Mathematische Annalen manuscript No. (will be inserted by the editor)

Positivity for equations involving polyharmonic operators with Dirichlet boundary conditions

Hans-Christoph Grunau^{1}, Guido Sweers^{2}

1 Mathematisches Institut, Universität Bayreuth, D-95440 Bayreuth, Germany 2 Vakgroep Algemene Wiskunde, Technische Universiteit Delft, Postbus 5031, 2600 GA Delft

Netherlands

Received: 13 April 1995 /Revised version: 6 March 1996

Mathematics Subject Classification (1991): 35J40; 35B50.

1 Introduction

Cranston, Fabes and Zhao ([26], [5]) established the uniform bound

sup ; \begin{matrix}x, y ∈ Ω \\ x ≠ y \end{matrix} \frac{∫_{Ω} G_{1,n} (x, z) G_{1,n} (z, y) dz}{G_{1,n} (x, y)} ≤ M < ∞,  (1)

where G_{1,n} (x, y) is the Green function for the Laplacian −Δ with Dirichlet boundary conditions on a Lipschitz domain Ω ⊂ ℝ^{n} with n ≥ 3 (see [27] for n = 2). This estimate was used in [23] and [18] to obtain positivity, uniformly with respect to f ≥ 0, for noncooperative elliptic systems as:

\{ \begin{matrix}−Δu & = & f − εv & in Ω, \\ −Δv & = & u & in Ω, \\ u = v & = & 0 & on ∂Ω, \end{matrix}

when ε ≥ 0 is small. In particular, Δ^{2}v + εv ≥ 0 in Ω, with v = Δv = 0 on ∂Ω, implies v ≥ 0 for ε small. In numerical experiments ([14]) for one dimension a similar behaviour was observed under Dirichlet boundary conditions v = \frac{∂}{∂n}v = 0.

In this paper we will derive a 3-G type theorem as in (1) but with G_{1,n} replaced by the Green function G_{m,n} for the m-polyharmonic operator with Dirichlet boundary conditions and with Ω replaced by the unit ball B in ℝ^{n} with n ≥ 1:

\{ \begin{matrix}(−Δ)^{m} u & = & f & in B, \\ D_{m}u & = & 0 & on ∂B, \end{matrix}  (2)

where

D_{m}u = (D^{k}u)_{k∈ℕ^{n},|k|≤m−1} .

We use a multi-index k ∈ ℕ^{n} with D^{k} = (\frac{∂}{∂x_{1}} )^{k_{1}} … (\frac{∂}{∂x_{n}} )^{k_{n}} and |k| = k_{1} +…+k_{n}. ℕ is the set of nonnegative integers. Notice that the Dirichlet boundary condition may also be written as

(u, \frac{∂}{∂r} u, \frac{∂^{2}}{∂r^{2}} u, …, \frac{∂^{m−1}}{∂r^{m−1}} u) = 0 on ∂B.

Boggio in [4] (see also [9], [10]) showed positivity of G_{m,n} on the ball B. Here we will establish two-sided pointwise estimates for G_{m,n} and estimates from above for derivatives of G_{m,n}. These estimates enable us to prove 3-G Theorem type results.

For these results we will only use the ball in ℝ^{n} as a domain. In contrary to the Laplacian the Green function for the biharmonic or polyharmonic operator is not positive for arbitrary domains. Hence the estimate of G_{m,n} from below (by a positive function) necessarily restricts the possible domains. The question whether or not the estimates from above remain true on arbitrary smooth and bounded domains, is open as far as the authors know. Counterexamples of sign changing superpolyharmonic functions can be found in [6], [7, p. 275], [17] and [22].

The 3-G results are used in three directions. First, one can prove that on B also the inverse of polyharmonic operators that are perturbed by small lower order terms, are positivity preserving. That is, if the coefficients a_{k} in

(−Δ)^{m} u + ∑_{|k|<2m} a_{k} (·) D^{k}u = f,  (3)

are sufficiently small, f positive implies that the solution u of the Dirichlet problem for (3) is positive. Secondly, as in [24], we obtain positivity results for systems of these operators. Thirdly, estimates for these Green functions on the unit ball in ℝ^{n} can be used to obtain local maximum principles for solutions of differential inequalities in arbitrary domains. Results in this last direction will be studied in a forthcoming paper [11]. As a consequence we will show in [11] existence of classical solutions of higher order semilinear elliptic equations for a much wider class as in [9]-[10].

2 Estimates for G_{m,n}

We will use the following notations.

i. B = {x ∈ ℝ^{n}; |x| < 1} with n ≥ 1. ii. s ∧ t = min (s, t) and s^{+} = max (s, 0) for s, t ∈ ℝ. iii. G_{m,n} is the Green operator for (2), that is, u = G_{m,n}f with

(G_{m,n}f) (x) = ∫_{B} G_{m,n} (x, y) f (y) dy.

iv. For x, y ∈ B:

\begin{aligned}[xy] & = & |x − y| , \\ [XY] & = & ||x| y − \frac{x}{|x|} | , \\ d (x) & = & 1 − |x| , \\ Θ (x, y) & = & [XY]^{2} − [xy]^{2} . \end{aligned}

Positivity for polyharmonic Dirichlet problems  3

Note that [XY] = \sqrt{|x|^{2} |y|^{2} − 2 x · y + 1} = [Y X] and that d (x) is the distance of x to the boundary. Moreover

Θ (x, y) = (1 − |x|^{2}) (1 − |y|^{2}) .  (4)

Hence [XY] > [xy] for x, y ∈ B. Lemma 2.1 (Boggio) [4, p. 126] The Green function G_{m,n} is as follows:

G_{m,n} (x, y) = k_{m,n} |x − y|^{2m−n} ∫_{1}^{[XY]/[xy]}\frac{(v^{2} − 1)^{m−1}}{v^{n−1}} dv  (5)

with k_{m,n} some fixed positive constants. Remark 1. For m = 2 one finds

G_{2,n}(x, y) =

= c_{n} \begin{cases}[xy]^{3} − [XY]^{3} + \frac{3}{2} & Θ (x, y) & [XY] & for n = 1, \\ [xy]^{2} (2 log [xy] − 2 log [XY]) + & Θ (x, y) & for n = 2, \\ − [xy] + [XY] − \frac{1}{2} & Θ (x, y) & [XY]^{−1} & for n = 3, \\ −2 log [xy] + 2 log [XY] − & Θ (x, y) & [XY]^{−2} & for n = 4, \\ [xy]^{−n+4} − [XY]^{−n+4} − \frac{n−4}{2} & Θ (x, y) & [XY]^{−n+2} & for n ≥ 5, \end{cases}

with

c_{n} = (2 |(n − 4) (n − 2)| S_{n})^{−1} for all n ∉ {2, 4} ,
c_{n} = (8S_{n})^{−1} for n ∈ {2, 4} .

S_{n} is the area of the surface of the unit ball in ℝ^{n} (S_{1} = 2).

Starting from the explicit formula we will prove a two-sided estimate. But to simplify the statements we will first define an ordering and a related equivalency relation for nonnegative functions. Definition 2.2 Let f, g ≥ 0. We call f ∼ g on Ω, iff there are c_{1}, c_{2} > 0 such that

c_{1}g (x) ≤ f (x) ≤ c_{2}g (x) for all x ∈ Ω.

We will say f ⪯ g on Ω, iff there is c > 0 such that f (x) ≤ c g (x) for all x ∈ Ω.

Proposition 2.3 On B^{2} (that is (x, y) ∈ B^{2}) we have the following. i. For 2m < n :

G_{m,n} (x, y) ∼ |x − y|^{2m−n} (1 ∧ \frac{d (x)^{m} d (y)^{m}}{|x − y|^{2m}} ) .

ii. For 2m = n :

G_{m,n} (x, y) ∼ log (1 + \frac{d (x)^{m} d (y)^{m}}{|x − y|^{2m}} ) .

iii. For 2m > n :

G_{m,n} (x, y) ∼ (d (x) d (y))^{m− \frac{1}{2} n} (1 ∧ \frac{d (x)^{\frac{1}{2} n} d (y)^{\frac{1}{2} n}}{|x − y|^{n}} ) .

Proposition 2.4 Let k ∈ ℕ^{n} with |k| ≤ 2m. Then on B^{2} we have the following.

i. For |k| ≥ 2m − n and n odd, or, |k| > 2m − n and n even:

(a) if |k| ≤ m then

|D_{x}^{k}G_{m,n} (x, y)| ⪯ |x − y|^{2m−n−|k|} (1 ∧ \frac{d (x)^{m−|k|} d (y)^{m}}{|x − y|^{2m−|k|}} ) ;

(b) if |k| ≥ m then

|D_{x}^{k}G_{m,n} (x, y)| ⪯ |x − y|^{2m−n−|k|} (1 ∧ \frac{d (y)^{m}}{|x − y|^{m}} ) .

ii. For |k| = 2m − n and n even: (a) if |k| ≤ m (that is m ≤ n) then

|D_{x}^{k}G_{m,n} (x, y)| ⪯ log (2 + \frac{d (y)}{|x − y|}) (1 ∧ \frac{d (x)^{m−|k|} d (y)^{m}}{|x − y|^{2m−|k|}} ) ;

(b) if |k| ≥ m (that is m ≥ n) then

|D_{x}^{k}G_{m,n} (x, y)| ⪯ log (2 + \frac{d (y)}{|x − y|}) (1 ∧ \frac{d (y)^{m}}{|x − y|^{m}} ) .

iii. For |k| ≤ 2m − n and n odd, or, |k| < 2m − n and n even: (a) if |k| ≤ m − \frac{1}{2}n then

|D_{x}^{k}G_{m,n} (x, y)| ⪯ d (x)^{m− \frac{1}{2} n−|k|} d (y)^{m− \frac{1}{2} n} (1 ∧ \frac{d (x)^{\frac{1}{2} n} d (y)^{\frac{1}{2} n}}{|x − y|^{n}} );

(b) if m − \frac{1}{2}n ≤ |k| ≤ m then

|D_{x}^{k}G_{m,n} (x, y)| ⪯ d (y)^{2m−n−|k|} (1 ∧ \frac{d (x)^{m−|k|} d (y)^{n−m+|k|}}{|x − y|^{n}} );

(c) if m ≤ |k| then

|D_{x}^{k}G_{m,n} (x, y)| ⪯ d (y)^{2m−n−|k|} (1 ∧ \frac{d (y)^{n−m+|k|}}{|x − y|^{n−m+|k|}} ) .

Remark 2. The Green function for the Laplacian (n > 2, m = 1) satisfies the estimates above on arbitrary bounded domains with C^{2,γ}-smooth boundary. This result (among others) was established by K. O. Widman in [25].

Remark 3. For p, q ≥ 0 one finds on B^{2}

(1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) ∼ (1 ∧ \frac{d (x)}{|x − y|})^{p} (1 ∧ \frac{d (y)}{|x − y|})^{q} ,  (6)

and if p + q > 0

log (1 + \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) ∼ log (2 + \frac{d (y)}{|x − y|}) (1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) .  (7)

These results follow from Lemma 3.2 viii. and ix.

Remark 4. From Lemma 3.3 it follows that for all three cases in Proposition 2.4 iii. we have

|D_{x}^{k}G_{m,n} (x, y)| ⪯ d (x)^{m− \frac{1}{2} n−|k|} d (y)^{m− \frac{1}{2} n} (1 ∧ \frac{d (x)^{\frac{1}{2} n} d (y)^{\frac{1}{2} n}}{|x − y|^{n}} ) .  (8)

In the case that m − \frac{1}{2}n ≤ |k| ≤ m both estimates (Prop. 2.4 iii.(b) and (8)) are equivalent as can be seen from Lemma 3.3 ii. When |k| > m the estimate in (8) is less sharp as the one in Prop. 2.4 iii. (c).

3 Proving the Green function estimates

We will repeatedly distinguish the cases [xy] ≥ \frac{1}{2} [XY] and [xy] ≤ \frac{1}{2} [XY]. The following Lemma will help us doing so.

