To end this section, we give a sketch of proof for the exterior problem, Theorem1.4.
We take the hyperbolic case as an example.
Theorem 4.2 Assume x : M → Hn+1\BHR is a compact immersed stable capillary hypersurface outside the hyperbolic ball BHR with constant mean curvature H ≥ 0 and constant contact angleθ ∈(0, π). Then x is totally umbilical.
Proof In this case, the differences occur thatx = −sinhRN¯ and the termq in the stability inequality (2.3) is given by
q= − 1
sinθcothR+cotθh(μ, μ).
By checking the proof of Proposition4.4, we see the Minkowski formula is
M
n(Va−sinhRcosθg(Y¯ a, ν))d A=
M
Hg(X¯ a, ν)d A.
We take the test function to be
ϕa=n(Va−sinhRcosθg(Y¯ a, ν))−Hg(X¯ a, ν). (4.47) Then
Mϕad A = 0. Also, by checking the proof of Proposition 4.5, we see that
¯∇μϕa =qϕaalong∂M. From Proposition4.6,ϕain (4.47) still satisfies (4.40). Then the proof is exactly the same as the interior problem, Theorem4.1.
5 Heintze–Karcher–Ros type inequality and Alexandrov theorem LetK =0 or ±1. Denote byM¯n+1(K)the space form with sectional curvatureK. As in previous section, we use the Poincaré ball model(Bn+1,g¯H)forM¯n+1(−1)and the model(Rn+1,g¯S)forM¯n+1(1).
In this section we consider anisometric embedding x : M → ¯Mn+1(K)into a ball B in a space form with free boundary, ı.e.,θ =π/2. To unify the notation, we use B to mean the unit ballBn+1 in the Euclidean case, the ballBHR with radius R
(R∈(0,∞))in the hyperbolic case and the ballBSRwith radiusR(R∈(0, π))in the spherical case. We denote =x(M). LetBbe decomposed by into two connected components. We choose one and denote it by. Denote byT the part of∂lying on
∂B. Thus,∂= ∪T.
We also unify the following notations:
V0=
andXais conformal vector field in (3.1), (4.5) and (4.45) in each case respectively.
We first prove other Minkowski type formulae.
Proposition 5.1 Let x : M → ¯Mn+1(K)be an embedded smooth hypersurface into B which meets B orthogonally. Letσk,k=1, . . . ,n be the k-th mean curvatures, i.e., the elementary symmetric functions acting on the principal curvatures. Then
Remark 5.1 Formula (5.2) is still true ifxis only an immersion.
Proof Due to the perpendicularity condition,μ = ¯N. Since Xa ⊥ ¯N along∂B, we seeXa⊥μalong∂ . From the conformal property, we have
divg¯Xa=(n+1)Va. Integrating it overand using Stokes’ theorem, we have
(n+1)
This is (5.1). Denote by XaT the tangential projection of Xa on . From above we know thatXTa ⊥μalong∂ . Let{eα}nα=1be an orthonormal frame on . From the conformal property, we have that
1 2
∇α(XaT)β + ∇β(XaT)α =Vagαβ−hαβg(X¯ a, ν). (5.3)
(cf. (3.5)). DenoteTk−1(h)= ∂σ∂hk the Newton transformation. MultiplyingTkαβ−1(h) In fact, sinceμis a principal direction ofh, it is also a principal direction of the Newton tensorTk−1ofh, which implies thatTk−1(XaT, μ)=0. The proof is completed.
Next we prove a Heintze–Karcher–Ros type inequality. In order to prove the Heintze–Karcher–Ros type inequality, we need a generalized Reilly formula, which has been proved by Qiu and Xia [48], Li and Xia [33,34].
Theorem 5.1 [33,48]Letbe a bounded domain in a Riemannian manifold(M¯,g¯) with piecewise smooth boundary∂. Assume that∂is decomposed into two smooth pieces∂1and∂2with a common boundary. Let V be a non-negative smooth Remark 5.2 The formula (5.4) here is a bit different with that in [33]. We do not do integration by parts on∂in the last step of the proof as [33,48].
Theorem 5.2 Let x : M → ¯Mn+1(K)be an embedded smooth hypersurface into B with∂ ⊂∂B. Assume lies in a half ball
Ba+= {Va ≥0} = {x,a ≥0}.
If has positive mean curvature, then
Moreover, equality in(5.5)holds if and only if is a spherical cap which meets∂B orthogonally.
Proof Recall∂= ∪T, whereT is the boundary part lying in∂B. See Figure1.
Let f be a solution of the mixed boundary value problem
⎧⎨
⎩
¯ f +K(n+1)f =1 in, f =0 on ,
VafN¯ − f(Va)N¯ =0 onT. (5.6) Since the existence of (5.6) has its own interest, we give a proof in Appendix A. From the Appendix we have f ∈W1,2()satisfying (5.6) in the weak sense, i.e., f =0 on follows from standard linear elliptic PDE theory.
From the fact ¯∇2Va= −K Vag, we see¯
¯Va+K(n+1)Va=0, ¯Vag¯− ¯∇2Va+VaRic=0. (5.8) By using Green’s formula, (5.6) and (5.8), we have
Using Hölder’s inequality for the RHS of (5.9), we have
Next, we use formula (5.4) in our situation with V = Va. Because of (5.8) and which implies that term II vanishes. We take the hyperbolic case for instance. First,
∂Bis umbilical inHn+1with all principal curvatures cothR = 12R+RR2R. Second, since Thus (5.12) follows for the hyperbolic case. For other two cases (5.12) follows simi-larly.
