Question

Writing \(f^{\prime \prime \prime}=\left(f^{\prime \prime}\right)^{\prime}\), consider the third order differential equation $$ f^{\prime \prime \prime}-a_{1} f^{\prime \prime}-a_{2} f^{\prime}-a_{3} f=0 $$ a. Show that \(\left[\begin{array}{l}f_{1} \\ f_{2} \\\ f_{3}\end{array}\right]\) is a solution to the system $$ \begin{array}{l} \left\\{\begin{array}{l} f_{1}^{\prime}= & a_{1} f_{1}+f_{2} \\ f_{2}^{\prime}= & a_{2} f_{1}+f_{3} \\ f_{3}^{\prime}=a_{3} f_{1} \end{array}\right. \\ \text { that is }\left[\begin{array}{l} f_{1}^{\prime} \\ f_{2}^{\prime} \\ f_{3}^{\prime} \end{array}\right]=\left[\begin{array}{lll} a_{1} & 1 & 0 \\ a_{2} & 0 & 1 \\ a_{3} & 0 & 0 \end{array}\right]\left[\begin{array}{l} f_{1} \\ f_{2} \\ f_{3} \end{array}\right] \end{array} $$ b. Show further that if \(\left[\begin{array}{l}f_{1} \\ f_{2} \\\ f_{3}\end{array}\right]\) is any solution to this system, then \(f=f_{1}\) is a solution to Equation 3.15 . where \(a_{1}, a_{2},\) and \(a_{3}\) are real numbers. Let \(f_{1}=f, f_{2}=f^{\prime}-a_{1} f\) and \(f_{3}=f^{\prime \prime}-a_{1} f^{\prime}-a_{2} f^{\prime \prime}\) Remark. A similar construction casts every linear differential equation of order \(n\) (with constant coefficients) as an \(n \times n\) linear system of first order equations. However, the matrix need not be diagonalizable, so other methods have been developed.

Short answer

\([f_1, f_2, f_3]^T\) satisfies the matrix system and makes \(f=f_1\) solve the differential equation.

Step-by-step solution

Define Terms

Let \( f_1 = f \), \( f_2 = f' - a_1 f \), and \( f_3 = f'' - a_1 f' - a_2 f \). We need to show \( \begin{bmatrix} f_1' \ f_2' \ f_3' \end{bmatrix} = \begin{bmatrix} a_1 & 1 & 0 \ a_2 & 0 & 1 \ a_3 & 0 & 0 \end{bmatrix} \begin{bmatrix} f_1 \ f_2 \ f_3 \end{bmatrix} \).

Compute Derivatives

Calculate the derivatives \( f_1' = f' \), \( f_2' = (f' - a_1 f)' = f'' - a_1 f' \), and \( f_3' = (f'' - a_1 f' - a_2 f)' = f''' - a_1 f'' - a_2 f' \).

Substitute into Matrix Equation

Substitute the definitions and derivatives into the matrix equation. For \( f_1' \), \( f' = a_1 f_1 + f_2 \). For \( f_2' \), substitute \( f_1'' \) into \( f''' = f_3' + a_1 f_2 + a_2 f_1 \). For \( f_3' \), note \( f''' = a_3 f_1 \) by definition.

Verify the Matrix Representation

Check: \( f_1' = a_1 f_1 + f_2 \), \( f_2' = a_2 f_1 + f_3 \), \( f_3' = a_3 f_1 \) matches: \( \begin{bmatrix} a_1 & 1 & 0 \ a_2 & 0 & 1 \ a_3 & 0 & 0 \end{bmatrix} \begin{bmatrix} f_1 \ f_2 \ f_3 \end{bmatrix} \). They represent equivalent systems.

Confirm Conditions for f

To solve part (b), verify if \( f_1 = f \) satisfies \( f''' - a_1 f'' - a_2 f' - a_3 f = 0 \) using \( f_1' \), \( f_2' \), and \( f_3' \). By substitution, show that each \( f_i' \) condition satisfies the third-order differential equation.

Key concepts

Linear System of First-Order Equations

Understanding the concept of a linear system of first-order equations is crucial when dealing with higher-order differential equations. These systems break down complex problems into simpler, more manageable parts. In this context, transforming a third-order differential equation into a system of first-order equations allows you to utilize a structured approach to solve it.

A first-order system typically consists of several equations that relate the derivatives of functions with each other. For example, in the given exercise, you use the functions \( f_1 \), \( f_2 \), and \( f_3 \) as part of a vector that helps break down the original third-order problem. By expressing these functions as a system of first-order equations, each derivative becomes one part of the overall solution set.

This method is particularly useful because first-order systems have a variety of solution techniques and are generally easier to solve using computational tools. It also highlights the interconnected nature of the variables and how they can be systematically analyzed.

Matrix Representation

Matrix representation plays a pivotal role in simplifying and solving systems of equations, especially when dealing with differential equations. By writing the system of first-order equations from the exercise in matrix form, you streamline the process of handling multiple equations.

In matrix notation, a system of linear equations can be expressed compactly as \( \mathbf{A} \mathbf{x} = \mathbf{b} \). For the exercise, the matrix \( \mathbf{A} \) is \( \begin{bmatrix} a_1 & 1 & 0 \ a_2 & 0 & 1 \ a_3 & 0 & 0 \end{bmatrix} \), and the vector form for derivatives becomes \( \begin{bmatrix} f_1' \ f_2' \ f_3' \end{bmatrix} \). This representation is powerful because it allows for the use of linear algebra techniques to find solutions.

With matrix representation, you can apply a range of methods like eigenvalue analysis, which provides insights into the behavior of the system over time. It also facilitates the use of numerical solvers, which can handle large systems efficiently, ultimately aiding in the understanding and solving of complex differential equations.

Differential Equations with Constant Coefficients

Differential equations with constant coefficients are among the most common types found in mathematical modeling. They offer significant advantages because their solutions tend to follow systematic patterns, allowing for extensive analytical solutions.

The third-order differential equation in the exercise has constant coefficients \( a_1 \), \( a_2 \), and \( a_3 \). This means these values do not change as the function progresses, simplifying the process of finding a general solution. For these types of equations, particular techniques like the characteristic equation method are often employed.

The process involves finding the roots of the characteristic polynomial, which corresponds to the matrix derived from the system of first-order equations. Based on the nature of these roots (real or complex), you derive solutions that describe the behavior of the system comprehensively. In practical applications, these solutions are invaluable in predicting system dynamics like oscillations, decay, or growth, depending on the context of the differential equation.