Positive powers of the Laplacian in the half-space under Dirichlet boundary conditions

We present explicit formulas for solutions to nonhomogeneous boundary value problems involving any positive power of the Laplacian in the half-space. For non-integer powers the operator becomes nonlocal and this requires a suitable extension of Dirichlet-type boundary conditions. A key ingredient in our proofs is a point inversion transformation which preserves harmonicity and allows us to use known results for the ball. We include uniqueness statements, regularity estimates, and describe the growth or decay of solutions at infinity and at the boundary.


INTRODUCTION
In this paper we study explicit formulas for solutions to nonhomogeneous boundary value problems for any positive power s > 0 of the Laplacian (−∆) s in the half-space R N + := {x ∈ R N : x 1 > 0}. Such explicit formulas play a prominent role in Liouville-type theorems, scaling arguments, and in the study of qualitative properties (monotonicity, symmetry, etc.) of solutions to nonlinear equations, see [16,17,24].
In our results, we consider a consistent extension of Dirichlet boundary conditions to the higher-order fractional setting given by where σ ∈ (0, 1], k ∈ N, and x = (x 1 , z) ∈ R N + . Note that for σ = 1, the traces (1.1) reduce to the usual (inward) normal derivatives associated to Dirichlet boundary conditions for the polyharmonic operator.
(1.2) If σ = 1 then (−∆) s = (−∆) m+1 is the usual iterated Laplacian, that is, and, if σ ∈ (0, 1), (−∆) s = (−∆) m+σ becomes a nonlocal operator and can be evaluated pointwisely using finite differences where c N,s is a positive normalization constant whose precise value can be found in (1.22) below. The integral (1.3) is finite whenever u is locally C 2s+α and belongs to L 1 s (see the notation section below for definitions), but it cannot be computed explicitly in general and this is one of the main difficulties in the nonlocal setting. For more information on the pointwise evaluation (1.3), see [3]. Since (−∆) s is a nonlocal operator for noninteger s, in this case one can prescribe data also on the complement of the half-space. In the following we use (x 1 ) α + to denote the function 0 if x 1 ≤ 0 (also if α < 0) and x α 1 if x 1 > 0. Our first result shows that the nonlocal Poisson kernel for the half-space is given by Γ s (x, y) := (−1) m γ N,σ (x 1 ) s + (−y 1 ) s |x − y| N for x ∈ R N and y ∈ R N \R N + , (1.4) where γ N,σ is a positive normalization constants, see (1.20) below. Observe that the kernel Γ s alternates sign depending on the parity of m. The case m = 0, i.e., s ∈ (0, 1), is remarked in [8, equation (3.40)].
Theorem 1.1. Let s ∈ (0, ∞)\N, r > 0, g ∈ L 1 (R N ) such that g has compact support in R N \R N + and let u : R N → R be given by (1.5)

6)
and there is some C(N, g, s) = C > 0 such that Moreover, u is the unique solution of (1.6) satisfying (1.7). If, in addition, g = 0 is nonnegative then there is C > 0 depending on g, N, and s such that (1.8) In particular, u given by (1.5) is a solution of (1. 2) with f = h k = 0 for all k ∈ {0, . . . , m}. Note also that, by (1.8), u can be negative even if the data of the problem is nonnegative. The assumption that g = 0 in {x ∈ R N : x 1 > −r} is needed to guarantee integrability in (1.5); this assumption can be weakened, but data which is nonzero at ∂ R N + necessarily requires a different kernel, see for example [4,Theorem 1.6] for the ball case.
Our next result concerns the Green function for (−∆) s in R N + , which is defined by Then u ∈ C 2s+β (R N + ) is a solution of (−∆) s u = f in R N + ,

10)
and there is some C(N, f , s) = C > 0 such that Moreover, u is the unique solution of (1.10) satisfying (1.11).
