1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

478 Answers to Odd-Numbered Exercises

37. a.u=−r

2

4

+c 1 ln(r)+c 2 ;

b.u=−(ln(r))

2

2

+c 1 ln(r)+c 2.

39.V(x,y)∼=a 0 =^1

a

∫a

0

f(x)dxify> 5 L(becausee−λ^1.^5 L∼=0).

41. The solution isθ(X,Y)=1. The low Biot number,B=0, means a very
    large value for conductivityκ, so very little or no cooling takes place.

Chapter 5

Section 5.1

1. ∂

(^2) u
∂x^2

+∂

(^2) u
∂y^2

=^1

c^2

∂^2 u

∂t^2

,0<x<a,0<y<b,0<t,

u(x, 0 ,t)=0, u(x,b,t)=0, 0 <x<a,0<t,

u( 0 ,y,t)=0, u(a,y,t)=0, 0 <y<b,0<t,

u(x,y, 0 )=f(x,y), ∂∂ut(x,y, 0 )=g(x,y),0<x<a,0<y<b.

3. ∂

(^2) u
∂x^2

+∂

(^2) u
∂y^2

+∂

(^2) u
∂z^2

=^1

c^2

∂^2 u

∂t^2

.

Section 5.2

1.

∫c

0

∂^2 u

∂z^2

dz=∂u

∂z

∣∣

∣∣

c

0

=0 by Eq. (12).

3.W′′+( 2 h/bκ)(T 2 −W)=0, 0 <x<a, W( 0 )=T 0 , W(a)=T 1.

W(x)=T 2 +Acosh(μx)+Bsinh(μx),whereμ^2 = 2 h/bκ,

A=T 0 −T 2 ,B=(T 1 −T 2 −Acosh(μa))/sinh(μa).

5.∇^2 u=^1 k∂∂ut,0<x<a,0<y<b,0<t,

∂u

∂x(^0 ,y,t)=0, u(a,y,t)=T^0 ,0<y<b,0<t,

u(x, 0 ,t)=T 0 ,

∂u

∂y(x,b,t)=0,^0 <x<a,0<t,

u(x,y, 0 )=f(x,y),0<x<a,0<y<b.