Introduction and statement of the main result
Let Ω \Omega Ω M M M M = S 2 , H 2 M=\mathbb{S}^{2},\mathbb{H}^{2} M = S 2 , H 2 1 , − 1 1,-1 1 , − 1 u u u
{ − Δ u = f ( u ) , in Ω , u = 0 , on ∂ Ω . (1) \begin{cases}
-\Delta u=f(u), & \text{in }\Omega,\\
u=0, & \text{on }\partial\Omega. \tag*{(1)}
\end{cases} { − Δ u = f ( u ) , u = 0 , in Ω , on ∂ Ω. (1) Here f : [ 0 , + ∞ ) → R f:[0,+\infty)\to\mathbb{R} f : [ 0 , + ∞ ) → R C 1 C^{1} C 1 f ( 0 ) ≥ 0 f(0)\ge0 f ( 0 ) ≥ 0 positive solutions of (1) that are semi-stable , meaning that the (linear) Schrödinger operator
− Δ − f ′ ( u ) I (2) -\Delta-f'(u)I \tag*{(2)} − Δ − f ′ ( u ) I (2) is non-negative in Ω \Omega Ω
Two classical problems that fall in this setting are the torsion problem , i.e.,
f ( s ) = 1 (3) f(s)=1 \tag*{(3)} f ( s ) = 1 (3) and the Dirichlet eigenvalue problem (when we consider the first eigenfunction ), i.e.,
f ( s ) = λ 1 s , (4) f(s)=\lambda_{1}s, \tag*{(4)} f ( s ) = λ 1 s , (4) where λ 1 \lambda_{1} λ 1 Ω \Omega Ω
2020 Mathematics Subject Classification. 58J32, 35J61, 58J20, 58J05.
The second author acknowledges support of INdAM - GNAMPA.
The third author acknowledges support of the project “Perturbation problems and asymptotics for elliptic differential equations: variational and potential theoretic methods” funded by the European Union – Next Generation EU and by MUR-PRIN-2022SENJZ3 and from the project “Analisi Geometrica e Teoria Spettrale su varietà Riemanniane ed Hermitiane” of the INdAM GNSAGA.
The first author acknowledges support of INdAM - GNAMPA Project, CUP E53C25002010001.
The first author acknowledges support of the project e.INS- Ecosystem of Innovation for Next Generation Sardinia (cod. ECS 00000038) funded by the Italian Ministry for Research and Education (MUR) under the National Recovery and Resilience Plan (NRRP) - MISSION 4 COMPONENT 2, “From research to business” INVESTMENT 1.5, “Creation and strengthening of Ecosystems of innovation” and construction of “Territorial R&D Leaders” and of the project Next Generation EU-CUP-J55F21004240001, DM 737-2021, risorse 2022 – 2023.
Discussion
The study of critical points of solutions of elliptic equations is a vast topic. Typical questions involve estimates on the number of critical points and their location. Some introductory surveys on the subject are [15 , 26 ] . The majority of available results concern the torsion problem (3) and the eigenvalue problem (4) , especially for convex domains in R n \mathbb{R}^n R n f f f
The first Dirichlet eigenfunction
This is perhaps the most studied case. The uniqueness of the critical point in the case of convex Euclidean domains is well-known [1 , 4 , 7 ] . This result (as most of the results for problems (3) and (4) ) is obtained in a somehow indirect way, namely it is a consequence of the fact that the first eigenfunction is log-concave, joined with the interior regularity properties of the eigenfunction.
For convex domains of the sphere S n \mathbb{S}^n S n [24 ] , again as a consequence of log-concavity estimates on the eigenfunction (see also [8 , 18 ] ). In the case of convex domains of S 2 \mathbb{S}^2 S 2 π 2 \frac{\pi}{2} 2 π [16 ] . Log-concavity of the eigenfunction for convex domains in surfaces of positive curvature has been studied in [21 ] . A new point of view has been taken in [22 ] , where log-concavity estimates are investigated when a given metric undergoes a conformal deformation.
Concerning the hyperbolic space H n \mathbb{H}^n H n n = 2 n = 2 n = 2 [16 ] that the first eigenfunction on horoconvex domains has exactly one, non-degenerate critical point. The result seems to be optimal in the sense that there exist examples of convex domains (which are not horoconvex) for which the first eigenfunction has at least two critical points [30 ] , or for which the level sets are non-convex [3 ] . A log-concavity estimate has been recently proved [22 ] in the case of horoconvex domains of H 2 \mathbb{H}^2 H 2
If we leave the realm of convex domains, estimates on the number of critical points for the first Dirichlet eigenfunction become more difficult or even unavailable. In the case of manifolds, we mention [11 , 12 ] where the authors are able to find metrics on a given Riemannian manifold for which the first eigenfunction has an arbitrary number of critical points.
The torsion function
Surprisingly enough, there are fewer results on the critical points of the torsion functions compared to the first eigenfunction. A seminal paper is [27 ] where the author proves the convexity of the level lines of the torsion function on planar convex domains, and the uniqueness and non-degeneracy of the critical point. On the other hand, if we drop the convexity assumption, we have no bounds on the number of critical points, even if a domain is close in a suitable sense to a convex one [14 ] . The result of [27 ] has been extended to any dimension in [23 ] .
In [16 ] it has been proved that the torsion function has a unique, non-degenerate critical point for horoconvex domains of H 2 \mathbb{H}^2 H 2 π 2 \frac{\pi}{2} 2 π [22 ] the authors prove the quasi-concavity of the torsion function for convex spherical domains of small diameter (less than 2 arctan ( 1 / 5 ) 2\arctan(1/5) 2 arctan ( 1/5 ) 1 2 \frac{1}{2} 2 1 n ≥ 3 n \ge3 n ≥ 3
In [14 ] the authors exhibit a family of star-shaped planar domains for which the torsion function has an arbitrary number of critical points. These domains are “not too far” from being convex.
General non-linearity
For a general non-linearity f f f [13 ] imply that for a domain in R n \mathbb{R}^{n} R n (1) is unique and non-degenerate. Again, the statement on the critical point is an indirect result which follows from the symmetry properties of a solution derived in [13 ] .
A second relevant result is [6 ] where the authors confine to strictly convex planar domains, dropping symmetry conditions, and requiring positivity and semi-stability of solutions. Under these hypotheses, they prove uniqueness and non-degeneracy of the critical point. Strict convexity has been relaxed in [9 ] . We remark that in [6 , 9 ] the study of the critical points has been attacked directly . Roughly speaking, the combinatorial properties of critical points of smooth functions on differentiable manifolds are related to the topology of the manifold though its Euler characteristic . However, classical theorems in differential topology (as the Poincaré-Hopf Theorem, see Theorem 2.1 ) imply that the topological information on the manifolds gives an information on a signed sum of the critical points, the sign depending on the nature of the critical points. With no other information, the research for estimates on the number of critical points is a very difficult task. Moreover, the difficulties of taking a direct approach increase considerably in dimension n ≥ 3 n \ge3 n ≥ 3 [15 , 26 ] for a deeper discussion on this point.
For domains of H 2 \mathbb{H}^{2} H 2 S 2 \mathbb{S}^{2} S 2 [16 ] . Namely, in [16 ] it has been proved that positive, semi-stable solutions of (1) have a unique-non degenerate critical point in the following cases:
Ω \Omega Ω H 2 \mathbb{H}^{2} H 2
Ω \Omega Ω S 2 \mathbb{S}^{2} S 2 π 2 \frac{\pi}{2} 2 π
The results of [16 ] are in the spirit of [6 ] , in the sense that they descend from an application of the Poincaré-Hopf Theorem to the counting of the index of a suitable vector field related to ∇ u \nabla u ∇ u
The results of [16 ] are quite good in the case of H 2 \mathbb{H}^{2} H 2 [3 , 30 ] , however the restriction on the diameter for domains in S 2 \mathbb{S}^{2} S 2 H 2 \mathbb{H}^{2} H 2 S 2 \mathbb{S}^{2} S 2 [16 ] to a class of domains larger than horoconvex domains.
Statement of the main results
In order to state the main results of the paper we recall the definition of uniformly star-shaped domain.
Definition 1.1. We say that a smooth domain Ω \Omega Ω S 2 \mathbb{S}^{2} S 2 H 2 \mathbb{H}^{2} H 2 uniformly star-shaped with respect to some point P ∈ Ω P \in\Omega P ∈ Ω
for any p ∈ ∂ Ω p \in\partial\Omega p ∈ ∂ Ω P P P p p p Ω \Omega Ω
there exists δ > 0 \delta> 0 δ > 0 ⟨ ν , e ⃗ r ⟩ ≥ δ \langle\nu,\vec{e}_{r}\rangle\ge\delta ⟨ ν , e r ⟩ ≥ δ ∂ Ω \partial\Omega ∂ Ω ν \nu ν ∂ Ω \partial\Omega ∂ Ω e ⃗ r \vec{e}_{r} e r r r r P P P
In the case of S 2 \mathbb{S}^{2} S 2 − P ∉ Ω ‾ -P \notin\overline{\Omega} − P ∈ / Ω
The first result concerns spherical domains . In the spherical case it is customary to denote the radial coordinate with respect to a point P P P θ \theta θ r r r
Theorem 1.2. Let Ω \Omega Ω S 2 \mathbb{S}^2 S 2 P ∈ Ω P \in\Omega P ∈ Ω ν \nu ν ∂ Ω \partial\Omega ∂ Ω k Ω k_{\Omega} k Ω ∂ Ω \partial\Omega ∂ Ω ν \nu ν
cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ > 0 on ∂ Ω , (5) \cos(\theta)k_{\Omega}+\sin(\theta)\langle\nu,\vec{e}_{\theta}\rangle>0 \quad\text{on }\partial\Omega, \tag*{(5)} cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e θ ⟩ > 0 on ∂ Ω , (5) where θ \theta θ P P P e ⃗ θ \vec{e}_{\theta} e θ (1) has a unique non-degenerate critical point, which is a maximum.
Finally (5) is sharp in the sense that, for any fixed n ∈ N n \in\mathbb{N} n ∈ N b 0 > 0 b_0>0 b 0 > 0 { Ω b } b ∈ ( 0 , b 0 ) \{\Omega_b\}_{b\in(0,b_0)} { Ω b } b ∈ ( 0 , b 0 )
lim b → 0 ( cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ ) ≥ 0 on ∂ Ω b (6) \lim_{b\to0}\left(\cos(\theta)k_{\Omega_b}+\sin(\theta)\langle\nu,\vec{e}_{\theta}\rangle\right)\geq0 \quad\text{on }\partial\Omega_b \tag*{(6)} b → 0 lim ( cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e θ ⟩ ) ≥ 0 on ∂ Ω b (6) and the solution u b u_b u b
{ − Δ u = 1 , in Ω b , u = 0 , on ∂ Ω b . (7) \begin{cases}
-\Delta u=1, & \text{in }\Omega_b,\\
u=0, & \text{on }\partial\Omega_b. \tag*{(7)}
\end{cases} { − Δ u = 1 , u = 0 , in Ω b , on ∂ Ω b . (7) admits at least n n n b ∈ ( 0 , b 0 ) b\in(0,b_0) b ∈ ( 0 , b 0 )
The geometric condition (5) involves both the geodesic curvature of the boundary and the constant δ \delta δ i i ii ii (5) takes into account the possibility that a domain exceeds the equator. Note that all convex domains satisfy the hypothesis of Theorem 1.2 . In Section 6 we provide examples of non-convex domains and domains exceeding the equator satisfying the assumptions of Theorem 1.2 .
The second result concerns hyperbolic domains . It extends the result of [16 ] to class of domains larger than horoconvex domains.
