Quasilinear parabolic problems with nonlinear boundary conditions

for each s ∈ [0, 1]. This follows from the mixed derivative theorem. The space in (3.37)
embeds into BUC(J × R+; X) if 1/p < sα and 1/p < 2 − 2s, i.e. if 1/pα < s < 1 − 1/2p.
This shows that the compatibility condition (iv) is necessary in case α > 2/(2p − 1).
We proceed with the determination of the regularity of φ|t=0 in case α > 2/(2p − 1).
Observe that this inequality is equivalent to 1/p < α(1 − 1/2p). In light of Theorem
3.2.1, we have the embedding
B sγ
pp(J; X) ∩ Lp(J; DA(γ, p)) ↩→ C(J; DA(γ − 1
p s , p)), (3.38)
if γ, γ − 1/ps ∈ (0, 1) and sγ ∈ (1/p, 2). Thus, by taking s = α and γ = 1 − 1/2p, we see
that φ|t=0 ∈ DA(1 − 1/2p − 1/pα, p). Hence property (vii) is established.
It remains to show (vi) and (viii). Let α > 1 + 3/(2p − 1), which in particular means
that α > 1 + 1/p. Exploiting (3.37) with s = 1/α yields
which in turn entails
∂tu ∈ H α−1
p
(J; Lp(R+; X)) ∩ Lp(R; H
∂tu ∈ H (α−1)s
p
(J; H
1
2(1− α )(1−s)
p
1
2(1− α )
p
(R+; X)),
(R+; X)),
for each s ∈ [0, 1], by the mixed derivative theorem. This space is embedded into
BUC(J ×R+; X) whenever (α−1)s > 1/p and 2(1−1/α)(1−s) > 1/p, i.e. if 1/p(α−1) <
s < 1 − α/2p(α − 1). An easy calculation shows that existence of such an s is equivalent
to α > 1 + 3/(2p − 1), which we just assumed. Therefore the trace ∂tu(0, 0) exists, and
we see that the compatibility condition (vi) is necessary.
Last but not least, we restrict ∂tf|t=0 to some finite interval J1 = [0, y0], which results
in a function that belongs to
B
1 1
2(1− − α pα )
pp (J1; X) ∩ Lp(J1; DA(1 − 1 1
α − pα , p)),
owing to property (v). Then we apply again (3.38), this time with γ = 1 − 1/α − 1/pα
and s = 2. This is possible on account of α > 1 + 3/(2p − 1). We immediately see that
∂tf|t=0, y=0 enjoys the regularity claimed in (viii). Hence the necessity part of Theorem
3.5.1 as well as the additional properties (vii) and (viii) are established.
We turn now to the sufficiency part of Theorem 3.5.1. Let the data f and φ be given
such that the conditions (i)-(vi) are satisfied. We will build the solution u ∈ Z of (3.27)
as a sum of two functions in Z.
At first we will construct a function u1 ∈ Z such that
u1 − a ∗ ∂ 2 yu1 + a ∗ Au1 = f, t ∈ J, y > 0, (3.39)
i.e. u1 solves the first equation of (3.27). For this purpose we extend the function f
w.r.t. y to all of R in such a way that (i),(iii),(v) are fulfilled with R+ replaced by
R. Let Y := Lp(R; X) and Λ ∈ BIP(Y ) be the operator defined in (3.28). Then we
have f|t=0 ∈ DΛ(1 − 1/pα, p), if α > 1/p, and ∂tf|t=0 ∈ DΛ(1 − 1/α − 1/pα, p) in case
α > 1 + 1/p, owing to (3.29) and (3.30). Thus, by Theorem 3.1.4, the Volterra equation
admits a unique solution v1 in the regularity class
v + a ∗ Λv = f, t ∈ J, (3.40)
H α p (J; Lp(R; X)) ∩ Lp(J; H 2 p(R; X)) ∩ Lp(J; Lp(R; DA)).
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