Problem 18 - Advanced Engineering Mathematics

Question

In Problems, 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: \( \lambda = n^2\pi^2 - 1 \). Eigenfunctions: \( y(x) = C_1 \cos(n\pi x) \).

Step 1: Write the Differential Equation

Start by analyzing the given differential equation \( y'' + (\lambda + 1)y = 0 \). This is a second-order linear homogeneous differential equation, where \( \lambda \) is the eigenvalue parameter we need to find.

Step 2: Solve the Characteristic Equation

The characteristic equation for the differential equation \( y'' + (\lambda + 1)y = 0 \) is \( r^2 + (\lambda + 1) = 0 \), which simplifies to \( r^2 = -(\lambda + 1) \). Thus, the roots are \( r = \pm i \sqrt{\lambda + 1} \).

Step 3: Find the General Solution

The general solution for the differential equation with imaginary roots \( \pm i \beta \) is \( y(x) = C_1 \cos(\beta x) + C_2 \sin(\beta x) \), where \( \beta = \sqrt{\lambda + 1} \). Thus, \( y(x) = C_1 \cos(\sqrt{\lambda + 1} x) + C_2 \sin(\sqrt{\lambda + 1} x) \).

Step 4: Apply the First Boundary Condition

Apply the boundary condition \( y'(0) = 0 \). Differentiate the general solution: \( y'(x) = -C_1 \sqrt{\lambda + 1} \sin(\sqrt{\lambda + 1} x) + C_2 \sqrt{\lambda + 1} \cos(\sqrt{\lambda + 1} x) \). Set \( y'(0) = 0 \): \( C_2 \sqrt{\lambda + 1} = 0 \), thus \( C_2 = 0 \).

Step 5: Apply the Second Boundary Condition

With \( C_2 = 0 \), the solution simplifies to \( y(x) = C_1 \cos(\sqrt{\lambda + 1} x) \). Apply the second boundary condition \( y'(1) = 0 \), meaning \( y'(1) = -C_1 \sqrt{\lambda + 1} \sin(\sqrt{\lambda + 1}) = 0 \). This implies \( \sin(\sqrt{\lambda + 1}) = 0 \).

Step 6: Determine Eigenvalues

From \( \sin(\sqrt{\lambda + 1}) = 0 \), \( \sqrt{\lambda + 1} = n\pi \) for \( n = 0, 1, 2, \ldots \) Therefore, \( \lambda + 1 = n^2\pi^2 \) and \( \lambda = n^2\pi^2 - 1 \). These are the eigenvalues.

Step 7: Write Eigenfunctions

For \( \lambda = n^2\pi^2 - 1 \), the corresponding eigenfunctions are \( y(x) = C_1 \cos(n\pi x) \). These functions, for \( n = 0, 1, 2, \ldots \), form the eigenfunctions for the problem.

Key Concepts

Boundary-Value Problem

A boundary-value problem in mathematics involves finding a solution to a differential equation that satisfies certain specified values, known as boundary conditions, at the boundaries of the interval. In the case of our exercise, the problem presents a second-order linear differential equation with specific conditions at two points, usually at the beginning and end of an interval.
If you've ever solved a puzzle only to find a piece missing, you know it doesn't feel complete. Boundary-value problems ensure all the pieces fit together, giving us a complete solution by specifying how our solution should behave at the limits of our domain.

Second-Order Linear Differential Equations

Second-order linear differential equations are equations that involve the second derivative of a function, along with the function itself and its first derivative. They are essential because they appear frequently in engineering, physics, and other disciplines to model a wide range of phenomena.
Our exercise is about understanding these equations, where the equation given is homogeneous, meaning there's no additional term unlinked to the function itself. These equations can often be solved by finding their characteristic equation, which helps identify the behavior of the solutions.

Characteristic Equation

The characteristic equation is obtained from a linear differential equation and is vital for determining the nature of its solutions. By substituting possible exponential solutions into the differential equation, we derive a polynomial equation - the characteristic equation.
For our second-order differential equation, the characteristic equation was used to find the roots, which guided us to the general solution. Imaginary roots indicate that the solution will involve sine and cosine functions. The process of deriving the characteristic equation is a critical step in solving second-order linear differential equations and sets the stage for identifying specific solutions corresponding to given boundary conditions.

Boundary Conditions

Boundary conditions are the specified values that a solution to a differential equation must satisfy at the boundaries of the interval. These conditions are crucial because they help determine the specific solution from the general solution.
In the given problem, the boundary conditions were applied twice: at the beginning and the end of the interval. The conditions ensure that our solution is not just generally, but exactly tailored to fit our problem's requirements.
They provide the constraints that transform an infinite set of potential solutions into the precise solution that fits within the context of the problem, allowing us to identify the eigenvalues and eigenfunctions accurately.