3.2. Some Canonical Problems Having Analytic Solutions
59
Therefore,
L -1 DexP( -aJb 2 + P)J
= I 2~exp[ -(:: + b
2 . ) ]d'
= e-ab r
t
_a_ exp [_ (a - 2b.)2]d'
J02~
4.
= e-ab r
t
(a + 2b. + a - 2b')ex p [ (a - 2b.)2]d'.
Jo 4~ 4~
4.
Dividing the right-hand side of the above equation into two integrations, and
making the following variable substitutions into each integral, respectively:
~_a-2b. (=a+2b.,
- f o '
fo
we obtain,
Substituting this equation into Eq. (3.2.12) and using Eq. (3.2.13), we finally
get
C
(VX){ [ x
] [x - .JV 2 + 4Ä.D t]
C(x, t) = 2 0 exp 2D L exp -2D;..JV2 + 4Ä.DL erfc
2.jiJ;J
L
[ x ] [x + .JV
2
+ 4Ä.D tJ}
+ exp 2D L .JV 2 + 4Ä.DL erfc
2.jiJ;J
L
•
(3.2.14)
This is the solution of Problem 1. In case of Ä. = 0, i.e., there is no radioactive
decay, the solution is reduced to
C(x,t) = ~o {erfc[;Jv'3] + exp(~)erfc[;~]}. (3.2.15)
Problem 2. Continuous Injection of a Radioactive Tracer into an Infinite
Horizontal Sand Column
Assume that there is a uniform one-dimensional flow with an average pore
velo city V in the sand column. Before t = 0 there was no radioactive tracer