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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

Any fool can criticize, complain, and condemn —and most fools do. But it takes character and self-control to be understanding and forgiving, Dale Carnegie

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

General Form of a Linear First-Order System

A linear first-order system 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:

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 analyze the following system:

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

We will solve this system using the eigenvalue and eigenvector method, also known as the exponential ansatz. 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.

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

In a previous solution approach, we used substitution to reduce the system to a second-order differential equation. The solution was found to be:

$\begin{cases} x = C_1e^{-t} + C_2e^{-6t} \\ y = \frac{1}{2}C_1e^{-t} - 2C_2e^{-6t} \end{cases}$

Solving the System Using the Eigenvalue and Eigenvector Method

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

To streamline the solution process, we can express the system in matrix form. 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 = $(\begin{smallmatrix}-2 &2\\2 &-5\end{smallmatrix})$ contains the coefficients of the system. The vector X = $(\begin{smallmatrix}x\\ y\end{smallmatrix})$represents the dependent variables.

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

The eigenvalue and eigenvector method involves finding solutions of the form: $X(t) = ve^{λt}$ where:

Formulating the Eigenvalue Problem

Assume a solution of the form: $X(t) = (\begin{smallmatrix}x\\ y\end{smallmatrix}) = ve^{λt} = (\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})e^{λt}$

Substituting into the original system X’ = AX: (🚀) $X’ = λve^{λt} = Ave^{λt}$

Cancelling the exponential term eλt (since it’s never zero): λv = Av. This equation is the matrix eigenvalue equation.

Setting Up the Characteristic Equation. To find the eigenvalues λ, we solve the characteristic equation: det(A -λI) = 0 where I is the identity matrix.

Compute A -λI = $(\begin{smallmatrix}-2-λ & 2\\2 &-5-λ\end{smallmatrix})$

Calculate the determinant: $det(A -λI) = (-2-λ)(-5-λ)-4 = 0 ↭[\text{Simplify}] (2+λ)(5+λ)-4 = 0 ↭ λ^2+7λ + 6 = 0$.

Solving for λ using the quadratic formula, $λ = \frac{-7±\sqrt{7^2-4·1·6}}{2·1} = \frac{-7±\sqrt{25}}{2}$.

Finding the Corresponding Eigenvectors

The roots are λ1 = -1, λ2 = -6.

For λ1 = -1. Substitute λ = 1 into the (A−λI)v=0:

$(\begin{smallmatrix}-2-(-1) & 2\\2 &-5-(-1)\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = (\begin{smallmatrix}-1 & 2\\2 &-4\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$

This yields the system:

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

The second equation is a multiple of the first, so we can choose a2 = 1, which gives a1 = 2. The eigenvector corresponding to λ = 1 is: $(\begin{smallmatrix}a_1\\a_2\end{smallmatrix})=(\begin{smallmatrix}2\\ 1\end{smallmatrix})$. The solution associated with λ = 1 is: $(\begin{smallmatrix}2\\ 1\end{smallmatrix})e^{-t}$

For λ2 = 6. Substitute λ = 6 into (A-λI)v = 0, we aim to calculate a1, a2.

$(\begin{smallmatrix}-2-(-6) & 2\\2 &-5-(-6)\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = (\begin{smallmatrix}4 & 2\\2 & 1\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$

This yields the system:

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

The second equation is a constant multiple (2) of the first one. Typically, we make a1 = 1, then a1 = -2. The eigenvector corresponding to λ = -6 is: $(\begin{smallmatrix}a_1\\a_2\end{smallmatrix})=(\begin{smallmatrix}1\\ -2\end{smallmatrix})$. The solution associated with λ = -6 is: $(\begin{smallmatrix}1\\ -2\end{smallmatrix})e^{-6t}$

Using the eigenvalues and eigenvectors, the general solution to the system is a linear combination of the solutions corresponding to each eigenvalue: X(t) = $C_1v_1e^{dλ_1t} + C_2v_2e^{dλ_2t}$.

