Question

Show that if the Wronskian of the \(n\) -times continuously differentiable functions \(\left\\{y_{1}, y_{2}, \ldots, y_{n}\right\\}\) has no zeros in \((a, b)\), then the differential equation obtained by expanding the determinant $$ \left|\begin{array}{ccccc} y & y_{1} & y_{2} & \cdots & y_{n} \\ y^{\prime} & y_{1}^{\prime} & y_{2}^{\prime} & \cdots & y_{n}^{\prime} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y^{(n)} & y_{1}^{(n)} & y_{2}^{(n)} & \cdots & y_{n}^{(n)} \end{array}\right|=0 $$ in cofactors of its first column is normal and has \(\left\\{y_{1}, y_{2}, \ldots, y_{n}\right\\}\) as a fundamental set of solutions on \((a, b)\).

Short answer

Answer: From the given problem, we can conclude that the set of functions \(\left\{y_{1}, y_{2}, \ldots, y_{n}\right\}\) is linearly independent and forms a fundamental set of solutions to the differential equation obtained by expanding the determinant. We also conclude that the resulting differential equation is normal on the interval \((a, b)\).

Step-by-step solution

Calculate the determinant

First, we will calculate the determinant of the given matrix. The matrix is: $$ \left|\begin{array}{ccccc} y & y_{1} & y_{2} & \cdots & y_{n} \\\ y^{\prime} & y_{1}^{\prime} & y_{2}^{\prime} & \cdots & y_{n}^{\prime} \\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\ y^{(n)} & y_{1}^{(n)} & y_{2}^{(n)} & \cdots & y_{n}^{(n)} \end{array}\right| $$

Expand the determinant in cofactors of its first column

To expand the determinant in cofactors of its first column, we will consider the elements of the first column and the corresponding minors. The expression will be: $$ \left|\begin{array}{c|ccc} y & y_{1} & y_{2} & \cdots & y_{n} \\\ \hline y^{\prime} & y_{1}^{\prime} & y_{2}^{\prime} & \cdots & y_{n}^{\prime} \\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\ y^{(n)} & y_{1}^{(n)} & y_{2}^{(n)} & \cdots & y_{n}^{(n)} \end{array}\right| = yW\left(y_{1}, y_{2}, \ldots, y_{n}\right) - y'W\left(y_{1}', y_{2}', \ldots, y_{n}'\right) + \dotsb + (-1)^{n} y^{(n)}W\left(y_{1}^{(n-1)}, y_{2}^{(n-1)}, \ldots, y_{n}^{(n-1)}\right) = 0 $$ where \(W(y_1, y_2, \ldots, y_n)\) represents the Wronskian of the given functions.

Show that the resulting differential equation is normal

Since the Wronskian has no zeros in \((a, b)\), it implies that the first term in the determinant is nonzero. Therefore, the differential equation is of the form: $$ yW\left(y_{1}, y_{2}, \ldots, y_{n}\right) = \text{polynomial in } y, y', \ldots, y^{(n)} $$ This equation is normal because it is of order \(n\) and its leading coefficient (the coefficient of \(y^{(n)}\)) is nonzero.

Prove that \(\left\{y_{1}, y_{2}, \ldots, y_{n}\right\}\) form a fundamental set of solutions

Since the Wronskian of these functions has no zeros in \((a, b)\), it means that they are linearly independent. Moreover, the functions \(\left\{y_{1}, y_{2}, \ldots, y_{n}\right\}\) satisfy the given differential equation, as we can see by substituting them into the equation one by one. Therefore, they form a fundamental set of solutions to the given differential equation. In conclusion, we have shown that the differential equation obtained by expanding the determinant in cofactors of its first column is normal and has \(\left\{y_{1}, y_{2}, \ldots, y_{n}\right\}\) as a fundamental set of solutions on \((a, b)\).

Key concepts

Wronskian determinant

The concept of the Wronskian determinant is a crucial tool in the study of differential equations. It's named after the Polish mathematician Józef Hoene-Wroński. The Wronskian helps to determine whether a set of solutions is linearly independent, which is essential for confirming that solutions form a 'fundamental set' for a differential equation.

To calculate the Wronskian for a set of functions, you arrange the functions and their derivatives into a matrix, and the determinant of this matrix is the Wronskian. Mathematically, for functions \(y_1, y_2, \ldots, y_n\), their Wronskian, denoted as \(W(y_1, y_2, \ldots, y_n)\), is the determinant of a matrix containing these functions and their derivatives up to the \(n-1\)th order. The calculation of a Wronskian involves basic operations like addition, multiplication, and most notably, differentiation of the functions involved.

If the Wronskian is non-zero on an interval, it implies that the functions are linearly independent on that interval. This is a key aspect since linear independence of solutions ensures that they span the solution space of the differential equation, allowing for more complex solutions to be built from these basic 'building blocks'.

Fundamental set of solutions

A 'fundamental set of solutions' is a concept directly linked with the Wronskian determinant when dealing with linear differential equations. Essentially, it refers to a set of solutions to a differential equation that are linearly independent and span the entire solution space for that equation.

The term 'fundamental' comes from the fact that any solution to a linear homogeneous differential equation can be expressed as a linear combination of the solutions in this set. To be considered a fundamental set, the functions included must satisfy two conditions. First, they must be solutions to the differential equation in question. Second, their Wronskian must be non-zero in the interval of interest, ensuring their linear independence.

In practice, finding a fundamental set of solutions is like discovering the building blocks that can be combined to form any possible solution to the differential equation. It's a powerful concept because once you have this set, you have a complete understanding of the solution space and can tackle a wide range of related problems.

Normal differential equation

A 'normal differential equation' is a type of ordinary differential equation with certain structural features. Specifically, a normal differential equation is one in which the highest order derivative is present and multiplied by a non-zero function. This function, which multiplies the highest order derivative, is known as the leading coefficient.

For an \(n\)th order differential equation, the normality condition means that the coefficient of the \(y^{(n)}\) term (the \(n\)th derivative of \(y\)) is not equal to zero. Why does this matter? Well, it's a matter of ensuring that the equation is well-behaved and doesn’t have singularities or undefined points within the interval of interest.

In the context of the exercise, the fact that the Wronskian does not have zeros implies that the leading coefficient isn't zero, confirming the normalcy of the differential equation. This property is crucial because it means that there is a consistent relationship between the degree of the equation and the behavior of the solutions, allowing analysts to apply standard techniques to solve it.

Cofactor expansion

Cofactor expansion is a method used in linear algebra to calculate the determinant of a matrix. This technique is essential when working with the Wronskian determinant in differential equations. In the context of the exercise provided, the differential equation is formed by expanding the determinant of a matrix using cofactor expansion along its first column.

The cofactor expansion involves two main components for each entry of the matrix: the minor and the cofactor itself. The minor of an element is the determinant of a smaller matrix created by deleting the row and column of that element. The cofactor is then calculated by attaching a sign to the minor according to its position, which is given by \( (-1)^{i+j} \) where \( i \) and \( j \) are the row and column indices of the element. By summing the products of each element in a row or column with its corresponding cofactor, one obtains the determinant of the entire matrix.

For differential equations, utilizing cofactor expansion can lead to an expression where the function and its derivatives are tied to the Wronskian of other functions. This links the concepts of determinants in linear algebra to the analysis of differential equations, showcasing the interconnectedness of mathematical disciplines.