h - Eldred Lee

BARD COLLEGE AT SIMON’S ROCK
FUNDAMENTALS OF INTERMEDIATE ELECTROMAGNETISM
MATH365T – PARTIAL DIFFERENTIAL EQUATIONS
This Project is Dedicated to Dr. Bernard Clark Musselman II
BY
ELDRED LEE
GREAT BARRINGTON, MA
15 MAY 2014
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Introduction
The purpose of this project is to show the ability to apply the knowledge of partial
differential equations and ordinary differential equations. I have decided to solve three problems
that are essential for intermediate electromagnetism that need to be done using the separation of
variables. The problems can be seen on the “Special Techniques” section of Introduction to
Electrodynamics by David J. Griffiths. In the first two problems, with the initial boundary values
given in each problem, I will be developing general equations for necessary potentials. The third
problem will involve applications of ordinary differential equations and power series. There will
be separations of variables in 2D and 3D Cartesian coordinates in the first two problems and
separation of variables in 3D spherical coordinates with azimuthal symmetry. Some steps might
have been skipped due to the desire to avoid the lengthiness of the project. MathType was used
to write the mathematical expressions in this project.
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Problems
1. A – Develop a general formula for the potential within the pipe & B – Find the potential
explicitly, for the case Vo (y) = Vo (a constant): A rectangular pipe, running parallel to the zaxis (from -∞ to +∞), has three grounded metal sides, at y = 0, y = a, and x = 0. The fourth
side, at x = b is maintained at a specific potential Vo (y) .
2. Find the potential inside the box: A cubical box (side of length a) consists of five metal
plates, which are welded together and grounded. The top is made of a separate sheet of metal,
insulated from the others, and held at a constant potential Vo .
3. Derive P3 (x) from the Rodrigues formula, and check that P3 (cosθ ) satisfies the angular
equation for l = 3.
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Problem 1
Separation of Variables in 2D Cartesian Coordinates
Diagram:
Three thick parallelogram-like lines are the three
grounded metal sides
In this problem, a rectangular pipe expends from –a to +a parallel to z-axis. The three
metal plates, as stated in the problem, are at y = 0, y = a, and x = 0. And at x = b, the plate is
maintained at constant potential Vo (y) . The potential can be denoted by V (x, y) and by
inspection, we see that there is no z-dependence. Thus, the Laplacian is independent of z. The
∂2 V ∂2 V
Laplacian becomes 2 + 2 = 0 . The boundary conditions are the following:
∂ X ∂Y
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In this problem, we need to look for the product solution of the form of V(x,y) = X(x)Y(y). The
first step is to do the separation of variables. When rewritten, the Laplacian equation becomes
X ''Y + Y '' X = 0 , thus becoming
X '' Y ''
+
= 0 . The sum of the x-component and the y-component
X Y
is zero; thus, this could also mean that the sum of a fixed constant and the negation of that fixed
constant is zero. Squaring the constant will make the ODE’s to be easier to solve. The constants,
in this case, will be k 2 and −k 2 . Thus, the separation of variables yields the following second
order linear homogeneous ordinary differential equations: X '' = k 2 X and Y '' = −k 2Y , which can
be written as X ''− k 2 X = 0 and Y ''+ k 2Y = 0 . The general solutions of these two ODEs turn out
to be X(x) = Aekx + Be− kx and Y (y) = C sin ky + D cos ky .
After finding the general solutions to the ordinary differential equations, it is time to use
the initial boundary conditions to find the four unknown constants. Going back to the equation
V(x,y) = X(x)Y(y), the potential becomes V (x, y) = (Aekx + Be− kx )(C sin ky + D cos ky) . Using the
boundary condition a), D becomes zero. Using the boundary condition c) and the value of
constant D, B = -A. Using the boundary condition b) and the values of constants B and D,
0 = C sin ka , which means that ka is an integer multiple of π in order to satisfy the boundary
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condition. Therefore, ka = nπ , which also means that k =
V (x, y) = (Ae
nπ x
a
− Ae
function sinh x =
−
nπ x
a
)(C sin
nπ
. With b), the potential is
a
nπ y
) . In addition, we know that the hyperbolic trigonometric
a
e x − e− x
, thus making the potential with boundary condition b)
2
V (x, y) = 2AC sinh(
nπ x
nπ y
)sin(
).