That (1.9) is the Green function for (−∆) s in R N + was known for s ∈ (0, 1) ∪ N, see [16, equation (3.1)] and [17,Remark 2.28]. Note that G s is a positive kernel, and therefore the half-space enjoys a positivity preserving property. For more information on maximum principles for (−∆) s , see [4,5]. We also point out that the Green function is not uniquely determined, since one can always add suitable harmonic functions (see Proposition 1.4 below); however, (1.9) is often referred to as the Green function for the half-space (because of its relationship with the ball's Green function via the Kelvin transform), and in the following we use this convention as well.
Finally, we introduce the boundary Poisson kernels for the half-space given by where α l,k are normalization constants, see (1.20) below, and ⌊a⌋ denotes the integer part of a. For s ∈ N, the kernels (1.12) were introduced in [13] using the equivalent expression In [13] a representation formula for a restricted set of functions is shown using (1.13) and it is also stated the conjecture that these kernels provide pointwise solutions for nonhomogeneous Dirichlet polyharmonic problems (under some smoothness and growth assumptions on the data, see [13,Satz 3]). Our next theorem is new even in the local case and shows, in particular, that the conjecture in [13,Satz 3] is true (at least) for compactly supported data. In the following x ′ = 0 if N = 1.
See also Theorem 4.13 below for more estimates on each of the kernels E k,s . We remark that the kernels E m−1,s and E m,s are connected via the trace operators with the Green function G s , see (4.40) and (4.41) below; however, the relationship between E k,s and G s is not so simple for k ≤ m − 2, see Remark 4.16. These identities are relevant to treat more general domains and for integration by parts formulas, for which only partial results are currently known in the fractional setting, see [20,26]. We also note that a boundary kernel for the half-space in the case s ∈ (0, 1) can be found in [8, equation (3.38)].
In general, one can find s-harmonic functions with much larger growth at infinity than those constructed in Theorem 1.3, as the next proposition shows. Proposition 1.4. Let m ∈ N 0 , σ ∈ (0, 1], and s = m + σ > 0. Then Observe that (x 1 ) σ −1 + is large or singular at ∂ R N + in the sense that it diverges as x 1 → 0. This is also the case of u as in Theorem 1.3 if h 0 = 0. These large s-harmonic functions are a purely nonlocal effect which can also be observed in other domains [1,2,8,19]. The proof of Proposition 1.4 is done first in dimension one N = 1 using several pointwise manipulations, and then extended to higher-dimensions using the following result.
for all x ∈ U.
In particular, u is s-harmonic in U if and only if v is s-harmonic in V .
A direct calculation (see Lemma 4.7 below) shows that the s-harmonic functions x k+σ −1 1 have the following relationship to the summands in (1.12): let σ ∈ (0, 1), m ∈ N, and k ∈ {0, . . ., m}, then In particular, this shows the strong influence that compactly supported data has on the growth or decay of solutions at infinity. The fact that our results are presented for compactly supported functions is not only for a better presentation, but also because general data yield a more complex problem and even for the Laplacian s = 1 this is an active research topic, see for example [28]. To mention one difficulty, without compact support bounds such as (1.7), (1.11), or (1.16) may not hold, and our uniqueness argument (see Lemma 3.4 below) cannot be applied; in fact, without growth assumptions uniqueness does not hold, by Proposition 1.4.
One of the key ingredients in our proofs is the following point inversion transformation.
The next proposition states that K s preserves s-harmonicity.
Proposition 1.6. For s, c > 0, v ∈ R N let κ and K s as in (1.17).
As a consequence, if U ⊂ R N \ {−v} is an open set and u ∈ L 1 s , then u is distributionally s-harmonic in U if and only if K s u is distributionally s-harmonic in κ(U). Proposition 1.6 can be deduced from [12,Lemma 3], where covariance under Möbius transformations is studied using a unique continuation argument. Here we present a different proof of Proposition 1.6 based on induction, which could be of independent interest.
Observe that, if c = 1 and v = 0, then κ is the usual Kelvin transform which maps B\{0} to R N \B conformally and vice versa; whereas, if c = 2 and v = e 1 , then κ maps B to the half-space R N + and ∂ B\{−e 1 } to ∂ R N + . Proposition 1.6 was known for s ∈ (0, 1), see for example [10,15], and the case s ∈ N is classical.
