Quasilinear parabolic problems with nonlinear boundary conditions
Quasilinear parabolic problems with nonlinear boundary conditions Quasilinear parabolic problems with nonlinear boundary conditions
Lemma 4.2.1 Let 1 < p < ∞, s1, s2 ∈ (0, 1), 0 < T2, 0 ≤ T1 ≤ T2, and Ω be an arbitrary domain in R n . Then |mf| Y T 2 ≤ |m| Y T 1 ∩L∞ |f| Y T 1 ∩L∞ + 2|m| Y T 1 , T 2 ∩L∞ |f| Y T 2 ∩L∞ (4.26) for all m, f ∈ Y T2 ∩ L∞([0, T2] × Ω). We turn now to products where one factor might be unbounded. Such a constellation arises for example when s1 < 1/p. We confine ourselves to the case Ω = R n . By means of extension and restriction, the subsequent multiplication property can be transferred to domains with sufficiently smooth boundary. Suppose f ∈ Y T2 and m ∈ C r1 ([0, T2]; C(R n )) ∩ C([0, T2]; C r2 (R n )) =: M T2 for some si < ri < 1, i = 1, 2. Letting J be a subinterval of [0, T2], we then estimate [mf] Lp(J;B s 2 pp(R n )) where � � I(m, f) ≤ ( J R n � = ( � � J Rn Rn ≤ |m| s L∞(J×Ω)[f] Lp(J;B 2 pp(Rn )) � B1(y) =: I1(m, f) + I2(m, f). |m(t, x)f(t, x) − m(t, y)f(t, y)| p |m(t, x) − m(t, y)| p |f(t, y)| p |x − y| n+s2p |x − y| n+s2p + I(m, f), dx dy dt) 1 � p + ( � J Rn � dx dy dt) 1 p · · ·) 1 p R n \B1(y) We put [m] C(J; Cr2 (Rn )) = supt∈J, x, y∈Rn |m(t, x) − m(t, y)| |x − y| −r2 . By hypothesis 1 − s2/r2 > 0. If η ∈ [0, 1 − s2/r2), then (1 − η)r2 − s2 > 0, and we obtain I1(m, f) ≤ (2|m| L∞(J×Rn )) η [m] 1−η C(J; Cr2 (Rn )) ( � � � |f(t, y)| p dx dy dt 1 ) p |x − y| n−((1−η)r2−s2)p Further, Hence [mf] Y T 2 2 J Rn B1(y) ≤ C(p, r2, s2, n)|m| η L∞(J×Rn ) [m]1−η C(J; Cr2 (Rn )) |f| Lp(J×Rn ). � � I2(m, f) ≤ 2|m| L∞(J×Rn )( J Rn � Rn \B1(y) ≤ C(p, s2, n)|m| L∞(J×Rn )|f| Lp(J×Rn ). |f(t, y)| p dx dy dt |x − y| n+s2p ≤ [mf] Lp([0,T1];B s2 pp(Rn )) + [mf] Lp([T1,T2];B s2 pp(Rn )) ≤ C(|m| L∞([0,T1]×Rn ) + [m] C([0,T1]; Cr2 (Rn )) )(|f| Lp([0,T1]×Rn ) + [f] T Y 1 ) 2 +|m| L∞([T1,T2]×Rn � ) [f] T Y 2 + C(1 + [m] C([T1,T2]; C 2 r2 (Rn )) )|f| Lp([0,T2]×Rn ) Let now [m] Cr1 (J; C(Rn )) = supt, τ∈J, x∈Rn |m(t, x) − m(τ, x)| |t − τ| −r1 . To estimate , we take (4.23) with Ω = Rn as starting point, apply Fubini’s theorem, and [mf] T Y 2 1 use the same estimation techniques as for [mf] T Y 2 2 [mf] Y T 2 1 to the result ≤ |m| L∞([0,T1]×Rn )[f] T Y 1 + C1[m] Cr1 ([0,T1]; C(R 1 n )) |f| Lp([0,T1]×Rn ) +2|m| L∞([T1,T2]×Rn )([f] T Y 2 + C2[m] Cr1 ([T1,T2]; C(R 1 n )) |f| Lp([0,T2]×Rn )). 66 ) 1 p � .
