Problem 18 - Advanced Engineering Mathematics

Question

The given vectors are solutions of a system \(\mathbf{X}^{\prime}=\mathbf{A X}\). Determine whether the vectors form a fundamental set on the interval \((-\infty, \infty)\). $$ \mathbf{X}_{1}=\left(\begin{array}{r} 1 \\ -1 \end{array}\right) e^{t}, \mathbf{X}_{2}=\left(\begin{array}{l} 2 \\ 6 \end{array}\right) e^{t}+\left(\begin{array}{r} 8 \\ -8 \end{array}\right) t e^{t} $$

Step-by-Step Solution

Answer

Yes, the vectors form a fundamental set on \((- fty, \infty)\).

Step 1: Define the Fundamental Set Condition

For two vectors to form a fundamental set of solutions, they must be linearly independent on the interval \((-fty, fty)\) and solve the differential system \(\mathbf{X}'=\mathbf{A} \mathbf{X}\). Linear independence can be determined using the Wronskian.

Step 2: Write Vectors in Component Form

Rewrite the given solutions in component form.\[ \mathbf{X}_{1} = \begin{pmatrix} 1 e^{t} \ -1 e^{t} \end{pmatrix} \]\[ \mathbf{X}_{2} = \begin{pmatrix} 2 e^{t} + 8t e^{t} \ 6 e^{t} - 8t e^{t} \end{pmatrix} \]

Step 3: Find the Wronskian

Calculate the Wronskian, \(W(t)\), using the determinant:\[ W(t) = \begin{vmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \end{vmatrix} = \begin{vmatrix} 1 & 2 + 8t \ -1 & 6 - 8t \end{vmatrix} \]Find the determinant:\[ W(t) = (1)(6 - 8t) - (-1)(2 + 8t) \]Simplify:\[ W(t) = 6 - 8t + 2 + 8t = 8 \]

Step 4: Analyze the Wronskian

The Wronskian \(W(t) = 8\) is non-zero (constant) for all \(t \in (-\infty, \infty)\). This implies that the two solutions \(\mathbf{X}_{1}\) and \(\mathbf{X}_{2}\) are linearly independent.

Step 5: Conclude the Fundamental Set Verification

Since the Wronskian is non-zero for all \(t\), \(\mathbf{X}_{1}\) and \(\mathbf{X}_{2}\) form a fundamental set of solutions for the differential system \(\mathbf{X}' = \mathbf{A} \mathbf{X}\) on the interval \((-fty, \infty)\).

Key Concepts

Linear Independence

Linear independence is a crucial concept when dealing with solutions to differential equations. It helps in determining whether a set of functions, like vectors in our case, provides distinct, meaningful solutions over the given interval. When vectors are linearly independent, it means none of the vectors can be expressed as a combination of the others in the set. This is vital because linear combinations would mean that one vector doesn't add additional information, making it redundant.

In mathematical terms, a set of vectors \( \{\mathbf{X}_{1}, \mathbf{X}_{2}, ..., \mathbf{X}_{n}\} \) is said to be linearly independent if the only solution to the equation:

• \( c_{1}\mathbf{X}_{1} + c_{2}\mathbf{X}_{2} + ... + c_{n}\mathbf{X}_{n} = 0 \)

is \( c_{1} = c_{2} = ... = c_{n} = 0 \).

This principle ensures each vector contributes unique information to the solution of the system, something that is verified using another mathematical tool called the Wronskian.

Wronskian

The Wronskian is a determinant used in the context of differential equations to test whether a set of functions are linearly independent over an interval. It provides a clear mathematical test — if the Wronskian is non-zero at any point in the interval, the functions are independent.

For vectors \( \mathbf{X}_{1} \) and \( \mathbf{X}_{2} \), the Wronskian \( W(t) \) is calculated using the determinant of a matrix formed by the component functions:

• \( W(t) = \begin{vmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \end{vmatrix} \)

This determinant should not be zero over the range under consideration.

In the given exercise, the computation gave us \( W(t) = 8 \), a constant that is not zero. Thus, \( \mathbf{X}_{1} \) and \( \mathbf{X}_{2} \) are independent over \((-\infty, \infty)\), fulfilling part of the requirement for being a fundamental set.

Fundamental Set of Solutions

A fundamental set of solutions of a differential equation is a collection of solutions that span the solution space for the system. For a system involving vectors, like \( \mathbf{X}_{1} \) and \( \mathbf{X}_{2} \), forming a fundamental set means any solution to the system can be expressed as a combination of these vectors.

The criteria for establishing a fundamental set includes:

• The solutions must solve the differential equation, \( \mathbf{X}' = \mathbf{A} \mathbf{X} \).
• The vectors must be linearly independent (as confirmed by a non-zero Wronskian).

For this exercise, both conditions are satisfied since the vectors are independent and solve the system across the entire interval \((-\infty, \infty)\).

This ensures that \( \mathbf{X}_{1} \) and \( \mathbf{X}_{2} \) indeed provide a complete and versatile solution set, giving students confidence in applying these concepts in solving differential equations.