Q29E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The equation

(1-x²)y"-2xy'+n(n+1)y=0

where n is an unspecified parameter is called Legendre’s equation. This equation appears in applications of differential equations to engineering systems in spherical coordinates.

(a) Find a power series expansion about x=0 for a solution to Legendre’s equation. 

(b) Show that for a non negative integer there exists an n^th degree polynomial that is a solution to Legendre’s equation. These polynomials upto a constant multiples are called Legendre polynomials.

(c) Determine the first three Legendre polynomials (upto a constant multiple).

Step-by-Step Solution

Answer

(a) The power series expansion is y(x)=a₀ [1+Σ_k=1^∞  (-1)^k [n(n-2)(n-4)...(n-2k+2)(n+1)(n+3)...(n+2k-1)]/(2k)! x^2k] +a₁ [x+Σ_k=1^∞  (-1)^k [(n-1)(n-3)...(n-2k+1)(n+2)(n+4)...(n+2k)]/(2k+1)! x^2k+1]. .

(b) We showed that for a nonnegative integer there exists a Legendre polynomial.

(c) The first three Legendre polynomials are P₀(x)=1, P₁(x)=x, P₂(x)=1-3/2x².

Define power series expansion:

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a recurrence relation for the coefficients. 

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable. 

It is generally given by the formula,

y(x)=Σ_n=0^∞ a_nx

Find the first sum substitute:

Given,

(1-x²)y"-2xy'+n(n+1)y=0

Let,

y(x)=Σ_k=0^∞ a_kx^k

Taking derivative of the above equation,

y'(x)=Σ_k=1^∞ ka_kx^k-1

y"(x)=Σ_k=2^∞ k(k-1)a_kx^k-2

Replace in the equation.

(1-x²)Σ_k=2^∞ k(k-1)a_kx^k-2 -2x Σ_k=1^∞ ka_kx^k-1 +n(n+1)Σ_k=0^∞ a_kx^k =0

Σ_k=2^∞ k(k-1)a_kx^k-2 -Σ_k=2^∞ k(k-1)a_kx^k-2 Σ_k=1^∞ ka_kx^k +n(n+1)Σ_k=0^∞ a_kx^k =0 

The first sum substitute is,

Σ_k=0^∞ (m+2)(m+1)a_m+2x^m -Σ_k=2^∞ k(k-1)a_kx^k -2Σ_k=1^∞ ka_kx^k+n(n+1)Σ_k=0^∞ a_kx^k =0

(2)(1)a₂+(3)(2)a₃+Σ_k=2^∞ (m+2)(m+1)a_m+2x^m - Σ_k=2^∞ k(k-1)a_kx^k - (2)(1) a₁x  -2Σ_k=1^∞ ka_kx^k + n(n+1) (a₀+a₁x) +n(n+1)Σ_k=0^∞ a_kx^k =0 

2a₂+n(n+1)a₀ + (6a₃-2a₁+n(n+1)a₁) x+Σ_k=2^∞  [(k+2)(k+1)a_k+2+(-k²+k-2k+n(n+1))a_k] x^k =0

2a₂+n(n+1)a₀ + (6a₃+(n-1)(n+2)a₁) x+Σ_k=2^∞  [(k+2)(k+1)a_k+2+(n(n+1)-k(k+1)) a_k] x^k =0

Hence the first sum substitute is 2a₂+n(n+1)a₀ + (6a₃+(n-1)(n+2)a₁) x+Σ_k=2^∞  [(k+2)(k+1)a_k+2+(n(n+1)-k(k+1)) a_k] x^k =0.

Find the general solution:

Coefficients must equal zero.

2a₂+n(n+1)a₀=0

a₂= -n(n+1)/2 a₀

6a₃+(n-1)(n+2)=0

a₃= -(n-1)(n+2)/6 a₁

(k+2)(k+1)a_k+2+(n(n+1)-k(k+1)) a_k=0

a_k+2= -[n(n+1)-k(k+1)]/(k+2)(k+1) a_k

= -[(n+k+1)(n-k)]/(k+2)(k+1) a_k

a₄=a₂₊₂

= -[(n+3)(n-2)]/(4)(3) a₂

=(-1)² [(n-2)n(n+1)(n+3)]/4! a₀

a₅=a₃₊₂

= -[(n+4)(n-3)]/(5)(4) a₃

=(-1)² [(n-3)(n-1)(n+2)(n+3)]/5! a₀

Now you can find general formula for coefficients.

a_2k=(-1)^k [n (n-2)(n-4)...(n-2k+2) (n+1)(n+3)...(n+2k-1)]/2k!   k≥1

a_2k+1 = (-1)^k [(n-1)(n-3)...(n-2k+1) (n+2)(n+4)...(n+2k)]/(2k+1)!   k≥1

Hence the general solution is,

y(x)=a₀ [1+Σ_k=1^∞ (-1)^k [n (n-2)(n-4)...(n-2k+2) (n+1)(n+3)...(n+2k-1)]/2k! ] x^2k +a₁ [x+Σ_k=1^∞ (-1)^k [(n-1)(n-3)...(n-2k+1) (n+2)(n+4)...(n+2k)]/(2k+1)! ] x^2k+1  k≥1

(b) Find the Legendre polynomials:

Let’s find the radius of convergence for this series.

|a_k+2/a_k|<1

|(n+k+1) (n-k)/(k+2)(k+1)| <1

This is satisfied for choosing k=n. Therefore the series cut in specific integers n and n+1 produce polynomials called Legendre polynomials.

Define Legendre polynomial:

Let’s find Legendre polynomial for n=0.

P₀(x)=1

For n=1, k can be 1 or zero.

P₁(x)=x

For n=2, k can be.

P₂(x)=1+(-1) (2+1)/[(2)(1)!] x²

=1-3/2 x²

The Legendre polynomials are P₀(x)=1, P₁(x)=x, and P₂(x)=1-3/2 x² .