Notes "orthogonal" January 2017
1
Expansion in orthogonal functions
To obtain a more useful form of the Green’s function, we’ll want to expand in
orthogonal functions that are (relatively) easy to integrate. We begin with
some basics and we’ll see how to obtain solutions for the potential in boundary
value problems with no charges inside the volume. Then we can move on to
compute the Green’s function.
1.1
Sturm-Liouville theory
See Lea Chapter 8 sections 1 and 2. Please review this material carefully.
1.2
General method for finding the potential
1. Choose coordinates so that the boundaries of the region correspond to
constant-coordinate surfaces. Then write the defining equation
∇2 Φ = 0
in terms of the chosen coordinates, called u, v, and w, separate variables,
and solve to determine the eigenfunctions. You’ll get equations something
like:
Dv V
Du U
= − (α + β) ;
= α;
U
V
where Du is a differential operator of the form
f (u)
Dw W
=β
W
d2
d
+ h (u)
+ g (u)
du2
du
and similarly for Dv and Dw .
2
Note which boundary has a non-zero value of potential. This boundary should be defined by the condition u = constant for one of the coordinates, u. You should choose your separation constants so that the
sets of eigenfunctions in the other two coordinates v and w are orthogonal
functions.
3
Use the boundaries on which Φ = 0 to determine the eigenfunctions
wW
).
V and W, and the eigenvalues, (which I’ll call α = DVv V and β = DW
If the boundaries are at a finite distance the eigenvalues will be countable
and can be labelled with an integer: αm . Otherwise they will form a
continuous set.
1
4
The last eigenfunction is determined from the differential equation in
the final coordinate, u:
γ=
Du U
= −α − β
U
At this point you should have a general solution of the form:
[
Φ (u, v, w) =
Amn Umn (u : −αm − β n ) Vm (v : αm ) Wn (w : β n )
m,n
where the coefficients Amn are still undetermined.
5
Now use the final, non-zero, boundary condition, together with orthogonality of the eigenfunctions Vm and Wn , to find the coefficients Amn .
[
Φ (u0 , v, w) =
Amn Umn (u0 : −αm − β n ) Vm (v : αm ) Wn (w : β n )
m,n
= known function of v, w
1.3
Rectangular coordinates
See Lea §8.2 and Griffiths Ch 3 §3, Jackson §2.9, or check here:
http://www.physics.sfsu.edu/~lea/courses/ugrad/360notes7.PDF
1.3.1
Rectangular 2-D problems with non-zero potential on more
than one side.
Since the general method allows for non-zero potential on only one side, we
have to solve these problems by superposition. Suppose we have a rectangular
region measuring a × b with potential V on the sides at x = 0 and y = b, the
sides at x = a and y = 0 being grounded. We solve two problems, each with
three sides grounded, one having potential V at x = 0 and one having potential
V at y = b. Then we add the results.
The solution is
Φ = Φ1 + Φ2
2
where (Lea Example 8.1)
∞
(2m+1)πy
(2m + 1) πx sinh
4V [
1
a
sin
Φ2 =
π m=0 2m + 1
a
sinh (2m+1)πb
a
and we obtain Φ2 by interchanging a and b, and noting that the potential is
zero at x = a and y = 0, so x → y and y → a − x
Φ1 =
∞
(2n+1)π(a−x)
(2n + 1) πy sinh
4V [ 1
b
sin
π n=0 2n + 1
b
sinh (2n+1)πa
b
With a = 2b the potential looks like this. (The variables are x/a and y/a, and
the contours are at fixed values of Φ/V.)
0.5
0.4
0.3
y
0.2
0.1
0
0.2
0.4
x
0.6
0.8
1
Blue 0.75, black, 0.5, red 0.25, dashed 0.9
1.3.2
Continuous set of eigenvalues
If one dimension of our rectangle becomes infinite, then the set of eigenvalues
is no longer countable but becomes a continuous set. Suppose we let b → ∞ in
the example above. With Φ (x, y) = X (x) Y (y) , Laplace’s equation takes the
form
Y
X
= k2 = −
X
Y
. The solutions to the differential equation that satisfy the boundary conditions
Φ = 0 at y = 0 and at x = a but Φ non-zero at x = 0 are of the form
sin ky sinh k(a − x)
But we no longer have an upper boundary in y to determine the values of k.
