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  Proof
First we will examine the behavior of Vo in a neighborhood of α1 = α1s on L+ . When α1
c
approaches α1s, then (Q Y) approaches zero. Since e3 approaches e1, eq 118 is reduced to:
e1(wo - ν1 )ρ10Vo = 1 - Y T1 .
~
˙~
(137)
Also eq 120 is reduced to
˙~
Y T1E3 (T1) = E4 (T1)
(138)
where E3 and E4 are defined as
1 - eoµ
0
E3 (T1) = 1 + (Ko / δo )(η / α o )∫
ds
(139)
K1(1 + e4µ)2
T1
1- µ
0
E4 (T1) = eo 1 + (Ko / δo )(η / α o )∫
ds.
(140)
T1 K1(1 + e4µ)
2
˙~
Since 0 < eo < 1, so E3 > E4 > 0 or Y T1 < eo < 1. Hence we find that Vo is positive in a
neighborhood of α1 = α1s. Suppose that there is a point on L+ where Vo < 0. Since Vo (Q) is
c
+
+
continuous on Lc for T1 < Tσ, one can find a point on Lc such that Vo vanishes. However,
this contradicts Prop. 1. Therefore, Vo must be positive on L+ . Since eo > 0 and Vo > 0 on L+ ,
c
c
~ > 0 on L+ (α , α ) and on C . u
in eq 120 E2 (T1) > 0 and E1 (T1) > 0 by eq 101. Therefore, T1
1s
o
s
c
When eo > 0, a solution with negative fo may exist by eq 96. We will study such a possi-
bility below.
Proposition 8
+
In Sp fo is positive if σ < σc.
Proof
Suppose that there exists a solution T1 = Tp of Problem P on L+ (α1s , αo) such that fo
c
(Tp) = 0 or Y (Tp) = eo (Tp)Q. From eq 101 and 102 we obtain
[
]
T ′(s,Tp ) = -(1/ η) α o + s2 s3µ(s,Tp )Q
(141)
f (s,Tp ) = s2µ(s,Tp )Q.
(142)
Tp is a solution of the following equation given as
σ = g1{T1: fo = 0} = g3 (T1)
(143)
where g3 is defined as
e4µ(s, T1)
^
0 k2 (s)
0
g3 (T1) = ∫
ds - (η / s3 )∫
ds
(144)
T1 K1(s)[1 + e4µ(s, T1)]
^
T1 k1(s)
^
where e4 is defined as
e4 = (s2 s3 / α o ) Q = s2 s3Y / (α oeo ).
(145)
^
It is easy to find
18
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