物理-數(shù)學方法課程

Mathematical

PhysicsCylindrical

FunctionsCylindrical

FunctionsFundamental

PropertiesEigenvalue

ProblemSymmetric

Cylindrical

ProblemsGeneral

Cylindrical

ProblemsConclusionFundamental

PropertiesCylindrical

Functions

of

order

mDefinition：x2

y"xy'(x2

m2

)

y

02k

mx1)k(k

m

1)(!2k

0

kJm

(

x)

mm

()xJcos

mx

m

()xJsin

mx()xNHm

(x)

Jm

(x)

i

Nm

(x)Special

solution

ofClassification：Bessel

FunctionofordermannFunctionofordermHankel

Function

ofordermFundamental

PropertiesGraphs

of

Cylindrical

FunctionsBessel

Functionsann

FunctionsProperties

of

Cylindrical

FunctionsSymmetryFor

m

N,

Zm(-x)

=(-1)m

Zm(x)Asymptotic

PropertiesNull

pointsRecurrence

FormulasBesselFunctionsann

FunctionsAsymptotic

PropertiesAs x

→0,

wehave

：

2Jm0N0

(

x)

2

ln

x

,

Nm0J

0

(

x)

1,As x

→∞,we

have：2exp[

i(H

(

x)

exp[i(

xH

( )

sin(

x

cos(

x

1m

x

2

xm

2

x

2

xJm

(

x)

Nm

(

x)

Null

Points

of Bessel

Functions2 4From

the

asymptotic

formula,

oneobtainscos(

x

1

m

1

)

0

x

1

m

1

(n

1

)2

4

2

4

2x(m)

(n

1

m

1

)nThe

positive

zerosappear

alterna

y.1

1

1

1

10xx(0)

x(1)

)2(

x

m)(

x

m1)(According

to

the

graph：There

are

infini y

many

positive

zeros.0

x(m)

x(m)

x(m)

x(m)

x(m)

1

2

3

n

n1The positive

zeroincreases

with

the

orderm.1

1

2

2

30x()m

x

m1)(

x()m

x

m1)(

x()mRecurrence

Formulas[xm

Zm

(

x)]'

xm

Zm1(

x)[xm

Zm

(

x)]'

xm

Zm1(

x)Basic

recurrence

formulasCorollary

2Zm

'mZm

/

x

Zm1Zm

'mZm

/

x

Zm1Corollary

12Zm

'

Zm1

Zm12mZm

/

x

Zm1

Zm1Proof

of

the

recurrence

formula

22k

mx(1)kk

0

k!(k

m

1)Jm

(

x)

21mm(

x2k

)'2k

mk

0

k!(k

m

1)(1)k[J

(

x)

/

x

]'

212k

12k

mxk

1

k!(k

m

1)2k(1)k

212k

12k

m1

(1)k

1xk

1

(k

1)!(k

m

1)l

k

1

lx/

x212l

1m

m2l

m1l

0

l!(l

m

1

1)(1)

Jm1(x)

/

xmApplications

of

the

recurrence

formulas[

x

m

Zm

]'

x

m

Zm1

Zk

xk

1[

x1k

Zk

1

]'[

x

m

Zm

]'

x

m

Zm1

Z0

x

1[

xZ1

]'

xn

Jmdx

xn

xm1[x1m

Jm1]'

dx

xn

Jm1

(n

m

1)

xn1Jm1dx

xn

J0dx

xn

x1[xJ1]'

dx

xn

J1

(n

1)

xn1J1dx

xn

J1

(n

1)xn1J0

(n

1)2

xn2

J0dx

xm

Jm1dx

xm

Jm

c

x3J0dx

x3J1

2x2

J0

4

xJ

0dx

xn

Jmdx

xn

Jm1

(n

m

1)

xn1Jm1dx

xJ0dx

xJ1

xn

J0dx

xn

J1

(n

1)xn1J0

(n

1)2

xn2

J0dxEx.1Ex.2

J1dx

J0Ex.4Ex.3

x2

J1dx

x2

J0

2

xJ

0dxEx.5

xJ

2dx

xJ1

2

J1dx

xm

Jm1dx

xm

Jm

cApplications

of

the

recurrence

formulasEigenvalue

problems

of

Bessel’s

Eq.Theresolution

of

rotational

symmetric

cylindrical

problemsGeneral

eigenvalue

problemsEigenvalue

problemsEigenvalues

andeigenfunctionsOrthogonality

and

completenessTypical

eigenvalue

problemsFiniteness

and

boundary

condition

of

the

1st

kindFiniteness

and

boundary

condition

of

the

2nd

kindBoundaryconditions

ofthe1st

kindut

a22uT

'a2k

2T

0u

T

(t)R()eimu

|t0

f

()eimnT

An

exp(

k

2a2t)R

Cn

Jm

(kn

)

