1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

482 Answers to Odd-Numbered Exercises

(A superscriptkin parentheses meanskth derivative.) The right-hand

side looks like the binomial theorem. In the case at hand, at most two

terms are not zero.

11.bn=







0(nodd),

−(− 1 )n/^2 ( 2 n+ 1 )

(n+ 2 )(n− 1 )

1 · 3 · 5 ···(n− 1 )

2 · 4 · 6 ···n (neven).

Section 5.10

1. The solution is as given in Eq. (5) with coefficients as shown in Eq. (7).
   The integration yields (see Section 5.9)

b 0 =^12 ;bn=0 forneven;b 1 = 3 /4and

bn=

(− 1 )(n−^1 )/^2

2 ·cn

1 · 3 · 5 ···(n− 2 )

2 · 4 · 6 ···(n− 1 )·

2 n+ 1

n+ 1 ,n=^3 ,^5 ,^7 ,....

3.u(φ,t)=T−

∑

bnPn(cos(φ))exp(−(μ^2 +n(n+ 1 ))kt/R^2 ),whereμ^2 =

γ^2 R^2 ,nis odd, andbnis as at the end of Part B, withT 0 =T.

5. The eigenfunctions are as in Part C, except thatnmust be odd in order to
   satisfy the boundary condition.


6. The nodal surfaces are: a sphere atρ= 0 .634 and two naps of a cone given
   byφ= 0. 304 πandρ= 0. 696 π.

Chapter 5 Miscellaneous Exercises

1.u(x,y,t)=

∑∞

m= 1

amsin(μmy)exp

(

−μ^2 mkt

)

+

∑∞

n= 1

amncos(λnx)sin(μmy)exp

(

−

(

μ^2 m+λ^2 n

)

kt

)

,

μm=mπb,λn=nπ/a,

am=T^1 −cosmπ(mπ),amn=^4 πT 3 (cos(nπ)−^1 n)( (^2) m^1 −cos(mπ)).
3.u(a/ 2 ,b/ 2 ,t)=

∑∞

n= 1

bmnsin

(nπ

2

)

sin

(mπ

2

)

exp

(

−

(

λ^2 n+μ^2 m

)

kt

)

,

whereλn=nπ/a,μm=mπ/b,and

bmn=^4 πT 2 (^1 −cos(mπ ))(mn^1 −cos(nπ)).