數(shù)學(xué)物理方程習(xí)題解答案

理解習(xí)題一1，驗(yàn)證下面兩個(gè)函數(shù)：u)ln

1y

u(x,y

x

siny

uxxu(,

12yx

y

)

(

y

)

x

xy

x(22)(x2y2)

y

(2

)

x

yy

x

y2y22(y2)(x2y2)2

x22(22)2(x)

u(,ln

12y

xx

yy

uysinyusinx

x

uyxx

xux,uyyyyuxsinxsinu)

x

sinxxyy

uf()()

uuuxyxy

f

證明：因?yàn)閡f()g(y)ugy)x

f(xy

ufxyuuu(x)(y)xy

g

()f

x)

u(x,)(

xy)

uuxx

(y)f(

)

ufx

u

xx

f

uxyyyyff

-4

1

2

u(,y)fx),)fy)122

f,1

2

u(,(3)fy)1

4，試導(dǎo)出均勻等截面的彈性桿作微小縱振動(dòng)的運(yùn)動(dòng)方程（略去空氣的阻力和桿的重量解性桿的假設(shè)直于桿的每一個(gè)截面上的每一點(diǎn)受力與位移的情形都是相同的，取桿的左端截面的形心為原點(diǎn)，桿軸為

[,x

(x,(xt)

(x

(xt(x(t)

SE(x

(x)(x(x)

xx2tt13t2ttx2tx11

(x,t

(xt)xtsExxxsx()[x]

x

E

u2

EEldQ()dSdt1C,Ku

x2

2

1

t

(u1

ldxdtx12Q2

t

2K(t)dtQ

t

K(x)dtQ12

x

K(t)K(u4tx

xxctxtxt11tttttxxctxtxt11tttttt

)(t)21tx

tx

c

2

t

[

c(u]44

x,x,,121

2

k4k1(u)c

t

(,yz)

u(xyzt)

u

(y)dsdt

u

dt

D(xz)

m

tt

D[(D)()()]

t

D(

)dxdydzdt

t1

t

2

[u(yzt)(y,z,t)]2t

dtdxdydzt

D()dxdydzdtdtdxdydzt

tl0xtl0xt

[D(

)]dxdydzdtt12D()

l

x

x

utt

u,0xxx

u

x

0,

x

u

,uttt

l

度quxttxx

(l)2

u

x

u

xx

qtu

t

l)2

,0xl處ux00u(x

u2u2ut

a

2

kc

b

2

4k

uu

xt

,()

u(xt)1

u)2

u,0x,t,0xttttt0,u(x),tttttt2t),tuu1xx

u1

2

,0x,0,ttxxu(xt1tt),t)x

Lu,iii

fi

u

u

x,

u

2

uiitt

u

xx

i

u1tu1x

tttu1

0,t)x

uu

2t2x

x),1ttu2

(2

u

t

t

t

()

uu12

u(xt)1

u)2

2ut2(xt,0,ttxxttuxx

u1

2

2,t,0,txxt0

Lu,iii

f

f

f

u

t

u

x,t

2uuLufiiii

u1

2

2ufx,)txx

ii

ut

ux

u

ut

x

u

t

tt

u12習(xí)題二1，

用分離變量法解齊次弦振動(dòng)方程Utt

U

xx

0

,0)U(,0)l),Ut)U(l,t)tU(,0)xU(x,0)t)(lt)tx

U(xsin(

l

U(xx(l),U(0,t)(l,tt

X(x

t)

U(,t)X()T()XxTX

TaT(t)(x)

X(xTt)

T

a

tX

X(xU(0,)T()U(lt)(l)T()

t)

X(l)

kl2kkkl2kk

X(x)

X)(xX(0)(l)

r2r

()

X(0)(l)

A

1ee

BX(x)

AB

rr

X

sin

A

sin

(

0)k

1,2,L)

k

l

),k

X

k

Bl

T

T

r

2

rl

l

k

tsinll

tL

UXk

k

kk

ktsintlll

,kL

U

k

llsinndx[(ktllsinndx[(ktsinktsint]sinuacos

kktt]sinlll

,kLak

u

t

sin

l

nl

nll

,L[l]0

nl

0

lnl2ll0l

a

n

ak

ll2l

bk

2lk43k0l4a

u

l

(2k(2ksinsin4llk

1(2l

]

k0,1,2L

X

k

k

l

x,LTAkk

11kll

tk0,1,2L

[cos

kk2lll

bk

axk2ksinnlaxk2ksinnl12l2ll

dx

1

3bk

l21

3

1kk2ll

u[nn

ntsintxlll3nnsin,nl0ll0,nkl(l)dxl1l0lnk4

kL

ucos

3lsinlla

1k

sin

sinll2,l0,t(1)l0<x<l,Ut)U(l,t0,t

U0<x<l,t0,t(xx0<x<l,U(0,(lt)tU(,t)X()T()

X(x)

X)TaT()(x)(t)X

X(l)

X(x

X)X(0)(l

(l)sindxk(l)sindxk

()cos

Bsin

A

sin

k

l

)2,k1,2,

X(x)Bkk

l

,kL

T

aT(t)k

t)

al

)2t)k

TtA

l

t

,1,2,L

UTkkk

k

l

)t

l

k1,2,U(,t)k

k

aek

k

l

)t

l

k

lk2ll0ll0l4l[1k(k

]8l2,k(k3

L

u

8l(2n3

e

nl

]t

(2nl

k

l

)

2

k0,1,2,L

X(x)k

l

,kLu(t)ekk

k

l

)t

lkldxll

k

l

[1k],kL

k4l,

,0,1,2

tttt

u()

l(2n2

e

l

]t

nl

0,0y)0y(x(xb)x

u,y))12

(,)1

,y(0,)u(,y)0y(x(xxb)xux)2

x,yy)f(ya)0y(xx,)21uy[ash(n/)

n)g)sindshsinaau(y)

