Quasilinear parabolic problems with nonlinear boundary conditions
Quasilinear parabolic problems with nonlinear boundary conditions Quasilinear parabolic problems with nonlinear boundary conditions
the natural regularity space of φ in (4.33), is equivalent to | · |Yl = | · | Lp([0,Tl]×R n ) + [ · ] Y T l 1 see (4.21) and (4.22) for the definition of [ · ] Y T l 1 , [ · ] T Y l . 2 + [ · ] T Y l , 2 Taking now for (g, φ) in (4.33) the right-hand sides of (4.5) and (4.31), application of the solution operator L ij H, i+1 yields where and (I − S ij H )ϕjv =L ij � ϕjf + k ∗ Cj(·, x, Dx)v H, i+1 ϕjh + γyCH, j(·, x ′ )v S ij Hw = Lij H, i+1 � j k ∗ (A# (Ti, xj, Dx) − A j # (·, x, Dx))w γy(B j # (Ti, x ′ j , Dx) − B j # (·, x′ , Dx))w � =: h ij H (f, h, v), (4.34) CH, j(t, x) = B#(t, x ′ , Dx)ϕj − ϕjb0, (4.35) γy denoting the trace operator at y = 0. One can then show an analogue to Claim 2 (cf. the proof of Theorem 4.1.1) asserting in particular existence of a small η0 > 0 which is such that whenever δ = maxi |Ti+1 − Ti|, η ≤ η0, the equation (I − S ij H )w = hij H (f, h, v) admits a unique solution w =: (I − S ij H )|−1 Zi+1(ϕjV −1 H, i (f, h))hij H (f, h, v) in Zi+1(ϕjV −1 H, i (f, h)) for all (f, h) ∈ ΞH, i+1, v ∈ Zi+1(V −1 H, i (f, h)), i = 0, . . . , M −1, and (f, h), i ≥ 1, refers to the solution of (4.30) on the time-interval [0, Ti], which is already known in the (i + 1)th time step. Further, Z1(ϕjV −1 H, 0 (f, h)) := {ϕjw : w ∈ Z1, ∂m t w|t=0 = ∂m t f|t=0, if α > m + 1/p, m = 0, 1}. The proof of these properties is similar to that of Claim 2 above, which is why we only consider the part that involves estimates on the boundary. Let w ∈ Zi+1(0). If p > n + 1 + 2/α, that is Y ↩→ BUC([0, T ] × Rn ), Lemma 4.2.1 yields j = 0, . . . , Nb. Here V −1 H, i |γy(B j # (Ti, x ′ j, Dx) − B j # (t, x′ , Dx))w|Yi+1 = |(b(Ti, x ′ j) − b j (·, ·)) · ∇x ′γyw|Yi+1 ≤ C|b(Ti, x ′ j) − b j (·, ·)| (Y T i , T i+1 ∩L∞) n| |∇x ′γyw| (Yi+1∩L∞) n ≤ C1|b(Ti, x ′ j) − b j (·, ·)| T (Y i , Ti+1 ∩L∞) n|w|Zi+1 =: κ1|w|Zi+1 , where the constant C1 does not depend on i, j. Similarly, if p ≤ n + 1 + 2/α, Lemma 4.2.2 shows that |γy(B j # (Ti, x ′ j, Dx) − B j # (t, x′ , Dx))w|Yi+1 ≤ ≤ C|(b(Ti, x ′ j)−b j (·, ·))| L∞([Ti,Ti+1]×R n ) n(1+[b(Ti, x ′ j)−b j (·, ·)] (M T i , T i+1 ) n)|w|Zi+1 ≤ C2η|w|Zi+1 =: κ2|w|Zi+1 , again with a constant not depending on i, j. In view of (4.24) and (4.25), it is clear that κ1, κ2 tend to zero if η and δ do so. This, together with Theorem 4.2.2 and an estimate 70 � ,
