chapter 3 boundary-value problems: part iivvanchur/phys5511/chapter3.pdf · 2017-05-01 · chapter...

Chapter 3

Boundary-Value Problems:

Part II

Problem Set #3: 3.1, 3.13, 3.17 (Due Monday March. 11th)

3.1 Spherical Coordinates

Spherical coordinates are used when boundary conditions have spherical sym-metry. The Laplace equation in spherical coordinates takes the followingform,

∇2Φ =1

r

∂2

∂r2(rΦ) +

1

r2 sin θ

∂

∂θ

(

sin θ∂Φ

∂θ

)

+1

r2 sin2 θ

∂2Φ

∂φ2= 0. (3.1)

To use the separation of variables method we can plug in the following ansatz

Φ (r, θ,φ) =U(r)

rP (θ)Q(φ) (3.2)

into (3.1) to yield

P (θ)Q(φ)1

r

d2

dr2U(r)+

U(r)

rQ(φ)

1

r2 sin θ

d

dθ

(

sin θd

dθP (θ)

)

+U(r)

rP (θ)

1

r2 sin2 θ

d2

dφ2Q(φ) = 0

(3.3)or after multiplying by r3 sin2 θ

U(r)P (θ)Q(φ) ,

r2 sin2 θ

U

d2

dr2U +

sin θ

P

d

dθ

(

sin θd

dθP

)

= −1

Q

d2

dφ2Q. (3.4)

Since the right hand side depends only on φ and the left hand side does notthere must be some constant m2 such that

−1

Q

d2

dφ2Q = m2. (3.5)

24

CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 25

andr2 sin2 θ

U

d2

dr2U +

sin θ

P

d

dθ

(

sin θd

dθP

)

= m2 (3.6)

Equation (3.5) has solutions

Q = Cm eimφ (3.7)

where m must be an integer for Q to be single valued.Similarly we can separate variables θ and r in (3.6) to get

r2

U

d2

dr2U =

m2

sin2 θ−

1

P sin θ

d

dθ

(

sin θd

dθP

)

(3.8)

orr2

U

d2

dr2U = l(l + 1) (3.9)

and

m2

sin2 θ−

1

P sin θ

d

dθ

(

sin θd

dθP

)

= l(l + 1) (3.10)

for some constant l(l + 1). These two equation can be rewritten as

d2

dr2U = l(l + 1)

U

r2(3.11)

and

1

sin θ

d

dθ

(

sin θd

dθP

)

=

(

m2

sin2 θ− l(l + 1)

)

P (3.12)

Equation (3.11) can be solve using polynomial ansatz

U ∝ rα (3.13)

whereα(α− 1) = l(l + 1) (3.14)

or

α =

l + 1

−l(3.15)

Thus, the most general solution of (3.11) is given by

U = Alrl+1 +Blr

−l. (3.16)

To determine l we switch to (3.12) rewritten in terms of x = cos θ,

d

dx

(

(1− x2)d

dxPl

)

=

(

m2

1− x2− l(l + 1)

)

Pl. (3.17)

This is the generalized Legendre equation whose solutions are the associatedLegendre functions.


3.2 Legendre polynomials

For starters let us consider a special casem = 0 corresponding to the ordinaryLegendre equation

d

dx

(

(1− x2)d

dxPl

)

+ l(l + 1)Pl = 0 (3.18)

If we assume the following expansion of

Pl(x) = xα∞∑

j=0

ajxj (3.19)

where a0 = 0, then we obtain

∞∑

j=0

[

(α + j) (α + j − 1) ajxα+j−2 − ((α+ j) (α + j + 1)− l (l + 1)) ajx

α+j]

=(3.20)

α(α− 1)a0xα−2 + (1 + α)αa1x

α−1 +

+∞∑

j=0

[(α + j + 2) (α+ j + 1) aj+2 − ((α + j) (α+ j + 1)− l (l + 1)) aj] xα+j = 0.

Since the coefficient of all of the powers of x have to vanish separately, wefind

α (α− 1) = 0 for a0 = 0

α (α + 1) = 0 for a1 = 0 (3.21)

and

aj+2 =(α + j) (α + j + 1)− l (l + 1)

(α+ j + 2) (α + j + 1)aj. (3.22)

The two equation (3.21) are redundant and the only condition that followsfrom them is α = 0 or 1. Then the expansion (3.19) has either odd (forα = 1) or even (for α = 0) powers of x.

