UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof.

Steven Errede

LECTURE NOTES 7
LAPLACE’S EQUATION
As we have seen in previous lectures, very often the primary task in an electrostatics problem
is e.g. to determine the electric field E ( r ) of a given stationary/static charge distribution
– e.g. via Coulomb’s Law:
Charge density
ρ ( r ′) ẑ
Field Point
Source r = r − r′ P
Point(s)
S r

v′
Volume r′
element dτ ′ Ο ŷ
in volume v′

x̂
1 rˆ r r − r′
E (r ) = ∫ r ρ ( r ′) dτ ′ r = r − r′ rˆ = =
4πε o v′
2
r r − r′

(x − xs ) + ( y p − ys ) + ( z p − zs )
2 2 2
r = r − r′ = p

Oftentimes ρ ( r ′ ) is complicated, and analytic calculation of E ( r ) is painful / tedious (or just

plain hard). (Numerical integration on a computer is likely faster/easier. . . )

Oftentimes it is easier to first calculate the potential V ( r ) , and then use E ( r ) = −∇V ( r )
1 1
Here: V ( r ) = ∫ r ρ ( r ′) dτ ′
4πε o v′
But even doing this integral analytically often can be very challenging. . .

Furthermore, often in problems involving conductors, ρ ( r ′ ) may not apriori (i.e. beforehand)
be known! Charge is free to move around, and often only the total free charge Qfree is controlled
/ known in the problem.
In such cases, it is usually better to recast the problem in DIFFERENTIAL form, using Poisson’s
equation:
ρ (r )
∇i E ( r ) = −∇i∇V ( r ) = −∇ 2V ( r ) =
εo
ρ (r )
Or: ∇ 2V ( r ) = − ⇐ Poisson’s Equation
εo

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 1

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Poisson’s equation, together with the boundary conditions associated with the value(s) allowed
for V ( r ) e.g. on various conducting surfaces, or at r = ∞, etc. enables one to uniquely determine
V ( r ) (we’ll see how / why shortly. . .).

The Poisson equation is an inhomogeneous second-order differential equation – its solution

consists of a particular solution for the inhomogeneous term (RHS of Poisson’s Equation) plus
the general solution for the homogeneous second-order differential equation:

∇ 2V ( r ) = 0 ⇐ Laplace’s Equation

commensurate with the boundary conditions for the specific problem at hand.

Very often, in fact, we are interested in finding the potential V ( r ) in a charge-free region,
containing no electric charge, i.e. where ρ ( r ′ ) = 0.

If ρ ( r ′ ) = 0, then ∇ 2V ( r ) = 0 and the TRIVIAL solution is V ( r ) = 0 ∀ r , which is boring /

useless!

We seek physically meaningful / non-trivial solutions V ( r ) ≠ 0 that satisfy ∇ 2V ( r ) = 0 and the

boundary conditions on V ( r ) for a given physical problem.

Now, before we go any further on this discussion, let’s back up a bit and take a (very) broad
generalized MATHEMATICAL view (or approach) to find V ( r ) .

First, let’s simplify the discussion, by talking about one-dimensional problems:

If ρ ( x ) = 0 , Laplace’s Equation in one-dimension becomes (in rectangular/Cartesian

coordinates):
d 2V ( x )
∇ V (r ) = 0
2
⇒ =0 ⇐ Note the total (not partial) derivative with regards to x.
dx 2
Integrating this equation (both sides) once, we have:

d 2V ( x ) d ⎛ dV ( x ) ⎞ ⎛ dV ⎞ dV ( x )
∫ 2
d x
dx = ∫ ⎜
dx ⎝ dx ⎠
⎟ dx = ∫ d ⎜
⎝ dx
⎟=
⎠ dx
= ∫ dx = m = 1st constant of integration

dV ( x )
Then: ∫ dx
dx = ∫ mdx = m ∫ dx

∫ dV ( x ) = V ( x ) = mx + b ← 2
nd
Or: constant of integration

d 2V ( x )
So: V ( x) = b + mx (equation for a straight line) is the general solution for = 0.
dx 2
y-intercept slope

2 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Depending on the boundary conditions for the problem, e.g. suppose V ( x = 5 ) = 0 Volts and
V ( x = 1) = 4 Volts, then together, these two boundary conditions uniquely specify what b and m
must be – we have two equations, and two unknowns (m & b) – solve simultaneously:

V ( x ) = b + mx ← equation for a straight line

y-intercept slope

V ( x = 5 ) = 0 = b + 5m → b = −5m
V ( x = 1) = 4 = b + 1m → 4 = −5m + 1m = −4m
∴ V ( x ) = 5 − 1x or: m = −1 and b = 5.
V ( x ) = 5 − 1x is the equation of a straight line for this problem.

5 b = 5 is y-intercept

4
V(x)
3

2 slope m = −1

0 1 2 3 4 5 x
x−a x x+a

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 3

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

General features of 1-D Laplace’s Equation ∇ 2V ( x ) = 0 and potential V ( x ) :

1. From above one-dimensional case V ( x ) = b + mx (general solution = straight line eqn.)

we can see that:

V ( x ) is the average of V ( x + a ) and V ( x − a ) i.e. V ( x ) = 1

2 {V ( x + a ) + V ( x − a )}
⇒ Laplace’s Equation is a kind of averaging instruction

The solutions of V ( x ) are as “boring” as possible, but fit the endpoints (boundary conditions)
properly.
This may be “obvious” in one-dimension, but it is also true / also holds in 2-D and 3-D cases of
∇ 2V ( r ) = 0 .

2. ∇ 2V ( r ) tolerates / allows NO local maxima or minima – extrema must occur at endpoints

i.e. ∇ 2V ( r ) = 0 requires the second spatial derivative(s) of V ( r ) to be zero.
- Not a proof, because e.g. ∃ fcns ( x ) where the second derivative vanishes other than at
endpoints - e.g. f ( x ) = x 4 (has a minimum at x = 0).

Laplace’s Equation in Two Dimensions (in Rectangular/Cartesian Coordinates)

∂ 2V ∂ 2V
If V = V ( x, y ) then ∇ 2V = 0 ⇒ + =0
∂x 2 ∂y 2

n.b. now have partial derivatives of V ( r ) .

⎛ ∂2 ∂2 ⎞
Because ∇ 2V = 0 = ⎜ 2 + 2 ⎟ V ( x, y ) now contains partial derivatives, the general solution
⎝ ∂x ∂y ⎠
does not contain just two arbitrary constants or any finite number - ∃ an infinite number of
possible solutions (in general)
– the most general solution is a linear combination of harmonic functions (sine and cosine
functions of x and y in rectangular coordinates and other functions (Bessel Functions) in
cylindrical coordinates).

Nevertheless, V ( x, y ) will still wind up being the average value of V around a point ( x, y )
within a circle of radius R centered on the point ( x, y ) .

