Quasilinear parabolic problems with nonlinear boundary conditions

Here the operator D is pseudo-sectorial in X, and we assume that D A 1/2 ↩→ DD. As
above we seek a solution in
Z = H α p (J; Lp(R+; X)) ∩ Lp(J; H 2 p(R+; X)) ∩ Lp(J; Lp(R+; DA)).
Before we state the theorem concerning (3.45) we note that any function v in the space
Z automatically satisfies v|y=0 ∈ Lp(J; DD) and
Dv(·, 0) ∈ B
1 1
α( − 2 2p )
pp
(J; X) ∩ Lp(J; DA( 1
In fact, if v ∈ Z, then it follows, by Theorem 3.5.1, that
v|y=0 ∈ Lp(J; DA(1 − 1
2p , p)),
which entails the first claim, for we have the embeddings
DA(1 − 1
2p , p) ↩→ D A 1 2
2
↩→ DD.
1 − 2p , p)). (3.46)
As for (3.46), we see with the aid of the mixed derivative theorem (Proposition 2.3.2)
that for w := A 1/2 v,
w ∈ H α
2
p (J; Lp(R+; X)) ∩ Lp(J; H 1 p(R+; X)) ∩ Lp(J; Lp(R+; D A 1 2 )). (3.47)
We extend w w.r.t. t to all of R such that (3.47) holds true, with J replaced by R.
As above, set Y = Lp(R; X) and define the operator F by (3.32). Then we know that
F is invertible and lies in the class BIP(Y ) with power angle θF < π/2. Further,
w ∈ H 1 p(R+; Y ) ∩ Lp(R+; DF ). Thus, Theorem 3.4.1 yields w|y=0 ∈ DF (1 − 1/p, p),
which by restriction to t ∈ J implies
A 1
2 v(·, 0) ∈ B
1 1
α( − 2 2p )
pp (J; X) ∩ Lp(J; DA( 1 1
2 − 2p , p)),
in view of (3.33). Assertion (3.46) now follows on account of D A 1/2 ↩→ DD.
Theorem 3.5.2 Let p ∈ (1, ∞), X be a Banach space of class HT , a ∈ K 1 (α, θa) with
α ∈ (0, 2) \
� 1
p , 2
p−1
, 1 + 1
p
�
, and A ∈ BIP(X) with power angle θA. Let further D be a
pseudo-sectorial operator in X such that D ∈ BIP(R(D)) with power angle θD. Suppose
that A and D commute in the resolvent sense, and that D A 1/2 ↩→ DD. Assume further
the angle conditions θa + θA < π and θD < π − max{θa, θA}/2. Then the problem (3.45)
has a unique solution in Z if and only if the data f and φ are subject to the following
conditions.
(i) f ∈ H α p (J; Lp(R+; X));
(ii) φ ∈ B
1 1
α( − 2 2p )
pp (J; X) ∩ Lp(J; DA( 1 1
2 − 2p , p));
2
2− pα
pp
(iii) f|t=0 ∈ B (R+; X) ∩ Lp(R+; DA(1 − 1
1
pα , p)), if α > p ;
(iv) f|t=0, y=0 ∈ D(D) and − ∂yf|t=0, y=0 + Df|t=0, y=0 = φ|t=0, if α > 2
p−1 ;
(v) ∂tf|t=0 ∈ B
1 1
2(1− − α pα )
pp
If this is the case, then additionally
(R+; X) ∩ Lp(R+; DA(1 − 1
α
1
1
− pα , p)), if α > 1 + p .
(vi) f|t=0, y=0 ∈ DA(1 − 1 1
2p − pα , p), if α > 2
2p−1 ;
(vii) Df|t=0, y=0, φ|t=0 ∈ DA( 1 1 1
2 − 2p − pα , p), if α > 2
p−1 .
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