+
neighborhood of Cs in Sp and Vo (Q) is continuous on lp (α o ) , Vo > 0 on lp (α o ) . Therefore,
there exists a unique solution T1 of Problem P on lp (α o ) such that fo > 0. When fo = 0, Vo > 0
because from eq 92 and 93 we find
Q - Y = e1(Q - fo ) = e1Q > 0. u
(180)
We will define Lp by eq 172 where Tp is the unique solution in Prop. 13. Then the line Lp
+
divides the region Sp into two regions, Sx and Spp defined by eq 173 and 174, respectively.
Proposition 13 implies that there exists a unique solution in Spp such that fo > 0. We will
search a solution T1 of Problem P in Sx such that T1 < Tσ ≤ Tx and fo < 0. It is easy to find
from eq 180 that Vo > 0 when fo < 0.
When fo < 0, we will study the behavior of Wo (T) for Tp ≤ T ≤ Tx. When T = Tp, fo = 0.
From eq 176 we obtain
(s, Tp )
0
Wo (Tp ) = (s2 η / α o )Q ∫
ds > 0.
(181)
Tp K1(s)[1 + e4(s, Tp )]
Using eq 121, we obtain:
˙
Wo (T ) = g1(T ) - g2 (T ) = E1(T ).
(182)
˙
˙
Because of eq 106 and 107 we find
Wo (T ) > 0
for Tp ≤ T ≤ Tx .
(183)
We may write eq 183 as
Wo (Tσ -) > 0
for Tp < Tσ ≤ Tx .
(184)
We will consider L+ defined by eq 162. We have found that there exists a unique solution
T1 of Problem P for α1s ≤ α1 ≤ α1p. As α1 increases from α1s to α1p, T1 increases from Ts to Tp
and fo decreases from fs to zero. Because of eq 184 there exists a unique solution T1 of Prob.
Proposition 14
There exists a unique solution T1 of Problem P on L+ with α1p < α1 such that Tp < T1 < Tσ
and fo < 0 if σ ≥ σx.
As stated above there may exist a solution T1 of Problem P such that T1 > Tσ in Case 2.
But we are not certain of the existence of such a solution.
SOLUTION IN Sp
-
We will consider a line Lc defined as
L- (α1s , α o ) = {(α1 , α o ) : α1 > α1s and α oc > α o > 0}
(185)
c
-
-
where (α1s, αo) ∈ Cs and (α1, αoc) is on the line Lc (Fig. 2b). It is clear that Lc belongs to Sp
where eo < 0.
23
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