数学物理方程第二版(谷超豪)第二章答案

数学物理方程第二版谷超豪

1 Ù9D §

à°7

2008c12 9F

8¹

12345

9D §9Ù½)¯K ÑÐ> ¯K ©lCþ{…ܯK

4 n!½)¯K) 5Ú-½5) ìC5

1481215

19D §9Ù½)¯K Ñ

~1.1 þ![\ » l,b §3Ó ¡þ §Ý´ Ó ,\ L¡Ú± 0 u)9 ,¿Ñl5Æ

dQ=k1(u u1)dSdt.

b \ —Ý ρ§'9 c,9D Xê k,Á Ñd §Ýu÷v §.):

\¶ x¶, \ u[x,x+ x] ã 9þ²ï.ü mlý

dQ1= k1(u u1)πl x;

ü mlx?,x+ x?6\ 9þ

πl2 u

,dQ2= k(x)(x,t)·

4

uπl2

dQ3=k(x+ x)(x+ x,t)·,

41

¡6\ 9þ

数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

ü m6\(x,x+ x) 9þ

uπl2

dQ=dQ1+dQ2+dQ3=k(x)· x k1(u u1)πl x.

x 4nþ,l t1 t26\ u[x1,x2]\ã 9þ

t2 x2

uπl2

k(x) k1(u u1)πldxdt.4t1x1

3ùã mS[x1,x2]\ãS :§Ýlu(x,t1)C u(x,t2),ÙáÂ9þ

x2 t2 x22

πl2 uπl

cρ(u(x,t2) u(x,t1))dx=cρdxdt.

44x1t1x1 â9þÅð,¿5¿ x1,x2,t1,t2 ?¿5, ¤¦ § 1 u4k1 u=k(x) (u u1).cρcρl~1.2Á í *ÑL§¤÷v © §.):

N(x,y,z,t)L«3 t,(x,y,z):?*ÑÔ ßÝ,D(x,y,z) *

N

dSdt.ÑXê,3á mãdtS,ÏLá -¡¬dS þ

dm= D(x,y,z)

Ïdl t1 t26\« (Γ L¡) þ

t2 t2

N

D(x,y,z)dSdt=div(DgradN)dxdydzdt.

t1Γt1 , ,l t1 t2, ¥TÔ O\

[N(x,y,z,t2) N(x,y,z,t1)]dxdydz=

t1

t2

N

dtdxdydz. â þÅð,¿5¿ ,t1,t2 ?¿5, ¤¦ §

N N N N=D+D+D.~1.3¬(·Yè)SÜ;õX9þ,¡ Yz9,3§ Ó Åì Ñ, 9 ÝÚ§¤;õ Ys9¤ '.±Q(t)L«§3ü NÈ¥¤; 9þ,Q0 Щ ¤; 9þ,KdQ= βQ,Ù¥β ~ê.qb ¬ '9 c,—Ý ρ,9D Xê k,¦§3 Ó §Ýu÷v §.à°7

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2

数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

):

¬S:(x,y,z)3 t §Ý u(x,y,z,t),w,

dQ = βQ, dt Q(t)=Q0e βt. Q(0)=Q0,

´ t1 t2 ,¬S? « ¥ 9þ O\ ul Ü6\ 9þ9¬¥ Yz9 Ú,=

t2

u

cρdtdxdydz=(Q(t1) Q(t2))dxdydz+t1

t2

u u uk+k+kdxdydzdtt1

t2

dQ

= dtdxdydz+

dtt1

t2

u u uk+k+kdxdydzdt.t1

5¿ t1,t29 ?¿5,k

u1 u u uβ=k+k+k+Q0e βt.cρcρ

~1.4 þ! ?3± ~ê§Ýu0 0 ¥,Áy:3~>6 ^e §Ý÷v © §

k 2uk1P0.24i2r u

= (u u0)+,cρ2cρωcρω

Ù¥i9r©OL« N >69>{,PL«î ¡ ± ,ωL«î ¡ ¡È, k1L« éu0 9 Xê.):

