1. 3. Boundary Value Problems II 3A. The Spherical Harmonics
Consider problems in 3D written in spherical coordinates
It is useful to find a set of complete, orthonormal functions for the angular part
Found in quantum mechanics class
Such a set of functions are called the spherical harmonics
Note that l runs from 0 to 
And m runs from –l to l
Any function of angle can be written in terms of them
They are orthonormal
And complete

2. Properties of Spherical Harmonics
 dependence:
 dependence
Pl|m| is a polynomial of order l – |m|
Complex conjugate:
Parity: changing x  –x
In angles,    –  and   – 
Value at  = 0:
Explicit form can be found in many sources
Look them up, use Maple program, or do whatever is most useful to you

3. The Spherical Harmonics

4. Spherical Harmonics and Legendre Functions
Legendre functions are a set of n’th order polynomials
They are orthogonal on the interval (–1, 1)
They are “normalized” by the condition
Compare to the normalization condition
Recall these functions have no  dependence
Recall that Yl0 is an l’th order polynomial in cos
It is clear that Yl0 and Pl are almost the same thing

5. Spherical Harmonics and the Laplacian
Spherical harmonics satisfy
Compare to the Laplacian
Suppose we write a function in the form:
Then the Laplacian acting on this will give

6. Problems with No Charge in a Region
Suppose we have no charge in a region, so
Then we must have:
Need two independent solutions, since 2nd order differential equation
Let’s try clm ~ r
Solutions are  = l or  = –l – 1
Therefore
If we want finite at 0, don’t include B
If we want zero at , don’t include A

7. Spheres with Boundary Conditions
Suppose we have a sphere, radius a, potential given on surface
We want potential inside or outside
The general solution inside will be
Match at r = a:
To find coefficients, multiply by Yl'm'* and integrate over angles
So we have:
Similarly:

8. Sample Problem 2.3 Redux
The potential on the surface of a sphere of radius a centered at the origin is given by  = Ez. Find the potential outside the sphere. x z
Write potential as
We then note
So we have
Our answer is
Where

9. Sample Problem 3.1 (1)
The potential on the surface of a sphere of radius a centered at the origin is given by  = V for cos > 0 and  = –V for cos < 0 Find the potential outside the sphere.
Our answer is
Where
Recall that
Integral vanishes unless m = 0
Parity tells you
Only odd l contributes, for which

10. Sample Problem 3.1 (2)
The potential on the surface of a sphere of radius a centered at the origin is given by  = V for cos > 0 and  = –V for cos < 0 Find the potential outside the sphere.
If needed, that last integral can actually be done in closed form
Put it all together
For r/a large, terms rapidly converge

11. 3B. The Free Green’s Function
Setting it up
Dirac delta in 3D in spherical coordinates:
You can tell this is right, because:
The Green’s function in free space satisfies:
Use completeness to write right hand side in terms of spherical harmonics

12. Factoring G into Radial and Angular Functions
The Green’s function, like any function, can be written in the form
When we put this in the Laplacian, we get
Matching this with above, we have
We know solution when right side vanishes for r  r'
To make it finite at r = 0 and at r = , must have

13. Factoring G into Radial and Angular Functions
The Green’s function, like any function, can be written in the form
When we put this in the Laplacian, we get
Matching this with above, we have
We know solution when right side vanishes for r  r'
To make it finite at r = 0 and at r = , must have

14. Matching at the Boundary
Multiply this by r, then integrate from r = r' –  to r = r' + , where  is small
First term integrated by fundamental theorem of calculus
Second term is small
Right side is trivial
Substitute in:
Since finite discontinuity in derivative, function must be continuous

15. Putting it All Together
Substitute the second formula in the first:
We therefore have have
In summary,

16. Multipole Expansion
Green’s functions are used to calculate the potential from an arbitrary charge distribution
For example, let’s find the potential outside a localized distribution
So r > r'
Define the multipole moments of the distribution qlm by the expression
Note that it is easy to show that
Then the potential is given by
Leading terms are large r are given by small l

