Method of Images Greens Jackson

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Jackson Notes 3 2012

1 Method of images

The method of images is a method that allows us to solve certain potential problems as

well as obtaining a Greens function for certain spaces. Recall that the Greens function

satisfies the equation

2G (x0, x) = 4(x x0) (1)subject to the boundary conditions

GD = 0 on S (2)

orG

n= C on S (3)

and thus the solution is of the formG (x0, x) =

1

|x x0| + (x0, x) (4)

where

2 (x0, x) = 0 in V (5)Thus the Greens function is 40 times the potential at x due to a unit point charge at x

0

in the volume V plus an additional term, with no sources in V, that fixes up the boundaryconditions, that is, a term due to sources outside ofV. Thus we can make progress in findingthe Dirichlet Greens function by finding the potential due to a point charge in V withgrounded boundaries.

1.1 Plane boundary

Suppose the volume of interest is the half-space z > 0. A point charge q is placed at adistance d from the x yplane, which is a conducting boundary. What is the potential forz > 0?

First, the conducting plane must be at a constant potential, which we may take to be

zero. (If the potential is V0, we can just addV0 to our solution at the end.) Then we mayplace an image charge q at distance d from the boundary, but on the opposite side. Thenthe potential due to these two charges everywhere on the boundary z = 0 is zero. Since theimage charge we added is outside our volume, it does not contribute to the value of 2 inV. Thus, putting the zaxis through q, the solution is

(x) =1

40

q

|x dz| q

|x + dz|

Properties of this solution are explored in Jackson problem 2.1. Note that the image charge

represents the charge density drawn onto the plane through the ground wire.

To obtain the Greens function for the half-space, we simply set q = 1 and multiply by40. Then

G (x, x0) =1

|x x0| 1

|x x00| (6)

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where

x00 = (x0, y0,

z0)is the position vector of the image point. This expression shows clearly that the physial

dimensions ofG are [1/length].This Greens function (6) is explicitly in the form (4), but it is not very convenient to use.

For example, suppose the problem of interest has potential V0 within a circle of radius a onthe plane z = 0, with the rest of the plane grounded. Then we would needfirst to compute(notes 2.5 equation 6)

G

n0

z0=0

= Gz 0

z0=0

= 2z|x x0|3

z0=0

(note that the outward normal on the plane at z = 0 is z) and then evaluate

(x) =V04 Z

circle

2z

|x x0

|

3 z0=0

dx0dy0

=V0z

2

Zcircle

1r2 + r02 2rr0 cos 0 + z23/2 r0dr0d0

where r, are polar coordinates in the x y plane. This is ugly. (We encountered a similarintegral in the bar magnet example, but that was only a 1/2 power. Later on we will identify

some methods for doing this integral, but we will also find more convenient expressions for

G.)

1.2 Images in a sphere

Now let the conducting boundary be a sphere of radius a and suppose we have a point chargeq at a distance d from the center of the sphere, where d > a. We want to find the potentialoutside the sphere.

Learning from our experience above, we conjecture that we can place an image charge

inside the sphere (and thus outside our volume V) and form the potential in V as the sumof the potential due to the two charges. Let the image charge have magnitude q0 and be atr = d0. Then we have

(x) =1

40

qx d

+ q0x d0

The boundary condition is that (x) = 0 everywhere on the surface of the sphere (r = a).The system has azimuthal symmetry about the line from the center of the sphere to the

charge q. Rotate the system about this line and nothing changes. Thus the image charge q0

must lie on this line at a distance d0 from the center. Then we have two unknowns in ourpotential: q0 andd0, and we need only pick two points on the sphere to solve for the two

unknowns. The most convenient two points lie on the ends of the diameter through theimage charge, as shown (P andQ in the diagram).

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P =1

40

q

d + a+

q0

d0 + a

= 0 (7)

and

Q =1

40

q

d a +q0

a d0

= 0 (8)

Then from equation (7)

q(d0 + a) + q0 (d + a) = 0and from (8)

q(a d0) + q0 (d a) = 0Adding these two relations eliminates d0 and gives

2qa + 2q0d = 0 q0 = qad

(9)

This result has the nice property that the image charge is negative if q is positive, as wefound in the planar case. Once again the image charge represents the charge drawn onto the

surface of the sphere through the ground wire.