a
a
Part A
The first part of the problem is to develop a general formula for the potential within the
pipe. To begin this part of the problem, the last initial boundary condition will be applied. In
V (x, y) = 2AC sinh(
nπ x
nπ y
)sin(
) , we know that 2AC is a constant, which we can combine and
a
a
make them into one constant. Griffiths uses Cn as the combined constant in his textbook. The
separation of variables gives us an infinite set of solutions for each n; thus, we have to use a
summation notation to write down the potential. The potential, using the summation notation,
∞
becomes V (x, y) = ∑ Cn sinh(
n=1
nπ x
nπ y
)sin(
) . Now, with this, we finally apply the last boundary
a
a
∞
condition d). Thus, V (b, y) = ∑ Cn sinh(
n=1
nπ b
nπ y
)sin(
) = Vo . What Griffiths calls it a “Fourier’s
a
a
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trick” has to be used in order to solve for Cn . Following is the method to find out Cn in the
∞
following Fourier sine series using the “Fourier’s trick”: V (0, y) = ∑ Cn sin(
n=1
Cn =
nπ y
) = Vo (y) ,
a
2 a
nπ y
Vo (y)sin(
)dy .
∫
0
a
a
∞
Thus, in this particular problem, when V (b, y) = ∑ Cn sinh(
n=1
Cn sinh(
nπ b
nπ y
)sin(
) = Vo ,
a
a
a
nπ b 2 a
nπ y
2
nπ y
) = ∫ Vo (y)sin(
)dy ; therefore, Cn =
Vo (y)sin(
)dy .
∫
0
0
nπ b
a
a
a
a
asinh(
)
a
Thus, the general formula for the potential becomes
nπ x
nπ y
)sin(
) a
∞ 2sinh(
nπ y
a
a
V (x, y) = ∑
Vo (y)sin(
)dy .
∫
0
nπ b
a
n=1
asinh(
)
a
Part B
This part of the problem asks us to find the potential explicitly for the case Vo (y) = Vo (a
constant). Given Vo (y) = Vo , we can conclude that Cn =
2
asinh(
a
nπ b
)
a
Vo ∫ sin(
0
nπ y
)dy . Thus, by
a
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integrating this expression, Cn =
2Vo
a
2Vo
1− cos nπ
(− cos nπ + 1) =
. However,
nπ b nπ
nπ b
n
π
asinh(
)
sinh(
)
a
a
the constant is different when n is odd and when n is even. The solution becomes the following:
Thus, the constant is zero whenever n is even. Cn =
∞
plug this back into V (x, y) = ∑ Cn sinh(
n=1
4Vo
whenever n is odd. When we
nπ b
nπ sinh(
)
a
nπ x
nπ y
)sin(
) , the potential becomes
a
a
nπ x
nπ y
sinh(
)sin(
)
4Vo
a
a
V (x, y) =
.
∑
nπ b
π n=odd
nsinh(
)
a
The answers to Problems 1 are the following:
nπ x
nπ y
)sin(
) a
nπ y
a
a
Part A: V (x, y) = ∑
Vo (y)sin(
)dy
∫
0
nπ b
a
n=1
asinh(
)
a
nπ x
nπ y
sinh(
)sin(
)
4Vo
a
a
Part B: V (x, y) =
∑
nπ b
π n=odd
nsinh(
)
a
∞
2sinh(
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Problem 2
Separation of Variables in 3D Cartesian Coordinates
Diagram:
In this problem, the top of the box (z-direction) is made of a separate sheet of metal,
insulated from the others, and held at a constant potential Vo . There are six boundary conditions
to this problem:
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⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
a)
V =0 x=0
b)
V =0 x=a
c)
V =0 y=0
d)
V =0 y=a
e)
V =0 z=0
f ) V = Vo z = a
This time, because of the z dependence, the Laplacian depends on z, therefore making this
problem a separation of variables in three-dimensional Cartesian coordinates. The Laplacian
∂2 V ∂2 V ∂2 V
becomes 2 + 2 + 2 = 0 . Similar to the first problem, we need to look for the product
∂ X ∂Y ∂ Z
solution of the form of V(x,y) = X(x)Y(y)Z(z). The first step is to do the separation of variables.
