KELVIN TRANSFORM FOR GRUSHIN OPERATORS AND CRITICAL SEMILINEAR EQUATIONS

ROBERTO MONTI AND DANIELE MORBIDELLI

Abstract. We study positive entire solutions u = u(x, y) of the critical equation

Δ_{x}u + (α + 1)^{2}|x|^{2α}Δ_{y}u = −u^{\frac{Q+2}{Q−2}}  in ℝ^{n} = ℝ^{m} × ℝ^{k}  (1)

where (x, y) ∈ ℝ^{m} × ℝ^{k}, α > 0 and Q = m + k(α + 1).

In the first part of the paper, exploiting the invariance of the equation with respect to a suitable conformal inversion, we prove a “spherical symmetry” result for solutions.

In the second part, we show how to reduce the dimension of the problem using a hyperbolic symmetry argument. Given any positive solution u of (1), after a suitable scaling and a translation in the variable y, the function v(x) = u(x, 0) satisfies the equation

div_{x}(p∇_{x}v) − qv = −pv^{\frac{Q+2}{Q−2}} , |x| < 1,  (2)

with a mixed boundary condition. Here, p and q are appropriate radial functions. In the last part, we prove that if m = k = 1 the solution of (2) is unique and that for m ≥ 3 and k = 1 problem (2) has unique solution in the class of x-radial functions.

1. Introduction and results

In this paper, we study entire positive solutions of the semilinear equation

Δ_{x}u + (α + 1)^{2}|x|^{2α}Δ_{y}u = −u^{\frac{Q+2}{Q−2}}  in ℝ^{n} = ℝ^{m} × ℝ^{k},  (1.1)

where Δ_{x} and Δ_{y} are Laplace operators in the variables x ∈ ℝ^{m} and y ∈ ℝ^{k}, respectively. Here, α > 0 is a positive real number and Q = m + k(α + 1) is the appropriate “dimension”. The partial differential operator L := Δ_{x} + (α + 1)^{2}|x|^{2α}Δ_{y} is known (at least for α = 1) as Grushin operator. The power \frac{Q+2}{Q−2} in the nonlinear term is the corresponding critical exponent. We prove symmetry and uniqueness results for entire positive solutions to equation (1.1).

If α = 0, then Q = n and (1.1) is the Yamabe equation in ℝ^{n}. A positive solution u yields a Riemannian metric ds^{2} = u^{\frac{4}{n−2}} |dz|^{2} conformal to the standard one in ℝ^{n} and with constant scalar curvature equal to 4\frac{n−1}{n−2}. Positive solutions are radial functions about some point in ℝ^{n}. This was proved by Gidas, Ni and Nirenberg in [GNN] under some assumptions on the behavior at infinity of solutions and by Caffarelli, Gidas and Spruck [CGS] in the general case (see also [CL] and [LZha] for simpler proofs).

Date: 19th April 2004.

2000 Mathematics subject classification. Primary 35H20. Secondary 34B15.

In the case α = 1, the nonlinear equation (1.1) already appeared in connection with the Cauchy-Riemann Yamabe problem solved by Jerison and Lee in [JL1, JL2]. The model space for this problem is the Heisenberg group ℂ^{n} × ℝ and in this setting the Yamabe equation for Webster scalar curvature is

Δ_{b}u = −u^{\frac{Q+2}{Q−2}} , (z, t) ∈ ℂ^{n} × ℝ = ℝ^{2n} × ℝ, Q = 2n + 2,  (1.2)

where Δ_{b} is the Heisenberg sublaplacian. If the function u is radial in the variable z, u = u(|z|, t), then Δ_{b}u = Δ_{z}u + 4|z|^{2}∂_{tt}u is a Grushin operator with α = 1. In this radial case, Jerison and Lee found in [JL1] a method for solving equation (1.1) with α = 1, m even integer and k = 1. This method has been also generalized to more general choices of m and k by Garofalo and Vassilev in [GV]. These techniques, however, seem very much to depend on the choice α = 1 and the problem of finding explicit solutions of (1.1) for any α > 0 is, to the authors’ knowledge, still open.

There are two main motivations for our interest in equation (1.1). First, a typical example of weakly pseudoconvex domain in the complex space is the generalized Siegel domain Ω_{p} = {(z_{1}, z_{2}) ∈ ℂ^{2} : Im(z_{2}) > |z_{1}|^{2p}} with p > 1, the case p = 1 being strictly pseudoconvex. Under a radiality assumption in the variable z_{1}, the natural boundary sublaplacian on ∂Ω_{p} takes the form of a Grushin operator with α > 1. Thus, understanding equation (1.1) seems to be the first step in the study of semilinear equations with geometric relevance at the boundary of weakly pseudoconvex domains.

Secondly, equation (1.1) also results as Euler equation for the following Sobolev inequality

(∫_{ℝ^{n}} |u|^{\frac{2Q}{Q−2}} dxdy)^{\frac{Q−2}{2Q}} ≤ C (∫_{ℝ^{n}}(|∇_{x}u|^{2} + (α + 1)^{2}|x|^{2α}|∇_{y}u|^{2})dxdy)^{1/2} , (1.3)

for functions in the Sobolev space D^{1}(ℝ^{n}), the completion of C_{0}^{∞}(ℝ^{n}) in the seminorm appearing in the right hand side. Inequality (1.3) follows from the Poincaré inequality in [FL] and from the representation formula in [FLW]. The search for extremal functions, our original motivation for this problem, naturally leads to the Grushin semilinear critical equation. The problem of extremal functions in the case α = 1 and for suitable m, k has been discussed by Beckner in [B].

Equation (1.1) with α > 0 is not invariant under x-translations. This introduces a new difficulty which does not appear in the case α = 0. In particular, the classical moving planes method no longer works if α > 0. We can, however, develop a moving spheres method exploiting the conformal invariance of equation (1.1). We introduce a spherical inversion and a Kelvin transform preserving the equation and we prove a “spherical” symmetry for solutions.

For z = (x, y) ∈ ℝ^{m} × ℝ^{k}, let

∥z∥ = (|x|^{2(α+1)} + |y|^{2})^{\frac{1}{2(α+1)}} .

The function Γ(z) = ∥z∥^{2−Q} solves the equation LΓ(z) = 0, z ≠ 0. Actually, Γ is a constant multiple of the fundamental solution for L with pole at the origin. Formulae representing the fundamental solution with pole at arbitrary points of ℝ^{n} have been computed by Beals, Greiner and Gaveau in [BGG].

The “norm” ∥z∥ is 1–homogeneous for the group of anisotropic dilations (x, y) ↦ δ_{λ}(x, y) = (λx, λ^{α+1}y), λ > 0. Using such dilations, a spherical inversion I can be defined by

I(z) = δ_{∥z∥^{−2}}(z), z ≠ 0.

In the case α = 0, the map I becomes the Möbius inversion z ↦ \frac{z}{|z|^{2}} .

The inversion I is a conformal map in the following sense:

Theorem 1. Let D_{α} = (∇_{x}, (α + 1)|x|^{α}∇_{y}). Then, for any u ∈ C^{1}(ℝ) and z ≠ 0,

|D_{α}(u ∘ I)(z)|^{2} = |J_{I}(z)|^{2/Q}|(D_{α}u)(I(z))|^{2}  (1.4)

where J_{I} = det \frac{∂I}{∂z} is the Jacobian of I.

The Jacobian of I satisfies |J_{I}(z)| = ∥z∥^{−2Q} and therefore |J_{I}(z)|^{\frac{Q−2}{2Q}} = Γ(z) is an L-harmonic function (see Lemma 2.2). This generalizes the analogous Euclidean phenomenon with the appropriate “dimension” Q (see [IM, Sec. 2.7]). Equation (1.4) can be seen as a Cauchy-Riemann system and it can be also described from a “metric” point of view. Consider the (singular) Riemannian metric ds^{2} = |dx|^{2} +(α+ 1)^{−2}|x|^{−2α}|dy|^{2} (this actually generates the subriemannian distance of the operator L studied by Franchi and Lanconelli [FL]). Then Theorem 1 essentially says that the map I is conformal in the metric ds^{2} (see the precise statement in Theorem 2.3).

The conformal inversion I induces the following Kelvin transform. Given a function u : ℝ^{n} → ℝ, define u^{∗} : ℝ^{n} \ {0} → ℝ by

u^{∗}(z) = Γ(z)u(I(z)), z ≠ 0.

Using the conformal property (1.4), we prove that the Kelvin transform preserves equation (1.1).

Theorem 2. If u ∈ C^{2}(ℝ^{n}) is a positive entire solution to equation (1.1), then u^{∗} is a solution of the same equation in ℝ^{n} \ {0}.

The function u^{∗} has, a priori, a singularity at z = 0. This singularity, however, turns out to be removable, as a byproduct of Theorem 3 below.

Concerning the notion of inversion, Korányi [K] seems to have been the first to introduce a Kelvin transform in the Heisenberg group. More recently, the existence of a “conformal inversion” has played a substantial role in the classification of the so called Heisenberg type groups (see the work of Cowling, Dooley, Korányi and Ricci [CDKR]).

A function u can be scaled according to the rule δ_{λ}u(z) = λ^{\frac{Q}{2} −1}u (δ_{λ}(z)), λ > 0. Then u solves (1.1) if and only if the scaled function δ_{λ}u does. Adapting some ideas of Li and Zhang [LZha] and using a suitable Hopf Lemma for L we prove the following

Theorem 3. Let u ∈ C^{2}(ℝ^{n}) be a positive solution of equation (1.1). Then for a suitable λ > 0, the function δ_{λ}u satisfies δ_{λ}u = (δ_{λ}u)^{∗}.

Any translated function u^{(b)}(x, y) := u(x, y + b), b ∈ ℝ^{k}, solves (1.1), if u does. Then, Theorem 3 applies to any solution u^{(b)} and, by means of the family of the spherical identities so obtained, the solution u can be determined, at least on the subspace x = 0, by an argument due, in the Euclidean case, to Li and Zhu [LZhu]. Precisely, if u is a positive solution with u = u^{∗}, then there exists y_{0} ∈ ℝ^{k} such that

u(0, y) = u(0, y_{0})(1 + |y − y_{0}|^{2})^{− \frac{Q−2}{2(α+1)}} , y ∈ ℝ^{k}.  (1.5)

This identity and Theorem 3 yield a symmetry result for solutions which is best described in the hyperbolic geometry.

To any function u = u(x, y) associate a function U of the variables ξ ∈ ℝ^{m} and η ∈ ℝ^{k} by letting

U(ξ, η) = |ξ|^{\frac{Q−2}{2(α+1)}} u (|ξ|^{\frac{1}{α+1}} \frac{ξ}{|ξ|}, η) .  (1.6)

In order to explain this functional change, fix for a moment m = 1. In this case, if u solves (1.1), then U solves the equation

Δ_{H}U + \frac{Q(Q − 2)}{4(α + 1)^{2}}U = −\frac{1}{(α + 1)^{2}} U^{\frac{Q+2}{Q−2}} , ξ > 0,  (1.7)

where Δ_{H} = ξ^{2}Δ+(1−k)ξ∂_{ξ}, with Δ = ∂_{ξ}^{2}+∑_{i=1}^{k} ∂_{η_{i}}^{2}, is the (k+1)-dimensional hyperbolic Laplacian (see Proposition 3.2). Incidentally, notice that the Kelvin transform for the Grushin operator corresponds, via (1.6), to the hyperbolic isometry ζ ↦ \frac{ζ}{|ζ|^{2}} (see Remark 3.3). Equation (1.7) is invariant under the group of hyperbolic isometries. This suggests that symmetry of solutions to (1.7), if any, should be of hyperbolic type.

Now let m ≥ 1 again. For any v ∈ ℝ^{m} with |v| = 1 and for any r ∈ (0, 1) define the k-dimensional sphere

Σ(v, r) = \{(x, y) ∈ ℝ^{n} : \frac{x}{|x|} = v, \frac{(1 − |x|)^{2} + |y|^{2}}{4|x|} = \frac{r^{2}}{1 − r^{2}} \} .

This is a hyperbolic sphere in the halfspace {(tv, y) : t > 0, y ∈ ℝ^{k}} centered at t = 1, y = 0. Theorem 4. Let u ∈ C^{2}(ℝ^{n}) be a positive solution to equation (1.1) such that u = u^{∗} and y_{0} = 0 in (1.5). Then the function U in (1.6) is constant on Σ(v, r) for any v with |v| = 1 and r ∈ (0, 1).

