Quasilinear parabolic problems with nonlinear boundary conditions
Quasilinear parabolic problems with nonlinear boundary conditions Quasilinear parabolic problems with nonlinear boundary conditions
and ψ1, ψ2 are subject to the compatibility conditions and ψ1|t=0 = 2 3 γ∇x · v0 − 4 3 γ∂yw0, (5.62) ψ2|t=0 = −γ∇x · v0 − γ∂yw0, (5.63) ∂tψ2|t=0 = −(∇x · hv + ∂yhw)|t=0, if η > 0. (5.64) All in all we have established necessity of (N3) (5.14), (5.16), (5.17), (5.55), (5.57) − (5.64). That these conditions are also sufficient for the existence of a unique pair (v, w) ∈ Z solving (5.12) and satisfying (5.49), is shown in the following. Suppose that (N3) is fulfilled. We first investigate the regularity of q0. Using assumptions (5.61)-(5.64) we see that with ψ1 ∈ Y, ψ2 ∈ Yκ and where Therefore q0 = 1 2 2A(b − 3a) ∗ (ψ2 − γp1) + ( 1 2ψ1 + 1 3ψ2 + γ∂yw1), 0Yκ := 0B ψ2 − γp1 ∈ 0Yκ , 1 2 ψ1 + 1 3 ψ2 + γ∂yw1 ∈ 0Y, δb2 (1− 1 p )+κ pp δb2 (1− q0 ∈ 0B 1 p ) pp (J; Lp(R n )) ∩ 0H κ p (J; B (J; Lp(R n )) ∩ Lp(J; B 1 1− p pp (R n )). 1 1− p pp (R n )), and so, by the same conclusions as in the previous case, we find that e −Gy δb2 +κ p0 ∈ 0Hp (J; Lp(R n+1 + )) ∩ 0H κ p (J; H 1 p(R n+1 + )). Next we look at p1. From (5.14),(5.55),(5.57),(5.58), and (5.60) it follows by Theorem 3.3.1 that (5.27) has a unique solution φp in the space H δa p (J; H −1 p (R n+1 )) ∩ H κ p (J; H 1 p(R n+1 )). So we can argue as in the case κ ∈ (0, 1/p) to see that (5.12) admits a unique solution (v, w) ∈ Z with (5.49). Theorem 5.3.3 Let 1 < p < ∞, and suppose that the kernels a �= 0 and b are of type (E). Let δa and δb denote the regularization order of a and b, respectively, and assume that κ = δa−δb > 1/p. Suppose further that δa �= 2 p−1 as well as p(2δa−δb) �= 2+δb+2p.Then (5.12) has a unique solution (v, w) ∈ Z satisfying (5.49) if and only if the data are subject to the conditions (N3). 94
Chapter 6 Nonlinear Problems 6.1 Quasilinear problems of second order with nonlinear boundary conditions Let Ω be a bounded domain in R n with C 2 boundary Γ which decomposes as Γ = ΓD∪ΓN with dist(ΓD, ΓN) > 0. Let further J0 = [0, T0] be a compact time-interval and U0 ⊂ R, U1 ⊂ R n be nonempty open convex sets. With the functions a : J0×Ω×U0×U1 → R n×n , f, g : J0 × Ω × U0 × U1 → R, b D : J0 × ΓD × U0 → R, and b N : J0 × ΓN × U0 × U1 → R, we put A(u)(t, x) = −a(t, x, u(t, x), ∇u(t, x)), t ∈ J0, x ∈ Ω, F (u)(t, x) = f(t, x, u(t, x), ∇u(t, x)), t ∈ J0, x ∈ Ω, G(u)(t, x) = g(t, x, u(t, x), ∇u(t, x)), t ∈ J0, x ∈ Ω, BD(u)(t, x) = b D (t, x, u(t, x)), t ∈ J0, x ∈ ΓD, BN(u)(t, x) = b N (t, x, u(t, x), ∇u(t, x)), t ∈ J0, x ∈ ΓN, where ∇ = ∇x refers to the spatial variables, and u : J0 × Ω → R is a C(J0; C 1 (Ω)) function subject to u(t, x) ∈ U0 and ∇u(t, x) ∈ U1, for all t ∈ J0, x ∈ Ω. Let further k ∈ BVloc(R+) ∩ K 1 (1 + α, θ) with k(0) = 0, θ < π and α ∈ [0, 1). Then the problem under consideration reads as ⎧ ⎪⎨ ⎪⎩ ∂tu + dk ∗ (A(u) : ∇ 2 u) = F (u) + dk ∗ G(u), t ∈ J0, x ∈ Ω BD(u) = 0, t ∈ J0, x ∈ ΓD BN(u) = 0, t ∈ J0, x ∈ ΓN u|t=0 = u0, x ∈ Ω. (6.1) Our goal is to prove unique existence of a local strong solution, more precisely, we are looking for an interval J = [0, T ] with 0 < T ≤ T0 and a unique solution u of (6.1) on J in the space Z T := H 1+α p (J; Lp(Ω)) ∩ Lp(J; H 2 p(Ω)). This will be achieved under appropriate assumptions by means of maximal Lp-regularity of a linear problem related to (6.1) and the contraction mapping principle. To fix notation, we denote the independent variables by t ∈ J0, x ∈ Ω, ξ ∈ U0, and η ∈ U1. For what is to follow we need the spaces X T = Lp(J; Lp(Ω)), X T 1 = H α p (J; Lp(Ω)), (6.2) 95
