On the L p - L q decay estimate for the Stokes equations with free boundary conditions in an exterior domain

Abstract

This paper deals with the $L_{p}$ - $L_{q}$ decay estimate of the $C^{0}$ analytic semigroup ${T (t)}_{t ⩾ 0}$ associated with the perturbed Stokes equations with free boundary conditions in an exterior domain. The problem arises in the study of free boundary problem for the Navier–Stokes equations in an exterior domain. We proved that $\begin{matrix} {‖ \nabla^{j} T (t) f ‖}_{L_{p}} ⩽ C_{p, q} t^{- \frac{j}{2} - \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q}} (j = 0, 1) \end{matrix}$ provided that $1 < q ⩽ p ⩽ \infty$ and $q \neq \infty$ . Compared with the non-slip boundary condition case, the gradient estimate is better, which is important for the application to proving global well-posedness of free boundary problem for the Navier–Stokes equations. In our proof, it is crucial to prove the uniform estimate of the resolvent operator, the resolvent parameter ranging near zero.

Keywords

Exterior domains Stokes equations free boundary problem without surface tension

1. Introduction

Let Ω be an exterior domain in the N-dimensional Euclidean space $R^{N}$ ( $N ⩾ 3$ ). Let Γ be the boundary of Ω, which is assumed to be a $C^{2}$ hypersurface. Let $n =^{⊤} (n_{1}, \dots, n_{N})$ be the unit outer normal to Γ, where $^{⊤} M$ denotes the transposed M. We consider the equations: $\begin{matrix} (1) & \{\begin{matrix} \partial_{t} u - J^{- 1} Div \tilde{S} (u, p) = 0, \tilde{div} u = 0 & in Ω \times (0, \infty), \\ \tilde{S} (u, p) n = 0 & on Γ \times (0, \infty), \\ u |_{t = 0} = u_{0} & in Ω . \end{matrix} \end{matrix}$ Here, $\tilde{S} (u, p)$ and $\tilde{div} u$ are given as follows: Let $A = A (x)$ be an $N \times N$ matrix whose $(i, j)$ components are $a_{i j} = a_{i j} (x)$ , and let $J (x)$ be a given function. Let $\begin{matrix} (2) & \begin{array}{l} {\tilde{D}}_{i j} (u) = \sum_{k = 1}^{N} (a_{k j} \frac{\partial u_{i}}{\partial x_{k}} + a_{k i} \frac{\partial u_{j}}{\partial x_{k}}) for u =^{⊤} (u_{1}, \dots, u_{N}), \\ {\tilde{S}}_{i j} (u, p) = \sum_{k = 1}^{N} J a_{j k} (μ {\tilde{D}}_{i k} (u) - δ_{i k} p), \tilde{div} u = \sum_{j, k = 1}^{N} J a_{k j} \frac{\partial u_{j}}{\partial x_{k}}, \end{array} \end{matrix}$ where μ is a positive constant describing the viscosity coefficient. And then, $\tilde{S} (u, p)$ is the $N \times N$ matrix whose $(i, j)$ element is ${\tilde{S}}_{i j} (u, p)$ . As for the remaining symbols, $\partial_{t} = \partial / \partial t$ , t being time variable, and $Div K$ denotes the N-vector whose i component is $\sum_{j = 1}^{N} \partial_{j} K_{i j}$ , where K is any $N \times N$ matrix valued function whose $(i, j)$ component is $K_{i j}$ .

Let R be a fixed positive constant such that $Ω^{c} \subset B_{R / 2}$ , where for any $D \subset R^{N}$ , $D^{c}$ denotes the complement of D in $R^{N}$ , that is, $D^{c} = R^{N} ∖ D$ , and $B_{L} = {x \in R^{N} ∣ | x | < L}$ for any $L > 0$ . Throughout the paper, we assume that $a_{j k} = a_{j k} (x)$ and $J = J (x)$ are of the forms: $a_{j k} (x) = δ_{j k} + b_{j k} (x)$ and $J (x) = 1 + J_{0} (x)$ , where $b_{i j} (x)$ and $J_{0} (x)$ are functions in $H_{r}^{1} (Ω)$ with $N < r < \infty$ such that $\begin{matrix} (3) & \begin{matrix} b_{i j} (x) = 0, J_{0} (x) = 0 for | x | ⩾ 2 R, \\ {‖ (b_{i j}, J_{0}) ‖}_{L_{\infty} (Ω)} + {‖ \nabla (b_{i j}, J_{0}) ‖}_{L_{r} (Ω)} ⩽ σ \end{matrix} \end{matrix}$ with some small positive number σ. Here and hereafter, $δ_{i j}$ denote the Kronecker delta symbols, that is, $δ_{i i} = 1$ and $δ_{i j} = 0$ for $i \neq j$ . Since σ is chosen small enough eventually, we may assume that $0 < σ < 1$ . Moreover, we assume that $\begin{matrix} (4) & \sum_{j, k = 1}^{N} J a_{k j} \frac{\partial u_{j}}{\partial x_{k}} = \sum_{j, k = 1}^{N} \frac{\partial}{\partial x_{k}} (J a_{k j} u_{j}) . \end{matrix}$ From (4) it follows that $\begin{matrix} (5) & \tilde{div} u = div u + \sum_{j, k = 1}^{N} (b_{k j} + J_{0} a_{k j}) \frac{\partial u_{j}}{\partial x_{k}} = div u + \sum_{j, k = 1}^{N} \frac{\partial}{\partial x_{k}} ((b_{k j} + J_{0} a_{k j}) u_{j}), \end{matrix}$ where $div u = \sum_{j = 1}^{N} \partial_{j} u_{j}$ for $u =^{⊤} (u_{1}, \dots, u_{N})$ . Equation (1) with (4) is obtained by applying the transformation $y = L (x) : = x + w (x)$ to the Stokes equations: $\begin{matrix} (6) & \{\begin{matrix} \partial_{t} v - Div S (v, q) = 0, div v = 0 & in \tilde{Ω} \times (0, \infty), \\ S (v, q) \tilde{n} = 0 & on \tilde{Γ} \times (0, \infty), \\ v |_{t = 0} = v_{0} & in \tilde{Ω}, \end{matrix} \end{matrix}$ where $\tilde{n}$ is the unit outer normal to $\tilde{Γ}$ , $\begin{array}{l} (7) & \begin{array}{l} S (v, q) = μ D (v) - q I, D (v) = \nabla v +^{⊤} \nabla v, \\ \tilde{Ω} = {y = L (x) ∣ x \in Ω}, \tilde{Γ} = {y = L (x) ∣ x \in Γ} . \end{array} \end{array}$ Moreover, if $w (x)$ is an N-vector of functions in $H_{r}^{1} (Ω)$ with $N < r < \infty$ such that $supp w \subset B_{2 R}$ and $‖ w ‖_{L_{\infty} (Ω)} + ‖ \nabla w ‖_{L_{r} (Ω)} ⩽ σ$ , then the assumption (3) is satisfied.

In this paper, we prove the $L_{p}$ - $L_{q}$ decay estimate of solutions to Eq. (1), which plays an important role to prove global well-posedness of free boundary problem for the Navier–Stokes equations in an exterior domain; [19].

The $L_{p}$ - $L_{q}$ decay estimate for the Stokes equations has been extensively studied in the non-slip boundary condition case, [3–5,7–11,13,14]. If we denote the Stokes semigroup for the exterior problem with non-slip boundary condition by ${D (t)}_{t ⩾ 0}$ , then we know that $\begin{matrix} (8) & {‖ D (t) f ‖}_{L_{p}} ⩽ C_{p, q} t^{- \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q}}, {‖ \nabla D (t) f ‖}_{L_{p}} ⩽ C_{p, q} t^{- min (\frac{1}{2} + \frac{N}{2} (\frac{1}{q} - \frac{1}{p}), \frac{N}{2 q})} ‖ f ‖_{L_{q}} \end{matrix}$ for every $t > 1$ , provided that $1 < q ⩽ p ⩽ \infty$ and $q \neq \infty$ . On the other hand, if we denote the Stokes semigroup for the exterior problem with free boundary conditions by ${N (t)}_{t ⩾ 0}$ , Shibata and Shimizu [20] proved that $\begin{matrix} (9) & {‖ \nabla^{j} N (t) f ‖}_{L_{p}} ⩽ C_{p, q} t^{- \frac{j}{2} - \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q}} \end{matrix}$ for every $t > 0$ and $j = 0, 1$ , provided that $1 < q ⩽ p ⩽ \infty$ and $q \neq \infty$ . The difference appears in the gradient estimate, that is, we have a better decay estimate in the free boundary condition case, because of the null force at the boundary; [12]. In this paper, we show that semigroup associated with Eq. (1) has the same decay properties as in (9). In all the papers mentioned above concerning (8) and (9), it is crucial to prove the local energy estimate, which, combined with the $L_{p}$ - $L_{q}$ decay estimate for the Stokes semigroup in $R^{N}$ , leads to (8) and (9). But, our approach in this paper is completely different and the crucial step is to prove the uniform estimate of the resolvent operator in a sector $Σ_{ϵ}$ with some $ϵ \in (0, π / 2)$ , where $\begin{matrix} (10) & Σ_{ϵ} = {λ \in C ∖ {0} ∣ | λ | ⩽ π - ϵ} . \end{matrix}$

To state our main results precisely, at this point we explain the symbols used in this paper. $C$ , $R$ , and $N$ denote the set of all complex numbers, real numbers and natural numbers, respectively. Let $N_{0} = N \cup {0}$ . For any multi-index $α = (α_{1}, \dots, α_{N}) \in N_{0}^{N}$ we set $\partial_{x}^{α} h = \partial_{1}^{α_{1}} \dots \partial_{N}^{α_{N}} h$ with $\partial_{i} = \partial / \partial x_{i}$ . Especially, for scalar, θ, and N-vector, $u =^{⊤} (u_{1}, \dots, u_{N})$ , functions and $n \in N_{0}$ , we set $\nabla^{n} θ = (\partial_{x}^{α} θ ∣ | α | = n)$ and $\nabla^{n} u = (\nabla^{n} u_{j} ∣ j = 1, \dots, N)$ . In particular, $\nabla^{0} θ = θ$ , $\nabla^{1} θ = \nabla θ$ , $\nabla^{0} u = u$ and $\nabla^{1} u = \nabla u$ . We use bold small letters to denote N-vectors and bold capital letters to denote $N \times N$ matrix. For $a =^{⊤} (a_{1}, \dots, a_{N})$ , $a |_{i} = a_{i}$ and $(K_{i j})$ denote the $N \times N$ matrix whose $(i, j)$ component is $K_{i j}$ . Let $δ_{i j}$ be the Kronecker delta symbol, that is, $δ_{i i} = 1$ and $δ_{i j} = 0$ for $i \neq j$ . Thus, $I = (δ_{i j})$ . For any N-vectors a and b with $a |_{i} = a_{i}$ and $b |_{i} = b_{i}$ , let $a \cdot b = ⟨ a, b ⟩ = \sum_{j = 1}^{N} a_{j} b_{j}$ . For any N-vector a, let $Π_{0} [a] = a - ⟨ a, n ⟩ n$ . Given $1 < q < \infty$ , we set $q^{'} = q / (q - 1)$ . For $L > 0$ , let $B_{L} = {x \in R^{N} ∣ | x | < L}$ and $S_{L} = {x \in R^{N} ∣ | x | = L}$ . Throughout the paper, $R > 0$ is a fixed constant such as $O = R^{N} ∖ Ω \subset B_{R / 2}$ and κ is an element in $C_{0}^{\infty} (R^{N})$ such that $κ (x) = 1$ for $x \in B_{R}$ and $κ (x) = 0$ for $x \notin B_{2 R}$ . For $0 < R ⩽ L$ , let $Ω_{L} = Ω \cap B_{L}$ and for $R ⩽ L_{1} < L_{2}$ , let $D_{L_{1}, L_{2}} = {x \in R^{N} ∣ L_{1} ⩽ | x | ⩽ L_{2}}$ . For any domain G in $R^{N}$ , let $L_{q} (G)$ , $H_{q}^{m} (G)$ , and $B_{q, p}^{s} (G)$ be the usual Lebesgue space, Sobolev space, and Besov space on G, while $‖ \cdot ‖_{L_{q} (G)}$ , $‖ \cdot ‖_{H_{q}^{m} (G)}$ , and $‖ \cdot ‖_{B_{q, p}^{s} (G)}$ denote their norms, respectively. We write $W_{q}^{0} (G) = L_{q} (G)$ , and $W_{q}^{s} (G) = B_{q, p}^{s} (G)$ . For a Banach space X with norm $‖ \cdot ‖_{X}$ , let $X^{d} = {(f_{1}, \dots, f_{d}) ∣ f_{i} \in X (i = 1, \dots, d)}$ , while the norm of $X^{d}$ is written by $‖ \cdot ‖_{X}$ for short, which is defined by $‖ f ‖_{X} = \sum_{j = 1}^{d} ‖ f_{j} ‖_{X}$ for $f = (f_{1}, \dots, f_{d}) \in X^{d}$ . Let $\begin{array}{l} {\hat{H}}_{q, 0}^{1} (Ω) = {θ \in L_{q, loc} (Ω) ∣ \nabla θ \in L_{q} {(Ω)}^{N}, θ |_{Γ} = 0}, H_{q, 0}^{1} (Ω) = {θ \in H_{q}^{1} (Ω) ∣ θ |_{Γ} = 0}, \\ {\hat{H}}_{q}^{1} (R^{N}) = {θ \in L_{q, loc} (R^{N}) ∣ \nabla θ \in L_{q} {(R^{N})}^{N}}, H_{q, Γ}^{1} (Ω_{5 R}) = {θ \in H_{q}^{1} (Ω_{5 R}) ∣ θ |_{Γ} = 0}, \\ {\dot{H}}_{q, 0}^{m} (G) = {u \in H_{q}^{m} (G) ∣ \partial_{x}^{α} u |_{\partial G} = 0 (| α | ⩽ m - 1)}, \\ {\dot{H}}_{q, a}^{m} (G) = {u \in H_{q, 0}^{m} (G) | \int_{G} u d x = 0}, \\ L_{q, 5 R} (D) = {f \in L_{q} (D) ∣ f (x) = 0 for | x | > 5 R}, D \in {R^{N}, Ω}, \end{array}$ where L is a positive number such that $O \subset B_{L}$ , G is a bounded domain with boundary $\partial G$ , and $m \in N$ . Let $H_{q, 0}^{0} (G) = L_{q} (G)$ . Let ${(u, v)}_{G} = \int_{G} u \cdot v d x$ and let ${(u, v)}_{\partial G} = \int_{\partial G} u \cdot v d ω$ , where $d ω$ denotes the surface element on $\partial G$ . For $1 ⩽ p ⩽ \infty$ , $L_{p} ((a, b), X)$ and $H_{p}^{m} ((a, b), X)$ denote the usual Lebesgue space and Sobolev space of X-valued functions defined on an interval $(a, b)$ , while $‖ \cdot ‖_{L_{p} ((a, b), X)}$ and $‖ \cdot ‖_{H_{p}^{m} ((a, b), X)}$ denote their norms, respectively. Let $C_{0}^{\infty} (G)$ be the set of all $C^{\infty}$ functions whose supports are compact and contained in G. For two Banach spaces X and Y, $X + Y = {x + y ∣ x \in X, y \in Y}$ , $L (X, Y)$ denotes the set of all bounded linear operators from X into Y and $L (X, X)$ is simply written by $L (X)$ . Let $‖ \cdot ‖_{L (X, Y)}$ be the operator norm for $L (X, Y)$ . For any domain U in $C$ , $Hol (U, L (X, Y))$ denotes the set of all $L (X, Y)$ -valued holomorphic functions defined on U. Let $R_{L (X, Y)} ({T (λ) ∣ λ \in U})$ be the $R$ bound of operator family $T (λ) \in Hol (U, L (X, Y))$ . Let $Σ_{ϵ, λ_{0}} = {λ \in Σ_{ϵ} ∣ | λ | ⩾ λ_{0}}$ and $Λ_{ϵ, λ_{0}} = {λ \in Σ_{ϵ} ∣ | λ | ⩽ λ_{0}}$ . For any Lebesgue measurable set D in $R^{N}$ , $| D |$ denotes the Lebesgue measure of D. Let $F$ and $F_{ξ}^{- 1}$ denote the Fourier transform and the Fourier inverse transform defined by $\begin{matrix} F [f] (ξ) = \int_{R^{N}} e^{- i x \cdot ξ} f (x) d x, F_{ξ}^{- 1} [g (ξ)] (x) = \frac{1}{{(2 π)}^{N}} \int_{R^{N}} e^{i x \cdot ξ} g (ξ) d ξ \end{matrix}$ with $x = (x_{1}, \dots, x_{N})$ and $ξ = (ξ_{1}, \dots, ξ_{N}) \in R^{N}$ . Finally, the letter C and $C_{a, b, c, \dots}$ denotes generic constants and generic constants depending on $a, b, c, \dots$ , respectively. The value of constants C and $C_{a, b, c, \dots}$ may change from line to line.

Let $\begin{array}{l} \tilde{\nabla} φ & =^{⊤} (\sum_{k = 1}^{N} a_{k 1} \frac{\partial φ}{\partial x_{k}}, \dots, \sum_{k = 1}^{N} a_{k N} \frac{\partial φ}{\partial x_{k}}) =^{⊤} A \nabla φ, \end{array}$ and then, the perturbed solenoidal space ${\tilde{J}}_{q} (Ω)$ is defined by $\begin{matrix} (11) & {\tilde{J}}_{q} (Ω) = {f \in L_{q} {(Ω)}^{N} ∣ {(f, J \tilde{\nabla} φ)}_{Ω} = 0 for any φ \in {\hat{H}}_{q^{'}, 0}^{1} (Ω)}, \end{matrix}$ where $q^{'} = q / (q - 1)$ . The following theorem is the main result of this paper.

Theorem 1.
Assume that $N ⩾ 3$ . Let $N < r < \infty$ and $1 < q ⩽ r$ . Then, there exists a $σ > 0$ such that if assumption ( 3 ) holds, then there exists a $C^{0}$ analytic semi-group ${T (t)}_{t ⩾ 0}$ associated with problem ( 1 ) such that for any $u_{0} \in {\tilde{J}}_{q} {(Ω)}^{N}$ , $u = T (t) u_{0}$ is a unique solution of problem ( 1 ) with some pressure term $p \in L_{p} ((0, T), H_{q}^{1} (Ω) + {\hat{H}}_{q}^{1} (Ω))$ possessing the estimates: $\begin{matrix} (12) & {‖ \nabla^{j} T (t) u_{0} ‖}_{L_{p} (Ω)} ⩽ C_{p, q} t^{- \frac{j}{2} - \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ u_{0} ‖_{L_{q} (Ω)} \end{matrix}$ for every $t > 0$ and $j = 0, 1$ , provided that $1 < q ⩽ p ⩽ \infty$ and $q \neq \infty$ .

To prove Theorem 1, we consider the resolvent problem: $\begin{matrix} (13) & \{\begin{matrix} λ u - J^{- 1} Div \tilde{S} (u, p) = f, \tilde{div} u = 0 & in Ω, \\ \tilde{S} (u, p) n = 0 & on Γ . \end{matrix} \end{matrix}$ Let $Σ_{ϵ} = {λ \in C ∖ {0} ∣ | arg λ | < π - ϵ}$ . To prove Theorem 1, it is crucial to prove Theorem 2.
Assume that $N ⩾ 3$ . Let $N < r < \infty$ , $1 < q ⩽ r$ and $0 < ϵ < π / 2$ . Then, there exists a small constant $σ > 0$ such that if the assumption ( 3 ) holds, then there exist operator families $S (λ)$ and $P (λ)$ with $\begin{matrix} S (λ) \in Hol (Σ_{ϵ}, L (L_{q} {(Ω)}^{N}, H_{q}^{2} {(Ω)}^{N})), P (λ) \in Hol (Σ_{ϵ}, L (L_{q} {(Ω)}^{N}, H_{q}^{1} (Ω) + {\hat{H}}_{q, 0}^{1} (Ω))) \end{matrix}$ such that for any $λ \in Σ_{ϵ}$ , $f \in L_{q} {(Ω)}^{N}$ , $u = S (λ) f$ and $p \in P (λ) f$ are unique solutions of resolvent problem ( 13 ).