17. Sample Problem 3.2 (1)
A localized charge distribution takes the form Find the first two non-vanishing contributions to the multipole expansion, and the potential at large r.
The multipole moments are:
In spherical coordinates:
Recall that
Integral vanishes unless m = 0, in which case the  integral just yields 2
Parity can be used to argue that the  integral vanishes unless l is odd
Radial integral can be done explicitly:
So we have:
There’s a reason we have Maple:
> for l from 1 to 11 by 2 do integrate(spherharm(l,0)* sin(theta)*(Pi/2-theta),theta=0..Pi) end do;

18. Sample Problem 3.2 (2)
A localized charge distribution takes the form Find the first two non-vanishing contributions to the multipole expansion, and the potential at large r.
We find
The potential is given by
If you prefer,

19. Green’s Function Inside a Sphere
Suppose we had a grounded conducting hollow sphere with charge inside
We already know the Green’s function (from method of images):
First term we already know how to write:
For the second term, multiply by a/r and replace r by a2/r
Note that a2/r is always larger than r', so we have
Put it together:

20. Sample Problem 3.3 (1)
A hollow conducting sphere of radius a has a line charge  along the z-axis. Find the potential everywhere
This is tricky, since the line charge is at  = 0 and  = 
The value of  is irrelevant, since we are on the z-axis
We want the charge density to integrate to 2a
In spherical coordinates, we would have
Check it:
Now use the Green’s function:

21. Sample Problem 3.3 (2)
A hollow conducting sphere of radius a has a line charge  along the z-axis. Find the potential everywhere
Work on everything in the blue box
Only even l contribute

22. Sample Problem 3.3 (3)
A hollow conducting sphere of radius a has a line charge  along the z-axis. Find the potential everywhere
So final answer is:

23. 3C. Green’s Function in a Box Using a Complete Set of Modes
Green’s Function for a 3d box of dimensions a  b  c
Green’s function must satisfy
And must also satisfy
Any function that vanishes on boundary must take the form
The same with delta function
Substitute into our equation:
Since our functions are independent:

24. Finding the Green’s Function
We can find the B’s using the orthonormality of the sine functions
Put it all together and we have the Green’s functions:

25. Sample Problem 3.4
A cubical box of side a has potential zero on the surface and a surface charge density  at z = a/2 . Find the potential everywhere.
Charge density is
Potential is

26. Sample Problem 3.5
A cubical box of side a has potential zero on the surface and uniform charge  inside. Find the potential everywhere.
Potential is

27. 3D. Solving Problems in a Cylinder
Azimuthal Dependance
For problems in cylinders, we will work in cylindrical coordinates
Write functions as
Laplacian in cylindrical coordinates:
Helpful to have complete orthonormal functions for these variables
For : it is naturally periodic, since  = 0 is the same as  = 2
We know a complete set of functions like this:
These are complete and (almost) orthonormal:
They are also nice because
Any function of  can be written in terms of these
The constants fm can be found from
Can also use functions

28. Z - Dependance
We might also need some complete functions in the z-direction
If there is no restriction on z, this might be
Complete and almost orthonormal
These are also helpful because
If instead, we have a problem where function vanishes at z = 0 and z = c, then we can use as our complete functions
Again, almost orthonormal
This leaves us to find “nice” basis of functions for  dependence

29. Bessel Functions
Let’s try to find functions R() that satisfy the equation
Divide this equation by k2
Define a new variable
Then
This is called Bessel’s Equation
For each value of m, two linearly independent solutions:
J is called a Bessel function and N is called a Neumann function
General solution of original equation is therefore

30. Properties of Bessel Functions
Nice series expression for the Bessel functions:
The Jm’’s are well behaved at the origin, the Nm’s are not
At large x, they oscillate
They have an infinite number of roots
Maple is happy to compute the functions
Maple can also find roots for you
> BesselJ(m,x);BesselY(m,x)
> BesselJZeros(m,n);