The K s transformation allows us to establish a link between (1.2) and its equivalent problem on balls, for which solutions and representation formulas are known [4]. However, the case of higher-order boundary Poisson kernels is rather delicate, since K s does not map directly the trace operator of the ball to that of the half-space; in particular, these kernels require subtle and intricate combinatorial identities, which are inherent to higher-order problems.
We remark that Poisson kernels for the half-space associated to a large class of (local) higher-order elliptic operators satisfying general boundary conditions were obtained in [6]. Due to their generality, the formulas from [6] are much less explicit than the ones presented in Theorem 1.3.
The paper is organized as follows. In Section 2 we collect some useful results concerning integration by parts, properties of the trace operator (1.1), distributional solutions, and the explicit formula for the Green function in a ball. Section 3 is devoted to properties of point-inversion transformations and the proof of Proposition 1.6. Finally, Section 4 contains the proofs of Theorems 1.1, 1.2, 1.3, Proposition 1.4, and Lemma 1.5; this section is divided in 4 subsections dedicated respectively to the nonlocal Poisson kernel, the Green function, one-dimensional s-harmonic functions, and the boundary Poisson kernels.
1.1. Notation. For m ∈ N 0 and U ⊂ R N open we write C m,0 (U) to denote the space of m-times continuously differentiable functions in U and, for σ ∈ (0, 1] and s = m + σ , we write C s (U) := C m,σ (U) to denote the space of functions in C m,0 (U) whose derivatives of order m are (locally) σ -Hölder continuous in U or (locally) Lipschitz continuous in U if σ = 1. We denote by C s (U) the set of functions u ∈ C s (U) such that We use u + := max{u, 0} to denote the positive part of u and (x 1 ) β + , β ∈ R, to denote the function to denote derivatives with respect to x, whenever they exist in some appropriate sense.
, (1.20) where Γ is the standard Gamma function. Furthermore, for a set A ⊂ R N we let |A| denote its Ndimensional Lebesgue measure. For x ∈ R N we sometimes write . We also use the standard multi-index notation: We use δ j,k to denote a Kronecker delta, that is, δ j,k = 1 if j = k and δ j,k = 0 if j = k. We often use the identity [14, equation (16) page 10] The constant in (1.3) is a positive normalization constant given by which allows the following relationship: if F ( f ) denotes the Fourier transform of f , then see, for example, [3,Theorem 1.9]. Finally, in dimension one (N = 1), the boundary integral is meant in

PRELIMINARY RESULTS AND DEFINITIONS
2.1. Integration by parts formula. We use the following integration by parts formula from [3].
We now show some frequently used properties of the trace operator D k+σ −1 defined in (1.1). Observe that, by definition, for any j ∈ N 0 , holds. Moreover, we have the following lemma.
By assumption, f ∈ C k ([0, ∞)) and Hence, by Taylor's theorem and then For sufficiently smooth functions, u is pointwisely s-harmonic if and only if u is distributionally sharmonic.
Proof. The result follows from Lemma 2.1 and the fundamental Lemma of calculus of variations.
The Green function of (−∆) s in the unitary ball B (see [2,12]) is given by where k N,s is as in (1.20).

POINT INVERSIONS
Proof. Identity (3.1) follows from differentiating the function with respect to t and evaluating at t = 1. Identity (3.3) follows from (1.23). To show (3.2) we use the Fourier transform of u, denoted by F (u). First, let ξ ∈ R N \{0} and note that and (3.2) follows by applying the inverse Fourier transform. Proof. Since , we assume without loss of generality that v = 0 and c = 1. We show (3.6) by induction. For s ∈ (0, 1) the claim is known (see, for example, [25, Proposition A.1]) and for s = 1 the claim follows from a direct computation. Let u ∈ C ∞ c (R N \{0}), s > 0, and assume as induction hypothesis that , and since Using (3.2) and the linearity of K s , we deduce that as claimed.
As a consequence, u is distributionally s-harmonic in U if and only if K s u is distributionally s-harmonic in κ(U), as claimed.
3.1. Uniqueness. Now we can use K s (as in (1.17) with c = 2 and v = e 1 ) to establish uniqueness of solutions to homogeneous problems in R N + .