(Effectively, Φ2 in the previous problem has gone to zero because of the sinh kb
3
in the denominator, leaving Φ1 with an undetermined k.) However, since the
eigenfunction is an even function of k, we need only positive values of k. Thus
the solution is
] ∞
A (k) sin ky sinh [k(a − x)] dk
(1)
Φ (x, y) =
0
To find the function A (k) , we use the known potential V (y) at x = 0 :
] ∞
A (k) sin ky sinh ka dk
Φ (0, y) = V (y) =
0
We now make use of the orthogonality of the functions sin ky by multiplying
both sides by sin k y and integrating over y :
] ∞
] ∞] ∞
V (y) sin k y dy =
A (k) sin ky sinh ka dk sin k y dy
0
0
On the RHS, we have
] ∞
sin ky sin k y dy =
0
=
=
=
0
]
1 ∞
[cos (k − k ) y − cos (k + k ) y] dy
2 0
]
r
3
3
3
1 ∞ q i(k−k3 )y
e
+ e−i(k−k )y − ei(k+k )y − e−i(k+k )y dy
4 0
]
1 ∞
{exp [i (k − k ) y] − exp [i (k + k ) y]} dy
4 −∞
π
[δ (k − k ) − δ (k + k )]
2
where we used Lea eqn 6.16. Thus
] ∞
]
π ∞
V (y) sin k y dy =
A (k) sinh ka [δ (k − k ) − δ (k + k )] dk
2 0
0
Since k and k are both positive, only the first delta function contributes, and
this integral reduces to
] ∞
π
V (y) sin k y dy = A (k ) sinh k a
2
0
and thus, dropping the primes,
A (k) =
2
π
]
∞
V (y)
0
4
sin ky
dy
sinh ka
For example, if V (y) = V0 e−y/h , then
]
2 ∞
sin ky
dy
A (k) =
V0 e−y/h
π 0
sinh ka
∞
(ik−1/h)y
1 V0
e(−ik−1/h)y
e
=
−
πi sinh ka
ik − h1
−ik − h1 0
1 V0
1
−1
=
−
πi sinh ka ik − h1
ik + h1
2
1 V0
2ih k
=
2
πi sinh ka k h2 + 1
and inserting this value into (1), we get
] V0 ∞
2h2 k2
sin ky sinh [k(a − x)]
dk
Φ (x, y) =
2
2
π 0
k h +1
k
sinh ka
Analysis: We can see right away that our result is dimensionally correct and
real. Also, since x ≤ a, k(a − x) ≤ ka and thus
sinh [k(a − x)]
≤1
sinh ka
for all k, so the integral converges. I couldn’t do this integral analytically, so
let’s compute
some values in the case h = a:
Φ a2 , a2 /V0 = π2 0.314 45 = 0.200
Φ a2 , a /V0 = π2 0.281 18 = 0.179
Φ a2 , 2a /V0 = π2 0.119 04 = 7.578 × 10−2
The potential decreases rapidly for y > a
1.3.3
A 3-D problem
Find the potential inside a cubical box of sidek a with grounded
walls, except
2x
2 l
for the side at z = a which has potential V0 1 − a − 1 . This potential
increases from zero at the walls to V0 at x = a/2,as shown in the diagram.
1
0.8
0.6
ential
0.4
0.2
0
0.2
0.4
x/a
5
0.6
0.8
1
There is no charge inside the box, so the potential satisfies Laplace’s equation:
∇2 Φ = 0
∂ Φ ∂ Φ ∂ 2Φ
+
+
= 0
∂x2
∂y2
∂z 2
2
2
Try separating variables:
Φ = XY Z
Then
X Y Z + XY Z + XY Z = 0
or
Y
Z
X
+
+
=0
X
Y
Z
For equation (2) to hold for all values of x, y, z we must have:
(2)
Y
Z
X
= k1 ,
= k2 ,
= k3 and k1 + k2 + k3 = 0
X
Y
Z
The potential is zero at x = 0 and at x = a, so we need k1 to be negative,
k1 = −α2 , which gives the solutions X = sin αx and X = cos αx, which have
multiple zeros. We choose the sine function to make X (0) = 0, and then choose
α = nπ/a to make X (a) = 0. A similar reasoning gets Y = sin (mπy/a) . Then
π2
k3 = n2 + m2 2
a
and to make Z (0) = 0 the appropriate solution for Z is the sinh .
Φ (x, y, z) =
s
x
y
πz sin mπ
sinh
Anm sin nπ
n2 + m2
a
a
a
n,m=1
∞
[
Finally we evaluate the potential at z = a :
%
2 &
∞
s
[
mπy
nπx
2x
V0 1 −
−1
sin
sinh
Anm sin
n2 + m2 π
=
a
a
a
n,m=1
We make use of the orhogonality of the sines by multiplying both sides by
sin (n πx/a) sin (m πy/a) and integrating over x and y :
2 &
] a] a %
t
m πy
2x
n πx
a2
3
3
−1
sin
dxdy = An m
sinh (n )2 + (m )2 π
V0 1 −
sin
a
a
a
4
0
0
Now we drop the primes on n and m . The integral over y is straighforward.