11Kaq, ¶ x¶,3 t1 t20u[x1,x2] ã 9

þO\ :l ٧ܩ6\ 9þ,lý¡6\ 9þ±9>6ÏL[x1,x2]ùã ) 9þ Ú,= t2 x2 t2 x2 x2 t2 ui2rkωdxdt k1P(u u0)dxdt+0.24dxdt.t1x1t1x1x1t1Ïd â9þ²ïÒ §Ý÷v §

uk 2uk1P0.24i2r

= (u u0)+.cρ2cρωcρω

à°7

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3

数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

~1.5 ÔNL¡ ýé§Ý u,d § .Ë Ñ 9þ dA}-À [ù(Stefan-Boltzmann)½Æ 'uu4,=

dQ=σu4dSdt.

b ÔNÚ± 0 m k9Ë vk9D ,qb ÔN± 0 ýé§Ý ® ¼êf(x,y,z,t),¦d TÔN9D ¯K >.^ .):

>.þ ¡È dS.3dt mS,²>. 6Ñ 9þ (k

k

dT Ë Ü0 9þ

σu4dSdt.

Ü0 ÏLT Ë ÔNL¡ 9þ

σf4dSdt.

â9þ²ïk

k

¤¦>.^

k

u

dSdt=σu4dSdt σf4dSdt. u

=σ(u4 f4). u

dSdt.9D Xê)

2Ð> ¯K ©lCþ{

~2.1^©lCþ{¦e ½)¯K ):

ut=a2uxx(t>0,0<x<π),

u(0,t)=ux(π,t)=0(t>0), u(x,0)=f(x)(0<x<π).):

u(x,t)=X(x)T(t),K

T +λa2T=0,

X +λX=0,

X(0)=X (π)=0.

à°7htqi2008@http://4

数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂ

2

1

λk=k+,k=0,1,2,...

2

∞ 1122

u(x,t)=Cke (k+)atsink+x.

2k=0

dЩ^

1

f(x)=Cksink+x

2k=0

π

12

f(ξ)sink+ξdξ. Ck=

02

∞

ìÀ Æ%°©

~2.2^©lCþ{¦)9D § Ð> ¯K:

ut=uxx(t>0,0<x<1), 0<x≤1, x, u(x,0)= 1 x,1<x<1, u(0,t)=u(1,t)=0(t>0).):

u(x,t)=

Ck=2

∞ k=1

Cke kπtsinkπx.

22

ξsinkπξdξ+

1

1

k=2n, 4kπ 0,n=22sin= n=0,1,2,...4( 1) k2,k=2n+1, (2n+1)22

(1 ξ)sinkπξdξ

u(x,t)=

∞ n=0

4( 1)n (2n+1)2π2t

esin(2n+1)πx.

(2n+1)22

~2.3XJk Ý l þ![ ,Ù± ±9üàx=0,x=lþ ý9,Щ§Ý©O u(x,0)=f(x),¯± §Ý©ÙXÛ?…y² f(x) u~êu0 ,ðku(x,t)=u0.):

2 u=auxx, t

ux|x=0=ux|x=l=0, u|=f(x).t=0

à°7

22 kπkπ

u(x,t)=Ckexp 2a2tcosx

llk=0

∞

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数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂ

ìÀ Æ%°©

l

l

C=

1kπ

0l

f(ξ)dξ,

C0=

2kl

f(ξ)cos

0l

ξdξ(k 0)f(x)≡u0 C0=u0,Ck=0(k 0) u(x,t)≡u0.

~2.43« t>0,0<x<l¥¦)Xe ½)¯K:

ut=a2uxx β(u u0), u(0,t)=u(l,t)=u0, u(x,0)=f(x),Ù¥α,β,u0þ ~ê,f(x) ® ¼ê.):

-u=u0+v(x,t)e βt,K 'uv Xe½)¯K:

vt=a2vxx,v(0,t)=v(l,t)=0,v(x,0)=f(x) u0.

)

∞ v(x,t)=Ck2π2a2

kπ

kexp k=1

l2tsinlx,

Ù¥

l

C2

k=

l

(f(ξ) ukπ0)sin

lξdξ=f2uk0+0

kπ(( 1)k 1),l

f=

2

kl

f(ξ)sin

kπ

0l

ξdξ. k

∞ 22 u(x,t)=ukπa2

kπ

0+fkexp 2t βtsinx

k=1

ll ∞4u0

(2k+1)2π2a2 (2k+1)πk=0

(2k+1)πexp l2t βtsinlx.à°7htqi2008@http://6

数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

~2.5 Ý l þ![\ Щ§Ý 0 C,à:x=0 ±~§u0, 3x=lÚý¡þ,9þ ±uÑ ± 0 ¥ ,0 §Ý 0 C,d \þ §Ý©Ù¼êu(x,t)÷veã½)¯K:

ut=a2uxx b2u,

u(0,t)=u0,(ux+Hu)|x=l=0, u(x,0)=0.Á¦Ñu(x,t).):