Moreover, $\begin{matrix} (14) & \begin{array}{l} {‖ S (λ) f ‖}_{L_{p} (Ω)} ⩽ C_{p, q} | λ |^{- 1 + \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (Ω)}, \\ {‖ \nabla S (λ) f ‖}_{L_{p} (Ω)} ⩽ C_{p, q} | λ |^{- \frac{1}{2} + \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (Ω)}, \\ {‖ (\nabla^{2} S (λ) f, \nabla P (λ) f) ‖}_{L_{q} (Ω)} ⩽ C_{q} max (1, | λ |^{- \frac{1}{2} + \frac{N}{2} (\frac{1}{q} - \frac{1}{\tilde{p}})}) ‖ f ‖_{L_{q} (Ω)} \end{array} \end{matrix}$ for any $λ \in Σ_{ϵ}$ and $f \in L_{q} {(Ω)}^{N}$ with some constant $C_{p, q}$ and $C_{q}$ provided that $1 < q ⩽ p ⩽ \infty$ and $N (1 / q - 1 / p) < 1$ . Here, $\tilde{p}$ is chosen in such a way that $\tilde{p} = \infty$ when $N < q ⩽ r$ , and ${(\tilde{p})}^{- 1} = q^{- 1} - N^{- 1} + δ_{0}$ with some constant $δ_{0}$ satisfying the condition: $0 < δ_{0} < N^{- 1} + p^{- 1} - q^{- 1}$ when $1 < q ⩽ N$ .

2. Preliminaries

2.1. Bogovskiĭ’s lemma

To keep the divergence free condition in using the cut-off technique, we use the following lemma due to Bogovskiiĭ [1,2] (cf. also Galdi [6]).

Lemma 3.
Let $1 < q < \infty$ , $m \in N_{0}$ and let G be a bounded domain whose boundary $\partial G$ is a hypersurface of $C^{m + 1}$ class. Then, there exists a linear operator $B : H_{q, 0}^{m} (G) \to H_{q, 0}^{m + 1} {(G)}^{N}$ having the following properties:
There exists a $ρ \in C_{0}^{\infty} (G)$ such that $ρ ⩾ 0$ , $\int_{G} ρ d x = 1$ , and $div B [f] = f - ρ \int_{Ω} f d x$ . In particular, if $\int_{G} f d x = 0$ , then $div B [f] = f$ .

We have the estimate: $\begin{matrix} {‖ B [f] ‖}_{H_{q}^{k + 1} (G)} ⩽ C_{q, k, G} ‖ f ‖_{H_{q}^{k} (G)} (k = 0, \dots, m) . \end{matrix}$

If $f = \partial g / \partial x_{i}$ with some $g \in H_{q, 0}^{m + 1} (G)$ , then $\begin{matrix} {‖ B [f] ‖}_{H_{q}^{k} (G)} ⩽ C_{q, k, G} ‖ g ‖_{H_{q}^{k} (G)} (k = 0, \dots, m) . \end{matrix}$

Remark 4.
Since $B [f] \in H_{q, 0}^{m + 1} (G)$ , the 0 extension of $B [f]$ to $R^{N}$ is also written by $B [f]$ . Thus, $B [f] \in H_{q}^{m + 1} {(R^{N})}^{N}$ and $supp B [f] \subset G$ .

To apply Lemma 3, we use the following lemma.
Lemma 5.
Let $1 < q < \infty$ and let $2 R < L_{1} < L_{2} < L_{3} < L_{4} < 5 R$ . Let χ be a function in $C^{\infty} (R^{N})$ such that $χ (x) = 1$ for $x \in B_{L_{2}}$ and $χ (x) = 0$ for $x \notin B_{L_{3}}$ . If $v \in H_{q}^{2} {(G)}^{N}$ , $G \in {R^{N}, Ω, Ω_{5 R}}$ , satisfies $div v = 0$ in $D_{L_{1}, L_{4}}$ , then $(\nabla χ) \cdot v \in {\dot{H}}_{q, a}^{2} (D_{L_{2}, L_{3}})$ .

To prove Lemma 5, we need the following lemma. Lemma 6.
Let $1 < q < \infty$ and let $2 R < L_{1} < L_{2} < L_{3} < L_{4} < L_{5} < L_{6} < 5 R$ . Let χ be a function in $C^{\infty} (R^{N})$ such that $supp \nabla χ \subset D_{L_{3}, L_{4}}$ . If $u \in H_{q}^{2} (G)$ , $G \in {R^{N}, Ω, Ω_{5 R}}$ , satisfies $div u = 0$ in $D_{L_{1}, L_{6}}$ , then there exists a $v \in H_{q}^{2} {(R^{N})}^{N}$ such that $supp v \subset D_{L_{2}, L_{5}}$ , $div v = 0$ in $R^{N}$ and $(\nabla χ) \cdot v = (\nabla χ) \cdot u$ in $R^{N}$ .
Proof.
Let $A_{0}, A_{1}, \dots, A_{5}$ and $B_{0}, B_{1}, \dots, B_{5}$ be numbers such that $L_{2} = A_{5} < A_{4} < A_{3} < A_{2} < A_{1} < A_{0} < L_{3} < L_{4} < B_{0} < B_{1} < B_{2} < B_{3} < B_{4} < B_{5} = L_{5}$ . Let $φ \in C_{0}^{\infty} (R^{N})$ such that $φ (x) = 1$ for $A_{2} < | x | < B_{2}$ and $φ (x) = 0$ for $| x | < A_{3}$ or $| x | > B_{3}$ . Note that $φ (x) = 1$ on $supp \nabla χ$ . Let $E = D_{A_{4}, A_{1}} \cup D_{B_{1}, B_{4}}$ . Since $div u = 0$ in $D_{L_{1}, L_{6}}$ and $supp φ \subset D_{A_{3}, B_{3}}$ , we have $div (φ u) = (\nabla φ) \cdot u$ in $D_{A_{4}, B_{4}}$ and $div (φ u) = (\nabla φ) \cdot u \in {\dot{H}}_{q}^{2} (E)$ . Moreover, we have $\begin{array}{l} \int_{E} (\nabla φ) \cdot u d x & = \int_{D_{A_{4}, B_{4}}} (\nabla φ) \cdot u d x = \int_{D_{A_{4}, B_{4}}} div (φ u) d x \\ = - \int_{S_{A_{4}}} \frac{x}{| x |} \cdot (φ u) d τ + \int_{S_{B_{4}}} \frac{x}{| x |} \cdot (φ u) d τ = 0, \end{array}$ which leads to $(\nabla φ) \cdot u \in {\dot{H}}_{q, a}^{3} (E)$ . By Lemma 3, $v = φ u - B [(\nabla φ) \cdot u]$ has the properties: $v \in H_{q}^{2} (R^{N})$ , $supp v \subset D_{A_{3}, B_{3}}$ , and $div v = 0$ in $R^{N}$ . Moreover, $(\nabla χ) \cdot u = (\nabla χ) \cdot v$ in $R^{N}$ , because $φ = 1$ on $supp \nabla χ$ and $B [(\nabla φ) \cdot u] = 0$ on $supp \nabla χ$ . □
Proof of Lemma 5.
By Lemma 6, there exists a $w \in H_{q}^{2} {(R^{N})}^{N}$ such that $(\nabla χ) \cdot v = (\nabla χ) \cdot w$ in $R^{N}$ , $supp w \subset B_{L_{4}}$ and $div w = 0$ in $R^{N}$ . Then, the assertion follows from the following observation: $\begin{array}{l} \int_{D_{L_{2}, L_{3}}} (\nabla χ) \cdot v d x & = \int_{D_{L_{2}, L_{3}}} (\nabla χ) \cdot w d x = \int_{B_{L_{4}}} (\nabla χ) \cdot w d x \\ = \int_{B_{L_{4}}} div (χ w) d x = \int_{S_{L_{4}}} \frac{x}{| x |} \cdot (χ w) d τ = 0, \end{array}$ because $χ |_{S_{L_{4}}} = 0$ . This completes the proof of Lemma 5. □

2.2. Resolvent estimate in a bounded domain

In this subsection, we consider the following equations: $\begin{matrix} (15) & \{\begin{matrix} λ u - J^{- 1} Div \tilde{S} (u, p) = f, & \tilde{div} u = g in Ω_{5 R}, \\ \tilde{S} (u, p) n |_{Γ} = h, & u |_{S_{5 R}} = 0, \end{matrix} \end{matrix}$ and prove the following theorem.

Theorem 7.
Let $N < r < \infty$ be the exponent given in ( 3 ). Let $1 < q < \infty$ , $0 < ϵ < π / 2$ and $λ_{1} > 0$ . Let $\begin{matrix} Λ_{ϵ, λ_{1}} = {λ \in Σ_{ϵ} ∣ | λ | ⩽ λ_{1}} . \end{matrix}$ Assume that $max (q, q^{'}) ⩽ r$ . Then, there exists a $σ > 0$ such that if the assumption ( 3 ) holds, then for any $λ \in Λ_{ϵ, λ_{1}} \cup {0}$ , $f \in L_{q} (Ω_{5 R})$ , $g \in H_{q}^{1} (Ω_{5 R})$ and $h \in H_{q}^{1} {(Ω_{5 R})}^{N}$ , problem ( 15 ) admits unique solutions $u \in H_{q}^{2} {(Ω_{5 R})}^{N}$ and $p \in H_{q}^{1} (Ω_{5 R})$ possessing the estimate: $\begin{matrix} (16) & \begin{matrix} ‖ u ‖_{H_{q}^{2} (Ω_{5 R})} + ‖ p ‖_{H_{q}^{1} (Ω_{5 R})} ⩽ C {‖ f ‖_{L_{q} (Ω_{5 R})} + ‖ h ‖_{H_{q}^{1} (Ω_{5 R})} + ‖ g ‖_{H_{q}^{1} (Ω_{5 R})}} \end{matrix} \end{matrix}$ for some constant $C > 0$ depending solely on ϵ and $λ_{1}$ .

To prove Theorem 7, first we consider the weak Dirichlet problem in $Ω_{5 R}$ , which is formulated as follows: $\begin{matrix} (17) & {(\nabla u, \nabla φ)}_{Ω_{5 R}} = {(f, \nabla φ)}_{Ω_{5 R}} for any φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R}) . \end{matrix}$ As was proved in the appendix of Shibata [18], we have
Theorem 8.
Let $1 < q < \infty$ . Then, for any $f \in L_{q} {(Ω_{5 R})}^{N}$ , problem ( 17 ) admits a unique solution $u \in H_{q, Γ}^{1} (Ω_{5 R})$ possessing the estimate: $\begin{matrix} (18) & ‖ u ‖_{H_{q}^{1} (Ω_{5 R})} ⩽ C ‖ f ‖_{L_{q} (Ω_{5 R})} . \end{matrix}$

To prove Theorem 7, next we consider the equations: $\begin{matrix} (19) & \{\begin{matrix} λ u - Div S (u, p) = f, & div u = g in Ω_{5 R}, \\ S (u, p) n |_{Γ} = h, & u |_{S_{5 R}} = 0 . \end{matrix} \end{matrix}$ For $g \in H_{q}^{1} (Ω_{5 R})$ , by the Poincarés inequality $\begin{matrix} | {(g, φ)}_{Ω_{5 R}} | ⩽ ‖ g ‖_{L_{q} (Ω_{5 R})} ‖ φ ‖_{L_{q^{'}} (Ω_{5 R})} ⩽ C ‖ g ‖_{L_{q} (Ω_{5 R})} ‖ \nabla φ ‖_{L_{q} (Ω_{5 R})} \end{matrix}$ for any $φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R})$ . Thus, by the Hahn–Banach theorem, there exists a $g \in L_{q} (Ω_{5 R})$ such that $\begin{matrix} {(g, φ)}_{Ω_{5 R}} = {(g, \nabla φ)}_{Ω_{5 R}} for any φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R}), ‖ g ‖_{L_{q} (Ω_{5 R})} ⩽ C ‖ g ‖_{L_{q} (Ω_{5 R})} . \end{matrix}$ Thus, by Theorem 1.5 in Shibata [16], we have
Theorem 9.
Let $1 < q < \infty$ and $0 < ϵ < π / 2$ . Then, there exists a $λ_{0} > 0$ such that for any $f \in L_{q} {(Ω_{5 R})}^{N}$ , $g \in H_{q}^{1} (Ω_{5 R})$ , and $h \in H_{q}^{1} {(Ω_{5 R})}^{N}$ , problem ( 19 ) with $λ = λ_{0}$ admits unique solutions $u \in H_{q}^{2} (Ω_{5 R})$ and $p \in H_{p}^{1} (Ω_{5 R})$ possessing the estimate: $\begin{array}{l} ‖ u ‖_{H_{q}^{2} (Ω_{5 R})} + ‖ p ‖_{H_{q}^{1} (Ω_{5 R})} ⩽ C {‖ f ‖_{L_{q} (Ω_{5 R})} + {‖ (g, h) ‖}_{H_{q}^{1} (Ω_{5 R})}} \end{array}$ for some constant C depending on ϵ and $λ_{0}$ .

By using Theorem 9, we prove
Theorem 10.
Let $1 < q < \infty$ , $0 < ϵ < π / 2$ and $λ_{1} > 0$ . Then, for any $λ \in Λ_{ϵ, λ_{1}} \cup {0}$ , $f \in L_{q} {(Ω_{5 R})}^{N}$ , $g \in H_{q}^{1} {(Ω_{5 R})}^{N}$ , and $h \in H_{q}^{1} {(Ω_{5 R})}^{N}$ , problem ( 19 ) admits unique solutions $u \in H_{q}^{2} (Ω_{5 R})$ and $p \in H_{q}^{1} (Ω_{5 R})$ possessing the estimate: $\begin{matrix} (20) & ‖ u ‖_{H_{q}^{2} (Ω_{5 R})} + ‖ p ‖_{H_{q}^{1} (Ω_{5 R})} ⩽ C {‖ f ‖_{L_{q} (Ω_{5 R})} + {‖ (g, h) ‖}_{H_{q}^{1} (Ω_{5 R})}} \end{matrix}$ for some constant C depending solely on ϵ, $λ_{1}$ and R.
Proof.
Let $λ_{0} > 0$ be the number given in Theorem 9 and let $v \in H_{q}^{2} (Ω_{5 R})$ and $q \in H_{q}^{1} (Ω_{5 R})$ be solutions of the equations $\begin{matrix} \{\begin{matrix} λ_{0} v - Div S (v, q) = f, & div v = g in Ω_{5 R}, \\ S (v, q) n |_{Γ} = h, & v |_{S_{5 R}} = 0, \end{matrix} \end{matrix}$ which possess the estimates: $\begin{matrix} (21) & ‖ v ‖_{H_{q}^{2} (Ω_{5 R})} + ‖ q ‖_{H_{q}^{1} (Ω_{5 R})} ⩽ C {‖ f ‖_{L_{q} (Ω_{5 R})} + {‖ (g, h) ‖}_{H_{q}^{1} (Ω_{5 R})}} . \end{matrix}$ Let $u = v + w$ and $p = q + r$ in (15), and then w and $r$ are solution of the equations: $\begin{matrix} (22) & \{\begin{matrix} λ w - Div S (w, r) = g, & div w = 0 in Ω_{5 R}, \\ S (w, r) n |_{Γ} = 0, & w |_{S_{5 R}} = 0 \end{matrix} \end{matrix}$ with $g = - (λ - λ_{0}) v$ . By (21), $\begin{matrix} (23) & ‖ g ‖_{L_{q} (Ω_{5 R})} ⩽ (λ_{0} + λ_{1}) C {‖ f ‖_{L_{q} (Ω_{5 R})} + {‖ (g, h) ‖}_{H_{q}^{1} (Ω_{5 R})}} \end{matrix}$ for any $λ \in Λ_{ϵ, λ_{1}} \cup {0}$ . In the sequel, we consider (22) for $λ \in Λ_{ϵ, λ_{1}} \cup {0}$ . First, we introduce the second Helmholtz decomposition. Let us consider the variational problem: $\begin{matrix} (24) & {(\nabla ψ, \nabla φ)}_{Ω_{5 R}} = {(h, \nabla φ)}_{Ω_{5 R}} for any φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R}), \end{matrix}$ subject to $ψ = h$ on Γ. By Theorem 8, the unique existence of solutions to problem (24) holds. Thus, one can define an operator $N : L_{q} {(Ω_{5 R})}^{N} \times W_{q}^{1 - 1 / q} (Γ) \to H_{q}^{1} (Ω_{5 R})$ by $N (h, h) = ψ$ which satisfies the estimate $\begin{matrix} {‖ N (h, h) ‖}_{H_{q}^{1} (Ω_{5 R})} ⩽ C (‖ h ‖_{L_{q} (Ω_{5 R})} + ‖ h ‖_{W_{q}^{1 - 1 / q} (Γ)}) . \end{matrix}$ Given $g \in L_{q} (Ω_{5 R})$ , let $Q_{1} g = N (g, 0)$ , and $P_{1} g = g - \nabla Q_{1} g$ . And then, $\begin{matrix} g = P_{1} g + \nabla Q_{1} g, ‖ P_{1} g ‖_{L_{q} (Ω_{5 R})} + ‖ Q_{1} g ‖_{H_{q}^{1} (Ω)} ⩽ C ‖ g ‖_{L_{q} (Ω_{5 R})} . \end{matrix}$ Especially, $\begin{matrix} {(P_{1} g, \nabla φ)}_{Ω_{5 R}} = 0 for any φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R}), Q_{1} g |_{Γ} = 0 . \end{matrix}$ This is called the second Helmholtz decomposition of g in $Ω_{5 R}$ . Inserting this decomposition into the right hand side in (22), we have $\begin{matrix} \{\begin{matrix} λ w - Div S (w, r + Q_{1} g) = P_{1} g, & div w = 0 in Ω_{5 R}, \\ S (w, r + Q_{1} g) n |_{Γ} = 0, & w |_{S_{5 R}} = 0, \end{matrix} \end{matrix}$ where we have used the fact that $Q_{1} g |_{Γ} = 0$ , and by (23) $\begin{matrix} ‖ P_{1} g ‖_{L_{q} (Ω_{5 R})} + ‖ Q_{1} g ‖_{H_{q}^{1} (Ω_{5 R})} ⩽ C {‖ f ‖_{L_{q} (Ω_{5 R})} + {‖ (g, h) ‖}_{H_{q}^{1} (Ω_{5 R})}} . \end{matrix}$ Thus, it suffices to solve (22) under the assumptions that $g \in J_{q} (Ω_{5 R})$ and that (23) holds, where $\begin{matrix} J_{q} (Ω_{5 R}) = {g \in L_{q} {(Ω_{5 R})}^{N} ∣ {(g, \nabla φ)}_{5 R} = 0 for any φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R})} . \end{matrix}$ Given $w \in H_{q}^{2} (Ω_{5 R})$ , let $\begin{matrix} K (w) = N (Div (μ D (w)) - \nabla div w, ⟨ μ D (w) n, n ⟩ - div w) . \end{matrix}$ Namely, $ψ = K (w)$ is a unique solution of the variational equation: $\begin{matrix} (25) & {(\nabla ψ, \nabla φ)}_{Ω_{5 R}} = {(Div (μ D (w)) - \nabla div w, \nabla φ)}_{5 R} for any φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R}), \end{matrix}$ subject to $ψ = ⟨ μ D (w) n, n ⟩ - div w$ on Γ. We have $\begin{matrix} {‖ K (w) ‖}_{H_{q}^{1} (Ω_{5 R})} ⩽ C ‖ \nabla w ‖_{H_{q}^{1} (Ω_{5 R})} . \end{matrix}$ Let $w \in H_{q}^{2} (Ω_{5 R})$ and $r \in H_{q}^{1} (Ω_{5 R})$ be solutions of problem (22) for $g \in J_{q} (Ω)$ , and then for any $φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R})$ $\begin{array}{l} 0 & = {(g, \nabla φ)}_{Ω_{5 R}} = λ {(w, \nabla φ)}_{Ω_{5 R}} - {(Div (μ D (w)) - \nabla div w, \nabla φ)}_{Ω_{5 R}} + {(\nabla r, \nabla φ)}_{Ω_{5 R}} \\ = {(\nabla (r - K (w)), \nabla φ)}_{Ω_{5 R}}, \end{array}$ because $div w = 0$ . Moreover, $r - K (w) = 0$ on Γ because $div w = 0$ . Thus, the uniqueness implies that $r = K (w)$ .

