Sketching Solutions of Homogeneous Linear System II

Recall

Quantity has its own quality, Joseph Stalin

Now I am become Death, the destroyer of worlds, Robert Oppenheimer

Differential equations

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, $\frac{dy}{dx}$

First-Order Linear Ordinary Differential Equations (ODEs)

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 (x₀, y₀) and
• Its partial derivative with respect to y, denoted as $\frac{∂f}{∂y}$, is also continuous near (x₀, y₀).

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.

Sketching Solutions of a Homogeneous Linear System with Complex Eigenvalues

Consider the system of differential equations: $\begin{cases} x’ = -x -y \\ y’ = 2x -3y \end{cases}$

Our goal is to solve this system, find the general solution, and sketch the trajectories of the solutions to understand the system’s behavior over time.

Step 1: Representing the System in Matrix Form. We can represent the system in matrix form as: x’ = Ax, where x =$(\begin{smallmatrix}x’\\ y’\end{smallmatrix})$, A =$(\begin{smallmatrix}-1 & -1\\ 2 & -3\end{smallmatrix})$

Step 2: Finding the Eigenvalues of Matrix A

To solve this system, we need to find the eigenvalues λ and eigenvectors v of the matrix A. The eigenvalues are found by solving the characteristic equation: det(A −λI), where I is the identity matrix and λ represents the eigenvalues.

To find the eigenvalues λ of matrix A, we solve the characteristic equation: |A - λI| = $\vert\begin{smallmatrix}-1-λ & -1\\ 2 & -3-λ\end{smallmatrix}\vert = (−1−λ)(−3−λ) + 2 = (λ+1)(λ+3) +2 = λ^2 + 4λ + 3 + 2 = λ^2 + 4λ + 5 = 0 $

Solving the Characteristic Equation: $λ^2 + 4λ + 5 = 0 ⇒[\text{Using the quadratic formula}] λ = \frac{-4±\sqrt{-4}}{2} = -2±i$

Step 3: Finding the Eigenvectors

We need to solve: (A −λI)v = 0

• λ₁ = -2 + i The matrix A - λI is: $\vert\begin{smallmatrix}−1−(−2+i) & -1\\ 2 & −3−(−2+i)\end{smallmatrix}\vert = \vert\begin{smallmatrix}1-i & -1\\ 2 & -1-i\end{smallmatrix}\vert$

For λ = -2 + i, we solve $\vert\begin{smallmatrix}1-i & -1\\ 2 & -1-i\end{smallmatrix}\vert(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$, which gives: (1-i)a₁ -a₂ = 0 ⇒ a₂ = (1-i)a₁.

From the first equation a₂ = (1-i)a₁. Substitute a₂ into the second equation: 2a₁ + (-1-i)a₁ = 0 ↭ $2a_1 + (-1-i)(1-i)a_1 = 0 ↭ 2a_1 +(-1 +i -i -1)a_1 = 0 ↭ 2a_1-2a_1=0↭ 0 = 0$ ⇒[The equation is always true, we can choose a₁ = 1 for simplicity] $a_2 = (1-i)$

Thus, the eigenvector corresponding to λ₁=−2+i is: $\vec{v_1} = (\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = (\begin{smallmatrix}1\\1-i\end{smallmatrix})$

Step 4: Constructing the General Solution

The corresponding solution to the system can be written as $x = e^{λt}v = e^{(-2+i)t}(\begin{smallmatrix}1\\1-i\end{smallmatrix}) = e^{-2t}e^{it}((\begin{smallmatrix}1\\1\end{smallmatrix})+ i(\begin{smallmatrix}0\\ -1 \end{smallmatrix})) = e^{-2t}(cos(t) + isin(t))((\begin{smallmatrix}1\\1\end{smallmatrix})+ i(\begin{smallmatrix}0\\ -1 \end{smallmatrix})) $

so that we get respectively for the real and imaginary parts of x

$x_1 = e^{-2t}((\begin{smallmatrix}1\\1\end{smallmatrix})cos(t) - (\begin{smallmatrix}0\\ -1\end{smallmatrix})sin(t))$

