Method of Applied Math

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Method of Applied Math

Lecture 2: Legendre’s Equation and Gamma function

Sujin Khomrutai, Ph.D.

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Legendre’s equation

Legendre’s eqn Legendre’s eqn EX 1.

Series Sol Legendre func Rodrigues’ form EX 2.

EX 3.

Gamma function Properties

EX 4.

EX 5.

EX 6.

EX 7.

Plan

✔ Legendre’s equation.

✔ Gamma function

✔ Applications

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Legendre’s equation

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

Definition For a number n , the equation of the form (1 − x 2 )y ′′ − 2xy ′ + n(n + 1)y = 0

is called the Legendre equation of order n.

• n = 0: (1 − x 2 ) y ′′ − 2 xy ′ = 0

• n = 1: (1 − x 2 ) y ′′ − 2 xy ′ + 2 y = 0

• n = 2: (1 − x 2 )y ′′ − 2xy ′ + 6y = 0

• n = 3: (1 − x 2 )y ′′ − 2xy ′ + 12y = 0

In most phenomena, n is a non-negative integer. So we restrict to

that n ∈ { 0, 1, 2, . . . , } .

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Example 1: (Electrostatics)

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

Example. A two metallic spherical caps are placed so that the

upper part stayed at a constant potential 110 V and the lower

part is grounded. The electrostatic potential at any point in the

space can be calculated by solving the Legendre’s equation.

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Series solutions of Legendre’s equation

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

Power series method. a = 0 is an ordinary point. Set solution y to the Legendre’s equation

(1 − x 2 )y ′′ − 2xy ′ + n(n + 1)y = 0, as

y =

∞

X

k=0

b k x k

y ′ =

∞

X

k =0

b k +1 (k + 1)x k

y ′′ =

∞

X

k=0

b k+2 (k + 2)(k + 1)x k .

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Series solutions of Legendre’s equation

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

The first term

(1 − x 2 )y ′′ = y ′′ − x 2 y ′′

=

∞

X

k =0

b k +2 ( k + 2)( k + 1) x k −

∞

X

k =0

b k +2 ( k + 2)( k + 1) x k +2

=

∞

X

k=0

b k+2 (k + 2)(k + 1)x k −

∞

X

j =2

b j j (j − 1)x j

= b 2 · 2 · 1 + b 3 · 3 · 2x +

∞

X

k =2

[b k +2 (k + 2)(k + 1) − b k k (k − 1)] x k

where we have use shifting and splitting.

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EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

The second term 2xy ′ = 2x

∞

X

k=0

b k +1 (k + 1)x k

=

∞

X

k =0

2b k+1 (k + 1)x k +1

=

∞

X

k =1

2b k kx k Third term

n ( n + 1) y = n ( n + 1)

∞

X

k =0

b k x k =

∞

X

k =0

n ( n + 1) b k x k

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EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

Thus

b 2 · 2 · 1 + b 3 · 3 · 2x +

∞

X

k =2

[ b k +2 ( k + 2)( k + 1) − b k k ( k − 1)] x k

−

∞

X

k=1

2b k kx k +

∞

X

k=0

n(n + 1)b k x k = 0

∴ [2b 2 + n(n + 1)b 0 ] + [6b 3 − 2b 1 + n(n + 1)b 1 ]x +

∞

X

k=2

[b k+2 (k + 2)(k + 1) − b k k (k − 1)

− 2b k k + n(n + 1)b k ]x k = 0

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EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

Finally, we get

[2b 2 + n(n + 1)b 0 ] + [6b 3 + (n − 1)(n + 2)b 1 ]x +

∞

X

k =2

[( k + 2)( k + 1) b k +2 − ( n − k )( n + k + 1) b k ] x k

Apply the fact: P ∞

k=0 c k (x − a) k = 0 ⇔ c k = 0 for all k : b 2 = − n(n + 1)

2 b 0

b 3 = − ( n − 1)( n + 2)

6 b 1

b k+2 = − ( n − k )( n + k + 1)

(k + 2)(k + 1) b k ∀ k = 2 , 3 , . . . .

