Quasilinear parabolic problems with nonlinear boundary conditions

Then (4.18) has a unique solution in the space Z if and only if the data f and g are
subject to the subsequent conditions.
(i) f ∈ H α p (J; Lp(R n+1
+ ));
(ii) g ∈ B
1
α(1− 2p )
pp
2
2− pα
pp
(J; Lp(R n )) ∩ Lp(J; B
(iii) f|t=0 ∈ B (R n+1
+ ), if α > 1
p ;
(iv) f|t=0, y=0 = g|t=0, if α > 2
2p−1 ;
1 1
2(1− − α pα )
pp
1
2− p
pp (R n ));
(v) ∂tf|t=0 ∈ B (R n+1
+ ), if α > 1 + 1
p ;
(vi) ∂tf|t=0, y=0 = ∂tg|t=0, if α > 1 + 3
2p−1 .
Proof. By means of a variable transformation of the form ¯x = QT ΛQx, where Q is a
rotation matrix and Λ is diagonal with Λii = 1/ √ λi (λ1, . . . , λn+1 denoting the positive
eigenvalues of the matrix a), problem (4.18) can be reduced to a problem of the same
structure but with a = In+1. The assertion follows then from Theorem 3.5.1 applied to
X = Lp(Rn ) and the operator A = −∆x ′, which belongs to BIP(X) and has power angle
0, and from the fact that all function spaces occurring in that theorem are preserved
under the above variable transformation. �
The corresponding result for (4.19) reads
Theorem 4.2.2 Let 1 < p < ∞, n ∈ N, a ∈ Sym{n + 1}, and b ∈ R n . Suppose
k ∈ K 1 (α, θ), where θ < π and α ∈ (0, 2) \
� 1
p , 2
p−1
, 1 + 1
p
�
. Assume that a is positive
definite.
Then (4.19) possesses a unique solution in the space Z if and only if the functions
f and h satisfy the following conditions.
(i) f ∈ H α p (J; Lp(R n+1
+ ));
(ii) h ∈ B
1 1
α( − 2 2p )
pp
2
2− pα
pp
(J; Lp(R n )) ∩ Lp(J; B
1
1− p
pp (R n ));
(iii) f|t=0 ∈ B (R n+1
+ ), if α > 1
p ;
(iv) −∂yf|t=0, y=0 + b · ∇x ′f|t=0, y=0 = h|t=0, if α > 2
p−1 ;
(v) ∂tf|t=0 ∈ B
1 1
2(1− − α pα )
pp
(R n+1
+ ), if α > 1 + 1
p .
Proof. Use the variable transformation described in the proof of Theorem 4.2.1 and
normalize the coefficient in front of the normal derivative to reduce (4.19) to a problem
of the same structure with a = In+1. The assertion is then a consequence of Theorem
3.5.1 applied to X = Lp(R n ), A = −∆x ′, and D = b · ∇x ′ (b ∈ Rn ). Note that
D ∈ BIP(R(D)) with power angle θD ≤ π/2 (cp. Prüss [65, Section 3]) and thus
θD < π − θ/2 = π − max{θ, θA}/2 showing the second angle condition in Theorem 3.5.1.
�
4.2.2 Pointwise multiplication
Let 1 < p < ∞, s1, s2 ∈ (0, 1), 0 < T2, 0 ≤ T1 ≤ T2, and Ω be an arbitrary domain in
R n . We are interested in pointwise multipliers for the intersection space
Y T2 := Y T2
1
∩ Y T2
2 := B s1
pp([0; T2]; Lp(Ω)) ∩ Lp([0, T2]; B s2
pp(Ω))
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