On the other hand, if $w \in H_{q}^{2} (Ω_{5 R})$ is a unique solution to problem: $\begin{matrix} (26) & \{\begin{matrix} λ w - Div S (w, K (w)) = g in Ω_{5 R}, \\ S (w, K (w)) n |_{Γ} = 0, w |_{S_{5 R}} = 0, \end{matrix} \end{matrix}$ then for any $φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R})$ $\begin{array}{l} 0 & = {(g, \nabla φ)}_{Ω_{5 R}} \\ = λ {(w, \nabla φ)}_{Ω_{5 R}} - {(\nabla div w, \nabla φ)}_{Ω_{5 R}} - {(Div (μ D (w)) - \nabla div w, \nabla φ)}_{Ω_{5} R} + {(\nabla K (w), \nabla φ)}_{Ω_{5 R}} \\ = - {(λ div w, φ)}_{Ω_{5 R}} - {(\nabla div w, \nabla φ)}_{Ω_{5 R}}, \end{array}$ subject to $div w = 0$ on Γ, because $\begin{matrix} 0 = ⟨ S (w, K (w)) n, n ⟩ = ⟨ μ D (w) n, n ⟩ - K (w) = div w on Γ . \end{matrix}$ Thus, $u = div w \in H_{q, Γ}^{1} (Ω_{5 R})$ satisfies the weak Dirichlet–Neumann problem: $\begin{matrix} λ {(u, φ)}_{Ω} + {(\nabla u, \nabla φ)}_{Ω} = 0 for any φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R}) . \end{matrix}$ We know that $u = 0$ for $λ \in Σ_{ϵ} \cup {0}$ , that is, $div w = 0$ . Thus, w and $r = K (w)$ are unique solution of problem (22) provided that $g \in J_{q} (Ω_{5 R})$ .

In the sequel, we consider problem (26) instead of problem (22). Let $\begin{array}{l} D_{q} (A) = {u \in J_{q} (Ω_{5 R}) \cap H_{q}^{2} (Ω_{5 R}) ∣ S (u, K (u)) n |_{Γ} = 0, u |_{S_{5 R}} = 0}, \\ A u = Div S (u, K (u)) for u \in D_{q} (A) . \end{array}$ Since $C_{0}^{\infty} (Ω_{5 R}) \subset H_{q^{'}, Γ}^{1} (Ω_{5 R})$ , we have $div u = 0$ in $Ω_{5 R}$ provided $u \in J_{q} (Ω) \cap H_{q}^{2} (Ω)$ , and so A maps $D_{q} (A)$ into $J_{q} (Ω_{5 R})$ . By Theorem 9, $λ_{0}$ lies in the resolvent set $ρ (A)$ of A, and so the resolvent ${(λ_{0} - A)}^{- 1}$ is a bounded linear operator from $J_{q} (Ω_{5 R})$ onto $D_{q} (A)$ . Since $H_{q}^{2} (Ω_{5 R})$ is compactly embedded into $L_{q} (Ω_{5 R})$ as follows from the Rellich compactness theorem, ${(λ_{0} - A)}^{- 1}$ is a compact operator from $J_{q} (Ω_{5 R})$ into itself. Since $λ - A = λ - λ_{0} + λ_{0} - A = (I - (λ_{0} - λ) {(λ_{0} - A)}^{- 1}) (λ_{0} - A)$ , what λ lies in $ρ (A)$ follows from the invertibility of $I - (λ_{0} - λ) {(λ_{0} - A)}^{- 1}$ . Thus, in view of the Riesz–Schauder theory (Fredholm alternative principle), it suffices to prove that the kernel of $I - (λ_{0} - λ) {(λ_{0} - A)}^{- 1}$ is trivial to prove that $λ \in ρ (A)$ . Let $w \in J_{q} (Ω_{5 R})$ satisfy $(I - (λ_{0} - λ) {(λ_{0} - A)}^{- 1}) w = 0$ . Especially, $w = (λ_{0} - λ) {(λ_{0} - A)}^{- 1} w \in D_{q} (A)$ , and so $\begin{array}{l} (λ - A) w & = (λ_{0} - λ) (λ - A) {(λ_{0} - A)}^{- 1} w = (λ_{0} - λ) (λ - λ_{0} + λ_{0} - A) {(λ_{0} - A)}^{- 1} \\ = (λ_{0} - λ) {(λ - λ_{0}) {(λ_{0} - A)}^{- 1} + I} w = (λ_{0} - λ) (I - (λ_{0} - λ) {(λ_{0} - A)}^{- 1}) w = 0 . \end{array}$ Thus, the uniqueness of the equations (26) implies that $λ \in ρ (A)$ . Let $w \in D_{q} (A)$ satisfy the equation $(λ - A) w = 0$ in $Ω_{5 R}$ . First, we consider the case that $2 ⩽ q < \infty$ . In this case, $w \in D_{q} (A) \subset D_{2} (A)$ , and so $\begin{array}{l} 0 & = {((λ - A) w, w)}_{Ω_{5 R}} = λ ‖ w ‖_{L_{2} (Ω_{5 R})}^{2} - {(Div S (w, K (w)), w)}_{Ω_{5 R}} \\ = λ ‖ w ‖_{L_{2} (Ω_{5 R})}^{2} - {(S (w, K (w)) n, w)}_{Γ} - {(S (w, K (w)) ω, w)}_{S_{5 R}} \\ + \frac{μ}{2} {‖ D (w) ‖}_{L_{2} (Ω_{5 R})} - {(K (w), div w)}_{Ω_{5 R}} \\ = λ ‖ w ‖_{L_{2} (Ω_{5 R})}^{2} + \frac{μ}{2} {‖ D (w) ‖}_{L_{2} (Ω_{5 R})}^{2} . \end{array}$ Taking the real part and the imaginary part, we have $\begin{matrix} (Im λ) ‖ w ‖_{L_{2} (Ω_{5 R})} = 0, (Re λ) ‖ w ‖_{L_{2} (Ω_{5 R})} + \frac{μ}{2} {‖ D (w) ‖}_{L_{2} (Ω_{5 R})} = 0 . \end{matrix}$ When $Im λ \neq 0$ , we have $w = 0$ . On the other hand, when $Im λ = 0$ , $λ ⩾ 0$ because $λ \in Σ_{ϵ} \cup {0}$ . Thus, $D (w) = 0$ , and so by the first Korn inequality and $w |_{S_{5 R}} = 0$ , we have $w = 0$ . Therefore, we have $C ∖ (- \infty, 0) \subset ρ (A)$ . Especially, $Λ_{ϵ, λ_{1}} \cup {0}$ is a compact set, and so we have $\begin{matrix} {‖ {(λ - A)}^{- 1} g ‖}_{H_{q}^{2} (Ω_{5 R})} ⩽ C ‖ g ‖_{L_{q} (Ω_{5 R})} \end{matrix}$ for any $λ \in Λ_{ϵ, λ_{1}} \cup {0}$ with some constant C depending solely on $λ_{0}$ and ϵ. When $1 < q < 2$ , the uniqueness follows from the existence for the dual problem, and so we also have $C ∖ (- \infty, 0) \subset ρ (A)$ . This completes the proof of Theorem 10. □
Proof of Theorem 7.
In view of the assumptions (3) and (4), problem (15) is written as follows: $\begin{matrix} (27) & \{\begin{matrix} λ u - Div S (u, p) - μ (A_{1} \nabla^{2} u + A_{2} \nabla u) + A_{3} \nabla p = f, & in Ω_{5 R}, \\ div u + A_{4} \nabla u = g & in Ω_{5 R}, \\ (S (u, p) n + μ A_{5} \nabla u + A_{6} p) n |_{Γ} = h, u |_{S_{5 R}} = 0 . \end{matrix} \end{matrix}$ Here, $A_{1}, \dots, A_{5}$ are some matrices such that $A_{1}, \dots, A_{5}$ vanish for $| x | ⩾ 2 R$ and satisfy the estimate: $\begin{matrix} (28) & {‖ (A_{1}, A_{3}, A_{4}, A_{5}, A_{6}) ‖}_{L_{\infty} (Ω)} ⩽ C σ, {‖ (\nabla A_{1}, A_{2}, \nabla A_{3}, \nabla A_{4}, \nabla A_{5}, \nabla A_{6}) ‖}_{L_{r} (Ω)} ⩽ C σ \end{matrix}$ with some positive constant C independent of σ.

Let $D \in {Ω_{5 R}, Ω, R^{N}}$ . For any $q \in (1, r]$ , we have $\begin{matrix} (29) & ‖ f g ‖_{L_{q} (D)} ⩽ C ‖ f ‖_{H_{q}^{1} (D)} ‖ g ‖_{L_{r} (D)} \end{matrix}$ for any $f \in H_{q}^{1} (D)$ and $g \in L_{r} (D)$ with some constant C depending on D, q and r. In fact, when $r = q$ , by the Sobolev imbedding theorem $\begin{matrix} ‖ f g ‖_{L_{q} (D)} ⩽ ‖ f ‖_{L_{\infty} (D)} ‖ g ‖_{L_{r} (D)} ⩽ C ‖ f ‖_{W_{q}^{1} (D)} ‖ g ‖_{L_{r} (D)}, \end{matrix}$ because $q = r > N$ . When $1 < q < r$ , let s be an exponent such that $1 / q = 1 / r + 1 / s$ . In this case, $N (1 / q - 1 / s) = N / r < 1$ , and so by Hölder’s inequality and the Sobolev imbedding theorem, $\begin{matrix} ‖ f g ‖_{L_{q} (D)} ⩽ ‖ f ‖_{L_{s} (D)} ‖ g ‖_{L_{r} (D)} ⩽ C ‖ f ‖_{W_{q}^{1} (D)} ‖ g ‖_{L_{r} (D)} . \end{matrix}$ Thus, we have (29).

By (29) and (28), $\begin{array}{l} {‖ μ (A_{1} \nabla^{2} u + A_{2} \nabla u) - A_{3} \nabla p ‖}_{L_{q} (Ω_{5 R})} ⩽ C σ (‖ u ‖_{H_{q}^{2} (Ω_{5 R})} + ‖ p ‖_{H_{q}^{1} (Ω_{5 R})}), \\ {‖ (A_{4} \nabla u, A_{5} \nabla u, A_{6} p) ‖}_{H_{q}^{1} (Ω_{5 R})} ⩽ C σ (‖ u ‖_{H_{q}^{2} (Ω_{5 R})} + ‖ p ‖_{H_{q}^{1} (Ω_{5 R})}), \end{array}$ for any $u \in H_{q}^{2} (Ω_{5 R})$ and $p \in H_{q}^{1} (Ω_{5 R})$ . Let $H = H_{q}^{2} {(Ω_{5 R})}^{N} \times H_{q}^{1} (Ω_{5 R})$ . Given $(v, q) \in H$ and $λ \in Λ_{ϵ, λ_{0}} \cup {0}$ , let u and $p$ be solutions of the equations: $\begin{matrix} (30) & \{\begin{matrix} λ u - Div S (u, p) = f + μ (A_{1} \nabla^{2} v + A_{2} \nabla v) - A_{3} \nabla q & in Ω_{5 R}, \\ div u = g + A_{4} \nabla v & in Ω_{5 R}, \\ S (u, p) n |_{Γ} = h - μ A_{5} \nabla v - A_{6} p |_{Γ}, u |_{S_{5 R}} = 0 . \end{matrix} \end{matrix}$ By Theorem 10, $(u, p) \in H$ exists uniquely. If we define the map $Φ : H \to H$ by $Φ (v, q) = (u, p)$ , then by (28), (29) and Theorem 10, we can find a small $σ > 0$ uniformly with respect to $λ \in Λ_{ϵ, λ_{0}} \cup {0}$ such that Φ is a contraction map. Thus, by the Banach fixed point theorem, we have Theorem 7, which completes the proof of Theorem 7 □

Finally, we consider the following weak Dirichlet problem: $\begin{matrix} (31) & {(\tilde{\nabla} u, \tilde{\nabla} φ)}_{Ω_{5 R}} = {(f, \tilde{\nabla} φ)}_{Ω_{5 R}} for any φ \in H_{q^{'}, Γ}^{1} (Ω_{5 R}) . \end{matrix}$ Under the assumption (3), using Theorem 8 and employing the same argument as in the last part of the proof of Theorem 7, we have Theorem 11.
Let $1 < q < \infty$ . Then, there exists a $σ > 0$ such that if the assumption ( 3 ) holds, then for any $f \in L_{q} {(Ω_{5 R})}^{N}$ , problem ( 31 ) admits a unique solution $u \in H_{q, Γ}^{1} (Ω_{5 R})$ possessing the estimate: $\begin{matrix} (32) & ‖ u ‖_{H_{q}^{1} (Ω_{5 R})} ⩽ C ‖ f ‖_{L_{q} (Ω_{5 R})} . \end{matrix}$

3. Resolvent problem in the whole space

In this section, we consider the resolvent problem for the Stokes equations: $\begin{matrix} (33) & λ u - Div S (u, p) = f, div u = 0 in R^{N} . \end{matrix}$ Let $\begin{matrix} (34) & R_{0} (λ) f (x) = F_{ξ}^{- 1} [\frac{P (ξ) \hat{f} (ξ)}{λ + | ξ |^{2}}] (x), Π f (x) = F_{ξ}^{- 1} [\frac{i ξ \cdot \hat{f} (ξ)}{| ξ |^{2}}] (x), \end{matrix}$ where $P (ξ) = {(δ_{i j} - ξ_{i} ξ_{j} / | ξ |^{2})}_{i, j = 1, \dots, N}$ , and then $u = R_{0} (λ) f$ and $p = Π f + c$ , where c is any constant, satisfy the equations (33). Moreover, the following theorem holds, which is used in later sections.

Theorem 12.
Let $1 < q < \infty$ and $0 < ϵ < π / 2$ . Then, the following assertions hold.
$R_{0} (λ) \in Hol (Σ_{ϵ}, L (L_{q} {(R^{N})}^{N}, H_{q}^{2} {(R^{N})}^{N}))$ and $\begin{matrix} (35) & {‖ (λ R_{0} (λ) f, λ^{1 / 2} \nabla R_{0} (λ) f, \nabla^{2} R_{0} (λ) f, \nabla Π f) ‖}_{L_{q} (R^{N})} ⩽ C_{ϵ} ‖ f ‖_{L_{q} (R^{N})} \end{matrix}$ for any $λ \in Σ_{ϵ}$ and $f \in L_{q} {(R^{N})}^{N}$ with some constant $C_{ϵ}$ depending solely on ϵ.

Let $\begin{matrix} L_{q, 5 R} {(R^{N})}^{N} = {f \in L_{q} {(R^{N})}^{N} ∣ f (x) = 0 (| x | > 5 R)} . \end{matrix}$ Then, there exists an operator $R_{0} (0) : L_{q, 5 R} {(R^{N})}^{N} \to H_{q, loc}^{2} {(R^{N})}^{N}$ such that for any $f \in L_{q, 5 R} {(R^{N})}^{N}$ , $u = R_{0} (0) f$ is a solution of the equations: $\begin{matrix} (36) & - Div S (u, p) = f, div u = 0 in R^{N} \end{matrix}$ with $p = Π f + c$ for any constant c. Moreover, $\begin{matrix} (37) & \begin{matrix} {‖ R_{0} (0) f ‖}_{H_{q}^{2} (B_{L})} ⩽ C_{L, R, ϵ} ‖ f ‖_{L_{q} (R^{N})}, \\ {‖ R_{0} (λ) f - R_{0} (0) f ‖}_{H_{q}^{2} (B_{L})} ⩽ C_{L, R, ϵ} | λ |^{1 / 4} ‖ f ‖_{L_{q} (R^{N})}, \\ R_{0} (0) f = c_{1 N} \int_{R^{N}} \frac{f_{i} (y)}{| x - y |^{N - 2}} d y + c_{2 N} \sum_{j = 1}^{N} \int_{R^{N}} \frac{(x_{i} - y_{i}) (x_{j} - y_{j}) f_{j} (y)}{| x - y |^{N}} d y, \\ Π f = c_{3 N} \sum_{i = 1}^{N} \int_{R^{N}} \frac{(x_{i} - y_{i}) f_{i} (y)}{| x - y |^{N}} d y \end{matrix} \end{matrix}$ for any $f =^{⊤} (f_{1}, \dots, f_{N}) \in L_{q, 5 R} {(R^{N})}^{N}$ and $L > 0$ , where $c_{1 N}$ , $c_{2 N}$ and $c_{3 N}$ are some constants depending on N.

Let $1 < q ⩽ p ⩽ \infty$ , $q \neq \infty$ and $N (1 / q - 1 / p) < 1$ . Let $f \in L_{q} {(R^{N})}^{N}$ and $λ \in Λ_{ϵ, 1 / 2}$ . Then, we have $\begin{matrix} (38) & {‖ \nabla^{i} R_{0} (λ) f ‖}_{L_{p} (R^{N})} ⩽ C | λ |^{\frac{i}{2} - 1 + \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (R^{N})} (i = 0, 1) . \end{matrix}$

Let $1 < q ⩽ p ⩽ \infty$ , $q \neq \infty$ and $N (1 / q - 1 / p) < 1$ . Let $f \in L_{q, 5 R} {(R^{N})}^{N}$ and $λ \in Λ_{ϵ, 1 / 2}$ . Then, we have the following estimations:

If $p (N - 1) > N$ , then we have $\begin{matrix} (39) & {‖ \nabla^{i} R_{0} (λ) f ‖}_{L_{p} (R^{N})} ⩽ C | λ |^{\frac{i - 1}{2}} ‖ f ‖_{L_{q} (R^{N})} (i = 0, 1) . \end{matrix}$

If $p (N - 1) ⩽ N$ , then choosing $0 < γ < 1$ in such a way that $(N - 1 + γ) p > N$ , we have $\begin{matrix} (40) & {‖ \nabla^{i} R_{0} (λ) f ‖}_{L_{p} (R^{N})} ⩽ C | λ |^{\frac{i - 1}{2} - \frac{γ}{2}} ‖ f ‖_{L_{q} (R^{N})} (i = 0, 1) . \end{matrix}$

Proof.
(1) Since $\begin{matrix} (41) & | λ + | ξ |^{2} | ⩾ (sin \frac{ϵ}{2}) (| λ | + | ξ |^{2}) \end{matrix}$ for any $λ \in Σ_{ϵ}$ , by the Fourier multiplier theorem of Marcinkiewicz–Mikhlin–Hörmander type we have (35) immediately.

(2) Let $φ_{0} (ξ)$ be a function in $C_{0}^{\infty} (R^{N})$ such that $φ_{0} (ξ) = 1$ for $| ξ | ⩽ 1$ and $φ_{0} (ξ) = 0$ for $| ξ | ⩾ 2$ . Let $\begin{array}{l} S_{0} (λ) f = F_{ξ}^{- 1} [\frac{φ_{0} (ξ) P (ξ) F [f] (ξ)}{λ + | ξ |^{2}}], S_{\infty} (λ) f = F_{ξ}^{- 1} [\frac{(1 - φ_{0} (ξ)) P (ξ) F [f] (ξ)}{λ + | ξ |^{2}}] \end{array}$ for any $λ \in Σ_{ϵ}$ with $| λ | ⩽ 1 / 2$ . Since $\begin{matrix} (42) & | \partial_{ξ}^{β} \frac{(1 - φ_{0} (ξ)) ξ^{α} P (ξ)}{λ + | ξ |^{2}} | ⩽ C_{β} | ξ |^{- | β |} (α \in N_{0}^{N}, | α | ⩽ 2) \end{matrix}$ for any $β \in N_{0}^{N}$ and $λ \in Σ_{ϵ} \cup {0}$ with $| λ | ⩽ 1 / 2$ , we see that $\begin{array}{l} S_{\infty} (0) f = F_{ξ}^{- 1} [\frac{(1 - φ_{0} (ξ)) P (ξ) F [f] (ξ)}{| ξ |^{2}}] \in H_{q}^{2} {(R^{N})}^{N}, \\ {‖ S_{\infty} (0) f ‖}_{H_{q}^{2} (R^{N})} ⩽ C_{ϵ} ‖ f ‖_{L_{q} (R^{N})}, \\ {‖ S_{\infty} (λ) f - S_{0} (0) f ‖}_{H_{q}^{2} (R^{N})} ⩽ C_{ϵ} | λ | ‖ f ‖_{L_{q} (R^{N})} \end{array}$ for any $λ \in Σ_{ϵ}$ with $| λ | ⩽ 1 / 2$ and $f \in L_{q} {(R^{N})}^{N}$ .

