The Laplace Equation
Chris Olm and Johnathan Wensman
December 3, 2008
Introduction (Part I)


We are going to be solving the Laplace equation in
the context of electrodynamics
Using spherical coordinates assuming azimuthal
symmetry
– Could also be solving in Cartesian or cylindrical
coordinates
– These would be applicable to systems with corresponding
symmetry

Begin by using separation of variables
– Changes the system of partial differential equations to
ordinary differential equations

Use of Legendre polynomials to find the general
solution
Introduction (Part II)

We will then demonstrate how to
apply boundary conditions to the
general solution to attain particular
solutions
– Explain and demonstrate using “Fourier’s
Trick”
– Analyzing equations to give us a workable
solution
The Laplace Equation

Cartesian coordinates
 2V  2V  2V


0
x 2
y 2
z 2
– V is potential
– Harmonic!

Spherical coordinates
1   2 V 
1
 
V 
1
 2V
0
r
 2
 sin 
 2
2
2
2
  r sin  
r r  r  r sin   
– r is the radius
–  is the angle between the z-axis and the vector we’re
considering
–  is the angle between the x-axis and our vector
Azimuthal Symmetry


Assuming azimuthal symmetry
simplifies the system
In this case, decoupling V from Φ
1   2 V 
1
 
V 
1
 2V
0
r

 sin 

  r 2 sin 2   2
r 2 r  r  r 2 sin   
becomes
  2 V 
1  
V 
r

 sin 
0
r  r  sin   
 
Potential Function


Want our function in terms of r and θ
So
V (r , )  R(r )( )
– Where R is dependent on r
– Θ is dependent on θ
ODEs!

Plugging in we get:
1 d  2 dR 
1
d 
d 
r

 sin 
0
R dr 
dr   sin  d 
d 
– These two ODEs
that must be equal and
.
opposite:
1 d  2 dR 
r
k
R dr  dr 
↓
d  2 dR 
r
  kR
dr  dr 
1
d 
d 
sin


  k
 sin  d 
d 
↓
d 
d 
sin


  k sin 
d 
d 
General Solution for R(r)

Assume
R  rl
– Plugging in we get
d 2
d 2
d l 1
(r * R' ) 
(r * lr l 1 ) 
lr  l (l  1)r l  kr l
dr
dr
dr
So
l (l  1)  k
– We can deduce that the equation is solved
when k=l or k=-(l+1)
– So our general solution for R(r) is
R(r )  Ar  B
l
1
r l 1
General Solution for Θ(θ)

Legendre polynomials
– The solutions to the Legendre differential
equation, where l is an integer
– Orthogonal
– Most simply derived using Rodriques’s
l
formula:
1 d 
Pl ( x) 

2
l
[(
x

1
)
],0  l  


l
2 l!  dx 
In our case x=cosθ so
( )  Pl (cos  )
Θ(θ) Part 2

Let l=0:
d 
d
d
 d 
 d
sin  (0)  0
P0 (cos  )  
(1)  
 sin 
 sin 
d 
d
d

d

d




 0(0  1) P0 (cos  ) sin(  )  0
00

Let l=3:
3


d 
d
d  1 
d
 d 
 (cos 2   1) 3   
P3 (cos  )  
sin 
 3 
 sin 
d 
d
d  2 3!  d cos( ) 
 d 
 

↘

d
d

d 5
3
d 
3

 15

3
 sin 
 sin    cos 2  sin   sin   
 cos   cos    
d  2
2
2
  d 
 2


↘

3
 15

 3(3  1) P3 (cos  )  12  cos 2  sin   sin  
2
 2

↘

□
d  15
3 2 
2
2
3
3
  cos  sin   sin    15(cos  sin   cos  sin  )  3 sin  cos 
d  2
2


↓
 18  5 cos 2  sin   sin   15(cos 3  sin   cos  sin 3  )  3 sin  cos 
□
The General Solution for
V

