Regularity Estimates for the Oblique Derivative Problem on Non-smooth Domains I

Pengfei Guan^{∗} Department of Mathematics McMaster University Hamilton, Ontario L8S 4K1 Canada

Eric Sawyer^{†} Department of Mathematics and Statistics McMaster University Hamilton, Ontario L8S 4K1 CANADA

1 Introduction

The oblique derivative problem was first posed by Poincaré in 1910 ([Poin]):

\{ \begin{matrix}P u & = & f & in Ω \\ ℓ^{→} u & = & g & on ∂Ω \end{matrix}  (1)

where P is an elliptic second order differential operator on R^{n}, Ω is a bounded domain in R^{n} and ℓ^{→} is a unit vector field on R^{n}. The prototypical example is the Neumann problem where P is the Laplacian △ and ℓ^{→}=n^{→}, the unit outward normal to ∂Ω. In this case the existence, uniqueness and regularity for (1) are explicitly known under very weak smoothness requirements on the structures P, Ω and ℓ^{→}. Roughly speaking, a solution u exists if and only if ∫_{∂Ω} g = ∫_{Ω} f, is unique up to constants, and is 2 degrees smoother than f and 1 degree smoother (plus \frac{1}{p} from the extension to Ω) than g. If ℓ^{→} is everywhere transversal to ∂Ω, then (1) is an elliptic boundary value problem and the existence, uniqueness and regularity results are the same as those for the Neumann problem, except that existence requires finitely many compatibility conditions, and uniqueness holds only up to a finite dimensional space (see [ADN]).

∗Supported in part by NSERC grant OGP0046732 and a Sloan Fellowship.

†Research supported in part by NSERC grant OGP0005149

When ℓ^{→} is permitted to become tangent to ∂Ω, the situation is significantly different. For example, as shown in [Bic], if P = △, Ω is the unit ball in R^{3} and ℓ^{→}= \frac{∂}{∂x}, then any harmonic function u = u (x, y, z) that is independent of x satisfies (1) with f = 0 and g = 0, thus exhibiting an infinite dimensional null space. On the other hand, infinitely many compatibility conditons are required for existence in the same problem but with Ω taken to be the complement of the unit ball. These examples suggest, as established in [Hor] and [EK], that the values of u should be prescribed on the manifold Γ on which ℓ^{→}=T^{→} +a n^{→} changes from pointing into Ω to pointing out of Ω (i.e. a changes sign from − to +) as ℓ^{→} crosses Γ in the direction of T^{→}, and that a jump discontinuity should be accepted on the manifold where a changes sign from + to −. Thus in the event Γ ≠ ϕ, we consider the problem

\{ \begin{matrix}P u & = & f & in Ω \\ ℓ^{→} u & = & g & on ∂Ω \\ u & = & h & on Γ \end{matrix} .  (2)

In 1969, Egorov and Kondrat’ev ([EK]) showed that in the case where the structures P, ∂Ω, ℓ^{→} and Γ are smooth, then u in (2) gains 1 derivative from f and 0 derivatives (plus \frac{1}{p} from the extension to Ω) from g. Moreover, they showed that the gain of 0 from g was sharp, but left open the sharpness of the gain from f. Subsequently, the question of completely characterizing the gain from g was solved in the smooth case. In his book [Ego], Egorov showed that u gains \frac{1}{k+1} derivatives (plus \frac{1}{p} from the extension to Ω) from

g in the L^{p}-Sobolev scale of spaces, if and only if ℓ^{→} is of type k on ∂Ω. As for the gain from f, Guan [Gu] and Smith [Smi] showed in 1990 that u actually has the elliptic gain of 2 derivatives from f in Hölder and L^{p}-Sobolev spaces respectively provided ℓ^{→} is of finite type on ∂Ω. More recently, the elliptic gain of 2 from f in Hölder spaces was extended to general smooth ℓ^{→} by the first author (see below where this is proved for less regular structures). The gain from f was characterized completely in [GS] in 1993. The result is that u gains 2 − ϵ derivatives from f for every ϵ > 0, and achieves the elliptic gain of 2 derivatives from f in the L^{p} Besov-Sobolev scale of spaces if and only if ℓ^{→} is of finite type in ∂Ω \Γ^{∗} and satisfies the A_{p}^{∓}condition on Γ (see subsection 1.1)

This paper is a sequel to [GS], in which we considered (2) for smooth P, ∂Ω, ℓ^{→} and Γ. However, in dealing with nonlinear versions of this problem, it is necesssary to consider the case where P and ℓ^{→} in (2) have limited smoothness. Thus the purpose of this paper is to consider the same problem, but with nonsmooth structure, namely where P has leading coefficients in C^{λ}, Ω has C^{λ+3} boundary ∂Ω, ℓ^{→}=T^{→} +a n^{→} is a C^{λ+2} unit vector field, and Γ is a C^{λ+2} manifold. Of course, the methods used in [GS] in the smooth case carry over for λ sufficiently large depending on the dimension n, but we will obtain optimal results for λ > 0 in any dimension. Here C^{d} denotes the usual Lipschitz space of continuous functions whose derivatives of order [d] are bounded (when d is an integer), or satisfy a Hölder condition of order d−[d] (when d is not an integer). See Theorems 9 and 10 below for applications to the regularity and existence of a nonlinear problem. More general nonlinear oblique derivative problems will be considered in a future paper.

In the case of limited smoothness, the techniques in [GS] cannot be used for the reduction to the boundary nor for the pseudodifferential calculus. For example, in the reduction to the boundary, we use here the solution to the Neumann problem to construct an approximate Poisson operator, and for the pseudodifferential calculus we use the method of symbol splitting. Several new features arise that were not present in our previous paper [GS]. For example:

• The gain from f in the L^{p} Besov-Sobolev scale of spaces can be strictly less than 2, and this is characterized by fractional variations on the A_{p}^{∓} condition.

• The gain from g in the L^{p} Besov-Sobolev scale of spaces takes the form ν + \frac{1}{p}, where ν

can be any number in [0, \frac{1}{3}] ∪ \{\frac{1}{2}\}, and this is characterized by a generalized “type” condition on ∂Ω.

• The gain from h in the L^{p} Besov-Sobolev scale of spaces, which had not been considered at all in [GS], takes the form \frac{γ}{p} + \frac{1}{p}, where γ can be any number in [0, \frac{1}{3}] ∪ \{\frac{1}{2}\}, and this is characterized by a generalized “type” condition near Γ.

• In addition to the L^{p} Besov-Sobolev scale of spaces, we obtain analogous results for the Hölder scale of spaces, Λ^{s} (Ω).

• In the elliptic setting, the solution u to (2) can be taken in Λ^{s} (Ω) for s all the way up to λ + 2, and up to λ + 2 − ϵ in H_{p}^{s} (Ω) (recall the leading coefficients of P are only in C^{λ}). This realization of the smoothness “cap” in the case of limited smoothness is achieved with the aid of symbol smoothing.

• In this paper, we need to construct a left parametrix (in [GS] we used the left parametrix from [MS]).

Our approach to solving (2) can be summed up as:

1. reduction of (2) to a pseudodifferential equation on ∂Ω.

2. construction of a parametrix for the boundary problem and reduction to the operators K, K and T:

\begin{matrix}Kf(x, t) = ∫_{R^{n}} e^{ix·ξ}e^{− ∫_{0}^{t} a(x,θ)Q(x,θ,ξ)dθ}f̂(ξ)dξ, \\ Kf(x, t) = ∫_{R^{n}} e^{ix·ξ} ∫_{0}^{t} e^{− ∫_{t′}^{t} a(x,θ)Q(x,θ,ξ)dθ}f^{∼}(ξ, t′)dt′dξ, \\ T f(x, t) = ∫_{R^{n}} e^{ix·ξ} ∫_{0}^{t} a(x, t′)Q(x, t′, ξ)e^{− ∫_{t′}^{t} a(x,θ)Q(x,θ,ξ)dθ}f^{∼}(ξ, t′)dt′dξ, \end{matrix} (3)

for x ∈ R^{n} and t ∈ (−1, 1).

3. reduction of boundedness properties of T to weighted norm inequalities for the Hardy operator, and corresponding estimates for K and K.

The first two points will be taken up in the next section, while the third point is part of the subject matter of the adjoining paper [GSII], concerned with mapping properties of the special classes of pseudodifferential operators that arise in the oblique derivative problem.

1.1 Statement of theorems

Let H_{p}^{s}(R^{n+2}), B_{p}^{s,p}(R^{n+2}) and Λ^{s} (R^{n+2}) be the L_{p} - Sobolev, Besov and Hölder spaces respectively on R^{n+2}. Let Ω be a connected open set in R^{n+2} with C^{λ+3} boundary ∂ Ω = Σ. Let

P (x, D) = ∑_{i,j=1}^{n+2} a_{ij}(x)\frac{∂}{∂x_{i}} \frac{∂}{∂x_{j}} + ∑_{i=1}^{n} b_{i}(x)\frac{∂}{∂x_{i}} + c(x)

be an elliptic second order real differential operator on R^{n+2} with C^{λ} coefficients, i.e. there are C, δ > 0 such that

C|ξ|^{2} ≥ ∑_{i,j=1}^{n+2} a_{ij}(x)ξ_{i}ξ_{j} ≥ δ|ξ|^{2}, x, ξ ϵ R^{n+2}

where a_{ij}, b_{i} and c are C^{λ} on R^{n+2}. Let ℓ⃗(x) be a real C^{λ+2} unit vector field in R^{n+2} with

ℓ⃗|_{Σ} = T⃗ + a(x)n⃗,

where n⃗ is the unit outward normal to Σ at x and T⃗ is tangential to Σ. Finally let N = {x ∈ Σ : a(x) = 0} ≠ ϕ. We assume throughout the following restriction (R_{int}) on T⃗ :

No integral curve of T⃗ in Σ lies in N for an infinite interval of time.  (4)

Furthermore, we suppose that ℓ⃗ satisfies one of the following two cases:

Case (I) a(x) ≥ 0 for all x ∈ Σ or a(x) ≤ 0 for all x ∈ Σ.

Case (II) a(x) takes on both positive and negative values and there is an n-dimensional C^{λ+2} submanifold Γ (possibly with several components) contained in N and open subsets Σ^{+} and Σ^{−} of Σ such that Σ is a disjoint union of Σ^{+}, Σ^{−} and Γ and

(i) a(x) ≥ 0 for x ∈ Σ^{+}

(ii) a(x) ≤ 0 for x ∈ Σ^{−}

(iii) ∂Σ^{+} = ∂Σ^{−} = Γ

(iv) ℓ⃗ is transversal to Γ and points in the direction of Σ^{+}.

For case (I) we consider the problem

\{ \begin{matrix}P u & = & f & in Ω \\ ℓ^{→} u & = & g & on ∂Ω \end{matrix} ,  (5)

and for case (II) the problem is

\{ \begin{matrix}P u & = & f & in Ω \\ ℓ^{→} u & = & g & on ∂Ω \\ u & = & h & on Γ \end{matrix} .  (6)

We always assume that either u ∈ Λ^{s} (Ω) for s > 1, or that u ∈ H_{p}^{s} (Ω) for s > 1 + \frac{1}{p}, so

that the traces of ℓ^{→} u on ∂Ω and u on Γ exist.

To define the A_{p,α} conditions, we will use the flow for T⃗ on Σ. Given x ∈ Σ, let γ⃗ (x, s) denote the integral curve of T⃗ through x, i.e.

\{ \begin{matrix}\frac{∂}{∂s}γ⃗ (x, s) = T⃗ (γ⃗ (x, s)) , s real, \\ γ⃗ (x, 0) = x. \end{matrix}  (7)

We will also need some more notation. Let y ∈ N. Define I_{y} to be the largest closed interval containing 0 (which may be {0} itself) such that γ⃗ (y, I_{y}) ⊂ N, i.e. a (γ⃗ (y, s)) = 0 for s ∈ I_{y}. Then F_{y} = γ⃗ (y, I_{y}) = {γ⃗ (y, s) : s ∈ I_{y}} is the longest segment of integral curve of T⃗ through y that lies in N and is thus bounded by restriction (R_{int}) (see (4)). We refer to F_{y} as the fibre of N at y. Set Γ^{∗} = ∪_{y∈Γ}F_{y}.

Definition 1 The vector field ℓ⃗ (or equivalently a) satisfies the A_{p,α}^{∓} condition at the fibre F_{y}, y ∈ Γ, if there are constants r > 0, R^{−} < 0 < R^{+}, such that a (γ⃗ (x, R^{−})) ≠ 0 and a (γ⃗ (x, R^{+})) ≠ 0 for x ∈ Γ, |x − y| < r and both of the following conditions hold:

[\frac{1}{∫_{σ}^{β} a (γ⃗ (x, t)) dt} ∫_{σ}^{β} a (γ⃗ (x, t))^{p′} dt]^{p−1} ≤ C \frac{1}{γ − β} [∫_{β}^{γ} a (γ⃗ (x, t)) dt]^{1−pα}  (8)

for all x ∈ Γ, |x − y| < r and all 0 < σ < β < γ < R^{+} with ∫_{σ}^{β} a (γ⃗ (x, t)) dt = ∫_{β}^{γ} a (γ⃗ (x, t)) dt, and also

[\frac{1}{∫_{β}^{γ} |a (γ⃗ (x, t))| dt} ∫_{β}^{γ} |a (γ⃗ (x, t))|^{p′} dt]^{p−1} ≤ C \frac{1}{β − σ} [∫_{σ}^{β} |a (γ⃗ (x, t))| dt]^{1−pα} (9)

for all x ∈ Γ, |x − y| < r and all R^{−} < σ < β < γ < 0 with ∫_{σ}^{β} |a (γ⃗ (x, t))| dt = ∫_{β}^{γ} |a (γ⃗ (x, t))| dt. If the above holds for all y ∈ Γ, we say that ℓ⃗ satisfies the A_{p,α}^{∓} condition on Γ.