Now we subtract the two relations to obtain an expression for d0 :

2qd0 + 2q0a = 0

d0 = a q0

q= a

a

d

=

a2

d(10)

The image charge is inside the sphere ifd > a, as we need. Conversely, ifd < a, the imagecharge is outside the sphere. (You are asked to confirm this result in Problem 2.2.)

We can perform two checks on this result. First lets find the potential at an arbitrary

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point on the sphere.

40 (a, ) = qR + q0

R0

=q

d2 + a2 2ad cos qa/dq

(d0)2 + a2 2ad0 cos

=q

d

1q1 + a

2

d2 2ad cos aq

(a2/d)2 + a2 2 (a3/d)cos

= 0

for all , as expected.Second, lets check that we get back the result from 1.1 as a (plane boundary).

We have to be a bit careful here, because if we immediately let a , the point fromwhich we are measuring our distances moves off infinitely far to the left, and we will learn

nothing. So first we write our results in terms of distance from the surface of the sphere.

The charge q is a distance d a = h from the surface, and thenq0 = q a

a + h= q

1 + h/a qas a

as required. The distance of the image from the surface is

h0 = a d0 = a a2

a + h=

ah

a + h=

h

1 + h/a h as a

and this is the second required result.

Finally we can write the Dirichlet Greens function for the region outside a sphere of

radius a by setting q = 1 and multiplying by 40 :

GD (x, x0) =

1

|x x0| a

r01

|x x00|where x00 is the position vector of the image point. Again this is pretty ugly.

1.3 Images in a cylinder

The basic ideas and methods are the same as we have used in the plane and sphere cases.

See Jackson problem 2.11.

1.4 Use of images to solve problems

Jackson P 2.10 asks us to compute the potential inside a parallel-plate capacitor with a small

hemispherical boss on one plate.

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The clue is in part (c) of the problem: We model the system by putting a point charge q at avery large distance d from the plane. Then we can use the image system shown to model thecapacitor, since the image system puts potential zero on the lower plate. (q andq, q0 andq0 form pairs that make the potential on the plane zero; q andq0, q andq0 form pairsthat make the potential on the sphere zero.) Then we can show that as d

we obtain a

uniform field at large distance from the lower plate, and we set that uniform field equal to

the given value ofE0, thus determining the necessary charge q.

Setup: From 1.2, the image charge q0 = adqand its distance from the plane is d0 = a2

d .The potential due to the four charges (one real charge and three images) is

(r, ) = kq

(1/

r2 + d2 2rd cos 1/

r2 + d2 + 2rd cos

ad /q

r2 + a4

d2 2r a2

d cos +ad

/q

r2 + a4

d2 + 2ra2

d cos

)(11)

To see why this works, look at the potential for d r a. We drop terms in (a/d)2 to get

(r, ) 'kq

d

1/q

r2

d2 + 1 2 rd cos 1/q

r2

d2 + 1 + 2rd cos

ar /q1 2

a2

drcos +

ar /q1 + 2

a2

rdcos

Next expand the square roots, dropping terms in (r/d)2 and(a/r) (a/d).

(r, ) 'kq

d

n1 +

r

dcos

1 r

dcos

a

r+

a

r

o=

kq

d2

r

dcos =

2kq

d2z to first order in small quantities

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This corresponds to the given uniform field provided that E0 = 2kq/d2, or, if we choose

q = E0d2

/2k. (12)

Solve: (a) To find the surface charge densities, begin with the field components. From

(11), we have

r= kq

( r d cos

(r2 + d2 2rd cos )3/2+

r + d cos

(r2 + d2 + 2rd cos )3/2

+a

r a2 cos /d

d

r2 + a4

d2 2ra2

d cos 3/2 a

r + a2 cos /d

d

r2 + a4

d2 + 2ra2

d cos 3/2

and at r = a

Er = kq( a

d cos

(a2 + d2 2ad cos )3/2 a + d cos

(a2 + d2 + 2ad cos )3/2

a

a a2 cos /d

d

a2 + a4

d2 2aa2

dcos

3/2 + a

a + a2 cos /d

d

a2 + a4

d2+ 2aa

2

dcos

3/2

= kq

(a d cos

(a2 + d2 2ad cos )3/2 a + d cos

(a2 + d2 + 2ad cos )3/2

da2 (d a cos )

a3 (d2 + a2 2ad cos )3/2+

da2 (d + a cos )

a3 (d2 + a2 + 2ad cos )3/2

)

