Page 117 - Fundamentals of Computational Geoscience Numerical Methods and Algorithms

5.1 Mathematical Background of Chemical Dissolution Front Instability Problems 105

5.1.2.3 The Second Special Case (Perturbation Solutions
for an Unstable State)
When a reactive transport system represented by the above-mentioned second spe-
cial problem is stable, the planar dissolution front remains planar, even though
both small perturbations of the dissolution front and the feedback effect of
porosity/permeability change are simultaneously considered in the analysis. How-
ever, when the reactive transport system is unstable, the planar dissolution front
can change from a planar shape into a complicated one. The instability of the
above-mentioned second special problem can be determined using a linear stabil-
ity analysis (Chadam et al. 1986, 1988, Ortoleva et al. 1987). The main purpose
of conducting such a linear stability analysis is to determine the critical condition
under which the chemical dissolution front of the reactive transport system becomes
unstable.
If a small time-dependent perturbation is added to the planar dissolution front,
then the total solution of the system is equal to the summation of the base solution
and the perturbed solution of the system.

S(ξ, y,τ) = ξ − δ exp(ωτ) cos(my), (5.52)

p total (ξ, y,τ) = p(ξ, τ) + δ ˆ p(ξ)exp(ωτ) cos(my), (5.53)

ˆ
C total (ξ, y,τ) = C(ξ, τ) + δC(ξ)exp(ωτ) cos(my), (5.54)
where ω is the growth rate of the perturbation; m is the wavenumber of the perturba-
tion; δ is the amplitude of the perturbation and δ<< 1 by the definition of a linear
stability analysis.
Since S(ξ, y,τ) is a function of coordinates ξ and y, the following derivatives
exist mathematically:

∂ ∂S ∂ ∂
= = , (5.55)
∂ξ ∂ξ ∂S ∂ξ
ξ S

∂ ∂S ∂ ∂ ∂S ∂ ∂
= + = + , (5.56)
∂y ∂y ∂S ∂y ∂y ∂ξ ∂y
ξ S S S
2 2
∂ ∂
= , (5.57)
∂ξ 2 ξ ∂ξ 2 S

∂ ∂ S ∂ ∂S ∂ ∂S ∂ ∂
2 2 2 2 2 2
= + + 2 + . (5.58)
2
∂y 2 ξ ∂y ∂ξ ∂y ∂ξ 2 ∂y ∂ξ∂y ∂y 2 S