Hassani S. Mathematical Methods: For Students of Physics and Related Fields

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26.5 Properties of Legendre Polynomials 627

Substituting these expressions in the DE, the reader may check that

(1 −z

)

− 2z

+ n(n +1)P

n +1

(2πi)

(ξ

− 1)

[nξ

− 2(n +1)ξz + n +2]

(ξ − z)

n+3

dξ.

The reader may also verify that

(ξ

− 1)

[nξ

− 2(n +1)ξz + n +2]

(ξ − z)

n+3

dξ



(ξ

− 1)

n+1

(ξ − z)

n+2



so that the integrand is the derivative of a function. Since the contour of

integration is closed, the lower and upper limits of integration coincide and

the integral vanishes. So, the Rodrigues formula indeed yields Legendre poly-

nomials.

Example 26.5.4.

As an illustration of the use of the Rodrigues formula, let us

evaluate the integral

I =

−1

(x) dx for k ≤ n.

The procedure is to replace P

(x) by the RHS of Equation (26.44) and integrate by

parts repeatedly. After one integration by parts, we get

I =

−1



− 1)



 

n−1

− 1)



−1

−

−1

k−1

n−1

− 1)

The ﬁrst term on the RHS of the second line is zero because each diﬀerentiation

reduces the power of (x

− 1)

by at most one unit. So after n −1 diﬀerentiations,

we get a sum of terms each having (x

−1) raised to various powers, with the lowest

power being one. All these terms vanish at x = 1 as well as at x = −1. Continuing

the integration by parts, we get

I =

−kx

k−1

n−2

− 1)



−1

  

=0 for same reason as above

(−1)

−1

k(k −1)x

k−2

n−2

− 1)

dx.

After k integrations by part, we obtain

I =

(−1)

−1

n−k

− 1)

because after k diﬀerentiations x

yields k!. Now, if k<n, the integral vanishes for

the same reason as above. If n = k, no diﬀerentiation will be left, and we have

I =

(−1)

−1

− 1)

dx.

628 Laplace’s Equation: Spherical Coordinates

Problem 26.14 shows how to evaluate the ﬁnal integral and obtain

−1

− 1)

dx =(−1)

2n+1

(n!)

(2n +1)!

Therefore,

I =

(−1)

−1

− 1)

dx =

n+1

(n!)

(2n +1)!

We summarize the above derivation

−1

(x) dx =

⎧

⎪

⎨

⎪

⎩

0ifk<n,

n+1

(n!)

(2n +1)!

if k = n.

(26.45)

If instead of x

we have a general polynomial of order k in x with n>k,the

integral will still vanish.



The result of the preceding example is summarized as

Box 26.5.1. Any polynomial of degree less than n is orthogonal to P

26.6 Expansions in Legendre Polynomials

The orthogonality of Legendre polynomials—as the orthogonality of the Fourier

trigonometric functions—makes them very useful for expansion of functions

deﬁned in the interval (−1, +1). Let f(x) be such a function. Then we write

f(x)=

∞



n=0

(x) (26.46)

and seek to ﬁnd c

.Butc

can be obtained by multiplying both sides of

the series by P

(x) and integrating from −1to+1. OntheLHS,weget



−1

f(x)P

(x) dx,andontheRHS

−1



∞



n=0

(x)



(x) dx =

∞



n=0

−1

(x)P

(x) dx



 

[2/(2n+1)]δ

by (26.43)

= c

2m +1

Equating the RHS and the LHS, we obtain

2m +1

−1

f(x)P

(x) dx or c

2n +1

−1

f(x)P

(x). (26.47)

Equations (26.46) and (26.47) give a procedure for expanding an arbitrary

function deﬁned in the interval (−1, +1) in terms of Legendre polynomials.

26.6 Expansions in Legendre Polynomials 629

If f (x) happens to be a polynomial of degree k, then it can be written

as a ﬁnite sum of Legendre polynomials of degree k and less. In fact, for

f(x)=x

,wehave

2n +1

−1

(x)dx =0 for n>k

by Box 26.5.1. Thus the coeﬃcients in the sum (26.46) beyond k are all zero.