-u(x,t)=e btv(x,t)+ψ(x),K ψ(x)÷v

b2

ψ 2ψ=0,

a

2

ψ(0)=u0,(ψ +Hψ)|x=l=0

,v(x,t)÷v

vt=a2vxx,

v|x=0=(vx+Hv)|x=l=0, v(x,0)= ψ(x).

bch

b(l x)

´

ψ(x)=

bch

bl

a

+aHsh+aHsh

b(l x) blau0.

'uv(x,t) ½)¯K ëì áP49^©lCþ{¦).

~2.6 » a .² ,ÙL¡ý9,3 ±>.þ ±~§u0, 3 »>.þ ±~§u1,¦ -ðG (= mtÃ' G ) §Ý©Ù.):

d½)¯K

2u1 u1 2u

++=0,2rr22

u(a,θ)=u0,0<θ<π,u(r,0)=u(r,π)=u1,0≤r≤a.

C u(r,θ)=v(r,θ)+u1,Kv÷v

2v1 v1 2v

++=0,2rr22

à°7

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数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

v(a,θ)=u0 u1,0<θ<π,v(r,0)=v(r,π)=0,0≤r≤a.

^©lCþ{¦),-v(r,θ)=R(r)Θ(θ),“\ §9>.^ k

r2R +rR λR=0,

Θ +λΘ=0, Θ(0)=Θ(π)=0,

λk=k2(k=1,2,...).

dA ¯K) λk=k2(k=1,2,...),

Θk=Aksinkθ,

Rk=Bkrk+Ckr k.

d) k.5 Ck=0,¤±

v=

∞Akrksinkθ.

k=1

“\ ±þ >.^ k

∞Akaksinkθ=u0 u1.

k=1

u´

π

A2k=ak

(uu0 u1)sinkθdθ=

2(0 u1)

0akkπ

[1 ( 1)k].nþ

u(r,θ)=u ∞4(u1+

0 u1) r 2n+1n=0

(2n+1)πa

sin(2n+1)θ.

3…ܯK

~3.1¦eã¼ê Fp“C :

(1)e ηx2

(η>0);(2)e a|x|(a>0);(3)

x

1(a22)k,

(a22)k

(a>0,k g,ê).

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数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

):(1)

∞

F[e

ηx

2

∞

]=

e

ηx

2

e

iλx

dx=e

λ2e

η(x+iλ)

2

dx=

e λ

2. ∞

∞

(2)

∞

∞

F[e

a|x|

]=

e

a|x|

cosλxdx i

e a|x|sinλxdx

∞

∞

∞

=2

e axcosλxdx=

2a

22

.(3)|^3ê½nÚXeFourierC 5 O ,

F[ ixf(x)]=

d

dλ

F[f].~3.2y²: f(x)3( ∞,∞)þýé È ,F[f] ëY¼ê.):

P f(λ)=F[f(x)],K

| f(λ+h) f(λ)|= ∞ (e ihx 1)e iλx f(x)dx ∞

∞ ∞

≤|e ihx 1|·|f(x)|dx≤2

|f(x)|dx.

∞

∞

dué?¿ x∈R,k

limh→0

|e ihx 1|=0.

∞

limh→0

| f(λ+h) f(λ)|≤lim

h→0

|e ihx 1|·|f(x)|dx=0.

∞

~3.3^Fp“C {¦)n‘9D § …ܯK

u t=a2(uxx+uyy+uzz),

u|t=0= (x,y,z).):é §ÚЩ^ ?1FourierC ( áP56),P

u(λ1,λ2,λ3,t)=F[u(x,y,z,t)], (λ1,λ2,λ3)=F[ (x,y,z)],

d u22

dt

= a2(λ21+λ2+λ3) u, u|t=0= .à°7

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数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

)þãODE

u= (λ1,λ2,λ3)e a(λ1+λ2+λ3)t.