On the other hand, by (41) $\begin{matrix} | \frac{φ_{0} (ξ) ξ^{α} P (ξ)}{λ + | ξ |^{2}} | ⩽ \frac{C}{| ξ |^{2}} \in L_{1} (B_{2}) \end{matrix}$ for any $λ \in Σ_{ϵ} \cup {0}$ , because $N ⩾ 3$ . So, $\begin{array}{l} S_{0} (0) f = F_{ξ}^{- 1} [\frac{φ_{0} (ξ) P (ξ) F [f] (ξ)}{| ξ |^{2}}] \in H_{q}^{2} {(B_{L})}^{N}, \\ \begin{matrix} {‖ S_{0} (0) f ‖}_{H_{q}^{2} (B_{L})} & ⩽ | B_{L} |^{1 / q} {‖ S_{0} (0) f ‖}_{H_{\infty}^{2} (R^{N})} ⩽ C_{L} \int_{| ξ | ⩽ 2} \frac{| F [f] (ξ) |}{| ξ |^{2}} d ξ \\ ⩽ C_{L} ‖ f ‖_{L_{1} (R^{N})} ⩽ C_{L, R} ‖ f ‖_{L_{q} (R^{N})} \end{matrix} \end{array}$ for any $f \in L_{q, 5 R} {(R^{N})}^{N}$ and $L > 0$ . Since $\begin{matrix} | \frac{1}{λ + | ξ |^{2}} - \frac{1}{| ξ |^{2}} | ⩽ \frac{C_{ϵ}}{| ξ |^{2}}, | \frac{1}{λ + | ξ |^{2}} - \frac{1}{| ξ |^{2}} | = \frac{| λ |}{| (λ + ξ |^{2}) | ξ |^{2}} ⩽ \frac{C_{ϵ} | λ |}{| ξ |^{4}} \end{matrix}$ as follows from (41), we have $\begin{matrix} | \frac{1}{λ + | ξ |^{2}} - \frac{1}{| ξ |^{2}} | ⩽ \frac{C_{ϵ} | λ |^{1 / 4}}{| ξ |^{3 / 2} | ξ |} = \frac{C_{ϵ} | λ |^{1 / 4}}{| ξ |^{5 / 2}}, \end{matrix}$ and so $\begin{array}{l} {‖ S_{0} (λ) f - S_{0} (0) f ‖}_{H_{q}^{2} (B_{L})} & ⩽ | B_{L} |^{1 / q} {‖ S_{0} (λ) f - S_{0} (0) f ‖}_{H_{\infty}^{2} (R^{N})} \\ ⩽ C_{L, ϵ} | λ |^{1 / 4} \int_{| ξ | ⩽ 2} \frac{| F [f] (ξ) |}{| ξ |^{5 / 2}} d ξ ⩽ C_{L, R, ϵ} | λ |^{1 / 4} ‖ f ‖_{L_{q} (B_{5 R})} \end{array}$ for any $λ \in Σ_{ϵ}$ with $| λ | ⩽ 1 / 2$ and $f \in L_{q, 5 R} {(R^{N})}^{N}$ . Note that $R_{0} (0) f = S_{0} (0) f + S_{\infty} (0) f$ .

From (34) it follows that $\begin{array}{l} i th component of R_{0} (0) f = F_{ξ}^{- 1} [\frac{P (ξ) \hat{f} (ξ)}{| ξ |^{2}}] |_{i} = {(- Δ)}^{- 1} f_{i} - \sum_{j = 1}^{N} \frac{\partial^{2}}{\partial x_{i} \partial x_{j}} {(- Δ)}^{- 2} f_{j}, \\ Π f = F [\frac{- i ξ \cdot \hat{f}}{| ξ |^{2}}] = \sum_{j = 1}^{N} \frac{\partial}{\partial x_{j}} {(- Δ)}^{- 1} f_{j} . \end{array}$ Here, ${(- Δ)}^{- 1}$ is the fundamental solution of $- Δ$ , which is given by $c_{N} | x |^{- (N - 2)}$ with some constant $c_{N}$ depending on N when $N ⩾ 3$ . Thus, using $Δ | x |^{s} = s (N + s - 2) | x |^{s - 2}$ ( $s \neq 0$ ) and $Δ log | x | = (N - 1) | x |^{- 2}$ , we have the third and fourth formulas in (37), which completes the proof of the assertion (2).

(3) Let $S_{0} (λ)$ and $S_{\infty} (λ)$ be the same operators as in the proof of (2) above. By (42) and the Fourier multiplier theorem of Marcinkiewicz–Mikhlin–Hörmander type, we have $\begin{matrix} (43) & {‖ S_{\infty} (λ) f ‖}_{H_{q}^{2} (R^{N})} ⩽ C_{q} ‖ f ‖_{L_{q} (R^{N})} . \end{matrix}$ Thus, by the Sobolev imbedding theorem and (43) $\begin{matrix} (44) & {‖ S_{\infty} (λ) f ‖}_{H_{p}^{1} (R^{N})} ⩽ C_{p, q} ‖ f ‖_{L_{q} (R^{N})} \end{matrix}$ provided that $1 < q ⩽ p ⩽ \infty$ , $q \neq \infty$ , and $N (1 / q - 1 / p) < 1$ .

Next, we consider $S_{0} (λ) f$ . Let $\begin{matrix} K_{λ}^{0} (x) = F_{ξ}^{- 1} [φ_{0} (ξ) P (ξ) {(λ + | ξ |^{2})}^{- 1}] (x), K_{λ}^{1} (x) = F_{ξ}^{- 1} [i ξ φ_{0} (ξ) P (ξ) {(λ + | ξ |^{2})}^{- 1}] (x), \end{matrix}$ and then $S_{0} (λ) f = K_{λ}^{0} * f$ and $\nabla S_{0} (λ) f = K_{λ}^{1} * f$ , where $f * g$ denotes the convolution of f and g, i.e., $\begin{matrix} f * g (x) = \int_{R^{N}} f (x - y) g (y) . \end{matrix}$ To estimate $K_{λ}^{0} (x)$ and $K_{λ}^{1} (x)$ , we use the following theorem (cf. [20, Theorem 2.3]). Theorem 13.
Let B be a Banach space and $| \cdot |_{B}$ its norm. Let α be a number $> - N$ and set $α = n + σ - N$ where n is a non-negative integer and $0 < σ ⩽ 1$ . Let $f (ξ)$ be a function in $C^{\infty} (R^{N} ∖ {0}, B)$ such that $\begin{matrix} \partial_{ξ}^{γ} f (ξ) \in L_{1} (R^{N}, B) for | γ | ⩽ n, {| \partial_{ξ}^{γ} f (ξ) |}_{B} ⩽ C_{γ} | ξ |^{α - γ} for any ξ \in R^{N} and γ \in N_{0}^{N} . \end{matrix}$ Let $g (x) = \int_{R^{N}} e^{- i x \cdot ξ} f (ξ) d ξ$ , and then $\begin{matrix} {| g (x) |}_{B} ⩽ C_{N, α} (max_{| γ | ⩽ n + 2} C_{γ}) | x |^{- (N + α)} for any x \in R^{N} ∖ {0} \end{matrix}$ where $C_{N, α}$ is a constant depending solely on N and α.

By (41) $\begin{matrix} (45) & | \partial_{ξ}^{β} (φ_{0} (ξ) P (ξ) {(λ + | ξ |^{2})}^{- 1}) | ⩽ C_{α, ϵ} {(| λ | + | ξ |^{2})}^{- 1} | ξ |^{- | β |} \end{matrix}$ for any $β \in N_{0}^{N}$ , and so $\begin{matrix} (46) & | \partial_{ξ}^{β} (ξ_{j} φ_{0} (ξ) P (ξ) {(λ + | ξ |^{2})}^{- 1}) | ⩽ C_{β, ϵ} | ξ |^{- 1 - | β |} or C_{β, ϵ} | λ |^{- 1 / 2} | ξ |^{- | β |} . \end{matrix}$ Thus, by Theorem 13, (45) and (46), $\begin{array}{l} (47) & 2 | K_{λ}^{i} (x) | ⩽ C_{ϵ} | x |^{- (N - 1)} | λ |^{\frac{i - 1}{2}} for any x \in R^{N} ∖ {0}, \\ (48) & | K_{λ}^{i} (x) | ⩽ C_{ϵ} | x |^{- N} | λ |^{\frac{i - 2}{2}} for any x \in R^{N} ∖ {0} \end{array}$ for $i = 0, 1$ . By (47) and (48), we have $\begin{matrix} (49) & {‖ K_{λ}^{i} ‖}_{L_{r} (R^{N})} ⩽ C_{r, ϵ} | λ |^{(\frac{i}{2} - 1 + \frac{N}{2}) - \frac{N}{2 r}} \end{matrix}$ provided that $(N - 1) r < N$ . In fact, $\begin{array}{l} \int_{R^{N}} {| K_{λ}^{i} (x) |}^{r} d x & ⩽ C_{r, ϵ} (\int_{| x | ⩽ | λ |^{- 1 / 2}} | x |^{- r (N - 1)} d x | λ |^{(\frac{i - 1}{2}) r} + \int_{| x | ⩾ | λ |^{- 1 / 2}} | x |^{- r N} d x | λ |^{(\frac{i - 2}{2}) r}), \end{array}$ from which (49) follows immediately. Thus, in case of $p = \infty$ , by the Hölder inequality and (49) $\begin{matrix} {‖ \nabla^{i} S_{0} (λ) f ‖}_{L_{\infty} (R^{N})} ⩽ {‖ K_{λ}^{i} ‖}_{L_{q^{'}} (R^{N})} ‖ f ‖_{L_{q} (R^{N})} ⩽ C_{q, ϵ} | λ |^{(\frac{i}{2} - 1 + \frac{N}{2}) - \frac{N}{2 q^{'}}} ‖ f ‖_{L_{q} (R^{N})} \end{matrix}$ for $i = 0, 1$ with $q^{'} = q / (q - 1)$ . In fact, $q^{'} (N - 1) < N$ for $q \in (N, \infty)$ , and so we can use (49). Since $(\frac{i}{2} - 1 + \frac{N}{2}) - \frac{N}{2 q^{'}} = \frac{i}{2} - 1 + \frac{N}{2 q}$ , we have (38) with $p = \infty$ .

In the case of $1 < p < \infty$ , choosing q and r in such a way that $1 + 1 / p = 1 / r + 1 / q$ , by the Young inequality and (49), we have $\begin{matrix} {‖ \nabla^{i} S_{0} (λ) f ‖}_{L_{p} (R^{N})} ⩽ {‖ K_{λ}^{i} ‖}_{L_{r} (R^{N})} ‖ f ‖_{L_{q} (R^{N})} ⩽ C_{p, q, ϵ} | λ |^{(\frac{i}{2} - 1 + \frac{N}{2}) - \frac{N}{2 r}} ‖ f ‖_{L_{q} (R^{N})} \end{matrix}$ for $i = 0, 1$ . In fact, $r (N - 1) < N$ follows from $1 + 1 / p = 1 / r + 1 / q$ and $N (1 / q - 1 / p) < 1$ , and so we can use (49). Since $(\frac{i}{2} - 1 + \frac{N}{2}) - \frac{N}{2 r} = \frac{i}{2} - 1 + \frac{N}{2} (\frac{1}{q} - \frac{1}{p})$ , we have (38).

(4) Recall that $R_{0} (λ) f = S_{0} (λ) f + S_{\infty} (λ) f$ . Since $N (1 / q - 1 / p) < 1$ , by the Sobolev imbedding theorem and (43), $\begin{matrix} (50) & {‖ \nabla^{i} S_{\infty} (λ) f ‖}_{L_{p} (R^{N})} ⩽ C {‖ S_{\infty} (λ) f ‖}_{H_{q}^{2} (R^{N})} ⩽ C ‖ f ‖_{L_{q} (R^{N})} (i = 0, 1) . \end{matrix}$ To estimate $S_{0} (λ) f$ , we first consider the case that $p (N - 1) > N$ . By (47), $\begin{matrix} \int_{| x | ⩾ 1} {| K_{λ}^{i} (x) |}^{p} d x ⩽ C_{ϵ} \int_{| x | ⩾ 1} | x |^{- (N - 1) p} d x | λ |^{\frac{i - 1}{2} p} ⩽ C_{p, ϵ} | λ |^{\frac{i - 1}{2} p} . \end{matrix}$ On the other hand, by (45) with $β = 0$ , $\begin{matrix} | K_{λ}^{0} (x) | ⩽ \int_{R^{N}} | φ_{0} (ξ) P (ξ) {(λ + | ξ |^{2})}^{- 1} | d ξ ⩽ C_{ϵ} \int_{| ξ | ⩽ 2} | ξ |^{- 2} d ξ < \infty, \end{matrix}$ because $N ⩾ 3$ , and by (46) with $β = 0$ , $\begin{matrix} | K_{λ}^{1} (x) | ⩽ \int_{R^{N}} | i ξ φ_{0} (ξ) P (ξ) {(λ + | ξ |^{2})}^{- 1} | d ξ ⩽ C_{ϵ} \int_{| ξ | ⩽ 2} | ξ |^{- 1} d ξ < \infty . \end{matrix}$ Thus, we have $\begin{matrix} {‖ K_{λ}^{1} ‖}_{L_{p} (R^{N})} ⩽ C_{ϵ, p} | λ |^{\frac{i - 1}{2}} \end{matrix}$ for $λ \in Σ_{ϵ, 1 / 2}$ and $i = 0, 1$ , and so by the Young inequality $\begin{matrix} (51) & {‖ \nabla^{i} S_{0} (λ) f ‖}_{L_{p} (R^{N})} ⩽ {‖ K_{λ}^{i} ‖}_{L_{p} (R^{N})} ‖ f ‖_{L_{1} (R^{N})} ⩽ C_{p, ϵ} | λ |^{\frac{i - 1}{2}} ‖ f ‖_{L_{1} (R^{N})} . \end{matrix}$ Since $supp f \subset B_{5 R}$ , we have $\begin{matrix} (52) & ‖ f ‖_{L_{1} (R^{N})} ⩽ C_{q, R} ‖ f ‖_{L_{q} (R^{N})}, \end{matrix}$ which, combined with (50) and (51), leads to (39).

Next, we consider the case that $p (N - 1) ⩽ N$ . Let γ be any number such that $0 < γ < 1$ and $(N - 1 + γ) p > N$ . By (47) and (48), $\begin{matrix} | K_{λ}^{i} (x) | ⩽ C_{ϵ} | x |^{- (N - 1 + γ)} | λ |^{\frac{i - 1}{2} - \frac{γ}{2}} . \end{matrix}$ Since $(N - 1 + γ) p > N$ , we have $\begin{matrix} \int_{| x | ⩾ 1} {| K_{λ}^{i} (x) |}^{p} d x ⩽ C_{p, ϵ} | λ |^{(\frac{i - 1}{2} - \frac{γ}{2}) p} . \end{matrix}$ On the other hand, by (45) with $β = 0$ , $\begin{matrix} | K_{λ}^{0} (x) | ⩽ \int_{R^{N}} | φ_{0} (ξ) P (ξ) {(λ + | ξ |^{2})}^{- 1} | d ξ ⩽ \int_{| ξ | ⩽ 2} | ξ |^{- 2} d ξ < \infty, \end{matrix}$ because $N ⩾ 3$ , and by (46) with $β = 0$ , $\begin{matrix} | K_{λ}^{1} (x) | ⩽ \int_{R^{N}} | i ξ φ_{0} (ξ) P (ξ) {(λ + | ξ |^{2})}^{- 1} | d ξ ⩽ \int_{| ξ | ⩽ 2} | ξ |^{- 1} d ξ < \infty . \end{matrix}$ Thus, $\begin{matrix} {‖ K_{λ}^{i} ‖}_{L_{p} (R^{N})} ⩽ C_{q, ϵ} | λ |^{\frac{i - 1}{2} - \frac{γ}{2}}, \end{matrix}$ and so by the Young inequality $\begin{matrix} {‖ \nabla^{i} S_{0} (λ) f ‖}_{L_{p} (R^{N})} ⩽ {‖ K_{λ}^{i} ‖}_{L_{p} (R^{N})} ‖ f ‖_{L_{1} (R^{N})}] ⩽ C_{p, ϵ} | λ |^{\frac{i - 1}{2} - \frac{γ}{2}} ‖ f ‖_{L_{1} (R^{N})}, \end{matrix}$ which, combined with (50) and (52), leads to (40). This completes the proof of Theorem 12. □

4. Resolvent problem in an exterior domain; $λ \neq 0$ case

In this section we consider the resolvent problem: $\begin{matrix} (53) & λ u - J^{- 1} Div \tilde{S} (u, p) = f, \tilde{div} u = 0 in Ω, \tilde{S} (u, p) n |_{Γ} = 0 \end{matrix}$ for any $λ \in Σ_{ϵ}$ with $| λ | ⩾ δ$ for any $δ > 0$ . First, we note the following fact.

Lemma 14.
If $u \in H_{q}^{2} {(Ω)}^{N}$ satisfies $\tilde{div} u = 0$ in Ω, then $\begin{matrix} {(u, J \tilde{\nabla} φ)}_{Ω} = 0 for any φ \in {\hat{H}}_{q^{'}, 0}^{1} (Ω) . \end{matrix}$
Proof.
Let ψ be a function in $C_{0}^{\infty} (R^{N})$ such that $ψ (x) = 1$ for $| x | ⩽ 1$ and $ψ (x) = 0$ for $| x | ⩾ 2$ and let $ψ_{L} (x) = ψ (x / L)$ . Since $\tilde{div} (ψ_{L} u) = (\nabla ψ_{L}) \cdot u$ for large L as follows from the fact that $J a_{k j} = δ_{k j}$ for $| x | ⩾ 2 R$ , we have $\begin{matrix} {(ψ_{L} u, J \tilde{\nabla} φ)}_{Ω} = - {((\nabla ψ_{L}) \cdot u, φ)}_{Ω} for any φ \in {\hat{H}}_{q^{'}, 0}^{1} (Ω) . \end{matrix}$ Extending φ by zero outside of Ω, we may assume that $φ \in {\hat{H}}_{q^{'}}^{1} (R^{N})$ , and so there exists a constant c such that $\begin{matrix} {‖ (φ - c) | x |^{- 1} ‖}_{L_{q^{'}} (R^{N})} ⩽ C ‖ \nabla φ ‖_{L_{q^{'}} (R^{N})} . \end{matrix}$ Since $div u = 0$ for $| x | ⩾ 2 R$ , by Lemma 5, $\int_{Ω} (\nabla ψ_{L}) \cdot u d x = 0$ . Thus. we have $\begin{array}{l} | {((\nabla ψ_{L}) \cdot u, φ)}_{Ω} | & = | {((\nabla ψ_{L}) \cdot u, φ - c)}_{Ω} | ⩽ C L^{- 1} \int_{L}^{2 L} | φ (x) - c | | u (x) | d x \\ ⩽ C \int_{L}^{2 L} \frac{| φ (x) - c |}{| x |} | u (x) | d x ⩽ {‖ (φ - c) | x |^{- 1} ‖}_{L_{q^{'}} (D_{L, 2 L})} ‖ u ‖_{L_{q} (D_{L, 2 L})} \\ ⩽ C ‖ \nabla φ ‖_{L_{q^{'}} (R^{N})} ‖ u ‖_{L_{q} (D_{L, 2 L})} \to 0 as L \to \infty . \end{array}$ This completes the proof of Lemma 14. □

The purpose of this section is to prove
Theorem 15.
Assume that $N ⩾ 3$ . Let $N < r < \infty$ be the exponent appearing in ( 3 ). Let $1 < q < \infty$ and $δ > 0$ . Assume that $max (q, q^{'}) ⩽ r$ . Then, there exists a $σ > 0$ such that if the assumption ( 3 ) holds, then for any $λ \in Σ_{ϵ, δ}$ and $f \in L_{q} {(Ω)}^{N}$ , problem ( 53 ) admits unique solutions $u \in H_{q}^{2} {(Ω)}^{N}$ and $p \in H_{q}^{1} (Ω) + {\hat{H}}_{q, 0}^{1} (Ω)$ possessing the estimate: $\begin{array}{l} {‖ (| λ | u, | λ |^{1 / 2} \nabla u) ‖}_{L_{q} (Ω)} + ‖ u ‖_{H_{q}^{2} (Ω)} + ‖ \nabla p ‖_{L_{q} (Ω)} ⩽ C_{δ} ‖ f ‖_{L_{q} (Ω)} \end{array}$ with some constant $C_{δ}$ depending solely on ϵ and δ.

