Quasilinear parabolic problems with nonlinear boundary conditions
Quasilinear parabolic problems with nonlinear boundary conditions Quasilinear parabolic problems with nonlinear boundary conditions
which is necessary for v to be a solution of (4.40) on [0, Ti+1]. Let Zi+1 = H α p ([0, Ti+1]; Lp(Ω)) ∩ Lp([0, Ti+1]; H 2 p(Ω)) for i = 0, . . . , M − 1, and set as usual Z1(v 0 ) = {w ∈ Z1 : ∂ m t w|t=0 = ∂ m t f|t=0, if α > m + 1/p, m = 0, 1}, as well as Zi+1(v i ) := {w ∈ Zi+1 : w| [0,Ti] = v i } for i > 0, where v i denotes the unique solution of (4.40) on [0, Ti] determined in the ith time step. In similar fashion as in the proofs of Theorem 4.1.1 and Theorem 4.2.3, one can now show that for each i = 0, . . . , M − 1, the mapping G leaves Zi+1(v i ) invariant and is a strict contraction, provided that δ is small enough. In fact, one uses the estimates derived in the above proofs together with the diffeomorphism property of the coordinate transformations ¯x = ˜ ϑj(x). Observe that we have uniform bounds for |Θj| and |Θ −1 j | w.r.t. i and j, as only the spatial variables are transformed and by compactness of Γ. So for sufficiently small δ, equation (4.52) admits a unique solution v =: Qi+1(f, g, h) in the space Zi+1(v i ). To see that this function indeed solves (4.40) on [0, Ti+1], one can again argue as in the proofs of Theorem 4.1.1 and Theorem 4.2.3. Here one has to consider the linear operators Ki+1 defined by + = N� � N1 Ki+1(f1, f2, f3) = (K (1) i+1 (f1, f2, f3), 0 , K (3) i+1 (f1, f2, f3)) � [A, ψj] h F ij(f1, Qi+1(f1, f2, f3)) + j=0 N2 � j=N1+1 [A, ψj] h j=N2+1 N ij (f1, f3, Qi+1(f1, f2, f3)), 0 , γΓN j=N2+1 [A, ψj] h D ij (f1, f2, Qi+1(f1, f2, f3)) N� [B, ψj] h N � ij (f1, f3, Qi+1(f1, f2, f3)) , for triples (f1, f2, f3) in the product space of the regularity classes of f, g, h whose temporal traces at t = 0 coincide with those of f, g, h (including the first temporal derivative) whenever these traces exist. Since Ki+1 contains only terms of lower order, one can prove existence of a triple (f1, f2, f3) satisfying (f1, f2, f3)+Ki+1(f1, f2, f3) = (f, g, h), provided that δ is sufficiently small. By simple computations, one shows then that Qi+1(f1, f2, f3) solves (4.40) on [0, Ti+1]. Finally, uniqueness implies v = Qi+1(f, g, h) = Qi+1(f1, f2, f3). This establishes the sufficiency part. We turn now to necessity. Suppose that v ∈ Z solves (4.40). Clearly, ϕjv ∈ Z for all j = 0, . . . , N, which in turn implies Θ −1 j (ϕjv) ∈ Hα p (J; Lp(R n+1 + )) ∩ Lp(J; H2 p(R n+1 + )) for all j = N1 + 1, . . . , N. Observe further that all those terms on the right-hand sides of (4.48), (4.50), and the first equation of (4.46) (i = M − 1) which involve the function v have the regularity desired for the corresponding inhomogeneity (f, g resp. h) on J × Rn+1 resp. J × R n+1 + . In view of the Theorems 4.1.1, 4.2.3, 4.2.4, and the diffeomorphism property of the variable transformations ¯x = ˜ ϑj(x), it thus follows that the functions ϕjf, ϕjg, and ϕjh enjoy the desired regularity on J ×Ω, J ×ΓD, and J ×ΓN, respectively, for each j. On account of �N j=0 ϕj(x) ≡ 1 on Ω, we eventually obtain the desired regularity for the data f, g, h themselves. The compatibility conditions can be seen in the same way. � 78
Chapter 5 Linear Viscoelasticity In this chapter we shall study a linear parabolic problem of second order which arises in the theory of viscoelasticity. In comparison to the problems investigated in the previous chapter, it has two new challenging features: (1) it is a vector-valued problem, and (2) it contains two independent kernels. As before we shall characterize unique existence of the solution in a certain class of optimal Lp-regularity in terms of regularity and compatibility conditions on the given data. The chapter is organized as follows. At first we recall the model equations from linear viscoelasticity, here following the presentation given in Prüss [63, Section 5]. In the second part we state the problem and discuss the assumptions on the kernels. The third and main part of this chapter is devoted to the thorough investigation of a half space case of the problem. For a derivation of the fundamental equations of continuum mechanics and of linear viscoelasticity, we further refer to the books by Christensen [12], Gurtin [41], Pipkin [62], and Gripenberg, Londen, Staffans [39]. 