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Applications of Legendre Polynomials
PHYS 500 - Southern Illinois University
October 19, 2016
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 1 / 11
Generating Function of Legendre Polynomials
Recall
The nth Legendre Polynomial Pn(x) is the only bounded polynomialsolution to Legendre’s Equation (1− x2)P ′′ − 2xP ′ + n(n + 1)P = 0 thatsatisfies Pn(1) = 1.
Our goal is to find yet another way to represent the Legendre PolynomialsPn(x).
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 2 / 11
Generating Function of Legendre Polynomials
Recall
The nth Legendre Polynomial Pn(x) is the only bounded polynomialsolution to Legendre’s Equation (1− x2)P ′′ − 2xP ′ + n(n + 1)P = 0 thatsatisfies Pn(1) = 1.
Our goal is to find yet another way to represent the Legendre PolynomialsPn(x).
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 2 / 11
Generating Function of Legendre Polynomials
Generating Function
For any |h| < 1, define the function
Ψ(x , h) =1√
1− 2xh + h2.
Define the function y = 2xh − h2 and expand Ψ(y) = (1− y)−1/2 inpowers of y about the point y = 0.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 3 / 11
Generating Function of Legendre Polynomials
Generating Function
For any |h| < 1, define the function
Ψ(x , h) =1√
1− 2xh + h2.
Define the function y = 2xh − h2 and expand Ψ(y) = (1− y)−1/2 inpowers of y about the point y = 0.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 3 / 11
Generating Function of Legendre Polynomials
Ψ =1√
1− y
= 1 +1
2y +
1
2(
3
2
1
2)y2 +
1
3!(
5
2
3
2
1
2)y3 + · · ·
= 1 + xh + h2(3
2x2 − 1
2) + · · ·
= p0(x) + hp1(x) + h2p2(x) + · · ·
Ψ(x , h) =∞∑n=0
hnpn(x).
Theorem.
The functions pn(x) are the Legendre Polynomials Pn(x).
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 4 / 11
Generating Function of Legendre Polynomials
Ψ =1√
1− y
= 1 +1
2y +
1
2(
3
2
1
2)y2 +
1
3!(
5
2
3
2
1
2)y3 + · · ·
= 1 + xh + h2(3
2x2 − 1
2) + · · ·
= p0(x) + hp1(x) + h2p2(x) + · · ·
Ψ(x , h) =∞∑n=0
hnpn(x).
Theorem.
The functions pn(x) are the Legendre Polynomials Pn(x).
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 4 / 11
Generating Function of Legendre Polynomials
Proof
We will show that the pn(x) satisfy pn(1) = 1 and that they satisfyLegendre’s equation.
Ψ(1, h) =1√
1− 2h + h2=
1
1− h= 1 + h + h2 + · · ·
=∞∑n=0
hnpn(1).
Since this holds for all |h| < 1, we must have pn(1) = 1.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 5 / 11
Generating Function of Legendre Polynomials
Proof
We will show that the pn(x) satisfy pn(1) = 1 and that they satisfyLegendre’s equation.
Ψ(1, h) =1√
1− 2h + h2=
1
1− h= 1 + h + h2 + · · ·
=∞∑n=0
hnpn(1).
Since this holds for all |h| < 1, we must have pn(1) = 1.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 5 / 11
Generating Function of Legendre Polynomials
Proof
We will show that the pn(x) satisfy pn(1) = 1 and that they satisfyLegendre’s equation.
Ψ(1, h) =1√
1− 2h + h2=
1
1− h= 1 + h + h2 + · · ·
=∞∑n=0
hnpn(1).
Since this holds for all |h| < 1, we must have pn(1) = 1.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 5 / 11
Generating Function of Legendre Polynomials
Proof
Next, we compute that
(1− x2)∂2Ψ
∂x2− 2x
∂Ψ
∂x+ h
∂2
∂h2(hΨ) = 0.
Proof
Substitute Ψ(x , h) =∑∞
n=0 hnpn(x) and compare the coefficients. It
yields:(1− x2)p′′l (x)− 2xp′l(x) + l(l + 1)p(x) = 0.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 6 / 11
Generating Function of Legendre Polynomials
Proof
Next, we compute that
(1− x2)∂2Ψ
∂x2− 2x
∂Ψ
∂x+ h
∂2
∂h2(hΨ) = 0.
Proof
Substitute Ψ(x , h) =∑∞
n=0 hnpn(x) and compare the coefficients. It
yields:(1− x2)p′′l (x)− 2xp′l(x) + l(l + 1)p(x) = 0.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 6 / 11
Application: Expansion of Electromagnetic Potential
Consider a charge q located at position R from the origin. We want tocompute the potential at some other position r. Let the polar angle θ bethe angle between r and R.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 7 / 11
Application: Expansion of Electromagnetic Potential
Solution 1
Recall that Gauss’s law says ∇2Φ(r , θ, φ) = −ρ(r ,θ,φ)ε0
. For all r < R, thecharge density ρ is zero:
∇2Φ =
[1
r2∂
∂r
(r2∂
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂
∂θ
)]Φ(r , θ) = 0.
