vec{a}}{l}$ where a is the linear acceleration along the circular path, leading to: $m\vec{a} = \vec{F}$. The forces acting on the pendulum are:

**The force of gravity**. This acts in the direction of the center of the Earth, given by -mgsin(θ) where g is the acceleration due to gravity. The negative sign indicates that gravity acts to restore the pendulum to the equilibrium position (θ = 0).**The damping force**. Opposes the velocity of the pendulum. It is proportional to the angular velocity θ’, represented by -c_{1}lθ’ where c_{1}is the damping coefficient (units of kg·m^{2}/s).

Thus, Newton’s equation becomes: $ml·θ’’ = -mgsin(θ) -c_1lθ’$

By dividing by ml, we simplify this to: $θ’’ + \frac{c_1}{m}θ’+\frac{g}{l}sin(θ) = 0$. This is a **non-linear second-order differential equation**. To make the notation more compact, we define: $c = \frac{c_1}{m}$ (the damping ratio), $k = \frac{g}{l}$ (the natural frequency of the pendulum).

This yields the equation: θ’’ + cθ’ + ksin(θ) = 0. This is a non-linear second-order differential equation due to the sin(θ) term.

To analyze this system, we convert it into a first-order system by introducing a new variable w = θ' (the angular velocity). Then, w’ = -ksin(θ) -cθ’ = -ksin(θ) -cw

This leads to the following system of equations:

$\begin{cases} θ’ = w \\ w’ = -ksin(θ) -cw \end{cases}$

This is now a system of **two first-order equations in the variables θ and w.**

We will now examine the case of under-damping, where c = 1 and k = 2. The system becomes:

$\begin{cases} θ’ = w \\ w’ = -2sin(θ) -w \end{cases}$

Step 1: Find Critical Points

Critical points (equilibrium points) are points where both derivatives are zero, i.e., where the system is in equilibrium θ’ = 0 and w’ = 0. Setting the right-hand sides of the equations to zero:

$\begin{cases} f(x, y) = 0 \\ g(x, y) = 0 \end{cases}$

In our example, we need to solve:

$\begin{cases} w = 0 \\ -2sin(θ)-w = 0 \end{cases}$

Simplify the second equation: -2sin(θ)-w = 0 ⇒ w = -2sin(θ). Since w = 0, too (first equation), this leads to: -2sin(θ) = 0. Solutions for θ = nπ, n ∈ ℤ.

Thus, the critical points are: (θ, w) = (nπ, 0), n ∈ ℤ. These correspond to the pendulum being at rest (w = θ’= 0) at various positions: θ = 0, ±π, ±2π, ±3π, ···

Physically, there are two notable critical points:

- (θ, w) = (0, 0), corresponding to a stable equilibrium point.
- (θ, w) = (π, 0), corresponding to an unstable equilibrium point.

Step 2: Linearizing the System Near the Critical Points

To analyze the behavior near each critical point, we linearize the system. This could involve approximating the non-linear functions with linear ones by taking their Taylor expansion around the critical point. Near θ = 0, we can use the approximation sin(θ) ≈ θ. The system then becomes:

$\begin{cases} θ’ = w \\ w’ = -2θ -w\end{cases}$

This is a linear system and can be written in matrix form as:

$(\begin{smallmatrix}θ’\\ w’\end{smallmatrix}) = (\begin{smallmatrix}0 & 1\\ -2 & -1\end{smallmatrix})(\begin{smallmatrix}θ\\ w\end{smallmatrix})$

To determine the behavior of the system, we calculate the eigenvalues of the matrix. The characteristic equation is:

$det(\begin{smallmatrix}0-λ & 1\\ -2 & -1-λ\end{smallmatrix}) = (−λ)(−1−λ)−(−2)(1)=λ(λ+1)−(−2)= λ^2 + λ +2 = 0 ↭ λ = \frac{-1±\sqrt{1-8}}{2} = \frac{-1±\sqrt{-7}}{2}=\frac{-1±i\sqrt{7}}{2}$.

The eigenvalues are complex with a negative real part indicating a spiral sink.

- The negative real part ($\frac{-1}{2}$) indicates
**exponential decay**, meaning that**trajectories spiral inward toward the critical point**as time progresses. - The imaginary part ($\frac{\sqrt{7}}{2}$) causes the oscillatory (spiraling) behavior.

Thus, near (θ, w)= (0, 0), the system exhibits a stable spiral or spiral sink. Trajectories spiral inward in a clockwise or counterclockwise fashion, depending on initial conditions, representing damped oscillations of the pendulum returning to the equilibrium position.

