Download ) ( , z

Solving the Laplace equation—separation of variables
1. Cartesian coordinates
 ∂2
∂2
∂2 
∇ Φ =  2 + 2 + 2 Φ = 0
∂y
∂z 
 ∂x
2
The separation of variables in Cartesian coordinates is done as follows. Let
Φ ( x, y , z ) = X ( x )Y ( y ) Z ( z )
where X, Y, and Z, are functions of x, y, and z, respectively. The Laplace equation now takes the
form
1 d2X
1 d 2Y
1 d 2Z
+
+
=0
X ( x) dx 2 Y ( y ) dy 2 Z ( z ) dz 2
Each of the terms is a function of a different coordinate, and therefore each term must be a
constant, otherwise, we will not get 0 on the right hand side. This means that
1 d2X
= C1
X dx 2
1 d 2Y
= C2
Y dy 2
1 d 2Z
= C3
Z dz 2
with
C1 + C 2 + C 3 = 0
Moreover, we cannot have all three constants positive, or all three negative.
When the constant is positive, the general solution is (for example, for X(x)),
X ( x) = Ae kx + Be − kx = A' cosh(kx) + B' sinh(kx) ,
where A, B, and ,
k 2 = C1
or alternatively A', B', and
k 2 = C1
are to be determined from
boundary conditions.
1
When the constant is negative, the general solution is
X ( x) = Ae ikx + Be − ikx = A' cos(kx) + B' sin(kx)
and, again, A, B, and
k 2 = −C1 , or alternatively A', B', and k 2 = −C1
are to be determined
from boundary conditions.
Let us examine several specific examples to see how the method works.
Exercise: Find the potential within an infinite pipe placed along the z-direction, whose crosssection is a square of size a × a . Let the coordinate origin be at the center of the square. The
walls of the pipe along y are held at potential V, those along x are grounded.
Solution: Firstly, we write the boundary conditions in the form
a
Φ ( x, y = ± ) = V ,
2
and
Φ( x = ±
a
, y) = 0 .
2
It is clear that the potential does not depend on z, and therefore the Laplace equation becomes
1 d2X
1 d 2Y
+
=0
X ( x) dx 2 Y ( y) dy 2
and clearly we have
1 d2X
= −C
X dx 2
1 d 2Y
=C
Y dy 2
Let us take C>0. The general solution for X is then
X ( x) = Ae ikx + Be − ikx = A' cos(kx) + B' sin(kx)
with
C = k 2 . The general solution for Y is
Y ( y ) = Ce ky + De − ky = C ' cosh(ky) + D' sinh(ky)
2
again with
C = k 2 . Now we apply the boundary conditions. Since
a
Φ ( x, y = ± ) = V ,
2
we find (remember that cosh is an even function, and sinh is an odd
function)
V = C ' cosh(ka / 2) + D' sinh(ka / 2) = C ' cosh(ka / 2) − D' sinh(ka / 2)
Clearly we must have D'=0, and therefore the general solution for Y is
Y ( y ) = C ' cosh(ky )
Now we take the other boundary condition,
Φ( x = ±
a
, y ) = 0 . This one gives (cos is an even
2
function and sin is an odd function)
0 = A' cos(ka / 2) + B' sin(ka / 2) = A' cos(ka / 2) − B' sin(ka / 2)
so we can choose B'=0 and
k = (2n + 1)
π
a
where n is an integer. Combining all these results, the general solution for the potential is
∞
πx 
πy 


Φ ( x, y ) = ∑ Cn cos (2n + 1)  cosh (2n + 1) 
a
a


n =0
Now we need to determine all the coefficients
a
Φ ( x, y = ± ) = V ,
2
C n . We use again the boundary condition
to write
πx 
π


