σ ≥ σx or Tσ ≤ Tx. First we will study Case 1. In Case 1 if a solution of Problem P exists on
~
L+, then Vo > 0 and T1 > 0 on L+ by the same reasoning as used in the proof of Proposition 7.
Proposition 11
There exists a unique solution T1 < Tx of Problem P on L+ if σ < σx.
Proof
For a given point (α1, αo) on L+, we will search a solution of Problem P under the as-
sumption that Vo > 0 at this point. Because of eq 105, 106 and 101 for a unique solution T1 to
exist we must have
Wo (Tx) > 0.
(164)
Using eq 99 and 100 we obtain
Wo (Tx ) = α oδo / (s3Ko ) + σ - σc > 0.
(165)
Hence, there exists a unique solution T1 < Tx of Problem P on L+ if Vo > 0. But if such a
solution exists, then Vo > 0. Therefore, there exists a unique solution of Problem P. u
Proposition 12
There exists a unique solution T1 = Tp of Problem P on L+ such that fo (Tp) = 0 if σ < σx.
Proof
It is easy to see that Tp is also a solution of eq 143 of Prop. 8. In this case σ ≥ σc. Hence,
there exists a unique solution Tp of eq 143. Since fo = 0, Vo > 0 by eq 90. We will show that
this solution is located on L+. We will denote the point such that T1 = Tp by (α1p, αo) and the
point on Cs by (α1s, αo). Since fo = 0 or Y eoQ = 0 at (α1p, αo), we obtain
[
}]
{
α1p = (1/ k1 ) ko + LK2 (Tp ) / ηeo (Tp ) α o
(166)
where eo (Tp) is given as
[
]
eo (Tp ) = s2 1 - (s3 / η)K2 (Tp ) .
(167)
From eq 69 we obtain
α1s = (1/ k1)[ko + LK2 (Ts ) / η]α o .
(168)
Using eq 166 and 168, we obtain
[
][
]
α1p - α1s = K2 (Tp ) - eo (Tp )K2 (Ts ) L / k1ηeo (Tp ) .
(169)
~
Since T1 > 0 on L+, K2(Tp) > K2(Ts). Also 1 > eo (Tp ) > 0 . Hence, α1p > α1s. Therefore, the
point (α1p, αo) is on L+. u
The unique solution T1 of Proposition 12 satisfies eq 143. We may write eq 143 and 145 as
σ = g3 (Tp)
(170)
{
}
e4 (Tp ) = s2 s3K2 (Tp ) / ηeo (Tp ) .
(171)
^
21
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