e2 (T1) = (e3 / e1 )Y - o (Q - Y).
˙
(119)
Differentiating eq 98 with respect to Q, we obtain
~
E1(T1)T1 = E2 (T1)
(120)
where E1 and E2 are defined as:
E1(T1) = g1(T1) - g2 (T1)
(121)
˙
˙
1 - (s,T1)
0
E2 (T1) = (eo / e1 )(δo / Ko ) + (η / α o )∫
ds.
(122)
T1 K (s)[1 + e (s,T )]2
1
4
1
Now we will begin our search of solutions of eq 92, 93 and 98 in Sp with a special case in
which eo vanishes. For the sake of brevity we will refer the problem of eq 92, 93 and 98 to as
Problem P hereafter.
Proposition 5
There exists a unique solution of Problem P such that eo = 0 and Vo > 0 if the following
condition holds true:
K2 (s) - (η / s3 )
0
σ<∫
ds.
(123)
K1(s)
Tx
Proof
When eo = 0, then T1 = Tx and from eq 93 we obtain
fo = Y = α o / s3 .
(124)
Using eq 101, 102 and 124, we will reduce eq 98 to
K2 (s) - (η / s3 )
0
α oδo / (s3Ko ) = ∫
ds - σ.
(125)
K1(s)
Tx
Since eq 123 holds true, for a given σ there exists a unique and positive αo (σ) that satisfies
eq 125.
We will denote αo (σ) by αoc , and consider a line Lc (α1s,αoc) defined as
Lc (α1s , α oc ) = {(α1 , α oc ) : α1 ≥ α1s }.
(126)
Our aim is to show Vo > 0 on Lc. From eq 122 we find that E2 (T1) vanishes if and only if T1
= Tx or eo = 0. From eq 111 and 113 we find that g1(Tx ) > 0 and g2 (Tx ) < 0; hence E1 (Tx) > 0.
˙
˙
~
~
Therefore, T1 vanishes in this case. Using eq 118, we find that Vo is positive. Therefore, Vo
is positive on Lc except at a point (α1s,αoc) where Vo vanishes. u
We will define go (σ) as
K2 (s) - (η / s3 )
0
go (σ) = ∫
ds - σ
(127)
K1(s)
Tx
or we may write go as
16
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