The motion of the pendulum will involve spirals that gradually shrink in size, representing the damping effect. The negative real part leads to the shrinking, while the oscillations correspond to the spiraling.

Consider (θ, w) = (1, 0), $(\begin{smallmatrix}θ’\\ w’\end{smallmatrix}) = (\begin{smallmatrix}0 & 1\\ -2 & -1\end{smallmatrix})(\begin{smallmatrix}θ\\ w\end{smallmatrix}) = (\begin{smallmatrix}0 & 1\\ -2 & -1\end{smallmatrix})(\begin{smallmatrix}1\\ 0\end{smallmatrix}) = (\begin{smallmatrix}0\\ -2\end{smallmatrix})$

This show that the motion is clockwise. Near (0, 0), the system exhibits spiral sinks, with trajectories spiraling inward in a clockwise direction. (Refer to Figure iii for a visual representation and aid in understanding it)

Behavior Around (π, 0). At (π, 0) we cannot approximate sin(θ) as θ. We calculate the Jacobian matrix:

For a system:

$\begin{cases} θ^′ = f(θ, w) \\ w^′ = g(θ, w) \end{cases}$

The Jacobian matrix at a critical point (θ_{0}, w_{0}) is:

J(θ, w) = $(\begin{smallmatrix}\frac{∂f}{∂θ} & \frac{∂f}{∂w}\\ \frac{∂g}{∂θ} & \frac{∂g}{∂w}\end{smallmatrix})\bigg|_{θ_0, w_0}$ =

$(\begin{smallmatrix}f_θ & f_w\\ g_θ & g_w\end{smallmatrix})\bigg|_{θ_0, w_0}$, and this is the matrix of the linearized system.

$(\begin{smallmatrix}\frac{∂w}{∂θ} & \frac{∂w}{∂w}\\ \frac{∂(-2sin(θ)-w)}{∂θ} & \frac{∂(-2sin(θ)-w)}{∂w}\end{smallmatrix}) = (\begin{smallmatrix}0 & 1\\ -2cos(θ) & -1\end{smallmatrix})$

Evaluate at (0, 0): $J_{(0, 0)} = (\begin{smallmatrix}0 & 1\\ -2 & -1\end{smallmatrix})$. This is consistent with the previous result.

Evaluate at (π, 0): $J_{(π, 0)} = (\begin{smallmatrix}0 & 1\\ 2 & -1\end{smallmatrix})$

Now, we find the eigenvalues of the matrix J(π, 0): $det(\begin{smallmatrix}0-λ & 1\\ 2 & -1-λ\end{smallmatrix}) = (−λ)(−1−λ)−(2)(1)=λ(λ+1)−2= λ^2 + λ -2 = 0 ↭ (λ +2)(λ - 1) = 0$

Solving this quadratic equation gives the eigenvalues: λ = 1, λ = -2. One positive eigenvalue ( λ = 1) and one negative eigenvalue (λ = −2) indicates a saddle point at (π, 0).

λ = 1, $(J-λI)\vec{v} = \vec{0} ↭ (\begin{smallmatrix}-1 & 1\\ 2 & -2\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = \vec{0}$

$\begin{cases} -a_1 + a_2 = 0 \\ 2a_1 -2a_2 = 0 \end{cases}$

-a_{1} + a_{2} = 0⇒ a_{2} = a_{1} =[Simplicity] 1. The second equation is redundant. Eigenvector: $(\begin{smallmatrix}1\\ 1\end{smallmatrix})$. The solution is $(\begin{smallmatrix}1\\ 1\end{smallmatrix})e^{t}$

λ = -2, $(J-λI)\vec{v} = \vec{0} ↭ (\begin{smallmatrix}2 & 1\\ 2 & 1\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = \vec{0}$

$\begin{cases} 2a_1 + a_2 = 0 \\ 2a_1 + a_2 = 0 \end{cases}$

2a_{1} + a_{2} = 0⇒ a_{2} = -2a_{1}. Eigenvector: $(\begin{smallmatrix}1\\ -2\end{smallmatrix})$. The solution is $(\begin{smallmatrix}1\\ -2\end{smallmatrix})e^{-2t}$

The eigenvalues are real and have opposite signs. This indicates a saddle point at (π, 0):

- It represents the pendulum inverted at the top (θ = π).
**Unstable Manifold**. The positive eigenvalue λ = 1 corresponds to an unstable direction $\vec{v_1}=(\begin{smallmatrix}1\\ 1\end{smallmatrix})$ (trajectories move away from the critical point (π, 0)).**Stable Manifold**. The negative eigenvalue λ = −2 corresponds to a stable direction $\vec{v_2}=(\begin{smallmatrix}1\\ -2\end{smallmatrix})$ (trajectories move toward the critical point (π, 0)).- Trajectories approach the critical point along one direction and diverge along another. Physically, the inverted pendulum position is unstable; any slight perturbation causes it to fall away from θ = π.