V = ∑ C n cos (2n + 1)  cosh (2n + 1)  .
a
2


n =0
∞
To proceed, we use the identity
3
a/2
1
1
dx cos(πx / a ) cos(' πx / a ) = δ '
∫
a −a / 2
2
which holds when both
and
'
are odd. Then
 πx 
 π 
V = ∑ C cos
 cosh 

 a 
 2 
gives
1
2
' πx 1
' π
' π
sin
dxV cos
= C ' cosh
=V
∫
2
2
' π
2
a −a / 2
a
a/2
Combining all the results, the solution for the potential within the pipe is
(−1) n
πx 
πy 
π



Φ ( x, y ) =
cos(2n + 1)  cosh (2n + 1)  / cosh (2n + 1) 
∑
a
2
2
π n =0 2 n + 1 


4V
∞
Exercise: Find the potential within a stripe located in the region
0 ≤ x ≤ a , 0 ≤ y ≤ ∞ . The edge at y=0 is held at a potential V, all other three edges are
grounded.
Solution: This is again a two-dimensional problem, so we again have
1 d2X
= −C
X dx 2
1 d 2Y
=C
Y dy 2
However, now we must choose C to be positive, in order to satisfy the boundary condition at
y → ∞ . To see this we write the general solution for Y(y) as
Y ( y ) = Ce ky + De − ky
We see that in order to satisfy that boundary condition, we must have C=0 and k>0. Hence the
general solution of the function Y in this case is
Y ( y ) = De − ky
4
Let us now take the general solution of the function X,
X ( x) = A' cos( kx) + B ' sin( kx)
Since the boundary conditions are that the potential vanishes (for any y) at x=0 and x=a, we
ka = nπ
must have A'=0 and
becomes
, where n is an integer. Collecting all results, the general solution
∞
Φ ( x, y ) = ∑ An e − nπy / a sin (nπx / a )
n =1
Now it remains to find the coefficients
An . To this end, we use the remaining boundary
condition, which says that when y=0, the potential is V. This one gives
∞
V = ∑ An sin (nπx / a )
n =1
Now we use the identity
 nπ   n 'π  a
x  sin 
x  = δ nn '
a   a  2
a
∫ dx sin 
0
to obtain
a
An ' =
2
a
∫ dxV
sin( n ' π x / a ) = V
0
Therefore, the coefficients
An
(
a
1 − (1) n '
πn'
)
are non zero only when n is odd. Consequently, the solution
reads
Φ ( x, y ) =
4V
π
1
∑ ne
− nπy / a
sin (nπx / a )
oddn
One may even calculate this sum. Let us denote
Z ≡ e ( iπ / a )( x + iy )
5
and then we can write the result in the form
1 Z n − Z *n 4V
Φ ( x, y ) =
=
Im ∑ Z n
∑
π nodd n
2i
π
nodd
4V
The result is then
Φ ( x, y ) =
 sin (πx / a ) 
  1 + Z  2V

Imln
arctan
 =
π
  1 − Z  π
 sinh (πy / a ) 
4V
Exercise: Find the potential within a rectangular box, of dimension (a,b,c) in the (x,y,z)
directions, respectively. All faces of the box are grounded, except the one at z=c which is held
at potential V(x,y) [V(x,y) is a given function].
Solution: This is a three-dimensional problem. We therefore start from
1 d2X
= C1
X dx 2
1 d 2Y
= C2
Y dy 2
1 d 2Z
= C3
Z dz 2
C1 + C 2 + C 3 = 0
We first determine the sign of the three constants from the boundary conditions. X(x) should
vanish at x=0 and at x=a. Therefore
X ( x) = A sin αx
with
α=
nπ
a
, i.e.,
C1 = −α 2
mπ
b
, i.e.,
C2 = − β 2
Y(y) should vanish at y=0 and y=b. Therefore
Y ( y ) = B sin βy
It follows that
C3 = α 2 + β 2
with
β=
and is positive. Since the potential vanishes also at z=0, we must
choose
(
)
Z ( z ) = C sinh α 2 + β 2 z .
6
The resulting general solution is therefore
Φ ( x, y , z ) =
∞
∑A
nm
sin (α n x )sin (β m y )sinh (γ nm z ),
n , m =1
α n = ( nπ ) / a
β m = (mπ ) / b
 n2
2
a
  m2
 +  2
 b
γ nm = π 