$x_2 = e^{-2t}((\begin{smallmatrix}1\\1\end{smallmatrix})sin(t) + (\begin{smallmatrix}0\\ -1\end{smallmatrix})cos(t))$

The general solution is a real-valued function, so we consider the real and imaginary parts separately. The real part provides one solution, and the imaginary part provides another. Therefore, the general solution is: $c_1[((\begin{smallmatrix}1\\1\end{smallmatrix})cos(t) - (\begin{smallmatrix}0\\ -1 \end{smallmatrix})sin(t))]e^{-2t} + c_2[((\begin{smallmatrix}1\\1\end{smallmatrix})sin(t) + (\begin{smallmatrix}0\\ -1 \end{smallmatrix})cos(t))]e^{-2t}$

The real question is what $[((\begin{smallmatrix}1\\1\end{smallmatrix})cos(t) - (\begin{smallmatrix}0\\ -1\end{smallmatrix})sin(t)), ((\begin{smallmatrix}1\\1\end{smallmatrix})sin(t) + (\begin{smallmatrix}0\\ -1 \end{smallmatrix})cos(t))]]$ look like? As a curve, this is bounded, periodic, the exponential decay e^-2t causes the amplitude of the trajectories to decrease over time, resulting in a spiral inward toward the origin. The system describes a spiral sink, meaning that as time progresses, the solutions are spiraling trajectories that move inward toward the origin.

How do you know that it goes around counterclockwise and not clockwise? By calculating the vector at (1, 0) from the system’s velocity field (our original system equations), we obtained (-1, 2), therefore the motion is counterclockwise (Refer to Figure v for a visual representation and aid in understanding it).

Given x=1, y=0: x’ = −x−y = −1−0 = −1 (motion to the left). y′ = 2x−3y = 2(1)−0 = 2 (motion upwards). Since x′ < 0 and y′ > 0, the motion is up and to the left from the point (1, 0), indicating a counterclockwise rotation around the origin. ​

• Sketch the following homogeneous linear system

A = $(\begin{smallmatrix}-2 & 3\\ -3 & -2\end{smallmatrix}).$ We find the eigenvalue λ by solving the characteristic equation: det(A -λI) = 0 ↭ $det (\begin{smallmatrix}-2-λ & 3\\ -3 & -2-λ\end{smallmatrix}) = 0↭ (-2-λ)(-2-λ)-(3)(-3) = 0↭ (λ+2)^2+9 = 0↭ (λ+2)^2 = -9↭ λ+2 = ±3i$. Thus, the eigenvalues are: λ = -2 ± 3i

The eigenvalues are complex conjugates with a real part Real(λ) = -2 < 0, which is negative. Since the real part of the eigenvalues is negative, the trajectories will spiral inward towards the origin.

The presence of the imaginary part ±3i indicates that the trajectories will be spirals (oscillatory bounded behaviour), causing the trajectories to spiral rather than move directly inward.

To determine the direction of the spiral, you can compute the action of the matrix on a standard basis vector, such as $(\begin{smallmatrix}1\\ 0\end{smallmatrix})$

$(\begin{smallmatrix}-2 & 3\\ -3 & -2\end{smallmatrix})(\begin{smallmatrix}1\\ 0\end{smallmatrix}) = (\begin{smallmatrix} -2 \\ -3 \end{smallmatrix})$. This vector points downwards and to the left, indicating a clockwise rotation if you follow the arrow.

