Question

Assume the characteristic equation of \(y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{1} y^{\prime}+a_{0} y=0\) has distinct roots \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\). It can be shown that the Vandermonde \(^{8}\) determinant has the value $$ \left|\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ \lambda_{1} & \lambda_{2} & & \lambda_{n} \\ \lambda_{1}^{2} & \lambda_{2}^{2} & & \lambda_{n}^{2} \\ \vdots & & & \vdots \\ \lambda_{1}^{n-1} & \lambda_{2}^{n-1} & \cdots & \lambda_{n}^{n-1} \end{array}\right|=\prod_{i, j=1 \atop i>i}^{n}\left(\lambda_{i}-\lambda_{j}\right) $$ Use this fact to show that \(\left\\{e^{\lambda_{1} t}, e^{\lambda_{2} t}, \ldots, e^{\lambda_{n} t}\right\\}\) is a fundamental set of solutions.

Short answer

Based on the given exercise, the Wronskian of the set of functions \(\{e^{\lambda_1 t}, e^{\lambda_2 t}, \dots, e^{\lambda_n t}\}\), using the Vandermonde determinant, is found to be $$W(\{e^{\lambda_1 t}, e^{\lambda_2 t}, \dots, e^{\lambda_n t}\}) = \left( \prod_{i, j=1 \atop i>j}^{n}\left(\lambda_i-\lambda_j\right) \right) \cdot \left( e^{\lambda_1 t}\cdot e^{\lambda_2 t} \cdot \cdots \cdot e^{\lambda_n t}\right)$$ which is nonzero due to distinct roots \(\lambda_i\) and non-zero exponential functions. Therefore, the set is a fundamental set of solutions.

Step-by-step solution

Write down the Wronskian of the given set of functions

The Wronskian of a set of functions is the determinant of a matrix formed by the functions and their derivatives. So, for the given set of functions \(\{e^{\lambda_1 t}, e^{\lambda_2 t}, \dots, e^{\lambda_n t}\}\), the Wronskian matrix is: $$ W(\{e^{\lambda_1 t}, e^{\lambda_2 t}, \dots, e^{\lambda_n t}\}) = \left| \begin{array}{cccc} e^{\lambda_1 t} & e^{\lambda_2 t} & \cdots & e^{\lambda_n t} \\ \lambda_1 e^{\lambda_1 t} & \lambda_2 e^{\lambda_2 t} & \cdots & \lambda_n e^{\lambda_n t} \\ \vdots & \vdots & & \vdots \\ \lambda_1^{n-1} e^{\lambda_1 t} & \lambda_2^{n-1} e^{\lambda_2 t} & \cdots & \lambda_n^{n-1} e^{\lambda_n t} \end{array} \right| $$

Factor out the exponentials

Since each row in the Wronskian matrix has a common exponential factor, we can factor it out: $$ W(\{e^{\lambda_1 t}, e^{\lambda_2 t}, \dots, e^{\lambda_n t}\}) = \left| \begin{array}{cccc} 1 & 1 & \cdots & 1 \\ \lambda_1 & \lambda_2 & \cdots & \lambda_n \\ \lambda_1^2 & \lambda_2^2 & \cdots & \lambda_n^2 \\ \vdots & \vdots & & \vdots \\ \lambda_1^{n-1} & \lambda_2^{n-1} & \cdots & \lambda_n^{n-1} \end{array} \right| \cdot \left( e^{\lambda_1 t}\cdot e^{\lambda_2 t} \cdot \cdots \cdot e^{\lambda_n t}\right) $$ This is now just the Vandermonde determinant multiplied by a product of exponential functions.