4 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

The Method of Relaxation - Iterative Computer Algorithm for Finding V ( x, y ) :

1
V ( x, y ) = V ( r ) dl
2π R circle∫C of
radius R
centered
on (x,y)

- Start with V ( x, y ) as specified on boundary (fixed)

- Choose reasonable “interpolated” values of V ( x, y ) (from boundary) on interior ( x, y )
points away from the boundaries.
- 1st pass reassigns V ( x, y ) = average value at interior point ( x, y ) of its nearest
neighbors.
- 2nd pass repeats this process . .
- 3rd pass repeats this process . . .
- etc. ….

After few iterations, V ( x, y ) of nth iteration settles down, e.g. when:

iteration n iteration n −1

ΔV ( x, y ) = Vn ( x, y ) − Vn −1 ( x, y ) ≤ tolerance

then QUIT iterating, V ( x, y ) is determined after nth iteration is “good enough”.

V ( x, y ) again will have no local maxima or minima – all extrema will occur on boundaries.
∇ 2V ( x, y ) = 0 has solution V ( x, y ) which is the most featureless function – as smooth as
possible.

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 5

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Laplace’s Equation in Three Dimensions

Can’t draw this on 2-D sheet of paper (because now this is a 4 dimensional problem!), but:

V ( x, y, z ) = V ( r ) = average value of V over a spherical surface of radius R centered on r .

1
i.e. V (r ) = ∫ Vda
4π R 2 sphere at r
of radius R

Again V ( r ) will have no local maxima or minima

- all extrema must occur at boundaries of problem (see work-through proof in Griffiths, p. 114)
- The average potential produced by a collection of charges, averaged over a sphere of
radius R is equal to the value of the potential at the center of that sphere!

Boundary Conditions on the Potential V ( r )

Dirichlet Boundary Conditions on V ( r ) :

V ( r ) itself is specified (somewhere) on the boundary - i.e. the value of V ( r ) is specified
(somewhere) on the boundary.

Neumann Boundary Conditions on V ( r ) :

The normal derivative of V ( r ) is specified somewhere on the boundary - i.e.
∇V ( r )inˆ = − E ⊥ ( r ) is specified somewhere on the boundary.

6 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Uniqueness Theorem(s):

Suppose we have two solutions of Laplace’s equation, V1 ( r ) and V2 ( r ) , each satisfying the
same boundary condition(s), i.e. the potentials V1 ( r ) and V2 ( r ) are specified on the boundaries.
We assert that the two solutions can at most differ by a constant. (n.b. Only differences in the
scalar potential V ( r ) are important / physically meaningful!)

Proof: Consider a closed region of space with volume v which is exterior to n charged
conducting surfaces S1, S2, S3. . . Sn that are responsible for generating the potential V.
The volume v is bounded (outside) by the surface S.

Charged Conducting Surfaces closed region of

S1 volume v , bounded
by enclosing surface S.
S2 S3

S4 v, S

Suppose we have two solutions V1(r) and V2(r) both satisfying ∇ 2V ( r ) = 0 i.e. ∇ 2V1 ( r ) = 0
and ∇ 2V2 ( r ) = 0 in the charge-free region(s) of the volume v.

V1(r) and V2(r) satisfy either Dirichlet boundary conditions or satisfy Neumann boundary
conditions ∇V ( r )inˆ on the surfaces S1, S2, S3. . . Sn. We also demand that V(r) be finite at r = ∞.

Let us define: VΔ ( r ) ≡ V1 ( r ) − V2 ( r ) = difference in the two potential solutions at the point r .

Since both ∇ 2V1 ( r ) = 0 and ∇ 2V2 ( r ) = 0 then:

∇ 2VΔ ( r ) = ∇ 2 (V1 ( r ) − V2 ( r ) ) = ∇ 2V1 ( r ) − ∇ 2V2 ( r ) = 0 Note that: ∇ 2V ( r ) = ∇i ∇V ( r ) ( )

separately separately
=0 =0

The potentials Vi =1,2 are uniquely specified on charged (equipotential) surfaces S1, S2, S3, . . . Sn
in the volume v .

( )
Now apply the divergence theorem to the quantity VΔ ∇VΔ ; we also define: EΔ ( r ) ≡ −∇VΔ ( r )

∫ ∇i(V ∇V ) dτ =
v
Δ Δ
S+
∫ (V ∇V )idA =
Δ Δ
S+
∫ ( )
VΔ − EΔ idA
S1 + S2 + S3 +... Sn S1 + S2 + S3 +... Sn

Volume integral
over enclosing Surface integral over
volume v ALL surfaces in v

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 7

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Then:
− ∫ ( ) ( )
VΔ EΔ idA = − ∫ VΔ EΔ idAS − ∫ VΔ EΔ idAS1 − ∫ VΔ EΔ idAS2 − ... − ∫ VΔ EΔ idASn ( ) ( ) ( )
S+ S S1 S2 Sn
S1 + S2 + S3 +... Sn

Recognizing that:
1. The conducting surfaces S1, S2, S3, . . . Sn, are equipotentials.
Thus: VΔ ( r ) = V1 ( r ) − V2 ( r ) (= a constant on surfaces S1, S2, S3, . . . Sn) must = 0 at/on those surfaces!!!
2. The volume v is arbitrary, so let’s choose volume v → ∞, and thus surface area S → ∞ as well.
3. ∫ EΔ idAi = Φ Ei = electric flux through ith surface.
Si

4. VΔ ( r → ∞ ) = V1 ( r → ∞ ) − V2 ( r → ∞ ) (= constant on surface S → ∞) must = 0

because V1 ( r → ∞ ) = V2 ( r → ∞ ) .

∴ ∫ (
∇i VΔ ∇VΔ dτ = − VΔS ) ∫ EΔ idAS − VΔS1 ∫ EΔ idAS1 − VΔS2 ∫E
Δ idAS2 − ... − VΔSn ∫E
Δ idASn
v =0 S = 0 S1 = 0 S2 = 0 Sn
all space all space
= Φ SE1 = Φ SE2 = Φ En
S
= Φ SE

Thus: ∫ (
∇i VΔ ∇VΔ dτ = 0 )
v
all space
=∇VΔ i ∇VΔ

However, using the identity ∇i VΔ ∇VΔ = VΔ ( ∇ 2VΔ ) + ∇VΔ ( ) ( )

2

( ) VΔ ( ∇ 2VΔ ) dτ + ∫ ( ∇V )
2
Then:
v
∫ ∇i VΔ ∇VΔ dτ = ∫
v v
Δ dτ = 0
all space all space =0 all space

∫ ( ∇V ) ∫ ( ∇V i∇V ) dτ = 0
2
= Δ dτ = Δ Δ
v v
all space mathematically all space
≥0
can be

∫ ( ∇V ) ( ) = ( ∇V ( r )i∇VΔ ( r ) ) = 0 .
2 2
The only way Δ dτ = 0 is iff (i.e. if and only if) the integrand ∇VΔ ( r ) Δ
V
mathematically
≥0

( ) = ( ∇V ( r )i∇VΔ ( r ) ) = 0 , then: ∇VΔ ( r )

2
If ∇VΔ ( r ) Δ itself must be = 0

( i.e. A ( r )i A ( r ) = 0 ⇒ A ( r ) ≡ 0 ) for all points ( r ) in volume v.

