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Ch11.3 (Ch12.2e) Neumann functions, Bessel Functions of the

2nd Kind講者： 許永昌 老師

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ContentsDefinitionOther formsRecurrence RelationWronskian FormulasExample: Coaxial Wave Guides

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Definition ( 請預讀 P699~P700)

Neumann Functions:Why do we need to define a new solution for

Bessel Eq.?When n, Jn and J-n are independent to each

other; however, J-n=(-1)nJn when n. We need to find a second solution. (Ch 9.6)

The asymptotic series of Jn(z)~

we want to find a solution Nn(z)~

cos.

sin

J x J xN x

2 1cos

2 2z

z

2 1

sin2 2

zz

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Series Form ( 請預讀 P700)

1)

2

0

2

cos 1'

sin

11 2 21! 2 ! !

1 12 1 2ln

2 ! 2 2 1

limn

n n n nn

n

n ns s

n

s

s s n

n

J x J x J x J xN x L Hospital rule

x xx

s n n s n s n

n sx x xJ x

s n s x

0

2 21

0 0

21

0

1

1

1 1 12 1 2 1ln 1 1

2 ! 2 1 ! ! 2

1 ! 11 2 1ln

! 2 2 !

n

s

s ss n s nn

ns s

ss nn

ns

s n

s n

s nx x xJ x n s s

s x s n s n s

n s x xJ x

s s n

2

0

1 1 .! 2

s n

s

xn s s

s

,

1 1 1sin 1 .

1 1n s

n s

Where

s n d dn s n s n s

s n dn s n dn

Reference: http://mathworld.wolfram.com/BesselFunctionoftheSecondKind.html

Q: How about Nn?

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Other Forms ( 請預讀 P701)

From the contour integral representation of Hankel function

10 0

2 2 2

2

cosh2

0 0

0 20 1

1exp1 2

1exp cosh , 0 ,

2 2cos cosh sin cosh ,

cos2 2cos cosh . cosh

1

ie

i i is i

i

i ix s

x t tH dti t

ix s ds t ie ie e iei

e ds x s i x s dsi i

xtN x s ds dt t s

t

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Recurrence relations ( 請預讀 P702)

1 11

1 11

1 1

1 1cos,

sin1 1 cos,

sincos

,sin

22

.2

t

J JN

J JN

J JN

g J J Jx N N N

xJ J J

x

1 11

1 11

1 11 1cos

,1 1sin

cos,

sincos

,sin

2 '2 ' .

2 'x

J JN

J JN

J JN

g J J JN N N

J J J

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' ' .A

W J J J Jx

Wronskian Formulas ( 請預讀 P702)

Wronskian (See Ch9.6):

Therefore, Find An from x0.

1 1

Note: '' ' 02 2

' 1exp exp .

'

x x

y Py Qy

y y AW B Pdt B dt

y y t x

1 1 2, ,

! 2 ! 2' '

! !2' , ' ,

! 2 !

2sin' ' .

xJ x J

xJ J J J

xxJ x J

x x x

W J x J x J x J xx

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Wronskian Formulas (continue)

''

1 1' sin'

sin sin

2' ' . 1

J JJ J W

WN N J J

J x N x N x J xx

1 1

1 1

21 ,

2. 2

J N N N J Jx x x

J N N Jx

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Wronskian Formulas (continue)

2sin' ' . 3J x J x J x J x

x

1 1

1 1

2sin3 ,

2sin. 4

J J J J J Jx x x

J J J Jx

1 1

1 1

2sin3 ,

2sin. 5

J J J J J Jx x x

J J J Jx

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Example ( 請預讀 P703~P704)

Coaxial Wave Guide:

Conditions:For TM mode (Hz=0 everywhere)zE=ikE,

Under these two conditions, we can get

Ez=0 on the boundary ( Faraday’s Law)

Therefore, we just consider the PDE for Ez.

2

22

1E E Ec

22 2

2 2, .t t z

ikE E k

c

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Example (continue)

PDE: -2Ez=(w/c)2Ez.

Separation Variables:

Boundary condition: Ez(a)= Ez(b)=0

2 22

1 1. 2z zE E

.i t ikz inz n n n n

n

E e e a J b N e

0,

0.

n n n n

n n n n

a J a b N a

a J b b N b

Therefore, g is quantized.If g and w are provided,

cut off frequency of this wave guide.

c

2 22 .kc

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Example (for TE mode) (continue)

Boundary conditions:nE=0 on the boundary (Faraday’s Law)Br=0 on the boundary (Gauss Law of magnetic

field or E=iwB.)For TE mode:

Ez=0 everywhererHz=0 on the boundary (Ampere’s Law).PDE: -2Hz=(w/c)2Hz.

22 2

2 2, .t t z

ikH H k

c

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Example (for TEM mode) (continue)

For TEM mode: Ez=Bz=0 everywhere.

If we use =t+ê3z for Maxwell Eqs., we will get

Therefore, there is no TEM mode for a hollow cylindrical wave guide, but TEM mode can be survived in a coaxial cylindrical wave guide.

222 0

0

, .

, 2 .0,

ˆ ˆ, & are constants.0.

ikzz t t t t

t t t t

tt t t t

E E E e E x yc

free source D electrostatic caseH E

E a bH E

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Homework11.3.2 (12.2.2e)11.3.3 (12.2.3e)

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NounsNeumann function: Nn(x) or Yn(x).P699