Quasilinear parabolic problems with nonlinear boundary conditions

where C0 is independent of δ (u ∈ Zi+1(0)!) and ε > 0 can be chosen arbitrary small,
cf. (4.12) for the last term. Theorem 4.2.1, the results in Section 4.2.2 on pointwise
multiplication, and (4.38) allow us now to estimate
Nb �
j=0
|γyCH, j(t, x)(v − ¯v)|Yi+1 ≤ C2(Nb + 1)|γy(v − ¯v)|Yi+1 ≤ C2(Nb + 1)C3|v − ¯v| 0Z 1/2
i+1
≤ C4(C0|kα/2| L1(0,δ) + ε + Cε|k| L1(0,δ))|v − ¯v|Zi+1 , (4.39)
the constants C0, C4 being independent of δ. In view of (4.37) and (4.39), it is now
apparent that GH is a strict contraction for δ sufficiently small. Consequently, for such
δ, (4.36) possesses a unique solution v =: QH, i+1(f, h) in the space Zi+1(V −1
i (f, h)). In
other words, QH, i+1 is a left inverse for the operator
VH, i+1 := � I + k ∗ A(·, x, Dx), γyB(t, x ′ , Dx) � :
Zi+1(v i ) → Ξi+1(Viv i ) × Yi+1(γyB(t, x ′ , Dx)v i ).
Here the symbols Ξi+1(ψ) and Yi+1(ψ) have to be understood like the corresponding one
for Z defined in (4.3).
To show that VH, i+1 is a surjection, we proceed as in the proof of Theorem 4.1.1.
Define the linear operator KH, i+1 by means of
� Nb �
= k ∗
j=0
KH, i+1(g, gb) = (K (1)
H, i+1 (g, gb), K (2)
H, i+1 (g, gb))
[A#(·, x, Dx),ψj](I − S ij
H )|−1
Zi+1(ϕjV −1 hij
H, i (g,gb)) H (g, gb, QH, i+1(g, gb))
Nb+N �
+k ∗ [A#(·, x, Dx),ψj](I − S ij
j=Nb+1
λ )|−1
Zi+1(ϕjV −1
H, i (g,gb)) hij (g, gb, QH, i+1(g, gb)),
Nb �
γy (B#(t, x
j=0
′ , Dx)ψj)(I − S ij
H )|−1
Zi+1(ϕjV −1 hij
H, i (g,gb)) H (g, gb, QH, i+1(g, gb))
Observe that KH, i+1 maps pairs (g, gb) ∈ Ξi+1 × Yi+1 satisfying the compatibility condition
(iv) into 0Ξi+1 × 0Yi+1. Indeed, if α > 2/(p − 1), the function
w := (I − S ij
H )|−1
Zi+1(ϕjV −1 hij
H, i (g,gb)) H (g, gb, QH, i+1(g, gb))
has temporal trace w|t=0 = ϕjg|t=0. Since ψj ≡ 1 on an open set Vj containing supp ϕj,
it follows that B#(t, x ′ , Dx)ψj ≡ 0 on supp ϕj. Hence (K (2)
H, i+1 (g, gb))|t=0 = 0.
The commutators [A#(t, x, Dx), ψj] are differential operators of first order, while multiplying
pointwise by B#(t, x ′ , Dx)ψj can be regarded as an operator of order zero. So
we see that by choosing δ small enough, the mapping (g, gb) ↦→ (f, h) − KH, i+1(g, gb) becomes
a strict contraction in the space {(g, gb) ∈ Ξi+1×Yi+1 : ∂ m t g|t=0 = ∂ m t f|t=0, if α >
m + 1/p, m = 0, 1; gb|t=0 = h|t=0, if α > 2/(p − 1)}; thus for such δ we find a pair
(g, gb) ∈ Ξi+1 × Yi+1 satisfying
(g, gb) + KH, i+1(g, gb) = (f, h).
72
�
.