Rewriting the Laplacian becomes X ''YZ + Y '' XZ + Z '' XY = 0 which becomes
X '' Y '' Z ''
+
+
= 0 . We now assign constants to each component of the Laplacian. The sum of the
X Y
Z
three components is zero once again. The first two components will be assigned negative squared
constants and the last component will be the negation of sum of the two to make the Laplacian
zero. Thus,
X ''
Y ''
Z ''
= −k 2 ,
= −l 2 , and
= (k 2 + l 2 ) . The rearrangements of these separated
X
Y
Z
variables become the following second-order linear homogeneous ordinary differential
equations: X ''+ k 2 X = 0 , Y ''+ l 2Y = 0 , and Z "− (k 2 + l 2 )Z = 0 . The general solutions of these
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three ordinary differential equations are X(x) = Asin kx + B cos kx , Y (y) = C sinly + D cosly , and
Z(z) = Ee
k 2 +l 2 z
+ Fe−
k 2 +l 2 z
Now, we apply the boundary conditions. By applying the first boundary condition, we
know that the constant B becomes zero. By applying the second boundary condition, just like one
of the boundary conditions in problem 1, 0 = Asin ka , which means that ka is an integer multiple
of π in order to satisfy the boundary condition. Therefore, ka = nπ , which also means that
k=
nπ
. The third and fourth boundary conditions are similar to the first and second boundary
a
conditions, which means that the constant D becomes zero and la is an integer multiple of π in
order to satisfy the boundary condition. Thus, la = mπ , which also means that l =
mπ
. By
a
applying the fifth boundary condition, E + F = 0, thus E = -F. Now, we plug in the expressions
for k and l to Z(z) = Ee
k 2 +l 2 z
+ Fe−
k 2 +l 2 z
as well as substituting E into -F. Thus solving this
expression yields the following:
Z(z) = Ee
= Ee
k 2 +l 2 z
⎪⎧
= E ⎨(e
⎪⎩
⎧ π
= E ⎨e a
⎩
(
k 2 +l 2 z
+ Fe−
− Ee−
k 2 +l 2 z
nπ 2 mπ 2
) +(
) z
a
a
n 2 +m 2 z
−e
k 2 +l 2 z
= E(e
) − (e
−
− (
π 2 2
n +m z
a
k 2 +l 2 z
− e−
nπ 2 mπ 2
) +(
) z
a
a
k 2 +l 2 z
)
⎪⎫
)⎬
⎪⎭
⎧ π
⎫
⎪ (e
⎬ = 2E ⎨
⎭
⎪
⎩
n 2 +m 2
z
a
−e
2
− π n 2 +m 2
z
a
⎫
)⎪
⎬
⎪
⎭
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e x − e− x
Just like in Problem 1, we have the hyperbolic trigonometric function, sinh x =
. Thus,
2
z⎤
⎡
the expression simplifies to the following: Z(z) = 2E sinh ⎢π n 2 + m 2 ⎥ .
a⎦
⎣
Now, we need to find the potential. Putting together X(x), Y(y), and Z(z), there has to be
two summations because there are two unspecified unknowns n and m. The constants A, C, and
2E are going to be absorbed into one constant Cn,m . Thus, the potential becomes
∞
∞
V (x, y, z) = ∑ ∑ Cn,m sin(
n=1 m=1
nπ x
mπ y
z
)sin(
)sinh(π n 2 + m 2 ) . We apply the final boundary
a
a
a
∞
∞
condition and the potential becomes Vo = ∑ ∑ Cn,m sinh(π n 2 + m 2 )sin(
n=1 m=1
nπ x
mπ y
)sin(
) Now,
a
a
we use the “Fourier’s trick.” In this case, because there is a double sum, there has to be a double
integral. Thus, the Fourier’s trick yields
a a
2
nπ x
mπ y
Cn,m sinh(π n 2 + m 2 ) = ( )2 Vo ∫ ∫ sin(
)sin(
)dx dy . The integral becomes zero when
0 0
a
a
a
either n or m is even and when both of them are odd, the integral becomes
16Vo
⎛ 2⎞
⎛ a(1− cos(nπ )) ⎞ ⎛ a(1− cos(mπ )) ⎞
.