One can show that the series expansion converges for x = 1 only if itterminates. Since α and j are positive integers or zero the relation (3.22)terminates only if l is a positive integer or zero. Then only one of the twoseries terminates when

(jmax + α)(jmax + α + 1)− l(l + 1)

(jmax + α+ 2)(jmax + α + 1)= 0 ⇒

jmax = l for α = 0

jmax = l − 1 for α = 1(3.23)


Thus,

Pl(x) =

∑l/2j=0 a2jx

2j where aj+2 =j(j+1)−l(l+1)(j+2)(j+1) aj for even l

∑(l−1)/2j=0 a2jx2j+1 where aj+2 =

(j+1)(j+2)−l(l+1)(j+3)(j+2) aj for odd l.

(3.24)For example, if we set Pl(1) = 1, then

P0(x) = 1

P1(x) = x

P2(x) =1

2(−1 + 3x2)

P3(x) =1

2(−3x+ 5x3)... (3.25)

We note (without proving) that the Legendre Polynomial have the fol-lowing properties

Pl(x) =1

2ll!

dl

dxl(x2 − 1)l (Rodrigues Formula) (3.26)

Pl+1(x) =2l + 1

l + 1Pl(x)−

l

l + 1Pl−1(x) (Recurrence Relation #1) (3.27)

Pl(x) =1

2l + 1

d

dx(Pl+1(x)− Pl−1(x)) (Recurrence Relation #2) (3.28)

∫ 1

−1

Pl(x) Pl′(x) =2

2l + 1δll′ (Orthogonality condition) (3.29)

The last property implies that the Legendre polynomials form a completeset of orthogonal function on the interval (−1, 1) and any function can beexpanded as

f(x) =∞∑

l=0

AlPl(x) where Al =2l + 1

2

∫ 1

−1

f(x)Pl(x)dx. (3.30)

Then the general solution of the Laplace equation in spherical coordinatesfor the case of azimuthal symmetry (i.e. m = 0) is given by

Φ(r, θ,φ) =∞∑

l=0

(Alrl +Blr

−l−1)Pl(cos θ). (3.31)

3.3 Azimuthal symmetry

Consider a sphere of radius a with potential V (θ) on its surface.To find thepotential everywhere inside the sphere we note that the potential should not


blow up at r = 0 and thus the Bl’s in (3.31) must vanish. Then our problemreduces to finding the appropriate coefficients in

Φ(r, θ,φ) =∞∑

l=0

AlrlPl(cos θ) (3.32)

for boundary conditions

V (θ) =∞∑

l=0

AlalPl(cos θ). (3.33)

Multiplying both sides of (3.33) by Pl′(cos θ) sin θ and integrating over θ leadsto

∫ π

0

V (θ)Pl′(cos θ) sin θdθ =

∞∑

l=0

Alal

∫ π

0

Pl(cos θ)Pl′(cos θ) sin θdθ =

∞∑

l=0

Alal

∫ 1

−1

Pl(x)Pl′(x)dx = Al′al′ 2

2l′ + 1(3.34)

or

Al =2l + 1

2al

∫ π

0

V (θ)Pl(cos θ) sin θdθ. (3.35)

For instance the potential inside of two hemispherical shells of radius aheld at different potentials, i.e.

V (θ) =

+V for 0 ≤ θ < π/2

−V for π/2 < θ ≤ π,(3.36)

can be estimated from

Φ(r, θ) = V

[

3

2

r

aP1(cos θ)−

7

8

(r

a

)3P3(cos θ) +

11

16

(r

a

)5P5(cos θ)...

]

.

(3.37)For the potential outside of the sphere (r/a)l would be replaced by (a/r)l+1.In fact this solution could have been obtained from an already known solution(2.29) for θ = 0. Since

Φ(r, 0) =∞∑

l=0

(Alrl +Blr

−l−1) (3.38)


and

Φ(r, 0) = V

(

1−r2 − a2

r√r2 + a2

)

=V√π

∞∑

j=1

(−1)j−1

(

2j − 12

)

Γ(

j − 12

)

j!

(a

r

)2j

(3.39)we get

Φ(r, θ) =V√π

∞∑

j=1

(−1)j−1

(

2j − 12

)

Γ(

j − 12

)

j!

(a

r

)2jP2j−1 (cos θ) (3.40)

which can be shown to be equivalent to (3.37).Another important problem with azimuthal symmetry that can be solved

by the same method is the potential due to a unit point charge at x′ placedanywhere on z axis. Then, the Taylor series expansion for x also on the zaxis is given by

1

|x− x′|=

1

|r − r′|=

1

r>

∞∑

l=0

rl<rl>

(3.41)

where r< (or r>) is the smaller (or larger) of r and r′, and thus,

1

|x− x′|=

∞∑

l=0

rl<rl+1>

Pl(cos θ). (3.42)