Lemma 3.1 If [xy] ≥ \frac{1}{2} [XY], then

d (x) d (y) ≤ 3 |x − y|^{2} ,  (9)
max (d (x) , d (y)) ≤ 3 |x − y| .  (10)

If [xy] ≤ \frac{1}{2} [XY], then

\frac{3}{4} |x − y|^{2} ≤ \frac{3}{16} [XY]^{2} ≤ d (x) d (y) ,  (11)

\frac{1}{4}d (x) ≤ d (y) ≤ 4d (x) .  (12)

Proof. If [xy] ≥ \frac{1}{2} [XY], then d (x) d (y) ≤ [XY]^{2} − [xy]^{2} ≤ \frac{3}{4} [XY]^{2} which implies (9). The estimate in (10) follows from

d (x)^{2} ≤ d (x) (d (y) + |x − y|) ≤

≤ 3 |x − y|^{2} + d (x) |x − y| ≤

≤ 4 |x − y|^{2} + \frac{1}{4}d (x)^{2}

and a similar estimate for d (y).

If [xy] ≤ \frac{1}{2} [XY], then d (x) d (y) ≥ \frac{1}{4} ([XY]^{2} − [xy]^{2}) ≥ \frac{3}{16} [XY]^{2} which implies (11) and moreover

d (x) ≤ d (y) + |x − y| ≤ d (y) + \sqrt{\frac{4}{3}d (x) d (y)} ≤

≤ d (y) + \frac{1}{\sqrt{3}} (d (x) + d (y)) ≤ \frac{3}{5}d (x) + \frac{8}{5}d (y)

implies d (x) ≤ 4d (y). Similarly d (y) ≤ 4d (x) and (12) follows.

The following equivalencies will be used several times.

Lemma 3.2 On ℝ^{+} × ℝ^{+} (for s, t ∈ ℝ^{+}) we have:

i.  min (\frac{1}{s}, \frac{1}{t}) ∼ \frac{1}{s + t},
ii.  max (s, t) ∼ s + t.

On B^{2} (for x, y ∈ B) we have with p, q ≥ 0 fixed:

iii.  Θ (x, y) ∼ d (x) d (y) ,
iv.  [XY] ∼ d (x) + d (y) + |x − y| ,
v.  1 ∧ \frac{d (y)}{|x − y|} ∼ 1 ∧ \frac{d (y)}{d (x)} ∧ \frac{d (y)}{|x − y|},
vi.  1 ∧ \frac{d (x) d (y)}{|x − y|^{2}} ∼ \frac{d (y)}{d (x)} ∧ \frac{d (x)}{d (y)} ∧ \frac{d (x) d (y)}{|x − y|^{2}},

vii.  1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ∼ 1 ∧ \frac{d (x)^{p}}{|x − y|^{p}} ∧ \frac{d (y)^{q}}{|x − y|^{q}} ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ,

viii.  1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ∼ (1 ∧ \frac{d (x)}{|x − y|})^{p} (1 ∧ \frac{d (y)}{|x − y|})^{q} .

On B^{2} (for x, y ∈ B) we have with p, q ≥ 0 and p + q > 0 fixed:

ix. log (1 + \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) ∼ log (2 + \frac{d (y)}{|x − y|}) (1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) .

Remark 1. The statements in viii., ix. imply (6,7) in the remark following Proposition 2.4.

Proof. i. and ii. are straightforward; iii. follows from (4). iv. For the first step we use

[XY]^{2} = |x|^{2} |y|^{2} − 2 x · y + 1 ≥ (1 − |x| |y|)^{2} ≥ (1 − |x|)^{2} = d (x)^{2} .

With a similar estimate by d (y) and with (4) we find left ≥ \frac{1}{3}right. By (4) we also find

[XY] ≤ \sqrt{|x − y|^{2} + 4d (x) d (y)} ≤
≤ |x − y| + 2\sqrt{d (x) d (y)} ≤ |x − y| + d (x) + d (y) .

v. We use Lemma 3.1 to find the following. If [xy] ≥ \frac{1}{2} [XY], then \frac{d(y)}{|x−y|} ⪯ \frac{d(y)}{d(x)} and

if [xy] ≤ \frac{1}{2} [XY], then 1 ⪯ \frac{d(y)}{d(x)}. vi. The sixth equivalency is proven as follows. Since min (t, t^{−1}) ≤ 1 on ℝ^{+}, one

finds left⪰right. As in v. we find by Lemma 3.1 if [xy] ≥ \frac{1}{2} [XY], then

\frac{d (x) d (y)}{|x − y|^{2}} ⪯ \frac{d (x) d (y)}{d (x)^{2}} and \frac{d (x) d (y)}{|x − y|^{2}} ⪯ \frac{d (x) d (y)}{d (y)^{2}},

and if [xy] ≤ \frac{1}{2} [XY], then

1 ⪯ \frac{d (y)}{d (x)} and 1 ⪯ \frac{d (x)}{d (y)}.

vii. By Lemma 3.1 if [xy] ≥ \frac{1}{2} [XY] then

\frac{d (x)^{p}}{|x − y|^{p}} ∧ \frac{d (y)^{q}}{|x − y|^{q}} ⪰ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ,

and if [xy] ≤ \frac{1}{2} [XY] then

\frac{d (x)^{p}}{|x − y|^{p}} ∧ \frac{d (y)^{q}}{|x − y|^{q}} ⪰ 1.

viii. One uses again Lemma 3.1. If [xy] ≥ \frac{1}{2} [XY] then

(1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) ∼ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ∼ (1 ∧ \frac{d (x)}{|x − y|})^{p} (1 ∧ \frac{d (y)}{|x − y|})^{q} .

If [xy] ≤ \frac{1}{2} [XY] then

(1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) ∼ 1 ∼ (1 ∧ \frac{d (x)}{|x − y|})^{p} (1 ∧ \frac{d (y)}{|x − y|})^{q} .

ix. Lemma 3.1 implies if [xy] ≥ \frac{1}{2} [XY] that

log (1 + \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) ∼ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ∼ (1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) ∼

∼ log (2 + \frac{d (y)}{|x − y|}) (1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) .

If [xy] ≤ \frac{1}{2} [XY] then by Lemma 3.1

log (1 + \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) ∼ log (1 + \frac{d (y)^{p+q}}{|x − y|^{p+q}} ) ∼

∼ log ((2 + \frac{d (y)}{|x − y|})^{p+q}) ∼ log (2 + \frac{d (y)}{|x − y|}) ∼

∼ log (2 + \frac{d (y)}{|x − y|}) (1 ∧ \frac{d (x)^{p} d (y)^{q}}{|x − y|^{p+q}} ) .

Lemma 3.3 Let r, s, t ≥ 0.

i. If s + t ≥ 2r, then on B^{2} we have

(1 ∧ \frac{d (x)^{t} d (y)^{s}}{|x − y|^{t+s}} ) ⪯ (\frac{d (y)}{d (x)})^{r−t} (1 ∧ \frac{d (x)^{r} d (y)^{r}}{|x − y|^{2r}} ) .  (13)

ii. If s + t = 2r, then on B^{2} we have

(1 ∧ \frac{d (x)^{t} d (y)^{s}}{|x − y|^{t+s}} ) ∼ (\frac{d (y)}{d (x)})^{r−t} (1 ∧ \frac{d (x)^{r} d (y)^{r}}{|x − y|^{2r}} ) .  (14)

By symmetry one may interchange x, t with y, s.

Proof of Lemma 3.3. We will use Lemma 3.1 several times.

• The case [xy] ≤ \frac{1}{2} [XY]. We have

(1 ∧ \frac{d (x)^{t} d (y)^{s}}{|x − y|^{t+s}} ) ∼ 1 ∼ (\frac{d (y)}{d (x)})^{r−t} ∼

∼ (\frac{d (y)}{d (x)})^{r−t} (1 ∧ \frac{d (x)^{r} d (y)^{r}}{|x − y|^{2r}} ) .

• The case [xy] ≥ \frac{1}{2} [XY]. We have

(1 ∧ \frac{d (x)^{t} d (y)^{s}}{|x − y|^{t+s}} ) ∼ \frac{d (x)^{t} d (y)^{s}}{|x − y|^{t+s}} =

= (\frac{d (y)}{d (x)})^{r−t} \frac{d (x)^{r} d (y)^{r}}{|x − y|^{2r}} \frac{d (y)^{s+t−2r}}{|x − y|^{s+t−2r}} ⪯

⪯ (\frac{d (y)}{d (x)})^{r−t} \frac{d (x)^{r} d (y)^{r}}{|x − y|^{2r}} ∼

∼ (\frac{d (y)}{d (x)})^{r−t} (1 ∧ \frac{d (x)^{r} d (y)^{r}}{|x − y|^{2r}} ) .

Note that when s + t = 2r we may replace ⪯ by = in the previous estimate.

Before proving both propositions of the previous section we will introduce two convenient transformations of (5). For the case that [xy] ≥ \frac{1}{2} [XY] we will use the integral transformation with s = 1 − v^{−2} and we obtain the following expression for G_{m,n}:

G_{m,n} (x, y) = \frac{1}{2}k_{m,n} |x − y|^{2m−n} ∫_{s=0}^{A_{xy}}s^{m−1} (1 − s)^{\frac{1}{2} n−m−1} ds,  (15)

where

A_{xy} = \frac{[XY]^{2} − [xy]^{2}}{[XY]^{2}} .  (16)

We set

f_{m,n} (t) = ∫_{s=0}^{t} s^{m−1} (1 − s)^{\frac{1}{2} n−m−1} ds.  (17)

By straightforward calculus it follows that for t ∈ [0, 1], if m ≥ j > 0,

|f_{m,n}^{(j)} (t)| ⪯ t^{m−j} (1 − t)^{\frac{1}{2} n−m−j} (t^{j−1} + (1 − t)^{j−1}) ⪯
⪯ t^{m−j} (1 − t)^{\frac{1}{2} n−m−j} ,  (18)

and if m < j

|f_{m,n}^{(j)} (t)| ⪯ (1 − t)^{\frac{1}{2} n−m−j} (t^{m−1} + (1 − t)^{m−1}) ⪯ (1 − t)^{\frac{1}{2} n−m−j} .  (19)

For the case that [xy] ≤ \frac{1}{2} [XY] we use the transformation with w = v^{−1} and (5) becomes:

G_{m,n} (x, y) = k_{m,n} |x − y|^{2m−n}  ∫_{w=[xy]/[XY]}^{1} w^{n−2m−1} (1 − w^{2})^{m−1} dw.  (20)

We also set

g_{m,n} (t) = ∫_{w=t}^{1} w^{n−2m−1} (1 − w^{2})^{m−1} dw.  (21)

Proof of Proposition 2.3.

First we prove Proposition 2.3 i. which follows rather directly from (15). Indeed, if n > 2m, then one finds by distinguishing the cases 0 ≤ t ≤ \frac{1}{2} and \frac{1}{2} ≤ t ≤ 1 that f_{m,n} in (17) satisfies

f_{m,n} (t) ∼ t^{m} on [0, 1]

and hence by (15), using respectively Lemma 3.2 iii./iv., i. and vi., that

G_{m,n} (x, y) ∼ |x − y|^{2m−n} (A_{xy})^{m} ∼

∼ |x − y|^{2m−n} (\frac{d (x) d (y)}{d (x)^{2} + d (y)^{2} + |x − y|^{2}} )^{m} ∼

∼ |x − y|^{2m−n} (min (\frac{d (y)}{d (x)}, \frac{d (x)}{d (y)}, \frac{d (x) d (y)}{|x − y|^{2}}))^{m} ∼

∼ |x − y|^{2m−n} min (1, (\frac{d (x) d (y)}{|x − y|^{2}})^{m}) .

For Proposition 2.3 ii. and iii. we have to distinguish two cases.

• The case [xy] ≥ \frac{1}{2} [XY]. Then A_{xy} ≤ \frac{3}{4} and the formula in (15) implies

G_{m,n} (x, y) ∼ |x − y|^{2m−n} (A_{xy})^{m} ∼

∼ |x − y|^{2m−n} \frac{d (x)^{m} d (y)^{m}}{[XY]^{2m}} ∼

∼ \frac{d (x)^{m} d (y)^{m}}{[XY]^{n}} ∼

∼ \frac{d (x)^{m} d (y)^{m}}{(d (x) + d (y) + |x − y|)^{n}} ∼

∼ d (x)^{m− \frac{1}{2} n} d (y)^{m− \frac{1}{2} n} (min (\frac{d (x)}{d (y)}, \frac{d (y)}{d (x)}, \frac{d (x) d (y)}{|x − y|^{2}}))^{\frac{1}{2} n} ∼

∼ d (x)^{m− \frac{1}{2} n} d (y)^{m− \frac{1}{2} n} min (1, (\frac{d (x) d (y)}{|x − y|^{2}})^{\frac{1}{2} n}) .