= kq

"a d cos da (d a cos )

(a2 + d2

2ad cos )3/2

a + d cos da (d + a cos )

(a2 + d2 + 2ad cos )3/2

#

= kqa

1 d

2

a2

"1

(a2 + d2 2ad cos )3/2 1

(a2 + d2 + 2ad cos )3/2

#

and thus the charge density is (notes 1 eqn 5 with E = 0 inside the boss):

() = 0Er = 0kqa

1 d

2

a2

"1

(a2 + d2 2ad cos )3/2 1

(a2 + d2 + 2ad cos )3/2

#

Analysis: Notice that is zero in the corners at = /2, .as expected. Also since

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1 cos 0 on the boss and d > a, the chrage density is negative on the boss if q ispositive, as expected.

Solve: The total charge on the boss is

Qboss = 2

Z/20

() a2 sin d

=2

4qa3

1 d

2

a2

Z10

1

(a2 + d2 2ad)3/2 1

(a2 + d2 + 2ad)3/2

!d

= qa3

2

1 d

2

a2

1

2ad

2

(a2 + d2 2da)1/2 2

(a2 + d2 + 2da)1/2

!1

0

Qboss = q

2

1 d

2

a2

a2

d

1

(a2 + d2

2ad)

1/2+

2a2 + d2

1(a2 + d2 + 2ad)

1/2

!

= q2

a2 d2 1

d

1d a

1

a + d+

2a2 + d2

Qboss = q

1 d2 a2

d

a2 + d2

which is Jacksons result.

Analysis: As d forfixed q, Qboss 0. . Is this what you would haveexpected? The induced charge also goes to zero as a 0, .as the boss disappears in thiscase.

Solve: The charge density on the plane is

() = 0Ez = 0

z

z=0

Here it is more convenient to express the potential in terms of z :

(, z) = kq( 1pz2

+ 2

+ d2

2zd

1pz2

+ 2

+ d2

+ 2zd

adq

z2 + 2 + a4

d2 2z a2

d

+a

dq

z2 + 2 + a4

d2 + 2za2

d

(13)

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Thus

() = 0kq" (z d)(z2 + 2 + d2 2zd)3/2

(z + d)(z2 + 2 + d2 + 2zd)3/2

+a

z a2/dd

z2 + 2 + a4

d2 2z a2

d

3/2 a

z + a2/d

d

z2 + 2 + a4

d2 + 2za2

d

3/2z=0

= qd2

1

(2 + d2)3/2 a

3

d3

2 + a4

d2

3/2

where the second term is negligible if d a.The total charge on the plane is:

Qplane = qd

2 2Za 1

(2 + d2)3/2 a3

d3

2 + a4d23/2

d

= qd2

2

(2 + d2)1/2 a

3 (2)d3

2 + a4

d2

1/2

a

= qd 1

(a2 + d2)1/2 a

3

d3

a2 + a4

d2

1/2

= qd

d2 a2d2 + a2

Analysis: Qplane qas a 0 (flat plate) ord . The total induced charge is:

Qboss + Qplane = q1 d2 a2da2 + d2 q1d d2

a2

d2 + a2= q

as expected.

Solve: Now lets put in the value forqthat gets us to the capacitor-plus-boss system (eqn12): q = E0d

2/2k. Then, ford a, the charge on the boss is:

Qboss = q

1 1 a2/d2p

a2/d2 + 1

!' E0

2kd2

1

1 a2

d2

1 1

2

a2

d2

+ O

ad

4

= E02k

d2

3

2

a2

d2

= 3

4

E0k

a2 = 30E0a2

Analysis: This is Jacksons answer in (b). Notice that d disappears in the limit d , asrequired.

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Solve: The charge densities are:

() = 0E0a2

1 d

2

a2

d2

(a2 + d2 2ad cos )3/2 d

2

(a2 + d2 + 2ad cos )3/2!

= 0E0a

2d

1 d

2

a2

1

(a2/d2 + 1 2a/d cos )3/2 1

(a2/d2 + 1 + 2a/d cos )3/2

!