3.4 Fields in a Conical Hole

Consider a conical hole in a conductor that has angle β relative z axis. Tofind the potential inside the hole we are restricted to the region of interestin spherical coordinates: 0 < θ < β and 0 < φ < 2π. The problem has anazimuthal symmetry, but the solutions are not given by Legendre polynomialssince we can no longer expand around θ = π/2 (or x = 0 in (3.18)). For thisparticular problem it makes sense to look for an expansion around θ = 0 (orx = 1 in (3.18)), and thus it is convenient to change variables

y =1

2(1− x) (3.43)

d

dy= −2

d

dx(3.44)

in (3.18) to rewrite the equation as

d

dy

(

y(1− y)d

dyPν

)

+ ν(ν + 1)Pν = 0. (3.45)


We have switched from subscript l to ν to emphasize the fact that the func-tions Pν(y) are not Legendre polynomials. Just like before we can plug in aseries expansion

Pν(y) = yα∞∑

j=0

ajyj (3.46)

into (3.45) to deduce that vanishing of a coefficient of the lowest power of yrequires α = 0 and the recursion relation between successive coefficients is

aj+1 =(j − ν) (j + ν + 1)

(j + 1)2aj . (3.47)

Moreover, if we choose a normalization corresponding to a0 then we obtainour final solution

Pν(y) = 1 +(−ν)(ν + 1)

1!1!y +

(−ν)(−ν + 1)(ν + 1)(ν + 2)

2!2!y2 + ... (3.48)

or

Pν(x) = 1+(−ν)(ν + 1)

1!1!

(

x− 1

2

)

+(−ν)(−ν + 1)(ν + 1)(ν + 2)

2!2!

(

x− 1

2

)2

+....

(3.49)These are the so-called Legendre functions of the first kind and order ν,where ν ′s are not necessarily integers. Their values are determined from theboundary conditions Φ(θ = β) = 0 corresponding to

Pν(cos β) = 0, (3.50)

and the most general solution is given by

Φ(r, θ,φ) = A0 +∞∑

k=1

AkrνkPνk(cos θ) where Pνk(cos β) = 0. (3.51)

Deep inside the hole (r ≈ 0) of a grounded conductor (A0 = 0) only theleading term would dominate, i.e.

Φ(r, θ,φ) = Arν1Pν1(cos θ), (3.52)

where

ν1 =

2.405β − 1

2 for small β

2 ln(

2π−β

)

for large β.(3.53)


3.5 Spherical Harmonics

Let us now consider the Laplace equation in spherical coordinates withoutazimuthal symmetry which boils down to solving the generalized Legendreequation for m = 0

d

dx

(

(1− x2)d

dxPl

)

=

(

m2

1− x2− l(l + 1)

)

Pl. (3.54)

It can be shown that for the equation to be finite between 1 and −1 theparameter l must be positive integer and

m = −l,−l + 1, ...., l − 1, l. (3.55)

Then the solutions of (3.54) are given by

Pml (x) =

(−1)m(1− x2)m/2 dm

dxmPl(x) for m ≥ 0

(−1)m (l−m)!(l+m)!Pl(x) for m < 0

(3.56)

Another definition of Pml ’s is obtained from the Rodrigues formula (3.26)

and is valid for positive as well as negative m,

Pml (x) =

(−1)m

2ll!(1− x2)m/2 dl+m

dxl+m(x2 − 1)l. (3.57)

The associated Legendre function form a set of orthogonal functions onthe interval (−1, 1),

∫ 1

−1

Pml (x)Pm

l′ (x)dx = δll′2

2l + 1

(l +m)!

(l −m)!. (3.58)

It is now convenient to define a set of orthonormal functions

Ylm(θ,φ) =

√

2l + 1

4π

(l −m)!

(l +m)!Pml (cos θ)eimφ (3.59)

withYl,−m(θ,φ) = (−1)mY ∗

lm(θ,φ). (3.60)

which also satisfy the orthogonality and normalization conditions

∫ 2π

0

dφ

∫ π

0

dθ sin θ Y ∗l′m′(θ,φ)Ylm(θ,φ) = δll′δmm′ (3.61)


and the completeness relation

∞∑

l=0

l∑

m=−l

Y ∗lm(θ

′,φ′)Ylm(θ,φ) = δ(φ− φ′)δ(cos θ − cos θ′). (3.62)

These function are known as spherical harmonics as they represent the angular-dependent part of the solutions of the Laplace equation in spherical coordi-nates (θ,φ). For example,

Y00 =1√4π

Y11 = −√

3

8πsin θ eiφ

Y10 =

√

3

8πcos θ

Y1,−1 =

√

3

8πsin θ e−iφ, (3.63)

Note that for m = 0 the spherical harmonic reduce to the Legendre polyno-mials

Yl0(θ,φ) =

√

2l + 1

4πPl(cos θ) (3.64)

and for m = l

Ymm(θ,φ) =(−1)m

2mm!