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EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

Let us consider the case n = 3. Recurrence equations are b 2 = − 3 · 4

2 b 0 , b 3 = − 2 · 5 3! b 1 b k +2 = − (3 − k)(3 + k + 1)

(k + 2)(k + 1) b k ∀ k ≥ 2 k = 2 : b 4 = − 1 · 6

4 · 3 b 2 = 1 · 6 · 3 · 4 4! b 0 k = 3 : b 5 = − 0 · 7

5 · 4 b 3 = 0 k = 4 : b 6 = − ( − 1) · 8

6 · 5 b 4 = − ( − 1) · 8 · 1 · 6 · 3 · 4

6! b 0

k = 5 : b 7 = − ( − 2) · 9

7 · 6 b 5 = 0

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EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

So in the case n = 3, we obtain the solution y = b 0

1 − 3 · 4

2! x 2 + 1 · 6 · 3 · 4

4! x 4 + · · ·

+ b 1

x − 2 · 5 3! x 3

This is the general solution with fundamental solutions y 1 = 1 − 3 · 4

2! x 2 + 1 · 6 · 3 · 4

4! x 4 + · · · an infinite series y 2 = x − 2 · 5

3! x 3 = − 5

3 x 3 + x a polynomial degree 3

For a general n ∈ { 0, 1, 2, . . . } , the Legendre equation has one

polynomial solution of degree n and an infinite series solution.

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Legendre functions

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

Theorem. For n ∈ { 0 , 1 , 2 , . . . } , the general solutions to the Legendre’s equation

(1 − x 2 )y ′′ − 2xy ′ + n(n + 1)y = 0 is

y = C 1 P n ( x ) + C 2 Q n ( x )

where P n is a polynomial of degree n, Q n an infinite series.

• P n = the Legendre polynomial.

• Q n = the Legendre function of the second kind.

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Legendre functions

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

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Legendre functions

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

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Rodrigues’ formula

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

The Legendre’s equation with n = 0 is (1 − x 2 )y ′′ − 2xy ′ = 0.

A polynomial of degree n = 0 is constant. It can be easily check that y = 1 is a solution. So we put

P 0 ( x ) = 1 .

Rodrigues’ formula If n ∈ { 1, 2, 3, . . . } then P n ( x ) = 1

2 n n!

d n

dx n ( x 2 − 1) n

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Example 2

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

EX. Verify Rodigues’ solution formula for n = 1 , 2 , 3.

n = 1 : P 1 ( x ) = 1 2

d

dx ( x 2 − 1) = x (1 − x 2 )x ′′ − 2x · x ′ + 2x = 0

(1 − x 2 ) · 0 − 2x · 1 + 2x = 0 True n = 2 : P 2 (x) = 1

2 2 · 2!

d 2

dx 2 (x 2 − 1) 2

= 1 8

d 2

dx 2 ( x 4 − 2 x 2 + 1) = 3

2 x 2 − 1 2 (1 − x 2 )( 3

2 x 2 − 1

2 ) ′′ − 2 x ( 3

2 x 2 − 1

2 ) ′ + 6( 3

2 x 2 − 1

2 ) = 0 (1 − x 2 ) · 3 − 2 x · 3 x + (9 x 2 − 3) = 0

(3 − 3x 2 ) − 6x 2 + (9x 2 − 3) = 0 True

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Example 2

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

n = 3 : P 3 (x) = 1 2 3 · 3!

d 3

dx 3 (x 2 − 1) 3

= 1 48

d 3

dx 3 ( x 6 − 3 x 4 + 3 x 2 − 1)

= 5

2 x 3 − 3 2 x (1 − x 2 )( 5

2 x 3 − 3

2 x) ′′ − 2x( 5

2 x 3 − 3

2 x) ′ + 12( 5

2 x 3 − 3

2 x) = 0 (1 − x 2 )(15x) − 2x( 15

2 x 2 − 3

2 ) + (30x 3 − 18x) = 0

(15x − 15x 3 ) − (15x 3 − 3x) + (30x 3 − 18x) = 0 True

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Example 3

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

EX. Solve the Legendre’s equation (1 − x 2 )y ′′ − 2xy ′ + 2y = 0.

