Quasilinear parabolic problems with nonlinear boundary conditions

3.2 A general trace theorem
Let X be a Banach space of class HT , p ∈ (1, ∞), J = [0, T ] or R+, and A be an
R-sectorial operator in X with arbitrary R-angle φR A < π. Further suppose γ ∈ [0, 1/p),
s > 1/p − γ, and
u ∈ Z := H s+γ
p (J; X) ∩ H γ p (J; DA). (3.17)
Our first aim is to prove that u(0) ∈ DA(1 + γ/s − 1/ps, p). To do so, note that, by the
mixed derivative theorem, we have the embedding
H s+γ
p (J; X) ∩ H γ p (J; DA) ↩→ H (1−θ)s+γ
p (J; DAθ), θ ∈ (0, 1). (3.18)
We now distinguish two cases.
Case 1: 1/p − γ < s < 2(1/p − γ). By hypothesis and (3.18), we see that
u ∈ H s+γ
p (J; X) ∩ H (1−θ)s+γ
p (J; DAθ), θ ∈ (0, 1).
If we put κ = (1 − θ)s + γ, B = Aθ , and a(t) = e−ttθs−1 , t > 0, then we are in the
situation of Theorem 3.1.4, provided that κ < 1/p and θsπ/2 + φR B < π. But these
two conditions are fulfilled with θ = 1/2. In fact, in case of θ = 1/2, we estimate
κ = s/2 + γ < (1/p − γ) + γ = 1/p, as well as
θs π
2 + φR sπ 1
B ≤ 4 + 2 φR π π
A < 2 + 2
= π,
where the inequality φR B ≤ φR A /2 follows from Proposition 2.2.1. Further,
1 + κ 1 2γ 2
θs − θsp = 2 + s − ps ,
and so, by Theorem 3.1.4 and Theorem 2.2.2, we obtain
u(0) ∈ DB(2 + 2γ
s
2 − ps , p) = D A 1 2γ 2
(2 +
2 s − ps , p) = DA(1 + γ 1
s − ps , p).
Case 2: s ≥ 2(1/p − γ). Here, we look at the regularity with respect to the operator
A. By hypothesis and (3.18),
u ∈ H (1−θ)s+γ
p (J; D A θ) ∩ H γ p (J; DA), θ ∈ (0, 1).
Therefore, if we choose the base space Xθ = D(A θ ), then we get
u ∈ H (1−θ)s+γ
p (J; Xθ) ∩ H γ p (J; D A 1−θ), θ ∈ (0, 1).
Again, we want to apply Theorem 3.1.4. So we have to ensure that 1/p < (1 − θ)s + γ
and (1 − θ)sπ/2 + (1 − θ)φR A < π. But these conditions are satisfied for
θ = 1 − η
� �
1
s p − γ
with an arbitrary η ∈ (1, 2/(1 + 1/p − γ)]. We namely have in this case
� �
1
(1 − θ)s + γ = η p − γ + γ > 1
p
as well as
(1 − θ)s π
2 + (1 − θ)φR A < η
� �
1
p − γ · π
� � � �
η 1
1
2 + s p − γ π ≤ η p − γ · π η
2 + 2 π ≤ π,
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