To prove Theorem 15 for large λ, we consider equations: $\begin{matrix} (54) & λ u - Div S (u, p) = f, div u = g = div g in Ω, S (u, p) n |_{Γ} = h . \end{matrix}$ If we consider the weak Dirichlet problem: $\begin{matrix} (55) & {(\nabla u, \nabla φ)}_{Ω} = {(f, \nabla φ)}_{Ω} for any φ \in {\hat{H}}_{q^{'}, 0}^{1} (Ω), \end{matrix}$ then as was proved in Shibata [18, Theorem 7 in Section 3] (cf. also Pruess and Simonett [15]), problem (55) admits a unique solution $u \in {\hat{H}}_{q, 0}^{1} (Ω)$ possessing the estimate $\begin{matrix} (56) & ‖ \nabla u ‖_{L_{q} (Ω)} ⩽ C ‖ f ‖_{L q (Ω)} . \end{matrix}$ Thus, applying Theorem 1.6 in Shibata [16] (cf. also Shibata [17]), we have
Theorem 16.
Let $N < r < \infty$ be the exponent appearing in ( 3 ). Let $1 < q < \infty$ , and assume that $max (q, q^{'}) ⩽ r$ . Then, there exists a $λ_{0} ⩾ 1$ such that for any $λ \in Σ_{ϵ, λ_{0}}$ , $f \in L_{q} {(Ω)}^{N}$ , $g \in H_{q}^{1} (Ω)$ , $g \in L_{q} {(Ω)}^{N}$ , and $h \in H_{q}^{1} (Ω)$ , problem ( 54 ) admits unique solutions $u \in H_{q}^{2} {(Ω)}^{N}$ and $p \in H_{q}^{1} (Ω) + {\hat{H}}_{q, 0}^{1} (Ω)$ possessing the estimate: $\begin{array}{l} {‖ (| λ | u, | λ |^{1 / 2} \nabla u) ‖}_{L_{q} (Ω)} + ‖ u ‖_{H_{q}^{2} (Ω)} + ‖ \nabla p ‖_{L_{q} (Ω)} \\ ⩽ C {‖ f ‖_{L_{q} (Ω)} + | λ | ‖ g ‖_{L_{q} (Ω)} + | λ |^{1 / 2} {‖ (g, h) ‖}_{L_{q} (Ω)} + {‖ (g, h) ‖}_{H_{q}^{1} (Ω)}} \end{array}$ with some constant C depending solely on ϵ and $λ_{0}$ .

Using Theorem 16 and employing the same argument as in the last part of the proof of Theorem 7 in Section 2.2, we have
Theorem 17.
Let $N < r < \infty$ be the exponent appearing in ( 3 ). Let $1 < q < \infty$ , and assume that $max (q, q^{'}) ⩽ r$ . Let $λ_{0}$ be the same constant as in Theorem 16 . Then, there exists a σ depending on ϵ and $λ_{0}$ such that if the assumption ( 3 ) holds, then for any $λ \in Σ_{ϵ, λ_{0}}$ , and $f \in L_{q} {(Ω)}^{N}$ , problem ( 53 ) admits unique solutions $u \in H_{q}^{2} {(Ω)}^{N}$ and $p \in H_{q}^{1} (Ω) + {\hat{H}}_{q, 0}^{1} (Ω)$ possessing the estimate: $\begin{array}{l} {‖ (| λ | u, | λ |^{1 / 2} \nabla u) ‖}_{L_{q} (Ω)} + ‖ u ‖_{H_{q}^{2} (Ω)} + ‖ \nabla p ‖_{L_{q} (Ω)} ⩽ C ‖ f ‖_{L_{q} (Ω)} \end{array}$ with some constant C depending solely on ϵ and $λ_{0}$ .

Let $Ξ_{ϵ, δ} = {λ \in Σ_{ϵ} ∣ δ ⩽ | λ | ⩽ λ_{0}}$ . In what follows, we consider Eq. (53) in the case that $λ \in Ξ_{ϵ, δ}$ . If we consider the equations: $\begin{matrix} (57) & \{\begin{matrix} λ u - J^{- 1} Div \tilde{S} (u, p) = g, & \tilde{div} u = 0 in Ω_{5 R}, \\ \tilde{S} (u, p) n |_{Γ} = 0, & u |_{S_{5 R}} = 0, \end{matrix} \end{matrix}$ then, by Theorem 7 we know that for $g \in L_{q} {(Ω_{5 R})}^{N}$ and $λ \in Ξ_{ϵ, δ}$ , Eq. (57) admits unique solutions $u \in H_{q}^{2} {(Ω_{5 R})}^{N}$ and $p \in H_{q}^{1} (Ω_{5 R})$ . Let $R_{1} (λ)$ and $Π_{1} (λ)$ be operators acting on $g \in L_{q} {(Ω_{5 R})}^{N}$ defined by $u = R_{1} (λ) g$ and $p = Π_{1} (λ) g$ . By Theorem 10, we know that $\begin{matrix} (58) & {‖ R_{1} (λ) g ‖}_{H_{q}^{2} (Ω_{5 R})} + {‖ Π_{1} (λ) g ‖}_{H_{q}^{1} (Ω_{5 R})} ⩽ C_{ϵ, λ_{0}} ‖ g ‖_{L_{q} (Ω_{5 R})} \end{matrix}$ for any $λ \in Ξ_{ϵ, δ}$ and $g \in L_{q} (Ω_{5 R})$ .

On the other hand, let $R_{0} (λ)$ and Π be the operators defined in (33) in Section 3. Let c be any constant. By Theorem 12, $u = R_{0} (λ) f$ and $p = Π_{0} f + c$ satisfy the Stokes equations: $\begin{matrix} (59) & (λ - Δ) u + \nabla p = f, div u = 0 in R^{N} \end{matrix}$ and the estimate: $\begin{matrix} (60) & {‖ (| λ | u, | λ |^{1 / 2} \nabla u, \nabla^{2} u, \nabla p) ‖}_{L_{q} (R^{N})} ⩽ C_{ϵ} ‖ f ‖_{L_{q} (R^{N})} \end{matrix}$ for any $λ \in Σ_{ϵ}$ with some constant $C_{ϵ}$ depending solely on ϵ.

Let $φ (x)$ be a function in $C_{0}^{\infty} (R^{N})$ such that $φ (x) = 1$ for $x \in B_{3 R}$ and $φ (x) = 0$ for $x \notin B_{4 R}$ . Given $f \in L_{q} {(Ω)}^{N}$ , let $f_{0}$ be the zero extension of f to the outside of Ω, that is, $f_{0} (x) = f (x)$ for $x \in Ω$ and $f_{0} (x) = 0$ for $x \notin Ω$ . On the other hand, let $r_{R}$ be the restriction operator from Ω to $Ω_{5 R}$ , that is, $r_{R} f = f |_{Ω_{5 R}}$ . Let $\begin{array}{l} R_{2} (λ) f = (1 - φ) R_{0} (λ) f_{0} + φ R_{1} (λ) r_{R} f + B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f)], \\ Π_{2} (λ) f = (1 - φ) (Π_{0} f_{0} + c (f)) + φ Π_{1} (λ) r_{R} f, \end{array}$ where $c (f)$ is a constant chosen in such a way that $\begin{matrix} (61) & \int_{D_{3 R, 4 R}} (Π_{0} f_{0} + c (f) - Π_{1} (λ) r_{R} f) d x = 0 . \end{matrix}$ Since $\tilde{div} u = div u$ for $x \notin B_{2 R}$ , we have $div (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f) = 0$ for $x \in Ω_{5 R} \cap B_{2 R}^{c}$ , and so by Lemma 3 and Lemma 5, we see that $\begin{matrix} (62) & \begin{matrix} B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f)] \in H_{q}^{3} {(R^{N})}^{N}, \\ supp B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f)] \subset D_{3 R, 4 R}, \\ div B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f)] = (\nabla φ) \cdot (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f), \\ {‖ B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f)] ‖}_{H_{q}^{3} (R^{N})} ⩽ C ‖ f ‖_{L_{q} (Ω)} . \end{matrix} \end{matrix}$ Note that $J^{- 1} Div \tilde{S} (u, p) = Div S (u, p)$ and $\tilde{div} u = div u$ for $x \notin B_{2 R}$ , and then setting $u = R_{2} (λ) f$ and $p = Π_{2} (λ) f$ , we have $\begin{matrix} (63) & λ u - J^{- 1} Div \tilde{S} (u, p) = f + T_{0} (λ) f, \tilde{div} u = 0 in Ω, \tilde{S} (u, p) n |_{Γ} = 0 \end{matrix}$ with $\begin{array}{l} T_{0} (λ) f & = 2 μ (\nabla φ) : \nabla (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f) + μ (Δ φ) (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f) \\ + λ B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f)] - μ Δ B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f)] \\ - (\nabla φ) (Π_{0} f_{0} + c (f) - Π_{1} (λ) r_{R} f) . \end{array}$ Here, $a : A$ is the N-vector whose ith component is $\sum_{j = 1}^{N} a_{j} A_{i j}$ for $a =^{⊤} (a_{1}, \dots, a_{N})$ and $A = {(A_{i j})}_{i, j = 1, \dots, N}$ .

Using Lemma 3, we see that $T_{0} (λ) \in L (L_{q} {(Ω)}^{N}, H_{q}^{1} {(Ω)}^{N})$ and $supp T_{0} (λ) f \subset D_{3 R, 4 R}$ , where $D_{L, M} = {x \in R^{N} ∣ L ⩽ | x | ⩽ M}$ with $L < M$ . By the Rellich compactness theorem, $T_{0} (λ)$ is a compact operator from $L_{q} {(Ω)}^{N}$ into itself, and therefore to prove the invertibility of $I + T_{0} (λ)$ , it suffices to prove that the kernel of $I + T_{0} (λ)$ is trivial. So, let f be an element of $L_{q} {(Ω)}^{N}$ such that $(I + T_{0} (λ)) f = 0$ . Our task is to prove that $f = 0$ . What $f = - T_{0} (λ) f$ implies that $\begin{matrix} (64) & supp f \subset D_{3 R, 4 R} . \end{matrix}$ Let $u = R_{2} (λ) f$ and $p = Π_{2} (λ) f$ , and then by (63) $u \in H_{q}^{2} {(Ω)}^{N}$ and $p \in H_{q}^{1} (Ω) + {\hat{H}}_{q, 0}^{1} (Ω)$ satisfy the homogeneous equations: $\begin{matrix} (65) & λ u - J^{- 1} Div \tilde{S} (u, p) = 0, \tilde{div} u = 0 in Ω, \tilde{S} (u, p) n |_{Γ} = 0 . \end{matrix}$ For a while, we assume that $\begin{matrix} (66) & u = 0, p = 0 . \end{matrix}$ By (66), we have $\begin{matrix} (67) & \begin{matrix} (1 - φ) R_{0} (λ) f_{0} + φ R_{1} (λ) r_{R} f + B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - R_{1} (λ) r_{R} f)] = 0, \\ (1 - φ) (Π_{0} f_{0} + c (f)) + φ Π_{1} (λ) r_{R} f = 0 \end{matrix} \end{matrix}$ in Ω. Since $φ = 1$ for $x \in B_{3 R}$ and $φ (x) = 0$ for $x \notin B_{4 R}$ , by (67) and (62) $\begin{matrix} (68) & \begin{matrix} R_{0} (λ) f_{0} = 0, Π_{0} f_{0} + c (f) = 0 for x \notin B_{4 R}, \\ R_{1} (λ) r_{R} f = 0, Π_{1} (λ) r_{R} f = 0 for x \in Ω_{3 R} . \end{matrix} \end{matrix}$ Let $\begin{matrix} w = \{\begin{matrix} R_{1} (λ) r_{R} f & for x \in D_{3 R, 5 R}, \\ 0 & for x \in B_{3 R}, \end{matrix} q = \{\begin{matrix} Π_{1} (λ) r_{R} f & for x \in D_{3 R, 5 R}, \\ 0 & for x \in B_{3 R}, \end{matrix} \end{matrix}$ and then, $w \in H_{q}^{2} {(B_{5 R})}^{N}$ and $q \in H_{q}^{1} (B_{5 R})$ . Moreover, by (64) w and $q$ satisfy the equations: $\begin{matrix} (69) & λ w - Div S (w, q) = h, div w = 0 in B_{5 R}, w |_{S_{5 R}} = 0, \end{matrix}$ where h is the zero extension of f, that is, $h (x) = f (x)$ for $x \in D_{3 R, 4 R}$ and $h (x) = 0$ for $x \notin D_{3 R, 4 R}$ , because $Div S (w, q) = J^{- 1} Div \tilde{S} (w, q)$ and $\tilde{div} w = div w$ for $x \notin B_{2 R}$ . On the other hand, setting $\tilde{w} = (R_{0} (λ) f_{0}) |_{B_{5 R}}$ and $\tilde{q} = (Π_{0} f_{0} + c (f)) |_{B_{5 R}}$ , from (68) and (59) it follows that $\tilde{w} \in H_{q}^{2} {(B_{5 R})}^{N}$ , $\tilde{q} \in H_{q}^{1} (B_{5 R})$ and they satisfy the equations (69), which leads to $\tilde{w} = w$ and $\tilde{q} = q + d$ with some constant d in $B_{5 R}$ . Especially, it follows from (61) that $\int_{D_{3 R, 4 R}} (\tilde{q} - q) d x = 0$ , and so $d = 0$ . Thus, we have $\tilde{q} = q$ in $Ω_{5 R}$ too. Noting that $R_{1} (λ) r_{R} f = w$ and $Π_{1} (λ) r_{R} f = q$ on $Ω_{R}$ , we have $R_{1} (λ) r_{R} f - R_{0} (λ) f_{0} = 0$ and $Π_{1} (λ) r_{R} f - (Π_{0} f_{0} + c (f)) = 0$ on $Ω_{5 R}$ . Inserting these relations into (67) yields that $\begin{array}{l} 0 = R_{0} (λ) f_{0} + φ (R_{1} (λ) r_{R} f - R_{0} (λ) f_{0}) = R_{0} (λ) f_{0}, \\ 0 = Π_{0} f_{0} + φ (Π_{1} (λ) r_{R} f - Π_{0} f_{0}) = Π_{0} f_{0} \end{array}$ in $Ω_{5 R}$ . By (59), $f_{0} = 0$ in $Ω_{5 R}$ , that is, $f = 0$ in $Ω_{5 R}$ , which, combined with (64), leads to $f = 0$ . Thus, to prove the invertibility of $I + T_{0} (λ)$ , it is sufficient to prove that u and $p$ satisfy the assumption (66).

First, we consider the case that $2 ⩽ q < \infty$ . Let ψ be a function in $C^{\infty} (R^{N})$ such that $ψ (x) = 1$ for $| x | ⩾ 4 R$ and $ψ (x) = 0$ for $| x | ⩽ 3 R$ . Since $div u = \tilde{div} u = 0$ for $x \notin B_{2 R}$ , by Lemma 3 and Lemma 5 $\begin{matrix} (70) & \begin{matrix} B [(\nabla ψ) \cdot u] \in H_{q}^{3} {(R^{N})}^{N}, \\ supp B [(\nabla ψ) \cdot u] \subset D_{3 R, 4 R}, \\ div B [(\nabla ψ) \cdot u] = (\nabla ψ) \cdot u . \end{matrix} \end{matrix}$ Let $v = ψ u - B [(\nabla ψ) \cdot u]$ and $q = ψ p$ , and then by (65) and (70), we see that $v \in H_{q}^{2} {(R^{N})}^{N}$ , $q \in {\hat{H}}_{q}^{1} (R^{N})$ , and v and $q$ satisfy the equations: $\begin{matrix} (71) & λ v - Div S (v, q) = g, div v = 0 in R^{N}, \end{matrix}$ with $g = - λ B [(\nabla ψ) \cdot u] + μ Div D (B [(\nabla ψ) \cdot u]) + (\nabla ψ) p - h$ , where $h =^{⊤} (h_{1}, \dots, h_{N})$ and $\begin{matrix} h_{i} = \sum_{j = 1}^{N} \frac{\partial}{\partial x_{j}} (\frac{\partial ψ}{\partial x_{i}} u_{j} + \frac{\partial ψ}{\partial x_{j}} u_{i}) + \sum_{j = 1}^{N} \frac{\partial ψ}{\partial x_{j}} D_{i j} (u) . \end{matrix}$ Note that $g \in L_{q} (R^{N})$ and $supp g \subset B_{4 R}$ . Since $q ⩾ 2$ , $g \in L_{2} {(R^{N})}^{N} \cap L_{q} {(R^{N})}^{N}$ , and so setting $z = R_{0} (λ) f$ and $r = Π_{0} g$ , by Theorem 12 and (34), we see that $z \in H_{q}^{2} {(R^{N})}^{N} \cap H_{2}^{2} {(R^{N})}^{N}$ , $r \in {\hat{H}}_{q}^{1} (R^{N}) \cap {\hat{H}}_{2}^{1} (R^{N})$ , z and $r$ satisfy the equations (71), and $| r (x) | ⩽ C | x |^{- (N - 1)}$ when $| x | ⩾ 5 R$ with some constant C. Since $λ \in Σ_{ϵ}$ , the uniqueness in the $L_{q} (R^{N})$ framework implies that $v = z$ and $q = r + c$ with some constant c. Thus, we have $\begin{matrix} (72) & u \in H_{2}^{2} {(Ω)}^{N}, \nabla p \in L_{2} {(Ω)}^{N}, sup_{| x | ⩾ 5 R} | p (x) - c | ⩽ C | x |^{- (N - 1)} \end{matrix}$ with some constant C. Let ω be a function in $C_{0}^{\infty} (R)$ such that $ω (t) = 1$ for $| t | ⩽ 1$ and $ω (t) = 0$ for $| t | ⩾ 2$ , and let $ω_{L} (x) = ω (| x | / L)$ . Note that $supp \nabla ω_{L} (x) \subset D_{L, 2 L}$ and $| \nabla ω_{L} (x) | ⩽ C | x |^{- 1}$ on $D_{L, 2 L}$ . Let $L > 5 R$ , and then by (65) and (72) $\begin{array}{l} 0 & = {(λ u, J u ω_{L})}_{Ω} - {(Div \tilde{S} (u, p), u ω_{L})}_{Ω} \\ = {(λ u, J u ω_{L})}_{Ω} + \sum_{i, j, k = 1}^{N} {(J a_{k j} (μ {\tilde{D}}_{i j} (u) - δ_{i j} p), \frac{\partial}{\partial ξ_{k}} (u_{i} ω_{L}))}_{Ω} \\ = {(λ u, J u ω_{L})}_{Ω} + \frac{μ}{2} {(J \tilde{D} (u), \tilde{D} (u) ω_{L})}_{Ω} - {(ω_{L} p, \tilde{div} u)}_{Ω} \\ + \sum_{i, j, k = 1}^{N} μ {(a_{k j} {\tilde{D}}_{i j} (u), u_{i} \frac{\partial ω_{L}}{\partial x_{k}})}_{Ω} - \sum_{i, k = 1}^{N} {(a_{k i} p, u_{i} \frac{\partial ω_{L}}{\partial x_{k}})}_{Ω} . \end{array}$ Since $u \in H_{2}^{2} {(Ω)}^{N}$ , we have $\begin{array}{l} lim_{L \to \infty} {(λ u, J u ω_{L})}_{Ω} = λ {(J u, u)}_{Ω}, \\ lim_{L \to \infty} {(J \tilde{D} (u), \tilde{D} (u) ω_{L})}_{Ω} = {(J \tilde{D} (u), \tilde{D} (u))}_{Ω}, \\ lim_{L \to \infty} \sum_{i, j, k = 1}^{N} {(a_{k j} {\tilde{D}}_{i j} (u), u_{i} \frac{\partial ω_{L}}{\partial x_{k}})}_{Ω} = 0 . \end{array}$ Since $a_{k i} = δ_{k i}$ on $supp \frac{\partial ω_{L}}{\partial x_{k}}$ , we have $\begin{matrix} \sum_{i, k = 1}^{N} {(a_{k i} p, u_{i} \frac{\partial ω_{L}}{\partial x_{k}})}_{Ω} = {(p, (\nabla ω_{L}) \cdot u)}_{Ω} = {(p - c, (\nabla ω_{L}) \cdot u)}_{Ω} + c \int_{Ω} (\nabla ω_{L}) \cdot u d x . \end{matrix}$ By (72), $\begin{array}{l} | {(p - c, (\nabla ω_{L}) \cdot u)}_{Ω} | & ⩽ C \int_{D_{L, 2 L}} | x |^{- (N - 1)} | u (x) | d x \\ ⩽ C {\int_{D_{L, 2 L}} | x |^{- 2 (N - 1)} d x}^{1 / 2} ‖ u ‖_{L_{2} (D_{L, 2 L})} \to 0 as L \to \infty, \end{array}$ because $2 (N - 1) ⩾ N + 1$ as follows from the assumption: $N ⩾ 3$ . Since $div u = 0$ for $| x | ⩾ 2 R$ as follows from $\tilde{div} u = 0$ in Ω, by Lemma 5 $\int_{Ω} (\nabla ω_{L}) \cdot u d x = 0$ . Since $\tilde{div} u = 0$ in Ω, summing up the observations above, we have $\begin{matrix} λ {(u, J u)}_{Ω} + \frac{μ}{2} {(\tilde{D} (u), J \tilde{D} (u))}_{Ω} = 0 . \end{matrix}$ Since $λ \in Ξ_{ϵ, δ}$ , we have $u = 0$ . By the first equation in (65), $(^{⊤} A) \nabla p = 0$ in Ω, and so $\nabla p = 0$ in Ω. Thus, $p$ is a constant. But, by the boundary condition in (65), $(^{⊤} A n) p = 0$ on Γ, which implies that $p = 0$ . Thus, u and $p$ satisfy the assumption (66), from which we conclude that Theorem 15 holds in the case that $2 ⩽ q < \infty$ .