In the last three steps we used respectively Lemma 3.2 iv., i. and vi. Notice that, if 2m = n, (9) implies

d (x)^{m− \frac{1}{2} n} d (y)^{m− \frac{1}{2} n} min (1, (\frac{d (x) d (y)}{|x − y|^{2}})^{\frac{1}{2} n}) ∼

∼ log (1 + (\frac{d (x) d (y)}{|x − y|^{2}})^{\frac{1}{2} n}) .

• The case [xy] ≤ \frac{1}{2} [XY]. Now it follows from (20) that we find

G_{m,n} (x, y) ∼ |x − y|^{2m−n} ∫_{w=[xy]/[XY]}^{1} w^{n−2m−1}dw ∼

∼ \{ \begin{matrix}if n = 2m : & log ([XY] / [xy]) ; \\ if n < 2m : & [XY]^{2m−n} . \end{matrix}

For n = 2m one uses log (1 + a) ∼ log (1 + a^{\frac{1}{2} n}) for a ∈ [\frac{3}{4}, ∞), and continues by

log (\frac{[XY]}{[xy]}) ∼ log (1 + \frac{[XY]^{2} − [xy]^{2}}{[xy]^{2}} ) ∼

∼ log (1 + \frac{d (x) d (y)}{|x − y|^{2}}) ∼ log (1 + (\frac{d (x) d (y)}{|x − y|^{2}})^{\frac{1}{2} n}) .

In the last step we used (11). For n < 2m one has, using respectively Lemma 3.2 iv., (11) and (12), and again (11)

[XY]^{2m−n} ∼ (d (x) + d (y) + |x − y|)^{2m−n} ∼
∼ (d (x) d (y))^{m− \frac{1}{2} n} ∼

∼ (d (x) d (y))^{m− \frac{1}{2} n} min (1, (\frac{d (x) d (y)}{|x − y|^{2}})^{\frac{1}{2} n}) .

Proposition 2.4 will be a consequence of the following lemmas. Lemma 3.4 We have on Ω_{1} := \{(x, y) ∈ B^{2}; [xy] ≥ \frac{1}{2} [XY]\} that i. if |k| ≤ m

|D_{x}^{k}G_{m,n} (x, y)| ⪯ |x − y|^{2m−n−|k|} (\frac{d (x)}{[XY]})^{m−|k|} (\frac{d (y)}{[XY]})^{m};

ii. if |k| ≥ m

|D_{x}^{k}G_{m,n} (x, y)| ⪯ |x − y|^{2m−n−|k|} (\frac{d (y)}{[XY]})^{m} .

Proof of Lemma 3.4. We will use (15). With the assumption [xy] ≥ \frac{1}{2} [XY] it is sufficient to estimate f_{m,n} (t) on [0, \frac{3}{4}] since 0 ≤ A_{xy} ≤ \frac{3}{4} for all (x, y) ∈ Ω_{1}. We find by (17-19) that

|f_{m,n}^{(j)} (t)| ⪯ t^{max(m−j,0)} on [0, \frac{3}{4}] for all j ≥ 0.  (22)

Since

A_{xy} = \begin{matrix}(1 − |x|^{2}) & (1 − |y|^{2}) \\ [XY] & 2 \end{matrix},

|D_{x}^{p} [XY]| ⪯ [XY]^{1−|p|} ,

and 1 − |x|^{2} ≤ 2d (x) ⪯ [XY] by Lemma 3.2 iv., it follows that on B^{2}

|D_{x}^{p}A_{xy}| ⪯ \frac{d (y)}{[XY]^{|p|+1}} .  (23)

By a tedious application of the chain rule we find that |D_{x}^{k}G_{m,n} (x, y)| ⪯

⪯ ∑_{p≤k} |D_{x}^{k−p} |x − y|^{2m−n}| |D_{x}^{p}f_{m,n} (A_{xy})| ⪯

⪯ ∑_{p≤k} |D_{x}^{k−p} |x − y|^{2m−n}| ∑_{j=0}^{|p|} |f_{m,n}^{(j)} (A_{xy})| ∑_{∑{j \atop i=1} p^{(i)}=p ; 1≤|p^{(i)}|} ∏_{i=1}^{j} |D_{x}^{p^{(i)}}A_{xy}| =  (24)

where the last sum is 1 for j = 0. Note that (23) implies

∑_{∑{j \atop i=1} p^{(i)}=p ; 1≤|p^{(i)}|} ∏_{i=1}^{j} |D_{x}^{p^{(i)}}A_{xy}| ⪯ \frac{d (y)^{j}}{[XY]^{|p|+j}} .

12  H.-Ch. Grunau and G. Sweers

Using respectively (22), Lemma 3.2 iii. and iv., and again iv., we continue by

(24) ⪯ ∑_{p≤k} |x − y|^{2m−n−|k|+|p|} ∑_{j=0}^{|p|} (A_{xy})^{max(m−j,0)} \frac{d (y)^{j}}{[XY]^{j+|p|}} ⪯

⪯ ∑_{p≤k} |x − y|^{2m−n−|k|+|p|} ∑_{j=0}^{|p|} (\frac{d (x) d (y)}{[XY]^{2}})^{max(m−j,0)} \frac{d (y)^{j}}{[XY]^{j+|p|}} ⪯

⪯ ∑_{p≤k} |x − y|^{2m−n−|k|+|p|} (\frac{d (x)}{[XY]})^{max(m−|p|,0)} (\frac{d (y)}{[XY]})^{m} [XY]^{−|p|} ⪯

⪯ |x − y|^{2m−n−|k|} (\frac{d (x)}{[XY]})^{max(m−|k|,0)} (\frac{d (y)}{[XY]})^{m} .

We used \frac{d(x)}{[XY]} ≤ 1, \frac{d(y)}{[XY]} ≤ 1 and [xy] ∼ [XY]. Lemma 3.5 We have on Ω_{2} := \{(x, y) ∈ B^{2}; [xy] ≤ \frac{1}{2} [XY]\} that i. if |k| > 2m − n

|D_{x}^{k}G_{m,n} (x, y)| ⪯ |x − y|^{2m−n−|k|};

ii. if |k| = 2m − n

|D_{x}^{k}G_{m,n} (x, y)| ⪯ log (\frac{[XY]}{[xy]}) for n even,

|D_{x}^{k}G_{m,n} (x, y)| ⪯ 1 for n odd;

iii. if |k| < 2m − n

|D_{x}^{k}G_{m,n} (x, y)| ⪯ [XY]^{2m−n−|k|} .

Proof of Lemma 3.5. First we will derive some estimates for g_{m,n} defined in (21). If n > 2m then for all j ≥ 0

|g_{m,n}^{(j)} (t)| ⪯ 1 on [0, 1] .

If n = 2m = 2 then

g_{1,2} (t) = − log (t)  (25)

and if n = 2m > 2 then

g_{m,2m} (t) = − log (t) + P_{2m−2} (t) ,

where P_{2m−2} is some polynomial of degree 2m − 2. Hence if n = 2m we get

|g_{m,2m} (t)| ⪯ − log t on [0, \frac{1}{2}] .  (26)

and for j > 0 that

|g_{m,2m}^{(j)} (t)| ⪯ t^{−j} on [0, \frac{1}{2}] .  (27)

Which implies that we obtain for n ≥ 2m and |p| ≥ 1:

|D_{x}^{p}g_{m,n} (\frac{[xy]}{[XY]})| ⪯

⪯ ∑_{j=1}^{|p|} |g_{m,n}^{(j)} (\frac{[xy]}{[XY]})| ∑_{{p^{(1)}+...+p^{(j)}=p \atop |p^{(i)}|≥1}} ∏_{i=1}^{j} |D_{x}^{p^{(i)}} (\frac{[xy]}{[XY]})| ⪯

⪯ ∑_{j=1}^{|p|} (\frac{[xy]}{[XY]})^{−j} ∑_{{p^{(1)}+...+p^{(j)}=p \atop |p^{(i)}|≥1}} ∏_{i=1}^{j} (\frac{[xy]^{1−|p^{(i)}|}}{[XY]} ) ⪯

⪯ |x − y|^{−|p|} .  (28)

Next we distinguish the cases n > 2m, n = 2m and n < 2m.

• The case n > 2m. Inequality (28) implies that

|D_{x}^{k}G_{m,n} (x, y)| ⪯

⪯ ∑_{p≤k} |D_{x}^{k−p} |x − y|^{2m−n}| |D_{x}^{p}g_{m,n} (\frac{[xy]}{[XY]})| ⪯

⪯ ∑_{p≤k} |x − y|^{2m−n−|k|+|p|} |x − y|^{−|p|} ⪯

⪯ |x − y|^{2m−n−|k|} .

• The case n = 2m. If |k| = 0 the result follows from (25) and (26). If |k| ≥ 1 we obtain by (28) that

|D_{x}^{k}G_{m,n} (x, y)| ∼ |D_{x}^{k}g_{m,n} (\frac{[xy]}{[XY]})| ⪯ |x − y|^{−|k|} .

• The case n < 2m. Integration by part yields

g_{m,n} (t) = \frac{1}{2m − n} t^{n−2m} (1 − t^{2})^{m−1} − \frac{2m − 2}{2m − n} g_{m−1,n} (t) =

= ∑_{i=1}^{ℓ} \frac{α_{m,n,i}}{2 (m − i)} t^{n−2m+2i−2} (1 − t^{2})^{m−i} − α_{m,n,ℓ} g_{m−ℓ,n} (t)  (29)

with

α_{m,n,i} = (−1)^{i−1} \begin{matrix}( & {m−1 \atop i} & ) \\ ( & {m− \frac{1}{2} n \atop i} & ) \end{matrix},

for all ℓ ≤ m − \frac{1}{2}n if n is even and for all ℓ < m if n is odd. Define

H_{m,n,ℓ} (x, y) = ∑_{i=1}^{ℓ} \frac{α_{m,n,i}}{2 (m − i)} [xy]^{2i−2} [XY]^{2m−n−2i+2} (A_{xy})^{m−i}

14  H.-Ch. Grunau and G. Sweers

and we have

G_{m,n} (x, y) = H_{m,n,ℓ} (x, y) − α_{m,n,ℓ} [xy]^{2m−n} g_{m−ℓ,n} (\frac{[xy]}{[XY]}) .

From

|D_{x}^{k} [xy]^{2i−2}| = 0 for k > 2i − 2,

since 2i − 2 is even and positive, and (23), we get for ℓ < m that

|D_{x}^{k}H_{m,n,ℓ} (x, y)| ⪯

⪯ ∑_{i=1}^{ℓ} ∑_{p≤k} |D_{x}^{k−p} |x − y|^{2i−2}| |D_{x}^{p} ([XY]^{2m−n−2i+2} (A_{xy})^{m−i})| ⪯

⪯ ∑_{i=1}^{ℓ} ∑_{{p≤k \atop |k|+2−2i≤|p|}} |x − y|^{2i−2−|k|+|p|} [XY]^{2m−n−2i+2−|p|} ⪯

⪯ [XY]^{2m−n−|k|}  (30)

Hence H_{m,n,ℓ} (x, y) for ℓ < m is the innocent part. The estimate for the derivatives of [xy]^{2m−n} g_{m−ℓ,n} ([xy] / [XY]) will be crucial.

We distinguish further the cases n is even and n odd. •• n even. We use (29) with ℓ = m − \frac{1}{2}n and we get that

|D_{x}^{k} ([xy]^{2m−n} g_{\frac{1}{2} n,n} (\frac{[xy]}{[XY]}))| ⪯

⪯ ∑_{{p≤k \atop |k|−2m+n≤|p|}} |D_{x}^{k−p} |x − y|^{2m−n}| |D_{x}^{p}g_{\frac{1}{2} n,n} (\frac{[xy]}{[XY]})| =  (31)

A next distinction will be |k| ≤ 2m − n and |k| > 2m − n.

• • • n even and |k| ≤ 2m − n. Using (28) and (26) we get

(31) ⪯ |x − y|^{2m−n−|k|} g_{\frac{1}{2} n,n} (\frac{[xy]}{[XY]}) + ∑_{{p≤k \atop 1≤|p|}} |x − y|^{2m−n−|k|+|p|} |x − y|^{−|p|} ⪯

⪯ |x − y|^{2m−n−|k|} log (\frac{[XY]}{[xy]}) + |x − y|^{2m−n−|k|} ⪯

⪯ |x − y|^{2m−n−|k|} log (\frac{[XY]}{[xy]}) .