We found separable solutions Φ(r , θ) = Rn(r)Pn(θ). The general solutionis given by
Φ(r , θ) =∞∑n=0
anRn(r)Pn(cos θ),
where Rn(r) = Arn + Br−n−1 and Pn(cos θ) is the nth LegendrePolynomial.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 8 / 11
Application: Expansion of Electromagnetic Potential
Solution 1
Recall that Gauss’s law says ∇2Φ(r , θ, φ) = −ρ(r ,θ,φ)ε0
. For all r < R, thecharge density ρ is zero:
∇2Φ =
[1
r2∂
∂r
(r2∂
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂
∂θ
)]Φ(r , θ) = 0.
We found separable solutions Φ(r , θ) = Rn(r)Pn(θ). The general solutionis given by
Φ(r , θ) =∞∑n=0
anRn(r)Pn(cos θ),
where Rn(r) = Arn + Br−n−1 and Pn(cos θ) is the nth LegendrePolynomial.
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 8 / 11
Application: Expansion of Electromagnetic Potential
For Rn(r) = Arn + Br−n−1, finite solution at r = 0 requires B = 0. Hence
Φ(r , θ) =∞∑n=0
anrnPn(cos θ).
To determine the constants an, we need boundary conditions.
When θ = 0, we must recover the potential of a point charge:
Φ(r , 0) =∞∑n=0
anrn =
q
4πε0
1
R − r
=q
4πε0(
1
R+
r
R2+
r2
R3+ · · · )
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 9 / 11
Application: Expansion of Electromagnetic Potential
For Rn(r) = Arn + Br−n−1, finite solution at r = 0 requires B = 0. Hence
Φ(r , θ) =∞∑n=0
anrnPn(cos θ).
To determine the constants an, we need boundary conditions.
When θ = 0, we must recover the potential of a point charge:
Φ(r , 0) =∞∑n=0
anrn =
q
4πε0
1
R − r
=q
4πε0(
1
R+
r
R2+
r2
R3+ · · · )
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 9 / 11
Application: Expansion of Electromagnetic Potential
For Rn(r) = Arn + Br−n−1, finite solution at r = 0 requires B = 0. Hence
Φ(r , θ) =∞∑n=0
anrnPn(cos θ).
To determine the constants an, we need boundary conditions.
When θ = 0, we must recover the potential of a point charge:
Φ(r , 0) =∞∑n=0
anrn =
q
4πε0
1
R − r
=q
4πε0(
1
R+
r
R2+
r2
R3+ · · · )
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 9 / 11
Application: Expansion of Electromagnetic Potential
Therefore an = q4πε0
1rn+1 . The full solution is
Φ(r , θ) =q
4πε0
∞∑n=0
rn
Rn+1Pn(cos θ).
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 10 / 11
Application: Expansion of Electromagnetic Potential
Solution 2
Go directly from the potential function Φ = q4πε0d
, where d = |R− r|.
The Law of Cosines gives
d = |R− r| =√R2 − 2Rr cos θ + r2 = R
√1− 2
r
Rcos θ +
( r
R
)2.
Change of variables h = rR and x = cos θ. Then
Φ =q
4πε0RΨ(x , h).
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 11 / 11
Application: Expansion of Electromagnetic Potential
Solution 2
Go directly from the potential function Φ = q4πε0d
, where d = |R− r|.
The Law of Cosines gives
d = |R− r| =√R2 − 2Rr cos θ + r2 = R
√1− 2
r
Rcos θ +
( r
R
)2.
Change of variables h = rR and x = cos θ. Then
Φ =q
4πε0RΨ(x , h).
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 11 / 11
Application: Expansion of Electromagnetic Potential
Solution 2
Go directly from the potential function Φ = q4πε0d
, where d = |R− r|.
The Law of Cosines gives
d = |R− r| =√R2 − 2Rr cos θ + r2 = R
√1− 2
r
Rcos θ +
( r
R
)2.
Change of variables h = rR and x = cos θ. Then
Φ =q
4πε0RΨ(x , h).
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 11 / 11
Application: Expansion of Electromagnetic Potential
As proven above, this can be expressed in terms of Legendre Polynomialsas
Φ =q
4πε0R
∞∑n=0
hnPn(x)
=q
4πε0
∞∑n=0
rn
Rn+1Pn(cos θ).
PHYS 500 - Southern Illinois University Applications of Legendre Polynomials October 19, 2016 12 / 11