On the nonstationary Stokes system in a cone: Asymptotics of solutions at infinity

J. Math. Anal. Appl. 486 (2020) 123821

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

On the nonstationary Stokes system in a cone: Asymptotics of solutions at infinity Vladimir Kozlov a , Jürgen Rossmann b,∗ a b

Institute of Mathematics, Linköping University, SE-58183 Linköping, Sweden Institute of Mathematics, University of Rostock, D-18051 Rostock, Germany

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 24 August 2018 Available online 3 January 2020 Submitted by P.G. Lemarie-Rieusset

The paper deals with the Dirichlet problem for the nonstationary Stokes system in a cone. The authors obtain existence and uniqueness results for solutions in weighted Sobolev spaces and study the asymptotics of the solutions at infinity. © 2020 Elsevier Inc. All rights reserved.

Keywords: Nonstationary Stokes system Conical points Asymptotics

0. Introduction Although the stationary Stokes system in domains with nonsmooth boundaries is well studied (see, e.g., the papers [3,6–8] for Lipschitz domains and the papers [4,15,20–22,25] for domains with corners and edges), there are only few papers dealing with the nonstationary Stokes system in domains with singular boundary points. One of them is our recent paper [18] for the problem in a 3-dimensional cone, the 2-dimensional case was studied in [24]. The present paper is a continuation of [18] and deals with the problem ∂u − Δu + ∇p = f, −∇ · u = g in K × (0, ∞), ∂t u(x, t) = 0 for x ∈ ∂K, t > 0, u(x, 0) = 0 for x ∈ K,

(1) (2)

where K is a 3-dimensional cone with vertex at the origin. In [18], we obtained existence, uniqueness and regularity results for solutions of the problem (1), (2) in weighted Sobolev spaces, where the weight function is a power of the distance from the vertex of the cone. However, the results in [18] are not optimal, for example, in the case that the cone K is contained in a half-space. One goal (but not the main goal) of the present paper is to improve the results for this particular case. * Corresponding author. E-mail addresses: [email protected] (V. Kozlov), [email protected] (J. Rossmann). https://doi.org/10.1016/j.jmaa.2019.123821 0022-247X/© 2020 Elsevier Inc. All rights reserved.

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When considering solutions of the problem (1), (2), the question on the behavior of the solutions both in a neighborhood of the vertex of the cone and at infinity arises. The major part of the present paper deals with the asymptotics of the solution at infinity. The asymptotics near the vertex of the cone is the subject of a forthcoming paper. Concerning the behavior at infinity, the results for the Stokes system are completely different from those for the heat equation and other parabolic problems given in [5], [9–12], [16,17]. We show in this paper that the solution is a finite sum of singular terms and a (more regular) remainder, where the singular terms depend on the eigenvalues of the Beltrami operator δ with Neumann boundary conditions on the intersection ∂Ω of ∂K with the unit sphere S 2 . In the case of the heat equation and other parabolic problems, such singular terms do not appear. Note that the eigenvalues of two different operator pencils appear in solvability and regularity results for the nonstationary Stokes system. Besides the eigenvalues of the Neumann problem for the Beltrami operator, one has to consider the eigenvalues of the pencil generated by the corresponding stationary problem (see Theorem 3.1). In analogous results for other parabolic problems (see [9–11]), only the eigenvalues of the second operator pencil play a role. The paper consists of three sections. Sections 1 and 2 are concerned with the parameter-depending problem su ˜ − Δ˜ u + ∇˜ p = f˜,

−∇ · u ˜ = g˜ in K,

u ˜ = 0 on ∂K,

(3)

which arises after the Laplace transformation with respect to the time t. Here, s is a complex number, Re s ≥ 0, s = 0. Section 1 deals with the solvability of this problem in weighted Sobolev spaces. In Subsections 1.1–1.3, we recall the main results of [18]. In particular, an existence and uniqueness result for solutions of this problem in the space Eβ2 (K) × Vβ1 (K) was obtained in [18]. Here Vβl (K) denotes the weighted Sobolev space of all functions (vector-functions) with finite norm uVβl (K) =

 

 2 1/2 r2(β−l+|α|) ∂xα u(x) dx ,

(4)

K |α|≤l

while Eβl (K) is the weighted Sobolev space with the norm uEβl (K) =

  

2 1/2  r2β + r2(β−l+|α|) ∂xα u(x) dx ,

(5)

K |α|≤l

r = |x| denotes the distance of the point x from the vertex of the cone. As was shown in [18], there are two neighboring β-intervals for which an existence and uniqueness result in the space Eβ2 (K) × Vβ1 (K) holds, namely the intervals 1 1 − λ1 < β < 2 2

and

 1 1 3 < β < min μ2 + , λ1 + . 2 2 2

(6)

Here, λ1 and μ2 are positive numbers depending on the cone. More precisely, λ1 is the smallest positive eigenvalue of the operator pencil L(λ) generated by the Dirichlet problem for the stationary Stokes system, while μ2 is the smallest positive eigenvalue of the operator pencil N (λ) generated by the Neumann problem for the Laplacian, respectively (μ2 (μ2 + 1) is the smallest positive eigenvalue of the operator −δ with Neumann boundary conditions, see Subsection 1.3). The eigenvalue λ1 is not greater than 1 since λ = 1 is always an eigenvalue of the pencil L(λ). In the case λ1 < 1, the inequalities (6) for β are sharp. However, the existence and uniqueness result given in [18] can be improved in the case that λ1 is equal to 1 and simple. This is done in Subsection 1.4. For example, the eigenvalue λ = 1 is the smallest positive eigenvalue and simple if K\{0} is contained in a half-space α1 x1 + α2 x2 + α3 x3 > 0. In this case we obtain the following

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weaker conditions on β, under which an existence and uniqueness result in the space Eβ2 (K) × Vβ1 (K) holds (see Theorem 1.2):     5 1 1 − min μ2 + 1, Re λ2 < β − < min μ2 , Re λ2 + 1 , β = ± , β = . 2 2 2

(7)

Here λ2 is the eigenvalue of L(λ) with smallest real part > 1. The uniqueness of the solution holds even for −μ2 − 12 < β < Re λ2 + 32 (see Lemma 1.5). Section 2 deals with the behavior of a solution (˜ u, p˜) ∈ Eβ2 (K) × Vβ1 (K) of the problem (3) at infinity. If 0 1 1 ∗ for example f˜ ∈ Eγ (K), g˜ ∈ Vγ (K) ∩ (V−γ (K)) ,  1 1 3 1 − λ1 < β < < γ < min μ2 + , λ1 + , 2 2 2 2

 g˜ dx = 0, K

then it follows from Lemma 1.3 of the present paper that u ˜ ∈ Eγ2 (K) and p˜ ∈ Vγ1 (K). This is not true, if the integral of g˜ over K is not equal to zero. Then we can represent (˜ u, p˜) as a sum of singular terms and a remainder (˜ v , q˜) ∈ Eγ2 (K) × Vγ1 (K). For γ < min( 32 , μ2 + 12 ), we obtain the decomposition 

 (−1)     (−1) v , q˜) u ˜, p˜ = η |s|r2 c1 (s) u0 (x, s), p0 (x) + (˜ (−1)

(−1)

with the formulas (31), (32) for u0 , p0 and c1 , where η is a smooth function on (0, ∞), η(r) = 0 for r < 1/2 and η(r) = 1 for r > 1. In the case γ > μ2 + 12 , additional singular terms appear, i.e., we obtain a decomposition   (−j,k) (−j,k)    (˜ u, p˜) = η |s|r2 cj,k (s) u0 , p0 + (˜ v , q˜), j,k

 (−j,k) (−j,k)  are singular functions depending on the eigenvalues of the Beltrami operator with where u0 , p0 Neumann boundary conditions (see Theorems 2.1 and 2.2). In Section 3, we consider the time-dependent problem (1), (2). The results of Section 1 enable us to obtain solvability results in weighted Sobolev spaces and regularity results for the solutions. Partially, these results can be found in our paper [18]. In the present paper, we weaken the conditions on the weight parameter β for the case that λ1 = 1 is the smallest positive eigenvalue of the pencil L(λ) and simple. In particular, there exists a unique solution (u, p) ∈ Wβ2,1 (Q) × L2 (R+ , Vβ1 (K)) for arbitrary f ∈ L2 (R+ , Vβ0 (K)), g ∈ 1 L2 (R+ , Vβ1 (K)), ∂t u ∈ L2 (R+ , (V−β (K))∗ ) if 12 − λ1 < β < 12 . Here, Wβ2,1 (Q) is the space of all u ∈ L2 (R+ , Vβ2 (K)) such that ∂t u ∈ L2 (R+ , Vβ0 (K)). By means of the results of Section 2, we describe the asymptotics of this solution at infinity. We prove that the velocity u is a finite sum of terms t  S

(j,k)

(x, t) =

 (j,k)  Ku (x, y, t − τ ) f (y, τ ) + Hu(j,k) (x, y, t − τ ) g(y, τ ) dy dτ

0 K (j,k)

(j,k)

and a remainder v ∈ Wγ2,1 (Q), γ > 12 , and derive point estimates for the kernels Ku and Hu derivatives. An analogous representation holds for the pressure p (see Theorem 3.4).

and their

1. Solvability of the parameter-depending problem Let Ω be a subdomain of the unit sphere S 2 with smooth (of class C 2,α ) boundary ∂Ω and let K = x ∈ R3 : ω = x/|x| ∈ Ω} be a cone with vertex at the origin. We consider the boundary value problem

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

4

s u − Δu + ∇p = f, −∇ · u = g in K,

u = 0 on ∂K\{0}.

(8)

Here, s is an arbitrary complex number, Re s ≥ 0. This section is concerned with the existence and uniqueness of solutions in the space Eβ2 (K) × Vβ1 (K). 1.1. Weighted Sobolev spaces on the cone For nonnegative integer l and real β, we define the weighted Sobolev spaces Vβl (K) and Eβl (K) as the sets of all functions (or vector functions) with finite norms (4) and (5), respectively. Note that the spaces Vβl (K) and Eβl (K) can be also defined as the closures of C0∞ (K\{0}) with respect to the above norms. ◦

1 β (K)

Furthermore, we define V

◦

and E 1β (K) as the spaces of all functions u ∈ Vβ1 (K) and u ∈ Eβ1 (K), ◦

◦

−1 respectively, which are zero on ∂K\{0}. The dual spaces of V 1β (K) and E 1β (K) are denoted by V−β (K) −1 and E−β (K), respectively. Note that the norm

u =



  1/2 r2β |u|2 + |∇u|2 dx

K ◦

is equivalent to the Eβ1 (K)-norm in E 1β (K). 1.2. The operator of the problem (8) Obviously, the mapping Eβ2 (K) × Vβ1 (K) (u, p) → f = su − Δu + ∇p ∈ Eβ0 (K) is continuous for arbitrary real β and complex s. Furthermore, the operator div realizes a continuous mapping ◦

from Eβ2 (K)∩ E 1β (K) into the space  1 ∗  1 ∗ Xβ1 (K) = Eβ1 (K) ∩ V−β (K) = Vβ1 (K) ∩ V−β (K) (see [18, Section 2.1]). We denote the operator  2  ◦ Eβ (K)∩ E 1β (K) × Vβ1 (K) (u, p)   → su − Δu + ∇p, −∇ · u ∈ Eβ0 (K) × Xβ1 (K) of the problem (8) by Aβ . Note that the integral of g over K exists if g ∈ Xβ1 (K), ˜ 1 (K) if R(Aβ ) ⊂ Eβ0 (K) × X β

1 2

< β < 52 , and that

5 1 <β< 2 2

(R(Aβ ) denotes the range of the operator Aβ ), where ˜ 1 (K) = g ∈ X 1 (K) : X β β

 g(x) dx = 0



K

˜ 1 (K) can be also defined as the closure of the set all if 12 < β < 52 (see [18, Lemma 2.12]). The space X β g ∈ C0∞ (K\{0}) satisfying the condition

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

5

 g(x) dx = 0

(9)

K

in Xβ1 (K). However, for β < Lemma 1.1. If β < Xβ1 (K).