With (30) we find if |k| = 2m − n that

|D_{x}^{k}G_{m,n} (x, y)| ⪯ log (\frac{[XY]}{[xy]}) ,

and if |k| < 2m − n then

|x − y|^{2m−n−|k|} log (\frac{[XY]}{[xy]}) ⪯ [XY]^{2m−n−|k|}

which implies

|D_{x}^{k}G_{m,n} (x, y)| ⪯ [XY]^{2m−n−|k|} .

• • • n even and |k| > 2m − n. With (28) we get

(31) ⪯ ∑_{{p≤k \atop |k|−2m+n≤|p|}} |x − y|^{2m−n−|k|+|p|} |x − y|^{−|p|} ⪯

⪯ |x − y|^{2m−n−|k|} .

•• n odd. If n = 1 we proceed straightforwardly from Boggio’s formula:

G_{m,1} (x, y) =

= c_{m,1} |x − y|^{2m−1} ∫_{v=1}^{[XY]/[xy]}(v^{2} − 1)^{m−1} dv =

= ∑_{q=0}^{m−1} α_{q} [XY]^{2m−2q−1} [xy]^{2q} + α_{m} [xy]^{2m−1} .

Hence it follows that

|D_{x}^{k}G_{m,n} (x, y)| ⪯

⪯ ∑_{q=0}^{m−1} ∑_{p≤k} |D_{x}^{k−p} [XY]^{2m−2q−1}| |D_{x}^{p} [xy]^{2q}| + |D_{x}^{k} [xy]^{2m−1}| ⪯

⪯ ∑_{q=0}^{m−1} ∑_{{p≤k \atop |p|≤2q}} [XY]^{2m−2q−1−|k|+|p|} [xy]^{2q−|p|} + [xy]^{2m−1−|k|} ⪯

⪯ [XY]^{2m−1−|k|} + [xy]^{2m−1−|k|} .

If |k| ≤ 2m−1, then [XY]^{2m−1−|k|}+[xy]^{2m−1−|k|} ⪯ [XY]^{2m−1−|k|} and if |k| > 2m−1, then [XY]^{2m−1−|k|} + [xy]^{2m−1−|k|} ⪯ [xy]^{2m−1−|k|}.

If n > 1 is odd we use (29) with ℓ = m − \frac{1}{2} (n − 1) to find

G_{m,n} (x, y) =

= H_{m,n,m− \frac{1}{2} (n−1)} (x, y) − α_{m,n,m− \frac{1}{2} (n−1)} [xy]^{2m−n} g_{\frac{1}{2} n− \frac{1}{2} ,n} (\frac{[xy]}{[XY]}) .

Notice that, since n − 3 is nonnegative and even,

g_{\frac{1}{2} n− \frac{1}{2} ,n} (t) = ∫_{t}^{1} (1 − w^{2})^{\frac{n−3}{2}} dw

is a polynomial in [xy] / [XY]. We find that

|D_{x}^{k} ([xy]^{2m−n} g_{\frac{1}{2} n− \frac{1}{2} ,n} (\frac{[xy]}{[XY]}))| ⪯ |x − y|^{2m−n−|k|} .  (32)

Hence, combining (30) and (32) we find that

|D_{x}^{k}G_{m,n} (x, y)| ⪯ [XY]^{2m−n−|k|} + |x − y|^{2m−n−|k|} ,

which implies if |k| ≥ 2m − n that

|D_{x}^{k}G_{m,n} (x, y)| ⪯ |x − y|^{2m−n−|k|} ,

and if |k| ≤ 2m − n that

|D_{x}^{k}G_{m,n} (x, y)| ⪯ [XY]^{2m−n−|k|} .

Proof of Proposition 2.4. Again we separate the two cases.

• The case [xy] ≥ \frac{1}{2} [XY]. We find by (10) that on Ω_{1}:

\frac{d (x)}{[XY]} ∼ \frac{d (x)}{[xy]} ⪯ 1 and \frac{d (y)}{[XY]} ∼ \frac{d (y)}{[xy]} ⪯ 1  (33)

and

log (2 + \frac{d (y)}{|x − y|}) ∼ 1.

Since t ∼ (1 ∧ t) for t ∈ [0, M] and M fixed, we find that Lemma 3.4 i. implies Proposition 2.4 i./ii. (a) and that Lemma 3.4 ii. implies Proposition 2.4 i./ii. (b). It remains to prove Proposition 2.4 iii.

If m − \frac{1}{2}n ≥ |k| it follows from Lemma 3.4 i. and (33) that

|D_{x}^{k}G_{m,n} (x, y)| ⪯ |x − y|^{2m−n−|k|} \frac{d (x)^{m−|k|}}{|x − y|^{m−|k|}} \frac{d (y)^{m}}{|x − y|^{m}} =

= d (x)^{m− \frac{1}{2} n−|k|} d (y)^{m− \frac{1}{2} n} \frac{d (x)^{\frac{1}{2} n} d (y)^{\frac{1}{2} n}}{|x − y|^{n}} ⪯

⪯ d (x)^{m− \frac{1}{2} n−|k|} d (y)^{m− \frac{1}{2} n} (1 ∧ \frac{d (x)^{\frac{1}{2} n} d (y)^{\frac{1}{2} n}}{|x − y|^{n}} ) .

which implies Proposition 2.4 iii.(a).

If m ≥ |k| ≥ m − \frac{1}{2}n it follows from Lemma 3.4 i. and (33) that

|D_{x}^{k}G_{m,n} (x, y)| ⪯ |x − y|^{2m−n−|k|} \frac{d (x)^{m−|k|}}{|x − y|^{m−|k|}} \frac{d (y)^{m}}{|x − y|^{m}} =

= d (y)^{2m−n−|k|} \frac{d (x)^{m−|k|} d (y)^{n−m+|k|}}{|x − y|^{n}} ⪯

⪯ d (y)^{2m−n−|k|} (1 ∧ \frac{d (x)^{m−|k|} d (y)^{n−m+|k|}}{|x − y|^{n}} )

which implies Proposition 2.4 iii.(b).

If 2m − n ≥ |k| ≥ m it follows from Lemma 3.4 ii. and (33) that

|D_{x}^{k}G_{m,n} (x, y)| ⪯ |x − y|^{2m−n−|k|} \frac{d (y)^{m}}{|x − y|^{m}} =

= d (y)^{2m−n−|k|} \frac{d (y)^{n−m+|k|}}{|x − y|^{n−m+|k|}} ⪯ d (y)^{2m−n−|k|} (1 ∧ \frac{d (y)^{n−m+|k|}}{|x − y|^{n−m+|k|}} ) ,

which implies Proposition 2.4 iii.(c).

• The case [xy] ≤ \frac{1}{2} [XY]. From (11-12) we find that on Ω_{2}

\frac{d (x)}{|x − y|} ⪰ 1 and \frac{d (y)}{|x − y|} ⪰ 1.

Hence Lemma 3.5 i. implies Proposition 2.4 i. (a) and (b). Since (11) shows that on Ω_{2}

[XY] ⪯ d (x) and [XY] ⪯ d (y) ,

Lemma 3.5 iii. implies Proposition 2.4 iii. (a), (b) and (c). For Proposition 2.4 ii. notice that on Ω_{2}

log (\frac{[XY]}{[xy]}) ⪯ log (2 + \frac{d (y)}{|x − y|}) ∼

∼ log (2 + \frac{d (y)}{|x − y|}) (1 ∧ \frac{d (x)^{(m−|k|)^{+}} d (y)^{m}}{|x − y|^{(m−|k|)^{+}+m}} ) ,

implying both (a) and (b).

4 Results of 3-G type

Proposition 4.1 For k ∈ ℕ^{n} with |k| ≤ 2m the following holds on B^{3} (x, y, z ∈ B).

i. If 2m ≥ |k| > 2m − n, then:

\frac{G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)|}{G_{m,n} (x, y)} ⪯ |x − z|^{2m−n−|k|} + |z − y|^{2m−n−|k|};

ii. if |k| = 2m − n with n even, then

\frac{G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)|}{G_{m,n} (x, y)} ⪯ log (\frac{3}{|x − z|}) + log (\frac{3}{|z − y|});

iii. if either |k| = 2m − n with n odd, or |k| < 2m − n, then:

\frac{G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)|}{G_{m,n} (x, y)} ⪯ 1.

Corollary 4.2 There exists M ∈ ℝ^{+}, depending on n, m only, such that for k ∈ ℕ^{n} with |k| ∈ [0, 2m − 1] the following holds. Let f ∈ L^{p} (B) , p > 1, with f ≥ 0 and we have

|(G_{m,n}D^{k}G_{m,n}f) (x)| ≤ M (G_{m,n}f) (x) for almost all x ∈ B.

Proof of Corollary 4.2. Note that, due to the additional assumption |k| ≤ 2m − 1, all right hand sides in Proposition 4.1 are uniformly integrable. That means there exists M ∈ ℝ^{+} such that for all x, y ∈ B

∫_{z∈B} \frac{G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)|}{G_{m,n} (x, y)}dz ≤ M.

Hence, since |k| ≤ 2m − 1 by dominated convergence, see also [8, Lemma 4.1], and Fubini-Tonelli

|(G_{m,n}D^{k}G_{m,n}f) (x)| =
= |∫_{z∈B} G_{m,n} (x, z) D_{z}^{k} ∫_{y∈B} G_{m,n} (z, y) f (y) dy dz| ≤

≤ ∫_{y∈B} ∫_{z∈B} G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)| dz f (y) dy ≤

≤ M ∫_{y∈B} G_{m,n} (x, y) f (y) dy = M (G_{m,n}f) (x) .

Before proving Proposition 4.1 we need a technical lemma.

Lemma 4.3 On B^{3} (x, y, z ∈ B) one finds that

i. T (x, y, z) := \frac{(1 ∧ \frac{d(x)d(z)}{|x−z|^{2}}) (1 ∧ \frac{d(z)d(y)}{|y−z|^{2}})}{(1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})} ⪯ 1;

ii. Q (x, y, z) := \frac{(1 ∧ \frac{d(x)d(z)}{|x−z|^{2}}) (1 ∧ \frac{d(y)}{|y−z|})}{(1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})} ⪯ 1 + \frac{|x − y|}{|x − z|};

iii. L (x, y, z) := \begin{matrix}log & (1 + \frac{d(x)d(z)}{|x−z|^{2}}) & (1 ∧ \frac{d(y)}{|y−z|}) \\ log & (1 + \frac{d(x)d(y)}{|x−y|^{2}}) \end{matrix} ⪯ 1 + \frac{|x − y|}{|x − z|};

iv. R (x, y, z) := \frac{|x − y|}{|x − z| |z − y|} ≤ \frac{1}{|x − z|} + \frac{1}{|z − y|}.

Proof of Lemma 4.3. First we suppose that [xy] ≤ \frac{1}{2} [XY]. Hence Lemma 3.1 implies

\frac{d(x)d(y)}{|x−y|^{2}} ⪰ \frac{d(x)}{|x−y|} ⪰ 1 and the estimates in i. and ii. follow from (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}}) ∼ 1. The estimate in iii. uses

\begin{matrix}log & (1 + t) \\ log & (1 + s) \end{matrix} ≤ 1 + \frac{t}{s} for t, s ∈ ℝ^{+}.  (34)

Indeed, let s ≥ 0 and one finds if α ≥ 1 that log (1 + αs) ≤ α log (1 + s) by the concavity of log; if α ∈ [0, 1] one finds log (1 + αs) ≤ log (1 + s); (34) follows with αs = t. From Lemma 3.2 ix. we find

log (1 + \frac{d (x) d (y)}{|x − y|^{2}}) ∼ log (2 + \frac{d (x)}{|x − y|}) ∼ log (1 + \frac{d (x)}{|x − y|}) ⪰ 1

and

log (1 + \frac{d (x) d (z)}{|x − z|^{2}}) ⪯ log (2 + \frac{d (x)}{|x − z|}) ⪯ 1 + log (1 + \frac{d (x)}{|x − z|}) .

