456 Answers to Odd-Numbered Exercises
Section 2.3
1.w(x,t)=−π^2 (T 0 +T 1 )sin
(
πx
a
)
exp
(
−π
(^2) kt
a^2
)
−^2
π
(T
0 −T 1
2
)
sin
( 2 πx
a
)
exp
(
−^4 π
(^2) kt
a^2
)
−···
- The partial differential equation is
∂^2 U
∂ξ^2
=∂U
∂τ
, 0 <ξ < 1 , 0 <τ.
5.w(x,t)=
∑∞
n= 1
bnsin
(
nπx
a
)
exp
(
−n^2 π^2 kt/a^2
)
,bn=T 02 (^1 −cos(nπ))
πn
.
7.w(x,t)as in the answer to Exercise 5, withbn=
2 βa
π ·
1
n.
- a.v(x)=C 1 ;
b.∂w∂t =D∂
(^2) w
∂x^2 ,0<x<a,0<t,
w( 0 ,t)=0, w(a,t)=0, 0 <t,
w(x, 0 )=C 0 −C 1 ;
c.C(x,t)=C 1 +
∑∞
n= 1
bnsin
(nπx
a
)
exp
(
−n^2 π^2 kt/a^2
)
,
bn=(C 0 −C 1 )
2 ( 1 −cos(nπ))
πn ;
d.t=−a
2
Dπ^2 ln
(
π
40
)
;
e.t= 6444 s= 107 .4min.
Section 2.4
1.a 0 =T 1 /2,an= 2 T 1 (cos(nπ)− 1 )/(nπ)^2.
3.u(x,t)asgiveninEq.(9),withλn=nπ/a,a 0 =T 0 /2, andan=
4 T 0 (2cos(nπ/ 2 )− 1 −cos(nπ ))/n^2 π^2.
- a. The general solution of the steady-state equation isv(x)=c 1 +c 2 x.
The boundary conditions arec 2 =S 0 ,c 2 =S 1 ; thus there is a solution
ifS 0 =S 1. If heat flux is different at the ends, the temperature cannot
approach a steady state. IfS 0 =S 1 ,thenv(x)=c 1 +S 0 x,c 1 undefined.