In other words, by the principle of superposition, the general solution to the system is: $(\begin{smallmatrix}x\\y\end{smallmatrix}) = \tilde{c_1}(\begin{smallmatrix}2\\ 1\end{smallmatrix})e^{-t}+ \tilde{c_2}(\begin{smallmatrix}1\\ -2\end{smallmatrix})e^{-6t}$. This matches the solution obtained via substitution, but presented in a clearer vector form, the difference is that $\frac{c_1}{2} = \tilde{c_1}, c_2 = \tilde{c_2}$. The constants C1 and C2 are determined by initial conditions.

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.

Consider the following homogeneous system of two coupled first-order linear differential equations with constant coefficients: $\begin{cases} x’ = ax + by \\ y’ = cx + dy \end{cases}$

where:

Matrix Representation

$(\begin{smallmatrix}x\\ y\end{smallmatrix})’ = (\begin{smallmatrix}a & b\\c & d\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix})$ (🚀) 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) ↭ $\vec{x’} = A\vec{x}$

We begin by assuming that the solutions for x(t) and y(t) take the form of exponential functions: $X(t) = ve^{λt} = (\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})e^{λt}$ where

Substituting the Trial Solution $X(t) = ve^{λt}$ into the System X’ = AX. leads to: $λve^{λt} = Ave^{λt}$. Since eλt is never zero, we can divide both sides by eλt. This equation is the eigenvalue equation: $Av = λv$.

This trial solution assumes that the solution grows (or decays) exponentially with time at a rate determined by λ.

Formulating and Solving the Eigenvalue Problem

To find the eigenvalues λ and eigenvectors v, we solve the characteristic equation derived from the eigenvalue equation.

The characteristic equation is obtained by ensuring that the system Av = λv has non-trivial solutions (i.e., v ≠ 0): det(A -λI) = 0 where I is the identity matrix.

The determinant is: det(A -λI) = $(\begin{smallmatrix}a-λ & b\\c & d-λ\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (a-λ)(d-λ) -bc = 0 ↭ λ^2 +(a+d)λ + (ad -bc) = 0$. This is a quadratic equation in λ, known as the characteristic equation.

The eigenvalues λ1 and λ2 are the roots of the characteristic equation: $λ^2 -(a+d)λ + (ad -bc) = 0$

Using the quadratic formula: $λ = \frac{(a+d)±\sqrt{(a+d)^2-4·1·(ad-bc)}}{2·1}$. These eigenvalues determine the nature of the system’s solutions (e.g., exponential growth, decay, oscillations).

For each eigenvalue λi, we find the corresponding eigenvector vi by solving:

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

$(\begin{smallmatrix}a-λ_i & b\\c &d-λ_i\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$.

$\begin{cases} (a -λ_i)a_1 + ba_2 \\ ca_1 + (d-λ_i)a_2 = 0 \end{cases}$

This gives us a linear system for a1 and a2 that we can solve to find the components of the corresponding eigenvector vi associated with each eigenvalue λi.

Once we have the eigenvalues and eigenvectors, the general solution to the system is a linear combination of the solutions corresponding to each eigenvalue: $\vec{x(t)} = c_1\vec{v_1}e^{λ_1} + c_2\vec{v_2}e^{λ_2}$ where $\vec{v_1} = (\begin{smallmatrix}a_{11}\\a_{12}\end{smallmatrix})$ and $\vec{v_2} = (\begin{smallmatrix}a_{21}\\a_{22}\end{smallmatrix})$ are the eigenvectors corresponding to the eigenvalues λ1 and λ2 respectively, and c1 and c2 are arbitrary constants determined by initial conditions.

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

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.

To find the eigenvalues, we compute the characteristic polynomial, which requires calculating det(A − λI).

Compute A -λI = $(\begin{smallmatrix}2-λ & 3\\ 4 & -2-λ\end{smallmatrix})$

Calculate the determinant: det(A -λI) = $det(\begin{smallmatrix}2-λ & 3\\ 4 & -2-λ\end{smallmatrix}) = (2-λ)(-2-λ)-12 = λ^2-2λ+2λ-4-12 = λ^2-16$.

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

Solving for λ: $λ^2-16 = 0 ↭ (λ-4)(λ+4) = 0$. The roots (eigenvalues) are λ1 = 4, λ2 = -4.