√

(2m+ 1)(2m)!

4πeimφ sinm θ. (3.65)

Since the spherical harmonics form a complete orthonormal set of func-tion, an arbitrary (square-integrable) function on a unit sphere can be ex-panded as,

g(θ,φ) =∞∑

l=0

l∑

m=−l

AlmYlm(θ,φ), (3.66)

where the coefficients are given by

Alm =

∫

dΩY ∗lm(θ,φ)g(θ,φ), (3.67)

and the most general solution to the Laplace equation can also be expandedin spherical harmonics as

Φ(r, θ,φ) =∞∑

l=0

l∑

m=−l

(

Almrl +Blmr

−(l+1))

Ylm(θ,φ). (3.68)


It will be useful to prove the so-called addition theorem for sphericalharmonics which expresses a Legendre polynomial of order l and angle γ interms of spherical harmonics of angles (θ,φ) and (θ′,φ′) where

cos γ = cos θ cos θ′ + sin θ sin θ′ cos(φ− φ′). (3.69)

Clearly, the expansion (3.66) implies that

Pl(cos γ) =∞∑

l′=0

l′∑

m=−l′

Al′m(θ′,φ′)Yl′m(θ,φ). (3.70)

If θ′ = 0, corresponding to (θ′,φ′) lying on the z axis, then cos γ = cos θ andthen Pl(cos γ) must satisfy

1

sin θ

d

dθ

(

sin θd

dθPl(cos γ)

)

+ l(l + 1)Pl(cos γ) = 0 (3.71)

and, since the Laplacian operator ∇2ψ = 1r2

1sin θ

ddθ

(

sin θ ddθψ)

is a scalar whichmust be invariant under rotations, we have

∇2Pl(cos γ) +l(l + 1)

r2Pl(cos γ) = 0 (3.72)

for arbitrary θ′. Therefore Pl(cos γ) must be expressible in terms of sphericalharmonics of order l, i.e.

Pl(cos γ) =l∑

m=−l

Am(θ′,φ′)Ylm(θ,φ) (3.73)

with coefficients

Am(θ′,φ′) =

∫

dΩY ∗lm(θ,φ)Pl(cos γ). (3.74)

To evaluate the integral of (3.74) we first expand Y ∗lm(θ,φ) in spherical

harmonic for spherical coordinates (γ, β), i.e.

Y ∗lm(θ,φ) =

l∑

m′=−l

Bmm′Ylm′(γ, β) (3.75)

where

Bmm′ =

∫

dΩγ,βY∗lm(θ,φ)Y

∗lm′(γ, β) (3.76)


and in particular

Bm0 =

∫

dΩγ,βY∗lm(θ,φ)Y

∗l0(γ, β) =

=

∫

dΩγ,βY∗lm(θ,φ)

√

2l + 1

4πPl (cos γ) (3.77)

and thus

Bm0(θ′,φ′) =

√

2l + 1

4πAm(θ

′,φ′). (3.78)

However in the limit γ → 0 equation (3.75) reducs to

Y ∗lm(θ,φ) =

l∑

m′=−l

Bmm′Ylm′(0, β)

=l∑

m′=−l

Bmm′δm′0

√

2l + 1

4π

= Bm0

√

2l + 1

4π(3.79)

By combining (3.78) and (3.79) we get

Am(θ′,φ′) =

4π

2l + 1Y ∗lm(θ

′,φ′).

which can now be substituted into (3.73) to obtain the addition theorem forspherical harmonics

Pl(cos γ) =4π

2l + 1

l∑

m=−l

Y ∗lm(θ

′,φ′)Ylm(θ,φ). (3.80)

This result can, for example, be used to express the potential at x due to aunit point charge at x′. By substituting (3.80) into (3.42) we obtain

1

|x− x′|= 4π

∞∑

l=0

l∑

m=−l

1

2l + 1

rl<rl+1>

Y ∗lm(θ

′,φ′)Ylm(θ,φ). (3.81)

3.6 Bessel functions

The cylindrical coordinates (ρ,φ, z) are used when the boundary surface has acylindrical symmetry, but the boundary conditions are not uniform along the


z axis. The Laplace equation in cylindrical coordinates takes the followingform

∇2Φ =1

ρ

∂

∂ρ

(

ρ∂

∂ρΦ

)

+1

ρ2∂2Φ

∂φ2+∂2Φ

∂z2= 0. (3.82)

As usual we apply the separation of variables technique and substitute

Φ(ρ,φ, z) = R(ρ)Q(φ)Z(z) (3.83)

to obtain the three ordinary differential equations

d2Z

dz2− k2Z = 0 (3.84)

d2Q

dφ2+ ν2Q = 0 (3.85)

d2R

dρ2+

1

ρ

dR

dρ+

(

k2 −ν2

ρ2

)

R = 0. (3.86)

Solutions of the first two equations (3.84) and (3.85) are given by

Z(z) =

A0 +B0z for k = 0

Akekz +Bke−kz for k = 0(3.87)

Q(φ) =

C0 +D0φ for ν = 0

Cνeiνφ +Dνe−iνφ for ν = 0(3.88)

and the solutions of (3.86) will be derived using series expansion. As we willsee shortly that by choosing a real k the solutions will be given in terms ofBessel function, when a chose of imaginary k would have led to solutions interms of modified Bessel functions.