Dn

Nm

(kn

)n1imu

Tn

(t)Rn

()en1n

nA

R

()f

()

(R'

)'

m2

R

k

2

R

0The

resolution

of

rotationalsymmetric

cylindrical

problemGeneral

eigenvalueproblemEigenvalue

problems：S

L

type

boundary

conditionsx2

y"

xy'm2

y

x2

y

0,

0

x

kb2mR

k

2

R

0(R')'Sturm

Liouvi

le

type

boundary

conditionsLet x

=

k

ρ,

y(x)=

R(ρ),

then

wehave：Eigenvalues

and

eigenfunctionsT eral

solution

of

the

universe

eq.

isy(

x)

AJ

m

(

x)

BNm

(

x)Fromboundary

conditions,

weget0nk0

(m)

0kn0

(m)

/

b

0The

eigenfunctions

areRn

()

yn

(x)

Cn

Jm

(

(m)

/

b)

Dn

Nm

(

(m)

/

b),

n

0，1,2,3,n

nOrthogonality

and

completeness0nbRn

(

)Pl

(

)dx

n,l

(N

m

)2NormOrthogonalitynbnR2

(

)d0m

2(N

)

Completenessn

nf

R

()n1f

()

Teralized

Fourier

coefficients

arenbnnf

(

)R

(

)d

1

0(

N

m

)2f

Finiteness

andboundaryconditionof

the

1st

kindTheeigenvalue

problem

isnTheeigenvalues

and

eigenfunctions

arekn

x(m)

/

b,

x(m)

is

the

nth

positive

zero

of

J

(

x)n

n

mRn

(

)

yn

(

x)

Jm

(

x(m)

/

b),

n

1,2,3,2mR

|

b

0R

k

2

R

0,

b(R'

)'The

orthogonality

and

norm

：20nb(

x(m)

)n

m1

n(N

m

)2

1

b2

J

2Rn

(

)Pl

(

)dx

n,l

(N

m

)2bnnf

(

)Rn

(

)d

1

0(N

m

)2f

f

J

(x(m)n

n

n

m

n

/

b)

n1

n1f R

()

f

()

Finiteness

andboundaryconditionof

the

1st

kindCompleteness

：Fourier’s

expansion

coefficients：n

mn

n(

x)xdxf

(

x

/

k

)Jk

(N

)1knb02

m

2

mnn(

x)xdxx(

m

)0f

(bx

/

x(m)

)Jb2(

x(m)

)2

(N

m

)2n

nFiniteness

andboundaryconditionof

the

1st

kindnbnn

1

0(

N

m

)2

(

c)R

(

)

df

f

J

(x(m)n

m

nn

n

/

b)

n1

n1f

R

()

(

c)

1cR

(c)n(N

m

)2nEx.

1：Expand

f

=

δ(ρ-c)in

[0,b]in

ageneralized

Fourierseries

of

the

Bessel

functions

Jm.2(m)nm1

b2

J

2

(x(m)

)/

b)cJ

(cxm1

nnmbnn

R

(

)

d10(

N

m

)2f

f

J

(x(m)n

m

nn

nm

/

b)

n1

n1f

R

()

mnxm

J

(

x)xdxx(

m

)0bm2(

x(m)

)m2

(N

m

)2n

nnx(m)

J2bm(

x(m)

)m1

n0(

m

)nxx

J

(

x)m1m1bm2(

x(m)

)m2

(N

m

)2n

nnx

x(m)

/

bFiniteness

and

boundary

condition

of

the

1st

kindEx.