Example 26.6.1.

We want to ﬁnd the Legendre expansion of a function f(x)

deﬁned as

f(x)=

⎧

⎨

⎩

if 0 <x≤ 1,

−V

if − 1 ≤ x<0.

To ﬁnd the coeﬃcients of expansion, we use Equation (26.47):

2n +1

−1

f(x)P

(x) dx

2n +1

−1

f(x)





=−V

(x) dx +

2n +1

f(x)





=+V

(x) dx (26.48)

2n +1



−

−1

(x) dx +

(x) dx



In the ﬁrst integral of the last line, we make the substitution x = −y so that

−1

(x) dx =

(−y)(−dy)=

(−y) dy =(−1)

(x) dx,

where we used (26.35) and, in the last equality, we changed the dummy variable of

integration from y to x (Section 3.2). Inserting this in (26.48), we obtain

2n +1

[1 − (−1)

]

(x) dx,

2n +1

⎧

⎨

⎩

0ifn is even



2k+1

(x) dx if n =2k +1

wherewehavewrittentheoddn as 2k +1 for k =0, 1,....

It remains to evaluate the integral of a Legendre polynomial of odd order in the

interval (0, 1). To this end, we use the Rodrigues formula:

2k+1

(x) dx =

2k+1

(2k +1)!

2k+1

− 1)

2k+1

(2k +1)!

− 1)

2k+1



2k+1

(2k +1)!

(

− 1)

2k+1



x=1

−

− 1)

2k+1



x=0

)

630 Laplace’s Equation: Spherical Coordinates

The ﬁrst term gives zero because there is no suﬃcient number of diﬀerentiations to

get rid of all factors of (x

− 1). For the second term, we note that (x

− 1)

2k+1

a polynomial in x whose derivatives of various orders consist of powers of x.These

powers will give zero at x = 0 except for the constant term (of zeroth power). So,

let us use binomial expansion for (x

− 1)

2k+1

which is equal to −(1 − x

)

2k+1

− 1)

2k+1



x=0

= −

2k+1



j=0

(2k +1)!

j!(2k +1− j)!

(−x

)



x=0

= −

2k+1



j=0

(2k +1)!

j!(2k +1− j)!

(−1)







x=0

whose constant term is obtained when k = j, all the other terms of the sum will

vanish either because of too many diﬀerentiations (when j<k, we end up diﬀeren-

tiating constants) or too few diﬀerentiations (when j>k,apowerofx will remain

which evaluates to zero at x = 0). Therefore,

− 1)

2k+1



x=0

= −

(2k +1)!

k!(k +1)!

(−1)







x=0

(2k +1)!

k!(k +1)!

(−1)

k+1

(2k)!

and

2k+1

(x) dx = −

2k+1

(2k +1)!



(2k +1)!

k!(k +1)!

(−1)

k+1

(2k)!



(−1)

(2k)!

2k+1

k!(k +1)!

(26.49)

Finally, we can write the coeﬃcient c

2k+1

2(2k +1)+1

2k+1

(x) dx =

(−1)

(4k + 3)(2k)!

2k+1

k!(k +1)!

with c

=0forevenn. The ﬁnal expansion series can now be given:

f(x)=

⎧

⎨

⎩

if 0 <x≤ 1

−V

if − 1 ≤ x<0

= V

∞



k=0

(−1)

(4k + 3)(2k)!

2k+1

k!(k +1)!

2k+1

(x)

= V

(x) −

(x)+

(x) −···



Example 26.6.2. We can easily obtain the Legendre expansion of the Dirac delta

function. The expansion coeﬃcients are given by

2n +1

−1

f(x)P

(x) dx =

2n +1

−1

δ(x)P

(x) dx =

2n +1

(0).

From Equation (26.39) and the discussion preceding it we can ﬁnd all values of

(0). Substituting these values in the above equation, we conclude that c

=0if

n is odd, and

4k +1



(−1)

(2k)!

(k!)



It now follows thatLegendre

expansion of the

Dirac delta

function

δ(x)=

∞



k=0

(−1)

(4k + 1)(2k)!