⎜⎝ ⎟⎠ Vo ⎜⎝
⎟⎠ ⎜⎝
⎟⎠ , which becomes 2
a
nπ
mπ
π nm
2
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⎧ 0
⎪
a a
2 2
nπ x
mπ y
Essentially, ( ) Vo ∫ ∫ sin(
)sin(
)dx dy = ⎨ 16V
0 0
o
a
a
a
⎪ 2
π
nm
⎩
Therefore, Cn,m =
∞
∞
m = even
n & m = odd
⎞
16Vo ⎛
1
. Thus, after plugging in the constant back to
⎜
2
π nm ⎝ sinh(π n 2 + m 2 ) ⎟⎠
V (x, y, z) = ∑ ∑ Cn,m sin(
n=1 m=1
n or
nπ x
mπ y
z
)sin(
)sinh(π n 2 + m 2 ) , the potential becomes
a
a
a
2
2 z
sinh(
π
n
+
m
)
16Vo
1
nπ x
mπ y
a
V (x, y, z) = 2 ∑ ∑
sin(
)sin(
)
.
π n=odd m=odd nm
a
a
sinh(π n 2 + m 2 )
The answer to Problem 2 is
2
2 z
16Vo
1
nπ x
mπ y sinh(π n + m a )
V (x, y, z) = 2 ∑ ∑
sin(
)sin(
)
.
π n=odd m=odd nm
a
a
sinh(π n 2 + m 2 )
Problem 3
Separation of Variables in Spherical Coordinates (Azimuthal Symmetry)
In this problem, in order to derive P3 (x) from the Rodrigues formula, it necessary to
undergo the separation of variables in spherical coordinates. The problem will have azimuthal
symmetry, which means that the potential is independent of φ .
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Laplace’s equation in three dimensions is the following:
1 ∂ ⎛ 2 ∂V ⎞
1
∂ ⎛
∂V ⎞
1
∂2 V
r
+
sin
θ
+
= 0 . As in the first two problems, we need
⎜
⎟
⎜
⎟
r 2 ∂r ⎝ dr ⎠ r 2 sin θ ∂θ ⎝
∂θ ⎠ r 2 sin 2 θ ∂φ 2
to look for the solution in the form of V (r,θ ,φ ) = R(r)P(θ )Φ(φ ) . However, after applying the
azimuthal symmetry, φ dependence disappears. Thus, we need to look for the solution in the
form of V (r,θ ) = R(r)P(θ ) . The Laplacian becomes
∂ ⎛ 2 ∂V ⎞
1 ∂ ⎛
∂V ⎞
⎜⎝ r
⎟⎠ +
⎜⎝ sin θ
⎟ = 0.
∂r
dr
sin θ ∂θ
∂θ ⎠
Putting the Laplacian into V (r,θ ) = R(r)P(θ ) , the equation becomes
P
∂ ⎛ 2 ∂R ⎞
R ∂ ⎛
∂P ⎞
⎜⎝ r
⎟⎠ +
⎜⎝ sin θ
⎟ = 0 . Now, dividing the whole equation by R(r)P(θ ) ,
∂r
dr
sin θ ∂θ
∂θ ⎠
1 ∂ ⎛ 2 ∂R ⎞
1
∂ ⎛
∂P ⎞
⎜⎝ r
⎟⎠ +
⎜⎝ sin θ
⎟ = 0.
R ∂r
dr
P sin θ ∂θ
∂θ ⎠
The two components, R (radial) and P (angular), are set equal to the constant l(l+1) and
-l(l+1) (just like Cartesian coordinates, one has to be negative and one has to be positive to make
the sum of the two components zero). Thus,
1 d ⎛ 2 dR ⎞
⎜r
⎟ = l(l + 1) and
R dr ⎝ dr ⎠
1
d ⎛
dP ⎞
d ⎛ 2 dR ⎞
⎜⎝ sin θ
⎟⎠ = −l(l + 1) . Thus, the radial component becomes
⎜r
⎟ = l(l + 1)R .
P sin θ dθ
dθ
dr ⎝ dr ⎠
The general solution of the radial component can be figured out with the following. Let
R(r) = r q .