Definition 2 The vector field ℓ⃗ (or equivalently a) satisfies the A_{p,α}^{=} condition at the fibre F_{y}, y ∈ (Σ^{+} ∩ N)\Γ^{∗}, if there are constants r > 0, R^{−} < 0 < R^{+}, such that a (γ⃗ (x, R^{−})) ≠ 0 and a (γ⃗ (x, R^{+})) ≠ 0 for x ∈ (Σ^{+} ∩ N) \ Γ^{∗}, |x − y| < r and (8) holds for all x ∈ (Σ^{+} ∩ N) \ Γ^{∗}, |x − y| < r and all R^{−} < σ < β < γ < R^{+} with ∫_{σ}^{β} a (γ⃗ (x, t)) dt = ∫_{β}^{γ} a (γ⃗ (x, t)) dt. If the above holds for all y ∈ (Σ^{+} ∩ N) \ Γ^{∗}, we say that ℓ⃗ satisfies the A_{p,α}^{=} condition on (Σ^{+} ∩ N) \ Γ^{∗}.

Definition 3 The vector field ℓ⃗ (or equivalently a) satisfies the A_{p,α}^{‡} condition at the fibre F_{y}, y ∈ (Σ^{−} ∩ N)\Γ^{∗}, if there are constants r > 0, R^{−} < 0 < R^{+}, such that a (γ⃗ (x, R^{−})) ≠ 0 and a (γ⃗ (x, R^{+})) ≠ 0 for x ∈ (Σ^{−} ∩ N) \ Γ^{∗}, |x − y| < r and (9) holds for all x ∈ (Σ^{−} ∩ N) \ Γ^{∗}, |x − y| < r and all R^{−} < σ < β < γ < R^{+} with ∫_{σ}^{β} |a (γ⃗ (x, t))| dt = ∫_{β}^{γ} |a (γ⃗ (x, t))| dt. If the above holds for all y ∈ (Σ^{−} ∩ N) \ Γ^{∗}, we say that ℓ⃗ satisfies the A_{p,α}^{‡} condition on (Σ^{−} ∩ N) \ Γ^{∗}.

Definition 4 The vector field ℓ⃗ (or equivalently a) satisfies the (T_{ν}) condition on Σ if

\begin{matrix}β − α ≤ C (∫_{α}^{β} |a (γ⃗ (x, t))| dt)^{ν} \\ for all α < β, and x ∈ Σ. \end{matrix}

Definition 5 The vector field ℓ⃗ (or equivalently a) satisfies the (P_{γ}) condition on Γ if

\begin{matrix}|β| ≤ C |∫_{0}^{β} a (γ⃗ (x, t)) dt|^{γ} \\ for all β ∈ R, and x ∈ Γ. \end{matrix}

Remark We remark that for the purposes of the above five definitions, the family of integral curves γ⃗ (x, t) can be replaced by any smooth family γ⃗ (x, t) satisfying the following condition, weaker than (7):

\{ \begin{matrix}∥\frac{∂}{∂s}γ⃗ (x, t) − T⃗ (γ⃗ (x, t))∥ ≤ C |a (x, t)| , \\ γ⃗ (x, 0) = x. \end{matrix}  (10)

This is a simple consequence of Lemma 6.38 of [GS], which required only that the curves be C^{1} and that the vector fields be Lipschitz.

We can now state our main theorems. Fix 0 < μ_{∞} < min \{1 − δ, \frac{λ+2}{2(λ+3)}\} and set

μ_{p} = min \{1 − δ − \frac{1}{p(λ+3)}, \frac{λ+2}{2(λ+3)}\} for 1 < p < ∞ where δ = max \{\frac{1}{2}, \frac{1}{λ+1}\}. We point out that the choice of δ is dictated by (25) below, and that μ_{p} arises in Lemma 19 in [GSII].

Theorem 6 Suppose that P, Ω and ℓ⃗ are as at the beginning of this subsection with λ > 0.

(i) If ℓ⃗ satisfies case (I) and the T_{ν} condition on Σ for some ν ≥ 0, then for all 1 < s ≤ λ+2, there is a subspace F of finite codimension in Λ^{s−2}(Ω)× Λ^{s−ν}(∂Ω) such that for (f, g) ∈ F, there exists u satisfying (5) and u ∈ Λ^{s}(Ω). Moreover, if u ∈ Λ^{s′}(Ω) for some s′ > 1 satisfies (5) with f, g as above, then u ∈ Λ^{s}(Ω), and there is C_{s} such that

∥u∥_{Λ^{s}(Ω)} ≤ C_{s} (∥f∥_{Λ^{s−2}(Ω)} + ∥g∥_{Λ^{s−ν}(∂Ω)} + ∥u∥_{Λ^{s−μ∞}(Ω)}) .

Furthermore, if the zero order term c (x) in the operator P is negative, then for every (f, g) ∈ Λ^{s−2}(Ω)× Λ^{s−ν}(∂Ω), there is a unique solution u ∈ Λ^{s}(Ω). (ii) If ℓ⃗ satisfies case (II) and the T_{ν} condition on Σ for some ν ≥ 0, then for all 1 < s ≤ λ + 2, there is a subspace F of finite codimension in Λ^{s−2}(Ω)× Λ^{s−ν}(∂Ω)× Λ^{s}(Γ) such that for (f, g, h) ∈ F, there exists u satisfying (6) and u ∈ Λ^{s}(Ω). Moreover, if u ∈ Λ^{s′}(Ω) for some s′ > 1 satisfies (6) with f, g, h as above, then u ∈ Λ^{s}(Ω), and there is C_{s} such that

∥u∥_{Λ^{s}(Ω)} ≤ C_{s} (∥f∥_{Λ^{s−2}(Ω)} + ∥g∥_{Λ^{s−ν}(∂Ω)} + ∥h∥_{Λ^{s}(Γ)} + ∥u∥_{Λ^{s−μ∞}(Ω)}) .

Furthermore, if the zero order term c (x) in the operator P is nonpositive, then for every (f, g, h) ∈ Λ^{s−2}(Ω)× Λ^{s−ν}(∂Ω)× Λ^{s}(Γ), there is a unique solution u ∈ Λ^{s}(Ω).

At this point we could immediately obtain a regularity theorem for a semilinear oblique derivative problem. However, in order to handle a fully nonlinear equation with semilinear oblique derivative in the next subsection, we need the following result which establishes an additional gain from g when it is multiplied by a tangential derivative of a.

Theorem 7 Suppose that P, Ω, and ℓ⃗ are as at the beginning of this subsection with λ > 0, and let D = ∑_{k=1}^{n+2} b_{k} (x) \frac{∂}{∂x_{k}} ∈ C^{λ+2} (\overline{Ω}) be a vector field tangent to ∂Ω for x ∈ ∂Ω. If for some s > 0, u ∈ Λ^{s}(Ω) and satisfies

\{ \begin{matrix}P u & = & 0 & in Ω \\ ℓ^{→} u & = & (Da) g_{1} & on ∂Ω \end{matrix}  (11)

in case (I) and

\{ \begin{matrix}P u & = & 0 & in Ω \\ ℓ^{→} u & = & (Da) g_{1} & on ∂Ω \\ u & = & 0 & on Γ \end{matrix}  (12)

in case (II) with g_{1} ∈ Λ^{λ− \frac{1}{4} +1} (∂Ω) as above, then u ∈ Λ^{λ+1}(Ω), and there is C_{λ} such that

∥u∥_{Λ^{λ+1}(Ω)} ≤ C_{λ} (∥g_{1}∥_{Λ^{λ− \frac{1}{4} +1}(∂Ω)} + ∥u∥_{Λ^{λ+1−μ∞}(Ω)}) .

Now we state the Sobolev space analogue.

Theorem 8 Let 1 < p < ∞ and α, ν ≥ 0. Suppose that P, Ω, and ℓ⃗ are as at the beginning of this subsection with \frac{λ}{λ+1} > α. (A) Suppose ℓ⃗ satisfies the A_{p,α}^{∓} condition on Γ, A_{p,α}^{=} on (Σ^{+} ∩ N)\Γ^{∗}, A_{p,α}^{‡} on (Σ^{−} ∩ N)\ Γ^{∗}, and also the T_{ν} condition on Σ. Then:

(i) Assume ℓ⃗ satisfies case (I) above, and let 1 + \frac{1}{p} < s < λ + 2 − α. There is a subspace F

of finite codimension in H_{p}^{s−2+α}(Ω)× B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω) such that for (f, g) ∈ F, there exists u satisfying (5) and u ∈ H_{p}^{s}(Ω). Moreover, if u ∈ H_{p}^{s′}(Ω) for some s′ > 1 + \frac{1}{p} satisfies (5) with f, g as above, then u ∈ H_{p}^{s}(Ω), and there is C_{s} such that

∥u∥_{H_{p}^{s}(Ω)} ≤ C_{s} (∥f∥_{H_{p}^{s−2+α}(Ω)} + ∥g∥_{B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω)} + ∥u∥_{H_{p}^{s−μp}(Ω)}) .

Furthermore, if the zero order term c (x) in the operator P is negative, then for every

(f, g) ∈ H_{p}^{s−2+α}(Ω)× B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω), there is a unique solution u ∈ Λ^{s}(Ω). (ii) Assume ℓ⃗ satisfies case (II) above and in addition, the P_{γ} condition on Γ for some γ ≥ 0, and let 1 + \frac{1}{p} < s < λ + 2 − α. There is a subspace F of finite codimension

in H_{p}^{s−2+α}(Ω)× B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω)× B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ) such that for (f, g, h) ∈ F, there exists u satisfying (6) and u ∈ H_{p}^{s}(Ω). Moreover, if u ∈ H_{p}^{s′}(Ω) for some s′ > 1 + \frac{1}{p} satisfies (6) with f, g, h as above, then u ∈ H_{p}^{s}(Ω), and there is C_{s,α} such that

∥u∥_{H_{p}^{s}(Ω)} ≤ C_{s,α} (∥f∥_{H_{p}^{s−2+α}(Ω)} + ∥g∥_{B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω)} + ∥h∥_{B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ)} + ∥u∥_{H_{p}^{s−μp}(Ω)}) .

Furthermore, if the zero order term c (x) in the operator P is nonpositive, then for every

(f, g, h) ∈ H_{p}^{s−2+α}(Ω)× B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω)× B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ), there is a unique solution u ∈ Λ^{s}(Ω). Finally, the above statements are true when α = \frac{1}{p(λ+3)}, since in this case the A_{p,α}^{∓}, A_{p,α}^{=} and A_{p,α}^{‡} conditions automatically follow from the smoothness of a. (B) Conversely, suppose ℓ⃗ satisfies case I or case II and λ > 0 satisies μ_{p} > 0. Then:

(i) If for some 1 + \frac{1}{p} < s < λ + 2 − α, α ≥ 0, ϵ > 0 and for every f ∈ H_{p}^{s−2+α}(Ω), there is an approximate solution u ∈ H_{p}^{s}(Ω) (with g = 0, h = 0), i.e. such that

∥P u − f∥_{H_{p}^{s−2+α+ϵ}(Ω)} ≤ C_{s,ϵ}∥f∥_{H_{p}^{s−2+α}(Ω)},
∥ℓ⃗u∥_{B_{p}^{s− \frac{1}{p} ,p}(∂Ω)}  ≤ C_{s} (∥f∥_{H_{p}^{s−2+α}(Ω)} + ∥u∥_{H_{p}^{s−1}(Ω)}) ,
∥u∥_{B_{p}^{s− \frac{1}{p} ,p}(Γ)}  ≤ C_{s} (∥f∥_{H_{p}^{s−2+α}(Ω)} + ∥u∥_{H_{p}^{s−1}(Ω)}) ,

then ℓ⃗ satisfies the A_{p,α}^{∓} condition on Γ, A_{p,α}^{=} on (Σ^{+} ∩ N)\Γ^{∗}, and A_{p,α}^{‡} on (Σ^{−} ∩ N)\Γ^{∗}.

