1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

474 Answers to Odd-Numbered Exercises

5. Check zero boundary conditions by substituting. Atx=a,find

Ancosh(μna)=^2 b

∫b

0

Sycos(μny)dy.

7.w(x,y)=

∑∞

n= 1

ancosh(λny)cos(λnx). From the condition aty=b,

ancosh(λnb)=^2

a

∫a

0

Sb

a

(x−a)cos(λnx)dx.

9.w(x,y)=

∑∞

n= 1

cnsinh(λny)+ansinh(λn(b−y))

sinh(λnb) sin(λnx),

an=cn=−

2

a

∫a

0

Hx(a−x)sin(λnx)dx=− 2 Ha^2

1 −cos(nπ)

n^3 π^3.

11. 12A+ 2 C=−K,12E+ 2 C=−K. There are many solutions.

Section 4.4

1.an=^2 a

∫a

0

f(x)sin

(

nπx

a

)

dx.

3.A(μ)=

2

π

∫∞

0

g 2 (y)sin(μy)dy.

5. a.u(x,y)=

∑

cncos(λnx)exp(−λny),λn=( 2 n− 1 )π/ 2 a,

cn= 4 (− 1 )n+^1 /π( 2 n− 1 );

b.u(x,y)=

∫∞

0

B(λ)cosh(λx)sin(λy)dλ,B(λ)=^2 λ

π(λ^2 + 1 )cosh(λa)

;

c.u(x,y)=

∫∞

0

A(λ)cos(λy)sinh(λx)dλ,A(λ)=

2sin(λb)

πλsinh(λa).

7.u(x,y)=

∑∞

1

bnsin(λnx)exp(−λny)

+

∫∞

0

(

A(μ)sinh(μx)

sinh(μa)

+B(μ)sinh(μ(a−x))

sinh(μa)

)

sin(μy)dμ,

λn=nπ/a,bn= 2 ( 1 −cos(nπ ))/nπ,A(μ)=B(μ)= 2 μ/π(μ^2 + 1 ).

Also see Exercise 8.

9. a.u(x,y)=π^2

∫∞

0

1 −cos(λa)

λ sin(λx)

sinh(λy)

sinh(λb)dλ;