Research Article Monotonicity of Harnack Inequality for Positive Invariant Harmonic Functions

Hindawi Publishing Corporation Journal of Applied Mathematics and Stochastic Analysis Volume 007, Article ID 397, pages doi:0.55/007/397 Research Article Monotonicity of Harnack Inequality for Positive Invariant Harmonic Functions Yifei Pan and Mei Wang Received 3 February 007; Revised 6 June 007; Accepted 7 October 007 A monotonicity property and a refined estimate of Harnack inequality are derived for positive solutions of the Weinstein equation. Copyright 007 Y. Pan and M. Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.. Introduction Let B n ={x R n : x < }, n, be the unit ball in R n, S n = B n. Consider the Weinstein equation in B n of the form Δ λ u = ( x { x u 4 x j j + λ j } u n x j + λ x j λ u = 0, (. where u = u(x, x B n, λ R. In this paper, we prove a monotonicity property and a refined estimate of Harnack inequality for positive solutions of (.. The differential operator Δ λ in (. is a natural extension of the Laplacian operator (λ = 0. If T is a Möbius transformation from B n onto B n and T (x denotes the Jacobian matrix, then for every solution u of (.inb n, the function dett (x (n λ/n u ( T(x (. is also a solution of (., as proved by Akın and Leutwiler []. More precisely, Δ λ { dett (x (n λ/n u ( T(x } = dett (x (n λ/n Δ λ u ( T(x (.3

Journal of Applied Mathematics and Stochastic Analysis for twice differentiable functions u(x inb n []. Therefore, Δ λ is also called invariant Laplacian and the solutions of (. are called invariant harmonic functions. The Dirichlet problem for (. and its half-space counterpart are challenging and interesting, as summarized in Liu and Peng [], where the authors also pointed out that invariant harmonic functions do not possess good boundary regularity in general. The classical Harnack inequality gives scale invariant bounds for nonnegative (nonpositive harmonic functions in the plane. Harnack-type inequalities have been an important tool in the general theoryof harmonicfunctionsand partial differential equations, on which Kassmann [3] provided a through introduction and survey of the development and applications. In this paper, we give an estimate of Harnack inequality bounds for any two points in B n based on a monotonicity property of positive invariant harmonic functions. In this section, we state the main results. The proofs are provided in the subsequent sections. For positive solutions of (., Theorem. describes a monotonicity property, Theorem. gives bounds for a Harnack-type inequality for two points on the same ray, and Theorem.3 extends the estimates for the Harnack-type inequality to any two points in B n. Two interesting special cases one on harmonic functions and another on the Laplace-Beltrami operator associated with the Poincaré metric are stated as corollaries. To study the properties of solutions of Δ λ u = 0, one often needs to distinguish the cases of λ / and λ< /. Throughout this paper, we denote δ = δ(λ = λ +λ +λ = 0, +λ, λ< for λ R. (.4 Theorem.. Let u(x, x B n, be a positive solution of Δ λ u = 0, λ R.Thenforζ S n, the function is decreasing for 0 r< and the function is increasing for 0 r<. ( r n δ +λ δ u(rζ (.5 ( + r ( + r n δ +λ δ u(rζ (.6 ( r Theorem.. Let u(x, x B n, be a positive solution of Δ λ u = 0, λ R.Thenforζ S n and 0 r r<, r λ+ δ ( +r n δ +r λ+ δ ( r u(r r +r n δ ζ u(rζ u(r +r r ζ. (.7 Notice that case λ = 0(δ = 0 gives the classical Harnack Inequality.