This homogeneous linear system is stable because the trajectories spiral inward towards the origin. The negative real part of the eigenvalues (Re(λ)=−2) causes the trajectories to move towards the origin, and the presence of the imaginary part (±3i) indicates that the trajectories are spirals. The system exhibits a stable spiral inward toward the origin in a clockwise direction (Refer to Figure i for a visual representation and aid in understanding it)

• Sketch the following homogeneous linear system

A = $(\begin{smallmatrix}2 & 3\\ -3 & 2\end{smallmatrix})$. Compute the characteristic equation $det(A-λI) = det((\begin{smallmatrix}2-λ & 3\\ -3 & 2-λ\end{smallmatrix})) = 0 ↭ (2-λ)^2-(-3)(3) = 0 ↭ (2-λ^2) = -9↭ 2-λ = ±3i$. Thus, the eigenvalues are: λ = 2 ± 3i

The real part of the eigenvalue Real(λ) = 2 > 0, which is positive, and an imaginary part Im(λ) = ±3. Since the real part of the eigenvalues is positive, the trajectories will spiral outward, away from the origin, moving away from the origin due to exponential growth.

The presence of the imaginary part ±3i (oscillatory bounded behaviour) indicates that the trajectories will be spirals.

To determine the direction of the spiral, you can compute the action of the matrix on a standard basis vector, such as $(\begin{smallmatrix}1\\ 0\end{smallmatrix})$

$(\begin{smallmatrix}2 & 3\\ -3 & 2\end{smallmatrix})(\begin{smallmatrix}1\\ 0\end{smallmatrix}) = (\begin{smallmatrix} 2 \\ -3 \end{smallmatrix})$. This vector points downwards and to the right, indicating a clockwise rotation if you follow the arrow.

This homogeneous linear system is unstable because the trajectories spiral outwards away from the origin. The positive real part of the eigenvalues (Re(λ) = 2) causes the trajectories to move away from the origin, and the presence of the imaginary part (±3i) indicates that the trajectories are spirals. The system exhibits an unstable spiral (trajectories spiral outward from the origin) in a clockwise direction (Refer to Figure ii for a visual representation and aid in understanding it)

• Sketch the following homogeneous linear system

A = $(\begin{smallmatrix}0 & 1\\ -5 & 0\end{smallmatrix})$

Step 1: Finding the Eigenvalues

We start by calculating the eigenvalues λ of A by solving the characteristic equation: det(A -λI) = 0 ↭ $det(\begin{smallmatrix}-λ & 1\\ -5 & -λ\end{smallmatrix}) = (−λ)(−λ)−(1)(−5)= λ^2+5 = 0$

Solve for λ: $λ^2 = -5 ↭ λ = ±\sqrt{5}i$, so the eigenvalues are purely imaginary.

Eigenvalues that are purely imaginary indicate oscillatory behavior in the system. Since the eigenvalues have no real component, there will be no exponential growth or decay, meaning the trajectories neither spiral inward nor outward but form closed orbits around the origin. This type of system is known as a center, and the trajectories are expected to be circles or ellipses centered at the origin.

Step 2: Finding the Eigenvector for $λ = +\sqrt{5}i$

To find the eigenvector corresponding to $λ = ±\sqrt{5}i$, we solve: (A−λI)v = 0

$v = (\begin{smallmatrix}-\sqrt{5}i & 1\\ -5 & -\sqrt{5}i\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = (\begin{smallmatrix}0\\ 0\end{smallmatrix})$

This gives the system of equations:

$\begin{cases} -\sqrt{5}ia_1 + a_2 = 0 \\ -5a_1 -\sqrt{5}a_2 = 0 \end{cases}$

From the first equation, we get: $a_2 = \sqrt{5}ia_1$. Let a₁ = 1, then $a_2 = \sqrt{5}ia_1 = \sqrt{5}i$. So, an eigenvector corresponding to $λ = \sqrt{5}i$ is v = $(\begin{smallmatrix}1\\ \sqrt{5}i\end{smallmatrix})$