Evaluate the Vandermonde determinant

According to the given exercise, the Vandermonde determinant is: $$ \left|\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ \lambda_1 & \lambda_2 & \cdots & \lambda_n \\ \lambda_1^2 & \lambda_2^2 & \cdots & \lambda_n^2 \\ \vdots & \vdots & & \vdots \\ \lambda_1^{n-1} & \lambda_2^{n-1} & \cdots & \lambda_n^{n-1} \end{array}\right| = \prod_{i, j=1 \atop i>j}^{n}\left(\lambda_i-\lambda_j\right) $$

Evaluate the Wronskian and check for non-zero value

Substituting the Vandermonde determinant into the Wronskian equation, we get: $$ W(\{e^{\lambda_1 t}, e^{\lambda_2 t}, \dots, e^{\lambda_n t}\}) = \left( \prod_{i, j=1 \atop i>j}^{n}\left(\lambda_i-\lambda_j\right) \right) \cdot \left( e^{\lambda_1 t}\cdot e^{\lambda_2 t} \cdot \cdots \cdot e^{\lambda_n t}\right) $$ Since the roots \(\lambda_i\) are distinct, the products \(\left(\lambda_i-\lambda_j\right)\) are always nonzero. Moreover, exponential functions are never equal to zero, so their product is nonzero as well. Therefore, the Wronskian is nonzero: $$ W(\{e^{\lambda_1 t}, e^{\lambda_2 t}, \dots, e^{\lambda_n t}\}) \neq 0 $$

Conclusion

We've shown that the Wronskian of the given set of functions, \(\{e^{\lambda_1 t}, e^{\lambda_2 t}, \dots, e^{\lambda_n t} \}\), is nonzero. This proves that the set is a fundamental set of solutions.

Key concepts

Vandermonde Determinant

The Vandermonde determinant is an important concept in linear algebra, particularly useful in solving systems of equations and analyzing polynomial roots. This determinant is uniquely defined for a matrix where each row is in a geometric progression.

For instance, consider a matrix where the first row contains all ones, and each subsequent row contains powers of a sequence of distinct variables, \( \lambda_i \), with the power increasing by one for each row. The expression for the Vandermonde determinant of such a matrix is given by the product of the differences between each pair of the variables \( \lambda_i \). Specifically, in the case of distinct roots \( \lambda_1, \lambda_2, \ldots, \lambda_n \) for some characteristic equation, we have the following expression for the determinant:

\[\prod_{i, j=1 \atop i>j}^{n}(\lambda_i-\lambda_j)\]
What's fascinating about the Vandermonde determinant is that if the variables are all distinct, the determinant is non-zero. This is because the determinant is a product of differences, and with distinct variables, none of these differences will be zero. Therefore, the matrix is invertible, which has implications for the uniqueness and existence of solutions to linear systems.

Characteristic Equation

A characteristic equation is a fundamental tool in the study of differential equations and linear algebra. It is typically derived from a linear differential equation or from the determinant of a matrix subtracted by a scalar multiple of the identity matrix (when calculating eigenvalues). In the context of differential equations, the characteristic equation is obtained by translating the differential equation into an algebraic equation.

For a linear homogeneous differential equation, like \(y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{1}y'+a_{0} y=0\), the solutions are intimately connected with the roots of the corresponding characteristic equation. If the roots, denoted by \(\lambda_1, \lambda_2, \ldots, \lambda_n\), are distinct, they provide a basis for the general solution to the differential equation. The exponential functions \(e^{\lambda_i t}\) become candidate solutions whose linear combination can be tweaked to satisfy initial conditions or particular scenarios.

Fundamental Set of Solutions

A fundamental set of solutions refers to a set of functionally independent solutions to a differential equation from which any solution to the differential equation can be constructed through linear combinations. For a linear homogeneous differential equation with constant coefficients, such a set is created by taking exponential functions where the exponents are the distinct roots of the characteristic equation.

In the given exercise, the set \(\{e^{\lambda_1 t}, e^{\lambda_2 t}, \ldots, e^{\lambda_n t}\}\) suggests that each function \(e^{\lambda_i t}\) is associated with a distinct root \(\lambda_i\) of the characteristic equation. Thanks to the non-zero Wronskian determinant, these solutions are linearly independent and hence, form a fundamental set. This non-zero condition ensures that the solution set spans the solution space and provides the maximum number of independent solutions equal to the order of the differential equation—solidifying the set's 'fundamental' status.

The beauty of this set lies in its ability to express any specific solution to the original equation, making the study of differential equations manageable and organized, and providing a structured path to find general solutions.