1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

Chapter 2 457

c.A=(S 1 −S 0 )/a,B=S 0 .IfS 0 =S 1 ,then∂∂ut=kAfor allt.

7.φ′′+λ^2 φ=0, 0<x<a,

φ( 0 )=0,φ(a)=0.

Solution:φn=sin(λnx),λn=nπ/a(n= 1 , 2 ,...).

9. The series

∑∞

n= 1

|An(t 1 )|converges.

11. No.u( 0 ,t)is constant ifut( 0 ,t)=0.

Section 2.5

1.v(x,t)=T 0.

3.u(x,t)=T 0 +

∑∞

n= 1

bnsin(λnx)exp

(

−λ^2 nkt

)

,λn=( 2 n− 1 )π/ 2 a,

bn=

8 T(− 1 )n+^1

π^2 ( 2 n− 1 )^2 −

4 T 0

π( 2 n− 1 ).

5. The steady-state solution isv(x)=T 0 −Tx(x− 2 a)/ 2 a^2. The transient
   satisfies Eqs. (5)–(8) with

g(x)=T 0 −v(x)=Tx(x−^2 a)

2 a^2

.

7.u(x,t)=T 0 +

∑∞

n= 1

cncos(λnx)exp

(

−λ^2 nkt

)

,

λn=( 2 n− 1 )π/ 2 a, cn=

4 (T 1 −T 0 )(− 1 )n+^1

π( 2 n− 1 ).

9.u(x,t)=T 1 cos(πx/ 2 a)exp

(

−

(

π

2 a

) 2

kt

)

.

11. The graph ofGin the interval 0<x< 2 ais made by reflecting the graph
    ofgin the linex=a(likeanevenextension).


12. a.u(x,t)=T 0 +

∑∞

n= 1

bnsin(λnx)exp

(

−λ^2 nkt

)

,λn=( 2 n− 1 )π/ 2 a,

bn=

1

a

∫ 2 a

0

g(x)sin

(nπx

2 a

)

dx.