Theorem 1.3. Let Ω \Omega Ω H 2 \mathbb{H}^2 H 2 P ∈ Ω P\in\Omega P ∈ Ω ν \nu ν ∂ Ω \partial\Omega ∂ Ω k Ω k_{\Omega} k Ω ∂ Ω \partial\Omega ∂ Ω ν \nu ν
cosh ( r ) k Ω − sinh ( r ) ⟨ ν , e ⃗ r ⟩ > 0 on ∂ Ω , (8) \cosh(r)k_{\Omega}-\sinh(r)\langle\nu,\vec{e}_{r}\rangle>0 \quad\text{on }\partial\Omega, \tag*{(8)} cosh ( r ) k Ω − sinh ( r ) ⟨ ν , e r ⟩ > 0 on ∂ Ω , (8) where r r r P P P e ⃗ r \vec{e}_{r} e r (1) has a unique non-degenerate critical point, which is a maximum. Finally (8) is sharp in the sense that, for any fixed n ∈ N n\in\mathbb{N} n ∈ N b 0 > 0 b_0>0 b 0 > 0 { Ω b } b ∈ ( 0 , b 0 ) \{\Omega_b\}_{b\in(0,b_0)} { Ω b } b ∈ ( 0 , b 0 )
lim b → 0 ( cosh ( r ) k Ω b − sinh ( r ) ⟨ ν , e ⃗ r ⟩ ) ≥ 0 on ∂ Ω b (9) \lim_{b\to0}\left(\cosh(r)k_{\Omega_b}-\sinh(r)\langle\nu,\vec{e}_{r}\rangle\right)\geq0 \quad\text{on }\partial\Omega_b \tag*{(9)} b → 0 lim ( cosh ( r ) k Ω b − sinh ( r ) ⟨ ν , e r ⟩ ) ≥ 0 on ∂ Ω b (9) and the solution u b u_b u b
{ − Δ u = 1 , in Ω b , u = 0 , on ∂ Ω b . (10) \begin{cases}
-\Delta u=1, & \text{in }\Omega_b,\\
u=0, & \text{on }\partial\Omega_b. \tag*{(10)}
\end{cases} { − Δ u = 1 , u = 0 , in Ω b , on ∂ Ω b . (10) admits at least n n n b ∈ ( 0 , b 0 ) b\in(0,b_0) b ∈ ( 0 , b 0 )
The geometric condition (8) , as in the spherical case, involves both the geodesic curvature of the boundary and the constant δ \delta δ (8) (hence we recover the result in [16 ] ); moreover, condition (8) is satisfied also by certain convex but not horoconvex domains, see Section 6.
The geometric conditions (5) and (8) have the common form
V k Ω − ∂ ν V > 0 (11) V k_{\Omega}-\partial_{\nu}V>0 \tag*{(11)} V k Ω − ∂ ν V > 0 (11) with
V = { cos ( θ ) in S 2 1 in R 2 cosh ( r ) in H 2 V =
\begin{cases}
\cos(\theta) & \text{in } \mathbb{S}^{2} \\
1 & \text{in } \mathbb{R}^{2} \\
\cosh(r) & \text{in } \mathbb{H}^{2}
\end{cases} V = ⎩ ⎨ ⎧ cos ( θ ) 1 cosh ( r ) in S 2 in R 2 in H 2 Now, let ( M , g ) (M,g) ( M , g ) n n n V V V ( M , g , V ) (M,g,V) ( M , g , V ) static triple if
Δ V g − D 2 V + V Ric ≡ 0 in M \Delta V g - D^{2}V + V\operatorname{Ric} \equiv0 \text{ in } M Δ V g − D 2 V + V Ric ≡ 0 in M The space forms S 2 , R 2 , H 2 \mathbb{S}^{2},\mathbb{R}^{2},\mathbb{H}^{2} S 2 , R 2 , H 2 V = cos ( θ ) , 1 , cosh ( r ) V=\cos(\theta),1,\cosh(r) V = cos ( θ ) , 1 , cosh ( r ) V V V P P P V V V static potential for the manifold. The study of Riemannian manifolds admitting static potentials has deep roots in Riemannian geometry and in particular, in general relativity. In this context, it has been pointed out that one natural notion of convexity for a domain Ω \Omega Ω κ i ≥ ∂ ν V V \kappa_i \ge\frac{\partial_\nu V}{V} κ i ≥ V ∂ ν V ∂ Ω \partial\Omega ∂ Ω κ i \kappa_i κ i i = 1 , … , n − 1 i=1,\ldots,n-1 i = 1 , … , n − 1 V > 0 V>0 V > 0 ∂ Ω \partial\Omega ∂ Ω static convexity , in the work of Brendle and Wang [5 ] on the Gibbons-Penrose inequality in Schwarzschild spacetime. Static convexity has since appeared naturally in many other related contexts, for instance in the Minkowski and Alexandrov-Fenchel type inequalities [25 ] , and in the study of geometric flows [19 ] , where static convexity is shown to be preserved along suitable curvature flows in hyperbolic space.
Condition (11) can be also interpreted as follows. Suppose that V > 0 V>0 V > 0 Ω \Omega Ω V κ Ω − ∂ ν V V\kappa_\Omega-\partial_\nu V V κ Ω − ∂ ν V ∂ Ω \partial\Omega ∂ Ω V − 2 g V^{-2}g V − 2 g g g g
For a domain Ω ⊂ H 2 \Omega\subset\mathbb{H}^{2} Ω ⊂ H 2 (8) is telling that the geodesic curvature of ∂ Ω \partial\Omega ∂ Ω cosh ( r ) − 2 \cosh(r)^{-2} cosh ( r ) − 2
On the other hand, if Ω ⊂ S 2 \Omega\subset\mathbb{S}^{2} Ω ⊂ S 2 (5) amounts to the positivity of the geodesic curvature of ∂ Ω \partial\Omega ∂ Ω Ω \Omega Ω S + 2 = { θ < π / 2 } \mathbb{S}^{2}_{+}=\{\theta<\pi/2\} S + 2 = { θ < π /2 } Ω ∩ S + 2 \Omega\cap\mathbb{S}^{2}_{+} Ω ∩ S + 2 S − 2 ∖ Ω ‾ \mathbb{S}^{2}_{-}\setminus\overline{\Omega} S − 2 ∖ Ω S − 2 = { θ > π / 2 } \mathbb{S}^{2}_{-}=\{\theta>\pi/2\} S − 2 = { θ > π /2 }
Equivalently, when V > 0 V>0 V > 0 (11) can be read as the positivity of a weighted curvature. More precisely, on an n n n ( M , g , V d v g ) (M,g,V\,\mathrm{d}v_g) ( M , g , V d v g ) Σ \Sigma Σ H H H H − ∂ ν V V H-\frac{\partial_\nu V}{V} H − V ∂ ν V [20 ] . In our case, the weight is the static potential of the manifold.
Condition (11) suggests that the results of Theorems 1.2 and 1.3 may be extended to static convex domains in static triples. However, such an extension turns out to be far from straightforward.
The second part of the statements of Theorems 1.2 and 1.3 shows that the geometric conditions (5) and (8) , which ensure the uniqueness of the critical point of u u u f ≡ 1 f\equiv1 f ≡ 1 (1) ) and construct examples of solutions to (1) in a family of geodesically star-shaped domains of M = S 2 M=\mathbb{S}^{2} M = S 2 H 2 \mathbb{H}^{2} H 2 (6) and (9) .
This type of examples first appeared in [14 ] and [9 ] , where the authors demonstrated the optimality of the convexity assumption on Ω \Omega Ω [6 ] when Ω ⊂ R 2 \Omega\subset\mathbb{R}^{2} Ω ⊂ R 2 [14 ] was later extended in [10 ] to the setting of a general d d d M M M f ( u ) f(u) f ( u )
Here, we construct more explicit examples of solutions to (1) , which preserve all the key features of those in [14 ] and [10 ] , directly within S 2 \mathbb{S}^{2} S 2 H 2 \mathbb{H}^{2} H 2 [2 ] .
We remark that the fact that the limiting configuration is built around a geodesic appears to be a necessary condition for this type of construction, and we believe that it holds in a more general setting. We intend to investigate this phenomenon in future papers.
Strategy of the proof
We give here some ideas about the proof of Theorem 1.2 1.2 1.2 spherical case . The proof of Theorem 1.3 1.3 1.3
The proof of Theorem 1.2 1.2 1.2 [16 ] , however it differs in many aspects. As in [16 ] , the most important tool that we use is the Poincaré-Hopf Theorem, which relates the index of the zeros of any vector field V V V Theorem 2.1 . In our setting, we are dealing with contractible sets, hence Theorem 2.1 gives
∑ i Ind p i V = 1 , (12) \sum_i \operatorname{Ind}_{p_i} V = 1, \tag*{(12)} i ∑ Ind p i V = 1 , (12) where the sum runs on the isolated zeros of V V V (12) with V = ∇ u V = \nabla u V = ∇ u u u u ∇ u \nabla u ∇ u [16 ] , we consider here an auxiliary vector field V V V (12) . We then relate the zeros of V V V ∇ u \nabla u ∇ u
The vector field V V V K 1 , K 2 K_{1}, K_{2} K 1 , K 2 ∇ u \nabla u ∇ u K 1 , K 2 K_{1}, K_{2} K 1 , K 2 (17) and (18) ). Extrinsically, K 1 , K 2 K_{1}, K_{2} K 1 , K 2 S 2 \mathbb{S}^{2} S 2 z ∂ x − x ∂ z z\partial_x-x\partial_z z ∂ x − x ∂ z z ∂ y − y ∂ z z\partial_y-y\partial_z z ∂ y − y ∂ z R 3 \mathbb{R}^{3} R 3
The proof of Theorem 1.2 1.2 1.2
i) We first observe that critical points of u u u p p p u u u Z : Ω → R Z:\Omega\to\mathbb{R} Z : Ω → R p p p p p p Z Z Z Z Z Z p p p (5) in Theorem 1.2 1.2 1.2 Z Z Z ∂ Ω \partial\Omega ∂ Ω Z Z Z − Δ − f ′ ( u ) -\Delta-f'(u) − Δ − f ′ ( u ) 0 0 0 Ω \Omega Ω − Δ − f ′ ( u ) -\Delta-f'(u) − Δ − f ′ ( u ) Ω \Omega Ω P P P Ω \Omega Ω
ii) The previous step allows to prove that the critical points of u u u V V V V V V (17) , (18) ). The proof of the second implication is in the same spirit of point i ) i) i ) Z Z Z [16 ] : the vector field V V V u u u
iii) We relate the index of p p p ∇ u \nabla u ∇ u p p p V V V V V V ∇ u \nabla u ∇ u Ind p V = 1 \operatorname{Ind}_{p}V=1 Ind p V = 1 Ind p V = 1 \operatorname{Ind}_{p}V=1 Ind p V = 1 V V V
iv) Finally we prove that condition (1.5) in Theorem 1.2 implies ⟨ V , ν ⟩ > 0 \langle V,\nu\rangle>0 ⟨ V , ν ⟩ > 0 ∂ Ω \partial\Omega ∂ Ω V V V
1 ≤ # { critical points of u } ≤ ∑ i Ind p i V = χ ( Ω ) = 1 , 1 \le\#\{\text{critical points of }u\} \le\sum_i \operatorname{Ind}_{p_i}V=\chi(\Omega)=1, 1 ≤ # { critical points of u } ≤ i ∑ Ind p i V = χ ( Ω ) = 1 , which gives the uniqueness of the zero of V V V u u u
Some of the previous points make use of ideas introduced by Cabrè and Chanillo in [6 ] .
The present paper is organized as follows. In § 2 we collect a few preliminary results on vector fields on the sphere, on the Poincaré-Hopf Theorem and on the behavior of solutions of elliptic PDE’s near singular zeros. The proof of the first part of Theorem 1.2 (concerning the uniqueness of the critical point under condition (1.5)) is presented in § 3 , while the proof of the first part of Theorem 1.3 (concerning the uniqueness of the critical point under condition (1.8)) is presented in § 4 . In § 5 we prove the optimality of conditions (1.5) and (1.8), completing the proofs of Theorems 1.2 and 1.3 . Finally, in § 6 we discuss some classes of domains for which Theorems 1.2 and 1.3 apply. In particular, in Subsection 6.4 we provide information on the location of the critical point for convex spherical domains of diameter larger than π / 2 \pi/2 π /2
Preliminaries and notation
In this section we collect a few preliminary results and we fix the notation.
