Quasilinear parabolic problems with nonlinear boundary conditions

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Let u1 be the restriction of v1 to y ∈ R+. Then clearly u1 ∈ Z, and u1 satisfies (3.39). Next we put φ1 := u1|y=0. Due to the necessity part of Theorem 3.5.1 (conditions (ii),(iv),(vi)), we see that φ1 ∈ B 1 α(1− 2p ) pp (J; X) ∩ Lp(J; DA(1 − 1 2p , p)), and f|t=0, y=0 = φ1|t=0, if α > 2 2p−1 , as well as ∂tf|t=0, y=0 = ˙ φ1|t=0, in case α > 1 + 3/(2p − 1). Therefore we deduce that φ − φ1 ∈ 0B 1 α(1− 2p ) pp (J; X) ∩ Lp(J; DA(1 − 1 , p)). Consider now the problem � v − a ∗ ∂ 2 yv + a ∗ Av = 0, t ∈ J, y > 0, v(t, 0) = φ − φ1, t ∈ J. 2p (3.41) Define this time A as the natural extension of A to Y1 := Lp(J; X). Let further B ∈ BIP(Lp(J; X)) be the inverse Volterra operator from Corollary 2.8.1 associated with the kernel a. Then B is invertible, A and B belong to BIP(Y1), their resolvents commute, and θA + θB ≤ θA + θa < π. Therefore, by Theorem 2.3.1, the operator B + A with domain D(B) ∩ D(A) = 0H α p (J; X) ∩ Lp(J; DA) is invertible and contained in BIP(Y1) with power angle θB+A ≤ max{θa, θA}. Moreover, the operator F1 := √ B + A is also invertible and belongs to BIP(Y1) with power angle θF1 ≤ max{θa, θA}/2 < π/2. We can now rewrite (3.41) as −v ′′ (y) + F 2 1 v(y) = 0, y > 0, v(0) = φ − φ1. (3.42) In the same way as above for the operator F , one can show that and 1 DF1 (1 − p , p) = 0B 1 DF1 (2 − p , p) = 0B 1 1 α( − 2 2p ) pp (R; X) ∩ Lp(R; DA( 1 1 2 − 2p , p)) (3.43) 1 α(1− 2p ) pp (J; X) ∩ Lp(J; DA(1 − 1 , p)). (3.44) Compare this with (3.33) and (3.34). Thus we have φ − φ1 ∈ DF1 (2 − 1/p, p), which implies existence of a solution u2 of (3.41) in the space 0H α p (J; Lp(R+; X)) ∩ Lp(J; H 2 p(R+; X)) ∩ Lp(J; Lp(R+; DA)), by virtue of Theorem 3.4.2. Finally let u := u1 + u2. Then u ∈ Z, and by construction u is a solution of problem (3.27). Uniqueness follows from Theorem 3.4.2. To see this, rewrite (3.27) with zeros on the right-hand side as (3.42) with initial value zero. This completes the proof of Theorem 3.5.1. � There is a corresponding result for the problem � u − a ∗ ∂ 2 yu + a ∗ Au = f, t ∈ J, y > 0, −∂yu(t, 0) + Du(t, 0) = φ(t), t ∈ J. 52 2p (3.45)
Here the operator D is pseudo-sectorial in X, and we assume that D A 1/2 ↩→ DD. As above we seek a solution in Z = H α p (J; Lp(R+; X)) ∩ Lp(J; H 2 p(R+; X)) ∩ Lp(J; Lp(R+; DA)). Before we state the theorem concerning (3.45) we note that any function v in the space Z automatically satisfies v|y=0 ∈ Lp(J; DD) and Dv(·, 0) ∈ B 1 1 α( − 2 2p ) pp (J; X) ∩ Lp(J; DA( 1 In fact, if v ∈ Z, then it follows, by Theorem 3.5.1, that v|y=0 ∈ Lp(J; DA(1 − 1 2p , p)), which entails the first claim, for we have the embeddings DA(1 − 1 2p , p) ↩→ D A 1 2 2 ↩→ DD. 1 − 2p , p)). (3.46) As for (3.46), we see with the aid of the mixed derivative theorem (Proposition 2.3.2) that for w := A 1/2 v, w ∈ H α 2 p (J; Lp(R+; X)) ∩ Lp(J; H 1 p(R+; X)) ∩ Lp(J; Lp(R+; D A 1 2 )). (3.47) We extend w w.r.t. t to all of R such that (3.47) holds true, with J replaced by R. As above, set Y = Lp(R; X) and define the operator F by (3.32). Then we know that F is invertible and lies in the class BIP(Y ) with power angle θF < π/2. Further, w ∈ H 1 p(R+; Y ) ∩ Lp(R+; DF ). Thus, Theorem 3.4.1 yields w|y=0 ∈ DF (1 − 1/p, p), which by restriction to t ∈ J implies A 1 2 v(·, 0) ∈ B 1 1 α( − 2 2p ) pp (J; X) ∩ Lp(J; DA( 1 1 2 − 2p , p)), in view of (3.33). Assertion (3.46) now follows on account of D A 1/2 ↩→ DD. Theorem 3.5.2 Let p ∈ (1, ∞), X be a Banach space of class HT , a ∈ K 1 (α, θa) with α ∈ (0, 2) \ � 1 p , 2 p−1 , 1 + 1 p � , and A ∈ BIP(X) with power angle θA. Let further D be a pseudo-sectorial operator in X such that D ∈ BIP(R(D)) with power angle θD. Suppose that A and D commute in the resolvent sense, and that D A 1/2 ↩→ DD. Assume further the angle conditions θa + θA < π and θD < π − max{θa, θA}/2. Then the problem (3.45) has a unique solution in Z if and only if the data f and φ are subject to the following conditions. (i) f ∈ H α p (J; Lp(R+; X)); (ii) φ ∈ B 1 1 α( − 2 2p ) pp (J; X) ∩ Lp(J; DA( 1 1 2 − 2p , p)); 2 2− pα pp (iii) f|t=0 ∈ B (R+; X) ∩ Lp(R+; DA(1 − 1 1 pα , p)), if α > p ; (iv) f|t=0, y=0 ∈ D(D) and − ∂yf|t=0, y=0 + Df|t=0, y=0 = φ|t=0, if α > 2 p−1 ; (v) ∂tf|t=0 ∈ B 1 1 2(1− − α pα ) pp If this is the case, then additionally (R+; X) ∩ Lp(R+; DA(1 − 1 α 1 1 − pα , p)), if α > 1 + p . (vi) f|t=0, y=0 ∈ DA(1 − 1 1 2p − pα , p), if α > 2 2p−1 ; (vii) Df|t=0, y=0, φ|t=0 ∈ DA( 1 1 1 2 − 2p − pα , p), if α > 2 p−1 . 53
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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 52 and 53: Our next objective is to show neces
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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