1 2

1 2

and β > 52 , the following assertion holds analogously to [24, Lemma 2].

or β > 52 , then the set of all g ∈ C0∞ (K\{0}) satisfying the condition (9) is dense in

Let A∗β denote the adjoint operator of Aβ . This operator is defined as a continuous mapping  ∗  ∗  ∗ ◦ 0 E−β (K) × Xβ1 (K) → Eβ2 (K)∩ E 1β (K) × Vβ1 (K) .  ∗  ∗ 1 Here, Xβ1 (K) = V−β (K) + Vβ1 (K) . Since the constant function g = 1 is an element of the space  1 ∗ Xβ (K) for 12 < β < 52 , the kernel of A∗β contains the pair (v, q) = (0, c) with constant c in the case 1 5 ∗ 2 < β < 2 . Other constant elements of ker Aβ do not exist. The kernel of the operator Aβ contains no constant elements except (u, p) = (0, 0). 1.3. Bijectivity of the operator Aβ We introduce the following operator pencils L(λ) and N (λ) generated by the Dirichlet problem for the stationary Stokes system and the Neumann problem for the Laplacian in the cone K, respectively. For every complex λ, we define the operator L(λ) as the mapping 

◦

1 W (Ω) × L2 (Ω)

 →

U P

  r2−λ − Δrλ U (ω) + ∇rλ−1 P (ω) ∈ W −1 (Ω) × L2 (Ω),   −r1−λ ∇ · rλ U (ω)

where r = |x| and ω = x/|x|. The properties of the pencil L are studied, e.g., in [14]. In particular, it is ¯ and −1 − λ are simultaneously eigenvalues of the pencil L(λ) or not. The known that the numbers λ, λ eigenvalues in the strip −2 ≤ Re λ ≤ 1 are real, and the numbers 1 and −2 are always eigenvalues of the pencil L(λ). If Ω is contained in a half-sphere, then λ = 1 and λ = −2 are the only eigenvalues in the interval [−2, 1] (cf. [14, Theorem 5.5.5]). We denote the eigenvalues with positive real part by λj , j = 1, 2, . . ., while λ−j = −1 − λj are the eigenvalues with negative real part, · · · ≤ Re λ−2 < λ−1 < −1 < 0 < λ1 < Re λ2 ≤ · · · . Here, 0 < λ1 ≤ 1 and −2 ≤ λ−1 < −1. Note that the eigenvalues λj and λ−j have the same geometric and algebraic multiplicities. The operator N (λ) is defined as N (λ) U =



− δU − λ(λ + 1)U ,

∂U   ∂n ∂Ω

for U ∈ W 2 (Ω).

As is known (see e.g. [14, Section 2.3]), the eigenvalues of this pencil are real, and generalized eigenfunctions do not exist. The spectrum contains, in particular, the simple eigenvalues μ1 = 0 and μ−1 = −1 with the eigenfunction φ1 = const. The interval (−1, 0) is free of eigenvalues. Let μj , j = 1, 2, . . ., be the nonnegative eigenvalues, and let μ−j = −1 − μj be the negative eigenvalues of the pencil N (λ),

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V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

· · · < μ−2 < −1 = μ−1 < μ1 = 0 < μ2 < · · · . Obviously, μj and μ−j are the solutions of the equation μ(μ + 1) = −Λj , where Λj is the jth eigenvalue of the operator −δ with Neumann boundary condition on ∂Ω. By [18, Theorem 2.1], the operator Aβ has closed range and its kernel has finite dimension for Re s ≥ 0, s = 0, if and only if the line Re λ = −β + 1/2 does not contain eigenvalues of the pencil L(λ) and the number −β − 1/2 is not an eigenvalue of the pencil N (λ). Furthermore, the following result was proved in [18, Lemma 2.10, Theorems 2.3–2.5]. Theorem 1.1. Suppose that Re s ≥ 0 and s = 0. 1) If −μ2 − 1/2 < β < λ1 + 3/2, β = − 12 , then Aβ is injective. 2) If 12 − λ1 < β < 12 , then Aβ is an isomorphism onto Eβ0 (K) × Xβ1 (K), and the estimate uVβ2 (K) + |s| uVβ0 (K) + pVβ1 (K)   1 (K))∗ ≤ c f Vβ0 (K) + gVβ1 (K) + |s| g(V−β

(10)

is valid for every solution (u, p) ∈ Eβ2 (K) × Vβ1 (K) of the problem (8).   ˜ 1 (K), and the 3) If 12 < β < min μ2 + 12 , λ1 + 32 , then Aβ is an isomorphism onto the space Eβ0 (K) × X β 2 1 estimate (10) is valid for every solution (u, p) ∈ Eβ (K) × Vβ (K) of the problem (8). For the proof of the last theorem, the authors used the following regularity assertion. Lemma 1.2. Suppose that (u, p) ∈ Eβ2 (K) × Vβ1 (K) is a solution of the problem (8), where Re s ≥ 0, s = 0, f ∈ Eβ0 (K) ∩ Eγ0 (K) and g ∈ Xβ1 (K) ∩ Xγ1 (K). We assume that one of the following two conditions is satisfied: (i) β < γ and the interval −γ − 1/2 ≤ λ ≤ −β − 1/2 does not contain eigenvalues of the pencil N (λ), (ii) β > γ and the strip −β + 1/2 ≤ Re λ ≤ −γ + 1/2 is free of eigenvalues of the pencil L(λ). Then u ∈ Eγ2 (K), p ∈ Vγ1 (K). If β < 12 < γ, then the interval −γ − 1/2 ≤ λ ≤ −β − 1/2 contains the eigenvalue μ−1 = −1 of the pencil N (λ), and Lemma 1.2 is not applicable. Then the following regularity assertion can be easily deduced from Lemma 1.2 and [18, Lemma 2.13]. Lemma 1.3. Let (u, p) ∈ Eβ2 (K) × Vβ1 (K) be a solution of the problem (8), where Re s ≥ 0, s = 0, f ∈ Eβ0 (K) ∩ Eγ0 (K),

g ∈ Xβ1 (K) ∩ Xγ1 (K),

− λ1 < β, γ < 12 + min(μ2 , λ1 + 1) and β, γ = 12 . In the case β < condition (9). Then u ∈ Eγ2 (K) and p ∈ Vγ1 (K). 1 2

1 2

< γ, we assume that g satisfies the

1.4. The case that the eigenvalue λ1 is equal to 1 and simple   The condition 12 − λ1 < β < min μ2 + 12 , λ1 + 32 , β = 12 for the bijectivity of the operator Aβ in Theorem 1.1 is sharp if λ1 < 1 (cf. [18, Lemmas 2.14, 2.15, 2.17]). As was mentioned above, the number λ = 1 is always an eigenvalue of the pencil L(λ) with the corresponding constant eigenvector (0, 1). In the

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case that Ω is contained in a half-sphere, the eigenvalues λ = 1 and λ = −2 are simple (have geometric and algebraic multiplicity 1), and all other eigenvalues lie outside the strip −2 ≤ Re λ ≤ 1 ([14, Theorem 5.5.5]). We assume in this subsection that (as in the just described case), the smallest positive eigenvalue of the pencil L(λ) is λ1 = 1 and that this eigenvalue is simple. Since the strip −2 ≤ Re λ ≤ 1 contains only real eigenvalues of the pencil L(λ) (cf. [14, Theorem 5.3.1]), it follows then that · · · ≤ Re λ−2 < −2 = λ−1 < λ1 = 1 < Re λ2 ≤ · · · . Then the results of Theorem 1.1 can be improved. Lemma 1.4. Suppose that Re s ≥ 0, s = 0, λ1 = 1 and that λ1 is a simple eigenvalue of the pencil L(λ). Then Aβ is an isomorphism onto Eβ0 (K) × Xβ1 (K) if max(−μ2 − 12 , 12 − Re λ2 ) < β < − 12 . Proof. Suppose that max(−μ2 − 12 , 12 −Re λ2 ) < β < − 12 . Then the kernel of Aβ is trivial (see Theorem 1.1) and the range of Aβ is closed (see [18, Theorem 2.1]). Thus, it suffices to show that the problem (8) is solvable in Eβ2 (K) × Vβ1 (K) for arbitrary f ∈ C0∞ (K\{0}), g ∈ C0∞ (K\{0}). By Theorem 1.1, there exists a solution (u, p) ∈ Eγ2 (K) × Vγ1 (K), where − 12 < γ < 12 . We show that u ∈ Eβ2 (K) and p − c ∈ Vβ1 (K) for a certain constant c. Let ζ be a two times continuously differentiable function with compact support in K, ζ = 1 in a neighborhood of the vertex of the cone K, and let η = 1 − ζ. Then ζ(u, p) ∈ Vγ2 (K) × Vγ1 (K), −∇ · (ζu) = ζg − u · ∇ζ ∈ Vβ1 (K) and −Δ(ζu) + ∇(ζp) = F , where 0 F = ζf − sζu + ζΔu − Δ(ζu) + p∇ζ ∈ Vγ0 (K) ∩ Vγ−2 (K). 0 If γ − 2 ≤ β ≤ γ, then Vγ0 (K) ∩ Vγ−2 (K) ⊂ Vβ0 (K), and we conclude from well-known regularity results for solutions of elliptic boundary value problems (see, e.g., [23, Chapter 3, Theorem 5.5]) that ζu ∈ Vβ2 (K) and ζp − c ∈ Vβ1 (K) since λ1 = 1 is the only eigenvalue of the pencil L(λ) in the strip 12 − γ ≤ Re λ ≤ 12 − β. Obviously, η(u, p) ∈ Eβ2 (K) × Vβ1 (K). Thus, u ∈ Eβ2 (K) and p − c ∈ Vβ1 (K). If β < γ − 2, then we conclude 2 1 0 0 first that ζu ∈ Vγ−2 (K) and ζp − c ∈ Vγ−2 (K). In this case, we conclude that F ∈ Vγ−4 (K) ∩ Vγ−2 (K). 2 Applying again the regularity result in [23, Chapter 3, Theorem 5.5], we obtain ζu ∈ Vβ (K), ζp −c ∈ Vβ1 (K) if γ − 4 ≤ β ≤ γ. In this way, after finitely many steps, we get u ∈ Eβ2 (K) and p − c ∈ Vβ1 (K) if max(−μ2 − 12 , 12 − Re λ2 ) < β < − 12 . Obviously, the pair (u, p − c) is also a solution of the problem (8). Thus, it is shown that (8) is solvable in the space Eβ2 (K) × Vβ1 (K) for arbitrary f ∈ C0∞ (K\{0}) and g ∈ C0∞ (K\{0}). The proof of the lemma is complete. 

The last lemma allows us to improve the first assertion of Theorem 1.1 if λ1 = 1 and λ1 is a simple eigenvalue. Lemma 1.5. Suppose that Re s ≥ 0, s = 0, λ1 = 1 and that λ1 is a simple eigenvalue of the pencil L(λ). Furthermore, we assume that −μ2 − 12 < β < Re λ2 + 32 and β = − 12 . Then the operator Aβ is injective. Proof. By Theorem 1.1, the operator Aβ is injective for −μ2 − 12 < β < 52 , β = − 12 . We show that Aβ is injective for 52 ≤ β < δ + 52 , where δ = min(μ2 , Re λ2 − 1). Suppose that the kernel of Aβ is not trivial for one β in the interval 52 ≤ β < δ + 52 . As was shown in [18] (see Formula (34) in [18]), there is the relation ker A∗γ ⊃ ker Aβ if − β ≤ γ ≤ 2 − β. Thus, the kernel of A∗γ is not trivial for −β ≤ γ ≤ 2 − β. However the interval [−β, 2 − β] has a nonempty intersection with the interval (max(−μ2 − 12 , 12 − Re λ2 ), − 12 ) since 2 − β > −μ2 − 12 and 2 − β > 12 − Re λ2 for β < δ + 52 . This contradicts Lemma 1.4. Consequently, the kernel of Aβ is trivial for 52 ≤ β < δ + 52 .

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V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

Furthermore, it follows from Lemma 1.2 that ker Aβ ⊂ ker Aγ = {0} if Re λ−2 < 12 − β < 12 − γ < −2 = λ−1 . This proves the lemma. 

5 2

<γ <δ+

5 2

≤ β < Re λ2 + 32 , i.e.,

Finally, we obtain the following result if μ2 > 2. Lemma 1.6. Suppose that Re s ≥ 0, s = 0, λ1 = 1 and that λ1 is a simple eigenvalue of the pencil L(λ). Furthermore, we assume that μ2 > 2 and 52 < β < min(μ2 + 12 , Re λ2 + 32 ). Then Aβ is an isomorphism onto Eβ0 (K) × Xβ1 (K). Proof. By Lemma 1.5, the operator Aβ is injective. Furthermore, under the given condition on β, the number − 12 − β is not an eigenvalue of the pencil N (λ) (since μ−2 = −μ2 − 1 < − 12 − β < −3 < μ−1 = −1), and the line Re λ = 12 − β is free of eigenvalues of the pencil L(λ) (since Re λ−2 = −Reλ2 − 1 < 12 − β < −2 = λ−1 ). By [18, Theorem 2.1], the range of Aβ is closed. Let f ∈ C0∞ (K\{0}), g ∈ C0∞ (K\{0}), and let g satisfy the condition (9). Then by Theorem 1.1, there exists a solution (u, p) ∈ Eγ2 (K) × Vγ1 (K), where 12 < γ < 52 . Since the interval − 21 − β ≤ λ ≤ − 12 − γ does not contain eigenvalues of the pencil N (λ), it follows from Lemma 1.2 that u ∈ Eβ2 (K) and p ∈ Vβ1 (K). Consequently, the range of the operator Aβ contains the set of all pairs (f, g) of C0∞ (K\{0})-functions satisfying the condition (9). By Lemma 1.1, the set of all g ∈ C0∞ (K\{0}) satisfying the condition (9) is dense in Xβ1 (K). Thus, the range of the operator Aβ is the set Eβ0 (K) × Xβ1 (K). The proof is complete.  We can summarize the results of Theorem 1.1 and Lemmas 1.4, 1.6 as follows. Theorem 1.2. Suppose that Re s ≥ 0, s = 0, λ1 = 1 and that λ1 is a simple eigenvalue of the pencil L(λ). Then the following assertions are true. 1) If max(−μ2 − 12 , 12 − Re λ2 ) < β < 12 , β = − 12 , or 52 < β < 12 + min(μ2 , Re λ2 + 1), then Aβ is an isomorphism onto Eβ0 (K) × Xβ1 (K). ˜ 1 (K). 2) If 12 < β < min(μ2 + 12 , 52 ), then Aβ is an isomorphism onto the space Eβ0 (K) × X β Note that the operator Aβ is not Fredholm for the values ± 21 and 52 of β. Indeed, for β = − 12 , we have 1 1 1 1 1 5 2 − β = λ1 and − 2 − β = μ1 . If β = 2 , then − 2 − β = μ−1 , while 2 − β = λ−1 if β = 2 . Thus, the conditions 1 5 of [18, Theorem 2.1] are not satisfied for β = ± 2 and β = 2 . Analogously to [18, Lemmas 2.14, 2.15, 2.17], one can also show that the bounds max(−μ2 − 12 , 12 − Re λ2 ) and min(μ2 + 12 , Re λ2 + 32 ) for β in the last theorem are sharp. 2. Behavior of solutions of the parameter-depending problem at infinity Suppose that f ∈ Eβ0 (K) and g ∈ Xβ1 (K), where 12 − λ1 < β < 12 . Then by Theorem 1.1 there exists a unique solution (u, p) ∈ Eβ2 (K) × Vβ1 (K) of the problem (8). By Lemma 1.3, this solution belongs to the space Eγ2 (K) ×Vγ1 (K) if f ∈ Eγ0 (K), g ∈ Xγ1 (K) and β < γ < 12 . However, this is not true in general if γ > 12 . We show that then the solution is a sum of some singular terms and a remainder (v, q) ∈ Eγ2 (K) × Vγ1 (K). 2.1. Special solutions of the parameter-depending problem In the sequel, let ν(x) denote the distance of the point x from the boundary ∂K. Obviously, the function ν is positively homogeneous of degree 1. In the neighborhood ν(x) < δ|x| of the boundary ∂K with sufficiently small δ, the function ν is two times continuously differentiable and satisfies the equality |∇ν| = 1. Furthermore, ∇ν(x) is orthogonal to ∂K at any point x ∈ ∂K\{0}. For an arbitrary vector function v in the neighborhood ν(x) < δ|x| of ∂K, we define