Vector fields
In this paper we consider as ambient manifolds the sphere S 2 \mathbb{S}^{2} S 2 H 2 \mathbb{H}^{2} H 2 1 1 1 − 1 -1 − 1 S 2 = { ( x , y , z ) ∈ R 3 : x 2 + y 2 + z 2 = 1 } \mathbb{S}^{2}=\{(x,y,z)\in\mathbb{R}^{3}:x^{2}+y^{2}+z^{2}=1\} S 2 = {( x , y , z ) ∈ R 3 : x 2 + y 2 + z 2 = 1 } S 2 \mathbb{S}^{2} S 2 R 3 \mathbb{R}^{3} R 3 S 2 \mathbb{S}^{2} S 2 R 3 \mathbb{R}^{3} R 3 ⟨ ⋅ , ⋅ ⟩ \langle\cdot,\cdot\rangle ⟨ ⋅ , ⋅ ⟩ S 2 \mathbb{S}^{2} S 2 H 2 \mathbb{H}^{2} H 2
Killing vector fields
We recall that a Killing vector field K K K M M M K K K
⟨ ∇ X K , Y ⟩ + ⟨ ∇ Y K , X ⟩ = 0 (13) \langle\nabla_XK,Y\rangle+\langle\nabla_YK,X\rangle=0 \tag*{(13)} ⟨ ∇ X K , Y ⟩ + ⟨ ∇ Y K , X ⟩ = 0 (13) for all vector fields X , Y X,Y X , Y K K K div K = 0 \operatorname{div}K=0 div K = 0 K K K K K K u : M → R u:M\to\mathbb{R} u : M → R
Δ ⟨ K , ∇ u ⟩ = ⟨ K , ∇ Δ u ⟩ , (14) \Delta\langle K,\nabla u\rangle=\langle K,\nabla\Delta u\rangle, \tag*{(14)} Δ ⟨ K , ∇ u ⟩ = ⟨ K , ∇Δ u ⟩ , (14) that is, K K K [29 ] for more information on Killing vector fields.
We consider on S 2 \mathbb{S}^{2} S 2 K 1 K_{1} K 1 K 2 K_{2} K 2 z ∂ x − x ∂ z z\partial_{x}-x\partial_{z} z ∂ x − x ∂ z z ∂ y − y ∂ z z\partial_{y}-y\partial_{z} z ∂ y − y ∂ z R 3 \mathbb{R}^{3} R 3 S 2 \mathbb{S}^{2} S 2 K i K_{i} K i S 2 \mathbb{S}^{2} S 2 ( θ , ϕ ) ∈ [ 0 , π ] × [ 0 , 2 π ] (\theta,\phi)\in[0,\pi]\times[0,2\pi] ( θ , ϕ ) ∈ [ 0 , π ] × [ 0 , 2 π ] S 2 \mathbb{S}^{2} S 2 θ \theta θ ( 0 , 0 , 1 ) (0,0,1) ( 0 , 0 , 1 ) K 1 , K 2 K_{1},K_{2} K 1 , K 2
K 1 = cos ( ϕ ) ∂ θ − cot ( θ ) sin ( ϕ ) ∂ ϕ K_{1}=\cos(\phi)\partial_{\theta}-\cot(\theta)\sin(\phi)\partial_{\phi} K 1 = cos ( ϕ ) ∂ θ − cot ( θ ) sin ( ϕ ) ∂ ϕ ( 0 , ± 1 , 0 ) (0,\pm1,0) ( 0 , ± 1 , 0 )
K 2 = sin ( ϕ ) ∂ θ + cot ( θ ) cos ( ϕ ) ∂ ϕ K_{2}=\sin(\phi)\partial_{\theta}+\cot(\theta)\cos(\phi)\partial_{\phi} K 2 = sin ( ϕ ) ∂ θ + cot ( θ ) cos ( ϕ ) ∂ ϕ ( ± 1 , 0 , 0 ) (\pm1,0,0) ( ± 1 , 0 , 0 )
We consider the Poincaré disk model for H 2 \mathbb{H}^{2} H 2 D D D R 2 \mathbb{R}^{2} R 2 ( x , y ) (x,y) ( x , y ) 4 ( 1 − x 2 − y 2 ) 2 ( d x 2 + d y 2 ) \frac{4}{(1-x^{2}-y^{2})^{2}}(dx^{2}+dy^{2}) ( 1 − x 2 − y 2 ) 2 4 ( d x 2 + d y 2 ) H 2 \mathbb{H}^{2} H 2 K 1 K_{1} K 1 K 2 K_{2} K 2
K 1 = 1 − x 2 + y 2 2 ∂ x − x y ∂ y K_{1}=\frac{1-x^{2}+y^{2}}{2}\partial_{x}-xy\partial_{y} K 1 = 2 1 − x 2 + y 2 ∂ x − x y ∂ y D D D x = 0 x=0 x = 0 ( ± 1 , 0 ) (\pm1,0) ( ± 1 , 0 )
K 2 = − x y ∂ x + 1 + x 2 − y 2 2 ∂ y K_{2}=-xy\partial_{x}+\frac{1+x^{2}-y^{2}}{2}\partial_{y} K 2 = − x y ∂ x + 2 1 + x 2 − y 2 ∂ y D D D y = 0 y=0 y = 0 ( 0 , ± 1 ) (0,\pm1) ( 0 , ± 1 )
Orientations and Hodge star operator
We choose on S 2 ⊂ R 3 \mathbb{S}^{2}\subset\mathbb{R}^{3} S 2 ⊂ R 3 N N N S 2 \mathbb{S}^{2} S 2 N = ( x , y , z ) N=(x,y,z) N = ( x , y , z ) R 2 \mathbb{R}^{2} R 2 ( 0 , 0 , − 1 ) (0,0,-1) ( 0 , 0 , − 1 ) p ∈ S 2 p\in\mathbb{S}^{2} p ∈ S 2 ( e ⃗ 1 , e ⃗ 2 ) (\vec e_{1},\vec e_{2}) ( e 1 , e 2 ) T p S 2 T_{p}\mathbb{S}^{2} T p S 2 ( d x 1 , d x 2 ) (dx_{1},dx_{2}) ( d x 1 , d x 2 ) T p ∗ S 2 T_{p}^{*}\mathbb{S}^{2} T p ∗ S 2 ⋆ \star ⋆ ⋆ d x 1 = d x 2 \star dx_{1}=dx_{2} ⋆ d x 1 = d x 2 ⋆ d x 2 = − d x 1 \star dx_{2}=-dx_{1} ⋆ d x 2 = − d x 1 V V V V ⊥ V^{\perp} V ⊥
V ⊥ : = ⋆ V . (15) V^{\perp}:=\star V. \tag*{(15)} V ⊥ := ⋆ V . (15) Extrinsically, V ⊥ V^{\perp} V ⊥ V V V π / 2 \pi/2 π /2 N N N
V ⊥ = N × V , V^{\perp}=N\times V, V ⊥ = N × V , where × \times × R 3 \mathbb{R}^{3} R 3
Analogously, we choose on H 2 \mathbb{H}^{2} H 2 D ⊂ R 2 D\subset\mathbb{R}^{2} D ⊂ R 2 ( ∂ x , ∂ y ) (\partial_{x},\partial_{y}) ( ∂ x , ∂ y ) V V V H 2 \mathbb{H}^{2} H 2 V ⊥ V^{\perp} V ⊥ V V V π 2 \frac{\pi}{2} 2 π
Poincaré-Hopf Theorem
We state here the Poincaré-Hopf Theorem which relates the index of a vector field with the Euler characteristic of the underlying manifold. We recall that for a vector field V V V Ind p i V \operatorname{Ind}_{p_i}V Ind p i V p i p_i p i V V V S n − 1 \mathbb{S}^{n-1} S n − 1 V V V p i p_i p i
Theorem 2.1 (Poincaré-Hopf). Let M M M n n n V V V M M M p i p_i p i ∂ M ≠ ∅ \partial M\ne\varnothing ∂ M = ∅ ⟨ V , ν ⟩ > 0 \langle V,\nu\rangle>0 ⟨ V , ν ⟩ > 0 ν \nu ν ∂ M \partial M ∂ M
∑ i Ind p i V = χ ( M ) , \sum_i \operatorname{Ind}_{p_i}V=\chi(M), i ∑ Ind p i V = χ ( M ) , where χ ( M ) \chi(M) χ ( M )
We refer e.g., to [28 ] for precise definitions and for a proof of Theorem 2.1 . We conclude this subsection by noting that if a vector field V V V M M M p p p det ( ∂ x j V i ( 0 ) ) ≠ 0 \det(\partial_{x_j}V_i(0)) \ne0 det ( ∂ x j V i ( 0 )) = 0 Ind p V = sign det ( ∂ x j V i ( 0 ) ) \operatorname{Ind}_p V = \operatorname{sign}\det(\partial_{x_j}V_i(0)) Ind p V = sign det ( ∂ x j V i ( 0 )) ( x 1 , … , x n ) (x_1,\ldots,x_n) ( x 1 , … , x n ) p p p 0 ∈ R n 0 \in\mathbb{R}^n 0 ∈ R n V = ∑ i = 1 n V i ∂ x i V = \sum_{i=1}^n V_i\partial_{x_i} V = ∑ i = 1 n V i ∂ x i
Nodal sets of solutions of elliptic PDE's
We recall here the following classical result on the behavior of solutions to elliptic equations on domains of S 2 \mathbb{S}^2 S 2 H 2 \mathbb{H}^2 H 2 q q q v ( q ) = 0 v(q)=0 v ( q ) = 0 ∇ v ( q ) = 0 \nabla v(q)=0 ∇ v ( q ) = 0 [17 ] .
Theorem 2.2. Let v v v − Δ v = a v -\Delta v = av − Δ v = a v Ω ⊂ S 2 , H 2 \Omega\subset\mathbb{S}^2,\mathbb{H}^2 Ω ⊂ S 2 , H 2 a a a q ∈ Ω q \in\Omega q ∈ Ω v ( q ) = ∇ v ( q ) = 0 v(q)=\nabla v(q)=0 v ( q ) = ∇ v ( q ) = 0 q q q v − 1 ( 0 ) v^{-1}(0) v − 1 ( 0 )
Proof of Theorem 1.2 (uniqueness of the critical point)
In this section we will prove Theorem 1.2 in several steps. We start by noting that by hypothesis f ( 0 ) ≥ 0 f(0) \ge0 f ( 0 ) ≥ 0 ∣ ∇ u ∣ ≠ 0 |\nabla u| \ne0 ∣∇ u ∣ = 0 ∂ Ω \partial\Omega ∂ Ω u > 0 u>0 u > 0 Ω \Omega Ω ∂ Ω \partial\Omega ∂ Ω
ν = − ∇ u ∣ ∇ u ∣ . (16) \nu= -\frac{\nabla u}{|\nabla u|}. \tag*{(16)} ν = − ∣∇ u ∣ ∇ u . (16) By assumption Ω \Omega Ω P ∈ Ω P \in\Omega P ∈ Ω P P P ( θ , ϕ ) ∈ [ 0 , π ] × [ 0 , 2 π ] (\theta,\phi) \in[0,\pi] \times[0,2\pi] ( θ , ϕ ) ∈ [ 0 , π ] × [ 0 , 2 π ] θ \theta θ P P P
The auxiliary vector field V
We introduce now the following vector field:
V = − V ~ ⊥ , (17) V=-\widetilde{V}^{\perp}, \tag*{(17)} V = − V ⊥ , (17) where
V ~ = ⟨ K 1 , ∇ u ⟩ ∇ ⟨ K 2 , ∇ u ⟩ − ⟨ K 2 , ∇ u ⟩ ∇ ⟨ K 1 , ∇ u ⟩ . (18) \widetilde{V}=\langle K_1,\nabla u\rangle\nabla\langle K_2,\nabla u\rangle-\langle K_2,\nabla u\rangle\nabla\langle K_1,\nabla u\rangle. \tag*{(18)} V = ⟨ K 1 , ∇ u ⟩ ∇ ⟨ K 2 , ∇ u ⟩ − ⟨ K 2 , ∇ u ⟩ ∇ ⟨ K 1 , ∇ u ⟩ . (18) Recall that K 1 , K 2 K_1,K_2 K 1 , K 2 ⊥ \perp ⊥ (15) .
Critical points of u are non degenerate. There are no critical points on the equator.