This essentially says that a solution u satisfying the assumptions in Theorem 4 is determined by its values on the m dimensional ball {|x| < 1}×{0} ⊂ ℝ^{m} ×ℝ^{k}. Then, it is natural to look for an equation satisfied by the function v(x) := u(x, 0). In fact, it turns out that the dimension of the original equation (1.1) can be reduced and the function v(x) must solve the problem

\{ \begin{matrix}div_{x}(p∇_{x}v) − qv = −pv^{\frac{Q+2}{Q−2}} & |x| < 1 \\ v > 0 & |x| ≤ 1 \\ \frac{∂v}{∂ν} + (\frac{Q}{2} − 1)v = 0 & |x| = 1, \end{matrix}  (1.8)

where p(x) = (1 − |x|^{2(α+1)})^{k} and q(x) = k(α + 1)(Q − 2)(1 − |x|^{2(α+1)})^{k−1}|x|^{2α}. Here, ν denotes the exterior normal to the unit ball in ℝ^{m}. The boundary condition in (1.8) is produced by the reduction procedure. The partial differential equation in (1.8) is of variational type and it is clearly related to the following Sobolev-Hardy inequality

( ∫_{B} |v|^{\frac{2Q}{Q−2}} (1 − |x|)^{k} dx)^{\frac{Q−2}{Q}} ≤ C ∫_{B} \{|∇_{x}v|^{2}(1 − |x|)^{k} + |v|^{2}|x|^{2α}(1 − |x|)^{k−1}\} dx

for functions v ∈ C^{1}(B̄), where B = {x ∈ ℝ^{m} : |x| < 1}. We shall discuss this inequality in a future study.

The last part of the paper is devoted to the study of uniqueness. The existence of solutions to (1.1) can be proved by Lions’ concentration-compactness method and we shall not address the problem, here. It is natural to conjecture that the solution of (1.1) is unique up to scaling and translations in y. This statement reduces to proving uniqueness for problem (1.8). In the case m > 1, we expect the solution v to be x-radial. The proof of this property seems to require new ideas and we have not yet been able to achieve it.

We can, however, prove the conjecture in the following case.

Theorem 5. Let m = k = 1 and α > 0. Up to a scaling and a vertical translation, there exists a unique positive solution u ∈ C^{2}(ℝ^{2}) of equation (1.1).

If m = k = α = 1 (so that Q = 3), the function in ℝ^{2}

u(x, y) = \frac{1}{((1 + x^{2})^{2} + y^{2})^{1/4}}

solves equation (1.1). By Theorem 5, this is the unique solution, up to a scaling and a vertical translation. Beckner proved in [B] that this function is an extremal for the Sobolev inequality (1.3) in the plane.

In order to prove Theorem 5, we show that problem (1.8) with m = k = 1 and Q = α + 2 has unique solution. The problem is

\{ \begin{matrix}(pu′)′ − qu + pu^{\frac{Q+2}{Q−2}} = 0 & in (−1, 1) \\ u > 0 & in [−1, 1] \\ αu(x) + 2xu′(x) = 0 & x = ±1, \end{matrix}  (1.9)

where p(x) = (1 − |x|^{2(α+1)}), q(x) = α(α + 1)|x|^{2α}.

The main step in the proof is showing that solutions are even functions.

Theorem 6. If u ∈ C^{1}([−1, 1])∩C^{2}(−1, 1) is a solution to problem (1.9) then u′(0) = 0 and u is even.

The uniqueness then follows by estimating how many times two even solutions can intersect (see the proof of Theorem 4.5).

If m > 1, assuming the function v to be x-radial transforms problem (1.8) into a problem in one variable. We prove that the solution is unique and thus we have the following Theorem 7. Let m ≥ 3, k = 1, n = m + 1 and α > 0. Up to a scaling and a vertical translation, there exists, in the class of x-radial functions, a unique positive solution u ∈ C^{2}(ℝ^{n}) of problem (1.1).

The problem of uniqueness of (radial) solutions of nonlinear PDEs is in general quite delicate. In our study of problem (1.8), after several attempts with different techniques, we have found out the way how to adapt the energy method of Kwong and Li [KL]. Unfortunately, the arguments are not simple and moreover, in the case m = 2 and k = 1, there is a clearly technical problem and the proof does not work (see Remark 4.2). In the perspective of the present work, it remains an interesting question to find a more natural and less complicated proof of Theorems 6 and 7.

A short description of the paper is now in order. In section 2, we study the conformal inversion, we prove Theorems 1 and 2, and we finally establish Theorem 3 and its corollary, identity (1.5). In section 3, we study the hyperbolic symmetry of solutions, we prove Theorem 4 and we show how the reduced problem (1.8) can be obtained. Section 4 is devoted to uniqueness results.

Notation. If a, b are vectors in ℝ^{d} for some d ∈ ℕ, we denote by ⟨a, b⟩ = ∑_{j=1}^{d} a_{j}b_{j} the standard inner product and by |a| = ⟨a, a⟩^{1/2} the Euclidean norm. ∇_{x} and ∇_{y} denote Euclidean gradients with respect to the variables x ∈ ℝ^{m} and y ∈ ℝ^{k}. Moreover, D_{α} = (∇_{x}, (α + 1)|x|^{α}∇_{y}) denotes the Grushin gradient and

div_{α}(f, g) = ∑_{i=1}^{m} \frac{∂f_{i}}{∂x_{i}} + (α + 1)|x|^{α} ∑_{j=1}^{k} \frac{∂g_{j}}{∂y_{j}} , (f, g) ∈ C^{1}(ℝ^{n}; ℝ^{m} × ℝ^{k})

is the Grushin divergence. With this notation, Lu = div_{α}D_{α}u. Finally, we let 2^{∗} = \frac{2Q}{Q−2}, so that \frac{Q+2}{Q−2} = 2^{∗} − 1.

2. Kelvin transform and spherical symmetry of solutions

In this section, we study the Kelvin transform for the operator L and we prove the main spherical symmetry result for solutions to problem (1.1). For z ∈ ℝ^{n} and λ > 0, we let

∥z∥ = (|x|^{2(α+1)} + |y|^{2})^{\frac{1}{2(α+1)}} and δ_{λ}(z) = (λx, λ^{α+1}y).

Proposition 2.1. The singular function

Γ(z) = ∥z∥^{2−Q},  Q = m + k(α + 1).  (2.1)

satisfies LΓ = 0 in ℝ^{n} \ {0}.

Proof. The proof is a short computation. Letting N(z) = |x|^{2(α+1)} + |y|^{2} and β = \frac{2−Q}{2(α+1)}, we have, for z ≠ 0, ∇_{x}Γ = 2(α + 1)βN^{β−1}|x|^{2α}x and ∇_{y}Γ = 2βN^{β−1}y. Then

Δ_{x}Γ = div_{x}∇_{x}Γ = 2(α + 1)βN^{β−2}|x|^{2α}{2(α + 1)(β − 1)|x|^{2(α+1)} + (2α + m)N},
Δ_{y}Γ = div_{y}∇_{y}Γ = 2βN^{β−2}{2(β − 1)|y|^{2} + kN},

and thus

LΓ = 2(α + 1)βN^{β−1}|x|^{2α}{Q + 2(α + 1)(β − 1) + 2α}.

Since Q + 2(α + 1)(β − 1) + 2α = 0, this shows that LΓ(z) = 0 for z ≠ 0. □

Define the inversion I : ℝ^{n} \ {0} → ℝ^{n} \ {0} by

I(z) = δ_{∥z∥^{−2}}(z), z ≠ 0.

Clearly, I^{2} is the identity. In the next propositions, we prove some basic properties of I. We denote by J_{I}(z) = det \frac{∂I(z)}{∂z} the Jacobian of I at the point z ≠ 0.

Lemma 2.2. For all z ≠ 0 we have |J_{I}(z)| = Γ(z)^{\frac{2Q}{Q−2}} = ∥z∥^{−2Q}.

Proof. We give a sketch of the proof. Let Φ(z) = ∥z∥ and consider a relatively open set A ⊂ {Φ = 1}. The coarea formula and a dilation argument yield for t > 0

∫_{δ_{t}(A)} \frac{dH^{n−1}(z)}{|∇Φ(z)|} = t^{Q−1}μ(A), where μ(A) = ∫_{A} \frac{dH^{n−1}(z)}{|∇Φ(z)|},  (2.2)

where H^{n−1} denotes the (n − 1)-dimensional Hausdorff measure in ℝ^{n}.

Now fix r > 0 and for δ > 0 define Ω_{δ} = {δ_{t}(z) : z ∈ A, r < t < r + δ}. The inverted set is I(Ω_{δ}) = {δ_{t}(z) : z ∈ A, r < 1/t < r + δ}. By the coarea formula and by (2.2),

|Ω_{δ}| = ∫_{r}^{r+δ} ∫_{δ_{t}(A)} \frac{1}{|∇Φ|}dH^{n−1}dt = μ(A) ∫_{r}^{r+δ} t^{Q−1}dt,

and analogously, |I(Ω_{δ})| = μ(A) ∫_{1/(r+δ)}^{1/r} t^{Q−1}dt.

Note that |I(Ω_{δ})|/|Ω_{δ}| does not depend on A. If z ∈ ℝ^{n} is a point such that ∥z∥ = r > 0 then, by the area formula, |J_{I}(z)| = lim_{δ→0} \frac{|I(Ω_{δ})|}{|Ω_{δ}|} = r^{−2Q} = ∥z∥^{−2Q}. The proof is concluded. □

The next theorem and the following corollary describe a remarkable conformality property of I.

Theorem 2.3. Let z = (x, y) ∈ ℝ^{n} with x ≠ 0, and |ζ|_{z} = \sqrt{|ξ|^{2} + (α + 1)^{−2}|x|^{−2α}|η|^{2}}, ζ = (ξ, η) ∈ ℝ^{n}. Then

lim_{ζ→z} \frac{|I(ζ) − I(z)|_{I(z)}}{|ζ − z|_{z}} = |J_{I}(z)|^{1/Q}.  (2.3)

Proof. Fix z ∈ ℝ^{n} with x ≠ 0. Define I_{x}(z) ∈ ℝ^{m} and I_{y}(z) ∈ ℝ^{k} by the relation I(z) = (I_{x}(z), I_{y}(z)), and let N(z) = |x|^{2(α+1)} + |y|^{2}. Then

|I(ζ) − I(z)|_{I(z)}^{2} = |I_{x}(ζ) − I_{x}(z)|^{2} + (α + 1)^{−2}|I_{x}(z)|^{−2α}|I_{y}(ζ) − I_{y}(z)|^{2}

= |\frac{ξ}{N(ζ)^{\frac{1}{α+1}}} − \frac{x}{N(z)^{\frac{1}{α+1}}} |^{2} + \frac{N(z)^{\frac{2α}{α+1}}}{(α + 1)^{2}|x|^{2α}} |\frac{η}{N(ζ)} − \frac{y}{N(z)}|^{2}

= N(z)^{− \frac{2}{α+1}} \{|ξ(\frac{N(ζ)}{N(z)})^{− \frac{1}{α+1}} − x|^{2} + \frac{1}{(α + 1)^{2}|x|^{2α}} |η(\frac{N(ζ)}{N(z)})^{−1} − y|^{2}\}.

By a Taylor expansion of the function N(ζ) at the point z,

\frac{N(ζ)}{N(z)} = 1 + \frac{Ω(z, ζ)}{N(z)} + o(|z − ζ|),  (2.4)

where we let Ω(z, ζ) = {2(α + 1)|x|^{2α}⟨x, ξ − x⟩ + 2⟨y, η − y⟩}. Notice that Ω(z, ζ) = O(|z − ζ|). Therefore

\begin{matrix}(\frac{N(ζ)}{N(z)} & )^{−1} = 1 − \frac{Ω(z, ζ)}{N(z)} + o(|z − ζ|), \\ (\frac{N(ζ)}{N(z)} & )^{−1/(α+1)} = 1 − \frac{Ω(z, ζ)}{(α + 1)N(z)} + o(|z − ζ|). \end{matrix}

In the following, N replaces N(z). By Lemma 2.2, N^{− \frac{2}{α+1}} = |J_{I}(z)|^{2/Q} and thus

|J_{I}(z)|^{−2/Q}|I(ζ)−I(z)|_{I(z)}^{2} = |ξ − x − \frac{Ω(z, ζ)}{(α + 1)N} ξ|^{2}

+ \frac{1}{(α + 1)^{2}|x|^{2α}} |η − y − \frac{Ω(z, ζ)}{N}η|^{2} + o(|z − ζ|^{2})

= |ζ − z|_{z}^{2} − \frac{2⟨ξ − x, ξ⟩}{(α + 1)N}Ω(z, ζ) + \frac{|ξ|^{2}}{(α + 1)^{2}N^{2}} Ω(z, ζ)^{2}

− \frac{2⟨η − y, η⟩}{(α + 1)^{2}|x|^{2α}N} Ω(z, ζ) + \frac{|η|^{2}}{(α + 1)^{2}|x|^{2α}N^{2}} Ω(z, ζ)^{2}

+ o(|z − ζ|^{2})

= |ζ − z|_{z}^{2} + Ω(z, ζ)R(z, ζ),

where R is defined by the last equality. The proof of the theorem will be completed as soon as we prove that R(z, ζ) = o(|ζ − z|). Using the explicit form of Ω,

R(z, ζ) = \frac{2}{N^{2}(α + 1)^{2}} [ − (α + 1)N⟨ξ − x, ξ⟩ + |ξ|^{2}(α + 1)|x|^{2α}⟨x, ξ − x⟩ + |ξ|^{2}⟨y, η − y⟩

− \frac{N}{|x|^{2α}} ⟨η − y, η⟩ + (α + 1)|η|^{2}⟨x, ξ − x⟩ + \frac{|η|^{2}}{|x|^{2α}} ⟨y, η − y⟩]

= \frac{2}{N^{2}(α + 1)}⟨ξ − x, ξ⟩( − (|x|^{2(α+1)} + |y|^{2}) + |x|^{2α}|ξ|^{2} + |η|^{2})

+ \frac{2}{N^{2}(α + 1)^{2}} ⟨η − y, y⟩(|ξ|^{2} − \frac{|x|^{2(α+1)} + |y|^{2}}{|x|^{2α}} + \frac{|η|^{2}}{|x|^{2α}} ) + o(|z − ζ|).

