On a conditioned Brownian motion and a maximum principle on the disk

A. Dall’Acqua, H.-Ch. Grunau, G.H. Sweers

April 15, 2003

Abstract

Bounds for the 3G-expression ∫_{Ω} G (x, z) G (z, y) dz/G (x, y) play a fundamental role in potential theory. Here G (x, y) is the Green function for the Laplace problem with zero Dirichlet boundary conditions on Ω. The 3G-formula equals 𝔼_{x}^{y} (τ_{Ω}), the expected lifetime for a Brownian motion starting in x ∈ Ω̄, that is killed on exiting Ω and conditioned to converge to and to be stopped at y ∈ Ω̄. Although it was shown by probabilistic methods for bounded (simply connected) 2d-domains that if x ∈ ∂Ω then the supremum of y ↦ 𝔼_{x}^{y} (τ_{Ω}) is assumed for some y at the boundary, the analogous question remained open for x in the interior. Here we are able to give an answer in the case that B ⊂ ℝ^{2} is the unit disk. The dependence of this quantity on the positions of x and y is investigated and it is shown that indeed 𝔼_{x}^{y} (τ_{B}) is maximized on B̄^{2} by opposite boundary points. The result gives also an answer to a number of questions related to the best constant for the positivity preserving property of some elliptic systems. In particular it confirms a relation with a ‘sum of inverse eigenvalues’ that was conjectured in [11].

1 Introduction

Let B = {x ∈ ℝ^{2} : |x| < 1} denote the unit disk and set

G_{B} (x, y) = \frac{1}{4π} log (\frac{|x|^{2} |y|^{2} − 2x · y + 1}{|x|^{2} − 2x · y + |y|^{2}}) for x, y ∈ B.

This function G_{B} is the Green function for

\{ \begin{matrix}−Δu = f & in B, \\ u = 0 & on ∂B, \end{matrix}  (1)

that is, the solution of (1) is given by u (x) = ∫_{B} G_{B} (x, y) f (y) dy. We will show that for every y ∈ B the function x ↦ H (x, y) (≡ 𝔼_{x}^{y} (τ_{B}) for Brownian motion normalized for −Δ) given by

H (x, y) = ∫_{B} \frac{G_{B} (x, z) G_{B} (z, y)}{G_{B} (x, y)}dz  (2)

maximum principle.

2000 Mathematics Subject Classification. Primary 60J65, 35J55; Secondary 35B50.

Key words and phrases. Lifetime of a conditioned brownian motion, positivity for elliptic systems,

is increasing away from y along the hyperbolic geodesics and along the curves of a complementary family. See Theorem 1 and Figure 1 below. As a consequence we will find that x ↦ H (x, y) has no interior maximum and we will even pinpoint the location of the maximum at the boundary.

Our aim in studying this problem was to supply an answer to some questions left open in [2], [9] and in [10], [11]. After explaining the background we will come back to this in section 1.3.

1.1 The link between analysis and probability

The model problem for the positivity preserving property of systems of second order elliptic boundary value problems that are coupled in a noncooperative way is

\{ \begin{matrix}−Δu = f − λv & in Ω, \\ −Δv = f & in Ω, \\ u = v = 0 & on ∂Ω, \end{matrix}  (3)

where Ω is a bounded set in ℝ^{n} and λ ∈ ℝ^{+}. One knows, at least for Ω that satisfy some boundary regularity, that there exists λ_{c} (Ω) ∈ (0, ∞) such that for all f ≥ 0 the solution u satisfies u ≥ 0 if and only if λ ≤ λ_{c} (Ω) . See [11], [12] and [15]. Since the solution u of (3) equals

u (x) = ∫_{y∈Ω} G_{Ω} (x, y) (1 − λ ∫_{z∈Ω} \frac{G_{Ω} (x, z) G_{Ω} (z, y)}{G_{Ω} (x, y)}dz) f (y) dy,

one can show that

λ_{c} (Ω)^{−1} = sup_{x,y∈Ω} ∫_{z∈Ω} \frac{G_{Ω} (x, z) G_{Ω} (z, y)}{G_{Ω} (x, y)}dz,  (4)

where G_{Ω} is the Green function for the Laplace problem with zero Dirichlet boundary condition on Ω. For rather general elliptic problems Cranston, Fabes and Zhao in [4] showed that the right hand side of (4) is finite. For the Laplacian such a bound has been obtained by Cranston in [6] for n ≥ 3 and with McConnell in [5] for n = 2.

The link between (4) and probability theory is:

𝔼_{x}^{y} (τ_{Ω}) = ∫_{z∈Ω} \frac{G_{Ω} (x, z) G_{Ω} (z, y)}{G_{Ω} (x, y)}dz,  (5)

where 𝔼_{x}^{y} (τ_{Ω}) is the expectation of the lifetime of a Brownian motion starting in x, conditioned to converge to and to be stopped at y and to be killed on exiting Ω.

The famous result from [5] states that there is a c > 0 such that

𝔼_{x}^{y} (τ_{Ω}) ≤ c|Ω| for all Ω ⊂ ℝ^{2},

where |Ω| is the Lebesgue measure of Ω.

Some details for identity (5). A Brownian motion that starts in x ∈ Ω and is killed on ∂Ω has transition density given by p_{Ω}(t, x, y) and has expected lifetime given by

E_{x}(τ_{Ω}) = ∫_{Ω} G_{Ω}(x, z)dz.

To consider Brownian motion that is conditioned to exit Ω through Γ ⊂ ∂Ω and stopped at leaving Ω, we use the transition density p_{Ω}^{h}(t, x, z) = p_{Ω}(t, x, z)\frac{h(z)}{h(x)} where h is the solution of

−Δh = 0  in Ω,
h = 0  on ∂Ω \ Γ,
h = 1  on Γ.

