Question

Suppose \(\left\\{y_{1}, y_{2}, \ldots, y_{n}\right\\}\) is a fundamental set of solutions of $$ P_{0}(x) y^{(n)}+P_{1}(x) y^{(n-1)}+\cdots+P_{n}(x) y=0 $$ on \((a, b),\) and let $$ \begin{aligned} z_{1} &=a_{11} y_{1}+a_{12} y_{2}+\cdots+a_{1 n} y_{n} \\ z_{2} &=a_{21} y_{1}+a_{22} y_{2}+\cdots+a_{2 n} y_{n} \\ & \vdots \vdots \vdots \\ z_{n} &=a_{n 1} y_{1}+a_{n 2} y_{2}+\cdots+a_{n n} y_{n}, \end{aligned} $$ where the \(\left\\{a_{i j}\right\\}\) are constants. Show that \(\left\\{z_{1}, z_{2}, \ldots, z_{n}\right\\}\) is a fundamental set of solutions of \((\mathrm{A})\) if and only if the determinant $$ \left|\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{array}\right| $$ is nonzero.HINT: The determinant of a product of \(n \times n\) matrices equals the product of the deter-

Short answer

Question: Show that a set of solutions \(z_1, z_2, \ldots, z_n\) of a differential equation form a fundamental set of solutions if and only if the determinant of the matrix formed by the constants \(a_{ij}\), where \(z_i = \sum_{j=1}^{n} a_{ij}y_j\), is nonzero.

Step-by-step solution

Show that if the determinant is nonzero, then \(\{z_1, z_2, \ldots, z_n\}\) is a fundamental set of solutions

Assume that the determinant of the matrix formed by the constants \(a_{ij}\) is nonzero. We have \(n\) linear independent solutions \(y_1, y_2, \ldots, y_n\) for the given differential equation. Now, we write the solution \(z_i\) as a linear combination of the solution \(y_j\): $$z_i = \sum_{j=1}^{n} a_{ij} y_j$$ To find a fundamental set of solutions, we need to make sure the Wronskian of the set \(\{z_1, z_2, \ldots, z_n\}\) remains nonzero.

Compute the Wronskian of \(\{z_1, z_2, \ldots, z_n\}\)

Compute the Wronskian \(W(z_1, z_2, \ldots, z_n)\): $$ W(z_1, z_2, \ldots, z_n) = \left|\begin{matrix} z_1 & z_2 & \cdots & z_n \\ z_1' & z_2' & \cdots & z_n' \\ \vdots & \vdots & \ddots & \vdots \\ z_1^{(n-1)} & z_2^{(n-1)} & \cdots & z_n^{(n-1)} \end{matrix}\right| $$ Use the linear combination for \(z_i\) and replace it in the determinant expression: $$ W(z_1, z_2, \ldots, z_n) = \left|\begin{matrix} \sum_{j=1}^n a_{1j}y_j & \sum_{j=1}^n a_{2j}y_j & \cdots & \sum_{j=1}^n a_{nj}y_j \\ \sum_{j=1}^n a_{1j}y_j' & \sum_{j=1}^n a_{2j}y_j' & \cdots & \sum_{j=1}^n a_{nj}y_j' \\ \vdots & \vdots & \ddots & \vdots \\ \sum_{j=1}^n a_{1j}y_j^{(n-1)} & \sum_{j=1}^n a_{2j}y_j^{(n-1)} & \cdots & \sum_{j=1}^n a_{nj}y_j^{(n-1)} \end{matrix}\right| $$

Show that the determinant is nonzero

A property of the determinant is that the determinant of a product of matrices equals the product of the determinants of the individual matrices. So, we have: $$ W(z_1, z_2, \ldots, z_n) = \left|\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{matrix}\right| \cdot W(y_1, y_2, \ldots, y_n) $$ As \(y_1, y_2, \ldots, y_n\) form a fundamental set of solutions, their Wronskian \(W(y_1, y_2, \ldots, y_n)\) is nonzero. Since we assumed that the determinant of the matrix formed by the constants \(a_{ij}\) is nonzero, we get \(W(z_1, z_2, \ldots, z_n) \ne 0\). Thus, \(z_1, z_2, \ldots, z_n\) form a fundamental set of solutions.

