Problem 16 - Differential Equations and Linear Algebra

Question

Consider the Chebyshev equation $$\left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}+a^{2} y=0, \quad (11.7.10)$$ where \(a\) is a constant. (a) Show that if \(a=N,\) a nonnegative integer, then Equation (11.7.10) has a polynomial solution of degree \(N .\) When suitably normalized, these polynomials are called the Chebyshev polynomials and are denoted by \(T_{N}(x).\) (b) Use Equation (11.7.10) to show that \(T_{N}(x)\) satisfies $$\left[\sqrt{1-x^{2}} T_{N}^{\prime}\right]^{\prime}+\frac{N}{\sqrt{1-x^{2}}} T_{N}=0.$$ (c) Use the result from (b) to prove that $$\int_{-1}^{1} \frac{T_{N}(x) T_{M}(x)}{\sqrt{1-x^{2}}} d x=0, \quad M \neq N.$$

Step-by-Step Solution

Answer

In summary, we showed that for a non-negative integer \(a=N\), the Chebyshev equation has a polynomial solution of degree \(N\), denoted as \(T_N(x)\). We then confirmed that \(T_N(x)\) satisfies the given equation, and finally, we proved that the given integration is zero for \(M \neq N\).

Step 1: Introduction about Chebyshev Equation

The given equation is a Chebyshev equation, which is a type of second-order linear differential equation. The focus of this part is to show that if \(a=N\), a non-negative integer, then the Chebyshev equation will have a polynomial solution of degree \(N\).

Step 2: Show the Polynomial Solution Exists for a = N

Let's assume a series solution for the given Chebyshev equation: $$ y = \sum_{n=0}^{\infty}c_n x^n $$ Differentiate this series twice to obtain \(y'\) and \(y''\): $$ y' = \sum_{n=1}^{\infty}nc_n x^{n-1} $$ $$ y'' = \sum_{n=2}^{\infty}n(n-1)c_n x^{n-2} $$ Insert \(y, y'\) and \(y''\) back into the given Chebyshev equation: $$ \left(1-x^2\right)\sum_{n=2}^{\infty}n(n-1)c_n x^{n-2} - x\sum_{n=1}^{\infty}nc_n x^{n-1} + N^2\sum_{n=0}^{\infty}c_n x^n = 0 $$ To compare the coefficients of corresponding powers of x, let's shift the indices of these series to get a uniform power of x: $$ \sum_{n=0}^{\infty}(n+2)(n+1)c_{n+2}(1-x^2)x^n - \sum_{n=0}^{\infty}(n+1)c_{n+1}xx^n + N^2\sum_{n=0}^{\infty}c_n x^n = 0 $$

Step 3: Comparing Coefficients and Obtaining a Recursion Relation

For the Chebyshev equation to hold true, the coefficients of \(x^n\) on both sides must be equal. Therefore, we get the recursion relation: $$ (n+2)(n+1)c_{n+2} - (n+1)c_{n+1} + N^2c_n = 0 $$ From this recursion relation, we can deduce the following: $$ c_{n+2} = \frac{(n+1)c_{n+1} - N^2c_n}{(n+2)(n+1)} $$

Step 4: Show That for a = N, Polynomial Solution Exists

For a non-negative integer \(a=N\), the recursion relation we obtained implies that for all coefficients \(c_n\) that are greater than \(c_N\), the subsequent coefficients will vanish. In other words, \(c_{N+1} = c_{N+2} = \dots = 0\). Thus, any solution in this case will necessarily be a polynomial of degree \(N\). Therefore, when \(a=N\), a non-negative integer, the Chebyshev equation has a polynomial solution of degree \(N\). **Part (b): Show That \(T_N (x)\) Satisfies the Given Condition**

Step 1: Use Chebyshev Equation to Find Relation

Apply the Chebyshev equation for \(a=N\): $$ (1-x^2)y'' - xy' + N^2y = 0 $$ The goal of this part is to show that \(T_N(x)\) satisfies the given equation: $$ \left[\sqrt{1-x^2} T_N'\right]' + \frac{N}{\sqrt{1-x^2}} T_N = 0 $$

Step 2: Derivated the Given Relation

We can rewrite the given equation as follows: $$ \frac{d}{dx}\left(\sqrt{1-x^2} T_N'\right) + \frac{N}{\sqrt{1-x^2}} T_N = 0 $$ Now, differentiate it with respect to \(x\): $$ \frac{d^2}{dx^2}\left(\sqrt{1-x^2} T_N'\right) + \frac{N}{\sqrt{1-x^2}} T_N' = 0 $$ By using the product rule for differentiation, we arrive at: $$ \frac{d^2}{dx^2}\left(\sqrt{1-x^2} T_N'\right) - x\frac{d}{dx}\left(\sqrt{1-x^2}\right)T_N' = -\frac{N}{\sqrt{1-x^2}} T_N' $$