a
] a
− cos mπy/a mπy
1 − (−1)m
dy =
sin
=a
a
mπ/a
mπ
0
0
6
The result is zero unless m is odd. Let u = 2x/a − 1, to get (for m odd),
]
s
u+1
a2
2a 1
a
du = Anm sinh π n2 + m2
V0 1 − u2 sin nπ
mπ −1
2
2
4
] 1
s
nπ
nπu
nπ
nπu
4V0
cos
+ cos
sin
du = Anm sinh π n2 + m2(3)
1 − u2 sin
mπ −1
2
2
2
2
When we integrate over u the sin nπu/2 term gives zero because we have an
odd integrand over an even range. Then sin nπ/2 is zero unless n is odd. The
remaining integrand is even, so we can use half the range. We do integation by
parts:
+1
] 1
] 1
2 1 − u2
nπu nπu
4
nπu
2
du =
sin
du
1 − u cos
u sin
+
2
nπ
2 nπ 0
2
0
0
The integrated term is zero. We do another integration by parts to get
$
#
+1
] 1
] 1
nπu
4
nπu 2
nπu
−2
2
1 − u cos
du = 2
cos
du
+
cos
u
2
nπ
nπ
2 0
nπ 0
2
−1
16
2
nπ
nπu +1
= −
−
cos
sin
2
nπ
2 0
(nπ)2
nπ
32
=
3 sin 2
(nπ)
where the first term is zero for n odd. So the left hand side of (3) is
π
128
4V0 32
sin2 n = V0
for m and n odd, and zero otherwise
3
mπ (nπ)
2
mn3 π4
Thus for m odd and n odd
Anm = V0
128
1
√
mn3 π 4 sinh n2 + m2 π
and finally we have the potential
Φ (x, y, z) =
128
V0
π4
∞
[
∞
[
m=1,odd n=1,odd
√
mπy sinh n2 + m2 πz
nπx
1
a
√
sin
sin
mn3
a
a sinh n2 + m2 π
Analysis: The result is dimensionally correct, and converges nicely.
The plot shows the series with values of m up to m = 11 and n up to n = 5.
The black line is the potential as a function of z for x = y = a/2,the green line
for x = y = a/4,and the red line for x = a/2, y = a/4 . The potential increases
faster near the center of the box, as expected, since the potential on the top
side is maximum at x = a/2. .
7
Phi/V0 0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
At z = a/2, y = a/2 the potential versus x/a looks like this:
0.14
0.12
0.1
0.08
hi/V0
0.06
0.04
0.02
0
0.2
0.4
x/a
0.6
0.8
The potential is symmetric about x = a/2, as expected.
8
1
0.9
1.0
z/a
2
Complex potential
Recall that both the real and imaginary parts of an analytic function satisfy
Laplace’s equation in two dimensions (Lea Chapter 2 section 4) . We may use
this fact to solve 2-d potential problems.
2.0.4
Potential in a wedge.
Suppose the region of interest is defined by the angular wedge 0 ≤ θ ≤ α.
Further suppose the potential is zero on the conducting boundaries at θ = 0
and θ = α. Then the analytic function
π
π f (z) = z π/α = rπ/α eiθπ/α = rπ/α cos θ + i sin θ
α
α
has imaginary part
π
θ
α
that satisfies the boundary conditions. If α = π/m for some integer m, then
f (z) is analytic everywhere. If α is an arbitrary real number, then f (z) may
have a branch point at the origin, but we may choose the branch cut so that
f (z) is still analytic in our region everywhere except AT the origin. Thus
v (r, θ) also satisfies Laplace’s equation in the volume, and so it is a solution of
the correct form. In fact the function f (z) = z nπ/α for any n has the same nice
properties. Thus the potential in a wedge-shaped region with opening angle α
and conducting boundaries at potential V0 is described by the complex potential
v (r, θ) = rπ/α sin
Φ (z) = A + iV0 + Σn an z nπ/α
The imaginary part is
V0 +
+∞
[
an rnπ/α sin
n=−∞
nπ
θ
α
which is the potential in the region. The coefficients an must be chosen to
satisfy any remaining boundary conditions in r. Compare with Jackson’s eqn.
2.72 which he derives using separation of variables in plane polar coordinates.
(You should read this derivation.)
If the origin is included within our region, the sum is over positive n only,
so that the potential remains finite. Then the potential near the origin (small
r) is dominated by the first (n = 1) term, and the field near the origin has
components
k π π l
π
θ r̂ + cos
θ θ̂
E = −∇Φ = −a1 rπ/α−1 sin
α
α
α
Thus E → 0 as r → 0 if π/α > 1 but E → ∞ as r → 0 if π/α < 1. The field
is small in a hole (α < π) but very large near a spike (α > π).
9
The diagrams show the cases of α = π/2 and 3π/2. For α = π/2, the
equipotential surfaces are given by
r2 sin 2θ = constant = 2xy
Thus the equipotentials are hyperbolae in this case.
y 12
y5
10
4
8
6
3
4
2
2
1
-6 -4 -2
-2
0
-4
0
1
2
3
4
5
x
equipotentials and field lines for α = π/2
2
4
6
-6
α = 3π/2
We may use conformal mapping to find the potential in a 2-d boundary value
problem with more complex boundaries. See Lea Ch 2 section 2.8. We choose
the mapping to simplify the boundary shape, solve the simpler problem, and
then map back.
10
8 10 12
x