Proposition 5

e2 (T1) = (e3 / e1 )Y - µ o (Q - Y).

(119)

Differentiating eq 98 with respect to Q, we obtain

E1(T1)T1 = E2 (T1)

(120)

where E1 and E2 are defined as:

E1(T1) = g1(T1) - g2 (T1)

(121)





1 - µ(s,T1)

E2 (T1) = (eo / e1 )(δo / Ko ) + (η / α o )∫

ds.

(122)

T1 K (s)[1 + e µ(s,T )]2









Now we will begin our search of solutions of eq 92, 93 and 98 in Sp with a special case in

which eo vanishes. For the sake of brevity we will refer the problem of eq 92, 93 and 98 to as

Problem P hereafter.

Proposition 5

There exists a unique solution of Problem P such that eo = 0 and Vo > 0 if the following

condition holds true:

K2 (s) - (η / s3 )

σ<∫

ds.

(123)

K1(s)

Proof

When eo = 0, then T1 = Tx and from eq 93 we obtain

fo = Y = α o / s3 .

(124)

Using eq 101, 102 and 124, we will reduce eq 98 to

K2 (s) - (η / s3 )

α oδo / (s3Ko ) = ∫

ds - σ.

(125)

K1(s)

Since eq 123 holds true, for a given σ there exists a unique and positive αo (σ) that satisfies

eq 125.

We will denote αo (σ) by αoc , and consider a line Lc (α1s,αoc) defined as

Lc (α1s , α oc ) = {(α1 , α oc ) : α1 ≥ α1s }.

(126)

Our aim is to show Vo > 0 on Lc. From eq 122 we find that E2 (T1) vanishes if and only if T1

= Tx or eo = 0. From eq 111 and 113 we find that g1(Tx ) > 0 and g2 (Tx ) < 0; hence E1 (Tx) > 0.

Therefore, T1 vanishes in this case. Using eq 118, we find that Vo is positive. Therefore, Vo

is positive on Lc except at a point (α1s,αoc) where Vo vanishes. u

We will define go (σ) as

K2 (s) - (η / s3 )

go (σ) = ∫

ds - σ

(127)

K1(s)

or we may write go as

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