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 foruis
u(r, t)=
α
t
exp
(
−
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 obtain
2 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 timetis
bρs
∫∞
0
u(r, t)2πr dr=2πbρsα
∫∞
0
r
t
exp
(
−
r^2
2 βt
)
dr
=2πbρsαβ.
The excess heat is therefore independent oftand so must be equal to the total heat
inputE, implying that
α=
E
2 πbρsβ
=
E
4 πbk
.
(iii) The total heat flow past a circle of radiusris
− 2 πrbk
∫∞
0
∂u(r, t)
∂r
dt=− 2 πrbk
∫∞
0
E
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.