Near (π,0), the system exhibits a saddle point, with trajectories diverging in one direction and converging in another (Refer to Figure iv for a visual representation and aid in understanding it).

Step 3. Plot trajectories around each critical point and then add some more. (Refer to Figure v for a visual representation and aid in understanding it)

The global behavior demonstrates the periodic nature of the pendulum system (sin(θ) is periodic with period 2π), with repeated critical points at multiples of θ, θ = nπ alternating between stable spirals (damped oscillations) and saddle points (more complex behavior).

Analyzing non-linear autonomous systems like the lightly damped pendulum involves:

**Identifying critical points**where the system is in equilibrium.**Linearizing the system**near these points to approximate behavior.**Calculating eigenvalues and eigenvectors**to determine stability and dynamics.**Interpreting physical behavior**in terms of the mathematical results.**Sketching phase portraits**to visualize the system’s trajectories and gain insight into the complex behaviors that non-linear systems can exhibit, such as oscillations, damping, and instability.

In ecological modeling, predator-prey systems are essential for understanding the interactions between two species: one being the prey and the other the predator. These models help predict population dynamics and can assist and inform policy makers to design and implement conservation and management strategies.

Consider the following system of differential equations: $\begin{cases} x’ = x(3-x)-xy \\ y’ = y(1-y) + xy \end{cases}$ where:

- x(t) represents the prey population at time t.
- y(t) represents the predator population at time t.
- Without the interaction term xy, both populations would exhibit logistic growth: The prey has a carrying capacity of 3. The predator has a carrying capacity of 1 🚀.
- The term x-y in x′ represents the decrease in prey due to predation.
- The term +xy in y′ represents the increase in predators due to consuming prey.

🚀 Definition: Carrying capacity is the maximum population size of a biological species that an environment can sustain indefinitely, given the availability of resources such as food, habitat, water, and other necessities.

It is symbolized by K in population models. It represents the equilibrium point where the population stabilizes due to resources become limited and other limiting factors like competition, predation (when considering prey-predator interactions), disease, and waste accumulation slow down growth.

The logistic growth equation models how a population grows rapidly at first and then slows as it approaches the carrying capacity. For a population N(t): $N’ = rN(1-\frac{N}{K})$ where

- N(t) is the population at time t; r is the intrinsic growth rate; K is the carrying capacity; and N′ is the rate of change of the population.
**Initial Exponential Growth**: When N is much less than K (N « K), the term $(1-\frac{N}{K})≈1$, so the population grows exponentially.**Growth Rate Decreases**: As N approaches K, $(1-\frac{N}{K})$ decreases, slowing the growth.**Stabilization**: The population stabilizes at N = K, where growth stops (N′ = 0).

**Application to the Prey-Predator Model**

- Prey Equation. x′ = x(3−x). This is a logistic growth equation. Carrying Capacity: K = 3. The prey population grows until it reaches a maximum sustainable size of 3 units (thousands, millions, etc., depending on the context).
- Predator Equation: y′=y(1−y). This is also a logistic growth equation. Carrying Capacity: K = 1. The predator population grows until it reaches a maximum sustainable size of 1 unit.

Step 1: **Finding the Critical Points**

Critical points occur where the populations do not change over time, i.e., where x′ = 0 and y′= 0.

x’ = x(3-x)-xy = 0 ⇒x = 0 or x(3-x) = xy ↭ y = 3 -x. y’ = y(1-y) + xy = 0 ⇒ y = 0 or y(1-y) = -xy ↭ x = y -1

Therefore, the critical points are:

- (0, 0)
- x = 0, y = x + 1: (0, 1).
- y = 0, x = 3 - y: (3, 0).
- y = 3 - x, x = y - 1 = 3 - x -1 = 2 - x ↭ 2x = 2 ↭ x = 1, y = 3 - x = 2: (1, 2).

Step 2: **Linearizing the System Near Each Critical Point.**

To understand the behavior near each critical point, we linearize the system by computing the Jacobian matrix.