Next, we consider the case that $1 < q < 2$ . Let f be any element of $L_{q^{'}} {(Ω)}^{N}$ . Let $v \in H_{q^{'}}^{2} {(Ω)}^{N}$ and $q \in H_{q^{'}}^{1} (Ω) + {\hat{H}}_{q^{'}, 0}^{1} (Ω)$ be solutions of the equations: $\begin{matrix} λ v - J^{- 1} Div \tilde{S} (v, q) = f, \tilde{div} v = 0 in Ω, \tilde{S} (v, q) n |_{Γ} = 0 . \end{matrix}$ Since $2 < q^{'} < \infty$ , the existence of such v and $q$ are guaranteed by the part of Theorem 15 already proved. Let $q = q_{1} + q_{2}$ with $q_{1} \in H_{q^{'}}^{1} (Ω)$ and $q_{2} \in {\hat{H}}_{q^{'}, 0}^{1} (Ω)$ . Since $v \in H_{q}^{2} {(Ω)}^{N}$ and $\tilde{div} v = 0$ , by Lemma 14, ${(J \tilde{\nabla} q_{2}, v)}_{Ω} = 0$ . And, $\tilde{S} (v, q) n = \tilde{S} (v, q_{1}) n$ on Γ because of $q_{2} = 0$ on Γ. Thus, $\begin{array}{l} {(f, J u)}_{Ω} & = λ {(v, J u)}_{Ω} - {(Div \tilde{S} (v, q), u)}_{Ω} = λ {(v, J u)}_{Ω} - {(Div \tilde{S} (v, q_{1}), u)}_{Ω} \\ = λ {(v, J u)}_{Ω} - {(\tilde{S} (v, q_{1}) n, u)}_{Γ} + \frac{μ}{2} {(J \tilde{D} (v), \tilde{D} (u))}_{Ω} - {(q_{1}, \tilde{div} u)}_{Ω} \\ (73) & = λ {(u, J v)}_{Ω} + \frac{μ}{2} {(J \tilde{D} (u), \tilde{D} (v))}_{Ω} . \end{array}$ Let $p = p_{1} + p_{2}$ with $p \in H_{q^{'}}^{1} (Ω)$ and $p_{2} \in {\hat{H}}_{q^{'}, 0}^{1} (Ω)$ . Since $v \in {\tilde{J}}_{q^{'}} (Ω)$ and $\tilde{S} (u, p_{1}) n = \tilde{S} (u, p) n = 0$ on Γ, analogously to (73) we have $\begin{array}{l} 0 & = {(λ u - J^{- 1} Div \tilde{S} (u, p), J v)}_{Ω} = λ {(u, J v)}_{Ω} - {(Div \tilde{S} (u, p_{1}), J v)}_{Ω} \\ = λ {(u, J v)}_{Ω} + \frac{μ}{2} {(J \tilde{D} (u), \tilde{D} (v))}_{Ω}, \end{array}$ which, combined with (73), leads to ${(f, J u)}_{Ω} = 0$ for any $f \in L_{q^{'}} (Ω)$ . Thus, we have $u = 0$ . By the equation in (53), we have $J \tilde{\nabla} p = 0$ , which yields that $p$ is a constant. But, by the boundary condition in (53), $(^{⊤} A n) p = 0$ on Γ, from which we can conclude that $p = 0$ . Therefore, u and $p$ satisfy the assumption (66) in the case that $1 < q < 2$ . This completes the proof of Theorem 15.
5. Resolvent problem near $λ = 0$ in an exterior domain; compactly supported data case

In this section, we consider the corresponding resolvent problem: $\begin{matrix} (74) & λ u - J^{- 1} Div \tilde{S} (u, p) = f, \tilde{div} u = 0 in Ω, S (u, p) n |_{Γ} = 0 \end{matrix}$ near $λ = 0$ . Let $\begin{array}{l} L_{q, R} (Ω) = L (L_{q, 5 R} {(Ω)}^{N}, H_{q}^{2} {(Ω_{5 R})}^{N}), \\ L_{q, 5 R} {(Ω)}^{N} = {f \in L_{q} {(Ω)}^{N} ∣ f (x) = 0 for x \notin Ω_{5 R}} . \end{array}$ The theorem which follows is the main result in this section.

Theorem 18.
Let $1 < q < \infty$ , $0 < ϵ < π / 2$ , and $R > R_{0} + 3$ . Then, there exist positive constants σ and d such that if the assumption ( 3 ) holds, then there exist solution operators $S (λ)$ and $P (λ)$ with $\begin{array}{l} S (λ) \in Hol (Λ_{ϵ, d}, L (L_{q, 5 R} {(Ω)}^{N}, H_{q}^{2} {(Ω)}^{N})), \\ P (λ) \in Hol (Λ_{ϵ, d}, L (L_{q, 5 R} {(Ω)}^{N}, H_{q}^{1} (Ω) + {\hat{H}}_{q, 0}^{1} (Ω))) \end{array}$ such that for any $f \in L_{q, 5 R} {(Ω)}^{N}$ and $λ \in Λ_{ϵ, d}$ , $u = S (λ) f$ and $p = P (λ) f$ are unique solutions of the equations ( 74 ) possessing the estimate: $\begin{matrix} {‖ (λ S (λ) f, λ^{1 / 2} \nabla S (λ) f, \nabla^{2} S (λ) f, \nabla P (λ) f) ‖}_{L_{q} (Ω)} ⩽ C ‖ f ‖_{L_{q} (Ω)} . \end{matrix}$

Moreover, let $1 < q ⩽ p ⩽ \infty$ , $q \neq \infty$ , $N (1 / q - 1 / p) < 1$ , and then for any $f \in L_{q, 5 R} {(Ω)}^{N}$ and $λ \in Λ_{ϵ, d}$ , the following estimations hold:
In the case that $p (N - 1) > N$ , $\begin{matrix} (75) & {‖ \nabla^{i} S (λ) f ‖}_{L_{p} (Ω)} ⩽ C_{p, q} | λ |^{\frac{i - 1}{2}} ‖ f ‖_{L_{q} (Ω)} (i = 0, 1) . \end{matrix}$

In the case that $p (N - 1) ⩽ N$ , choosing $γ \in (0, 1)$ in such a way that $(N - 1 + γ) p > N$ , we have $\begin{matrix} (76) & {‖ \nabla^{i} S (λ) f ‖}_{L_{p} (Ω)} ⩽ C_{p, q, γ} | λ |^{\frac{i - 1}{2} - \frac{γ}{2}} ‖ f ‖_{L_{q} (Ω)} (i = 0, 1) . \end{matrix}$

First, we construct a solution of the equations (74) near $λ = 0$ for $f \in L_{q, 5 R} {(Ω)}^{N}$ . Let $f_{0}$ be the zero extension of f to $Ω^{c}$ , that is, $f_{0} (x) = f (x)$ for $x \in Ω$ and $f_{0} (x) = 0$ for $x \notin Ω$ and let $r_{5 R} f$ is the restriction of f to $Ω_{5 R}$ . Let $u \in H_{q}^{2} {(Ω_{5 R})}^{N}$ and $p \in H_{q}^{1} (Ω_{5 R})$ be unique solutions of the equations: $\begin{matrix} (77) & \begin{array}{l} - J^{- 1} Div \tilde{S} (u, p) = g, \tilde{div} u = 0 in Ω_{5 R}, \\ \tilde{S} (u, p) n |_{Γ} = 0, u |_{S_{5 R}} = 0 . \end{array} \end{matrix}$ Let A and B be operators acting on $g \in L_{q} (Ω_{5 R})$ defined by $\begin{matrix} (78) & u = A g, p = B g, \end{matrix}$ where u and $p$ are unique solutions of the equations (77). Let $R_{0} (λ)$ and Π be the operator defined in (34). For any constant c, $u = R_{0} (λ) f$ and $p = Π f + c$ solve the equations (33), and so we choose a constant $c (f)$ in such a way that $\begin{matrix} (79) & \int_{Ω_{5 R}} (Π f_{0} + c (f) - B r_{5 R} f) d x = 0 . \end{matrix}$ Let φ be a function in $C_{0}^{\infty} (R^{N})$ such that $φ (x) = 1$ for $| x | ⩽ (3 + 1 / 3) R$ and $φ (x) = 0$ for $| x | ⩾ (3 + 2 / 3) R$ and then we define operators $Φ (λ)$ and Ψ acting on $f \in L_{q, 5 R} {(Ω)}^{N}$ by $\begin{matrix} (80) & \begin{array}{l} Φ (λ) f = (1 - φ) R_{0} (λ) f_{0} + φ A r_{5 R} f + B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - A r_{5 R} f)], \\ Ψ f = (1 - φ) (Π f_{0} + c (f)) + φ B r_{5 R} f . \end{array} \end{matrix}$ Since $div R_{0} (λ) f_{0} = 0$ in $R^{N}$ , $div A r_{5 R} f = 0$ for $x \in D_{2 R, 5 R}$ , by Lemma 5, $\begin{matrix} (81) & (\nabla φ) \cdot (R_{0} (λ) f_{0} - A r_{5 R} f) \in {\dot{H}}_{q, a}^{2} (D_{3 + 1 / 3) R, (3 + 1 / 2) R} . \end{matrix}$ Thus, by Lemma 3, $B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - A r_{5 R} f)]$ are well-defined and possess the following properties: $\begin{matrix} (82) & \begin{matrix} B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - A r_{5 R} f)] \in H_{q}^{3} {(R^{N})}^{N}, \\ {‖ B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - A r_{5 R} f)] ‖}_{H_{q}^{3} (R^{N})} ⩽ C ‖ f ‖_{H_{q}^{2} (Ω)}, \\ supp B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - A r_{5 R} f)] \subset D_{3 R, 4 R}, \\ div B [(\nabla φ) \cdot R_{0} (λ) f_{0} - A r_{5 R} f)] = (\nabla φ) \cdot (R_{0} (λ) f - A r_{5 R} f) in R^{N} . \end{matrix} \end{matrix}$ Especially, $div Φ (λ) f = 0$ for $x \notin B_{2 R}$ , and therefore noting that $\tilde{div} w = div w$ and $J^{- 1} Div \tilde{S} (w, r) = Div S (w, r)$ for $x \notin B_{2 R}$ , we have $\begin{matrix} (83) & \begin{array}{l} λ Φ (λ) f - J^{- 1} Div \tilde{S} (Φ (λ) f, Ψ f) = f + S (λ) f, \tilde{div} Φ (λ) f = 0 in Ω, \\ \tilde{S} (Φ (λ) f, Ψ f) n |_{Γ} = 0, \end{array} \end{matrix}$ with $\begin{array}{rcl} (84) & \begin{matrix} S (λ) f & = λ φ A r_{5 R} f + λ B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - A r_{5 R} f)] + μ Div D (B [(\nabla φ) \cdot (R_{0} (λ) f_{0} - A r_{5 R} f)]) \\ + μ T_{1} (\nabla φ) \nabla (R_{0} (λ) f_{0} - A r_{5 R} f_{0}) + μ T_{2} (\nabla^{2} φ) (R_{0} (λ) f_{0} - A r_{5 R} f_{0}), \end{matrix} \end{array}$ where $T_{1} (\nabla φ) \nabla g$ and $T_{2} (\nabla^{2} φ) g$ are defined by $\begin{array}{l} i th component of T_{1} (\nabla φ) \nabla g = \frac{\partial φ}{\partial x_{i}} div g + \sum_{j = 1}^{N} \frac{\partial φ}{\partial x_{j}} \frac{\partial g_{i}}{\partial x_{j}} + \sum_{j = 1}^{N} \frac{\partial φ}{\partial x_{j}} D_{i j} (g), \\ i th component of T_{2} (\nabla^{2} φ) g = \sum_{j = 1}^{N} \frac{\partial^{2} φ}{\partial x_{i} \partial x_{j}} g_{j} + (Δ φ) g_{i} \end{array}$ for $g =^{⊤} (g_{1}, \dots, g_{N})$ .

We prove Lemma 19.
Let $1 < q < \infty$ . Then, the inverse of the operator $I + S (0)$ exists in $L (L_{q, 5 R} {(Ω)}^{N})$ .
Proof.
Let $R_{0} (0)$ be the operator given in Theorem 12 and let $\begin{matrix} (85) & \begin{array}{l} Φ (0) f = (1 - φ) R_{0} (0) f_{0} + φ A r_{5 R} f + B [(\nabla φ) \cdot (R_{0} (0) f_{0} - A r_{5 R} f)], \\ S (0) f = μ Div D (B [(\nabla φ) \cdot (R_{0} (0) f_{0} - A r_{5 R} f)]) \\ + T_{1} (\nabla φ) \nabla (R_{0} (0) f_{0} - A r_{5 R} f) + T_{2} (\nabla^{2} φ) (R_{0} (0) f_{0} - A r_{5 R} f)) . \end{array} \end{matrix}$ By (36), $\begin{matrix} (86) & - J^{- 1} Div \tilde{S} (Φ (0) f, Ψ f) = f + S (0) f, \tilde{div} Φ (0) f = 0 in Ω, \tilde{S} (Φ (0) f, Ψ f) n |_{Γ} = 0 . \end{matrix}$ By Lemma 3, Lemma 5 and Lemma 6, we have $\begin{matrix} (87) & \begin{matrix} B [(\nabla φ) \cdot (R_{0} (0) f_{0} - A r_{5 R} f)] \in H_{q}^{3} {(R^{N})}^{N}, \\ {‖ B [(\nabla φ) \cdot (R_{0} (0) f_{0} - A r_{5 R} f)] ‖}_{H_{q}^{3} (R^{N})} ⩽ C ‖ f ‖_{H_{q}^{2} (Ω)}, \\ supp B [(\nabla φ) \cdot (R_{0} (0) f_{0} - A r_{5 R} f)] \subset D_{3 R, 4 R}, \\ div B [(\nabla φ) \cdot (R_{0} (0) f_{0} - A r_{5 R} f)] = (\nabla φ) \cdot (R_{0} (0) f - A r_{5 R} f) in R^{N} . \end{matrix} \end{matrix}$ Since $S (0) f \in H_{q}^{1} (Ω)$ and $supp S (0) f \subset D_{3 R, 4 R}$ as follows from (87), by the Rellich compactness theorem $S (0)$ is a compact operator from $L_{q, 5 R} {(Ω)}^{N}$ into itself. Thus, to prove the invertibility of $I + S (0)$ , it suffices to prove that the kernel of $I + S (0)$ is trivial. Let f be an element of $L_{q, 5 R} {(Ω)}^{N}$ such that $(I + S (0)) f = 0$ . Since $f = - S (0) f$ , we have $\begin{matrix} (88) & supp f \subset D_{3 R, 4 R} . \end{matrix}$ Let $u = Φ (0) f$ and $q = Ψ f$ , and then $u \in H_{q, loc}^{2} {(Ω)}^{N}$ and $q \in {\hat{H}}_{q}^{1} (Ω)$ . Moreover, by (86), u and $q$ satisfy the homogeneous equations: $\begin{matrix} (89) & - J^{- 1} Div \tilde{S} (u, q) = 0, \tilde{div} u = 0 in Ω, \tilde{S} (u, q) n |_{Γ} = 0 . \end{matrix}$ First, we assume that the uniqueness of solutions to (89) holds, that is, $\begin{matrix} (90) & u = 0, q = 0 . \end{matrix}$ And then, by (80) and (85), $\begin{matrix} (91) & (1 - φ) R_{0} (0) f_{0} + φ A r_{5 R} f + B_{0} = 0, (1 - φ) (Π f_{0} + c (f)) + φ B r_{5 R} f = 0 \end{matrix}$ with $B_{0} = B [(\nabla φ) \cdot (R_{0} (0) f_{0} - A r_{5 R} f)]$ . Since $supp B_{0} \subset D_{3 R, 4 R}$ as follows from (87) and since $φ = 1$ for $| x | ⩽ 3 R$ and $φ (x) = 0$ for $| x | ⩾ 4 R$ , we have $\begin{matrix} (92) & \begin{array}{l} R_{0} (0) f_{0} = 0, Π f_{0} + c (f) = 0 for | x | > 4 R, \\ A r_{5 R} f = 0, B r_{5 R} f = 0 for | x | < 3 R . \end{array} \end{matrix}$ Let $\begin{matrix} w = \{\begin{matrix} A r_{5 R} f & for x \in Ω_{5 R}, \\ 0 & for x \notin Ω, \end{matrix} r = \{\begin{matrix} B r_{5 R} f & for x \in Ω_{5 R}, \\ 0 & for x \notin Ω, \end{matrix} \end{matrix}$ and then by (92) and the fact that $Div S (w, q) = J^{- 1} Div \tilde{S} (A r_{5 R} f, B r_{5 R} f) = r_{5 R} f$ and $div w = \tilde{div} A r_{5 R} f = 0$ for $x \in D_{2 R, 5 R}$ , we see that $w \in H_{q}^{2} {(B_{5 R})}^{N}$ and $r \in H_{q}^{1} (B_{5 R})$ satisfy the Stokes equations: $\begin{matrix} (93) & - Div S (w, r) = f_{0}, div w = 0 in B_{5 R}, w |_{S_{5 R}} = 0, \end{matrix}$ because $supp f \subset D_{3 R, 4 R}$ (cf. (88)). On the other hand, by (92) we see that $R_{0} (0) f_{0}$ and $Π f_{0} + c (f)$ also satisfy the equations (93). The uniqueness of solutions for Eq. (93) implies that $w = R_{0} (0) f_{0}$ and $r = Π f_{0} + c (f) + d$ in $B_{5 R}$ , d being a constant. By (79), we have $\begin{matrix} \int_{Ω_{5 R}} d d x = \int_{Ω_{5 R}} (r - (Π f_{0} + c (f))) d x = \int_{Ω_{5 R}} (B r_{5 R} f - (Π f_{0} + c (f))) d x = 0, \end{matrix}$ and so $d = 0$ . Thus, we have $R_{0} (0) f_{0} = A r_{5 R} f$ and $Π f_{0} + c (f) = B r_{5 R} f$ in $Ω_{5 R}$ , which, combined with (91), leads to $B_{0} = 0$ , and therefore $\begin{array}{l} 0 = R_{0} (0) f_{0} + φ (A r_{5 R} f - R_{0} (0) f_{0}) = R_{0} (0) f_{0}, \\ 0 = Π f_{0} + c (f) + φ (B r_{5 R} f - (Π f_{0} + c (f))) = Π f_{0} + c (f) \end{array}$ in $Ω_{5 R}$ . Thus, $\begin{matrix} f_{0} = - Div S (R_{0} (0) f_{0}, Π f_{0} + c (f)) = 0 in Ω_{5 R}, \end{matrix}$ which, combined with (88), leads to $f = 0$ . From this we conclude that ${(I + S (0))}^{- 1}$ exists as an element of $L (L_{q, 5 R} {(Ω)}^{N})$ . In particular, if we define w and $r$ by $\begin{matrix} (94) & w = Φ (0) {(I + S (0))}^{- 1} f, r = Ψ {(I + S (0))}^{- 1} f, \end{matrix}$ then $w \in H_{q, loc}^{2} (Ω)$ and $r \in H_{q, loc}^{1} (Ω)$ satisfy the equations: $\begin{matrix} (95) & J^{- 1} Div \tilde{S} (w, r) = f, \tilde{div} w = 0 in Ω, \tilde{S} (w, r) n |_{Γ} = 0, \end{matrix}$ provided that $f \in L_{q, 5 R} {(Ω)}^{N}$ .

