
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
69
operator γ ◦ F
1
has a fixed point u
1
. Then u
1
(x, t; v
0
), where v
0
is the
fixed point of the operator Υ, is a solution of (22), (23), (24). To complete
the proof we must show that u
1
∈ [z, Z].We show that u
1
(x, t) ≥ z(x, t)
for every (x, t) ∈ D. Suppose, on the contrary, that there exists (ζ, τ ) ∈ D
such that u
1
(ζ, τ) < z (ζ, τ) . Letting w (x, t) = u
1
(x, t)−z(x, t), we see that
w (ζ, τ) < 0. w achieves a negative minimum at some point (ζ
0
, η
0
) ∈ D.
By the strong maximum principle (see
16
) we have either (ζ
0
, η
0
) ∈ Γ or
(ζ
0
, η
0
) = (ζ
0
, 0) with ζ
0
∈ Ω. Since w (x, t) ≥ 0 for (x, t) ∈ Γ, we see that
min
D
w (x, t) = w (ζ
0
, 0) < 0.
We now proceed as in the proof of Lemma 3.1 to complete the proof.
Similarly, we show that u
1
(x, t) ≤ Z(x, t) for every (x, t) ∈ D. Now, for
u ∈ [z, Z] we have that δ (u) = u. Hence F
1
(x, t, u) and F (x, t, u) coincide.
Consequently, u
1
is a solution of the original problem.
Acknowledgements: The first draft of this paper was written while
the author was visiting the Division of Applied Mathematics at Brown
University through a Grant from The Arab Fund For Economic And Social
Development, Kuwait. He is grateful to both Institutions and Professor
John Mallet-Paret for making the visit possible. Also, he wishes to thank
King Fahd University of Petroleum and Minerals for its constant support.
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