In the last equality, we replaced ⟨ξ − x, x⟩ with ⟨ξ − x, ξ⟩ (and the same we did for η) and we consequently added an o(|z − ζ|). Now the claim follows, because both the round brackets in the last two lines tend to zero as ζ → z. □

Corollary 2.4. Let u, v ∈ C^{1}(ℝ^{n}). Then for z ≠ 0

⟨D_{α}(u ∘ I)(z), D_{α}(v ∘ I)(z)⟩ = |J_{I}(z)|^{2/Q}⟨D_{α}u(I(z)), D_{α}v(I(z))⟩.  (2.5)

Proof. We first prove the corollary for x ≠ 0. Then the proof for any z ≠ 0 follows by continuity. Notice that for x ≠ 0

|D_{α}u(z)| = lim sup_{ζ→z} \frac{|u(ζ) − u(z)|}{|ζ − z|_{z}}.  (2.6)

The inequality “≥” follows from |u(ζ) − u(z)| ≤ |D_{α}u(z)||ζ − z|_{z} + o(|ζ − z|), and equality is achieved by choosing ζ_{ε} = z + ε(∇_{x}u(z), (α + 1)^{2}|x|^{2α}∇_{y}u(z)) and letting ε → 0.

By Theorem 2.3 and (2.6) we get

|D_{α}(u ∘ I)(z)| = lim_{ζ→z} \frac{|I(ζ) − I(z)|_{I(z)}}{|ζ − z|_{z}} lim sup_{ζ→z} \frac{|u(I(ζ)) − u(I(z))|}{|I(ζ) − I(z)|_{I(z)}}
= |J_{I}(z)|^{1/Q}|D_{α}u(I(z))|.

Developing this identity for the function u ∘ I + v ∘ I, we find (2.5). □

Let u : ℝ^{n} → ℝ be a function. The Kelvin transform u^{∗} : ℝ^{n} \ {0} → ℝ of u is defined by

u^{∗}(z) = Γ(z)u(I(z)), z ≠ 0.  (2.7)

Theorem 2.5. If u ∈ C^{2}(ℝ^{n}) is a positive solution of (1.1), then u^{∗} is a solution of the same equation in ℝ^{n} \ {0}.

Proof. We first prove that if u ∈ C^{1}(ℝ^{n}) and φ ∈ C_{0}^{1}(ℝ^{n} \ {0}), then

∫_{ℝ^{n}}⟨D_{α}u, D_{α}φ⟩dz = ∫_{ℝ^{n}}⟨D_{α}u^{∗}, D_{α}φ^{∗}⟩dz.  (2.8)

In order to prove (2.8), let v = u ∘ I and ψ = φ ∘ I. Then

⟨D_{α}u^{∗}, D_{α}φ^{∗}⟩ = ⟨D_{α}(Γv), D_{α}(Γψ)⟩

= Γ^{2}⟨D_{α}v, D_{α}ψ⟩ + vψ|D_{α}Γ|^{2} + Γ⟨D_{α}(ψv), D_{α}Γ⟩

= Γ^{2}⟨D_{α}v, D_{α}ψ⟩ + div_{α}(vψΓD_{α}Γ),

because LΓ = 0 in ℝ^{n} \ {0}, by Proposition 2.1. On the other hand, by (2.5)

⟨D_{α}v(z), D_{α}ψ(z)⟩ = |J_{I}(z)|^{2/Q}⟨D_{α}u(I(z)), D_{α}φ(I(z))⟩.

By Lemma 2.2, Γ(z)^{2} = |J_{I}(z)|^{\frac{Q−2}{Q}}, and then

∫_{ℝ^{n}}⟨D_{α}u^{∗}, D_{α}φ^{∗}⟩dz = ∫_{ℝ^{n}} (Γ^{2}⟨D_{α}v, D_{α}ψ⟩ + div_{α}(vψΓD_{α}Γ))dz

= ∫_{ℝ^{n}} Γ^{2}⟨D_{α}v, D_{α}ψ⟩dz = ∫_{ℝ^{n}}⟨D_{α}u(I(z)), D_{α}φ(I(z))⟩|J_{I}(z)|dz

= ∫_{ℝ^{n}}⟨D_{α}u(z), D_{α}φ(z)⟩dz.

The proof of (2.8) is concluded.

In order to conclude the proof of the theorem, it suffices to note that

∫_{ℝ^{n}}(u^{∗})^{\frac{Q+2}{Q−2}} φ^{∗}dz = ∫_{ℝ^{n}} u(I(z))^{\frac{Q+2}{Q−2}} φ(I(z))^{∗}|J_{I}(z)|dz = ∫_{ℝ^{n}} u^{\frac{Q+2}{Q−2}} φ dz

for all φ ∈ C_{0}^{∞}(ℝ^{n} \ {0}). □

Now, we are going to apply the moving spheres method to equation (1.1). Given a function u : ℝ^{n} → ℝ and λ > 0, define the function δ_{λ}u(z) = λ^{\frac{Q}{2} −1}u (δ_{λ}(z)). Then let

u_{λ}(z) = (δ_{λ^{2}}u)^{∗}(z) = (\frac{∥z∥}{λ})^{2−Q}u (\frac{λ^{2}x}{∥z∥^{2}} , \frac{λ^{2(α+1)}y}{∥z∥^{2(α+1)}} ) , z ≠ 0, λ > 0.

If Lu = −u^{2^{∗}−1}, 2^{∗} = \frac{2Q}{Q−2}, then δ_{λ}u and u_{λ} solve the same equation. The statement concerning δ_{λ}u is a simple computation. The statement concerning u_{λ} is a consequence of Theorem 2.5.

The knowledge of the singular solution (2.1) easily provides the following “characteristic” Hopf Lemma.

Lemma 2.6 (Hopf–type Lemma). Let v ∈ ℝ^{k} with |v| = 1, t ∈ ℝ, Ω = {(x, y) ∈ ℝ^{n} : ⟨y, v⟩ > t}, and (0, y_{0}) ∈ ∂Ω. If a function u ∈ C^{2}(Ω) ∩ C^{1}(Ω̄) satisfies u > 0 in Ω, u(0, y_{0}) = 0 and Lu ≤ 0 in Ω, then ⟨∇_{y}u(0, y_{0}), v⟩ > 0.

Proof. The proof is a short computation. Let y_{1} = y_{0} + v, z_{0} = (0, y_{0}), z_{1} = (0, y_{1}) and z = (x, y). The point z_{1} = z_{0} + (0, v) belongs to Ω. The function Λ(z) = Γ(z − z_{1}) − Γ(z_{0} − z_{1}) satisfies LΛ(z) = 0 for z ≠ z_{1}, Λ(z) = 0 for ∥z − z_{1}∥ = γ_{1} := ∥z_{0} − z_{1}∥ and Λ(z) = 1 for ∥z − z_{1}∥ = γ_{0}, for a suitable γ_{0} ∈ (0, γ_{1}).

Let R = {γ_{0} < ∥z − z_{1}∥ < γ_{1}} ⊂ Ω. If ε > 0 is small enough, the function u_{ε}(z) = u(z)−εΛ(z) is strictly positive on ∥z −z_{1}∥ = γ_{1}. Moreover, u_{ε}(z) = u(z) ≥ 0 on ∥z − z_{1}∥ = γ_{0}. Since u_{ε} ≥ 0 on ∂R and Lu_{ε}(z) = Lu(z) − εLΛ(z) = Lu(z) ≤ 0 on R, by the weak Maximum Principle it follows that u ≥ εΛ on R. Thus, using u(0, y_{0}) = 0, we find

⟨∇_{y}u(0, y_{0}), v⟩ = lim_{t→0} \frac{u(0, y_{0} + tv)}{t} ≥ ε lim_{t→0} \frac{Λ(0, y_{0} + tv)}{t}

= ε lim_{t→0} \frac{1}{t}\{|y_{0} − y_{1} + tv|^{\frac{2−Q}{α+1}} − |y_{0} − y_{1}|^{\frac{2−Q}{α+1}} \}

= ε lim_{t→0} \frac{1}{t}\{(1 − t)^{\frac{2−Q}{α+1}} − 1\} = ε\frac{Q − 2}{α + 1} > 0.

This ends the proof. □

Theorem 2.7. Let u ∈ C^{2}(ℝ^{n}) be a positive solution of equation (1.1). Then there exists λ > 0 such that u = u_{λ}.

Proof. The proof is articulated in three steps. Here, the scheme of the elegant proof by Li and Zhang in [LZha, Sec. 2] can be adapted without significant changes.

Step 1. If u ∈ C^{2}(ℝ^{n}) is a positive solution of (1.1) then there exists λ_{0} > 0 such that

u_{λ}(z) ≤ u(z), λ ∈ (0, λ_{0}), ∥z∥ ≥ λ.

Step 2. Define

λ̄ = sup{λ_{0} > 0 : u_{λ}(z) ≤ u(z), ∥z∥ ≥ λ, 0 < λ ≤ λ_{0}}.  (2.9)

If λ̄ < +∞ then u ≡ uλ̄ on ℝ^{n} \ {0}.

Step 3. For b ∈ ℝ^{k}, let u^{(b)}(z) = u(z + (0, b)), z ∈ ℝ^{n}, and let λ̄_{b} be defined as in (2.9) relatively to u^{(b)}. If there exists b ∈ ℝ^{k} such that λ̄_{b} = +∞, then u ≡ 0.

Proof of Step 1. Let z ∈ ℝ^{n} with ∥z∥ ≤ 1 and for λ > 0 define the function φ(λ) = λ^{\frac{Q−2}{2}} u(δ_{λ}(z)). It is easy to check that

φ′(λ) = λ^{\frac{Q−4}{2}} [\frac{Q − 2}{2}u(δ_{λ}(z)) + λ⟨∇_{x}u(δ_{λ}(z)), x⟩ + (α + 1)λ^{α+1}⟨∇_{y}u(δ_{λ}(z)), y⟩].

Since lim_{λ→0} [...] = \frac{Q−2}{2}u(0) > 0 uniformly in ∥z∥ ≤ 1, we have φ′(λ) > 0 for λ ∈ (0, λ_{1}) and ∥z∥ ≤ 1, where λ_{1} is small enough. Then, for any z with ∥z∥ ≤ 1 and 0 < λ < λ_{1}, we have φ(λ_{1}) − φ(λ) > 0, by the mean value theorem. Letting μ = λ/λ_{1}, this yields u(z) ≥ μ^{\frac{Q−2}{2}} u(δ_{μ}(z)), for ∥z∥ ≤ λ_{1} and 0 < μ < 1. Thus, choosing μ = λ^{2}/∥z∥^{2} < 1,

(\frac{λ}{∥z∥})^{Q−2}u(δ_{λ^{2}∥z∥^{−2}}(z)) ≤ u(z), 0 < λ ≤ ∥z∥ ≤ λ_{1}.

Now note that

u(z) − (\frac{λ_{1}}{∥z∥})^{Q−2} min_{∥ζ∥=λ_{1}} u(ζ) ≥ 0,  ∥z∥ ≥ λ_{1}.  (2.10)

This follows from Bony’s Maximum Principle [Bo]. Indeed, the function in the left hand side is superharmonic for the operator L and nonnegative on ∥z∥ = λ_{1} and as ∥z∥ → ∞.

Fix now λ_{0} ∈ (0, λ_{1}) such that

λ_{0}^{Q−2} max_{∥z∥≤λ_{0}} u(z) ≤ λ_{1}^{Q−2} min_{∥z∥=λ_{1}} u(z).

