Problem 16 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

Eigenvalues: \( \lambda = \frac{n^2}{4} \), Eigenfunctions: \( y_n(x) = B \sin\left(\frac{n}{2} x\right) \).

Step 1: Understand the differential equation

The boundary-value problem given is a second-order linear differential equation: \[ y'' + \lambda y = 0, \] with boundary conditions \( y(-\pi) = 0 \) and \( y(\pi) = 0 \). Our goal is to find the eigenvalues \( \lambda \) and the corresponding eigenfunctions \( y(x) \).

Step 2: Solve the characteristic equation

Assume a trial solution of the form \( y(x) = e^{rx} \). Substituting into the differential equation gives the characteristic equation \[ r^2 + \lambda = 0, \] which implies \( r = \pm i\sqrt{\lambda} \).

Step 3: Formulate the general solution

Since the roots are complex, the general solution of the differential equation is: \[ y(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x), \] where \( A \) and \( B \) are constants.

Step 4: Apply boundary conditions

Apply the first boundary condition \( y(-\pi) = 0 \): \[ A \cos(-\pi\sqrt{\lambda}) + B \sin(-\pi\sqrt{\lambda}) = 0. \]Secondly, apply the condition \( y(\pi) = 0 \): \[ A \cos(\pi\sqrt{\lambda}) + B \sin(\pi\sqrt{\lambda}) = 0. \]

Step 5: Solve the system of equations

The two boundary conditions form a system that generally allows solutions only for certain values of \( \lambda \). By setting these equations in the standard matrix form, we observe that the determinant, \[ \begin{vmatrix} \cos(-\pi\sqrt{\lambda}) & \sin(-\pi\sqrt{\lambda}) \ \cos(\pi\sqrt{\lambda}) & \sin(\pi\sqrt{\lambda}) \end{vmatrix} = 0, \] which leads to the condition \[ \sin(2\pi\sqrt{\lambda})=0. \]

Step 6: Find eigenvalues

The condition \( \sin(2\pi\sqrt{\lambda})=0 \) implies \[ 2\pi\sqrt{\lambda} = n\pi, \] where \( n \) is an integer. Solving for \( \lambda \) gives \[ \lambda = \left( \frac{n}{2} \right)^2 = \frac{n^2}{4}. \]

Step 7: Determine eigenfunctions

For each eigenvalue \( \lambda = \frac{n^2}{4} \), the corresponding eigenfunction is \[ y_n(x) = A \cos\left(\frac{n}{2} x\right) + B \sin\left(\frac{n}{2} x\right). \] Using the boundary conditions, we find that \( A = 0 \), hence, the eigenfunctions are \[ y_n(x) = B \sin\left(\frac{n}{2} x\right). \]

Key Concepts

Boundary-Value Problems

Boundary-value problems involve differential equations where specific conditions, known as boundary conditions, are set at different points, often at the endpoints of an interval. In the given exercise, these boundary conditions are specified at \( x = -\pi \) and \( x = \pi \). These problems require solutions that satisfy both the differential equation and these boundary conditions.

Boundary-value problems can be contrasted with initial value problems, where the conditions are specified at a single point. In practical applications:

• Boundary-value problems often model physical phenomena like heat conduction, wave motion, or structural deflection, where the conditions are determined by the physical setup and geometry.
• Satisfying the boundaries is essential because the solution that meets these criteria usually represents a real and meaningful physical state.

These problems can have multiple solutions or none, and are critical in fields like engineering and physics.

Differential Equations

Differential equations describe how a certain quantity changes over time or space. They form the basis for modeling continuous processes and can appear in various forms, such as ordinary (ODEs) or partial (PDEs). In this exercise, the equation \( y'' + \lambda y = 0 \) is a second-order linear ordinary differential equation (ODE).

Key aspects of differential equations include:

• Order: The order of a differential equation is determined by the highest derivative. Here, it's second-order due to \( y'' \).
• Linearity: If a differential equation is linear, it implies solutions can typically be superimposed to form new solutions.
• Solution Methods: Solving these equations involves integrating the derivative terms to find a general solution, often using characteristic equations and trial solutions.

In the exercise, interpreting \( \lambda \) as a parameter leads us to find specific solutions (eigenfunctions) that meet the boundary conditions, making it crucial to solving the boundary-value problem.

Characteristic Equation

The characteristic equation is a fundamental tool for solving linear differential equations, particularly those with constant coefficients. It helps to identify the general form of solutions for such equations.

For a differential equation like \( y'' + \lambda y = 0 \), we assume solutions of the form \( y(x) = e^{rx} \). Substituting this into the differential equation yields the characteristic equation:

• \( r^2 + \lambda = 0 \)

By solving for \( r \), we find \( r = \pm i\sqrt{\lambda} \), indicating that the solutions are oscillatory, involving sine and cosine functions:

• The general solution is \( y(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) \).

The characteristic equation thus provides a pathway to deduce the nature of the solutions—whether they are exponential, oscillating, or a combination. It also draws a direct connection between the differential equation and its possible solutions, especially useful for systems exhibiting wave-like behaviors, as in this problem with sine functions as solutions.