This is a so-called Doob’s conditioned Brownian motion, see [7, Part 2, Chap. X]. The expected lifetime is given by

E_{x}^{h}(τ_{Ω}) = ∫_{Ω} G_{Ω}(x, z)\frac{h(z)}{h(x)}dz.  (6)

We want to consider the expectation for the time that Brownian motion spends going from x to y and staying inside Ω. This can be approximated by the expected lifetime for the following conditioned Brownian motion. One considers the domains Ω_{ε} = Ω \ B_{ε}(y) and the functions h_{y,ε} such that

−Δh_{y,ε} = 0  in Ω,
h_{y,ε} = 1  on ∂B_{ε}(y),
h_{y,ε} = 0  on ∂Ω,

with the expected lifetime given by (6) replacing h by h_{y,ε} and G_{Ω} by G_{Ω_{ε}}. The expectation of the time we are interested in becomes the expected lifetime of the Brownian motion starting at x and conditioned to leave Ω \ {y} at {y}. This is now given by

E_{x}^{y}(τ_{Ω\{y}}) = lim_{ε→0} E_{x}^{h_{y,ε}}(τ_{Ω_{ε}}).  (7)

For x and y in the interior, using that

\frac{h_{y,ε}(z)}{h_{y,ε}(x)} → \frac{G_{Ω}(z, y)}{G_{Ω}(x, y)},

and that in dimension n > 1 it holds G_{Ω_{ε}} → G_{Ω}, identity (5) follows from (6) and (7). In the particular case of y ∈ ∂Ω a similar procedure leads to

𝔼_{x}^{y} (τ_{Ω}) = ∫_{z∈Ω} G_{Ω} (x, z) \frac{K_{Ω} (y, z)}{K_{Ω} (y, x)}dz,  (8)

where K_{Ω} (y, ·) is the Poisson kernel for y ∈ ∂Ω, namely the function such that u (x) = ∫_{y∈∂Ω} K_{Ω} (x, y) g (y) dσ_{y} solves

\{ \begin{matrix}−Δu = 0 & in Ω, \\ u = g & on ∂Ω. \end{matrix}

For sufficiently regular domains the expression in (8) is a continuous extension of (5) to Ω × Ω̄. Note that in the above we have used the analyst’s −Δ instead of −\frac{1}{2}Δ.

1.2 Notation and main result

Since the remainder is concerned with the unit disk in ℝ^{2} we will skip the subscript Ω and write G (x, y) = G_{B} (x, y) etc. In 2 dimensions the direct relation between conformal maps and Green functions is best exploited using ℂ instead of ℝ^{2}. For the sake of clear notation we will use boldface for the complex alternative:

for x ∈ ℝ^{2} set x = x_{1} + ix_{2},
for h : ℝ^{2} → ℝ^{2} set h (x) = h_{1} (x) + ih_{2} (x) .

The explicit expressions of the Green function and of the Poisson kernel in the disk can now be written as

G (x, y) = \frac{1}{4π} log (\begin{matrix}|ȳx − 1| & 2 \\ |x − y| & 2 \end{matrix} ) , where x, y ∈ B,

K (x, y) = \frac{1}{2π} \frac{1 − |y|^{2}}{|x − y|^{2}} , where x ∈ ∂B, y ∈ B.

By using dominated convergence and taking limits one can extend the definition of H in (2) up to the closure B̄ × B̄. The complete definition of H then reads:

H (x, y) = \{ \begin{matrix}\frac{∫_{B} G (x, z) G (z, y) dz}{G (x, y)} & if x, y ∈ B with x ≠ y, \\ 0 & if x = y ∈ B̄, \\ \frac{∫_{B} K (x, z) G (z, y) dz}{K (x, y)} & if x ∈ ∂B, y ∈ B, \\ \frac{∫_{B} K (y, z) G (z, x) dz}{K (y, x)} & if x ∈ B, y ∈ ∂B, \\ π |x − y|^{2} ∫_{B} K (x, z) K (y, z) dz & if x, y ∈ ∂B with x ≠ y. \end{matrix}

This function H lies in C (B̄ × B̄) and is strictly positive on B̄^{2}\ {(x, x) ; x ∈ B̄}. The

only delicate part is the case x = y ∈ ∂B, for which we refer to formula (12) below.

A precise formulation of the result is the following:

Theorem 1 For all y ∈ B̄ the function x ↦ H (x, y) is

i. increasing along ‘the hyperbolic geodesics through y’ in increasing euclidean distance;

ii. increasing along the orthogonal trajectories of ‘the hyperbolic geodesics through y’ in increasing euclidean distance.

Remark 1.1 For B ‘the hyperbolic geodesics through y’ are the circles through y that intersect ∂B perpendicular. The orthogonal trajectories are again circles. See Figure 1.

Remark 1.2 For y ∈ ∂B part i of Theorem 1 has been proved by Griffin, McConnell and Verchota in [9].

x ↦ H(x, y)

Figure 1: The geodesics through y in green (light) and the orthogonal trajectories in red (dark).

Corollary 2 One directly finds that:

i. sup_{x∈B̄} H (x, y) = H (−y/ |y| , y) for any y ∈ B̄\ {0};

ii. sup_{x∈B̄} H (x, 0) = H (e, 0) with e = (1, 0);

iii. and sup_{x,y∈B̄} H (x, y) = H (−e, e) .

Remark 2.1 Since the problem has a rotational symmetry one finds that e above might be replaced by any a ∈ ∂B.

