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Sketching Solutions of Homogeneous Linear System

Recall

The first principle is that you must not fool yourself – and you are the easiest person to fool, Richard Feynman

Behind this mask there is more than just flesh. Beneath this mask there is an idea… and ideas are bulletproof, Alan Moore

Fourier Series

Differential equations

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

  1. 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.
  2. Constants. Fixed numerical values that do not change.
  3. 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., $\frac{dy}{dx} = 3x +5y$):

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) .

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:

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

Fourier Series

Sketching Solutions of a 2x2 Homogeneous Linear System with Constant Coefficients

This model considers the competition between Massachusetts and New Hampshire in terms of their tourism advertising budgets. The variables x(t) and y(t) represent deviations from their average advertising budgets at time t, where:

The underlying idea is that if one state spends more than usual on tourism in a given year, there’s pressure to spend less in the future to balance other needs (like education, military, or social services, e.g. -x, -3y). Conversely, if a state spends less than average, there’s pressure from local businesses and entrepreneurs to increase tourism advertising. The states are also in competition with each other to attract tourists, which introduces interaction terms between x(t) and y(t) (e.g., b and c).

$\begin{cases} x’ = -x +by \\ y’ = cx -3y \end{cases}$

The given system of differential equations represents a model for competition, potentially between two entities such as nations in a conflict or companies in a market. Here, x and y represent the states of the two competing entities, and b and c are positive constants that describe the interaction between them.

The system of differential equations modeling this dynamic is:

$\begin{cases} x’ = -x +2y \\ y’ = -3y \end{cases}$

This can be rewritten in matrix form as x’ = Ax where x = $(\begin{smallmatrix}x\\y\end{smallmatrix})$ and A = $(\begin{smallmatrix}-1 & 2\\ 0 & -3\end{smallmatrix})$

Finding Eigenvalues and Eigenvectors

To solve this system, we first find the eigenvalues and eigenvectors of the matrix A. The characteristic equation is derived from the determinant of A −λI, where I is the identity matrix and λ represents the eigenvalues.

To find the eigenvalues, we solve the characteristic equation: |A - λI| = $\vert\begin{smallmatrix}-1-λ & 2\\ 0 & -3-λ\end{smallmatrix}\vert =$

= $(-1-λ)(-3-λ)-2·0 = 0 ↭ λ^2+4λ+3 = 0 ↭[\text{Factoring the quadratic}] (λ+3)(λ+1) = 0 ↭[\text{Solving for Eigenvalues}] λ = -3, λ = -1$

We now find the eigenvectors corresponding to each eigenvalue.

$(A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

For λ = -3, $(\begin{smallmatrix}2 & 2\\0 &0\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$.

This leads to the system of equations:

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

The eigenvector corresponding to λ1 = -3 is $\vec{α_1} = (\begin{smallmatrix}1\\ -1\end{smallmatrix})$

$(A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

For λ -1, we solve $(\begin{smallmatrix}0 & 2\\0 &-2\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$, which gives:

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

The eigenvector corresponding to λ2 = -1 is $\vec{α_2} = (\begin{smallmatrix} 1\\ 0\end{smallmatrix})$

The general solution to the system is a linear combination of the solutions corresponding to each eigenvalue (the eigenvectors multiplied by their exponential terms): $(\begin{smallmatrix}x\\y\end{smallmatrix}) = C_1α_1e^{λ_1t} + C_2α_2e^{λ_2t} =[\text{Substituting the eigenvalues and eigenvectors}] C_1(\begin{smallmatrix}1\\ -1\end{smallmatrix})e^{-3t} + C_2(\begin{smallmatrix} 1\\ 0\end{smallmatrix})e^{-t}$

This corresponds to the system: $\begin{cases} x(t) = c_1e^{-3t}+ c_2e^{-t} \\ y(t) = -c_1e^{-3t} \end{cases}$

To sketch the solutions, we examine the behavior for different values of C1 and C2 (c1=±1, c2=0 or c1= 0, c2 = ± 1), and then fill in the rest.

Sketching Solutions of 2x2 Homogeneous Linear System

Thus, all trajectories (starting from different initial conditions) converge towards the origin as t→∞, indicating that the origin is a sink node, a stable equilibrium point. In other words, both advertising budgets eventually return to their average levels due to the negative exponential terms. The mathematical analysis shows that the system is stable and converges to equilibrium.