If ∇VΔ ( r ) = 0 for all points r in volume v , then VΔ ( r ) = (same) constant at all points in
volume v. ∴ VΔ ( r ) = V1 ( r ) − V2 ( r ) = constant at all points in volume v.

8 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Dirichlet Boundary Conditions (V specified on surfaces S1, S2, S3, . . ., Sn)

If V1(r) and V2(r) are specified on the surfaces S1, S2, S3, . . ., Sn in the volume v enclosed by
surface S (Dirichlet boundary conditions), then: VΔ ( r ) = V1 ( r ) − V2 ( r ) = 0
(i.e. the problem is over-determined).
must = 0!

∴ VΔ ( r ) = 0 throughout the volume v and V1 ( r ) = V2 ( r ) throughout the volume v.

i.e. the two solutions V1(r) and V2(r) for ∇ 2V ( r ) = 0 are identical – there is only one unique solution.

Neumann Boundary Conditions ( E ⊥ specified on surfaces S1, S2, S3, . . ., Sn)

If ∇V1 inˆ = − E1⊥ and ∇V2 inˆ = − E2⊥ are specified on the surfaces S1, S2, S3, . . ., Sn in the volume v
enclosed by surface S (Neumann boundary conditions), then ∇VΔ ( r ) = ∇V1 ( r ) − ∇V2 ( r ) = 0 at
all points in volume v and ∇VΔ inˆ = 0.

Then VΔ ( r ) = V1 ( r ) − V2 ( r ) = constant, but is not necessarily = 0 !!!

Here, solutions V1(r) and V2(r) can differ, but only by a constant Vo.
e.g. V1 ( r ) = V2 ( r ) + Vo ⇒ problem is NOT over-determined for V ( r ) .
( E ( r ) is over-determined / unique, but not V ( r ) ).

Physical Example:

The Parallel Plate Capacitor: E = ΔV/d =100 V/m + 100 V

E d=1m
0V

Or: E = ΔV/d = 100 V/ m +500 V

E d=1m
+400 V

ΔV = 100 V in both cases – thus E-field is same/identical in both cases!

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 9

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

If we instead specify the charge densit(ies) ρ ( r ) within the volume v (see figure below), then we
also have a uniqueness theorem for the electric field associated with Poisson’s equation
(∇i E ( r ) = −∇ 2V ( r ) = ρ ( r ) ε o . )

Charged Conducting Surfaces closed region of

Q1 volume v , bounded
S1 by enclosing surface S.
S2 Q2 Q3 S3

Q4 S4 v, S
ρ ( r ) specified

Suppose there are two electric fields E1 ( r ) and E2 ( r ) , both satisfying all of the boundary
conditions of this problem. Both obey Gauss’ law in differential and integral form everywhere
within the volume v:

∇i E1 ( r ) = ρ ( r ) ε o and: ∇i E2 ( r ) = ρ ( r ) ε o

1 1
∫ E1 ida =
εo
Qiencl and: ∫ E2 ida =
εo
Qiencl
i th conducting i th conducting
surface , Si surface , Si

At the outer boundary (enclosing surface S) we also have:

1 1
∫ E1 ida = ∫ E2 ida =
encl encl
Qtot and: Qtot
S εo S εo

We define the difference in electric fields: EΔ ( r ) ≡ E1 ( r ) − E2 ( r ) which, in the region between

the conductors, obeys ∇i EΔ ( r ) = ∇i E1 ( r ) − ∇i E2 ( r ) = ρ ( r ) ε o − ρ ( r ) ε o = 0 , and obeys
1 1
∫ Si
EΔ ida = ∫ E1 ida − ∫ E2 ida =
Si Si εo
Qiencl −
εo
Qiencl = 0 over each boundary surface Si.

Even though we do not know how the charge Qi on the ith conducing surface Si is distributed,
we do know that each surface Si is an equipotential, hence the scalar potential VΔ ≡ V1 − V2 on
each surface is at least a constant on each surface Si (n.b. VΔ may not necessarily be = 0, since in
general V2 may not in general be equal to V1 on each/every surface Si).

10 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

( )
Using Griffith’s product rule # 5: ∇i fA = f ∇i A + Ai ∇f , then: ( ) ( )
( )
∇i VΔ EΔ = VΔ ∇i EΔ + EΔ i ∇VΔ( ) ( )
However, in the region between conductors, we have shown (above) that ∇i EΔ ( r ) = 0 ,
( ) (
and EΔ ≡ −∇VΔ , hence: ∇i VΔ EΔ = EΔ i ∇VΔ = − EΔ i EΔ = − EΔ2 . )
If we integrate this relation over the entire volume v (with associated enclosing surface S):

∫ ∇i(V E ) dτ = ∫
v Δ Δ all S
VΔ EΔ ida = − ∫ EΔ2 dτ
v

Note that the surface integral covers all boundaries of the region in question – the enclosing outer
surface S and all of the Si inner surfaces associated with the i conductors. Since VΔ is a constant
on each surface, it can be pulled outside of the surface integral (n.b. if the outer surface S is at
infinity, then for localized sources of charge, VΔ ( r = ∞ ) = 0 ). Thus:

VΔ ∫ EΔ ida = − ∫ EΔ2 dτ
all S v

But since we have shown above that ∫ Si

EΔ ida = 0 for each surface Si, then ∫ all S
EΔ ida = 0 .

∫ E dτ = 0 . Note that the integrand E ( r ) = E ( r )i E ( r ) is always non-negative.

2 2
Therefore: Δ Δ Δ Δ
v

Hence, in general, the only way that this integral can vanish is if EΔ ( r ) ≡ E1 ( r ) − E2 ( r ) = 0
everywhere, thus, we must have E1 ( r ) = E2 ( r ) .

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 11

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Solving Laplace’s Equation ( ∇ 2V ( r ) = 0 ) in 3-D, 2-D and 1-D Situations

In general, when solving the potential V ( r ) problems in 3 (or less) dimensions, first note the
symmetries associated with the problem. Then, if you have:

Rectangular Solve Rectangular

Cylindrical Symmetry ⇒ Problem Cylindrical Coordinates
Spherical Using Spherical

In 2-D and 3-D problems, the general solutions to ∇ 2V ( r ) = 0 are the harmonic functions (an ∞-
series solution, in principle) e.g. of sines and cosines, Bessel functions, or Legendre
Polynomials and/or Spherical Harmonics.

The boundary conditions / symmetries will select a subset of the ∞-solutions.

We will now work through derivations of finding solutions to Laplace’s Equations in

3-dimensions in rectangular (i.e. Cartesian) coordinates, cylindrical coordinates, and spherical
coordinates. We will also use / show the method of separation of variables.

