Homogeneous Linear Systems with Constant Coefficients

Recall

One who asks a question is a fool for a minute; one who does not remains a fool forever, Chinese proverb

Linear First-Order Systems: The Matrix Approach

A general linear first-order system of two variables has the form:

$\begin{cases} x' = a(t)x + b(t)y + r_1(t) \\\\ y' = c(t)x + d(t)y + r_2(t) \end{cases}$ where:

• $a(t), b(t), c(t)$, and $d(t)$ are functions of time $t$ (or constants in simpler cases).
• $r_1(t)$ and $r_2(t)$ represent external forcing functions or inputs.
• $x(t_0) = x_0$ and $y(t_0) = y_0$ are the initial conditions.

Homogeneous Linear Systems with Constant Coefficients

Let’s explore homogeneous linear systems of first-order ordinary differential equations (ODEs) with constant coefficients. Specifically, we will return to our egg-cooling temperature model:

$\begin{cases} x' = -2x + 2y \\\\ y' = 2x - 5y \end{cases}$ where:

• We have swapped $T_1$ and $T_2$ for standard variables $x$ and $y$ for simplicity.
• x(t) and y(t) are the dependent variables.
• t is the independent variable (typically representing time).
• x’ = $\frac{dx}{dt}$ and y’ = $\frac{dy}{dt}$ denote the first derivatives of x(t) and y(t) with respect to t.

The given system of ODE’s is a homogeneous linear system because:

• The equations are perfectly linear with respect to $x$ and $y$.
• There are no standalone forcing terms depending only on $t$.

There are no constant or forcing terms (i.e., terms that do not depend on x or y).

In a previous section, we solved this exact system using the substitution method (converting it into a single second-order ODE). Our solution was: $\begin{cases} x = C_1e^{-t} + C_2e^{-6t} \\\ y = \frac{1}{2}C_1e^{-t} - 2C_2e^{-6t} \end{cases}$

While substitution works for 2x2 systems, it becomes computationally messy (nightmarish) for larger systems. Instead, we will solve this using the Eigenvalue and Eigenvector method. This approach transforms the system into an algebraic problem, allowing us to find the general solution by identifying the system’s eigenvalues and corresponding eigenvectors.

Solving the System Using Eigenvalues

$\begin{cases} x' = -2x + 2y \\\\ y' = 2x - 5y \end{cases}$

To streamline the solution process, we can express the system in matrix form: $X' = AX$. This facilitates the application of linear algebra techniques, particularly those involving eigenvalues and eigenvectors.

$(\begin{smallmatrix}x\\\ y\end{smallmatrix})' = (\begin{smallmatrix}-2 & 2\\ 2 & -5\end{smallmatrix})(\begin{smallmatrix}x\\\ y\end{smallmatrix})$ where:

• The matrix A = $(\begin{smallmatrix}-2 & 2\\ 2 &-5\end{smallmatrix})$ is the constant coefficient matrix.
• The vector X = $(\begin{smallmatrix}x\\ y\end{smallmatrix})$ is the state vector.

Thus, the system can be succinctly written as: X’ = AX. The goal is to find the general solution to this system.

Because the derivative of an exponential function is proportional to itself, we assume a solution of the form: $X(t) = ve^{λt}$ where:

• $\lambda$ is an unknown scalar (the eigenvalue).
• $\mathbf{v} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ is an unknown constant vector (the eigenvector).

Substitute this guess into the original system $X' = AX$:

1. The derivative of our guess is $X' = \lambda \mathbf{v} e^{\lambda t}$.
2. The right side is $AX = A(\mathbf{v} e^{\lambda t})$.

Equating them: $\lambda \mathbf{v} e^{\lambda t} = A \mathbf{v} e^{\lambda t}$

Since $e^{\lambda t}$ is never zero, we can divide both sides by it: $\lambda \mathbf{v} = A \mathbf{v}$

Simplify, $A \mathbf{v} - \lambda \mathbf{v} = 0 \implies (A - \lambda I)\mathbf{v} = 0$

This is the standard matrix eigenvalue equation! Our calculus problem has been completely reduced to finding the eigenvalues ($\lambda$) and eigenvectors ($\mathbf{v}$) of matrix $A$.