Hence it follows that

L (x, y, z) ⪯ \frac{1 + log (1 + \frac{d(x)}{|x−z|})}{log (1 + \frac{d(x)}{|x−y|})} ⪯ 1 + \begin{matrix}log & (1 + \frac{d(x)}{|x−z|}) \\ log & (1 + \frac{d(x)}{|x−y|}) \end{matrix} ⪯ 1 + \frac{|x − y|}{|x − z|}.

If [xy] ≥ \frac{1}{2} [XY], then

(1 ∧ \frac{d (x) d (y)}{|x − y|^{2}}) ∼ log (1 + \frac{d (x) d (y)}{|x − y|^{2}}) ∼ \frac{d (x) d (y)}{|x − y|^{2}}

and we distinguish the following cases.

• The case |x − z| ≥ \frac{1}{2} |x − y|. Using Lemma 3.2 vi. we get

T (x, y, z) ⪯ \frac{|x − y|^{2}}{d (x) d (y)} \frac{d (x) d (z)}{|x − z|^{2}} \frac{d (y)}{d (z)} ≤ 4,  (35)

and using Lemma 3.2 v. we obtain

Q (x, y, z) ⪯ \frac{|x − y|^{2}}{d (x) d (y)} \frac{d (x) d (z)}{|x − z|^{2}} \frac{d (y)}{d (z)} ≤ 4.  (36)

Similarly we obtain that

L (x, y, z) ⪯ \frac{|x − y|^{2}}{d (x) d (y)} \frac{d (x) d (z)}{|x − z|^{2}} \frac{d (y)}{d (z)} ≤ 4.  (37)

• The case |x − z| ≤ \frac{1}{2} |x − y|. Then |y − z| ≥ |y − x| − |x − z| ≥ \frac{1}{2} |x − y| and one proceeds for i. as in (35) using T (x, y, z) = T (y, x, z). In order to prove ii. we proceed by using Lemma 3.2 vii.:

Q (x, y, z) ⪯ \frac{|x − y|^{2}}{d (x) d (y)} \frac{d (x)}{|x − z|} \frac{d (y)}{|y − z|} ≤ 2\frac{|x − y|}{|x − z|}.

Similarly we find by Lemma 3.2 ix.,viii. and again ix.:

L (x, y, z) ⪯ \frac{|x − y|^{2}}{d (x) d (y)} log (1 + \frac{d (x)}{|x − z|}) \frac{d (y)}{|y − z|} ≤ 2\frac{|x − y|}{|x − z|}.

Note that the estimate in iv. is straightforward.

Proof of Proposition 4.1.

• The case n > 2m. One finds by Proposition 2.3, Proposition 2.4 i., respectively,

(6) and Lemma 4.3 that

\frac{G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)|}{G_{m,n} (x, y)} ⪯

⪯ \begin{matrix}|x − y| & n−2m & (1 ∧ \frac{d(x)^{m}d(z)^{m}}{|x−z|^{2m}} ) & (1 ∧ \frac{d(z)^{^{(m−|k|)}^{+}}d(y)^{m}}{|y−z|^{(m−|k|)^{+}+m}} ) \\ (1 ∧ \frac{d(x)^{m}d(y)^{m}}{|x−y|^{2m}} ) & |x − z| & n−2m & |z − y| & n−2m+|k| \end{matrix} =

= \begin{matrix}|x − y| & n−2m \\ |x − z| & n−2m & |z − y| & n−2m+|k| \end{matrix} (T (x, y, z))^{(m−|k|)^{+}} (Q (x, y, z))^{m∧|k|} ⪯

⪯ \begin{matrix}|x − y| & n−2m \\ |x − z| & n−2m & |z − y| & n−2m+|k| \end{matrix} (1 + (\frac{|x − y|}{|x − z|})^{m∧|k|}) =

= \frac{1}{|z − y|^{|k|}} R (x, y, z)^{n−2m} + \frac{1}{|z − y|^{(|k|−m)^{+}}} R (x, y, z)^{n−2m+(m∧|k|)} ⪯

⪯ \frac{1}{|z − y|^{|k|} |x − z|^{n−2m}} + \frac{1}{|z − y|^{|k|+n−2m}} +

+\frac{1}{|z − y|^{(|k|−m)^{+}} |x − z|^{n−2m+(m∧|k|)}} ⪯

⪯ |x − z|^{2m−n−|k|} + |z − y|^{2m−n−|k|} .  (38)

• n = 2m and 0 < |k| ≤ 2m. By using Lemma 3.2 ix. we have

\frac{G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)|}{G_{m,n} (x, y)} ⪯

⪯ \begin{matrix}log & (1 + \frac{d(x)d(z)}{|x−z|^{2}}) & (1 ∧ \frac{d(x)^{m−1}d(z)^{m−1}}{|x−z|^{2m−2}} ) \\ log & (1 + \frac{d(x)d(y)}{|x−y|^{2}}) & (1 ∧ \frac{d(x)^{m−1}d(y)^{m−1}}{|x−y|^{2m−2}} ) \end{matrix}·

· |y − z|^{2m−n−|k|} (1 ∧ \frac{d (z)^{(m−|k|)^{+}} d (y)^{m}}{|y − z|^{(m−|k|)^{+}+m}} ) =

= \begin{matrix}|x − y| & n−2m \\ |x − z| & n−2m & |z − y| & n−2m+|k| \end{matrix} ·

· (T (x, y, z))^{(m−|k|)^{+}} L (x, y, z) (Q (x, y, z))^{(m∧|k|)−1} .

Since L and Q can be bounded similarly (see Lemma 4.3 ii. and iii.) we may continue as in (38).

• n = 2m and |k| = 0. We find by using Lemma 3.2 ix., the estimate for T (x, y, z) and the symmetry of G_{m,n} (x, y) that

\frac{G_{m,n} (x, z) G_{m,n} (z, y)}{G_{m,n} (x, y)} ⪯

⪯ log (2 + \frac{d(x)}{|x−z|}) (1 ∧ \frac{d(x)^{m}d(z)^{m}}{|x−z|^{2m}} ) log (2 + \frac{d(y)}{|y−z|}) (1 ∧ \frac{d(z)^{m}d(y)^{m}}{|y−z|^{2m}} ) ; max (log (2 + \frac{d(x)}{|x−y|}) , log (2 + \frac{d(y)}{|x−y|})) (1 ∧ \frac{d(x)^{m}d(y)^{m}}{|x−y|^{2m}} ) ⪯

⪯ log (2 + \frac{d(x)}{|x−z|}) log (2 + \frac{d(y)}{|y−z|}) ; max (log (2 + \frac{d(x)}{|x−y|}) , log (2 + \frac{d(y)}{|x−y|})) =  (39)

If |x − z| ≥ \frac{1}{2} |x − y| then log (2 + \frac{d(x)}{|x−z|}) ⪯ log (2 + \frac{d(x)}{|x−y|}) and we continue by

(39) ⪯ log (2 + \frac{d (y)}{|y − z|}) ⪯ log (\frac{3}{|y − z|}) .

If |x − z| ≤ \frac{1}{2} |x − y| then |y − z| ≥ |y − x| − |x − z| ≥ \frac{1}{2} |x − y| and we conclude by

(39) ⪯ log (2 + \frac{d (x)}{|x − z|}) ⪯ log (\frac{3}{|x − z|}) .

• n < 2m and 0 ≤ |k| < 2m − n, or, n < 2m and |k| = 2m − n and n odd. Note that by Lemma 3.3 (see also (8)) we may estimate as follows:

\frac{G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)|}{G_{m,n} (x, y)} ⪯

⪯ \begin{matrix}(d (x) d (z)) & m− \frac{1}{2} n & (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}}) & \frac{1}{2} n \\ (d (x) d (y)) & m− \frac{1}{2} n & (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}}) & \frac{1}{2} n \end{matrix} ·

·d (z)^{m− \frac{1}{2} n−|k|} d (y)^{m− \frac{1}{2} n} (1 ∧ \frac{d (z) d (y)}{|y − z|^{2}})^{\frac{1}{2} n} =

= d (z)^{2m−n−|k|} (T (x, y, z))^{\frac{1}{2} n} ⪯ 1

• n < 2m and |k| = 2m − n with n even. We use Lemma 3.3. Note that |k| = 2m − n implies m − |k| = n − m. We have

\frac{G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)|}{G_{m,n} (x, y)} ⪯

⪯ \begin{matrix}(d (x) d (z)) & m− \frac{1}{2} n & (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}}) & \frac{1}{2} n & log & (2 + \frac{d(y)}{|z−y|}) & (1 ∧ \frac{d(z)^{^{(m−|k|)}^{+}}d(y)^{m}}{|y−z|^{(m−|k|)^{+}+m}} ) \\ (d (x) d (y)) & m− \frac{1}{2} n & (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}}) & \frac{1}{2} n \end{matrix} ⪯

⪯ \frac{d (z)^{m− \frac{1}{2} n} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} n} log (2 + \frac{d(y)}{|z−y|}) (\frac{d(y)}{d(z)})^{m− \frac{1}{2} n} (1 ∧ \frac{d(z)d(y)}{|y−z|^{2}})^{\frac{1}{2} n}}{d (y)^{m− \frac{1}{2} n} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{\frac{1}{2} n}} =

= log (2 + \frac{d (y)}{|z − y|}) (T (x, y, z))^{\frac{1}{2} n} ⪯ log (\frac{3}{|z − y|}) .

• n < 2m and 2m − n < |k| ≤ 2m − \frac{1}{2}n. We find that

\frac{G_{m,n} (x, z) |D_{z}^{k}G_{m,n} (z, y)|}{G_{m,n} (x, y)} ⪯

⪯ \begin{matrix}(d (x) d (z)) & m− \frac{1}{2} n & (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}}) & \frac{1}{2} n & |z − y| & 2m−n−|k| & (1 ∧ \frac{d(z)^{^{(m−|k|)}^{+}}d(y)^{m}}{|y−z|^{(m−|k|)^{+}+m}} ) \\ (d (x) d (y)) & m− \frac{1}{2} n & (1 ∧ ^{\frac{d(x)^{\frac{1}{2} n}d(y)^{\frac{1}{2} n}}{|x−y|^{n}}} ) \end{matrix} =

= |z − y|^{2m−n−|k|} \frac{(\frac{d(z)}{d(y)})^{m− \frac{1}{2} n} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} n} (1 ∧ \frac{d(z)^{^{(m−|k|)}^{+}}d(y)^{m}}{|y−z|^{(m−|k|)^{+}+m}} )}{(1 ∧ ^{\frac{d(x)^{\frac{1}{2} n}d(y)^{\frac{1}{2} n}}{|x−y|^{n}}} )} = (40)

By Lemma 3.3 it follows that

(1 ∧ \frac{d (y) d (z)}{|y − z|^{2}})^{m− \frac{1}{2} n} ∼ (1 ∧ \frac{d (z)}{|y − z|})^{2m−n} (\frac{d (y)}{d (z)})^{m− \frac{1}{2} n} .  (41)

By Lemma 3.2 and (41) it follows that

(1 ∧ \frac{d (z)^{(m−|k|)^{+}} d (y)^{m}}{|y − z|^{(m−|k|)^{+}+m}} ) ∼

\begin{matrix}∼ (1 ∧ \frac{d (z) d (y)}{|y − z|^{2}})^{2m− \frac{1}{2} n−|k|} (1 ∧ \frac{d (y)}{|y − z|})^{|k|+n−2m} · \\ · (1 ∧ \frac{d (z)}{|y − z|})^{(|k|−m)^{+}} (1 ∧ \frac{d (z)}{|y − z|})^{n−2m} (1 ∧ \frac{d (y) d (z)}{|y − z|^{2}})^{m− \frac{1}{2} n} ⪯ \end{matrix}

⪯ (1 ∧ \frac{d (z) d (y)}{|y − z|^{2}})^{2m− \frac{1}{2} n−|k|} (1 ∧ \frac{d (y)}{|y − z|})^{|k|+n−2m} (\frac{d (y)}{d (z)})^{m− \frac{1}{2} n} ,  (42)

and using (42) we continue by

(40) ⪯ |z − y|^{2m−n−|k|} ·

·\frac{(1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} n} (1 ∧ \frac{d(z)d(y)}{|y−z|^{2}})^{2m− \frac{1}{2} n−|k|} (1 ∧ \frac{d(y)}{|y−z|})^{|k|+n−2m}}{(1 ∧ ^{\frac{d(x)^{\frac{1}{2} n}d(y)^{\frac{1}{2} n}}{|x−y|^{n}}} )} =

= |z − y|^{2m−n−|k|} (T (x, y, z))^{2m− \frac{1}{2} n−|k|} (Q (x, y, z))^{|k|+n−2m} ⪯

⪯ |z − y|^{2m−n−|k|} (1 + \frac{|x − y|}{|x − z|})^{|k|+n−2m} ⪯

⪯ |z − y|^{2m−n−|k|} + |x − z|^{2m−n−|k|} .

