[
]
-1
e1 = 1 - eo = d1 d2 1 + (ko η)-1 Ly .
(94)
Now the problem of finding a solution with positive Vo is reduced to that of finding T1 < 0
that satisfies eq 47, 92 and 93.
It follows from eq 92 and 93 that there are two possible types of solutions satisfying one
of the following conditions given as
eo ≤ 0 or eo > 0 and Y ≥ eoQ, then fo ≥ 0
(95)
eo > 0 and Y < eoQ, then fo < 0.
(96)
Since s3 is a positive number and K2 (T) is an increasing and continuous function if σ = 0,
we will define Tx < 0 as
K2 (Tx ) = η / s3 , σ = 0.
(97)
Using eq 92 and 93, we will write eq 47 as
g1(T1) = g2 (T1)
(98)
where g1 and g2 are defined as
g1(T1) = σ + (δo/Ko)fo(T1)
(99)
0
0
f (s, T1)
K2 (s)
g2 (T1) = ∫
ds + ∫
ds
(100)
K1(s)T ′(s, T1)
K1(s)
T1
T1
and T′ and f are given as
[
]
T ′(s,T1 ) = -η-1 α o + s2 s3(s,T1 ){Q - Y(T1 )} / e1(T1)
(101)
f (s,T1 ) = fo (T1) + s2(s,T1 ){Q - Y(T1)} / e1(T1)
(102)
fo ( T1 ) = [Y(T1) - eo (T1)Q]/ e1(T1)
(103)
(s,T1 ) = [wo - ν(s) ]/[wo - ν(T1) ].
(104)
It is noted that g1 and g2 may be discontinuous at T1 = Tσ due to the singularity of K2 (T1).
Below we will study the properties of g1 (T1) and g2 (T1) that will be used later for existence
proofs.
Proposition 4
For given αo and α1, g1 (T1) and g2 (T1) have the following properties:
g1 (Tm) < g2 (Tm)
(105)
g1(T1) > 0 for Tm < T1 < 0
(106)
˙
g2 (T1) < 0 for Tm < T1 < 0 if eo > 0 and Vo > 0
˙
(107)
where a dot denotes differentiation with respect to T1.
14
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