Gradient estimates for the constant mean curvature equation in hyperbolic space
aa r X i v : . [ m a t h . DG ] D ec Gradient estimates for the constant meancurvature equation in hyperbolic space
Rafael L´opez
Departamento de Geometr´ıa y Topolog´ıaInstituto de Matem´aticas (IEMath-GR)Universidad de Granada18071 Granada, Spain
Abstract
We establish gradient estimates for solutions to the Dirichlet problem forthe constant mean curvature equation in hyperbolic space. We obtain theseestimates on bounded strictly convex domains by using the maximum princi-ples theory of Φ-functions of Payne and Philippin. These estimates are thenemployed to solve the Dirichlet problem when the mean curvature H satisfies H <
AMS 2010 Classification Number: 35J62, 35J25, 35J93, 35B38, 53A10Keywords: hyperbolic space, Dirichlet problem, maximum principle, critical point
In this paper we consider the Dirichlet problem for the constant mean curvatureequation on a domain of a horosphere in three-dimensional hyperbolic space H . Inorder to fix the terminology, we consider the upper halfspace model of H , that is,1 = { ( x , x , x ) ∈ R : x > } endowed with the hyperbolic metric g = g /x , g being the Euclidean metric. After a rigid motion of H , a horosphere can beexpressed as a horizontal plane P a of equation x = a , a >
0. Let Ω be a domain of P a , where we identify Ω with its orthogonal projection Ω × { } on the plane x = 0.We study the Dirichlet problemdiv Du p | Du | ! = − u p | Du | − H ! in Ω (1) u = a > ∂ Ω , (2)where u > H ∈ R is a constant and D and div denotethe gradient and the divergence operators in the Euclidean plane R . The graphΣ u = { ( x, u ( x )) : x ∈ Ω } , x = ( x , x ), represents a surface in H with constantmean curvature H computed with respect to the upwards orientation. The studyof the solutions of the Dirichlet problem (1)-(2) depends strongly of the relationbetween H and the value 1, the modulus of the sectional curvature − H . Forexample, if H <
H > u lies above the horosphere P a (respectivelybelow P a ) and the geometric behaviour of Σ u in both cases is completely different:let us observe that in hyperbolic geometry, the translations along the x -coordinateare not isometries of H .In this article, we will use the theory of maximum principles developed by Payneand Philippin to obtain estimates of the gradient for a solution of (1)-(2). We deriveestimates of the gradient | Du | in terms of C bounds of u . Theorem 1.1.
Let Ω ⊂ R be a bounded strictly convex domain. Let u be a solutionof (1)-(2) and denote u M = sup Ω u and C = 11 − H u M a · If ≤ H < or if H < with u M < r H − H a, (3) then | Du | ≤ p C − (1 + HC ) HC in Ω . (4)2f we have estimates for the gradient of solutions of (1)-(2), it is natural toaddress the problem of the existence of solutions of the Dirichlet problem. In thecontext of the hyperbolic space, the results of existence require some assumptionon the convexity of the domain Ω. If 0 ≤ H <
1, the convexity of ∂ Ω is enoughto ensure the existence of a solution of (1)-(2): see [9, 14]. However, if
H <
0, themere convexity of Ω does not ensure the existence of solutions and it is requiredstronger convexity. More exactly, the solvability of the Dirichlet problem (1)-(2)was proved if the curvature κ of ∂ Ω satisfies − k < H < Theorem 1.2.
Let Ω ⊂ R be a bounded strictly convex domain. Let R be thediameter of ∂ Ω . If − ≤ H < satisfies R < − − H + 2 r HH − , (5) then there exists a unique solution of (1)-(2). We notice that we need to assume that the diameter of Ω is small in relationwith the value of H but, in contrast, it is not necessary strong convexity of ∂ Ω andwe allow that the existence of regions of ∂ Ω whose curvature is closed to 0.Theorems 1.1 and 1.2 will be proved in §
3. In the proof of these results, weneed to show the uniqueness of critical points of solutions of (1)-(2). Althoughthis may be expected because the resemblance of (1) with other quasilinear ellipticequations, as for example, the capillary equation ([1, 2, 6]) or the singular minimalsurface equation ([12]), we prove this uniqueness only in the range of values
H < §
2. Finally we prove in § u M of a solution of (1)-(2) when H < u are obtained by comparing u with known solutions of (1),as for example, radial solutions. However, our result establishes a lower estimate ofthe value u at the critical point in terms of the curvature of ∂ Ω and H : see Theorem4.1. 3 Uniqueness of critical points
The first result in this paper establishes, under some hypothesis, the uniquenessof critical points of a solution of the Dirichlet problem. The topic on the numberof critical points of solutions for elliptic equations is a subject of high interest andthe literature is very extensive, especially related with the question of the convexityof level sets of solutions of elliptic equations. In the context of the constant meancurvature equation in Euclidean space, and if the domain is convex, Sakaguchi provedthe existence of a unique critical point assuming Dirichlet boundary condition orNeumann boundary condition ([19]). In this paper we address this problem for theconstant mean curvature equation in hyperbolic space when
H < Theorem 2.1.