In terms of variable x = kρ equation (3.86) takes the form of the Besselequation

1

x

d

dx

(

xdR(x)

dx

)

+

(

1−ν2

x2

)

R(x) = 0. (3.89)

If we substitute expansion

R(x) =∞∑

j=0

ajxj+α, (3.90)

where a0 = 0, into (3.89) then

∞∑

j=0

ajxj+α−2

(

(j + α)2 − ν2)

+∞∑

j=0

ajxj+α = 0. (3.91)


Since coefficient of every power of x must vanish independently we get

a0(α2 − ν2) = 0 (3.92)

a1((1 + α)2 − ν2) = 0 (3.93)

and a recursive relation

aj+2 = −1

(j + 2 + α)2 − ν2aj. (3.94)

It follows thatα = ±ν (3.95)

and

a2j+1 = 0 (3.96)

a2j = −1

4j(j + α)a2j−2. (3.97)

If we choose

α0 =1

2αΓ(α+ 1)(3.98)

then the recursion relation leads to th following solutions

Jν(x) =(x

2

)ν∞∑

j=0

(−1)j

j!Γ(j + ν + 1)

(x

2

)2j

(3.99)

J−ν(x) =(x

2

)−ν∞∑

j=0

(−1)j

j!Γ(j − ν + 1)

(x

2

)2j(3.100)

known as Bessel functions of the first kind of order ±ν. These two solutionsare linearly independent only when ν is not an integer and otherwise

J−ν(x) = (−1)νJν(x). (3.101)

In either case one can construct another linearly independent solutions usingthe Neumann function (or Bessel function of the second kind)

Nν(x) ≡Jν(x) cos(νπ)− J−ν(x)

sin(νx). (3.102)

Alternatively, one can use the Hankel functions (or Bessel function of thethird kind) defined as linear combinations of Bessel and Neumann functions,

H(1)ν (x) ≡ Jν(x) + iNν(x) (3.103)

H(1)ν (x) ≡ Jν(x)− iNν(x) (3.104)


which also form a fundamental set of solutions to the Bessel equation (3.89).

We note that all of these functions, i.e. Ων = Jν , Nν , H(1)ν or H(2)

ν satisfy veryuseful recursion formulas,

Ων−1(x) + Ων+1(x) =2ν

xΩν(x) (3.105)

Ων−1(x)− Ων+1(x) = 2d

dxΩν(x). (3.106)

One can use Bessel function to expand an arbitrary function on the interval0 ≤ ρ ≤ a into Fourier-Bessel series

f(ρ) =∞∑

n=1

AνnJν(xνnρ

a) (3.107)

where xνn are the roots ofJν(x) = 0 (3.108)

and

Aνn =2

a2J2ν+1(xνn)

∫ a

0

ρf(ρ)Jν

(

xνnρ

a

)

. (3.109)

Although the expansion in terms of√ρJν(xνn

ρa) is very useful for homoge-

neous Dirichlet boundary conditions an alternative expansion in terms of√ρJν(yνn

ρa) , where yνn are the roots of

d

dxJν(x) = 0, (3.110)

is useful for homogeneous Neumann boundary conditions. Some other ex-pansions are given by

∞∑

n=0

anJν+n(x) Neumann series (3.111)

∞∑

n=0

anJν+n((ν + n)x) Kapteyn series (3.112)

∞∑

n=1

anJν(n+ x) Schlomilch series. (3.113)


3.7 Cylindrical Coordinates

By summing over all of the solutions to the Laplace equation of the form(3.83) we obtain

Φ(ρ,φ, z) = (A00 +B00z)(C00 +D00φ)(E00 + F00 log ρ) + (3.114)

+∑

ν =0

(Aν0 +Bν0z)(Cν0eiνφ +Dν0e

−iνφ)(Eν0ρν + Fν0ρ

−ν) +

+∑

k =0

(A0kekz +B0ke

−kz)(C0k +D0kφ)(E0kJ0(kρ) + F0kN0(kρ)) +

+∑

k =0

∑

ν =0

(Aνkekz +Bνke

−kz)(Cνkeiνφ +Dνke

−iνφ)(EνkJν(kρ) + FνkNν(kρ)).

