444 Answers to Odd-Numbered Exercises
7.u(r)= 325 + 104 ( 0. 25 −r^2 )/4;u( 0 )=950.
9.u(x)=T 0 +AL^2 ( 1 −e−x/L).Section 0.5
1.G(x,z)={
z(a−x)/(−a), 0 <z≤x,
x(a−z)/(−a), x≤z<a.3.G(x,z)={
cosh(γz)sinh(γ (a−x))/(−γcosh(γa)), 0 <z≤x,
cosh(γx)sinh(γ (a−z))/(−γcosh(γa)), x≤z<a.5.G(ρ,z)=
(c−ρ)/ρ
−c/z^2, 0 ≤z<ρ,
(c−z)/z
−c/z^2,ρ≤z<c.7.G(x,z)=
sinh(γz)e−γx
−γ, 0 <z≤x,sinh(γx)e−γz
−γ, x≤z.9.u(ρ)=(ρ^2 −c^2 )/6.11.u(x)=∫a0G(x,z)f(z)dz=∫x0z(a−x)
−a f(z)dz+∫axx(a−z)
−a f(z)dz.
There are two cases:
(i)x≤a/2, sou(x)=∫aa/ 2x(a−z)
−a dz;
and
(ii)x>a/2, sou(x)=∫axx(a−z)
−a dz.Results:u(x)={
−ax/ 8 , 0 <x<a/2,
−x(a−x)^2 / 2 a, a/ 2 <x<a.- (i) At the left boundary,x=l<z, so the second line of Eq. (17) holds.
The boundary condition (2) is satisfied byvbecause it is satisfied byu 1.
At the right boundary, use the first line of Eq. (17).
(ii) Atx=z,bothlinesofEq.(17)givethesamevalue.
(iii)v′(z+h)−v′(z−h)=u 1 (z)u′ 2 (z+h)−u′ 1 (z−h)u 2 (z)
W(z).
Ashapproaches 0, the numerator approachesW(z).