Proof
When T1 = Tm, then Y = 0, eo = s2 and e1 = 1 s2. From eq 101 through 103, we obtain
fo = s2 Q/ (1 s2) < 0
(108)
f / T ′ = s2 η(1 - )Q / (e1α o + s2 s3Q) ≥ 0.
(109)
It follows from eq 108 that g1 (Tm) < σ.
Since Tm < Tσ by eq 40, we will write eq 100 as
Tσ K2 (s)
f (s, T1)
0
g2 (Tm ) = σ + ∫
ds + ∫
(110)
ds.
Tm K1(s)T ′(s, T1)
Tm K1(s)
Since the second and the third terms in the right side of eq 110 are positive, we find that g2
(Tm) > σ; eq 105 holds true.
Differentiating eq 99 with respect to T1, we obtain
g1(T1) = (δo / Ko ) (α o / η) (e3 / e1 )y
2 ˙
˙
(111)
where e3 is defined as
e3 = e1 + s2 s3 (Q - Y) / α o = [(1 - s2 )α o + s2 s3Q] / α o > 0.
(112)
It follows from eq 111 that 106 holds true.
Differentiating eq 100 with respect to T1, we obtain
0
g2 (T1) = -(η / α o )(e3 / e1 )∫
W (s,T1)
(113)
˙
2
2
ds
K1(s)[(1 + e4(s,T1)]2
T1
where e4 and W are defined as
e4 (T1) = (s2 s3 / α o )(Q - Y) / e1
(114)
W (s,T1) = [α o (1 - s2) + s2 s3Q]Y + eoα o(Q - Y).
˙
(115)
˙
Since Vo > 0, by eq 92, Q Y > 0. Differentiating eq 104 with respect to T1, we obtain
(s,T ) = [wo - ν(T1)]-1 ν(T1) = o (T1) ≥ 0.
(116)
˙
˙
Since eo > 0, so W (s,T1) > 0. Hence, eq 107 holds true. u
˙
˙
It is noted that g1(T1) and g2 (T1) may be discontinuous at T1 = Tσ due to the singularity
of K2 (T1). We will consider a straight line L (α1s, αo) in Sp defined as
L(α1s, αo) = [(α1, αo): α1 ≥ α1s]
(117)
where (α1s , α o ) ∈Cs . We will study the behavior of T1 and Vo on L(α1s , α o ). Differentiating
eq 92 with respect to Q, we obtain
e1(wo - ν1 )ρ30Vo = 1 - e2T1
~
~
(118)
where a tilde denotes differentiation with respect to Q for a given αo and e2 is defined as
15
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