As an example we can consider a cylinder with radius a and height L withboundary conditions

Φ(ρ,φ, z) =

⎧

⎪

⎨

⎪

⎩

0 for ρ = a

0 for z = 0

V (ρ,φ) for z = L.

(3.115)

If we are to solve the Laplace equation for the interior of the cylinder, thenthe solutions which blow up at ρ = 0 must vanish. Moreover the solutionsmust be matched at φ = 0 and φ = 2π. This leads to

Φ(ρ,φ, z) = (A00 +B00z) + (3.116)

+∑

ν =0

(Aν0 +Bν0z)(Cν0eiνφ +Dν0e

−iνφ)ρν +

+∑

k =0

(A0kekz +B0ke

−kz)(C0k +D0kφ)J0(kρ) +

+∑

k =0

∑

ν =0

(Aνkekz +Bνke

−kz)(Cνkeiνφ +Dνke

−iνφ)Jν(kρ).

The boundary condition Φ(ρ,φ, z = 0) = 0 implies that

0 = A00 + (3.117)

+∑

ν =0

Aν0(Cν0eiνφ +Dν0e

−iνφ)ρν +

+∑

k =0

(A0k +B0k)(C0k +D0kφ)J0(kρ) +

+∑

k =0

∑

ν =0

(Aνk +Bνk)(Cνkeiνφ +Dνke

−iνφ)Jν(kρ),


or

A00 = 0 (3.118)

Av0 = 0

A0k = −B0k

Aνk = −Bνk

and

Φ(ρ,φ, z) = B00z + (3.119)

+∑

ν =0

z(Cν0eiνφ +Dν0e

−iνφ)ρν +

+∑

k =0

(ekz − e−kz)(C0k +D0kφ)J0(kρ) +

+∑

k =0

∑

ν =0

(ekz − e−kz)(Cνkeiνφ +Dνke

−iνφ)Jν(kρ)

=∑

ν

z(Cν0eiνφ +Dν0e

−iνφ)ρν

∑

k =0

∑

ν

sinh(kz)(Cνkeiνφ +Dνke

−iνφ)Jν(kρ)

The boundary condition Φ(ρ = a,φ, z) = 0 implies that

0 =∑

ν

z(Cν0eiνφ +Dν0e

−iνφ)aν (3.120)

+∑

k =0

∑

ν

sinh(kz)(Cνkeiνφ +Dνke

−iνφ)Jν(ka)

or

Cν0 = 0 (3.121)

Dν0 = 0

and ifJm(kmna) = 0 (3.122)

defines kmn’s from its roots kmna then

Φ(ρ,φ, z) =∞∑

m=0

∞∑

n=1

sinh(kmnz) (Amn sin(mφ) +Bmn cos(mφ))Jν(kmnρ).

(3.123)


The remaining boundary condition fixes the coefficients Amn and Bmn using(2.42) and (3.109),

Amn =2

sinh(kmnL)πa2J2m+1(kmna)

∫ 2π

0

dφ

∫ a

0

dρ ρV (ρ,φ)Jm(kmnρ) sin(mφ)

Bmn =2

sinh(kmnL)πa2J2m+1(kmna)

∫ 2π

0

dφ

∫ a

0

dρ ρV (ρ,φ)Jm(kmnρ) cos(mφ)

with a correction B0n → 12B0n due to 1

2 coefficient in (2.41).How would the solution (3.123) change if a → ∞ ?

3.8 Spherical expansion of Green’s functions

If the problem involves the distributions of charges AND boundary condi-tions, then it is important to be able to express the Green functions usingexpansion appropriate for the boundary conditions. For example, considera sphere of radius a, with a specified potential V (φ, θ) at its surface, and apoint charge. This problem was already solved using the method of Green’sfunctions,

Φ(x) =1

4πϵ0

∫

ρ(x′)GDd3x′ −

1

4π

∫

Φ∂

∂n′GDda′, (3.124)

such that GD = 0 at the surface of the sphere, i.e.

GD(x,x′) =

1

|rx− r′x′|−

1

| r′ra x− ax′|. (3.125)

It was already shown (3.81) that

1

|rx− r′x′|=

4π∑∞

l=0

∑lm=−l

12l+1

rl

r′l+1Y ∗lm(θ

′,φ′)Ylm(θ,φ) for r < r′

4π∑∞

l=0

∑lm=−l

12l+1

r′l

rl+1Y ∗lm(θ

′,φ′)Ylm(θ,φ) for r′ < r(3.126)

and the expansion of (3.125) in spherical harmonics is give by,

GD(x,x′) =

⎧

⎨

⎩

4π∑∞

l=0

∑lm=−l

12l+1

(

rl

r′l+1 − a2l+1

(rr′)l+1

)