The remaining boundary condition is that at z=c, the potential is some given function V(x,y).
Therefore, we determine the coefficients
V ( x, y ) =
∞
∑A
nm
An
from the equation
sin (α n x )sin (β m y )sinh (γ nm c )
n , m =1
A double use of the trigonometric delta-function condition,
 nπ 
 n 'π  a
dx
sin
x
sin
x  = δ nn '



∫0
 a

 a
 2
a
will give
a
Anm
b
4
=
dx dyV ( x, y ) sin (α n x )sin (β m y )
ab sinh (γ nm c ) ∫0 ∫0
which completes the solution.
2. Circular symmetry
We start with an example.
7
Exercise: An infinite cylindrical shell (of radius a), parallel to the z-axis, is held at potential V0
for
π / 2 < ϕ < 3π / 2 , and the other half is held at − V0 . Find the potential in the entire space.
Solution: Clearly the potential is independent of z, and therefore we need the Laplace equation in
two dimensions. It is also clear from the symmetry of the problem that one should choose polar
coordinates,
Φ ≡ Φ( ρ ,ϕ )
and the Laplace equation is
1 ∂  ∂Φ  1 ∂ 2 Φ
∇ Φ=
ρ
+
=0
ρ ∂ρ  ∂ρ  ρ 2 ∂ϕ 2
2
We write
Φ(ρ , ϕ ) = R( ρ )Ψ (ϕ )
and find
ρ d  dR  1 d 2 Ψ
ρ
+
=0
R dρ  dρ  Ψ dϕ 2
The dependence on the angle
ϕ
must be unique, namely,
Ψ (ϕ + 2π ) = Ψ (ϕ )
and therefore
the general solution for this function is
Ψ (ϕ ) = A cos nϕ + B sin nϕ
so that the Laplace equation becomes
1 d  dR  n 2
ρ
−
R=0
ρ dρ  dρ  ρ 2
The possible solutions of this equation are
ρ n , ρ − n , ln ρ , (n = 0)
Therefore, the general solution is
8
∞
[
]
Φ(ρ , ϕ ) = a0 ln ρ + ∑ ( An ρ n + Bn ρ − n )(Cn cos(nϕ ) + Dn sin (nϕ )
n =1
Let us now return to our problem, and write down the general solution for the potential inside
and outside the cylinder, keeping mind that it cannot diverge. This means that we have
n
ρ
Φ in ( ρ , ϕ ) = ∑   ( An cos nϕ + Bn sin nϕ )
n =1  a 
n
a
Φ out ( ρ , ϕ ) = ∑   (Cn cos nϕ + Dn sin nϕ )
n =1  ρ 
We now apply the boundary conditions, taking these two potentials at
2π
∫ dϕ cos mϕ cos nϕ = πδ
ρ =a.
Since
2π
∫ dϕ sin mϕ sin nϕ = πδ
n ,m
0
n ,m
0
2π
∫ dϕ cos mϕ sin nϕ = 0
0
it is straightforward to write all coefficients in terms of the integrals
2π
∫ dϕΦ(a,ϕ ) cos mϕ
2π
and
0
∫ dϕΦ(a,ϕ ) sin mϕ
0
Then, by the specific boundary conditions of this problem, we need
π /2
[
∫
2π
+
∫
π
0
3 /2
2π
3π / 2
3π / 2
−
]dϕ sin mϕ = 0
∫
π
/2
and
π /2
[
∫
0
+
∫
3π / 2
−
∫ ]dϕ cos mϕ =
π /2
4
π
sin m
m
2
9
It therefore turns out that the coefficients
Bn , Dn are all zero and (since the potential must be
continuous at the shell)
An = Cn =
4V0
( −1)
π (2 + 1)
which completes the solution
 (−1)  ρ  2+1

