Quasilinear parabolic problems with nonlinear boundary conditions

It can be written as
where
l(z, ξ) =
ζ(z, ξ) :=
We first study the function k defined by
k(z, τ) =
1
+
â(z)τ 2
�
ζ + 4(1 − κ)√ �
ζ + 1
√ √ (ζ + 2)
ζ + 1 + κζ + 1
−1 , (5.45)
1
, κ(z) :=
â(z)|ξ| 2
4(1 − κ(z))
â(z)τ 2 + 1 + �
� 1
� 1
â(z)τ 2 + 1
1
( ˆb(z)+ 4
3
â(z)
ˆ b(z) + 4
3 â(z).
â(z))τ 2 + 1
, (z, τ) ∈ Σ π
2 +η × Ση.
Remember that both â and ˆ b are analytic functions in C \ R−, since a and b are assumed
to be of type (E).
Lemma 5.3.1 There exist c > 0, η > 0 such that
��
��� 1
|k(z, τ)| ≥ c
â(z)τ 2
� �
�
�
� + 1
, (z, τ) ∈ Σ π
2 +η × Ση. (5.46)
Proof. Let ν = 1/(â(z)τ 2 ). Assume for the moment that z is fixed with arg(z) ∈ [0, π/2)
and τ ∈ (0, ∞). Then we have arg(1/ � da(z)) ∈ [0, θa] and arg(1/ � db(z)) ∈ [0, θb]. Now we
examine two cases.
Case 1: arg(1/ � da(z)) ≥ arg(1/ � db(z)). It follows that
� � �
� � �
1
1 ω
1
arg ≤ arg + ≤ arg , ∀ω ≥ 0.
�db(z) �da(z) �db(z) �da(z)
Thus we have
arg(κ) = arg
�
�da(z)
�db(z) + 4
3 � da(z)
�
⎛
= arg ⎝
1
�db(z)
1 4
�da(z)
+ 3
1
�db(z)
⎞
⎠ ≤ 0
as well as arg(1 − κ) ≥ 0. Moreover, it is easy to see that arg(1 − κ) ≤ θa and |κ| < 1.
From arg(1/ � da(z)) ∈ [0, θa] and arg(z) ∈ [0, π/2) we infer that arg(ν) ∈ [0, π/2 + θa).
Since arg(κ) ≤ 0, we have arg(κν) ≤ arg(ν). This together with |κν| < |ν| and arg(ν) ≥ 0
implies arg(κν + 1) ≤ arg(ν + 1). Therefore, arg(1 + � (κν + 1)/(ν + 1)) ≤ 0. On the
other hand we have arg(κν) ∈ [0, π/2 + θa), by definition of κ and the inequality
�
0 ≤ arg ( � db(z) + 4
3 � da(z)) −1�
� �
1
≤ arg .
�da(z)
Thus arg(1 + � (κν + 1)/(ν + 1)) ∈ (−(π/4 + θa/2), 0]. By writing
k(z, τ) = ν + 4(1 − κ(z))
89
�
1 +
� �−1 κν + 1
ν + 1