Assume first that the critical point p p p θ ≠ π 2 \theta\ne\frac{\pi}{2} θ = 2 π p p p ∇ u ( p ) = 0 \nabla u(p)=0 ∇ u ( p ) = 0 D 2 u ∣ p ( X , ⋅ ) = 0 D^2u|_p(X,\cdot)=0 D 2 u ∣ p ( X , ⋅ ) = 0 X ∈ T p S 2 X \in T_p\mathbb{S}^2 X ∈ T p S 2 X = K 1 ( p ) X=K_1(p) X = K 1 ( p ) K 1 , K 2 K_1,K_2 K 1 , K 2 p p p X X X p p p a K 1 + b K 2 aK_1+bK_2 a K 1 + b K 2 a 2 + b 2 = 1 a^2+b^2=1 a 2 + b 2 = 1 K 1 K_1 K 1 K 2 K_2 K 2 P P P Z = ⟨ K 1 , ∇ u ⟩ Z=\langle K_1,\nabla u\rangle Z = ⟨ K 1 , ∇ u ⟩
− Δ Z = f ′ ( u ) Z -\Delta Z=f'(u)Z − Δ Z = f ′ ( u ) Z in Ω \Omega Ω (14) , since K 1 K_1 K 1 Z ( p ) = 0 Z(p)=0 Z ( p ) = 0 ∇ Z ( p ) = 0 \nabla Z(p)=0 ∇ Z ( p ) = 0
Assume now that p p p ∇ u ( p ) = 0 \nabla u(p)=0 ∇ u ( p ) = 0 K 1 ( p ) = 0 K_1(p)=0 K 1 ( p ) = 0 P P P Z = ⟨ K 1 , ∇ u ⟩ Z=\langle K_1,\nabla u\rangle Z = ⟨ K 1 , ∇ u ⟩ − Δ Z = f ′ ( u ) Z -\Delta Z=f'(u)Z − Δ Z = f ′ ( u ) Z Z ( p ) = 0 Z(p)=0 Z ( p ) = 0 ∇ Z ( p ) = 0 \nabla Z(p)=0 ∇ Z ( p ) = 0
We will prove in Subsection 3.7 that under assumption (5) Z Z Z ∂ Ω \partial\Omega ∂ Ω Z Z Z Ω \Omega Ω p p p Z = 0 Z=0 Z = 0 Z Z Z ∂ Ω \partial\Omega ∂ Ω Z = 0 Z=0 Z = 0 ω ⊊ Ω \omega\subsetneq\Omega ω ⊊ Ω ω ‾ \overline{\omega} ω ∂ Ω \partial\Omega ∂ Ω Z Z Z Z = 0 Z=0 Z = 0 ∂ ω \partial\omega ∂ ω Z Z Z ω \omega ω − Δ Z = f ′ ( u ) Z -\Delta Z=f'(u)Z − Δ Z = f ′ ( u ) Z ω \omega ω − Δ − f ′ ( u ) -\Delta-f'(u) − Δ − f ′ ( u ) Ω \Omega Ω u u u (2) .
We conclude that critical points of u u u θ = π 2 \theta=\frac{\pi}{2} θ = 2 π
The zeros of V not on the equator coincide with the zeros of ∇ u \nabla u ∇ u
In this subsection we prove that ∇ u \nabla u ∇ u V V V Ω ∖ { θ = π 2 } \Omega\setminus\{\theta=\frac{\pi}{2}\} Ω ∖ { θ = 2 π } ∇ u \nabla u ∇ u V ~ \widetilde V V ∇ u ( p ) = 0 \nabla u(p)=0 ∇ u ( p ) = 0 V ~ ( p ) = 0 \widetilde V(p)=0 V ( p ) = 0 p p p V ~ ( p ) = 0 \widetilde V(p)=0 V ( p ) = 0 ∇ u ( p ) ≠ 0 \nabla u(p)\ne0 ∇ u ( p ) = 0
Let
K : = a K 2 + b K 1 (19) K:=aK_2+bK_1 \tag*{(19)} K := a K 2 + b K 1 (19) where
a = ⟨ K 1 ( p ) , ∇ u ( p ) ⟩ , b = − ⟨ K 2 ( p ) , ∇ u ( p ) ⟩ . a=\langle K_1(p),\nabla u(p)\rangle,\qquad b=-\langle K_2(p),\nabla u(p)\rangle. a = ⟨ K 1 ( p ) , ∇ u ( p )⟩ , b = − ⟨ K 2 ( p ) , ∇ u ( p )⟩ . Note that, since ∇ u ( p ) ≠ 0 \nabla u(p)\ne0 ∇ u ( p ) = 0 K 1 , K 2 K_1,K_2 K 1 , K 2 p p p ( a , b ) ≠ ( 0 , 0 ) (a,b)\ne(0,0) ( a , b ) = ( 0 , 0 ) K K K
Define Z = ⟨ K , ∇ u ⟩ Z=\langle K,\nabla u\rangle Z = ⟨ K , ∇ u ⟩
− Δ Z = f ′ ( u ) Z -\Delta Z=f'(u)Z − Δ Z = f ′ ( u ) Z since K K K Z ( p ) = 0 Z(p)=0 Z ( p ) = 0
∇ Z ( p ) = V ~ ( p ) = 0. \nabla Z(p)=\widetilde V(p)=0. ∇ Z ( p ) = V ( p ) = 0. We are led to a contradiction as in Subsection 3.2.
The vector field V satisfies ⟨ V , ν ⟩ > 0 \langle V,\nu\rangle>0 ⟨ V , ν ⟩ > 0 ∂ Ω \partial\Omega ∂ Ω
We start by noting that
⟨ V , ν ⟩ ∣ ∂ Ω = − 1 ∣ ∇ u ∣ ⟨ V , ∇ u ⟩ = − 1 ∣ ∇ u ∣ ⟨ V ⊥ , ∇ ⊥ u ⟩ = − 1 ∣ ∇ u ∣ ⟨ V ~ , ∇ ⊥ u ~ ⟩ , \left.\langle V,\nu\rangle\right|_{\partial\Omega}
=-\frac{1}{|\nabla u|}\langle V,\nabla u\rangle
=-\frac{1}{|\nabla u|}\langle V^\perp,\nabla^\perp u\rangle
=-\frac{1}{|\nabla u|}\langle\widetilde V,\widetilde{\nabla^\perp u}\rangle, ⟨ V , ν ⟩ ∣ ∂ Ω = − ∣∇ u ∣ 1 ⟨ V , ∇ u ⟩ = − ∣∇ u ∣ 1 ⟨ V ⊥ , ∇ ⊥ u ⟩ = − ∣∇ u ∣ 1 ⟨ V , ∇ ⊥ u ⟩ , where we have used the fact that ( V ~ ⊥ ) ⊥ = − V ~ (\widetilde V^\perp)^\perp=-\widetilde V ( V ⊥ ) ⊥ = − V ∇ ⊥ u \nabla^\perp u ∇ ⊥ u ( ∇ u ) ⊥ (\nabla u)^\perp ( ∇ u ) ⊥
Now we can write
− ⟨ V ~ , ∇ ⊥ u ~ ⟩ = ⟨ K 1 , ∇ u ⟩ D K 2 ( ∇ u , ∇ ⊥ u ) − ⟨ K 1 , ∇ u ⟩ D 2 u ( K 2 , ∇ ⊥ u ) − ⟨ K 2 , ∇ u ⟩ D K 1 ( ∇ u , ∇ ⊥ u ) + ⟨ K 2 , ∇ u ⟩ D 2 u ( K 1 , ∇ ⊥ u ) (20) -\langle\widetilde V,\widetilde{\nabla^\perp u}\rangle
=\langle K_1,\nabla u\rangle DK_2(\nabla u,\nabla^\perp u)-\langle K_1,\nabla u\rangle D^2u(K_2,\nabla^\perp u)
-\langle K_2,\nabla u\rangle DK_1(\nabla u,\nabla^\perp u)+\langle K_2,\nabla u\rangle D^2u(K_1,\nabla^\perp u) \tag*{(20)} − ⟨ V , ∇ ⊥ u ⟩ = ⟨ K 1 , ∇ u ⟩ D K 2 ( ∇ u , ∇ ⊥ u ) − ⟨ K 1 , ∇ u ⟩ D 2 u ( K 2 , ∇ ⊥ u ) − ⟨ K 2 , ∇ u ⟩ D K 1 ( ∇ u , ∇ ⊥ u ) + ⟨ K 2 , ∇ u ⟩ D 2 u ( K 1 , ∇ ⊥ u ) (20) = ⟨ K 1 , ∇ u ⟩ D K 2 ( ∇ u , ∇ ⊥ u ) − ⟨ K 2 , ∇ u ⟩ D K 1 ( ∇ u , ∇ ⊥ u ) − ⟨ K 1 , ∇ u ⟩ D 2 u ( K 2 , ∇ ⊥ u ) + ⟨ K 2 , ∇ u ⟩ D 2 u ( K 1 , ∇ ⊥ u ) . (21) =\langle K_1,\nabla u\rangle DK_2(\nabla u,\nabla^\perp u)-\langle K_2,\nabla u\rangle DK_1(\nabla u,\nabla^\perp u)
-\langle K_1,\nabla u\rangle D^2u(K_2,\nabla^\perp u)+\langle K_2,\nabla u\rangle D^2u(K_1,\nabla^\perp u). \tag*{(21)} = ⟨ K 1 , ∇ u ⟩ D K 2 ( ∇ u , ∇ ⊥ u ) − ⟨ K 2 , ∇ u ⟩ D K 1 ( ∇ u , ∇ ⊥ u ) − ⟨ K 1 , ∇ u ⟩ D 2 u ( K 2 , ∇ ⊥ u ) + ⟨ K 2 , ∇ u ⟩ D 2 u ( K 1 , ∇ ⊥ u ) . (21) where in the second inequality we have just re-arranged the terms.