Proof. Let u be as stated and note that Moreover, by adjusting the constant C, we also have Then, using the monotonicity of the Kelvin transform, we have for x ∈ B\{−e 1 }, Since w is s-harmonic in B, by Proposition 1.6, and satisfies that w = 0 in R N \B and |w(x)| ≤ Cδ (x) s−1+α for x ∈ B, we have that w ≡ 0, by [4,Theorem 1.5]. Then u(x) = 2 N−2s K s (w)(x) = 0 for x ∈ R N \{−e 1 }, and the statement follows.
, K s and κ be given by (1.17) with v = e 1 , c = 2, and let γ N,σ as in (1.20). Note that K s g ∈ L 1 s , by Lemma 3.3, and K s g = 0 on R N \ B 1+r (0) for somer ∈ (0, 1). Hence, by [4, Theorem 1.1], the function v : Recall that the absolute value of the Jacobian for y → κy is 2 N |x + v| −2N and, for x, y ∈ R N \ {−e 1 }, and, by Lemma 3.3, u = K s v ∈ L 1 s . By Proposition 1.6 and Lemma 2.3, it follows that (−∆) s u(x) = 0 for every x ∈ R N + . Finally, to see that u ∈ C s 0 (R N + ) and (1.7), note that |v(x)| ≤ C(1 − |x| 2 ) s for x ∈ B and for some C > 0 (depending on g, N, and s); but then, by linearity, and thus |u(x)| ≤ C4 s |x + e 1 | −N x s 1 for x ∈ R N + which implies (1.7). Note that u is the unique solution of (1.6) satisfying (1.7), by Lemma 3.4.
We now argue (1.8). Assume in addition that g is nonnegative and let u = K s v in R N \{−e 1 } where v is given by (4.1), which implies that (−1) m u > 0 in R N + , since γ N,σ > 0 by (1.20) and K s preserves positivity. The upper bound in (1.8) follows form (4.3) and the lower bound can be argued similarly since, for x ∈ B, Similarly, as in the proof of Proposition 1.6 it follows that which shows that G U is a Green function for (−∆) s in U. Proof. Let κ = 2 x+e 1 |x+e 1 | 2 − e 1 (i.e. c = 2, v = e 1 ), then κ(B) = R N + = {x ∈ R N : x 1 > 0} and the claim follows from Proposition 4.1 and a simple direct calculation.

4.3.
One-dimensional s-harmonic functions. In this subsection we show Lemma 1.5 and Proposition 1.4. We mention that a similar result to Lemma 1.5 is remarked in [23, proof of Theorem 3] for s ∈ (0, 1).
To close this subsection, we show that the s-harmonic functions in Proposition 1.4 can be obtained using suitable boundary kernels. These kernels play a prominent role in the next subsection where we construct solutions of the nonhomogeneous problem (1.15).

28)
and sup Proof. Observe that, for x ∈ R N + , Let R > 0 be such that supp(h) ⊂ B R (0) ⊂ R N−1 and L as in (4.17). Then, by Lemma 4.10, we have that sup x, x∈L, Fix z ∈ B R (0) and note that, for x ∈ L 0 , we have |x − z| ≥ 1 and therefore by (4.30),(4.31), and the proof is finished.
We are ready to show the main theorem of this section. Then and, for k ∈ {0, . . . , m}, Proof. If N = 1 and k ∈ {0, . . . , m}, then, by Lemma 4.9 (with j = k), we have for which is s-harmonic, by Proposition 1.4, and (4.33), (4.34), (4.35) are clearly satisfied. Now, let N ≥ 2, k ∈ {0, . . ., m}, and note that Lemma 4.12 implies that each summand is (s − i)-harmonic, which in turn implies that v k is pointwisely s-harmonic in R N + (as we argued in (4.13)). We now show (4.35). Observe that, by a change of variables, But then, (4.35) follows from (1 + |y ′ | 2 ) N+m−k 2 and the fact that h is uniformly bounded. The claim (4.34) follows from Lemma 4.11 and the fact that, for x ∈ R N + , Finally, we show (4.33). Fix x ′ ∈ ∂ R N + and let j ∈ {k, . . . , m}, then, by (4.36), = lim (4.37) = lim Observe that the integral in (4.38) is finite, by Lemma 4.8, and the interchange between derivative and integral in (4.37) can be justified as in the proof of Lemma 4.11 using that h is compactly supported. If there is some γ i odd, then by a change of variables, So we may assume that j − k is even. Then, by Lemmas 4.8 and 4.9, Then, since D s−2 z = 0 on ∂ R N + , Iterating this argument, we obtain but then z ≡ 0 in R N , by Lemma 3.4, and the uniqueness follows.
