Problem 18 - Advanced Engineering Mathematics

Question

In Problems 11-20, find the eigenvalues and eigenfunctions for the given boundary-value problem. $$ y^{\prime \prime}+(\lambda+1) y=0, y^{\prime}(0)=0, y^{\prime}(1)=0 $$

Step-by-Step Solution

Answer

Eigenvalues are \( n^2\pi^2 - 1 \); eigenfunctions are \( \cos(n\pi x) \).

Step 1: Understand the Problem

We are given a differential equation \( y'' + (\lambda + 1) y = 0 \) with boundary conditions \( y'(0) = 0 \) and \( y'(1) = 0 \). Our task is to find eigenvalues \( \lambda \) and eigenfunctions \( y(x) \) for this problem.

Step 2: Assume a Solution Form

Assume a solution of the form \( y(x) = A \cos(kx) + B \sin(kx) \), where \( k^2 = \lambda + 1 \). This form is chosen because it solves \( y'' + k^2y = 0 \).

Step 3: Apply the Boundary Condition at x = 0

The condition \( y'(0) = 0 \) implies \( y'(x) = -Ak \sin(kx) + Bk \cos(kx) \). At \( x = 0 \), we get \( 0 = Bk \). Therefore, \( B = 0 \).

Step 4: Simplify the Solution

With \( B = 0 \), the solution simplifies to \( y(x) = A \cos(kx) \).

Step 5: Apply the Boundary Condition at x = 1

The condition \( y'(1) = 0 \) requires \( y'(1) = -Ak \sin(k) = 0 \). Since \( A eq 0 \) for a non-trivial solution, it follows that \( \sin(k) = 0 \).

Step 6: Find the Values of k

The solutions to \( \sin(k) = 0 \) are \( k = n\pi \) where \( n \) is an integer.

Step 7: Determine Eigenvalues

Given \( k^2 = \lambda + 1 \), we find \( \lambda = n^2\pi^2 - 1 \) as the eigenvalues.

Step 8: Find Corresponding Eigenfunctions

The eigenfunctions corresponding to \( \lambda_n = n^2\pi^2 - 1 \) are \( y_n(x) = A\cos(n\pi x) \). Since any constant multiple is an eigenfunction, we often choose \( A = 1 \) for simplicity.

Key Concepts

Boundary-Value Problem

A boundary-value problem is a type of differential equation coupled with a set of additional constraints, or "boundary conditions." These conditions specify the behavior of the solution at the boundaries of the domain. In this case, we are given a second-order differential equation:

• The differential equation: \( y'' + (\lambda + 1) y = 0 \)
• Boundary conditions imposed at \( x = 0 \) and \( x = 1 \): \( y'(0) = 0 \) and \( y'(1) = 0 \)

Boundary conditions like these are crucial as they determine the allowable solutions to the equation. Unlike initial value problems where the solution is sought from a given starting point, boundary-value problems identify solutions that fit behavior constraints at two or more points.
This approach is often seen in physical situations like quantum mechanics and structural engineering, where physical constraints need adherence at the borders of a studied domain.

Differential Equations

Differential equations involve derivatives, which represent how a quantity changes. Here, we work with a second-order differential equation, which involves the second derivative of a function. The problem we are dealing with is expressed as:

• Equation: \( y'' + (\lambda + 1) y = 0 \)

The goal is to determine functions that satisfy this equation, which describe phenomena with respect to an independent variable, often time or space.
The solutions depend on the conditions applied, and may take various forms: polynomial, exponential, or as here, trigonometric, depending on the nature of the equation. These mathematical expressions make sense of physical dynamics, like oscillations or growth rates. Our equation suggests oscillatory solutions, which lead us to consider trigonometric functions for solutions.

Trigonometric Functions

Trigonometric functions, including sine and cosine, naturally emerge when solving differential equations related to periodic or oscillatory behavior. In this exercise, we assume a trigonometric form for the solution:

• Assumed form: \( y(x) = A \cos(kx) + B \sin(kx) \)

Such a form is suitable due to the resemblance of our equation to the simple harmonic motion equation, \( y'' + k^2y = 0 \), associated with sinusoidal properties of physical systems like springs and pendulums.
By applying the boundary conditions, which are derivatives at specific points, the trigonometric components simplify our work:

• At \( x=0 \), the sine term is eliminated since it must cancel out for \( y'(0) = 0 \).
• At \( x=1 \), ensuring \( y'(1) = 0 \) leads us to require \( \sin(k) = 0 \), dictating values of \( k \).

Eigenvalue Problems

An eigenvalue problem is a special type of boundary-value problem that appears in various fields including physics and engineering. It involves determining specific values (eigenvalues) and corresponding functions (eigenfunctions) that solve a differential equation under given conditions. For the stated equation:

• Eigenvalues \( \lambda \) are parameters for which the differential equation admits non-trivial solutions.
• Eigenfunctions \( y(x) \) are the corresponding solutions to the equation with these eigenvalues.

The solutions need to satisfy both the differential equation and the boundary constraints, leading to discrete eigenvalues. Here, after applying boundary conditions and solving \( \sin(k) = 0 \), we find:

• Eigenvalues: \( \lambda = n^2\pi^2 - 1 \), where \( n \) is an integer.
• Eigenfunctions: \( y_n(x) = A\cos(n\pi x) \), with typically \( A = 1 \) chosen for simplicity.

Eigenvalues and eigenfunctions play a crucial role in vibrational analysis, stability assessment, and other areas where identifying natural frequencies and modes is essential.