Question

(a) Convert the scalar equation $$ P_{0}(t) y^{(n)}+P_{1}(t) y^{(n-1)}+\cdots+P_{n}(t) y=F(t) $$ into an equivalent \(n \times n\) system $$ \mathbf{y}^{\prime}=A(t) \mathbf{y}+\mathbf{f}(t) $$ (b) Suppose (A) is normal on an interval \((a, b)\) and \(\left\\{y_{1}, y_{2}, \ldots, y_{n}\right\\}\) is a fundamental set of solutions of $$ P_{0}(t) y^{(n)}+P_{1}(t) y^{(n-1)}+\cdots+P_{n}(t) y=0 $$ on \((a, b)\). Find a corresponding fundamental matrix \(Y\) for $$ \mathbf{y}^{\prime}=A(t) \mathbf{y} $$ on \((a, b)\) such that $$ y=c_{1} y_{1}+c_{2} y_{2}+\cdots+c_{n} y_{n} $$ is a solution of \((\mathrm{C})\) if and only if \(\mathrm{y}=Y \mathrm{c}\) with $$ \mathbf{c}=\left[\begin{array}{c} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \end{array}\right] $$ is a solution of (D).

Short answer

In summary, to convert the given scalar differential equation into an equivalent \(n \times n\) system: (a) We first introduced new variables for higher-order derivatives and then represented the original equation as a system of first-order differential equations. The resulting system was then expressed in the form \(\mathbf{y}' = A(t) \mathbf{y} + \mathbf{f}(t)\). (b) Given a fundamental set of solutions for the homogeneous system, we created a fundamental matrix \(Y(t)\) and demonstrated that it corresponds to the given fundamental set of solutions by showing that \(\mathbf{y} = Y \mathbf{c}\).

Step-by-step solution

(a) Converting the scalar equation into an equivalent \(n \times n\) system

To convert the scalar equation into a system of first-order differential equations, we will introduce new variables for the higher-order derivatives. Define the following variables: $$ y_1 = y, $$ $$ y_2 = y^{(1)}, $$ $$ \cdots $$ $$ y_n = y^{(n-1)}. $$ Now, we can express the system as a first-order differential equation system as: $$ \begin{cases} y_1' = y_2, \\ y_2' = y_3, \\ \cdots \\ y_{n-1}' = y_n, \\ y_n' = \frac{1}{P_0(t)}[-P_1(t)y_n - P_2(t)y_{n-1} - \cdots - P_n(t)y - F(t)] \end{cases} $$ Let \(\mathbf{y}=\left[\begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array}\right]\), and \(A(t)\) be an \(n \times n\) matrix defined by $$ A(t) = \left[\begin{matrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ -\frac{P_n(t)}{P_0(t)} & -\frac{P_{n-1}(t)}{P_0(t)} & -\frac{P_{n-2}(t)}{P_0(t)} & \cdots & -\frac{P_1(t)}{P_0(t)} \end{matrix}\right]. $$ With the defined \(\mathbf{y}\) and \(A(t)\), the system can be rewritten as: $$ \mathbf{y}' = A(t) \mathbf{y} + \mathbf{f}(t), $$ where \(\mathbf{f}(t) = \left[\begin{array}{c}0 \\ 0 \\ \vdots \\ - \frac{F(t)}{P_0(t)} \end{array}\right]\).