• n < 2m and 2m − \frac{1}{2}n < |k| ≤ 2m. Similar as in the previous case we obtain the estimate in (40). By Lemma 3.2 we get

(1 ∧ \frac{d (z)^{(m−|k|)^{+}} d (y)^{m}}{|y − z|^{(m−|k|)^{+}+m}} ) ⪯ (\frac{d (y)}{d (z)})^{m− \frac{1}{2} n} (1 ∧ \frac{d (y)}{|y − z|})^{\frac{1}{2} n} ,

and we continue by

(40) ⪯ |y − z|^{2m−n−|k|} \frac{(1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} n} (1 ∧ \frac{d(y)}{|y−z|})^{\frac{1}{2} n}}{(1 ∧ ^{\frac{d(x)^{\frac{1}{2} n}d(y)^{\frac{1}{2} n}}{|x−y|^{n}}} )} =

= |z − y|^{2m−n−|k|} (Q (x, y, z))^{\frac{1}{2} n} ⪯

⪯ |z − y|^{2m−n−|k|} (1 + (\frac{|x − y|}{|x − z|})^{\frac{1}{2} n}) =

= |z − y|^{2m−n−|k|} + |z − y|^{2m− \frac{1}{2} n−|k|} (R (x, y, z))^{\frac{1}{2} n} ⪯

⪯ |z − y|^{2m−n−|k|} + |z − y|^{2m− \frac{1}{2} n−|k|} |x − z|^{− \frac{1}{2} n} ⪯

⪯ |z − y|^{2m−n−|k|} + |x − z|^{2m−n−|k|} .

In the last step we used that 2m − n − |k| ≤ −\frac{1}{2}n.

5 Positivity for related equations and systems

5.1 More general equations

Consider the equation

\{ \begin{matrix}((−Δ)^{m} + A) u & = & f & in B, \\ D_{m}u & = & 0 & on ∂B, \end{matrix}  (43)

with

A = ∑_{|α|<2m} a_{α} (x) D^{α}  (44)

and a_{α} ∈ C \binom{}{B̄}. The boundary condition is as in (2). The operator A is a lower order perturbation of (−Δ)^{m}.

In this section we will fix some p > 1. The operator G_{m,n} is well defined on L^{p} (B) and we may use the regularity theory of [1].

Theorem 5.1 There exists ε_{0} > 0 such that, if ∥a_{α}∥_{∞} ≤ ε_{0} for all α with |α| < 2m, then the following holds.

i. For all f ∈ L^{p} (B) there exists a solution u ∈ W^{2m,p} (Ω) ∩ W_{0}^{m,p} (Ω) of (43). ii. Moreover, if f ∈ L^{p} (B) and 0 ≢ f ≥ 0 in B, then the solution of (43) satisfies

u > 0 in B.

Remark 1. Let Ω be a simply connected two dimensional domain and let h : B → Ω be a bijection such that h (x_{1} + ix_{2}) = h_{1} (x_{1}, x_{2}) + i h_{2} (x_{1}, x_{2}) is a holomorphic mapping. Then Δ (u ∘ h) = \frac{1}{2} |∇h|^{2} (Δu) ∘ h. We write g (x) = 2 |(∇h) (x)|^{−2}. If ∂Ω is sufficiently smooth, then a Theorem of Kellogg-Warschawski (see [20]) implies that h is sufficiently smooth and that there exist c_{i} > 0 such that c_{1} ≤ |(∇h) (x)| ≤ c_{2}. The elliptic problem (−Δ)^{m} u = f in Ω with Dirichlet boundary condition can be transformed to

(−g (·) Δ)^{m} (u ∘ h) = f ∘ h in B,

which can also be written as

((−Δ)^{m} + A) (u ∘ h) = g^{−m} (f ∘ h) in B,

for some A as in (44). Since 0 ≢ f ≥ 0 yields 0 ≢ g^{−m} (f ∘ h) ≥ 0 one finds that domains that are close to the disk still have a positive Green function G_{m,n,Ω}. Close to the disk means ∥h − I∥_{C^{2m−1}\binom{}{B̄}} sufficiently small. For example this holds for an ellipse that is close to a circle, see [12].

The theorem will be proven in two steps. We may rewrite (43) as

u = −G_{m,n}Au + G_{m,n}f

and then formally

u = (I + G_{m,n}A)^{−1} G_{m,n}f.

First we will justify that (I + G_{m,n}A)^{−1} is well defined. The next step will be to prove the estimate (I + G_{m,n}A)^{−1} G_{m,n} ≥ cG_{m,n}.

Definition 5.2 By S ≥ T we mean that (Sf) (x) ≥ (T f) (x) for all f ∈ L^{p} (B) with f ≥ 0.

Lemma 5.3 Set W_{D}^{2m,p}(B) := W^{2m,p}(B) ∩ W_{0}^{m,p}(B). If ∥a_{α}∥_{∞} ≤ η_{1} and η_{1} is sufficiently small, then

(I + G_{m,n}A)^{−1} ∈ L (W_{D}^{2m,p}(B); W_{D}^{2m,p}(B)) .

Proof. It is sufficient to prove that ∥G_{m,n}A∥ < 1. Set

κ_{n,m} = ∑_{|k|<2m} 1 = \binom{2m − 1 + n}{2m − 1}  (45)

and we obtain that

∥Au∥_{p} ≤ ∑_{|α|<2m} ∥a_{α}∥_{∞} ∥D^{α}u∥_{p} ≤ κ_{n,m} η_{1} ∥u∥_{W^{2m,p}} .

By [1] it follows that G_{m,n}Au ∈ W_{D}^{2m,p}(B) and that G_{m,n} : L^{p} (B) → W_{D}^{2m,p}(B) is bounded. For η_{1} small enough we have ∥G_{m,n}A∥ < 1.

Lemma 5.4 Let M be as Corollary 4.2, ∥a_{α}∥_{∞} ≤ η. Then

G_{m,n}AG_{m,n} ≤ Mηκ_{n,m} G_{m,n},

with κ_{n,m} as in (45).

The proof is straightforward from Corollary 4.2. By symmetry we also have

−G_{m,n}AG_{m,n} ≤ Mηκ_{n,m} G_{m,n}.

Lemma 5.5 If ∥a_{α}∥_{∞} ≤ η_{2} and η_{2} is sufficiently small, then there is c > 0 such that

(I + G_{m,n}A)^{−1} G_{m,n} ≥ c G_{m,n}.

Proof of Lemma 5.5. Assume that f ∈ L^{p} (B). By [1] it follows that G_{m,n}f ∈ W_{D}^{2m,p}(B). The solution u of (43) is the limit in W_{D}^{2m,p}(B) of {u_{ν}} where

u_{ν} = ∑_{i=0}^{ν} (−G_{m,n}A)^{i} G_{m,n}f.  (46)

Hence there is a subsequence, again denoted by {u_{ν}}, such that u_{ν} → u pointwise. Let us denote A_{z} = ∑_{|α|<2m} a_{α} (z) D_{z}^{α} and G = G_{m,n}. By Lemma 5.4 we find, assuming ∥a_{α}∥_{∞} ≤ η, that for f ≥ 0

|((G_{m,n}A)^{i} G_{m,n}f) (x)| =

= |∫_{z_{1}} G (x, z_{1}) A_{z_{1}} ∫_{z_{2}} G (z_{1}, z_{2}) A_{z_{2}} ⋯ ∫_{y} G (z_{i}, y) f (y) dydz_{i} … dz_{2}dz_{1}| =

\begin{matrix}= | ∫_{y} ∫_{z_{i}} … ∫_{z_{2}} ∫_{z_{1}} \frac{G (x, z_{1}) A_{z_{1}}G (z_{1}, z_{2})}{G (x, z_{2})} \frac{G (x, z_{2}) A_{z_{2}}G (z_{2}, z_{3})}{G (x, z_{3})} ⋯ \\ … \frac{G (x, z_{i}) A_{z_{i}}G (z_{i}, y)}{G (x, y)}G (x, y) f (y) dz_{1}dz_{2} … dz_{i}dy| ≤ \end{matrix}

\begin{matrix}≤ ∫_{y} ∫_{z_{i}} … ∫_{z_{2}} ∫_{z_{1}} |\frac{G (x, z_{1}) A_{z_{1}}G (z_{1}, z_{2})}{G (x, z_{2})}| dz_{1} |\frac{G (x, z_{2}) A_{z_{2}}G (z_{2}, z_{3})}{G (x, z_{3})}| dz_{2} ⋯ \\ … |\frac{G (x, z_{i}) A_{z_{i}}G (z_{i}, y)}{G (x, y)}| dz_{i}G (x, y) f (y) dy ≤ \end{matrix}

≤ (κ_{n,m}ηM)^{i} (∫_{y} G (x, y) f (y) dy) .  (47)

Take η_{1} such that C = κ_{n,m}η_{1}M < 1 and it follows from (46) that

|u_{ν} (x) − (G_{m,n}f) (x)| ≤ |(∑_{i=1}^{ν} (−G_{m,n}A)^{i} G_{m,n}f) (x)| ≤

≤ ∑_{i=1}^{ν} |((G_{m,n}A)^{i} G_{m,n}f) (x)| ≤

≤ \frac{C}{1 − C} (G_{m,n}f) (x) ,

which implies

u_{ν} (x) ≥ \frac{1 − 2C}{1 − C} (G_{m,n}f) (x) .  (48)

Hence we find for 0 ≤ f ≢ 0 that u_{ν} > 0 in B for C < \frac{1}{2}. Since C does not depend on ν it follows that u satisfies (48).

Proof of Theorem 5.1. For ε_{0} = η_{1} as in Lemma 5.3 it follows that we have a solution in W_{D}^{2m,p} (Ω) of (43) for all f ∈ L^{p} (Ω). Indeed, G_{m,n}f ∈ W_{D}^{2m,p} (Ω) by [1] and Lemma 5.3 implies u = (I + G_{m,n}A)^{−1} G_{m,n}f ∈ W_{D}^{2m,p} (Ω). For ε_{0} = η_{2} (< η_{1}) as in Lemma 5.5 we find that u ≥ cG_{m,n}f > 0.

5.2 Systems with the same polyharmonic operator

Consider the system

\{ \begin{matrix}Lu & = & Au + (f, 0, …, 0)^{T} & in B, \\ Du & = & 0 & on ∂B, \end{matrix}  (49)

with u a vector-function with k components and where we have

i. L is a diagonal k × k-matrix of polyharmonic operators: (L)_{ii} = (−Δ)^{m}; ii. D is a diagonal k × k-matrix of the corresponding Dirichlet boundary condition:

(D)_{ii} = D_{m}; iii. A is a (full) matrix of lower order coupling terms:

A_{i,j} = ∑_{|α|<2m} a_{α}^{ij} D^{α},

with a_{α}^{ij} ∈ C \binom{}{B̄}.

The question is the following. What kind of (small) nonzero coupling is allowed such that one has u_{1} > 0 whenever 0 ≢ f ≥ 0?

The last equation of (49), (−Δ)^{m} u_{k} = ∑_{j=1}^{k} A_{k,j}u_{j}, is ’solved’ by

u_{k} = T_{k} ∑_{j=1}^{k−1} A_{k,j} u_{j}

where

T_{k} = (I − G_{m} A_{k,k})^{−1} G_{m}

The operator is T_{k} is well defined for ∥a_{α}^{kk}∥_{∞} small. Moreover, for every ε > 0 there is δ > 0 such that if ∥a_{α}^{kk}∥_{∞} < δ then

(1 − ε) G_{m} ≤ T_{k} ≤ (1 + ε) G_{m}.

We obtain a system with k − 1 terms, of which the coupling terms are

A_{i,j}^{(k−1)} = A_{i,j} + A_{i,k}T_{k}A_{k,j}.

