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Q29E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 450
Fundamentals Of Differential Equations And Boundary Value Problems

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

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Short Answer

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 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

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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=0∞ akxk

Taking derivative of the above equation,

y'(x)=Σk=1∞ kakxk-1

y"(x)=Σk=2∞ k(k-1)akxk-2

Replace in the equation.

(1-x2)Σk=2∞ k(k-1)akxk-2 -2x Σk=1∞ kakxk-1 +n(n+1)Σk=0∞ akxk =0

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

The first sum substitute is,

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

(2)(1)a2+(3)(2)a3+Σk=2∞ (m+2)(m+1)am+2xm - Σk=2∞ k(k-1)akxk - (2)(1) a1x -2Σk=1∞ kakxk + n(n+1) (a0+a1x) +n(n+1)Σk=0∞ akxk =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 .

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