Problem 21 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

Eigenvalues are \( \lambda_n = (n\pi)^4 \) with eigenfunctions \( y_n(x) = \cos(n\pi x) \).

Step 1: Substitute Eigenvalue Assumption

Assume \( \lambda = \alpha^4 \). Therefore, the differential equation changes into \( y^{(4)} - \alpha^4 y = 0 \). We will solve this new equation.

Step 2: Solve Characteristic Equation

For the differential equation \( y^{(4)} - \alpha^4 y = 0 \), the characteristic equation is \( r^4 - \alpha^4 = 0 \). Factoring gives \( (r^2 - \alpha^2)(r^2 + \alpha^2) = 0 \). The roots are \( r = \pm \alpha, \pm i\alpha \).

Step 3: Construct General Solution

Using the roots of the characteristic equation, the general solution is \( y(x) = C_1 e^{\alpha x} + C_2 e^{-\alpha x} + C_3 \cos(\alpha x) + C_4 \sin(\alpha x) \).

Step 4: Apply Boundary Condition at x=0

First, apply \( y(0) = 0 \). This leads to \( C_1 + C_2 + C_3 = 0 \). Next, apply \( y''(0) = 0 \). Compute \( y''(x) = \alpha^2 (C_1 e^{\alpha x} + C_2 e^{-\alpha x} - C_3 \cos(\alpha x) - C_4 \sin(\alpha x)) \) and set \( y''(0) = 0 \), giving \( \alpha^2 (C_1 + C_2 - C_3) = 0 \).

Step 5: Solve Boundary Conditions at x=1

Apply \( y(1) = 0 \). Substitute into the general solution: \( C_1 e^{\alpha} + C_2 e^{-\alpha} + C_3 \cos(\alpha) + C_4 \sin(\alpha) = 0 \). Finally, \( y''(1) = 0 \) gives \( \alpha^2 (C_1 e^{\alpha} + C_2 e^{-\alpha} - C_3 \cos(\alpha) - C_4 \sin(\alpha)) = 0 \).

Step 6: Simplify and Solve for Coefficients

Solving the system of equations from boundary conditions, one finds that non-trivial solutions require that either certain terms vanish or cancel. Specifically, \( \sin(\alpha) = 0 \) for the sine terms to allow non-zero coefficients. Thus, \( \alpha = n \pi \) for integer \( n \).

Step 7: Determine Eigenvalues and Eigenfunctions

Given \( \alpha = n \pi \), the eigenvalues are \( \lambda = (n \pi)^4 \). The corresponding eigenfunctions derived from the non-zero coefficients satisfying the conditions are \( y_n(x) = C_3 \cos(n \pi x) \), due to the earlier simplification.

Key Concepts

Differential Equations

Differential equations are mathematical equations connecting a function with its derivatives. They allow us to model how quantities change, which can be about motion, growth, or other processes. These equations come in various forms like ordinary differential equations (ODEs) and partial differential equations (PDEs).

In our exercise, we encounter a fourth-order differential equation, specifically: - \( y^{(4)} - \lambda y = 0 \) Here, we deal with an ordinary differential equation as it involves derivatives with respect to only one variable, \(x\). This particular equation is pivotal because it describes not just simple motion but more complex behaviors, which can turn up in mechanics and physics.

Solving differential equations involves finding the 'general solution,' which includes all possible solutions that satisfy the equation itself. In approaching the solution, constants will be determined by additional information, often through initial conditions or boundary conditions.

Boundary-value Problems

Boundary-value problems (BVPs) are a specific type of differential equation problem where the solution is determined by the values of the solution at the boundaries of the domain. These problems often arise when you're dealing with physical situations, indicating constraints at the edges, like the ends of a string.

The key points about BVPs are: - They require conditions at the boundary of the interval, which provides constraints to narrow down the infinite solutions.

In our exercise, the boundary conditions are given as: - \( y(0) = 0 \) - \( y''(0) = 0 \) - \( y(1) = 0 \) - \( y''(1) = 0 \) These conditions specify the values or requirements of the solution and its derivatives at specific points (here \(x=0\) and \(x=1\)). Applying these at the solutions help us find a unique answer that fits all the boundary requirements.

Eigenfunctions

Eigenfunctions are essential in mathematics and physics because they give information about the system's behavior under certain conditions. When you solve a boundary-value problem, like in this exercise, you find these eigenfunctions which are special types of solutions.

For our problem, the eigenfunction is linked to the general solution derived from solving the differential equation. Here, our solution form is - \( y(x) = C_1 e^{\alpha x} + C_2 e^{-\alpha x} + C_3 \cos(\alpha x) + C_4 \sin(\alpha x) \)

The eigenfunctions come from finding the specific constants \(C_1, C_2, C_3,\) and \(C_4\) such that the boundary conditions are satisfied without resulting in a trivial (zero) solution. In this context, due to boundary conditions, we find that - \( y_n(x) = C_3 \cos(n \pi x) \) represents the family of eigenfunctions for the problem. These are valuable as they describe possible modes of vibration or behavior within the constraints of the problem.

Characteristic Equation

Characteristic equations are a central concept when dealing with differential equations, especially in finding eigenvalues and eigenfunctions. These equations help simplify solving differential equations by breaking them into a form that is easier to manage. This method involves transforming the differential equation into an algebraic one using the solutions of its characteristic equation.

For our fourth-order differential equation \( y^{(4)} - \lambda y = 0 \), assuming \( \lambda = \alpha^4 \), the characteristic equation becomes: - \( r^4 - \alpha^4 = 0 \) Solving this involves finding the roots, resulting in solutions: - \( r = \pm \alpha, \pm i\alpha \) These roots help form the general solution. Characteristic equations are advantageous because they turn a problem with derivatives into one reminiscent of pure algebra, much simpler to approach and solve. The roots of the characteristic equation typically lead to expressions involving exponents and trigonometric functions, combining to give a comprehensive solution for the differential equation.