analogous to (4.17), shows that S ij −1 H : Zi+1(ϕjVi (f, h)) → Zi+1 is contractive if both η and δ are sufficiently small. Assuming this, we thus obtain, aside from (4.32), ϕjv = (I − S ij H )|−1 Zi+1(ϕjV −1 H, i (f, h))hij H (f, h, v), j = 0, . . . , Nb. Multiplying these equations by ψj and summing over all j then results in Nb � v = GH(v) := j=0 + ψj(I − S ij H )|−1 Zi+1(ϕjV −1 H, i (f, h))hij H Nb+N � j=Nb+1 (f, h, v) ψj(I − S ij )| −1 Zi+1(ϕjV −1 H, i (f, h))hij (f, v), (4.36) a fixed point equation for v ∈ Zi+1(v i ). Since GH leaves this space invariant, the contraction principle is applicable, provided that GH is a strict contraction. To verify that this can be arranged by selecting δ sufficiently small, we let v, ¯v ∈ Zi+1(v i ) and estimate with the aid of Theorem 4.1.1 and 4.2.2 |GH(v)−GH(¯v)|Zi+1 = � Nb � � � j=0 Nb+N � + j=Nb Nb+N � ≤ C( j=0 ψj(I − S ij H )|−1 Zi+1(0) Lij H, i+1 ψj(I − S ij )| −1 Zi+1(0) Lij � � k ∗ Cj(·, x, Dx)(v − ¯v) γyCH, j(·, x)(v − ¯v) i+1k ∗ Cj(·, � � x, Dx)(v − ¯v) � Zi+1 |Cj(·, x, Dx)(v − ¯v)|Xi+1 + Nb � |γyCH, j(t, x)(v − ¯v)|Yi+1 ), with C > 0 not depending on δ. By extension to J × Rn+1 , estimate (4.14), and restriction to J × R n+1 + , we obtain for the first sum Nb+N � j=0 |Cj(·, x, Dx)(v − ¯v)|Xi+1 ≤ C1(ε + Cε|k| L1(0,δ))|v − ¯v|Zi+1 , (4.37) where ε > 0 can be chosen arbitrary small and C1, Cε > 0 do not depend on δ. Turning to the second sum, we introduce the space which is normed by 0Z 1/2 i+1 = 0H α 2 |w| 0Z 1/2 i+1 j=0 p (J; Lp(R n+1 + )) ∩ Lp(J; H 1 p(R n+1 = |Bk w|Xi+1 α/2 + |∇xw| n+1 X , i+1 + )), where k α/2(t) = t α/2−1 , t > 0, and Bk α/2 = (k α/2∗) −1 in Xi+1. Suppose that u ∈ Zi+1(0). By causality, B 2 k α/2 u| [0,Ti] = 0, and so we have |u| 0Z 1/2 = |(kα/2χ [0,Ti+1−Ti]) ∗ B i+1 2 k u|Xi+1 α/2 + |∇xu| n+1 Xi+1 ≤ |kα/2| L1(0,Ti+1−Ti)|B 2 k u|Xi+1 α/2 + ε|∇2xu| + Cε|u|Xi+1 ≤ C0|k α/2| L1(0,δ)|Bku|Xi+1 + ε|∇2 xu| + Cε|k| L1(0,δ)|Bku|Xi+1 ≤ (C0|k α/2| L1(0,δ) + ε + Cε|k| L1(0,δ))|u|Zi+1 ∀u ∈ Zi+1(0), (4.38) 71
- 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 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: One can then construct functions a
- 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
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analogous to (4.17), shows that S ij<br />
−1<br />
H : Zi+1(ϕjVi (f, h)) → Zi+1 is contractive if both η<br />
and δ are sufficiently small.<br />