Y ∗lm(θ


4π∑∞

l=0

∑lm=−l

12l+1

(

r′l

rl+1 − a2l+1

(rr′)l+1

)

Y ∗lm(θ

′,φ′)Ylm(θ,φ) for r′ < r,

(3.127)for the exterior of the sphere and

GD(x,x′) =

⎧

⎨

⎩

4π∑∞

l=0

∑lm=−l

12l+1

(

rl

r′l+1 − (rr′)l

a2l+1

)

Y ∗lm(θ


4π∑∞

l=0

∑lm=−l

12l+1

(

r′l

rl+1 − (rr′)l

a2l+1

)

Y ∗lm(θ

′,φ′)Ylm(θ,φ) for r′ < r.

(3.128)


for the interior of the sphere.For illustration let us find potential inside a sphere of radius b at potential

V (θ,φ) and a charge Q on a uniformly charged ring of radius a < b in thexy plane with center at x = y = 0, i.e.

ρ(x′) =Q

2πa2δ(r′ − a)δ(cos θ′). (3.129)

From (3.127)

∂GD

∂n′ =

[

∂GD

∂r′

]

r′=b

= −4π∞∑

l=0

l∑

m=−l

rl

bl+2Y ∗lm(θ

′,φ′)Ylm(θ,φ) (3.130)

and the second term in (3.124) is

−1

4π

∫

Φ∂

∂n′GDda′ =

∞∑

l=0

l∑

m=−l

rl

bl+2Ylm(θ,φ)

∫

dΩ′V (θ′,φ′)Y ∗lm(θ

′,φ′).

(3.131)To calculate the first term in (3.124) we also use the expansion (3.127) and(3.129)

1

4πϵ0

∫

ρ(x′)GDd3x′ =

Q

2πa2ϵ0

∫ 2π

0

dφ′∞∑

l=0

l∑

m=−l

1

2l + 1

(

rl<r>l+1

−(r>r<)

l

b2l+1

)

Y ∗lm(π/2,φ

′)Ylm(θ,φ)

(3.132)where r< (or r>) is the smaller (or larger) of r and a. But because of theazimuthal symmetry of the potential due to the ring only m = 0 terms wouldsurvive,

1

4πϵ0

∫

ρ(x′)GDd3x′ =

= dφ′∞∑

l=0

1

2l + 1

(

rl<r>l+1

−(r>r<)

l

b2l+1

)

Y ∗l0(π/2,φ

′)Yl0(θ,φ)

=Q

4πa2ϵ0

∞∑

l=0

(

rl<r>l+1

−(r>r<)

l

b2l+1

)

Pl(0)Pl(cos θ) (3.133)

and the final answer is

Φ(x) = (3.134)

=Q

4πa2ϵ0

∞∑

l=0

(

rl<r>l+1

−(r>r<)

l

b2l+1

)

Pl(0)Pl(cos θ) +

+∞∑

l=0

l∑

m=−l

rl

bl+2Ylm(θ,φ)

∫

dΩ′V (θ′,φ′)Y ∗lm(θ

′,φ′). (3.135)


3.9 Cylindrical expansion of Green’s functions

For the boundary value problems with cylindrical symmetry it is useful toexpand the Green’s function in cylindrical coordinates. This corresponds tosolving the Laplace equation in cylindrical coordinates, i.e.

∇2G(x,x′) = −4π

ρδ(ρ− ρ′)δ(φ− φ′)δ(z − z′), (3.136)

where

δ(z − z′) =1

2π

∫ ∞

−∞dk eik(z−z′) =

1

π

∫ ∞

0

dk cos(k(z − z′)) (3.137)

δ(φ− φ′) =1

2π

∞∑

m=−∞

eim(φ−φ′). (3.138)

The Green’s function can be expanded as

G(x,x′) =1

2π2

∞∑

m=−∞

∫ ∞

0

dk eim(φ−φ′) cos(k(z − z′)) gm(k, ρ, ρ′), (3.139)

where the functions gm(k, ρ, ρ′) must satisfy

1

ρ

d

dρ

(

ρdgmdρ

)

−(

k2 +m2

ρ2

)

gm = −4π

ρδ(ρ− ρ′) (3.140)

For ρ = ρ′ the solutions of (3.140) are the modified Bessel functions

Im(kρ) ≡ i−mJm(ikρ) (3.141)

Km(kρ) ≡π

2iν+1H(1)

m (ikρ). (3.142)

Then the most general solution is a linear combination of (3.141) and (3.142),

ψ1(kρ) = aIm(kρ) + bKm(kρ) satisfies b.c. for ρ < ρ′ (3.143)