Observe that These kernels are connected via the trace operators with the Green function G s , see (4.40) and (4.41) below; however, the relationship between E k,s and G s is not so simple for k ≤ m − 2, see Remark 4.16.
Lemma 4.14. Let m ∈ N 0 , σ ∈ (0, 1], s = m + σ , x ∈ R N , and z ∈ ∂ R N + , then E m,s (x, z) and, if m ≥ 1, where we used that |x −z| = |x − z| for z ∈ ∂ R N + , z 1 = 0. But then equation (4.41) follows, since On the other hand, for any z ∈ ∂ R N + we can also compute where we have used an equivalent expression for G s to the one in (1.9). Let us first remark that 2 ω N = 4 s Γ(s) 2 k N,s (see (1.20)). Then, as y → z.
Thus, we have  where G s is as in (1.9), then Lemma 4.14 shows that E m,s is the Martin kernel for the half-space (up to a constant). (ii) Let s > 1, y ∈ R N + , and P s−1 as in (4.42), then P s−1 (x, y) = ω N 2 ∂ R N + E m−1,s−1 (x, z)E m,s (y, z) dz for x ∈ R N + , where E k,s are given in (1.12). A similar relationship was observed in the ball, see [4]. (iii) As in the ball case, the kernel Γ s satisfies a recurrence formula (see [4,Proposition 3.1]). Let m ∈ N, σ ∈ (0, 1), s = m + σ , and Γ s as in (1.4).Then Γ s (x, y) = Γ s−1 (x, y) − ∂ R N + E m,s (x, z)D s−1 Γ s−1 (z, y) dz for x ∈ R N + , y ∈ R N \ R N + .
(iv) As a consequence of Theorem 1.2 and Lemma 4.14 one can deduce a higher-order Hopf Lemma for the homogeneous Dirichlet problem as in [17,Theorem 5.7] and [4,Corollary 1.9] where the case of a ball is considered and the positivity of the trace D s u on the boundary of the domain can be obtained if (−∆) s u is positive. Note however that in the half-space one requires additional growth assumptions at infinity for the solution (such as (1.7), (1.11), or (1.16)). Without these assumptions, the function x s 1 (which is s-harmonic by Proposition 1.4) would be a counterexample.
(v) Similar arguments to those presented for the half-space can be used to deduce the kernels for the complement of the ball B c := {x ∈ R N : |x| > 1} using the Kelvin transform (K s as in (1.17) with c = 1 and v = 0). Here, similarly as in the ball case (see [4]), for k ∈ N 0 the boundary trace operator for B c is given by The associated Green function and nonlocal Poisson kernels are see also [9, equation (8.6)] and [22, equation (1.6.11)] for the case s ∈ (0, 1). Boundary Poisson kernels for B c can also be obtained via the Kelvin transform using the boundary Poisson kernels for the ball given in [4]. for j odd.
Actually, a closer look at integration by parts formulas [24,Lemma 8] suggests that, for j ∈ {0, . . . , m}, the function u j : R N + → R given by u j (x) = ∂ R N + K j,m+1 (x, y)h(y) dy is a solution of (−∆) m+1 u j = 0 in R N + and, for k ∈ {0, . . . , m}, (−∆) k 2 u j = δ j,k h for k even, ∂ x 1 (−∆) k−1 2 u j = δ j,k h for k odd on ∂ R N + . The relationship between E k,s and G s does not seem to be simple in general and therefore a generalization of our results to more general domains is not immediate.