2k+1

(k!)

(x).



26.7 Physical Examples 631

26.7 Physical Examples

The most common physical problems involving Laplace’s equation are those

from electrostatics in empty space, and steady-state heat transfer. In each

case, a surface is held at some (not necessarily uniform) potential or tem-

perature and the potential or temperature is sought in regions away from

the surface. In the present context, these surfaces are typically (portions of)

spheres.

Example 26.7.1.

Two solid heat-conducting hemispheres of radius a, separated

by a very small insulating gap, form a sphere. The two halves of the sphere are two solid

heat-conducting

hemispheres held

at temperatures

and −T

in contact—on the outside—with two (inﬁnite) heat baths at temperatures T

and

−T

[Figure 26.1(a)]. We want to ﬁnd the temperature distribution T (r, θ, ϕ)inside

the sphere. We choose a spherical coordinate system in which the origin coincides

with the center of the sphere and the polar axis is perpendicular to the equatorial

plane. The hemisphere with temperature T

is assumed to constitute the northern

hemisphere.

Since the problem has azimuthal symmetry, T is independent of ϕ,andwecan

immediately write the general solution from Equation (26.29). However, since the

origin is in the region of interest, we need to exclude all negative powers of r.This

is accomplished by setting all the B coeﬃcients equal to zero. Thus, we have

T (r, θ)=

∞



n=0

(cos θ). (26.50)

It remains to calculate the constants A

. This is done by noting that

T (a, θ)=

⎧

⎪

⎨

⎪

⎩

if 0 ≤ θ<

−T

<θ≤ π.

In terms of u =cosθ, this is written as

T (a, u)=

⎧

⎨

⎩

−T

if − 1 ≤ u<0,

if 0 <u≤ 1.

Substituting this in Equation (26.50), we obtain

T (a, θ)=

⎧

⎨

⎩

−T

if − 1 ≤ u<0

if 0 <u≤ 1

∞



n=0



≡c

(u), (26.51)

which—except for using u instead of x—is entirely equivalent to the expansion of

Example 26.6.1, where we found that even coeﬃcients are absent and

2k+1

≡ A

2k+1

(−1)

(4k + 3)(2k)!

2k+1

k!(k +1)!

Finding A

2k+1

from this equation and inserting the result in (26.50) yields

T (r, θ)=T

∞



k=0

(−1)

(4k + 3)(2k)!

2k+1

k!(k +1)!





2k+1

(cos θ), (26.52)

where we have substituted cos θ for u.



632 Laplace’s Equation: Spherical Coordinates

(b)(a)

−T

−V

Figure 26.1: (a) Two heat-conducting hemispheres held at two diﬀerent temperatures.

(b) Two electrically conducting hemispheres held at two diﬀerent potentials. The upper

hemispheres have the polar angle range 0 ≤ θ<π/2 or 0 < cos θ ≤ 1,andthelower

hemispheres have the range π/2 <θ≤ π or −1 ≤ cos θ<0.

Example 26.7.2. Consider two electrically conducting hemispheres of radius a

two electrically

conducting

hemispheres held

at potentials V

and −V

separated by a small insulating gap at the equator. The upper hemisphere is held

at potential V

and the lower one at −V

as shown in Figure 26.1(b). We want to

ﬁnd the potential at points outside the resulting sphere. Since the potential must

vanish at inﬁnity, we expect the ﬁrst term in Equation (26.29) to be absent, i.e.,

= 0. To ﬁnd B

, substitute a for r in (26.29), and let cos θ ≡ u. Then,

Φ(a, u)=

∞



k=0

k+1



≡c

(u),

where

Φ(a, u)=

−V

if − 1 <u<0,

if 0 <u<1.

The calculation of the coeﬃcients is identical to that of Example 26.6.1. Thus,

=0forevenk and

2m+1

2m+2

=(−1)

(4m + 3)(2m)!

2m+1

(m +1)!m!

2m+1

(−1)

(4m + 3)(2m)!

2m+1

m!(m +1)!