We remark that the constant C2 stems from finding an upper bound for the integral � T2 T1 |t − τ|−β dt, τ ∈ [0, T2], where β < 1 is a fixed number. Thus if T1 and T2 vary within the set [0, T ], C2 can be chosen to be independent of those numbers. Set and |m| M T 1 = |m| L∞([0,T1]×R n ) + [m] C r 1 ([0,T1]; C(R n )) + [m] C([0,T1]; C r 2 (R n )) [m] M T 1 , T 2 = [m] C r 1 ([T1,T2]; C(R n )) + [m] C([T1,T2]; C r 2 (R n )) . Then our observations can be summarized as follows. Lemma 4.2.2 Suppose that 1 < p < ∞, 0 < T , 0 ≤ T1 ≤ T2 ≤ T , and 0 < si < ri < 1 for i = 1, 2. Let Y T2 s1 = Bpp([0; T2]; Lp(Rn )) ∩ Lp([0, T2]; Bs2 pp(Rn )), and M T2 r1 = C ([0, T2]; C(Rn )) ∩ C([0, T2]; Cr2 (Rn )). Then there exists a constant C > 0 not depending on T1 and T2, such that � |mf| Y T2 ≤ C (4.27) for all m ∈ M T2 and f ∈ Y T2 . 4.2.3 Variable coefficients � |m| M T 1 |f| Y T 1 + |m| L∞([T1,T2]×R n )(1 + [m] M T 1 , T 2 )|f| Y T 2 The goal of this subparagraph is to extend the results proven in Subsection 4.2.1 to variable coefficients. Besides we no longer stick to differential operators consisting only of the principal part but consider general operators of second (Volterra equation) respectively first order (boundary condition). To start with, recall the notation R n+1 + = {(x′ , y) ∈ Rn+1 : x ′ ∈ Rn , y > 0}. Define the operators A(t, x, Dx) and B(t, x ′ , Dx) by and A(t, x, Dx) = −a(t, x) : ∇ 2 x + a1(t, x) · ∇x + a0(t, x), t ∈ J, x ∈ R n+1 + , B(t, x ′ , Dx) = −∂y + b(t, x ′ ) · ∇x ′ + b0(t, x ′ ), t ∈ J, x ′ ∈ R n , (4.28) respectively. We are concerned with the two separate problems � v + k ∗ A(·, x, Dx)v = f, t ∈ J, x ∈ R n+1 + , v = g, t ∈ J, x ′ ∈ R n , y = 0 � v + k ∗ A(·, x, Dx)v = f, t ∈ J, x ∈ R n+1 + , B(t, x ′ , Dx)v = h, t ∈ J, x ′ ∈ R n , y = 0 and seek, as in Subsection 4.2.1, unique solutions in the regularity class Z = H α p (J; Lp(R n+1 + )) ∩ Lp(J; H 2 p(R n+1 + )). Concerning (4.29), we have the following result. , (4.29) , (4.30) Theorem 4.2.3 Let 1 < p < ∞, J = [0, T ], n ∈ N, and k ∈ K 1 (α, θ), where θ < π, and α ∈ (0, 2) \ � 1 p , 2 2p−1 1 3 , 1 + p , 1 + 2p−1 � . Suppose a ∈ Cul(J × R n+1 + , Sym{n + 1}), a1 ∈ L∞(J × R n+1 + , Rn+1 ), a0 ∈ L∞(J × R n+1 + ), and assume further that there exists c0 > 0 such that a(t, x)ξ · ξ ≥ c0|ξ| 2 , t ∈ J, x ∈ Rn+1 , ξ ∈ Rn+1 . Then (4.29) has a unique solution in the space Z if and only if the data f and g are subject to the conditions (i)-(vi) stated in Theorem 4.2.1. 67
- 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 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: endowed with the norm | · | Y T 2
- 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)
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- 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
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- 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