Repeating such an argument for the last equation of the newly obtained system, which corresponds to the one but last equation of (49), we find

u_{k−1} = T_{k−1} ∑_{j=1}^{k−2} A_{k−1,j}^{(k−1)} u_{j}

where

T_{k−1} = (I − G_{m} A_{k−1,k−1}^{(k−1)})^{−1} G_{m}.

The operator is T_{k−1} is well defined for ∥a_{α}^{kk}∥_{∞} , ∥a_{α}^{k,k−1}∥_{∞} , ∥a_{α}^{k−1,k}∥_{∞} and ∥a_{α}^{k−1,k−1}∥_{∞}

small. Moreover, for every ε > 0 there is δ > 0 such that if ∥a_{α}^{ij}∥_{∞} < δ for i, j ∈ {k − 1, k} then

(1 − ε) G_{m} ≤ T_{k−1} ≤ (1 + ε) G_{m}.

Etc. In an iterative way one defines A^{(k)} = A for ℓ = k and for ℓ ∈ {1, …, k − 1} :

\begin{aligned}T_{ℓ} & = & (I − G_{m} A_{ℓ,ℓ}^{(ℓ)})^{−1} G_{m} \\ A_{i,j}^{(ℓ−1)} & = & A_{i,j}^{(ℓ)} + A_{i,ℓ}^{(ℓ)}T_{ℓ}A_{ℓ,j}^{(ℓ)} & 1 ≤ i, j ≤ ℓ − 1. \end{aligned}

Finally we obtain

u_{1} = T_{1}f.

One proves the following.

Lemma 5.6 If all ∥a_{α}^{ij}∥_{∞} are sufficiently small, then we find that the operator T_{1} ∈

L (L^{p} (B) ; W_{D}^{2m,p} (B)) is well defined. Moreover, for all ε > 0 there is δ > 0, such

that if ∥a_{α}^{ij}∥_{∞} < δ for all i, j ≤ k, |α| < 2m, then

(1 − ε) G_{m} ≤ T_{1} ≤ (1 + ε) G_{m}.

The next theorem is a direct corollary.

Theorem 5.7 There exists ε_{0} > 0 such that, if ∥a_{α}^{ij}∥_{∞} ≤ ε_{0} for all i, j ≤ k and all α with |α| < 2m, then the following holds. If f ∈ L^{p} (Ω) with 0 ≢ f ≥ 0 in B, then the solution of (49) satisfies u_{1} > 0 in B.

28  H.-Ch. Grunau and G. Sweers

5.3 Systems with different polyharmonic operators In general there is no hope to find

G_{m,n}AG_{ℓ,n} ≤ c G_{m,n}  (50)

for arbitrary m and ℓ. However, if m ≤ ℓ such a result holds. Lemma 5.8 Suppose ℓ ≥ m > 0, |k| < 2ℓ. Then we have on B × B × B:

\frac{G_{m,n}(x, z) |D_{z}^{k}G_{ℓ,n}(z, y)|}{G_{m,n}(x, y)} ⪯ |x − z|^{1−n} + |y − z|^{1−n}.

Proof. We use Proposition 2.4 in the following weak form

|D_{z}^{k}G_{ℓ,n}(z, y)| ⪯ |z − y|^{1−n} (1 ∧ \frac{d(y)^{ℓ}}{|z − y|^{ℓ}} ) ⪯

⪯ |z − y|^{1−n} (1 ∧ \frac{d(y)^{m}}{|z − y|^{m}} ) = |z − y|^{2m−n−(2m−1)} (1 ∧ \frac{d(y)^{m}}{|z − y|^{m}} ) .

Now we may use the proof of Proposition 4.1 replacing |D_{z}^{k}G_{ℓ,n}(z, y)| formally by

|Dk̃ ; zG_{m,n}(z, y)| for some k̃, |k̃| = 2m − 1.

Corollary 5.9 Suppose that ℓ ≥ m > 0 and let A = ∑_{|α|<2ℓ} a_{α} D^{α} with a_{α} ∈ C(B̄). Then there is c > 0 such that (50) holds.

This allows us to generalize the result of the previous section. Theorem 5.10 Consider (49) with (L)_{ii} = (−Δ)^{m_{i}} and (D)_{ii} = D_{m_{i}} and suppose that m = m_{1} ≤ m_{i} for all i ≥ 1. Assume that A_{i,j} = ∑_{|α|<2m_{j}} a_{α}^{ij} D^{α}. There exists ε_{0} > 0 such that, if ∥a_{α}^{ij}∥_{∞} ≤ ε_{0} for all i, j and α, then the following holds. If f ∈ L^{p} (Ω) with 0 ≢ f ≥ 0 in B, then the solution of (49) satisfies u_{1} > 0 in B.

Notice that polyharmonic equations with some non Dirichlet boundary conditions can be taken care of this way. For instance the boundary value problem

{ (−Δ)^{m_{1}+m_{2}+...+m_{k}} u = f in B, D_{m_{1}}Δ^{m_{2}+m_{3}+...+m_{k}}u = 0 on ∂B, ⋮ D_{m_{k−2}}Δ^{m_{k−1}+m_{k}}u = 0 on ∂B, D_{m_{k−1}}Δ^{m_{k}}u = 0 on ∂B, D_{m_{k}}u = 0 on ∂B,  (51)

can be transformed to a system as in (49) with different polyharmonic operators on the diagonal.

6 Only small perturbations

For second order boundary value problems the following holds. Let u ∈ W_{D}^{2,p} (Ω) satisfy

\{ \begin{matrix}−Δu = au + f & in Ω, \\ u = 0 & on ∂Ω, \end{matrix}

where Ω is a bounded sufficiently regular domain in ℝ^{n}. Then there exists a first eigenvalue λ_{1} such that

– if a < λ_{1} then 0 ≢ f ≥ 0 implies u > 0;

– if a = λ_{1} then 0 ≢ f ≥ 0 implies nonexistence of a solution u;

– if a > λ_{1} then 0 ≢ f ≥ 0 implies u ≱ 0.

For the Dirichlet problem with the Laplacian replaced by a polyharmonic operator in general domains there is no such structure. What will remain is the following. Consider

\{ \begin{matrix}(−Δ)^{m} u = au + f & in Ω, \\ D_{m}u = 0 & on ∂Ω, \end{matrix}  (52)

with Ω a bounded domain in ℝ^{n} with ∂Ω ∈ C^{2m,γ} for some γ > 0. Let λ_{1,m,n} denote the first eigenvalue:

λ_{1,m,n} = inf_{u∈W_{0}^{m,2}(Ω)} \frac{∫_{Ω} |∇Δ^{\frac{m−1}{2}} u (x)|^{2} dx}{∫_{Ω} |u (x)|^{2} dx} for m odd;

λ_{1,m,n} = inf_{u∈W_{0}^{m,2}(Ω)} \frac{∫_{Ω} |Δ^{\frac{m}{2}} u (x)|^{2} dx}{∫_{Ω} |u (x)|^{2} dx} for m even.

Let Φ_{1,m,n} denote a corresponding eigenfunction which is normalized by

max_{x∈Ω} Φ_{1,m,n} (x) = 1.

To simplify notation let ((−Δ)^{m} − a)^{inv} denote the Green operator corresponding to the Dirichlet problem (52): u = ((−Δ)^{m} − a)^{inv} f. These operators

((−Δ)^{m} − a)^{inv} : L^{p} (Ω) → W_{D}^{2m,p} (Ω)

are well defined for a ∈ C \binom{}{Ω̄} with a < λ_{1,m,n}.

Theorem 6.1 Suppose that there is a ∈ C \binom{}{Ω̄} with a < λ_{1,m,n} in Ω̄, such that

((−Δ)^{m} − a)^{inv} : L^{p} (Ω) → W_{D}^{2m,p} (Ω) ⊂ L^{p} (Ω) is positivity preserving.

Then for all b ∈ C \binom{}{Ω̄} with a ≤ b < λ_{1,m,n} in Ω̄, we find that

((−Δ)^{m} − b)^{inv} : L^{p} (Ω) → W_{D}^{2m,p} (Ω) ⊂ L^{p} (Ω) is positivity preserving.

Moreover, ((−Δ)^{m} − a)^{inv} being strongly positive implies ((−Δ)^{m} − b)^{inv} to be strongly positive.

Remark 1. An operator T : L^{p} (Ω) → L^{p} (Ω) is called strongly positive iff 0 ≢ f ≥ 0 implies T f > 0.

Lemma 6.2 Let λ < λ_{1,m,n} and suppose that ((−Δ)^{m} − λ)^{inv} : L^{p} (Ω) → W_{D}^{2m,p} (Ω) ⊂ L^{p} (Ω) is strongly positive. Then Φ_{1,m,n} is unique and Φ_{1,m,n} > 0.

Remark 2. If ((−Δ)^{m} − λ)^{inv} can be represented by an integral operator, then this lemma is a consequence of a theorem by Jenč (see [21, page 337]).

Lemma 6.3 Let m > 1. For λ ≪ 0 the operator ((−Δ)^{m} − λ)^{inv} : L^{p} (Ω) → W_{D}^{2m,p} (Ω) ⊂ L^{p} (Ω) is not positivity preserving.

Remark 3. The result in Lemma 6.3 seems to be folklore, as we learnt from Bernis [3], but we have not been able to locate a reference for this fact.

In the following corollary we will suppose that u ∈ W_{D}^{2m,p} (Ω) and f ∈ L^{p} (Ω) are as in (52).

Corollary 6.4 Let m > 1 and suppose that Φ_{1,m,n} > 0. Then there is −∞ < λ_{c} ≤ λ_{1,m,n}, depending on m, n and Ω, such that for all a ∈ C \binom{}{Ω̄} the following holds.

i. If a < λ_{c}, then

0 ≢ f ≥ 0 ⇒ u ≰ 0,

but not positivity preserving:

∃ 0 ≢ f ≥ 0 with u ≥ 0;
∃ 0 ≢ f ≥ 0 with u ≱ 0.

ii. If λ_{c} ≤ a < λ_{1,m,n}, then positivity preserving:

f ≥ 0 ⇒ u ≥ 0.

iii. If a = λ_{1,m,n}, then for 0 ≢ f ≥ 0 there is no solution. iv. If a > λ_{1,m,n}, then positivity killing:

0 ≢ f ≥ 0 ⇒ u ≱ 0.

Remark 4. If m = 1 then by standard results we find that the same holds with λ_{c} = −∞.

Remark 5. The fact that ((−Δ)^{m} − λ)^{inv} is not positivity preserving for λ ≪ 0 implies that the corresponding parabolic initial boundary value problem doesn’t preserve positive initial values. See also [2].

Remark 6. By the example in [17] one finds that for m > 1 a first eigenfunction Φ_{1,m,n} may change sign in strictly convex, arbitrarily smooth domains. If Ω is the unit ball then Φ_{1,m,n} > 0. We expect Φ_{1,m,n} to be positive on domains that are close to a ball. Such a result holds for n = 2, see [12].

Proof of Corollary 6.4. First we prove iv. Let a > λ_{1,m,n} and suppose that we have 0 ≢ f ≥ 0 and 0 ≢ u ≥ 0. Then, like in the case of the Laplacian:

0 < ∫_{Ω} fΦ_{1,m,n}dx = ∫_{Ω} ((−Δ)^{m} u − au) Φ_{1,m,n}dx =

= ∫_{Ω} u (λ_{1,m,n} − a) Φ_{1,m,n}dx ≤ 0,  (53)

which is a contradiction.

iii. also follows by (53).

ii. From Theorem 6.1 one finds that the set of λ in ℝ for which the operator ((−Δ)^{m} − λ)^{inv} : L^{p} (Ω) → W_{D}^{2m,p} (Ω) ⊂ L^{p} (Ω) is positivity preserving is an interval [λ_{c}, λ_{1,m,n}) or (λ_{c}, λ_{1,m,n}). Lemma 6.3 yields that λ_{c} > −∞ when m > 1. Since

(λ ↦ ((−Δ)^{m} − λ)^{inv}) : ℝ → L (L^{p} (Ω) ; W_{D}^{2m,p} (Ω))

is continuous except at the eigenvalues, we find that the interval is left closed.

i. The first claim again follows by (53). From the previous argument we find that ((−Δ)^{m} − λ)^{inv} is not positivity preserving for any λ < λ_{c}. Now suppose that a ≤ λ < λ_{c}. If ((−Δ)^{m} − a)^{inv} is positivity preserving then by Theorem 6.1 ((−Δ)^{m} − λ)^{inv} would be positivity preserving, which is a contradiction. The other claim follows by taking u = Φ_{1,m,n}. Indeed, Φ_{1,m,n} > 0 and ((−Δ)^{m} − a) Φ_{1,m,n} = (λ_{1,m,n} − a) Φ_{1,m,n} > 0.