1.3 Earlier related results

Critical numbers related to (4) have been studied before in a number of papers. Caristi and Mitidieri in [2] considered the radially symmetric case (in any dimension n), that is, system (3) for radially symmetric functions and hence −Δ replaced by −r^{1−n} \frac{∂}{∂r} (r^{n−1} \frac{∂}{∂r} ). They showed that the corresponding H_{radial} (r, s) is maximal for (r, s) being extremal which means r = 0 and s = 1 or vice versa. The critical number that they find for this radial case is as follows:

sup_{r,s∈[0,1]} H_{radial} (r, s) = \frac{1}{2n}.

In the one-dimensional case they also considered \frac{∂^{2}}{∂x^{2}} + c without assuming symmetry.

Maximal lifetime on the disk. Griffin, McConnell and Verchota in [9] considered H for general simply connected 2-dimensional domains Ω but fixed y ∈ ∂Ω. Two of their main results for such Ω are

sup_{x∈Ω̄,y∈∂Ω} H (x, y) = sup_{x,y∈∂Ω} H (x, y)

and that (with our ‘analytic’ normalization)

sup_{x,y∈Ω̄} H (x, y) ≤ \frac{1}{2π} |Ω| .

For Ω = B they sharpen for y ∈ ∂B :

sup_{x∈B̄,y∈∂B} H (x, y) ≤ 2 log 2 − 1 = \frac{2 log 2 − 1}{π} |B| .

The numerical values are \frac{1}{2π} = .159155... and \frac{2 log 2−1}{π} = .12296.... Our result improves the last estimate by

sup_{x,y∈B̄} H (x, y) = sup_{x∈B̄,y∈∂B} H (x, y) ≤ 2 log 2 − 1,

thereby giving an estimate for the lifetime inequality on a disk with a small hole which is sharper than 1/(2π) (which corresponds to 1/π in [9, Remark 5.7]).

Domain optimization. In [10] Kawohl and coauthor showed that the disk does not give the smallest bound for H among all convex planar sets of equal area. Indeed, they considered a sector-like domain S, with |S| = |B|, and proved that:

sup_{x∈S̄,y∈∂S} H (x, y) < sup_{x∈B̄,y∈∂B} H (x, y) .

The question remains open if

sup_{x,y∈S̄} H (x, y) < sup_{x,y∈B̄} H (x, y) ?  (9)

In the present paper we show that sup_{x∈B̄,y∈∂B} H (x, y) = sup_{x,y∈B̄} H (x, y) holds. We expect the

last identity to hold for all planar domains Ω. Let us put it as a conjecture.

Conjecture 3 If Ω is a (simply connected) planar domain, then

sup_{x,y∈Ω} H (x, y) = sup_{x,y∈∂Ω} H (x, y) .

The obvious consequence of this conjecture is (9). We want to remark that such a result is not likely to hold on a manifold. Consider for example the surface of a ball with a small hole near the pole, see Fig.2. Taking y near the north pole one expects the maximum of H to be attained near the south pole.

Relation with eigenvalues In one dimension critical numbers for sign-changing in (3) were studied by Schröder [14]. The precise result was revisited in [11]. Due to the fact that in one dimension the boundary consists of isolated points one recovers an eigenvalue problem for the critical number.

A relation between that critical number and the Dirichlet eigenvalues in an interval I ⊂ ℝ is

sup_{x,y∈I} H (x, y) = ∑_{k=1}^{∞} \frac{1}{λ_{k}} = 2 ∑_{k=1}^{∞} \frac{(−1)^{k−1}}{λ_{k}} .  (10)

Note that for the unit interval I = (0, 1) these eigenvalues are λ_{k} = π^{2}k^{2},

Figure 2: Sphere with a small hole near the north pole.

For the disk one finds

sup_{x,y∈B} H (x, y) = 4 ∑_{ν=1}^{∞} (−1)^{ν−1} ∑_{k=1}^{∞} \frac{m_{ν,k}}{λ_{ν,k}} ,  (11)

where λ_{ν,k} is the eigenvalue for the eigenfunction with k − 1 circular nodal lines and ν radial nodal lines, and where m_{ν,k} is the multiplicity, that is, m_{ν,k} = 1 for ν = 0 and m_{ν,k} = 2 for ν ≥ 1. The numbers for the two right hand sides above can be found in [11].

1.4 Scheme for the proofs

In section 2 we will consider the case where one of the points lies on the boundary. As mentioned before the case with one point at the boundary has been previously studied by Griffin, McConnell and Verchota in [9]. We will need a more precise characterization of H and in doing so we will recover some of their results. Instead of using power series in ℂ our basic tools will be conformal mappings, a monotonicity result for a convolution (see Proposition 4) and the maximum principle.

Since the function under consideration is symmetric, H(x, y) = H(y, x), the behaviour of x ∈ B ↦ H(x, y) with y ∈ ∂B can be used for the behaviour of x ∈ ∂B ↦ H(x, y) with y ∈ B. Using such a result on the boundary and by several applications of the maximum principle one is able to transfer a inequality valid on the boundary to the interior. This is done in section 3 and will lead to our main result.

Most of the steps consist of deriving estimates for some tailor-made functions. Since all these technicalities might blur the line of arguments we hope to clarify our approach by complementing each intermediate result for a increasing direction of x ↦ H(x, y) (or a related function) by a sketch.

2 The proof for one point lying on the boundary

Assuming y ∈ ∂B we may suppose without loss of generality that y = e = (1, 0). The numerator ∫_{B} K (e, z) G (z, x) dz equals:

E (x) := −\frac{1 − xx̄}{8π} (\frac{log (1 − x)}{x} + \frac{log (1 − x̄)}{x̄} + 1) for x ∈ B̄\ \{e\} and E (e) = 0.

