Quasilinear parabolic problems with nonlinear boundary conditions

Combining (5.38) with (5.39), setting
and using (5.37) leads to
q0 := 1
2A(1 ∗ gw) − 1 2
2A(b − 3a) ∗ γp1 + γ∂yw1
q0 = p0 + 1 2
2A(b − 3a) ∗ p0 + γ∇x · vp
= Aa ∗ p0+ 1
2
= 1
2
= 1
2
Further,
A(b + 4
3
A(b − 2
3
a) ∗ p0−∆xF −2
1
4 A(b + 3a) ∗ p0 − ∆xF −2
1
(5.40)
−2
−2 1
a) ∗ p0+∆xF 1 p0−∆xF 1 A(b + 3a) ∗ (F −G)(F +G)−1p0 �
1
−Aa ∗ (F +G) + A(b +
� A(b − 2
3
3 a) ∗ (F −G)� (F +G) −1 p0
a) ∗ F − A(b + 4
3 a) ∗ G� (F + G) −1 p0.
2Ka ∗ q0 = � I + 2DnF −2
1 [K(b − 2
3 a) ∗ F − G](F + G)−1� p0
= � (A + 2Dn + 2DnK(b − 2
3 a)∗)F + AG� (F + G) −1 F −2
1 p0
= � A + [2Dn + 2DnK(b − 2
3a)∗]F (F + G)−1� F −2
1 p0
= � A + [2Dn(K(b + 4
3a)∗) + 2DnK(b − 2
3a)∗]F (F + G)−1� F −2
1 p0
= � A + DnK(4b + 4
3a) ∗ F (F + G)−1� F −2
1 p0 = Lp0, (5.41)
where the operator L is defined by
L = � A + DnK(4b + 4
3 a) ∗ F (F + G)−1� F −2
1 . (5.42)
We claim now that q0 ∈ 0Y . Indeed, in virtue of (5.18),(5.19), there exist ψ1, ψ2 ∈ Y
such that
and
q0 = 1
2 (ψ1 + Ab ∗ ψ2) − 1 2
2A(b − 3a) ∗ γp1 + γ∂yw1
= 1
2 (ψ1 + A(b − 2
3a) ∗ ψ2) + 1
3ψ2 − 1 2
2A(b − 3a) ∗ γp1 + γ∂yw1
= 1 2
2A(b − 3a) ∗ (ψ2 − γp1) + ( 1
2ψ1 + 1
3ψ2 + γ∂yw1),
ψ1|t=0 = 2
3 γ∇x · v0 − 4
3 γ∂yw0, ψ2|t=0 = −γ∇x · v0 − γ∂yw0 (5.43)
in case p > 1 + 2/δa. But from (5.43) and the definition of p1 and w1, we deduce that
ψ2 − γp1 , 1
2 ψ1 + 1
3 ψ2 + γ∂yw1 ∈ 0Y.
Hence the claim follows, because A(b∗) ∈ B(0Y ).
From q0 ∈ 0Y and K(a∗) ∈ B(0Y ) we conclude further that K(a ∗ q0) ∈ 0Y . That is,
to solve (5.41) for p0, we have to show that L has a bounded inverse on 0Y . To achieve
this, we shall use, aside from extension and restriction, the joint (causal) H∞ (Σ π
2 +η ×Ση)
- calculus (0 < η < π/2) of the pair (∂t, Dn) in Lp(R+ × Rn ), cf. Example 2.4.1.
For this purpose we look at the symbol l(z, ξ) of L (in Lp(R+ × Rn )). Taking the
Laplace-transform in t and the Fourier-transform in x we obtain for l(z, ξ) :
⎛
⎜
l(z, ξ) = ⎝
1
+
â(z)|ξ| 2
� 1
â(z)|ξ| 2 + 1
4ˆb(z)+ 4
3 â(z)
ˆ 4
b(z)+ 3 â(z)
�
1
â(z)|ξ| 2 + 1 + �
1
( ˆb(z)+ 4 + 1
â(z))|ξ|2 3
88
⎞
�
�−1 ⎟ 1
⎠ + 2 . (5.44)
â(z)|ξ| 2