Laplace’s Equation ∇ 2V ( x, y, z ) = 0 and Potential Problems with Rectangular Symmetry

(Rectangular / Cartesian coordinates)
In Three Dimensions: Solve Laplace’s equation in rectangular / Cartesian coordinates:

⎛ ∂2 ∂2 ∂2 ⎞ ∂ 2V ∂ 2V ∂ 2V
∇ 2V ( x, y, z ) = ⎜ 2 + 2 + 2 ⎟ V ( x, y, z ) = 2 + 2 + 2 = 0
⎝ ∂x ∂y ∂z ⎠ ∂x ∂y ∂z

The solutions of ∇ 2V = 0 in rectangular coordinates are known as harmonic functions (i.e. sines
and cosines) (→ Fourier Series Solutions).

It is usually (but not always) possible to find a solution to the Laplace Equation, ∇ 2V = 0 which
also satisfies the boundary conditions, via separation of variables technique, i.e. try a product
solution of the form:

V ( x, y , z ) = X ( x ) Y ( y ) Z ( z )

where: X (x) x
Y (y) are functions only of y respectively.
Z (z) z

∂ 2V ( x, y, z ) ∂ 2V ( x, y, z ) ∂ 2V ( x, y, z )
Then: ∇ 2V ( x, y, z ) = 0 ⇒ + + =0
∂x 2 ∂y 2 ∂z 2

But: V ( x, y, z ) = X ( x ) Y ( y ) Z ( z )

12 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

∂2 X ( x)Y ( y ) Z ( z ) ∂2 X ( x)Y ( y ) Z ( z ) ∂2 X ( x )Y ( y ) Z ( z )
Thus: + + =0
∂x 2 ∂y 2 ∂z 2
∂2 X ( x) ∂ 2Y ( y ) ∂2Z ( z )
= Y ( y) Z ( z) + X ( x) Z ( z ) + X ( x)Y ( y ) =0
∂x 2 ∂y 2 ∂z 2

Now divide both sides of the above equation by X ( x ) Y ( y ) Z ( z ) :

1 ∂2 X ( x) 1 ∂ Y ( y)
2
1 ∂ Z ( z)
2

Then: = + + =0
X ( x ) ∂x 2 Y ( y ) ∂y 2 Z ( z ) ∂z 2

1 ∂2 X ( x) 1 ∂ Y ( y)
2
1 ∂ Z ( z)
2

But: = + + = 0 i.e. C1 + C2 + C3 = 0
X ( x ) ∂x 2 Y ( y ) ∂y 2 Z ( z ) ∂z 2
fcn(x) only fcn(y) only fcn(z) only
independent of y, z independent of x, z independent of x, y
= C1 = C2 = C3

True for all points (x, y, z)

in volume v of problem.
The only way the above equation can be true for all points (x, y, z) in volume v is if:

1 ∂2 X ( x) d 2 X ( x)
= constant C1 ⇒ − C1 X ( x ) = 0 #1
X ( x) ∂x 2 dx 2
1 ∂ Y ( y) d 2Y ( y )
2

= constant C 2 ⇒ − C2Y ( y ) = 0 #2
Y ( y ) ∂y 2 dy 2
1 ∂ Z ( z) d 2Z ( z )
2

= constant C3 ⇒ − C3 Z ( z ) = 0 #3
Z ( z ) ∂z 2 dz 2
Note total derivatives now!!!
Subject to the constraint: C1 + C2 + C3 = 0
Can now solve 3 ORDINARY 1-D differential equations, #1–3, which are subject to C1 + C2 + C3 = 0,
PLUS the specific Dirichlet / Neumann boundary conditions for the problem on either V (x, y, z)
or ∇V ( x, y, z )inˆ at surfaces for this 3-D problem.

Essentially, we have replaced the 3-D problem with three 1-D problems, and the constraint:
C1 + C2 + C3 = 0.

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 13

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

* If one has a 2-D rectangular coordinate problem ( ∇ 2V ( x, y ) = 0 ) , then: V (x, y) = X(x)Y(y) (only).

1 d 2 X ( x) d 2 X ( x)
= C1 ⇒ − C1 X ( x ) = 0
X ( x ) dx 2 dx 2
1 d Y ( y) d 2Y ( y )
2

= C2 ⇒ − C2Y ( y ) = 0
Y ( y ) dy 2 dy 2

Subject to the constraint: C1 + C2 = 0, i.e. C1 = −C2.

Plus BC’s: either on V (x, y) or ∇V ( x, y )inˆ for the 2-D problem.

* If one has a 1-D rectangular coordinate problem ( ∇ 2V ( x ) = 0 ) , then: V(x) = X(x) (only).

d 2V ( x ) d 2 X ( x) 1 d X ( x)
2

= 0 ⇒ = 0 ⇒ = 0 = C1
dx 2 dx 2 X ( x) dx 2
d 2 X ( x)
= 0 ⇒ X ( x) = V ( x) = ax + b is the 1-D general solution.
dx 2

For 1-D problem ( ∇ 2V ( x ) = 0 ) , only need to solve one ordinary differential equation subject to
dV ( x)
the constraint C1 = 0 and BC’s on either V(x) or .
dx

14 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

The General Solution V ( x, y, z ) = X ( x ) Y ( y ) Z ( z ) for ∇ 2V ( x, y, z ) = 0

in Rectangular Coordinates

Since we have the constraint C1 + C2 + C3 = 0, at least one of the Ci’s (i = 1, 2 or 3) must be less
than zero.

Let us “choose” C1 = −α 2 , C2 = − β 2 , C3 = γ 2

Then: C1 + C2 + C3 = 0
−α 2 − β 2 + γ 2 = 0 or: α2 + β2 = γ 2

The boundary conditions on the surfaces will define α and β , and hence define γ .

IMPORTANT NOTE:
The geometry (x – y – z) of the problem and the boundary conditions dictate whether:
C1 > 0 or C1 < 0
C2 > 0 or C2 < 0
C3 > 0 or C3 < 0

i.e. have sine / cosine type solutions vs. sinh / cosh (or e x , e− x ) type solutions for x, y, z.

Then the General Solution is (for above choice of C1 = −α 2 , C2 = − β 2 , C3 = γ 2 ):

∞
V ( x, y , z ) = ∑A
m,n=0
mn sin (α n x ) sin ( β m y ) sinh ( γ mn z ) so we also have the additional series solutions:
could be could be could be
cos cos cosh = α n2 + β m2

∞
+ ∑ Bmn cos (α n x ) cos ( β m y ) sinh ( γ mn z )
m,n=0
= α n2 + β m2
∞
+ ∑ Cmn sin (α n x ) sin ( β m y ) cosh ( γ mn z )
m,n=0
= α n2 + β m2
∞
+ ∑ Dmn cos (α n x ) cos ( β m y ) cosh ( γ mn z )
m,n=0
= α n2 + β m2

1 x −x 1 x −x
n.b. cosh( x) =
2
(e + e ) sinh( x) =
2
(e − e )
1 ix − ix 1
n.b. sin( x) =
2i
(e − e ) cos( x) = ( eix + e − ix )
2
i ≡ −1
− ix
n.b. eix = cos( x) + i sin( x) e = cos( x) − i sin( x)
n.b. e x = cosh( x) + sinh( x) e − x = cosh( x) − sinh( x)

The BC’s and symmetries will determine which of the coefficients Amn, Bmn, Cmn, Dmn = 0.