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1 d ⎛ 2 dR ⎞
⎜r
⎟ = l(l + 1)
R dr ⎝ dr ⎠
d ⎛ 2 dR ⎞
⎜r
⎟ = l(l + 1)R
dr ⎝ dr ⎠
dR
= qr q−1
dr
dR
r2
= qr q+1
dr
d ⎛ 2 dR ⎞
q
⎜⎝ r
⎟⎠ = (q + 1)qr = q(q + 1)R = l(l + 1)R
dr
dr
The first solution to this is R = r l and the second solution has
q = −(l + 1)
. Thus,
−l = q + 1
q(q + 1) = −(l + 1)(−l) = l(l + 1) . The solution of the radial equation becomes R(r) = Ar l +
The angular component is
B
.
r l+1
1
d ⎛
dP ⎞
⎜⎝ sin θ
⎟ = −l(l + 1) . Finding the general solution
P sin θ dθ
dθ ⎠
to the angular component is complicated. The following is the separation of variables for angular
component. We will also derive the Rodrigues formula using Legendre’s differential equation.
Note that the polynomial solution of Pl (x) of the Legendre equation that will be obtained during
the work is a Legendre polynomial, if Pl (1) = 1 . Pl (1) = 1 , is the definition of Legendre
polynomial.
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We first let l(l + 1) = k . Then the angular equation becomes
There will be a change of variables by letting x(θ ) = cosθ , thus
of P as a function of θ through the variable x. Thus,
The angular equation then becomes
1
d ⎛
dP ⎞
⎜⎝ sin θ
⎟ = −k .
P sin θ dθ
dθ ⎠
d
1 d
=−
. We can think
dx
sin θ dθ
dP dP dx dP
dP
=
=
(− sin θ ) =
(− 1− x 2 ) .
dθ dx dθ dx
dx
d
dP
((1− x 2 ) ) + kP = 0 . We now guess the power series
dx
dx
∞
P = ∑ an x n , thus the first derivative and the second derivative become
n=0
dP ∞
= ∑ nan x n−1 and
dx n=1
d 2P ∞
= ∑ n(n − 1)an x n−2 respectively. When we plug these into the new angular equation, we
2
dx
n=2
obtain an equation in the form of the Legendre’s differential equation. Legendre’s differential
equation is generally in the form of (1− x 2 )
d2y
dy
− 2x + l(l + 1)y = 0 . By plugging in the
2
dx
dx
d 2P
dP
guessed power series and first and second derivatives, we obtain (1− x ) 2 − 2x
+ kP = 0 ,
dx
dx
2
where k = l(l + 1) . Simplifying this equation gives us
∞
∑ n(n − 1)a x
n
n=2
n−2
∞
∞
∞
n=1
n=1
n=0
− ∑ n(n − 1)an x n − 2∑ nan x n + k ∑ nan x n . In general, for x s , where s is a
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positive integer, as+2 = as
s(s + 1) − k
s(s + 1) − l(l + 1)
= as
. The right hand side of this equation
(s + 2)(s + 1)
(s + 2)(s + 1)
is 0 when s=l. Thus, after rearranging this equation, it becomes
as =
−(s + 2)(s + 1)
l(l − 1)
as+2 (s ≤ l − 2) . By letting s = l − 2 , we obtain al−2 = −
al .
(l − s)(l + s + 1)
2(2l − 1)
We have the recurrence relation as =
−(s + 2)(s + 1)
as+2 s = 0,1,2... . The Legendre
(l − s)(l + s + 1)
differential equation, as written above, is (1− x 2 )
d 2P
dP
− 2x
+ kP = 0 , and when s=l, we
2
dx
dx
obtain al+2 , al+4 , al+6 ... al+2 m for all positive integers m. The two linearly independent solutions
of the Legendre differential equation are the following:
P1 = 1−
⎛ (l − 2n + 2)(l + 2n − 1) ⎞ 2n
l(l + 1) 2
x ++ (−1)n ⎜
⎟⎠ x +
⎝
2!
(2n)!
P2 = x −
⎛ (l − 2n + 1)(l + 2n) ⎞ 2n+1
(l − 1)(l + 2) 3
x ++ (−1)n ⎜
+
⎟⎠ x
⎝
3!
(2n + 1)!
