Quasilinear parabolic problems with nonlinear boundary conditions

To see the converse direction, suppose the triple (v, w, p) satisfies (5.21), (5.22), and
(5.23). Let q = −∇x·v−∂yw and deduce from (5.21), by performing the same calculation
as above, that
⎧
⎨
⎩
∂tq − da ∗ (∆xq + ∂2 yq) − (db + 1
3da) ∗ (∆xp + ∂2 yp) = −∇x · fv − ∂yfw (J × R n+1
+ )
γq = γp (J × R n )
q|t=0 = −∇x · v0 − ∂yw0 (R n+1
+ ).
(5.25)
Then we replace the right-hand side of the first and third equation of (5.25) by the
corresponding terms in (5.23), to discover
⎧
⎨
⎩
∂t(q − p) − da ∗ (∆x + ∂ 2 y)(q − p) = 0 (J × R n+1
+ )
γ(q − p) = 0 (J × R n )
(q − p)|t=0 = 0 (R n+1
+ ).
(5.26)
So by uniqueness of the solution of (5.26), we find q = p, that is, (5.20) is established.
System (5.12) now follows immediately.
Observe that if we once know the boundary value γp, then the pressure p is uniquely
determined by (5.23). With p being known, (v, w) can then be obtained via (5.21). So
we have to find a formula for γp involving only the given data.
Step 2. To approach our goal, we continue by extending the functions ψf := −∇x · fv −
∂yfw and ψ0 := −∇x · v0 − ∂yw0 to all of R w.r.t. y so that the new functions (again
denoted by ψf and ψ0) lie in the corresponding regularity classes on J × R n+1 , that is
• ψf ∈ H δa−1
p (J; H −1
p (R n+1 ));
• ψf |t=0 ∈ B
• ψ0 ∈ B
2
1− pδa
pp
2 2
1− − δa pδa
pp
(R n+1 ).
(R n+1 ), if δa > 1 + 1
p ;
We then consider the problem
� ∂tq − dk ∗ (∆xq + ∂ 2 yq) = ψf , t ∈ J, x ∈ R n , y ∈ R,
q|t=0 = ψ0, x ∈ R n , y ∈ R,
(5.27)
on the space X−1 := H −1
p (R n+1 ). By integration we see that (5.27) is equivalent to the
Volterra equation
q(t) + (k ∗ Λq)(t) = (1 ∗ ψf )(t) + ψ0, t ∈ J, (5.28)
where Λ = Dn+1 with domain D(Λ) = H 1 p(R n+1 ). One readily verifies that
• 1 ∗ ψf + ψ0 ∈ H δa
p (J; X−1);
• ψ0 ∈ (X−1, D(Λ)) 1−1/pδa, p;
• ψf (0) ∈ (X−1, D(Λ)) 1−1/δa−1/pδa, p, if δa > 1 + 1
p .
So according to Theorem 3.1.4, (5.27) admits a unique solution φp in the space
H δa
p (J; H −1
p (R n+1 )) ∩ Lp(J; H 1 p(R n+1 )).
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