Let Ω ⊂ R be a bounded strictly convex domain and let H ∈ R . If H < , then a solution u of (1)-(2) has exactly one critical point, which coincideswith the point where u attains its global maximum. We prove this result as a consequence of the following arguments.A first step consists in proving the existence of at least one critical point of asolution u of (1)-(2). When H ≤
0, this is achieved by the Hopf maximum principle.Indeed, because the right-hand side of (1) is non positive, the minimum of u isattained at some boundary point, proving u > a in Ω. Since Ω is bounded, thefunction u has a global maximum at some interior point.This argument fails if 0 < H <
1. For this range of values of H (also if H ≤
0) wewill use a comparison principle based in the standard theory of quasilinear ellipticequations ([7, Th. 10.1]). In our context, it can be formulated as follows: if twosurfaces Σ and Σ have a common interior point p and with constant mean curvature H and H , respectively, with respect to the orientations that coincide at p , if Σ lies above Σ around p , then H ≥ H (the same conclusion holds if p is a commonboundary point with tangent boundaries at p ): see [11, p. 194]. Lemma 2.2.
Suppose Ω ⊂ R is a bounded domain. If H < , then a solution of(1)-(2) satisfies u > a in Ω .Proof. The proof is by contradiction. Suppose that there exists x ∈ Ω such that u attains the minimum value, u ( x ) ≤ a . Let p = ( x , u ( x )). For b < u ( x ), consider4he the horosphere P b of equation x = b , whose mean curvature is 1 with respectto the upwards orientation. Then we move up P b by letting b ր ∞ , until the firsttouching point with Σ u at b = u ( x ). Then the horosphere P b touches Σ u at p ,which is an interior point of both Σ u and P b . As Σ u lies above P b , we arrive acontradiction with the comparison principle.Once proved this lemma, we follow with the proof of Theorem 2.1. Denote u k = ∂u/∂x k , k = 1 ,
2, and consider the summation convention of repeated indices.Equation (1) can be expressed as(1 + | Du | )∆ u − u i u j u ij + 2(1 + | Du | ) u − H (1 + | Du | ) / u = 0 . Denote v k = u k , k = 1 ,
2, and we differentiate the above identity with respect to x k obtaining: (cid:0) (1 + | Du | ) δ ij − u i u j (cid:1) v kij + 2 (cid:18) u i ∆ u − u j u ij + 2 u i u − Hu i u (1 + | Du | ) / (cid:19) v ki − | Du | ) u (1 − H p | Du | ) v k = 0 (6)for k = 1 , δ ij is the Kronecker delta. Equation (6) is a linear ellipticequation in v k . We need to apply the Hopf Maximum Principle ([7, Th. 3.5]) to thisequation. Then we have to know that the term of v k is non positive, or equivalently,1 − H p | Du | ≥ H ≤
0, this inequality is clear. If 0 < H <
1, one needs to estimate | Du | in termsof H . For this, we prove the next lemma, which is implicitly contained in the proofof the main result in [13]. Lemma 2.3.
