Quasilinear parabolic problems with nonlinear boundary conditions

Quasilinear parabolic problems with nonlinear boundary conditions Quasilinear parabolic problems with nonlinear boundary conditions

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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

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

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