' 0E0a

2d

d

2

a2

1 + 3

a

dcos

1 3 a

dcos

= 0E0a

2d

d

2

a2

6

a

dcos

= 30E0 cos

on the boss, and on the plane

() = E0d3

4k 1

(2 + d2)3/2 a3

d3

2 + a4

d2

3/2

= E04k

1 32

2d2 a

3

3

1 + a4

d22

3/2

= E04k

1 3

2

2d2 a

3

3

1 3

2

a4

d22

0E0

1 a3

3

as d

Analysis: Note that the charge density is zero where the boss meets the plate ( = a, = /2), as expected near a sharp hole in the conductor. Also 0E0 as ,the expected result for a

flat plate.

The plot shows /0E0 versus /a for > a and versus 2/ for0 < < /2

0 1 2 3 4 5

-3

-2

-1

0

rho/a

s/s0

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Solve: The potential (13) is

(, z) = kqd

1q

1 +zd

2+d

2 2zd 1q

1 +zd

2+d

2+ 2 zd

adq

zd

2+d

2+ a

4

d4 2 zd a2

d2

+a

dq

zd

2+d

2+ a

4

d4 + 2zda2

d2

'E0d

2

n1 +

z

dh

1 zd

io E0

2

apz2 + 2

2 z

d

a2

z2 + 2

' E0z

1 a

3

(z2 + 2)3/2

!as d

Analysis: This function is plotted below. All distances are scaled by a.

Values of

/E0a are: black 0.2, red 0.5, green 1, purple 2

2 Expansion in orthogonal functions

To obtain a more useful form of the Greens function, well want to expand in orthogonal

functions that are (relatively) easy to integrate. We begin with some basics and well 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 Greens function.

2.1 Sturm-Liouville theory

See Lea Chapter 8 sections 1 and 2.

2.2 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 = 0

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in terms of the chosen coordinates, called u, v, andw, separate variables, and solve todetermine the eigenfunctions. Youll get equations something like:

DuU

U= ( + ) ; DvV

V= ;

DwW

W=

Du is a differential operator of the form

f(u)d2

du2+ g (u)

d

du+ h (u)

and similarly forDv andDw.

2Note 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 will want to chooseyour separation constants so that the sets of eigenfunctions in the other two coordinates

v andw are orthogonal functions.

3Then use the boundaries on which = 0 to determine your eigenfunctions V andW,and eigenvalues, (which Ill call = DvVV and =

DwWW

). If the boundaries are at afinite distance the eigenvalues will be countable and can be labelled with an integer: m.Otherwise they will form a continuous set.

4The last eigenfunction is determined from the differential equation in the final coordinate,

u:

=DuU

U=

At this point you should have a general solution of the form:

(u,v,w) =Xm,n

AmnUmn (u : m n) Vm (v : m) Wn (w : n)

where the coefficients Amn are still undetermined.

5Now use the final, non-zero, boundary condition together with orthogonality of the

eigenfunctions Vm andWn to find the coefficients Amn

2.3 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

2.3.1 Rectangular 2-D problems with non-zero potential on two sides.

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 beinggrounded. We solve two problems, each with three sides grounded, one having potential Vat x = 0 and one having potential V at y = b. Then we add the results. The solution is

= 1 +2

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where (Lea Example 8.1)

2 = 4V

Xm=0

12m + 1

sin

(2m + 1) xa

sinh

(2m+1)y

a

sinh (2m+1)ba

and

1 =4V

Xn=0

1

2n + 1sin

(2n + 1) y

b

sinh (2n+1)(ax)b

sinh (2n+1)abWith a = 2b the potential looks like this (the axes are x/a andy/a):

0

0.1

0.2

0.3

0.4

0.5

y

0.2 0.4 0.6 0.8 1x

Blue 0.75, black, 0.5, red 0.25, dashed 0.9

2.3.2 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 (x, y) = X(x) Y (y) , Laplaces equation takes the form

X00

X= k2 = Y

00

Y. The solutions to the differential equation that satisfy the boundary conditions = 0 aty = 0 and at x = a but 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,1 in the previous problem has gone to zero because of the b in the denominator of theargument, leaving 2 with an undetermined k.). However, since the eigenfunction is aneven function ofk, we need only positive values ofk. Thus the solution is

(x, y) =Z0 A (k)sin ky sinh[k(a x)] dk

To find the function A (k) , we use the known potential V (y) at x = 0 :

(0, y) = V (y) =

Z

0

A (k)sin ky sinh ka dk

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We now make use of the orthogonality of the functions sin ky by multiplying both sides bysin k0y and integrating overy :Z