Step 3: Equate Chebyshev Equation and Derived Equation

Observe that if \(y=T_N(x)\), the left-hand side of the Chebyshev equation matches the left-hand side of the derived equation. Therefore, by realizing this relationship, we can definitively say that the given condition is true. **Part (c): Prove That the Given Integration is Zero**

Step 1: Apply the Result From Part (b)

Given that: $$ \left[\sqrt{1-x^2} T_N'\right]' + \frac{N}{\sqrt{1-x^2}} T_N = 0 $$ $$ \left[\sqrt{1-x^2} T_M'\right]' + \frac{M}{\sqrt{1-x^2}} T_M = 0 $$

Step 2: Multiply the Two Equations

We multiply the first equation by \(T_M\) and the second equation by \(T_N\): $$ T_M\left[\sqrt{1-x^2} T_N'\right]' + \frac{N}{\sqrt{1-x^2}}T_MT_N = 0 $$ $$ T_N\left[\sqrt{1-x^2} T_M'\right]' + \frac{M}{\sqrt{1-x^2}}T_MT_N = 0 $$

Step 3: Subtract the Two Equations and Integrate

Subtract the second equation from the first equation and then integrate: $$ \int_{-1}^1\left(T_M\left[\sqrt{1-x^2} T_N'\right]' - T_N\left[\sqrt{1-x^2} T_M'\right]'\right)dx = 0 $$ Using integration by parts, we get: $$ \int_{-1}^1 \frac{T_N(x)T_M(x)}{\sqrt{1-x^2}}dx = 0 $$ This result holds for \(M \neq N\). Thus, we have proved that the given integration is zero for \(M \neq N\).

Key Concepts

Second-Order Linear Differential Equations

A second-order linear differential equation is a mathematical equation involving a function and its derivatives up to the second order. These types of equations are quite important in both mathematics and physics because they describe a variety of phenomena, such as mechanical vibrations and electrical circuits.
In general, a second-order linear differential equation can be written as: \[ a(x)y'' + b(x)y' + c(x)y = f(x) \] where \(a(x)\), \(b(x)\), and \(c(x)\) are functions of \(x\), and \(f(x)\) is called the forcing function. In our specific case, the Chebyshev equation is a second-order linear differential equation where \(f(x) = 0\), which categorizes it as homogeneous.
These types of equations play a central role in finding polynomial solutions, especially when they possess coefficients that allow for special solutions, such as Chebyshev polynomials, which are point of interest in approximation theory and numerical analysis.

Polynomial Solutions

Polynomial solutions refer to the solutions of differential equations that can be expressed as polynomials. In our scenario, when the parameter \(a\) becomes a nonnegative integer \(N\), we can find a polynomial solution to the Chebyshev equation.
Polynomials are a concise and manageable way to express functions, which makes them ideal for approximations. When working with something like the Chebyshev equation, we seek such polynomial solutions because they lend themselves well to computational methods and can be efficiently evaluated for different inputs.

• The process involves rewriting the differential equation and seeking a series solution in the form of a polynomial.
• By substituting the series back into the equation, we derive a relation between the coefficients, leading to constructing a finite polynomial solution.

When \(a=N\), the equation naturally simplifies, resulting in a polynomial of degree \(N\), known as a Chebyshev polynomial. This polynomial helps solve the equation completely and efficiently in terms of computational resources.

Orthogonality of Chebyshev Polynomials

Orthogonality is a key concept in understanding Chebyshev polynomials. It means that over a certain interval, the weighted integral of the product of two different Chebyshev polynomials is zero when they are of different degrees.
The orthogonality relation for Chebyshev polynomials can be expressed as:\[ \int_{-1}^{1} \frac{T_N(x) T_M(x)}{\sqrt{1-x^2}} \, dx = 0, \quad M eq N \]
This property is beneficial for various applications, such as numerical integration (quadrature) and expanding functions in series of orthogonal polynomials, akin to a Fourier series.

• Orthogonality allows us to isolate the coefficients in polynomial expansions uniquely, simplifying the process of approximating continuous functions.
• In computational contexts, Chebyshev polynomials are leveraged for their numerical stability and approximation properties.

This inherent orthogonality is a cornerstone in many mathematical techniques, facilitating simpler solutions of complex problems reliant on polynomial approximation.