Φ in (ρ , ϕ ) =
  cos(2 + 1)ϕ 

∑
π =0  2 + 1  a 

2 +1

4V0 ∞  (−1)  R 
Φ out (ρ , ϕ ) =


cos(
2
+
1
)
ϕ


∑
π =0  2 + 1  ρ 

4V0
∞
Exercise: An infinite cylindrical shell (of radius a), parallel to the z-axis, is held at potential
V (ϕ ) (a given function). Find the potential inside the cylinder.
Solution: Again, the potential is independent on z. The general form for the potential inside the
cylinder may be written in the form
 ρ  n

Φ(ρ , ϕ ) = ∑   An einϕ + cc 
n =1 

 a 
∞
(
)
(cc means complex conjugate). Now we use
1
2π
2π
∫ dϕe
in ϕ
= δ n ,0
0
to obtain from the boundary condition
1
An =
2π
2π
∫ dϕV (ϕ )e
−inϕ
0
The solution is therefore given in the form
10
2π
1
ρ n in (ϕ −ϕ ')
d
V
'
(
'
)
e
+ cc
ϕ
ϕ
∫
an
n =1 2π 0
∞
Φ (ρ , ϕ ) = ∑
Since this is a geometric sum, we find
Φ (ρ , ϕ ) =
1
2π
2π
∫ dϕ 'V (ϕ ' )
0
2
ρ
cos(ϕ − ϕ ') − 2
ρ2
a2 = 1
2π
ρ2
ρ
1 + 2 − 2 cos(ϕ − ϕ ')
a
a
a
1−
2π
∫ dϕ 'V (ϕ ' )
0
1+
ρ2
a
2
−2
ρ2
a2
ρ
a
cos(ϕ − ϕ ')
In the last step, we have omitted a constant term from the potential.
3. Laplace equation in cylindrical coordinates
This geometry is a continuation of the previous examples, but now the potential depends also on
the coordinate z. The Laplacian in cylindrical coordinates is
1 ∂  ∂Φ  1 ∂ 2Φ ∂ 2Φ
∇ Φ=
+
=0
ρ
+
ρ ∂ρ  ∂ρ  ρ 2 ∂ϕ 2 ∂z 2
2
As before, we separate variables,
Φ ( ρ , ϕ , z ) = R ( ρ ) Ψ (ϕ ) Z ( z )
and then, in general,
d 2Z
− k 2Z = 0
2
dz
d 2Ψ
+ν 2 Ψ = 0
2
dϕ
d 2 R 1 dR  2 ν 2 
+
+  k − 2  R = 0
dρ 2 ρ dρ 
ρ 
11
At this stage,
k ,ν
are unknown constants. However, we will be interested in problems in which
Ψ (ϕ + 2π ) = Ψ (ϕ ) and therefore ν = n ,
where n is an integer. Clearly we know the general solution for the differential equations of Z
and Ψ . What about the function R? Let us change the variables of that equation, so that it will
the dependence on the angle is uniquely defined,
become dimensionless, ρ = x / k . (x is now a dimensionless variable.) The third equation then
reads (R=R(x))
d 2 R 1 dR  n 2 
+
+ 1 −  R = 0 .
dx 2 x dx  x 2 
This is the Bessel equation, and its solutions are the Bessel functions (documented in books).
There are two types of these functions. We will deal only with the first kind, which is wellbehaved at small arguments. The Bessel functions of the first type are usually denoted by J,
R( x) ≡ J n ( x), with J n ( x) = (−1) n J −n ( x)
and they have many nice properties (all documented in books). Generally speaking, the Bessel
function is a complicated polynom. Let us now see using examples how we solve the Laplace
equation in cylindrical coordinates.
Exercise: A cylinder of radius a and height L, is placed parallel to the z axis. Its basis at z=0 is
grounded, and so is its face at ρ = a . The basis at z=L is held at a potential V ( ρ , ϕ ) (a given
function). We need to find the potential everywhere within the cylinder.
Solution: Let us first consider the general solutions of the three equations,