31. Bessel Functions With Boundary Conditions
Our Bessel functions satisfy:
If we want them to be well-behaved at the origin, only want to use Jm’s
If it is forced to vanish at some radius a, then
We know the roots of Jm are xmn, so
We therefore want to write functions of  in terms of these functions:
Are these functions orthogonal?
Not terribly hard to show that this vanishes unless n = n'
Harder to actually find the integral

32. Bessel Functions Are Complete and Orthogonal
Given any function on a circle, we write it in polar coordinates:
We then write this function in terms of the azimuthal modes:
If the function vanishes at  = a, we expand the g’s in terms of Bessel functions
We can always find the coefficients Amn:

33. Cylinder with Ends Specified (1)
Consider a cylinder radius a and height c with no charge inside
Potential is zero on lateral surface, specified on top and bottom
We write the potential everywhere in the form
Substitute into Laplace’s equation:
Bessel functions satisfy Bessel’s equation:
Therefore:

34. Cylinder with Ends Specified (2)
Only way this can be satisfied is if
General solution of this equation is
So we have:
Must still match on the bottom and the top

35. Matching the End Conditions
We can use orthogonality to get the coefficients for the bottom
Look back three slides
In a similar manner, for the top we have
Two equation in two unknowns
Solve for mn and mn and substitute in

36. Sample Problem 3.4 (1)
A cylinder of radius a and height 2a has potential  = 0 on the lateral boundary and  = V on the ends. Find the potential everywhere, and particularly at the center
Center at origin
Potential is
On the ends, this implies
Use ortho- gonality to find
Subtract the two equations:
Therefore:

37. Sample Problem 3.4 (2)
A cylinder of radius a and height 2a has potential  = 0 on the lateral boundary and  = V on the ends. Find the potential everywhere, and particularly at the center
The  integral is zero unless m = 0
For final integral, let y = /a
Make Maple do the work
> for i to 10 do evalf(2*int(y*BesselJ(0,BesselJZeros(0,i)*y),
y=0..1)/BesselJ(1,BesselJZeros(0,i))^2) end do;

38. Sample Problem 3.4 (3)
A cylinder of radius a and height 2a has potential  = 0 on the lateral boundary and  = V on the ends. Find the potential everywhere, and particularly at the center
To get the value at the origin, set z = 0 and  = 0
J0(0) = 1 and cosh(0) = 1
Add seven terms

39. Cylinder with Lateral Surface Specified
What if the potential were specified on the sides instead?
We want complete functions that vanish on z = 0 and z = c
And still periodic on 
This suggests using modes
Write a complete set of functions
Substitute into Laplace’s equation
Since these are complete orthonormal functions on  and z, we must have

40. Modified Bessel Functions
Almost like Bessel’s Equation but sign on last term is wrong sign
The solutions are actually modified Bessel functions
They are essentially Bessel functions with imaginary arguments
The Km’s diverge at the origin, so don’t use them
It then has to match on the boundary  = a
Orthogonality lets you find the A’s

41. Sample Problem 3.5 (1)
A cylinder of radius a and height 2a has potential  = 0 on the top and bottom and  = V on the lateral surface. Find the potential everywhere, and particularly at the center
Bottom at z = 0
Potential is
On the lateral surface
Use orthogonality to find
Therefore:

42. Sample Problem 3.5 (2)
A cylinder of radius a and height 2a has potential  = 0 on the top and bottom and  = V on the lateral surface. Find the potential everywhere, and particularly at the center
At center, we have
I0(0) = 1, and sine alternates
Let Maple do the work
> add(4*evalf((-1)^n/BesselI(0,Pi*(n+1/2))/(2*n+1)/Pi),n=0..6);

43. Sample Problem 3.5b
A cylinder of radius a and height 2a has potential  = V on all surfaces. Find the potential at the center
The hard way
We previously solved it for just the ends or just the sides:
By superposition, the potential for this problem is the sum of these two problems
The easy way:
The solution of Poisson’s equation is:
Sometimes, clever approaches can give you the answer much more quickly
You could have done the first problem by subtraction