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 15

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

We solve for the non-zero coefficients Apq, Bpq, Cpq and Dpq by taking inner products.
i.e. we multiply V ( x, y, z ) = ∑ ( stuff ) by e.g. sin(α p x) sin( β q y ) to project out the p-qth
component (i.e. we use the orthogonality properties of the individual terms in sin( ) and cos( )
Fourier Series.) and then integrate over the relevant intervals in x and y:
e.g.
xo yo
∫ ∫
0 0
V ( x, y ) sin(α p x) sin( β q y )dxdy
⎧⎪⎛ ∞ ⎞
⎨⎜ ∑ Amn sin(α n x) sin( β m y ) sinh(γ mn z ) *sin(α p x) sin( β q y ) ⎟
xo yo
=∫ ∫
⎪⎩⎜⎝ m, n =0 ⎟
0 0
= constant here ⎠
⎛ ∞ ⎞
+ ⎜ ∑ Bmn cos(α n x) cos( β m y ) sinh(γ mn z ) *sin(α p x) sin( β q y ) ⎟
⎜ m, n =0 ⎟
⎝ = constant here ⎠
⎛ ∞ ⎞
+ ⎜ ∑ Cmn sin(α n x) sin( β m y ) cosh(γ mn z ) *sin(α p x) sin( β q y ) ⎟
⎜ m, n =0 ⎟
⎝ = constant here ⎠
⎛ ∞ ⎞ ⎫⎪
+ ⎜ ∑ Dmn cos(α n x) cos( β m y ) cosh(γ mn z ) *sin(α p x) sin( β q y ) ⎟ ⎬ dxdy
⎜ m,n =0 ⎟⎪
⎝ = constant here ⎠⎭

Fourier Functions: orthonormality properties of sin ( ) and cos ( ):

sin(α n x) sin(α p x)dx = ∼∼ δ np ( == 10 for n≠p)

xo
∫
for n = p
0
some
constant

cos (α n x ) sin (α p x ) dx = 0
xo
∫ 0

Kroenecker δ -function: δ np ( == 10 for n≠p)

for n = p

So all terms in above Σ’s vanish, except for a single term (in each sum) – that for the Apq /Bpq
/Cpq /Dpq coefficient!!! The BC’s will e.g. kill off 3 out of remaining 4 non-zero terms, thus only
one term survives…

Suppose only the Apq coefficient survives. Its analytic form is now known for all integers p and q.

Then the analytic form of 3-D potential V ( x, y, z ) is now known – it is an infinite series solution
of the form:
∞
V ( x, y , z ) = ∑A
m,n=0
mn sin (α n x ) sin ( β m y ) sinh ( γ mn z )
= α n2 + β m2

16 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Laplace’s Equation ∇ 2V ( ρ , ϕ , z ) = 0
And Potential Problems with Cylindrical Symmetry (Cylindrical Coordinates)

ẑ r r = ρ + z = ρρˆ + zzˆ r = ρ 2 + z2

Ο ŷ ∇ 2V ( ρ , ϕ , z ) = 0
1 ∂ ⎛ ∂V
⎞ 1 ∂ 2V ∂ 2V
ϕ ρ = ρ
⎟+ 2 + 2 =0
ρ ∂ρ ⎜⎝ ∂ρ
⎠ ρ ∂ϕ
2
∂z
∂ V 1 ∂V 1 ∂ V ∂ 2V
2 2
ϕ̂ = 2+ + + + =0
∂ρ ρ ∂ρ ρ 2 ∂ϕ 2 ∂z 2
x̂ ρ̂

Again, we use the separation of variables technique:

V ( ρ , ϕ , z ) = R ( ρ ) Q (ϕ ) Z ( z ) ⇒ ∇ 2V = 0 ⇒ yields 3 ordinary differential equations:

d 2Z ( z )
2
− k 2 Z ( z ) = 0 ⇒ Z ( z ) = e ± kz
dz
d Q (ϕ )
2

+ν 2Q (ϕ ) = 0 ⇒ Q (ϕ ) = e ± iνϕ
dϕ 2

d 2R ( ρ ) 1 dR ( ρ ) ⎛ 2 ν 2 ⎞
+ + ⎜k − 2 ⎟ R(ρ) = 0
dρ2 ρ dρ ⎝ ρ ⎠

Note(s):
1.) k is arbitrary without imposing boundary conditions.
2.) k appears in both Z(z) and R(ρ) equations.
3.) In order for Q(φ) to be single-valued (i.e. Q (ϕ ) = Q (ϕ + 2π ) ), ν must be an integer!

d 2R ( x) 1 dR ( x ) ⎛ ν 2 ⎞
Let x ≡ kp Then: + + ⎜1 − 2 ⎟ R ( x ) = 0 ⇐ Bessel’s Equation
dx 2 x dx ⎝ x ⎠
∞
R ( x ) = xα ∑ a j x j ⇐ Power Series Solution α = ±ν
j =0

1
a2 j ≡ − a2 j − 2 for j = 0, 1, 2, 3, . . . .
4 j ( j +α )

All odd powers of xj have vanishing coefficients, i.e. a1 = a3 = a5 = a2j+1 = 0

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 17

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Coefficients a2j expressed in terms of a0:

⎡ ( −1) j Γ (α + 1) ⎤ ( −1)
j

a2 j = ⎢ 2 j ⎥ a0 = 2( j +α )
⎢⎣ 2 j !Γ ( j + α + 1) ⎥⎦ 2 j !Γ ( j + α + 1)
1
where a0 = α Γ ( x ) = Gamma Function
2 Γ (α + 1)

There exist TWO solutions of the Radial Equation (i.e. Bessel’s Equation):
They are:
Bessel Functions of 1st kind, of order ±ν :
ν
( −1)
j 2j
∞
⎛x⎞ ⎛x⎞
J +ν ( x ) = ⎜ ⎟
⎝2⎠
∑ ⎜ ⎟
j = 0 j !Γ ( j + ν + 1) ⎝ 2 ⎠
These series converge for

−ν
( −1)
j 2j
⎛x⎞ ∞ ⎛x⎞
J −ν ( x ) = ⎜ ⎟ ∑ ⎜ ⎟ all values of x.
⎝ 2 ⎠ j =0 j !Γ ( j −ν + 1) ⎝ 2 ⎠

If ν is not an integer (which is not the case here), then the J ±ν ( x ) form a pair of
n.b. linearly independent solutions to the 2nd order Bessel’s Equation:
R ( x ) = Aν Jν ( x ) + A−ν J −ν ( x ) for ν ≠ integer

However, note that if ν = integer (which is the case for us here) then the Bessel functions
Jν ( x ) and J −ν ( x ) are NOT linearly independent!!