Putting together R(r), Θ(θ) and
summing over all l
V (r , )  R(r )( )
becomes

Bl 

l
V (r , )    Al r  l 1 Pl (cos )
r 
l 0 
Applying the general
solution
Example 1

The potential is specified on a hollow
sphere of radius R
V( R , )  V0 ()
What is the potential on the inside of
the sphere?
Applying boundary
conditions and intuition

We know it must take the form

B 

V( r , )    A l r l  l l1 Pl (cos )
r 
l0 

And on the surface of the sphere must be V0,
also all Bl must be 0, so we get

V( r , θ)   A l r l Pl cosθ 
l 0

The question becomes are there any Al
which satisfy this equation?
Yes!
But doing so is tricky

First we note that the Legendre
polynomials are a complete set of
orthogonal functions
– This has a couple consequences we can
exploit
0 if l'  l









P
x
P
x
dx

P
cos

P
cos

sin

d


 2
l'
1 l l'
0 l
 2l  1 if l' =l
1

Applying this property

We can multiply our general solution by
Pl’’(cos θ) sin θ and integrate (Fourier’s Trick)




A l r l   Pl cos Pl ' cos sin d  Vo θ  Pl cos sin  d
0
0


2
A l' r
  Vo Pl ' cos sin  d
2l '1 0
l'

2l  1
Al 
Vo Pl cos sin  d
2r l 0
Solving a particular case
V0 ()  sin 2  / 2

We could plug this into our equation giving

2l  1
Al 
sin 2  / 2Pl cos sin  d
l 
2R 0

In scientific terms this is unnecessarily
cumbersome (in layman's terms this is a hard
integral we don’t want or need to do)
The better way


Instead let’s use the half angle formula to rewrite
our potential as
V0 
1
1  cos
2
V0 
1
P0 cos  P1 cos
2
Plugging THIS into our equation gives

2l  1 1
P0 cos  P1 cosPl cos sin  d
Al 
l 
2r 0 2

Now we can practically read off the values of Al
Getting the final answer
A0 

1
2
A1  
1
2R
Plugging these into our general solution we
get

V( r , θ)   A l r l Pl cosθ 
l 0
1 0
1
 r P0 cosθ  
rP1 cosθ 
2
2R

1
r

1  cosθ 
2 R

Example 2


Very similar to the first example
The potential is specified on a hollow
sphere of radius R
V( R , )  V0 ()

What is the potential on the outside of the
sphere?
Proceeding as before

Must be of the form

B 

V( r , )    A l r l  ll1 Pl (cos )
r 
l0 

All Al must be 0 this time, and again at the
surface must be V0, so

Vr ,   
l 0

Bl
P cos 
l 1 l
r
and
V( R , )  Vo ()
Fourier’s Trick again

By applying Fourier’s Trick again we
can solve for Bl

Bl ' 2
  Vo Pl ' cos sin  d
l ' 1
r 2l '1 0

2l  1
Bl  r l 1
Vo Pl cos sin  d

2 0
 As far as we can solve without a specific potential
Conclusion


By solving for the general solution we can
easily solve for the potential of any system
easily described in spherical coordinates
This is useful as the electric field is the
gradient of the potential
– The electric field is an important part of
electrostatics
References





1)Griffiths, David. Introduction to Electrodynamics. 3rd ed.
Upper Saddle River: Prentice Hall, 1999.
2)Blanchard, Paul, Robert Devaney, and Glen Hall. Differential
equations. 3rd ed. Belmont: Thomson Higher Education, 2006.
3) White, J. L., “Mathematical Methods Special Functions
Legendre’s Equation and Legendre Polynomials,”
http://www.tmt.ugal.ro/crios/Support/ANPT/Curs/math/s8/s8l
egd/s8legd.html, accessed 12/2/2008.
4) Weisstein, Eric W. "Laplace's Equation--Spherical
Coordinates." From MathWorld--A 5) Wolfram Web Resource.
http://mathworld.wolfram.com/LaplacesEquationSphericalCoor
dinates.html, accessed 12/2/2008