ψ2(kρ) = a′Im(kρ) + b′Km(kρ) satisfies b.c. for ρ′ < ρ (3.144)

but due to symmetry (ρ↔ ρ′) we must have

gm(k, ρ, ρ′) = ψ1(kρ<)ψ2(kρ>). (3.145)


Then (3.140) gives us

∫ ρ′+ϵ

ρ′−ϵ

d

(

ρdgmdρ

)

−∫ ρ′+ϵ

ρ′−ϵ

(

k2ρ+m2

ρ

)

gm dρ = −4π

∫ ρ′+ϵ

ρ′−ϵ

δ (ρ− ρ′) dρ

[

ρdgmdρ

]ρ=ρ′+ϵ

ρ=ρ′−ϵ

= −4π

[

dgmdρ

]

ρ=ρ′+ϵ

−[

dgmdρ

]

ρ=ρ′−ϵ

= −4π

ρ′

k (ψ1 (kρ′)ψ′

2 (kρ′)− ψ′

1 (kρ′)ψ2 (kρ

′)) = −4π

ρ′

kW [ψ1 (kρ′) ,ψ2 (kρ

′)] = −4π

ρ′(3.146)

where W [ψ1,ψ2] is the Woronskian of ψ1 and ψ2.As an aside, consider the Sturm-Liouville equation

d

dx

[

p(x)dy(x)

dx

]

+ (q(x) + λr(x)) y(x) = 0 (3.147)

with two solutions y1(x) and y2(x) for some λ. Then,

d

dx[p(x)W (y1,y2)] =

d

dx[p(x) (y1(x)y

′2(x)− y′1(x)y2(x))] = (3.148)

= y1(x) (p(x)y′2(x))

′ − y2(x)(p(x)y′1(x))

′ = 0(3.149)

and thus

W (y1, y2) ∝1

p(x). (3.150)

In our problem p = ρ and

W (ψ1(kρ),ψ2(kρ)) = −4π

kρ(3.151)

for all ρ. For the solution gm(k, ρ, ρ′) to be finite at ρ = 0 and to vanish atρ = ∞

ψ1(kρ) = AIm(kρ) (3.152)

ψ2(kρ) = Km(kρ) (3.153)

and the constant A = 4π determined from (3.151) since

W (Im)x), Km(x)) = −1

x. (3.154)


Therefore the expansion of the Green’s function in cylindrical coordinatesis given by

1

|x− x′|=

2

π

∞∑

m=−∞

∫ ∞

0

dk eim(φ−φ′) cos(k(z − z′))Im(kρ<)Km(kρ>) (3.155)

=4

π

∫ ∞

0

dk cos(k(z − z′))

(

1

2I0(kρ<)K0(kρ>) +

∞∑

m=1

cos(m(φ− φ′))Im(kρ<)Km(kρ>)

)

If we let x′ → 0, then only m = 0 term is finite and

1√

ρ2 + z2=

2

π

∫ ∞

0

cos(kz)K0(kρ)dk (3.156)

If we let z′ → 0 then the Green’s function can be obtain by two methods.We can either replace ρ2 → R2 = ρ2 + ρ′2 − 2ρρ′ cos(φ − φ′) in (3.156) orwe can derive it directly from (3.155). By equating the integrands of bothexpressions we obtain,

K0(k√

ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′)) = I0(kρ<)K0(kρ>)+2∞∑

m=1

cos(m(φ−φ′))Im(kρ<)Km(kρ>)

(3.157)which must hold for all z. Then one can show by expanding around k ∼ 0that in polar coordinates the Green’s function is given by

log

(

1√

ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′)

)

= 2 log

(

1

ρ>

)

+∞∑

m=1

1

m

(

ρ<ρ>

)m

cos(m(φ−φ′)).

(3.158)For example, the potential of a line of charge with linear charge density λ is

Φ(x) =1

4πϵ0

∫

ρe(x′)G(x,x′)d3x′ =

1

4πϵ0

∫ 2π

0

dφ′∫ +∞

−∞dz′∫ ∞

0

ρ′dρ′(

λδ(ρ′)

2πρ′

)

G(x,x′)

=λ

π2ϵ0

∫ +∞

−∞dz′∫ ∞

0

dk cos(k(z − z′))

(

1

2I0(0)K0(kρ)

)

=λ

2π2ϵ0

∫ ∞

0

dk

(∫ +∞

−∞dz′′ cos(kz′′)

)

K0(kρ)

=λ

2πϵ0

∫ ∞

−∞dk δ(k)K0(kρ)

= limk→0

λ

2πϵ0

(

− log

(

kρ

2

))

= −λ log (ρ)

2πϵ0+ const. (3.159)

chapter 3 boundary-value problems: part iivvanchur/phys5511/chapter3.pdf · 2017-05-01 · chapter...

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