Quasilinear parabolic problems with nonlinear boundary conditions

Remarks 3.1.2 (i) It is not difficult to see that the second assertion of Theorem 3.1.2
remains true if we allow κ to lie in [1/p, 1 + 1/p).
(ii) Let κ = 0, A be invertible, and assume that the kernel a admits in addition the
estimate C ≤ s −α |â(s)|, s > 0, for some constant C > 0. Then by taking J = R+ in the
above lines and by employing this additional estimate, one obtains the implications
x ∈ DA(1 − 1
pα , p) ⇒ S(·)x ∈ Lp(R+; DA), if α > 1
p ;
y ∈ DA(1 − 1 1
α − pα , p) ⇒ (1 ∗ S)(·)y ∈ Lp(R+; DA), if α > 1 + 1
p ,
see also Prüss [64, Theorem 6].
(iii) The proof of Theorem 3.1.2 is inspired by the estimates derived in the proof of
Theorem 6 in Prüss [64].
We next want to show that these conditions on the traces are also necessary. In the
following theorem, again, the operator A is only sectorial. For technical reasons, we
have to assume maximal regularity of the solutions of (3.4) and (3.5), respectively, on
the whole halfline. Later on we shall extend this result to problems on compact timeintervals
where the operator A is R-sectorial.
Theorem 3.1.3 Let X be a Banach space of class HT , p ∈ (1, ∞), and A a sectorial
operator in X with spectral angle φA. Suppose that κ ∈ [0, 1/p), a ∈ K 1 (α, θa) with
α ∈ (1/p − κ, 2), α + κ �= 1 + 1/p. Further let ω ≥ 0 and assume that θa + φA < π. Then
and if α + κ > 1 + 1/p,
e−ω· AS(·)x ∈ Hκ p (R+; X) ⇒ x ∈ DA(1 + κ 1
α − pα , p), (3.11)
e−ω· A(1 ∗ S)(·)y ∈ Hκ p (R+; X) ⇒ y ∈ DA(1 + κ 1 1
α − α − pα , p). (3.12)
Proof. The main idea of the proof is to use Proposition 1 in [64], which says that for
every function g in Lp(R+; X), the Laplace transform ˆg(λ) exists for Re λ > 0, and with
p −1 + q −1 = 1 the estimate
� ∞
0
|ˆg(λ)| p λ p−2 dλ ≤ |g| p
Lp(R+;X) Γ(1/q)p
(3.13)
holds true.
We first show implication (3.11). Let Bκ be defined as in the proof of Theorem
3.1.2. Suppose that g(t) := Bκ(e −ωt AS(t)x), t ≥ 0, is contained in Lp(R+; X). Then
g is Laplace transformable, according to the proposition mentioned above. From the
resolvent equation for S(·), S(t)x + (a ∗ AS)(t)x = x, t ≥ 0, it follows by the convolution
theorem that
ˆg(λ) = λκ
λ + ω A(1 + âω(λ)A) −1 x, Re λ > 0.
Here, aω(t) := a(t)e −ωt , t ≥ 0. Therefore, using (3.13) we obtain
� ∞
η := ( λ1+κ− 1
p
λ + ω |A(1 + âω(λ)A) −1 p dλ
x|X) ≤ C|g|p
λ Lp(R+;X) . (3.14)
0
From Lemma 2.6.2 we know that aω ∈ K 1 (α, θa). So, in similar manner as in the proof
of Theorem 3.1.2, we can derive the following estimate from the resolvent equation for
A and parabolicity of (3.4):
|A(|λ| α + A) −1 x| ≤ (C1 + C2(|âω(λ)| |λ| α ) −1 ) |A(1/âω(λ) + A) −1 x|, Re λ > 0,
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