数学物理方程第二版(谷超豪)第二章答案
发布时间：2024-11-10
数学物理方程第二版谷超豪
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~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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数学物理方程第二版谷超豪
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):
¬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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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
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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
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~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://
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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
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数学物理方程第二版(谷超豪)第二章答案.doc
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