Let Ω ⊂ R be a bounded strictly convex domain of R and let 0, that is, H − p | Du | ≤ , which yields (8).As a consequence of Lemma 2.3, the Hopf Maximum Principle for equation (6)implies that if v k takes a non-negative maximum in Ω or a non positive minimumin Ω, then v k must be a constant function ([7, Th. 3.5]). We point out also thatthe function u is analytic by standard theory ([3, 15]), and the same holds for thefunctions v k .For each θ ∈ R , let (cos θ, sin θ ) be a vector of the unit circle S . Since (6) is alinear equation on v k , the function v ( θ ) = Du · (cos θ, sin θ ) = v cos θ + v sin θ (10)also satisfies (6). Denote n the outward unit normal vector of ∂ Ω. Because u isconstant along ∂ Ω, we have ( v , v ) = Du = ( Du · n ) n along ∂ Ω, that is,( v , v ) = ∂u∂n n . From (10), v ( θ ) = ∂u∂ n n · (cos θ, sin θ ) along ∂ Ω . 6n the other hand, since u is constant along ∂ Ω, the Strong Maximum Principle ofHopf ([7, Th. 3.6]) implies that any outward directional derivative on ∂ Ω is negativeand thus, ∂u∂ n < ∂ Ω . Fix θ ∈ R . Since ∂ Ω is strictly convex, the map n : ∂ Ω → S is one-to-one. Itfollows that there exist exactly two points of ∂ Ω where n ( s ) is orthogonal to thefixed direction (cos θ, sin θ ). By the definition of v ( θ ), the function v ( θ ) vanishesalong ∂ Ω at exactly two points.We now follow the argument given by Philippin in [17] to prove the uniqueness ofthe critical points. By completeness, we give it briefly. The proof is by contradiction,and suppose that there are at least two critical points of u in Ω. Let P and P betwo critical points which are fixed in the sequel. The argument consists into thefollowing steps.1. The function v ( θ ) is not constant in Ω because v ( θ ) only has two zeroes along ∂ Ω.2. As a consequence, the critical points of v ( θ ) are isolated point of Ω because v ( θ ) is analytic.3. Let N θ = v ( θ ) − ( { } ) be the nodal set of v θ . Because v ( θ ) is analytic, stan-dard theory asserts that near to a critical point of v ( θ ), the function v ( θ ) isasymptotic to a harmonic homogeneous polynomial ([3]). Following Cheng[4], N θ is diffeomorphic to the nodal set of the homogeneous polynomial thatapproximates, in particular, N θ is formed by a set of regular analytic curvesat regular points, the so-called nodal lines. On the other hand, in a neighbourof a critical point, the nodal lines form an equiangular system.We claim that there does not exist a closed component of N θ contained in Ω.This is because if N θ encloses a subdomain Ω ′ of Ω where v ( θ ) = 0 along ∂ Ω ′ ,the maximum principle implies that v ( θ ) is identically 0 in Ω ′ , a contradiction.4. We prove that N θ is formed exactly by one nodal line. Suppose by contradic-tion that there are two nodal lines L and L . Because both L an L are notclosed, then the arcs L and L end at the boundary points where v ( θ ) van-ishes, being both points the two end-points of L i . Since Ω is simply-connected,7hen the two arcs L and L enclose at least one subdomain Ω ′ ⊂ Ω where v ( θ ) vanishes along ∂ Ω ′ . This is impossible by the maximum principle.5. As a conclusion, the nodal set N θ is formed exactly by one arc. We now givean orientation to the arc N θ for all θ . The chosen orientation in N θ is that wefirst pass through P and then through P . With respect to this orientation,we are ordering the two boundary points where v ( θ ) vanishes. More precisely,denote by P ( θ ) the initial point of N θ , which after passing P and then P ,finishes at the other boundary point, which is denoted by Q ( θ ).6. Let us consider θ varying in the interval [0 , π ]. We observe that by the def-inition of v ( θ ) in (10), the functions v (0) and v ( π ) coincides up to the sign,that is, v (0) = − v ( π ) and thus the nodal lines N and N π coincide as setsof