Page 177 - Physical Principles of Sedimentary Basin Analysis

6.15 Cooling sills and dikes 159

where c 2 is the second integration constant. The first boundary condition, (η=0) = 1,
implies that c 2 = 1. The second boundary condition, (η →∞) = 0, then implies that

∞ 2

c 1 e −x dx =−1. (6.222)
0
∞ −x 2 √ √
Tables of definite integrals give that e dx = π/2 and therefore c 1 =−2/ π,
0
which shows that (η) = 1 − erf(η) = erfc(η) is the solution (6.202). The variable η is
called a similarity variable, and it reduces the partial differential equation into an ordinary
differential equation, which in this case is straightforward to integrate.
Cooling of the semi-infinite half-space leads to the same solution in a slightly different
manner. We have that T s < T 0 and the dimensionless temperature is therefore

T (z, t) − T s
= (η) or T (z, t) = T 0 + (T s − T 0 ) (η), (6.223)
T 0 − T s
because we want the dimensionless temperature to be a number from 0 to 1. The
boundary conditions for cooling of the semi-infinite half-space are

(η = 0) = 0 and (η →∞) = 1. (6.224)

The expression for (η) becomes the same as in equation (6.221), where the first of the two
boundary conditions in (6.224) leads to c 2 = 0. The second of these boundary conditions
√
leads to c 1 = 2/ π, and the function is therefore (η) = erf(η), which yields the
solution (6.205). This solution is as we have already seen equal to the solution (6.202).

Exercise 6.20 The surface temperature is suddenly increased by 10 C. How much time is
◦
needed for the temperature to increase by 1 Cat0.1 m, 1 m and 10 m?
◦
Exercise 6.21 The surface temperature is increased from the initial temperature T 0 to T s =
T 0 + T . Find the depth where the temperature has increased by T/2 as a function of
time. Hint: erfc(0.4769) = 0.5.

Exercise 6.22 Use definition (6.203) of the complementary error function and defini-
tion (6.204) of the error function and show that

√
π ∞ −x 2
erfc(η) = e dx. (6.225)
2 η

6.15 Cooling sills and dikes
A sill is a tabular igneous intrusion that is oriented parallel to the planar structure of the
surrounding rock, and a dike is a near-vertical planar igneous intrusion that cuts through the
bedding of the country rock, see Figure 6.26. These intrusions are assumed to be emplaced
◦
instantaneously with a high temperature (∼1000 C). The thermal impact on the coun-
try rock from such intrusions is found by solving the temperature equation (6.201) with