If ∥z∥ ≥ λ_{1} and λ ∈ (0, λ_{0}), we have ∥δ_{λ^{2}∥z∥^{−2}}(z)∥ = \frac{λ^{2}}{∥z∥} ≤ λ_{0}. Thus

(\frac{λ}{∥z∥})^{Q−2}u(δ_{λ^{2}∥z∥^{−2}}(z)) ≤ (\frac{λ_{0}}{∥z∥})^{Q−2} max_{∥ζ∥≤λ_{0}} u(ζ) ≤ (\frac{λ_{1}}{∥z∥})^{Q−2} min_{∥ζ∥=λ_{1}} u(ζ) ≤ u(z).

Proof of Step 2. Let w_{λ} = u − u_{λ}. Then w_{λ}(z) = 0 for ∥z∥ = λ. Moreover, if 0 < λ ≤ λ̄,

w_{λ}(z) ≥ 0, Lw_{λ}(z) ≤ 0, for ∥z∥ ≥ λ.

If wλ̄ ≡ 0 on {∥z∥ ≥ λ̄} the proof is finished. Assume this does not hold. Then, by the Maximum Principle it must be wλ̄ > 0 on ∥z∥ > λ̄. By the elliptic Hopf Lemma and by Lemma 2.6, for all z with ∥z∥ = λ̄ we have ⟨ν(z), ∇wλ̄(z)⟩ > 0, where ν(z) is the unit normal at the point z to the surface {ζ ∈ ℝ^{n} : ∥ζ∥ = ∥z∥}.

By continuity, for some r > λ̄ suitably close to λ̄ and for all λ ∈ [λ̄, r] we have

w_{λ}(z) > 0,  λ < ∥z∥ ≤ r.  (2.11)

We are going to extend this inequality on ∥z∥ ≥ r, at least for λ ∈ [λ̄, λ̄ + ε_{0}) for a small ε_{0} < r − λ̄.

Using the Maximum Principle as in the proof of (2.10) we get

u(z) − uλ̄(z) ≥ (\frac{r}{∥z∥})^{Q−2} min_{∥ζ∥=r}(u(ζ) − uλ̄(ζ)) > 0, ∥z∥ ≥ r.

Then

u(z) − u_{λ}(z) ≥ uλ̄(z) − u_{λ}(z) + (\frac{r}{∥z∥})^{Q−2} min_{∥ζ∥=r}(u(ζ) − uλ̄(ζ)), ∥z∥ ≥ r.

Now

|uλ̄(z) − u_{λ}(z)| = \frac{1}{∥z∥^{Q−2}} |λ̄^{Q−2}u(δ_{λ̄^{2}∥z∥^{−2}}(z)) − λ^{Q−2}u(δ_{λ^{2}∥z∥^{−2}}(z))|.

If ∥z∥ ≥ r, the points δ_{λ̄^{2}∥z∥^{−2}}(z) and δ_{λ^{2}∥z∥^{−2}}(z) are in a compact set. Then, by uniform continuity, there exists ε_{0} > 0 such that for ∥z∥ ≥ r and λ ∈ (λ̄, λ̄ + ε_{0}) we have

|λ̄^{Q−2}u(δ_{λ̄^{2}∥z∥^{−2}}(z)) − λ^{Q−2}u(δ_{λ^{2}∥z∥^{−2}}(z))| ≤ \frac{1}{2}r^{Q−2} min_{∥ζ∥=r}(u(ζ) − uλ̄(ζ)).

Therefore, for all z with ∥z∥ ≥ r and λ ∈ (λ̄, λ̄ + ε_{0}) it holds

w_{λ}(z) = u(z) − u_{λ}(z) ≥ \frac{1}{2}(\frac{r}{∥z∥})^{Q−2} min_{∥ζ∥=r}(u(ζ) − uλ̄(ζ)) > 0,

which, together with (2.11), contradicts the definition of λ̄ and finishes the proof of Step 2.

Proof of Step 3. If u is a solution, then u^{(b)}(z) = u(z + (0, b)) is still a solution and there exists a maximal λ̄_{b} > 0 such that

(u^{(b)})_{λ}(z) ≤ u^{(b)}(z), λ ∈ (0, λ̄_{b}), ∥z∥ ≥ λ.

We first show that if λ̄_{b} = +∞ for some b, then λ̄_{b} = +∞ for all b ∈ ℝ^{k}. Let b be such that λ̄_{b} = +∞. Letting z_{b} = z − (0, b), we have

u(z) ≥ (\frac{λ}{∥z_{b}∥})^{Q−2}u(δ_{λ^{2}∥z_{b}∥^{−2}}(z_{b}) + (0, b))

for all λ > 0 and ∥z_{b}∥ ≥ λ. For any fixed λ > 0, it follows that lim inf_{∥z∥→∞} ∥z∥^{Q−2}u(z) ≥

λ^{Q−2}u(0, b) > 0. Letting λ → +∞, this implies

lim_{∥z∥→∞} ∥z∥^{Q−2}u(z) = +∞.  (2.12)

Assume now that there is a ≠ b such that λ̄_{a} < +∞. Then, by Step 2, it is u^{(a)} ≡ (u^{(a)})λ̄_{a}, i.e.

u(z) ≡ (\frac{λ̄_{a}}{∥z_{a}∥})^{Q−2}u(δ_{λ̄_{a}^{2}∥z_{a}∥^{−2}}(z_{a}) + (0, a)), z_{a} = z − (0, a) ≠ 0.

This gives lim_{∥z∥→∞} ∥z∥^{Q−2}u(z) = λ̄_{a}^{Q−2}u(0, a) < +∞, which contradicts (2.12). Now we know that λ̄_{b} = +∞ for any b. Let

g_{λ,b}(z) := u(z + (0, b)) − (\frac{λ}{∥z∥})^{Q−2}u(δ_{λ^{2}∥z∥^{−2}}(z) + (0, b))

for λ > 0, ∥z∥ ≥ λ, b ∈ ℝ^{k}. Since g_{∥z∥,b}(z) = 0 and g_{∥z∥,b}(δ_{r}(z)) ≥ 0, r ≥ 1, it follows that

\frac{d}{dr}g_{∥z∥,b}(δ_{r}(z)) ≥ 0, for r = 1.

This, after a short computation, furnishes

2⟨∇u(z + (0, b)), (x, (α + 1)y)⟩ + (Q − 2)u(z + (0, b)) ≥ 0, z ∈ ℝ^{n}, b ∈ ℝ^{k},

or equivalently

2⟨∇u(z), (x, (α + 1)(y − b))⟩ + (Q − 2)u(z) ≥ 0, z ∈ ℝ^{n}, b ∈ ℝ^{k}.

Dividing by |b| and letting |b| → ∞ with b/|b| → b_{0}, |b_{0}| = 1, we get

⟨∇_{y}u(z), b_{0}⟩ ≤ 0, for |b_{0}| = 1, z ∈ ℝ^{n}.

This implies ∇_{y}u ≡ 0 and therefore u does not depend on y and ultimately solves

Δ_{x}u(x) = −u(x)^{\frac{Q+2}{Q−2}} , x ∈ ℝ^{m}.

This implies u ≡ 0. If m ≥ 3, this is a well known result on entire subcritical semilinear equations (see, for instance, [CL]). Note that Q > m + 1 > m and thus \frac{Q+2}{Q−2} < \frac{m+3}{m−1} < \frac{m+2}{m−2}. If m = 2, the previous argument still works, only add one mute variable to ℝ^{2}. If m = 1, the function u = u(x) is positive and strictly concave on the real line and this is not possible.

This ends the proof of Claim 3 and of the theorem. □

In the classical moving spheres method, the solution of the Yamabe equation in ℝ^{n} can be determined explicitly on the whole space (see [LZhu]). Here, the argument provides the explicit form of the solution only on the set {x = 0}.

Corollary 2.8. Let u ∈ C^{2}(ℝ^{n}) be a positive solution of equation (1.1) such that u = u^{∗}. Then there exists y_{0} ∈ ℝ^{k} such that for all y ∈ ℝ^{k}

u(0, y) = u(0, y_{0})(1 + |y − y_{0}|^{2})^{\frac{2−Q}{2(α+1)}} .  (2.13)

Proof. The assumption u = u^{∗} reads

u(z) = ∥z∥^{2−Q}u (δ_{1/∥z∥^{2}}(z)) ,  (2.14)

where z = (x, y).

For a fixed b ∈ ℝ^{k}, define u^{(b)}(z) = u(z + (0, b)). Clearly, Lu^{(b)} = −(u^{(b)})^{2^{∗}−1} and hence, by Theorem 2.7, there exists λ_{b} > 0 such that u^{(b)} = (u^{(b)})_{λ_{b}}, that is,

u(z + (0, b)) = (∥z∥/λ_{b})^{2−Q}u (δ_{λ_{b}^{2}/∥z∥^{2}}(z) + (0, b)) .

Letting z_{b} = z − (0, b) for all z, this identity becomes

u(z) = (∥z_{b}∥/λ_{b})^{2−Q}u (δ_{λ_{b}^{2}/∥z_{b}∥^{2}}(z_{b}) + (0, b)) ,  (2.15)

for all z ≠ (0, b). Multiplying (2.15) by ∥z∥^{Q−2} and letting ∥z∥ → ∞, we find

\begin{aligned}u^{∗}(0) = lim_{∥z∥→∞} ∥z∥^{Q−2}u(z) & = λ_{b}^{Q−2} lim_{∥z∥→∞} (∥z_{b}∥/∥z∥)^{2−Q}u (δ_{λ_{b}^{2}/∥z_{b}∥^{2}}(z_{b}) + (0, b)) \\ & = λ_{b}^{Q−2}u(0, b), \end{aligned}

and using u(0) = u^{∗}(0) we get

λ_{b}^{Q−2} = \frac{u(0, 0)}{u(0, b)}.  (2.16)

¿From (2.14) and (2.15) we also have

∥z∥^{2−Q}u (δ_{1/∥z∥^{2}}(z)) = (∥z_{b}∥/λ_{b})^{2−Q} u (δ_{λ_{b}^{2}/∥z_{b}∥^{2}}(z_{b}) + (0, b)) .

Now let f(y) = u(0, y). Setting x = 0 in the last identity and using (2.16) we obtain

|y|^{\frac{2−Q}{α+1}} f (\frac{y}{|y|^{2}} ) = \frac{f(0)}{f(b)}|y − b|^{\frac{2−Q}{α+1}} f (\frac{λ_{b}^{2(α+1)}(y − b)}{|y − b|^{2}} + b) ,

and by a first order Taylor approximation with |y| → ∞,

\begin{aligned}|y|^{\frac{2−Q}{α+1}} & \{f(0) + ⟨∇f(0), \frac{y}{|y|^{2}} ⟩ + o (\frac{1}{|y|})\} = \\ & = \frac{f(0)}{f(b)}|y − b|^{\frac{2−Q}{α+1}} \{f(b) + λ_{b}^{2(α+1)} ⟨∇f(b), \frac{y − b}{|y − b|^{2}} ⟩ + o (\frac{1}{|y − b|})\} . \end{aligned} (2.17)

The function f has a maximum point y_{0} ∈ ℝ^{k}, because u is infinitesimal at infinity. Without loss of generality, we can assume that y_{0} = 0 and ∇f(0) = 0. Using again (2.16) and rearranging terms in (2.17), we get

f(0)^{− \frac{2(α+1)}{Q−2}} \{1 − (\frac{|y|}{|y − b|})^{\frac{Q−2}{α+1}} \} = (\frac{|y|}{|y − b|})^{\frac{Q−2}{α+1}} f(b)^{− \frac{Q+2α}{Q−2}} ⟨∇f(b), \frac{y − b}{|y − b|^{2}} ⟩ + o (\frac{1}{|y|}) .

We multiply this identity by y_{i}, i = 1, ..., k, and let y_{i} → ∞. Notice that

lim_{y_{i}→+∞} y_{i} \{1 − (\frac{|y|}{|y − b|})^{\frac{Q−2}{α+1}} \} = −\frac{Q − 2}{α + 1}b_{i},

and

lim_{y_{i}→+∞} y_{i} ⟨∇f(b), \frac{y − b}{|y − b|^{2}} ⟩ = ∂_{i}f(b).

Whence,

f(0)^{− \frac{2(α+1)}{Q−2}} ∇(1 + |b|^{2}) = −\frac{2(α + 1)}{Q − 2}f(b)^{− \frac{Q+2α}{Q−2}} ∇f(b) = ∇(f(b)^{− \frac{2(α+1)}{Q−2}} ),

where ∇ is the gradient with respect to b. This finally gives for b ∈ ℝ^{k}

f(b) = f(0)(1 + |b|^{2})^{− \frac{Q−2}{2(α+1)}} .

This is (2.13) with y_{0} = 0. □

3. Grushin and hyperbolic symmetry

In this section, we prove radial symmetry properties of solutions to equation (1.1). After a suitable functional change, such solutions become radial functions in the hyperbolic space.