Y. Pan and M. Wang 3 Theorem.3. Let u(x, x B n, be a positive solution of Δ λ u = 0, λ R. Letξ,ξ S n and 0 r r <. Then { } u r ξ f λ r, r exp gλ r,ξ,ξ u ( { } f λ r,r exp gλ r,ξ r ξ,ξ (.8 with λ+ δ +r n δ r f λ r,r =, +r r π g λ r,ξ,ξ = ξ ξ n +λ δ r. r (.9 Remarks.4. The sharpness of the bounds in the above theorems are discussed in the proofs of Lemma.4, Theorems. and.3. The equality case can be achieved in Theorem.. Only trivial equalities are attained in Theorem.3. Case λ = 0 corresponds to harmonic functions, wherein Theorem.3 canbestatedas the following. Corollary.5. Let u(x be a positive harmonic function in B n. ξ,ξ S n,0 r r <. Then f { } u r ξ r, r exp g r u f { } r,r exp g r, (.0 r ξ where f ( +r n r r,r =, +r r g π r = g r,ξ,ξ = ξ ξ nr. r (. Notice that when r = r = x, r = 0, (.8becomes r u(x n ( + r u(0 +r n, (. ( r the classical Harnack inequality in B n. Case λ = n/ corresponds to the Laplace-Beltrami operator Δ n/ associated with the Poincaré metric. In this case, Theorem.3 has the following form. Corollary.6. Let u(x, x B n, be a positive solution of Δ n/ u = 0. Letξ,ξ S n and 0 r r <. Then C u r ξ u C, (.3 r ξ where C = C ( ( +r r,r,ξ,ξ = r n exp{ π ξ +r r ξ (n r }. (.4 r

4 Journal of Applied Mathematics and Stochastic Analysis. Proof of Theorem. Our proofs depend on an integral representation for positive solutions of Weinstein equations in B n derived by Leutwiler [4] based on earlier work of Huber [5] and Brelot-Collin and Brelot [6]. We state Leutwiler s representation as a theorem below (with a slightly different parametrization. Theorem. (The representation theorem, Leutwiler [4, Theorem 3.]. Each positive solution of (. admits the unique representation where μ is a positive measure on S n. ( x +λ δ u(x = dμ(η, (. S n x η n+λ δ According to the representation theorem, every positive solution of Δ λ u = 0canbe identified with its integral representation (. and the corresponding positive measure μ.notice thatwhen λ /,the integrand of (. is the Poisson kernel P λ (x,η = ( x +λ x η n+λ. (. In the sequel, we will use the representation formula (. in terms of the Poisson kernel for positive solutions of (. for the case λ /. The solutions of (. with λ < / is related to that of λ> / by a corresponding principle, also proved by Leutwiler [4]. We state a special case of the correspondence principle as a lemma. Lemma. (The correspondence principle, Leutwiler [4, Lemma 3.4]. If u(x, x B n, is a solution of Δ λ u = 0, λ< /, then Δ λ ũ = 0 with ũ(x = ( x (+λ u(x, λ = ( + λ >. (.3 We need the following two lemmas for the proof of Theorem.. Lemma.3. Let x R n, x =r, ζ S n.ifλ /, then ( n +λ (n λ r ( r λ x ζ n+λ ( r +λ r x ζ n+λ ( n +λ +(n λ r ( r λ x ζ n+λ. (.4 Proof. Write x = x η = rη, η ζ = n i= η i ζ i.since r x ζ = ( x rη ζ + = (r η ζ, r r x ζ n+λ = ( x ζ (n+λ/ = (n +λ x ζ n+λ (r η ζ, r (.5

Y. Pan and M. Wang 5 we have ( r +λ r x ζ n+λ = ( + λ( r λ ( r x ζ n+λ ( r +λ ( / r x ζ n+λ x ζ (n+λ = ( + λ( r λ r x ζ n+λ ( r +λ (n +λ x ζ n+λ (r η ζ x ζ (n+λ = ( + λ( r λ r x ζ ( r +λ (n +λ(r η ζ x ζ n+λ+. (.6 To prove the right-side inequality in Lemma.3,it sufficesto show ( + λr x ζ ( r (n +λ(r η ζ ( n +λ +(n λ r x ζ, (.7 which is equivalent to (n +λ ( r (r η ζ (n +λ( + r x ζ. (.8 Since λ /, n +λ>0, the above becomes or which, after a simplification, is equivalent to ( r (r η ζ ( + r x ζ, (.9 ( r(r η ζ r rη ζ +, (.0 η ζ. (. The inequality is true since ζ,η S n. To prove the left-side inequality in Lemma.3,it suffices to show that ( + λr x ζ ( r (n +λ(r η ζ ( n +λ (n λ r x ζ, (. which is equivalent to which is, after a simplification, (n +λ ( r (r η ζ (n +λ( r x ζ, (.3 η ζ, (.4 true since ζ,η S n. This completes the proof of Lemma.3.

