Quasilinear parabolic problems with nonlinear boundary conditions

elongs to H∞ (Σ π
2 +η × Ση). Hence the associated operator is bounded in Lp(R+ × Rn ),
by the joint H∞ (Σ π
2 +η × Ση) - calculus of the pair (∂t, Dn). By causality, extension
and restriction, it is then clear that L−1 ∈ B(Lp(J × Rn )). In view of [F −1 , L−1 ] = 0
we also have L−1 ∈ B(D(F )), where D(F ) = 0H δa/2
p (J; Lp(Rn )) ∩ Lp(J; H1 p(Rn )). Since
0Y = (Lp(J; Lp(Rn )), D(F )) 1−1/p, p, by real interpolation it follows that L−1 ∈ B(0Y ).
In sum we have proved
Theorem 5.3.1 Let 1 < p < ∞, and suppose that the kernels a �= 0 and b are of type
(E). Let δa and δb denote the regularization order of a and b, respectively, and assume
that δa ≤ δb. Suppose further that δa /∈ { 2 1
p−1 , 1 + p }. Then (5.12) has a unique solution
(v, w) ∈ Z if and only if the conditions (N1) are satisfied.
5.3.3 The case 0 < δa − δb < 1/p
Let κ = δa − δb (= α − β). Suppose that (v, w) ∈ Z solves (5.12) and satisfies in addition
According to Corollary 2.8.1, the latter implies
p = −∇x · v − ∂yw ∈ H κ p (J; H 1 p(R n+1
+ )). (5.49)
(db ∗ ∇xp, db ∗ ∂yp) ∈ (0H δa−1
p (J; Lp(R n+1
+ )))n+1 .
In light of (5.21), we therefore obtain again necessity of (5.13) and (5.15). In the same
way as in the case δa < δb, we further see that conditions (5.14),(5.16), and (5.17) are
necessary. Concerning gw we deduce from (5.22) that
where
gw = da ∗ ψ1 + db ∗ ψ2, with ψ1 ∈ Y, ψ2 ∈ Yκ, (5.50)
δb2 (1−
Yκ = B
1
p )+κ
pp
(J; Lp(R n )) ∩ H κ p (J; B
Observe as well that we have the compatibility conditions
1
1− p
pp (R n )).
ψ1|t=0 = 2
3γ∇x · v0 − 4
3γ∂yw0, if p > 1 + 2 , (5.51)
δa
ψ2|t=0 = −γ∇x · v0 − γ∂yw0, if p > 2+δb . (5.52)
2κ+δb
Finally, from (5.49) and (5.23) there emerge the two conditions
∇x · v0 + ∂yw0 ∈ B
(∇x · fv + ∂yfw)t=0 ∈ B
2κ
1+ − δb 2
pδb pp
2κ
1+ − δb 2
− δb 2
pδb pp
(R n+1
+ ), (5.53)
(R n+1
+ ), if α > 1
where ∇x· and ∂y have to be understood in the distributional sense.
In sum we have shown necessity of
(N2) (5.13), (5.14), (5.15), (5.16), (5.17), (5.50) − (5.54).
p , (5.54)
Turning to the converse, we suppose that all conditions in (N2) are fulfilled. Let us
look first at q0. In virtue of (5.50),(5.51), and (5.52), we see that
q0 = 1 2
2A(b − 3a) ∗ (ψ2 − γp1) + ( 1
2ψ1 + 1
3ψ2 + γ∂yw1),
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