Quasilinear parabolic problems with nonlinear boundary conditions
Quasilinear parabolic problems with nonlinear boundary conditions Quasilinear parabolic problems with nonlinear boundary conditions
Proof. We begin with the necessity part. Suppose we have a function u ∈ Z which is a solution of (3.45). Then u is also a solution of the problem � u − a ∗ ∂ 2 yu + a ∗ Au = f, t ∈ J, y > 0, u(t, 0) = ϕ(t), t ∈ J, where ϕ := u|y=0. Therefore, conditions (i),(iii) and (v) are necessary, by the first part of the proof of Theorem 3.5.1, where the case α = 2/(2p − 1) is admissible, too. Moreover, we see that f|t=0, y=0 ∈ DA(1 − 1/2p − 1/pα, p), if α > 2/(2p − 1), i.e. property (vi) is fulfilled. To show condition (ii), we proceed similarly as in the proof of Theorem 3.5.1. We extend u w.r.t. t to all of R such that u ∈ Z1 (see (3.31) for the definition), u|y=0 ∈ Lp(R; DD), and Du(·, 0) ∈ B 1 1 α( − 2 2p ) pp (R; X) ∩ Lp(R; DA( 1 1 2 − 2p , p)). Let Y := Lp(R; X) and define D as the natural extension of D to this space. Let F be as in (3.32). Then the space DF (1 − 1/p, p) is given by (3.33). Thus we see that u ∈ H 2 p(R+, Y ) ∩ Lp(R+; D F 2), u|y=0 ∈ D(D), and Du|y=0 ∈ DF (1 − 1/p, p). Now consider u as the solution of the problem −u ′′ (y) + F 2 u(y) = g, y > 0, −u ′ (0) = ¯ φ − Du(0), with some g ∈ Lp(R+; Y ) and an extension ¯ φ of φ on the real line. Since θF ≤ max{θa, θA}/2 < π, it then follows from Theorem 3.4.3 that ¯ φ−Du(0) ∈ DF (1−1/p, p), i.e. ¯ φ ∈ DF (1 − 1/p, p), which after restriction to t ∈ J yields condition (ii). Finally, we have to prove (iv) and (vii). By the mixed derivative theorem, it follows from u ∈ Z that and further ∂yu ∈ H α 2 p (J; Lp(R+; X)) ∩ Lp(J; H 1 p(R+; X)), ∂yu ∈ H αs 2 p (J; H1−s p (R+; X)), for all s ∈ [0, 1]. This space embeds into BUC(J ×R+; X) if αs/2 > 1/p and 1−s > 1/p, i.e. in case 2/pα < s < 1 − 1/p. Such an s exists if and only if α > 2/(p − 1) as a short computation shows. Thus in this situation the trace (∂yu)(0, 0) ∈ X exists. Further, and B 1 1 α( − 2 2p ) pp Du|y=0, φ ∈ B (J; X) ∩ Lp(J; DA( 1 1 1 α( − 2 2p ) pp (J; X) ∩ Lp(J; DA( 1 1 2 − 2p , p)) 2 1 1 1 1 − 2p , p)) ↩→ C(J; DA( 2 − 2p − pα , p)), thanks to Theorem 3.2.1 and the remark before Theorem 3.5.2. Therefore, by the closedness of D, f|t=0, y=0 ∈ D(D) and −∂yf|t=0, y=0 + Df|t=0, y=0 = φ|t=0, where each term in this equation lies in the space DA(1/2 − 1/2p − 1/pα, p). Hence, conditions (iv) and (vii) are proved. This completes the necessity part of the proof. 54
We come now to sufficiency. Suppose we are given the data f and φ which fulfill the conditions (i)-(v). We proceed similarly as in the proof of Theorem 3.5.1. Firstly, let u1 ∈ Z be a function which satisfies (3.39). Such a function was constructed in the proof of Theorem 3.5.1. Put φ1 := −∂yu1|y=0 + Du1|y=0 and consider the problem � v − a ∗ ∂2 yv + a ∗ Av = 0, t ∈ J, y > 0, (3.48) −∂yv(t, 0) + Dv(t, 0) = φ(t) − φ1(t), t ∈ J. Owing to the necessity part of Theorem 3.5.2 and the above remark concerning (3.46), φ1 is well-defined and lies in the same regularity class as the data φ. In addition, we see that (φ−φ1)|t=0 = 0, if α > 2/(p−1), by construction of φ1 and thanks to condition (iv). That is, the compatibility condition is satisfied. To solve (3.48), we let Y1 = Lp(J; X), D be the natural extension of D to this space, and define the operator F1 as in the paragraph following problem (3.41). Then (3.48) can be rewritten as −v ′′ (y) + F 2 1 v(y) = 0, y > 0, −v ′ (0) + Dv(0) = φ − φ1. (3.49) Observe that D is pseudo-sectorial in Y1 and D ∈ BIP(R(D)) with power angle θD ≤ θD. In view of (3.43), φ − φ1 ∈ DF1 (1 − 1/p, p). Further, we have θD + θF1 ≤ θD + max{θa, θA}/2 < π. By Theorem 3.4.3, problem (3.48) thus admits a unique solution u2 in the space with 0H α p (J; Lp(R+; X)) ∩ Lp(J; H 2 p(R+; X)) ∩ Lp(J; Lp(R+; DA)) Du2(·, 0) ∈ DF1 (1 − 1/p, p) = 0B 1 1 α( − 2 2p ) pp (J; X) ∩ Lp(J; DA( 1 1 2 − 2p , p)). Finally put u := u1 + u2. Then u clearly possesses the desired regularity and solves (3.45). Uniqueness follows by Theorem 3.4.3. In fact, rewrite (3.45) with zero data as (3.49) with zeros on the right-hand side. Then it is apparent that the zero-function is the only solution of (3.45) in Z. � Remarks 3.5.1 (i) Theorem 3.5.1 and Theorem 3.5.2 are also true, if the kernel a is of the form a = b+dk∗b, where b is like a above and k ∈ BVloc(R+) with k(0) = k(0+) = 0. (ii) The proofs of the two foregoing theorems are inspired by Prüss [65]. In the case a ≡ 1 Theorem 3.5.1 is equivalent with a version of [65, Thm. 5] for compact J, while Theorem 3.5.2, roughly speaking, can be regarded as an extension of [65, Thm. 6], at least in the case D A 1/2 ↩→ DD. 55
- Page 5 and 6: Chapter 1 Introduction The present
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- Page 27 and 28: Using (2.19) for aω and bω yields
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- Page 31 and 32: Example 2.8.1 For J = [0, T ] and a
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- Page 46 and 47: 3.2 A general trace theorem Let X b
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We come now to sufficiency. Suppose we are given the data f and φ which fulfill the<br />
<strong>conditions</strong> (i)-(v). We proceed similarly as in the proof of Theorem 3.5.1.<br />
Firstly, let u1 ∈ Z be a function which satisfies (3.39). Such a function was constructed<br />
in the proof of Theorem 3.5.1. Put φ1 := −∂yu1|y=0 + Du1|y=0 and consider<br />
the problem �<br />
v − a ∗ ∂2 yv + a ∗ Av = 0, t ∈ J, y > 0,<br />
(3.48)<br />
−∂yv(t, 0) + Dv(t, 0) = φ(t) − φ1(t), t ∈ J.<br />
Owing to the necessity part of Theorem 3.5.2 and the above remark concerning (3.46),<br />
φ1 is well-defined and lies in the same regularity class as the data φ. In addition, we see<br />
that (φ−φ1)|t=0 = 0, if α > 2/(p−1), by construction of φ1 and thanks to condition (iv).<br />
That is, the compatibility condition is satisfied. To solve (3.48), we let Y1 = Lp(J; X),<br />
D be the natural extension of D to this space, and define the operator F1 as in the<br />
paragraph following problem (3.41). Then (3.48) can be rewritten as<br />
−v ′′ (y) + F 2 1 v(y) = 0, y > 0, −v ′ (0) + Dv(0) = φ − φ1. (3.49)<br />
Observe that D is pseudo-sectorial in Y1 and D ∈ BIP(R(D)) <strong>with</strong> power angle θD ≤<br />
θD. In view of (3.43), φ − φ1 ∈ DF1 (1 − 1/p, p). Further, we have θD + θF1 ≤ θD +<br />
max{θa, θA}/2 < π. By Theorem 3.4.3, problem (3.48) thus admits a unique solution u2<br />
in the space<br />
<strong>with</strong><br />
0H α p (J; Lp(R+; X)) ∩ Lp(J; H 2 p(R+; X)) ∩ Lp(J; Lp(R+; DA))<br />
Du2(·, 0) ∈ DF1 (1 − 1/p, p) = 0B<br />
1 1<br />
α( − 2 2p )<br />
pp (J; X) ∩ Lp(J; DA( 1 1<br />
2 − 2p , p)).<br />
Finally put u := u1 + u2. Then u clearly possesses the desired regularity and solves<br />
(3.45).<br />
Uniqueness follows by Theorem 3.4.3. In fact, rewrite (3.45) <strong>with</strong> zero data as (3.49)<br />
<strong>with</strong> zeros on the right-hand side. Then it is apparent that the zero-function is the only<br />
solution of (3.45) in Z. �<br />
Remarks 3.5.1 (i) Theorem 3.5.1 and Theorem 3.5.2 are also true, if the kernel a is of<br />
the form a = b+dk∗b, where b is like a above and k ∈ BVloc(R+) <strong>with</strong> k(0) = k(0+) = 0.<br />
(ii) The proofs of the two foregoing theorems are inspired by Prüss [65]. In the case<br />
a ≡ 1 Theorem 3.5.1 is equivalent <strong>with</strong> a version of [65, Thm. 5] for compact J, while<br />
Theorem 3.5.2, roughly speaking, can be regarded as an extension of [65, Thm. 6], at<br />
least in the case D A 1/2 ↩→ DD.<br />
55