The Jacobian matrix J(x,y) is: J(x, y) = $(\begin{smallmatrix}\frac{∂f}{∂x} & \frac{∂f}{∂y}\\ \frac{∂g}{∂x} & \frac{∂g}{∂y}\end{smallmatrix})$, where: f(x, y) = x(3−x) −xy = 3x -x^{2}-xy and g(x,y) = y(1−y) + xy = y -y^{2} + 3y.

J(x, y) = $(\begin{smallmatrix}3−2x−y & −x\\ y & 1−2y+x\end{smallmatrix})$

Step 3: **Evaluating the Jacobian at Each Critical Point**

J(0, 0) = $(\begin{smallmatrix}3−0−0 & −0\\ 0 & 1−0+0\end{smallmatrix}) = (\begin{smallmatrix}3 & 0\\ 0 & 1\end{smallmatrix})$

J(0, 1) = $(\begin{smallmatrix}3−0−1 & −0\\ 1 & 1−2⋅1+0\end{smallmatrix}) = (\begin{smallmatrix}2 & 0\\ 1 & -1\end{smallmatrix})$

J(3, 0) = $(\begin{smallmatrix}3−2⋅3−0 & −3\\ 0 & 1−0+3\end{smallmatrix}) = (\begin{smallmatrix}-3 & -3\\ 0 & 4\end{smallmatrix})$

J(1, 2) = $(\begin{smallmatrix}3−2⋅1−2 & −1\\ 2 & 1−2⋅2+1\end{smallmatrix}) = (\begin{smallmatrix}-1 & -1\\ 2 & -2\end{smallmatrix})$

Step 4: **Computing Eigenvalues and Eigenvectors**

- J(0, 0) = $(\begin{smallmatrix}3 & 0\\ 0 & 1\end{smallmatrix})$. Eigenvalues: λ
_{1}= 3, λ_{2}= 1.

Computing Eigenvectors. We solve $(J-λI)\vec{v} = \vec{0}$ for each eigenvalue.

λ_{1} = 3, $(\begin{smallmatrix}0 & 0\\ 0 & -2\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$. Equation $-2a_2 = 0⇒ a_2 = 0,\text{Eigenvector: } \vec{v_1} = (\begin{smallmatrix}1\\ 0\end{smallmatrix})$ (x-axis direction)

λ_{2} = 1, $(\begin{smallmatrix}2 & 0\\ 0 & 0\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$. Equation $2a_1 = 0⇒ a_1 = 0,\text{Eigenvector: } \vec{v_2} = (\begin{smallmatrix}0\\ 1\end{smallmatrix})$ (y-axis direction)

Both eigenvalues are positive, indicating that any small perturbation will cause the system to move away from this point. Trajectories move away from (0,0) along the x-axis and y-axis. This is not a stable state; any small introduction of individuals (prey or predators) leads to population growth. Classification: Unstable Node.

Biological Meaning: **If both populations are near zero, they will grow rapidly away from zero** (Refer to Figure i for a visual representation and aid in understanding it).

- J(0, 1) = $(\begin{smallmatrix}2 & 0\\ 1 & -1\end{smallmatrix})$. Characteristic Equation: (2−λ)(−1−λ) = 0. Eigenvalues: λ
_{1}= 2, λ_{2}= -1.

λ_{1} = 2, $(\begin{smallmatrix}0 & 0\\ 1 & -3\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$. Equation $a_1-3a_2 = 0⇒ a_1 = 3a_2,\text{Eigenvector: } \vec{v_1} = (\begin{smallmatrix}3\\ 1\end{smallmatrix})$

λ_{2} = -1, $(\begin{smallmatrix}3 & 0\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$. Equation $3a_1 = 0⇒ a_1 = 0,\text{Eigenvector: } \vec{v_2} = (\begin{smallmatrix}0\\ 1\end{smallmatrix})$

One positive (λ_{1} = 2) and one negative (λ_{2} = -1) eigenvalue. Unstable direction: $\vec{v_1} = (\begin{smallmatrix}3\\ 1\end{smallmatrix})$. Stable direction: $ \vec{v_2} = (\begin{smallmatrix}0\\ 1\end{smallmatrix})$ (y-axis).

Prey population is zero, and the predator population is at its carrying capacity of 1. Predators cannot sustain themselves without prey.

Trajectories approach (0, 1) along the stable direction (y-axis). Trajectories move away along the unstable direction. Represents an unstable equilibrium where the prey population is zero.

The introduction of prey leads to dynamics moving away from this point. The trajectory typically moves toward the stable spiral at (1,2) if prey are introduced.