Taking account of the above information in (5.11), we obtain n
Combining (5.10) and (5.13), we conclude (5.5).
We are remained to consider the equality case. If a spherical cap which meets
∂Borthogonally, the Minkowski formula (5.1) implies that equality in (5.5) holds, for has constant mean curvature. Conversely, if equality in (5.5) holds, then equality in (5.11) holds, which implies that ¯∇2f+K fg¯=0 holds in. Restricting this equation on , in view of f =0 on we know that must be umbilical. Thus it is a spherical cap andis the intersection of two geodesic balls. It is easy to show the contact angle
must be π2. Indeed, we have an explicit form for f:
This implies that these two intersecting geodesic balls are perpendicular. The proof is
completed.
As an application we give an integral geometric proof of the Alexandrov Theorem, which was obtained by Ros and Souam in [53] by using the method of moving plane.
Our proof has the same flavor of Reilly [49] and Ros [51], see also [33,48].
Theorem 5.3 Let x : M → ¯Mn+1(K)be an embedded smooth CMC hypersurface into B which meets B orthogonally. Assume lies in a half ball. Then is either a spherical cap or part of totally geodesic hyperplane.
Remark 5.3 The condition that lies in a half ball cannot be removed because there are other embedded CMC hypersurfaces, like the Denaulay hypersurfaces in a ball which meets∂Borthogonally which does not lie in a half ball.
Proof We take the hyperbolic case for instance.
We claim first that the constant mean curvature H is non-negative. To prove this claim, let the totally geodesic hyperplane{x,a =0}move upward alongadirection along the totally geodesic foliation ofHn+1, until it touches at some pointpat a first time. It is clear thatH=H(p)≥0. IfH =0, then the boundary point lemma or the interior maximum principle implies that must be some totally geodesic hyperplane.
Next we assumeH >0. In this case the two Minkowski formulae (5.1) and (5.2) yield
The above equation means, for the constant mean curvature hypersurface , the Heintze–Karcher–Ros inequality is indeed an equality. By the classification of equal-ity case in (5.5), we conclude that must be a spherical cap. The proof is completed.
Using the higher order Minkowski formulae (5.2) and the Heintze–Karcher–Ros inequality (5.5), we can also prove the rigidity when has constant higher order mean curvatures or mean curvature quotients as Ros [51] and Koh and Lee [30].
Theorem 5.4 Let x:M → ¯Mn+1(K)be an isometric immersion into a ball with free boundary. Assume that lies in a half ball.
(i) Assume x is an embedding and has nonzero constant higher order mean curva-turesσk,1≤k≤n. Then is a spherical cap.
(ii) Assume x has nonzero constant curvature quotient, i.e., σk
σl =const., σl >0, 1≤l<k≤n. Then is a spherical cap.
Note that in Theorem5.4(ii), we do not need assume the embeddedness ofx, since in the proof we need only use the higher order Minkowski formulae (5.2) (without use of the Heintze–Karcher–Ros inequality), which is true for immersions, see Remark5.1.
On the other hand, the condition of embeddedness may not be removed in Theorem 5.4(i) in view of Wente’s counterexample. The proof is similar, we leave it to the interested reader.
Appendix A. Existence of weak solution of (5.6)
In this Appendix we discuss the existence of weak solution of (5.6), namely
⎧⎨
⎩
¯ f +K(n+1)f =1 in, f =0 on , VafN¯ − f(Va)N¯ =0 onT
(A.1)
in the weak sense (5.7). Since our Robin boundary condition has a different sign, i.e.
(Va)N¯
Va <0, we can not apply known results about the existence for mixed boundary problem, for example, [37]. Here we use the Fredholm alternative theorem. In order to use it, we have to show that
⎧⎨
⎩
φ¯ +K(n+1)φ=0 in, φ=0 on , VaφN¯ −φ(Va)N¯ =0 onT
(A.2)
has only the trivial solutionφ =0 inW1,2(). For a general domainwe do not know how to prove it. Nevertheless, we can prove it for domains under the conditions given in Theorem5.2.
Proposition A.1 Let x : M → ¯Mn+1(K)be an embedded smooth hypersurface into B with∂ ⊂∂B. Assume lies in a half ball Ba+= {Va≥0} = {x,a ≥0}.If has positive mean curvature, then(A.2)has only the trivial solutionφ=0.
Proof Letφ∈W1,2()be a weak solution of (A.2), i.e.,φ=0 on and
[ ¯g(¯∇φ, ¯∇ϕ)−K(n+1)φϕ]d=
T φϕd A,
for allϕ ∈ W1,2()withϕ =0 on . The classical elliptic PDE theory gives the regularityφ ∈ C∞(\).¯ Now we can use the Reilly type formula, (5.4), as in the proof of Theorem5.2. Replacing f byφin (5.4), using (5.12), we get
−
Va| ¯∇2φ+Kφg|¯ 2d=
VaH(φν)2d A.
SinceVaandHis positive, it follows that
¯∇2φ+Kφg¯=0 in (A.3)
andφν =0 on . From (A.3) and Remark4.2, we seeφmust be of form φ=
n+1
i=0
biVi,
forbi ∈R,i =0,1, . . .n+1. By checking the boundary conditionVaφN¯−φ(Va)N¯ = 0 onT, we seeb0=0. Moreover, sinceφ=0 on , must be the totally geodesic hyperplane through the origin if one ofbi = 0, which is a contraction to having
positive mean curvature. We get the assertion.
With this Proposition one can use the Fredholm alternative to get a unique weak solution of (A.1).
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