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
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., $\frac{dy}{dx} = 3x +5y$):
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:
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:
Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x0, y0), meaning that it satisfies the initial condition y(x0) = y0.
This theorem ensures that under these conditions, the solution exists and is unique near x = x0.
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 such systems by:
To illustrate these concepts, we will apply them 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:
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), $k = \frac{g}{l}$ (relates to 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 (lightly damped case), 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:
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.
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 gradually to the equilibrium position (to rest at the bottom).
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 (θ0, w0) is:
J(θ, w) = $(\begin{matrix}\frac{∂f}{∂θ} & \frac{∂f}{∂w}\\ \frac{∂g}{∂θ} & \frac{∂g}{∂w}\end{matrix})\bigg|_{θ_0, w_0}$ =
$(\begin{matrix}f_θ & f_w\\ g_θ & g_w\end{matrix})\bigg|_{θ_0, w_0}$, and this is the matrix of the linearized system.
$(\begin{matrix}\frac{∂w}{∂θ} & \frac{∂w}{∂w}\\ \frac{∂(-2sin(θ)-w)}{∂θ} & \frac{∂(-2sin(θ)-w)}{∂w}\end{matrix}) = (\begin{matrix}0 & 1\\ -2cos(θ) & -1\end{matrix})$
Evaluate at (0, 0): $J_{(0, 0)} = (\begin{matrix}0 & 1\\ -2 & -1\end{matrix})$. 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, $(J-λI)\vec{v} = \vec{0} ↭ (\begin{smallmatrix}-1 & 1\\ 2 & -2\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = \vec{0}$
$\begin{cases} -a_1 + a_2 = 0 \\ 2a_1 -2a_2 = 0 \end{cases}$
-a1 + a2 = 0⇒ a2 = a1 =[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, $(J-λI)\vec{v} = \vec{0} ↭ (\begin{smallmatrix}2 & 1\\ 2 & 1\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = \vec{0}$
$\begin{cases} 2a_1 + a_2 = 0 \\ 2a_1 + a_2 = 0 \end{cases}$
2a1 + a2 = 0⇒ a2 = -2a1. 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). The general solution is $c_1(\begin{smallmatrix}1\\ 1\end{smallmatrix})e^{t}+c_2(\begin{smallmatrix}1\\ -2\end{smallmatrix})e^{-2t}$
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:
In ecological modeling, predator-prey systems are critical tools for understanding the interactions between two species: one being the prey and the other the predator. These models help researchers predict population dynamics and can assist and inform policy makers and conservationist 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:
🚀 Definition: The carrying capacity is the maximum population size of species that an environment can sustain indefinitely, given the availability of resources such as food, habitat, water, and other environmental conditions.
It is commonly denoted 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, slowing down growth. Eventually, the population stabilizes at K, where growth stops because births and deaths balance out.
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
Application to the Prey-Predator Model
Step-by-Step Analysis
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, meaning neither population is changing at that instant.
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:
Step 2: Linearizing the System Near Each Critical Point.
To understand how the system behaves near these equilibrium points, we use linearization. This involves finding the Jacobian matrix, which is the matrix of first-order partial derivatives of the system.
The Jacobian matrix J(x,y) is: J(x, y) = $(\begin{matrix}\frac{∂f}{∂x} & \frac{∂f}{∂y}\\ \frac{∂g}{∂x} & \frac{∂g}{∂y}\end{matrix})$, where: f(x, y) = x(3−x) −xy = 3x -x2-xy and g(x,y) = y(1−y) + xy = y -y2 + xy.
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
We determine the stability of each critical point by finding the eigenvalues of the Jacobian matrices at those points. The sign of the real parts of the eigenvalues tells us if the equilibrium is stable (eigenvalues have negative real parts) or unstable (at least one eigenvalue has a positive real part).
Computing Eigenvectors. We solve $(J-λI)\vec{v} = \vec{0}$ for each eigenvalue.
λ1 = 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)
λ2 = 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. This equilibrium is an unstable node.
Biological Meaning: If both populations are near zero, they will grow rapidly away from zero. Neither species remains extinct; they tend to increase from this point. Refer to Figure i for a visual representation and aid in understanding it.
λ1 = 2, $(\begin{smallmatrix}0 & 0\\ 1 & -3\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$. Equation $a_1-3a_2 = 0⇒ a_1 = 3a_2,\text{Eigenvector: } \vec{v_1} = (\begin{smallmatrix}3\\ 1\end{smallmatrix})$
λ2 = -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 (λ1 = 2) and one negative eigenvalue (λ2 = -1) eigenvalue. This indicates a saddle point, which is unstable. Unstable direction: $\vec{v_1} = (\begin{smallmatrix}3\\ 1\end{smallmatrix})$. Stable direction: $ \vec{v_2} = (\begin{smallmatrix}0\\ 1\end{smallmatrix})$ (y-axis).
(0, 1), prey population is zero and the predator population is at its carrying capacity of 1. Predators cannot sustain themselves without prey, so small deviations cause the system to leave this point. If prey are introduced, the system moves away from (0, 1).
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.
As it was previously stated, 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.
λ1 = -3, $(\begin{smallmatrix}0 & -3\\ 0 & 7\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$. Equation $-3a_2 = 0⇒ a_2 = 0,\text{Eigenvector: } \vec{v_1} = (\begin{smallmatrix}1\\ 0\end{smallmatrix})$
λ2 = 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 (λ2 = 4) and one negative (λ1 = -3) eigenvalue. This is another saddle point, which is unstable. 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.
Biological Meaning: The prey population is at its maximum carrying capacity (K = 3) with no predators. Any introduction of predators or slight changes or perturbations will destabilize this equilibrium and drive the system moves away from (3, 0) toward (the stable spiral at) (1, 2).
Computing eigenvectors for complex eigenvalues involves dealing with complex arithmetic. However, for phase portraits, we can focus on the qualitative behavior:
Biological Meaning: Both populations will oscillate around this point (1, 2) and slowly spiral into this equilibrium, 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 | $(\begin{matrix}3 & 0\\ 0 & 1\end{matrix})$ | $(\begin{matrix}2 & 0\\ 1 & -1\end{matrix})$ | $(\begin{matrix}-3 & -3\\ 0 & 4\end{matrix})$ | $(\begin{matrix}-1 & -1\\ 2 & -2\end{matrix})$ |
λ1, λ2 | 3, 1 | 2, -1 | -3, 4 | $\frac{-3±\sqrt{7}i}{2}$ |
v1, v2 | $\vec{v_1} = (\begin{matrix}1\\ 0\end{matrix}), \vec{v_2} = (\begin{matrix}0\\ 1\end{matrix})$ | $\vec{v_1} = (\begin{matrix}3\\ 1\end{matrix}), \vec{v_2} = (\begin{matrix}0\\ 1\end{matrix})$ | $\vec{v_1} = (\begin{matrix}1\\ 0\end{matrix}), \vec{v_2} = (\begin{matrix}3\\ -7\end{matrix})$ | |
Unstable node | Saddle (unstable) | Saddle (unstable) | Spiral stable |