FORMATION FACTOR BOUNDS

To obtain some useful bounds, I again consider the form of (1)  

$$G(g_1,g_2) = \frac{g_1}{F_1} + \frac{g_2}{F_2} + \int_0^\infty \frac{dx {\cal G}(x)}{\frac{1}{g_1} + \frac{x}{g_2}}.$$ (4)

For reasons that will become apparent I want to compare the values of G(g₁+2g₀,g₂+2g₀) and G(g₁,g₂)+2g₀, where g₀ can take any positive value, but g₀ is limited in the negative range by the limitations that both g₁+2g₀ and g₂+2g₀ must always be nonnegative. A straightforward, but somewhat tedious calculation shows that  

$$\begin{array} {l} G(g_1+2g_0,g_2+2g_0) - G(g_1,g_2) - 2g_0 =... ...cal G}(x)}{(1+x)(g_2+xg_1)[g_2+xg_1+2(1+x)g_0]}\, .\end{array}$$ (5)

The right hand side of this equation is always positive whenever g₀ >0 and $g_1 \ne g_2$. It vanishes when g₀ = 0 or g₁ = g₂. If g₁ < g₂, then for negative values of the parameter g₀, allowed values of g₀ lie in the range $0 \gt 2g_0 \ge -g_1$. For such values of g₀, the right hand side of (5) is strictly negative.

The limiting case obtained by taking $2g_0 \to -g_1$ is most useful because, in this limit, $G(g_1+2g_0,g_2+2g_0) \to (g_2-g_1)/F_2$-- thus eliminating the unknown functional ${\cal G}(x)$ from this part of the expression. Then, (5) shows that  

$$G(g_1,g_2) \ge g_1 + \frac{g_2-g_1}{F_2} \equiv S_2(g_1,g_2),$$ (6)

which is a general lower bound on G(g₁,g₂) without any further restrictions on the measurable quantities $g_1 \le g_2$, and F₂.

A second bound can be obtained (again in the limit 2g₀ = -g₁) by noting that  

$$\int_0^\infty \frac{dx x{\cal G}(x)}{(1+x)(g_2+xg_1)} \le \int_0^\infty \frac{dx {\cal G}(x)}{g_2+xg_1},$$ (7)

and then recalling that  

$$\int_0^\infty \frac{dx {\cal G}(x)}{g_2+xg_1} = \frac{1}{g_1g_2}\left[G(g_1,g_2) - \frac{g_1}{F_1} - \frac{g_2}{F_2}\right].$$ (8)

Substituting (7) into (5) produces an upper bound on G(g₁,g₂). By subsequently substituting (8) and then rearranging the result, the final bound is  

$$G(g_1,g_2) \le g_2 + \frac{g_1-g_2}{F_1} \equiv S_1(g_1,g_2).$$ (9)

Comparing (6) and (9), I see consistency requires that  

$$g_1 + \frac{g_2-g_1}{F_2} \le g_2 + \frac{g_1-g_2}{F_1}$$ (10)

must be true. Rearranging this expression gives the condition  

$$0 \le (g_2-g_1)\left(1 - \frac{1}{F_1} - \frac{1}{F_2}\right),$$ (11)

the validity of which I need to check. In the limit g₁ = g₂ = 1, a sum rule follows from (4), and from this I have:  

$$1 - \frac{1}{F_1} - \frac{1}{F_2} = \int_0^\infty\frac{dx {\cal G}(x)}{1+x} \ge 0.$$ (12)

This shows explicitly that (11) is always satisfied as long as $g_2 \ge g_1$. If this inequality $g_2 \ge g_1$does not hold, then the sense of the bounding inequalities is changed, so the expressions for the upper and lower bounds trade places.

When $g_2 = \hbox{const}$ and g₁ varies (as would be expected in a series of thermal conductivity experiments with different fluids in the same porous medium), then (6) and (9) are both straight lines that cross at g₁ = g₂. The general bounds are therefore  

$$\min(S_1,S_2) \le G(g_1,g_2) \le \max(S_1,S_2),$$ (13)

where S₁ and S₂ were defined in (6) and (9).