Problem 11 - Advanced Engineering Mathematics

Question

(a) The differential equation is $$\left(1+x^{2}\right) y^{\prime \prime}+2 x y^{\prime}+\frac{\lambda}{1+x^{2}} y=0.$$ Letting \(x=\tan \theta\) we have \(\theta=\tan ^{-1} x\) and $$\begin{aligned} \frac{d y}{d x} &=\frac{d y}{d \theta} \frac{d \theta}{d x}=\frac{1}{1+x^{2}} \frac{d y}{d \theta} \\ \frac{d^{2} y}{d x^{2}} &=\frac{d}{d x}\left[\frac{1}{1+x^{2}} \frac{d y}{d \theta}\right]=\frac{1}{1+x^{2}}\left(\frac{d^{2} y}{d \theta^{2}} \frac{d \theta}{d x}\right)-\frac{2 x}{\left(1+x^{2}\right)^{2}} \frac{d y}{d \theta} \\\ &=\frac{1}{\left(1+x^{2}\right)^{2}} \frac{d^{2} y}{d \theta^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}} \frac{d y}{d \theta}. \end{aligned}$$ The differential equation can then be written in terms of \(y(\theta)\) as $$\begin{array}{c} \left(1+x^{2}\right)\left[\frac{1}{\left(1+x^{2}\right)^{2}} \frac{d^{2} y}{d \theta^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}} \frac{d y}{d \theta}\right]+2 x\left[\frac{1}{1+x^{2}} \frac{d y}{d \theta}\right]+\frac{\lambda}{1+x^{2}} y \\\ =\frac{1}{1+x^{2}} \frac{d^{2} y}{d \theta^{2}}+\frac{\lambda}{1+x^{2}} y=0 \end{array}$$ or $$\frac{d^{2} y}{d \theta^{2}}+\lambda y=0.$$ The boundary conditions become \(y(0)=y(\pi / 4)=0 .\) For \(\lambda \leq 0\) the only solution of the boundary-value problem is \(y=0 .\) For \(\lambda=\alpha^{2}>0\) the general solution of the differential equation is \(y=c_{1} \cos \alpha \theta+c_{2} \sin \alpha \theta\) The condition \(y(0)=0\) implies \(c_{1}=0\) so \(y=c_{2} \sin \alpha \theta .\) Now the condition \(y(\pi / 4)=0\) implies \(c_{2} \sin \alpha \pi / 4=\) 0\. For \(c_{2} \neq 0\) this condition will hold when \(\alpha \pi / 4=n \pi\) or \(\lambda=\alpha^{2}=16 n^{2},\) where \(n=1,2,3, \ldots .\) These are the eigenvalues with corresponding eigenfunctions \(\sin 4 n \theta=\sin \left(4 n \tan ^{-1} x\right),\) for \(n=1,2,3, \dots\) (b) An orthogonality relation is $$\int_{0}^{1} \frac{1}{x^{2}+1} \sin \left(4 m \tan ^{-1} x\right) \sin \left(4 n \tan ^{-1} x\right) d x=0, \quad m \neq n.$$

Step-by-Step Solution

Answer

Eigenvalues are \( \lambda = 16n^2 \), eigenfunctions are \( \sin(4n\tan^{-1} x) \).

Step 1: Convert Variables

We start by substituting \( x = \tan \theta \), hence \( \theta = \tan^{-1} x \). This transformation helps simplify the differential equation.

Step 2: Differentiate with Respect to \( \theta \)

We first find the derivatives: \( \frac{dy}{dx} = \frac{1}{1+x^2} \frac{dy}{d\theta} \) and \( \frac{d^2y}{dx^2} = \frac{1}{(1+x^2)^2} \frac{d^2y}{d\theta^2} - \frac{2x}{(1+x^2)^2} \frac{dy}{d\theta} \).

Step 3: Substitute Derivatives into the Differential Equation

Replace the derivatives in the equation \( (1+x^2) y'' + 2x y' + \frac{\lambda}{1+x^2} y = 0 \). Simplifying gives \( \frac{d^2y}{d\theta^2} + \lambda y = 0 \).

Step 4: Solve the Simplified Differential Equation

The equation \( \frac{d^2y}{d\theta^2} + \lambda y = 0 \) has the general solution \( y = c_1\cos \alpha \theta + c_2\sin \alpha \theta \), where \( \alpha^2 = \lambda \).