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

vν = v · ∇ν

9

and vτ = v − vν ∇ν.

Obviously vτ · ∇ν = 0 near ∂K. Now, let μ be an eigenvalue of the pencil N (λ), and let φ be an eigenfunction corresponding to this eigenvalue. Then the function p0 (x) = rμ φ(ω)

(11)

is a solution of the Neumann problem Δp0 = 0 in K, ∇p0 · ∇ν = 0 on ∂K\{0}. Furthermore, let χ be a two times continuously differentiable function on (0, ∞) such that χ(r) = 1 for 2r < δ and χ(r) = 0 for r > δ. We define ν    √ u0 (x, s) = s−1 v (0) (x) − χ e−ν s vτ(0) (x) , r

where v (0) = −∇p0

(12)

√ and s is the square root of s with positive real part. More precisely, u0 (x, s) is defined by (12) for ν(x) < δ|x| and u0 (x, s) = s−1 v (0) (x) if ν(x) > δ|x|. Lemma 2.1. Let p0 and u0 be the functions (11) and (12), respectively. Then (s − Δ) u0 + ∇p0 = χ

ν  r

e−ν

√ s

f0 + R1 , −∇ · u0 = χ

ν  r

e−ν

√ s

g0 + R2 ,

where f0 = s−1/2



− vτ(0) Δν − 2

3 (0)   ∂ν ∂vτ + s−1/2 Δ vτ(0) , g0 = s−1 ∇ · vτ(0) , ∂xj ∂xj j=1

R1 , R2 are continuous in K and satisfy the estimates √ s/3

|R1 | ≤ c s−1 rμ−3 e−δr Re

,

√ s/3

|R2 | ≤ c s−1 rμ−2 e−δr Re

(13)

with a constant c independent of s and r. Furthermore, u0 = 0 on ∂K\{0}. The remainders R1 and R2 in Lemma 2.1 are √

ν  R1 = s−1 Δ, χ( ) e−ν s vτ(0) r

and R2 = s−1 e−ν

√ s

ν vτ(0) · ∇χ( ). r

Here [Δ, χ] denotes the commutator of Δ and χ. Since ∇χ( νr ) is equal to zero for 2ν < δr, the terms R1 , R2 (0)

satisfy the estimate (13). Obviously, the terms vτ Δν,

(0) ∂ν ∂vτ ∂xj ∂xj

(0)

and ∇ · vτ

are positively homogeneous of

(0) Δ vτ

degree μ − 2, while is positively homogeneous of degree μ − 3. It follows from the last lemma that     (s − Δ) u0 + ∇p0  + |s|1/2 ∇ · u0  ≤ c |s|−1/2 rμ−2 for r > |s|−1/2 . Next, we construct functions UN and PN with the leading terms u0 and p0 , respectively, such that     (s − Δ) UN + ∇PN  + |s|1/2 ∇ · UN  ≤ c |s|−(N +1)/2 rμ−N −2 (1 + | log r|k )

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

10

for r > |s|−1/2 , where k ≤ N . We introduce the polynomials P1 (ν) = −k!

k  νj j=0

, P2 (ν) =

j!

k+1 k−1  k!  (2ν)j νj (ν) = −k! (k − j + 1) , P 3 k 2 + 2 j=1 j! j! j=0

for integer k ≥ 0 (P3 = 0 if k = 0) which satisfy the equalities P1 (ν) − P1 (ν) = ν k , P2 (ν) − 2P2 (ν) = −ν k , P3 (ν) − P3 (ν) = P1 (ν) − 2P1 (ν). Then the following lemma holds. Lemma 2.2. Let f and g be positively homogeneous of degree μ in the neighborhood ν(x) < δ|x| of ∂K. Then the functions u = s−1/2 e−ν and p = e−ν

√  s

 √ √ P1 (ν s) g ∇ν + P2 (ν s) fτ

√  s

 √ √ P3 (ν s) g + P1 (ν s) fν satisfy the equations

(s − Δ) u + ∇p =

√

s e−ν

√ s

 √ k  (ν s) f + R1 ,

∇ · u = e−ν

√ s

 √ k  (ν s) g + R2

in the neighborhood ν(x) < δ|x| of ∂K, where R1 , R2 have the form R1 = s−1/2 rμ−1

k+1 

k+1  √ √ (ν s)j aj (ω) + s−1 rμ−2 (ν s)j bj (ω),

j=0

R2 = s−1/2 rμ−1

k+1 

j=1

√ (ν s)j cj (ω).

j=0

Furthermore, uτ = 0 and uν = s−1/2 P1 (0) g on ∂K\{0}. Proof. Since P2 (0) = 0, we have uτ = 0 and uν = s−1/2 P1 (0) g on ∂K\{0}. One easily checks that √ √   √ √ √ ∇ · u = −e−ν s P1 (ν s) g + s−1/2 e−ν s ∇ · P1 (ν s) g ∇ν + P2 (ν s) fτ  √  √   √ √ = e−ν s (ν s)k g + s−1/2 P1 (ν s) ∇ · g ∇ν + s−1/2 P2 (ν s)∇ · fτ .

  Here, the functions ∇ · g ∇ν and ∇ · fτ are homogenous of degree μ − 1. Furthermore, ∇p =

√

s e−ν

√  s

  √ √ √  (ν s)k fν ∇ν + P3 (ν s) − P3 (ν s) g∇ ν + R

√ √ √ √ √ where R = s−1/2 P1 (ν s) ∇fν + s−1/2 P3 (ν s) ∇g. Using the equalities (s − Δ) e−ν s = s e−ν s Δν and Δq(ν) = q  (ν) + q  (ν) Δν, we get

(s − Δ) u = e−ν

√ s

  √ √ Δν P1 (ν s) g∇ν + P2 (ν s) fτ

3   √ √ ∂ν ∂  P1 (ν s) g∇ν + P2 (ν s) fτ ∂xj ∂xj j=1 √   √ √ −s−1/2 e−ν s Δ P1 (ν s) g∇ν + P2 (ν s) fτ  √  √  √ √ √  = s e−ν s (ν s)k fτ + 2P1 (ν s) − P1 (ν s) g∇ν + R ,

+2 e−ν

√ s

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

11

where R is equal to −1/2

s

3   √ √  √ ∂ν ∂fτ   P2 (ν s) − P2 (ν s) fτ Δν + 2 − s−1 P2 (ν s) Δfτ ∂x ∂x j j j=1

3   √ √ ∂ν ∂(g∇ν)  −s−1/2 (ν s))k gΔν ∇ν + 2 − s−1 P1 (ν s) Δ(g∇ν) . ∂xj ∂xj j=1

Consequently, (s − Δ) u + ∇p =

√

s e−ν

√ s

 √ k  (ν s) f + R + R .

This proves the lemma.  Next, we construct special singular functions with the leading part (u0 , p0 ) defined by (11) and (12). Lemma 2.3. Let μ be an eigenvalue of the pencil N (λ) with the eigenfunction φ, and let p0 and u0 be the functions (11) and (12), respectively. There exist functions uj , pj of the form uj (x, s) = s−(j+2)/2 rμ−j−1

j 

(log r)k Φj,k (ω),

k=0

pj (x, s) = s−j/2 rμ−j

j 

(log r)k Ψj,k (ω),

k=0

j = 1, 2, . . ., and functions wj , qj of the form wj (x, s) = s−(j+2)/2 rμ−j−1 e−ν

√ s

j  j  √ (ν s)i (log r)k Φi,j,k (ω), i=0 k=0

qj (x, s) = s−(j+1)/2 rμ−j−1 e−ν

√ s

j  j  √ (ν s)i (log r)k Ψi,j,k (ω) i=0 k=0

such that the functions

UN =

N  j=0

uj − χ

N ν  

r

wj ,

PN =

j=1

N  j=0

pj − χ

N ν  

r

qj

j=1

have the following properties for N = 0, 1, 2, . . .. 1) There are the representations (s − Δ) UN + ∇PN = χ

ν  r

f + R1 ,

∇ · UN = χ

ν  r

g + R2 ,

where f and g are finite sums of terms of the form s(1−d)/2 rμ−d e−ν

√ s

√ (ν s)j (log r)k Φ(ω),

(14)

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

12

and s−d/2 rμ−d e−ν

√ s

√ (ν s)j (log r)k Ψ(ω),

(15)

respectively, with nonnegative integers d ≥ N + 2 and j, k ≤ N . The remainders R1 , R2 vanish for 2ν < δr and for ν > δr and satisfy the estimate √  N   |R1 | + |s|1/2 |R2 | ≤ c |s|(1−μ)/2 1 +  log |s| e−δr Re s/3

(16)

for r > |s|−1/2 . 2) The function UN satisfies the boundary condition UN = 0 on ∂K\{0}. Proof. For the case N = 0, we refer to Lemma 2.1. We assume that we already constructed the functions UN , PN for a certain N ≥ 0. Then (s − Δ) UN + ∇PN = χ

ν  r

(f1 + f2 ) + R1 ,

∇ · UN = χ

ν  r

(g1 + g2 ) + R2 .

Here f1 , g1 are finite sums of terms of the form (14), (15), respectively, with d = N +2 and integers j, k ≤ N , while f2 , g2 are functions of the same form with d ≥ N + 3 and j, k ≤ N . The remainders R1 , R2 satisfy (16). By Lemma 2.2, there exist functions v, q of the form v(x, s) = s−(N +3)/2 rμ−N −2 e−ν

+1  N  √ N s

√ (ν s)j (log r)k Φj,k (ω),

j=0 k=0 +1  N  √ N −(N +2)/2 μ−N −2 −ν s

q(x, s) = s

r

e

√ (ν s)j (log r)k Ψj,k (ω)

j=0 k=0

in the neighborhood ν(x) < δ|x| of ∂K such that (s − Δ) v + ∇q = f1 + f3 ,

∇ · v = g1 + g3 ,

where f3 and g3 are finite sums of terms of the form (14), (15), respectively, with d ≥ N + 3 and integers j ≤ N + 1, k ≤ N . Furthermore,

vτ = 0,

vν = s−(N +3)/2 rμ−N −2

N 

(log r)k Φ0,k (ω) · ∇ν

k=0

Consequently, there exists a solution p of the Neumann problem Δp = 0 in K,

∇p · ∇ν = −s vν on ∂K\{0}

which has the form p (x, s) = s−(N +1)/2 rμ−N −1

N +1 

(log r)k ψk (ω)

k=0

on ∂K\{0}.