Definition 3.1. Given a function u = u(x, y) with x ∈ ℝ^{m} and y ∈ ℝ^{k}, define the function U = U(ξ, η) by

U(ξ, η) = |ξ|^{β}u (|ξ|^{\frac{1}{α+1}} \frac{ξ}{|ξ|}, η) ,  β = \frac{Q − 2}{2(α + 1)}.  (3.1)

We let U = T (u) and u = T^{−1}(U).

In the case m = k = α = 1, this function change was introduced by Beckner [B] in his study of extremal functions for the Sobolev inequality in the Grushin plane.

In order to explain the functional transformation T, we choose m = 1 and we introduce the hyperbolic space. Let H = {ζ = (ξ, η) ∈ ℝ × ℝ^{k} : ξ > 0} be the n = k + 1 dimensional hyperbolic spaces with the metric ds_{H}^{2} = ξ^{−2}(dξ^{2} + |dη|^{2}). Denote by d_{H} the hyperbolic distance on H. The hyperbolic gradient is D_{H}U = ξ∇U and the natural measure is dμ = ξ^{−k−1}dξdη. The Laplace-Beltrami operator Δ_{H} is defined by the integration by parts formula

∫_{H}⟨D_{H}U, D_{H}Φ⟩dμ = − ∫_{H} ΦΔ_{H}U dμ

for test functions Φ ∈ C_{0}^{∞}(H).

We briefly recall the unit ball model for the hyperbolic space. Let B = {z = (x, y) ∈ ℝ × ℝ^{k} : |z| < 1} be the n = k + 1 dimensional unit ball endowed with the metric ds_{B}^{2} = \frac{4(dx^{2}+|dy|^{2})}{(1−|z|^{2})^{2}}. It is known that the Möbius map S : B → H defined by S(x, y) = (−1, 0) + 2\frac{(x+1,y)}{|(x+1,y)|^{2}} is a hyperbolic isometry between B and H. Moreover, S takes the geodesic spheres {(x, y) ∈ B : x^{2} + |y|^{2} = r^{2}}, r < 1, onto the spheres

\{(ξ, η) ∈ H : \frac{(1 − ξ)^{2} + |η|^{2}}{4ξ} = \frac{r^{2}}{1 − r^{2}} \}.

The critical semilinear equation (1.1) for the Grushin operator is related to a semilinear equation in the hyperbolic space.

Proposition 3.2. Let m = 1. If u is a positive solution to equation (1.1) in ℝ^{+} ×ℝ^{k}, then U = T (u) is a solution to the equation in H

Δ_{H}U + \frac{Q(Q − 2)}{4(α + 1)^{2}}U = −\frac{1}{(α + 1)^{2}} U^{\frac{Q+2}{Q−2}} ,  (3.2)

where Δ_{H} is the hyperbolic Laplacian.

Proof. The proof is an easy computation. Integrating equation (1.1) against a test function φ ∈ C_{0}^{∞}(ℝ^{+} × ℝ^{k}), we obtain

∫_{x>0} \{⟨D_{α}u, D_{α}φ⟩ − u^{\frac{Q+2}{Q−2}} φ\}dz = 0.  (3.3)

¿From (3.1), we find ∂_{x}u = (α+1)ξ^{\frac{α}{α+1}} [−βξ^{−β−1}U+ξ^{−β}∂_{ξ}U] and ∇_{y}u = ξ^{−β}∇_{η}U(ξ, η), where ξ = x^{α+1} and β = \frac{Q−2}{2(α+1)}. Analogous formulas hold for the test function Φ = T (φ).

Performing the change of variable ξ = x^{α+1} and dz = dxdy = \frac{1}{α+1}ξ^{− \frac{α}{α+1}} dξdη, we obtain

∫_{x>0}⟨D_{α}u, D_{α}φ⟩dz = (α + 1) ∫_{ξ>0} \{β^{2}UΦ − βξ∂_{ξ}[UΦ] + ξ^{2}⟨∇_{ξ,η}U, ∇_{ξ,η}Φ⟩\}\frac{dξdη}{ξ^{k+1}} .

We used the relation \frac{α}{α+1} − 2β = 1 − k. After an integration by parts of the term ξ^{−k}∂_{ξ}[UΦ] and introducing hyperbolic gradient and measure, we get

\begin{aligned}∫_{x>0}⟨D_{α}u, D_{α}φ⟩dz & = (α + 1) ∫_{H} \{ − \frac{Q(Q − 2)}{4(α + 1)^{2}}UΦ + ⟨D_{H}U, D_{H}Φ⟩\}dμ \\ & = −(α + 1) ∫_{H} \{\frac{Q(Q − 2)}{4(α + 1)^{2}}U + Δ_{H}U\}Φ dμ. \end{aligned}

An even simpler computation shows that

∫_{x>0} u^{\frac{Q+2}{Q−2}} φ dz = \frac{1}{α + 1} ∫_{H} U^{\frac{Q+2}{Q−2}} Φ dμ.

Comparing the last two formulas with (3.3), we get the proof of the proposition. □

Remark 3.3. Equation (3.2) is invariant under hyperbolic isometries. Indeed, via the functional transformation T, translations u(x, y) ↦ u(x, y + b), b ∈ ℝ^{k}, correspond to translations (hyperbolic isometries) U(ξ, η) ↦ U(ξ, η + b), and dilations u(x, y) ↦ λ^{\frac{Q}{2} −1}u(λx, λ^{α+1}y), λ > 0, correspond to the hyperbolic isometries U(ξ, η) ↦ U(λ^{α+1}ξ, λ^{α+1}η). This observation suggests how to construct the Kelvin transform u^{∗} introduced in (2.7). Consider the hyperbolic isometry of (H, d_{H}), (ξ, η) = ζ ↦ \frac{ζ}{|ζ|^{2}} and for U : H → ℝ let

U^{†}(ξ, η) = U(\frac{ξ}{ξ^{2} + |η|^{2}} , \frac{η}{ξ^{2} + |η|^{2}} ).

Then it holds T (u^{∗}) = (T (u))^{†} for any u : ℝ^{+} × ℝ^{k} → ℝ. Indeed, since T (u)(ξ, η) = U(ξ, η) = ξ^{\frac{Q−2}{2(α+1)}} u(ξ^{\frac{1}{α+1}} , η),

(T (u))^{†}(ξ, η) = U^{†}(ξ, η) = (\frac{ξ}{ξ^{2} + |η|^{2}} )^{\frac{Q−2}{2(α+1)}} u((\frac{ξ}{ξ^{2} + |η|^{2}} )^{\frac{1}{α+1}} , \frac{η}{ξ^{2} + |η|^{2}} )),

and therefore

T^{−1}(U^{†})(x, y) = x^{− \frac{Q−2}{2}} U^{†}(x^{α+1}, y)

= x^{− \frac{Q−2}{2}} (\frac{x^{α+1}}{x^{2(α+1)} + |y|^{2}} )^{\frac{Q−2}{2(α+1)}} u(\frac{x}{[x^{2(α+1)} + |y|^{2}]^{\frac{1}{α+1}}} , \frac{y}{x^{2(α+1)} + |y|^{2}} )

= ∥z∥^{2−Q}u(δ_{∥z∥^{−2}}(z)).

Then, the Kelvin transform in the Grushin space origins from a hyperbolic reflection. The construction not only produces the correct form for the inversion z ↦ δ_{1/∥z∥^{2}}(z), but it also yields the form of the singular solution Γ(z) = ∥z∥^{2−Q} for L appearing in the definition of u^{∗}.

Now we prove the main hyperbolic symmetry theorem. Let m, k ≥ 1 and for v ∈ ℝ^{m} with |v| = 1 consider the “halfspace”

H_{v} = {(x, y) ∈ ℝ^{m} × ℝ^{k} : x = tv, t > 0}.

H_{v} carries a natural structure of k + 1 dimensional hyperbolic space. We use the coordinates (t, y) on H_{v}: by abuse of notation, (t, y) ∈ H_{v} stands for (tv, y) ∈ H_{v}. The metric here is ds^{2} = t^{−2}(dt^{2} + |dy|^{2}).

For any v ∈ ℝ^{m} with |v| = 1 and for any r ∈ (0, 1) define the k–dimensional sphere

Σ(v, r) = \{(t, y) ∈ H_{v} : \frac{(1 − t)^{2} + |y|^{2}}{4t} = \frac{r^{2}}{1 − r^{2}} \} .  (3.4)

Let B = {z = (x, y) ∈ ℝ × ℝ^{k} : |z| < 1} be the unit ball endowed with the hyperbolic metric. The map S_{v} : B → H_{v}, defined by

S_{v}(z) = \frac{1 − x^{2} − |y|^{2}}{(1 + x)^{2} + |y|^{2}} (v, 0) + \frac{2}{(1 + x)^{2} + |y|^{2}} (0, y)  (3.5)

is an isometry and it transforms the spheres {z ∈ ℝ^{k+1} : |z| = r}, r ∈ (0, 1), into the spheres (3.4). Thus, Σ(v, r) is a hyperbolic sphere in H_{v} centered at t = 1 and y = 0.

Introduce the class of functions u : ℝ^{n} → ℝ

S = {u : U = T (u) is constant on each Σ(v, r), |v| = 1 and r ∈ (0, 1)} .

Theorem 3.4. Let m, k ≥ 1 and n = m + k. If u ∈ C^{2}(ℝ^{n}) is a positive solution to equation (1.1) with u = u^{∗} and y_{0} = 0 in (2.13), then u ∈ S.

Proof. Let z = (x, y) ∈ ℝ^{n}, b ∈ ℝ^{k}, z_{b} = z −(0, b). By (2.15) in the proof of Corollary 2.8, there exists λ_{b} > 0 such that

\begin{aligned}u(z) & = (∥z_{b}∥/λ_{b})^{2−Q}u (δ_{λ_{b}^{2}/∥z_{b}∥^{2}}(z_{b}) + (0, b)) \\ & = (\frac{∥(x, y − b)∥}{λ_{b}})^{2−Q}u (\frac{λ_{b}^{2}x}{∥(x, y − b)∥^{2}} , \frac{λ_{b}^{2(α+1)}y}{∥(x, y − b)∥^{2(α+1)}} + b) . \end{aligned}  (3.6)

Moreover, by (2.16), λ_{b} is determined by u(0)λ_{b}^{2−Q} = u(0, b), and this, by (2.13) with y_{0} = 0, gives

λ_{b} = (1 + |b|^{2})^{\frac{1}{2(α+1)}} .

Let ζ = (ξ, η), ζ_{b} = ζ − (0, b) and |ζ_{b}| = (|ξ|^{2} + |η − b|^{2})^{1/2}. By definition (3.1) and (3.6), we have

U(ζ) = |ξ|^{\frac{Q−2}{2(α+1)}} (\frac{|ζ_{b}|^{\frac{1}{α+1}}}{λ_{b}} )^{2−Q} u (\frac{λ_{b}^{2}|ξ|^{\frac{1}{α+1}}}{|ζ_{b}|^{\frac{2}{α+1}}} \frac{ξ}{|ξ|}, \frac{λ_{b}^{2(α+1)}}{|ζ_{b}|^{2}} (η − b) + b) ,

and using u(x, y) = |x|^{1− \frac{Q}{2}} U(|x|^{α}x, y), we finally get

U(ζ) = U (\frac{(1 + |b|^{2})ξ}{|ζ_{b}|^{2}} , \frac{(1 + |b|^{2})(η − b)}{|ζ_{b}|^{2}} + b) , b ∈ ℝ^{k}.  (3.7)

In order to prove the theorem it suffices to choose m = 1 and consider the case ξ > 0. Let H and B be the hyperbolic halfspace and ball, respectively. The map I_{b} : ℝ^{n} \ {(0, b)} → ℝ^{n} \ {(0, b)} given by

I_{b}(ξ, η) = (\frac{(1 + |b|^{2})ξ}{|ζ_{b}|^{2}} , \frac{(1 + |b|^{2})(η − b)}{|ζ_{b}|^{2}} + b)

is a spherical inversion with respect to the sphere

Σ_{b} = {(ξ, η) ∈ ℝ × ℝ^{k} : ξ^{2} + |η − b|^{2} = 1 + |b|^{2}}.

Clearly, (1, 0) ∈ Σ_{b} for any b ∈ ℝ^{k}. Let Σ_{b}^{+} = Σ_{b} ∩ {ξ > 0}. Since the choice of b is arbitrary, the function U is symmetric with respect to any reflection fixing the point (1, 0). This clearly means that U is radial.