Step 1: Formulating the Characteristic Equation

To find the eigenvalues $\lambda$, we require that the matrix $(A - \lambda I)$ has a determinant of zero (otherwise, the only solution is the trivial vector $\mathbf{v} = 0$).

$$\det(A - \lambda I) = 0 \implies \begin{vmatrix} -2-\lambda & 2 \\ 2 & -5-\lambda \end{vmatrix} = 0$$

Calculate the determinant ($ad - bc = 0$): $(-2-\lambda)(-5-\lambda) - (2)(2) = 0$

Expand and simplify: $(10 + 2\lambda + 5\lambda + \lambda^2) - 4 = 0 \implies \lambda^2 + 7\lambda + 6 = 0$

Solving this quadratic equation yields our two eigenvalues: $(\lambda + 1)(\lambda + 6) = 0 \implies \lambda_1 = -1, \quad \lambda_2 = -6$ or alternatively $λ = \frac{-7±\sqrt{7^2-4·1·6}}{2·1} = \frac{-7±\sqrt{25}}{2} = \frac{-7±5}{2}$

Step 2: Finding the Corresponding Eigenvectors

Now we find the specific direction vector $\mathbf{v}$ associated with each eigenvalue by plugging $\lambda$ back into $(A - \lambda I)\mathbf{v} = 0$.

For $\lambda_1 = -1$: $\begin{pmatrix} -2-(-1) & 2 \\ 2 & -5-(-1) \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$$\begin{pmatrix} -1 & 2 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

This gives us a redundant system of equations (The second equation is a multiple -(ii) = -2(i)- of the first, which means we did it right!): $-a_1 + 2a_2 = 0 \implies a_1 = 2a_2$

We can pick any non-zero value for $a_2$. If we choose $a_2 = 1$, then $a_1 = 2$. Eigenvector 1: $\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$. Fundamental Solution 1: $X_1(t) = \begin{pmatrix} 2 \\ 1 \end{pmatrix}e^{-t}$

For $\lambda_2 = -6$: $\begin{pmatrix} -2-(-6) & 2 \\ 2 & -5-(-6) \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$$\begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

This yields the equation: $2a_1 + a_2 = 0 \implies a_2 = -2a_1$ (the second equation is a constant multiple (2) of the first one).

If we choose $a_1 = 1$, then $a_2 = -2$. Eigenvector 2: $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$. Fundamental Solution 2: $X_2(t) = \begin{pmatrix} 1 \\ -2 \end{pmatrix}e^{-6t}$

Step 3: Constructing the General Solution

By the Principle of Superposition, the general solution to a linear system is any linear combination of its fundamental solutions.

The General Vector Solution: $X(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t}$

Substituting our results: $\begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 \begin{pmatrix} 2 \\ 1 \end{pmatrix}e^{-t} + c_2 \begin{pmatrix} 1 \\ -2 \end{pmatrix}e^{-6t}$ where $c_1$ and $c_2$ are arbitrary constants determined by initial conditions.

Reconciling with the Substitution Method

If we break this vector equation back into a system of scalar equations, we get: $\begin{cases} x(t) = 2c_1e^{-t} + c_2e^{-6t} \\ y(t) = 1c_1e^{-t} - 2c_2e^{-6t} \end{cases}$ Notice how this perfectly matches the solution we obtained earlier via substitution! The only difference is how the arbitrary constants are scaled. If we map $C_1 = 2c_1$ and $C_2 = c_2$, the two solution sets are mathematically identical. The eigenvalue method simply presents it in a much cleaner, more structurally insightful vector form.

General Homogeneous Linear Systems with Constant Coefficients

In the study of ordinary differential equations (ODEs), understanding systems of linear differential equations is crucial for modeling and analyzing a wide range of physical, biological, and engineering phenomena.

A general linear first-order system of two variables has the form: $\begin{cases} x' = ax + by \\\\ y' = cx + dy \end{cases}$

where:

• $x(t)$ and $y(t)$ are the dependent state variables.
• $t$ is the independent variable (typically representing time).
• $x' = \frac{dx}{dt}$ and $y' = \frac{dy}{dt}$ denote the first time derivatives of x(t) and y(t) with respect to t.* $a, b, c$, and $d$ are constant coefficients that mathematically characterize the system’s behavior.

Matrix Representation

We can express the system elegantly in matrix form as: $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$

Or, more succinctly: $X' = AX$ where the matrix A = $(\begin{smallmatrix}a & b\\ c &d\end{smallmatrix})$ contains the coefficients from the original system of equations. The vector X = $(\begin{smallmatrix}x\\ y\end{smallmatrix})$ represents the system’s state at any time t.

X′ = AX is the system in matrix form. Our goal is to find the general solution for x(t) and y(t), that is, for the vector $X(t)$.

