Proposition 15
-
There exists at least one solution T1 of Problem P on Lc such that Tx < T1 < Ts < Tσ, Vo > 0
and fo > 0.
Proof
-
Since eo < 0 in this case, we must search a solution T1 on Lc such that T1 > Tx. First we
examine eq 98 when T1 = Tx. Since αo < αoc in this case, using eq 154, we find that g1 (Tx) <
g2 (Tx).
We will assume that Vo > 0 and fo > 0 for the time being. From eq 99 and 100 we obtain:
Wo (Ts ) = ao (Ts ) fo - a1(Ts ) + (η / α o )a2 (Ts )
(186)
where ao is defined by eq 177, and a1 and a2 are defined as
K2 (s)
Tσ
a1(T ) = ∫
(187)
ds
K1(s)
T
(s,T )
0
a2 (T ) = (s2 / e1)(Q - Y)∫
(188)
ds.
K1(s)[1 + e4(s,T )
T
Since Ts is the solution of eq 98 when Vo = 0 (Q = Y), we obtain
ds
0
Wo (Ts ) = (δo / Ko + (η / α o )∫
f - a1(Ts ) = 0 if Vo = 0
Ts K1(s) s
(189)
where fs is given by eq 68. Using eq 189, we will reduce eq 186 to
Wo (Ts ) = (-1/ e1 )(δo / Ko )E4 (Ts )(Q - Y)
(190)
where E4 is defined by eq 140. It follows from eq 190 that g1 (Ts) > g2 (Ts). Since g1 and g2 are
continuous functions of T1 for T1 < Tσ, there exists at least one solution T1 of Prob. P on L-
c
such that Tx < T1 < Ts < Tσ if Vo > 0 and fo > 0.
~
When such a solution exists, from eq 137 and 138 we find that T1 < 0 and Vo > 0 in a
-
neighborhood of α1 = α1s in Sp . By the similar reasoning to that used in the proof of Prop-
osition 7, we find that Vo > 0 on L- . Since Vo > 0 and T1 < Tσ on L- , hence by Proposition 10
c
c
fo > 0. u
It is clear that a solution of Proposition 15 is unique if g2 (T1) ≤ 0 on L- . We will study
˙
c
˙
the sign of g2 (T1) below. We will write eq 113 as
g2 (T1) = -(1/ α o )(e3 / e1 )W1(T1)
(191)
˙
2
where W1 is defined as
0
W (s,T1)
W1(T1) = (η / α o )∫
(192)
ds.
K1(s)(1 + e4)2
T1
˙
The sign of g2 (T1) depends on that of W1(T1). We will write eq 192 as
W1(T1) = w1αo + w2α1.
(193)
w1 and w2 are defined as
w1(T1) = ya3 + (e1e5 - y)a4 > 0
˙
˙
(194)
24
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