Classification: Saddle Point (Unstable).

Biological Meaning: The prey population is zero while the predator population is at its carrying capacity, which is unsustainable. Small changes will cause the system to move away from this point.

- J(3, 0) = $(\begin{smallmatrix}-3 & -3\\ 0 & 4\end{smallmatrix})$. Characteristic Equation: (−3−λ)(4−λ)= 0. Eigenvalues: λ
_{1}= -3, λ_{2}= 4.

λ_{1} = -3, $(\begin{smallmatrix}0 & -3\\ 0 & 7\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$. Equation $-3a_2 = 0⇒ a_2 = 0,\text{Eigenvector: } \vec{v_1} = (\begin{smallmatrix}1\\ 0\end{smallmatrix})$

λ_{2} = 4, $(\begin{smallmatrix}-7 & -3\\ 0 & 0\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$. Equation $-7a_1 -3a_2 = 0⇒ 3a_2 = -7v_1,\text{Eigenvector: } \vec{v_2} = (\begin{smallmatrix}3\\ -7\end{smallmatrix})$

Classification: One positive (λ_{2} = 4) and one negative (λ_{1} = -3) eigenvalue. Stable direction: $\vec{v_1} = (\begin{smallmatrix}1\\ 0\end{smallmatrix})$ (x-axis). Unstable direction: $\vec{v_2} = (\begin{smallmatrix}3\\ -7\end{smallmatrix})$

Trajectories approach (3, 0) along the stable direction (x-axis). Prey population is at its carrying capacity of 3, with no predators present. Trajectories move away along the unstable direction. Represents an unstable equilibrium where the predator population is zero. Saddle Point (Unstable).

Biological Meaning: The prey population is at its maximum without predators. Introduction of predators or slight changes will destabilize this equilibrium. Typically, the system moves away from (3, 0) toward (1, 2).

- J(1, 2) = $(\begin{smallmatrix}-1 & -1\\ 2 & -2\end{smallmatrix})$. Characteristic Equation: det(J−λI) = λ
^{2}+3λ +4 = 0. Eigenvalues: $λ = \frac{-3±\sqrt{(-3)^2-4·4}}{2} = \frac{-3±i\sqrt{7}}{2}$ Complex eigenvalues with a negative real part.

Computing eigenvectors for complex eigenvalues involves dealing with complex arithmetic. However, for phase portraits, we can focus on the qualitative behavior:

- The negative real part ($\frac{-3}{2}$) indicates that trajectories
**spiral inward toward the critical point**(1, 2). - The imaginary part ($\frac{\sqrt{7}}{2}$) indicates oscillatory behavior.
- Classification: The system exhibits a stable spiral at (1, 2). Populations oscillate while approaching equilibrium.

Biological Meaning: The populations will oscillate around this point, eventually settling into a stable coexistence. It is the only stable equilibrium, where both species coexist in a balanced state.

Many initial conditions lead to trajectories that spiral into (1, 2). Populations may oscillate before stabilizing due to the predator-prey interactions. Some initial conditions may lead to extinction of one or both species or one species reaching its carrying capacity without the other stabilizing.

Critical points | (0, 0) | (0, 1) | (3, 0) | (1, 2) |
---|---|---|---|---|

J | $(\begin{smallmatrix}3 & 0\\ 0 & 1\end{smallmatrix})$ | $(\begin{smallmatrix}2 & 0\\ 1 & -1\end{smallmatrix})$ | $(\begin{smallmatrix}-3 & -3\\ 0 & 4\end{smallmatrix})$ | $(\begin{smallmatrix}-1 & -1\\ 2 & -2\end{smallmatrix})$ |

λ_{1}, λ_{2} |
3, 1 | 2, -1 | -3, 4 | $\frac{-3±\sqrt{7}i}{2}$ |

v_{1}, v_{2} |
$\vec{v_1} = (\begin{smallmatrix}1\\ 0\end{smallmatrix}), \vec{v_2} = (\begin{smallmatrix}0\\ 1\end{smallmatrix})$ | $\vec{v_1} = (\begin{smallmatrix}3\\ 1\end{smallmatrix}), \vec{v_2} = (\begin{smallmatrix}0\\ 1\end{smallmatrix})$ | $\vec{v_1} = (\begin{smallmatrix}1\\ 0\end{smallmatrix}), \vec{v_2} = (\begin{smallmatrix}3\\ -7\end{smallmatrix})$ | |

Unstable node | Saddle (unstable) | Saddle (unstable) | Spiral stable |

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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