We consider (21) first. Writing K i = ⟨ K i , ∇ u ∣ ∇ u ∣ ⟩ ∇ u ∣ ∇ u ∣ + ⟨ K i , ∇ ⊥ u ∣ ∇ u ∣ ⟩ ∇ ⊥ u ∣ ∇ u ∣ K_i=\langle K_i,\frac{\nabla u}{|\nabla u|}\rangle\frac{\nabla u}{|\nabla u|}+\langle K_i,\frac{\nabla^\perp u}{|\nabla u|}\rangle\frac{\nabla^\perp u}{|\nabla u|} K i = ⟨ K i , ∣∇ u ∣ ∇ u ⟩ ∣∇ u ∣ ∇ u + ⟨ K i , ∣∇ u ∣ ∇ ⊥ u ⟩ ∣∇ u ∣ ∇ ⊥ u (21) , and collecting ∣ ∇ u ∣ 2 |\nabla u|^2 ∣∇ u ∣ 2 (21) simplifies to
− ∣ ∇ u ∣ 2 ( ⟨ K 1 , ∇ u ∣ ∇ u ∣ ⟩ ⟨ K 2 , ∇ ⊥ u ∣ ∇ u ∣ ⟩ − ⟨ K 2 , ∇ u ∣ ∇ u ∣ ⟩ ⟨ K 1 , ∇ ⊥ u ∣ ∇ u ∣ ⟩ ) D 2 u ( ∇ ⊥ u ∣ ∇ u ∣ , ∇ ⊥ u ∣ ∇ u ∣ ) . (22) -|\nabla u|^2\left(\left\langle K_1,\frac{\nabla u}{|\nabla u|}\right\rangle\left\langle K_2,\frac{\nabla^\perp u}{|\nabla u|}\right\rangle-\left\langle K_2,\frac{\nabla u}{|\nabla u|}\right\rangle\left\langle K_1,\frac{\nabla^\perp u}{|\nabla u|}\right\rangle\right)D^2u\left(\frac{\nabla^\perp u}{|\nabla u|},\frac{\nabla^\perp u}{|\nabla u|}\right). \tag*{(22)} − ∣∇ u ∣ 2 ( ⟨ K 1 , ∣∇ u ∣ ∇ u ⟩ ⟨ K 2 , ∣∇ u ∣ ∇ ⊥ u ⟩ − ⟨ K 2 , ∣∇ u ∣ ∇ u ⟩ ⟨ K 1 , ∣∇ u ∣ ∇ ⊥ u ⟩ ) D 2 u ( ∣∇ u ∣ ∇ ⊥ u , ∣∇ u ∣ ∇ ⊥ u ) . (22) We have that
D 2 u ( ∇ ⊥ u ∣ ∇ u ∣ , ∇ ⊥ u ∣ ∇ u ∣ ) = Δ u − ∂ ν ν 2 u = k Ω ∂ ν u = − k Ω ∣ ∇ u ∣ D^{2}u\left(\frac{\nabla^{\perp}u}{|\nabla u|},\frac{\nabla^{\perp}u}{|\nabla u|}\right)=\Delta u-\partial_{\nu\nu}^{2}u=k_{\Omega}\partial_{\nu}u=-k_{\Omega}|\nabla u| D 2 u ( ∣∇ u ∣ ∇ ⊥ u , ∣∇ u ∣ ∇ ⊥ u ) = Δ u − ∂ ν ν 2 u = k Ω ∂ ν u = − k Ω ∣∇ u ∣ while
⟨ K 1 , ∇ u ∣ ∇ u ∣ ⟩ ⟨ K 2 , ∇ ⊥ u ∣ ∇ u ∣ ⟩ − ⟨ K 2 , ∇ u ∣ ∇ u ∣ ⟩ ⟨ K 1 , ∇ ⊥ u ∣ ∇ u ∣ ⟩ \left\langle K_{1},\frac{\nabla u}{|\nabla u|}\right\rangle\left\langle K_{2},\frac{\nabla^{\perp}u}{|\nabla u|}\right\rangle-\left\langle K_{2},\frac{\nabla u}{|\nabla u|}\right\rangle\left\langle K_{1},\frac{\nabla^{\perp}u}{|\nabla u|}\right\rangle ⟨ K 1 , ∣∇ u ∣ ∇ u ⟩ ⟨ K 2 , ∣∇ u ∣ ∇ ⊥ u ⟩ − ⟨ K 2 , ∣∇ u ∣ ∇ u ⟩ ⟨ K 1 , ∣∇ u ∣ ∇ ⊥ u ⟩ is the signed area of the parallelogram formed by K 1 , K 2 K_{1},K_{2} K 1 , K 2 ( ∇ u , ∇ ⊥ u ) (\nabla u,\nabla^{\perp}u) ( ∇ u , ∇ ⊥ u ) K 1 , K 2 K_{1},K_{2} K 1 , K 2 positive on the upper hemisphere and negative in the lower hemisphere, while this area vanishes at the equator. We have proved that (21) can be written as
∣ ∇ u ∣ 3 A ( K 1 , K 2 ) k Ω . |\nabla u|^{3}A(K_{1},K_{2})k_{\Omega}. ∣∇ u ∣ 3 A ( K 1 , K 2 ) k Ω . An explicit computation allows to establish that A ( K 1 , K 2 ) = cos ( θ ) A(K_{1},K_{2})=\cos(\theta) A ( K 1 , K 2 ) = cos ( θ )
Now we proceed to study (20) . To do so, we write (if K 1 ≠ 0 K_{1}\ne0 K 1 = 0
D K 2 ( ∇ u , ∇ ⊥ u ) = D K 2 ( ⟨ ∇ u , K 1 ∣ K 1 ∣ ⟩ K 1 ∣ K 1 ∣ + ⟨ ∇ u , K 1 ⊥ ∣ K 1 ∣ ⟩ K 1 ⊥ ∣ K 1 ∣ , ⟨ ∇ ⊥ u , K 1 ∣ K 1 ∣ ⟩ K 1 ∣ K 1 ∣ + ⟨ ∇ ⊥ u , K 1 ⊥ ∣ K 1 ∣ ⟩ K 1 ⊥ ∣ K 1 ∣ ) = ⟨ ∇ u , K 1 ∣ K 1 ∣ ⟩ ⟨ ∇ ⊥ u , K 1 ⊥ ∣ K 1 ∣ ⟩ D K 2 ( K 1 ∣ K 1 ∣ , K 1 ⊥ ∣ K 1 ∣ ) + ⟨ ∇ u , K 1 ⊥ ∣ K 1 ∣ ⟩ ⟨ ∇ ⊥ u , K 1 ∣ K 1 ∣ ⟩ D K 2 ( K 1 ⊥ ∣ K 1 ∣ , K 1 ∣ K 1 ∣ ) = ( ⟨ ∇ u , K 1 ∣ K 1 ∣ ⟩ 2 + ⟨ ∇ u , K 1 ⊥ ∣ K 1 ∣ ⟩ 2 ) D K 2 ( K 1 ∣ K 1 ∣ , K 1 ⊥ ∣ K 1 ∣ ) = ∣ ∇ u ∣ 2 D K 2 ( K 1 ∣ K 1 ∣ , K 1 ⊥ ∣ K 1 ∣ ) . \begin{aligned}
DK_{2}(\nabla u,\nabla^{\perp}u) \\
&=DK_{2}\left(\left\langle\nabla u,\frac{K_{1}}{|K_{1}|}\right\rangle\frac{K_{1}}{|K_{1}|}+\left\langle\nabla u,\frac{K_{1}^{\perp}}{|K_{1}|}\right\rangle\frac{K_{1}^{\perp}}{|K_{1}|},\left\langle\nabla^{\perp}u,\frac{K_{1}}{|K_{1}|}\right\rangle\frac{K_{1}}{|K_{1}|}+\left\langle\nabla^{\perp}u,\frac{K_{1}^{\perp}}{|K_{1}|}\right\rangle\frac{K_{1}^{\perp}}{|K_{1}|}\right) \\
&=\left\langle\nabla u,\frac{K_{1}}{|K_{1}|}\right\rangle\left\langle\nabla^{\perp}u,\frac{K_{1}^{\perp}}{|K_{1}|}\right\rangle DK_{2}\left(\frac{K_{1}}{|K_{1}|},\frac{K_{1}^{\perp}}{|K_{1}|}\right) \\
&+\left\langle\nabla u,\frac{K_{1}^{\perp}}{|K_{1}|}\right\rangle\left\langle\nabla^{\perp}u,\frac{K_{1}}{|K_{1}|}\right\rangle DK_{2}\left(\frac{K_{1}^{\perp}}{|K_{1}|},\frac{K_{1}}{|K_{1}|}\right) \\
&=\left(\left\langle\nabla u,\frac{K_{1}}{|K_{1}|}\right\rangle^{2}+\left\langle\nabla u,\frac{K_{1}^{\perp}}{|K_{1}|}\right\rangle^{2}\right)DK_{2}\left(\frac{K_{1}}{|K_{1}|},\frac{K_{1}^{\perp}}{|K_{1}|}\right) \\
&=|\nabla u|^{2}DK_{2}\left(\frac{K_{1}}{|K_{1}|},\frac{K_{1}^{\perp}}{|K_{1}|}\right).
\end{aligned} D K 2 ( ∇ u , ∇ ⊥ u ) = D K 2 ( ⟨ ∇ u , ∣ K 1 ∣ K 1 ⟩ ∣ K 1 ∣ K 1 + ⟨ ∇ u , ∣ K 1 ∣ K 1 ⊥ ⟩ ∣ K 1 ∣ K 1 ⊥ , ⟨ ∇ ⊥ u , ∣ K 1 ∣ K 1 ⟩ ∣ K 1 ∣ K 1 + ⟨ ∇ ⊥ u , ∣ K 1 ∣ K 1 ⊥ ⟩ ∣ K 1 ∣ K 1 ⊥ ) = ⟨ ∇ u , ∣ K 1 ∣ K 1 ⟩ ⟨ ∇ ⊥ u , ∣ K 1 ∣ K 1 ⊥ ⟩ D K 2 ( ∣ K 1 ∣ K 1 , ∣ K 1 ∣ K 1 ⊥ ) + ⟨ ∇ u , ∣ K 1 ∣ K 1 ⊥ ⟩ ⟨ ∇ ⊥ u , ∣ K 1 ∣ K 1 ⟩ D K 2 ( ∣ K 1 ∣ K 1 ⊥ , ∣ K 1 ∣ K 1 ) = ( ⟨ ∇ u , ∣ K 1 ∣ K 1 ⟩ 2 + ⟨ ∇ u , ∣ K 1 ∣ K 1 ⊥ ⟩ 2 ) D K 2 ( ∣ K 1 ∣ K 1 , ∣ K 1 ∣ K 1 ⊥ ) = ∣∇ u ∣ 2 D K 2 ( ∣ K 1 ∣ K 1 , ∣ K 1 ∣ K 1 ⊥ ) . Here we have used the fact that D K 2 DK_{2} D K 2 ⟨ U ⊥ , V ⟩ = − ⟨ U , V ⊥ ⟩ \langle U^{\perp},V\rangle=-\langle U,V^{\perp}\rangle ⟨ U ⊥ , V ⟩ = − ⟨ U , V ⊥ ⟩ U ⊥ ⊥ = − U U^{\perp\perp}=-U U ⊥⊥ = − U U , V U,V U , V K 2 ≠ 0 K_{2}\ne0 K 2 = 0
D K 1 ( ∇ u , ∇ ⊥ u ) = ∣ ∇ u ∣ 2 D K 1 ( K 2 ∣ K 2 ∣ , K 2 ⊥ ∣ K 2 ∣ ) . DK_{1}(\nabla u,\nabla^{\perp}u)=|\nabla u|^{2}DK_{1}\left(\frac{K_{2}}{|K_{2}|},\frac{K_{2}^{\perp}}{|K_{2}|}\right). D K 1 ( ∇ u , ∇ ⊥ u ) = ∣∇ u ∣ 2 D K 1 ( ∣ K 2 ∣ K 2 , ∣ K 2 ∣ K 2 ⊥ ) . Hence (20) is rewritten as
∣ ∇ u ∣ 2 ⟨ F , ∇ u ⟩ |\nabla u|^{2}\langle F,\nabla u\rangle ∣∇ u ∣ 2 ⟨ F , ∇ u ⟩ where
F = D K 2 ( K 1 ∣ K 1 ∣ , K 1 ⊥ ∣ K 1 ∣ ) K 1 − D K 1 ( K 2 ∣ K 2 ∣ , K 2 ⊥ ∣ K 2 ∣ ) K 2 , F=DK_{2}\left(\frac{K_{1}}{|K_{1}|},\frac{K_{1}^{\perp}}{|K_{1}|}\right)K_{1}-DK_{1}\left(\frac{K_{2}}{|K_{2}|},\frac{K_{2}^{\perp}}{|K_{2}|}\right)K_{2}, F = D K 2 ( ∣ K 1 ∣ K 1 , ∣ K 1 ∣ K 1 ⊥ ) K 1 − D K 1 ( ∣ K 2 ∣ K 2 , ∣ K 2 ∣ K 2 ⊥ ) K 2 , which is a combination of the two Killing fields K 1 , K 2 K_{1},K_{2} K 1 , K 2 K 1 K_{1} K 1 K 2 K_{2} K 2
Now, we need to use the explicit expressions of K 1 , K 2 K_{1},K_{2} K 1 , K 2 K 1 = z ∂ x − x ∂ z K_{1}=z\partial_{x}-x\partial_{z} K 1 = z ∂ x − x ∂ z K 2 = z ∂ y − y ∂ z K_{2}=z\partial_{y}-y\partial_{z} K 2 = z ∂ y − y ∂ z ∣ K 1 ∣ 2 = x 2 + z 2 |K_{1}|^{2}=x^{2}+z^{2} ∣ K 1 ∣ 2 = x 2 + z 2 ∣ K 2 ∣ 2 = y 2 + z 2 |K_{2}|^{2}=y^{2}+z^{2} ∣ K 2 ∣ 2 = y 2 + z 2