Fourier_C

u(x,y,z,t)= (x,y,z) F 1[e a(λ1+λ2+λ3)t],

F 1[e a(λ1+λ2+λ3)t] 122

a2(λ21+λ2+λ3)tei(λ1x+λ2y+λ3z)dλdλdλe=123

(2π)3R3

2

1x+y2+z2

=.exp 4a2t(2a3

u(x,y,z,t)=

1

(ξ,η,ζ)e

R3

(x ξ)

2+(y η)2+(z ζ)2

4at

2

2

2

2

2

2

2

2

2

2

2

2

(2a3

dξdηdζ.

~3.4y²(3.29)¤L« ¼ê÷v àg §(3.15)±9Щ^ (3.16).):

aq áy²È© —Âñ5.

~3.5¦)9D §(3.17) …ܯK,® (1)u|t=0=sinx,(2)^òÿ{¦) k. þ 9D §(3.17),b

u(x,0)= (x)(0<x<∞), u(0,t)=0.):(1)

∞ ∞

√(x ξ)2112 4atu(x,t)= (ξ)edξ= (x 2aξ)e ξdξ ∞2a ∞

∞

√122

sin(x 2aξ)e ξdξ=e atsinx.= ∞(2)é (x) Ûòÿ,=

x≥0, (x),

Φ(x)= ( x),x<0.

¦)XeCauchy¯K

2 ut=auxx,

u|t=0=Φ(x),htqi2008@http://

10

à°7

数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

( x>0)

u(x,t)=

1

∞

2a (x ξ)

2

Φ(ξ)e4at

dξ

=1 ∞

∞2a (ξ)e (x ξ)2 0

(x ξ)2 4atdξ+ ( ξ)e

4atdξ=a1 0 ∞

∞

(ξ)e x2+ξ24atshxξ

2a2tdξ.

~3.6y²¼ê

v(x,y,t;ξ,η,τ)=

14πa2(t τ)

e (x ξ)2+(y η)

24a(t τ)éuCþ(x,y,t)÷v §

vt=a2(vxx+vyy),

éuCþ(ξ,η,τ)÷v §

vτ+a2(vξξ+vηη)=0.

):

éL ª¦ = y.

~3.7y²:XJu1(x,t),u2(y,t)©O´eãü ½)¯K ):

u

=a2 2u, u=a2 2u, u1|t=0= 1(x); u2|t=0= 2(y),Ku(x,y,t)=u1(x,t)u2(y,t)´½)¯K u2 2u 2

u =a+,u|t=0= 1(x) 2(y) ).):

u u1 u2 2u 2

u =u2+u1=a21+u12

2 2=a

2 2

(u1u2) 2(u1u2) 2+2=a2 2u 2u 2+2

,

u|t=0=(u1u2)|t=0= 1(x) 2(y).

à°7

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数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

~3.8 Ñe 9D §…ܯK) L ª:

u 2u2 2u =a+, n u|=αi(x)βi(y). t=0

i=1

):

dU\ n þK(J½ A^FourierC )

n

(x ξ)2+(y η)21 ∞∞

αi(ξ)βi(η)exp dξdη.u(x,y,t)=2

4ai=1 ∞ ∞4a2t

~3.9 y ‘9D §…ܯK

u2 2u +=a u|t=0= (x,y)) L ª

1

u(x,y,t)=

4πa2t

):

∞ ∞

2u,

∞ ∞

(ξ,η)e

(x ξ)2+(y η)2

4at

dξdη.

ì áP58-59 y² {?1 y.

44 n!½)¯K) 5Ú-½5

~4.1y² §ut=a2uxx+cu(c≥0)ä)| X>.^ Ð> ¯K) 5Ú-½5.):

C v(x,t)=u(x,t)e ct,Kv(x,t)÷v §vt=a2vxx,…k

|v|x=α|=|ue ct|x=α|≤B,|v|x=β|=|ue ct|x=β|≤B,|v|t=0|=|u|t=0|≤M.

â9D § 4 nk

|v(x,t)|≤max{M,B},

é?Ût>0

|u(x,t)|=|v(x,t)ect|≤max{Mect,Bect}.

