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Lecture 1 Sujin Khomrutai – 1 / 27
Method of Applied MathLecture 2: Legendre’s Equation and Legendre
Polynomials
Sujin Khomrutai, Ph.D.
Legendre’s equation
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 / 27
Definition Let n be a constant. The ODE 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 theory, n can be any real number, however, real phenomenashow non-negative integer n. We restrict to the latter case.
Example 1: (Electrostatics)
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 / 27
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
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 / 27
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
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 / 27
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
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 / 27
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
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 / 27
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
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 / 27
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: power series = 0 ⇔ coefficients = 0
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.
Series solutions of Legendre’s equation
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 / 27
Simplify the presentation, n = 3. Recurrence equations are
b2 = −3 · 4
2b0, b3 = −
2 · 5
3!b1
bk+2 = −(3− k)(3 + k + 1)
(k + 2)(k + 1)bk ∀ k ≥ 2
k = 2 : b4 = −1 · 6
4 · 3b2 =
1 · 6 · 3 · 4
4!b0
k = 3 : b5 = −0 · 7
5 · 4b3 = 0
k = 4 : b6 = −(−1) · 8
6 · 5b4 = −
(−1) · 8 · 1 · 6 · 3 · 4
6!b0
k = 5 : b7 = −(−2) · 9
7 · 6b5 = 0
Series solutions of Legendre’s equation
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 / 27
The general solution to Legendre’s equation with n = 3 is
y = b0
[
1−3 · 4
2!x2 +
1 · 6 · 3 · 4
4!x4 + · · ·
]
+ b1
[
x−2 · 5
3!x3
]
The fundamental solutions
y1 = x−2 · 5
3!x3 a polynomial degree 3
y2 = 1−3 · 4
2!x2 +
1 · 6 · 3 · 4
4!x4 + · · · an infinite series.
It is true in general, when n is a non-negative integer.
Legendre functions
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 / 27
Theorem. Let n ∈ {0, 1, 2, . . .}. The general solutions to theLegendre’s equation
(1− x2)y′′ − 2xy′ + n(n+ 1)y = 0
can be expressed as
y = C1Pn(x) + C2Qn(x)
where Pn(x) is a polynomial of degree n and Qn is an infiniteseries.
• Pn = the Legendre polynomial or the Legendre function ofthe first kind.
• Qn = the Legendre function of the second kind.
Legendre functions
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 / 27
Legendre functions
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 / 27
Rodrigues’ formula
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 / 27
The Legendre polynomial in the case n = 0 can be taken
P0(x) = 1.
In fact, the Legendre’s equation with n = 0 is
(1− x2)y′′ − 2xy′ = 0,
which has y = 1 as a solution. For n ≥ 1, we have:
Rodrigues’ formula If n ∈ {1, 2, 3, . . .} then
Pn(x) =1
2nn!
dn
dxn(x2 − 1)n
Example 2
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 / 27
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
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 / 27
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
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 / 27
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
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 / 27
The factorials are
0! = 1! = 1
2! = 2 · 1 = 2
3! = 3 · 2 · 1 = 6,
...
n! = n(n− 1)(n− 2) · · · 3 · 2 · 1 (n ≥ 4)
In above n must be a non-negative integer, i.e. 0, 1, 2, . . .
The Gamma function is the function defined for real numbersas well and it gives the factorials for non-negative integers.
The Gamma function
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 / 27
Definition The Gamma function Γ(x) is defined by
• If x > 0,
Γ(x) =
∫
∞
0
e−ttx−1 dt.
• If x < 0 and x 6= −1,−2, . . .,
Γ(x) =Γ(x+ n)
(x+ n− 1)(x+ n− 2) · · · (x+ 1)x
where n is a positive integer such that x+ n > 0.
The Gamma function
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 / 27
The graph of Gamma function is as shown below
Properties of Gamma function
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 / 27
(1) Γ(1) = 1. For x > 0,
Γ(x+ 1) = xΓ(x)
(2) Generally, for x > 0 and a positive integer n,
Γ(x+ n) = (x+ n− 1) · · · (x+ 1)xΓ(x) (n ≥ 2).
E.g. Γ(x+2) = (x+1)xΓ(x),Γ(x+3) = (x+2)(x+1)xΓ(x),etc.
(3) For an integer n,
Γ(n+ 1) = n!.
Properties of Gamma function
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 / 27
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
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 / 27
(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
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 / 27
EX. Evaluate
Γ(1) + Γ(4),Γ(2.8)
Γ(0.8).
Example 5
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 / 27
EX. Evaluate
Γ(−2.5)
in terms of Γ(1.5).
Example 6
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 / 27
EX. Express
1
(ν + 1)(ν + 2) · · · (ν + n)
as a ratio of Gamma function.
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