1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

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

)

−···

3. 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.

9. 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.

5. 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.