PDES: GENERAL AND PARTICULAR SOLUTIONS
An infrared laser delivers a pulse of (heat) energyEto a pointPon a large insulated
sheet of thicknessb, thermal conductivityk, specific heatsand densityρ. The sheet is
initially at a uniform temperature. Ifu(r, t)is the excess temperature a timetlater, at a
point that is a distancer(b)fromP, then show that a suitable expression foruisu(r, t)=α
texp(
−
r^2
2 βt)
, (20.37)
whereαandβare constants. (Note that we userinstead ofρto denote the radial coordinate
in plane polars so as to avoid confusion with the density.)
Further,(i)show thatβ=2k/(sρ);(ii)demonstrate that the excess heat energy in the
sheet is independent oft, and hence evaluateα; and(iii)prove that the total heat flow past
any circle of radiusrisE.The equation to be solved is the heat diffusion equation
k∇^2 u(r,t)=sρ∂u(r,t)
∂t.
Since we only require the solution forrbwe can treat the problem as two-dimensional
with obvious circular symmetry. Thus only ther-derivative term in the expression for∇^2 u
is non-zero, giving
k
r∂
∂r(
r∂u
∂r)
=sρ∂u
∂t, (20.38)
where nowu(r,t)=u(r, t).
(i) Substituting the givenexpression (20.37) into (20.38) we obtain2 kα
βt^2(
r^2
2 βt− 1
)
exp(
−
r^2
2 βt)
=
sρα
t^2(
r^2
2 βt− 1
)
exp(
−
r^2
2 βt)
,
from which we find that (20.37) is a solution, providedβ=2k/(sρ).
(ii) The excess heat in the system at any timetisbρs∫∞
0u(r, t)2πr dr=2πbρsα∫∞
0r
texp(
−
r^2
2 βt)
dr=2πbρsαβ.The excess heat is therefore independent oftand so must be equal to the total heat
inputE, implying thatα=E
2 πbρsβ=
E
4 πbk.
(iii) The total heat flow past a circle of radiusris− 2 πrbk∫∞
0∂u(r, t)
∂rdt=− 2 πrbk∫∞
0E
4 πbkt(
−r
βt)
exp(
−
r^2
2 βt)
dt=E
[
exp(
−
r^2
2 βt)]∞
0=E for allr.As we would expect, all the heat energyEdeposited by the laser will eventually flow past
a circle of any given radiusr.