If ν = m = integer (0, 1, 2, 3, …), then J − m ( x ) = ( −1) J m ( x )

m

∴ ⇒ We must find another linearly independent solution for R(x) when ν = m = integer

It is “customary” to replace J ± ν ( x ) by just Jν ( x ) and another function Nν ( x )

(called Neumann Functions)
Jν ( x ) cos (νπ ) − J −ν ( x )
Where: Nν ( x ) = Bessel Function of 2nd kind ≡
sin (νπ )
NOTE: Nν ( x ) is divergent (i.e singular) at x → 0

Complex Bessel Functions = Bessel Functions of 3rd kind = Hankel Functions

Hankel Functions are complex linear combinations of Jν ( x ) and Nν ( x ) (Bessel Functions of 1st
and 2nd kind respectively). They are defined as follows:

Hν(1) ( x ) ≡ Jν ( x ) + iNν ( x ) The Hankel Functions Hν(1) ( x ) and Hν( 2) ( x ) also form a
Hν( 2) ( x ) ≡ Jν ( x ) − iNν ( x ) fundamental set/basis of solutions to the Bessel equation.

18 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

The General Solution for ∇ 2V ( ρ , ϕ , z ) = 0 in Cylindrical Coordinates:

V ( ρ , ϕ , z ) = R ( ρ ) Q (ϕ ) Z ( z )

cosh(kmnz) is also allowed

∞
V ( ρ ,ϕ , z ) = ∑ J (k
m,n =0
m mn ρ ) sinh ( kmn z ) ⎡⎣ Amn sin ( mϕ ) + Bmn cos ( mϕ ) ⎤⎦

cosh(kmnz) is also allowed

∞
+ ∑ N (k
m,n =0
m mn ρ ) sinh ( kmn z ) ⎡⎣Cmn sin ( mϕ ) + Dmn cos ( mϕ ) ⎤⎦

Apply ALL boundary conditions on surfaces

(and also impose for r = ∞,that V(r = ∞) = finite! {If r = ∞ is part of the problem!})

Note that sometimes we want V ( r ) only inside some finite region of space, e.g. coaxial
capacitor – if so, then don’t have to worry about r = ∞ solutions being finite – an example – the
Coaxial Capacitor:
End View of a Coaxial Capacitor

If the r = 0 region is an excluded region in the problem, then must include (i.e. allow) the Nν ( x )
solutions (singular at x = k ρ = 0 )!!!

If r = 0 region is included in problem, then ALL coefficients Cmn = Dmn ≡ 0 (for all m, n),
if V ( ρ , ϕ , z ) is finite @ r = 0 .

Using/imposing BC’s on surfaces, orthogonality conditions on sines, cosines, Jν ( x ) , Nν ( x ) ,

etc. can find / determine values for all Amn, Bmn, Cmn, Dmn coefficients!!

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 19

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

2-Dimensional Circular Symmetry

Laplace’s Equation in (Circular) Cylindrical Coordinates

ρ
ϕ x

∇ 2V ( ρ , ϕ ) = 0 V ( ρ , ϕ ) = R ( ρ ) Q (ϕ )
Again try product solution
Potential, V ( ρ , ϕ ) is independent of z
(e.g. infinitely long coaxial cable)

1 ∂ ⎛ ∂V ⎞ 1 ∂ 2V
∇ V ( ρ ,ϕ ) =
2
ρ ⎟+ 2 =0
ρ ∂ρ ⎜⎝ ∂ρ ⎠ ρ ∂ϕ
2

Get:
ρ d ⎛ dR ( ρ ) ⎞ 1 d Q (ϕ )
2

⎜ρ ⎟ = C1 = − Let C1 = k2
R(ρ) dρ ⎝ dρ ⎠ Q (ϕ ) d ϕ 2
d ⎛ dR ( ρ ) ⎞ 2 d 2 Q (ϕ )
Then: ρ ⎜ρ ⎟ − k R ( ρ ) = 0 and + k 2 Q (ϕ ) = 0
dρ ⎝ dρ ⎠ dϕ 2

Require all solutions Q (ϕ ) to be single-valued, i.e. Q (ϕ ) = Q (ϕ + 2π )

because must have V (ϕ ) = V (ϕ + 2π ) .

Solutions for Q (ϕ ) are of the form:

Q (ϕ ) = A cos kϕ + B sin kϕ
Q (ϕ ) = Q (ϕ + 2kπ ) requires k = integer = 0, ± 1, ± 2, ± 3, … ± n …
d 2 Q (ϕ )
+ n 2Q (ϕ ) = 0 ⇒ Qn (ϕ ) = An cos ( nϕ ) + Bn sin ( nϕ )
dϕ 2

singular @ ρ → ∞ singular @ ρ = 0
d ⎛ dR ( ρ ) ⎞ 2
ρ ⎜ρ ⎟ − n R ( ρ ) = 0 ⇒ Rn ( ρ ) = Cn ρ + Dn ρ
−n
n
for n ≥ 1 (i.e. n = 1, 2, 3, . . .)
dρ ⎝ dρ ⎠
Ro ( ρ ) = Co + Do ln ( ρ ) for n = 0 only

20 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

General Solution for ∇ 2V ( ρ , ϕ ) = 0 in Two Dimensions:

Cylindrical (a.k.a. Zonal) Harmonics
∞
V ( ρ , ϕ ) = V0 + V1 ln ( ρ ) + ∑ ⎡⎣ an ρ n cos ( nϕ ) + bn ρ − n cos ( nϕ ) + cn ρ n sin ( nϕ ) + d n ρ − n sin ( nϕ ) ⎤⎦
n =1

Again, apply BC’s on all relevant surfaces, impose V ( r → ∞ ) = finite, etc. – these will dictate /
determine all coefficients, V0, V1, an, bn, cn and dn.

i.e. Solve for V0, V1, an, bn, cn and dn by applying all boundary conditions, V ( r → ∞ ) = finite,
and using orthogonality conditions / properties:
2π ρo
an = " A " ∫ dϕ ∫ d ρ ρ V ( ρ , ϕ )ρ n cos ( nϕ ) dA = ρ d ρ dϕ
0 0
2π ρo
bn = " B " ∫ dϕ ∫ d ρ ρ V ( ρ , ϕ )ρ − n cos ( nϕ )
0 0
2π ρo
cn = " C " ∫ dϕ ∫ d ρ ρ V ( ρ , ϕ )ρ n sin ( nϕ )
0 0
2π ρo
d n = " D " ∫ dϕ ∫ d ρ ρ V ( ρ , ϕ )ρ − n sin ( nϕ )
0 0
“A”, “B”, “C”, “D” are appropriate normalization factors (we will discuss later).

