Lecture 1 Sujin Khomrutai – 1 / 28

Method of Applied MathLecture 2: Legendre’s Equation and Gamma

function

Sujin Khomrutai, Ph.D.

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.

Lecture 1 Sujin Khomrutai – 2 / 28

Plan

✔ Legendre’s equation.✔ Gamma function✔ Applications

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.

Lecture 1 Sujin Khomrutai – 3 / 28

Definition For a number n, the equation of the form

(1− x2)y′′ − 2xy′ + n(n+ 1)y = 0

is called the Legendre equation of order n.

• n = 0: (1− x2)y′′ − 2xy′ = 0

• n = 1: (1− x2)y′′ − 2xy′ + 2y = 0

• n = 2: (1− x2)y′′ − 2xy′ + 6y = 0

• n = 3: (1− x2)y′′ − 2xy′ + 12y = 0

In most phenomena, n is a non-negative integer. So we restrict tothat n ∈ {0, 1, 2, . . . , }.

Example 1: (Electrostatics)

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.

Lecture 1 Sujin Khomrutai – 4 / 28

Example. A two metallic spherical caps are placed so that theupper part stayed at a constant potential 110 V and the lowerpart is grounded. The electrostatic potential at any point in thespace can be calculated by solving the Legendre’s equation.

Series solutions of 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.

Lecture 1 Sujin Khomrutai – 5 / 28

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

(1− x2)y′′ − 2xy′ + n(n+ 1)y = 0,

as

y =∞∑

k=0

bkxk

y′ =∞∑

k=0

bk+1(k + 1)xk

y′′ =∞∑

k=0

bk+2(k + 2)(k + 1)xk.

Series solutions of 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.

Lecture 1 Sujin Khomrutai – 6 / 28

The first term

(1− x2)y′′ = y′′ − x2y′′

=∞∑

k=0

bk+2(k + 2)(k + 1)xk −∞∑

k=0

bk+2(k + 2)(k + 1)xk+2

=∞∑

k=0

bk+2(k + 2)(k + 1)xk −∞∑

j=2

bjj(j − 1)xj

= b2 · 2 · 1 + b3 · 3 · 2x

+∞∑

k=2

[bk+2(k + 2)(k + 1)− bkk(k − 1)]xk

where we have use shifting and splitting.

Series solutions of 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.

Lecture 1 Sujin Khomrutai – 7 / 28

The second term

2xy′ = 2x∞∑

k=0

bk+1(k + 1)xk

=∞∑

k=0

2bk+1(k + 1)xk+1

=∞∑

k=1

2bkkxk

Third term

n(n+ 1)y = n(n+ 1)∞∑

k=0

bkxk =

∞∑

k=0

n(n+ 1)bkxk

Series solutions of 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.

Lecture 1 Sujin Khomrutai – 8 / 28

Thus

b2 · 2 · 1 + b3 · 3 · 2x

+∞∑

k=2

[bk+2(k + 2)(k + 1)− bkk(k − 1)]xk

−∞∑

k=1

2bkkxk +

∞∑

k=0

n(n + 1)bkxk = 0

∴ [2b2 + n(n+ 1)b0] + [6b3 − 2b1 + n(n+ 1)b1]x

+∞∑

k=2

[bk+2(k + 2)(k + 1)− bkk(k − 1)

− 2bkk + n(n+ 1)bk]xk = 0

Series solutions of 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.

Lecture 1 Sujin Khomrutai – 9 / 28

Finally, we get

[2b2 + n(n+ 1)b0] + [6b3 + (n− 1)(n+ 2)b1]x

+∞∑

k=2

[(k + 2)(k + 1)bk+2 − (n− k)(n+ k + 1)bk]xk

Apply the fact:∑

∞

k=0 ck(x− a)k = 0 ⇔ ck = 0 for all k:

b2 = −n(n+ 1)

2b0

b3 = −(n− 1)(n+ 2)

6b1

bk+2 = −(n− k)(n+ k + 1)

(k + 2)(k + 1)bk ∀ k = 2, 3, . . . .

Series solutions of 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.

Lecture 1 Sujin Khomrutai – 10 / 28

Let us consider the case n = 3. Recurrence equations are

b2 = −3 · 42

b0, b3 = −2 · 53!

b1

bk+2 = −(3− k)(3 + k + 1)

(k + 2)(k + 1)bk ∀ k ≥ 2

k = 2 : b4 = −1 · 64 · 3b2 =

1 · 6 · 3 · 44!

b0

k = 3 : b5 = −0 · 75 · 4b3 = 0

k = 4 : b6 = −(−1) · 86 · 5 b4 = −(−1) · 8 · 1 · 6 · 3 · 4

6!b0

k = 5 : b7 = −(−2) · 97 · 6 b5 = 0

Series solutions of 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.

Lecture 1 Sujin Khomrutai – 11 / 28

So in the case n = 3, we obtain the solution

y = b0

[

1− 3 · 42!

x2 +1 · 6 · 3 · 4

4!x4 + · · ·

]

+ b1

[

x− 2 · 53!

x3

]

This is the general solution with fundamental solutions

y1 = 1− 3 · 42!

x2 +1 · 6 · 3 · 4

4!x4 + · · · an infinite series

y2 = x− 2 · 53!

x3 = −5

3x3 + x a polynomial degree 3

For a general n ∈ {0, 1, 2, . . .}, the Legendre equation has onepolynomial solution of degree n and an infinite series solution.

