Quasilinear parabolic problems with nonlinear boundary conditions

3.3 More time regularity for Volterra equations
This section deals with the question under what conditions the solution of equation (3.1)
lies in the space
H α+κ
p (J; X) ∩ H κ p (J; DA),
where, in contrast to Section 3.1, we assume κ ∈ (1/p, 1 + 1/p) and α + κ < 2 + 1/p.
Theorem 3.3.1 Let X be a Banach space of class HT , p ∈ (1, ∞), J a compact timeinterval
[0, T ], 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). Further let κ ∈ (1/p, 1 + 1/p), α + κ < 2 + 1/p and
α + κ �= 1 + 1/p. Assume the parabolicity condition 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
(i) f(0) ∈ D(A);
(ii) f = h + (1 ∗ a)Af(0), with h ∈ H α+κ
p
case α + κ > 1 + 1/p.
(J; X), and ˙ h(0) ∈ DA(1 + κ 1 1
α − α − pα , p) in
Proof. We first show the necessity part. Suppose that u ∈ Z solves (3.1). Then, κ > 1/p
entails α + κ > 1/p. Thus, the trace x := u(0) ∈ X exists. Furthermore we see that
Au ∈ C(J; X). So, by the closedness of A, x ∈ D(A) and we infer from (3.1) that
u(t) + (a ∗ A(u − x))(t) = f(t) − (1 ∗ a)(t)Ax, t ∈ J.
From A(u − x) ∈ 0H κ p (J; X), it follows that a ∗ A(u − x) ∈ 0H α+κ
p (J; X), in view of
Corollary 2.8.1. In addition, 1 ∗ a is absolutely continuous on J and vanishes at t = 0.
Hence, f(0) = x as well as f = h+(1∗a)Af(0), where h is defined by h = u+a∗A(u−x) ∈
Hα+κ p (J; X). This proves (i) and the first part of (ii).
To verify the second condition in (ii), suppose that α + κ > 1 + 1/p. Owing to
a ∗ A(u − x) ∈ 0H α+κ
p (J; X), we have ˙ h(0) = ˙u(0). We now consider two cases. In
case of κ ∈ (1/p, 1), we can argue as in the proof of Theorem 3.1.4, Case 3, to see that
˙u(0) ∈ DA(1 + κ/α − 1/α − 1/pα, p). If κ ∈ [1, 1 + 1/p), then
˙u ∈ H α+κ−1
p
(J; X) ∩ H κ−1
p (J; DA),
and we obtain the desired regularity of ˙u(0) with the aid of Theorem 3.1.4, applied to
the equation
˙u(t) + (a ∗ A ˙u)(t) = g(t), t ∈ J,
with some g ∈ Hα+κ−1 p (J; X). Hence, the necessity part is complete.
We now want to prove the converse. Uniqueness follows immediately from Theorem
3.1.1. Concerning sufficiency, we suppose validity of (i) and (ii) and distinguish two
cases.
If α+κ < 1+1/p, we set u = x+u1, where u1 is defined as solution of v+a∗Av = h−x
on J. Condition (i) ensures that the constant function w(t) = x, t ∈ J lies in Z. From
the second condition, we deduce h(0) = f(0) = x and h − x ∈ 0H α+κ
p (J; X). So, due to
Remark 3.1.1(ii), we obtain u1 ∈ 0H α+κ
p (J; X) ∩ 0H κ p (J; DA). Thus u ∈ Z. Further,
u + a ∗ Au = (x + (1 ∗ a)Ax) + (h − x) = f,
by condition (ii). Hence, u ∈ Z is indeed the solution of (3.1).
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