To realize more concretely this fact, consider the map S : B → H defined in (3.5) with v ∈ ℝ, v = 1. S takes the “plane” π_{b} = {(x, y) ∈ B : x + ⟨b, y⟩ = 0} onto the halfsphere Σ_{b}^{+}, and S(0) = (1, 0). By Theorem 4.3.7 in [R], the points S(ζ) and S(I_{b}(ζ)) in B are symmetric with respect to the plane π_{b} for any ζ ∈ H. Therefore, by (3.7), the function U_{B} : B → ℝ defined by U_{B}(x, y) = U(S(x, y)) is symmetric with respect to the plane π_{b}. Since b ∈ ℝ^{k} is arbitrary, the function U_{B} is radial about the origin. Now, the claim follows from the fact that S transforms the spheres {(x, y) ∈ B : x^{2} + |y|^{2} = r^{2}}, r ∈ (0, 1), into the spheres (3.4). □

Let again m = 1 and n = k + 1. Theorem 3.4 has the following corollary, which is actually a hyperbolic symmetry result. Consider the class of functions

A = \{U ∈ C^{2}(H) : u(x, y) = |x|^{\frac{2−Q}{2}} U(|x|^{α+1}, y) ∈ C^{2}(ℝ^{n})\} .

Corollary 3.5. If U ∈ A is a positive solution to equation (3.2), then U is a radial function about some point in H for the hyperbolic metric.

The condition u ∈ C^{2}(ℝ^{n}) prescribes a suitable vanishing behavior of U on ∂H (i.e. at infinity). It is not clear whether this condition is precise. However, any attempt to prove directly Corollary 3.5 without requiring any similar condition must face the difficult task to apply a Hopf Lemma at boundary points which could be in principle at infinity. In our case, this tool is provided by Lemma 2.6.

The following theorem shows how to reduce equation (1.1) to a lower dimensional one.

Theorem 3.6. Let m, k ≥ 1 and n = m + k. If u ∈ C^{2}(ℝ^{n}) is a positive solution to equation (1.1) with u = u^{∗} and y_{0} = 0 in (2.13), then the function v(x) = u(x, 0), x ∈ ℝ^{m}, is a solution of the problem

\{ \begin{matrix}div_{x}(p∇_{x}v) − qv = −pv^{2^{∗}−1} & |x| < 1 \\ v > 0 & |x| ≤ 1 \\ \frac{∂v}{∂ν} + (\frac{Q}{2} − 1)v = 0 & |x| = 1, \end{matrix}  (3.8)

where p(x) = (1 − |x|^{2(α+1)})^{k} and q(x) = k(α + 1)(Q − 2)(1 − |x|^{2(α+1)})^{k−1}|x|^{2α}.

Proof. Let x ∈ ℝ^{m} be a point such that 0 < |x| < 1. By Theorem 3.4, the function y ↦ u(x, y) is radial, and therefore ∇_{y}u(x, 0) = 0. Then, for any i = 1, ..., k

\frac{∂^{2}u}{∂y_{i}^{2}}(x, 0) = lim_{ε→0} \frac{2}{ε^{2}} (u(x, εe_{i}) − u(x, 0)),

where e_{i} = (0, ..., 1, ..., 0) ∈ ℝ^{k} with 1 in the i–th coordinate.

Let U = T (u) and let ξ = |x|^{α}x. By Theorem 3.4, for any ε > 0 there is a unique point ξ_{ε} ∈ ℝ^{m} of the form ξ_{ε} = tξ with t ∈ (0, 1) and such that U(ξ, εe_{i}) = U(ξ_{ε}, 0). By (3.4), ξ_{ε} is determined by the condition

\frac{(1 − |ξ_{ε}|)^{2}}{|ξ_{ε}|} = \frac{(1 − |ξ|)^{2} + ε^{2}}{|ξ|} ,

which gives

|ξ_{ε}| = \frac{1}{2|ξ|}(1 + |ξ|^{2} + ε^{2} − \sqrt{(1 + |ξ|^{2} + ε^{2})^{2} − 4|ξ|^{2}}).

Letting φ(ε) = |ξ\sqrt{ε}|, we get φ(0) = |ξ| and φ′(0) = |ξ|/(|ξ|^{2} − 1). From (3.1) we find

u(x, εe_{i}) = \frac{1}{|ξ|^{β}} U(ξ, εe_{i}) = \frac{1}{|ξ|^{β}} U(ξ_{ε}, 0) = (\frac{|ξ_{ε}|}{|ξ|})^{β} u(|ξ_{ε}|^{\frac{1}{α+1}} \frac{x}{|x|}, 0).

Therefore,

\frac{∂^{2}u}{∂y_{i}^{2}}(x, 0) = lim_{ε→0} \frac{2}{ε} (\frac{φ(ε)^{β}}{|ξ|^{β}} u(φ(ε)^{\frac{1}{α+1}} \frac{x}{|x|}, 0) − u(x, 0))
= \frac{2}{|ξ|^{β}} \frac{d}{dε} (φ(ε)^{β}u(φ(ε)^{\frac{1}{α+1}} \frac{x}{|x|}, 0))|_{ε=0}

= −\frac{1}{(α + 1)(1 − |x|^{2(α+1)})}((Q − 2)u(x, 0) + 2⟨∇_{x}u(x, 0), x⟩).

The left hand side is a continuous function on |x| ≤ 1, and thus it must be

(Q − 2)u(x, 0) + 2⟨∇_{x}u(x, 0), x⟩ = 0, for |x| = 1.

Moreover,

Lu(x, 0) = Δ_{x}u(x, 0) − \frac{k(α + 1)|x|^{2α}}{1 − |x|^{2(α+1)}} ((Q − 2)u(x, 0) + 2⟨∇_{x}u(x, 0), x⟩).

Multiplying the equation Lu(x, 0) = −u(x, 0)^{2^{∗}−1} by p(x) = (1 − |x|^{2(α+1)})^{k} and letting q(x) = k(α + 1)(Q − 2)(1 − |x|^{2(α+1)})^{k−1}|x|^{2α}, we finally get

div_{x}(p(x)∇_{x}u(x, 0)) − q(x)u(x, 0) = −p(x)u(x, 0)^{2^{∗}−1}.

The proof is concluded. □

4. Uniqueness of solutions

In this section we prove the uniqueness Theorems 5 and 7. By Theorem 3.6, after a scaling and a vertical translation, the problem of uniqueness for (1.1) is reduced to the problem of uniqueness of solutions to (1.8). In the case m = k = 1, the function u = u(x) with x ∈ (−1, 1) solves problem (1.9). In case m > 1, for a radial function u = u(|x|) = u(r), r ∈ (0, 1), the partial differential equation in (1.8) becomes an ordinary equation. Adopting the notation

\begin{matrix}⟨r⟩ = 1 − |r|^{2(α+1)}, s(r) = r^{m−1}, \\ p(r) = ⟨r⟩^{k}, q(r) = k(α + 1)(Q − 2)⟨r⟩^{k−1}r^{2α}, \end{matrix}  (4.1)

this ordinary equation takes the form

(psu′)′ − qsu + psu^{2^{∗}−1} = 0,  r ∈ (0, 1).  (4.2)

In order to study our uniqueness problem, we discuss a functional change suitable for the “energy method” introduced by Kwong and Li in [KL]. The discussion of the functional change is presented for the radial case m > 1, but with straightforward adaptations it also works for m = 1, where no radiality is assumed (but the relevant interval is (−1, 1) instead of (0, 1)).

Consider the following auxiliary functions

h(r) = (s(r)p(r))^{ϑ},  ϑ = \frac{1}{2} \frac{Q − 2}{Q − 1},  (4.3)

and

G(r) = h(r)^{2^{∗}−2}(c_{1}r^{2α}⟨r⟩^{−2} + c_{2}r^{−2}),  (4.4)

where c_{1} and c_{2} are constants depending on m, k, α, and precisely

\begin{aligned}c_{1} & = 2ϑk(α + 1)\{\frac{Q(m − 1)}{Q − 1} − Q + 2(α + 1)\}, \\ c_{2} & = ϑ(m − 1)\{1 − \frac{(m − 1)Q}{2(Q − 1)}\}. \end{aligned}  (4.5)

Associate to the function u the function z by letting

z(r) = h(r)u(r),  (4.6)

and introduce the energy

E(z(r)) = h(r)^{2^{∗}−2}z′(r)^{2} + \frac{2}{2^{∗}} z(r)^{2^{∗}} + G(r)z(r)^{2}.  (4.7)

The reason for introducing z, G and E is described by the following Proposition 4.1. If the function u solves equation (4.2), then the function z = hu solves the equation

\frac{d}{dr}E(z(r)) = G′(r)z(r)^{2},  r ∈ (0, 1).  (4.8)

Proof. The function z satisfies the equation

h^{\frac{1}{ϑ} −1}z′′ + (\frac{1}{ϑ} − 2)h^{\frac{1}{ϑ} −2}h′z′ + F z + h^{\frac{1}{ϑ} +1−2^{∗}}z^{2^{∗}−1} = 0,

where

F = (2 − \frac{1}{ϑ})h^{\frac{1}{ϑ} −3}h′^{2} − h^{\frac{1}{ϑ} −2}h′′ − qsh^{−1}.

Multiplying the equation by 2h^{2^{∗}−1− \frac{1}{ϑ}} z′ we get

(h^{2^{∗}−2}(z′)^{2})′ + (z^{2})′G + \frac{2}{2^{∗}} (z^{2^{∗}})′ = 0,

where

G = h^{2^{∗}−2}[(2 − \frac{1}{ϑ})(\frac{h′}{h} )^{2} − \frac{h′′}{h} − \frac{qs}{h^{1/ϑ}} ].

Using (4.3), it can be checked by a rather long computation (we omit the details here) that G is the function in (4.4) with c_{1} and c_{2} as in (4.5). □

Equation (4.8) will prove to be useful in comparing solutions of equation (4.2). Now we are going to discuss the sign of the function G′ in the case k = 1.

If m = k = 1, we have Q = α + 2 and 2^{∗} = \frac{2Q}{Q−2} = 2(α + 2)/α. Problem (1.8) becomes problem (1.9) with

p(x) = (1 − |x|^{2(α+1)}) and q(x) = α(α + 1)|x|^{2α},  |x| < 1.  (4.9)

Letting m = 1 and k = 1 in (4.4)–(4.6) we find

\begin{matrix}z(x) = p(x)^{\frac{α}{2(α+1)}} u(x),  G(x) = \frac{α^{2}|x|^{2α}}{(1 − |x|^{2(α+1)})^{\frac{2α}{α+1}}} , \\ E(z(x)) = p(x)^{\frac{2}{α+1}} z′(x)^{2} + \frac{2}{2^{∗}} z(x)^{2^{∗}} + G(x)z(x)^{2}, |x| < 1. \end{matrix}  (4.10)

Clearly, G′ > 0 on (0, 1), because G is increasing.

In the case m > 1 and k = 1, (4.4) becomes

G(r) = c_{1}r^{2α+ \frac{2(m−1)}{Q−1}} ⟨r⟩^{\frac{2}{Q−1} −2} + c_{2}r^{\frac{2(m−1)}{Q−1} −2}⟨r⟩^{\frac{2}{Q−1}} = G_{1}(r) + G_{2}(r),

where c_{1} and c_{2} are as in (4.5) with k = 1. If m ≥ 3 then both G_{1} and G_{2} are increasing. Indeed, for all m ≥ 1 and k = 1, the constant c_{1} and 2 − \frac{2}{Q−1} are both positive. Therefore the function G_{1} is increasing with G_{1}(0) = 0, and G_{1}(1) = +∞. The function G_{2} is identically zero if m = 1, because in this case c_{2} = 0. The exponent \frac{2(m−1)}{Q−1} − 2 is negative for all m ≥ 1. The constant c_{2} is positive as soon as m ≥ 3. Then, if m ≥ 3 the function G_{2} is increasing on (0, 1) and G_{2}(0) = −∞, G_{2}(1) = 0.

Thus, for m ≥ 3 and k = 1, G is increasing on (0, 1), G(0) = −∞ and G(1) = +∞. The singularities at 0 and 1 will be carefully examined.

Remark 4.2. For m = 2 and k = 1 we have c_{2} < 0. The function G fails to enjoy the properties needed in the uniqueness argument.

We need the following Lemma.

Lemma 4.3. Let u, v ∈ C^{2}(0, 1) ∩ C^{1}([0, 1]) be positive solutions of

\{ \begin{matrix}(psu′)′ − qsu + psu^{2^{∗}−1} = 0 \\ u′(1) + \frac{Q − 2}{2}u(1) = 0, \end{matrix}  (4.11)

where p, q, s are as in (4.1) and k = 1. Letting η = u(1)/v(1), we have u(x)−ηv(x) = O(1 − x)^{2} and u′(x) − ηv′(x) = O(1 − x) as x → 1.

Proof. The Wronskian w = uv′ − vu′ satisfies (psw)′ = (vu^{2^{∗}−1} − uv^{2^{∗}−1})ps. Since (psw)(1) = 0 we have

w(x) = \frac{1}{p(x)s(x)} ∫_{x}^{1}(v^{2^{∗}−2} − u^{2^{∗}−2})psuv dt.  (4.12)

The functions u and v are in C^{1}([0, 1]) and v > 0 on [0, 1]. Then u/v ∈ C^{1}([0, 1]) and

(\frac{u}{v})′ = \frac{u′v − uv′}{v^{2}} = −\frac{w}{v^{2}} .

