Quasilinear parabolic problems with nonlinear boundary conditions

with ψ1 ∈ Y, ψ2 ∈ Yκ and
where
Thus
ψ2 − γp1 ∈ 0Yκ , 1
2 ψ1 + 1
3 ψ2 + γ∂yw1 ∈ 0Y,
0Yκ := 0B
δb2 (1− 1
p )+κ
pp
δb2 (1−
q0 ∈ 0B
1
p )
pp
(J; Lp(R n )) ∩ H κ p (J; B
(J; Lp(R n )) ∩ Lp(J; B
1
1− p
pp (R n )).
1
1− p
pp (R n )),
which entails K(a∗q0) ∈ 0Yκ. In view of L −1 ∈ B(0Yκ) and 2K(a∗q0) = Lp0, it therefore
follows that p0 ∈ 0Yκ. Observe now that
p0 ∈ 0Yκ ⇔ e −Gy δb2 +κ
p0 ∈ 0Hp
(J; Lp(R n+1
+ )) ∩ Hκ p (J; H 1 p(R n+1
+ )).
Concerning p1, we proceed as in the case δa ≤ δb. According to Theorem 3.1.4, it
follows from (5.13),(5.53),(5.54) that (5.27) admits a unique solution φp in the space
which is embedded into
H δa
p (J; H −1
p (R n+1 )) ∩ H κ p (J; H 1 p(R n+1 )),
δb2 +κ
Hp
(J; Lp(R n+1
+ )) ∩ Hκ p (J; H 1 p(R n+1
+ )),
by the mixed derivative theorem. Consequently, due to (5.30),
as well as γp ∈ Yκ. From
we then deduce
δb2 +κ
p ∈ Hp
(J; Lp(R n+1
+ )) ∩ Hκ p (J; H 1 p(R n+1
+ )),
δb
2
δa (1− 2
Yκ ↩→ B
1
p )
pp
(1 − 1
p
δa 1
) + κ > 2 (1 − p )
(J; Lp(R n )) ∩ Lp(J; B
1
1− p
pp (R n )).
Hence, p and γp lie in the right regularity classes when (5.21) is solved for (v, w). Using
this fact, together with (5.13),(5.14),(5.15),(5.16), and (5.17), Theorem 3.5.2 yields
(v, w) ∈ Z.
Theorem 5.3.2 Let 1 < p < ∞, and suppose that the kernels a �= 0 and b are of
type (E). Let δa and δb denote the regularization order of a and b, respectively, and
assume that 0 < κ := δa − δb < 1/p. Suppose further that δa /∈ { 2 1
p−1 , 1 + p } as well as
p(2δa − δb) �= 2 + δb. Then (5.12) has a unique solution (v, w) ∈ Z satisfying (5.49) if
and only if the data are subject to the conditions (N2).
5.3.4 The case δa − δb > 1/p
Let κ > 1/p. Suppose that (v, w) ∈ Z solves (5.12) and satisfies in addition (5.49). Then
the latter implies
in particular
p ∈ C(J; H 1 p(R n+1
+ )),
p|t=0 = −∇x · v0 − ∂yw0 ∈ H 1 p(R n+1
+ ), (5.55)
92