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Lemma 4.2.1 Let 1 < p < ∞, s1, s2 ∈ (0, 1), 0 < T2, 0 ≤ T1 ≤ T2, and Ω be an<br />
arbitrary domain in R n . Then<br />
|mf| Y T 2 ≤ |m| Y T 1 ∩L∞ |f| Y T 1 ∩L∞ + 2|m| Y T 1 , T 2 ∩L∞ |f| Y T 2 ∩L∞ (4.26)<br />
for all m, f ∈ Y T2 ∩ L∞([0, T2] × Ω).<br />
We turn now to products where one factor might be unbounded. Such a constellation<br />
arises for example when s1 < 1/p. We confine ourselves to the case Ω = R n . By means<br />
of extension and restriction, the subsequent multiplication property can be transferred<br />
to domains <strong>with</strong> sufficiently smooth <strong>boundary</strong>.<br />
Suppose f ∈ Y T2 and m ∈ C r1 ([0, T2]; C(R n )) ∩ C([0, T2]; C r2 (R n )) =: M T2 for some<br />
si < ri < 1, i = 1, 2. Letting J be a subinterval of [0, T2], we then estimate<br />
[mf] Lp(J;B s 2<br />
pp(R n ))<br />
where<br />
� �<br />
I(m, f) ≤ (<br />
J<br />
R n<br />
�<br />
= (<br />
�<br />
�<br />
J Rn Rn ≤ |m| s<br />
L∞(J×Ω)[f]<br />
Lp(J;B 2<br />
pp(Rn ))<br />
�<br />
B1(y)<br />
=: I1(m, f) + I2(m, f).<br />
|m(t, x)f(t, x) − m(t, y)f(t, y)| p<br />
|m(t, x) − m(t, y)| p |f(t, y)| p<br />
|x − y| n+s2p<br />
|x − y| n+s2p<br />
+ I(m, f),<br />
dx dy dt) 1<br />
�<br />
p + (<br />
�<br />
J Rn �<br />
dx dy dt) 1<br />
p<br />
· · ·) 1<br />
p<br />
R n \B1(y)<br />
We put [m] C(J; Cr2 (Rn )) = supt∈J, x, y∈Rn |m(t, x) − m(t, y)| |x − y| −r2 . By hypothesis<br />
1 − s2/r2 > 0. If η ∈ [0, 1 − s2/r2), then (1 − η)r2 − s2 > 0, and we obtain<br />
I1(m, f) ≤ (2|m| L∞(J×Rn )) η [m] 1−η<br />
C(J; Cr2 (Rn )) (<br />
� � �<br />
|f(t, y)| p dx dy dt 1<br />
) p<br />
|x − y| n−((1−η)r2−s2)p<br />
Further,<br />
Hence<br />
[mf] Y T 2<br />
2<br />
J Rn B1(y)<br />
≤ C(p, r2, s2, n)|m| η<br />
L∞(J×Rn ) [m]1−η<br />
C(J; Cr2 (Rn )) |f| Lp(J×Rn ).<br />
� �<br />
I2(m, f) ≤ 2|m| L∞(J×Rn )(<br />
J Rn �<br />
Rn \B1(y)<br />
≤ C(p, s2, n)|m| L∞(J×Rn )|f| Lp(J×Rn ).<br />
|f(t, y)| p dx dy dt<br />
|x − y| n+s2p<br />
≤ [mf] Lp([0,T1];B s2 pp(Rn )) + [mf] Lp([T1,T2];B s2 pp(Rn ))<br />
≤ C(|m| L∞([0,T1]×Rn ) + [m] C([0,T1]; Cr2 (Rn )) )(|f| Lp([0,T1]×Rn ) + [f] T<br />
Y 1 )<br />
2<br />
+|m| L∞([T1,T2]×Rn �<br />
) [f] T<br />
Y 2 + C(1 + [m] C([T1,T2]; C<br />
2<br />
r2 (Rn )) )|f| Lp([0,T2]×Rn )<br />
Let now [m] Cr1 (J; C(Rn )) = supt, τ∈J, x∈Rn |m(t, x) − m(τ, x)| |t − τ| −r1 . To estimate<br />
, we take (4.23) <strong>with</strong> Ω = Rn as starting point, apply Fubini’s theorem, and<br />
[mf] T<br />
Y 2<br />
1<br />
use the same estimation techniques as for [mf] T<br />
Y 2<br />
2<br />
[mf] Y T 2<br />
1<br />
to the result<br />
≤ |m| L∞([0,T1]×Rn )[f] T<br />
Y 1 + C1[m] Cr1 ([0,T1]; C(R<br />
1<br />
n )) |f| Lp([0,T1]×Rn )<br />
+2|m| L∞([T1,T2]×Rn )([f] T<br />
Y 2 + C2[m] Cr1 ([T1,T2]; C(R<br />
1<br />
n )) |f| Lp([0,T2]×Rn )).<br />
66<br />
) 1<br />
p<br />
�<br />
.