(ii) If for some 1 + \frac{1}{p} < s < λ + 2 − α, ν ≥ 0, ϵ > 0 and for every g ∈ B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω), there is an approximate solution u ∈ H_{p}^{s}(Ω) (with f = 0, h = 0), i.e. such that

∥P u∥_{H_{p}^{s−2+ \frac{1}{p(λ+3)}} (Ω)}  ≤ C_{s} (∥g∥_{B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω)} + ∥u∥_{H_{p}^{s−1}(Ω)}) ,
∥ℓ⃗u − g∥_{B_{p}^{s−ν− \frac{1}{p} +ϵ,p}(∂Ω)} ≤ C_{s,ϵ}∥g∥_{B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω)},
∥u∥_{B_{p}^{s− \frac{1}{p} ,p}(Γ)}  ≤ C_{s} (∥g∥_{B_{p}^{s−ν− \frac{1}{p} ,p}(∂Ω)} + ∥u∥_{H_{p}^{s−1}(Ω)}) ,

then ℓ⃗ satisfies the T_{ν} condition on Σ.

(iii) If for some 1 + \frac{1}{p} < s < λ + 2 − α, γ ≥ 0, ϵ > 0 and for every h ∈ B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ), there is an approximate solution u ∈ H_{p}^{s}(Ω) (with f = 0, g = 0), i.e. such that

∥P u∥_{H_{p}^{s−2+ \frac{1}{p(λ+3)}} (Ω)} ≤ C_{s} (∥h∥_{B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ)} + ∥u∥_{H_{p}^{s−1}(Ω)}) ,
∥ℓ⃗u∥_{B_{p}^{s− \frac{1}{p} ,p}(∂Ω)}  ≤ C_{s} (∥h∥_{B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ)} + ∥u∥_{H_{p}^{s−1}(Ω)}) ,
∥u − h∥_{B_{p}^{s− \frac{γ}{p} − \frac{1}{p} +ϵ,p}(Γ)} ≤ C_{s,ϵ}∥h∥_{B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ)},

then ℓ⃗ satisfies the P_{γ} condition on Γ.

Remark 1 In part (A) of Theorem 8, we can weaken the requirement μ_{p} > 0 to simply 1 − δ − α > 0 - see the remark following Lemma 19 in [GSII].

Remark 2 In Theorems 6 and 8, the assumptions that ℓ⃗, Γ are C^{λ+2} and ∂Ω is C^{λ+3} can all be relaxed somewhat. More precisely, if we assume that ℓ⃗, Γ and ∂Ω are C^{λ′} with λ′ + \frac{1}{p} ≤ λ + 2 − α, and assume that there is a C^{λ′} flow for T⃗ in a neighbourhood of N (the assumption above that ∂Ω be in C^{λ+3} is used only to guarantee the existence of a C^{λ+2} flow), then Theorem 6 remains true for 1 < s ≤ λ′, and Theorem 8 remains true for 1 + \frac{1}{p} < s < λ′, but with the gain of \frac{1}{p(λ+3)} from f replaced by \frac{1}{p(λ′+1)}. The

proofs are the same. Furthermore, the above smoothness assumptions on ℓ⃗, Γ and g are needed merely in a neighbourhod of N. Away from N, only the usual elliptic smoothness is required.

Remark 3 If ℓ⃗ satisfies the A_{p}^{∓} condition on Γ, etc. as in part (A) of Theorem 8, and if N = Γ, then the a priori inequalities in part (A) hold with s = 2, α = 0 and μ_{p} = \frac{1}{2} provided P ∈ C^{0}, and ℓ⃗,Γ, ∂Ω ∈ C^{2}. This can be proved by locally approximating P by constant coefficient elliptic operators and using α = 0.

1.2 Applications to nonlinear problems

Here we give two nonlinear applications of Theorem 6 - first to the regularity of a fully nonlinear elliptic equation with a semilinear oblique derivative. In particular, we show that if the data and structures are all smooth, then so are the solutions.

Theorem 9 Let P (x, u, ξ, ζ) be a function of the variables (x, u, ξ, ζ) ∈ R^{n+2} ×R×R^{n+2} × R^{(n+2)^{2}}, g be a function on ∂Ω, and h (x, u) be a function of the variables (x, u) ∈ R^{n+2}×R. Suppose that P, ∂_{ζ}P ∈ C^{λ} and g, h ∈ C^{λ+2} for some noninteger λ > 0. Suppose further that u ∈ Λ^{s} (\overline{Ω}), s > 2, is such that both ∂_{ζ}P and ∂_{u}h are elliptic, i.e. for some C, c > 0,

\begin{matrix}C|η|^{2} ≥ ∑_{i,j=1}^{n+2} \frac{∂}{∂ζ_{ij}} P (x, u, ∇u, ∇^{2}u) η_{i}η_{j} ≥ c|η|^{2}, & x, η ∈ R^{n+2}, \\ C ≥ ∂_{u}h (x, u) ≥ c, & x ∈ R^{n+2}. \end{matrix}

(i) If ℓ⃗ satisfies case (I) above, and if in addition u satisfies

\{ \begin{matrix}P (x, u, ∇u, ∇^{2}u) & = & 0 & in Ω \\ ℓ^{→} u & = & g (x) & on ∂Ω \end{matrix} ,

then u ∈ Λ^{λ+2} (\overline{Ω}). (ii) If ℓ⃗ satisfies case (II) above, and if in addition u satisfies

\{ \begin{matrix}P (x, u, ∇u, ∇^{2}u) & = & 0 & in Ω \\ ℓ^{→} u & = & g (x) & on ∂Ω \\ h (x, u) & = & 0 & on Γ \end{matrix} ,

then u ∈ Λ^{λ+2} (\overline{Ω}). In particular, if the functions P, g, h and the manifolds Γ, ∂Ω and the vector field ℓ⃗ are all smooth, and if u is a Λ^{s} solution to (i) or (ii) as above with s > 2, then u is also smooth. Finally, if P is linear in the second order derivatives, i.e.

P u = ∑_{i,j=1}^{n+2} a_{ij} (x, u, ∇u) \frac{∂^{2}u}{∂x_{i}∂x_{j}} + c (x, u, ∇u) ,

then the above conclusions follow from the weaker assumption u ∈ Λ^{s} (\overline{Ω}), s > 1.

Proof: We prove only case (II), the other case being similar but simpler. So suppose u ∈ Λ^{s} (\overline{Ω}) with 2 < s < s + \frac{1}{4} ≤ λ + 2. We will show that u ∈ Λ^{s+ \frac{1}{4}} (\overline{Ω}). For this, let D = ∑_{k=1}^{n+2} b_{k} (x) \frac{∂}{∂x_{k}} ∈ C^{λ+2} (\overline{Ω}) be a vector field tangent to ∂Ω for x ∈ ∂Ω, and commuting with T^{→}, i.e.[T^{→}, D] = 0. Applying D to the nonlinear equation above we obtain

\{ \begin{matrix}∑_{i,j=1}^{n+2} \frac{∂}{∂ζ_{ij}} P (x, u, ∇u, ∇^{2}u) \frac{∂^{2}(Du)}{∂x_{i}∂x_{j}} & = & −DP (x, u, ∇u, ∇^{2}u) − ∂_{u}P (x, u, ∇u, ∇^{2}u) Du \\ & & − ∑_{k=1}^{n+2} \frac{∂}{∂ξ_{k}} P (x, u, ∇u, ∇^{2}u) D (\frac{∂}{∂x_{k}} u) \\ & & + ∑_{i,j=1}^{n+2} \frac{∂}{∂ζ_{ij}} P (x, u, ∇u, ∇^{2}u) [\frac{∂^{2}}{∂x_{i}∂x_{j}} , D] u & in Ω \\ ℓ^{→} (Du) & = & g (x) + [ ℓ^{→}, D] u & on ∂Ω \\ ∂_{u}h (x, u) (Du) & = & −Dh (x, u) & on Γ \end{matrix} .

Since [ ℓ^{→}, D] u = [an⃗, D] = −Da (x) n⃗u, this can be rewritten as

\{ \begin{matrix}∑_{i,j=1}^{n+2} a_{ij} (x) \frac{∂^{2}}{∂x_{i}∂x_{j}} (Du) & = & f (x) & in Ω \\ ℓ^{→} (Du) & = & g (x) + Da (x) g_{1} (x) & on ∂Ω \\ Du & = & h (x) & on Γ \end{matrix} ,

where a_{ij} = \frac{∂}{∂ζ_{ij}} P (x, u, ∇u, ∇^{2}u) ∈ Λ^{s−2} (Ω), f ∈ Λ^{s−2} (Ω), g ∈ Λ^{λ+2} (∂Ω), g_{1} = n⃗u ∈

Λ^{s−1} (∂Ω) and h = −\frac{Dh(x,u)}{∂_{u}h(x,u)} ∈ Λ^{min\{λ+1,s\}} ⊂ Λ^{s−1+ \frac{1}{4}} (Γ). By Theorem 6, there is w ∈

Λ^{s−1+ \frac{1}{4}} (\overline{Ω}) such that

\{ \begin{matrix}∑_{i,j=1}^{n+2} a_{ij} (x) \frac{∂^{2}}{∂x_{i}∂x_{j}} w & = & f (x) & in Ω \\ ℓ^{→} w & = & g (x) & on ∂Ω \\ w & = & h (x) & on Γ \end{matrix} ,

and by Theorem 7, Du − w ∈ Λ^{s−1+ \frac{1}{4}} (\overline{Ω}). Thus Du ∈ Λ^{s−1+ \frac{1}{4}} (\overline{Ω}). To obtain u ∈

Λ^{s+ \frac{1}{4}} (\overline{Ω}), we note that we have proved ∇Du ∈ Λ^{s−2+ \frac{1}{4}} (\overline{Ω}) for any tangential derivative D, and now using P (x, u, ∇u, ∇^{2}u) = 0 together with the implicit function theorem, we obtain ∇^{2}u ∈ Λ^{s−2+ \frac{1}{4}} (\overline{Ω}), and hence u ∈ Λ^{s+ \frac{1}{4}} (\overline{Ω}) as required. Finally, if P is linear in the second order derivatives, we needn’t differentiate the problem, but can immediately apply Theorem 6 provided s > 1.

Our second application of Theorem 6 is the following nonlinear existence theorem for small data.

Theorem 10 Let Ω and ℓ^{→} be as in Theorem 6. Suppose that λ > 0 is nonintegral and that

\begin{matrix}P (x, u, ξ, ζ) ∈ C^{λ} (\overline{Ω} × R × R^{n+2} × R^{(n+2)^{2}}) , \\ ∂_{u}P, ∂_{ξ}P, ∂_{ζ}P ∈ C^{λ} (\overline{Ω} × R × R^{n+2} × R^{(n+2)^{2}}) . \end{matrix}

Moreover, suppose P (x, 0, 0, 0) = 0 and that there is C > 0 such that

\begin{matrix}C|η|^{2} ≥ ∑_{i,j=1}^{n+2} \frac{∂}{∂ζ_{ij}} P (x, 0, 0, 0) η_{i}η_{j} ≥ C^{−1}|η|^{2}, \\ −C ≤ ∂_{u}P (x, 0, 0, 0) ≤ −C^{−1}, \end{matrix} for all x ∈ \overline{Ω}.

(i) If ℓ^{→} satisfies case (I), then there is c_{0} > 0 such that if ∥f∥_{Λ^{λ}(Ω)} and ∥g∥_{Λ^{λ+2}(∂Ω)} ≤ c_{0}, then there is u ∈ Λ^{λ+2} (Ω) satisfying

\{ \begin{matrix}P (x, u, ∇u, ∇^{2}u) & = & f (x) & in Ω \\ ℓ^{→} u & = & g & on ∂Ω \end{matrix} .

(ii) If ℓ^{→} satisfies case (II), then there is c_{0} > 0 such that if ∥f∥_{Λ^{λ}(Ω)}, ∥g∥_{Λ^{λ+2}(∂Ω)} and ∥h∥_{Λ^{λ+2}(Γ)} ≤ c_{0}, then there is u ∈ Λ^{λ+2} (Ω) satisfying

\{ \begin{matrix}P (x, u, ∇u, ∇^{2}u) & = & f (x) & in Ω \\ ℓ^{→} u & = & g & on ∂Ω \\ u & = & h & on Γ \end{matrix} .

Proof: We prove only part (ii), the first part being similar. Let

X = {w : w ∈ Λ^{λ+2} (Ω) and ℓ^{→} w ∈ Λ^{λ+2} (∂Ω)} ,

Y = Λ^{λ} (Ω) × Λ^{λ+2} (∂Ω) × Λ^{λ+2} (Γ) .

Then X and Y are Banach spaces with the obvious norms. Consider the map F : X → Y given by

F (w) = (P (x, w, ∇w, ∇^{2}w) , ℓ^{→} w, w |_{Γ}) .

By our hypotheses, F (0) = 0 and F has continuous first Fréchet derivatives. Furthermore, since ∂_{u}P (x, 0, 0, 0) is negative, Theorem 6 implies that F′ (0) is invertible. Thus the implicit function theorem (see e.g. [Sch]) shows that F is an open map near w = 0, and this completes the proof of the theorem.

In the next section, we reduce (2) to a pseudodifferential equation on the boundary, and then use our results on a special class of pseudodifferential operators (see the following paper [GSII] in this journal) to prove Theorems 6, 7 and 8.

