Problem 12 - Advanced Engineering Mathematics

Question

In Problems 11-20, find the eigenvalues and eigenfunctions for the given boundary-value problem. $$ y^{\prime \prime}+\lambda y=0, y^{\prime}(0)=0, y(\pi / 4)=0 $$

Step-by-Step Solution

Answer

The eigenvalues are \( \lambda_n = \left(\frac{(2n+1)\pi}{8}\right)^2 \), and the eigenfunctions are \( y_n(x) = A \cos\left(\frac{(2n+1)\pi}{8}x\right) \).

Step 1: Set up the Characteristic Equation

For the differential equation \( y'' + \lambda y = 0 \), we assume a solution of the form \( y = e^{rx} \). Substituting into the differential equation gives \( r^2 + \lambda = 0 \). Therefore, the characteristic equation is \( r = \pm i\sqrt{\lambda} \). This implies the general solution is \( y(x) = A \cos(\sqrt{\lambda}x) + B \sin(\sqrt{\lambda}x) \).

Step 2: Apply Boundary Condition at x = 0

Apply the first boundary condition \( y'(0) = 0 \). Compute the derivative: \( y'(x) = -A \sqrt{\lambda} \sin(\sqrt{\lambda}x) + B \sqrt{\lambda} \cos(\sqrt{\lambda}x) \). At \( x=0 \), \( y'(0) = 0 = -A \sqrt{\lambda} \sin(0) + B \sqrt{\lambda} \cos(0) \), so \( B \sqrt{\lambda} = 0 \). If \( \lambda eq 0 \), then \( B = 0 \).

Step 3: Apply Boundary Condition at x = \( \pi/4 \)

Use the second boundary condition \( y(\pi/4) = 0 \). Substitute into the solution: \( y(\pi/4) = A \cos(\sqrt{\lambda} \pi/4) = 0 \). For \( A eq 0 \), this implies \( \cos(\sqrt{\lambda} \pi/4) = 0 \).

Step 4: Find Eigenvalues

From \( \cos(\sqrt{\lambda} \pi/4) = 0 \), the cosine function is zero at odd multiples of \( \pi/2 \). Set \(\sqrt{\lambda} \pi/4 = (2n + 1)\pi/2\) for integer \( n \). Solving for \( \lambda \): \( \sqrt{\lambda} = (2n + 1)\pi/8 \). Hence, \( \lambda_n = \left(\frac{(2n+1)\pi}{8}\right)^2 \).

Step 5: Determine Eigenfunctions

For each eigenvalue \( \lambda_n \), the eigenfunction satisfies \( y(x) = A\cos\left(\frac{(2n+1)\pi}{8}x\right) \). The constant \( A \) can be determined based on normalization or initial conditions if specified. Otherwise, in the context of finding eigenfunctions, it remains arbitrary.

Key Concepts

Boundary-Value Problems

Boundary-value problems are a critical concept in differential equations where you're asked to find solutions that meet specific criteria at the boundaries of the domain. They require you to satisfy equations at two or more points, known as boundary conditions. For instance, consider a vibrating string fixed at both ends; finding its mode shapes involves solving boundary-value problems.

In the provided exercise, the boundary-value problem is articulated as a differential equation with two boundary conditions:

• \(y'(0) = 0\): The derivative of the solution at \(x=0\) must equal zero.
• \(y(\pi/4) = 0\): The solution itself must zero at \(x=\pi/4\).

Understanding the nature of these problems helps in comprehending both the mathematical behavior across a region and the specific boundaries where conditions are applied.

Differential Equations

Differential equations represent relationships involving rates of change and are fundamental in modeling various physical phenomena. They typically express the relation between a function and its derivatives, encapsulating how the function changes over space or time.

The differential equation in our problem, \(y'' + \lambda y = 0\), is a classical second-order linear differential equation. The goal is to look for solutions that also satisfy accompanying boundary conditions.

• The general solution approach involves characteristic equations and assumes solutions in convenient forms like exponentials or trigonometric functions, exploiting known derivatives.

Grasping how these equations translate into practical, real-world models can significantly enhance the learning experience.

Characteristic Equation

The characteristic equation is central to solving many differential equations, particularly linear second-order ones. It arises when we assume a potential solution form, often using the exponential function due to its desirable derivative properties.

In our problem, you substitute \(y = e^{rx}\) into the differential equation to derive the characteristic equation \(r^2 + \lambda = 0\). This enables finding general solutions related to sine and cosine when the solution involves complex roots:

• \(r = \pm i\sqrt{\lambda}\), suggesting oscillatory solutions.

These insights provide structure to exploring how varied solutions satisfy different types of boundary conditions.

Boundary Conditions

Boundary conditions specify the behavior of solutions at the boundaries of the domain, dictating the permissible solutions to differential equations in boundary-value problems.

For our exercise, two boundary conditions were applied:

• \(y'(0) = 0\): This requires the function's slope to be zero at \(x=0\), indicating a horizontal tangent and influencing the coefficients in our general solution.
• \(y(\pi/4) = 0\): This ensures the function is zero at \(x=\pi/4\), pinpointing the permissible eigenvalues as it confines the form of the solution.

When solving, you apply these conditions to constrain and ultimately determine the constants and parameters in the general solutions, ensuring they are valid within the problem's framework.