Since the functions involved in the differential equations are continuous, the resulting velocity field is continuous. This implies that small changes in the state (values of x and y) lead to small changes in their rates of change, resulting in smooth trajectories. Consequently, the trajectories do not exhibit abrupt changes; instead, they evolve smoothly over time.

$\begin{cases} x = c_1e^{-3t}+ c_2e^{-t} \\ y = -c_1e^{-3t} \end{cases}$

$\begin{cases} x = c_1e^{-3t}+ c_2e^{-t} \\ y = -c_1e^{-3t} \end{cases} → 0 \text{ as } t → \infty$.

Suppose t = 10, e-10 is indeed a small number, but it is certainly bigger than e-30. As t → ∞, the term involving e-3t decay more rapidly than those involving e-t. Therefore, the long-term behavior of the system is dominated by the term involving e-t.

Suppose t = -10, e10 is indeed a big number, but it is certainly smaller than e30. They are both big, but relatively speaking e-3t is the term that dominate as t → -∞. This means that after a long time, the behavior of x(t) and y(t) is primarily determined by the eigenvalue λ2 = -1, and x(t) ≈ C2e-t, while y(t) approaches zero more quickly due to the e-3t term.

This system models a situation where, despite different initial responses to changes in advertising budgets, both Massachusetts and New Hampshire eventually revert to their normal budgets over time. Massachusetts is influenced by New Hampshire (as indicated by the 2y term in the first equation), but New Hampshire does not respond to Massachusetts’ actions (since the second equation is independent of x). In other words, Massachusetts is more reactive in its tourism policies, while New Hampshire remains more insulated.

As time progresses, both states’ advertising budgets will return to equilibrium, indicating that peace (or balance) in tourism advertising is eventually restored.

A node is a specific type of equilibrium point (also called fixed point) where the solution trajectories converge towards the point (approach the point) as time progresses (t → ∞). When the node is located at the origin (0,0) and all trajectories in its vicinity converge towards this point over time, it is referred to as a sink or sink node. In this system, the origin is a sink node because all trajectories converge to it over time.

A source node is an equilibrium point where trajectories diverge or move away from the node as time progresses (i.e., as t→ ∞).

A = $(\begin{smallmatrix}-1 & 4\\ -2 & 5\end{smallmatrix})$, $\vec{x} = C_1(\begin{smallmatrix}2\\ 1\end{smallmatrix})e^{t} + C_2(\begin{smallmatrix} 1\\ 1\end{smallmatrix})e^{3t}$

For λ1 = 1 > 0, the eigenvector is $(\begin{smallmatrix}2\\ 1\end{smallmatrix})$. This means that the trajectory associated with this eigenvalue lies along the line y = $\frac{1}{2}x$. For λ2 = 3 > 0, the eigenvector is $(\begin{smallmatrix}1\\ 1\end{smallmatrix})$. This means that the trajectory associated with this eigenvalue lies along the line y = x.

Since both eigenvalues are positive, all trajectories will diverge or move away from the origin as t increases.

As t increases, the dominate term is $C_1(\begin{smallmatrix}1\\ -1\end{smallmatrix})e^{3t}$,causing rapid growth in the direction of $(\begin{smallmatrix}1\\ -1\end{smallmatrix})$. For any other initial point not lying on these lines, the trajectory will initially be influenced by both eigenvectors, but as t increases, the term e3t will dominate, causing the trajectory to align with the direction of $(\begin{smallmatrix}1\\\ -1\end{smallmatrix})$

Since all trajectories move away from the origin, the system is classified as unstable. Refer to Figure A for a visual representation and aid in understanding it.

Sketch the following homogeneous linear system

A = $(\begin{smallmatrix}-3 & 0\\ 3 & -2\end{smallmatrix})$, $\vec{x} = C_1(\begin{smallmatrix}1\\ -3\end{smallmatrix})e^{-3t} + C_2(\begin{smallmatrix} 0\\ 1\end{smallmatrix})e^{-2t}$

For λ1 = -3 < 0, the eigenvector is $(\begin{smallmatrix}1\\ -3\end{smallmatrix})$. This means that the trajectory associated with this eigenvalue lies along the line y = -3x. For λ2 = -2 < 0, the eigenvector is $(\begin{smallmatrix}0\\ 1\end{smallmatrix})$. This means that the trajectory associated with this eigenvalue lies along the line x = 0 (the y-axis).

