Question

Linear Independence For definiteness, consider Theorem \(3.8\) in the case \(n=3 .\) Let \(\left\\{y_{1}, y_{2}, y_{3}\right\\}\) be a fundamental set of solutions for \(y^{\prime \prime \prime}+p_{2}(t) y^{\prime \prime}+p_{1}(t) y^{\prime}+p_{0}(t) y=0\), where \(p_{0}, p_{1}, p_{2}\) are continuous on \(a

Short answer

Question: Prove that the given set of three fundamental solutions \(\{y_1, y_2, y_3\}\) to a third-order homogeneous linear differential equation of the form \(y^{\prime \prime \prime} + p_2(t) y^{\prime \prime} + p_1(t)y^\prime + p_0(t)y = 0\) is linearly independent. Answer: To prove that the set of solutions \(\{y_1, y_2, y_3\}\) is linearly independent, we showed that the only scalars \(k_1, k_2, k_3\) that can satisfy the linear combination \(k_1 y_1(t) + k_2 y_2(t) + k_3 y_3(t) = 0\) are all equal to zero. We evaluated the linear combination and its derivatives at an arbitrary point \(t_0\), plugged them into the differential equation, and formed a system of linear equations with the unknown coefficients. The determinant of the coefficient matrix, or Wronskian, indicated that the given set of solutions was a fundamental set, which implies linear independence. Therefore, we proved that the set \(\{y_1, y_2, y_3\}\) is linearly independent.

Step-by-step solution

Write the linear combination and its derivatives

Consider the linear combination of the three solutions: \(k_1 y_1(t) + k_2 y_2(t) + k_3 y_3(t)\). We will also consider its first and second derivatives with respect to \(t\): \((k_1 y_1'(t) + k_2 y_2'(t) + k_3 y_3'(t))\) and \((k_1 y_1''(t) + k_2 y_2''(t) + k_3 y_3''(t))\), where \(k_1, k_2, k_3 \in \mathbb{R}\).

Evaluate the expressions at \(t_0\)

Evaluate the expressions for the linear combination and its first and second derivatives at an arbitrary point \(t_0\) in the interval \((a, b)\): 1. \(k_1 y_1(t_0) + k_2 y_2(t_0) + k_3 y_3(t_0) = 0\) 2. \(k_1 y_1'(t_0) + k_2 y_2'(t_0) + k_3 y_3'(t_0) = 0\) 3. \(k_1 y_1''(t_0) + k_2 y_2''(t_0) + k_3 y_3''(t_0) = 0\)

Use the differential equation on the linear combination

Since \(y_1, y_2, y_3\) are solutions to the given third-order homogeneous linear differential equation, we know that for each \(i \in \{1, 2, 3\}\): \(y_i^{\prime \prime \prime}(t) + p_2(t)y_i^{\prime\prime}(t) + p_1(t)y_i^\prime(t) +p_0(t)y_i(t) = 0\) By plugging in the expressions for the linear combination of the solutions and their derivatives, we get: \(k_1(y_1^{\prime\prime\prime}(t_0) + p_2(t_0)y_1^{\prime\prime}(t_0) + p_1(t_0)y_1^\prime(t_0) + p_0(t_0)y_1(t_0)) + k_2(y_2^{\prime\prime\prime}(t_0) + p_2(t_0)y_2^{\prime\prime}(t_0) + p_1(t_0)y_2^\prime(t_0) + p_0(t_0)y_2(t_0)) + k_3(y_3^{\prime\prime\prime}(t_0) + p_2(t_0)y_3^{\prime\prime}(t_0) + p_1(t_0)y_3^\prime(t_0) + p_0(t_0)y_3(t_0)) = 0\)

Rewrite the equation and combine terms

Using the equations we obtained in Step 2, rewrite the equation from Step 3: \(k_1(-p_2(t_0)y_1^{\prime\prime}(t_0) - p_1(t_0)y_1^\prime(t_0) - p_0(t_0)y_1(t_0)) + k_2(-p_2(t_0)y_2^{\prime\prime}(t_0) - p_1(t_0)y_2^\prime(t_0) - p_0(t_0)y_2(t_0)) + k_3(-p_2(t_0)y_3^{\prime\prime}(t_0) - p_1(t_0)y_3^\prime(t_0) - p_0(t_0)y_3(t_0)) = 0\) Now, we can combine terms: \(k_1 y_1^{\prime\prime\prime}(t_0) + k_2 y_2^{\prime\prime\prime}(t_0) + k_3 y_3^{\prime\prime\prime}(t_0) = 0\)