Indeed, since z ↦ K (e, z) ∈ L^{p} (B) for p ∈ (1, 2) the Dirichlet problem for the Poisson equation −Δu = K (e, ·) in B with u = 0 on ∂B has a unique solution in W^{2,p} (B) ∩ W_{0}^{1,p} (B) by [8, Theorem 9.15]. Since G is the kernel for the solution operator from L^{p} (B) to W^{2,p} (B) ∩ W_{0}^{1,p} (B) this Dirichlet problem is solved by u (x) = ∫_{B} K (e, z) G (z, x) dz. Next one checks straightforwardly that E lies in W^{2,p} (B) ∩ C_{0} \binom{}{B̄} for p ∈ (1, 2) and by [1, Theorem IX.17] it follows that E ∈ W^{2,p} (B) ∩ W_{0}^{1,p} (B). Since −ΔE = −4\frac{∂}{∂x} \frac{∂}{∂x̄}E = K (e, ·) in B one finds E = u, the unique solution. The expression for E can also be deduced from an explicit formula for ∫_{B} G(x, z)G(z, y)dz with x, y ∈ B, which is given in [13].

Dividing E(x) by K(e, x) yields:

H (x, e) = −\frac{(1 − x) (1 − x̄)}{4} (\frac{log (1 − x)}{x} + \frac{log (1 − x̄)}{x̄} + 1) ,  (12)

for x ∈ B̄\ {e} and by continuity H (e, e) = 0. We remark that log denotes the analytic extension of the standard logarithm to ℂ\ (−∞, 0] and that the function x ↦ \frac{log(1−x)}{x} is extended by −1 for x = 0.

2.1 In the halfplane

We consider the conformal map from the ball B onto the halfplane ℝ^{+} × ℝ that maps (−1, 0) to (0, 0) and (0, 0) to (1, 0). This map is given by h (x) = \frac{1+x}{1−x}. Note that h(e) = ∞. We let X denote an element of ℝ^{+} × ℝ, or in complex notation X = X_{1} + iX_{2} ∈ ℝ^{+} + iℝ. The inverse of h is also a conformal map and is defined by h^{−1} (X) = \frac{X−1}{X+1}.

It follows from a property of conformal maps that

H (x, e) = ∫_{ℝ^{+}×ℝ} \frac{K (e, h^{−1}(Z))}{K (e, x)}G_{ℝ^{+}×ℝ} (Z, h (x)) |(h^{−1})′ (Z_{1} + iZ_{2})|^{2} dZ_{1}dZ_{2},

where G_{ℝ^{+}×ℝ}(X, Y) = \frac{1}{4π} log (1 + \frac{4X_{1}Y_{1}}{|X − Y |^{2}} ). Next, by defining the function

H̃ (X) := H (x, e) for X = h (x) ,

one finds

H̃ (X) = \frac{1}{4π} ∫_{ℝ^{+}×ℝ} \frac{Z_{1}}{X_{1}} log (1 + \frac{4X_{1}Z_{1}}{|X − Z|^{2}} ) \frac{4}{((1 + Z_{1})^{2} + Z_{2}^{2})^{2}} dZ_{1}dZ_{2}.

We will show that X_{2} ⟼ H̃ (X_{1}, X_{2}) is decreasing for X_{2} > 0. In doing that we need:

Proposition 4 Let f, g ∈ L^{2}(ℝ), f, g ≥ 0, f (t) = f (|t|) , g (t) = g (|t|) and f, g decreasing for t > 0. Then

t ↦ ∫_{ℝ} f (x) g (x + t) dx,  (13)

is decreasing on ℝ^{+}.

Proof. We suppose first that additionally g ∈ C_{0}^{∞} (ℝ). One has

\frac{∂}{∂t} ∫_{ℝ} f (x) g (x + t) dx = ∫_{−∞}^{+∞} f (x) g′ (x + t) dx

= ∫_{−∞}^{−t} f (x) g′ (x + t) dx + ∫_{−t}^{+∞} f (x) g′ (x + t) dx.

Using that g′ (x + t) = −g′ (−x − t) , one gets

\frac{∂}{∂t} ∫_{ℝ} f (x) g (x + t) dx = − ∫_{−∞}^{−t} f (x) g′ (−x − t) dx + ∫_{−t}^{+∞} f (x) g′ (x + t) dx.

Changing the coordinates one obtains

\frac{∂}{∂t} ∫_{ℝ} f (x) g (x + t) dx = ∫_{+∞}^{0} f (−y − t) g′ (y) dy + ∫_{0}^{+∞} f (y − t) g′ (y) dy

= ∫_{0}^{+∞} g′ (y) (f (y − t) − f (−y − t)) dy.

Now for t > 0, one has |y − t| < |−y − t|. Hence the function (13) is decreasing.

The preceding arguments yields the result also for g as in the hypothesis. We observe that such g may be approximated in L^{2}(ℝ) by (g_{k})_{k∈ℕ} ⊂ C_{0}^{∞}(ℝ) having the additional properties above. This is achieved by using an even and in positive x-direction decreasing mollifier in C_{0}^{∞}(ℝ).

Corollary 5 The relations

max_{X_{2}∈ℝ} H̃ (X_{1}, X_{2}) = H̃ (X_{1}, 0) and X_{2} \frac{∂}{∂X_{2}} H̃ (X_{1}, X_{2}) ≤ 0,

hold for every X_{1} ∈ [0, +∞).

Proof. For every X_{1} ∈ ℝ^{+}, one has

H̃ (X) = \frac{1}{π} ∫_{ℝ^{+}} \frac{Z_{1}}{X_{1}} ∫_{ℝ} log (1 + \frac{4X_{1}Z_{1}}{|X − Z|^{2}} ) \frac{1}{((1 + Z_{1})^{2} + Z_{2}^{2})^{2}} dZ_{2}dZ_{1}.