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

13

(see, e.g., [13, Lemma 6.1.13]). Here ψN +1 = 0 if μ − N − 1 is not an eigenvalue of the pencil N (λ). We set   ν  √ and u = s−1 v  − χ e−ν s vτ . r

v  = −∇p

Analogously to Lemma 2.1, we get the representations (s − Δ)u + ∇p = χ

ν 

F + R1 ,

r

∇ · u = χ

ν  r

G + R2 in K,

where F , G are finite sums of terms of the form (14) and (15), respectively, with d ≥ N +3, j = 0, k ≤ N +1, and R1 , R2 satisfy the estimate (16). Furthermore, uτ = 0, uν = s−1 vν = −s−1 ∇p · ∇ν = vν on ∂K\{0}. We consider the functions UN +1 = UN + u − χ

ν  r

v = UN + uN +1 − χ

ν  r

wN +1

and PN +1 = PN + p − χ

ν  r

q = PN + pN +1 − χ

ν  r

qN +1 ,

√

where uN +1 = s−1 v  , wN +1 = v + s−1 e−ν s vτ , pN +1 = p and qN +1 = q. Obviously, uN +1 , pN +1 , wN +1 and qN +1 have the desired form, and the function UN +1 satisfies the boundary condition UN +1 = 0 on ∂K\{0}. Furthermore, we obtain (s − Δ) UN +1 + ∇PN +1 = χ

ν  r

ν ν (f2 − f3 + F ) + R1 + R1 + [Δ, χ( )] v − q ∇χ( ) r r

and ∇ · UN +1 = χ

ν  r

ν (g2 − g3 + G) + R2 + R2 − v · ∇χ( ). r

This proves the lemma.  Remark 2.1. 1) In the case μ = 0, where u0 = 0 and p0 is constant, we can obviously set UN = u0 = 0 and PN = p0 for arbitrary N . 2) Logarithmic terms in the representations of UN and PN appear only in the case N > 0 if at least one of the numbers μ − 1, μ − 2, . . . , μ − N is an eigenvalue of the pencil N (λ). In fact, the logarithm appears at most with an exponent N  ≤ N , where N  is the number of eigenvalues of the pencil N (λ) in the set {μ − 1, μ − 2, . . . , μ − N }. In the following, η is a two times continuously differentiable function on (0, ∞), η(r) = 0 for r < 1/2, η(r) = 1 for r > 1. Furthermore, we define ηs (r) = η(|s| r2 ). Then the following assertion holds.

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

14

Lemma 2.4. Let μ be an eigenvalue of the pencil N (λ) with the eigenfunction φ. Furthermore, let UN , PN be the functions described in Lemma 2.3. Then   (s − Δ) (ηs UN ) + ∇(ηs PN )2 0

Vβ (K)

 2N    ≤ c |s|−β−μ−1/2 1 +  log |s|

(17)

for β + μ < N + 1 and  2   2 ∇ · (ηs UN )2 1  1 + |s| U ) ∇ · (η s N V (K) (V

−β

β

(K))∗

 2N    ≤ c |s|−β−μ−1/2 1 +  log |s|

for β + μ < N + 12 , where c is a constant independent of s. Here, N  is the number of eigenvalues of the pencil N (λ) in the set {μ − 1, μ − 2, . . . , μ − N }, and N  ≤ N  + 1. If μ − 1, μ − 2, . . . , μ − N are not eigenvalues of the pencil N (λ), then N  = N  = 0. Proof. By Lemma 2.3,   ηs (s − Δ) UN + ∇PN = ηs χ(r−1 ν) f + ηs R1 and ηs ∇ · UN = ηs χ(r−1 ν) g + ηs R2 , where f, g are sums of terms of the form (14) and (15), respectively, and R1 , R2 satisfy the estimate (16). Obviously, ηs χf 2V 0 (K) ≤ β

   ν 2 √    2(β+μ−N −2)  1 + | log r|2N e−ν Re s dx. ηs χ  r r

c |s|N +1

K

Here by N  , we denoted the number of eigenvalues of the pencil N (λ) in the set {μ − 1, μ − 2, . . . , μ − N } (cf. Remark 2.1). Since      √     ν(x) 2 −ν(x) Re √s χ ν(ω) 2 e−r ν(ω) Re s dω ≤ c |s|−1/2 r−1 , e dω = χ  r Ω

Ω

we get ηs χe−ν

√ s

 2N    f 2V 0 (K) ≤ c |s|−β−μ−1/2 1 +  log |s| β

if β + μ − N < 1.

The same estimate holds for ηs R1 . Thus, (17) holds. Analogously, we obtain ∇ · (ηs UN )2(V 1

−β (K))

∗

≤ ∇ · (ηs UN )2V 0

β+1 (K)

 2N    ≤ c |s|−β−μ−5/2 1 +  log |s|

and  2N    ∇ · (ηs UN )2V 1 (K) ≤ c |s|−β−μ−1/2 1 +  log |s| β

if β + μ < N . Suppose that N ≤ β + μ < N + 12 . Then  2N   1 +  log |s| ≤c , |s|β+μ+1/2 

∇ ·

(ηs UN +1 )2V 1 (K) β

+ |s| ∇ · 2

(ηs UN +1 )2(V 1 (K))∗ −β

where N  denotes the number of eigenvalues of the pencil N (λ) in the set {μ − 1, μ − 2, . . . , μ − N − 1}. Furthermore, one can easily show that

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

15

  ◦ ηs (UN +1 − UN ) = ηs uN +1 − χ(r−1 ν) wN +1 ∈ Eβ2 (K)∩ E 1β (K) and  2N     2   1 +  log |s| 2  ηs (UN +1 − UN )2 2 + |s| ηs (UN +1 − UN ) V 0 (K) ≤ c . Vβ (K) β |s|β+μ+1/2 This implies ∇ · (ηs UN +1 − ηs UN )2V 1 (K) + |s|2 ∇ · (ηs UN +1 − ηs UN )2(V 1

−β (K))

β

 2N    . ≤ c |s|−β−μ−1/2 1 +  log |s|

∗

Thus, the desired estimate for ∇ ·(ηs UN ) holds in the case β +μ < N + 12 . If N  = 0, i.e., μ −1, μ −2, . . . , μ −N are not eigenvalues of the pencil N (λ), then UN and PN do not contain logarithmic terms (see Remark 2.1) and we have   UN (x, s) = |s|−(μ+1)/2 UN |s|1/2 x, |s|−1 s .   ˆN +1 (x, s) = |s|−(μ+1)/2 UN +1 |s|1/2 x, |s|−1 s , i.e., U ˆN +1 (x, s) arises if we replace log r by We define U 1/2 log(|s| r) in the representation of UN +1 (x, s). Then 2 2 ˆN +1 )2 1 ˆ ∇ · (ηs U V (K) + |s| ∇ · (ηs UN +1 )(V 1

−β (K))

β

∗

≤ c |s|−β−μ−1/2 .

Furthermore,        ˆN +1 − UN = |s|−(μ+1)/2 uN +1 |s|1/2 x, |s|−1 s − χ ν wN +1 |s|1/2 x, |s|−1 s . U r One can easily show that   ˆN +1 − UN )2 2 ηs (U

Vβ (K)

  ˆN +1 − UN )2 0 + |s|2 ηs (U ≤ c |s|−β−μ−1/2 V (K) β

and, consequently, 2 2 ˆN +1 − UN )2 1 ˆ ∇ · ηs (U V (K) + |s| ∇ · ηs (UN +1 − UN )(V 1

−β (K))

β

∗

≤

c |s|β+μ+1/2

.

This proves the inequality    2 ∇ · (ηs UN )2 1 + |s|2 ∇ · (ηs UN )(V 1 V (K)

−β (K))

β

∗

≤ c |s|−β−μ−1/2

for the case N  = 0. The proof of the lemma is complete.  Using Lemma 2.4, we can construct special solutions (V, Q) of the problem (s − Δ) V + ∇Q = 0,

∇ · V = 0 in K,

V = 0 on ∂K,

(18)

with the leading term ηs (u0 , p0 ), where p0 , u0 are the functions (11), (12) with a nonnegative eigenvalue μ of the pencil N (λ).

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

16

Corollary 2.1. Let μ be a nonnegative eigenvalue of the pencil N (λ) with the eigenfunction φ, and let N be the smallest integer greater than μ − λ1 . Furthermore, let UN , PN be the functions described in Lemma 2.3. Then there exists a nontrivial solution (V, Q) of the problem (18) which has the form V = ηs UN + v, where (v, q) ∈ Eβ2 (K) × Vβ1 (K) with some β ∈

1 2

Q = ηs PN + q,

− λ1 ,

1 2



.

Proof. By Lemma 2.4, we have (s−Δ)(ηs UN )+∇(ηs PN ) ∈ Eβ0 (K) and ∇·(ηs UN ) ∈ Xβ1 (K) for β+μ < N + 12 . Moreover, ηS UN = 0 on ∂K\{0}. Suppose that 1 1 − λ1 < β < min(N − μ, 0) + . 2 2 Then by Theorem 1.1, there exists a pair (v, q) ∈ Eβ2 (K) × Vβ1 (K) such that (V, Q) = ηs (UN , PN ) + (v, q) / Vβ1 (K) since β > − 12 and μ ≥ 0. Hence Q = 0.  is a solution of (18). Obviously, ηs PN ∈ In the case μ = μ1 = 0, the vector function (V, Q) in Corollary 2.1 depends on s. In the case μ = 0, we have N = 0, U0 = u0 = 0 and P0 = p0 is a constant. If we choose v = 0, q = (1 − ηs ) p0 , we get the solution (V, Q) = (0, p0 ) of the problem (18). Obviously, q ∈ Vβ1 (K) for arbitrary β > − 12 . 2.2. Asymptotics of the solution Let μj , j = 1, 2, . . ., denote the nonnegative eigenvalues of the pencil N and let φj,k , k = 1, . . . , σj , be orthonormalized eigenfunctions corresponding to μj . Then the functions φ−j,k = φj,k are also eigenfunctions corresponding to the negative eigenvalues μ−j = −1 − μj . For arbitrary integer j = 0, and k = 1, . . . , σj , we set (cf. (11), (12)) (j,k)

p0

(j,k)

(x) = rμj φj,k (ω),

u0

(j,k)

 ν   √ (x, s) = s−1 v (j,k) (x) − χ e−ν s vτ(j,k) (x) , r (j,k)

(j,k)

(j,k)

(j,k)

where v (j,k) = −∇p0 . Furthermore, let un , pn , wn , and qn , n = 1, 2, . . ., be the functions described in Lemma 2.3 for the eigenvalue μ = μj and the eigenfunction φ = φj,k . For integer N ≥ 0 we define (cf. Lemma 2.3) (j,k)

UN

=

N  n=0

un(j,k) − χ

N ν  

r

wn(j,k) ,

(j,k)

PN

=

n=1

N  n=0

pn(j,k) − χ

N ν  

r

qn(j,k) .

n=1

Let μ−j = −1 − μj be a negative eigenvalue of N (λ) and let γ > μj + 12 . Then we denote by Mj,γ the smallest integer such that Mj,γ > γ − μj − 32 and set (−j,k)

U j,k,γ = UMj,γ ,

(−j,k)

P j,k,γ = PMj,γ

for j = 1, 2, . . . .

Lemma 2.5. Suppose that Re s ≥ 0, s = 0, and that (u, p) ∈ Eβ2 (K) × Vβ1 (K) is a solution of the problem (8) with the data f ∈ Eβ0 (K) ∩ Eγ0 (K), g ∈ Xβ1 (K) ∩ Xγ1 (K), where γ > β > − 12 . If the numbers β − 12 and γ − 12 are not eigenvalues of the pencil N (λ), then (u, p) = ηs (r)

σj   j∈Jβ,γ k=1

  cj,k (s) U j,k,γ , P j,k,γ + (v, q),

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

where v ∈ Eγ2 (K), q ∈ Vγ1 (K) and Jβ,γ denotes the set of all j such that β − then

17

1 2

< μj < γ − 12 . If |s| = 1,

  ≤ c f Eγ0 (K) + gXγ1 (K) + uEβ2 (K) + pVβ1 (K) .

(19)

vEγ2 (K) + qVγ1 (K) +

σj  

|cj,k (s)|

j∈Jβ,γ k=1

Proof. We assume first that γ ≤ β + 12 . Then − 12 < γ − μj − 1 < 0 for j ∈ Jβ,γ and consequently, Mj,γ = 0. It follows from (8) that  ∇p · ∇q dx = F, q

1 for all q ∈ V−β (K),

K

 where F, q =

  1 (K) for p ∈ Vβ1 (K). By [18, (f + Δu) · ∇q − s g q dx. Obviously, F is continuous on V−β

K 2 Lemma 2.5], the functional F is also continuous on V1−γ (K), and

  2 2 (K) , F (V1−γ (K))∗ ≤ c f Eγ0 (K) + gXγ1 (K) + uEβ if |s| = 1. Hence, it follows from [18, Lemma 2.6] that p=

σj  

(−j,k)

cj,k p0

(x) + q  ,

j∈Jβ,γ k=1 0 where q  ∈ Vγ−1 (K), 0 q  Vγ−1 (K) +

(−j,k)

Since (1 − ηs ) p0



  1 (K))∗ + F (V 2 |cj,k | ≤ c F (V−β ∗ . 1−γ (K))

0 ∈ Vγ−1 (K) for μj < γ − 12 , this implies

p(x) = ηs (r)

σj  

(−j,k)

cj,k p0

(x) + q(x)

j∈Jβ,γ k=1 0 with a remainder q ∈ Vγ−1 (K), where  0 0 qVγ−1 (K) ≤ q Vγ−1 (K) + c



|cj,k |

if |s| = 1. We define v(x) = u(x) − ηs (r)

σj  

(−j,k)

cj,k u0

(x).

j∈Jβ,γ k=1 (−j,k)

1 1 One easily checks that ηs u0 ∈ Vγ−1 (K) for j ∈ Jβ,γ . Using the imbedding Eβ2 (K) ⊂ Vβ1 (K) ∩Vβ−1 (K) ⊂ 1 1 Vγ−1 (K), we conclude that v ∈ Vγ−1 (K). For |s| = 1, we obtain the estimate

