Quasilinear parabolic problems with nonlinear boundary conditions

Theorem 3.5.1 Let p ∈ (1, ∞) and X be a Banach space of class HT . Suppose that
a ∈ K1 �
1
(α, θa) with α ∈ (0, 2) \ p , �
2 1 3
2p−1 , 1 + p , 1 + 2p−1 and A ∈ BIP(X) with power
angle θA. Assume further θa + θA < π. Then the problem (3.27) has a unique solution
in Z if and only if the data f and φ satisfy the following conditions.
(i) f ∈ H α p (J; Lp(R+; X));
(ii) φ ∈ B
1
α(1− 2p )
pp
2
2− pα
pp
(J; X) ∩ Lp(J; DA(1 − 1 , p));
(iii) f|t=0 ∈ B (R+; X) ∩ Lp(R+; DA(1 − 1
1
pα , p)), if α > p ;
(iv) f|t=0, y=0 = φ|t=0, if α > 2
2p−1 ;
(v) ∂tf|t=0 ∈ B
1 1
2(1− − α pα )
pp
2p
(R+; X) ∩ Lp(R+; DA(1 − 1
(vi) ∂tf|t=0, y=0 = ˙ φ|t=0, if α > 1 + 3
2p−1 .
If this is the case, then additionally
α
1
1
− pα , p)), if α > 1 + p ;
(vii) f|t=0, y=0, φ|t=0 ∈ DA(1 − 1 1
2p − pα , p), if α > 2
2p−1 ;
(viii) ∂tf|t=0, y=0, φ|t=0
˙ ∈ DA(1 − 1 1 1
3
2p − α − pα , p), if α > 1 + 2p−1 .
Proof. We begin with the necessity part. Suppose that u ∈ Z is a solution of (3.27).
Then clearly f = u − a ∗ ∂ 2 yu + a ∗ Au ∈ H α p (J; Lp(R+; X)), i.e. the first condition is
satisfied.
Next we extend u w.r.t. y to all of R by u(t, y) = 3u(t, −y) − 2u(t, −2y), y < 0,
resulting in a function (again denoted by u) that belongs to
H α p (J; Lp(R; X)) ∩ Lp(J; H 2 p(R; X)) ∩ Lp(J; Lp(R; DA)).
Define A as the natural extension of A to Y := Lp(R; X) and let G = −∂ 2 y with domain
D(G) = H 2 p(R; X). Then G is sectorial and belongs to the class BIP(Y ) with power
angle θG = 0. Since both operators commute in the resolvent sense, Theorem 2.3.1 yields
that
Λ := A + G (3.28)
with domain D(Λ) = D(A) ∩ D(G) is sectorial and belongs to BIP(Y ) with power angle
θΛ ≤ θA, in particular Λ ∈ RS(Y ) with R-angle φ R Λ ≤ θA, by Theorem 2.2.3. It now
follows from Theorem 3.1.4 that u|t=0 ∈ DΛ(1 − 1/pα, p), if α > 1/p. Proposition 2.3.1
gives
DΛ(1 − 1
pα , p) = DG(1 − 1
pα , p) ∩ DA(1 − 1
pα , p)
= B
2
2− pα
pp
(R; X) ∩ Lp(R; DA(1 − 1
pα , p)), (3.29)
which entails the third condition, by restriction to y ∈ R+. If α > 1+1/p, then Theorem
3.1.4 further yields ∂tu|t=0 ∈ DΛ(1 − 1/α − 1/pα, p). By employing Proposition 2.3.1
once more, we get
DΛ(1 − 1 1
α − pα
1
α
, p) = B2(1−
pp
Thus, after restriction to y ∈ R+ we arrive at condition (v).
1
− pα )
(R; X) ∩ Lp(R; DA(1 − 1 1
α − pα , p)). (3.30)
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