Proof of Lemma 6.2. First some preliminaries that can be found in [21]. (E, ≤) is a vector lattice if f, g ∈ E implies f ∨ g, f ∧ g ∈ E. (E, ∥·∥ , ≤) is called a Banach lattice if (E, ∥·∥) is a Banach space and if (E, ≤) is a vector lattice such that |f| ≤ |g| implies ∥f∥ ≤ ∥g∥. A subset A ⊂ E is called a lattice ideal if |f| ≤ |g| and g ∈ A implies f ∈ A. An operator S on a Banach lattice E is called irreducible if {0} and E are the only closed lattice ideals that are invariant under S.

One has that L^{p} (Ω) is a Banach lattice and that the operator

T := ((−Δ)^{m} − λ)^{inv} : L^{p} (Ω) → W_{D}^{2m,p} (Ω) ⊂ L^{p} (Ω)

is positive, compact and irreducible. By De Pagter’s Theorem ([19]) the spectral radius of T is positive. By the Krein-Rutman Theorem (see e.g. [21]) it follows that the first eigenvalue of T has a unique (up to multiplication) eigenfunction Φ and that Φ > 0. Since its first eigenvalue is (λ_{1,m,n} − λ)^{−1} we find that Φ satisfies (−Δ)^{m} Φ = λ_{1,m,n}Φ and hence that Φ_{1,m,n} = cΦ and Φ_{1,m,n} > 0.

Proof of Lemma 6.3. We have to show that ((−Δ)^{m} − λ)^{inv} is not positivity preserving for λ ≪ 0. We use different arguments than Bernis in [3].

If ((−Δ)^{m})^{inv} is not positivity preserving we are done. Hence we may assume that ((−Δ)^{m})^{inv} does preserve positivity. We construct a function a ∈ C \binom{}{Ω̄}, with a ≤ 0 such that ((−Δ)^{m} − a)^{inv} is not positivity preserving.

Set u = ((−Δ)^{m})^{inv} 1, which is hence positive by assumption. Since u ∈ W_{D}^{2m,n} (Ω) ⊂ C \binom{}{Ω̄} there will be some B_{ρ} (x_{0}) ⊂ Ω such that u ≥ ε > 0 in \overline{B_{ρ} (x_{0})}. Let χ ∈ C^{∞} (ℝ) be such that 0 ≤ χ ≤ 1, χ (r) = 0 for r > 1 and χ = 1 in r < \frac{1}{2}. We define

ũ (x) = χ (\frac{|x − x_{0}|}{ρ}) (16 ρ^{−2} |x − x_{0}|^{2} − 1) + (1 − χ (\frac{|x − x_{0}|}{ρ})) u (x) .

Then the following holds: there exists some δ ∈ ℝ^{+} such that

(−Δ)^{m} ũ \{ \begin{matrix}= & 0 & for |x − x_{0}| ≤ \frac{1}{2}ρ, \\ ≥ & −δ^{−1} & for \frac{1}{2}ρ ≤ |x − x_{0}| < ρ, \\ = & 1 & elsewhere in Ω, \end{matrix}

and

ũ \{ \begin{matrix}< & 0 & for |x − x_{0}| < \frac{1}{4}ρ, \\ ≥ & 0 & for \frac{1}{4}ρ ≤ |x − x_{0}| < \frac{1}{2}ρ, \\ ≥ & δ & for \frac{1}{2}ρ ≤ |x − x_{0}| < ρ, \\ ≥ & 0 & elsewhere in Ω. \end{matrix}

Next we define

a (x) = −δ^{−2} (1 − χ (2\frac{|x − x_{0}|}{ρ})) ,

and we find that

((−Δ)^{m} − a) ũ \{ \begin{matrix}= & 0 & for |x − x_{0}| < \frac{1}{4}ρ, \\ ≥ & 0 & for \frac{1}{4}ρ ≤ |x − x_{0}| < \frac{1}{2}ρ, \\ ≥ & −δ^{−1} + δ^{−2}δ & for \frac{1}{2}ρ ≤ |x − x_{0}| < ρ, \\ ≥ & 1 & elsewhere in Ω. \end{matrix}

We constructed a sign-changing function ũ with ((−Δ)^{m} − a) ũ ≥ 0 in Ω and D_{m}ũ = 0 on ∂Ω. Hence Theorem 6.1 yields that ((−Δ)^{m} − λ)^{inv} is not positivity preserving for λ ≤ −δ^{−2} = min a.

Proof of Theorem 6.1. First assume that p ≥ 2. Suppose that u ∈ W_{D}^{2m,p} satisfies

(−Δ)^{m} u − bu = f in Ω,  (54)

and we find that u solves

(−Δ)^{m} u − au = f + (b − a) u in Ω,

where (b − a) ≥ 0.

We will show that the nonlinear problem

\{ \begin{matrix}((−Δ)^{m} − a) u = f + (b − a) |u| & in Ω, \\ D_{m}u = 0 & on ∂Ω, \end{matrix}  (55)

has a solution u_{0}. If a solution u_{0} exists we will find that f + (b − a) |u_{0}| ≥ 0 implies

u_{0} = ((−Δ)^{m} − a)^{inv} (f + (b − a) |u_{0}|) ≥ 0.

The existence of u_{0} is established by minimizing the functional F on W_{0}^{m,2} (Ω) defined by

F (u) = \frac{1}{2} ∫_{Ω} (|Ψu|^{2} − au^{2} − 2fu − (b − a) |u| u) dx,

where

Ψu = \{ \begin{matrix}Δ^{\frac{m}{2}} u & if m is even, \\ ∇Δ^{\frac{m−1}{2}} u & if m is odd. \end{matrix}

We prove that F is coercive on W_{0}^{m,2} (Ω). Set

ε = 2λ_{1,m,n} ∧ min_{x∈Ω̄} (λ_{1,m,n} − b (x)) .

Then

F (u) ≥ \frac{1}{2} ∫_{Ω} (|Ψu|^{2} − au^{2} − 2fu − (b − a) u^{2}) dx ≥

≥ \frac{1}{2} ∫_{Ω} ((\frac{\frac{1}{2}ε}{λ_{1,m,n}} + \frac{λ_{1,m,n} − \frac{1}{2}ε}{λ_{1,m,n}}) |Ψu|^{2} − bu^{2} − 2fu) dx ≥

≥ \frac{1}{2} ∫_{Ω} (\frac{ε}{2λ_{1,m,n}} |Ψu|^{2} + (λ_{1,m,n} − \frac{1}{2}ε − b) u^{2} − 2fu) dx ≥

≥ \frac{1}{2} ∫_{Ω} (\frac{ε}{2λ_{1,m,n}} |Ψu|^{2} + \frac{1}{2}εu^{2} − 2fu) dx ≥

≥ \frac{ε}{4λ_{1,m,n}} ∫_{Ω} |Ψu|^{2} dx − \frac{1}{ε} ∫_{Ω} f^{2}dx.

Now let {u_{ν}}_{ν=1}^{∞} be a minimizing sequence in W_{0}^{m,2} (Ω). Since F (·) is sequentially weakly lower semicontinuous and coercive in W_{0}^{m,2} (Ω) there exists a weak limit in W_{0}^{m,2} (Ω), say u_{0}, and F (u_{0}) ≤ lim inf_{ν→∞} F (u_{ν}). Since u_{ν} ⇀ u_{0} in W_{0}^{m,2} (Ω) we find u_{ν} → u_{0} in L^{2} (Ω) . Using regularity theory and appropriate imbeddings a bootstrapping argument shows that u_{0} ∈ W_{D}^{2m,p} (Ω) . As we remarked before we find that u_{0} ≥ 0 and hence that u_{0} solves (52) with b instead of a. For p ∈ (1, 2) the claim follows by approximating f ∈ L^{p} (Ω) by f_{k} ∈ L^{2} (Ω) and using [1].

Acknowledgements This work was supported by the Deutsche Forschungsgemeinschaft.

References

1. S. Agmon, A. Douglis and L. Nirenberg, Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions I, Commun. Pure Appl. Math. 12, 623-727 (1959)

2. F. Bernis, Change of sign of the solutions to some parabolic problems, in: V. Lakshmikantham (ed.), Nonlinear Analysis and Applications, pp. 75-82, New York: Marcel Dekker 1987

3. F. Bernis, private communication

4. T. Boggio, Sulle funzioni di Green d’ordine m, Rend. Circ. Mat. Palermo 20, 97-135 (1905)

5. M. Cranston, E.Fabes and Z. Zhao, Conditional gauge and potential theory for the Schrödinger operator, Transactions Amer. Math. Soc. 307, 171-194 (1988)

6. R.J. Duffin, On a question of Hadamard concerning super-biharmonic functions, J. Math. Phys. 27, 253-258 (1949)

7. P.R. Garabedian, Partial Differential Equations, second edition, New York: Chelsea 1986

8. D. Gilbarg and N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, second edition, Berlin Heidelberg New York: Springer-Verlag 1983

9. H.-Ch. Grunau, Das Dirichletproblem für semilineare elliptische Differentialgleichungen höherer Ordnung, Thesis, Göttingen 1990

10. H.-Ch. Grunau, The Dirichlet problem for some semilinear elliptic differential equations of arbitrary order, Analysis 11, 83-90 (1991)

11. H.-Ch. Grunau and G. Sweers, Classical solutions for some higher order semilinear elliptic equations under weak growth conditions, Nonlinear Anal., to appear 12. H.-Ch. Grunau and G. Sweers, Positivity for perturbations of polyharmonic operators with

Dirichlet boundary conditions in two dimensions, Math. Nachr., to appear

13. M. Grüter and K.O. Widman, The Green function for uniformly elliptic equations, Manuscr. Math. 37, 303-342 (1982)

14. W. Gunsenheimer, Nichtnegativität Greenscher Funktionen für u^{(iv)} + u = f in Abhängigkeit vom zugrundeliegenden Definitionsbereich [−L, L], Arbeitsbericht, Universität Bayreuth, 1994

15. I.W. Herbst and Z. Zhao, Note on a 3G theorem (d = 2), in: E. Çinlar e.a. (eds.), Seminar on Stochastic Processes, 1988, pp. 183-184, Boston Basel Berlin: Birkhäuser 1989

16. H. Hueber and M. Sieveking, Uniform bounds for the quotients of Green functions on C^{1,1}- domains, Ann. Inst. Fourier 32, 105-117 (1982)

17. V.A. Kozlov, V.A. Kondrat’ev and V.G. Maz’ya, On sign variation and the absence of “strong” zeros of solutions of elliptic equations, Math. USSR Izvestiya 34, 337-353 (1990)

18. E. Mitidieri and G. Sweers, Weakly coupled elliptic systems and positivity, Math. Nachr. 173, 259-286 (1995)

19. B. de Pagter, Irreducible compact operators, Math. Z. 192, 149-153 (1986)

20. Ch. Pommerenke, Boundary behaviour of conformal maps, Berlin Heidelberg New York: Springer 1992

21. H.H. Schaefer, Banach lattices and positive operators, Berlin Heidelberg New York: Springer 1974

22. H.S. Shapiro and M. Tegmark, An elementary proof that the biharmonic Green function of an eccentric ellipse changes sign, SIAM Rev. 36, 99-101 (1994)

23. G. Sweers, Positivity in C \binom{}{Ω̄} for elliptic systems, Math. Z. 209, 251-271 (1992)

24. G. Sweers, Positivity for a strongly coupled elliptic systems by Green function estimates, J. Geometric Analysis 4, 121-142 (1994)

25. K.O. Widman, Inequalities for the Green function and boundary continuity of the gradient of solutions of elliptic differential equations, Math. Scand. 21, 17-37 (1967)

26. Z. Zhao, Green function for Schrödinger operator and conditioned Feynman-Kac gauge, J. Math. Anal. Appl. 116, 309-334 (1986)

27. Z. Zhao, Green functions and conditioned gauge theorem for a 2-dimensional domain, in: E. Çinlar e.a. (eds.), Seminar on Stochastic Processes, 1987, pp. 283-294, Boston Basel Berlin: Birkhäuser 1988