Quasilinear parabolic problems with nonlinear boundary conditions

In order to formulate the corresponding result for (4.30), we put s1 = α( 1 1
2 − 2p ), s2 =
1 − 1
p , as well as Y = Y1 ∩ Y2, where Y1 = B s1
pp(J; Lp(Rn )) and Y2 = Lp(J; Bs2 pp(Rn )).
Theorem 4.2.4 Let 1 < p < ∞, J = [0, T ], n ∈ N, and k ∈ K 1 (α, θ), where θ < π,
and α ∈ (0, 2) \
� 1
p , 2
p−1
, 1 + 1
p
�
. Let further a, a1, and a0 be as in Theorem 4.2.3.
Assume b ∈ Cul(J × R n ), as well as (b, b0) ∈ Y n+1 , if p > n + 1 + 2/α, and (b, b0) ∈
(C r1 (J; C(R n )) ∩ C(J; C r2 (R n ))) n+1 with some ri > si, i = 1, 2, otherwise.
Then (4.30) has a unique solution in the space Z if and only if the data f and h
satisfy the conditions (i)-(iii) and (v) stated in Theorem 4.2.2, as well as
(iv) B(0, x ′ , Dx)f|t=0, y=0 = h|t=0, if α > 2
p−1 .
We only prove Theorem 4.2.3. Theorem 4.2.4 can be established by means of the same
techniques; the proof is even much simpler, since one does not have to consider perturbations
on the boundary.
Proof of Thm. 4.2.4. We begin with the ”only if” part. Suppose v ∈ Z is a solution
of (4.30). By the regularity assumptions on A(t, x, Dx), it is evident that A(t, x, Dx)v ∈
Lp(J; Lp(R n+1
+ )), which in turn implies k ∗ A(t, x, Dx)v ∈ 0H α p (J; Lp(R n+1
+ )), according
to Corollary 2.8.1. Thus f ∈ Hα p (J; Lp(R n+1
+ )). Since ∂j t v|t=0 = ∂ j
t f|t=0 in case α >
j+1/p, j = 0, 1, Theorem 4.2.1 shows (iii) and (v). Another consequence of that theorem
is that v|y=0 ∈ Y . Also, ∇xv|y=0 ∈ Y n+1 , by Theorem 4.2.2. Since p > n + 1 + 2/α
entails Y ↩→ C(J × Rn ), the coefficients b and b0 exhibit just the regularity needed for
the application of Lemma 4.2.1 and Lemma 4.2.2, respectively. Hence condition (ii) is
necessary. Last but not least, the compatibility condition (iv) follows from the regularity
assumptions on b, b0, and the embedding Y ↩→ C(J; Lp(Rn )), which is valid whenever
α > 2/(p − 1), cp. the proof of Theorem 3.5.2 and that of Theorem 4.2.2.
We come now to the sufficiency part. Although problem (4.30) is far more complicated
than (4.1), the strategy we are going to follow is basically the same as in the proof
of Theorem 4.1.1. Unless stated otherwise and apart from trivial modifications, the
notation used here is adopted from that proof. Note that we decompose the boundary
differential operator as B(t, x ′ , Dx) = B#(t, x ′ , Dx) + b0(t, x ′ ).
Given η > 0, the assumptions on a and b permit us to choose a large ball Br0 (0) ⊂
Rn+1 such that
|a(t, x) − a(t, ∞)| B(R (n+1) 2 η
≤ , for all t ∈ J, x ∈ Rn+1
) +
2 , |x| ≥ r0,
as well as
|b(t, x ′ ) − b(t, ∞)| B(R n ,R) ≤ η
2 , for all t ∈ J, x′ ∈ R n , |x ′ | ≥ r0.
Put U0 = {x ∈ Rn+1 : |x| > r0}. We can cover Br0 (0) ∩ Rn+1 + by finitely many balls
Uj = Brj (xj), j = 1, . . . , Nb + N, and select a partition 0 =: T0 < T1 < . . . < TM−1 <
TM := T such that for all i = 0, . . . , M − 1, the following conditions are fulfilled:
∃x ′ j ∈ R n : Brj (xj) = Brj ((x′ j, 0)), 1 ≤ j ≤ Nb;
Brj (xj) ∩ {(x ′ , y) ∈ R n+1 : y = 0} = ∅, Nb + 1 ≤ j ≤ Nb + N;
|a(t, x)−a(Ti, xj)| B(R (n+1) 2 ) ≤η, t ∈ [Ti, Ti+1], x ∈ R n+1
+ ∩Brj (xj), 1≤j ≤Nb+N;
|b(t, x ′ )−b(Ti, x ′ j)| B(R n ,R) ≤η, t ∈ [Ti, Ti+1], (x ′ , 0) ∈ Brj (xj), 1≤j ≤Nb;
|a(t, ∞)−a(Ti, ∞)| B(R (n+1) 2 ) , |b(t, ∞)−b(Ti, ∞)| B(R n ,R) ≤ η
2 , t ∈ [Ti, Ti+1].
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