2 Transference of the oblique derivative problem to the boundary and proofs of theorems

In this section we reduce the oblique derivative problem to the boundary using the solution to the Neumann problem to consttruct an approximate Poisson operator. This is achieved largely via the method of symbol splitting. For example, if σ ∈ C^{λ}S_{1,δ}^{m}, then σ = σ^{♯} + σ^{♭} where σ^{♯} ∈ S_{1,γ}^{m} is smooth and σ^{♭} ∈ C^{λ}S_{1,γ}^{m−λ(γ−δ)} has better order. See Proposition 4 in [GSII]. In particular, this applies to the operator of multiplication by the function a (x), and we will repeatedly interchange a and a^{♯} whenever convenient. We will also make use of the boundedness results for special operators in [GSII].

2.1 The Neumann problem and a change of variables

We study the oblique derivative problem by reducing it to a pseudodifferential equation on the boundary. In an earlier paper (see section 3 of [GS]), we dealt with this problem in the case of smooth data, following the line of argument in [BFG]. In order to simplify the equation, we straightened the normal direction with respect to (g_{i,j}) (where P = ∑_{i,j} g_{i,j} \frac{∂}{∂_{x_{i}}} \frac{∂}{∂_{x_{j}}}

+ lower order terms) and then the tangential direction T^{→} to \frac{∂}{∂t} (where ℓ^{→}=T^{→} +a n^{→}). We shall do the same here, but since the structures are no longer smooth, the process must be modified.

First, in order to simplify the computation of ℓ^{→} on the boundary, we pick a constant c such that the Neumann problem

\{ \begin{matrix}P F + cF = & f & on Ω \\ n^{→} F = & 0 & on ∂Ω \end{matrix}

has a unique solution F, which we denote by F = Nf. Then, by the theory of elliptic boundary value problems, N : Λ^{s}(Ω) → Λ^{s+2}(Ω) is bounded for 0 ≤ s ≤ λ, and N : H_{p}^{s}(Ω) → H_{p}^{s+2}(Ω) is bounded for 0 ≤ s < λ. With v = u − Nf, the oblique derivative problem reduces to

\{ \begin{matrix}P v = & cNf & on Ω \\ ℓ^{→} v = & g + (Da) g_{1}− T^{→} Nf & on ∂Ω \end{matrix}  (13)

or

\{ \begin{matrix}P v = & cNf & on Ω \\ ℓ^{→} v = & g + (Da) g_{1}− T^{→} Nf & on ∂Ω \\ v = & h − Nf & on Γ. \end{matrix}  (14)

Since N : Λ^{s}(Ω) → Λ^{s}(Ω) is compact, our main theorem will follow if we can solve the following equations with the appropriate estimates:

\{ \begin{matrix}P v = & 0 & on Ω \\ ℓ^{→} v = & g + (Da) g_{1}− T^{→} Nf & on ∂Ω \end{matrix}  (15)

and

\{ \begin{matrix}P v = & 0 & on Ω \\ ℓ^{→} v = & g + (Da) g_{1}− T^{→} Nf & on ∂Ω \\ v = & h − Nf & on Γ \end{matrix} .  (16)

As in section 3 of [GS], we now change variables repeatedly to simplify calculations. As indicated there, we want to choose the coordinate charts such that a(x, t) ≠ 0 on the top and bottom of the t-interval (where \frac{∂}{∂t} = T^{→}) to ensure that singularities won’t propagate out of the local charts.

Now ℓ^{→}=T^{→} +a n^{→}. If a ≠ 0 on the boundary, the problem is elliptic, so the result is well-known. Suppose N = {a = 0} ≠ ϕ. Fix p ∈ N, and let γ be the integral curve of the vector field T^{→} through p with γ(0) = p. Then γ is C^{λ+2} and, by the finite length restriction (4), there are s_{1} < 0 < s_{2} such that | a(γ(s_{1})) |> 0, | a(γ(s_{2})) |> 0 and γ((s_{1}, s_{2})) is not self-intersecting. Here, s_{1}, s_{2} can be chosen so that a(γ(s)) is as small as we wish for s_{1} <s < s_{2} and so that either γ((s_{1}, s_{2}))∩Γ = ∅ or γ((s_{1}, s_{2}))∩Γ consists of a single point q and the segment of γ between p and q is completely in N . (Note that γ can’t intersect Γ in more than one point since once γ is in ∑^{+} (resp. ∑^{−}), it must remain there for all future time by assumptions (iii) and (iv) in case (II) of section 1.1. Suppose γ((s_{1}, s_{2})) ∩ Γ = {q}. Now T^{→} is transversal to Γ at q, and so by the theory of ODE, there is a small neighborhood V of q in ∂Ω, and t_{1} < 0 < t_{2} and a C^{λ+2} diffeomorphism Φ : I^{n} × (t_{1}, t_{2}) ⟶ V, such that Φ(x, ·) are the integral curves of the vector field T^{→} and Φ(0, 0) = q, Φ(·, 0) ⊂ Γ, | a(Φ(·, t_{1})) |> 0, | a((Φ (·, t_{2})) |> 0 . So, p ∈ V automatically as Φ(0, s) passes through p. On the other hand, if γ((s_{1}, s_{2})) ∩ Γ = ϕ, choose any hypersurface H of ∂Ω through p at which T^{→} is transversal. We can then find V and Φ as above with Φ(·, 0) ⊂ H,but V ∩ Γ = ∅.

By introducing geodesic normal coordinates with respect to ∂Ω, and denoting by r the normal variable, then for any local coordinates y_{1}, ..., y_{n+1} in ∂Ω, in particular for the charts (Φ, V) constructed above, we have

P = α\frac{∂^{2}}{∂r^{2}} + n+1 ; ∑_{k=1} b_{k} \frac{∂}{∂y_{k}} \frac{∂}{∂r} + n+1 ; ∑_{i,j=1} g_{ij} \frac{∂}{∂y_{i}} \frac{∂}{∂y_{j}} + lower order terms.  (17)

In summary, for every p ∈ ∂Ω, if a(p) = 0, we can find a neighborhood U of p in the boundary which has the following property:

\begin{matrix}U = I^{n} × I, I = (−1, 1), | a(x, −1) |≠ 0, | a(x, 1) |≠ 0, ∀x ∈ I^{n,}, \\ either U ∩ Γ = ∅ or U ∩ Γ = \{t = 0\}. \end{matrix}

In the above coordinate chart, P satisfies (17) with α > 0, and ℓ^{→}= \frac{∂}{∂t} −a\frac{∂}{∂r} . Furthermore,

the coefficients in P and ℓ^{→} have the same degree of smoothness (C^{λ} and C^{λ+2} respectively) as those in the original variables by the theory of ODE. By shrinking U if necessary, we may assume that a is as small as we wish on U. For points p ∈ ∂Ω, a(p) ≠ 0, we simply choose a neighbourhood U such that a ≠ 0 on Ū. Since ∂Ω is compact, there is finite collection U_{1}, ..., U_{k} of neighbourhoods as above covering ∂Ω such that each U_{j} either has the property (*) or a ≠ 0 on Ū_{j} . Extend U_{i} to be an open set in R^{n+2} such that U_{j} = I^{n} × I × I_{ϵ}, I_{ϵ} = (−ϵ, ϵ), U_{j} ∩ ∂Ω = I^{n} × I × {0}, and U_{j} ∩ Ω = {r > 0}. Let U_{0} = {r > δ }, so that if δ

is small enough, then ∪_{j=0}^{k} U_{j} ⊃ Ω̄.

2.2 Pseudodifferential equations on the boundary

On those U (we drop the index j for the convenience) with property (*), equations (15) and (16) now become

\{ \begin{matrix}P v & = & 0 & in & r > 0 \\ ℓ^{→} v & = & g̃ + (Da) g_{1}− T^{→} F & on & r = 0 \end{matrix}  (18)

and

\{ \begin{matrix}P v & = & 0 & in & r > 0 \\ ℓ^{→} v & = & g̃ + (Da) g_{1}− T^{→} F & on & r = 0 \\ v & = & h − F & on & t = r = 0. \end{matrix}  (19)

where

\begin{matrix}P = & \frac{∂^{2}}{∂r^{2}} + n+1 ; ∑_{k=1} b_{k} \frac{∂}{∂y_{k}} \frac{∂}{∂r} + n+1 ; ∑_{i,j=1} g_{ij} \frac{∂}{∂y_{i}} \frac{∂}{∂y_{j}} + lower order terms, \\ ℓ^{→}= & \frac{∂}{∂t} − a\frac{∂}{∂r} . \end{matrix}

Now, if w_{0} = w |_{r=0} and F ≠ f |_{∂Ω}, then w is uniquely determined by w_{0} via the Poisson operator Ẽ of P, namely w = Ẽw_{0}. Therefore, to solve (18) and (19), we need only calculate ℓ^{→} Ẽw_{0} and then solve for w_{0} on the boundary. The calculation is standard if the coefficients of P are smooth (see e.g. [BFG]), but otherwise some care must be exercised. Here, Ẽ is defined by the theory of the Dirichlet problem for elliptic operators (see e.g. [ADN]), but we will instead use an approximate version E defined as follows.

Write

\begin{aligned}P & = (\frac{∂^{2}}{∂r^{2}} + 2B(x, D_{x})\frac{∂}{∂r} + P_{1}(x, D_{x})) \\ & +B′(r, x, D_{x})\frac{∂}{∂r} + P′_{1}(r, x, D_{x}) + lower order terms, \end{aligned}

where B′(0, x, D_{x}) = P′_{1}(0, x, D_{x}) = 0. Now set δ = max \{\frac{1}{λ+1}, \frac{1}{2}\}. Using the symbol splitting in Proposition 4 in [GSII], we define an approximate Poisson operator by

(Eu_{0}) (r, x) = ∫ e^{ix·ξ}e^{−r[\sqrt{−P_{1}^{♯}(x,iξ)+B^{♯}(x,iξ)^{2}}+B^{♯}(x,iξ)]}û_{0}(ξ)dξ  (20)

where ξ = (ξ_{1}, ..., ξ_{n+1}) is the dual variable of x = (x_{1}, ..., x_{n+1}), and

B = B^{♯} + B^{♭}, P_{1} = P_{1}^{♯} + P^{♭},  (21)

with

B^{♯} = i ∑_{k=1}^{n+1} b_{k}^{♯} (x, ξ) ξ_{k}, P_{1}^{♯} = − ∑_{i,j=1}^{n+1} g_{i,j}^{♯} (x, ξ) ξ_{i}ξ_{j},  (22)

where

\begin{matrix}b_{k}^{♯}, g_{i,j}^{♯} ∈ S_{1,δ}^{0}, \\ b_{k}^{♭} = b_{k} − b_{k}^{♯}, g_{i,j}^{♭} = g_{i,j} − g_{i,j}^{♯} ∈ C^{λ}S_{1,δ}^{−λδ}, \end{matrix}

and where

g̃_{i,j}^{♯} = g_{i,j}^{♯} − b_{i}^{♯}b_{j}^{♯}  (23)

is positive definite for |ξ| large.

Since E is a Poisson integral, we have that E : Λ^{s}(∂Ω) → Λ^{s}(Ω) for 0 < s ≤ λ and

E : B_{p}^{s− \frac{1}{p} ,p}(∂Ω) → H_{p}^{s}(Ω) for 0 ≤ s < λ. We now compute P E:

P E = (\frac{∂^{2}}{∂r^{2}} + 2B(x, D_{x})\frac{∂}{∂r} + P_{1}(x, D_{x})) E
+B′(r, x, D_{x})\frac{∂}{∂r} E + P′_{1}(r, x, D_{x})E + lower order terms.
= (\frac{∂^{2}}{∂r^{2}} + 2B^{♯}(x, D_{x})\frac{∂}{∂r} + P_{1}^{♯}(x, D_{x})) E + B′(r, x, D_{x})\frac{∂}{∂r} E + P′_{1}(r, x, D_{x})E
+2 (B^{♭}(x, D_{x})\frac{∂}{∂r} + P_{1}^{♭}(x, D_{x})) E + lower order terms.

(24)

Since B′(0, x, D_{x}) = P′_{1}(0, x, D_{x}) = 0, the second, third and last terms in (24) are bounded

operators Λ^{s}(∂Ω) → Λ^{s−min\{λ,1\}}(Ω) for 0 < s ≤ λ + 1, B_{p}^{s− \frac{1}{p} ,p}(∂Ω) → H_{p}^{s−min\{λ,1\}}(Ω) for \frac{1}{p} < s < λ + 1. A direct calculation shows that the first term, (\frac{∂^{2}}{∂r^{2}} + 2B^{♯} \frac{∂}{∂r} + P_{1}^{♯}) E, is a

combination of bounded operators Λ^{s+1+δ}(∂Ω) → Λ^{s}(Ω) for 0 < s ≤ λ, B_{p}^{s− \frac{1}{p} +1+δ,p}(∂Ω) → H_{p}^{s}(Ω) for 0 ≤ s < λ. Also g_{i,j}^{♭}∂_{i}∂_{j} ∘ E maps Λ^{s+2−δλ}(∂Ω) → Λ^{s}(Ω), for 0 < s ≤ λ, and

B_{p}^{s+2−δλ− \frac{1}{p}} (∂Ω) → H_{p}^{s}(Ω), for 0 ≤ s < λ, 1 < p < ∞. Thus these error terms have order 1 + δ and 2 − δλ respectively. Note that our choice of δ equalizes these orders for λ ≤ 1. Summarizing,

P E = B_{1},  (25)

where

\begin{matrix}B_{1} : Λ^{s+1+δ}(∂Ω) → Λ^{s}(Ω), & for 0 < s ≤ λ, \\ B_{1} : B_{p}^{s+1+δ− \frac{1}{p}} (∂Ω) → H_{p}^{s}(Ω), & for 0 ≤ s < λ, 1 < p < ∞. \end{matrix}

Now we compute ℓ^{→} E |_{∂Ω}. We have

ℓ^{→} E |_{∂Ω}= L_{1}

where

\begin{aligned}L_{1} & =T^{→} +a\sqrt{P_{1}^{♯}(x, D_{x}) − B^{♯}(x, D_{x})^{2}} + aB^{♯}(x, D_{x}) \\ & =T^{→} +a\sqrt{∑_{i,j=1}^{n+1} g̃_{i,j}^{♯} (x, ξ) ξ_{i}ξ_{j}} + ia ∑_{k=1}^{n+1} b_{k}^{♯} (x, ξ) ξ_{k}, \end{aligned}  (26)

by (22) and (23).

