Quasilinear parabolic problems with nonlinear boundary conditions

and up solves
� Aup − ∆xup − ∂ 2 yup = fp, t ∈ J, x ∈ R n , y > 0
−γ∂yup + γDup = hp, t ∈ J, x ∈ R n .
(5.34)
Observe that u1 is determined by the data and does not depend on p0. Note further
that the compatibility condition is satisfied in either case. Theorem 3.5.2 yields u1 ∈ Z.
To summarize we see that step 2 shows the equivalence
as well as the implication
(5.21), (5.23) ⇔ (5.30), (5.32), (5.34), (5.35)
p0 ∈ 0Y ⇒ (v, w) ∈ Z. (5.36)
Step 3. We will now employ condition (5.22), together with (5.32),(5.34), to derive a
formula for p0.
To begin with, the function up can be written in the form
up(y)=e −F y (F + D) −1 hp + 1 −1
F
2
(cp. Prüss [65, p. 6]), which implies
� ∞
γup = (F + D) −1 hp + (F + D) −1
0
[e −F |y−s| + (F − D)(F + D) −1 e −F (y+s) ]fp(s) ds
� ∞
A short computation using the Fourier transform shows that
(F + D) −1 �
F + ((∇x∇x·) + Dn)F
=
−1 ∇x·
∇x
F
�
so we obtain
and furthermore
γvp = ∇xF −2
1 p0 − F F −2
1
+∇xF −2
1
γ∇x · vp = ∆xF −2
� ∞
0
� ∞
1 p0 − ∆xF F −2
1
+∆xF −2 1
1 A(b +
1 p0 − ∆xF −2
1
= ∆xF −2
On the one hand, we now have
0
0
e −F s fp(s) ds.
F −2
1 , F1 := (A + 2Dn) 1
2 ,
e −F s A(b + 1
3 a) ∗ ∇xe −Gs p0 ds+
e −F s A(b + 1
3 a) ∗ Ge−Gs p0 ds,
1 A(b + 3a) ∗ (F + G)−1p0+ 3a) ∗ G(F + G)−1p0 1 A(b + 3a) ∗ (F − G)(F + G)−1p0. (5.37)
−γ∂yw = −γ∂ywp − γ∂yw1 = γ∇x · vp + p0 − γ∂yw1. (5.38)
On the other hand, it follows from (5.22) that
−γ∂yw = 1
= 1
2A(1 ∗ gw) − 1 2
2A(b − 3a) ∗ γp
2A(1 ∗ gw) − 1 2
2A(b − 3a) ∗ γp1 − 1 2
2A(b − 3a) ∗ p0. (5.39)
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