Quasilinear parabolic problems with nonlinear boundary conditions

Note that φp only depends upon the data and the selected extension operator.
We now set
p0 := γp − γφp. (5.29)
Then clearly, γp can be determined immediately as soon as p0 is known, and vice versa.
Putting p1 := φp| n+1
R ∈ Z−1 := H
+
δa
p (J; H−1 p (R n+1
+ ))∩Lp(J; H1 p(R n+1
+ )), it further follows
from the construction of φp that (5.23) is equivalent to the identity
By the mixed derivative theorem, we have
Notice also that
Consequently
p = e −Gy p0 + p1. (5.30)
Z−1 ↩→ H δa 2
p (J; Lp(R n+1
+ )) ∩ Lp(J; H 1 p(R n+1
+ )).
e −Gy p0 ∈ 0H δa 2
p (J; Lp(R n+1
+ )) ∩ Lp(J; H 1 p(R n+1
+ ))
δa (1− 2
⇐⇒ p0 ∈ 0Y := 0B
1
p )
pp
(J; Lp(R n )) ∩ Lp(J; B
1
1− p
pp (Rn )).
p0 ∈ 0Y ⇒ p ∈ H δa 2
p (J; Lp(R n+1
+ )) ∩ Lp(J; H 1 p(R n+1
+ )).
According to Theorem 3.5.2, this regularity of p suffices to obtain (v, w) ∈ Z when (5.21)
is solved for this pair of functions. In fact, let
� � � � �
v
v0
fv − (db +
u = , u0 = , f =
w
w0
1
3da) ∗ ∇xp
fw − (db + 1
�
,
3da) ∗ ∂yp
� � �
�
A(1 ∗ gv)
0 −∇x
h =
, D =
.
γp
−∇x· 0
Then system (5.21) is equivalent to the following problem for u.
Setting
⎧
⎨
⎩
∂tu − da ∗ ∆xu − da ∗ ∂ 2 yu = f, t ∈ J, x ∈ R n , y > 0
−γ∂yu + γDu = h, t ∈ J, x ∈ R n
�
fv − (db +
f1 =
1
3da) ∗ ∇xp1
da) ∗ ∂yp1
fw − (db + 1
3
as well as
�
1 −A(b +
fp =
3a) ∗ ∇xe−Gyp0 A(b + 1
3a) ∗ Ge−Gyp0 the solution u can be written as
where u1 is defined by means of
⎧
⎨
⎩
u|t=0 = u0, x ∈ R n , y > 0.
� �
A(1 ∗ gv)
, h1 =
γp1
� �
0
, hp =
p0
�
,
�
,
(5.31)
u = u1 + up, (5.32)
∂tu1 − da ∗ ∆xu1 − da ∗ ∂ 2 yu1 = f1, t ∈ J, x ∈ R n , y > 0
−γ∂yu1 + γDu1 = h1, t ∈ J, x ∈ R n
u1|t=0 = u0, x ∈ R n , y > 0,
86
(5.33)