Quasilinear parabolic problems with nonlinear boundary conditions

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Our next objective is to show necessity of (ii). For this purpose we choose an extension of the solution u ∈ Z w.r.t. t to all of R (again denoted by u) which lies in the regularity class Z1 := H α p (R; Lp(R+; X)) ∩ Lp(R; H 2 p(R+; X)) ∩ Lp(R; Lp(R+; DA)). (3.31) Set A1 = I +A with domain D(A1) = D(A) and let A1 be the natural extension of A1 to Y := Lp(R; X). Then the operator A1 is invertible, and by Theorem 2.3.1 it belongs to the class BIP(X) with power angle θA1 ≤ θA; thus A1 is invertible as well and contained in BIP(Y ) with power angle θA1 ≤ θA. Further let B ∈ BIP(Y ) be the inverse Volterra operator from Theorem 2.8.1. Then the resolvents of A1 and B commute, and we have θB + θA1 ≤ θa + θA < π. This allows us to apply Theorem 2.3.1 to the pair (B, A1) in the space Y yielding that B + A1 with domain D(B) ∩ D(A1) = H α p (R; X) ∩ Lp(R; DA) is invertible and contained in BIP(Y ) with power angle θB+A1 ≤ max{θa, θA}. The function u ∈ Z1 now satisfies a problem of the form Bu − ∂ 2 yu + A1u = g, y > 0, t ∈ R, u(t, 0) = ϕ(t), t ∈ R, with g ∈ Lp(R; Y ) and some ϕ, which is an extension of φ to all of R. To determine the regularity of ϕ, we apply Theorem 3.4.2 to the invertible operator F := � B + A1, (3.32) which belongs again to BIP(Y ) and has power angle θF ≤ max{θa, θA}/2 < π/2. This results in ϕ ∈ DF (2 − 1/p, p). Due to Theorem 2.2.2 and Proposition 2.3.1, we have γ DF (γ, p) = D (B+A1) 1/2(γ, p) = DB+A1 ( Therefore which implies DF (1 − 1 1 2 p , p) = Bα( pp 1 − 2p ) DF (2 − 1 1 p , p) = Bα(1− pp Here we employ the embeddings B B 1 α(1− 2p ) pp 1 α(1− 2p ) pp 2p ) (R; X) ∩ Lp(R; DA(1 − 1 (R; X) ∩ Lp(R; DA(1 − 1 2 , p) = DB( γ 2 (R; X) ∩ Lp(R; DA( 1 γ , p) ∩ DA1 ( 2 , p), γ ∈ (0, 1). 2 1 − 2p , p)), (3.33) (R; X) ∩ Lp(R; DA(1 − 1 2p , p)). (3.34) 1 1 − 2 2p 2p , p)) ↩→ Bα( ) pp (R; D 1 ), (3.35) A 2 2p α 2 , p)) ↩→ Hp (R; DA( 1 1 2 − 2p , p)), (3.36) which follow from the mixed derivative theorem and real interpolation. Hence, we have shown that 1 α(1− 2p ϕ ∈ B ) (R; X) ∩ Lp(R; DA(1 − 1 , p)). pp By restriction to J, we see that the second condition in Theorem 3.5.1 is necessary. To derive condition (iv), we notice that u ∈ Z satisfies u ∈ H αs p (J; H 2(1−s) p (R+; X)), (3.37) 50 2p
for each s ∈ [0, 1]. This follows from the mixed derivative theorem. The space in (3.37) embeds into BUC(J × R+; X) if 1/p < sα and 1/p < 2 − 2s, i.e. if 1/pα < s < 1 − 1/2p. This shows that the compatibility condition (iv) is necessary in case α > 2/(2p − 1). We proceed with the determination of the regularity of φ|t=0 in case α > 2/(2p − 1). Observe that this inequality is equivalent to 1/p < α(1 − 1/2p). In light of Theorem 3.2.1, we have the embedding B sγ pp(J; X) ∩ Lp(J; DA(γ, p)) ↩→ C(J; DA(γ − 1 p s , p)), (3.38) if γ, γ − 1/ps ∈ (0, 1) and sγ ∈ (1/p, 2). Thus, by taking s = α and γ = 1 − 1/2p, we see that φ|t=0 ∈ DA(1 − 1/2p − 1/pα, p). Hence property (vii) is established. It remains to show (vi) and (viii). Let α > 1 + 3/(2p − 1), which in particular means that α > 1 + 1/p. Exploiting (3.37) with s = 1/α yields which in turn entails ∂tu ∈ H α−1 p (J; Lp(R+; X)) ∩ Lp(R; H ∂tu ∈ H (α−1)s p (J; H 1 2(1− α )(1−s) p 1 2(1− α ) p (R+; X)), (R+; X)), for each s ∈ [0, 1], by the mixed derivative theorem. This space is embedded into BUC(J ×R+; X) whenever (α−1)s > 1/p and 2(1−1/α)(1−s) > 1/p, i.e. if 1/p(α−1) < s < 1 − α/2p(α − 1). An easy calculation shows that existence of such an s is equivalent to α > 1 + 3/(2p − 1), which we just assumed. Therefore the trace ∂tu(0, 0) exists, and we see that the compatibility condition (vi) is necessary. Last but not least, we restrict ∂tf|t=0 to some finite interval J1 = [0, y0], which results in a function that belongs to B 1 1 2(1− − α pα ) pp (J1; X) ∩ Lp(J1; DA(1 − 1 1 α − pα , p)), owing to property (v). Then we apply again (3.38), this time with γ = 1 − 1/α − 1/pα and s = 2. This is possible on account of α > 1 + 3/(2p − 1). We immediately see that ∂tf|t=0, y=0 enjoys the