Thus we have reduced the solving of (18) and (19) to the solution of the pseudodifferential equations

L_{1}u = g̃− T^{→} F  (27)

and

\{ \begin{matrix}L_{1}u = g̃− T^{→} F \\ u |_{t=0}= h − F. \end{matrix} .  (28)

2.3 Microlocalization

Note that the pseudodifferential operator L_{1} is elliptic on the cone {| ξ_{n+1} |> λ′ | ξ′ |} for any λ′ > 0, and thus we can easily invert L_{1} in this cone. So it remains to investigate L_{1} on a cone {| ξ_{n+1} |< λ | ξ′ |}. As in [GS], our goal now is to eliminate ξ_{n+1} from underneath the root sign in (26). First we change notation. Set t = x_{n+1}, τ = ξ_{n+1}, x = (x_{1}, ..., x_{n}) , and ξ = (ξ_{1}, ..., ξ_{n}) (which used to be denoted ξ′) so that on the cone {| τ |< λ | ξ |} we have

\begin{matrix}L_{1}(x, t, ξ, τ) = iτ + a(t, x) \sqrt{g̃_{n+1,n+1}^{♯}τ^{2} + ∑_{j=1}^{n} g̃_{n+1,j}^{♯} (x, t, ξ, τ) ξ_{i}ξ_{j} + ∑_{i,j=1}^{n+1} g̃_{i,j}^{♯} (x, t, ξ, τ) ξ_{i}ξ_{j}} \\ +ia(t, x) ∑_{k=1}^{n+1} b_{k}^{♯} (x, t, ξ, τ) ξ_{k} + ia(t, x)b_{n+1}^{♯} (x, t, ξ, τ) τ. \end{matrix}

Now decompose a (x, t, ξ, τ) = a^{♯} (x, t, ξ, τ)+a^{♭} (x, t, ξ, τ) with a^{♯} ∈ S_{1,δ}^{0} and a^{♭} ∈ C^{λ+2}S_{1,δ}^{−δ(λ+2)}

where δ = max \{\frac{1}{λ+1}, \frac{1}{2}\}, and let

\begin{matrix}L̃_{1}(x, t, ξ, τ) = iτ + a^{♯} (x, t, ξ, τ) \sqrt{g̃_{n+1,n+1}^{♯}τ^{2} + ∑_{j=1}^{n} g̃_{n+1,j}^{♯} (x, t, ξ, τ) ξ_{i}ξ_{j} + ∑_{i,j=1}^{n+1} g̃_{i,j}^{♯} (x, t, ξ, τ) ξ_{i}ξ_{j}} \\ +ia^{♯} (x, t, ξ, τ) ∑_{k=1}^{n+1} b_{k}^{♯} (x, t, ξ, τ) ξ_{k} + ia^{♯} (x, t, ξ, τ) b_{n+1}^{♯} (x, t, ξ, τ) τ. \end{matrix}

Sinced a is small in U, we may assume that a^{♯} is also small there.

Lemma 11 L̃_{1}(x, t, ξ, τ) = L̃_{0}(x, t, ξ, τ)L̃ (x, t, ξ, τ) where

L̃ (x, t, ξ, τ) = iτ + ia^{♯}Q (x, t, ξ, a^{♯}) ,

with Q, (\frac{∂}{∂z} )^{k} Q (x, t, ξ, z) ∈ S_{1,δ}^{1}, c |ξ| ≤ ReQ ≤ c^{−1} |ξ| for some c > 0; and

L̃_{0} (x, t, ξ, τ) = 1 + a^{♯}Q_{0} (x, t, ξ, a^{♯}) ,

with Q_{0} ∈ S_{1,δ}^{1}.

Proof. Let z = a^{♯}and write

\begin{aligned}L̃_{1}(x, t, ξ, τ) & = iτ + z (\sqrt{g̃_{n+1,n+1}^{♯}τ^{2} + ∑_{j=1}^{n} g̃_{n+1,j}^{♯}ξ_{i}ξ_{j} + ∑_{i,j=1}^{n+1} g̃_{i,j}^{♯}ξ_{i}ξ_{j}} + iz ∑_{k=1}^{n+1} b_{k}^{♯}ξ_{k} + ib_{n+1}^{♯}τ) \\ & = f (x, t, ξ̃, τ̃, z) |ξ| , \end{aligned}

where ξ̃ = \frac{ξ}{|ξ|}, τ̃ = \frac{τ}{|ξ|} and |\frac{∂}{∂_{τ}}f (x, t, ξ̃, τ̃, z) |_{_{τ}=0}| ≥ 1 − c |z| for some constant c > 0. Since

z is small, \frac{∂}{∂_{τ}}f |_{_{τ}=0} and so by the Malgrange preparation theorem,

f (x, t, ξ̃, τ̃, z) = f_{0} (x, t, ξ̃, τ̃, z) (iτ̃ + g_{0} (x, t, ξ̃, z)) , f_{0} ≠ 0.

But f (x, t, ξ̃, τ̃, z) = 0 and so g_{0} (x, t, ξ̃, z) = zg_{1} (x, t, ξ̃, z) and |(\frac{∂}{∂z} )^{k} g| ≤ C_{k}. Now let

L̃ (x, t, ξ, τ) = |ξ| (iτ̃ + g_{0} (x, t, ξ̃, a^{♯})) = iτ + a^{♯} |ξ| g_{1} (x, t, ξ̃, a^{♯}) = iτ + a^{♯}Q

and

L̃_{0} (x, t, ξ, τ) = f_{0} (x, t, ξ̃, τ̃, a^{♯}) .

A simple computation shows that \frac{ReQ}{|ξ|} is bounded above and below by two positive constants and that f_{0} (x, t, ξ̃, τ̃, 0) = 1. This completes the proof of Lemma 11.

We now give the corresponding operator decomposition.

Lemma 12 Let L and L_{0} denote the operators with symbols iτ + aQ (x, t, ξ, a) and 1 + a^{♯}Q_{0} (x, t, ξ, a^{♯}) where Q and Q_{0} are as in Lemma 11 respectively. We have

L_{1} = L_{0} ∘ L + aB_{δ} + B_{−δ}  (29)

where B_{δ} ∈ \overline{O}_{(0,λ+2)}^{δ} and B_{−δ} ∈ \overline{O}_{(0,λ+2)}^{−δ}.

Proof. Since a^{♭} ∈ C^{λ+2}S_{1,δ}^{−δ−1}, L_{1} − L̃_{1} ∈ \overline{O}_{(0,λ+2)}^{−δ}. By Lemma 11 and Propostion 5 in [GSII],

\begin{matrix}L̃_{0} ∘ L̃ − L̃_{1} = (a^{♯}Q_{0}) ∘ (a^{♯}Q) − a^{♯}Q_{0}a^{♯}Q = (aQ_{0}) ∘ (a^{♯}Q) − aQ_{0}a^{♯}Q + B_{−δ} \\ = a \{Q_{0} ∘ (a^{♯}Q) − Q_{0}a^{♯}Q\} + B_{−δ} = aB_{δ} + B_{−δ}. \end{matrix}

On the other hand,

L̃ − L = a^{♯}Q (x, t, ξ, a^{♯}) − aQ (x, t, ξ, a) = a^{♭}B_{1} = B_{−δ}.

If we now write

\begin{matrix}L_{1} = L̃_{1} + (L_{1} − L̃_{1}) = L̃_{0} ∘ L̃ + (L̃_{1} − L̃_{0} ∘ L̃) + (L_{1} − L̃_{1}) \\ = L̃_{0} ∘ L + L̃_{0} ∘ (L̃ − L) + (L̃_{1} − L̃_{0} ∘ L̃) + (L_{1} − L̃_{1}) , \end{matrix}

the lemma follows.

2.4 A Parametrix

An operator K is called a left (respectively right) parametrix of an operator L if KL = I +S (respectively LK = I + S) where S is a smoothing operator. In this paper however, it will be more convenient to call K a left (or right) parametrix of L if KL (or LK) = I + S provided there is δ > 0 such that S is in \overline{O}_{I}^{−δ} for some appropriate interval I. As we shall show, this definition is sufficient to carry out the estimates we need.

In our local setting, the conditions in cases (I), and (II) in §1 can be restated as follows: (I) a(x, t) ≥ 0 for (x, t)ϵ I^{n} × I or a(x, t) ≤ 0 for (x, t)ϵ I^{n} × I; a(x, ±1) ≠ 0 for x ∈ I^{n}. (II) a(x, t) ≥ 0 for t ≥ 0, x ϵ I^{n} and a(x, t) ≤ 0 for t ≤ 0, x ϵ I^{n}; a(x, ±1) ≠ 0 for

x ∈ I^{n}.

First, we will construct a left parametrix K for L microlocally in the cone {|τ| < λ |ξ|} in cases (I) and (II). Then we use this to obtain a parametrix for L_{1}, and finally to obtain a parametrix and a priori estimates for the original problem (13) and (14).

By the choice of the coordinate charts, there exist c > 0, 1 > c′ > 0, such that |a(x, t)| ≥ c for |t| ≥ c′. Since ρ = ρ_{j} in the previous section is compactly supported in I^{n} × I, we may pick ρ^{∗} and ρ̃ ∈ C_{c}^{∞}(I^{n} × I), ρ^{∗} = 1 on supp ρ, and ρ̃ = 1 on supp ρ^{∗}. By making the change of variable t → −t if necessary, we may assume that a(x, t) ≥ 0 in case (I). Recall that L in Lemma 11 is given by

L(t, x, τ, ξ) = iτ + a(t, x)Q(x, t, ξ, a),

where c |ξ| ≤ |ReQ| ≤ c^{−1} |ξ|. Now, as in subsection 3.2, let

Ã(x, t, t′, ξ) = ∫_{t′}^{t} a(x, θ)Q(x, θ, ξ, a)dθ,
A(x, t, t′) = ∫_{t′}^{t} a(x, θ)dθ,
K(x, t, t′, ξ) = ρ̃(x, t)e^{−Ã(x,t,t′,ξ)}.

Define

\begin{aligned}Ku(x, t) & = (2π)^{− \frac{n}{2}} ∫_{t_{0}}^{t} ∫_{|ξ|≥1} e^{ix·ξ}K(x, t, t′, ξ)ũ^{∗}(ξ, t′)dξdt′, \\ T u(x, t) & = (2π)^{− \frac{n}{2}} ∫_{t_{0}}^{t} ∫_{|ξ|≥1} e^{ix·ξ}K(x, t, t′, ξ)a(x, t′)Q (x, t′, ξ, a (x, t′)) ũ^{∗}(ξ, t′)dξdt′, \\ Ku_{0}(x, t) & = −(2π)^{− \frac{n}{2}} ∫_{|ξ|≥1} e^{ix·ξ}ρ̃(x, t_{0})e^{−Ã(x,t,t_{0},ξ)}ũ_{0}^{∗}(ξ)dξ, \end{aligned}

(30)

where u^{∗} = ρ^{∗}u , ũ^{∗}(ξ, t′) = ∫_{R^{n}} e^{−ix·ξ}u^{∗}(x, t′)dx, u_{0}^{∗} (x) = ρ^{∗} (x, 0) u_{0} (x) , ũ_{0}^{∗}(ξ) = ∫_{R^{n}} e^{−ix·ξ}u_{0}^{∗}(x)dx, and t_{0} = −1 if a satisfies case (I), while t_{0} = 0 if a satisfies case (II).

Remark The choice of t_{0} is crucial. With the above choice, a(x, θ) keeps the same sign in (t′, t) for t′ ∈ (t_{0}, t), and Re (Ã(x, t, t′, ξ)) ≥ 0 and A(x, t, t′) ≥ 0 for t′ ∈ (t_{0}, t).

Claim If μ_{p} = min \{1 − δ − \frac{1}{p(λ+3)}, \frac{λ+2}{2(λ+3)}\} > 0, then the operator K is a left parametrix for L in Case I, and a left parametrix for L in Case II for functions u with u vanishing at t = 0 (i.e., KLu = u + Su if u |_{t=0}= 0).