Step 3: Constructing the Complex Solution

The complex solution is: $\vec{x}(t) = e^{λt}v = (\begin{smallmatrix} 1 \\ \sqrt{5}i\end{smallmatrix})e^{\sqrt{5}it} =[\text{Using Euler’s formula}] (\begin{smallmatrix} 1 \\ \sqrt{5}i\end{smallmatrix})(cos(\sqrt{5}t)+isin(\sqrt{5}t)) = ((\begin{smallmatrix} 1 \\ 0\end{smallmatrix})+(\begin{smallmatrix} 0 \\ \sqrt{5}\end{smallmatrix})i)(cos(\sqrt{5}t)+isin(\sqrt{5}t)) $

Step 4: Separating Real and Imaginary Parts to Form Real Solutions

The real and imaginary parts of x(t) give us two linearly independent solutions:

The real solutions are: $\vec{x_1} = (\begin{smallmatrix} 1 \\ 0\end{smallmatrix})cos(\sqrt{5}t)-(\begin{smallmatrix} 0 \\ \sqrt{5}\end{smallmatrix})sin(\sqrt{5}t) = (\begin{smallmatrix} cos(\sqrt{5}t) \\ -\sqrt{5}sin(\sqrt{5}t)\end{smallmatrix}). \vec{x_2} = (\begin{smallmatrix} 1 \\ 0\end{smallmatrix})sin(\sqrt{5}t)+(\begin{smallmatrix} 0 \\ \sqrt{5}\end{smallmatrix})cos(\sqrt{5}t) = (\begin{smallmatrix} sin(\sqrt{5}t) \\ \sqrt{5}cos(\sqrt{5}t)\end{smallmatrix})$

Therefore, the general solution can be written as a linear combination of these two real solutions. The general solution for a system with purely imaginary eigenvalues can be expressed as: $\vec{x}(t) = C_1(\begin{smallmatrix} cos(\sqrt{5}t) \\ -\sqrt{5}sin(\sqrt{5}t)\end{smallmatrix}) + C_2(\begin{smallmatrix} sin(\sqrt{5}t) \\ \sqrt{5}cos(\sqrt{5}t)\end{smallmatrix})$

Step 5: Interpreting the Solution and Phase Portrait

The eigenvalues are purely imaginary, which indicates that the system has oscillatory behavior and periodic motion, so there is no exponential growth or decay (λ has no real part). This results in closed trajectories that form circular or elliptical orbits around the origin.

The system represents a center, meaning that trajectories will be closed curves (circles or ellipses) around the origin. They neither converge to nor diverge from the origin. Instead, they form closed orbits. (Refer to Figure iii for a visual representation and aid in understanding it)

Step 6: Determining the Direction of Rotation

To determine whether the closed trajectories are rotating clockwise or counterclockwise, we examine the effect of A on a standard basis vector,, such as $(\begin{smallmatrix}1\\ 0\end{smallmatrix})$

$ (\begin{smallmatrix}0 & 1\\ -5 & 0\end{smallmatrix})(\begin{smallmatrix}1\\ 0\end{smallmatrix}) = (\begin{smallmatrix}0\\ -5\end{smallmatrix})$. This results, $(\begin{smallmatrix}0\\ -5\end{smallmatrix})$, indicates that the vector $(\begin{smallmatrix}1\\ 0\end{smallmatrix})$ is rotated into the vector $(\begin{smallmatrix}0\\ -5\end{smallmatrix})$, suggesting a clockwise rotation around the origin.

General Homogeneous Linear System

In the study of differential equations, particularly in advanced calculus like Calculus III or IV, homogeneous linear systems play a crucial role. These systems allow us to model and understand complex dynamical behaviors in various fields such as physics, engineering, and economics. Specifically, we focus on systems of the form: $\vec{x’} = A\vec{x}$ where:

• $\vec{x}(t)$ is a vector of unknown functions, $\vec{x} = (\begin{smallmatrix}x(t)\\ y(t)\end{smallmatrix})$.
• A is a constant 2x2 matrix, A = $(\begin{smallmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{smallmatrix})$
• $\vec{x’}(t)$ denotes the derivative of $\vec{x}(t)$ with respect to t.