regularity claimed in (viii). Hence the necessity part of Theorem 3.5.1 as well as the additional properties (vii) and (viii) are established. We turn now to the sufficiency part of Theorem 3.5.1. Let the data f and φ be given such that the conditions (i)-(vi) are satisfied. We will build the solution u ∈ Z of (3.27) as a sum of two functions in Z. At first we will construct a function u1 ∈ Z such that u1 − a ∗ ∂ 2 yu1 + a ∗ Au1 = f, t ∈ J, y > 0, (3.39) i.e. u1 solves the first equation of (3.27). For this purpose we extend the function f w.r.t. y to all of R in such a way that (i),(iii),(v) are fulfilled with R+ replaced by R. Let Y := Lp(R; X) and Λ ∈ BIP(Y ) be the operator defined in (3.28). Then we have f|t=0 ∈ DΛ(1 − 1/pα, p), if α > 1/p, and ∂tf|t=0 ∈ DΛ(1 − 1/α − 1/pα, p) in case α > 1 + 1/p, owing to (3.29) and (3.30). Thus, by Theorem 3.1.4, the Volterra equation admits a unique solution v1 in the regularity class v + a ∗ Λv = f, t ∈ J, (3.40) H α p (J; Lp(R; X)) ∩ Lp(J; H 2 p(R; X)) ∩ Lp(J; Lp(R; DA)). 51
Page 1 and 2: Gutachter: Quasilinear parabolic pr
Page 3 and 4: Contents 1 Introduction 3 2 Prelimi
Page 5 and 6: Chapter 1 Introduction The present
Page 7 and 8: to assume that m, c are bounded fun
Page 9 and 10: We give now an overview of the cont
Page 11 and 12: conditions of order ≤ 1. Sections
Page 13 and 14: Chapter 2 Preliminaries 2.1 Some no
Page 15 and 16: Clearly, φA ∈ [0, π) and φA
Page 17 and 18: Definition 2.2.3 Let X and Y be Ban
Page 19 and 20: µ ∈ Σφα }. Let N ∈ N, Tj
Page 21 and 22: We remark that a theorem of the Dor
Page 23 and 24: Here f(A, ·) ∈ H0(Σ π 2 +η; B
Page 25 and 26: Further, K ∞ (α, θa) := {a ∈
Page 27 and 28: Using (2.19) for aω and bω yields
Page 29 and 30: 2.7 Evolutionary integral equations
Page 31 and 32: Example 2.8.1 For J = [0, T ] and a
Page 33: We conclude this section by illustr
Page 36 and 37: kernel a. The operator B is inverti
Page 38 and 39: x := f(0) ∈ X exists and we are l
Page 40 and 41: with two positive constants C1, C2
Page 42 and 43: with two positive constants C1 and
Page 44 and 45: derivative theorem to this pair of
Page 46 and 47: 3.2 A general trace theorem Let X b
Page 48 and 49: 3.3 More time regularity for Volter
Page 50 and 51: Theorem 3.4.2 Suppose X is a Banach
Page 54 and 55: Let u1 be the restriction of v1 to
Page 56 and 57: Proof. We begin with the necessity
Page 59 and 60: Chapter 4 Linear Problems of Second
Page 61 and 62: The strategy for solving (4.1) is n
Page 63 and 64: Since ψj ≡ 1 on supp ϕj, we may
Page 65 and 66: Turning to (c), let g ∈ Ξi+1 and
Page 67 and 68: endowed with the norm | · | Y T 2
Page 69 and 70: We remark that the constant C2 stem
Page 71 and 72: One can then construct functions a
Page 73 and 74: analogous to (4.17), shows that S i
Page 75 and 76: Apply now V#, i+1 := I + k ∗ A#(
Page 77 and 78: Given a function v ∈ H 2 p(R n+1
Page 79 and 80: v is a solution of (4.40) on Ji+1 :
Page 81 and 82: Chapter 5 Linear Viscoelasticity In
Page 83 and 84: where δij denotes Kronecker’s sy
Page 85 and 86: problem ⎧ ⎪⎨ ⎪⎩ ∂tv −
Page 87 and 88: To see the converse direction, supp
Page 89 and 90: and up solves � Aup − ∆xup
Page 91 and 92: It can be written as where l(z, ξ)
Page 93 and 94: elongs to H∞ (Σ π 2 +η × Ση
Page 95 and 96: which allows us to write the first
Page 97 and 98: Chapter 6 Nonlinear Problems 6.1 Qu
Page 99 and 100: sufficiently small, say T ≤ T1
Page 101 and 102: (d) bD ∈ C(J0 × ΓD × U0), ∃C
Page 103 and 104:
which entails (6.14). Corresponding
Page 105 and 106:
substitution operators to be studie
Page 107 and 108:
for all t, τ ∈ J, ξ, η ∈ K,
Page 109 and 110:
to write where h2(t, τ, x) = h21(t
Page 111 and 112:
Lemma 6.2.3 Let 0 < s < s0 < 1, ρ
Page 113 and 114:
Bibliography [1] Albrecht, D.: Func
Page 115 and 116:
[54] Lunardi, A.: On the heat equat
Page 117 and 118:
kleines T mit Hilfe des Kontraktion
Page 119:
Personal Details Curriculum Vitae N
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Quasilinear parabolic problems with nonlinear boundary conditions

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