Quasilinear parabolic problems with nonlinear boundary conditions

with two positive constants C1 and C2. Employing this estimate as well as the inequality
λ/(λ + ω) ≥ C0 > 0, λ ≥ 1, we deduce that
η ≥ C p
� ∞
0 (
1
λκ− 1
p
|âω(λ)| |A(1/âω(λ) + A) −1 p dλ
x|)
λ
≥ C p
� 1
∞ α+κ−
λ p
0 (
C1|âω(λ)|λα |A(λ
+ C2
α + A) −1 p dλ
x|)
λ .
1
We now exploit the assumption lim sup µ→∞ |â(µ)| µ α < ∞ to get an upper bound
|âω(λ)| λα ≤ C3 < ∞, 1 ≤ λ < ∞, and thus arrive at
C|g| p
Lp(R+;X) ≥
� ∞
1
α+κ−
(λ p |A(λ
1
α + A) −1 p dλ
x|)
λ
� ∞ κ 1
1+ −
= (r α pα |A(r + A) −1 p dr
x|X)
rα ,
1
i.e. x ∈ DA(1 + κ/α − 1/pα, p).
The proof of (3.12) is similar. Suppose g(t) := Bκ(e −ωt A(1 ∗ S)(t)x), t ≥ 0, lies in
Lp(R+; X). Then g is Laplace transformable, on account of the proposition mentioned
at the beginning of the proof. Integrating the resolvent equation for S(·) yields with
S1 := 1 ∗ S the relation S1(t)x + (a ∗ AS1)(t)x = tx, t ≥ 0. Thus, by the convolution
theorem, we obtain that
ˆg(λ) =
Using (3.13) then yields
� ∞
( λ1+κ− 1
1
λ κ
(λ + ω) 2 A(1 + âω(λ)A) −1 x, Re λ > 0.
p
(λ + ω) 2 |A(1 + âω(λ)A) −1 x|X)
p dλ
λ
≤ C|g|p
Lp(R+;X) ,
in consequence of which we arrive at
� ∞
1
κ−1−
(λ p |A(1 + âω(λ)A) −1 p dλ
x|X)
λ ≤ ˜ C|g| p
Lp(R+;X) ,
1
thanks to the inequality λ/(λ + ω) ≥ C0 > 0, λ ≥ 1. By the same line of conclusions as
in the first part of the proof we then obtain
� ∞
(r
1
Hence x ∈ DA(1 + κ/α − 1/α − 1/pα, p). �
κ 1 1
1+ − − α α pα |A(r + A) −1 p dr
x|X)
rα ≤ ˜ C|g| p
Lp(R+;X) .
We have now got all important ingredients of the main theorem concerning (3.1) which
reads as follows.
Theorem 3.1.4 Let X be a Banach space of class HT , p ∈ (1, ∞), J a compact timeinterval
[0, T ] or R+, and A an R-sectorial operator in X with R-angle φR A . Suppose that
a belongs to K1 (α, θa) with α ∈ (0, 2) and that in addition a ∈ L1(R+) in case J = R+.
Further let κ ∈ [0, 1/p) and α + κ /∈ {1/p, 1 + 1/p}. Assume the parabolicity condition
40