21[bsh(n/)

nxfshsinbbb2,,tu0,txx,u0,ttt

U(X(x(t)

Xx

2

X

TaTt)(x)X

X(l

X(x

X)X(l)

dxkdxTkdxkdxTk

rr

(x

Be

A

1-1ee

B

X(x

X(x)

AB

A

X(x

rr

()cos

Bsin

sincos

B

1)k0,1,22

k

(kl

]

，k=0,1

X(xAk

(kl

u(t)(t)cos

(k2)l

k

(kl

]T(tk(x

(kl

x

(,0)tk

(k2)l

(0)kk

l(2)3l0ll(l0l(k2)(t)lkT(0)

]2(t)

Ck

(k(ktDll

t

Ck

xk4)sinxT(0)cosnxT(0)lxk4)sinxT(0)cosnxT(0)l22l(k3ll

l34824(2k3(2dx4l348k3k

nn

kk

(kl

tx,t)

l

k

24(4kk[t(4l2lk3(4kt(42l2l

xx]u0(0,)t)tx(x,0)(x,0)0xtnxL

u(,t)tnn

Tn

22T(n2nxl

)cosnxdx)cosnxdx

2

sinxnxdx

2

12

[sin(1)xsin(1)x]dx

1

)xdx)]0

1

11)])1

[cos(1)]

,nk4T

nttnttnn)nn(0),(0)nnT)Atn

TBn

n

0

T()cossinantn

Tsinant(0)ncos

T)n

n1,2,

Tt)cosant0,1,2,Ln(x,t)

(t)cosnn

cosnxn

x0

n

cosn

n

2

ant22uxtxxutxxx8cos())

(n2

,nL

(n2)u(,t)()cosT()cos2nn

(2n4

Tn

(2n4

]TCnnn

(2n4

]t

n9en9e

ue

n

]t

cos

n4

(2nCcos44nn

ue

34

)t

)t94442ttxxtx,t

cos,nL

u(x,)

Tcosnxn

n

Tan)n

2

T(t(t)nn

T)T(t)Tnnnn

an)]uen

ban)

]

cosnxn

n

Cnxxn

cos

[0,

]

nnk2nk

u

2

4

1k

(2ktx

ttxxnnllnnnllnnttttxxnnllnnnllnnttn1t2uusin,0xll0,txu0,xttt

l

n

u

n

T(n

l

u

nna[T(t)2T(t)]sinlllln2an()T(tl0,n2

T(0)n

Tn

naT(t))2(tT(0)(0)nn

T()n

2

Tn

nT(t))(t)sinT(0)nnt)2(ttl2Wl

t

1

d2nW)tnlWf(sinn1

ln

1a2all

ttnnn2bxxttnnn2bxxn

(t00

sinll

22a[sin]2ll

u

11n[sincos]sin42lll

0,,0yyuAyu0,xy

u

n

X(x)cosn

b

Xn

b

)n

u(0,y)

n

nyX(0)cosXb

0dyb0u(a,)(ann

b

AyXa)n

n

kLAb,nkk1,2L

Dn(x)ebLn

X(0)X(a)nn

0xnn()nnb,L

)n)n

0

x

n

n

b

4xak

kbk

b

(2k2ttutxt

A,

l

n

u(t)sinnn

l

T()

Ae

Ae

sinnn

l

fn

2nl2

Tnn

l

(0)n

lnTlnlllnl

l02l2)[1]ln

l

)2fnT(0)nnlT(0)

)

t0n[1]sin(2nsint0n[1]sin(2nsinlsinsinsin

Tn

nl

t(t))T()flT(0)(t))T(t)l

tT(

)f

)

t

n(t)lT(0)f(n

)2(t

t

enn

n)(tl

()

t

fe

l

)(t

)

d

(t)n

nl

)t

en

nl

)(t

)

dn

l

)t

f(n

l

)

2

[1

l

)t

]

Al22

n

()t4Tl(22)l

n

]t(2ll2ut,tttxx0,txttt

u

n

T()sinn

l

f(tnn

nl

f(t)n

l

0

sin

llllsin

n

]

121121T(0)nT

2)2Tln

]Ttnt

n

22

[1n

]

t0

sin

n(tl

l2na2)

(sin

l),l

,k0,1,2u

l2k2(222]

[sin

l(2(2ksin]sin(2kall,xxyy

11(ww)xy(2y2)224

2

1v2

1()0,01vsin224

[asinn

n

n

[annn

124

2

nnCnnCab

1,n240,n124

2u

112424

sin2124

)sin21xy12

)1()f

f

fu(

a[cosnsin2n

nsinnf0sinn20nDn

111

ff)cosnf)sin

nn12nn12ab

11n

ff

L

0

0

()(

u(

n

)

2

,nL

sinn

n1,2,

n

nBn

n

Bn

u(

nn

2

0

f

n

d

u

Axy

u

yye

f(,y)(,y(,y121fy)f(,y)2

2

xxyyxxxxxn2xxyyxxxxxn2an

f

f

u

f(xy)1

f(x,)2

111x4yy4Axy12121212

31[x12

4

Axyx

2

y

2

)

4

]

x

y

ybyy2v()

v

,yaa),0byy2y()sinnn

aana

4n3

[(

nnnnnnnnnnnach(2tttnnnnnnnnnnnach(2tttYab()()2

nYchn

nsha

A2A2an

BnYn2a

a

n

nchnnsina

2

n

ch

(2(2sin3ux,ttxxusint,txttt

l

t2sinlxvlv12

1t1td(t0)l1t1td(t0)llx11222l112vvvtttl