D K 2 ( K 1 , ⋅ ) = − x ∂ y , D K 1 ( K 2 , ⋅ ) = − y ∂ x . DK_{2}(K_{1},\cdot)=-x\partial_{y},\qquad DK_{1}(K_{2},\cdot)=-y\partial_{x}. D K 2 ( K 1 , ⋅ ) = − x ∂ y , D K 1 ( K 2 , ⋅ ) = − y ∂ x . We also check that K 1 ⊥ = − x y ∂ x + ( x 2 + z 2 ) ∂ y − y z ∂ z K_{1}^{\perp}=-xy\partial_{x}+(x^{2}+z^{2})\partial_{y}-yz\partial_{z} K 1 ⊥ = − x y ∂ x + ( x 2 + z 2 ) ∂ y − y z ∂ z
D K 2 ( K 1 ∣ K 1 ∣ , K 1 ⊥ ∣ K 1 ∣ ) = − x . DK_{2}\left(\frac{K_{1}}{|K_{1}|},\frac{K_{1}^{\perp}}{|K_{1}|}\right)=-x. D K 2 ( ∣ K 1 ∣ K 1 , ∣ K 1 ∣ K 1 ⊥ ) = − x . Analogously, since K 2 ⊥ = ( − y 2 − z 2 ) ∂ x + x y ∂ y + x z ∂ z K_{2}^{\perp}=(-y^{2}-z^{2})\partial_{x}+xy\partial_{y}+xz\partial_{z} K 2 ⊥ = ( − y 2 − z 2 ) ∂ x + x y ∂ y + x z ∂ z
D K 1 ( K 2 ∣ K 2 ∣ , K 2 ⊥ ∣ K 2 ∣ ) = y . DK_{1}\left(\frac{K_{2}}{|K_{2}|},\frac{K_{2}^{\perp}}{|K_{2}|}\right)=y. D K 1 ( ∣ K 2 ∣ K 2 , ∣ K 2 ∣ K 2 ⊥ ) = y . Therefore
F = − x K 1 − y K 2 = − x z ∂ x − y z ∂ y + ( x 2 + y 2 ) ∂ z . F=-xK_1-yK_2=-xz\partial_x-yz\partial_y+(x^2+y^2)\partial_z. F = − x K 1 − y K 2 = − x z ∂ x − y z ∂ y + ( x 2 + y 2 ) ∂ z . However this is not the most convenient way to look at F F F ( θ , ϕ ) ∈ [ 0 , π ] × [ 0 , 2 π ] (\theta,\phi)\in[0,\pi]\times[0,2\pi] ( θ , ϕ ) ∈ [ 0 , π ] × [ 0 , 2 π ] K 1 = cos ( ϕ ) ∂ θ − cot ( θ ) sin ( ϕ ) ∂ ϕ K_1=\cos(\phi)\partial_\theta-\cot(\theta)\sin(\phi)\partial_\phi K 1 = cos ( ϕ ) ∂ θ − cot ( θ ) sin ( ϕ ) ∂ ϕ K 2 = sin ( ϕ ) ∂ θ + cot ( θ ) cos ( ϕ ) ∂ ϕ K_2=\sin(\phi)\partial_\theta+\cot(\theta)\cos(\phi)\partial_\phi K 2 = sin ( ϕ ) ∂ θ + cot ( θ ) cos ( ϕ ) ∂ ϕ x = sin ( θ ) cos ( ϕ ) x=\sin(\theta)\cos(\phi) x = sin ( θ ) cos ( ϕ ) y = sin ( θ ) sin ( ϕ ) y=\sin(\theta)\sin(\phi) y = sin ( θ ) sin ( ϕ )
F = − sin ( θ ) ∂ θ . F=-\sin(\theta)\partial_\theta. F = − sin ( θ ) ∂ θ . We conclude that (20) can be re-written as
− ∣ ∇ u ∣ 2 sin ( θ ) ⟨ ∇ u , e ⃗ θ ⟩ = ∣ ∇ u ∣ 3 sin ( θ ) ⟨ ν , e ⃗ θ ⟩ . -|\nabla u|^2\sin(\theta)\langle\nabla u,\vec e_\theta\rangle=|\nabla u|^3\sin(\theta)\langle\nu,\vec e_\theta\rangle. − ∣∇ u ∣ 2 sin ( θ ) ⟨ ∇ u , e θ ⟩ = ∣∇ u ∣ 3 sin ( θ ) ⟨ ν , e θ ⟩ . We have proved that
⟨ V , ν ⟩ = ∣ ∇ u ∣ 2 ( cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ ) (23) \langle V,\nu\rangle=|\nabla u|^2\left(\cos(\theta)k_\Omega+\sin(\theta)\langle\nu,\vec e_\theta\rangle\right) \tag*{(23)} ⟨ V , ν ⟩ = ∣∇ u ∣ 2 ( cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e θ ⟩ ) (23) on ∂ Ω \partial\Omega ∂ Ω (5) .
The index of any zero of V is 1
There are two families of zeros of V V V V V V Ω ∖ { θ = π 2 } \Omega\setminus\{\theta=\frac{\pi}{2}\} Ω ∖ { θ = 2 π } ∇ u \nabla u ∇ u V V V Ω ∩ { θ = π 2 } \Omega\cap\{\theta=\frac{\pi}{2}\} Ω ∩ { θ = 2 π }
Now, we observe that the zeros p p p V V V Ω ∩ { θ = π 2 } \Omega\cap\{\theta=\frac{\pi}{2}\} Ω ∩ { θ = 2 π } ⟨ ∇ u ( p ) , e ⃗ θ ⟩ = 0 \langle\nabla u(p),\vec e_\theta\rangle=0 ⟨ ∇ u ( p ) , e θ ⟩ = 0 ⟨ K i ( p ) , ∇ u ( p ) ⟩ = 0 \langle K_i(p),\nabla u(p)\rangle=0 ⟨ K i ( p ) , ∇ u ( p )⟩ = 0 p p p ⟨ ∇ u ( p ) , e ⃗ θ ⟩ ≠ 0 \langle\nabla u(p),\vec e_\theta\rangle\ne0 ⟨ ∇ u ( p ) , e θ ⟩ = 0 K K K K 1 , K 2 K_1,K_2 K 1 , K 2 ⟨ K ( p ) , ∇ u ( p ) ⟩ = 0 \langle K(p),\nabla u(p)\rangle=0 ⟨ K ( p ) , ∇ u ( p )⟩ = 0 K ( p ) = 0 K(p)=0 K ( p ) = 0 Z = ⟨ K , ∇ u ⟩ Z=\langle K,\nabla u\rangle Z = ⟨ K , ∇ u ⟩ Z ( p ) = 0 Z(p)=0 Z ( p ) = 0 V ( p ) = 0 V(p)=0 V ( p ) = 0 ∇ Z ( p ) = 0 \nabla Z(p)=0 ∇ Z ( p ) = 0 − Δ Z − f ′ ( u ) Z = 0 -\Delta Z-f'(u)Z=0 − Δ Z − f ′ ( u ) Z = 0 Ω \Omega Ω
We have to compute the index of the zeros in any of such families.
Let p p p ∇ u \nabla u ∇ u V V V Ω ∖ { θ = π 2 } \Omega\setminus\{\theta=\frac{\pi}{2}\} Ω ∖ { θ = 2 π } ( e ⃗ 1 , e ⃗ 2 ) (\vec e_1,\vec e_2) ( e 1 , e 2 ) T p S 2 T_pS^2 T p S 2
∇ e ⃗ 1 V ( e ⃗ 1 ) = − ⟨ ∇ ⟨ ∇ u , K 1 ⟩ , e ⃗ 1 ⟩ ⟨ ∇ ⊥ ⟨ ∇ u , K 2 ⟩ , e ⃗ 1 ⟩ + ⟨ ∇ ⟨ ∇ u , K 2 ⟩ , e ⃗ 1 ⟩ ⟨ ∇ ⊥ ⟨ ∇ u , K 1 ⟩ , e ⃗ 1 ⟩ = ⟨ ∇ ⟨ ∇ u , K 1 ⟩ , e ⃗ 1 ⟩ ⟨ ∇ ⟨ ∇ u , K 2 ⟩ , e ⃗ 2 ⟩ − ⟨ ∇ ⟨ ∇ u , K 2 ⟩ , e ⃗ 1 ⟩ ⟨ ∇ ⟨ ∇ u , K 1 ⟩ , e ⃗ 2 ⟩ = − D 2 u ( K 2 , e ⃗ 1 ) D 2 u ( K 1 , e ⃗ 2 ) + D 2 u ( K 1 , e ⃗ 1 ) D 2 u ( K 2 , e ⃗ 2 ) . \begin{aligned}
\nabla_{\vec e_1}V(\vec e_1)
=-\langle\nabla\langle\nabla u,K_1\rangle,\vec e_1\rangle\langle\nabla^\perp\langle\nabla u,K_2\rangle,\vec e_1\rangle+\langle\nabla\langle\nabla u,K_2\rangle,\vec e_1\rangle\langle\nabla^\perp\langle\nabla u,K_1\rangle,\vec e_1\rangle\\
&=\langle\nabla\langle\nabla u,K_1\rangle,\vec e_1\rangle\langle\nabla\langle\nabla u,K_2\rangle,\vec e_2\rangle-\langle\nabla\langle\nabla u,K_2\rangle,\vec e_1\rangle\langle\nabla\langle\nabla u,K_1\rangle,\vec e_2\rangle\\
&=-D^2u(K_2,\vec e_1)D^2u(K_1,\vec e_2)+D^2u(K_1,\vec e_1)D^2u(K_2,\vec e_2).
\end{aligned} ∇ e 1 V ( e 1 ) = − ⟨ ∇ ⟨ ∇ u , K 1 ⟩ , e 1 ⟩ ⟨ ∇ ⊥ ⟨ ∇ u , K 2 ⟩ , e 1 ⟩ + ⟨ ∇ ⟨ ∇ u , K 2 ⟩ , e 1 ⟩ ⟨ ∇ ⊥ ⟨ ∇ u , K 1 ⟩ , e 1 ⟩ = ⟨ ∇ ⟨ ∇ u , K 1 ⟩ , e 1 ⟩ ⟨ ∇ ⟨ ∇ u , K 2 ⟩ , e 2 ⟩ − ⟨ ∇ ⟨ ∇ u , K 2 ⟩ , e 1 ⟩ ⟨ ∇ ⟨ ∇ u , K 1 ⟩ , e 2 ⟩ = − D 2 u ( K 2 , e 1 ) D 2 u ( K 1 , e 2 ) + D 2 u ( K 1 , e 1 ) D 2 u ( K 2 , e 2 ) . In the same way we see that
∇ e ⃗ 2 V ( e ⃗ 2 ) = − D 2 u ( K 2 , e ⃗ 1 ) D 2 u ( K 1 , e ⃗ 2 ) + D 2 u ( K 1 , e ⃗ 1 ) D 2 u ( K 2 , e ⃗ 2 ) \nabla_{\vec e_2}V(\vec e_2)=-D^2u(K_2,\vec e_1)D^2u(K_1,\vec e_2)+D^2u(K_1,\vec e_1)D^2u(K_2,\vec e_2) ∇ e 2 V ( e 2 ) = − D 2 u ( K 2 , e 1 ) D 2 u ( K 1 , e 2 ) + D 2 u ( K 1 , e 1 ) D 2 u ( K 2 , e 2 ) and
∇ e ⃗ 2 V ( e ⃗ 1 ) = ∇ e ⃗ 1 V ( e ⃗ 2 ) = 0. \nabla_{\vec e_2}V(\vec e_1)=\nabla_{\vec e_1}V(\vec e_2)=0. ∇ e 2 V ( e 1 ) = ∇ e 1 V ( e 2 ) = 0. Moreover, writing K i = ⟨ K i , e ⃗ 1 ⟩ e ⃗ 1 + ⟨ K i , e ⃗ 2 ⟩ e ⃗ 2 K_i=\langle K_i,\vec e_1\rangle\vec e_1+\langle K_i,\vec e_2\rangle\vec e_2 K i = ⟨ K i , e 1 ⟩ e 1 + ⟨ K i , e 2 ⟩ e 2 ∇ e ⃗ 1 V ( e ⃗ 1 ) \nabla_{\vec e_1}V(\vec e_1) ∇ e 1 V ( e 1 ) ∇ e ⃗ 2 V ( e ⃗ 2 ) \nabla_{\vec e_2}V(\vec e_2) ∇ e 2 V ( e 2 )
∇ e ⃗ 1 V ( e ⃗ 1 ) = ∇ e ⃗ 2 V ( e ⃗ 2 ) = ( ⟨ K 1 , e ⃗ 1 ⟩ ⟨ K 2 , e ⃗ 2 ⟩ − ⟨ K 2 , e ⃗ 1 ⟩ ⟨ K 1 , e ⃗ 2 ⟩ ) ( D 2 u ( e ⃗ 1 , e ⃗ 1 ) D 2 u ( e ⃗ 2 , e ⃗ 2 ) − D 2 u ( e ⃗ 1 , e ⃗ 2 ) 2 ) = A ( K 1 , K 2 ) det D 2 u = cos ( θ ) det D 2 u . \begin{aligned}
\nabla_{\vec e_1}V(\vec e_1)=\nabla_{\vec e_2}V(\vec e_2)
=\left(\langle K_1,\vec e_1\rangle\langle K_2,\vec e_2\rangle-\langle K_2,\vec e_1\rangle\langle K_1,\vec e_2\rangle\right)
\left(D^2u(\vec e_1,\vec e_1)D^2u(\vec e_2,\vec e_2)-D^2u(\vec e_1,\vec e_2)^2\right) \\
&=A(K_1,K_2)\det D^2u=\cos(\theta)\det D^2u.
