Page 167 - Physical Principles of Sedimentary Basin Analysis

6.12 Transient convective heat flow 149

where the prime denotes differentiation. This expression is rewritten as
U V V
=−Pe + , (6.174)
U V V
where the left-hand side of the equality is dependent only on ˆ t and the right-hand side is
dependent only on ˆz. The equality must therefore be equal to a constant, which is written
−k. The minus sign before k anticipates that it is a positive number. The equation for the
factor U becomes
U
=−k, (6.175)
U
and the solution for U is (see Exercise 6.5)

U(ˆ t) = U 0 e −k ˆ t , (6.176)
where U 0 is an arbitrary constant. We cannot allow U(ˆ t) to grow exponentially with time,
which explains why the constant k has to be a positive number. The equation for V is

V − Pe V + kV = 0, (6.177)
and it is factorized in the following way:

d d
− r 1 − r 2 V = 0 (6.178)
dˆz dˆz
where r 1,2 are the roots
1
2
r 1,2 = Pe ± Pe − 4k (6.179)
2
2
of the quadratic equation r − Per + k = 0. We see that functions of the form exp(r i ˆz),
i = 1, 2, are solutions of equation (6.178). The function V can therefore be written as any
linear combination of these two solutions

V (ˆz) = c 1 e r 1 ˆz + c 2 e r 2 ˆz . (6.180)
The boundary conditions for the transient solution require that V (ˆz=0) = 0 and V (ˆz=1) =
0, which is the same as c 1 + c 2 = 0 and c 1 exp(−D) + c 2 exp(D) = 0, respectively, where

2
D = (Pe/2) − k. The boundary conditions therefore require that c 2 =−c 1 and that
exp(2D) = 1, where the latter equation has the non-trivial solution D = iπn, with i =
√ 2 2 2 2
−1, and where n is any integer. From D =−(πn) it follows that k n = (Pe/2) +(πn)
1
and that r 1,2 = Pe ± iπn. There is not just one k, but one for each integer n. Since D is a
2
complex number it follows from the formula for the exponentiation of a complex number,
exp(iφ) = cos φ + i sin φ, where φ is real that
V n (ˆz) = c 3 exp(Pe ˆz/2) sin(πnˆz), (6.181)

where c 3 is an arbitrary constant and where the subscript n tells us that there is one V for
each integer n. (Only the complex part i sin φ fulfills the boundary conditions. Only this