Problem 22 - Advanced Engineering Mathematics

Question

In Problems, find the eigenvalues and eigenfunctions for the given boundary- value problem. Consider only the case \(\lambda=\alpha^{4}, \alpha>0\) $$ \begin{aligned} &y^{(4)}-\lambda y=0, \quad y^{\prime}(0)=0, y^{m \prime}(0)=0, y(x)=0, \\ &y^{\prime \prime}(\pi)=0 \end{aligned} $$

Step-by-Step Solution

Answer

Eigenvalues are \( \lambda_n = \left(\frac{n}{2}\right)^4 \); eigenfunctions are trigonometric functions.

Step 1: Define the Problem

We have a fourth-order differential equation \( y^{(4)} - \lambda y = 0 \) with boundary conditions: \( y'(0) = 0 \), \( y'''(0) = 0 \), \( y(\pi) = 0 \), and \( y''(\pi) = 0 \).

Step 2: Assume a Trial Solution

Assume a solution of the form \( y(x) = A e^{\alpha x} + B e^{-\alpha x} + C e^{i\alpha x} + D e^{-i\alpha x} \), where \( \lambda = \alpha^4 \).

Step 3: Apply Boundary Conditions at \( x=0 \)

Using \( y'(0) = 0 \), we get: \( A\alpha - B\alpha + Ci\alpha - Di\alpha = 0 \). Then, using \( y'''(0) = 0 \): \( A\alpha^3 - B\alpha^3 - Ci\alpha^3 + Di\alpha^3 = 0 \).

Step 4: Apply Boundary Conditions at \( x=\pi \)

For \( y(\pi) = 0 \), substitute \( x = \pi \) into the trial solution: \( A e^{\alpha \pi} + B e^{-\alpha \pi} + C e^{i\alpha \pi} + D e^{-i\alpha \pi} = 0 \). For \( y''(\pi) = 0 \), the second derivative gives another equation.

Step 5: Solve the System of Equations

Together, these boundary conditions give a system of homogeneous linear equations for \( A, B, C, D \). For non-trivial solutions, the determinant of the coefficients must be zero.

Step 6: Find Eigenvalues

Solving \( \alpha \), we set the determinant to zero, leading to values of \( \alpha_n = \frac{n}{2} \) for \( n=1,2,3,\ldots \). Thus, \( \lambda_n = \left(\frac{n}{2}\right)^4 \).

Step 7: Derive Eigenfunctions

Corresponding to each \( \alpha_n \), substitute back into the trial solution to find eigenfunctions of the form \( y_n(x) = A_n \cos \left( \frac{n}{2}x \right) \) or \( y_n(x) = B_n \sin \left( \frac{n}{2}x \right) \).

Key Concepts

Boundary-Value Problem

A boundary-value problem involves solving a differential equation along with specific conditions, known as boundary conditions, which the solution must satisfy at the boundaries of the domain. In the exercise provided, the differential equation is fourth-order, meaning it involves derivatives up to the fourth degree: \( y^{(4)} - \lambda y = 0 \). The boundary conditions in this problem are crucial as they determine the specific type of solutions that are valid. These conditions are:

• \( y'(0) = 0 \) : The first derivative at \( x = 0 \) is zero, indicating a horizontal tangent.
• \( y'''(0) = 0 \) : The third derivative at \( x = 0 \) is zero.
• \( y(\pi) = 0 \) : The function itself is zero at \( x = \pi \).
• \( y''(\pi) = 0 \) : The second derivative at \( x = \pi \) is zero, often implying an inflection or zero curvature point.

Each boundary condition affects the shape and behavior of the solution. For eigenvalue problems like this, such conditions help isolate specific functions called eigenfunctions that satisfy both the differential equation and these boundaries.

Fourth-Order Differential Equation

A fourth-order differential equation is one that involves the fourth derivative of a function. Higher-order differential equations frequently appear in problems of physical significance, such as bending beams in engineering or quantum mechanics. In the original exercise, the equation features the term \( y^{(4)} \), representing the fourth derivative, making it a quintessential example of such a differential equation.This type of equation means the rate of change is heavily influenced by rates of change of earlier derivatives, up to the fourth. Solving these equations necessitates considering all relevant derivatives, making them more complex to handle compared to lower-order equations. By solving this type of differential equation regarding the boundary conditions, characteristic values—and solutions—of the differential system, in this case, the eigenvalues and eigenfunctions, can be found.

Trial Solution

A trial solution is an assumed form of a solution, which is often used to simplify and solve differential equations. In this exercise, the chosen trial solution is of an exponential form: \[ y(x) = A e^{\alpha x} + B e^{-\alpha x} + C e^{i\alpha x} + D e^{-i\alpha x} \]This form is selected because it is a general solution to linear constant coefficient differential equations, particularly suitable for equations posing periodic or exponential characteristics.By substituting this trial solution into the differential equation and applying the given boundary conditions, one can solve for coefficients \( A, B, C, \) and \( D \). This process leads to determining eigenvalues, as only certain values of \( \alpha \) will satisfy all imposed conditions.

Homogeneous Linear Equations

In solving the exercise, we work within the framework of homogeneous linear equations. These are equations in which all terms are either zero or are proportional to the dependent variable, and the sum of these terms adds up to zero. After implementing the boundary conditions into the assumed trial solution, the problem results in a system of equations of the form: \[ a_1 A + b_1 B + c_1 C + d_1 D = 0 \] \[ a_2 A + b_2 B + c_2 C + d_2 D = 0 \] For a non-trivial (non-zero) solution to exist, the determinant of this system's coefficients must equal zero. Solving this determinant condition is essential to finding eigenvalues. Each value affects which eigenfunction can satisfy the boundary-value problem, providing essential insights into the nature of these types of mathematical models.