Proof. We have

KL = K ∘ ∂_{t} + K ∘ aQ.

Performing integration by parts in K ∘ ∂_{t} we get

KLu = u − Ru − Ku_{0} + O_{(−1,λ+2)}^{−∞},

where u_{0} = u (t_{0}, x) and

Ru = − ∫ ∫ e^{ix·ξ}a (t′, x) Q (t′, x, ξ, a) e^{−Ã(x,t,t′,ξ)}ũ^{∗}(t′, ξ)dξdt′ + K ∘ aQ

By Lemma 19 in [GSII], we have that R is in \overline{O}_{(−1,λ+2)}^{−μ_{p}}. We get

KLu = u + S′u − K ∘ u_{0},

where S′ is in O_{(−1,λ+2)}^{−μ_{p}}. In Case I, u_{0} = u (x, −1) = 0 since supp u is compact. This completes the proof of the claim.

Now we turn to constructing a left parametrix for L_{1}. Since L_{0} ∈ C^{λ+2}S_{1,δ}^{0} is elliptic, there is L_{0}^{−1} ∈ C^{λ+2}S_{1,δ}^{0} such that L_{0}^{−1} ∘ L_{0} = I + J where J is in \overline{O}_{(0,λ+2)}^{−2}. Now set K̃ = K ∘ L_{0}^{−1}. By the rough ψdo calculus and Lemma 12 (since a ∈ C^{λ+2}), we have

K̃ ∘ L_{1}u = K ∘ Lu + K ∘ aB_{δ}u + \overline{O}_{(0,λ+2)}^{−δ}u
= K ∘ Lu + (K ∘ a) B_{δ}u + \overline{O}_{(0,λ+2)}^{−δ}u
= u − Ru − Ku_{0} + \overline{O}_{(0,λ+2)}^{−μ_{p}}u,

where u_{0} = u |_{t=t_{0}}, since K ∘ a ∈ \overline{O}_{(0,λ+2)}^{\frac{1}{(λ+3)p} −1} and μ_{p} > 0. By Lemma 19 in [GSII],

K̃ ∘ L_{1}u = u − Ku_{0} + S̃u,

with S̃ ∈ \overline{O}_{(0,λ+2)}^{−μ_{p}}. Note that in case (I), u_{0} is smooth and so K̃ ∘ L_{1} = I + S̃.

Up to this point we have only worked on the cone {|τ| < |ξ′|} for L_{1} microlocally. We are now ready to obtain a left parametrix for the model problem in the coordinate chart:

L_{1}u = g − ∂_{t}F + (Da) g_{1} + aw  (31)

or

\{ \begin{matrix}L_{1}u & = g − ∂_{t}F + (Da) g_{1} + aw \\ u |_{t=0} & = h − F \end{matrix} .  (32)

Proposition 13 If u satisfies either (31) or (32) above with the assumption that g, F, g_{1}, u and w are smooth at t = −1 and t = 1, then

\begin{aligned}u & = K̃(φg) + K̃ ∘ (Da) φg + T ∘ L_{0}^{−1} (F) + K̃ ∘ aw \\ & +\overline{O}_{(0,λ+2)}^{0} (F) + \overline{O}_{(0,λ+2)}^{−μ_{p}} (u) + \overline{O}_{(0,λ+2)}^{−1} (w) + \overline{O}_{(0,λ+1)}^{−1} (g_{1}) + \overline{O}_{(0,λ+2)}^{−1} (g) , \end{aligned}

in case (I), and

\begin{aligned}ρu_{b} & = K̃(φg) + K̃ ∘ (Da) φg + T ∘ L_{0}^{−1} (F) + K̃ ∘ aw + K \{h + B_{1} (g) |_{t=0} +B_{2} (g_{1}) |_{t=0}\} \\ & +\overline{O}_{(0,λ+2)}^{0} (F) + \overline{O}_{(0,λ+2)}^{−μ_{p}} (u) + \overline{O}_{(0,λ+2)}^{−1} (w) + \overline{O}_{(0,λ+1)}^{−1} (g_{1}) + \overline{O}_{(0,λ+2)}^{−1} (g) , \end{aligned}

in case (II), where B_{1}, B_{2} ∈ \overline{O}_{(0,λ)}^{−1} (∂Ω).

Proof: Let ψ̃ ∈ C^{∞} (R^{n}) be homogeneous of degree zero on |ξ| ≥ 1, with supp ψ̃ ⊂ {|ξ_{n+1}| > λ′ |ξ′|}. If we apply ψ̃ to the above equations and commute ψ̃ with L_{1} we get

L_{1}ψ̃u = ψ̃g − ∂_{t}ψ̃F + ψ̃ (Da) g_{1} + ψ̃aw + \overline{O}_{(0,λ+2)}^{0}u.

Since L_{1} is elliptic on \{|ξ_{n+1}| > λ′ |ξ′|\}, there is L_{1}^{−1} ∈ \overline{O}_{(0,λ+1)}^{−1} such that L_{1}^{−1} ∘ L_{1} = I + \overline{O}_{(−1,λ+2)}^{−1} and L_{1} ∘ L_{1}^{−1} = I + \overline{O}_{(−1,λ+1)}^{−1} (again by the rough ψdo calculus). Apply L_{1}^{−1} to this equation to obtain

ψ̃u = L_{1}^{−1}ψ̃g − L_{1}^{−1}∂_{t}ψ̃F + L_{1}^{−1}ψ̃ (Da) g_{1} + L_{1}^{−1}ψ̃aw + \overline{O}_{(0,λ+2)}^{−1}u.  (33)

Now pick ψ, φ ∈ C^{∞} (R^{n}) homogeneous of degree zero on |ξ| ≥ 1, with supp ψ ⊂ {|ξ_{n+1}| > λ′ |ξ′|} and ψ + φ = 1 on |ξ| ≥ 1. Replacing ψ̃ by ψ in (33) we obtain

ψu = L_{1}^{−1}ψg − L_{1}^{−1}∂_{t}ψF + L_{1}^{−1}ψ (Da) g_{1} + L_{1}^{−1}ψaw + \overline{O}_{(0,λ+2)}^{−1}u.

Now we wish to identify ψu |_{t=0} in case (II). If we use the equation

∂_{t}ψu = −ψaQ̃u + ψg − ∂_{t}ψF + ψ (Da) g_{1} + ψaw,

integrate in the t variable from −1 to 0, and use the fact that evaluation of ψu, ψg etc. at t = −1 is smoothing, we obtain

ψu |_{t=0} = ∫_{−1}^{0} ψaQ̃u + ∫_{−1}^{0} ψg − ψF |_{t=0} + ∫_{−1}^{0} ψ (Da) g_{1} + ∫_{−1}^{0} ψaw + \overline{O}_{(0,λ+2)}^{−∞}F
= − ∫_{−1}^{0} ∂_{t} ∘ ∂_{t}^{−1}ψaQ̃u + ∫_{−1}^{0} ∂_{t} ∘ ∂_{t}^{−1}ψg − ψF |_{t=0} + ∫_{−1}^{0} ∂_{t} ∘ ∂_{t}^{−1}ψ (Da) g_{1}
+ ∫_{−1}^{0} ∂_{t} ∘ ∂_{t}^{−1}ψaw + \overline{O}_{(0,λ+2)}^{−∞}F
= −ψF |_{t=0} +\overline{O}_{(0,λ+1)}^{−1} (g_{1}) |_{t=0} +\overline{O}_{(0,λ+2)}^{−1} (g) |_{t=0}
− ∫_{−1}^{0} ∂_{t} ∘ a (∂_{t}^{−1} ∘ ψQ̃) u − ∫_{−1}^{0} ∂_{t} ∘ [∂_{t}^{−1}ψ, a] ∘ Q̃ ∘ ψ̃u
+ ∫_{−1}^{0} ∂_{t} ∘ a∂_{t}^{−1} ∘ ψw + ∫_{−1}^{0} ∂_{t} ∘ [∂_{t}^{−1}ψ, a] w + \overline{O}_{(0,λ+2)}^{−∞} (F, u, g, g_{1}) ,

where ψ̃ = 1 on supp ψ, supp ψ̃ ⊂ {|ξ_{n+1}| > λ′ |ξ′|}.

Now if we substitute (33) for ψ̃u in the above equation, and note that a (x, 0) = 0, we get

\begin{matrix}ψu |_{t=0}= −ψF |_{t=0} +\overline{O}_{(0,λ+1)}^{−1} (g_{1}) |_{t=0} +\overline{O}_{(0,λ+2)}^{−1} (g) |_{t=0} \\ +\overline{O}_{(0,λ+2)}^{−1} (F) |_{t=0} +\overline{O}_{(0,λ+2)}^{−2} (g, g_{1}, w, u) |_{t=0} . \end{matrix}  (34)

On the other hand, applying φ to the equations yields

L_{1}φu = φg − ∂_{t}φF + φ (Da) g_{1} + φaw + [φ, L_{1}] u
= φg − ∂_{t}φF + φ (Da) g_{1} + φaw + [φ, aQ̃] u
= φg − ∂_{t}φF + φ (Da) g_{1} + φaw + [φ, a] ∘ Q̃u + a [φ, Q̃] u.

By the rough ψdo calculus, [φ, Q̃] ∈ \overline{O}_{(0,λ+2)}^{δ}. Now apply K̃ to the above to obtain

\begin{aligned}φu & = K (φu |_{t=t_{0}}) + S̃u + K̃φg − K̃ ∘ ∂_{t}φF + K̃φ (Da) g_{1} + K̃φaw \\ & +K̃ ∘ [φ, a] ∘ Q̃u + K̃ ∘ a ∘ \overline{O}_{(0,λ+2)}^{δ} (u) + \overline{O}_{(0,λ+2)}^{−1} (u) . \end{aligned}

Next we commute ∂_{t} with L_{0}^{−1}, use integration by parts in K̃ ∘ ∂_{t}φF, commute Da with φ, commute a with L_{0}^{−1}, and use the fact that Ka ∈ \overline{O}_{(0,λ)}^{1− \frac{1}{p(λ+3)}} to arrive at

\begin{aligned}φu & = K (φu |_{t=t_{0}}) + K̃ψg − φF + T ∘ L_{0}^{−1}φF + K (L_{0}^{−1}φF |_{t=t_{0}}) + K̃aφw + K̃ ∘ [φ, a] ∘ Q̃u \\ & +K̃ ∘ [φ, a] w + K̃ ∘ (Da) ψg_{1} + \overline{O}_{(0,λ+2)}^{−μ_{p}} (u) + \overline{O}_{(0,λ+2)}^{0} (F) + \overline{O}_{(0,λ+1)}^{−1} (g_{1}) + \overline{O}_{(0,λ+2)}^{−1} (w) . \end{aligned}

In case (I), L_{0}^{−1}φF |_{t=t_{0}} is smoothing, while in case (II),

L_{0}^{−1}φF |_{t=0}= φF |_{t=0} +\overline{O}_{(0,λ+2)}^{−1} (F) |_{t=0} .

Claim K̃ ∘ [φ, a] ∈ \overline{O}_{(0,λ+2)}^{− \frac{δ}{4} −1}.

Proof. To begin, note that [φ, a] = φ ∘ (a^{♯} + a^{♭}) − aφ = (a^{♯})_{x} φ_{ξ} + \overline{O}_{(0,λ+2)}^{− \frac{δ}{4} −1}. We need to show that K̃ ∘ (a^{♯})_{x} ∈ \overline{O}_{(0,λ+2)}^{− \frac{δ}{4}}, and for this we write K̃ ∘ (a^{♯})_{x} = K̃ ∘ a_{x} − K̃ ∘ (a^{♭})_{x}, where K̃ ∘ a_{x} ∈ \overline{O}_{(0,λ+1)}^{− \frac{δ}{4}} by commuting a_{x} with L_{0}^{−1}. We thus get K̃ ∘ (a^{♯})_{x} ∈ \overline{O}_{(0,λ+1)}^{− \frac{δ}{4}} since a^{♭} ∈ \overline{O}_{(0,λ+2)}^{−1−δ}. To show K̃ ∘ (a^{♯})_{x} ∈ \overline{O}_{(0,λ+2)}^{− \frac{δ}{4}}, we apply D_{x} or D_{t} to K̃ ∘ (a^{♯})_{x} to get

\begin{matrix}D_{x} ∘ K̃ ∘ (a^{♯})_{x} = K̃ ∘ (a^{♯})_{x} ∘ D_{x} + \overline{O}(0,λ+1)^{\frac{1}{2}}, \\ D_{t} ∘ K̃ ∘ (a^{♯})_{x} = −KaQ ∘ L_{0}^{−1} ∘ (a^{♯})_{x} + L_{0}^{−1} ∘ (a^{♯})_{x} ∈ \overline{O}(0,λ+1)^{\frac{1}{2}}, \end{matrix}

and this completes the proof of the claim.