2m+2

Having found the coeﬃcients, we can write the potential:

Φ(r, θ)=V

∞



m=0

(−1)

(4m + 3)(2m)!

2m+1

m!(m +1)!





2m+2

2m+1

(cos θ), (26.53)

where cos θ has been restored. Equation (26.53) is the multipole expansion of the

potential of the two hemispheres. It is interesting to note that the monopole term

26.7 Physical Examples 633

(the term with a single power of r in the denominator) is absent. It follows from

Equation (10.33) that the total charge on the two spheres must be zero. This is

consistent with the symmetry of the problem from which we expect equal surface

charge densities of opposite signs on the two hemispheres.



Example 26.7.3. As yet another example of the solution of Laplace’s equation

in spherical coordinates, consider a grounded neutral conducting sphere of radius conducting sphere

in an originally

uniform electric

ﬁeld

a placed in an originally uniform electric ﬁeld E

which is assumed to be inﬁnite

in extent (see Figure 26.2). We want to ﬁnd the electrostatic potential everywhere

outside the sphere. Choosing the ﬁeld to be in the positive z-direction and placing

the center of the sphere at the origin, we will have a problem that is azimuthally

symmetric. The general solution is therefore given by Equation (26.29). The bound-

aries outside the sphere consist of the sphere itself as well as inﬁnity. The electric

ﬁeld at inﬁnity is the original uniform ﬁeld, because the ﬁeld due to the charges

induced on the sphere vanishes at inﬁnity. The potential of this ﬁeld (at inﬁnity)

can be deduced from

E = E

= −∇Φ ⇒ E

= −

∂Φ

∂z

∂Φ

∂x

=0=

∂Φ

∂y

Thus, the potential at inﬁnity is independent of x and y,andcanbewrittenas

Φ(r, θ)=−E

z = −E

r cos θ = −E

(cos θ)forr →∞.

As r →∞,theB terms in Equation (26.29) will go to zero, and we must have the

“limiting” equality

∞



k=0

(u) →−E

(u).

Figure 26.2: The electric ﬁeld in the vicinity of a sphere placed in an external uniform

ﬁeld will change, but the ﬁeld far away from the sphere will remain almost uniform.

We could express the gradient in terms of spherical coordinates, but, as the reader will

note, the initial manipulation is noticeably easier in the Cartesian coordinates.

634 Laplace’s Equation: Spherical Coordinates

The orthogonality of the Legendre polynomials requires the coeﬃcients on both sides

to be equal. This gives

=0 fork =0andk ≥ 2,A

= −E

The B’s are obtained by applying the boundary condition of the sphere itself,

namely the fact that it is grounded. This means that Φ(a, θ) = 0, or

0=A

(u)+

∞



k=0

k+1

(u)=



− E



(u)+

∞



k=2

k+1

(u).

Again orthogonality of the Legendre polynomials requires the coeﬃcient of each

polynomial to vanish. This yields

=0,B

= E

, and B

=0 for k ≥ 2.

Inserting all these coeﬃcients in Equation (26.29), we obtain

Φ(r, θ)=−E



r −



(cos θ). (26.54)

Because of the simplicity of the expression for potential, we can evaluate some

other physical quantities of interest. For example, the electric ﬁeld at all points in

space is

E = −∇Φ=−

∂Φ

∂r

−

∂Φ

∂θ

= −

∂Φ

∂r

= E



1+2



cos θ,

= −

∂Φ

∂r

= −E



1 −



sin θ,

=0.

This is the sum of the original uniform ﬁeld

(cos θ

− sin θ

)=E

and the ﬁeld due to the charges induced on the sphere

sph

(2 cos θ

+sinθ

which (see Example 16.2.1) is the ﬁeld of an electric dipole with dipole moment

p =

=4π

It is interesting to note that at r = a, the only nonvanishing component of

the ﬁeld is E

. This is consistent with the known fact that electrostatic ﬁelds

are perpendicular to conducting surfaces. Furthermore, this perpendicular ﬁeld is

related to the surface charge density by E = σ/

. Therefore,

σ = 



r=a

=3

cos θ,

indicating an accumulation of positive charge on the “upper” (right in the ﬁgure)

hemisphere and an identical distribution of negative charge on the “lower” (left in

the ﬁgure) hemisphere.