à°7htqi2008@http://12

数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

y 5 y²¯K

ut=a2uxx+cu,

u|x=α=u|x=β=0, u|=0t=0

k").¯¢þ,d M=B=0,Ïd|u(x,t)|≤0,=u(x,t)≡0.

y-½5, y²¯K

ut=a2uxx+cu,

u|x=α=η1(t),u|x=β=η2(t), u|=ε(x)t=0

η1(t),η2(t)Úε(x) ,)½ . t≤T ,

|η1(t)|<η,|η2(t)|<η,max|ε(x)|<ε,

Két≤T,α≤x≤β,¤á

|u(x,t)|≤max{ηect,εect}≤max{ηecT,εecT}.

d¯K´-½ .

~4.2|^y²9D §4 n {,y²÷v §uxx+uyy=0 ¼ê3k.4« þ ج L§3>.þ .):

u(x,y)3±Γ >. « þNÚ. Ä u34« þ ëY

5, u ½ ± M.qÏΓ´48,u3Γþ k m.eyM=m.

^ y{. u(x,y)3 S,:(x0,y0)

u(x0,y0)=M>m.

9ϼê

v(x,y)=u(x,y)+

M m

[(x x0)2+(y y0)2],24R

Ù¥R´± : ¥%! ¹« ».d k

v(x0,y0)=u(x0,y0)=M,

à°7

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数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

M m

[(x x0)2+(y y0)2]|Γ<m+(M m)=M.24R

vØ3>.Γþ ,§73 S,:(x1,y1) ,3ù:A

v|Γ=u|Γ+

vxx≤0,vyy≤0, vxx+vyy≤0.

k

, ¡

vxx+vyy=uxx+uyy+

—gñ.ÏdAkM=m.

M m

>04R2

~4.3y²Ð> ¯K

ut a2uxx=f(x,t),

u|x=0=µ1(t),(ux+hu)|x=l=µ2(t)(h>0), u|= (x)t=0 )u(x,t)3Rt1:{0≤t≤t1,0≤x≤l}¥÷v

u(x,t)≤

λt

eµ(t)12

eλt1max0,max (x),maxe λtµ1(t),,max(e λtf),

0≤t≤t10≤x≤lhRt1

Ù¥λ ?¿ ~ê.):v

C v(x,t)=e λtu,Ù¥λ ?¿ ~ê.du Ð> ¯K´ v÷

vt a2vxx+λv=e λtf(x,t), v λt v|=eµ(t),+hv=e λtµ2(t), x=01 x=l v|t=0= (x).

Äv3Rt1þ ,XJv(x,t)3Rt1þk ,K3 :kvt≥0,vxx≤0…v>0,?

v=

¤±

|u(x,t)|≤eλt1

1 λt1

[ef(x,t) (vt a2vxx)]≤e λtf(x,t).1

max(e λtf(x,t)).Rt1

2 ì áP62y²Ù§ Oª,= (Ø.à°7

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数学物理方程第二版谷超豪

5êÆÔn §6SK‘ùÂìÀ Æ%°©

5) ìC5

~5.1y²e 9D §Ð> ¯K

2 u auxx=0, t

u|x=0=u|x=l=0, u|= (x)t=0

) t→+∞ ê/P~u",Ù¥ ëY¼ê,… (0)= (l)=0.):

d½)¯K )

u(x,t)=

Ù¥

∞ k=1

Ake

k2π2a2

lt

sin

kπx,l

kπ2l

(x)sinxdx.Ak=

l0l

ìC5 d (x)∈C[0,l] ,é k,

|Ak|≤C1,

Ù¥C1 = k' ~ê.

∞ 2π22π2 a2π2(k2 1) at t at lll1+e|u(x,t)|≤C1 e≤Ce.

k=2

~5.2y²: (x,y) R2þ k.ëY¼ê,… ∈L1(R2) , ‘9D §…ܯK ), t→+∞ ,±t 1P~Ǫu".):

22

1 (x ξ)+(y η)

4at|u(x,y,t)|≤dξdη| (ξ,η)|e

4πa2tR2

1≤| (ξ,η)|dξdη=Ct 1.24πatR2

~5.3y²: (x,y,z) R3þ k.ëY¼ê,… ∈L1(R3) ,n‘9D §…ܯK ), t→+∞ ,±t 3/2P~Ǫu".):

2)2+(z ζ)21 (x ξ)+(y η4at|u(x,y,z,t)|≤| (ξ,η,ζ)|edξdηdζ(2a3R3

1

| (ξ,η,ζ)|dξdηdζ=Ct 3/2.≤3

(2aR3

à°7htqi2008@http://15