The Exponential Ansatz

We begin by assuming that the solutions for $x(t)$ and $y(t)$ take the form of exponential functions. This trial solution is known as the Exponential Ansatz: $X(t) = \mathbf{v} e^{\lambda t} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} e^{\lambda t}$ where

• $\lambda$ is an unknown scalar (eigenvalue) representing the exponential growth, decay, or oscillation rate.
• $\mathbf{v} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ is an unknown constant direction vector (the corresponding eigenvector).
• $e^{λt}$ represents the exponential growth or decay rate.

  This trial solution assumes that the system’s behavior can be described by exponential functions, which is a very common approach for linear systems with constant coefficients. Because the derivative of an exponential is simply a multiple of itself, it perfectly fits linear differential structures.

Formulating the Eigenvalue Problem

Substitute the trial solution $X(t) = \mathbf{v} e^{\lambda t}$ back into our matrix system $X' = AX$, we get: $(\mathbf{v} e^{\lambda t})' = A(\mathbf{v} e^{\lambda t})$

Taking the derivative of the left side yields $\lambda \mathbf{v} e^{\lambda t}$: $\lambda \mathbf{v} e^{\lambda t} = A \mathbf{v} e^{\lambda t}$

Since $e^{\lambda t}$ is strictly positive and never zero, we can divide both sides by it: $\lambda \mathbf{v} = A \mathbf{v}$

Rearranging this into a standard homogeneous linear algebra equation gives us: $A \mathbf{v} - \lambda \mathbf{v} = 0 \implies (A - \lambda I)\mathbf{v} = 0$ where $I$ is the identity matrix.

Solving the Characteristic Equation

To find the eigenvalues $\lambda$ and eigenvectors $\mathbf{v}$, we must ensure that the matrix equation $(A - \lambda I)\mathbf{v} = 0$ has non-trivial solutions (i.e., we don’t want the boring solution $\mathbf{v} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$).

For a non-trivial solution to exist, the determinant of the matrix $(A - \lambda I)$ must be exactly zero: $\det(A - \lambda I) = 0$

Let’s expand this determinant for our general $2 \times 2$ matrix $\det \begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix} = (a - \lambda)(d - \lambda) - bc = 0$

Expanding the binomials: $ad - a\lambda - d\lambda + \lambda^2 - bc = 0$

Simplifying, $\lambda^2 - (a+d)\lambda + (ad - bc) = 0$

💡 The Trace-Determinant Shortcut: Notice the coefficients of this polynomial!

1. $(a+d)$ is the sum of the main diagonal, known as the Trace of matrix $A$, denoted $Tr(A)$.
2. $(ad-bc)$ is exactly the Determinant of matrix $A$, denoted $\det(A)$.

Therefore, for any $2 \times 2$ system, the characteristic equation is always beautifully written or expressed as: $\boxed{\lambda^2 - Tr(A)\lambda + \det(A) = 0}$

The eigenvalues $\lambda_1$ and $\lambda_2$ are the roots of this characteristic equation.

Using the quadratic formula: $\lambda = \frac{(a+d) \pm \sqrt{(a+d)^2 - 4(ad-bc)}}{2}$.

These eigenvalues dictate the fundamental nature of the system’s solutions (e.g., real roots yield exponential growth/decay, complex roots yield sine/cosine oscillations).

Finding Eigenvectors and the General Solution

For each eigenvalue $\lambda_i$ we found, we substitute it back into the equation $(A - \lambda_i I)\mathbf{v}_i = 0$ to find its corresponding eigenvector $\mathbf{v}_i = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$.

$$\begin{pmatrix} a - \lambda_i & b \\ c & d - \lambda_i \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

This produces a redundant system of linear equations:

$$\begin{cases} (a - \lambda_i)a_1 + b a_2 = 0 \\ c a_1 + (d - \lambda_i)a_2 = 0 \end{cases}$$

We solve this system to find the components of the eigenvector $\mathbf{v}_i$.

Finally, by the Principle of Superposition, the general solution to the system is a linear combination of the fundamental solutions corresponding to each eigenvalue/eigenvector pair: $\boxed{X(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t}}$ where $c_1$ and $c_2$ are arbitrary constants determined by the system’s specific initial conditions.

Example: The Egg Cooling Model

Let’s apply this generalized theory to a specific homogeneous linear system with constant coefficients. Recall our egg-cooling temperature model:

$\begin{cases} x'_1 = 2x_1 + 3x_2 \\\ x'_2 = 4x_1 -2x_2 \end{cases}$

We can rewrite the system in matrix form as $X' = AX$, where: $X = (\begin{smallmatrix}x_1\\\ x_2\end{smallmatrix}), A = (\begin{smallmatrix}2 & 3\\\ 4 & -2\end{smallmatrix})$. The general solution will involve finding the eigenvalues and eigenvectors of matrix A.

