Quasilinear parabolic problems with nonlinear boundary conditions

θa + φR A < π. Then (3.1) has a unique solution in Z := Hα+κ p (J; X) ∩ Hκ p (J; DA) if and
only if the function f satisfies the subsequent conditions.
(i) f ∈ H α+κ
p (J; X);
(ii) f(0) ∈ DA(1 + κ 1
α − pα , p), if α + κ > 1/p;
(iii) f(0) ˙ ∈ DA(1 + κ 1 1
α − α − pα , p), if α + κ > 1 + 1/p.
Proof. We consider three cases with respect to the sum α + κ.
Case 1: α + κ < 1/p. Because here Hα+κ p (J; X) = 0H α+κ
p (J; X), we are in the
situation of Theorem 3.1.1.
Case 2: 1/p < α + κ < 1 + 1/p. We begin with the necessity part. Suppose
that u ∈ Z is a solution of (3.1). Then Au ∈ Hκ p (J; X) = 0H κ p (J; X), which entails
a ∗ Au ∈ 0H α+κ
p (J; X), thanks to Corollary 2.8.1. Thus f = u + a ∗ Au ∈ Hα+κ p (J; X),
i.e. condition (i) is proved.
To show the second condition we extend u in the case J = [0, T ] to a function v on
R+ such that
v ∈ ZR+ := Hα+κ p (R+; X) ∩ H κ p (R+; DA).
If J = R+, we simply set v := u. From a ∈ K 1 (α, θA), it follows by Lemma 2.6.2
that the kernel a1 defined by a1(t) = a(t)e −t , t ≥ 0, belongs to K 1 (α, θA), too. Further,
(R+; X). Therefore, g := v+a1∗Av ∈
Hα+κ p (R+; X) as well as g − g(0)e−· ∈ 0H α+κ
p (R+; X). Define now v1 by means of the
equation v1 + a1 ∗ Av1 = g − g(0)e−t , t ∈ R+. This makes sense owing to Theorem
3.1.1, which also yields v1 ∈ 0H α+κ
p (R+; X) ∩ Hκ p (R+; DA). Then v2 := v − v1 ∈ ZR+
a1 ∈ L1(R+). Thus we deduce that a1∗Av ∈ 0H α+κ
p
is the solution of the equation w + a1 ∗ Aw = g(0)e −t , t ∈ R+. Denoting the resolvent
of (3.1) by S(·) it is not difficult to see that v2(t) = e −t S(t)g(0), t ≥ 0. Consequently,
by Theorem 3.1.3, we get g(0) = v(0) = u(0) = f(0) ∈ DA(1 + κ/α − 1/pα, p). This
establishes condition (ii).
To prove the converse, suppose that the conditions (i) and (ii) are satisfied. Uniqueness
is a direct consequence of Theorem 3.1.1. Concerning existence, we construct a
solution of (3.1) in the following way. If J = [0, T ], we define u1, u2 ∈ Z as the solutions
of the problems w1 + a ∗ Aw1 = f − f(0), t ∈ J, and w2 + a ∗ Aw2 = f(0), t ∈ J,
respectively. These solutions exist and lie in Z, by virtue of Theorem 3.1.1 and Theorem
3.1.2. Thus u := u1 + u2 has the desired regularity and solves (3.1). In case
J = R+ let v1 be the solution of w1 + a ∗ Aw1 = f(0), t ∈ [0, 1]. Condition (ii) implies
v1 ∈ Hα+κ p ([0, 1]; X) ∩ Hκ p ([0, 1]; DA), by Theorem 3.1.2. We extend v1 to a function
u1 ∈ Z. Then g := u1 +a∗Au1 −f ∈ 0H α+κ
p (R+; X), and the solution u2 of the problem
w2 + a ∗ Aw2 = g, t ∈ R+, lies in Z, thanks to Theorem 3.1.1. Hence, u := u1 + u2 ∈ Z
is a solution of (3.1).
Case 3: α + κ > 1 + 1/p. We first prove the necessity part. Suppose that u ∈ Z is
a solution of (3.1). Then condition (i) can be derived as in the second case. Our next
objective is to show (iii). The idea behind the following argument is a reduction to a
situation of Case 2.
For this purpose we extend u to a function v ∈ Hα+κ p (R; X) ∩ Hκ p (R; DA). Define A
as the natural extension of A to Y := Lp(R; X) and let G := (I − D2 t ) α/2 with domain
D(G) = Hα p (R; X). Then the operators A, G are sectorial in Y with spectral angles
φA ≤ φR A and φG = 0. Thus, φA + φG < π. Furthermore the resolvents of A and G
commute, and the pair (G, A) is coercively positive. This allows us to apply the mixed
41