d 2Z
− k 2Z = 0
2
dz
d 2Ψ
+ n2Ψ = 0
2
dϕ
d 2 R 1 dR  2 n 2 
+
+  k − 2  R = 0
dρ 2 ρ dρ 
ρ 
The general solution of
Ψ (ϕ ) consists of trigonometric functions,
12
Ψ (ϕ ) = An sin nϕ + Bn cos nϕ
As for the equation for Z, it can have a trigonometric solution as well (for imaginary k) or
hyperbolic solution (for a real k). According to the boundary conditions, there is no reason to
expect that potential as function of z is oscillatory, and therefore k is any real constant, and the
general solution is
Z ( z ) = C (k ) sinh kz + D(k ) cosh kz
but since the potential vanishes at z=0, we choose
D(k ) = 0 . The solution for R is the Bessel
function,
R ( ρ ) ∝ J n ( kρ )
So far, k is an unknown function. However, the boundary conditions tell us that the potential
vanishes at
ρ = a , and therefore we need to have J n (ka) = 0 , or, in other words, that ka will
be a root of the Bessel function of order n. Each Bessel function has an infinite number of roots,
and therefore ka takes an infinite number of discrete values, all of them are roots of the n-th
Bessel function. Namely,
k nm a
is the m-th root of the n-th Bessel function. It follows that the
general solution of our problem is
∞
∞
Φ ( ρ , ϕ , z ) = ∑∑ J n (k nm ρ ) sinh( k nm z )( Amn sin nϕ + Bmn cos nϕ )
n =0 m =1
The Bessel functions have the property
1
1
∫ dαJν ( xν α ) Jν ( xν α )α = 2 Jν
n'
n
2
+1
( xνn )δ nn '
0
where
xνn
is the n-th root of the Bessel function
Jν (x ) , i.e., Jν ( xνn ) = 0.
We now use this property with the boundary condition at z=L to determine all the coefficients in
terms of the given function V ( ρ , ϕ ) . First we use the delta functions of the trigonometric
functions to obtain
∑
m
Amn sinh(k nm L) J n (k nm ρ ) =
1
π
2π
∫ dϕV ( ρ ,ϕ ) sin nϕ
0
13
∑
Bmn sinh(k nm L) J n (k nm ρ ) =
m
1
π
2π
∫ dϕV ( ρ , ϕ ) cos nϕ
0
and secondly we use the property of the Bessel function, which can be written in the form
ρ
a
∫ d ρ J ν ( xν
0
n'
a
) J ν ( xν n
ρ
a2 2
)ρ =
J ν +1 ( xν n )δ nn '
a
2
Then
2π
a
Amn
1
= 2 2
dϕ dρρJ n (k nm ρ )V ( ρ , ϕ ) sin nϕ
a πJ n+1 (k nm a) sinh(k nm L) ∫0 ∫0
Bmn
1
= 2 2
dϕ dρρJ n (k nm ρ )V ( ρ , ϕ ) cos nϕ
a πJ n+1 (k nm a) sinh(k nm L) ∫0 ∫0
2π
a
which completes the solution.
As an example, we take the case where the function V ( ρ , ϕ ) is simply a constant, V. Then
Amn = 0 and Bmn ∝ δ n 0 , and we need to consider only the n=0 Bessel function.
4. Laplace equation in spherical coordinates
The Laplacian in spherical coordinates is
1 ∂2
1
1
∂ 2Φ
∂ 
∂Φ 
∇ Φ(r , θ , ϕ ) =
(rΦ) + 2
=0
 sin θ
+
r ∂r 2
r sin θ ∂θ 
∂θ  r 2 sin 2 θ ∂ϕ 2
2
We will first consider problems in which the potential is independent of
ϕ.
14
Because of the form of the Laplacian, it is useful to write the variable separation in the form
Φ (r , θ ) =
U (r )
P (θ )
r
We also see that upon defining µ = cosθ , then
d
d
= − sin θ
dθ
dµ
1 d 
d 
d2
d
d 
d 
2
 (1 − µ 2 )