Since both eigenvalues are negative, all trajectories will converge to the origin as t increases. As t increases, t → ∞ the dominate term is $C_2(\begin{smallmatrix}1\\ 0\end{smallmatrix})e^{-2t}$ (because e-2t decays more slower than e-3t).

Along the line y = -3x and x = 0, trajectories will move towards the origin. For any other initial point not lying on these lines, the trajectory will initially be influenced by both eigenvectors, but as t increases, the term e-2t will dominate, causing the trajectory to align with the direction of $(\begin{smallmatrix} 0\\\ 1\end{smallmatrix})$ and converge to the origin.

Since all trajectories move towards the origin, the system is classified as stable. Refer to Figure B for a visual representation and aid in understanding it.

Sketch the following homogeneous linear system

A = $(\begin{smallmatrix}4 & 0\\ 2 & -1\end{smallmatrix})$, $\vec{x} = C_1(\begin{smallmatrix}5\\ 2\end{smallmatrix})e^{4t} + C_2(\begin{smallmatrix} 0\\ 1\end{smallmatrix})e^{-t}$

For λ1 = 4 > 0, the eigenvector is $(\begin{smallmatrix}5\\ 2\end{smallmatrix})$. This means that the trajectory associated with this eigenvalue lies along the line y = 25x. For λ2 = -1 < 0, the eigenvector is $(\begin{smallmatrix}0\\ 1\end{smallmatrix})$. This means that the trajectory associated with this eigenvalue lies along the line x = 0 (the y-axis).

Since λ1 = 4 > 0, trajectories associated with this eigenvalue will diverge from the origin as t increases. Since λ2 = -1 < 0, trajectories associated with this eigenvalue will converge to the origin as t increases.

The term $C_1(\begin{smallmatrix}5\\ 2\end{smallmatrix})e^{4t}$ grows exponentially, while $C_2(\begin{smallmatrix} 0\\ 1\end{smallmatrix})e^{-t}$ decays exponentially. Therefore, the e4t term will dominate the behaviour as t increases.

Along the lines y = 25x and x = 0, trajectories will move towards the origin. or any other initial point not lying on these lines, the trajectory will initially be influenced by both eigenvectors, but as t increases, the term e4t will dominate causing the trajectory to align with the direction of $(\begin{smallmatrix}5\\ 2\end{smallmatrix})$ and diverge form the origin (Refer to Figure C for a visual representation and aid in understanding it)

Given these features, the origin acts as a saddle point in the phase portrait of this system:

  1. Any initial point lying on the y-axis will move towards the origin along this line as t→∞, due to the decaying term e-t.
  2. Any initial point lying on y = 25x will move away from the origin as t→∞, since the term e4t grows exponentially.
  3. All trajectories (except those exactly on the y-axis) will eventually align with the direction of v1 and diverge from the origin.

Sketch the following homogeneous linear system

Let $\vec{x} = C_1(\begin{smallmatrix}1\\ -1\end{smallmatrix})e^{3t} + C_2(\begin{smallmatrix} 1\\ 0\end{smallmatrix})e^{t}$

It’s quite similar as before, However, as t increases, we go the other way. In this case, as t → ∞ the dominate term is $C_1(\begin{smallmatrix}1\\ -1\end{smallmatrix})e^{3t}$,causing rapid growth in the direction of $(\begin{smallmatrix}1\\ -1\end{smallmatrix})$. Conversely, as t → -∞, the dominate term is $C_2(\begin{smallmatrix} 1\\ 0\end{smallmatrix})e^{t}$, the system approaches the origin slowly along the direction of $(\begin{smallmatrix} 1\\ 0\end{smallmatrix})$ (Refer to Figure iii for a visual representation and aid in understanding it)

Sketching Solutions of 2x2 Homogeneous Linear System

In this system, we have an unstable node at the origin, where all trajectories diverge from the origin over time.

Consider the system of differential equations: $\begin{cases} x’ = -x +3y \\ y’ = 5x -3y \end{cases}$

Step 1: Representing the System in Matrix Form

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.

We can express the system in matrix form as: x’ = Ax where A = $(\begin{smallmatrix}-1 & 3\\ 5 & -3\end{smallmatrix})$ and x = $(\begin{smallmatrix}x\\ y\end{smallmatrix})$ This matrix equation encapsulates the system of differential equations.

To solve this system, we first find the eigenvalues and eigenvectors of the matrix A. The characteristic equation is derived from the determinant of A −λI, where I is the identity matrix and λ represents the eigenvalues.