    1 0 0 2 (K) + vVγ−1 |cj,k | . (K) + qVγ−1 (K) ≤ c q Vγ−1 (K) + uEβ

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

18

Furthermore, it follows from Lemma 2.4 that (s − Δ) v + ∇q ∈ Eγ0 (K) and ∇ · v ∈ Xγ1 (K). If |s| = 1, then    (s − Δ) v + ∇qEγ0 (K) + ∇ · vXγ1 (K) ≤ c f Vγ0 (K) + gXγ1 (K) + |cj,k | . Moreover, v = 0 on ∂K\{0}. Applying [18, Lemma 2.4], we conclude that v ∈ Eγ2 (K) and q ∈ Vγ1 (K). Furthermore, the estimate (19) holds if |s| = 1. This proves the theorem for the case γ ≤ β + 12 . Suppose the assertion of the theorem is true for some γ > β > − 12 and that f ∈ Eβ0 (K) ∩ Eγ0 (K) and g ∈ Xβ1 (K) ∩Xγ1 (K), where γ  is such that γ < γ  ≤ γ + 12 and γ  − 12 is not an eigenvalue of the pencil N (λ). Since Eβ0 (K) ∩ Eγ0 (K) ⊂ Eγ0 (K) and Xβ1 (K) ∩ Xγ1 (K) ⊂ Xγ1 (K), it follows from the induction hypothesis that (u, p) = ηs (r)

σj  

  cj,k U j,k,γ , P j,k,γ + (v  , q  ),

j∈Jβ,γ k=1

where v  ∈ Eγ2 (K) and q  ∈ Vγ1 (K). If |s| = 1, then   v  Eγ2 (K) + q  Vγ1 (K) ≤ c f Eγ0 (K) + gXγ1 (K) + uEβ2 (K) + pVβ1 (K) . The same estimate holds for |cj,k (s)|, j ∈ Jβ,γ , k = 1, . . . , σj . Consequently, σj  

(u, p) = ηs (r)

   cj,k U j,k,γ , P j,k,γ + (V, Q),

(20)

j∈Jβ,γ k=1

where (V, Q) = (v  , q  ) − ηs (r)

σj  

     cj,k U j,k,γ − U j,k,γ , P j,k,γ − P j,k,γ . Here, ηs (U j,k,γ − U j,k,γ ) ∈

j∈Jβ,γ k=1 j,k,γ 

(−j,k)

(−j,k)

since U −U contains only functions un and wn with index n ≥ Mj,γ + 1 > γ − μj .  Analogously, ηs (P j,k,γ − P j,k,γ ) ∈ Vγ1 (K). Thus, V ∈ Eγ2 (K), Q ∈ Vγ1 (K) and V = 0 on ∂K\{0}. For |s| = 1, we obviously get Eγ2 (K)

j,k,γ

  V Eγ2 (K) + QVγ1 (K) ≤ c f Eγ0 (K) + gXγ1 (K) + uEβ2 (K) + pVβ1 (K) . Let F = (s − Δ) V + ∇Q and G = −∇ · V . From Lemma 2.4 we conclude that F ∈ Vγ0 (K) and G ∈ Xγ1 (K). If |s| = 1, then F V 0 (K) + GX 1 (K) ≤ f V 0 (K) + gX 1 (K) + c γ

γ

γ

σj  

γ

|cj,k (s)|.

j∈Jβ,γ k=1

Therefore, by the first part of the proof, we have

(V, Q) = ηs (r)

σj  

 (−j,k) (−j,k)  cj,k u0 , p0 + (v, q),

j∈Jγ,γ  k=1

where v ∈ Eγ2 (K) and q ∈ Vγ1 (K). For |s| = 1, we have   vE 2 (K) + qV 1 (K) ≤ c F V 0 (K) + GX 1 (K) + V Eγ2 (K) + QVγ1 (K) . γ

γ

γ

γ

(21)

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

19

The same estimate holds for |cj,k (s)|, j ∈ Jγ,γ  . Combining (21) with (20), we get

(u, p) = η(r)

σj  

   cj,k U j,k,γ , P j,k,γ + (v, q).

j∈Jβ,γ  k=1

Moreover, the desired estimate for v, q and the coefficients cj,k holds if |s| = 1. Thus, the assertion of the lemma is true for all γ > β.  2.3. A formula for the coefficients Let j ≥ 1, i.e., μj ≥ 0. Then we denote by Mj the smallest integer greater than μj − λ1 and consider (j,k) (j,k) the vector functions (UMj , PMj ) introduced in the foregoing subsection. By Corollary 2.1, there exist solutions (V (j,k) , Q(j,k) ) of the problem (18) which have the form (j,k)

V (j,k) = ηs UMj + v (j,k) ,

(j,k)

Q(j,k) = ηs PMj + q (j,k) ,

(22)

where v (j,k) ∈ Eβ2 (K) and q (j,k) ∈ Vβ1 (K), 12 − λ1 < β < min(0, Mj − μj ) + 12 . For the eigenvalue μ1 = 0, the pair (V, Q) = (0, 1) is a solution of the form (22) (see Remark 2.1). We introduce the bilinear forms as (u, p, v, q) =

     u · sv − Δv + ∇q − p ∇ · v dx K

and A(u, p, v, q) = as (u, p, v, q) − as (v, q, u, p). Lemma 2.6. Suppose that (u, p) ∈ Eβ2 (K) × Vβ1 (K), (v, q) ∈ Eδ2 (K) × Vδ1 (K) and u = v = 0 on ∂K\{0}. If 0 ≤ β + δ ≤ 2, then A(u, p, v, q) = 0. Proof. Under the assumptions of the lemma, we have −Δu + ∇p ∈ Vβ0 (K), −Δv + ∇q ∈ Vδ0 (K), ∇ · u ∈ 0 0 Xβ1 (K) and ∇ · v ∈ Xδ1 (K). Using the imbeddings Eβ2 (K) ⊂ E−δ (K), Eδ2 (K) ⊂ V−β (K),  ∗ 1 1 1 0 Vβ1 (K) ⊂ V−δ (K) + V2−δ (K) ⊂ V−δ (K) + V1−δ (K) ⊂ Xδ1 (K)  ∗ and Vδ1 (K) ⊂ Xβ1 (K) for 0 ≤ β + δ ≤ 2, we conclude that  



(−Δu + ∇p) · v − (∇ · u) q dx =

K

 

 u · (−Δv + ∇q) − p ∇ · v dx.

K

This proves the lemma.  Lemma 2.7. If i, j are positive integers, μj < μi + 1 and γ > μi + 1/2, then   2μj + 1 (j,k) (j,k) δi,j δk,l A ηs UMj , ηs PMj , ηs U i,l,γ , ηs P i,l,γ = − s for k = 1, . . . , σj , l = 1, . . . , σi .

(23)

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

20

Proof. Let SR be the intersection of the cone K with the sphere |x| = R. Then the left-hand side of (23) is the limit of     x (j,k) (j,k) (j,k) (j,k) U i,l,γ · ∂r UMj − UMj · ∂r U i,l,γ + P i,l,γ UMj − PMj U i,l,γ · dσ R SR

  (j,k)   (j,k) as R → ∞. Since U i,l,γ · ∂r UMj  + UMj · ∂r U i,l,γ  ≤ cRμj −μi −4 for large R and μj − μi < 1, the left-hand side of (23) is equal to 



lim

R→∞ SR

 x (j,k) (j,k) P i,l,γ UMj − PMj U i,l,γ · dσ. R

We consider the leading terms (−i,l)

p0

(x) = r−1−μi φi,l (ω),

(−i,l)

u0

(−i,l)

of P i,l,γ and U i,l,γ , where v (−i,l) = −∇p0 (j,k) UMj ,

 ν   √ e−ν s vτ(−i,l) , = s−1 v (−i,l) − χ r (j,k)

, and the analogous leading terms p0

(j,k)

, u0

respectively. Obviously, 



(−i,l)

p0

(j,k)

u0

(j,k)

− p0

(−i,l) 

u0

·

x dσ = s−1 (AR + BR ), R

SR

where  AR = −



(−i,l)

p0

(j,k)

∇p0

(j,k)

− p0

SR

= −(μi + μj + 1) R

μj −μi

(−i,l) 

∇p0

·

x dσ R

 φi,l (ω) φj,k (ω) dω = −(2μj + 1) δi,j δk,l Ω

and  BR =

χ

ν  r

e−ν

√ s

  x (−i,l) (j,k) (j,k) p0 vτ − p0 vτ(−i,l) · dσ. R

SR

Here,  |BR | ≤ c

χ

ν  r

SR

= cR

μj −μi



e−ν Re

√ s μj −μi −2

r

dσ

√   χ ν(ω) e−R ν(ω) Re s dω ≤ c |s|−1/2 Rμj −μi −1 ,

Ω

i.e., BR → 0 as R → ∞ if μj < μi + 1. Analogously,  SR



 x (j,k) (j,k) P i,l,γ UMj − PMj U i,l,γ · dσ − R





(−i,l)

p0

(j,k)

u0

SR

tends to zero as R → ∞ if μj < μi + 1. This proves the lemma. 

(j,k)

− p0

(−i,l) 

u0

·

x dσ R

(j,k)

of PMj

and

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

21

Using the last two lemmas, we can prove the following theorem. Theorem 2.1. Suppose that Re s ≥ 0, s = 0, 12 − λ1 < β < 12 < γ < 32 , and that (u, p) ∈ Eβ2 (K) × Vβ1 (K) is a solution of the problem (8) with the data f ∈ Eβ0 (K) ∩ Eγ0 (K) and g ∈ Xβ1 (K) ∩ Xγ1 (K). If γ − 12 is not an eigenvalue of the pencil N (λ), then (u, p) = ηs

σj 

 (−j,k) (−j,k)  cj,k (s) u0 , p0 + (v, q),

(24)

j∈Jγ k=1

where v ∈ Eγ2 (K), q ∈ Vγ1 (K), Jγ is the set of all j such that 0 ≤ μj < γ − cj,k (s) = −

s 1 + 2μj





1 2

and

 f (x) · V (j,k) (x, s) + g(x) Q(j,k) (x, s) dx.

(25)

K

The remainder (v, q) and the coefficients cj,k satisfy the estimates v2Vγ2 (K)

+ |s|

2

v2Vγ0 (K)

+

q2Vγ1 (K)

+

σj 

|s|μj −γ+1/2 |cj,k (s)|2

j∈Jγ k=1



≤ c f 2Vγ0 (K) + g2Vγ1 (K) + |s|2 g2(V 1

−γ



(26)

(K))∗

with a constant c independent of s. Proof. If j ∈ Jγ and γ < 32 , then −1 < γ − μj − (−j,k)

U j,k,γ (x, s) = u0

3 2

< 0 and, consequently, Mj,γ = 0. This means that (−j,k)

(x, s) and P j,k,γ (x, s) = p0

(x) = r−μj −1 φj,k (ω)

for j ∈ Jγ . Furthermore, the number β − 12 is not an eigenvalue of the pencil N (λ) since these eigenvalues lie outside the interval (−1, 0). Therefore, the decomposition (24) follows from Lemma 2.5. Let V (j,k) , Q(j,k) be the functions (22). We prove the formula (25). Obviously, 



     (j,k) (j,k) f · V (j,k) + g Q(j,k) dx = as ηs UMj , ηs PMj , u, p + as v (j,k) , q (j,k) , u, p .

K

  Here v (j,k) ∈ Eδ2 (K) and q (j,k) ∈ Vδ1 (K) with some δ ∈ 12 − λ1 , 12 . From Lemma 1.2 it follows that (u, p) ∈ Eβ2  (K) × Vβ1 (K) with arbitrary β  , max(−δ, 12 − λ1 ) < β  < 12 . Hence by Lemma 2.6, the equality     as v (j,k) , q (j,k) , u, p = as u, p, v (j,k) , q (j,k)   holds. Since as u, p, V (j,k) , Q(j,k) = 0, it follows that 



   (j,k) (j,k) f · V (j,k) + g Q(j,k) dx = A ηs UMj , ηs PMj , u, p .

K

Obviously,  α  ∂x (ηs U (j,k) ) ≤ c rμj −|α|−1 | log r|Mj , Mj

 α  ∂x (ηs P (j,k) ) ≤ c rμj −|α| | log r|Mj Mj

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

22

(j,k)

(j,k)

2 for |α| ≤ 2. Therefore, ηs UMj ∈ E−γ (K) and ηs PMj   (j,k) (j,k) A ηs UMj , ηs PMj , v, q = 0. Consequently,



1 ∈ V−γ (K) if μj < γ − 12 , and Lemma 2.6 implies

σi     (j,k) (j,k) (−i,l) (−i,l)  f · V (j,k) + g Q(j,k) dx = ci,l A ηs UMj , ηs PMj , ηs u0 , ηs p0 . i∈Jγ l=1

K

Using Lemma 2.7, we obtain (25). We prove the estimate (26). First, let |s| = 1. Under the conditions of the theorem, we have M1 = 0 and Mj ≤ 1 for 0 < μj < 1. Since μj − 1 is not an eigenvalue of the pencil N (λ) for 0 < μj < 1, we get   ηs (r) U (j,k) (x, s) ≤ c rμj −1 for 0 ≤ μj < γ − 1 (cf. Remark 2.1) and Mj 2  2   (j,k)  ηs (r) f · UMj (x, s) dx ≤ c f 2Vγ0 (K) .