2：Expand

f

=

ρm

in

[0,b]

in

a

generalized

Fourier

seriesoftheBessel

functions

Jm.(0)2n

0

nn

nf

J

(x

/

b)n1n1f

R

()

dnbnn

R

(

)

1

20(

N

0

)2f

n00x3J

(

x)dxb4(

x(0)

)4

(N

0

)2n

nx(

0)1

002132(

0)nxn

1

nx

J(

x(0)

)4

1

b2

J

2

(

x(0)

)b4

4xJ

2x

J

x3J0dx

x3J1

2x2

J0

4xJ1(0)

3

(0)

(0)(0)

4

2

(0)n1

nn(

x

)

J

(

x

)2b2[(

x

)

4xn

]J1(

xn

)Finiteness

and

boundary

condition

of

the

1st

kindEx.3：Expand

f

=

ρ2

in

[0,b]

in

a

generalized

Fourier

seriesoftheBessel

functions

J0.Finiteness

and

boundary

condition

of

the

2nd

kindThe

eigenvalue

problem

isThe

eigenvalues

and

eigenfunctions

are0(1)0(0)0(0)

xy

(

x)

J

(y

(

x)

J

(J

'

(

x)

0,

n

n

0

n(m)m

nn

nmR

(

)

R

(

)

kn

(m)

/

b,

where

(m)

is

the

nth positive

root

ofn

n

/

b),

n

0,1,2,3,；

/

b),

n

1,2,3,;

m

02mR'|

b

0R

k

2

R

0,

b(R'

)'The

orthogonality

and

norm

：2200

nnm

nn

nnb(

N

0

)2

1

b2

/

J

2

(

(0)

)(

N

m

)2

1

b2

[1

(m

/

(m)

)2

]/

J

2

(

(m)

)Rn

(

)Pl

(

)d

n,l

(

N

m

)2Finiteness

and

boundary

condition

of

the

2nd

kindnbnn

1

0(

N

m

)2

(

c)R

(

)

df

(1)n

0

nn

nf

J

(x

/

b)

n0

n0f

R

()

(

c)

1cR

(c)n(N

m

)2n200

n(1)n1

b2

J

2

(

x(1)

)/

b)cJ

(cx

(0)

x(1)n

n

J

0

'

(

x)

J1(

x)Ex.

1：Expand

f

=

δ(ρ-c) in

[0,b]inageneralized

Fourierseries

of

the

Bessel

functions

J0.nbnn

1

0(

N

0

)21

R

(

)

df

(1)n

0

nn

nf

J

(x

/

b)

n0

n0f

R

()

1

nn

nxJ

(

x)dx00b2(

x(1)

)2

(N

0

)2x(1)

0(1)xxJ1(

x)0n(

x(1)

)2

(N

0

)2n

nb2nx

x(1)

/

b

1b

1

00(

N

0

)2001

R

(

)

df

b10021

b2

J

2

(0)11dFiniteness

and

boundary

condition

of

the

2nd

kindEx.2：Expand

f

=1 in

[0,b]in

ageneralized

Fourierseries

of

the

Bessel

functions

J0.Boundary

conditions

of

the

1st

kindThe

eigenvalue

problem

isBy

boundary

conditions,

one

gets

BmJm

(ka)

Nm

(ka)

A

0Jm

(kb)

N

(kb)2mR

|

a

R

|

b

0R

k

2

R

0,

a

b(R'

)'R(

)

AJ

m

(k

)

BNm

(k

)T eral

solution

isThe

condition

ofnon-zero

solution

isJm

(ka)

Nm

(ka)

0Jm

(kb)

Nm

(kb)Rotational

symmetricproblemsAxial

symmetric

problems

(the

case

of

m

=0)Heat

problemWave

problemSteady

problemRotational

symmetric

problemsHeat

problemWave

problemSteady

problemAxial

symmetric

heat

problemsbnn(b

)J

(k

)d

1

0

n2

20(N

0

)2根據(jù)完備性：A

ut

a22u,

b定解問題為：

2

2u

|

b

0,

u

|t0

f

b

定解問題有軸對稱性，相應的半通解為u

n1

An

exp(

a

kn

t)[Cn

J0

(kn

)

Dn

N0

(kn

)]

2

2u(0,t)有界，u(b,t)