For λ1 = 4. Substitute λ1 into (A -λI)v = 0

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

This yields the system:

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

The second equation is a multiple of the first so we can choose a1 = 3 and a2 = 2. The eigenvector corresponding to λ1 = 4 is $v_1 = (\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = (\begin{smallmatrix}3\\ 2\end{smallmatrix})$. The solution associated with λ1 = 4 is: $(\begin{smallmatrix}3\\ 2\end{smallmatrix})e^{4t}$

For λ2 = -4. Substitute λ2 into (A -λI)v = 0

$(\begin{smallmatrix}6 & 3\\ 4 & 2\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$

This yields the system:

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

The second equation is a multiple (23) of the first so we can choose a1 = 1 and a2 = -2. The eigenvector corresponding to λ1 = -4 is $v_1 = (\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = (\begin{smallmatrix}1\\ 2\end{smallmatrix})$. The solution associated with λ1 = 4 is: $(\begin{smallmatrix}1\\ 2\end{smallmatrix})e^{-4t}$

The general solution to the system is a linear combination of the solutions corresponding to each eigenvalue: X(t) = $C_1v_1e^{dλ_1t} + C_2v_2e^{dλ_2t} = C_1(\begin{smallmatrix}3\\ 2\end{smallmatrix})e^{4t} + C_2(\begin{smallmatrix}1\\ 2\end{smallmatrix})e^{-4t}$ where C1 and C2 are constants determined by initial conditions.

Repeated Real Eigenvalues

Consider a model where a fish tank is divided into three compartments, each maintained at different temperatures:

The compartments are well insulated from the outside environment, and heat flows only between these compartments. Over time, these compartments exchange heat, and our goal is to determine how the temperatures in each compartment evolve over time (Refer to Figure i for a visual representation and aid in understanding it).

Sketching Solutions of 2x2 Homogeneous Linear System

To represent this mathematically, let xi(t) represent the temperature in the i-th compartment at time t. The rate of change of temperature in each compartment depends on the temperature differences between neighboring compartments.

Setting Up the Differential System

We aim to calculate the variables xi, which represent the temperatures in each compartment as functions of time.

We assume that the heat flow from compartment i to compartment j is proportional to the difference in temperature between the compartments, following Newton's law of cooling. This means the rate at which heat flows is proportional to xj - xi.

Thus, the system of equations governing the temperature changes is.

Thus, for compartment 1, the net rate of heat flow into it is proportional to (x2 - x1) + (x3 - 11). Therefore, the differential equation for x1 is:

$x_1’ = a(x_3-x_1)+a(x_2-x_1) ↭[\text{This can be simplified to: }] x_1’ = -2ax_1+ax_2+ax_3$

There is symmetry in our differential system regarding the heat flow between compartments (each compartment interacts with the others in a similar way). For simplicity, we assume that a = 1, which leads to the following system of differential equations:

$\begin{cases} x_1’ = -2x_1+x_2+x_3 \\ x_2’ = x_1-2x_2+x_3 \\ x_2’ = x_1+x_2-2x_3 \end{cases}$

This system can be written in matrix form as: x’ = Ax where x = $(\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix})$ and A = $\Bigl( \begin{smallmatrix}-2 & 1 & 1\\ 1 & -2 & 1\\ 1 & 1 & -2\end{smallmatrix} \Bigr)$

We now aim to find the general solution to this system, which requires solving for the eigenvalues and eigenvectors of the matrix A.

To find the eigenvalues, we solve the characteristic equation: |A - λI| = $\Bigl \vert\begin{smallmatrix}-2-λ & 1 & 1\\ 1 & -2-λ & 1\\ 1 & 1 & -2-λ\end{smallmatrix}\Bigr \vert = -(λ+2)^3-3(-2-λ)+2 = 0 ↭−(8+12λ+6λ^2 +λ^3)−3(−2−λ)+2 = 0 ↭ (−8+6+2)+(−12λ+3λ)+(−6λ^2) −λ^3 =0 ↭\text{Simplify and multiply both sides by -1} λ^3 +6λ^2 +9λ = 0$

Factoring it: λ(λ +3)2 = 0. Eigenvalues: λ1 = 0, λ2, 3 = -3 (repeated root).