If we let l be a positive even integer, P1 is a polynomial of degree n. But if we let l be a positive
odd integer, P2 is a polynomial of degree n. A polynomial solution Pl (x) of the Legendre
d 2P
dP
differential equation, (1− x ) 2 − 2x
+ l(l + 1)P = 0 , is a Legendre polynomial whenever
dx
dx
2
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Pl (1) = 1 . As stated before the definition of Legendre polynomials is when Pl (1) = 1 . However,
the recurrence relation, as =
−(s + 2)(s + 1)
l(l − 1)
as+2 (s ≤ l − 2) , which became al−2 = −
al
(l − s)(l + s + 1)
2(2l − 1)
as stated above, will not have the solution of Pl (1) = 1 when it is plugged back into the power
∞
series guess, P = ∑ an x n . However, by choosing al = 1 when l = 0 and
n=0
al =
(2l)!
l = 1,2, 3 , we will later on obtain the result of Pl (1) = 1 . al = 1 when l = 0 and
2 l (l!)2
al =
(2l)!
l = 1,2, 3 are therefore chosen to fulfill the definition of Legendre polynomials.
2 l (l!)2
Therefore, after guessing, we obtain al−2n = (−1)n
[l/2]
Pl (x) = ∑ (−1)n
n=0
following:
(2l − 2n)!
and
2 n!(l − n)!(l − 2n)!
l
(2l − 2n)!
x l−2n leading to the Legendre polynomials such as the
2 n!(l − n)!(l − 2n)!
l
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P0 (x) = 1
P1 (x) = x
1
1
P2 (x) = (3x 2 − 1)
P3 (x) = (5x 3 − 3x) and so on. Notice that all of the Legendre
2
2
1
4
2
P4 (x) = (35x − 30x + 3)
8
polynomials are 1 whenever x=1. The guesses, al = 1 when l = 0 and al =
(2l)!
l = 1,2, 3
2 l (l!)2
confirm the definition of Legendre polynomials. The steps to obtain
al−2n
[l/2]
(2l − 2n)!
(2l − 2n)!
= (−1) l
and Pl (x) = ∑ (−1)n l
x l−2n are the following:
2 n!(l − n)!(l − 2n)!
2 n!(l − n)!(l − 2n)!
n=0
n
After guessing al =
(2l)!
,
2 l (l!)2
l(l − 1)
l(l − 1)(2l)!
al = −
2(2l − 1)
2(2l − 1)2 l (l!)2
l(l − 1)(2l)(2l − 1)(2l − 2)!
=−
2(2l − 1)2 l l(l − 1)!l(l − 1)(l − 2)!
al−2 = −
al−2
If we have al−4 , this becomes al−4 = −
when l − 2n ≥ 0 , al−2n = (−1)n
This simplifies to al−2 = −
(2l − 2)!
.
2 (l − 1)!(l − 2)!
l
(l − 2)(l − 3)
(2l − 4)!
al−2 = l
. Thus, in general,
4(2l − 3)
2 2!(l − 2)!(l − 4)!
(2l − 2n)!
, where m is a positive integer.
2 n!(l − n)!(l − 2n)!
l
Lee 20
We can now plug al−2n = (−1)n
[l/2]
Pl (x) = ∑ (−1)n
n=0
(2l − 2n)!
into the power series guess to obtain
2 n!(l − n)!(l − 2n)!
l
(2l − 2n)!
x l−2n , where [l / 2] is denoted as a nearest integer function.
2 n!(l − n)!(l − 2n)!
l
⎧
⎪⎪ l / 2 if l = even
Thus, ⎨ (l − 1)
if l = odd
⎪
⎪⎩ 2
. We now need to find the Rodrigues formula, which is a formula
[l/2]
to obtain Legendre polynomials. We first switch Pl (x) = ∑ (−1)n
n=0
[l/2]
Pl (x) = ∑ (−1)n
n=0
(2l − 2n)!
x l−2n to
2 n!(l − n)!(l − 2n)!
l
⎧ (2l − 2n)!
⎫
1
x (2l−2n)−l ⎬ . We now peel off a derivative,
⎨
2 n!(l − n)! ⎩ [(2l − 2n) − l]!
⎭
l
dl j
j!