Prove that the coefficients are zero

Notice that now we have a system of linear equations with the unknowns \(k_1\), \(k_2\), \(k_3\) and the matrices formed as follows: \(\begin{pmatrix} y_1(t_0) & y_2(t_0) & y_3(t_0) \\ y_1'(t_0) & y_2'(t_0) & y_3'(t_0) \\ y_1''(t_0) & y_2''(t_0) & y_3''(t_0) \end{pmatrix} \begin{pmatrix} k_1 \\ k_2 \\ k_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\) The determinant of the coefficient matrix, known as the Wronskian, is: \(W(y_1, y_2, y_3)(t_0) = \det \begin{pmatrix} y_1(t_0) & y_2(t_0) & y_3(t_0) \\ y_1'(t_0) & y_2'(t_0) & y_3'(t_0) \\ y_1''(t_0) & y_2''(t_0) & y_3''(t_0) \end{pmatrix}\) Since the given set of solutions is a fundamental set of solutions, their Wronskian is nonzero for every point in the considered range, including \(t_0\). Therefore, the only solution to the above linear system of equations is the trivial solution, i.e. \(k_1 = k_2 = k_3 = 0\). This means that the set \(\{y_1, y_2, y_3\}\) is linearly independent.

Key concepts

Wronskian

The Wronskian is a determinant used in the study of differential equations to determine if a set of solutions is linearly independent. It's named after the Polish mathematician Józef Maria Hoene-Wroński. For a set of functions, say \{y_1, y_2, y_3\}, the Wronskian is calculated by forming a matrix where each row consists of the function and its derivatives, all evaluated at a particular point.

• The first row contains the functions: \(y_1(t), y_2(t), y_3(t)\).
• The second row includes their first derivatives: \(y_1'(t), y_2'(t), y_3'(t)\).
• The third row has their second derivatives: \(y_1''(t), y_2''(t), y_3''(t)\).

Then, the determinant of this matrix, called the Wronskian, is denoted by:\[W(y_1, y_2, y_3)(t) = \det \begin{pmatrix} y_1(t) & y_2(t) & y_3(t) \ y_1'(t) & y_2'(t) & y_3'(t) \ y_1''(t) & y_2''(t) & y_3''(t) \end{pmatrix}\]If the Wronskian is non-zero at some point in the interval of interest, the functions are linearly independent. This makes the Wronskian a powerful tool for analyzing the solutions to differential equations, particularly in proving linear independence among solutions.

Fundamental Set of Solutions

A fundamental set of solutions is a collection of solutions to a linear homogeneous differential equation that can be used to express any solution of the differential equation through linear combinations. For a third-order differential equation, you’ll typically need three linearly independent solutions.
Here's why a fundamental set of solutions is essential:

• Completeness: Any solution to the differential equation can be written as a combination of the fundamental set.
• Linearly Independent: The solutions do not overlap; this means no solution in the set can be written as a combination of the others.
• Utility: Knowing a fundamental set simplifies solving the differential equation, especially for varying initial conditions.

In theory, if you have a fundamental set of solutions for a differential equation of order \(n\), you can find a solution for any particular initial condition in the form:\[ k_1 y_1(t) + k_2 y_2(t) + ... + k_n y_n(t) = y(t)\]where \(k_1, k_2, ..., k_n\) are constants. Thus, finding a fundamental set is a central task in solving differential equations.

Homogeneous Differential Equations

Homogeneous differential equations are a category of differential equations where every term is a multiple of the function or its derivatives. These do not have a standalone term that depends solely on the independent variable, such as a constant or function of time.
The general form for a linear homogeneous differential equation of order \(n\) is:\[ a_n(t)y^{(n)} + a_{n-1}(t)y^{(n-1)} + ... + a_1(t)y' + a_0(t)y = 0\]where all terms are proportional to either \(y\) or its derivatives.
Key characteristics of homogeneous differential equations include:

• They always pass through the origin, as setting \(y = 0\) leads to the simplest solution, \(y(t) = 0\).
• They allow for the principle of superposition, meaning the sum of solutions is also a solution.
• Solving such equations typically involves finding one or more linearly independent solutions.

Understanding these equations is crucial as they form the backbone of many theoretical and applied problems in physics and engineering. They showcase how the governing processes stay consistent when scaled or combined, which is vital for understanding complex natural phenomena.