(27)

K

  (j,k) Analogously, the estimate ηs (r) PMj (x, s) ≤ c rμj implies  2   (j,k)  ηs (r) g · PMj (x, s) dx ≤ c g2Vγ+1 0 (K) .

(28)

K

We consider the integral of f · v (j,k) + g q (j,k) over K. By Lemma 2.3, the pair (v (j,k) , q (j,k) ) is a solution of the Dirichlet problem for the system (j,k)

(j,k)

(s − Δ) v (j,k) + ∇q (j,k) = −(s − Δ) ηs UMj − ∇ηs PMj , (j,k)

−∇ · v (j,k) = ∇ · (ηs UMj ) in the space Eδ2 (K) × Vδ1 (K) with some δ in the interval

1 2

− λ1 < δ < 12 . Using Lemma 2.4, we get

  (s − Δ) v (j,k) + ∇q (j,k) 2 0 + ∇ · v (j,k) 2X 1 (K) ≤ c V (K) δ

δ

if δ < Mj − μj + 12 . Here, Mj − μj + Theorem 1.1,

1 2

>

1 2

− λ1 . Suppose that

1 2

− λ1 < δ < min(Mj − μj + 12 , 12 ). Then by

v (j,k) Eδ2 (K) + q (j,k) 2V 1 (K) ≤ c. δ

Since 0 < γ + δ < 2, we have r−2γ ≤ r2δ−4 if r ≤ 1 and r−2γ ≤ r2δ if r ≥ 1. Hence,   2     (j,k) 2 2δ−4 (j,k) 2 dx ≤ c f Vγ0 (K) r |v | dx + r2δ |v (j,k) |2 dx  f ·v K

K

K

≤ c f 2Vγ0 (K) v (j,k) 2E 2 (K) ≤ c f 2Vγ0 (K) . δ

Analogously, 2     g · q (j,k) dx ≤ c g2Xγ1 (K) . K

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

23

Thus,     2 cj,k (s)2 ≤ c f 2 0 for |s| = 1. + g 1 Eγ (K) Xγ (K)

(29)

Since the operator Aγ is injective for −μ2 − 1/2 < γ < λ1 + 3/2, we get   vEγ2 (K) + qVγ1 (K) ≤ c F Eγ0 (K) + GXγ1 (K) , where F =f−

σj 

  (−j,k) (−j,k) cj,k (s) (s − Δ) (ηs u0 ) + ∇(ηs p0 )

j∈Jγ k=1

and G=g+

σj 

(−j,k)

cj,k (s) ∇ · (ηs u0

).

j∈Jγ k=1

Using Lemma 2.4 and (29), we obtain   F Eγ0 (K) + GXγ1 (K) ≤ c f Eγ0 (K) + gXγ1 (K) . This proves (26) in the case |s| = 1. If s is an arbitrary number in the half-plane Re s ≥ 0, s = 0, we set x = |s|−1/2 y and define u ˆ(y) = u(x), pˆ(y) = |s|−1/2 p(x), fˆ(y) = |s|−1 f (x), gˆ(y) = |s|−1/2 g(x). Obviously, (ˆ u, pˆ) is a solution of the Dirichlet problem for the system  −1  |s| s − Δ u ˆ + ∇ pˆ = fˆ, −∇ · u ˆ = gˆ in K. Hence, u ˆ(y) = η(|y|2 )

σj 

cj,k (|s|−1 s) u0

(−j,k)

(30)

(y, |s|−1 s) + vˆ(y), and an analogous decomposition with a

j∈Jγ k=1

remainder qˆ(y) holds for pˆ(y). Here, (−j,k)

u0

(y, |s|−1 s) = |s|−μj /2 u0

(−j,k)

(−j,k)

(x, s), p0

(y) = |s|−(μj +1)/2 p0

(−j,k)

(x)

and ˆ v 2Eγ2 (K)

+

ˆ q 2Vγ1 (K)

σj      cj,k (|s|−1 s)2 ≤ c fˆ2 0 + g 2Xγ1 (K) . Eγ (K) + ˆ j∈Jγ k=1

In particular, we get p(x) = ηs (x)

σj 

|s|−μj /2 cj,k (|s|−1 s) p0

(−j,k)

(x) + |s|1/2 qˆ(|s|1/2 x).

j∈Jγ k=1

We conclude from this and from (24) that cj,k (s) = |s|−μj /2 cj,k (|s|−1 s) and q(x) = |s|1/2 qˆ(|s|1/2 x). Using the equalities

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

24

ˆ q 2Vγ1 (K) = |s|γ−1/2 q2Vγ1 (K) ,

fˆ2Eγ0 (K) = |s|γ−1/2 f 2Eγ0 (K)

and  ˆ g 2Xγ1 (K) = |s|γ−1/2 g2Vγ1 (K) + |s|2 g2(V 1

−γ

 (K))∗

,

we obtain (26). The proof is complete.  The set of all μj , j ∈ Jγ , contains the simple eigenvalue μ1 = 0. The corresponding singular functions in (24) are (−1)

p0

(x) =

|Ω|−1/2 , r

(−1)

u0

 ν   √ (x, s) = |Ω|−1/2 s−1 v − χ e−ν s vτ , r

(31)

where v = −∇ r−1 = r−3 x and vτ = v − (v · ∇ν)∇ν. Furthermore, the constant pair (0, |Ω|−1/2 ) is the (−1) (−1) solution (V (1) , Q(1) ) in Corollary 2.1 for the eigenvalue μ1 = 0. Consequently, the coefficient of (u0 , p0 ) in Theorem 2.1 is  c1 (s) = −s |Ω|−1/2 g dx. (32) K (−1)

(−1)

Hence, the term c1 (s) (u0 , p0 ) does not appear in the asymptotics of (u, p) if g satisfies (9). Then the condition on γ in the last theorem can be weakened. Theorem 2.2. Suppose that Re s ≥ 0, s = 0, and that (u, p) ∈ Eβ2 (K) × Vβ1 (K) is a solution of the problem (8) with the data f ∈ Eβ0 (K) ∩ Eγ0 (K) and g ∈ Xβ1 (K) ∩ Xγ1 (K), where 1 1 3 − λ1 < β < < γ < min(λ1 , μ2 ) + . 2 2 2 If g satisfies the condition (9) and γ − (u, p) = ηs (r)

1 2

is not an eigenvalue of the pencil N (λ), then 

σj 

 (−j,k) (−j,k)  cj,k (s) u0 , p0 + (v, q),

(33)

j∈Jγ \{1} k=1

where Jγ is the same set as in Theorem 2.1, v ∈ Eγ2 (K), q ∈ Vγ1 (K), and the coefficients cj,k are given by the formula (25). The remainder (v, q) and the coefficients cj,k satisfy the estimate (26) with a constant c independent of s. Proof. Let γ  ≤ γ and 12 < γ  < min(μ2 + 12 , λ1 + 32 ). Then Lemma 1.3 implies u ∈ Eγ2 (K) and p ∈ Vγ1 (K). The interval γ  − 12 < μ < γ − 12 contains the eigenvalues μj with index j ∈ Jγ , j = 1. For these eigenvalues, (−j,k) the inequalities −1 < γ − μj − 32 < 0 are satisfied. This means that Mj,γ = 0, U j,k,γ (x, s) = u0 (x, s) and (−j,k) j,k,γ 2 1 P (x, s) = p0 (x). Using Lemma 2.5, we get (33), where (v, q) ∈ Eγ (K) × Vγ (K). As in the proof of Theorem 2.1, we obtain   K

   (j,k) (j,k) (−i,l) (−i,l)  ci,l A ηs UMj , ηs PMj , ηs u0 , ηs p0 f · V (j,k) + g Q(j,k) dx = i,l

for j ∈ Jγ , j = 1, where the summation is extended over all i ∈ Jγ \{1}, l = 1, . . . , σi . Obviously, μj < μi + 1 for i, j ∈ Jγ \{1}. Therefore, Lemma 2.7 implies (25) for j ∈ Jγ , j = 1, k = 1, . . . , σj .

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

25

Analogously to the proof of Theorem 2.1, we obtain (27) and (28) for |s| = 1. Since v (j,k) ∈ Eδ2 (K) and q ∈ Vδ1 (K), 12 −λ1 < δ < min( 12 , Mj −μj + 12 , 2 −γ), we obtain (analogously to the proof of Theorem 2.1) (j,k)

       f · v (j,k) + g q (j,k) dx ≤ c f 2Vγ0 (K) + g2Xγ1 (K)  K

if |s| = 1. Since the operator Aγ is injective by Theorem 1.1, we obtain (26) in the case |s| = 1. In the case |s| = 1, the estimate (26) can be obtained by means of the transformation x = |s|−1/2 y as in the proof of Theorem 2.1.  At the end of the section, we estimate the functions V (j,k) and Q(j,k) in the formula (25) for the coefficients cj,k (s), j ∈ Jγ . In the case j = 1 (i.e., μj = 0), the functions V (j,k) = 0 and Q(j,k) = |Ω|−1/2 are constant. Since γ < min(λ1 , μ2 ) + 32 in Theorems 2.1 and 2.2, it suffices to consider the case 0 < μj < min(λ1 , μ2 ) + 1. Then Mj ≤ 1. More precisely, we have Mj = 0 if μj < λ1 and Mj = 1 if λ1 ≤ μj < λ1 + 1. Therefore, (j,k) (j,k) the functions UMj and PMj contain only logarithmic terms if μj ≥ λ1 and μj − 1 is an eigenvalue of the pencil N (λ) (see Remark 2.1). In the case 0 < μj < μ2 + 1, μj = 1, this is not possible since the intervals (−1, 0) and (0, μ2 ) are free of eigenvalues. In the next lemma, we obtain point estimates of the solutions of the problem (8). Let Nβl,σ (K) be the weighted Hölder space with the norm uN l,σ (K) = β

 |α|≤l

sup x∈K

 α  ∂x u(x) |x|l+σ−β−|α|

+



sup

x,y∈K |α|=l 2|x−y|<|x|

|x|β

|∂xα u(x) − ∂yα u(y) , |x − y|σ

where l is a nonnegative integer, β and σ are real numbers, 0 < σ < 1. Lemma 2.8. Let (u, p) ∈ Eδ2 (K) × Vδ1 (K) be a solution of the problem (8), where f (x) = 0, g(x) = 0 for 2|s| |x|2 < 1. Then the following assertions are true.

1 2

− λ1 < δ <

1 2

and

1) If 2|s| |x|2 < 1, then   α ∂x u(x)2 ≤ c |s|δ+λ1 −1/2 r2(λ1 −|α|) (1 + | log |sr2 ||)2d (f, g)2δ for |α| ≤ 2,   α ∂x p(x)2 ≤ c |s|δ+λ1 −1/2 r2(λ1 −|α|−1) (1 + | log |sr2 ||)2d (f, g)2δ for |α| ≤ 1, where the constant c is independent of f, g, s. Here 1 (K))∗ . (f, g)δ = f Vδ0 (K) + gVδ1 (K) + |s| g(V−δ

Furthermore, d = 1 only in the case when λ1 = 1 and there exists a generalized eigenvector corresponding to this eigenvalue. Otherwise, d = 0. 0,σ 1,σ 2) If f ∈ Eγ0 (K) ∩ Nγ+σ−1/2 (K), g ∈ Xγ1 (K) ∩ Nγ+σ−1/2 (K), γ ≥ 92 and 2sr2 > 1, then  α  ∂x u(x)2 ≤ c |s|γ+|α|−5/2 |x|−4 |||(f, g)|||2γ,σ for |α| ≤ 2,  α  ∂x p(x)2 ≤ c |s|γ−1/2 |x|−2−2|α| |||(f, g)|||2γ,σ for |α| ≤ 1  where |||(f, g)|||γ,σ = (f, g)γ + |s|−1 f N 0,σ

+ gN 1,σ γ+σ−1/2 (K) γ+σ−1/2 (K)

 and c is independent of f, g, s.

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

26

Proof. 1) We assume first that |s| = 1 and that χ = χ(r) is a smooth cut-off function equal to one near r = 0, χ(r) = 0 for 2r2 > 1. Since f and g are zero on the support of χ, we get −Δ(χu) + ∇(χp) = F, ∇ · (χu) = u · ∇χ in K, where F = −[Δ, χ] u + p ∇χ − sχu. Using the imbedding 0,σ Vδ2 (K) ⊂ Nδ+σ−1/2 (K)

(cf. [22, Lemma 3.6.2]) and regularity results for elliptic systems (see [1]), we conclude that F N 0,σ

δ+σ−1/2 (K)

+ u · ∇χN 1,σ

δ+σ−1/2 (K)

We may assume that the line Re λ = [19, Theorem 5.2, Corollary 5.1] that

5 2

  ≤ c uEδ2 (K) + pVβ1 (K) ≤ c (f, g)δ .

− δ is free of eigenvalues of the pencil L(λ). Then it follows from

(χu, χp) =

κj 

cj,k (uj,k , pj,k ) + (v, q),

j∈Iδ k=1

where Iδ is the set of all j such that 0 < Re λj < uj,k (x) = rλj

k  (log r)l l=0

l!

5 2

− δ, cj,k are constants, uj,k , pj,k are of the form

Φj,k−l (ω), pj,k (x) = rλj −1

k  (log r)l l=0

l!