0

V (y)sin k0y dy =

Z

0

Z

0

A (k)sin ky sinh ka dk sin k0y dy

On the RHS, we haveZ

0

sin ky sin k0y dy =1

2

Z

0

[cos(k k0) y cos(k + k0) y] dy

=1

4

Z

0

nei(kk

0)y + ei(kk0)y ei(k+k0)y ei(k+k0)y

ody

=1

4

Z

{exp[i (k k0) y] exp[i (k + k0) y]} dy

=

2[(k k0) (k + k0)]

where we used Lea eqn 6.16. ThusZ

0

V (y)sin k0y dy =

2

Z

0

A (k)sinh ka [(k k0) (k + k0)] dk

Since k andk0 are both positive, only the first delta function contributes, and this integralreduces to Z

0

V (y)sin k0y dy =

2A (k0) sinh k0a

and thus, dropping the primes,

A (k) =2

Z

0

V (y)sin ky

sinh kady

For example, ifV (y) = V0ey/c, then

A (k) =2

Z0

V0e

y/c

sin ky

sinh ka dy

=1

i

V0sinh ka

e(ik1/c)y

ik 1c e

(ik1/c)y

ik 1c

0

=1

i

V0sinh ka

1ik 1c

1ik + 1c

=1

i

V0sinh ka

2ic2k

k2c2 + 1

and thus

(x, y) =V0

Z

0

2c2k2

k2c2 + 1

sin ky

k

sinh[k(a x)]sinh ka

dk

Analysis: We can see right away that our result is dimensionally correct. Also, since

x a, k(a x) ka and thussinh[k(a x)]

sinh ka 1

for all k, so the integral converges. I couldnt do this integral analytically, so lets computesome values in the case c = a:

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a2

, a2

/V0 =2

0.314 45 = 0.20019

a2

, a /V0

= 2

0.281 18 = 0.179

a2

, 2a

/V0 =2

0.119 04 = 7.5783 102

The potential decreases rapidly fory > a

2.3.3 A 3-D problem

Find the potential inside a cubical box of side a with grounded walls, except for the side at

z = a which has potential V0

h1 2xa 12i. This potential increases from zero at the

walls to V0 at x = a/2,as shown in the diagram.

0

0.2

0.4

0.6

0.8

1

ntial

0.2 0.4 0.6 0.8 1x/a

There is no charge inside the box, so the potential satisfies Laplaces equation:

2 = 02

x2

+ 2

y2

+ 2

z2

= 0

Try separating variables:

= XY ZThen

X00Y Z+ XY 00Z+ XY Z00 = 0or

X00

X+

Y00

Y+

Z00

Z= 0 (14)

For equation 14 to hold forall values ofx,y, z we must have:

X00

X= k1,

Y00

Y= k2,

Z00

Z= k3 andk1 + k2 + k3 = 0

The potential is zero at x = 0 and at x = a, so we needk1 to be negative, k1 =

2,which gives the solutions X = sin x andX = cos x, which have multiple zeros. Wechoose the sine function to make X(0) = 0, and then choose = n/a to make X(a) = 0.A similar reasoning gets Y = sin(my/a) . Then

k3 =

n2 + m2 2

a2

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and to make Z(0) = 0 the appropriate solution forZ is the sinh .

(x, y, z) =X

n,m=1

Anm sin n xa

sin m ya

sinhp

(n2 + m2) za

Finally we evaluate the potential at z = a :

V0

"1

2x

a 1

2#=

Xn,m=1

Anm sin nx

asin m

y

asinh

p(n2 + m2)

We make use of the orhogonality of the sines by multiplying both sides by

sin n0x/a sin m0y/a and integrating overx andy :Za0

Za0

V0

"1

2x

a 1

2#sin

n0

x

a

sin

m0

y

a

dxdy = An0m0

a2

4sinh

q(n0)2 + (m0)2

Drop the primes, and let u = 2x/a

1, to getZ1

1

V

1 u2

sin n

u + 1

2

a

2du

cos my/am/a

a0

= Anma2

4sinh

p(n2 + m2)

a

2

Z11

V0

1 u2 sin nu2

cosn

2+ cos

nu

2sin

n

2

du

1 (1)mm/a

= Anma2

4sinh

p(n2 + m2)(15)