- Page 46 and 47: 3.2 A general trace theorem Let X b
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- Page 50 and 51: Theorem 3.4.2 Suppose X is a Banach
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- 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
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- 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
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- Page 87 and 88: To see the converse direction, supp
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- Page 93 and 94: elongs to H∞ (Σ π 2 +η × Ση
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- 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
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- Page 113 and 114: Bibliography [1] Albrecht, D.: Func
- Page 115 and 116: [54] Lunardi, A.: On the heat equat
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and ψ1, ψ2 are subject to the compatibility <strong>conditions</strong><br />
and<br />
ψ1|t=0 = 2<br />
3 γ∇x · v0 − 4<br />
3 γ∂yw0, (5.62)<br />
ψ2|t=0 = −γ∇x · v0 − γ∂yw0, (5.63)<br />
∂tψ2|t=0 = −(∇x · hv + ∂yhw)|t=0, if η > 0. (5.64)<br />
All in all we have established necessity of<br />
(N3) (5.14), (5.16), (5.17), (5.55), (5.57) − (5.64).<br />
That these <strong>conditions</strong> are also sufficient for the existence of a unique pair (v, w) ∈ Z<br />
solving (5.12) and satisfying (5.49), is shown in the following.<br />
Suppose that (N3) is fulfilled. We first investigate the regularity of q0. Using assumptions<br />
(5.61)-(5.64) we see that<br />
<strong>with</strong> ψ1 ∈ Y, ψ2 ∈ Yκ and<br />
where<br />
Therefore<br />
q0 = 1 2<br />
2A(b − 3a) ∗ (ψ2 − γp1) + ( 1<br />
2ψ1 + 1<br />
3ψ2 + γ∂yw1),<br />
0Yκ := 0B<br />
ψ2 − γp1 ∈ 0Yκ , 1<br />
2 ψ1 + 1<br />
3 ψ2 + γ∂yw1 ∈ 0Y,<br />
δb2 (1− 1<br />
p )+κ<br />
pp<br />
δb2 (1−<br />
q0 ∈ 0B<br />
1<br />
p )<br />
pp<br />
(J; Lp(R n )) ∩ 0H κ p (J; B<br />
(J; Lp(R n )) ∩ Lp(J; B<br />
1<br />
1− p<br />
pp (R n )).<br />
1<br />
1− p<br />
pp (R n )),<br />
and so, by the same conclusions as in the previous case, we find that<br />
e −Gy δb2 +κ<br />
p0 ∈ 0Hp<br />
(J; Lp(R n+1<br />
+ )) ∩ 0H κ p (J; H 1 p(R n+1<br />
+ )).<br />
Next we look at p1. From (5.14),(5.55),(5.57),(5.58), and (5.60) it follows by Theorem<br />
3.3.1 that (5.27) has a unique solution φp in the space<br />
H δa<br />
p (J; H −1<br />
p (R n+1 )) ∩ H κ p (J; H 1 p(R n+1 )).<br />
So we can argue as in the case κ ∈ (0, 1/p) to see that (5.12) admits a unique solution<br />
(v, w) ∈ Z <strong>with</strong> (5.49).<br />
Theorem 5.3.3 Let 1 < p < ∞, and suppose that the kernels a �= 0 and b are of type<br />
(E). Let δa and δb denote the regularization order of a and b, respectively, and assume that<br />
κ = δa−δb > 1/p. Suppose further that δa �= 2<br />
p−1 as well as p(2δa−δb) �= 2+δb+2p.Then<br />
(5.12) has a unique solution (v, w) ∈ Z satisfying (5.49) if and only if the data are subject<br />
to the <strong>conditions</strong> (N3).<br />
94