Hence defining

f (Z_{2}) = log (1 + _{\frac{4X_{1}Z_{1}}{(X_{1}−Z_{1})^{2}+Z_{2}^{2}}} ) ,
g (Z_{2}) = _{\frac{1}{((1+Z_{1})^{2}+Z_{2}^{2})^{2}}} ,

we can write

∫_{ℝ} log (1 + \frac{4X_{1}Z_{1}}{|X−Z|^{2}} ) _{\frac{1}{((1+Z_{1})^{2}+Z_{2}^{2})^{2}}} dZ_{2} = ∫_{ℝ} f (Z_{2} − X_{2}) g (Z_{2}) dZ_{2}.

Applying Proposition 4 one gets that the function H̃ is decreasing for X_{2} positive and increasing for X_{2} negative for every X_{1} ∈ ℝ^{+}. The claim follows using the regularity of the function. The case X_{1} = 0 goes similarly by proceeding to the limit.

Figure 3: Illustration of Corollary 5; arrows denote increasing directions of X ↦ H̃(X).

2.2 Back in the disk

Using the properties of conformal mapping, see [3, Sect. III.3], from the increasing direction of H̃ we get an increasing direction of H (x, e) . The lines h^{−1} ({X_{1} = k_{1}}) , varying k_{1} in ℝ^{+}, are circles inside the disk which are tangent to ∂B in (1, 0) . Hence, we have for every (x_{1}, x_{2}) that the function H is increasing in the direction

v_{(x_{1},x_{2})} = (−x_{2}, \frac{2x_{1} − x_{1}^{2} − 1 + x_{2}^{2}}{2 (1 − x_{1})} ), if x_{2} > 0,  (14)

and in the −v_{(x_{1},x_{2})}–direction, if x_{2} < 0. In particular we obtain that

x_{2} \frac{∂}{∂θ} H (x, e) := x_{2} (−x_{2} \frac{∂}{∂x_{1}} + x_{1} \frac{∂}{∂x_{2}} ) H (x, e) ≥ 0 when |x| = 1.  (15)

Here we write x_{1} = |x| cos θ and x_{2} = |x| sin θ.

x ↦ H(x, e)

Figure 4: The result of Corollary 5 transformed back to the disk; arrows denote increasing directions of x ↦ H(x, e).

Since we will proceed through properties of the differential equation that H satisfies let us fix the following formula.

Lemma 6 For a, b ∈ C^{2} with b ≠ 0 the following identity holds

−Δ (\frac{a}{b}) − 2\frac{∇b}{b} · ∇ (\frac{a}{b}) + \frac{−Δb}{b} (\frac{a}{b}) = \frac{−Δa}{b}.  (16)

Having e ∈ ∂B one finds −ΔK(x, e) = 0 and −Δ (∫_{B} G(x, z)K(z, e)dz) = K(x, e) in B and by (16) the function H satisfies:

−ΔH (x, e) − 2\frac{∇K (x, e)}{K (x, e)} · ∇H (x, e) = 1 when x ∈ B.

Let us consider the derivative with respect to the angle \frac{∂}{∂θ} H. Since \frac{∂}{∂θ} = x_{1} \frac{∂}{∂x_{2}} −x_{2} \frac{∂}{∂x_{1}} , we get

∇\frac{∂}{∂θ} H = ∇ ((x_{1} \frac{∂}{∂x_{2}} − x_{2} \frac{∂}{∂x_{1}} ) H) = (R + \frac{∂}{∂θ} ) ∇H,

with R = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) . Since \frac{∂}{∂θ} and Δ commute and since R is skew-symmetric, one

obtains that \frac{∂}{∂θ} H (x, e) satisfies

−Δ\frac{∂}{∂θ} H − 2∇ log(K) · ∇\frac{∂}{∂θ} H = −\frac{∂}{∂θ} ΔH − 2\frac{∇K}{K} · (R + \frac{∂}{∂θ} ) ∇H =

= \frac{∂}{∂θ} (−ΔH − 2\frac{∇K}{K} · ∇H) + 2 (\frac{∂}{∂θ} \frac{∇K}{K} ) · ∇H − 2\frac{∇K}{K} · R∇H
= 0 + 2 ((\frac{∂}{∂θ} + R) ∇ log K) · ∇H
= 2 (∇\frac{∂}{∂θ} log K) · ∇H.

By the symmetry one observes that \frac{∂}{∂θ} H(x, e) = 0 in \{x ∈ B : x_{2} = 0\}. Furthermore from (15) it follows that \frac{∂}{∂θ} H(x, e) ≥ 0 in \{x ∈ ∂B : x_{2} > 0\}. A priori one knows that x ↦ \frac{∂}{∂θ} H (x, e) is in C^{2}(B^{+}) ∩ C(B̄^{+} \ \{e\}) . It remains to study the behavior in e. Using the explicit formula of H(x, e) given by (12), we will prove the following:

Lemma 7 The following identity holds

lim_{\begin{matrix}x→e, \\ x∈B^{+} \end{matrix}} \frac{∂}{∂θ} H(x, e) = 0.

Proof. Since \frac{∂}{∂θ} = i (x\frac{∂}{∂x} − x̄\frac{∂}{∂x̄}), one gets

\frac{∂}{∂θ} H (x, e) = i\frac{(1 − x̄)}{4} (log (1 − x) + \frac{x}{x̄} log (1 − x̄) + x)

−i\frac{(1 − x) (1 − x̄)}{4} (−\frac{log (1 − x)}{x} − \frac{1}{1 − x})

−i\frac{(1 − x)}{4} (\frac{x̄}{x} log (1 − x) + log (1 − x̄) + x̄)

+i\frac{(1 − x) (1 − x̄)}{4} (−\frac{log (1 − x̄)}{x̄} − \frac{1}{1 − x̄})

= i\frac{log (1 − x)}{4x} (1 − 2x̄ + xx̄) − i\frac{log (1 − x̄)}{4x̄} (1 − 2x + xx̄) − i\frac{x̄ − x}{2}.