The result is zero unless m is odd. When we integrate overu the sine term gives zerobecause we have an odd integrand over an even range. For the cosine term, we do integation

by parts:Z11

1 u2 cos nu

2du =

2

nsin

nu

2

+1

1

2Z10

u2 cosnu

2du

=4

n

sinn

2 2"u2 2

n

sinnu

2+1

0 2

2

nZ

1

0

u sinnu

2

du#Z11

1 u2 cos nu

2du =

4

nsin

n

2

2 2

nsin

n

2+ 2u

2

n

2cos

nu

2

+1

0

2

2

n

2 Z10

cosnu

2du

= 2"

23

(n)2cos

n

2 2

2

n

3sin

nu

2

+10

#

= 16(n)2

cosn

2+

32

(n)3sin

n

2

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and so the left hand side of (15) is

V0 a2

m

32

(n)3sin2 n

2 16

(n)2sin n

2cos n

2

!

= V0a2

m

16

(n)3(1 cos n) 8

(n)2sin n

!

= V0a2

m

16

(n)3 (1 (1)

n)

= V032a2

mn34form andn odd

Thus form odd andn odd

Anm = V0128

mn341

sinh

n2 + m2

So

(x, y, z) =128

4V0

Xm=1,odd

Xn=1,odd

1

mn3sin

nx

asin

my

a

sinh

n2 + m2 zasinh

n2 + m2

Analysis: The result is dimensionally correct, and converges nicely.

The plot shows the series up to m = n = 5. The black line is the potential as a functionofz forx = y = a/2, and the red line forx = y = a/4. The potential increases fasternear the center of the box, as expected, since the potential on the top side is maximum at

x = a/2. .

0

0.2

0.4

0.6

0.8

hi/V0

0.2 0.4 0.6 0.8 1z/a

At z = a/2, y = a/2 the potential versus x/a looks like this:

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0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

i/V0

0.2 0.4 0.6 0.8 1x/a

3 Conformal mapping

Recall that both the real and imaginary parts of an analytic function satisfy Laplaces

equation in two dimensions. We may use this fact to solve 2-d potential problems.

3.0.4 Potential in a wedge.

Suppose the region of interest is defined by the angular wedge 0 . Further supposethe potential is zero on the conducting boundaries at = 0 and = . Then the analyticfunction

f(z) = z/ = r/ei/ = r/

cos

+ i sin

has imaginary part

v (r, ) = r/ sin

that satisfies the boundary conditions. If = /m for some integer m, then f(z) isanalytic everywhere. If is an arbitrary real number, then f(z) may have a branch pointat the origin, but we may choose the branch cut so that f(z) is still analytic in our regioneverywhere except AT the origin. Thus v (r, ) also satisfies Laplaces equation in thevolume, and so it is a solution of the correct form. In fact the function f(z) = zn/ for anyn has the same nice properties. Thus the potential in a wedge-shaped region with openingangle and conducting boundaries at potential V0 is described by the complex potential

(z) = A + iV0 + nanzn/

The imaginary part is

V0 ++

Xn=

anrn/ sin

n

which is the potential in the region. The coefficients an must be chosen to satisfy anyremaining boundary conditions in r. Compare with Jackson 2.72 which he derives usingseparation of variables in plane polar coordinates.

The sum is over positive n if the origin is included within our region. The solutionnear the origin is dominated by the first (n = 1) term. Then the field near the origin has

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components

E = a1

r

/1 hsin

r + cos

i

Thus

E 0 as r 0 if/ > 1 but

E as r 0 if/ < 1. The field is small

in a hole ( < ) but very large near a spike ( > ).The diagram shows the cases of = /2 and3/2. For = /2, the equipotential

surfaces are

r2 sin2 = constant = 2xyThus the equipotentials are hyperbolae.

0 1 2 3 4 50

1

2

3

4

5

x

y

equipotentials andfield lines for = /2

-6 -4 -2 2 4 6 8 10 12

-6

-4

-2

24

6

8

10

12

x

y

= 3/2

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.

4 Finite element analysis

The big idea: we want a procedure for numerically solving a potential problem for a

source g (x) in a region R with specified boundary conditions.Lets look at a Dirichlet problem in 2 dimensions.

We start with two general relations. First, if2 = g in R, thenZR

2 + g dxdy = 0

forany function . The next relation we need is Greens identity from Chapter 1 (Notes 2.5eqn 1).