Step 1: The Characteristic Equation

To find the eigenvalues, we must solve the characteristic equation $\det(A - \lambda I) = 0$.

Let’s calculate the determinant: $\det \begin{pmatrix} 2-\lambda & 3 \\ 4 & -2-\lambda \end{pmatrix} = (2-\lambda)(-2-\lambda) - (3)(4)=[\text{Simplify}] (\lambda^2 - 4) - 12 = \lambda^2 - 16$.

Thus, we obtain the characteristic equation: $λ^2-16 = 0$

💡 Using the Trace-Determinant Shortcut: $Tr(A) = 2 + (-2) = 0$. $\det(A) = (2)(-2) - (3)(4) = -4 - 12 = -16$. Formula: $\lambda^2 - Tr(A)\lambda + \det(A) = 0 \implies \lambda^2 - 0\lambda - 16 = 0$. It matches perfectly and saves us the binomial expansion!

Solving for λ: $\lambda^2 - 16 = 0 \implies (\lambda - 4)(\lambda + 4) = 0$. The roots (eigenvalues) are: $\lambda_1 = 4, \quad \lambda_2 = -4$

Step 2: Finding the Eigenvectors

Now we plug each eigenvalue back into the equation $(A - \lambda I)\mathbf{v} = 0$ to find its corresponding eigenvector.

For $\lambda_1 = 4$: $\begin{pmatrix} 2 - 4 & 3 \\ 4 & -2 - 4 \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$\begin{pmatrix} -2 & 3 \\ 4 & -6 \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

This yields the redundant system:

$$\begin{cases} -2a_1 + 3a_2 = 0 \\ 4a_1 - 6a_2 = 0 \end{cases}$$

The second equation is simply $-2$ times the first. Looking at $-2a_1 + 3a_2 = 0$, we can choose $a_1 = 3$ and $a_2 = 2$ to satisfy it.

• Eigenvector 1: $\mathbf{v}_1 = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$; $\lambda_1 = 4$.
• Fundamental Solution 1: $X_1(t) = \begin{pmatrix} 3 \\ 2 \end{pmatrix}e^{4t}$.

Similarly, for $\lambda_2 = -4$:

$$\begin{pmatrix} 2 - (-4) & 3 \\ 4 & -2 - (-4) \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$$$\begin{pmatrix} 6 & 3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

This yields the system:

$$\begin{cases} 6a_1 + 3a_2 = 0 \\ 4a_1 + 2a_2 = 0 \end{cases}$$

The second equation is a multiple ($\frac{2}{3}$) of the first. Simplifying the second equation gives $2a_1 + a_2 = 0 \implies a_2 = -2a_1$. If we choose $a_1 = 1$, then $a_2 = -2$.

• Eigenvector 2: $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$; $\lambda_2 = -4$
• Fundamental Solution 2: $X_2(t) = \begin{pmatrix} 1 \\ -2 \end{pmatrix}e^{-4t}$

Step 3: The General Solution

The general solution to the system is a linear combination of the fundamental solutions corresponding to each eigenvalue: $X(t) = C_1 \mathbf{v}_1 e^{\lambda_1 t} + C_2 \mathbf{v}_2 e^{\lambda_2 t}$

Substituting our calculated eigenvector and eigenvalues: $\begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix} = C_1 \begin{pmatrix} 3 \\ 2 \end{pmatrix}e^{4t} + C_2 \begin{pmatrix} 1 \\ -2 \end{pmatrix}e^{-4t}$ where $C_1$ and $C_2$ are arbitrary constants determined by the initial conditions.

Breaking this into a system of scalar equations:

$$\begin{cases} x_1(t) = 3C_1e^{4t} + C_2e^{-4t} \\ x_2(t) = 2C_1e^{4t} - 2C_2e^{-4t} \end{cases}$$

💡 Dynamical Systems Insight: The Saddle Point

Notice that our solution has two competing exponential terms: $e^{4t}$ and $e^{-4t}$.

• The $e^{-4t}$ term decays rapidly to $0$ as time goes forward. This represents an attractive, stable direction.
• The $e^{4t}$ term blows up to infinity as time goes forward. This represents a repulsive, unstable direction.

Because the system possesses one negative (stable) eigenvalue and one positive (unstable) eigenvalue, the origin $(0,0)$ is classified geometrically as a Saddle Point. Unless the initial conditions perfectly align with the stable eigenvector (forcing $C_1 = 0$), the unstable $e^{4t}$ term will eventually overpower the system and push the trajectory away toward infinity!

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