ψ(f, g) ≤ Mρ + |bξ(·, ·, w)|∞, m, f, g ∈ Σ, M being independent of T because (f − g)|t=0 = 0. Hence the assertion follows <strong>with</strong> µ(T ) = [w] 0,s (Y 1 1 ) m + T r−s1+ 1 p . � By repeating the above considerations <strong>with</strong> the roles of J and Ω being reversed, one obtains (under the corresponding assumptions, cf. (H4) in Section 6.1) the estimate � � ≤ C ρ + µ(T ) + |bξ(·, ·, w)|∞, m [b(·, ·, f) − b(·, ·, g)] Y 0,s 2 2 We come now to exponents greater than 1. |f − g| (Y 0,s 1 1 ∩Y 0,s2 2 ) m, f, g ∈ Σ. Lemma 6.2.2 Let w ∈ (Y 1,s1 1 ∩ Y 1,s2 2 ) m be a fixed K-valued function and ρ > 0. Let further Σ be as described above. Suppose bξ, btξ ∈ L∞(J × Ω × K, Rm ), bξξ ∈ L∞(J × Ω × K, Rm×m ), and assume that there exist CHL > 0, r1 ∈ (s1, 1), and Cb ∈ Lp(Ω) such that |btξ(t, x, ξ) − btξ(τ, x, η)| ≤ CHL(Cb(x)|t − τ| r1 + |ξ − η|), |bξξ(t, x, ξ) − bξξ(τ, x, η)| ≤ CHL(|t − τ| r1 + |ξ − η|), for all t, τ ∈ J, ξ, η ∈ K, and a.a. x ∈ Ω. Then there exists a constant C > 0 not depending on T and ρ such that � � ≤ C ρ + µ(T ) + |bξ(·, ·, w)|∞, m [b(·, ·, f) − b(·, ·, g)] Y 1,s 1 1 |f − g| (Y 1,s 1 1 ∩Y 1,s2 2 ) m, f, g ∈ Σ. Proof. For brevity we set Y = Y 1,s1 1 ∩ Y 1,s2 2 . Remember that we have the embedding Y ↩→ Cr (J; C( ¯ Ω)) for some r ∈ (s1, 1). Let now f, g ∈ Σ be arbitrary functions. Put h1(t, τ, x) = bt(t, x, f(t, x)) − bt(t, x, g(t, x)) − bt(τ, x, f(τ, x)) + bt(τ, x, g(τ, x)), h2(t, τ, x) = bξ(t, x, f(t, x)) · ft(t, x) − bξ(t, x, g(t, x)) · gt(t, x) for t, τ ∈ J, and a.a. x ∈ Ω. Then [b(·, ·, f) − b(·, ·, g)] 1,s Y 1 1 � T � T � ≤ 2� ( i=1 −bξ(τ, x, f(τ, x)) · ft(τ, x) + bξ(τ, x, g(τ, x)) · gt(τ, x) 0 0 Ω = ( � T � T 0 0 � Ω (|h1(t, τ, x) + h2(t, τ, x)|) p |t − τ| 1+s1p |hi(t, τ, x)| p 1 dx dτ dt) p =: I1 + I2. |t − τ| 1+s1p dx dτ dt) 1 p Concerning I1, we may use the estimates from the proof of Lemma 6.2.1, thereby obtaining � I1 ≤ C (ρ + |btξ(·, ·, w)|∞, m)[f − g] 0,s (Y 1 1 ) m � + (ρ + µ(T ))|f − g|∞, m � � ≤ C (ρ0 + |btξ|∞, m)µ(T )[f − g] (Cr 1 ) m + M(ρ + µ(T ))|f − g|Y m ≤ C(ρ + µ(T ))|f − g|Y m. The term I2 is more sophisticated. We employ the identity aA − bB − cC + dD = (a − b − c + d)D + (−a + b + c)(A − B − C + D) + 106
to write where h2(t, τ, x) = h21(t, τ, x) · gt(τ, x) + +(a − c)(A − B) + (a − b)(A − C) +(−bξ(t, x, f(t, x)) + bξ(t, x, g(t, x)) + bξ(τ, x, f(τ, x))) · h22(t, τ, x) +(bξ(t, x, f(t, x)) − bξ(τ, x, f(τ, x)) · (ft(t, x) − gt(t, x)) + +(bξ(t, x, f(t, x)) − bξ(t, x, g(t, x))) · (ft(t, x) − ft(τ, x)) =: I3(t, τ, x) + I4(t, τ, x) + I5(t, τ, x) + I6(t, τ, x), h21(t, τ, x) = bξ(t, x, f(t, x)) − bξ(t, x, g(t, x)) − bξ(τ, x, f(τ, x)) + bξ(τ, x, g(τ, x)), h22(t, τ, x) = ft(t, x) − gt(t, x) − ft(τ, x) + gt(τ, x), t, τ ∈ J, a.a. x ∈ Ω. The summand I3 can be estimated by mimicking the middle part of the proof of Lemma 6.2.1. Letting h23(t, τ, x) = f(t, x) − g(t, x) − f(τ, x) + g(τ, x) and r0 = min{r, r1} we get |h21(t,τ, x)| ≤ ≤ C(|h23(t, τ, x)| + |f − g|∞, m(|f(t, x) − f(τ, x)| + |g(t, x) − g(τ, x)| + |t − τ| r1 )) ≤ C(|t − τ| r [f − g] (C r 1 ) m + |f − g|∞, m(|t − τ| r ([f] (C r 1 ) m + [g] (C r 1 )m) + |t − τ|r1 )) ≤ C(|t − τ| r [f − g]Y m(1 + ρ + [w] (C r 1 )m) + [f − g]Y m|t − τ|r1 ) ≤ C|t − τ| r0 [f − g]Y m for all t, τ ∈ J, and a.a. x ∈ Ω. Thus, � T � T � |I3(t, τ, x)| ( 0 0 Ω p 1 dx dτ dt) p ≤ |t − τ| 1+s1p ≤ C[f − g]Y m( � T � |gt(τ, x)| 0 Ω p � T ( 0 dt ) dx dτ)1 p |t − τ| 1+(s1−r0)p |f − g|Y m ≤ Cµ(T )|f − g|Y m. ≤ CT r0−s1 [g]X m 1 Turning to I4, we immediately see that � T � T � ( 0 0 Ω As for I5, we estimate |I4(t, τ, x)| p 1 dx dτ dt) p ≤ M(ρ + |bξ(·, ·, w)|∞, |t − τ| 1+s1p m)[f − g] 1,s (Y1 ) m. |I5(t, τ, x)| ≤ |bξ(t, x, f(t, x)) − bξ(τ, x, f(t, x))||ft(t, x) − gt(t, x)|+ + |bξ(τ, x, f(t, x)) − bξ(τ, x, f(τ, x))||ft(t, x) − gt(t, x)| ≤ (|btξ|∞, m|t − τ| + |bξξ| ∞, m 2[f] (C r 1 ) m|t − τ|r )|ft(t, x) − gt(t, x)| ≤ C|t − τ| r |ft(t, x) − gt(t, x)| 107
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Gutachter: Quasilinear parabolic pr
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Contents 1 Introduction 3 2 Prelimi
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Chapter 1 Introduction The present
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to assume that m, c are bounded fun
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We give now an overview of the cont
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conditions of order ≤ 1. Sections
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Chapter 2 Preliminaries 2.1 Some no
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Clearly, φA ∈ [0, π) and φA
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Definition 2.2.3 Let X and Y be Ban
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µ ∈ Σφα }. Let N ∈ N, Tj
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We remark that a theorem of the Dor
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Here f(A, ·) ∈ H0(Σ π 2 +η; B
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Further, K ∞ (α, θa) := {a ∈
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Using (2.19) for aω and bω yields
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2.7 Evolutionary integral equations
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Example 2.8.1 For J = [0, T ] and a
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We conclude this section by illustr
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kernel a. The operator B is inverti
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x := f(0) ∈ X exists and we are l
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with two positive constants C1, C2
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with two positive constants C1 and
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derivative theorem to this pair of
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3.2 A general trace theorem Let X b
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3.3 More time regularity for Volter
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Theorem 3.4.2 Suppose X is a Banach
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Our next objective is to show neces
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Let u1 be the restriction of v1 to
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Proof. We begin with the necessity
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