Show that if \(\{z_1, z_2, \ldots, z_n\}\) is a fundamental set of solutions, the determinant is nonzero

Now assume that \(\{z_1, z_2, \ldots, z_n\}\) is a fundamental set of solutions, which means its Wronskian \(W(z_1, z_2, \ldots, z_n)\) is nonzero. Using the result from Step 3, we have: $$ W(z_1, z_2, \ldots, z_n) = \left|\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{matrix}\right| \cdot W(y_1, y_2, \ldots, y_n) $$ As \(W(z_1, z_2, \ldots, z_n) \ne 0\) and \(W(y_1, y_2, \ldots, y_n) \ne 0\), this implies that the determinant of the matrix formed by the constants \(a_{ij}\) must be nonzero. Thus, we have shown that the set \(\{z_1, z_2, \ldots, z_n\}\) is a fundamental set of solutions of the given differential equation if and only if the determinant of the matrix formed by the constants \(a_{ij}\) is nonzero.

Key concepts

Differential Equations

Differential equations are mathematical equations that describe how a particular quantity changes over time, often with respect to other variables. They are fundamental to modeling real-world systems in physics, engineering, biology, and economics, among other fields. A differential equation involves derivatives, which represent rates of change, and controls the behavior of a function that models a physical scenario.

For instance, if \( P_0(x), P_1(x), ..., P_n(x) \) are functions of \( x \) and \( y \) is an unknown function of \( x \) sought after, a general \( n \)th order linear homogeneous differential equation can be written as: \[ P_{0}(x) y^{(n)} + P_{1}(x) y^{(n-1)} + \cdots + P_{n}(x) y = 0. \] The solutions to these equations reveal the progression of the system described, and in the context of this exercise, a 'fundamental set of solutions' refers to a set of functions that solve the differential equation and possess a very important property, linear independence, which we will discuss next.

Wronskian

The Wronskian is a determinant used in the study of differential equations to determine whether a set of functions is linearly independent. The concept is named after the Polish mathematician Hoene Wronski. It has particular significance in the theory of ordinary differential equations, as it can provide insights about the existence and uniqueness of solutions based on initial conditions.

To calculate the Wronskian of a set of functions \( \{y_1, y_2, ..., y_n\} \) each of which is a solution to a differential equation, you would construct a matrix with each row being the derivatives of these functions up to the \( (n-1) \)th derivative, and then compute the determinant of this matrix. Formally, the Wronskian of \( y_1, y_2, \ldots, y_n \) is given by: \[ W(y_1, y_2, \ldots, y_n) = \left| \begin{matrix} y_1 & y_2 & \cdots & y_n \ y_1' & y_2' & \cdots & y_n' \ \vdots & \vdots & \ddots & \vdots \ y_1^{(n-1)} & y_2^{(n-1)} & \cdots & y_n^{(n-1)} \end{matrix} \right|. \] If the Wronskian is nonzero at a particular point, the functions are linearly independent at that point.

Linear Independence

Linear independence is a core concept in linear algebra that is relevant in many mathematical areas, including the theory of differential equations. A set of functions or vectors is said to be linearly independent if no one in the set can be written as a linear combination of the others. In simpler terms, in a linearly independent set, there is no redundancy; each function or vector adds unique information.

In the context of differential equations, the linear independence of solutions is crucial. Only when solutions are linearly independent can they span the solution space - meaning that they can be used to construct any possible solution to the differential equation. So, when dealing with the fundamental set of solutions, it's vital to confirm their linear independence; this is often accomplished through the use of the Wronskian.

Determinants

Determinants are mathematical objects that are used to perform various computations in linear algebra, such as finding the inverse of a matrix, as well as in calculus and differential equations for solving systems of linear equations. Determinants can be thought of as a scalar quantity that encapsulates certain properties of a square matrix. Importantly, the determinant can tell us if a matrix is invertible (non-singular) or not - a non-zero determinant implies an invertible matrix, whereas a zero determinant highlights a singular, or non-invertible, matrix.

The significance of the determinant in the context of our differential equations problem is profound. It allows verifying the linear independence of the solutions by connecting it to the Wronskian. If the determinant of the matrix formed by constants \( \{a_{ij}\} \)—which are known as the coefficients in the linear combinations of our solutions—is non-zero, then the set of solutions created by these combinations \( \{z_{1}, z_{2}, ..., z_{n}\} \) is guaranteed to also be linearly independent, constituting another fundamental set of solutions for the differential equation.