Assuming this, we thus obtain, aside from (4.32),<br />
ϕjv = (I − S ij<br />
H )|−1<br />
Zi+1(ϕjV −1<br />
H, i (f, h))hij H<br />
(f, h, v), j = 0, . . . , Nb.<br />
Multiplying these equations by ψj and summing over all j then results in<br />
Nb �<br />
v = GH(v) :=<br />
j=0<br />
+<br />
ψj(I − S ij<br />
H )|−1<br />
Zi+1(ϕjV −1<br />
H, i (f, h))hij H<br />
Nb+N �<br />
j=Nb+1<br />
(f, h, v)<br />
ψj(I − S ij )| −1<br />
Zi+1(ϕjV −1<br />
H, i (f, h))hij (f, v), (4.36)<br />
a fixed point equation for v ∈ Zi+1(v i ). Since GH leaves this space invariant, the contraction<br />
principle is applicable, provided that GH is a strict contraction.<br />
To verify that this can be arranged by selecting δ sufficiently small, we let v, ¯v ∈<br />
Zi+1(v i ) and estimate <strong>with</strong> the aid of Theorem 4.1.1 and 4.2.2<br />
|GH(v)−GH(¯v)|Zi+1 =<br />
� Nb<br />
�<br />
�<br />
�<br />
j=0<br />
Nb+N �<br />
+<br />
j=Nb<br />
Nb+N �<br />
≤ C(<br />
j=0<br />
ψj(I − S ij<br />
H )|−1<br />
Zi+1(0) Lij<br />
H, i+1<br />
ψj(I − S ij )| −1<br />
Zi+1(0) Lij<br />
� �<br />
k ∗ Cj(·, x, Dx)(v − ¯v)<br />
γyCH, j(·, x)(v − ¯v)<br />
i+1k ∗ Cj(·,<br />
�<br />
�<br />
x, Dx)(v − ¯v)<br />
� Zi+1<br />
|Cj(·, x, Dx)(v − ¯v)|Xi+1 +<br />
Nb �<br />
|γyCH, j(t, x)(v − ¯v)|Yi+1 ),<br />
<strong>with</strong> C > 0 not depending on δ. By extension to J × Rn+1 , estimate (4.14), and<br />
restriction to J × R n+1<br />
+ , we obtain for the first sum<br />
Nb+N �<br />
j=0<br />
|Cj(·, x, Dx)(v − ¯v)|Xi+1 ≤ C1(ε + Cε|k| L1(0,δ))|v − ¯v|Zi+1 , (4.37)<br />
where ε > 0 can be chosen arbitrary small and C1, Cε > 0 do not depend on δ.<br />
Turning to the second sum, we introduce the space<br />
which is normed by<br />
0Z 1/2<br />
i+1 = 0H α<br />
2<br />
|w| 0Z 1/2<br />
i+1<br />
j=0<br />
p (J; Lp(R n+1<br />
+ )) ∩ Lp(J; H 1 p(R n+1<br />
= |Bk w|Xi+1 α/2 + |∇xw| n+1<br />
X ,<br />
i+1<br />
+ )),<br />
where k α/2(t) = t α/2−1 , t > 0, and Bk α/2 = (k α/2∗) −1 in Xi+1. Suppose that u ∈ Zi+1(0).<br />
By causality, B 2 k α/2 u| [0,Ti] = 0, and so we have<br />
|u|<br />
0Z 1/2 = |(kα/2χ [0,Ti+1−Ti]) ∗ B<br />
i+1<br />
2 k u|Xi+1 α/2 + |∇xu| n+1<br />
Xi+1 ≤ |kα/2| L1(0,Ti+1−Ti)|B 2 k u|Xi+1 α/2 + ε|∇2xu| + Cε|u|Xi+1<br />
≤ C0|k α/2| L1(0,δ)|Bku|Xi+1 + ε|∇2 xu| + Cε|k| L1(0,δ)|Bku|Xi+1<br />
≤ (C0|k α/2| L1(0,δ) + ε + Cε|k| L1(0,δ))|u|Zi+1 ∀u ∈ Zi+1(0), (4.38)<br />
71
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