By the mean value theorem, there exists ξ ∈ (x, 1) such that

η − \frac{u(x)}{v(x)} = \frac{u(1)}{v(1)} − \frac{u(x)}{v(x)} = (\frac{u}{v})′(ξ)(1 − x) = −\frac{w(ξ)}{v(ξ)^{2}} (1 − x),

Now, since p(x) = O(1 − x), as x → 1, (4.12) and de l’Hôpital rule give w(ξ) = (1 − ξ)(γ_{1} + o(1)) for ξ → 1 and for some constant γ_{1} ∈ ℝ. Then u(x) − ηv(x) = (1 − x)^{2}(γ_{2} + o(1)) for x → 1 and for some new constant γ_{2} ∈ ℝ.

Integrating (4.11) for u and v on (x, 1) yields

u′(x) − ηv′(x) = \frac{1}{(ps)(x)} ∫_{x}^{1} [ − (u − ηv)qs + (u^{2^{∗}−1} − ηv^{2^{∗}−1})ps]dt.

Using s(x) → 1, \frac{p(x)}{1−x} → 2α + 2 and u(x) − ηv(x) = O(1 − x)^{2} as x → 1, de l’Hôspital gives

lim_{x→1} \frac{u′(x) − ηv′(x)}{1 − x} = \frac{1}{2}(u(1)^{2^{∗}−1} − ηv(1)^{2^{∗}−1}).

This finishes the proof of the lemma. □

Now we show that for m = k = 1 all solutions of (1.9) are even functions and satisfy u′(0) = 0. In the higher dimensional case, this is a consequence of the radiality assumption.

Theorem 4.4. If u ∈ C^{2}(−1, 1) ∩ C^{1}([−1, 1]) solves problem (1.9) then u′(0) = 0.

Proof. Assume by contradiction that u′(0) < 0. The function v(x) = u(−x) is a new solution to problem (1.9), because p and q in (4.9) are even functions. Clearly, v(0) = u(0) and v′(0) > 0.

We claim that u(x) < v(x) for all x ∈ (0, 1]. Indeed. Letting

r(x) = \frac{u′(x)}{u(x)} and R(x) = \frac{v′(x)}{v(x)},  (4.13)

we have r(0) < R(0). Assume by contradiction that there exists a point ξ ∈ (0, 1] such that u(ξ) = v(ξ), and let ξ be the smallest one. It cannot be ξ = 1 because the Cauchy problem with data at 1 has unique solution (this fact, although not completely trivial because of the singular terms, can be proved by the contraction principle). It must be u′(ξ) > v′(ξ) because, if u′(ξ) = v′(ξ), then u ≡ v, by uniqueness for the Cauchy problem with data at ξ. Therefore r(ξ) > R(ξ). Then, by continuity, there exists a point b ∈ (0, ξ) such that r(b) = R(b). Let b be the smallest one. Then r(x) < R(x) for x ∈ (0, b), that is u′/u < v′/v on the same interval. This condition is equivalent to (u/v)′ < 0 on (0, b), and thus the function u/v is strictly decreasing on this interval. Let

ω = \frac{u(b)}{v(b)}.  (4.14)

Since u(0) = v(0), we have ω ∈ (0, 1).

According to (4.10), define the functions z = p^{\frac{α}{2(α+1)}} u and ζ = p^{\frac{α}{2(α+1)}} v. Then ω = \frac{z(b)}{ζ(b)}. Moreover, since r(b) = R(b), we also have \frac{u′(b)}{v′(b)} = ω. Thus

\frac{z′(b)}{ζ′(b)} = \frac{(p(b)^{\frac{α}{2(α+1)}} )′ u(b) + p(b)^{\frac{α}{2(α+1)}} u′(b)}{(p(b)^{\frac{α}{2(α+1)}} )′ v(b) + p(b)^{\frac{α}{2(α+1)}} v′(b)} = ω.  (4.15)

Let E(z) be the energy associated with z as in (4.10). Integrating (4.8) over (0, b) and using G(0) = 0, p(0) = 1 and p′(0) = 0 (this ensures ζ(0) = u(0) and z′(0) = u′(0)), we get

p(b)^{\frac{2}{α+1}} (z′(b))^{2} + \frac{2}{2^{∗}} z(b)^{2^{∗}} + G(b)z(b)^{2} = u′(0)^{2} + \frac{2}{2^{∗}} u(0)^{2^{∗}} + ∫_{0}^{b} G′(x)z(x)^{2}dx. (4.16)

Recall now that u(0) = v(0) and u′(0)^{2} = v′(0)^{2} by definition of v. Subtract from (4.16) the same identity for ζ multiplied by ω^{2}, obtaining

\frac{2}{2^{∗}} (1 − ω^{2−2^{∗}})z(b)^{2^{∗}} = (1 − ω^{2})(u′(0)^{2} + \frac{2}{2^{∗}} u(0)^{2^{∗}}) + ∫_{0}^{b} G′(x)(z(x)^{2} − ω^{2}ζ(x)^{2})dx.

This is a contradiction. Indeed, the right hand side is strictly positive because G′ > 0 on (0, 1), z^{2} − ω^{2}ζ^{2} > 0 on (0, b) and ω ∈ (0, 1). On the other hand, 2^{∗} > 2 and therefore the left hand side is strictly negative. We have proved that u < v on (0, 1].

Next we show that r < R on the entire interval (0, 1). Let w = uv′ − vu′ and note that R − r = \frac{w}{uv}. We have to show that w > 0 on (0, 1). Since u and v satisfy (1.9), we have (pw)′ = (u^{2^{∗}−2} − v^{2^{∗}−2})puv. Integrating over (x, 1) and using p(1) = 0, we get for x ∈ (0, 1)

w(x) = \frac{1}{p(x)} ∫_{x}^{1}(v^{2^{∗}−2} − u^{2^{∗}−2})puv dt > 0,

and hence r < R on (0, 1).

Then, the function u/v is strictly decreasing on (0, 1). Let

η = \frac{u(1)}{v(1)} = \frac{u′(1)}{v′(1)}.

The last equality follows from the boundary conditions. By (4.10) and (4.8), we have for any x ∈ (0, 1)

E(z(x)) − η^{2}E(ζ(x)) = E(z(0)) − η^{2}E(ζ(0)) + ∫_{0}^{x} G′(t)(z(t)^{2} − η^{2}ζ(t)^{2})dt. (4.17)

We are going to let x → 1 in this identity. There is some singular term to estimate. We first show that the left hand side tends to zero. Recalling (4.10), compute

lim_{x→1} G(x)(z(x)^{2} − η^{2}ζ(x)^{2}) = α^{2} lim_{x→1}(1 − x^{2(α+1)})^{− \frac{α}{α+1}} (u(x)^{2} − η^{2}v(x)^{2}) = 0,

because p(x) = O(1 − x), u(x) − ηv(x) = O(1 − x)^{2} by Lemma 4.3, and 2 − \frac{α}{α+1} > 0. Note also that z(1) = ζ(1) = 0, which gives \frac{2}{2^{∗}} (z(1)^{2^{∗}} − η^{2}ζ(1)^{2^{∗}}) = 0. Moreover, it is also

lim_{x→1} p(x)^{\frac{2}{α+1}} (z′(x)^{2} − η^{2}ζ′(x)^{2}) = 0.  (4.18)

In order to prove (4.18), write z′ = h′u + hu′, where h(x) = p(x)^{\frac{α}{2(α+1)}} , so that h = O(1−x)^{\frac{α}{2(α+1)}} and h′ = O(1−x)^{− \frac{α+2}{2(α+1)}} , as x → 1. The same computation can be done for ζ. Inserting these estimates in the left hand side of (4.18) along with Lemma 4.3 proves the claim. Thus we have shown that lim_{x→1} E(z(x)) − η^{2}E(ζ(x)) = 0.

Now we analyze the right hand side in (4.17). Write again z′ = (p^{\frac{α}{2(α+1)}} )′u+p^{\frac{α}{2(α+1)}} u′. We have z′(0) = u′(0) and ζ′(0) = v′(0) = −u′(0), by v(x) = u(−x). The energies for z and ζ are equal at 0: E(z(0)) = E(ζ(0)) = \frac{2}{2^{∗}} u(0)^{2^{∗}} + u′(0)^{2}. Thus, letting x → 1 in (4.17) yields

0 = (1 − η^{2})(u′(0)^{2} + \frac{2}{2^{∗}} u(0)^{2^{∗}}) + ∫_{0}^{1} G′(x)(z(x)^{2} − η^{2}ζ(x)^{2})dx.  (4.19)

This is a contradiction, because the right hand side is strictly positive. Indeed, η ∈ (0, 1) and z/ζ = u/v > η on (0, 1). The proof is concluded. □

Thanks to Theorem 4.4, the proof of Theorem 5, or equivalently the uniqueness for problem (1.9), is reduced to the uniqueness for the following problem

\{ \begin{matrix}(pu′)′ − qu + pu^{2^{∗}−1} = 0 & in (0, 1) \\ u > 0 & in [0, 1] \\ u′(0) = 0 \\ αu(1) + 2u′(1) = 0. \end{matrix}  (4.20)

Thus the proof of both Theorems 5 and 7 is a consequence of the following Theorem 4.5. Let m ≠ 2 and k = 1. There exists a unique solution u ∈ C^{2}(0, 1) ∩ C^{1}([0, 1]) to the problem

\{ \begin{matrix}(psu′)′ − qsu + psu^{2^{∗}−1} = 0 & in (0, 1) \\ u > 0 & in [0, 1] \\ u′(0) = 0 \\ u′(1) + (\frac{Q}{2} − 1)u(1) = 0. \end{matrix}  (4.21)

Proof. Assume that there exist two different solutions u, v of Problem (4.21). By the uniqueness for the Cauchy problem for our equation with data at 0 or at 1, we can assume that u(0) ≠ v(0) and u(1) ≠ v(1). We will show that this assumption gives a contradiction. The argument of the proof is based on a Sturm-type comparison of u and v.

First we prove that u and v intersect at least once in (0, 1). Assume by contradiction that the solutions u and v satisfy u < v on (0, 1). Here we use only superlinearity. The Wronskian w = uv′ − vu′ satisfies the equation

(psw)′ = (u^{2^{∗}−2} − v^{2^{∗}−2})psuv.  (4.22)

Integrating (4.22) over (0, 1) and using (psw)(0) = (psw)(1) = 0 we get ∫_{0}^{1}(v^{2^{∗}−2} − u^{2^{∗}−2})psuv dx = 0. But this is not possible, because u < v on (0, 1).

Next, we prove that the solutions u and v must intersect at least twice in (0, 1). This is the most delicate part of the proof and it involves the fact that the function G in increasing. Assume by contradiction that u and v intersect only once in (0, 1). For example, assume that u(b) = v(b), u < v on (0, b) and u > v on (b, 1) for some b ∈ (0, 1). The function u/v is strictly increasing on (0, 1). Indeed, since (u/v)′ = −w/v^{2}, in order to prove this statement, we show that w(x) < 0 for all x ∈ (0, 1). Take x ∈ (0, 1). If x ≤ b, then integrating (4.22) yields

(psw)(x) − (psw)(0) = (psw)(x) = ∫_{0}^{x}(u^{2^{∗}−2} − v^{2^{∗}−2})psuv dt < 0,

and thus w(x) < 0 because u < v, on (0, b). If x > b, integrate (4.22) on (x, 1) and use the fact that u > v on (b, 1). We have proved that u/v is increasing.

Now let η = \frac{u(1)}{v(1)} and recall that G is increasing on (0, 1). Let z = s^{ϑ}p^{ϑ}u and ζ = s^{ϑ}p^{ϑ}v, where ϑ = \frac{Q−2}{2(Q−1)}, as in (4.3) and (4.6). Fix 0 < ε < σ < 1 and integrate (4.8) on (ε, σ) to get

E(z(σ)) − η^{2}E(ζ(σ)) = E(z(ε)) − η^{2}E(ζ(ε)) + ∫_{ε}^{σ} G′(t)[z(t)^{2} − η^{2}ζ(t)^{2}]dt. (4.23)

We now prove that: Claim 1. lim_{ε→0} E(z(ε)) − η^{2}E(ζ(ε)) ≤ 0.

Claim 2. lim_{σ→1} E(z(σ)) − η^{2}E(ζ(σ)) = 0.

As soon as the claims are proved, letting ε → 0 and σ → 1 in (4.23) gives a contradiction and we may conclude, as desired, that u and v intersect at least twice.

Indeed, the integral in the right hand side of (4.23) is negative because z(x)−ηζ(x) < 0 and G′(x) > 0 for all x ∈ (0, 1).