In what follows, we prove (90). Using (37), we have $\begin{matrix} (96) & \begin{array}{l} | u (x) | = | [R_{0} (0) f_{0}] (x) | ⩽ C_{R} | x |^{- (N - 2)} ‖ f_{0} ‖_{L_{1} (R^{N})}, \\ | \nabla u (x) | = | \nabla [R_{0} (0) f_{0}] (x) | ⩽ C_{R} | x |^{- (N - 1)} ‖ f_{0} ‖_{L_{1} (R^{N})}, \\ | q (x) - c (f) | = | [Π f_{0}] (x) | ⩽ C_{R} | x |^{- (N - 1)} ‖ f_{0} ‖_{L_{1} (R^{N})}, \end{array} \end{matrix}$ for $| x | ⩾ 6 R$ . Since $supp f_{0} \subset B_{5 R}$ , we have $‖ f_{0} ‖_{L_{1} (R^{N})} ⩽ | B_{5 R} |^{1 / q^{'}} ‖ f ‖_{L_{q} (Ω)}$ in (96). First, we prove (90) in the case that $2 ⩽ q < \infty$ . In this case, $u \in H_{2, loc}^{2} {(Ω)}^{N}$ and $q \in H_{2, loc}^{1} (Ω)$ . Let ψ be a function in $C_{0}^{\infty} (R^{N})$ such that $ψ (x) = 1$ for $| x | ⩽ 1$ and $ψ (x) = 0$ for $| x | ⩾ 2$ , and let $ψ_{L} (x) = ψ (x / L)$ . Multiplying (89) with $ψ_{L} J u$ , integrating the resultant formula on Ω and applying the divergence theorem of Gauß, we have $\begin{matrix} (97) & \begin{matrix} 0 & = {(J^{- 1} Div \tilde{S} (u, q), ψ_{L} J u)}_{Ω} \\ = - \frac{μ}{2} {(J \tilde{D} (u), ψ_{L} \tilde{D} (u))}_{Ω} - μ \sum_{i, j = 1}^{N} {({\tilde{D}}_{i j} (u), J a_{k j} \frac{\partial ψ_{L}}{\partial x_{k}} u_{j})}_{Ω} + \sum_{i, j = 1}^{N} {(q, J a_{j i} \frac{\partial ψ_{L}}{\partial x_{j}} u_{i})}_{Ω} . \end{matrix} \end{matrix}$ By (96), $\begin{array}{l} | {({\tilde{D}}_{i j} (u), J a_{k j} \frac{\partial ψ_{L}}{\partial x_{k}} u_{j})}_{Ω} | ⩽ C L^{- 1} \int_{D_{L, 2 L}} | \nabla u (x) | | u (x) | d x ⩽ C L^{- (N - 2)} ‖ f ‖_{L_{q} (Ω)}^{2} \to 0, \\ | {(q - c (f), J a_{j i} \frac{\partial ψ_{L}}{\partial x_{j}} u_{i})}_{Ω} | ⩽ C L^{- 1} \int_{D_{L, 2 L}} | q (x) - c (f) | | u (x) | d x ⩽ C L^{- (N - 2)} ‖ f ‖_{L_{q} (Ω)}^{2} \to 0 \end{array}$ as $L \to \infty$ , because $N ⩾ 3$ . Moreover, by Lemma 6, $\begin{matrix} \sum_{i, j = 1}^{N} {(c (f), J a_{j i} \frac{\partial ψ_{L}}{\partial x_{j}} u_{i})}_{Ω} = c (f) \int_{D_{L, 2 L}} (\nabla ψ_{L}) \cdot u d x = 0, \end{matrix}$ because $J = 1$ and $a_{j i} = δ_{j i}$ for $| x | ⩾ 2 R$ . Thus, letting $L \to \infty$ in (97), we have $\tilde{D} (u) = 0$ in Ω. If we consider the transformation: $y = A x$ , then $v (y) = u (A^{- 1} y)$ satisfies $D (v) = 0$ in $Ω_{\infty} = {y = A x ∣ x \in Ω}$ . This implies that v is a vector of first order polynomials of the form $A y + b$ with some constant anti-symmetric matrix $A$ and some constant N vector b. Since $u (x) = O (| x |^{- (N - 2)})$ as $| x | \to \infty$ , $v (y)$ also behaves like $| y |^{- (N - 2)}$ as $| y | \to \infty$ , which yields that $v (y) = 0$ . So, $u = 0$ . Inserting this fact into (89), we have $\nabla q = 0$ in Ω and $q |_{Γ} = 0$ , which implies that $q = 0$ . This completes the proof of (90) in the case that $2 ⩽ q < \infty$ .

Next, we consider the case that $1 < q < 2$ . Let g be any element of $L_{q^{'}, 5 R} (Ω)$ . Since $2 < q^{'} < \infty$ , by (94) we know the existence of $w \in H_{q^{'}, loc}^{2} {(Ω)}^{N}$ and $r \in H_{q^{'}, loc}^{1} (Ω)$ such that w and $r$ satisfy the equations: $\begin{matrix} (98) & - J^{- 1} Div \tilde{S} (w, r) = g, \tilde{div} w = 0 in Ω, \tilde{S} (w, r) n |_{Γ} = 0 . \end{matrix}$ Moreover, by (37) and (94) $\begin{matrix} (99) & \begin{array}{l} | w (x) | = | [R_{0} (0) {({(I + S (0))}^{- 1} g)}_{0}] (x) | ⩽ C_{R} | x |^{- (N - 2)} {‖ {({(I + S (0))}^{- 1} g)}_{0} ‖}_{L_{1} (R^{N})}, \\ | \nabla w (x) | = | \nabla [R_{0} (0) {({(I + S (0))}^{- 1} g)}_{0}] (x) | ⩽ C_{R} | x |^{- (N - 1)} {‖ {({(I + S (0))}^{- 1} g)}_{0} ‖}_{L_{1} (R^{N})}, \\ | r (x) - c_{g} | = | [Π {({(I + S (0))}^{- 1} g)}_{0}] (x) | ⩽ C_{R} | x |^{- (N - 1)} {‖ {({(I + S (0))}^{- 1} g)}_{0} ‖}_{L_{1} (R^{N})}, \end{array} \end{matrix}$ for $| x | ⩾ 6 R$ with some constant $c_{g}$ . Here, ${({(I + S (0))}^{- 1} g)}_{0}$ is the zero extension of ${(I + S (0))}^{- 1} g$ to $Ω^{c}$ . In (99), we use the estimate: $\begin{matrix} {‖ {({(I + S (0))}^{- 1} g)}_{0} ‖}_{L_{1} (R^{N})} ⩽ | B_{5 R} |^{1 / q^{'}} {‖ {(I + S (0))}^{- 1} g ‖}_{L_{q} (Ω_{5 R})} ⩽ | B_{5 R} |^{1 / q^{'}} ‖ g ‖_{L_{q} (Ω_{5 R})}, \end{matrix}$ because ${(I + S (0))}^{- 1} \in L (L_{q, 5 R} {(Ω)}^{N})$ . By (98) $\begin{array}{l} {(J ψ_{L} u, g)}_{Ω} = - {(ψ_{L} u, Div \tilde{S} (w, r))}_{Ω} & = \frac{μ}{2} {(ψ_{L} J \tilde{D} (u), \tilde{D} (w))}_{Ω} + \sum_{i, j, k = 1}^{N} {(J a_{k j} \frac{\partial ψ_{L}}{\partial x_{k}} u_{i}, {\tilde{D}}_{i j} (w))}_{Ω} \\ - \sum_{i, k = 1}^{N} {(J a_{k i} \frac{\partial ψ_{L}}{\partial x_{k}} u_{i}, r - c_{g})}_{Ω} + c_{g} \int_{R^{N}} (\nabla ψ_{L}) \cdot u d x \end{array}$ for large $L > 2 R$ , because $J = 1$ and $a_{k i} = δ_{k i}$ for $| x | > 2 R$ . Since u, $q$ satisfy (96), and w, $r$ satisfy (99), we have $\begin{matrix} lim_{L \to \infty} \sum_{i, j, k = 1}^{N} {(J a_{k j} \frac{\partial ψ_{L}}{\partial x_{k}} u_{i}, {\tilde{D}}_{i j} (w))}_{Ω} = 0, lim_{L \to \infty} \sum_{i, k = 1}^{N} {(J a_{k i} \frac{\partial ψ_{L}}{\partial x_{k}} u_{i}, r - c_{g})}_{Ω} = 0 . \end{matrix}$ Moreover, by Lemma 6 we have $\begin{matrix} \int_{R^{N}} (\nabla ψ_{L}) \cdot u d x = 0 . \end{matrix}$ Thus, letting $L \to \infty$ , we have $\begin{matrix} (100) & {(J u, g)}_{Ω} = \frac{μ}{2} {(J \tilde{D} (u), \tilde{D} (w))}_{Ω} . \end{matrix}$ On the other hand, by (89) $\begin{array}{l} 0 = - {(J^{- 1} Div \tilde{S} (u, q), J ψ_{L} w)}_{Ω} & = \frac{μ}{2} {(ψ_{L} J \tilde{D} (u), \tilde{D} (w))}_{Ω} + \sum_{i, j, k = 1}^{N} {(J a_{k j} \frac{\partial ψ_{L}}{\partial x_{k}} w_{i}, {\tilde{D}}_{i j} (u))}_{Ω} \\ - \sum_{i, k = 1}^{N} {(J a_{k i} \frac{\partial ψ_{L}}{\partial x_{k}} w_{i}, q - c (f))}_{Ω} + c (f) \int_{R^{N}} (\nabla ψ_{L}) \cdot w d x . \end{array}$ Using (96), (99) and Lemma 6 and letting $L \to \infty$ , we have $\begin{matrix} 0 = \frac{μ}{2} {(J \tilde{D} (u), \tilde{D} (w))}_{Ω}, \end{matrix}$ which, combined with (100), leads to ${(J u, g)}_{Ω} = 0$ . The arbitrariness of choice of g yields that $u = 0$ on $Ω_{5 R}$ . Then, by using the equations and the boundary conditions, we also have $q = 0$ on $Ω_{5 R}$ . Thus, $u \in H_{q, loc}^{2} (R^{N} ∖ B_{5 R})$ and $q \in H_{q, loc}^{1} (R^{N} ∖ B_{5 R})$ satisfy the homogeneous equations: $\begin{matrix} Div S (u, q) = 0, div u = 0 in B_{5 R}^{c}, u |_{S_{5 R}} = 0 . \end{matrix}$ Since the boundary of $R^{N} ∖ B_{5 R}$ is $S_{5 R}$ which is a smooth compact hypersurface, by the Sobolev imbedding theorem and a bootstrap argument, we see that $u \in H_{2, loc}^{2} (R^{N} ∖ B_{5 R})$ and $q \in H_{2, loc}^{1} (R^{N} ∖ B_{5 R})$ . Therefore, we have $u \in H_{2, loc}^{2} (Ω)$ and $q \in H_{2, loc}^{1} (Ω)$ . Since the estimate (96) holds, using the same argument as in the case that $2 ⩽ q < \infty$ , we have (90). This completes the proof of Lemma 19. □
Proof of Theorem 18.
In view of (83), we will represent solutions u and $p$ of (74) by $\begin{matrix} (101) & u = Φ (λ) {(I + S (λ))}^{- 1} f, p = Ψ {(I + S (λ))}^{- 1} f . \end{matrix}$ Since $\begin{array}{l} S (0) f - S (λ) f & = - λ φ A r_{5 R} f + λ B [(\nabla φ) \cdot A r_{5 R} f] + λ B [(\nabla φ) \cdot (R_{0} (0) f_{0} - R_{0} (λ) f_{0})] \\ - λ B [(\nabla φ) \cdot R_{0} (0) f_{0}] + μ Div D (B [(\nabla φ) \cdot (R_{0} (0) f_{0} - R_{0} (λ) f_{0})]) \\ + μ T_{1} (\nabla φ) \nabla (R_{0} (0) f_{0} - R_{0} (λ) f_{0}) + μ T_{2} (\nabla^{2} φ) (R_{0} (0) f_{0} - R_{0} (λ) f_{0}) \end{array}$ as follows from (84) and (85), by (37), (40) and Theorem 7, $‖ S (0) - S (λ) ‖_{L (L_{q} (Ω_{5 R}))} ⩽ C | λ |^{1 / 4}$ for any $λ \in Λ_{ϵ, 1 / 2}$ , and therefore $\begin{matrix} {‖ {(I + S (λ))}^{- 1} ‖}_{L (L_{q, 5 R} (Ω))} ⩽ 1 \end{matrix}$ for any $λ \in Σ_{ϵ, d}$ with some small constant $d > 0$ . Thus, letting $\begin{matrix} S (λ) = Φ (λ) {(I + S (λ))}^{- 1}, P (λ) = Ψ {(I + S (λ))}^{- 1}, \end{matrix}$ applying Theorem 12(4) and Theorem 7, and noting that $(i - 1) / 2 ⩽ 0$ and $(i - 1) / 2 - γ / 2 < 0$ for $i = 0, 1$ , we have Theorem 18. This completes the proof of Theorem 18. □

6. A proof of Theorem 2

As was seen in (27), the equations (13) are written as follows: $\begin{matrix} (102) & \{\begin{matrix} λ u - Div S (u, p) - μ (A_{1} \nabla^{2} u + A_{2} \nabla u) + A_{3} \nabla p = f & in Ω, \\ div u + A_{4} \nabla u = 0 & in Ω, \\ S (u, p) n + μ A_{5} \nabla u + A_{6} p = 0 & on Γ . \end{matrix} \end{matrix}$ Here, $A_{1}, \dots, A_{6}$ are the same matrices as in (27), which satisfy the condition: $supp A_{i} \subset B_{2 R}$ ( $i = 1, \dots, 6$ ) and the estimate (28). Let $f \in L_{q} {(Ω)}^{N}$ , and let $c (f)$ be a constant such that $\int_{Ω_{5 R}} (Π f_{0} + c (f)) d x = 0$ . And then, by Poincaré’s inequality, $\begin{matrix} (103) & {‖ Π f_{0} + c (f) ‖}_{H_{q}^{1} (Ω_{5 R})} ⩽ C ‖ f ‖_{L_{q} (Ω)} . \end{matrix}$ Let $u = R_{0} (λ) f_{0} + v$ and $p = Π f_{0} + c (f) + q$ in problem (13), and then by (102) $\begin{matrix} (104) & \{\begin{matrix} λ v - J^{- 1} Div \tilde{S} (v, q) = U_{0} (λ) f & in Ω, \\ \tilde{div} v = U_{1} (λ) f & in Ω, \\ \tilde{S} (v, q) n = U_{2} (λ) f & on Γ, \end{matrix} \end{matrix}$ with $\begin{array}{l} U_{0} (λ) f = μ (A_{1} \nabla^{2} R_{0} (λ) f_{0} + A_{2} \nabla R_{0} (λ) f_{0}) - A_{3} \nabla Π f_{0}, \\ U_{1} (λ) f = - A_{4} \nabla R_{0} (λ) f_{0}, \\ U_{2} (λ) f = - S (R_{0} (λ) f_{0}, Π f_{0} + c (f)) n - μ A_{5} \nabla R_{0} (λ) f_{0} - A_{6} (Π f_{0} + c (f)) . \end{array}$ Note that $supp U_{i} (λ) f \subset B_{2 R}$ ( $i = 0, 1, 2$ ). In the case that $N < q < \infty$ , by (29), (28), Theorem 12 with $p = \infty$ , and (103), we have $\begin{matrix} (105) & \begin{matrix} {‖ U_{0} (λ) f ‖}_{L_{q} (Ω_{5 R})} + {‖ (U_{1} (λ) f, U_{2} (λ) f) ‖}_{H_{q}^{1} (Ω_{5 R})} & ⩽ C ({‖ \nabla R_{0} (λ) f_{0} ‖}_{L_{\infty} (R^{N})} + ‖ f ‖_{L_{q} (Ω)}) \\ ⩽ C | λ |^{- \frac{1}{2} + \frac{N}{2 q}} ‖ f ‖_{L_{q} (Ω)} \end{matrix} \end{matrix}$ for any $λ \in Λ_{ϵ, 1 / 2}$ . On the other hand, in the case that $1 < q ⩽ N$ , choosing s in such a way that $s ⩾ p$ and $N (1 / q - 1 / s) < 1$ , by (29), (28), Theorem 12 with $p = s$ , and (103) we have $\begin{matrix} (106) & \begin{matrix} {‖ U_{0} (λ) f ‖}_{L_{q} (Ω_{5 R})} + {‖ (U_{1} (λ) f, U_{2} (λ) f) ‖}_{H_{q}^{1} (Ω_{5 R})} & ⩽ C ({‖ \nabla R_{0} (λ) f_{0} ‖}_{L_{s} (R^{N})} + ‖ f ‖_{L_{q} (Ω)}) \\ ⩽ C | λ |^{- \frac{1}{2} + \frac{N}{2} (\frac{1}{q} - \frac{1}{s})} ‖ f ‖_{L_{q} (Ω)} \end{matrix} \end{matrix}$ for any $λ \in Λ_{ϵ, 1 / 2}$ .

