Quasilinear parabolic problems with nonlinear boundary conditions

derivative theorem to this pair of operators acting on Y to obtain
1−κ
1−
|A α G 1−κ
α y|Y ≤ C|Ay + Gy|Y , for all y ∈ D(A) ∩ D(G), (3.15)
where C > 0 is a constant not depending on y. By definition, D(G 1−κ
α ) = H1−κ p (R; X).
This together with (3.15) implies
H α p (R; X) ∩ Lp(R; DA) ↩→ H 1−κ
p (R; D 1−
A 1−κ
α ).
1
− Thus, by the boundedness of DtG α in Lp(R; D 1−
A 1−κ
α ),
1
−
˙v = DtG α G 1−κ
α G κ
α v ∈ H α+κ−1
p
(R; X) ∩ Lp(R; D A 1− 1−κ
To determine the regularity of ˙v(0) we now use the necessity part of the second case.
Let b(t) = t α+κ−2 , t ≥ 0. Then b ∈ K 1 (α + κ − 1, (α + κ − 1)π/2). Since A is R-sectorial,
the operator C := A 1−(1−κ)/α is R-sectorial as well, and
φ R C
≤ (1 − 1−κ
α )φR A
≤ ( α+κ−1
α
)(π − α π
2
) = ( α+κ−1
α
α ).
)π − ( α+κ−1
2 )π,
by Proposition 2.2.1 and the parabolicity condition, combined with Remark 2.6.1. So
we see that θb + φR C < π. By the necessity part of Case 2, applied to the equation
˙v(t) + (b ∗ C ˙v)(t) = g(t), t ∈ [0, 1],
with some g ∈ Hα+κ−1 p ([0, 1]; X), we get ˙ f(0) = ˙v(0) ∈ DC(1 − 1/p(α + κ − 1), p). It
follows now, by Theorem 2.2.2, that ˙ f(0) lies in
D A 1− 1−κ
α
�
�
1 1 − p(α+κ−1) , p
�
= DA (1 − 1−κ
α )(1 − 1
�
= DA 1 + κ
�
1 1
α − α − pα , p .
p(α+κ−1)
�
), p
Hence condition (iii) is satisfied.
It remains to show (ii). Put J1 := [0, 1] in case J = R+ and J1 := J, otherwise.
Define w1 by means of
w1(t) + (a ∗ Aw1)(t) = t ˙
f(0), t ∈ J1.
Then, due to Theorem 3.1.2, it follows from condition (iii) that u|J1−w1 ∈ Hα+κ p (J1; X)∩
Hκ p (J1; DA). We extend u|J1 − w1 to a function v ∈ ZR+ and put h(t) := v(t) + (a1 ∗
Av)(t), t ∈ R+, where a1 is defined as in Case 2. As above, we see that h ∈ Hα+κ p (R+; X).
By construction, we have ˙v(0) = ˙ h(0) = 0, i.e. h−ψ(·)h(0) ∈ 0H α p (R+; X), where ψ(t) =
(1 + t)e−t , t ≥ 0. In fact, ψ ∈ Hα+κ p (R+), ψ(0) = 1, ψ(t) ˙ = −te−t , t ≥ 0, in particular
˙ψ(0) = 0. Define now the function v1 by means of v1+a1∗Av1 = h−ψh(0), t ∈ R+. This
is possible in view of Theorem 3.1.1, which also gives v1 ∈ 0H α+κ
p (R+; X)∩H κ p (R+; DA).
Consequently, v2 := v−v1 ∈ ZR+ , and v2 solves the equation w+a1∗Aw = ψh(0), t ∈ R+.
If S(·) denotes the resolvent for (3.1), then one verifies that
v2(t) = e −t S(t)h(0) + e −t (1 ∗ S)(t)h(0), t ≥ 0.
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