6 Journal of Applied Mathematics and Stochastic Analysis Lemma.4. Let u(x be a positive solution of Δ λ u = 0, λ /, x =r. Then n +λ (n λ r r u(x u(x n +λ +(n λ r r r u(x. (.5 Proof. By the representation theorem, u has the integral representation (. of the Poisson kernel with a positive measure μ on S n, ( x +λ u(x = S n x ζ n+λ dμ(ζ. (.6 Applying Lemma.3,wehave It follows that u(x r S n ( ( x +λ r x ζ n+λ dμ(ζ S n ( n +λ +(n λ r ( r λ x ζ n+λ dμ(ζ n +λ +(n λ r = r n +λ +(n λ r = ( ( x +λ = S n r x ζ n+λ S n r u(x. dμ(ζ ( x +λ x ζ n+λ dμ(ζ ( n +λ +(n λ r (.7 r u(x. (.8 The left-side inequality in Lemma.4 can be proved in the same manner. For the equality case, consider u y (x = ( x +λ / x y n+λ which is invariant harmonic in R n \{y} for y S n. A simple calculation shows that the equalities hold for u y (xwhenx = x y and x = x y, respectively. This completes the proof of Lemma.4. Now we prove Theorem.. Proof. First, consider the case λ /(δ = 0. Define ϕ(r = then ϕ (r n +λ +(n λ r ϕ(r = r, For x = rζ, x =r, ζ S n, denote ( rn ( + rn +λ, ψ(r = +λ, for 0 r<, (.9 ( + r ( r ψ (r n +λ (n λ r ψ(r = r. (.0 I(r,ζ = ϕ(ru(rζ, J(r,ζ = ψ(ru(rζ. (.

Y. Pan and M. Wang 7 To prove Theorem. for λ /, we need to show that I(r,ζ is decreasing and J(r,ζ is increasing in r.bylemma 4., d ϕ (r logi(r,ζ = dr ϕ(r + u(x u(x r n +λ +(n λ r n +λ +(n λ r r + r = 0. Therefore, log I(r,ζ is decreasing in r, and so is I(r,ζ. Similarly, (. d ψ (r logj(r,ζ = dr ψ(r + u(x u(x r n +λ (n λ r r ( n +λ (n λ r r = 0. Hence, J(r,ζ is increasing in r.wehaveprovedtheorem. for the case λ /. If Δ λ u = 0, λ< /, then satisfies (.3 ũ(x = ( x (+λ u(x (.4 Δ λ ũ = 0 with λ = ( + λ > (.5 by the correspondence principle (Lemma.. From the above results for λ> /, ( r n n ( r ( ũ(rζ = r (+λ u(rζ = ( r n λ + λ ( + r ( + r (+λ u(rζ (.6 is decreasing in r,and ( + r n n ( + r ( ũ(rζ = r (+λ u(rζ = ( + r n λ + λ ( r ( r (+λ u(rζ (.7 is increasing in r. Recallthatδ = 0forλ / andδ = +λ for λ< /, we have shown that is decreasing in r and ( r n δ +λ δ u(rζ (.8 ( + r ( + r n δ +λ δ u(rζ (.9 ( r is increasing in r for any λ R. This completes the proof of Theorem..