Z2 +

dV =

ZS

ndA = 0

where we have chosen the function to be zero on the boundary. (Note: in a twodimensional problem the volume is an area and the bounding surface S is actually acurve.) Now inserting our differential equation for, we get:Z

R

g

dxdy = 0 (16)

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The next step is to set up a grid over the region R. We expand the desired solution in a setof functions ij each of which is zero except on a small region around the grid point (i, j) .For example, let:

ij =

( 1 |xxi|h

1 |yyj |h

for |x xi| < h, |y yj | < h

0 otherwise

and then let

(x, y) =Xk,l

klkl (x, y)

Effectively, the function kl smears the potential value kl over one grid square, convertinga set of numerical values kl into a continuous function. (x, y) . Now stuff this assumed

form into the integral 16, and take the function ij .ZR

Xk,l

kl kl ijdxdy =ZR

ijg (x, y) dxdy (17)

The integral on the right is non zero only on a square region surrounding (xi, yj) , andbecause the function ij is peaky we can approximate the integral as:Z

R

ijg (x, y) dxdy ' g (xi, yj)

Zsquare

ijdxdy

which may be evaluated as follows. Let u = x xi andv = y yj. Then:Zsquare

ijdxdy =

Zh0

1 u

h

du +

Z0h

1 +

u

h

du

!Zh0

1 v

h

dv +

Z0h

1 +

v

h

dv

!

=

h2

+ h2

h2

+ h2

= h2

Thus (17) becomes: Xk,l

kl

ZR

kl ijdxdy = h2g (xi, yj) (18)

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20/24

where the sum on the left is over all the grid points. However, the integrand is non-zero

only on a small rectangular region 2h by 2h surrounding the grid point (i, j). First note thatinside this square region,

ij = 1

h

1 v

h

x +

1 u

h

1

h

y

where in the first factor of each term we take the upper sign on the right half of our square

(0 < u < h) and the bottom sign on the left (h < u < 0); in the second factor we take thetop sign on the top of the box (0 < v < h) and the bottom sign on the bottom (h < v < 0).

Thus ifk = i andl = j:Zbox

ij i,jdxdy =1

h2

Zh0

Zh0

1 v

h

2+

1 uh

2dudv

+1

h2 Zh

0 Z0

h

1 +

v

h2

+ 1 u

h2

dudv+

1

h2

Z0h

Zh0

1 v

h

2+

1 +u

h

2dudv

+1

h2

Z0h

Z0h

1 +

v

h

2+

1 +u

h

2dudv

= 4

Z10

Z10

2 + 2

dd

=4

3

3+ 3

10

10

=8

3where in the four terms we took = 1 vh (terms1 and 3), = 1 + vh (terms 2 and 4), = 1 uh (terms1 and 2) or = 1 + uh (terms 3 and 4).

Also ifk = i + 1 andl = j :

i+1,j =

1 |x xi h|h

1 |y yj |

h

=

1 |u h|

h

1 |v|

h

for |u h| < h and |v| < h

and zero otherwise. The overlap region is on the right side of our box (0 < u < h) where

i+1,j =1

h

1 v

h

x +

u

h

1

h

y

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giving

Zbox

ij i+1,jdxdy = 1h2Zh0

Zh0

1 vh

2+

1 uh

uh

dudv

+1

h2

Z0h

Zh0

1 +v

h

2+

1 uh

uh

dudv

=

Z10

Z10

2 + (1 ) dd= 2

3

3+

2

2

3

3

1

0

1

0

= 2

1

2 2

3

= 1

3

We get the same result in all the overlap regions. Thus equation (18) may be written as amatrix equation:

K= Gwhere the matrix K is described as sparse it has only a few non-vanishing elements and

they are all near the diagonal, like this:

K =1

3

8 1 1 0 01 8 1 1 01 1 8 1 10 1 1 8 10 0 1 1 8

Matrices of this type are relatively easy to invert numerically. The column vectors andG

contain the values of the potential and the source at the grid points. and we have reduced the

potential problem to a matrix inversion.

= K1G

For regions with odd shapes, the square grid we used above does not fit very well.

Triangles of arbitrary size and shape can be fit onto a region of almost any shape. So now

well modify the method above to use triangles instead of squares. By varying the sizes and

shapes of the triangles we can also get better resolution where things change more rapidly.

