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

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 2x2 Homogeneous Linear System with Constant Coefficients

Let’s analyze and sketch the phase portraits of several homogeneous linear systems. Each system is defined by a matrix A and its corresponding general solution.

This model considers the competition between Massachusetts and New Hampshire in terms of their tourism advertising budgets. We model how deviations from their average advertising budgets evolve over time. Let:

• x(t): Departure of Massachusetts’ tourism budget from its normal (average) level at time t.
• y(t): Departure of New Hampshire’s tourism budget from its normal (average) level at time t.

The underlying idea is that if a state spends more than usual on tourism in a given year, there’s pressure to spend less in the future to balance other needs (e.g., education, military, or social services; -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, introducing interaction terms between x(t) and y(t) (e.g., b and c).

We model the situation with the following system:

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

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:

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

• Massachusetts’ rate of change in advertising budget depends negatively on its own current deviation (−x) and positively on New Hampshire’s current deviation (+2y). This means if Massachusetts is above its normal budget, there’s pressure to decrease (−x), but if New Hampshire is above its normal budget, Massachusetts feels competitive pressure to increase (+2y).
• New Hampshire’s rate of change depends only on its own current deviation (−3y) and is independent of Massachusetts’ budget (x). This suggests that New Hampshire does not react to Massachusetts' actions.

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.

• Eigenvalue λ = -3

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

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 $\vec{α_1} = (\begin{smallmatrix}1\\ -1\end{smallmatrix})$

• Eigenvalue λ = -1

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

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

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

2a₂ = 0 ⇒ a₂ = 0. a₁ can be any real number. Let’s choose a₁ = 1 for simplicity.

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.

• e₁: c₁ = 1, c₂ = 0. In this case, x = e^-3t, y = -e^-3t. t = 0 ⇒ (x, y) = (1, -1). As t increases, it is always in the direction of $(\begin{smallmatrix}1\\\ -1\end{smallmatrix})$, meaning the trajectory lies along the line y = -x (approaching the origin faster due to the exponential term), but decreasing in length by $e^{-3t}$. The eigenvector $(\begin{smallmatrix}1\\ -1\end{smallmatrix})$ determines the direction on which it travels and the eigenvalue determine which way it goes. If the eigenvalue is negative (λ = -3), it is approaching the origin as t increases. Refer to Figure ii for a visual representation and aid in understanding it.

• e₂: c₁ = -1, c₂ = 0. Similarly, x = -e^-3t, y = e^-3t. t = 0 ⇒ (x, y) = (-1, 1). As t increases, it is always in the direction of $(\begin{smallmatrix}-1\\ 1\end{smallmatrix})$, meaning the trajectory lies along the line y = -x, but decreasing in length by $e^{-3t}$.

• e₃: c₁ = 0, c₂ = 1 ⇒ x = e^-t, y = 0. t = 0 ⇒ (x, y) = (1, 0). As t increases, t is always in the direction of $(\begin{smallmatrix}1\\ 0\end{smallmatrix})$, so the trajectory lies along the x-axis, moving towards the origin as t increases.

• e₄: c₁ = 0, c₂ = -1 ⇒ x = -e^-t, y = 0. t = 0 ⇒ (x, y) = (-1, 0). As t increases, t is always in the direction of $(\begin{smallmatrix}-1\\ 0\end{smallmatrix})$. The trajectory lies along the negative x-axis, also approaching the origin.

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 a fixed point) where the solution trajectories either converge towards the point (approach the point) or diverge away from it as as time progresses (t → ∞). Depending on the direction of the trajectories, nodes can be further classified as sinks or sources.

A sink node is an equilibrium point where all nearby trajectories in the phase plane move towards the node as time progresses (t → ∞). When a node is located at the origin (0,0) and all trajectories in its vicinity converge towards this point over time, it is specifically referred to as a sink or sink node. This behavior indicates that the system is stable, as perturbations from the equilibrium point will decay over time, bringing the system back to equilibrium.

A source node is an equilibrium point where all nearby trajectories in the phase plane move away from the node as time progresses (i.e., as t→ ∞). This behavior signifies that the system is unstable, as small perturbations from the equilibrium point will grow over time, causing the system to diverge from equilibrium.

• Sketch the following homogeneous linear system

Given 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 in the vicinity of the origin will diverge away from it as time progresses, making the origin an unstable node or source.

As t increases, the dominate term is the one associated with the larger eigenvalue in magnitude λ₂ = 3, i.e., $C_1(\begin{smallmatrix}1\\ -1\end{smallmatrix})e^{3t}$, will dominate the behavior of the system 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 e^3t will dominate, causing the trajectory to align with the direction of the eigenvector $(\begin{smallmatrix}1\\\ -1\end{smallmatrix})$, resulting in rapid growth along the line y = x.

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 in the vicinity of the origin will converge towards it as time progresses, making the origin a stable node or sink. As t increases, t → ∞ the term associated with the smaller eigenvalue in magnitude 𝜆₂ = − 2, i.e., $C_2(\begin{smallmatrix}1\\ 0\end{smallmatrix})e^{-2t}$ decays more slower than the term associated with 𝜆₁ = − 3. Therefore, the behavior of the system will be dominated by the e^-2t term.

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 = ²⁄₅x. 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 e^4t term will dominate the behaviour as t increases.

Along the lines y = ²⁄₅x and x = 0, trajectories will move away or towards the origin respectively. Any other initial point not lying on these lines or all other trajectories, will initially be influenced by both eigenvectors, but as t increases, the term e^4t 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. To sum up:

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 = ²⁄₅x will move away from the origin as t→∞, since the term e^4t 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 to the previous systems discussed. However, the behavior changes as t varies. In this case: the dominant term is $C_1(\begin{smallmatrix}1\\ -1\end{smallmatrix})e^{3t}$, causing rapid growth in the direction of $(\begin{smallmatrix}1\\ -1\end{smallmatrix})$. This implies that as time progresses forward, the trajectories move away from the origin along the line defined by y = −x.

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)

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

• Sketch solutions of a homogeneous linear system with constant coefficients

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

• Eigenvalue for λ = -6

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 $\vec{α_1} = (\begin{smallmatrix} 3\\ -5\end{smallmatrix})$

• Eigenvalue λ = 2

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

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 λ₂ = 2 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 $(\begin{smallmatrix} 1\\ 1\end{smallmatrix})$, which lies along the line y = x, 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 $(\begin{smallmatrix}3\\ -5\end{smallmatrix})$, which lies along the line y = −5/3x, corresponding to stable decay as t → -∞, indicating that trajectories are attracted to the origin along this direction in backward time.

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 the line spaned by $\vec{α_2}$ (the unstable eigenvector), which corresponds to y = x (C₁ = 0, C₂ = 1), y = -x (C₁ = 0, C₂ = -1) as t → ∞.
3. Stable direction: Trajectories approach the origin along the line spanned by $\vec{α_1} = (\begin{smallmatrix} 3\\ -5\end{smallmatrix})$ (C₂ = 0, C₁ = 1, -1), y = −5/3x, y = 5/3x. 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

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