8 Journal of Applied Mathematics and Stochastic Analysis 3. Proof of Theorem. The proof for Theorem. is based on Theorem. and the following lemma. Lemma 3.. Let f (r be a positive function on r [0,.Iffora,b R, then for 0 r r<, a + br r f (r f (r a br f (r, (3. r +r a ( r (b+a/ +r a ( r (b a/ +r r f (r f (r +r r f (r. (3. Proof. The integral Thus for 0 r r <, by (3., a br r dr = aln( + r+ (b aln( r + C. (3.3 r f ln f (r ln f (r (r = r f (r dr r r ( a br +r r dr ln +r a ( r r that is, the right-side inequality in (3. holds. Similarly, by the left-side of (3., r ln f (r ln f (r a + br ( +r r r dr ln +r hence the left-side inequality in (3.holds. Now we prove Theorem.. a ( r r (b a/, (3.4 (b+a/, (3.5 Proof. For λ /, u(rζ satisfies Lemma.4. Therefore, (3. holds with f (r = u(rζ, a = n +λ, b = n +λ +.Let0 r r<. Inequality (3.inLemma 3. implies +r n λ ( r λ+ +r n+λ ( r n+ +r r u(r ζ u(rζ +r r u(r ζ, (3.6 which is equivalent to r λ+ ( +r n +r λ+ ( r u(r r +r n ζ u(rζ u(r +r r ζ. (3.7 If Δ λ u = 0, λ< /, then ũ(x = ( x (+λ u(x satisfies Δ λ ũ = 0with λ = ( + λ > / according to the correspondence principle (Lemma.. From the above proof

Y. Pan and M. Wang 9 for λ> /, that is, r λ+( +r n +r λ+( r ũ(r r +r n ζ ũ(rζ ũ(r +r r ζ, (3.8 ( +r n λ ( r u(r n λ ζ u(rζ u(r ζ. (3.9 +r r Combining the cases λ / andλ< / completes the proof of Theorem.. The equality cases are achieved by the function u y (x = ( x +λ δ / x y n+λ δ at x = x y and x = x y, respectively. 4. Proof of Theorem.3 The proof of Theorem.3 is based on the following three lemmas. Lemma 4.. Let u(x, x B n, be a positive solution of Δ λ u = 0, λ R. Letϕ(t, t [0,] be the shortest arc on the great circle connecting ξ and ξ. Then d dt u( rϕ(t r (n +λ δϕ (t ( r u ( rϕ(t, r [0,]. (4. Proof. Direct calculation shows that d dt rϕ(t η n+λ δ = (n +λ δrϕ (t η rϕ(t η n+λ δ+. (4. Applying rϕ(t η rφ(t η r,wehave S n d dt rϕ(t η n+λ δ dμ(η (n +λ δrϕ (t η = S n rϕ(t η n+λ δ+ dμ(η n +λ δ r S n ϕ (t η rϕ(t η n+λ δ ( r dμ(η = r (n +λ δϕ (t ( r +λ δ ( r ( r +λ δ S n rϕ(t η n+λ δ dμ(η = r (n +λ δϕ (t ( r ( r +λ δ u( rϕ(t. (4.3

0 Journal of Applied Mathematics and Stochastic Analysis By Lebesgue s dominant convergence theorem, Lemma 4.. Let d dt u( rϕ(t ( d rϕ(t +λ δ = dt S n rϕ(t η n+λ δ dμ(η = ( r +λ δ d dt rϕ(t η S n r (n +λ δϕ (t ( r u ( rϕ(t. ξ i = ( 0,...,0,r cosθ i,r sinθ i S n, i =,, n+λ δ dμ(η θ θ π, ϕ(t = ( 0,...,0,r cosθ t,r sinθ t, θt = tθ +( tθ, t [0,]. (4.4 (4.5 Then ϕ (t π ξ ξ. (4.6 Proof. It suffices to prove for n =. For computation convenience, we use the complex plane notations in R. Denote ξ i = e θi ; ξ ξ = e iθ e iθ ( = exp i θ + θ = exp i θ + θ ( ( exp i θ θ (isin θ θ. ( exp i θ θ (4.7 Notice that sin x x π for x π, (4.8 so θ θ π implies Furthermore, e iθ e iθ θ θ = sin( π θ θ. (4.9 ϕ (t = d dt ( e itθ +i( tθ = ϕ(ti ( θ θ. (4.0