One observes that

lim_{\begin{matrix}x→e, \\ x∈B^{+} \end{matrix}} \frac{∂}{∂θ} H (x, e) = 0.

Hence we can say that \frac{∂}{∂θ} H (·, e) ∈ C^{2}(B) ∩ C(B̄) and that it satisfies the boundary value problem

\{ \begin{matrix}−Δ\frac{∂}{∂θ} H − 2\frac{∇K}{K} · ∇\frac{∂}{∂θ} H = 2∇\frac{∂}{∂θ} log K · ∇H & in B^{+}, \\ \frac{∂}{∂θ} H ≥ 0 & on ∂B^{+}. \end{matrix}  (17)

Proposition 8 The inequality x_{2} \frac{∂}{∂θ} H (x, e) ≥ 0 holds for all x ∈ B.

x ↦ H(x, e)

Figure 5: For y = e ∈ ∂B the function x ↦ H(x, y) is increasing along semicircles to the left.

Proof. Since K (x, e) = \frac{1 − |x|^{2}}{|x − e|^{2}} we have

\frac{∂}{∂θ} log K = − \begin{matrix}\frac{∂}{∂θ} & |x − e| & 2 \\ |x − e| & 2 \end{matrix} = −\begin{matrix}(x_{1} \frac{∂}{∂x_{2}} − x_{2} \frac{∂}{∂x_{1}} ) & ((x_{1} − 1)^{2} + x_{2}^{2}) \\ |x − e| & 2 \end{matrix}
= 2\frac{x_{2} (x_{1} − 1) − x_{1}x_{2}}{|x − e|^{2}} = \frac{−2x_{2}}{|x − e|^{2}} ,

and

∇\frac{∂}{∂θ} log K = ∇ (\frac{−2x_{2}}{|x − e|^{2}} ) = \frac{−2}{|x − e|^{2}} (0, 1) + \frac{2x_{2}}{|x − e|^{4}} (2 (x_{1} − 1) , 2x_{2})
= \frac{2}{|x − e|^{4}} (2x_{2} (x_{1} − 1) , 2x_{2}^{2} − |x − e|^{2})

= 2\frac{(2x_{2} (x_{1} − 1) , x_{2}^{2} − (x_{1} − 1)^{2})}{|x − e|^{4}}
= \frac{4 (1 − x_{1})}{|x − e|^{4}} (−x_{2}, \frac{2x_{1} − x_{1}^{2} − 1 + x_{2}^{2}}{2 (1 − x_{1})} ) .

We see that ∇\frac{∂}{∂θ} log K (x_{1}, x_{2}) has the direction of v_{(x_{1},x_{2})} as defined in (14) . Hence the term

∇\frac{∂}{∂θ} log K · ∇H,

is non-negative. Applying Theorem A to (17) the claim follows.

3 The proof for both points in the interior

3.1 Tangential directions

We consider now y in the interior. Without loss in generality we can suppose that y = (−s, 0) with s ∈ (0, 1). The case s = 0 gives the radial symmetric case which has been considered previously by Caristi and Mitidieri in [2].

Let us fix x at the boundary and consider H (x, y) . Let C_{s} = {y : |y| = s} . From the previous section it follows that the maximum of H (x, ·) in C_{s} is attained in y = −sx. This is equivalent to ask for y = (−s, 0) that x = (1, 0) . So using the symmetry of the problem we can say that

x_{2} \frac{∂}{∂θ} H (x, (−s, 0)) ≤ 0 when x ∈ ∂B.  (18)

x ↦ H(x, y) x ↦ H^{s}(x)

Figure 6: Using the symmetry between x and y we may conclude that for any y ∈ B, the function x ↦ H(x, y) (left) is increasing along ∂B from the nearest boundary point of y to the most distant boundary point. Putting y = (−s, 0) with s > 0 it means increasing to the right along ∂B. Also the function x ↦ H^{s}(x) (right) is increasing to the right along ∂B.

We consider the conformal map k_{s} from the disk onto the disk that maps y into 0. We have

k_{s} (x) = \frac{x + s}{1 + sx}.

Proceeding as before we will now study the function

H^{s} (x) := H (k_{s}^{−1}(x), k_{s}^{−1}(0)) ,

which due to the behaviour of conformal mappings transforms into

H^{s} (x) = ∫_{B} \frac{G (x, z) G (z, 0)}{G (x, 0)} |(k_{s}^{−1})′ (z_{1} + iz_{2})|^{2} dz_{1}dz_{2}.

We have k_{s}^{−1} (z) = \frac{z−s}{1−sz} and |(k_{s}^{−1})′ (z_{1} + iz_{2})|^{2} = \begin{matrix}(1−s^{2}) & ^{2} \\ |e−sz| & ^{4} \end{matrix} , hence

H^{s} (x) = ∫_{B} \frac{G (x, z) G (z, 0)}{G (x, 0)} \begin{matrix}(1 − s^{2}) & 2 \\ |e − sz| & 4 \end{matrix} dz_{1}dz_{2}.

One gets that for x ≠ 0 the function H^{s} satisfies

−ΔH^{s} (x) + \frac{2}{r |log r|} \frac{∂}{∂r} H^{s} (x) = \begin{matrix}(1 − s^{2}) & 2 \\ |e − sx| & 4 \end{matrix} .

Proposition 9 The inequality x_{2} \frac{∂}{∂θ} H^{s} (x) ≤ 0 holds for all x ∈ B.

Proof. By symmetry it is sufficient to consider the case x ∈ B^{+}. We consider the function Θ (x) := \frac{∂}{∂θ} H^{s} (x) or to be more specific

Θ (x) = x_{1} \frac{∂}{∂x_{2}} H^{s} (x) − x_{2} \frac{∂}{∂x_{1}} H^{s} (x) .

