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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# The equation(1-x2)y"-2xy'+n(n+1)y=0where 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 nth 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).

(a) The power series expansion is y(x)=a0 [1+Σk=1 (-1)k [n(n-2)(n-4)...(n-2k+2)(n+1)(n+3)...(n+2k-1)]/(2k)! x2k] +a1 [x+Σk=1 (-1)k [(n-1)(n-3)...(n-2k+1)(n+2)(n+4)...(n+2k)]/(2k+1)! x2k+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.

See the step by step solution

## 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)=Σn=0 anx

## Find the first sum substitute:

Given,

(1-x2)y"-2xy'+n(n+1)y=0

Let,

y(x)=Σk=0akxk

Taking derivative of the above equation,

y'(x)=Σk=1kakxk-1

y"(x)=Σk=2k(k-1)akxk-2

Replace in the equation.

(1-x2k=2k(k-1)akxk-2 -2x Σk=1kakxk-1 +n(n+1)Σk=0akxk =0

Σk=2k(k-1)akxk-2k=2k(k-1)akxk-2 Σk=1kakxk +n(n+1)Σk=0akxk =0

The first sum substitute is,

Σk=0 (m+2)(m+1)am+2xmk=2k(k-1)akxk -2Σk=1kakxk+n(n+1)Σk=0akxk =0

(2)(1)a2+(3)(2)a3k=2 (m+2)(m+1)am+2xm - Σk=2k(k-1)akxk - (2)(1) a1x -2Σk=1kakxk + n(n+1) (a0+a1x) +n(n+1)Σk=0akxk =0

2a2+n(n+1)a0 + (6a3-2a1+n(n+1)a1) x+Σk=2 [(k+2)(k+1)ak+2+(-k2+k-2k+n(n+1))ak] xk =0

2a2+n(n+1)a0 + (6a3+(n-1)(n+2)a1) x+Σk=2 [(k+2)(k+1)ak+2+(n(n+1)-k(k+1)) ak] xk =0

Hence the first sum substitute is 2a2+n(n+1)a0 + (6a3+(n-1)(n+2)a1) x+Σk=2 [(k+2)(k+1)ak+2+(n(n+1)-k(k+1)) ak] xk =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)ak+2+(n(n+1)-k(k+1)) ak=0

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

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

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.

a2k=(-1)k [n (n-2)(n-4)...(n-2k+2) (n+1)(n+3)...(n+2k-1)]/2k! k≥1

a2k+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+Σk=1 (-1)k [n (n-2)(n-4)...(n-2k+2) (n+1)(n+3)...(n+2k-1)]/2k! ] x2k +a1 [x+Σk=1 (-1)k [(n-1)(n-3)...(n-2k+1) (n+2)(n+4)...(n+2k)]/(2k+1)! ] x2k+1 k≥1

## (b) Find the Legendre polynomials:

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

|ak+2/ak|<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 .