It follows from eq 71 that Ts is a decreasing function of αo. Similarly it is easy to find that Ts
is a decreasing function of δo , αoδo and σ. u
Next we will study the region Si. For a given αo , we will consider a segment l(α o ) of a
^
^
straight line L(α ie , α o ) (Fig. 2a) defined as
^
l(α o ) = {(α1 , α o ) : α o = α o and α1e < α1 < α1s}
(72)
^
^
L(α1e , α o ) = {(α1 , α o ) : α o = α o , α1 > α1e }
(73)
^
^
where (α1e , αo ) ∈ Le and (α1s , αo ) ∈Cs . The flux fo (α1) on l(α o ) is given by eq 51 as
^
^
^
fo (α1) = (k1α1 - koα o ) / L.
^
(74)
The flux depends linearly on α1, fo (α1) > 0 on l(α o ), fo (α1e ) = 0 and fo (α1s ) = fs where fs
^
is given as
fs = K2 {Ts (α o )}(α o / η).
^
^
(75)
The P' (δ ) vanishes at the point (α1s , α o ). Suppose that P' (δ ) vanishes at some point on
^
l(αo ) , then there is no solution of eq 58 at that point by Prop. 1. We will seek solutions on
^
l(α o ) under the condition of P' (δ ) > 0.
^
Proposition 3
For a given α o , there exists a unique T1 on l(α o ) such that Ts < T1 < Tσ and eq 58 holds
^
^
true if P' (δ ) > 0.
Proof
For a given point (α1 , α o ) on l(α o ) we will write eq 58 as
^
^
h3 (α1) = h4 (T1 , α1)
(76)
where h3 and h4 are defined as
(
)
h3 (α1) = σ + δo / Ko fo (α1)
(77)
0
0
h4 (T1 , α1) = ∫ [K2 (s) / K1(s)]ds - (η / αo ) fo (α1)∫ [1/ K1(s)]ds.
^
(78)
T
T
1
1
Differentiating h4 (T1, α1) with respect to T1, we obtain
˙
h4 (T1 , α1) = -(η / αo )P′(δ-).
^
(79)
Since P' (δ ) > 0, h4(T1, α1) is a decreasing function of T1. When T1 approaches Ts (α o ) , we
^
will evaluate h4. Since Ts (α o ) is the solution of eq 62, from eq 63 and 64 we obtain
^
0
σ + (δo / Ko ) fs = ∫ [K2 (s) - K2 (Ts )][1/ K1(s)]ds.
(80)
T
s
Using eq 80, from eq 78 we obtain
0
[1/ K1(s)] ds.
h4 (Ts , α1) = σ + (δo / Ko ) fs + (η / α o )K1(Ts )P′(δ-)∫
(81)
^
Ts
It follows from eq 80 and 81 that h4(Ts, α1) > h3(α1). Also when T1 approaches Tσ, we have
12
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