Using eq 61, we will write eq 58 as
h1(T1) = h2(T1)
(62)
where h1 and h2 are defined as
h1(T1) = σ + (δo/Ko)K2(T1)(αo/η)
(63)
0
h2 (T1) = ∫ [K2 (s) - K2 (T1)][1/ K1(s)]ds.
(64)
T1
Since fo > 0, then T1 < Tσ. Using eq 37, we will reduce eq 64 to
{K (s) / K (s)}ds - K (T )∫ [1 / K (s)]ds.
T
0
σ
h2 (T1) = σ + ∫
(65)
2
1
2
1
1
T1
T1
Differentiating h1 (T1) and h2 (T1) with respect to T1, we obtain
˙
˙
h1(T1) = (δο/ Kο )K2 (T1)(αο /η)
(66)
[
]
0
˙
h2 (T1) = -K2 (T1)∫ 1/ K (s) ds
˙
(67)
1
T1
where a dot denotes differentiation with respect to T1. Therefore, from eq 63 and 65, h1 (Tm)
˙
˙
< h (T ) and h (T ) > h (T ). From eq 66 and 67 we find that h (T ) > 0 and h (T ) < 0 for
σ
σ
1
1
2
1
2
m
1
2
Tm < T1 < Tσ. Also h1(T1) and h2(T1) are continuous for T1 < Tσ. Therefore, there exists a
unique T1 such that T1 < Tσ and that eq 44 holds true. u
We will denote the unique T1 of Prop. 1 for a given αo by Ts (αo). It is easy to see that the
function Ts (αo) is continuous for T1 < Tσ. If we denote fo by fs, when T1 = Ts , then fs is given
as
fs = K2(Ts)(αo/η).
(68)
Substituting fo in eq 51 with fs, we will define a curve Cs in Sf (Fig. 2a) as
Cs = {(α1, αo):αo = k1[ko + η1LK2{Ts(αo)}]1α1}.
(69)
We also define the region Si bounded by Le and Cs as
{
}
Si = (α1 , α o ):(k1 / ko )α1 > α o > k1[ko + η-1LK2 {Ts (α o )}]-1α1 .
(70)
From eq 62, Ts depends on αo, δo, and σ for a given soil. We will show the nature of such
dependence below.
Proposition 2
The solution Ts of eq 62 is a decreasing function of αo, δo , αoδo and σ.
Proof
Differentiating eq 62 with respect to αo, we obtain
{
]}
[
T
0
K2 (Ts ) s δo (ηKo )-1α o + ∫ 1/ K1 (s) ds = -δo (ηKo )-1 K2 (Ts ).
˙
(71)
αo
Ts
11
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