Laplace’s Equation ∇ 2V ( r , ϑ , ϕ ) = 0 In Spherical Coordinates

ẑ V = V ( r ,ϑ , ϕ )

ϕ̂
r̂
r
θ θ
Ο ŷ

ϕ ϕ̂

x̂

∇ 2V ( r , ϑ , ϕ ) = 0
1 ∂ ⎛ 2 ∂V ⎞ 1 ∂ ⎛ ∂V ⎞ 1 ∂ 2V
= ⎜ r ⎟ + ⎜ sin θ ⎟ + =0
r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ∂θ ⎝ ∂θ ⎠ r 2 sin 2 θ ∂ϕ 2

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 21

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Again, try separation of variables / try product solution:

U (r )
V ( r ,ϑ , ϕ ) = P (θ ) Q (ϕ ) ⇐ of this form!!
r

d 2U ( r ) U ( r ) Q (ϕ ) d ⎛ dP (θ ) ⎞ U ( r ) P (θ ) d 2Q (ϕ )
P (θ ) Q (ϕ ) + ⎜ sin θ ⎟+ 2 2 =0
dr 2 r 2 sin θ dθ ⎝ dθ ⎠ r sin θ dϕ 2

Multiply by r 2 sin 2 θ / U ( r ) P (θ ) Q (ϕ ) :

⎡ 1 d 2U ( r ) 1 1 d ⎛ dP (θ ) ⎞ ⎤ 1 d Q (ϕ )
2

r 2 sin 2 θ ⎢ + 2 ⎜ sin θ ⎟⎥ + =0
⎢⎣U ( r ) dr r sin θ P (θ ) dθ dθ ⎠ ⎥⎦ Q (ϕ ) dϕ 2
2
⎝
function of r +θ only function of ϕ only

1 d 2 Q (ϕ ) d 2 Q (ϕ )
Now: = −m 2
⇒ + m 2 Q (ϕ ) = 0
Q (ϕ ) dϕ 2
dϕ 2

Solutions are of the form: Q (ϕ ) = e ± imϕ where m = integer = 0, 1, 2, 3, …

Since V ( r , ϑ , ϕ ) = V ( r , ϑ , ϕ ± 2π ) i.e. Q (ϕ ) = Q (ϕ ± 2π )
Then: Q (ϕ ) must be single-valued!
Thus:
⎡ 1 d 2U ( r ) 1 1 d ⎛ dP (θ ) ⎞ ⎤
r 2 sin 2 θ ⎢ + 2 ⎜ sin θ ⎟⎥ = +m
2

⎢⎣U ( r ) dr r sin θ P (θ ) dθ ⎝ dθ ⎠ ⎥⎦
2

1 d U (r ) 1 d ⎛ dP (θ ) ⎞
2
1 m2
= − ⎜ sin θ ⎟ +
U ( r ) dr 2 r 2 sin θ P (θ ) dθ ⎝ dθ ⎠ r 2 sin 2 θ

multiply above equation by r2:

r2 d U (r ) ⎛ dP (θ ) ⎞
2
1 1 d m2
=− ⎜ sin θ ⎟ + = −α (α ≥ 0 )
U ( r ) dr 2
sin θ P (θ ) dθ ⎝ d θ ⎠ sin 2
θ
function only of r function only of θ

must hold for any/all r and θ !!

d 2U ( r ) α 1 d ⎛ dP (θ ) ⎞ ⎡ m2 ⎤
∴ − 2 U (r ) = 0 and ⎜ sin θ +
⎟ ⎢α − ⎥ P (θ ) = 0
dr 2 r sin θ dθ ⎝ dθ ⎠ ⎣ sin 2 θ ⎦

let / define α ≡ ( + 1) where = integer = 0, 1, 2, 3, …

(Trust me, ☺ I know the answer . . .)

d 2U ( r ) ( + 1) 1 d ⎛ dP (θ ) ⎞ ⎡ m2 ⎤
∴ − U ( r ) = 0 and ⎜ sin θ ⎟+⎢ ( + 1) − ⎥ P (θ ) = 0
dr 2 r2 sin θ dθ ⎝ dθ ⎠ ⎣ sin 2 θ ⎦

22 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Now let x = cos θ x 2 = cos 2 θ = 1 − sin 2 θ

dx = d cos θ = sin θ dθ ∴ sin 2 θ = 1 − cos 2 θ = 1 − x 2
sin θ = 1 − x 2

1 d ⎛ sin 2 θ dP (θ ) ⎞ ⎡ m2 ⎤
Then: ⎜ ⎟ + ( + 1) − P (θ ) = 0
sin θ dθ ⎝ sin θ dθ ⎠ ⎢⎣ sin 2 θ ⎥⎦
⎡ m2 ⎤
d ⎛ 2 dP ( x ) ⎞
Becomes: ⎜(1 − x ) dx ⎟ ( ) 1 − x 2 ⎥ P ( x ) = 0 ⇐
+ ⎢ + 1 − Generalized Legendre' Equation
dx ⎝ ⎠ ⎢⎣ ( ) ⎥⎦
d 2U ( r ) ( + 1) U r = 0 are of the form:
General Solutions of the radial equation, 2
− ( )
dr r2
U ( r ) = Ar + Br − ( +1) (l + A + B) are determined by boundary conditions…

For m = 0 (azimuthally-symmetric problems – no φ-dependence) the general solution for

azimuthally-symmetric potential V ( r , θ ) is of the form:
∞
V ( r ,θ ) = ∑ ⎡⎣ A r + B r − ( +1)
⎤⎦ P ( cos θ )
=0
" ordinary " Legendre '
Polynomial of order

The coefficients A and B are determined by the boundary conditions

n.b. If ∃ no charges at r = 0, then B = 0 ∀ !!

Rodrigues’ Formula is useful for “ordinary” Legendre' Polynomials:

⎛ 1 ⎞d
P ( x) ≡ ⎜ ⎟
⎝ 2 ! ⎠ dx
( x 2 − 1)

The coefficients A and B can be found / determined by evaluating V ( r , θ ) on the conducting

surfaces in the problem, e.g. suppose we want to determine V ( r ) inside a conducting sphere of
radius r = a. Then on the surface of the conducting sphere at radius r = a (an equipotential!):
V ( r = a,θ ) = ∑ A a P ( cos θ ) = constant ⇐ Legendre' Series

n.b. inside conducting sphere, e.g. there are no charges at r = 0 ∴ B =0∀

In order to determine coefficients, take inner product:

⎛ ( 2 + 1) ⎞ π
A =⎜ ⎟ ∫0 V ( r = a, θ ) P ( cos θ ) sin θ dθ
⎝ 2a ⎠
normalization
factor

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 23

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Orthogonality condition on P ( x ) ’s:

−1 2 ⎪⎧δ l 'l = 0 for l ' ≠ l ⎪⎫
∫ P ′ ( x ) P ( x ) dx = δ′ ⎨ ⎬
−1
( 2 + 1) Kroenecker ⎪⎩δ l 'l = 1 for l ' = l ⎭⎪
δ -function

n.b. The P ( cos θ ) functions form a complete orthonormal basis set on the unit circle (r = 1)
for −1 ≤ cos θ ≤ 1 or: 0 ≤ θ ≤ π

“Ordinary” Legendre’ Polynomials P ( x ) ( x = cos θ ) defined on the interval −1 ≤ x ≤ 1 :

P0 ( x ) = 1
P1 ( x ) = x
1
P2 ( x ) =
2
( 3 x 2 − 1)
1
P3 ( x ) = ( 5 x 3 − 3 x )
2
1
P4 ( x ) = ( 35 x 4 − 30 x 2 + 3)
8
1
P5 ( x ) = ( 63 x 5 + 70 x 3 + 15 x )
8
…

Note: All P =even ( x ) functions are even functions of x: P =even ( − x ) = + P =even ( x )

All P =odd ( x ) functions are odd functions of x: P =odd ( − x ) = − P =odd ( x )
under x → −x reflection. Generally speaking, P ( − x ) = ( −1) P ( x ) .