Since Δ and \frac{∂}{∂θ} commute, one finds

−ΔΘ (x) = −\frac{∂}{∂θ} ΔH^{s} (x) = \frac{∂}{∂θ} [−\frac{2}{r| log r|} \frac{∂}{∂r} H^{s} (x) + \begin{matrix}(1 − s^{2}) & 2 \\ |e − sx| & 4 \end{matrix} ]

= −\frac{2}{r| log r|} \frac{∂}{∂r} Θ (x) − 4\begin{matrix}(1 − s^{2}) & 2 \\ |e − sx| & 6 \end{matrix} sx_{2}.

A priori we have that Θ ∈ C^{2}(B̄ \ {0}). It remains to study the behavior of Θ in 0. We have

\frac{∂}{∂θ} H^{s}(x) = \frac{1}{G(x, 0)} \frac{∂}{∂θ} R(x),

where R(x) satisfies

\{ \begin{matrix}−ΔR (x) = −\frac{(1−s^{2})^{2}}{4π|e−sx|^{4}} log |x|^{2} & in B, \\ R (x) = 0 & on ∂B. \end{matrix}  (19)

Since the right hand side of (19) is in L^{p}(B) for every p ∈ (1, +∞), one gets R ∈ W^{2,p}(B) and hence, using the Sobolev imbedding theorem it follows that

R ∈ C^{1,α}(B̄) for every α ∈ (0, 1).  (20)

Setting Ω = B_{\frac{1}{2}} (0), we have \frac{∂}{∂θ} R and G^{−1}(·, 0) ∈ C(Ω̄) (where we extend G^{−1}(·, 0) in 0 by 0). Hence Θ ∈ C^{2}(B^{+}) ∩ C^{0}(B̄^{+}).

Using (18) and the fact that H^{s} is symmetric in x_{2} = 0, we find that Θ(x) ≤ 0 on ∂B^{+}. We may summarize:

−ΔΘ (x) + \frac{2}{r| log r|} \frac{∂}{∂r} Θ (x) = −4\begin{matrix}(1−s^{2}) & ^{2} \\ |e−sx| & ^{6} \end{matrix} sx_{2} in B^{+},
Θ (x) ≤ 0  on ∂B^{+}.

The claim follows applying the maximum principle, see Theorem A.

x ↦ H^{s}(x)

Figure 7: A conformal mapping changed H(x, y) to H^{s}(x) and put y in the center. By Proposition 9 the mapping x ↦ H^{s}(x) is increasing to the right along all semicircles around 0.

3.2 Radial directions

In order to prove Theorem 1, it remains to show that the function H^{s} (x_{1}, 0) is increasing on the interval (0, 1) . We will show that the function H^{s} is increasing in radial direction by using the maximum principle. First we will show that the function H^{s} satisfies a zero Neumann boundary condition:

Lemma 10 The identity \frac{∂}{∂r} H^{s} (x) = 0 holds for all x ∈ ∂B.

Proof. We write

H^{s} (x) = \frac{R (x)}{G (x, 0)},

with R (x) = ∫_{B} G (x, z) G (z, 0) \begin{matrix}(1−s^{2}) & ^{2} \\ |e−sz| & ^{4} \end{matrix} dz_{1}dz_{2} and observe that R (x) = G (x, 0) = 0 for x ∈ ∂B. Moreover

−ΔR (x) = G (x, 0) \begin{matrix}(1 − s^{2}) & 2 \\ |e − sx| & 4 \end{matrix} and − ΔG (x, 0) = 0 for x ≠ 0, x ∈ B.

Since −Δ = −\frac{∂^{2}}{∂r^{2}} − \frac{1}{r} \frac{∂}{∂r} − \frac{1}{r^{2}} \frac{∂^{2}}{∂^{2}ϕ}, we find that at the boundary

−\frac{∂^{2}}{∂r^{2}} R (x) = \frac{∂}{∂r} R (x) ,  (21)
−\frac{∂^{2}}{∂r^{2}} G (x, 0) = \frac{∂}{∂r} G (x, 0) .  (22)

Using the series expansion near the boundary for R (x) and G (x, 0), we get for x ∈ ∂B:

lim_{B∋ξ→x} \frac{∂}{∂r} H^{s} (ξ) = lim_{B∋ξ→x} \frac{\frac{∂}{∂r} G (ξ, 0)}{G (ξ, 0)} ( \frac{\frac{∂}{∂r} R (ξ)}{\frac{∂}{∂r} G (ξ, 0)} − \frac{R (ξ)}{G (ξ, 0)})

= lim_{B∋ξ→x} \frac{1}{|ξ|−1} ( \frac{\frac{∂}{∂r} R(ξ)+(|ξ|−1) \frac{∂^{2}}{∂r^{2}} R(ξ)+..}{\frac{∂}{∂r} G(ξ,0)+(|ξ|−1) \frac{∂^{2}}{∂r^{2}} G(ξ,0)+..} − \frac{\frac{∂}{∂r} R(ξ)+ \frac{|ξ|−1}{2} \frac{∂^{2}}{∂r^{2}} R(ξ)+..}{\frac{∂}{∂r} G(ξ,0)+ \frac{|ξ|−1}{2} \frac{∂^{2}}{∂r^{2}} G(ξ,0)+..})

= \frac{1}{2} \frac{\frac{∂^{2}}{∂r^{2}} R (x) \frac{∂}{∂r} G (x, 0) − \frac{∂^{2}}{∂r^{2}} G (x, 0) \frac{∂}{∂r} R (x)}{(\frac{∂}{∂r} G (x, 0))^{2}},

which is zero by using (21) and (22).