\end{aligned} ∇ e 1 V ( e 1 ) = ∇ e 2 V ( e 2 ) = ( ⟨ K 1 , e 1 ⟩ ⟨ K 2 , e 2 ⟩ − ⟨ K 2 , e 1 ⟩ ⟨ K 1 , e 2 ⟩ ) ( D 2 u ( e 1 , e 1 ) D 2 u ( e 2 , e 2 ) − D 2 u ( e 1 , e 2 ) 2 ) = A ( K 1 , K 2 ) det D 2 u = cos ( θ ) det D 2 u . Since p p p θ = π 2 \theta=\frac{\pi}{2} θ = 2 π cos ( θ ) ≠ 0 \cos(\theta)\ne0 cos ( θ ) = 0 p p p V V V
We investigate now the second family of zeros of V V V p ∈ Ω ∩ { θ = π 2 } p\in\Omega\cap\{\theta=\frac{\pi}{2}\} p ∈ Ω ∩ { θ = 2 π } ⟨ ∇ u ( p ) , e ⃗ θ ⟩ = 0 \langle\nabla u(p),\vec e_\theta\rangle=0 ⟨ ∇ u ( p ) , e θ ⟩ = 0 V ( p ) = 0 V(p)=0 V ( p ) = 0 V V V p p p T p S 2 T_pS^2 T p S 2 ( e ⃗ θ , e ⃗ ϕ ) (\vec e_\theta,\vec e_\phi) ( e θ , e ϕ ) e ⃗ θ = ( 0 , 0 , − 1 ) \vec e_\theta=(0,0,-1) e θ = ( 0 , 0 , − 1 ) e ⃗ ϕ = ( − y p , x p , 0 ) \vec e_\phi=(-y_p,x_p,0) e ϕ = ( − y p , x p , 0 ) p = ( x p , y p , 0 ) p=(x_p,y_p,0) p = ( x p , y p , 0 ) S 2 S^2 S 2 ∇ u = ± ∣ ∇ u ∣ e ⃗ ϕ \nabla u = \pm|\nabla u|\vec e_\phi ∇ u = ± ∣∇ u ∣ e ϕ p p p D K i DK_i D K i
∇ e ⃗ θ V ( e ⃗ ϕ ) = ∇ e ⃗ ϕ V ( e ⃗ θ ) = 0 \nabla_{\vec e_\theta}V(\vec e_\phi)=\nabla_{\vec e_\phi}V(\vec e_\theta)=0 ∇ e θ V ( e ϕ ) = ∇ e ϕ V ( e θ ) = 0
∇ e ⃗ θ V ( e ⃗ θ ) = ∇ e ⃗ ϕ V ( e ⃗ ϕ ) = ± ∣ ∇ u ( p ) ∣ D 2 u ( p ) ( e ⃗ θ , e ⃗ ϕ ) \nabla_{\vec e_\theta}V(\vec e_\theta)=\nabla_{\vec e_\phi}V(\vec e_\phi)=\pm|\nabla u(p)|D^2u(p)(\vec e_\theta,\vec e_\phi) ∇ e θ V ( e θ ) = ∇ e ϕ V ( e ϕ ) = ± ∣∇ u ( p ) ∣ D 2 u ( p ) ( e θ , e ϕ )
Now, since ∣ ∇ u ( p ) ∣ ≠ 0 |\nabla u(p)|\ne0 ∣∇ u ( p ) ∣ = 0 D 2 u ( p ) ( e ⃗ θ , e ⃗ ϕ ) ≠ 0 D^2u(p)(\vec e_\theta,\vec e_\phi)\ne0 D 2 u ( p ) ( e θ , e ϕ ) = 0 V V V p p p
We show now that necessarily D 2 u ( p ) ( e ⃗ θ , e ⃗ ϕ ) ≠ 0 D^2u(p)(\vec e_\theta,\vec e_\phi)\ne0 D 2 u ( p ) ( e θ , e ϕ ) = 0 D 2 u ( p ) ( e ⃗ θ , e ⃗ ϕ ) = 0 D^2u(p)(\vec e_\theta,\vec e_\phi)=0 D 2 u ( p ) ( e θ , e ϕ ) = 0 ( e ⃗ θ , e ⃗ ϕ ) (\vec e_\theta,\vec e_\phi) ( e θ , e ϕ ) T p S 2 T_pS^2 T p S 2 D 2 u ( p ) D^2u(p) D 2 u ( p ) λ ∈ R \lambda\in\mathbb{R} λ ∈ R D 2 u ( p ) ( e ⃗ θ ) = λ e ⃗ θ D^2u(p)(\vec e_\theta)=\lambda\vec e_\theta D 2 u ( p ) ( e θ ) = λ e θ λ = 0 \lambda=0 λ = 0 λ ≠ 0 \lambda\ne0 λ = 0
Assume that λ = 0 \lambda=0 λ = 0 V V V p = ( 0 , 1 , 0 ) p=(0,1,0) p = ( 0 , 1 , 0 ) Z = ⟨ K 2 , ∇ u ⟩ Z=\langle K_2,\nabla u\rangle Z = ⟨ K 2 , ∇ u ⟩ p p p K 2 ( p ) = e ⃗ θ K_2(p)=\vec e_\theta K 2 ( p ) = e θ Z ( p ) = 0 Z(p)=0 Z ( p ) = 0 ∇ Z ( p ) = D 2 u ( p ) ( K 2 ) − D K 2 ( ∇ u ) = D 2 u ( p ) ( e ⃗ θ ) − D K 2 ( ∇ u ) = 0 \nabla Z(p)=D^2u(p)(K_2)-DK_2(\nabla u)=D^2u(p)(\vec e_\theta)-DK_2(\nabla u)=0 ∇ Z ( p ) = D 2 u ( p ) ( K 2 ) − D K 2 ( ∇ u ) = D 2 u ( p ) ( e θ ) − D K 2 ( ∇ u ) = 0 D K 2 ( ∇ u ) = 0 DK_2(\nabla u)=0 D K 2 ( ∇ u ) = 0 p p p ∇ u = ( ± ∣ ∇ u ∣ , 0 , 0 ) \nabla u=(\pm|\nabla u|,0,0) ∇ u = ( ± ∣∇ u ∣ , 0 , 0 ) D K 2 ( X ) = ( 0 , X 3 , − X 2 ) DK_2(X)=(0,X_3,-X_2) D K 2 ( X ) = ( 0 , X 3 , − X 2 ) X = ( X 1 , X 2 , X 3 ) X=(X_1,X_2,X_3) X = ( X 1 , X 2 , X 3 )
Assume that λ ≠ 0 \lambda\ne0 λ = 0 K = a K 1 + b K 2 K=aK_1+bK_2 K = a K 1 + b K 2 Z = ⟨ K , ∇ u ⟩ Z=\langle K,\nabla u\rangle Z = ⟨ K , ∇ u ⟩ Z ( p ) = 0 Z(p)=0 Z ( p ) = 0 ∇ Z ( p ) = 0 \nabla Z(p)=0 ∇ Z ( p ) = 0 − Δ Z − f ′ ( u ) Z = 0 -\Delta Z-f'(u)Z=0 − Δ Z − f ′ ( u ) Z = 0 Ω \Omega Ω a , b a,b a , b p = ( 0 , 1 , 0 ) p=(0,1,0) p = ( 0 , 1 , 0 ) ∇ u = ± ∣ ∇ u ∣ e ⃗ ϕ \nabla u=\pm|\nabla u|\vec e_\phi ∇ u = ± ∣∇ u ∣ e ϕ
∇ Z = D 2 u ( p ) ( a K 1 ( p ) + b K 2 ( p ) ) − a D K 1 ( p ) ( ∇ u ( p ) ) − b D K 2 ( p ) ( ∇ u ( p ) ) = ( ± ∣ ∇ u ∣ a + λ b ) e ⃗ θ . \begin{aligned}
\nabla Z={}&D^2u(p)(aK_1(p)+bK_2(p))-aDK_1(p)(\nabla u(p))-bDK_2(p)(\nabla u(p))\\
={}&(\pm|\nabla u|a+\lambda b)\vec e_\theta.
\end{aligned} ∇ Z = = D 2 u ( p ) ( a K 1 ( p ) + b K 2 ( p )) − a D K 1 ( p ) ( ∇ u ( p )) − b D K 2 ( p ) ( ∇ u ( p )) ( ± ∣∇ u ∣ a + λb ) e θ . where we have used the fact that at p p p K 1 = 0 K_1=0 K 1 = 0 K 2 = e ⃗ θ K_2=\vec e_\theta K 2 = e θ a = λ a=\lambda a = λ b = ∓ ∣ ∇ u ∣ b=\mp|\nabla u| b = ∓ ∣∇ u ∣
Application of Poincaré-Hopf and conclusion.