Thus Y. Pan and M. Wang ϕ (t = θ θ π e iθ e iθ π = ξ ξ. (4. Lemma 4.3. Let u(x, x B n, be a positive solution of Δ λ u = 0, λ R. Letξ,ξ S n. Then for r [0,, { exp π ξ ξ n +λ δ r ( r } u rξ π u ( exp{ rξ ξ ξ n +λ δ r ( r }. (4. Proof. For ξ,ξ S n, there exists a Möbius transformation T in R n such that T(S n = S n,andforallr [0,], T ( rξ i = ( 0,...,0,r cosθi,r sinθ i for i =,, θ θ π (4.3 with dett (x =, T ( rξ T ( rξ = rξ rξ, (4.4 and u(t(x is also a positive solution of (. with respect to the measure μ(t (x. Since Δ λ is invariant under such orthogonal transformation, we may assume without loss of generality that Let ξ i = ( 0,...,0,cosθ i,sinθ i for i =,, θ θ π. (4.5 ϕ(t = ( 0,...,0,r cosθ t,r sinθ t, θt = tθ +( tθ, t [0,]. (4.6 Then ϕ(0 = ξ, ϕ( = ξ.by(4.inlemma 4. and (4.6inLemma 4., (d/dtu ( rϕ(t u ( rϕ(t dt (d/dtu( rϕ(t u ( rϕ(t dt 0 0 n +λ δ r ( r 0 ϕ (t dt π ζ ζ n +λ δ r ( r. (4.7 Since we have ln u rξ u = ln u rϕ( rξ u ( rϕ(0 = 0 (d/dtu ( rϕ(t u dt, (4.8 rϕ(t ln u rξ u rξ π ξ ξ n +λ δ r ( r. (4.9

Journal of Applied Mathematics and Stochastic Analysis Therefore, π ξ ξ n +λ δ r ( r ln u rξ u π rξ ξ ξ n +λ δ r ( r. (4.0 This completes the proof of Lemma 4.3. The proof of Theorem.3 follows immediately. Proof. Write u r ξ u = u r ξ r ξ u u r ξ r ξ u. (4. r ξ Applying Lemma 4.3 and Theorem., Theorem.3 follows. The function u y (x = ( x +λ δ / x y n+λ δ at x = x y and x = x y gives the trivial equalities achieved in Theorem.. Acknowledgment The authors thank the editors and the referees for their valuable comments and suggestions. References [] Ö. Akın and H. Leutwiler, On the invariance of the solutions of the Weinstein equation under Möbius transformations, in Classical and Modern Potential Theory and Applications, K. GowriSankaran, J. Bliedtner, D. Feyel, M. Goldstein, W. K. Hayman, and I. Netuka, Eds., vol. 430 of NATO Advanced Science Institutes Series C: Mathematical and Physical Sciences, pp. 9 9, Kluwer Academic, Dordrecht, The Netherlands, 994. [] C. Liu and L. Peng, Boundary regularity in the Dirichlet problem for the invariant Laplacians on the unit real ball, Proceedings of the American Mathematical Society, vol. 3, no., pp. 359 368, 004. [3] M. Kassmann, Harnack inequalities: an introduction, Boundary Value Problems, vol. 007, Article ID 845, pages, 007. [4] H. Leutwiler, Best constants in the harnack inequality for the Weinstein equation, Aequationes Mathematicae, vol. 34, no. -3, pp. 304 35, 987. [5] A. Huber, On the uniqueness of generalized axially symmetric potentials, Annals of Mathematics, vol. 60, no., pp. 35 358, 954. [6] B. Brelot-Collin and M. Brelot, Représentation intégrale des solutions positives de l équation n L k (u = ( u/ xi +(k/x n ( u/ x n,(k constante réelle dans le demi-espace E(x n > 0, de R n, Bulletin de la Classe des Sciences. Académie Royale de Belgique, vol. 58, pp. 37 36, 97. Yifei Pan: Department of Mathematical Sciences, Indiana University-Purdue University Fort Wayne, Fort Wayne, IN 46805-499, USA Current address: School of Mathematics and Informatics, Jiangxi Normal University, Nanchang 33007, China Email address: pan@ipfw.edu Mei Wang: Department of Statistics, University of Chicago, Chicago, IL 60637, USA Email address: meiwang@galton.uchicago.edu