points. However, when θ runs in [0 , π ], the ends points of N interchangeits position when θ arrives to the value θ = π , that is, in the nodal line N π .Consequently, and according to the chosen orientation in N θ , P (0) = Q ( π )and P ( π ) = Q (0). Because all the arcs N θ pass first P and then P , thisinterchange of the end points between N and N π implies the existence of an-other nodal line for v ( π ). This is impossible by the item 4: this contradictioncompletes the proof of Theorem 2.1.We extend Theorem 2.1 in the limit case u = 0 along ∂ Ω. Corollary 2.4. Let Ω ⊂ R be a bounded strictly convex domain. Let H be a realnumber with H < . If u is a solution (1) and u = 0 along ∂ Ω , then u has a uniquecritical point.Proof. We consider positive values a sufficiently close to 0 so the set Ω a = { x ∈ Ω : u ( x ) > a } is strictly convex. Then Theorem 2.1 asserts the existence of aunique critical point, which must coincide for all a because Ω a ′ ⊂ Ω a if a < a ′ . Theargument finishes by letting a → 0. 8 Proof of theorems 1.1 and 1.2 In this section we apply the theory of the maximum principle developed by Payneand Philippin in [16] for some Φ-functions associated to equation (1). We introducethe notation employed there. Consider an equation of typediv( g ( q ) Du ) + ρ ( q ) f ( u ) = 0 , (11)where ρ, g > g is a C function of its argument and ρ and f are C functions. Here q = | Du | . We also assume that (11) satisfies the elliptic condition g ( ξ ) + 2 ξg ′ ( ξ ) > ξ > 0. We define the Φ-functionΦ( x ; α ) = Z q c g ( ξ ) + 2 ξg ′ ( ξ ) ρ ( ξ ) dξ + α Z uc f ( η ) dη, where α is a real parameter and c , c ∈ R . Here the functions g and ρ are evaluatedin q .We now prove Theorem 1.1. Proof of Theorem 1.1. For equation (1), we take c = 0, c = 1 and the functions g , ρ and f are g ( ξ ) = 1 √ ξ , ρ ( ξ ) = 1 √ ξ − H, f ( u ) = 2 u · (12)Following the theory of Payne and Philippin, we require that ρ is positive, which itis clear if H ≤ 0. On the other hand, in the range 0 < H < 1, the evaluation of ρ at q is non-negative by Lemma 2.3. A straight-forward computation of Φ( x ; α ) givesΦ( x ; α ) = log (1 + q )(1 − H p q ) u α ! , x ∈ Ω . When Ω is strictly convex, it is proved in [16, Corollary 1] that Φ( x ; 2) attainsits maximum at one critical point of u . By Theorem 2.1, we know that the function u has exactly one critical point, which we denote by O , and let u M = u ( O ), whichcoincides with the maximum value of u in Ω. Then we find1 + | Du | (1 − H p | Du | ) u ≤ − H ) u M , | Du | (1 − H p | Du | ) ≤ − H ) (cid:16) u M u (cid:17) ≤ − H ) (cid:16) u M a (cid:17) . (13)Recall the value C = u M / ((1 − H ) a ). It follows from (13) that(1 + HC ) p | Du | ≤ C. The inequality (4) is shown provided 1 + HC > 0. This inequality is immediate if0 ≤ H < 1. In case H < 0, the inequality 1 + HC > (cid:18) δ ij − u i u j | Du | (cid:19) Φ ij + W i Φ i ≥ α − (cid:16) H p q + ( α − q − (cid:17) u (1 + q ) , (14)where W i is a vector function uniformly bounded in Ω. In order to apply the FirstHopf Maximum Principle, we require that the right-hand side in (14) is non-negative.If α ∈ [0 , H ≤ < H < H p q + ( α − q − ≤ ( α − q ≤ . Following [16], we deduce that Φ( x ; α ) attains its maximum at some some boundarypoint for all α ∈ [0 , α = 0, we deduce the following result. Corollary 3.1. Let Ω ⊂ R be a bounded domain and let H ≤ . If u is a solutionof (1), max Ω | Du | = max ∂ Ω | Du | . The same holds when < H < if, in addition, Ω is strictly convex, and u = a > on ∂ Ω . roof. If we take α = 0 in (14), then there exists a boundary point P ∈ ∂ Ω suchthat 1 + | Du | (1 − H p | Du | ) ≤ q M (1 − H p q M ) , where q M = | Du | ( P ). It follows directly that | Du | ≤ q M , proving the result.From Theorem 1.1 and Corollary 3.1, we prove the existence result of Theorem1.2. Proof