Applications of Legendre Polynomials - Physics | SIUechitamb/Teaching/Phys500/Legendre... · 2016-10-20 · Generating Function of Legendre Polynomials Recall The nth Legendre Polynomial

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).



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).



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.



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.



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).



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).



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.



Proof


Ψ(1, h) =1√

1− 2h + h2=

1

1− h= 1 + h + h2 + · · ·

=∞∑n=0

hnpn(1).




Proof


Ψ(1, h) =1√

1− 2h + h2=

1

1− h= 1 + h + h2 + · · ·

=∞∑n=0

hnpn(1).




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.



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.


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.



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.



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.



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+ · · · )




Φ(r , θ) =∞∑n=0

anrnPn(cos θ).



Φ(r , 0) =∞∑n=0

anrn =

q

4πε0

1

R − r

=q

4πε0(

1

R+

r

R2+

r2

R3+ · · · )




Φ(r , θ) =∞∑n=0

anrnPn(cos θ).



Φ(r , 0) =∞∑n=0

anrn =

q

4πε0

1

R − r

=q

4πε0(

1

R+

r

R2+

r2

R3+ · · · )



Therefore an = q4πε0

1rn+1 . The full solution is

Φ(r , θ) =q

4πε0

∞∑n=0

rn

Rn+1Pn(cos θ).



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).



Solution 2




d = |R− r| =√R2 − 2Rr cos θ + r2 = R

√1− 2

r

Rcos θ +

( r

R

)2.


Φ =q

4πε0RΨ(x , h).



Solution 2




d = |R− r| =√R2 − 2Rr cos θ + r2 = R

√1− 2

r

Rcos θ +

( r

R

)2.


Φ =q

4πε0RΨ(x , h).



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).


Applications of Legendre Polynomials - Physics | SIUechitamb/Teaching/Phys500/Legendre... · 2016-10-20 · Generating Function of Legendre Polynomials Recall The nth Legendre Polynomial

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