(b) Finding the fundamental matrix Y

Given a fundamental set of solutions \(\left\\{y_{1}, y_{2}, \ldots, y_{n}\right\\}\) for the homogeneous part of the scalar equation, we need to find a corresponding fundamental matrix \(Y(t)\) where \(y=c_1 y_1+c_2 y_2+\cdots+c_n y_n\) is a solution of the system if and only if \(\mathbf{y}=Y \mathbf{c}\). For each \(y_k\) in the fundamental set, we will define \(\mathbf{y}_k = \left[\begin{array}{c} y_{k} \\ y_{k}^{(1)} \\ \cdots \\ y_{k}^{(n-1)} \end{array}\right]\). To find \(Y(t)\), we concatenate \(\mathbf{y}_1, \mathbf{y}_2, \ldots, \mathbf{y}_n\) and form the matrix: $$ Y(t) = \left[\begin{array}{c|c|c|c} \mathbf{y}_1 & \mathbf{y}_2 & \ldots & \mathbf{y}_n \end{array}\right] $$ Now, we will show that \(\mathbf{y}=Y \mathbf{c}\). Let \(y=c_1 y_1+c_2 y_2+\cdots+c_n y_n\). Then $$ \mathbf{y} = \left[\begin{array}{c} y \\ y^{(1)} \\ \vdots \\ y^{(n-1)} \end{array}\right] = c_1 \left[\begin{array}{c} y_1 \\ y_1^{(1)} \\ \vdots \\ y_1^{(n-1)} \end{array}\right] + c_2 \left[\begin{array}{c} y_2 \\ y_2^{(1)} \\ \vdots \\ y_2^{(n-1)} \end{array}\right] + \cdots + c_n \left[\begin{array}{c} y_n \\ y_n^{(1)} \\ \vdots \\ y_n^{(n-1)} \end{array}\right], $$ which can also be written as: $$ \mathbf{y} = Y \mathbf{c} = \left[\begin{array}{c|c|c|c} \mathbf{y}_1 & \mathbf{y}_2 & \ldots & \mathbf{y}_n \end{array}\right] \left[\begin{array}{c} c_1 \\ c_2 \\ \cdots \\ c_n \end{array}\right]. $$ Thus, the condition \(\mathbf{y}=Y \mathbf{c}\) is fulfilled, and the fundamental matrix \(Y(t)\) corresponds to the given fundamental set of solutions.

Key concepts

System of Linear Equations

In mathematics, a system of linear equations forms the basis of solving complex mathematical and real-world problems. When we talk about a system of linear equations, we're referring to multiple linear equations that we deal with together. A common goal is to find values for variables that satisfy all the equations in the system simultaneously.

When solving these systems, each equation is linear, meaning it has no exponents or non-linear terms, and it can be written in the form of a straight line when graphed. The system can have one, none, or infinitely many solutions depending on the relationships between the equations.

For example, converting a higher-order differential equation into a system of first-order linear differential equations allows us to analyze and solve it using linear algebra techniques. We define new variables corresponding to the derivatives, changing the problem's complexity but providing a structured way to find solutions more systematically.

Fundamental Matrix

The fundamental matrix is a powerful tool in the world of linear algebra and differential equations. It is specifically used when dealing with systems of linear differential equations. This concept ties to the notion of a fundamental set of solutions.

A fundamental matrix helps simplify and solve systems by organizing fundamental solutions into a single matrix. It essentially transforms a system of linear differential equations into a matrix form, with each column of the matrix representing a fundamental solution to the system.

To construct a fundamental matrix, you'll populate it with solutions from the fundamental set corresponding to each variable involved in the system. This organization allows you to easily express any solution of the system as a combination of solutions in the fundamental matrix.

Using a matrix this way not only simplifies calculations but also helps in exploring further properties of the solutions, such as stability and behavior over time.

First-Order Differential Equation Systems

First-order differential equation systems are equations involving derivatives of unknown functions, where the highest derivative is the first derivative. These systems are crucial in modeling many phenomena in fields such as physics, biology, and engineering.

In simpler terms, first-order systems are collections of equations that describe how things change. They provide a means to understand and predict the behavior of dynamic systems over time. To solve these systems, you'll often use techniques from linear algebra and calculus.

Transforming higher-order differential equations into first-order systems can involve defining a set of new variables, as seen in our exercise. This action reduces complex problems into manageable chunks, allowing for practical and efficient computation of solutions using well-established mathematical methods.

Analyzing first-order systems provides insights into the trajectory and stability of systems, helping predict long-term outcomes and responses to initial conditions or external influences.