Page 198 - Physical Principles of Sedimentary Basin Analysis

180 Heat flow

which gives
∂U z z 2
(z, t) = √ exp − . (6.288)
∂t 2 πκt 3 4κt

When ∂U/∂t is inserted into solution (6.285) we get

z t T surf (λ) z 2
T (z, t) = √ exp − dλ. (6.289)
2 πκ 0 (t − λ) 3/2 4κ(t − λ)
The remaining step towards the temperature solution (6.282) is a change in integration
variable from λ to
z z 2
μ(λ) = √ or t − λ = 2 (6.290)
2 κ(t − λ) 4κμ
√
3
2
which gives that dλ = (z /2κμ ) dμ. The new integration limits are μ(λ=0) = z/2 κt
and μ(λ=t) =∞.
Note 6.12 Linearly increasing surface temperature. The temperature solution (6.282)is

2c ∞ z 2 −μ 2
T (z, t) = √ (t − ) e dμ (6.291)
√
π z/2 κt 4κμ 2
when T surf (t) = ct. The next step is integration by parts

e e −μ 2
−μ 2 −μ 2
dμ =− − 2 e dμ (6.292)
μ 2 μ
and then use the definition of the complementary error function (6.203).

Exercise 6.32
t
(a) Let the temperature be written as T (z, t) = f (z, t − λ) dλ and show that
0
∂T t ∂ f
= f (z, t − λ)| λ=t + (z, t − λ) dλ. (6.293)
∂t 0 ∂t
(b) Insert temperature (6.285) from Duhamel’s theorem into the temperature equation and
show that

2
2
∂T ∂ T ∂U t ∂ ∂U ∂ U
−κ = T surf (t) (z, t − λ) + T surf (λ) − κ dλ. (6.294)
∂t ∂z 2 ∂t ∂t ∂t ∂z 2
λ=t 0
The factor ∂U/∂t is zero at t = 0 when z > 0, but it is undefined for t = 0 and z = 0. The
integral is zero because U is a solution of the temperature equation, and expression (6.285)
is therefore a solution of the temperature equation for z > 0.