It follows from eq 51 that fo vanishes on Le.The line Le divides S into two regions, and we
will denote one of them by Sm defined as
Sm = {(α1 , α o ):α o > (k1 / ko )α1}.
(53)
Therefore, melting takes place in Sm. We will exclude Sm from our discussion hereafter. For
a special case where αo = 0 and fo ≥ 0, from eq 49 we find
fo = -K1(T1)P′(δ-).
(54)
It follows from eq 54, P7 and 51 that α1 and fo also vanish. Now we will seek solutions with
Vo = 0 in the region Sf defined as
Sf = S - (Sm + Le ) = {(α1 , α o ):0 < α o < (k1 / ko )α1}.
(55)
Suppose that such solutions exist in Sf. Then, eq 26 and 27 are reduced to
ρ3 (δ+) = 0
(56)
-1
r(δ+) = d2 fo > 0.
(57)
It follows from eq 56 and 57 that an ice layer grows in the solutions
When Vo = 0 in Sf , eq 47 is reduced to
(
)
0
0
σ + δo / Ko fo = ∫ (K2 / K1)dT - (η / α o ) fo ∫ (K1 )-1 dT.
(58)
T1
T1
We will write eq 58 as
(
)
0
0
δ / K + (η / α ) (K )-1 dT f = (K / K ) - σ.
o∫
o ∫1 2
o
(59)
1
1
o
T1
T
Suppose that T1 ≥ Tσ , using eq 37, we obtain
0
∫T1 (K2 / K1)dT - σ = γ (Tσ - T1 ) ≤ 0,
if T1 ≥ Tσ .
(60)
It follows from eq 59 and eq 60 that fo ≤ 0. Therefore, if there exists T1 such that fo > 0 and eq
58 holds true, then T1 must be less than Tσ. Our next aim is to find points (α1, αo) in Sf with
T1 < Tσ and Vo = 0. Hereafter, we will assume that σ ≥ 0 and δo > 0 are given constants.
By eq P7 P′ (δ ) is nonnegative if Vo = 0. We will begin our study with a special case in
which P′ (δ ) vanishes.
Proposition 1
If P′ (δ ) vanishes for a given αo, then there exists a unique T1 such that Tm < T1 < Tσ and
that eq 58 holds true.
Proof
When P′ (δ ) vanishes, then fo is given as
fo = K2 (T1)(αo / η).
(61)
10
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