26.8 Problems 635

26.8 Problems

26.1. Show that by writing P (u) ≡ Θ(θ)—with u =cosθ—and using the

chain rule, the second equation of (26.2) becomes

−

sin θ

dθ



(1 −u

)



+ αP =0.

26.2. Choose a solution of the form u



∞

n=0

for the Legendre DE,

assume that a

and a

are both nonzero, and show that the only solution for

r is r =0.

26.3. Derive Equation (26.15).

26.4. Derive Equations (26.18), (26.19), and (26.20).

26.5. Derive Equations (26.21) and (26.22) and show that they can both be

written as (26.23).

26.6. Show by mathematical induction (or otherwise) that Equation (26.24)

satisﬁes P

(1) = 1.

26.7. Show that Legendre polynomials and the hypergeometric function are

related via (26.25) and (26.26).

26.8. Suppose that Q represents electric charge. Show that in (26.34) Q

is the total charge and Q

is the dot product of

and the electric dipole

moment.

26.9. (a) Change t to −t and u to −u, and show that the generating function

g(t, u) of Legendre polynomials does not change.

(b) Now substitute −t for t and −u for u in Equation (26.30) and compare the

resulting equation with (26.30) to derive the parity of Legendre polynomials.

26.10. (a) Show that (1/t)[ln(1 + t) −ln(1 −t)] is an even function of t.

(b) Use the Maclaurin expansion of ln(1 ± t) to derive the following series:

[ln(1 + t) −ln(1 −t)] = 2

∞



k=0

2k +1

26.11. (a) Show that P

(0) = 0 if n is odd.

(b) Show that for u = 0, Equation (26.38) yields

(0) = −

2n − 1

2n−2

(0).

(0) = (−1)

(2n −1)!!

(2n)!!

(0) = (−1)

(2n −1)!!

(2n)!!

Now use the result of Problem 11.1 to obtain the ﬁnal form of (26.39).

636 Laplace’s Equation: Spherical Coordinates

26.12. Suppose f (x)=



∞

k=0

(x). Show that

−1

[f(x)]

dx =

∞



m=0

2m +1

26.13. Show the following two equalities:

(1 −z

)

− 2z

+ n(n +1)

n +1

(2πi)

(ξ

− 1)

[nξ

− 2(n +1)ξz + n +2]

(ξ − z)

n+3

dξ

n +1

(2πi)

dξ



(ξ

− 1)

n+1

(ξ − z)

n+2



dξ.

26.14. In the integral



−1

− 1)

dx,letu =(x

− 1)

and dv = dx and

integrate by parts to show that

−1

− 1)

dx = −2n

−1

− 1)

dx.

Integrate by parts a few more times and show that

−1

− 1)

dx =(−2)

n(n −1) ...(n −m +1)

(2m −1)!!

−1

− 1)

n−m

dx.

Set m = n and, using the result of Problem 11.1, obtain the following ﬁnal

result:

−1

− 1)

dx =(−1)

2n+1

(n!)

(2n +1)!

26.15. Use the procedures of Example 26.5.4 and the previous problem to

show that for m ≥ n:

−1

(x) dx =

if m and n have opposite parities,

(m+n)/2+1

m!(

m+n

(m−n)!(m+n+1)!

if m and n have the same parities,

where having the same parity means being both even or both odd.

26.16. Show that



(x) dx =0ifk ≥ 1. Hint: Extend the interval of

integration to (−1, 1) and use the orthogonality of Legendre polynomials.

26.17. Find the Legendre expansion for the function f(x)=|x| in the interval

(−1, +1). Hint: Break up the integrals into two pieces, employ the recurrence

relation to express xP

(x)intermsofP

n−1

(x)andP

n+1

(x), and use the

result of Example 26.6.1.

26.18. (a) Find the total charge on the upper and lower hemispheres and on

the entire sphere of Example 26.7.3.

(b) Using p =





dq(r



), calculate the (induced) dipole moment of the sphere.