If 3-D spherical coordinate problem DOES have azimuthal / ϕ -dependence, then m 2 ≠ 0

in Associated Legendre' Equation (A.L.E.):

1 d ⎛ dP (θ ) ⎞ ⎡ m2 ⎤
⎜ sin θ ⎟+⎢ ( + 1) − ⎥ P (θ ) = 0 x = cos θ
sin θ dθ ⎝ dθ ⎠ ⎣ sin 2 θ ⎦

Solutions to A.L.E. are Associated Legendre' Polynomials (A.L.P.’s)

m
dm
Associated Legendre’ Polynomials: P m ( x ) ≡ ( −1) (1 − x 2 ) 2 P ( x)
m

dx m "ordinary "
Legendre '
Polynomial

m = ± integer ≠ 0
i.e. m = ±1, ±2, ±3,... but have a constraint on m !!! − ≤m≤+

i.e. m = − , − + 1, − + 2,... − 2, −1, 0, +1, +2, − 2, − 1,

24 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Also: P − m ( x ) = ( −1)
m ( − m )! m
P ( x)
( + m )!

Orthogonality condition for P m ( x ) for fixed m:

1 2 ( + m )!
∫ P m′ ( x ) P m ( x ) dx = δ′
−1
( 2 + 1) ( − m )! Kroenecker
δ -function

We now define normalized P (θ ) Q (ϕ ) functions known as Spherical Harmonics:

Y m (θ , ϕ ) ≡
(2 + 1) ( − m ) ! m
P ( cos θ ) eimϕ
4π ( + m ) ! = Qm (ϕ )

The Spherical Harmonics Y m (θ , ϕ ) form a complete orthonormal set of basis “vectors” on the
surface of the unit sphere (r = 1)

Note that Y − m (θ , ϕ ) = ( −1) Y *m (θ , ϕ )

m
complex conjugate

i.e. i → −i where i ≡ −1

Y m (θ , ϕ ) Normalization and Orthogonality Condition:

2π π
∫ dϕ ∫ sin θ dθ Y *′m′ (θ , ϕ ) Y m (θ , ϕ ) = δ ′ δ m′m
0 0
Ω= 4π
i.e. ∫ d ΩY *′m′ (θ , ϕ ) Y m (θ , ϕ ) = δ ′ δ m′m d Ω = sin θ dθ dϕ
Ω= 0

∞
Completeness’ Relation: ∑ ∑ Y (θ ,ϕ ) Y (θ ,ϕ ) = δ ( cos θ − cos θ ') δ (ϕ − ϕ ')
=0 m =−
*
m m

DIRAC δ -functions

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 25

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

Y m (θ , ϕ ) Spherical Harmonics

1
Use Y − m (θ , ϕ ) = ( −1) Y *m (θ , ϕ )
m
=0 Y00 =
4π
in order to obtain Y2− 2 , Y2−1 , Y3−3 , Y3− 2 , Y3−1 etc.
3
=1 Y11 = − sin θ eiϕ
8π
3
Y10 = − cos θ Note:
4π

Y m (θ , ϕ ) ≡
(2 + 1) ( − m ) ! m
P ( cos θ ) eimϕ
4π ( + m ) !

Y22 =
1 15
sin 2 θ e 2iϕ Y 0 (θ , ϕ ) =
(2 + 1)
P ( cos θ )
4 2π 4π
5
=2 Y21 = − sin θ cos θ eiϕ
8π
5 ⎛3 1⎞
Y20 = ⎜ cos θ − ⎟
2

4π ⎝ 2 2⎠

1 35
Y33 = − sin 3 θ e3iϕ
4 4π
1 105 2
=3 Y32 = sin θ cos θ e 2iϕ
4 2π
1 21
Y31 = − sin θ ( 5cos 2 θ − 1) eiϕ
4 4π
7 ⎛5 3 ⎞
Y30 = ⎜ cos θ − cos θ ⎟
3

4π ⎝ 2 2 ⎠
…. …. etc.

26 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois

2005 - 2008. All rights reserved.
 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede

General Solution for Laplace’s Equation ∇ 2V ( r , θ , ϕ ) in Spherical Polar Coordinates

∞ +
V ( r ,θ , ϕ ) = ∑ ∑ ⎡⎣ A m r + B m r −( +1)
⎤Y m (θ , ϕ )
⎦
=0 m =−

Coefficients A m and B m are determined by / from Boundary Conditions on spherical surface(s)

If V = V (θ , ϕ ) on surface (e.g. at r = a)
(i.e. no charge at r = 0 in problem → B m = 0 ∀ ,m )

∞ +
Then: V (θ , ϕ ) = ∑ ∑A m Y m (θ , ϕ ) on surface (r = a).
=0 m =−
4π
And: A m = ∫ d ΩY *m (θ , ϕ ) V (θ , ϕ ) on surface (r = a).
0

Note: V (θ = 0, ϕ ) = ∑
∞
(2 + 1)
A0
" north " =0 4π
pole

(2 + 1) 4π
d ΩP ( cos θ ) V (θ , ϕ )
A0 =
4π ∫ 0

General Comments: The method of separation of variables used in Laplace’s equation ∇ 2V = 0

in rectangular, cylindrical and spherical coordinates shows up again in Poisson’s Equation
ρ free
∇ 2V = − and also in the wave equation (valid for all classical wave phenomena)
εo
1 ∂ ψ (r,t )
2

∇ ψ (r,t ) + 2
2
= 0 and in Schrödinger’s wave equation Hψ = Eψ in Quantum
c ∂t 2

Mechanics problems. These equations will appear again and again, in one form or another for
E&M, Classical Mechanics, Quantum Mechanics courses as well as for Classical / Newtonian
Gravity problems…

For more detailed information e.g. on separation of variables and solutions to 3-D Wave
1 ∂ 2ψ
Equation ∇ 2ψ = − 2 2 in rectangular, cylindrical and spherical coordinates see Prof. S.
c ∂t
Errede lecture notes (Lecture IV – parts 1 & 2) on (sound) waves in 1-D, 2-D, 3D Physics 406
Acoustical Physics of Music website:

http://online.physics.uiuc.edu/courses/phys406/406_lectures.html

and also see/read his Fourier Analysis Lectures on this website, if interested.

©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 27

2005 - 2008. All rights reserved.