0,半通解化為u

n1

An

exp(

a

kn

t)J0

(kn

),

kn

xn

/

b

2

2

(0)由初始條件得：b

n1

An

J0

(kn

)2

2

Example

1半徑為b的無限長圓柱體，柱面上溫度為零，初始溫度分布為

f=b2

–ρ2，確定柱內溫度

u

的變化。Solution：以圓柱體的對稱軸為

z

軸，建立柱坐標。Axial

symmetric

heat

problemsA

Jbnn(k

)d

1

0

n20(

N

0

)2根據(jù)完備性：A

ut

a22u,

b定解問題為：

2u

|

b

0,

u

|t0

f

A定解問題有軸對稱性，相應的半通解為u

n0

An

exp(

a

kn

t)[Cn

J

0

(kn

)

Dn

N0

(kn

)]

2

2u(0,t)有界，u

(b,t)

0,半通解化為u

n0

An

exp(

a

kn

t)J

0

(kn

),

kn

xn

/

b

2

2

(1)由初始條件得：A

n0

An

J0

(kn

)2

Example

2半徑為b的無限長圓柱體，柱面上絕熱，初始溫度分布為

f=

Aρ2

，確定柱內溫度

u

的變化。Solution：以圓柱體的對稱軸為

z

軸，建立柱坐標。Axial

symmetric

wave

problems

bn(b

)J

(k

)d

1

0

n2

20(N

0

)2根據(jù)完備性：Bn

0,An

utt

a22u,

b定解問題為：

2

2u

|

b

0,

u

|t0

b

,

ut

|t0

0定解問題有軸對稱性，相應的半通解為u

(

A

cos

ak

t

B

sin

ak

t)[C

J

(k

)

D

N

(k

)]n1

n

n

n

n

n

0

n

n

0

nu(0,t)有界，u(b,t)

0,半通解化為u

n1

(

An

cos

aknt

Bn

sin

aknt)J

0

(kn

),

kn

xn

/

b

(0)b

n1

An

J

0

(kn

)

2

2

由初始條件得：

0

n1

Bn

akn

J

0

(kn

)Example

3半徑為b的圓形膜，邊緣固定，初始形狀是旋轉拋物面f=b2

–ρ2，初始速度為零，求膜的振動情況。Solution：以圓形膜的中心為原點，建立極坐標。Axial

symmetric

wave

problems0

n1n

n

0

nn1n

0

nB

ak

J

(k

)

(

c)

A

J

(k

)由初始條件得：bn

n

(

c)J

(k

)d

1

0

n0ak

(N

0

)2根據(jù)完備性：An

0,Bn

utt

a22u,

bu

|

b

0,

u

|t

0

0,

ut

|t

0

(

c)定解問題為：n1u

(

An

cos

aknt

Bn

sin

aknt)[Cn

J0

(kn

)

Dn

N0

(kn

)]定解問題有軸對稱性，相應的半通解為(0)n1n

n

n

n

0

n

n

nu

(

A

cos

ak

t

B

sin

ak

t)J

(k

),

k

x

/

bu(0,t)有界，u(b,t)

0,半通解化為Example

4半徑為b的圓形膜，邊緣固定，初始位移為零，初始速度為

f=δ(ρ-c)，求膜的振動情況。Solution：以圓形膜的中心為原點，建立極坐標。Axial

symmetric

steady

problemsbn

J

(k

)d

1

0

n20sinh

kn

L(N

0

)2n根據(jù)完備性：B

2u

|

b

0,

u

|z0

0,

u

|zL

定解問題為：定解問題有軸對稱性，相應的半通解為u

(

A

cosh

k

z

B

sinh

k

z)[C

J

(k

)

D

N

(k

)]n1

n

n

n

n

n

0

n

n

0

nu(0,z)有界，u(b,z)

0,半通解化為u

n1

(

An

cosh

kn

z

Bn

sinh

kn

z)J

0

(kn

),

kn

xn

/

b

(0)由下底條件得：0

A

J

(k

)

A

0n1

n

0

n

n由上底條件得：

n1

Bn

sinh(

kn

L)J

0

(kn

)2

Example

5半徑為b，高為L的圓柱體，下底和側面都保持零度，上底的溫度分布為ρ2，求柱內的穩(wěn)恒溫度分布。Solution：以圓柱體的軸為

z

軸，下底中心為原點，建立柱坐標。uzz

2u

0,

b,

0

z

LAxial

symmetric

steady

problemsn1n1nBn

sinh(

kn

L)J

0

(kn

)B

2

A

sinh(

kn

L)J

0

(kn

)A

由上下底條件得：nbn

1

0sinh

kn

L(N

0

)2nAJ

0

(kn

)d

,

B

根據(jù)完備性：A

2u

|

0,

u

|

B

,

u

|

A

b

z0

zL定解問題為：n1u

[

An

sinh

kn

z

Bn

sinh

kn

(L

z)][Cn

J0

(kn

)