Sol. We get a solution P 1 (x) = x. Use the reduction of order Q 1 ( x ) = P 1 ( x )

Z 1

(P 1 (x)) 2 e −

R

₋₂x 1−x2

dx

= x

Z 1

x 2 e − ln(1 − x

²

) dx

= x

Z 1

x 2 (1 − x 2 ) dx = x 2 ln

1 + x 1 − x

− 1

∴ y = C 1 x + C 2 x

2 ln

1 + x 1 − x

− 1

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The Gamma function

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

The factorials n ! are 0! = 1! = 1

2! = 2 · 1 = 2

3! = 3 · 2 · 1 = 6, ...

n! = n(n − 1)(n − 2) · · · 3 · 2 · 1 Observe that

Z ∞

0

e − t dt =

Z ∞

0

e − t tdt = 1,

Z ∞

0

e − t t 2 dt = 2 = 2!, Z ∞

e − t t 3 dt = 6 = 3!.

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The Gamma function

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

Definition For x > 0, the integral Z ∞

0

e − t t x − 1 dt is called the Gamma function denoted Γ(x).

If x < 0 and x 6∈ {− 1, − 2, . . . } we put

Γ(x) = Γ(x + n)

(x + n − 1)(x + n − 2) · · · (x + 1)x where n ∈ { 1 , 2 , . . . } satisfies x + n > 0.

Consecutive product formula

Γ(x + n) = (x + n − 1)(x + n − 2) · · · (x + 1)xΓ(x).

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The Gamma function

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

The graph of Gamma function is as shown below

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Properties of Gamma function

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

(1) Γ(1) = 1. For x > 0, Γ(x + 1) = xΓ(x)

(2) Generally, for n ∈ { 1, 2, . . . } we get

Γ(x + n) = (x + n − 1) · · · (x + 1)xΓ(x).

(3) In particular, if n ∈ { 0 , 1 , 2 , . . . } then

Γ( n + 1) = n ! .

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Properties of Gamma function

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

Proof. (1) For Γ(1), consider Γ(1) =

Z ∞

0

e − t t 1 − 1 dt =

Z ∞

0

e − t dt

= ( − e − t )

∞

t=0 = 0 − ( − e 0 ) = 1 . Next, for x > 0 consider

Γ(x + 1) =

Z ∞

0

e − t t (x+1) − 1 dt =

Z ∞

0

e − t t x dt

= ( − e − t t x )

∞

t =0 −

Z ∞

0

( − e − t xt x − 1 )dt (by parts)

= x

Z ∞

0

e − t t x − 1 dt = xΓ(x).

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Properties of Gamma function

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

(2) For n = 2, we use (1) ( x → x + 1):

Γ( x + 2) = Γ(( x + 1) + 1) = ( x + 1)Γ( x + 1) , and then use (1) one more time:

Γ(x + 1) = xΓ(x).

Thus

Γ( x + 2) = ( x + 1) x Γ( x )

The argument can be extended to n = 3, n = 4 , . . . .

(3) Use x = 1 in (2) and that Γ(1) = 1.

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Example 4

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

EX. Evaluate

Γ(1) + Γ(4) , Γ(2.8)

Γ(0.8) .

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Example 5

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

EX. Given that Γ(0.5) = √

π. Evaluate

Γ( − 2 . 5) .

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Example 6

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

EX. Express

1

( ν + 1)( ν + 2)( ν + 3)

as a ratio of the Gamma function.

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Example 7

EX 3.

EX 4.

EX 5.

EX 6.

EX 7.

EX. Evaluate the integral Z ∞

0

t 1.35 e − t dt.

Express the value in terms of the Gamma function.