Step 5: Apply Boundary Conditions \( y(0) = y(\pi/4) = 0 \)

Using \( y(0) = 0 \), solve to find \( c_1 = 0 \). With \( y \) simplified to \( y = c_2 \sin \alpha \theta \), use \( y(\pi/4) = 0 \) to solve for \( \alpha = 4n \), giving eigenvalues \( \lambda = 16n^2 \).

Step 6: Find Eigenfunctions

The eigenfunctions corresponding to these eigenvalues are \( \sin(4n\theta) = \sin(4n\tan^{-1} x) \) for \( n = 1, 2, 3, \ldots \).

Step 7: Verify Orthogonality

The orthogonality relation is \( \int_0^1 \frac{1}{x^2+1} \sin(4m\tan^{-1} x) \sin(4n\tan^{-1} x) \, dx = 0 \) for \( m eq n \), ensuring the orthogonality of the eigenfunctions.

Key Concepts

Boundary Value Problems

Boundary value problems are an essential topic in differential equations. They involve finding a function that satisfies a differential equation and adheres to specific conditions at the boundaries of the domain, often given as points. In our case, the boundary conditions are given as \( y(0) = 0 \) and \( y(\pi/4) = 0 \).

Such problems naturally arise in fields like physics and engineering, where determining the steady-state or equilibrium condition of a system is necessary.

• Boundary conditions are critical, as they determine the set of all possible solutions.
• Finding the correct solution involves satisfying both the differential equation and the boundary conditions simultaneously.
• Solutions to these problems are not unique and often depend heavily on the conditions set by the problem's boundaries.

Understanding boundary value problems lays the foundation for tackling more complex systems and their associated behaviors in practical scenarios.

Eigenvalues and Eigenfunctions

The solution of differential equations, especially in boundary value problems, frequently involves finding eigenvalues and eigenfunctions. In our exercise, the parameter \( \lambda \) plays a crucial role as it represents possible eigenvalues once we meet both the differential equation and boundary conditions. An initial transformation to \( \theta \) helped simplify the equation, leading to the general solution form.

To find the eigenvalues, we applied the given boundary conditions, resulting in an equation involving \( \alpha \), or \( 4n \), leading us to express eigenvalues as \( \lambda = 16n^2 \).

• Each \( n \) gives a distinct eigenvalue and its corresponding eigenfunction, \( \sin(4n\theta) \).
• Eigenvalues are used to characterize and solve differential equations bound by specific conditions.
• Related eigenfunctions represent oscillatory modes, which can correspond to natural frequencies or energy levels in physical systems.

Therefore, understanding the concept of eigenvalues and eigenfunctions is vital for solving complex differential equations efficiently.

Orthogonality

Orthogonality is a central property related to eigenfunctions in boundary value problems. In simpler terms, orthogonal functions are ‘perpendicular’ in the function space, meaning their inner product evaluates to zero. This property is crucial as it means the functions do not interfere with one another.

This exercise showed that different eigenfunctions, when paired like \( \sin(4m \tan^{-1} x) \) and \( \sin(4n \tan^{-1} x) \) with \( m eq n \), are orthogonal due to:
\[\int_0^1 \frac{1}{x^2+1} \sin(4m \tan^{-1} x) \sin(4n \tan^{-1} x) \, dx = 0\]

• Orthogonality is fundamental in separating solutions, particularly when expanding functions in terms of eigenfunctions.
• It reveals the independence of modes in a system, such as various frequencies in a physical problem.
• This property simplifies computational methods like Fourier series expansions.

Orthogonality ensures that complex problems can be decomposed into simpler, non-interacting components, making them easier to solve.

Variable Transformation

A crucial strategy in solving differential equations, especially those in boundary value contexts, involves variable transformation. This process helps simplify the equation, making it more manageable. In our differential equation, substituting \( x = \tan \theta \) transformed the original variable \( x \) into \( \theta \).

This change allowed us to reduce the complex differential equation to a standard second-order form;
\[ \frac{d^2y}{d\theta^2} + \lambda y = 0\]
Variable transformations often aim to:

• Simplify or linearize complex equations, lowering computational difficulty.
• Allow the use of standard solving techniques, which are well-documented and easier to apply.
• Better fit the problem to recognized boundary conditions.

In essence, variable transformation is a powerful method for reframing problems to draw on simpler, more familiar solutions.