Proof of Claim 1. Write explicitly (4.7) for k = 1. A short computation gives

E(z(x)) = x^{\frac{2(m−1)}{Q−1}} p(x)^{\frac{2}{Q−1}} (z′(x))^{2} + \frac{2}{2^{∗}} z(x)^{2^{∗}} + G(x)z(x)^{2}.  (4.24)

Now, by (4.4) for k = 1, we have

G(x) = p(x)^{\frac{2}{Q−1}} x^{\frac{2(m−1)}{Q−1}} (c_{1}x^{2α}p(x)^{−2} + c_{2}x^{−2}),  (4.25)

with c_{1} and c_{2} as in (4.5). Therefore, since c_{2} = 0 if m = 1, we have, as ε → 0

G(ε) = \begin{cases}O(ε^{2α}) & if m = 1, \\ O(ε^{\frac{2(m−1)}{Q−1} −2}) & if m ≥ 2. \end{cases}

Moreover, since z(ε) = u(ε)ε^{\frac{(m−1)(Q−2)}{2(Q−1)}} p(ε)^{\frac{Q−2}{2(Q−1)}} , we have z′(ε) = k_{1}(ε)u(ε)+k_{2}(ε)u′(ε), for functions k_{1} and k_{2} such that for ε → 0

k_{1} = \begin{cases}O(ε^{2α+1}) & if m = 1 \\ O(ε^{\frac{(m−1)(Q−2)}{2(Q−1)} −1}) & if m ≥ 2 \end{cases} and k_{2} = O(ε^{\frac{(m−1)(Q−2)}{2(Q−1)}} ).

We used p′(ε) = O(ε^{2α+1}).

We prove Claim 1 for m ≥ 3. In this case, inserting into (4.24) the asymptotic behavior of the terms appearing in it,

E(z(ε)) = ε^{\frac{2(m−1)}{Q−1}} (O(ε^{\frac{(m−1)(Q−2)}{2(Q−1)} −1})u(ε) + O(ε^{\frac{(m−1)(Q−2)}{2(Q−1)}} )u′(ε))^{2}

+ \frac{2}{2^{∗}} O(ε^{\frac{(m−1)(Q−2)}{2(Q−1)} 2^{∗}})u(ε)^{2^{∗}} + O(ε^{\frac{2(m−1)}{Q−1} −2})O(ε^{\frac{(m−1)(Q−2)}{Q−1}} )u(ε)^{2}

= ε^{\frac{Q(m−1)}{Q−1} −2}O(1) → 0

as ε → 0, because \frac{Q(m−1)}{Q−1} − 2 > 0 if m ≥ 3. The same holds for E(ζ(ε)). We prove Claim 1 for m = 1. In this case

\begin{aligned}E(z(ε)) & = (O(ε^{1+2α})u(ε) + O(1)u′(ε))^{2} \\ & + \frac{2}{2^{∗}} p(ε)^{\frac{2^{∗}(Q−2)}{2(Q−1)}} u(ε)^{2^{∗}} + O(ε^{2α})u(ε)^{2} → \frac{2}{2^{∗}} u(0)^{2^{∗}}, \end{aligned}

as ε → 0 (recall that u′(0) = 0). Therefore, if m = 1 we have

lim_{ε→0} E(z(ε)) − η^{2}E(ζ(ε)) = \frac{2}{2^{∗}} [u(0)^{2^{∗}} − η^{2}v(0)^{2^{∗}}] ≤ 0,

because u(0) ≤ v(0) and then \frac{u(0)^{2^{∗}}}{v(0)^{2^{∗}}} ≤ (\frac{u(0)}{v(0)})^{2} ≤ (\frac{u(1)}{v(1)})^{2} = η^{2}.

Proof of Claim 2. By (4.24),

\begin{aligned}E(z(x)) − η^{2}E(ζ(x)) & = x^{\frac{2(m−1)}{Q−1}} p(x)^{\frac{2}{Q−1}} [z′(x)^{2} − η^{2}ζ′(x)^{2}] + \frac{2}{2^{∗}} [z(x)^{2^{∗}} − η^{2}ζ(x)^{2^{∗}}] \\ & + G(x)[z(x)^{2} − η^{2}ζ(x)^{2}] := A_{1} + A_{2} + A_{3}. \end{aligned}

We first show that A_{3} → 0 as x → 1. By (4.25), G(x) = O(1)p(x)^{\frac{2}{Q−1} −2}, as x → 1. Thus

\begin{aligned}G(x)[z(x)^{2} − η^{2}ζ(x)^{2}] & = O(1 − x)^{\frac{2}{Q−1} −2}[z(x) − ηζ(x)][z(x) + ηζ(x)] \\ & = O(1 − x)^{− \frac{Q−2}{Q−1}} [u(x) − ηv(x)][u(x) + ηv(x)], \end{aligned}

because z(x) = x^{\frac{(m−1)(Q−2)}{2(Q−1)}} p(x)^{\frac{Q−2}{2(Q−1)}} u(x) = O(1 − x)^{\frac{Q−2}{2(Q−1)}} u(x) and the same for ζ. Then the last line tends to zero as x → 1, because u(x) − ηv(x) = O(1 − x)^{2} by Lemma 4.3 and 2 − \frac{Q−2}{Q−1} > 0.

The proof that A_{2} tends to zero is trivial, because both z(x) and ζ(x) tend to zero. In order to show that lim_{x→1−} A_{1} = 0, a more careful analysis is needed. Indeed

we have, as x → 1,

A_{1} = O(1 − x)^{\frac{2}{Q−1}} [z′(x) − ηζ′(x)][z′(x) + ηζ′(x)].

Since z(x) = x^{\frac{(m−1)(Q−2)}{2(Q−1)}} p(x)^{\frac{(Q−2)}{2(Q−1)}} u(x), we have z′(x) = k_{1}(x)u(x) + k_{2}(x)u′(x) with k_{1}(x) = O(1 − x)^{\frac{Q−2}{2(Q−1)} −1} and k_{2}(x) = O(1 − x)^{\frac{Q−2}{2(Q−1)}} , as x → 1. Therefore

z′(x) − ηζ′(x) = O(1 − x)^{\frac{Q−2}{2(Q−1)} −1}[u(x) − ηv(x)] + O(1 − x)^{\frac{Q−2}{2(Q−1)}} [u′(x) − ηv′(x)]

and z′(x) + ηζ′(x) = O(1 − x)^{\frac{Q−2}{2(Q−1)} −1}. Hence,

A_{1} = O(1 − x)^{\frac{Q}{Q−1} −2}[u(x) − ηv(x)] + O(1 − x)^{\frac{Q}{Q−1} −1}[u′(x) − ηv′(x)].

By Lemma 4.3, u(x) − ηv(x) = O(1 − x)^{2} and u′(x) − ηv′(x) = O(1 − x) as x → 1. This finishes the proof of Claim 2.

Thus far, we have proved that two solutions of problem (4.21) must intersect at least twice in (0, 1). Now an essentially similar argument provides the conclusion of the proof of the theorem. Indeed, assume by contradiction there exist two solutions u and v such that v > u on (0, a) and v < u on (a, b) and u(b) = v(b) for some 0 < a < b < 1. Integrating equation (4.22) we get

(psw)(x) = ∫_{0}^{x}(u^{2^{∗}−2} − v^{2^{∗}−2})psuv dt.

Since v > u on (0, a), then w(x) < 0 for x ∈ (0, a]. By continuity, w < 0 on (0, a + δ) for some positive δ. Since (\frac{v}{u})′ = \frac{w}{v^{2}} , the function \frac{v}{u} is strictly decreasing on (0, a + δ]. Moreover, since \frac{v}{u}(a) = \frac{v}{u}(b) = 1, there exists ξ ∈ (a, b) such that (\frac{v}{u})′(ξ) = 0 and (\frac{v}{u})′ < 0 on (0, ξ). Therefore v′(ξ)u(ξ) = v(ξ)u′(ξ), so that, letting τ = v(ξ)/u(ξ), we also have τ = v′(ξ)/u′(ξ).

As in (4.23), for ε ∈ (0, ξ) we have

E(ζ(ξ)) − τ^{2}E(z(ξ)) = E(ζ(ε)) − τ^{2}E(z(ε)) + ∫_{ε}^{ξ} G′(x)[ζ(x)^{2} − τ^{2}z(x)^{2}]dx. (4.26)

As in the proof of Claim 1

lim_{ε→0} E(ζ(ε)) − τ^{2}E(z(ε)) = δ_{m,1} \frac{2}{2^{∗}} [v(0)^{2^{∗}} − τ^{2}u(0)^{2^{∗}}],

where δ_{m,1} is the Kroneker symbol. This term is nonnegative, because v(0) > u(0) and τ ∈ (0, 1). The integral in (4.26) is also positive because G′ > 0 and ζ − τz > 0 on (0, ξ).

Thus, letting ε → 0 in the right hand side of (4.26), we get a positive limit. This is a contradiction because the left hand side is negative. Indeed, by ζ(ξ) − τz(ξ) = 0 and ζ′(ξ) − τz′(ξ) = 0 we obtain

E(ζ(ξ)) − τ^{2}E(z(ξ)) = \frac{2}{2^{∗}} [ζ(ξ)^{2^{∗}} − τ^{2}z(ξ)^{2^{∗}}] = \frac{2}{2^{∗}} z(ξ)^{2^{∗}}[\frac{ζ(ξ)^{2^{∗}}}{z(ξ)^{2^{∗}}} − \frac{ζ(ξ)^{2}}{z(ξ)^{2}} ] < 0.

The functions u and v cannot exist. Problem (4.21) has unique solution. □

References

[BGG] R. Beals, P. Greiner, B. Gaveau, Green’s functions for some highly degenerate elliptic operators, J. Funct. Anal. 165 (1999), 407–429.

[B] W. Beckner, On the Grushin operator and hyperbolic symmetry, Proc. Amer. Math. Soc. 129 (2001), 1233–1246

[Bo] J. M. Bony, Principe du maximum, inégalite de Harnack et unicité du problème de Cauchy pour les opérateurs elliptiques dégénérés, Ann. Inst. Fourier (Grenoble) 19 (1969) 277–304.

[CGS] L. A. Caffarelli, B. Gidas, J. Spruck, Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth, Comm. Pure Appl. Math. 42 (1989), 271– 297.

[CL] W. X. Chen, C. Li, Classification of solutions of some nonlinear elliptic equations, Duke Math. J. 63 (1991), 615–622.

[CDKR] M. Cowling, A. H. Dooley, A. Korányi, F. Ricci, H-type groups and Iwasawa decompositions, Adv. Math. 87 (1991), 1–41.

[FLW] B. Franchi, G. Lu, R. L. Wheeden, A relationship between Poincaré-type inequalities and representation formulas in spaces of homogeneous type, Internat. Math. Res. Notices (1996), no. 1, 1–14.

[FL] B. Franchi, E. Lanconelli, Hölder regularity theorem for a class of non uniformly elliptic operators with measurable coefficients, Ann. Scuola Norm. Sup. Pisa 10 (1983), 523–541.

[GV] N. Garofalo, D. Vassilev, Symmetry properties of positive entire solutions of Yamabe-type equations on groups of Heisenberg type, Duke Math. J. 106 (2001), 411–448.

[GNN] B. Gidas, W. M. Ni, L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Phys. 68 (1979), 209–243.

[IM] T. Iwaniec, G. Martin, Geometric Function Theory and Non–linear Analysis, Oxford Math. Monographs, 2001.

[JL1] D. Jerison, J. M. Lee, The Yamabe problem on CR manifolds, J. Differential Geom. 25 (1987), 167–197.

[JL2] D. Jerison, J. M. Lee, Extremals for the Sobolev inequality on the Heisenberg group and the CR Yamabe problem, J. Amer. Math. Soc. 1 (1988), 1–13.

[K] A. Korányi, Kelvin transforms and harmonic polynomials on the Heisenberg group, J. Funct. Anal. 49 (1982), 177–185.

[KL] M. K. Kwong, Y. Li, Uniqueness of radial solutions of semilinear elliptic equations, Trans. Amer. Math. Soc. 333 (1992), 339–363. M

[LZha] Y. Y. Li, L. Zhang, Liouville–type theorems and Harnack–type inequalities for semilinear elliptic equations, J. Anal. Math. 99 (2003), 27–87.

[LZhu] Y. Y. Li, M. Zhu, Uniqueness theorems through the method of moving spheres, Duke Math. J. 80 (1995), 383–417.

[R] J. Ratcliffe, Foundations of hyperbolic manifolds, Springer, New York, 1994.

Roberto Monti: Mathematisches Institut, Universität Bern, Sidlerstrasse 5, 3012 Bern, Switzerland

  E-mail address: roberto.monti@math-stat.unibe.ch

Daniele Morbidelli: Dipartimento di Matematica, Università di Bologna, Piazza di Porta San Donato, 5, 40127 Bologna, Italy

  E-mail address: morbidel@dm.unibo.it