=
 sin θ
 = (1 − µ ) 2 − 2 µ
sin θ dθ 
dθ 
d µ dµ 
dµ 
dµ
and
The Laplacian is now
∇ 2 Φ ( r ,θ ) = P ( µ )
1 d 2U U (r ) d 
dP 
 (1 − µ 2 )  = 0
+ 3
2
r dr
dµ 
r dµ 
We begin with the part which depends on r. We have
r 2 d 2U
= const.
U (r ) dr 2
It is very easy to see that the solution is either U (r ) ∝ r α +1 and/or U (r ) ∝ r −α and in both cases
the constant is const. = α (α + 1) . Such solutions imply that the r-dependence of the potential is
r α and/or r −1−α . Since we want the potential and its all derivatives to be nicely-behaved, we take
α = , where is an integer. This implies that the function P satisfies the equation
( + 1) P ( µ ) +
dP ( µ ) 
d 
 (1 − µ 2 ) =0
dµ 
dµ 
(We now have P since for every we will have a different function.) The last equation is the
Legendre equation, and its solutions are the Legendre polynomials. Those have many properties,
for example,
π
2δ k
∫−1dµPk (µ ) P (µ ) = ∫0 dθ sin θPk (θ ) P (θ ) = 2 + 1
1
1
P' k (0)
∫ dµP (µ ) = k (k + 1)
k
0
(P’ denotes the derivative.)
The general expression of the potential in spherical coordinates, when it is independent of
∞
[
ϕ , is
]
Φ ( r , θ ) = ∑ A r + B r − −1 P (cos θ )
=0
15
Exercise: A grounded sphere of radius a is placed in a uniform electric field, E 0 . Find the field
outside the sphere.
Solution: Let us take the external field along the z-direction, E 0 = zˆE 0 . The potential that
creates such a field is then Φ 0 = − zE 0 = − r cos θE 0 . The potential that we are looking for is
the solution of the Laplace equation, and has the form
+1
∞ 
r
a 
Φ ( r , θ ) = ∑  A   + B   P (cos θ )
 r  
=0 
 a
(Note that P0 (cosθ ) = 1 )
r → ∞ , this potential should tend to Φ 0 , and therefore we know that all A = 0 except for
A1 = − E0 a (Note that P1 (cosθ ) = cosθ ). At r=a the potential should vanish. Since the Legendre
As
polynomials is a complete orthogonal set, it means that the coefficient of each of them should
vanish separately. Hence
B = 0 , except for B1 = − A1 = E0 a . Therefore the potential is
3
a 3 cosθ
a
Φ ( r ,θ ) = E0
− E0 r cosθ = E 0   z − E 0 z
2
r
r
and the electric field is
ˆ
3( z / r )rˆ − zˆ 3 3 cos θrˆ − z
E (r ) = E0 + E0 a 3
=
E
+
E
a
0
0
r3
r3
The second term here is the electric field of a dipole given by
p = E0 a 3 zˆ
Does the field satisfy the boundary conditions at r=a? We can write the field as
 2a 3 
Er (r , θ ) = E0 1 + 3  cos θ
r 