Ψj,k−l (ω)

((Φj,k , Ψj,k ) are eigenvectors or generalized eigenvectors of the pencil L(λ) corresponding to the eigenvalue λj ), and vN 2,σ

δ+σ−1/2 (K)

+ qN 1,σ

δ+σ−1/2 (K)

+



|cj,k | ≤ c (f, g)δ .

j∈Iδ

If λ1 < 1, then u1,k and p1,k do not contain logarithmic terms. In the case λ1 = 1, the functions u1,k , p1,k contain log r (with power 1) if there exist generalized eigenvectors corresponding to this eigenvalue (see [14, Theorems 5.3.2 and 5.4.1]). Hence |cj,k uj,k | + r |cj,k pj,k | ≤ c rλ1 (1 + | log r|d ) (f, g)δ for r < 1. Since moreover  |α|≤2

sup rδ+|α|−5/2 |∂xα v(x)| +

x∈K

 |α|≤2

sup rδ+|α|−3/2 |∂xα q(x)| ≤ c (f, g)δ

x∈K

and δ < 12 , we obtain the desired estimate for |s| = 1, 2r2 < 1. If |s| is arbitrary, then the equations (30) are satisfied for the functions u ˆ, pˆ, fˆ, gˆ introduced in the proof of Theorem 2.1. Since fˆ(y) = 0 and gˆ(y) = 0 for 2|y|2 < 1, we obtain 2  α ∂y u ˆ(y) ≤ c |y|2(λ1 −|α|) (1 + | log |y|2 |)2d (fˆ, gˆ)2δ = c |y|2(λ1 −|α|) (1 + | log |y|2 |)2d |s|δ−1/2 (f, g)2δ for |α| ≤ 2, 2|y|2 < 1 and the analogous estimate for ∂yα pˆ(y). This proves the assertion 1).

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

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2) We start again with the case |s| = 1. Since f (x) and g(x) are zero for 2|x|2 < 1, we have (f, g)δ ≤ c (f, g)γ with a certain constant c. Thus, by Theorem 1.1 and Lemma 2.5, (u, p) admits the decomposition (u, p) = ηs (r)

σj 

  cj,k (s) U j,k,γ , P j,k,γ + (v, q),

j∈Jγ k=1

where vEγ2 (K) + qVγ1 (K) +

σj  

|cj,k | ≤ c (f, g)γ .

k=1 j∈Jγ

Since γ − 2 > δ and v(x) = u(x), q(x) = p(x) for 2|x|2 < 1, we get the same estimate for the norm of (v, q) 2 1 (K) × Vγ−2 (K). Obviously, in Vγ−2  α  ∂x ηs U j,k,γ  ≤ c r−2 for |α| ≤ 2,

 α  ∂x ηs P j,k,γ  ≤ c r−1−|α| for |α| ≤ 1.

Furthermore, it follows from Lemma 2.3 that (s − Δ)v + ∇qN 0,σ

γ+σ−1/2 (K)

+ ∇ · vN 1,σ

γ+σ−1/2 (K)

≤ c |||(f, g)|||γ,σ .

Since, moreover, vN 0,σ

γ+σ−1/2 (K)

≤ c vVγ2 (K) ≤ c (f, g)γ ,

we conclude from [19, Theorem 5.1, Corollary 5.1] that vN 2,σ

γ+σ−1/2 (K)

+ qN 1,σ

γ+σ−1/2 (K)

≤ c |||(f, g)|||γ,σ .

In particular,  α  ∂x v(x) ≤ c r−γ−|α|+5/2 |||(f, g)|||γ,σ for |α| ≤ 2 and   α ∂x q(x) ≤ c r−γ−|α|+3/2 |||(f, g)|||γ,σ for |α| ≤ 1. This proves the desired estimates for |s| = 1, 2r2 > 1. In the case |s| = 1, we obtain the estimate analogously to part 1) by means of the coordinate transformation x = |s|−1/2 y.  We estimate the functions V (j,k) and Q(j,k) by means of the last lemma. Lemma 2.9. Let μj be an eigenvalue of the pencil N , 0 < μj < min(λ1 , μ2 ) + 1, μj = 1. Then the following estimates are valid for 2|s| r2 < 1:  α (j,k)   ∂x V (x, s) ≤ c |s|(|α|−1−μj )/2 (|s|1/2 r)λ1 −|α| (1 + | log |sr2 ||)d for |α| ≤ 2,  α (j,k)   ∂x Q (x, s) ≤ c |s|(|α|−μj )/2 (|s|1/2 r)λ1 −|α|−1 (1 + | log |sr2 ||)d for |α| ≤ 1. Here d is the same number as in Lemma 2.8. For 2|s| r2 > 1, the estimates

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

28

 α (j,k)  ∂x V (x, s) ≤ c |s|(|α|−1−μj )/2 (|s|1/2 r)μj −1 for |α| ≤ 2,  α (j,k)  ∂x Q (x, s) ≤ c |s|(|α|−μj )/2 (|s|1/2 r)μj −|α| for |α| ≤ 1

(34) (35)

are valid. The same estimates with an additional factor (1 +| log |s||)m on the right-hand sides hold if μj = 1. Proof. We start with the case 2|s|r2 < 1. In this case, (V (j,k) , Q(j,k) ) coincides with (v (j,k) , q (j,k) ). Here (v (j,k) , q (j,k) ) is the solution of the problem (8), where (j,k)

(j,k)

f = −(s − Δ)(ηs UMj ) + ∇ (ηs PMj ), v (j,k) ∈ Eδ2 (K), q (j,k) ∈ Vδ1 (K), (j,k)

and PMj

1 2

(j,k)

g = ∇ · (ηs UMj ), (j,k)

− λ1 < δ < 12 . For 0 < μj < min(μ2 , λ1 ) + 1, μj = 1, the functions UMj

do not contain logarithmic factors. Hence, (f, g)2δ ≤ c |s|−δ−μj −1/2

by means of Lemma 2.4. In the case μj = 1, an additional factor (1 + | log |s||)2 appears on the right-hand side of the last estimate (cf. Remark 2.1). Using Lemma 2.8, we obtain the desired estimates for the case 2|s| r2 < 1. We consider the case 2|s|r2 > 1. Let γ ≥ 92 and let γ − 12 be not an eigenvalue of the pencil N (λ). Furthermore, let Nj > γ + μj − 12 . We consider the functions (j,k)

u(j,k) = V (j,k) − ηs UNj

 (j,k) (j,k)  = v (j,k) + ηs UMj − UNj

(j,k)

and p(j,k) = Q(j,k) − ηs PNj . The pair (u(j,k) , p(j,k) ) is a solution of (8) with (j,k)

(j,k)

f = −(s − Δ) (ηs UNj ) − ∇(ηs PNj ),

(j,k)

g = ∇ · (ηs UNj ).

By Lemma 2.4, f ∈ Eγ0 (K) and g ∈ Xγ1 (K). Hence, Lemma 2.8 implies  α (j,k) 2  ∂x u  ≤ c |s|γ+|α|−5/2 |x|−4 |||(f, g)|||2γ,σ for |α| ≤ 2, 2|s|r2 > 1. Using Lemma 2.3, one can easily show that  2m |||(f, g)|||2γ,σ ≤ c |s|−γ−μj −1/2 1 + | log |s|| with a certain integer m ≥ 0 (cf. Lemma 2.4). This implies  α (j,k)    ∂x u  ≤ c |s|(|α|−μj −1)/2 1 + | log |s|| m (|s|r2 )−1 for |α| ≤ 2, 2|s|r2 > 1. Since  α    ∂x (ηs U (j,k) (x, s)) ≤ c |s|(|α|−1−μj )/2 1 + | log |s|| m (|s|1/2 r)μj −1 Nj

for |α| ≤ 2, we obtain  α (j,k)   m  ∂x V (x, s) ≤ c |s|(|α|−μj −1)/2 1 + | log |s|| (|s|1/2 r)μj −1 for |α| ≤ 2, 2|s|r2 > 1. Analogously,

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

29

 α (j,k)    ∂x Q  ≤ c |s||α|−μj 1 + | log |s|| m (|s|1/2 r)μj −|α| for |α| ≤ 1, 2|s|r2 > 1. If μj = 1, then these estimates can be improved. As was noted above, (j,k) (j,k) (j,k) the functions UMj and PMj do not contain logarithmic terms in this case. Thus, UMj (x, s) = |s|−(μj +1)/2 UMj (|s|1/2 x, |s|−1 s) and PMj (x, s) = |s|−μj /2 PMj (|s|1/2 x, |s|−1 s). Consequently, (j,k)

(j,k)

(j,k)

V (j,k) (x, s) = |s|−(μj +1)/2 V (j,k) (|s|1/2 x, |s|−1 s) and Q(j,k) (x, s) = |s|−μj /2 Q(j,k) (|s|1/2 x, |s|−1 s). Since   α (j,k) ∂x V (x, s) ≤ c rμj −1 for |α| ≤ 2,

 α (j,k)  ∂ x Q (x, s) ≤ c rμj −|α| for |α| ≤ 1

if |s| = 1 and 2r2 > 1, we obtain (34) and (35). The proof is complete.  3. The time-dependent problem In this section, we consider the time-dependent problem (1), (2). We start with existence and uniqueness theorems which can be easily deduced from the results in Section 1. Using Theorems 2.1 and 2.2, we describe the behavior of the solutions at infinity. 3.1. Solvability results Let Q = K × R+ = K × (0, ∞). By Wβ2l,l (Q), we denote the weighted Sobolev space of all functions u = u(x, t) on Q with finite norm uW 2l,l (Q) =

l  ∞ 

β

0 k=0

∂tk u(·, t)2V 2l−2k (K) dt

1/2 .

β

  In particular, Wβ0,0 (Q) = L2 R+ , Vβ0 (K) and Wβ2,1 (Q) is the set of all u ∈ L2 (R+ , Vβ2 (K)) such that ◦

2l,l ∂t u ∈ L2 (R+ , Vβ0 (K)). The space W 2l,l β (Q) is the subspace of all u ∈ Wβ (Q) satisfying the condition ∂tk u|t=0 for x ∈ K, k = 0, . . . , l − 1. Note that ∂tk u(·, 0) ∈ Vβ2l−2k−1 (K) for u ∈ Wβ2l,l (Q), k = 0, . . . , l − 1 (see [10, Proposition 3.1]). By [10, Proposition 3.4], the Laplace transform realizes an isomorphism from ◦

2l,l ˜(x, s) for Re s > 0 with values in Eβ2l (K) and W β (Q) onto the space Hβ2l of all holomorphic functions u finite norm

˜ uHβ2l = sup γ>0

+∞ l  

|γ + iτ |2k ˜ u(·, γ + iτ )2V 2l−2k (K) dτ

−∞ k=0

1/2 .

β

The proof of the analogous result in nonweighted spaces can be found in [2, Theorem 8.1]. The following theorem is proved in [18, Theorem 3.1].       1 Theorem 3.1. Suppose that f ∈ L2 R+ , Vβ0 (K) , g ∈ L2 R+ , Vβ1 (K) , ∂t g ∈ L2 R+ , (V−β (K))∗ and g(x, 0) = 0 for x ∈ K, where   −λ1 + 1/2 < β < min μ2 + 1/2 , λ1 + 3/2 , In the case β > 1/2, we assume in addition that

β = 1/2.

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

30

 g(x, t) dx = 0

for almost all t.

(36)

K

  Then there exists a unique solution (u, p) ∈ Wβ2,1 (Q) × L2 R+ , Vβ1 (K) of the problem (1), (2) satisfying the estimate uW 2,1 (Q) + pL2 (R+ ,Vβ1 (K)) β   1 (K))∗ ) ≤ c f W 0,0 (Q) + gL2 (R+ ,Vβ1 (K)) + ∂t gL2 (R+ ,(V−β

(37)

β

with a constant c independent of f , g. Analogously, the following theorem can be proved by means of Theorem 1.2.       1 Theorem 3.2. Suppose that f ∈ L2 R+ , Vβ0 (K) , g ∈ L2 R+ , Vβ1 (K) , ∂t g ∈ L2 R+ , (V−β (K))∗ and g(x, 0) = 0. Furthermore, we assume that λ1 = 1, that λ1 is a simple eigenvalue of the pencil L(λ) and that β satisfies the inequalities (7). In the case 1/2 < β < 5/2, we assume in addition that g satisfies the   condition (36). Then there exists a unique solution (u, p) ∈ Wβ2,1 (Q) × L2 R+ , Vβ1 (K) of the problem (1), (2) satisfying the estimate (37). Furthermore, the following regularity assertion for the solution can be easily deduced from Lemma 1.3 (cf. [18, Theorem 3.2]).   Theorem 3.3. Suppose that (u, p) ∈ Wβ2,1 (Q) × L2 R+ , Vβ1 (K) is a solution of the problem (1), (2) with the data   f ∈ L2 R+ , Vβ0 (K) ∩ Vγ0 (K) ,

(38)

and     1 1 g ∈ L2 R+ , Vβ1 (K) ∩ Vγ1 (K) , ∂t g ∈ L2 R+ , (V−β (K))∗ ∩ (V−γ (K))∗ ,

(39)

where 12 − λ1 < β, γ < min(μ2 + 12 , λ1 + 32 ), β = 12 , γ = 12 . In the case γ > 12 , we assume in addition that g   satisfies the condition (36). Then u ∈ Wγ2,1 (Q) and p ∈ L2 R+ , Vγ1 (K) . 3.2. Asymptotics at infinity   We consider the solution (u, p) ∈ Wβ2,1 (Q) × L2 R+ , Vβ1 (K) of the problem (1), (2) with the data (38),   (39). If 12 − λ1 < β < γ < 12 , then it follows from Theorem 3.3 that u ∈ Wγ2,1 (Q) and p ∈ L2 R+ , Vγ1 (K) . Now, let 12 − λ1 < β < 12 < γ < 32 . We denote the Laplace transforms of u(x, t) and p(x, t) by u ˜(x, s) and p˜(x, s), respectively. Under the condition of Theorem 2.1, we get the decomposition (24) for (˜ u, p˜), i.e., σj 

(˜ u, p˜) = ηs

 (−j,k) (−j,k)  cj,k (s) u0 , p0 + (V, Q),

(40)

j∈Jγ k=1

where V ∈ Eγ2 (K), Q ∈ Vγ1 (K), Jγ denotes the set of all j such that 0 ≤ μj < γ − cj,k (s) = −

s 1 + 2μj

 K

1 2

and

  f˜(y, s) · V (j,k) (y, s) + g˜(y, s) Q(j,k) (y, s) dy.