Legendre functions

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.

Lecture 1 Sujin Khomrutai – 12 / 28

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

(1− x2)y′′ − 2xy′ + n(n+ 1)y = 0

is

y = C1Pn(x) + C2Qn(x)

where Pn is a polynomial of degree n, Qn an infinite series.

• Pn = the Legendre polynomial.

• Qn = the Legendre function of the second kind.

Legendre functions

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.

Lecture 1 Sujin Khomrutai – 13 / 28

Legendre functions

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.

Lecture 1 Sujin Khomrutai – 14 / 28

Rodrigues’ formula

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.

Lecture 1 Sujin Khomrutai – 15 / 28

The Legendre’s equation with n = 0 is

(1− x2)y′′ − 2xy′ = 0.

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

P0(x) = 1.

Rodrigues’ formula If n ∈ {1, 2, 3, . . .} then

Pn(x) =1

2nn!

dn

dxn(x2 − 1)n

Example 2

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.

Lecture 1 Sujin Khomrutai – 16 / 28

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

n = 1 : P1(x) =1

2

d

dx(x2 − 1) = x

(1− x2)x′′ − 2x · x′ + 2x = 0

(1− x2) · 0− 2x · 1 + 2x = 0 True

n = 2 : P2(x) =1

22 · 2!d2

dx2(x2 − 1)2

=1

8

d2

dx2(x4 − 2x2 + 1) =

3

2x2 − 1

2

(1− x2)(3

2x2 − 1

2)′′ − 2x(

3

2x2 − 1

2)′ + 6(

3

2x2 − 1

2) = 0

(1− x2) · 3− 2x · 3x+ (9x2 − 3) = 0

(3− 3x2)− 6x2 + (9x2 − 3) = 0 True

Example 2

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.

Lecture 1 Sujin Khomrutai – 17 / 28

n = 3 : P3(x) =1

23 · 3!d3

dx3(x2 − 1)3

=1

48

d3

dx3(x6 − 3x4 + 3x2 − 1)

=5

2x3 − 3

2x

(1− x2)(5

2x3 − 3

2x)′′ − 2x(

5

2x3 − 3

2x)′ + 12(

5

2x3 − 3

2x) = 0

(1− x2)(15x)− 2x(15

2x2 − 3

2) + (30x3 − 18x) = 0

(15x− 15x3)− (15x3 − 3x) + (30x3 − 18x) = 0 True

Example 3

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.

Lecture 1 Sujin Khomrutai – 18 / 28

EX. Solve the Legendre’s equation

(1− x2)y′′ − 2xy′ + 2y = 0.

Sol. We get a solution P1(x) = x. Use the reduction of order

Q1(x) = P1(x)

∫

1

(P1(x))2e−

∫−2x

1−x2dxdx

= x

∫

1

x2e− ln(1−x2)dx

= x

∫

1

x2(1− x2)dx =

x

2ln

(

1 + x

1− x

)

− 1

∴ y = C1x+ C2

[

x

2ln

(

1 + x

1− x

)

− 1

]

The Gamma function

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.

Lecture 1 Sujin Khomrutai – 19 / 28

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∫

∞

0

e−tdt =

∫

∞

0

e−ttdt = 1,

∫

∞

0

e−tt2dt = 2 = 2!,

∫

∞

0

e−tt3dt = 6 = 3!.

The Gamma function

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.

Lecture 1 Sujin Khomrutai – 20 / 28

Definition For x > 0, the integral

∫

∞

0

e−ttx−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).

The Gamma function

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.

Lecture 1 Sujin Khomrutai – 21 / 28

The graph of Gamma function is as shown below

Properties of Gamma function

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.

Lecture 1 Sujin Khomrutai – 22 / 28

(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!.

Properties of Gamma function

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.

Lecture 1 Sujin Khomrutai – 23 / 28

Proof. (1) For Γ(1), consider

Γ(1) =

∫

∞

0

e−tt1−1dt =

∫

∞

0

e−tdt

= (−e−t)∣

∣

∣

∞

t=0= 0− (−e0) = 1.

Next, for x > 0 consider

Γ(x+ 1) =

∫

∞

0

e−tt(x+1)−1dt =

∫

∞

0

e−ttxdt

= (−e−ttx)∣

∣

∣

∞

t=0−∫

∞

0

(−e−txtx−1)dt (by parts)

= x

∫

∞

0

e−ttx−1dt = xΓ(x).

Properties of Gamma function

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.

Lecture 1 Sujin Khomrutai – 24 / 28

(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)xΓ(x)

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

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

Example 4

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.

Lecture 1 Sujin Khomrutai – 25 / 28

EX. Evaluate

Γ(1) + Γ(4),Γ(2.8)

Γ(0.8).

Example 5

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.

Lecture 1 Sujin Khomrutai – 26 / 28

EX. Given that Γ(0.5) =√π. Evaluate

Γ(−2.5).

Example 6

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.

Lecture 1 Sujin Khomrutai – 27 / 28

EX. Express

1

(ν + 1)(ν + 2)(ν + 3)

as a ratio of the Gamma function.

Example 7

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.

Lecture 1 Sujin Khomrutai – 28 / 28

EX. Evaluate the integral

∫

∞

0

t1.35e−tdt.

Express the value in terms of the Gamma function.