Thus we have

\begin{aligned}φu & = K̃ψg − φF + T ∘ L_{0}^{−1}φF + K̃(Da) φg_{1} + K̃aφw \\ & +\overline{O}_{(0,λ+2)}^{−μ_{p}} (u) + \overline{O}_{(0,λ+2)}^{0} (F) + \overline{O}_{(0,λ+1)}^{−1} (g_{1}) + \overline{O}_{(0,λ+2)}^{−1} (w) , \end{aligned}

in case (I), and

\begin{aligned}φu & = K̃ψg − φF + K (Da) φg_{1} + K (φu |_{t=0} +φF |_{t=0}) + K̃aφw \\ & +\overline{O}_{(0,λ+2)}^{−μ_{p}} (u) + \overline{O}_{(0,λ+2)}^{0} (F) + \overline{O}_{(0,λ+1)}^{−1} (g_{1}) + \overline{O}_{(0,λ+2)}^{−1} (w) , \end{aligned}

in case (II). By (34),

(φu + φF) |_{t=0} = (u − ψu) |_{t=0} +φF |_{t=0} +smoothing term
= h − ψF |_{t=0} −ψu |_{t=0} +smoothing term
= h + \overline{O}_{(0,λ+2)}^{−1} (g) |_{t=0} +\overline{O}_{(0,λ+1)}^{−1} (g_{1}) |_{t=0}
+\overline{O}_{(0,λ+2)}^{−1} (F) |_{t=0} +\overline{O}_{(0,λ+2)}^{−2} (g, g_{1}u, w) |_{t=0} .

The Proposition now follows since K ∘ \overline{O}_{(0,λ+2)}^{−1} |_{t=0}∈ \overline{O}_{(0,λ+2− \frac{1}{p})}^{0} (∂Ω).

Now we return to the original equations (13) and (14) and obtain a left parametrix for them. As in subsection 2.1, \overline{Ω} ⊂ ∪_{j=0}^{k}U_{j} and ∂Ω ⊂ ∪_{j=1}^{k}U_{j}. Let ρ ∈ C^{λ+2} (Ω) ∩C_{c}^{∞} (U_{j})for some j ≥ 1 (elliptic theory handles the case j = 0) . Let U = U_{j}.

Proposition 14 Suppose that either u ∈ Λ^{s} (Ω) for some s > 1, or u ∈ H_{p}^{s} (Ω) for some s > 1 + \frac{1}{p}, where 1 < p < ∞. Let ρ, ρ̃ ∈ C_{c}^{∞} (U) with ρ̃ = 1 in a rectangular (relative to the coordinate system) subdomain Ũ of U with Ũ ⊃ supp ρ. If u satisfies (13), then

\begin{aligned}ρu_{b} & = ρK̃(ρ̃g) + ρK̃ ∘ (Da) ρ̃g + ρT ∘ L_{0}^{−1} (ρ̃Nf |_{∂Ω}) \\ & +\overline{O}_{(0,λ+2)}^{0} (ρ̃Nf |_{∂Ω}) + \overline{O}_{(0,λ+2)}^{−μ_{p}} (ρ̃u_{b}, ρ^{∗}u_{b}) + \overline{O}_{(0,λ+1)}^{−1} (ρ̃g_{1}) + \overline{O}_{(0,λ+2)}^{−1} (ρ̃g) , \end{aligned} (35)

while if u satisfies (14), then

\begin{aligned}ρu_{b} & = ρK̃(ρ̃g) + ρK̃ ∘ (Da) ρ̃g + ρT ∘ L_{0}^{−1} (ρ̃Nf |_{∂Ω}) \\ & −ρK \{ρ̃h + \overline{B}_{1} (ρ̃g) |_{Γ} +\overline{B}_{2} (ρ̃g_{1}) |_{Γ}\} \\ & +\overline{O}_{(0,λ+2)}^{0} (ρ̃Nf |_{∂Ω}) + \overline{O}_{(0,λ+2)}^{−μ_{p}} (ρ̃u_{b}, ρ^{∗}u_{b}) + \overline{O}_{(0,λ+1)}^{−1} (ρ̃g_{1}) + \overline{O}_{(0,λ+2)}^{−1} (ρ̃g) , \end{aligned} (36)

where u_{b} = u |_{∂Ω}, \overline{B}_{j} ∈ \overline{O}_{(0,λ+2)}^{−1}, j = 1, 2.

Proof: We prove (36), the proof for (35) being similar but easier. Let v = u − Nf, where N denotes the Neumann operator. We have

\{ \begin{matrix}P v = & 0 & on Ω \\ ℓ^{→} v = & g + (Da) g_{1}− T^{→} Nf |_{∂Ω} & on ∂Ω \\ v = & h − Nf & on Γ \end{matrix} .

We must prove (36) for v. Choose ρ^{∗} ∈ C_{c}^{∞} (U) with ρ^{∗} = 1 on a rectangular subdomain U^{∗} where U^{∗} ⊃ supp ρ, a ≠ 0 on the top and bottom of U^{∗}, and ρ̃ = 1 on supp ρ^{∗}. Then ρ^{∗}v satisfies

\{ \begin{matrix}P (ρ^{∗}v) = & 0 & on Ω \\ ℓ^{→} (ρ^{∗}v) = & ρ^{∗}g + (Da) ρ^{∗}g_{1} − ρ^{∗} T^{→} Nf |_{∂Ω} − ( ℓ^{→} ρ^{∗}) v & on ∂Ω \\ ρ^{∗}v = & ρ^{∗}h − ρ^{∗}Nf & on Γ \end{matrix} .

Now set ṽ = Eρ^{∗}v_{b} so that ℓ^{→} ṽ |_{∂Ω}= L_{1} (ρ^{∗}v_{b}) by the construction of E.

Meanwhile,

\begin{matrix}P (ρ^{∗}v − ṽ) = & [ρ^{∗}, P] v − B_{1} (ρ^{∗}v_{b}) & on Ω \\ ρ^{∗}v − ṽ = & 0 & on ∂Ω \end{matrix} ,  (37)

where B_{1} is as in (25). Solving the Dirichlet problem (37), we obtain

ρ^{∗}v − ṽ = G {[ρ^{∗}, P] v − B_{1} (ρ^{∗}v_{b})} ,

where G ∈ \overline{O}_{(1,λ+2)}^{−2} (Ω), P G = I, and G |_{∂Ω}= 0. Since T^{→} (ρ^{∗}v − ṽ) = 0 on ∂Ω, we conclude that

ℓ^{→} (ρ^{∗}v − ṽ) |_{∂Ω}= a\overrightarrow{n}G [ρ^{∗}, P] v − aB′_{0} (ρ^{∗}v_{b}) ,

where B′_{0} = (\overrightarrow{n}GB_{1}) |_{∂Ω}. So B′_{0} : Λ^{s+1−δλ} (∂Ω) → Λ^{s} (∂Ω) for 0 < s ≤ λ + 1 and

B′_{0} : B_{p}^{s− \frac{1}{p} +1−δλ,p} (∂Ω) → B_{p}^{s− \frac{1}{p} ,p} (∂Ω) for 0 ≤ s < λ + 1. But, for any ρ̃ ∈ C_{c}^{∞} (Ω) with ρ̃ = 1 on the support of ρ^{∗} we get

\begin{matrix}ρ^{∗}g + (Da) ρ^{∗}g_{1} − ρ^{∗} T^{→} Nf |_{∂Ω} − ( ℓ^{→} ρ^{∗}) v_{b} − L_{1}v_{b} \\ = −aB′_{0} (ρ^{∗}v_{b}) + aB̃_{0} (ρ̃v_{b}) , \end{matrix}

and

\overrightarrow{n}G [ρ^{∗}, P] ρ̃v |_{∂Ω}= B̃_{0} (ρ̃v_{b}) ,

with B̃_{0} ∈ \overline{O}_{(0,λ+1)}^{0} (∂Ω). By Proposition 13, with w = B′_{0} (ρ^{∗}v_{b}) + B̃_{0} (ρ^{∗}v_{b}),

ρ^{∗}v_{b} = K̃(ρ^{∗}g) + K̃ ∘ (Da) ρ^{∗}g + T ∘ L_{0}^{−1} (ρ^{∗}Nf |_{∂Ω}) − K̃( ℓ^{→} ρ^{∗}) v_{b} + (K̃ ∘ a) B̃_{0}ρ̃v_{b}

+K̃ ∘ a ∘ B′_{0} (ρ^{∗}v_{b}) − K \{ρ^{∗}h + \overline{B}_{1} (ρ^{∗}g) |_{Γ} +\overline{B}_{2} (ρ^{∗}g_{1}) |_{Γ} +\overline{B}_{1} (( ℓ^{→} ρ^{∗}) v_{b}) |_{Γ}\}

+\overline{O}_{(0,λ+2)}^{0} (ρ̃Nf |_{∂Ω}) + \overline{O}_{(0,λ+2)}^{−μ_{p}} (ρ^{∗}v_{b}) + \overline{O}_{(0,λ+1)}^{−1} (ρ̃g_{1}) + \overline{O}_{(0,λ+2)}^{−1} (ρ̃g) .

If we now multiply the above identity by ρ, commute a with L_{0}^{−1}, and note that K̃ ∘ a ∈

\overline{O}_{(0,λ+2)}^{−1+ \frac{1}{p(λ+3)}} (∂Ω). So we have K̃ ∘ a ∘ B̃_{0}, K̃ ∘ a ∘ B′_{0} ∈ \overline{O}_{(0,λ+2)}^{−μ_{p}} (∂Ω) and

\begin{matrix}|x − y| ≥ δ_{0} > 0, \\ ∫_{t′}^{t} a (x, θ) dθ ≥ δ_{0} > 0, \end{matrix}

for (x, t) ∈ supp ρ, (y, t′) ∈ supp ( ℓ^{→} ρ^{∗}) ∪ supp (ρ̃ − ρ^{∗}), we finally conclude

\begin{aligned}ρK̃(ρ̃ − ρ^{∗}) , ρ\overline{K}∘ ℓ^{→} ρ^{∗}, ρK ∘ (B_{1} ℓ^{→} ρ^{∗}) |_{Γ}, ρK ∘ (ρ̃ − ρ^{∗}) & ∈ \overline{O}_{(0,λ+2)}^{−1} (∂Ω) \\ ρT ∘ L_{0}^{−1} (ρ̃ − ρ^{∗}) & ∈ \overline{O}_{(0,λ+2)}^{0} (∂Ω) \end{aligned} ,

and (36) now follows easily.

We now consider a priori estimates for the original equations (13) and (14). Suppose that P, Ω, and \overrightarrow{ℓ}=\overrightarrow{T}+a\overrightarrow{n} are as at the beginning of subsection 1.5 with λ > 0. Consider the following problems:

\{ \begin{matrix}P u & = & f & in Ω \\ ℓ^{→} u & = & g + (Da) g_{1} & on ∂Ω \end{matrix} ,  (38)

and

\{ \begin{matrix}P u & = & f & in Ω \\ ℓ^{→} u & = & g + (Da) g_{1} & on ∂Ω \\ u & = & h & on Γ \end{matrix} .  (39)

Proposition 15 Suppose that P, Ω, and ℓ⃗ are as at the beginning of subsection 1.5 with λ > 0, let μ_{p} = min \{1 − δ + \frac{1}{p(λ+3)}, \frac{λ+2}{2(λ+3)}\} for 1 < p < ∞ and 0 < μ_{∞} < min \{1 − δ, \frac{λ+2}{2(λ+3)}\},

and let D = ∑_{k=1}^{n+2} b_{k} (x) \frac{∂}{∂x_{k}} ∈ C^{λ+2} (\overline{Ω}) be a vector field tangent to ∂Ω for x ∈ ∂Ω.

(i) Let 1 < s ≤ λ + 2. If ℓ⃗ satisfies case (I) and the T_{δ} condition for some δ ≥ 0, and if u ∈ Λ^{s′}(Ω) for some s′ > 1 satisfies (38), then u ∈ Λ^{s}(Ω), and there is C_{s} such that

∥u∥_{Λ^{s}(Ω)} ≤ C_{s} (∥u∥_{Λ^{s−μ∞}(Ω)} + ∥f∥_{Λ^{s−2}(Ω)} + ∥g∥_{Λ^{s−δ}(∂Ω)} + ∥g_{1}∥_{Λ^{s− \frac{1}{4}} (∂Ω)}) . (40)

(ii) Let 1 < s ≤ λ + 2. If ℓ⃗ satisfies case (II) and the T_{δ} condition for some δ ≥ 0, and if u ∈ Λ^{s′}(Ω) for some s′ > 1 satisfies (39), then u ∈ Λ^{s}(Ω), and there is C_{s} such that

∥u∥_{Λ^{s}(Ω)} ≤ C_{s} (∥u∥_{Λ^{s−μ∞}(Ω)} + ∥f∥_{Λ^{s−2}(Ω)} + ∥g∥_{Λ^{s−δ}(∂Ω)} + ∥g_{1}∥_{Λ^{s− \frac{1}{4}} (∂Ω)} + ∥h∥_{Λ^{s}(Γ)}) .