Dn

N0

(kn

)]定解問題有軸對稱性，相應的半通解為(0)n10

n

n

n[

An

sinh

kn

z

Bn

sinh

kn

(L

z)]Ju

(k

),

k

x

/

bu(0,z)有界，u(b,z)

0,半通解化為Example

6半徑為b，高為L的圓柱體，側面電勢保持為零，上底的電勢為A，下底的電勢分布為Bρ2，求柱內的電勢分布。Solution：以圓柱體的軸為

z

軸，下底中心為原點，建立柱坐標。uzz

2u

0,

b,

0

z

LRotational

symmetric

heatproblemsn

1

nA

J

(k

)n1由初始條件得：A

bnnAJ

(k

)d

1

1

n0(

N

1

)2根據(jù)完備性：A

u

|

b

0,

u

|t

0

A

cosut

a22u,

b定解問題為：2

2n1n

n

n

1

n

n

1

nA

exp(

a

k

t)[C

J

(k

)

D

N

(k

)]cosu

定解問題有轉動對稱性，相應的半通解為(1)2

2n1n

n1

nn

nA

exp(

a

k

t)Ju

(k

)

cos

,

k

x

/

bu(0,t)有界，u(b,t)

0,半通解化為Example

7半徑為b的無限長圓柱體，柱面上溫度為零，初始溫度分布為

f=Aρcos

φ，確定柱內溫度

u

的變化。Solution：以圓柱體的對稱軸為

z

軸，建立柱坐標。Rotational

symmetric

wave

problemsd

bn

J

(k

)

1

2

n20(N

2

)2根據(jù)完備性：Bn

0,An

utt

a22u,

b定解問題為：

2

2u

|

b

0,

u

|t0

（b

）sin

2

,

ut

|t0

0定解問題有轉動對稱性，相應的半通解為u

(

A

cos

ak

t

B

sin

ak

t)[C

J

(k

)

D

N

(k

)]sin

2n1

n

n

n

n

n

2

n

n

2

nu(0,t)有界，u(b,t)

0,半通解化為u

n1

(

An

cos

aknt

Bn

sin

aknt)J

2

(kn

)

sin

2

,

kn

xn

/

b

(2)

n1

An

J

2

(kn

)

2

由初始條件得：

0

n1

Bn

akn

J

2

(kn

)Example

8半徑為b的圓形膜，邊緣固定，初始形狀是

ρ2sin2φ，初始速度為零，求膜的振動情況。Solution：以圓形膜的中心為原點，建立極坐標。Rotational

symmetric

steady

problemsbAJ

(k

)d

1

1

n0sinh

kn

L(N

1

)2n根據(jù)完備性：An

0，Bn

uzz

2u

0,

b,

0

z

L定解問題為：u

|

b

0,

u

|z0

A

sin

,

u

|zL

0定解問題有轉動對稱性，相應的半通解為u

[

A

sinh

k

z

B

sinh

k

(L

z)][C

J

(k

)

D

N

(k

)]sin

n1

n

n

n

n

n

1

n

n

1

nu(0,z)有界，u(b,z)

0,半通解化為u

n1[

An

sinh

kn

z

Bnsinh

kn

(L

z)]J1(kn

)

sin

,

kn

xn

/

b

(1)

0

n1

An

sinh(

kn

L)J1(kn

)

由上下底條件得：

A

n1

Bn

sinh(

kn

L)J1(kn

)Example

9半徑為b，高為L的圓柱體，側面和上底保持零度，下底的溫度分布為Aρsinφ，求柱內的穩(wěn)恒溫度分布。Solution：以圓柱體的軸為

z

軸，下底中心為原點，建立柱坐標。General

Cylindrical

ProblemsProgramer

,

resolve

the

conditions

into

Fourierserieswith

respect

to

the

variable

.Then,

find

the

symmetric

cylindrical

solutions

foreach

condition

with