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

31

Let ψ be a C ∞ -function on (−∞, +∞) with support in the interval [0, 1] satisfying the conditions 1

1 tj ψ(t) dt = 0 for j = 1, . . . , N,

ψ(t) dt = 1, 0

0

˜ we denote the Laplace transform of ψ. The function ψ˜ is where N is an integer, 2N > 12 − γ − λ1 . By ψ, ˜ analytic in C and satisfies the conditions ψ(0) = 1, ψ˜(j) (0) = 0 for j = 1, . . . , N . Since sn ψ˜(j) (s) is the j  dn Laplace transform of the function (−1)j dt n t ψ(t) , it follows that  (j)  ψ˜ (s) ≤ cj,n |s|−n for every j ≥ 0 and n ≥ 0, where cj,n is independent of s, Re s ≥ 0. Thus, we can replace the function ˜ 2 ) and obtain ηs (r) = η(|s| r2 ) in (40) by 1 − ψ(sr u ˜(x, s) =

σj    (j,k)  ˜ ˜ (j,k) (x, y, s) g˜(y, s) dy + V, K (x, y, s) f˜(y, s) + H u

(41)

 ˜ (j,k) (x, y, s) · f˜(y, s) + H ˜ (j,k) (x, y, s) g˜(y, s) dy + Q, K p p

(42)

u

j∈Jγ k=1 K

p˜(x, s) =

σj    j∈Jγ k=1 K

where s 1 + 2μj s ˜ (j,k) (x, y, s) = − H u 1 + 2μj ˜ (j,k) (x, y, s) = − K u

 

 ˜ 2 ) u(−j,k) (x, s) ⊗ V (j,k) (y, s), 1 − ψ(sr 0

 ˜ 2 ) Q(j,k) (y, s) u(−j,k) (x, s) 1 − ψ(sr 0

and s 1 + 2μj s ˜ (j,k) (x, y, s) = − H p 1 + 2μj

˜ (j,k) (x, y, s) = − K p

 

 ˜ 2 ) p(−j,k) (x) V (j,k) (y, s), 1 − ψ(sr 0

 ˜ 2 ) p(−j,k) (x) Q(j,k) (y, s). 1 − ψ(sr 0

By Theorem 2.2, the representation (41), (42) holds also if γ < min(λ1 , μ2 ) + 32 and g satisfies the condition ˜ u(j,k) (x, y, s) is the Laplace transform of (36). The matrix K

Ku(j,k) (x, y, t) =

1 m ∂ 2πi t

+i∞ 

˜ (j,k) (x, y, s) ds est s−m K u

−i∞

(the integral is absolutely convergent if m ≥ 1). Analogously, the inverse Laplace transforms Kp , Hu and ˜ p, H ˜ u and H ˜ p are defined. We estimate Ku(j,k) , Kp(j,k) , Hu(j,k) and Hp(j,k) . Hp of K Lemma 3.1. Suppose that 0 ≤ μj < min(λ1 , μ2 ) + 1 and μj = 1. Then    α β (j,k) |x| −2−μj  |y| μj −1 ∂x ∂y Ku (x, y, t) ≤ c t−(3+|α|+|β|)/2 1 + √ 1+ √ t t √   |y| λ1 −|β|  |y| + t d √ × 1 + log for |α|, |β| ≤ 2, |y| |y| + t

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

32

  α (j,k)  |x| −2−μj  |y| μj −|β| ∂x Hu (x, y, t) ≤ c t−(4+|α|+|β|)/2 1 + √ 1+ √ t t √   |y| λ1 −|β|−1  |y| + t d √ × for |α| ≤ 2, |β| ≤ 1. 1 + log |y| |y| + t Here, d is the same number as in Lemma 2.8. Furthermore,   α β (j,k)  |x| −1−μj −|α|  |y| μj −1 ∂x ∂y Kp (x, y, t) ≤ c t−(4+|α|+|β|)/2 1 + √ 1+ √ t t √   |y| λ1 −|β|  |y| + t d √ × 1 + log for |α| ≤ 1, |β| ≤ 2, |y| |y| + t    α (j,k) |x| −1−μj −|α|  |y| μj −|β| ∂x Hp (x, y, t) ≤ c t−(5+|α|+|β|)/2 1 + √ 1+ √ t t √   |y| λ1 −|β|−1  |y| + t d √ × for |α|, |β| ≤ 1. 1 + log |y| |y| + t Proof. First note that all theorems of this paper are not only valid for Re s ≥ 0 but for all complex s = σeiθ , where σ > 0 and |θ| ≤ δ + π2 with a sufficiently small positive number δ. Therefore, one can replace the (1) (2) path of integration by the contour Γt,δ = Γt,δ ∪ Γt,δ , where Γt,δ = {s = t−1 eiθ : −δ − π/2 < θ < δ + π/2} and (1)

Γt,δ = {s = σe±i(δ+π/2) : σ > t−1 }. (2)

This means, we have Ku(j,k) (x, y, t) =

1 2πi

 ˜ (j,k) (x, y, s) ds. est K u Γt,δ

Obviously,  α  ˜ 2 )) u(−j,k) (x, s) ≤ c |s|−1+|α|/2 |x|−2−μj for |s| r2 ≥ 1, ∂x (1 − ψ(sr 0  α  ˜ 2 )) u(−j,k) (x, s) ≤ c |s|−1 |x|−2−μj −|α| |sr2 |N +1 for |s| r2 ≤ 1. ∂x (1 − ψ(sr 0 If 2N ≥ μj + |α|, this implies  α  ˜ 2 )) u(−j,k) (x, s) ≤ c |s|(μj +|α|)/2 (1 + |s|1/2 r)−2−μj . ∂x (1 − ψ(sr 0

Using Lemma 2.9, we obtain the inequality  |s|1/2 |y| λ1 −|β|  β (j,k)  ∂y V (y, s) ≤ c |s|(|β|−1−μj )/2 1 + |s|1/2 |y|  μj −1 1 + |s|1/2 |y| d  × 1 + log . 1 + |s|1/2 |y| 1/2 |s| |y| This directly yields

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

    |x| −2−μj  α β ˜ (j,k) (x, y, s) ds ≤ c t−(3+|α|+|β|)/2 1 + √ est K  ∂x ∂y u t (1)

Γt,δ

√  |y| + t d |y| μj −1  |y| λ1 −|β|  √ 1 + log × 1+ √ . |y| t |y| + t

Furthermore,  ∞    |x| −2−μj   α β st ˜ (j,k) e Ku (x, y, s) ds ≤ c 1 + √ e−σt sin δ σ (1+|α|+|β|)/2 ∂x ∂y t (2)

1/t

Γt,δ

 μj −1  σ 1/2 |y| λ1 −|β|  1 + σ 1/2 |y| d × 1 + σ 1/2 |y| 1 + log dσ. 1 + σ 1/2 |y| σ 1/2 |y| In the case |y| ≥

√

t, we have σ 1/2 |y| ≤ 1 + σ 1/2 |y| ≤ 2 σ 1/2 |y| for σ ≥ 1/t and, consequently,     α β ˜ (j,k) (x, y, s) ds ∂ est K ∂  x y u (2)

Γt,α



|x| −2−μj ≤c 1+ √ t

∞

 μj −1 e−σt sin δ σ (1+|α|+|β|)/2 σ 1/2 |y| dσ

1/t



|x| −2−μj −|α|  |y| μj −1 √ . = c t−(3+|α|+|β|)/2 1 + √ t t In the case |y| ≤

√

t, we get     |x| −2−μj  α β ˜ (j,k) (x, y, s) ds ≤ c 1 + √ ∂ est K (A + B), ∂  x y u t (2)

Γt,α

where −2 |y| 

λ1 −|β|  d  e−σt sin δ σ (1+|α|+|β|)/2 σ 1/2 |y| 1 + | log(σ 1/2 |y|)| dσ

A= 1/t

and ∞ B=

 μj −1 e−σt sin δ σ (1+|α|+|β|)/2 σ 1/2 |y| dσ.

|y|−2

Substituting σt = τ , we get A ≤ ct

−(3+|α|+|β|)/2

 |y| λ1 −|β| ∞ √ e−τ sin δ τ (1+λ1 +|α|)/2 t 1

  d |y|   × 1 +  log √  + log τ dτ t  d   |y| λ1 −|β|  |y|   ≤ c t−(3+|α|+|β|)/2 √ 1 +  log √  . t t

33

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

34

Furthermore, ∞ B≤

μj +1  e−σt sin δ σ (1+|α|+|β|)/2 σ 1/2 |y| dσ

|y|−2

∞ ≤

 |y| μj +1 μj +1  e−σt sin α σ (1+|α|+β|)/2 σ 1/2 |y| dσ = c t−(3+|α|+|β|)/2 √ . t

1/t

Hence,     α β ˜ (j,k) (x, y, s) ds ∂ ∂ est K  x y u (2)

Γt,δ

  |x| −2−μj  |y| λ1 −|β|  |y| d  √ ≤ c t−(3+|α|+|β|)/2 1 + √ 1 +  log √  t t t if |y| ≤

√

t. Thus both in the cases |y| ≥

√

t and |y| ≤

√

t, we obtain the estimate

    |x| −2−μj  α β ˜ (j,k) (x, y, s) ds ≤ c t−(3+|α|+|β|)/2 1 + √ ∂ ∂ est K  x y u t (2)

Γt,δ

√  |y| μj −1  |y| λ1 −|β|  |y| + t d √ × 1+ √ . 1 + log |y| |y| + t t (j,k)

This proves the estimate for the functions Ku gously by means of the inequality

. The other estimates of the lemma can be proved analo-

 α  ˜ 2 )) p(−j,k) (x) ≤ c |s|(μj +|α|+1)/2 (1 + |s|1/2 r)−1−μj −|α| , ∂x (1 − ψ(sr 0

and the estimate of Q(j,k) in Lemma 2.9.  Using Theorem 2.1 and Theorem 2.2, we obtain the following result.   Theorem 3.4. Let (u, p) ∈ Wβ2,1 (Q) × L2 R+ , Vβ1 (K) be a solution of the problem (1), (2) with the data (38), (39). We assume that 12 − λ1 < β < 12 < γ < 32 and that γ − 1/2 is not an eigenvalue of the pencil N (λ). Then u and p admit the decomposition σj 

u=

S (j,k) + v,

j∈Jγ k=1

p=

σj 

T (j,k) + q,

j∈Jγ k=1

where Jγ is the set of all j such that 0 ≤ μj < γ − 12 , t  S

(j,k)

(x, t) =



 Ku(j,k) (x, y, t − τ ) f (y, τ ) + Hu(j,k) (x, y, t − τ ) g(y, τ ) dy dτ,

0 K

t  T

(j,k)

(x, t) = 0 K



 Kp(j,k) (x, y, t − τ ) · f (y, τ ) + Hp(j,k) (x, y, t − τ ) g(y, τ ) dy dτ,

(43)

V. Kozlov, J. Rossmann / J. Math. Anal. Appl. 486 (2020) 123821

35

  (j,k) (j,k) (j,k) (j,k) and (v, q) ∈ Wγ2,1 (Q) × L2 R+ , Vγ1 (K) . Here, Ku , Hu , Kp and Hp satisfy the estimates of Lemma 3.1. Furthermore, the estimate vWγ2,1 (Q) + qL2 (R+ ,Vγ1 (K))    ≤ c f L2 R+ ,Vγ0 (K) + gL2 R+ ,Vγ1 (K) + ∂t gL2 R+ ,(V−γ 1 (K))∗ is valid with a constant c independent of f and g. The same result holds if 12 − λ1 < β < 12 < γ < min(λ1 , μ2 ) + 32 , the number γ − 1/2 is not an eigenvalue of the pencil N (λ) and g satisfies the condition (36). Then the set Jγ in (43) can be replaced by Jγ \{1}. References [1] S. Agmon, A. Douglis, L. Nirenberg, Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions. I, Comm. Pure Appl. Math. 12 (1959) 623–727. [2] M.S. Agranovich, M.I. Vishik, Elliptic problems with a parameter and parabolic problems of general type, Uspekhi Mat. Nauk 19 (3 (117)) (1964) 3–161. [3] R. Brown, Z. Shen, Estimates for the Stokes operator in Lipschitz domains, Indiana Univ. Math. 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