Step 2: Finding the Eigenvalues of Matrix A

To find the eigenvalues, we solve the characteristic equation: |A - λI| = $\vert\begin{smallmatrix}-1-λ & 3\\ 5 & -3-λ\end{smallmatrix}\vert$

$|A - λI| = 0 ↭ = $λ^2 +4λ -12 = 0 ⇒[\text{This factors as: }] (λ+6)(λ-2) = 0 ↭ λ = -6, λ = 2$

Step 3: Finding the Eigenvectors

We will find an eigenvector corresponding to each eigenvalue λ by solving (A −λI)v = 0 ↭ $(A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

For λ = -6, we solve $(\begin{smallmatrix}5 & 3\\5 &3\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$, this yields the system:

$\begin{cases} 5a_1 + 3a_2 = 0 \\ 5a_1 + 3a_2 = 0 \end{cases}$

Both equations are the same, indicating infinite solutions along a line. $a_1 = \frac{-3}{5}a_2$. Choose $a_2 = -5$ for simplicity: $a_1 = \frac{-3}{5}a_2 = \frac{-3}{5}·(-5) = 3$

The eigenvector corresponding to λ1 = -6 is $\vec{α_1} = (\begin{smallmatrix} 3\\ -5\end{smallmatrix})$

$(A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

For λ = 2, we solve $(\begin{smallmatrix}-3 & 3\\5 & -5\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$. This yields the system:

$\begin{cases} -3a_1 + 3a_2 = 0 \\ 5a_1 -5a_2 = 0 \end{cases}$

$-3a_1 + 3a_2 = 0 ⇒ a_1 = a_2.$ Choose a1 = a2 = 1.

The eigenvector corresponding to λ1 = -6 is $\vec{α_2} = (\begin{smallmatrix} 1\\ 1\end{smallmatrix})$

Step 4: Constructing the General Solution

The general solution to the system is a linear combination of the eigenvectors multiplied by exponential functions of their corresponding eigenvalues: $(\begin{smallmatrix}x\\y\end{smallmatrix}) = C_1α_1e^{λ_1t} + C_2α_2e^{λ_2t} =[\text{Substituting the eigenvalues and eigenvectors}] C_1(\begin{smallmatrix}3\\ -5\end{smallmatrix})e^{-6t} + C_2(\begin{smallmatrix} 1\\ 1\end{smallmatrix})e^{2t}$

$x(t) = 3C_1e^{-6t} + C_2e^{2t}, y(t) = -5C_1e^{-6t} + C_2e^{2t}$

Step 5: Analyzing the Behavior of the Solutions

For large positive t, the term $C_2(\begin{smallmatrix} 1\\ 1\end{smallmatrix})e^{2t}$ dominates because of the positive eigenvalue λ2 = 2. This means the trajectories diverge away from the origin in the direction of the eigenvector $(\begin{smallmatrix} 1\\ 1\end{smallmatrix})$, corresponding to unstable growth as t→∞.

For large negative t, the term $C_1(\begin{smallmatrix}3\\ -5\end{smallmatrix})e^{-6t}$ dominates because e-6t decays rapidly. The trajectories approach the origin along the direction of the eigenvector $(\begin{smallmatrix}3\\ -5\end{smallmatrix})$, corresponding to stable decay as t → -∞.

  1. The solution represents a saddle point at the origin (Refer to Figure iv for a visual representation and aid in understanding it).
  2. Unstable direction: Trajectories move away from the origin along the line spanned by $\vec{α_2} = (\begin{smallmatrix} 1\\ 1\end{smallmatrix})$ for t → ∞. Diverge from the origin along another direction (the unstable eigenvector) y = x (C1 = 0, C2 = 1), y = -x (C1 = 0, C2 = -1).
  3. Stable direction: Trajectories approach the origin along the line spanned by $\vec{α_1} = (\begin{smallmatrix} 3\\ -5\end{smallmatrix})$ for t → -∞. Approach the origin from one direction (along the stable eigenvector).
  4. The trajectories between these two directions exhibit curved, saddle-like behavior, approaching the stable manifold as t→−∞ and diverging along the unstable manifold as t→∞.

    The system represents a scenario where two companies are engaged in an advertising war, and the solution suggests that everyone ends up overspending in advertising. Companies are driven to either aggressive growth or rapid decline The saddle point nature implies that finding a balanced, sustainable strategy is challenging in this competitive environment

Sketching Solutions of 2x2 Homogeneous Linear System

Bibliography

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].
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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