Problem 21 - Advanced Engineering Mathematics

Question

In Problems 21 and 22, find the eigenvalues and eigenfunctions for the given boundary-value problem. Consider only the case \(\lambda=\alpha^{4}, \alpha>0\) $$ \begin{aligned} &y^{(4)}-\lambda y=0, \quad y(0)=0, \quad y^{\prime \prime}(0)=0, \quad y(1)=0 \\\ &y^{\prime \prime}(1)=0 \end{aligned} $$

Step-by-Step Solution

Answer

Eigenvalues are determined by boundary conditions leading to permissible \( \alpha \). Eigenfunctions correspond to these \( \alpha \) values.

Step 1: Write Down the Differential Equation

The given differential equation is \( y^{(4)} - \lambda y = 0 \). We are considering the case where \( \lambda = \alpha^4 \) with \( \alpha > 0 \). The boundary conditions are \( y(0) = 0 \), \( y''(0) = 0 \), \( y(1) = 0 \), and \( y''(1) = 0 \).

Step 2: Assume a Solution Form

Assume a solution of the form \( y(x) = e^{rx} \), since the characteristic equation for a fourth order linear differential equation with constant coefficients is \( r^4 - \lambda = 0 \).

Step 3: Solve the Characteristic Equation

Substituting the assumed form into the differential equation gives \( r^4 = \lambda = \alpha^4 \). Thus, the roots of the characteristic equation are \( r = \pm \alpha \) and \( r = \pm i \alpha \).

Step 4: Write the General Solution

The general solution based on the characteristic roots is \( y(x) = C_1 e^{\alpha x} + C_2 e^{-\alpha x} + C_3 e^{i\alpha x} + C_4 e^{-i\alpha x} \).

Step 5: Apply the Boundary Condition \( y(0) = 0 \)

Substituting \( x = 0 \) into the general solution gives \( C_1 + C_2 + C_3 + C_4 = 0 \).

Step 6: Apply the Boundary Condition \( y''(0) = 0 \)

Calculate \( y''(x) = C_1 \alpha^2 e^{\alpha x} + C_2 \alpha^2 e^{-\alpha x} - C_3 \alpha^2 e^{i\alpha x} - C_4 \alpha^2 e^{-i\alpha x} \). Substitute \( x = 0 \) to get \( C_1 + C_2 - C_3 - C_4 = 0 \).

Step 7: Apply the Boundary Condition \( y(1) = 0 \)

Substituting \( x = 1 \) into the general solution gives \( C_1 e^{\alpha} + C_2 e^{-\alpha} + C_3 e^{i\alpha} + C_4 e^{-i\alpha} = 0 \).

Step 8: Apply the Boundary Condition \( y''(1) = 0 \)

Substituting \( x = 1 \) into the expression for \( y''(x) \) and setting it to zero gives \( C_1 \alpha^2 e^{\alpha} + C_2 \alpha^2 e^{-\alpha} - C_3 \alpha^2 e^{i\alpha} - C_4 \alpha^2 e^{-i\alpha} = 0 \).

Step 9: Solve the System of Equations

Solve the system of equations based on the boundary conditions to find relations between the constants \( C_1, C_2, C_3, \) and \( C_4 \). This eventually gives conditions on \( \alpha \) that lead to non-trivial solutions.

Step 10: Find Eigenvalues and Eigenfunctions

The eigenvalues are obtained when the determinant of the system is zero, which will indicate the permissible \( \alpha \) values. Eigenfunctions correspond to these \( \alpha \) values and are constructed from the general solution satisfying the boundary conditions.

Key Concepts

Boundary-Value Problem

A boundary-value problem is a differential equation paired with a set of additional constraints known as boundary conditions. These conditions are specified at the boundaries of the interval in which the solution is defined. In simpler terms, you are solving a differential equation subject to certain prescribed values at the endpoints of the given interval.
For the given problem, the differential equation is a fourth-order equation with boundary conditions specified at points 0 and 1:

• At point 0: \( y(0) = 0 \) and \( y''(0) = 0 \)
• At point 1: \( y(1) = 0 \) and \( y''(1) = 0 \)

These conditions ensure the solution is unique and tailored to fit specific physical or geometric properties of a system.

Fourth Order Differential Equation

In mathematics and physics, higher-order differential equations are equations that involve derivatives of an order higher than the first. A fourth-order differential equation includes the fourth derivative of the function. Solving these equations can be intricate, as they often involve more complex methods than first-order or second-order equations.
For the problem at hand, the given differential equation is \( y^{(4)} - \lambda y = 0 \). Because it's fourth-order, this equation has four solutions (or roots), which we discover using techniques like the characteristic equation method. This equation arises in many areas of science and engineering, such as in beam theory where it describes the deflection of a beam under a load.

Characteristic Equation

The characteristic equation is a crucial tool for solving linear differential equations with constant coefficients. It is derived by assuming a solution of a certain form (usually exponential) and substituting it into the differential equation. This converts the differential equation into an algebraic equation.
For a fourth-order equation such as \( y^{(4)} - \lambda y = 0 \), we assume a solution of the form \( y(x) = e^{rx} \). Substituting this form transforms our differential equation into \( r^4 = \lambda \), known as the characteristic equation. Solving this allows us to find the roots \( r \), which give insights into the behavior of the solution. In this problem, the roots are \( \pm \alpha \) and \( \pm i\alpha \), reflecting oscillatory and exponential growth components of the potential solutions.

Homogeneous Boundary Conditions

Homogeneous boundary conditions simplify many boundary-value problems by setting the conditions equal to zero. This can mean the function and certain derivatives vanish at specified points, creating a straightforward approach to determine the form of the eigenfunctions.
In the given problem, all specified conditions are homogeneous:

• \( y(0) = 0 \)
• \( y''(0) = 0 \)
• \( y(1) = 0 \)
• \( y''(1) = 0 \)

These conditions lead to a clearer path for solving the differential equation because they simplify the mathematics involved. They often result in well-defined and discrete eigenvalues, as nontrivial solutions can only occur for particular parameter values.