1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

Chapter 7 487

17.f(t)=

∑∞

−∞

1

2 a

G

(inπ

a

)

einπt/a.

19.F(s)must be of the formF(s)=G(s)/H(s),whereG(s)is never infinite.

The solutions ofH(s)=0 must form an arithmetic sequence of purely

imaginary numbers, andH′(s)=0ifH(s)=0.

21.F(s)=(^1 −e

−πs) 2

s( 1 −e−^2 πs)=

1 −e−πs

s( 1 +e−πs).

23.F(s)=^1 +e

−πs

(s^2 + 1 )( 1 −e−πs).

25.u(x,t)=sin(ωx)sin(ωt)

sin(ω)

+

∑∞

n= 1

(− 1 )n+^12 ω

ω^2 −n^2 π^2

sin(nπx)sin(nπt).

27.U(x,s)= ω

s^2 +ω^2

emx,m=^1

2

−

√

1

4

+s.

29.α^2 =

1

2

(

1

4 ±

√(

1

4

) 2

+ω^2

)

.Sinceαmust be real, take the+sign.

Chapter 7

Section 7.1

1. 16(ui+ 1 − 2 uij+ui− 1 )=−1,i= 1 , 2 ,3,u 0 =0,u 4 =1. Solution:u 1 =
   11 /32,u 2 = 5 /8,u 3 = 27 /32.


2. 16(ui+ 1 − 2 ui+ui− 1 )−ui=−^12 i,i= 1 , 2 ,3,u 0 =0,u 4 =1. Solution:
   u 1 = 0 .285,u 2 = 0 .556,u 3 = 0 .800.


3. 16(ui+ 1 − 2 ui+ui− 1 )=^14 i,i= 0 , 1 , 2 ,3,u 0 − 2 (u 1 −u− 1 )=1,u 4 =0.
   Solution:u 0 = 0 .422,u 1 = 0 .277,u 2 = 0 .148,u 3 = 0 .051.
   7.n=3:u 1 = 4 .76,u 2 = 4 .24;n=4:u 1 = 6 .65,u 2 = 9 .14,u 3 = 5 .92. The
   actual solution,

u(x)=−sin(

√

10 x)

sin(

√

10 )

,

has a maximum of about 50. The boundary value problem is nearly

singular.