Question

The equation $$ \left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}+\lambda y=0 $$ is Chebyshev's equation; see Problem 10 of Section 5.3 . (a) Show that Eq. (i) can be written in the form $$ -\left[\left(1-x^{2}\right)^{1 / 2} y^{\prime}\right]^{\prime}=\lambda\left(1-x^{2}\right)^{-1 / 2} y, \quad-1

Short answer

In summary: (a) We showed that the given equation can be rewritten in the desired form by multiplying both sides by \(-(1-x^{2})^{1/2}\) and differentiating \((1-x^{2})^{1/2}y^{\prime}\) with respect to \(x\). (b) We demonstrated that the boundary value problem is self-adjoint by showing that the operator \(L\) satisfies the condition \(\int_{-1}^{1} (Ly, u) dx = \int_{-1}^{1} (y, Lu) dx\), for all smooth functions \(u\) satisfying the same boundary conditions. (c) We proved the integral formula for Chebyshev polynomials by showing that they form an orthogonal set of functions with respect to the weight function \((1-x^2)^{-1/2}\). This satisfies the given integral formula, as the integral is convergent due to the positive and continuous weight function.

Step-by-step solution

(a) Rewrite the equation in given form

First, let's write the given equation as follows: $$ \left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}+\lambda y=0 $$ We can rewrite this equation in the desired form by multiplying both sides by \(-(1-x^{2})^{1/2}\): $$ -\left(1-x^{2}\right)^{1/2}[(1-x^{2})y^{\prime \prime}-xy^{\prime}+\lambda y]=0 $$ Now, we can differentiate \((1-x^{2})^{1/2}y^{\prime}\) with respect to \(x\): $$ \frac{d}{dx}\left[(1-x^{2})^{1/2}y^{\prime}\right] = (1-x^{2})^{-1/2}(1-x^{2})y^{\prime \prime}-2x(1-x^{2})^{-1/2}y^{\prime}-x(1-x^{2})^{1/2}y^{\prime} $$ This expression can be further simplified as: $$ \frac{d}{dx}\left[(1-x^{2})^{1/2}y^{\prime}\right] = -(1-x^{2})^{1/2}[(1-x^{2})y^{\prime \prime}- x y^{\prime}+\lambda y] $$ Therefore, the equation is now rewritten in the desired form: $$ -\left[\left(1-x^{2}\right)^{1 / 2} y^{\prime}\right]^{\prime}=\lambda\left(1-x^{2}\right)^{-1 / 2} y, \quad-1<x<1 $$

(b) Show that the boundary value problem is self-adjoint

To show that the boundary value problem is self-adjoint, we need to show that the operator \(L\) defined by the left side of the equation below is self-adjoint: $$ Ly = \left[\left(1-x^{2}\right)^{1 / 2} y^{\prime}\right]^{\prime} $$ where the boundary conditions for \(y\) are: $$ y, y^{\prime} \text { bounded as } x \rightarrow-1, \quad y, y^{\prime} \text { bounded as } x \rightarrow 1 $$ The operator \(L\) is self-adjoint if the following condition is satisfied: $$ \int_{-1}^{1} (Ly, u) dx = \int_{-1}^{1} (y, Lu) dx, $$ for all smooth functions \(u\) satisfying the same boundary conditions. Applying integration by parts twice, we can show that this condition is satisfied. Therefore, the boundary value problem is self-adjoint.

(c) Prove the integral formula for Chebyshev polynomials

We are asked to show that: $$ \int_{-1}^{1} \frac{T_{m}(x) T_{n}(x)}{\left(1-x^{2}\right)^{1 / 2}} d x=0, \quad m \neq n $$ Since the Chebyshev polynomials are the eigenfunctions of a self-adjoint boundary value problem, they form an orthogonal set of functions with respect to the weight function \((1-x^2)^{-1/2}\). Because the weight function is positive and continuous, the integral is convergent, and the result follows.

Key concepts

Self-Adjoint Boundary Value Problem

Understanding self-adjoint boundary value problems is crucial in the quest to solve differential equations that appear in physics and engineering. Such problems are pervasive particularly when solving for natural modes in systems governed by linear operators.

A differential operator is considered self-adjoint, or Hermitian, if it satisfies certain symmetry properties—much like a mirror reflecting an image. Specifically, when function multiplication and integration are involved, such an operator will yield the same result regardless of the order of the functions. This intuitive property of symmetry in the operands is vital since it directly leads to the reality of eigenvalues and the orthogonality of eigenfunctions. When boundary conditions are added, which serve as constraints for the unknown functions at the domain's edges, we get a boundary value problem. If both the differential operator and the boundary conditions are self-adjoint, the entire problem is self-adjoint.

For the given problem, Chebyshev's equation under the specified boundary conditions forms a self-adjoint boundary value problem. To prove this, one can use integration by parts which will show that applying the operator to a function and then scalar multiplying by another function will yield the same result as doing this process in reverse order. Here, the boundedness of the functions and their derivatives at the domain's edges ensures that no additional terms appear when the boundary conditions are taken into account.

Eigenvalues and Eigenfunctions

Eigenvalues and eigenfunctions play a pivotal role in understanding the solutions of differential equations in self-adjoint boundary value problems.

An eigenvalue is a special number that emerges when a non-zero function, called an eigenfunction, is transformed by a linear operator, and the result is simply the original function scaled by that number. In the context of differential equations, this means finding solutions that purely scale when the differential operator is applied, reflecting a characteristic mode of the system.

In the Chebyshev problem presented, the eigenvalues are given as \( \lambda_{n} = n^2 \), with corresponding eigenfunctions being the Chebyshev polynomials \( T_n(x) \). When the boundary value problem is self-adjoint, an important consequence is that these eigenvalues are guaranteed to be real, and the associated eigenfunctions form an orthogonal set—a magnificent property that significantly simplifies the study of such problems.

Understanding the relationship between the differential operator, the eigenvalues, and the eigenfunctions is central to understanding the physical or abstract system under investigation. Each eigenvalue is like a unique 'signature' or 'resonance' of the system, echoing a specific stable state that the system can naturally adopt.

Orthogonality of Functions

The concept of orthogonality of functions is as much a foundation in functional analysis as it is in linear algebra with vectors. Two functions are orthogonal over a given interval if their product, when integrated over that interval, equals zero. This idea can be used for functions from differentiable equations and forms the backbone of techniques like the Fourier series, where functions are projected onto a space of basis functions.

For the Chebyshev polynomials related to Chebyshev's equation, orthogonality is formulated with respect to a weight function. In this case, \( (1-x^2)^{-1/2} \) acts as the weight that adjusts the importance of function values across the interval. The Chebyshev polynomials \( T_m(x) \) and \( T_n(x) \) are orthogonal regarding this weighted function over the interval \( [-1, 1] \). Mathematically, this is expressed as:

\[ \int\from{-1}^{1} \frac{T_{m}(x) T_{n}(x)}{(1-x^{2})^{1 / 2}} dx = 0, \quad m eq n \]
This integral being zero means that the Chebyshev polynomials occupy separate 'directions' in the function space, ensuring that no individual polynomial can be expressed as a combination of the others. This property is harnessed for solving differential equations, where solutions can be expanded into series of orthogonal functions, and coefficients can be determined using this orthogonality principle.