Quasilinear parabolic problems with nonlinear boundary conditions

and
Gκ(λ) = λ κ A(1 + â(λ)A) −1 , Reλ > 0, (3.9)
we then obtain
(tBκAS)ˆ(λ) = φ(λ)
λ2 Gκ(λ), Reλ > 0.
Inversion of the Laplace transform now yields for t > 0
tBκAS(t)x = 1
�
e
2πi Γ
λt =
(tBκAS)ˆ(λ)x dλ
1
� ∞
e
2π
(γ+iρ)t dρ
φ(γ + iρ) Gκ(γ + iρ)x
(γ + iρ) 2
= 1
2π
−∞
� ∞
−∞
e γt+iσ φ(γ + i σ
t ) Gκ(γ + i σ
t )x
t dσ
,
(γt + iσ) 2
where we used the change of variables σ = tρ. By 1-regularity of a and parabolicity of
(3.4) we get a bound |φ(λ)| ≤ C, for all Re λ > 0. Using this estimate and choosing
γ = 1
t we obtain
� ∞
dσ
|BκAS(t)x|X ≤ C |Gκ((1 + iρ)/t)x|X , t > 0.
1 + σ2 −∞
Taking the Lp-norm on the interval J = [0, T ] and applying the continuous version of
Minkowski’s inequality yields
� ∞ � T
|BκAS(·)x| Lp(J;X) ≤ C ( |Gκ((1 + iρ)/t)x| p dσ
X
dt)1/p .
1 + σ2 −∞
0
Now we employ the change of variables s = √ 1 + σ2 /t for the inner integral and enlarge
its interval of integration to get
� ∞ � ∞ 2
−
|BκAS(·)x| Lp(J;X) ≤ C ( (s p |Gκ( (1+iσ)s
√
1+σ2 )x|X) p ds) 1/p dσ
−∞
≤ C sup {(
σ∈R
1
T
� ∞
(s
1
T
− 1
√ 1+σ 2
p |Gκ( (1+iσ)s
(1 + σ2 1
1−
) 2p
p ds
)x|X)
s )1/p }.
This shows that BκAS(·)x ∈ Lp(J; X) whenever
� ∞ 1
κ−
η := sup {( (s p |â(
σ∈R
(1+iσ)s
√
1+σ2 )|−1 |A(1/â( (1+iσ)s
√
1+σ2 ) + A)−1 p ds
x|X)
s )1/p } < ∞. (3.10)
1
T
Now we have by the resolvent equation
� �−1 1
A + A
â(λ)
= A(|λ| α + A) −1 +
for Re λ > 0, thus using the parabolicity of (3.4)
|A(1/â(λ) + A) −1 x| ≤ |A(|λ| α + A) −1 x| +
�
+ |λ| α − 1
� � �−1 1
+ A A(|λ|
â(λ) â(λ) α + A) −1 ,
+| |λ| α â(λ) − 1| |(1 + â(λ)A) −1 | |A(|λ| α + A) −1 x|
≤ (C1 + C2|â(λ)| |λ| α ) |A(|λ| α + A) −1 x|,
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