of Theorem 1.2. The uniqueness of solutions is a consequence that the right-hand side of (1) is non-decreasing on u by Lemma 2.3 ([7, Th. 10.2]).For the existence of a solution u of (1)-(2), we apply a modified version of thecontinuity method to the family of Dirichlet problems parametrized by τ ∈ [0 , P τ : (cid:26) Q τ [ u ] = 0 in Ω u = a on ∂ Ω , where Q τ [ u ] = div Du p | Du | ! + 2 u p | Du | − τ H ! . See [7, Th. 11.4]. The graph Σ u τ of a solution of u τ of P τ is a graph on P a withconstant mean curvature τ H and boundary ∂ Ω. Let us observe that for the value τ = 0, there is a solution of P because ∂ Ω is convex ([13, 14]). As usual, let A be the subset of [0 , 1] consisting of all τ for which the Dirichlet problem P τ has a C ,α solution. The proof consists in showing that 1 ∈ A because standard regularityPDE results guarantee that any solution of Q τ [ u ] = 0 is smooth in Ω.First observe that A 6 = ∅ because 0 ∈ A . On the other hand, the set A is openin [0 , 1] because ∂Q τ [ u ] ∂u = − u p | Du | − τ H ! ≤ , since H < 0. 11inally, the main difficulty lies in proving that A is closed. This follows if we areable to derive a priori C and C estimate of u τ for every τ ∈ [0 , 1] and dependingonly on the initial data. In other words, we have to find a constant M , dependingonly on H , a and Ω, such that if u τ is a solution of P τ , thensup Ω | u τ | + sup Ω | Du τ | ≤ M. (15)See [7, Th. 13.8]. However, by using Theorem 1.1, it is enough if we find an upperbound for sup Ω | u τ | . We now use a geometric viewpoint of the solutions of P τ .Fix H ∈ R . After a dilation from the origin of R , which is an isometry of H ,we suppose a = 1. Then the diameter of ∂ Ω coincides with the Euclidean one. Let C R ⊂ P be the circumscribed circle of ∂ Ω, which has a radius equal to R . After ahorizontal translation in R , if necessary, we suppose that the centre of C R is (0 , , D R ⊂ P the disc bounded by C R , which contains Ω in its interior. Weknow that Σ u lies above the plane x = 1. On the disc D R , we are going to place anumbilical surface Σ w with the same mean curvature H and being a graph on D R .Indeed, and from the Euclidean viewpoint, Σ w is a spherical cap which is a graphof a function w on the disc D R . Then we prove that Σ u lies in the interior of thedomain determined by Σ w and the plane P , or in other words, u < w in Ω. Thiswill be proved by doing dilations p → tp , p ∈ R from the origin O of R . Afterthat, we have u M < w M , where u M and w M are the global maximum of u and w respectively. But now, we notice that w M depends only on the initial data, that is,from Ω, a and H .The first step is to show the existence of the surface Σ w . Consider (part of) theEuclidean sphere in R of radius m > w ( r ) = c + √ m − r , ≤ r ≤ R, where c = − mH, m = (1 − c ) + R , (16)0 < c < w ( R ) = 1. The mean curvature of Σ w is H with respect to theupwards orientation. If we see c as a parameter varying from 0 to 1, the value ofthe mean curvature of Σ w goes from 0 to − /R . It is not difficult to see that theright-hand side of (5) is less than 1 /H . Thus R < /H , that is, − /R < H .12efinitively, given H under the hypothesis of Theorem 1.2, we have assured theexistence of Σ w .We now do the argument of comparison between Σ u and Σ w by dilations. Bydilations of Σ w with respect to the origin O of R , namely, t Σ w and t > 1, wetake t sufficiently big so t Σ w does not meet Σ u . Then let t ց t Σ w with Σ u . Because an interior touching point is notpossible because both surfaces have the same (constant) mean curvature, then thefirst touching point occurs at t = 1, that is, when Σ w comes back to its initialposition and Σ w touches Σ u only at some boundary point of Ω. In particular, Σ u iscontained inside the domain determined by Σ w and the plane x = 