(41)

(iii) Let α ≥ 0, 1 + \frac{1}{p} < s < λ + 2 − α. Suppose ℓ⃗ satisfies case (I) above, and in addition satisfies the A_{p,α}^{∓} condition on Γ, A_{p,α}^{=} on (Σ^{+} ∩ N) \ Γ^{∗}, A_{p,α}^{‡} on (Σ^{−} ∩ N) \ Γ^{∗} for some α ≥ 0 and 1 < p < ∞. Assume moreover that δλ−α > 0. If u ∈ H_{p}^{s′}(Ω) for some s′ > 1+ \frac{1}{p} satisfies (38), then u ∈ H_{p}^{s}(Ω), and there is C_{s} such that

∥u∥_{H_{p}^{s}(Ω)} ≤ C_{s} (∥u∥_{H_{p}^{s−μp}(Ω)} + ∥f∥_{H_{p}^{s−2+σ}(Ω)} + ∥g∥_{B_{p}^{s−δ− \frac{1}{p} ,p}(∂Ω)} + ∥g_{1}∥_{B_{p}^{s− \frac{1}{4} − \frac{1}{p} ,p}(∂Ω)}) . (42)

(iv) Let α ≥ 0, 1 + \frac{1}{p} < s < λ + 2 − α. Suppose ℓ⃗ satisfies case (II) above, and in addition satisfies the A_{p,α}^{∓} condition on Γ, A_{p,α}^{=} on (Σ^{+} ∩ N) \ Γ^{∗}, A_{p,α}^{‡} on (Σ^{−} ∩ N) \ Γ^{∗} for some α ≥ 0 and 1 < p < ∞, and also the P_{γ} condition for some γ ≥ 0. If u ∈ H_{p}^{s′}(Ω) for some s′ > 1 + \frac{1}{p} satisfies (39), then u ∈ H_{p}^{s}(Ω), and there is C_{s} such that

∥u∥_{H_{p}^{s}(Ω)} ≤ C_{s} (∥u∥_{H_{p}^{s−μp}(Ω)} + ∥f∥_{H_{p}^{s−2+α}(Ω)} + ∥g∥_{B_{p}^{s−δ− \frac{1}{p} ,p}(∂Ω)} + ∥g_{1}∥_{B_{p}^{s− \frac{1}{4} − \frac{1}{p} ,p}(∂Ω)} + ∥h∥_{B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ)}) .

(43)

Proof: Note that

K̃a = K ∘ L_{0}^{−1} ∘ a = K ∘ a + K ∘ [L_{0}^{−1}, a] ,
[L_{0}^{−1}, a] = L_{0}^{−1} ∘ a − aL_{0}^{−1} = L_{0}^{−1} ∘ a^{♯} − a^{♯}L_{0}^{−1} + L_{0}^{−1} ∘ a^{♭} − a^{♭}L_{0}^{−1}
= (L_{0}^{−1})_{ξ} ∘ (a^{♯})_{x} + \overline{O}_{(0,λ+2)}^{−1−δ} ∈ \overline{O}_{(0,λ+2)}^{−1},

by the improved estimates in Propostion 4 in [GSII]. The Proposition now follows from Proposition 14, Theorems 31, 34, 37 and Lemmas 26, 27 in [GSII], and the fact that the T_{δ} condition implies the P_{δ} condition.

Note that Theorem 7 follows from the above proposition.

Proposition 16 If the zero order term c (x) in the differential operator P is negative (respectively nonnegative) and μ_{p} > 0, then the solution u to (38) (respectively (39)) is unique.

Proof: Suppose u_{1} and u_{2} are solutions in H_{p}^{s′}(Ω) for some s′ > 1 + \frac{1}{p}, and set u = u_{1} − u_{2}. Then u satisfies (38) or (39) with f = g = h = 0. Let 1 < s < min {λ + 2, s′ + μ_{p}}. From the a priori estimates in Proposition 15 we have u ∈ H_{p}^{s}(Ω). By the Sobolev embedding theorem, we now obtain u ∈ H_{q}^{s−μ_{p}}(Ω) for \frac{1}{q} ≥ \frac{1}{p} − \frac{μ_{p}}{n}. Applying the a priori estimates in Proposition 15 again, but for q in place of p, we conclude that u ∈ H_{q}^{s}(Ω). Alternating applications of the Sobolev embedding theorem and the a priori estimates a finite number of times leads to u ∈ Λ^{s}(Ω). Since s > 1, the standard argument using the Hopf lemma as in [GS] shows that u ≡ 0, so u_{1} = u_{2}.

Proposition 17 Suppose s > 1 + \frac{1}{p} and that P = ∑_{i,j}^{n+2} a_{ij}∂_{i}∂_{j} + ∑_{k}^{n+2} b_{k}∂_{k} + c where c (x) ≤ c_{0} < 0 for some constant c_{0}. Let ∥P ∥_{Λ^{λ}} ≡ ∑_{i,j}^{n+2} ∥a_{ij}∥_{Λ^{λ}} + ∑_{k}^{n+2} ∥b_{k}∥_{Λ^{λ}} + ∥c∥_{Λ^{λ}} and suppose that

∑_{i,j=1}^{n+2} a_{ij}ξ_{i}ξ_{j} ≥ c_{1} |ξ|^{2} for all ξ ∈ R^{n+2}.

If u satisfies (38) or (39), then the terms ∥u∥_{Λ^{s−μ∞}(Ω)} and ∥u∥_{H_{p}^{s−μp}(Ω)} can be omitted from

the inequalities (40) to (43) in Proposition 15 if we replace C_{s} by C (c_{0}, c_{1}, ∥P ∥_{Λ^{λ}} , Ω, \overrightarrow{ℓ}).

Proof: If for example

∥u∥_{H_{p}^{s}(Ω)} ≤ C (∥f∥_{H_{p}^{s−2+α}(Ω)} + ∥g∥_{B_{p}^{s−δ− \frac{1}{p} ,p}(∂Ω)} + ∥h∥_{B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ)})  (44)

fails, then there are sequences {P_{k}}, {u_{k}}, {f_{k}}, {g_{k}}, and {h_{k}} such that

∥P_{k}∥_{Λ^{λ}(Ω)} ≤ C′ and P_{k} is uniformly elliptic for all k,
∥u_{k}∥_{H_{p}^{s−μp}(Ω)} = 1 for all k,
∥f_{k}∥_{H_{p}^{s−2+α}(Ω)} + ∥g_{k}∥_{B_{p}^{s−δ− \frac{1}{p} ,p}(∂Ω)} + ∥h_{k}∥_{B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ)} → 0 as k → ∞.

¿From (43) we obtain that ∥u_{k}∥_{H_{p}^{s}(Ω)} ≤ C for all k, and so by compactness there is a subsequence, which we continue to denote by {u_{k}}, such that u_{k} converges in H_{p}^{s−ϵ}(Ω) to some function u ∈ H_{p}^{s−ϵ}(Ω). Also, there is an elliptic operator P such that P_{k} → P in Λ^{λ−ϵ}. However since P_{k} ∈ Λ^{λ}, a straightforward computation with a difference operator

shows that P ∈ Λ^{λ}. Now P_{k}u_{k} → P u in H_{p}^{s−ϵ−2}(Ω), ℓ⃗u_{k} |_{∂Ω}→ ℓ⃗u |_{∂Ω} in B_{p}^{s−ϵ−1− \frac{1}{p} ,p}(∂Ω)

and u_{k} |_{Γ}→ u |_{Γ} in B_{p}^{s−ϵ− \frac{2}{p} ,p}(Γ). Thus u satisfies (6) with f = g = g_{jk} = h ≡ 0 and ∥u∥_{H_{p}^{s−μp}(Ω)} = 1, contradicting Proposition 16. Thus (44) holds and the proof is complete.

Now we can give the proof of Theorems 6 and 8.

Proof (of Theorems 6 and 8). The proof of Theorem 6 is similar to the proof of part (A) of Theorem 8 given below. First suppose that P (x, 0) < 0. Fix 1 + \frac{1}{p} < s < λ + 2 − α

and f ∈ H_{p}^{s−2+α}(Ω), g ∈ B_{p}^{s−δ− \frac{1}{p} ,p}(∂Ω) and h ∈ B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ) (again, we consider only case (II)). Choose f_{k} ∈ C^{∞}(Ω), g_{k} ∈ C^{∞}(∂Ω), h_{k} ∈ C^{∞}(Γ) and P_{k} ∈ C^{∞} (Ω) such that

\begin{matrix}∥f − f_{k}∥_{H_{p}^{s−2+α}(Ω)} + ∥g − g_{k}∥_{B_{p}^{s−δ− \frac{1}{p} ,p}(∂Ω)} + ∥h − h_{k}∥_{B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ)} → 0 as k → ∞, \\ ∥P − P_{k}∥_{Λ^{λ}(Ω)} → 0 as k → ∞. \end{matrix}

Then by a theorem in [Win], there is u_{k} ∈ Λ^{λ+1} (Ω) satisfying

\{ \begin{matrix}P_{k}u_{k} & = & f_{k} & in Ω \\ ℓ^{→} u_{k} & = & g_{k} & on ∂Ω \\ u_{k} & = & h_{k} & on Γ \end{matrix} .

But the two earlier Propositions, together with the apriori estimates in Proposition 15, show that

∥u_{k}∥_{H_{p}^{s}(Ω)} ≤ C, for all k.  (45)

Thus given ϵ > 0 with 1 + \frac{1}{p} < s − ϵ, there is a subsequence, which we continue to denote by {u_{k}}, that converges to u in H_{p}^{s−ϵ} (Ω). It follows upon letting k → ∞ that

\{ \begin{matrix}P u & = & f & in Ω \\ ℓ^{→} u & = & g & on ∂Ω \\ u & = & h & on Γ \end{matrix} .

Finally, by (45), we obtain ∥u∥_{H_{p}^{s}(Ω)} ≤ C also.

For general P, choose a constant c sufficiently negative that the coefficient of the zero order term of P + c < 0. By the result just proved, we can solve the problem

\{ \begin{matrix}(P + c) u & = & f & in Ω \\ ℓ^{→} u & = & g & on ∂Ω \\ u & = & h & on Γ \end{matrix} ,

for u ∈ H_{p}^{s} (Ω) given that f ∈ H_{p}^{s−2+α}(Ω), g ∈ B_{p}^{s−δ− \frac{1}{p} ,p}(∂Ω) and h ∈ B_{p}^{s− \frac{γ}{p} − \frac{1}{p} ,p}(Γ). But then P u = f − cu where cu is smoother than f and functional analysis now completes the proof.

Finally, we consider the necessary assertions in part (B) of Theorem 8. If for some ϵ > 0 there is, for every f ∈ H_{p}^{s−2+α} (Ω), a function u ∈ H_{p}^{s} (Ω) with P u − f ∈ H_{p}^{s+ϵ} (Ω),

( ℓ^{→} +b) u |_{∂Ω}≡ g ∈ B_{p}^{s− \frac{1}{p} ,p} (∂Ω) and u |_{Γ}≡ h ∈ B_{p}^{s− \frac{1}{p} ,p} (∂Ω), then for any c ∈ R, we have

\{ \begin{matrix}(P + c) u & = & f + cu + (P u − f) & in Ω \\ ℓ^{→} u & = & g & on ∂Ω \\ u & = & h & on Γ \end{matrix} .

Now choose c sufficiently negative and let N denote the Neumann operator for P + c. Then by Proposition 14,

ρT ∘ L_{0}^{−1} (ρ̃N (f + cu + (P u − f))) ∈ B_{p}^{s− \frac{1}{p} ,p} (∂Ω) ,

since the remaining terms in (36) are automatically in this Besov space. Since T maps

B_{p}^{t,p} (∂Ω) to B_{p}^{t− \frac{1}{(3+λ)p} ,p} (∂Ω) by Theorem 31 in [GSII], we conclude that

ρT ∘ L_{0}^{−1} (ρ̃Nf) ∈ B_{p}^{s+ϵ− \frac{1}{(3+λ)p} − \frac{1}{p} ,p} (∂Ω)

if ϵ ≤ \frac{1}{(3+λ)p} − α since both u and P u − f are smoother than f by order ϵ. Thus, since the Neumann operator is surjective (for c sufficiently negative), and L_{0}^{−1} is elliptic on the cone

\{|τ| < λ |ξ|\}, we see that ρφ ∘ T ∘ φρ̃ is bounded from B_{p}^{s− \frac{1}{p} ,p} (U) to B_{p}^{s+ϵ− \frac{1}{(3+λ)p} − \frac{1}{p} ,p} (∂Ω) if ϵ ≤ \frac{1}{(3+λ)p} − α. So by Theorem 31 and Lemma 26 in [GSII],

\begin{aligned}ℓ^{→} satisfies & A_{p,β}^{∓} on Γ, \\ & A_{p,β}^{=} on (Σ^{+} ∩ N) \ Γ^{∗}, \\ & A_{p,β}^{‡} on (Σ^{−} ∩ N) \ Γ^{∗}, \end{aligned}  (46)

with β = \frac{1}{(3+λ)p} − ϵ. Now repeat this argument to obtain that ℓ^{→} satisfies (46) with β =

\frac{1}{(3+λ)p} − 2ϵ (provided of course that β ≤ α). In this way we finally see that ℓ^{→} satisfies (46) with β = α. This completes the proof of part (B)(i) of Theorem 8. The proofs of the two other assertions in part (B) are similar if we make use of Theorems 34 and 37 in [GSII] in place of Theorem 31. We do not repeat the details.

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