(x ) =
x j−l ( j ≥ l) , where (2l − 2m) = j in our case. Thus, the power series becomes
l
dx
( j − l)!
1
d l (2l−2n)
Pl (x) = ∑ (−1) l
x
, which then becomes
2 n!(l − n)! dx l
n=0
[l/2]
n
Pl (x) =
1 dl l
l!
(−1)n
x (2l−2n) after extending the upper limit of the summation from
l
l ∑
2 l! dx n=0
n!(l − n)!
[l / 2] to l. The additional terms introduced after the extension of the upper limit of the sum from
[l / 2] to l have zero derivative. Notice that the sum in Pl (x) =
1 dl l
l!
(−1)n
x (2l−2n) is
l
l ∑
2 l! dx n=0
n!(l − n)!
Lee 21
the binomial expansion of (x 2 − 1)l , thus obtaining the Rodrigues formula,
1 dl
Pl (x) = l
[(x 2 − 1)n ] .
l
2 l! dx
The Pl (x) component in the Legendre polynomial is defined by the Rodrigues formula,
l
1 ⎛ d⎞
which is Pl (x) = l ⎜ ⎟ (x 2 − 1)l . To derive P3 (x) from the Rodrigues formula, simply plug in
2 l! ⎝ dx ⎠
3 into l. Thus,
3
P3 (x) =
1 ⎛ d⎞ 2
1 d3 2
3
(x
−
1)
=
(x − 1)3
⎜ ⎟
2 3 l! ⎝ dx ⎠
48 dx 3
1 d2
1 d
⎡(x 2 − 1) + (5x 2 − 1) ⎤⎦
x(x 2 − 1)2 =
2
8 dx
8 dx ⎣
1
= ⎡⎣ 2x(5x 2 − 1) + 10x(x 2 − 1) ⎤⎦
8
5
3
= x3 − x
2
2
=
5
3
cos 3 θ − cosθ .
2
2
Now, in order to check that P3 (cosθ ) satisfies the angular equation, for l = 3, we do the
In the form of Legendre polynomial, this equation becomes P3 (cosθ ) =
following:
Lee 22
dP3
d ⎛5
3
3
⎤
⎞ ⎡⎛ 5 ⎞
3
2
=
⎜⎝ cos θ − cosθ ⎟⎠ = ⎢⎜⎝ ⎟⎠ 3cos θ (− sin θ ) − (− sin θ ) ⎥
dθ dθ 2
2
2
2
⎣
⎦
3
= − sin θ (5 cos 2 θ − 1)
2
sin θ
dP3
3
= − sin 2 θ (5 cos 2 θ − 1)
dθ
2
∂ ⎛
dP3 ⎞
3 ∂
⎡sin 2 θ (5 cos 2 θ − 1) ⎤⎦
⎜⎝ sin θ
⎟⎠ = −
∂θ
dθ
2 ∂θ ⎣
3
= − ⎡⎣(5 cos 3 θ − 1)2sin θ − sin 3 θ (10 cosθ ) ⎤⎦
2
1 ∂ ⎛
∂P3 ⎞
3
3
2
⎜ sin θ
⎟ = − ⎡⎣ 2(5 cos θ − 1) − 10sin θ cosθ ⎤⎦
sin θ ∂θ ⎝
∂θ ⎠
2
3
⎡5
⎤
= −3(4) ⎢ cos 3 θ − cosθ ⎥ = −l(l + 1)P3
2
⎣2
⎦
qed.
Lee 23
References
Fry, J. N. (n.d.). Legendre Polynomials: Rodrigues’ Formula and Recursion Relations.
Retrieved April 13, 2014, from http://www.phys.ufl.edu/~fry/6346/legendre.pdf
Griffiths, D. J. (1999). Chapter 3: Special Techniques. Introduction to Electrodynamics (3rd ed.,
pp. 127-136). Upper Saddle River, N.J.: Prentice Hall.
Moore, C. E. (n.d.). The Legendre Polynomials. Retrieved April 13, 2014, from https://
www.morehouse.edu/facstaff/cmoore/Legendre%20Polynomials.htm
Series Solutions of Differential Equations. (n.d.). Retrieved April 11, 2014, from http://
www.most.gov.mm/techuni/media/EM_03011_p2chap34.pdf

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