Quasilinear parabolic problems with nonlinear boundary conditions

Proof. We begin with the necessity part. Suppose we have a function u ∈ Z which is a
solution of (3.45). Then u is also a solution of the problem
� u − a ∗ ∂ 2 yu + a ∗ Au = f, t ∈ J, y > 0,
u(t, 0) = ϕ(t), t ∈ J,
where ϕ := u|y=0. Therefore, conditions (i),(iii) and (v) are necessary, by the first part of
the proof of Theorem 3.5.1, where the case α = 2/(2p − 1) is admissible, too. Moreover,
we see that f|t=0, y=0 ∈ DA(1 − 1/2p − 1/pα, p), if α > 2/(2p − 1), i.e. property (vi) is
fulfilled.
To show condition (ii), we proceed similarly as in the proof of Theorem 3.5.1. We
extend u w.r.t. t to all of R such that u ∈ Z1 (see (3.31) for the definition), u|y=0 ∈
Lp(R; DD), and
Du(·, 0) ∈ B
1 1
α( − 2 2p )
pp (R; X) ∩ Lp(R; DA( 1 1
2 − 2p , p)).
Let Y := Lp(R; X) and define D as the natural extension of D to this space. Let F
be as in (3.32). Then the space DF (1 − 1/p, p) is given by (3.33). Thus we see that
u ∈ H 2 p(R+, Y ) ∩ Lp(R+; D F 2), u|y=0 ∈ D(D), and Du|y=0 ∈ DF (1 − 1/p, p). Now
consider u as the solution of the problem
−u ′′ (y) + F 2 u(y) = g, y > 0, −u ′ (0) = ¯ φ − Du(0),
with some g ∈ Lp(R+; Y ) and an extension ¯ φ of φ on the real line. Since θF ≤
max{θa, θA}/2 < π, it then follows from Theorem 3.4.3 that ¯ φ−Du(0) ∈ DF (1−1/p, p),
i.e. ¯ φ ∈ DF (1 − 1/p, p), which after restriction to t ∈ J yields condition (ii).
Finally, we have to prove (iv) and (vii). By the mixed derivative theorem, it follows
from u ∈ Z that
and further
∂yu ∈ H α
2
p (J; Lp(R+; X)) ∩ Lp(J; H 1 p(R+; X)),
∂yu ∈ H αs
2
p (J; H1−s p (R+; X)),
for all s ∈ [0, 1]. This space embeds into BUC(J ×R+; X) if αs/2 > 1/p and 1−s > 1/p,
i.e. in case 2/pα < s < 1 − 1/p. Such an s exists if and only if α > 2/(p − 1) as a short
computation shows. Thus in this situation the trace (∂yu)(0, 0) ∈ X exists. Further,
and
B
1 1
α( − 2 2p )
pp
Du|y=0, φ ∈ B
(J; X) ∩ Lp(J; DA( 1
1 1
α( − 2 2p )
pp (J; X) ∩ Lp(J; DA( 1 1
2 − 2p , p))
2
1
1 1 1
− 2p , p)) ↩→ C(J; DA( 2 − 2p − pα , p)),
thanks to Theorem 3.2.1 and the remark before Theorem 3.5.2. Therefore, by the closedness
of D, f|t=0, y=0 ∈ D(D) and
−∂yf|t=0, y=0 + Df|t=0, y=0 = φ|t=0,
where each term in this equation lies in the space DA(1/2 − 1/2p − 1/pα, p). Hence,
conditions (iv) and (vii) are proved. This completes the necessity part of the proof.
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