462 Answers to Odd-Numbered Exercises
u(x,t)=√T^0
4 πkt∫a0exp(
−(x′−x) 2
4 kt)
dx′=T^0
2
[
erf(
a√−x
4 kt)
+erf(
√x
4 kt)]
.
- Interpretation:uis the temperature in a rod with insulation on the cylin-
drical surface and on the left end. At the right end, heat is being forced
into the rod at a constant rate (becauseq(a,t)=−κ∂∂ux(a,t)=−κS,so
heat is flowing to the left, into the rod). The accumulation of heat energy
accounts for the steady increase of temperature.
19.( 1 / 6 ka)u 3 −(a/ 6 k)u 1 satisfies the boundary conditions.
21.w(x,t)=−^2 u∂∂xu,whereu(x,t)=a 0 +∑
ancos(nπx)exp(
−n^2 π^2 t)
,
wherea 0 = 2(
1 −e−^1 /^2)
andan=^1 −e− 1 / (^2) cos(nπ)
1
4 +(nπ)^2
.
23.u 2 =T 0 β^2 V
β 1 +β 2,u 1 =T 0(
1 − β^1 V
β 1 +β 2)
,
whereV= 1 −exp(−(β 1 +β 2 )t)andβi=h/ci.25.u(ρ,t)=^1 ρ∑∞
n= 1bnsin(λnρ)exp(
−λ^2 nkt)
,
λn=nπ/a,bn=2
a∫a0ρT 0 sin(λnρ)dρ.27.v(x)=T 0 +Sx−S sinh(λx)
γcosh(γa).
- Ifλ=0, the differential equation isφ′′=0 with general solutionφ(x)=
c 1 +c 2 x. The boundary conditions requirec 2 =0butallowc 1 =0. Thus,
this value ofλpermits the existence of a nonzero solution, and therefore
λ=0isaneigenvalue. - ChooseB(ω)=^2 π
∫∞
0 f(t)sin(ωt)dt.Iffhas a Fourier integral represen-
tation, then this choice ofBwill makeu( 0 ,t)=f(t),0<t.- a. v(x)=−Ix/aK+c 1 +c 2 ( 1 −e−aKx/T),
c 1 =h 1 ,c 2 =(h 2 −h 1 +IL/aK)/( 1 −e−aKL/T).
b. ∂(^2) w
∂x^2
+μ∂w
∂x
=^1
k∂w
∂t,0<x<L,0<t,
w( 0 ,t)=0, w(L,t)=0, 0 <t,
w(x, 0 )=h 0 (x)−v(x),0<x<L,
whereμ=aK/T,k=T/S.