Let $w \in H_{q}^{2} (Ω_{5 R})$ and $r \in H_{q}^{1} (Ω_{5 R})$ be solutions of the equations: $\begin{matrix} (107) & \{\begin{matrix} λ w - J^{- 1} Div \tilde{S} (w, r) = - U_{0} (λ) f, & \tilde{div} w = - U_{1} (λ) f in Ω_{5 R}, \\ \tilde{S} (w, r) n |_{Γ} = - U_{2} (λ) f, & w |_{S_{5 R}} = 0 . \end{matrix} \end{matrix}$ By Theorem 7, we know the existence of w and $r$ possessing the estimate: $\begin{matrix} (108) & ‖ w ‖_{H_{q}^{2} (Ω_{5 R})} + ‖ r ‖_{H_{q}^{1} (Ω_{5 R})} ⩽ C ({‖ U_{0} (λ) f ‖}_{L_{q} (Ω_{5 R})} + {‖ (U_{1} (λ) f, U_{2} (λ) f) ‖}_{H_{q}^{1} (Ω_{5 R})}) . \end{matrix}$ Let φ be a function in $C_{0}^{\infty} (R^{N})$ such that $φ (x) = 1$ for $| x | ⩽ 3 R$ and $φ (x) = 0$ for $| x | ⩾ 4 R$ . Since $U_{1} (λ) f = 0$ and $\tilde{div} w = div w = 0$ for $x \in D_{2 R, 5 R}$ , by Lemma 5, $(\nabla φ) \cdot w \in {\dot{H}}_{q, a}^{3} (D_{2 R, 5 R})$ , and therefore $div (\nabla w - B [(\nabla φ) \cdot w]) = 0$ in $Ω_{5 R}$ . Let $y = φ w - B [(\nabla φ) \cdot w]$ and $θ = φ r$ , and then by (107) $\begin{matrix} (109) & \{\begin{matrix} λ y - J^{- 1} Div \tilde{S} (y, θ) = - U_{0} (λ) f + U_{5} (λ) f & in Ω, \\ \tilde{div} y = - U_{1} (λ) f & in Ω, \\ \tilde{S} (y, θ) n = - U_{2} (λ) f & on Γ \end{matrix} \end{matrix}$ with $\begin{matrix} (110) & \begin{matrix} U_{5} (λ) f & = J^{- 1} Div \tilde{S} (φ w, φ r) - φ J^{- 1} Div \tilde{S} (w, r) \\ + J^{- 1} Div \tilde{S} (B [(\nabla φ) \cdot w], 0) - λ B [(\nabla φ) \cdot w], \end{matrix} \end{matrix}$ because $φ = 1$ near Γ. Let s be the exponent given in (106) and let $\begin{matrix} (111) & \tilde{p} = \{\begin{matrix} \infty & for N < q < \infty, \\ s & for 1 < q ⩽ N . \end{matrix} \end{matrix}$ By (105), (106), (108) and Lemma 3, $\begin{matrix} (112) & \begin{array}{l} ‖ y ‖_{H_{q}^{2} (Ω)} + ‖ \nabla θ ‖_{L_{q} (Ω)} ⩽ C | λ |^{- \frac{1}{2} + \frac{N}{2} (\frac{1}{q} - \frac{1}{\tilde{p}})} ‖ f ‖_{L_{q} (Ω)}, \\ {‖ U_{5} (λ) f ‖}_{L_{q} (Ω_{5 R})} ⩽ C | λ |^{- \frac{1}{2} + \frac{N}{2} (\frac{1}{q} - \frac{1}{\tilde{p}})} ‖ f ‖_{L_{q} (Ω)} . \end{array} \end{matrix}$ Finally, let z and π be solutions of the equations: $\begin{matrix} (113) & λ z - J^{- 1} Div \tilde{S} (z, π) = - U_{5} (λ) f, \tilde{div} z = 0 in Ω, \tilde{S} (z, π) n |_{Γ} = 0 . \end{matrix}$

In the sequel, we assume that $λ \in Λ_{ϵ, d}$ and $1 < q ⩽ p ⩽ \infty$ , $q \neq \infty$ , $N (1 / q - 1 / p) < 1$ . Then, by Theorem 18 $\begin{matrix} (114) & {‖ (\nabla^{2} z, \nabla π) ‖}_{L_{q} (Ω)} ⩽ C {‖ U_{5} (λ) f ‖}_{L_{q} (Ω)}, \end{matrix}$ and moreover, $\begin{matrix} (115) & {‖ \nabla^{i} z ‖}_{L_{p} (Ω)} ⩽ C | λ |^{\frac{i - 1}{2}} {‖ U_{5} (λ) f ‖}_{L_{q} (Ω)} (i = 0, 1) \end{matrix}$ provided that $p (N - 1) > N$ , and $\begin{matrix} (116) & {‖ \nabla^{i} z ‖}_{L_{p} (Ω)} ⩽ C | λ |^{\frac{i - 1}{2} - \frac{γ}{2}} ‖ U_{5} (λ) f ‖ L_{q} (Ω) (i = 0, 1) \end{matrix}$ provided that $p (N - 1) ⩽ N$ . Here, γ is a chosen in such a way that $0 < γ < 1$ and $(N - 1 + γ) p > N$ . We may assume that $0 < d < 1 / 2$ .

In the case that $p (N - 1) > N$ , combining (112) and (115) yields that $\begin{matrix} {‖ \nabla^{i} z ‖}_{L_{p} (Ω)} ⩽ C | λ |^{\frac{i}{2} - 1 + \frac{N}{2} (\frac{1}{q} - \frac{1}{\tilde{p}})} ‖ f ‖_{L_{q} (Ω)} . \end{matrix}$ Since $p ⩽ \tilde{p}$ , we have $\begin{matrix} (117) & {‖ \nabla^{i} z ‖}_{L_{p} (Ω)} ⩽ C | λ |^{\frac{i}{2} - 1 + \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (Ω)} \end{matrix}$ for $i = 0, 1$ and $λ \in Λ_{ϵ, d}$ provided $p (N - 1) > N$ .

Next, we consider the case that $p (N - 1) ⩽ N$ . Choosing $γ > 0$ in such a way that $(N - 1 + γ) p > N$ , by (112) and (116) we have $\begin{matrix} (118) & {‖ \nabla^{i} z ‖}_{L_{p} (Ω)} ⩽ C | λ |^{\frac{i}{2} - 1 + \frac{N}{2} (\frac{1}{q} - \frac{1}{\tilde{p}}) - \frac{γ}{2}} ‖ f ‖_{L_{q} (Ω)} \end{matrix}$ for $i = 0, 1$ and $λ \in Σ_{ϵ, d}$ . We choose γ in such a way that $\begin{matrix} (119) & \frac{N}{2} (\frac{1}{q} - \frac{1}{\tilde{p}}) - \frac{γ}{2} = \frac{N}{2} (\frac{1}{q} - \frac{1}{p}), \end{matrix}$ that is, $γ = N / p - N / \tilde{p}$ . In the case that $N < q < \infty$ , $\tilde{p} = \infty$ , and so $γ = N / p$ . In this case, $(N - 1 + γ) p = (N - 1) p + N > N$ , and so we have $\begin{matrix} (120) & {‖ \nabla^{i} z ‖}_{L_{p} (Ω)} ⩽ C | λ |^{\frac{i}{2} - 1 + \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (Ω)} . \end{matrix}$ In the case that $1 < q ⩽ N$ , $\tilde{p} = s$ , and so $γ = N / p - N / s$ . The requirement : $(N - 1 + γ) p > N$ implies that $1 - 1 / N > 1 / s$ . On the other hand, s must satisfy the condition: $N (1 / q - 1 / s) < 1$ , which implies that $1 / q - 1 / N < 1 / s$ . Thus, choosing s in such a way that $1 / q - 1 / N < 1 / s < 1 - 1 / N$ , we have (119) as well as $(N - 1 + γ) p > N$ , and therefore we also have (120) provided $1 < q ⩽ N$ .

Moreover, by (114) and (112), $\begin{matrix} (121) & {‖ (\nabla^{2} z, \nabla π) ‖}_{L_{q} (Ω)} ⩽ C | λ |^{- \frac{1}{2} + \frac{N}{2} (\frac{1}{q} - \frac{1}{\tilde{p}})} ‖ f ‖_{L_{q} (Ω)} . \end{matrix}$ Since $u = R_{0} (λ) f_{0} - y - z$ and $p = Π f_{0} + c (f) + θ + π$ are unique solutions of the equations (13), recalling that $\tilde{p} ⩾ p$ , by Theorem 12, (112), Sobolev’s imbedding theorem, (117), (120) and (121), $\begin{matrix} (122) & \begin{array}{l} {‖ \nabla^{i} u ‖}_{L_{p} (Ω)} ⩽ C | λ |^{\frac{i}{2} - 1 + \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (Ω)} (i = 0, 1), \\ {‖ (\nabla^{2} u, \nabla p) ‖}_{L_{q} (Ω)} ⩽ C | λ |^{- \frac{1}{2} + \frac{N}{2} (\frac{1}{q} - \frac{1}{\tilde{p}})} ‖ f ‖_{L_{q} (Ω)} \end{array} \end{matrix}$ for any $λ \in Λ_{ϵ, d}$ . This completes the proof of Theorem 2.

7. A proof of Theorem 1

First, we consider problem (1) in the semigroup setting. For this we will eliminate the pressure term $p$ by using the solutions of the variational equation: $\begin{matrix} (123) & {(\tilde{\nabla} u, J \tilde{\nabla} φ)}_{Ω} = {(f, J \tilde{\nabla} φ)}_{Ω} for any φ \in {\hat{H}}_{q^{'}, 0}^{1} (Ω) . \end{matrix}$ Then, by (3), (55), and (56), we have

Theorem 20.

Let $N < r < \infty$ be the exponent appearing in ( 3 ). Let $1 < q ⩽ r$ . Then, there exists a $σ > 0$ such that if the assumption ( 3 ) holds, then for any $f \in L_{q} {(Ω)}^{N}$ , problem ( 123 ) admits a unique solution $u \in {\hat{H}}_{q, 0}^{1} (Ω)$ possessing the estimate: $\begin{matrix} ‖ \nabla u ‖_{L_{q} (Ω)} ⩽ C ‖ f ‖_{L_{q} (Ω)} . \end{matrix}$

By (28), $\begin{matrix} (124) & J^{- 1} Div \tilde{S} (u, p) = μ J^{- 1} Div (J A \tilde{D} (u)) - \tilde{\nabla} p, \tilde{div} u = \sum_{j, k = 1}^{N} \frac{\partial}{\partial x_{k}} (J a_{k j} u_{j}) . \end{matrix}$ Moreover, $\begin{matrix} (125) & ⟨ \tilde{S} (u, p) n, n ⟩ = \sum_{i, j, k = 1}^{N} J a_{j k} {\tilde{D}}_{i k} (u) n_{j} n_{i} - (\sum_{i, j = 1}^{N} J a_{i j} n_{i} n_{j}) p \end{matrix}$ with $n =^{⊤} (n_{1}, \dots, n_{N})$ . For any $u \in H_{q}^{2} {(Ω)}^{N}$ , let $K (u)$ be a unique solution of the equations: $\begin{matrix} (126) & \begin{matrix} {(\tilde{\nabla} K (u), J \tilde{\nabla} φ)}_{Ω} = {(μ J^{- 1} Div (J A \tilde{D} (u)) - \tilde{\nabla} \tilde{div} u, J \tilde{\nabla} φ)}_{Ω} for any φ \in {\hat{H}}_{q^{'}, 0}^{1} (Ω), \end{matrix} \end{matrix}$ subject to $\begin{matrix} (127) & K (u) = {(J \sum_{i, j = 1}^{N} a_{i j} n_{i} n_{j})}^{- 1} (\sum_{i, j, k = 1}^{N} J a_{j k} {\tilde{D}}_{i k} (u) n_{j} n_{i}) - \tilde{div} u on Γ . \end{matrix}$ By Theorem 124 and (29), we know the existence of the unique solution $K (u) \in {\hat{H}}_{q, 0}^{1} (Ω) + {\hat{H}}_{q, 0}^{1} (Ω)$ possessing the estimate: $\begin{matrix} (128) & {‖ \nabla K (u) ‖}_{L_{q} (Ω)} ⩽ C ‖ \nabla u ‖_{H_{q}^{1} (Ω)} . \end{matrix}$ Let ${\tilde{J}}_{q} (Ω)$ be the solenoidal space given in (11). Given $f \in {\tilde{J}}_{q} (Ω)$ , let $u \in H_{q}^{2} {(Ω)}^{N}$ and $p \in H_{q}^{1} (Ω) + {\hat{H}}_{q, 0}^{1} (Ω)$ be solutions of the equations: $\begin{matrix} (129) & \begin{matrix} λ u - J^{- 1} Div \tilde{S} (u, p) & = f, \tilde{div} u = 0 in Ω, \tilde{S} (u, p) n |_{Γ} = 0 . \end{matrix} \end{matrix}$ Using (124) and (126), we have $\begin{array}{l} 0 & = {(f, J \tilde{\nabla} φ)}_{Ω} = {(λ u - J^{- 1} Div \tilde{S} (u, p), J \tilde{\nabla} φ)}_{Ω} \\ = {(λ u, J^{⊤} A \nabla φ)}_{Ω} - {(\tilde{\nabla} \tilde{div} u, J \tilde{\nabla} φ)}_{Ω} - {(\tilde{\nabla} K (u), J \tilde{\nabla} φ)}_{Ω} + {(\tilde{\nabla} p, J \tilde{\nabla} φ)}_{Ω} \\ = - λ {(\tilde{div} u, φ)}_{Ω} - {(\tilde{\nabla} \tilde{div} u, J \tilde{\nabla} φ)}_{Ω} + {(\tilde{\nabla} p - K (u), J \tilde{\nabla} φ)}_{Ω} . \end{array}$ Since $\tilde{div} u = 0$ , we have $\begin{matrix} (130) & {(\tilde{\nabla} (p - K (u)), J \tilde{\nabla} φ)}_{Ω} = 0 for any φ \in {\hat{H}}_{q^{'}, 0}^{1} (Ω) . \end{matrix}$ Since $\tilde{S} (u, p) n |_{Γ} = 0$ , by (125) and (126) $\begin{matrix} p = {(\sum_{i, j = 1}^{N} J a_{i j} n_{i} n_{j})}^{- 1} (\sum_{i, j, k = 1}^{N} J a_{j k} {\tilde{D}}_{i k} (u) n_{j} n_{i}) = K (u) - \tilde{div} u on Γ . \end{matrix}$ Since $\tilde{div} u = 0$ , we have $p - K (u) = 0$ on Γ, which, combined with (130), leads to $p = K (u)$ . Thus, u is a unique solution of the equations: $\begin{matrix} (131) & \begin{matrix} λ u - J^{- 1} Div \tilde{S} (u, K (u)) & = f, \tilde{div} u = 0 in Ω, \tilde{S} (u, K (u)) n |_{Γ} = 0 . \end{matrix} \end{matrix}$ By Theorem 2, $\begin{matrix} (132) & u = S (λ) f, K (u) = P (λ) f . \end{matrix}$ Let $\begin{array}{l} A u = J^{- 1} Div \tilde{S} (u, K (u)) for u \in D_{q} (Ω), \\ D_{q} (Ω) = {u \in {\tilde{J}}_{q} (Ω) \cap H_{q}^{2} {(Ω)}^{N} ∣ \tilde{S} (u, K (u)) n |_{Γ} = 0} . \end{array}$ In view of Lemma 14, $u \in {\tilde{D}}_{q} (Ω)$ implies $\tilde{div} u = 0$ in Ω. By Theorem 2, $S (λ) f = {(λ I - A)}^{- 1} f$ for $f \in {\tilde{J}}_{q} (Ω)$ , and so $A$ generates a continuous analytic semi-group ${T (t)}_{t ⩾ 0}$ . In fact, ${T (t)}_{t ⩾ 0}$ is given by $\begin{matrix} (133) & T (t) f = \frac{1}{2 π i} \int_{Γ} e^{λ t} S (λ) f d λ \end{matrix}$ with suitable contour from $\infty e^{- i (π - ϵ)}$ to $\infty e^{i (π - ϵ)}$ in the sector $Σ_{ϵ}$ ( $ϵ \in (0, π / 2)$ ). Using the standard argument as in the theory of semi-group, we see easily that ${T (t)}_{t ⩾ 0}$ is a continuous analytic semi-group.

Moreover, by Theorem 2 and the analytic semigroup property, we have (12). In fact, when $1 < q ⩽ r$ , $1 < q ⩽ p ⩽ \infty$ and $N (1 / q - 1 / p) < 1$ , by Theorem 2 and the change of variable: $λ t = λ^{'}$ in (133), we have $\begin{matrix} {‖ \nabla^{i} T (t) f ‖}_{L_{p} (Ω)} ⩽ C_{q, p} t^{- \frac{i}{2} - \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (Ω)}, \end{matrix}$ for $i = 0, 1$ and $t > 0$ . Next, we consider the case $N (1 / q - 1 / p) ⩾ 1$ . If there exists a $q_{1}$ such that $1 < q < q_{1} ⩽ r$ and $N (1 / q - 1 / q_{1}) < 1$ and $N (1 / q_{1} - 1 / p) < 1$ , then using the semigroup property: $T (t) f = T (t / 2) T (t / 2) f$ , we have $\begin{array}{l} {‖ T (t) ‖}_{L_{p} (Ω)} & ⩽ C {(t / 2)}^{- \frac{N}{2} (\frac{1}{q_{1}} - \frac{1}{p})} {‖ T (t / 2) f ‖}_{L_{q_{1}} (Ω)} ⩽ C {(t / 2)}^{- \frac{N}{2} (\frac{1}{q_{1}} - \frac{1}{p})} t^{- \frac{N}{2} (\frac{1}{q} - \frac{1}{q_{1}})} ‖ f ‖_{L_{q} (Ω)} \\ = C {(t / 2)}^{- \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (Ω)} . \end{array}$ Repeated use of this argument finite times yields that $\begin{matrix} {‖ T (t) ‖}_{L_{p} (Ω)} ⩽ C t^{- \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (Ω)} \end{matrix}$ provided that $1 < q ⩽ r$ and $q ⩽ p ⩽ \infty$ . And then, choosing $q_{1}$ in such a way that $1 < q < q_{1} ⩽ r$ , $N (1 / q_{1} - 1 / p) < 1$ and $1 < q < q_{1}$ , $\begin{array}{l} {‖ \nabla T (t) f ‖}_{L_{p} (Ω)} & = {‖ \nabla T (t / 2) (T (t / 2) f) ‖}_{L_{q_{1}} (Ω)} \\ ⩽ C {(t / 2)}^{- \frac{1}{2} - \frac{N}{2} (\frac{1}{q_{1}} - \frac{1}{p})} {‖ T (t / 2) f ‖}_{L_{q_{1}} (Ω)} \\ ⩽ C {(t / 2)}^{- \frac{1}{2} - \frac{N}{2} (\frac{1}{q_{1}} - \frac{1}{p})} {(t / 2)}^{- \frac{N}{2} (\frac{1}{q} - \frac{1}{q_{1}})} ‖ f ‖_{L_{q} (Ω)} \\ = C {(t / 2)}^{- \frac{1}{2} - \frac{N}{2} (\frac{1}{q} - \frac{1}{p})} ‖ f ‖_{L_{q} (Ω)} . \end{array}$ Therefore, we have (12), which completes the proof of Theorem 1.

Finally we discuss the solution formula of the equations: $\begin{matrix} (134) & \{\begin{matrix} \partial_{t} u - J^{- 1} Div S (u, p) = f, & \tilde{div} u = 0 in Ω \times (0, T), \\ \tilde{S} (u, p) n |_{Γ} = 0, & u |_{t = 0} = 0 . \end{matrix} \end{matrix}$ Given $g \in L_{q} {(Ω)}^{N}$ , let $ψ \in {\hat{H}}_{q, 0}^{1} (Ω)$ be a unique solution of the variational problem: $\begin{matrix} {(\tilde{\nabla} ψ, J \tilde{\nabla} φ)}_{Ω} = {(g, J \tilde{\nabla} φ)}_{Ω} for any φ \in {\hat{H}}_{q^{'}, 0}^{1} (Ω) . \end{matrix}$ Let P and Q be operators acting on g defined by $\begin{matrix} P g = g - \tilde{\nabla} ψ, Q g = ψ . \end{matrix}$ We have $\begin{matrix} (135) & \begin{array}{l} P g \in {\tilde{J}}_{q} (Ω), Q g \in {\hat{H}}_{q, 0}^{1} (Ω), \\ g = P g + \tilde{\nabla} Q g, {‖ (P g, \nabla Q g) ‖_{L_{q} (Ω)} ⩽ C ‖ g ‖}_{L_{q} (Ω)} . \end{array} \end{matrix}$ Using P and Q, we rewrite (134) as follows: $\begin{matrix} (136) & \{\begin{matrix} \partial_{t} u - J^{- 1} Div S (u, p + Q f) = P f, & \tilde{div} u = 0 in Ω \times (0, T), \\ \tilde{S} (u, p + Q f) n |_{Γ} = 0, & u |_{t = 0} = 0, \end{matrix} \end{matrix}$ where we have used (124) and $Q f |_{Γ} = 0$ . Since $P f \in {\tilde{J}}_{q} (Ω)$ for any $t > 0$ , we have $\begin{matrix} u = \int_{0}^{t} T (t - s) (P f) (s) d s, p = K (u) - Q f . \end{matrix}$ Summing up, we have

Proposition 21.

Let $N < r < \infty$ be the exponent appearing in ( 3 ) and let $1 < q ⩽ r$ as well as $1 < p < \infty$ . Let $α > 0$ . Then, there exists a $σ > 0$ such that if the assumption ( 3 ) holds, then for any $f \in C^{α} ([0, T), L_{q} {(Ω)}^{N})$ , $\begin{matrix} u = \int_{0}^{t} T (t - s) (P f) (s) d s, p = K (u) - Q f \end{matrix}$ are unique solutions of problem ( 136 ). Here, $C^{α} ([0, T), L_{q} {(Ω)}^{N})$ denotes the set of all $L_{q} {(Ω)}^{N}$ valued Hölder continuous functions of order α defined on $[0, T)$ .

Footnotes

Acknowledgements

Partially supported by JSPS Grant-in-aid for Scientific Research (A) 17H0109 and Top Global University Project.

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