Quasilinear parabolic problems with nonlinear boundary conditions

problem
⎧
⎪⎨
⎪⎩
∂tv − da ∗ (∆xv + ∂2 yv) − (db + 1
∂tw − da ∗ ∆xw − (db + 4
3da) ∗ ∂2 yw − (db + 1
3da) ∗ (∇x∇x · v + ∂y∇xw) = fv (J × R n+1
+ )
3da) ∗ ∂y∇x · v = fw (J × R n+1
+ )
−da ∗ γ∂yv − da ∗ γ∇xw = gv (J × Rn )
−(db − 2
3 da) ∗ γ∇x · v − (db + 4
3 da) ∗ γ∂yw = gw (J × R n )
v|t=0 = v0 (R n+1
+ )
w|t=0 = w0 (R n+1
+ ),
(5.12)
where γ denotes the trace operator at y = 0. We seek a unique solution (v, w) of (5.12)
in the regularity class
Z := (H δa
p (J; Lp(R n+1
+ )) ∩ Lp(J; H 2 p(R n+1
+ )))n+1 .
5.3.1 The case δa ≤ δb: necessary conditions
In this and the following subsection, we assume that δa ≤ δb. On the basis of the results
in Section 3.5, we first derive necessary conditions for the existence of a solution (v, w)
of (5.12) in the space Z.
Suppose that we are given such a solution. By Corollary 2.8.1, it then follows imme-
diately that
(fv, fw) ∈ (H δa−1
p (J; Lp(R n+1
+ )))n+1 . (5.13)
Taking the temporal trace of (v, w) and (∂tv, ∂tw), respectively, at t = 0 gives according
to Theorem 3.5.2 (see also the results from Chapter 4)
and
in case δa > 1 + 1/p. Putting
(v0, w0) ∈ (B
(fv, fw)|t=0 ∈ (B
δa (1− 2
Y = B
1
p )
pp
1
2(1− pδa )
pp
(R n+1
1 1
2(1− − δa pδa )
pp
+ ))n+1 , (5.14)
(R n+1
(J; Lp(R n )) ∩ Lp(J; B
+ ))n+1
1
1− p
pp (R n ))
Theorem 3.5.2 further yields (γ∂yv, γ∇xw, γ∂yw, γ∇x · v) ∈ Y 2n+2 . So if we set
φ = −γ∂yv − γ∇xw,
then it follows from the first boundary condition in (5.12) that gv is of the form
(5.15)
gv = da ∗ φ, with φ ∈ Y n . (5.16)
δa (1− 2
As, in case p > 1 + 2/δa, we have the embedding B
1
p )
pp (J; Lp(Rn )) ↩→ C(J; Lp(Rn )),
we see that φ satisfies in addition the compatibility condition
φ|t=0 = −γ∂yv0 − γ∇xw0, if p > 1 + 2 . (5.17)
δa
Turning to the second boundary condition in (5.12), we put
ψ1 = 2
3 γ∇x · v − 4
3 γ∂yw, ψ2 = −γ∇x · v − γ∂yw.
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