Proposition 11 The inequality r \frac{∂}{∂r} H^{s}(x) ≥ 0 holds for all x ∈ B.

x ↦ H^{s}(x)

Figure 8: A conformal mapping changed H(x, y) to H^{s}(x) and, roughly said, it put y in the center. Here is the result from Proposition 11: the function x ↦ H^{s}(x) is radially increasing. The combination with Figure 7 and the inverse conformal mapping lead to Figure 1.

Proof. The function H^{s} satisfies

−ΔH^{s} (x) = \begin{matrix}(1 − s^{2}) & 2 \\ |e − sx| & 4 \end{matrix} + \frac{4}{|x|^{2} (log |x|^{2})}x · ∇H^{s} (x) .  (23)

Let us define Ξ (x) := r \frac{∂}{∂r} H^{s} (x) = x · ∇H^{s} (x). One has

−ΔΞ (x) = −2ΔH^{s} (x) − x_{1} \frac{∂}{∂x_{1}} ΔH^{s} (x) − x_{2} \frac{∂}{∂x_{2}} ΔH^{s} (x) = …

and by (23)

… = 2\begin{matrix}(1−s^{2}) & ^{2} \\ |e−sx| & ^{4} \end{matrix} + \frac{8}{|x|^{2}(log|x|^{2})}x · ∇H^{s} (x) + x · ∇ (\begin{matrix}(1−s^{2}) & ^{2} \\ |e−sx| & ^{4} \end{matrix} + \frac{4}{|x|^{2}(log|x|^{2})}x · ∇H^{s} (x))
= 2\begin{matrix}(1−s^{2}) & ^{2} \\ |e−sx| & ^{4} \end{matrix} + \frac{8}{|x|^{2}(log|x|^{2})}Ξ (x) + 4sx_{1} \begin{matrix}(1−s^{2}) & ^{2} \\ |e−sx| & ^{6} \end{matrix} (1 − sx_{1})

+ \frac{4x_{1}}{|x|^{2}(log|x|^{2})} \frac{∂}{∂x_{1}} Ξ (x) − \frac{8x_{1}^{2}}{|x|^{4}(log|x|^{2})}Ξ (x) − \frac{8x_{1}^{2}}{|x|^{4}(log^{2}|x|^{2})}Ξ (x)

− 4s^{2}x_{2}^{2} \begin{matrix}(1−s^{2}) & ^{2} \\ |e−sx| & ^{6} \end{matrix} + \frac{4x_{2}}{|x|^{2}(log|x|^{2})} \frac{∂}{∂x_{2}} Ξ (x) − \frac{8x_{2}^{2}}{|x|^{4} log|x|^{2}} Ξ (x) − \frac{8x_{2}^{2}}{|x|^{4}(log^{2}|x|^{2})}Ξ (x) ,

that gives

−ΔΞ (x) − \frac{4x·∇Ξ(x)}{|x|^{2} log|x|^{2}} + \frac{8Ξ(x)}{|x|^{2}(log^{2}|x|^{2})} = 2\begin{matrix}(1−s^{2}) & ^{2} \\ |e−sx| & ^{4} \end{matrix} (\frac{1 − s|x|^{2}}{|e − sx|^{2}} ) .  (24)

One sees that the right hand side of (24) is non-negative. Furthermore, since

Ξ(x) = \frac{r}{G(x, 0)} \frac{∂}{∂r} R(x) − \frac{R(x)}{(G(x, 0))^{2}} r \frac{∂}{∂r} G(x, 0),

with R ∈ C^{1,α}(B̄) (from (20)), one has that Ξ(0) = 0 and that Ξ is continuous in B. With help of the preceding Lemma 10 we get that Ξ ∈ C^{0}(B̄). Hence, summarizing we have

\{ \begin{matrix}−ΔΞ (x) − \frac{4}{|x|^{2} log|x|^{2}} x · ∇Ξ (x) + \frac{8Ξ(x)}{|x|^{2}(log^{2}|x|^{2})} ≥ 0 & in B \ \{0\}, \\ Ξ (x) = 0 & on ∂B ∪ \{0\}. \end{matrix}

The maximum principle stated in Theorem A finally yields Ξ ≥ 0 in B.

Appendix A: A version of the Maximum Principle

The maximum principle had to be repeatedly applied to differential operators of which the coefficients become singular on the boundary. We prefer to give the precise formulation of a maximum principle, which is adequate for this situation. For a proof, we refer to [8, Sect. 3.1].

Theorem A Suppose that Ω ⊂ ℝ^{n} is open, bounded and connected, and that b ∈ C(Ω; ℝ^{n}) and c ∈ C(Ω; ℝ) with c ≥ 0. Set L = −Δ + b · ∇ + c. If u ∈ C^{2}(Ω) satisfies

\{ \begin{matrix}Lu(x) ≥ 0 & for x ∈ Ω, \\ lim inf_{Ω∋x→x_{∂}} u(x) ≥ 0 & for x_{∂} ∈ ∂Ω, \end{matrix}

then u ≥ 0 in Ω.

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A. Dall’Acqua, Dept. of Applied Mathematical Analysis, ITS Faculty, Delft University of Technology, Pobox 5031 2600GA Delft The Netherlands email: a.dallacqua@ITS.TUDelft.nl

H.C. Grunau Inst. für Analysis und Numerik, Fakultät für Mathematik, Otto-von-Guericke-Universität Magdeburg, Postfach 4120 D-39016 Magdeburg, Germany email: Hans-Christoph.Grunau@mathematik.uni-magdeburg.de

G.H. Sweers Dept. of Applied Mathematical Analysis, ITS Faculty, Delft University of Technology, Pobox 5031 2600GA Delft The Netherlands email: G.H.Sweers@ITS.TUDelft.nl