 a3 
Eθ (r ,θ ) = − E0 1 − 3  sin θ
 r 
16
The field inside the sphere is zero, therefore the tangential component,
Eθ ( a , θ ) = 0 , as found
in our solution. The normal component of the field has a discontinuity, namely, the induced
surface charge density is
σ (θ ) = E0
3
cos θ
4π
But there is no induced net charge:
π
∫ dθ sin θσ (θ ) = 0
0
Exercise: A sphere of radius a carries surface charge density
σ 0 (θ ) . Find the electric field.
Solution: The solution of the Laplace equation in this case is in general
∞
r
Φ in ( r , θ ) = ∑ A   P (cos θ ), r ≤ a
a
=0
∞
a
Φ out ( r ,θ ) = ∑ B  
r
=0
+1
P (cos θ ), r ≥ a
Since the tangential components of the field are continuous at r=a, it follows that A = B
for all . As for the normal components, we have
 ∂Φ out ∂Φ in 
 − ∂r + ∂r  = 4πσ 0 (θ )
r =a
This gives
∞
∑ (2 + 1)
=0
π
1
We now use
A
P (cos θ ) = 4πσ 0 (θ )
a
2δ k
∫ dµP (µ ) P (µ ) = ∫ dθ sin θP (θ ) P (θ ) = 2 + 1
k
−1
k
0
17
to find
π
A = 2πa ∫ dθ sin θσ 0 (θ ) P (cos θ )
0
This completes the solution. For example, when σ 0 (θ ) = σ cos θ , then A1 = 4πaσ / 3 and all other
coefficients vanish.
Exercise: The potential on the surface of a sphere of radius a is given as function of θ , V (θ ) .
Find the potential inside the sphere.
Solution: The solution of the Laplace equation inside the sphere is
∞
r
Φ in ( r , θ ) = ∑ A   P (cos θ ), r ≤ a
a
=0
and the boundary condition gives
∞
V (θ ) = ∑ A P (cos θ )
=0
Therefore
π
2 + 1
A =
dθ sin θV (θ ) P (cos θ )
2 ∫0
5. Laplace equation in spherical coordinates
The Laplacian in spherical coordinates is
1 ∂2
1
∂ 
∂Φ 
1
∂ 2Φ
∇ Φ(r , θ , ϕ ) =
(rΦ) + 2
=0
 sin θ
+
r ∂r 2
r sin θ ∂θ 
∂θ  r 2 sin 2 θ ∂ϕ 2
2
Let us write
18
Φ=
U (r )
P (θ )Q (ϕ )
r
and the Laplace equation takes the form
 1 d 2U 1
d 
dP   1 d 2Q
1
r sin θ 
+
2
2
sin θ dθ   + Q dϕ 2 = 0
U
dr
P
r
θ
d
θ
sin


2
2
For the potential to be unique, we put
1 d 2Q
= −m 2
2
Q dϕ
with m being an integer. The derivation now is like the one above, with
r 2 d 2U
= ( + 1)
U (r ) dr 2
where is also an integer. This leaves us with
1 d 
dP(θ )  
m2 

 P(θ ) = 0
sin
+
(
+
1
)
−
θ


sin θ dθ 
dθ  
sin 2 θ 
or, alternatively, with µ = cos θ
d 
dP( µ )  
m2 
 P( µ ) = 0
 (1 − µ 2 )
 +  ( + 1) −
dµ 
dµ  
1 − µ 2 
The solutions of this equation are the associated Legendre polynomials,
Pm ( µ ) . The general
solution of the Laplace equation in spherical coordinates is therefore
∞
[
Φ ( r , θ , ϕ ) = ∑∑ Pm (cos θ )e imϕ Am r + Bm r −−1
]
=0 m
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It is customary to combine together the full dependence on the angles, by introducing the
spherical harmonic functions Ym (θ , ϕ )
( 2 + 1)( − m)! m
P (cos θ )e imϕ ,
4π ( + m)!
Ym (θ , ϕ ) =
where m takes the values:
and
− ,−( − 1),....,0,..., ( − 1), Y −m (θ ,ϕ ) = ( −1) m Y*m (θ ,ϕ ).
The general solution of the Laplace equation in spherical coordinates is then
∞
Φ (r ,θ , ϕ ) = ∑
∑Y
m
[
(θ , ϕ ) Am r + Bm r − −1
]
=0 m = − The coefficients A and B are found according to the general rule which states that if we have
the function
∞
g (θ , ϕ ) = ∑
∑A
Y (θ , ϕ )
m m
=0 m = − then
Am = ∫ dΩY*m (θ , ϕ ) g (θ , ϕ )
dΩ ≡ dϕ sin θdθ
20