1. Therefore, wededuce that the global maximum u M is less than the highest point of Σ w , namely, w M = c + m = m (1 − H ) and u M < m (1 − H ) . The above argument has been done for the value H , but it holds for τ H , τ ∈ [0 , H by τ H . We now prove the C estimates for the problems P τ .Fix − ≤ H < u τ the solution of P τ , τ ∈ [0 , u τ is τ H and τ H > H for τ ∈ [0 , τ ∈ [0 , u τ < w M = m (1 − H ) . (17)In order to use Theorem 1.1, and because H < a = 1, it suffices to prove m (1 − H ) < r H − H , (18)that is, m < √ H − H · (19)However, from (16), we deduce m = (1 + mH ) + R , which leads to m = H + p H + (1 − H ) R − H · By using (5), we conclude the desired inequality (19). Once we have obtained (18),Theorem 1.1 applies deducing an a priori estimate for | Du | . Hence, and together1317), we have proved the existence of M in (15). This completes the proof of Theorem1.2. Remark 3.2. We compare this result with Theorem 1.1 in [13]. In [13], the hy-pothesis requires that Ω is strongly convex in terms of the boundary data H , namely, κ > | H | . However in Theorem 1.2 we need that the domain Ω is strictly convex butit may contain regions where the curvature κ of ∂ Ω is close to . In contrast, thesize of the domain is small in relation to the value of H . In this section, for H < 1, we prove an estimate from below of the global maximumof a solution of (1)-(2). Theorem 4.1. Let Ω ⊂ R be a bounded strictly convex domain with curvature κ > . If H < and u is a solution of (1)-(2), then u M ≥ − Hκ , (20) where κ = max ∂ Ω κ . Firstly, we need to prove a minimum principle for the function Φ( x ; 1). The nextresult is inspired by other similar in the torsional creep problem ([17]). Proposition 4.2. Let Ω ⊂ R be a bounded strictly convex domain. Let H be areal number with H < . If u is a non-radial solution of (1)-(2), then the function Φ( x ; α ) attains its minimum value on ∂ Ω for any α ∈ [1 , .Proof. Following [16, inequality (2.15)], it was proved that if u is a solution of (1)-(2),then Φ( x ; α ) satisfies the next elliptic differential equation (cid:18) δ ij − u i u j | Du | (cid:19) Φ ij + ˜ W i Φ i = 2( α − α − (cid:16) − H p q + 1) + q (cid:17) ( q + 1) u , (21)14here ˜ W i is a vector function which is singular at the critical point of u . It is notdifficult to see that if α ∈ [1 , α − α − ≤ − H p q ) + q is alwaysnon-negative: this is immediate for H ≤ < H < 1, we use Lemma 2.3.By the Hopf Maximum Principle, and since the vector functions ˜ W i are singularat the critical points of u , we conclude that Φ( x ; α ) attains its minimum at theunique critical point of u or at a boundary point. Recall that by Theorem 2.1, thefunction u has exactly one critical point O .The proof of Proposition 4.2 finishes if we discard the case that the minimumoccurs at some critical point. The proof follows now the next steps.1. The function Φ( x ; α ) is not constant in Ω. The proof is by contradiction. If Φis constant, then the left-hand side of (21) is 0. If we see the right-hand sideof (21), the only possibility to be 0 is that α is 1 or 2. We prove that this isnot possible. We consider the case α = 1 because the argument for α = 2 issimilar. By the expression of Φ( x ; 1), we find that1 + | Du | (1 − H p | Du | ) u is constant, in particular, | Du | is constant along ∂ Ω. Since u = a along ∂ Ω,then ∂u/∂ n is constant along ∂ Ω. Then u is a solution of the Dirichlet problem(1)-(2) together the Neumann condition ∂u/∂ n = ct along ∂ Ω. A result ofSerrin establishes that Ω is a round disk and u is a radial function u = u ( r )([20]). This is a contradiction.2. After a change of coordinates, suppose that the critical point is O = (0 , u ( O ) = u ( O ) = 0. A new change of coordinates allows toassume u ( O ) = 0. Since u is a maximum of u , we have u ( O ) ≤ u ( O ) ≤ Claim: u ( O ) < u ( O ) < u ( O ) = 0 (the same argumentif u ( O ) = 0). Here we follow the same notation as in the proof of Theorem2.1. If the function v = u is constant in Ω, then u depends only on the15ariable x and the boundary condition (2) is impossible. Thus v is a nonconstant analytic function. Since v vanishes at O as well as v and v , thefunction v vanishes at O with a finite order m ≥ 1. Thus there exist at leasttwo nodal lines of v which form an equiangular system in a neighbour of O .We have proved in Theorem 2.1 that this is impossible because there existsexactly one nodal line of v .3. Finally we prove that Φ( x ; α ) can not attain its minimum at O . We know u ( O ) = u ( O ) = u ( O ) = 0. We need the first and second partial derivativesof Φ at O ∈ Ω. Following the notation employed in [16, p. 197], at the criticalpoint O we haveΦ i ( O , α ) = 0 , Φ ij ( O ; α ) = 2 g + 2 q g ′ ρ u ik u jk + αf u ij . Hence, and from (1),Φ ( O ; α ) = 21 − H u ( O ) + 2 αu ( O ) u ( O )Φ ( O ; α ) = 0Φ ( O ; α ) = 21 − H u ( O ) + 2 αu ( O ) u ( O ) . Because O is a minimum of Φ( x ; α ), we find that Φ ( O ; α ) ≥ ( O ; α ) ≥ 0. Since u ( O ) , u ( O ) < − H > − H u ( O ) + 2 αu ( O ) ≤ − H u ( O ) + 2 αu ( O ) ≤ . Then u ( O ) + u ( O ) = ∆ u ( O ) ≤ − α (1 − H ) u ( O ) · (22)Finally, equation (1) at O yields∆ u ( O ) = − − H ) u ( O ) · (23)16omparing (22) and (23), we conclude, α ≤ 1. Thus if α ∈ (1 , α = 1. Denote by O α the the minimum point of Φ( x ; α ).We have proved that O α lies in ∂ Ω for all α ∈ (1 , O must be a boundary point, because on the contrary, Φ( x ; α ) would beconstant for some parameter α ∈ (1 , α = 1 andthe proof of Proposition 4.2 is completed. Remark 4.3. In case that u is a radial solution, then u can be expressed as u ( r ) = − Hm + √ m − r , ≤ r ≤ R. It is not difficult to see that if we denote u ′ = u ′ ( r ) , then the functional Φ( x ; α ) = 1 + u ′ (1 − H √ u ′ ) u α is constant only when the parameter α is α = 1 .Proof of Theorem 4.1. First suppose that u is not a radial solution. By the proof ofProposition 4.2, we know that Φ( x ; 1) attains its minimum at some point Q ∈ ∂ Ω.Then if q M = | Du | ( Q ), we have1 + | Du | )(1 − H p | Du | ) u ≥ q M (1 − H p q M ) a . We evaluate this inequality at the only critical point O , obtaining (cid:18) u M a (1 − H ) (cid:19) ≥ q M (1 − H p q M ) · (24)On the other, ∂ Φ( Q ; 1) /∂ n ≤ Q is the minimum of Φ( x ; 1). If u n and u nn denote the first and second outward normal derivatives of u along ∂ Ω, by theexpression of Φ i (see [16, p. 197]), we deduce u n u nn (1 + u n )(1 − H p u n ) + u n u ≤ Q · (25)17n normal coordinates, and taking into account that u is constant along ∂ Ω, equation(1) along ∂ Ω becomes u nn (1 + u n ) / + κu n p u n = − u p u n − H ! . Combining this equation at Q with (25) and using that u n ≤ − a ≥ κ ( Q ) u n ( Q )1 − H p u n ( Q ) · Hence, and as κ ( Q ) ≤ κ , 1 a κ ≤ u n ( Q )(1 − H p u n ( Q )) · As | Du | = u n at Q , we obtain from (24) (cid:18) u M a (1 − H ) (cid:19) ≥ u n ( Q )(1 − H p u n ( Q )) ≥ a κ , proving the result.Suppose now that u is a radial solution. Then u ( r ) = c + √ m − r , where m > c = − Hm and 0 ≤ r ≤ R . Since m > R , u M = u (0) = (1 − H ) m > (1 − H ) R = 1 − Hκ · This proves the inequality (20) and completes the proof of Theorem 4.1. Acknowledgements The author has been partially supported by MEC-FEDER grant no. MTM2017-89677-P. 18 eferences [1] L. Barbu. On some estimates for a fluid surface in a short capillary tube. Appl.Math. Comp. (2013), 8192–8197.[2] L. Barbu, C. Enache. A minimum principle for a soap film problem in R . Z.Angew. Math. Phys. (2013), 321–328.[3] L. Bers. Local behavior of solutions of general elliptic equations. Comm. PureAppl. Math. (1955), 473–496.[4] S. 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