Finding Eigenvectors

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

$(\begin{smallmatrix}-2 & 1 & 1\\1 &-2 & 1 \\1 & 1 & -2\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix}) = 0$.

This leads to:

$\begin{cases} -2a_1 + a_2 + a_3 = 0 \\ a_1 -2a_2 + a_3 = 0 \\ a_1 + a_2 -2a_3 = 0 \end{cases}$

This is a non-independent solution, the determinant of the coefficients is zero. This system has infinitely many solutions.

Solution: $(\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix})e^{0t} = (\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix})$. This is a constant solution, and intuitively there is a trivial solution, all compartments share the same temperature, hence there is no reason why they should change their temperatures as time goes on. So, the problem itself suggest a solution $v_1 =(\begin{smallmatrix}1\\1\\1\end{smallmatrix})$

We can check that this solution is indeed a solution of the system of equations.

We solve $(A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0 ↭ (A+3I)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

$(\begin{smallmatrix}1 & 1 & 1\\1 &1 & 1 \\1 & 1 & 1\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix}) = 0$.

$\begin{cases} a_1 + a_2 + a_3 = 0 \\ a_1 + a_2 + a_3 = 0 \\ a_1 + a_2 + a_3 = 0 \end{cases}$

This a dependent system of equation, meaning they describe the same relationship between the variables. Futhermore, I just have one equation, but repeated two times, namely $a_1 + a_2 + a_3 = 0 ⇒a_3 = -(a_1+a_2)$

  1. If a1 = 1, a2 = 0 ⇒ a3 = -1, $v_2 = (\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix})$
  2. If a1 = 1, a2 = -1 ⇒ a3 = 0, $v_3 = (\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix})$

Two independent solutions are: $(\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix})e^{-3t}$ and $(\begin{smallmatrix}1\\ -1 \\ 0\end{smallmatrix})e^{-3t}$

The reader should be cautions, e.g., $(\begin{smallmatrix}0 \\ 1 \\ -1\end{smallmatrix})$ is not independent of the previous both: $(\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix}) - (\begin{smallmatrix}1\\ -1 \\ 0\end{smallmatrix}) = (\begin{smallmatrix}0 \\ 1 \\ -1\end{smallmatrix})$

The general solution is a linear combination of the eigenvectors: $(\begin{smallmatrix}x_1\\x_2\\x_3\end{smallmatrix}) = c_1(\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix})e^{-3t} + c_2(\begin{smallmatrix}1\\ -1 \\ 0\end{smallmatrix})e^{-3t}+c_3(\begin{smallmatrix}1\\1\\1\end{smallmatrix})$

Long-Term Behavior. As t → ∞, $(\begin{smallmatrix}x_1\\x_2\\x_3\end{smallmatrix}) → c_3(\begin{smallmatrix}1\\1\\1\end{smallmatrix})$. All solutions tend to the solution where all compartments reach the same temperature, corresponding to the eigenvalue λ = 0.

Understanding Repeated Eigenvalues

A complete eigenvalue (or a non-defective eigenvalue) is a repeated eigenvalue for which there exists a sufficient number of linearly independent eigenvectors to form a complete basis. This means that if λ is an eigenvalue with algebraic multiplicity m (the number of times it is repeated), there are m linearly independent eigenvectors associated with λ. As a result, you can construct a full set of independent solutions.

In our problem, λ = −3 is a complete eigenvalue because we found two linearly independent eigenvectors corresponding to its algebraic multiplicity of 2.

A defective eigenvalue (or a non-complete eigenvalue) is also a repeated eigenvalue, but in this case, there are not enough linearly independent eigenvectors to match its algebraic multiplicity.

The Spectral or Principal Axis Theorem

The Spectral or Principal Axis Theorem is a fundamental result in linear algebra and has significant implications in various fields such as physics, engineering, and mathematics. It states that for a real symmetric matrix, the eigenvalues are real, and there exists an orthogonal basis of eigenvectors.

The Spectral or Principal Axis Theorem states that if A is a real n × n symmetric matrix (i.e., A = AT), then:

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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