Our goal is to find the general solution to this system, which involves understanding the concepts of linear independence, the Wronskian determinant, and the fundamental matrix.

General Solution of the System

Theorem. The general solution to the system $\vec{x'} = A\vec{x}$ can be expressed as a linear combination of two linearly independent solutions: $\vec{x(t)} = c_1\vec{x_1} +c_2\vec{x_2}$ where c₁ and c₂ are constants, and $\vec{x_1}(t)$ and $\vec{x_2}(t)$ are linearly independent solutions of the differential equation.

1. These solutions $\vec{x_1}(t)$ and $\vec{x_2}(t)$ are called fundamental solutions because they form a basis for the solution space of the differential equation. This means that any solution $\vec{x}(t)$ can be written as a linear combination of these two solutions.
2. Two solutions are linearly independent if one is not a scalar multiple of the other. In other words, there are no constants k such that x₂(t) = kx₁(t) for all t. This property ensures that the solutions span the entire solution space of the differential equation.
3. Constants c₁ and c₂. These constants are determined by initial conditions or specific requirements of the problem. If we are given x(t₀), we can solve for c₁ and c₂ to satisfy the condition.

Determining Linear Independence: The Wronskian

To verify whether two solutions $\vec{x_1}$ and $\vec{x_2}$ are linearly independent, we use the Wronskian.

Definition. The Wronskian for two vector-valued functions is defined as the determinant of a matrix formed by placing the solutions side by side: $W(\vec{x_1}(t), \vec{x_2}(t)) := |\vec{x_1} \vec{x_2}| = det(\begin{smallmatrix}x_1(t) & x_2(t)\\ y_1(t) & y_2(t)\end{smallmatrix})$ where $\vec{x_1}(t) = (\begin{smallmatrix}x_1(t)\\y_1(t)\end{smallmatrix})$ and $\vec{x_2} = (\begin{smallmatrix}x_2(t)\\y_2(t)\end{smallmatrix})$

The Wronskian gives us valuable information about the relationship between the two solutions:

• If the Wronskian is identically zero (W(t) ☰ 0 for all t), then the two solutions $\vec{x_1}, \vec{x_2}$ are linear dependent. This means one solution is a scalar multiple of the other.
• If the Wronskian W(t) is never zero on an interval, (W(t) ≠ 0 for all t in an interval), then the solutions $\vec{x_1}, \vec{x_2}$ are linearly independent. This means they span the entire solution space, forming a base for the solution space.

  Note: In linear systems with constant coefficients, the Wronskian is either always zero or never zero. This is due to the uniqueness of solutions in linear differential equations.

Fundamental Matrix Solution

A powerful tool in solving linear systems is the fundamental matrix.

A fundamental matrix X(t) for the system $\vec{x'} = A\vec{x}$ is a 2 x 2 matrix whose columns are linearly independent solutions of the system, $X := [\vec{x_1} \vec{x_2}] = (\begin{smallmatrix}x_1(t) & x_2(t)\\\ y_1(t) & y_2(t)\end{smallmatrix})$. It provides a compact representation of the full solution space.

Each column of X(t) is one of the independent solutions, so the matrix X(t) contains all the information about the general solution to the system.

Properties:

1. Non-singular Determinant. The determinant of X(t), which is the Wronskian of the two solutions, is never zero. Since $\vec{x_1}(t)$ and $\vec{x_2}(t)$ are linearly independent, then det(X(t)) ≠ 0 for all t.
2. The matrix X(t) satisfies the original differential equation: X'(t) = AX(t).

This property follows from the fact that each column of X(t) is a solution to the differential equation

X’ = AX ↭ $[\vec{x_1’} \vec{x_2’}] = A[\vec{x_1} \vec{x_2}] =[\text{Simple matrix multiplication}] [A\vec{x_1} A\vec{x_2}] ↭ \vec{x_1’} = A\vec{x_1}, \vec{x_2’} = A\vec{x_2}$ and this is true because each colum of X(t) is a solution to the differential equation.

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