Tσ
K2 (s)
g3 (Tm ) = σ + ∫
ds > σ.
(146)
Tm K1(s)
Differentiating eq 144 with respect to T1, we obtain
e4
^˙
0
g3 (T1) = -(η / s3 )∫
ds.
(147)
T1 K1(s)(1 + e4)2
^
It follows from eq 147 that g3 (T1) < 0 . Since g3(T1) is continuous and Tm < T1 < Tx < Tσ, for
˙
a solution T1 = Tp of eq 143 to exist we must have
σ - g3 (Tx ) > 0.
(148)
When T1 approaches Tx, eo approaches zero and g3 (T1) approaches σc. Hence eq 148 is
reduced to
σ > σc.
(149)
+
It is clear that eq 149 does not hold in this case. Hence the solution Tp does not exist on Lc .
+
+
Since fo > 0 in a neighborhood of α1 = α1s on Lc and fo (Q) is continuous on Lc for T1 < Tσ,
+
+
fo is positive on Lc . Since αo is an arbitrary positive number, fo is positive in Sp if σ < σc. u
Proposition 9
+
There exists a unique solution of Problem P in Sp .
Proof
+
We will consider a line Lc defined by eq 135. In view of Proposition 4 we need to show
that g1 (Tx) > g2 (Tx). From eq 99 and 100 we obtain
g1(Tx ) = α oδo / (s3Ko )
(150)
g2 (Tx ) = σc .
(151)
For the sake of convenience we will define Wo (T ) for T < 0 as
Wo (T ) = g1 (T ) g2 (T ).
(152)
From eq 150 and 151 we obtain
Wo (Tx ) = αoδo / (s3Ko ) + σ - σc .
(153)
Using eq 132, we will reduce eq 153 to
Wo (Tx ) = δo (αo - αoc ) / (s3Ko ).
(154)
Since αo > αoc, we find that g1 (Tx) > g2 (Tx). Since g1 (T1) and g2 (T1) are continuous for T1 <
+
Tσ, there exists a unique solution of Problem P on Lc . Since αo is arbitrary, for any given
+
point (α1 αo) in Sp there exists a unique solution. u
When Vo > 0, unfrozen water may exist in R2 (T < T1) and the amount of unfrozen water
in R2 depends on T. Hence, the rate of heave depends on T and is given for T ≤ T1 as
19
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