5.1 Model equations Let Ω ⊂ R 3 be an open set with boundary ∂Ω of class C 2 . The set Ω shall represent a body, i.e. a solid or fluid material. Acting forces lead to a deformation of the body, displacing every material point x ∈ Ω at time t to the point x + u(t, x). The vector field u : R × Ω → R 3 is called the displacement field, or briefly displacement. The velocity v(t, x) of the material point x ∈ Ω at time t is then given by v(t, x) = ˙u(t, x), the dot indicating partial derivative w.r.t. t. The deformation of the body induces a strain E(t, x), which will depend linearly on the gradient ∇u(t, x), provided that the deformation is small enough. We will put E(t, x) = 1 2 (∇u(t, x) + (∇u(t, x))T ), t ∈ R, x ∈ Ω, (5.1) i.e. E is the symmetric part of the displacement gradient ∇u. The strain in turn causes stress in a way which has to be specified, expressing the properties of the material the body is made of. The stress is described by the symmetric tensor S(t, x). If ρ denotes the mass density and assuming that it is time independent, i.e. ρ(t, x) = ρ0(x), the balance of momentum law implies ρ0(x)ü(t, x) = div S(t, x) + ρ0(x)f(t, x), t ∈ R, x ∈ Ω, (5.2) 79
- 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 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: v is a solution of (4.40) on Ji+1 :
- 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
Chapter 5<br />
Linear Viscoelasticity<br />
In this chapter we shall study a linear <strong>parabolic</strong> problem of second order which arises in<br />
the theory of viscoelasticity. In comparison to the <strong>problems</strong> investigated in the previous<br />
chapter, it has two new challenging features: (1) it is a vector-valued problem, and (2)<br />
it contains two independent kernels. As before we shall characterize unique existence<br />
of the solution in a certain class of optimal Lp-regularity in terms of regularity and<br />
compatibility <strong>conditions</strong> on the given data.<br />
The chapter is organized as follows. At first we recall the model equations from<br />
linear viscoelasticity, here following the presentation given in Prüss [63, Section 5]. In<br />
the second part we state the problem and discuss the assumptions on the kernels. The<br />
third and main part of this chapter is devoted to the thorough investigation of a half<br />
space case of the problem.<br />
For a derivation of the fundamental equations of continuum mechanics and of linear<br />
viscoelasticity, we further refer to the books by Christensen [12], Gurtin [41], Pipkin<br />
[62], and Gripenberg, Londen, Staffans [39].<br />
5.1 Model equations<br />
Let Ω ⊂ R 3 be an open set <strong>with</strong> <strong>boundary</strong> ∂Ω of class C 2 . The set Ω shall represent<br />
a body, i.e. a solid or fluid material. Acting forces lead to a deformation of the body,<br />
displacing every material point x ∈ Ω at time t to the point x + u(t, x). The vector field<br />
u : R × Ω → R 3 is called the displacement field, or briefly displacement. The velocity<br />
v(t, x) of the material point x ∈ Ω at time t is then given by v(t, x) = ˙u(t, x), the dot<br />
indicating partial derivative w.r.t. t.<br />
The deformation of the body induces a strain E(t, x), which will depend linearly on<br />
the gradient ∇u(t, x), provided that the deformation is small enough. We will put<br />
E(t, x) = 1<br />
2 (∇u(t, x) + (∇u(t, x))T ), t ∈ R, x ∈ Ω, (5.1)<br />
i.e. E is the symmetric part of the displacement gradient ∇u.<br />
The strain in turn causes stress in a way which has to be specified, expressing the<br />
properties of the material the body is made of. The stress is described by the symmetric<br />
tensor S(t, x). If ρ denotes the mass density and assuming that it is time independent,<br />
i.e. ρ(t, x) = ρ0(x), the balance of momentum law implies<br />
ρ0(x)ü(t, x) = div S(t, x) + ρ0(x)f(t, x), t ∈ R, x ∈ Ω, (5.2)<br />
79