数学物理方程(王明新等，清华大学出版社) 课后习题答案
时间：2025-07-06
数学物理方程(王明新等，清华大学出版社)学习指导与课后题目解答
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数学物理方程(王明新等，清华大学出版社)学习指导与课后题目解答
111141416§1.1§1.2
..............................................................................................
§2.1§2.2
..............................................................................................
§3.1...............................................§3.2
...............................................
§4.1...............................................§4.2
...............................................
§5.1...............................................§5.2...............................................
§5.3
...........................................
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数学物理方程(王明新等，清华大学出版社)学习指导与课后题目解答
§1.1
—
,
Fu-
bini
,
:
1.1
Rn,f
′
,
f(x)dx=0,
′
f≡0.
1.2 Rn,f
v∈C0∞
( ),
f(x)v(x)dx=0,
f≡0.
1.3(Stokes
)
Rm
,
C1
m
div vdx=
v· ndS,
n
,dS
m=1,2,3
,
,Green
§1.2
1.1
l
,
(x=0)
,
.
ρ(x),
f0(x,t).
1.1
u
(
1.1)
[a,b],
[t1,t2]
1
v,
(1.1)
,
T(x),
u
[t1,t2],
[a,b]×
(1.1)
数学物理方程(王明新等，清华大学出版社)学习指导与课后题目解答
t=t2
t=t1
=
[a,b]×[t1,t2]
+
bρut|t=t2dx
b
ρut|t=t1dx=
t2 b
f0dxdt+
t2
[T(b)sinαb T(a)sinαa]dt.a
a
t1
a
t1
sinα=
ux
αuab=
x
1+u2x x=a
≈ux|x=a,sin1+u2x
x=b
≈ux|x=b,
b
t2
ρ(x)uttdtdx=
x,t2) ut(x,t1)]dx
a
t1
b
ρ(x)[ut(t2=
a
f0dxdt+
t2
[T(b)ux(b,t) T(a)ux(a,t)]dt=
t12 b
a
t1
bta
t1
t2f0dxdt+
b(T(x)ux)xdxdt.
t1
a
Fubini
a,b,t1,t2
ρ(x)utt=(T(x)ux)x+f0(x,t).
,
ρ(x)=ρ
,
T(x)=(l x)ρg,
f0(x,t)=0.
:
utt=g[(l x)ux]x.
1.2
,
,
ρ(x)=ρ0(
),T(x)=T0(
),f0(x,t)= γut,
,
.
(1.2)
ρ0utt=(T0ux)x γut.
a
(
(1.2)
(1.2)
γ>0
数学物理方程(王明新等，清华大学出版社)学习指导与课后题目解答
.
,
ρ(x).
,
S(x).
T(x)=ES(x)ux,
E.
.
u
f0(x,t).
[a,b]×[t1,t2]
ba
P=:
ρ(x)S(x)[ut(x,t2) ut(x,t1)]dx=u
I1
b
ρ(x)S(x)I2,
a
t2t1
uttdtdx.
u
I1+I2= P,
:
1.4
(
1.3
t2
I1=(x)ux|x=b ES(x)ux|x=a)dt=
(ESt1
t2 b
(ES(x)ux)xdxdt,I t1a
t22=
f0(x,t)dxdt.
1
bta
a,b,t1,t2
,
ρ(x)S(x)utt=(ES(x)ux)x+f0(x,t).,
ρ(x)=ρ(
),S(x)=S0(
).
ρutt=(Eux)x+f(x,t).
1.3),
h,
ρ
E
1
x
2
h
utt.
1.3
,
ρ(x)=ρ(
),
f0(x,t)=0.
:
S(x)=S0
1
x
h
2
ux
=ρ1
x
x(1.3)
E.
(1.3)
,
数学物理方程(王明新等，清华大学出版社)学习指导与课后题目解答
D,
[t1,t2],
D
D.
D
dS,dS
TdS n,
u
TdS n· u=T u· ndS.
t2T u· ndSdt=
t1
D
t2t1
D
·(T u)dxdydt.
f1,
t2dt.
t1
f1dxdyD
2
ρutt2dxdy
ρutdtdxdy.D
|
D
|t1dxdy=
tρuttD
t1
t2t1
t2ρuttdxdydt=
D
t1
D
·(T u)dxdydt+
t2
f1dxdydt,
t1D
D,t1,t2
ρutt= ·(T u)+f1,
utt=a2 u+f.
u(x,y,0)= (x,y),ut(x,y,0)=ψ(x,y),(x,y)∈ .
(1)u(x,y,t)=g(x,y,t),(x,y)∈ ,t≥0,(2)
u
n
=0,(x,y)∈ ,t≥0.1.6
,
(1)
(2)
(3)
,
E,
ρ,
S.
[a,b],
[t1,t2].
(b,t),
ESux(a,t)
ESuxESux(b,t) ESux(a,t),
t2
t2[ESux(b,t) ESux(a,t)]dt=(ESux)xdxdt.
t1
t1
b
a
bt1
t2
b
ρSutt1dx
ρSut,
a
|a
|t2dx.
t2b
b
t2
(ESux)xdxdt=
ρSutρSuttdtdx.
t1
a
a
|t2dx
b
ρSuta
|t1dx=
ba
t1
a,b,t1,t2
ρSutt=(ESux)x.
:
a,b
数学物理方程(王明新等，清华大学出版社)学习指导与课后题目解答
(1)u(0,t)=u(l,t)=0;(2)ux(0,t)=ux(l,t)=0;
(3)(αu βux)|x=0=(αu+βux)|x=l=0.1.7
r,
ρ
,
,
.
c,
k,
u0
,
k1.
u
[a,b]
V.
V
[t1,t2]
.
,[t1,t2]
V
[t1,t2]
V
V
.
[t1,t2]
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