1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

Chapter 4 473

7. a. See Eq. (11).an=0,cn= 200 ( 1 −cos(nπ ))/nπ;
   b.u(x,y)=u 1 (x,y)+u 2 (x,y),u 1 (x,y)is the solution to Part a,

u 2 (x,y)=

∑∞

n= 1

cnsinh(μnx)

sinh(μna)

sin(μny),

μn=nπ/b,cn= 200 ( 1 −cos(nπ ))/nπ.

c.u(x,y)=u 1 (x,y)+u 2 (x,y),where

u 1 (x,y)=

∑∞

n= 1

cnsinh(λny)

sinh(λnb)

sin(λnx),

u 2 (x,y)=

∑∞

n= 1

cnsinh(μnx)

sinh(μna)

sin(μny).

In both series,cn= 2 ab(− 1 )n+^1 /nπ. Also noteu(x,y)=xy.

Section 4.3

1. a.u(x,y)=1, but the form found by applying the methods of this sec-
   tion is

u(x,y)=

∑∞

n= 1

ansinh(λny)+sinh(λn(b−y))

sinh(λnb)

cos(λnx)

+

∑∞

n= 1

bncosh(μnx)

cosh(μna)

sin(μny),

where

λn=(^2 n−^1 )π

2 a

, an=

4sin

(( 2 n− 1 )π

2

)

π( 2 n− 1 )

,

μn=nbπ, bn=^2 (^1 −cosnπ(nπ)).

b.u(x,y)=y/b, and this is found by the methods of this section. In this

case, 0 is an eigenvalue.

c.^4

π

∑∞

1

(− 1 )n+^1 cos(λny)

( 2 n− 1 )

sinh(λn(a−x))

sinh(λna)

,λn=

(

2 n− 1

2

π

b

)

.

3.b 0 b=V 20 ,bnsinh(λnb)=^2 V^0 (cosn 2 (πnπ) 2 −^1 ).