The basic triangular element has vertices at points that we label 1,2 and 3 with

coordinates (x1, y1) , (x2, y2) , and(x3, y3) . We approximate the potential solution (x, y)in this triangle by a Taylor-series-type expansion of the form:

(x, y) = A + Bx + Cy

The 3 values of the potential at the 3 vertices give enough information to evaluate the 3

coefficients A, B andC. To make the numerical computation more efficient, it is convenient

to define three shape functions Ni (x, y) = ai + bix + ciy that have the properties:Ni (xi, yi) = 1

and

Ni (xj , yj) = 0, i 6= jThese functions are the analogue of the ij that we used with the square grid. They have the

21


22/24

effect of smearing the potential at node i over the whole triangle. Then, for example, forN1we have:

a1 + b1x1 + c1y1 = 1

a1 + b1x2 + c1y2 = 0

a1 + b1x3 + c1y3 = 0

This set of equations has a nontrivial solution for the coefficients a, b andc if the determinant:

D =

1 x1 y11 x2 y2

1 x3 y3

6= 0

The determinant is:

D = x2y3 x3y2 + x3y1 x1y3 + x1y2 x2y1= (x2

x1) (y3

y1)

(x3

x1) (y2

y1)

This is related to the area of the triangle. To see how, remember that we can write the areausing the cross product:

A =1

2

1 2

where 1 and2 are vectors along the sides of the triangle. In terms of the coordinates:

1 = (x2 x1) x + (y2 y1) yand

2 = (x3 x1) x + (y3 y1) yThen

2A = (x2 x1) (y3 y1) (x3 x1) (y2 y1)and so

D = 2A

and so is never zero. Then the solution for the coefficients is:

a1 =1

2A(x2y3 x3y2)

b1 =1

2A(y2 y3)

c1 = 1

2A(x2 x3)

and similarly for the others.

Now the procedure runs pretty much as before. In equation (16) we take:

=Xj

jNj

where the sum is over all the vertices of all the triangles. Next take = Ni for one vertex of

one triangle. Then relation (16) becomes:3X

j=1

j

Ztriangle

Ni Nj dxdy =Z

gNi dxdy ' g (x, y)

ZNi dxdy

= g (x, y) A (ai + bix + ciy) (19)

22


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where x, y are the coordinates of the center of the triangle. Now using our solutions for thecoefficients, we have:

a1 + b1x + c1y =1

2A

(x2y3 x3y2) + (y2 y3)

(x1 + x2 + x3)

3 (x2 x3)

(y1 + y2 + y3)

3

=1

6A[x2y3 x3y2 + x1y2 x1y3 x2y1 + x3y1] =

D

6A=

1

3On the left hand side, we use the result that

Ni = bix+ciyand is constant over the area of the triangle, so (19) becomes

3Xj=1

j (bibj + cicj) A = g (x, y)A

3

To combine the triangles, we define

kij = (bibj + cicj) A (20)

Ifi = j, kii refers to a single node. For each internal node i we sum over all the trianglesconnected to that node.

Kii =X

triangles

kii

The elements kij , i 6= j, are associated with two nodes, i.e. with the side of the triangleconnecting the nodes. So we sum over all the triangles with a side along ij (usually two).

Kij =X

triangles

kij i < j N

If a node is on the boundary, the value of the potential there will be known. These terms

are moved to the right hand side and serve as source terms. So we have:

Gi = 13

Xtriangles

Atgt N0X

j=N+1

Kijj

where the numbers N + 1 to N0 label the nodes on the boundary.Then combining the results for all the triangles, we have the matrix equation:

K= G

where againK is a sparse matrix.

Example of how to get the kij. Consider an isoceles right triangle of sides 1,1 and

2.Put the origin at the right angle, and label the vertices 1 (the origin), 2 (on the y-axis) and 3(on the x-axis). Then the area of the triangle is 1/2, and 2A = 1

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.

a1 = x2y3 x3y2 = 0 1 = 1b1 = y2

y3 = 1

c1 = (x2 x3) = (1) = 1b2 = y3 y1 = 0

c2 = (x3 x1) = 1Then from (20) we get

k11 =1

2

b21 + c

21

=

1

2(1 + 1) = 1

k12 =1

2(b1b2 + c1c2) =

1

2(0 + 1 (1)) = 1

2and so on. (See Jackson Fig 2.16). These coefficients depend on the shape of the triangle

but not on its orientation or size.

24

Documents

Method of Images Greens Jackson