We have proved that V V V ∇ u \nabla u ∇ u V V V Ind p V = 1 \operatorname{Ind}_pV=1 Ind p V = 1 p p p V ( p ) = 0 V(p)=0 V ( p ) = 0 ⟨ V , ν ⟩ > 0 \langle V,\nu\rangle>0 ⟨ V , ν ⟩ > 0 ∂ Ω \partial\Omega ∂ Ω
1 = χ ( Ω ) = ∑ p : V ( p ) = 0 Ind p V = ∑ p : V ( p ) = 0 1 ≥ ∑ p : ∇ u ( p ) = 0 1 = # { critical points of u } ≥ 1 , 1=\chi(\Omega)=\sum_{p:V(p)=0}\operatorname{Ind}_pV=\sum_{p:V(p)=0}1\ge\sum_{p:\nabla u(p)=0}1=\#\{\text{critical points of }u\}\ge1, 1 = χ ( Ω ) = p : V ( p ) = 0 ∑ Ind p V = p : V ( p ) = 0 ∑ 1 ≥ p : ∇ u ( p ) = 0 ∑ 1 = # { critical points of u } ≥ 1 , therefore V V V ∇ u \nabla u ∇ u Ω \Omega Ω V V V ∇ u \nabla u ∇ u
Remark 3.1. We have proved that, a posteriori, there are no points on Ω ∩ { θ = π 2 } \Omega\cap\{\theta=\frac{\pi}{2}\} Ω ∩ { θ = 2 π } ⟨ ∇ u , e ⃗ θ ⟩ = 0 \langle\nabla u,\vec e_\theta\rangle=0 ⟨ ∇ u , e θ ⟩ = 0
The function ⟨ K 1 , ∇ u ⟩ \langle K_1,\nabla u\rangle ⟨ K 1 , ∇ u ⟩ ∂ Ω \partial\Omega ∂ Ω
We are only left to prove that condition (1.5) implies that ⟨ K 1 , ∇ u ⟩ \langle K_1,\nabla u\rangle ⟨ K 1 , ∇ u ⟩ ∂ Ω \partial\Omega ∂ Ω ( θ , ϕ ) (\theta,\phi) ( θ , ϕ ) P P P θ \theta θ P P P
We consider a parametrization γ : [ 0 , ∣ ∂ Ω ∣ ] → S 2 ⊂ R 3 \gamma:[0,|\partial\Omega|]\to S^2\subset\mathbb{R}^3 γ : [ 0 , ∣ ∂ Ω∣ ] → S 2 ⊂ R 3 ∂ Ω \partial\Omega ∂ Ω
γ ( t ) : = ( sin ( θ ( t ) ) cos ( ϕ ( t ) ) , sin ( θ ( t ) ) sin ( ϕ ( t ) ) , cos ( θ ( t ) ) ) , \gamma(t):=(\sin(\theta(t))\cos(\phi(t)),\sin(\theta(t))\sin(\phi(t)),\cos(\theta(t))), γ ( t ) := ( sin ( θ ( t )) cos ( ϕ ( t )) , sin ( θ ( t )) sin ( ϕ ( t )) , cos ( θ ( t ))) , where t t t ∂ Ω \partial\Omega ∂ Ω
ϕ : [ 0 , ∣ ∂ Ω ∣ ] → [ 0 , 2 π ] \phi: [0,|\partial\Omega|] \to[0,2\pi] ϕ : [ 0 , ∣ ∂ Ω∣ ] → [ 0 , 2 π ] is a strictly increasing bijection: ϕ ′ ( t ) > 0 \phi'(t)>0 ϕ ′ ( t ) > 0 t ∈ [ 0 , ∣ ∂ Ω ∣ ] t \in[0,|\partial\Omega|] t ∈ [ 0 , ∣ ∂ Ω∣ ]
Since u = 0 u=0 u = 0 ∂ Ω \partial\Omega ∂ Ω u ( θ ( t ) , ϕ ( t ) ) = 0 u(\theta(t),\phi(t))=0 u ( θ ( t ) , ϕ ( t )) = 0 t ∈ [ 0 , ∣ ∂ Ω ∣ ] t \in[0,|\partial\Omega|] t ∈ [ 0 , ∣ ∂ Ω∣ ] θ ′ ( t ) ∂ θ u + ϕ ′ ( t ) ∂ ϕ u = 0 \theta'(t)\partial_\theta u+\phi'(t)\partial_\phi u=0 θ ′ ( t ) ∂ θ u + ϕ ′ ( t ) ∂ ϕ u = 0 ⟨ K 1 , ∇ u ⟩ = cos ( ϕ ) ∂ θ u − cot ( θ ) sin ( ϕ ) ∂ ϕ u \langle K_1,\nabla u\rangle=\cos(\phi)\partial_\theta u-\cot(\theta)\sin(\phi)\partial_\phi u ⟨ K 1 , ∇ u ⟩ = cos ( ϕ ) ∂ θ u − cot ( θ ) sin ( ϕ ) ∂ ϕ u t ∈ [ 0 , ∣ ∂ Ω ∣ ] t\in[0,|\partial\Omega|] t ∈ [ 0 , ∣ ∂ Ω∣ ] ⟨ K 1 , ∇ u ⟩ ∣ ∂ Ω = 0 \left.\langle K_1,\nabla u\rangle\right|_{\partial\Omega}=0 ⟨ K 1 , ∇ u ⟩ ∣ ∂ Ω = 0 t t t
{ θ ′ ( t ) ∂ θ u + ϕ ′ ( t ) ∂ ϕ u = 0 , cos ( ϕ ( t ) ) ∂ θ u − cot ( θ ( t ) ) sin ( ϕ ( t ) ) ∂ ϕ u = 0. \begin{cases}
\theta'(t)\partial_\theta u+\phi'(t)\partial_\phi u=0,\\
\cos(\phi(t))\partial_\theta u-\cot(\theta(t))\sin(\phi(t))\partial_\phi u=0.
\end{cases} { θ ′ ( t ) ∂ θ u + ϕ ′ ( t ) ∂ ϕ u = 0 , cos ( ϕ ( t )) ∂ θ u − cot ( θ ( t )) sin ( ϕ ( t )) ∂ ϕ u = 0. Since ∇ u ≠ 0 \nabla u\ne0 ∇ u = 0 ∂ Ω \partial\Omega ∂ Ω t ∈ [ 0 , ∣ ∂ Ω ∣ ] t\in[0,|\partial\Omega|] t ∈ [ 0 , ∣ ∂ Ω∣ ] t t t
F ( t ) : = θ ′ ( t ) cot ( θ ( t ) ) sin ( ϕ ( t ) ) + ϕ ′ ( t ) cos ( ϕ ( t ) ) = 0. F(t):=\theta'(t)\cot(\theta(t))\sin(\phi(t))+\phi'(t)\cos(\phi(t))=0. F ( t ) := θ ′ ( t ) cot ( θ ( t )) sin ( ϕ ( t )) + ϕ ′ ( t ) cos ( ϕ ( t )) = 0. To simplify the notation, we shall omit from now on the dependence in t t t F F F [ 0 , ∣ ∂ Ω ∣ ] [0,|\partial\Omega|] [ 0 , ∣ ∂ Ω∣ ] K 1 K_1 K 1 ∂ Ω \partial\Omega ∂ Ω
We note that
F ( 0 ) = ϕ ′ ( 0 ) > 0 , F ( ϕ − 1 ( π ) ) = − ϕ ′ ( ϕ − 1 ( π ) ) < 0. F(0)=\phi'(0)>0,\qquad F(\phi^{-1}(\pi))=-\phi'(\phi^{-1}(\pi))<0. F ( 0 ) = ϕ ′ ( 0 ) > 0 , F ( ϕ − 1 ( π )) = − ϕ ′ ( ϕ − 1 ( π )) < 0. Therefore F F F t t t F ( t ) = 0 F(t)=0 F ( t ) = 0
ϕ ′ ( t ) F ′ ( t ) = − sin ( ϕ ) ϕ ′ 3 − sin ( ϕ ) ϕ ′ θ ′ 2 sin 2 ( θ ) + cot ( θ ) cos ( ϕ ) ϕ ′ 2 θ ′ + cot ( θ ) sin ( ϕ ) ϕ ′ ′ θ ′ + cos ( ϕ ) ϕ ′ ϕ ′ ′ . \begin{aligned}
\phi'(t)F'(t) \\
&=-\sin(\phi)\phi'^3-\frac{\sin(\phi)\phi'\theta'^2}{\sin^2(\theta)}+\cot(\theta)\cos(\phi)\phi'^2\theta'+\cot(\theta)\sin(\phi)\phi''\theta'+\cos(\phi)\phi'\phi''.
\end{aligned} ϕ ′ ( t ) F ′ ( t ) = − sin ( ϕ ) ϕ ′3 − sin 2 ( θ ) sin ( ϕ ) ϕ ′ θ ′2 + cot ( θ ) cos ( ϕ ) ϕ ′2 θ ′ + cot ( θ ) sin ( ϕ ) ϕ ′′ θ ′ + cos ( ϕ ) ϕ ′ ϕ ′′ . Now, for a t t t F ( t ) = 0 F(t)=0 F ( t ) = 0 ϕ ′ cos ( ϕ ) = − cot ( θ ) sin ( ϕ ) θ ′ \phi'\cos(\phi)=-\cot(\theta)\sin(\phi)\theta' ϕ ′ cos ( ϕ ) = − cot ( θ ) sin ( ϕ ) θ ′ F ( t ) = 0 F(t)=0 F ( t ) = 0
ϕ ′ ( t ) F ′ ( t ) = − sin ( ϕ ) ( ϕ ′ ( ϕ ′ 2 + 1 + cos 2 ( θ ) sin 2 ( θ ) θ ′ 2 ) + cot ( θ ) ( θ ′ ϕ ′ ′ − ϕ ′ θ ′ ′ ) ) . \phi'(t)F'(t)=-\sin(\phi)\left(\phi'\left(\phi'^2+\frac{1+\cos^2(\theta)}{\sin^2(\theta)}\theta'^2\right)+\cot(\theta)(\theta'\phi''-\phi'\theta'')\right). ϕ ′ ( t ) F ′ ( t ) = − sin ( ϕ ) ( ϕ ′ ( ϕ ′2 + sin 2 ( θ ) 1 + cos 2 ( θ ) θ ′2 ) + cot ( θ ) ( θ ′ ϕ ′′ − ϕ ′ θ ′′ ) ) . Assume that
ϕ ′ ( ϕ ′ 2 + 1 + cos 2 ( θ ) sin 2 ( θ ) θ ′ 2 ) + cot ( θ ) ( θ ′ ϕ ′ ′ − ϕ ′ θ ′ ′ ) ≠ 0 (24) \phi'\left(\phi'^2+\frac{1+\cos^2(\theta)}{\sin^2(\theta)}\theta'^2\right)+\cot(\theta)(\theta'\phi''-\phi'\theta'')\ne0 \tag*{(24)} ϕ ′ ( ϕ ′2 + sin 2 ( θ ) 1 + cos 2 ( θ ) θ ′2 ) + cot ( θ ) ( θ ′ ϕ ′′ − ϕ ′ θ ′′ ) = 0 (24) for all t t t t = 0 t=0 t = 0 t = ϕ − 1 ( π ) t=\phi^{-1}(\pi) t = ϕ − 1 ( π ) F F F F ′ ( t ) < 0 F'(t)<0 F ′ ( t ) < 0 F F F [ 0 , ϕ − 1 ( π ) ] [0,\phi^{-1}(\pi)] [ 0 , ϕ − 1 ( π )] F ′ ( t ) > 0 F'(t)>0 F ′ ( t ) > 0 F F F [ ϕ − 1 ( π ) , ∣ ∂ Ω ∣ ] [\phi^{-1}(\pi),|\partial\Omega|] [ ϕ − 1 ( π ) , ∣ ∂ Ω∣ ] F F F [ 0 , ϕ − 1 ( π ) ] [0,\phi^{-1}(\pi)] [ 0 , ϕ − 1 ( π )] [ ϕ − 1 ( π ) , ∣ ∂ Ω ∣ ] [\phi^{-1}(\pi),|\partial\Omega|] [ ϕ − 1 ( π ) , ∣ ∂ Ω∣ ]
Remark 3.2. We remark that the argument is invariant under rotations of the domain around P P P K 1 K_1 K 1 K 2 K_2 K 2 K 1 , K 2 K_1,K_2 K 1 , K 2
We prove now that (5) implies (24) . We consider the above arc-length parametrization γ \gamma γ ∂ Ω \partial\Omega ∂ Ω
γ ′ ( t ) = ( cos ( θ ( t ) ) cos ( ϕ ( t ) ) θ ′ ( t ) − sin ( θ ( t ) ) sin ( ϕ ( t ) ) ϕ ′ ( t ) , cos ( θ ( t ) ) sin ( ϕ ( t ) ) θ ′ ( t ) + sin ( θ ( t ) ) cos ( ϕ ( t ) ) ϕ ′ ( t ) , − sin ( θ ( t ) ) θ ′ ( t ) ) . \gamma'(t)=\left(\cos(\theta(t))\cos(\phi(t))\theta'(t)-\sin(\theta(t))\sin(\phi(t))\phi'(t),\right.
\left.\cos(\theta(t))\sin(\phi(t))\theta'(t)+\sin(\theta(t))\cos(\phi(t))\phi'(t),-\sin(\theta(t))\theta'(t)\right). γ ′ ( t ) = ( cos ( θ ( t )) cos ( ϕ ( t )) θ ′ ( t ) − sin ( θ ( t )) sin ( ϕ ( t )) ϕ ′ ( t ) , cos ( θ ( t )) sin ( ϕ ( t )) θ ′ ( t ) + sin ( θ ( t )) cos ( ϕ ( t )) ϕ ′ ( t ) , − sin ( θ ( t )) θ ′ ( t ) ) . and
∣ γ ′ ( t ) ∣ 2 = θ ′ ( t ) 2 + sin 2 ( θ ( t ) ) ϕ ′ ( t ) 2 = 1 |\gamma'(t)|^2=\theta'(t)^2+\sin^2(\theta(t))\phi'(t)^2=1 ∣ γ ′ ( t ) ∣ 2 = θ ′ ( t ) 2 + sin 2 ( θ ( t )) ϕ ′ ( t ) 2 = 1 We fix an orientation for γ \gamma γ ( γ ′ ( t ) , N ( t ) , ν ( t ) ) (\gamma'(t),N(t),\nu(t)) ( γ ′ ( t ) , N ( t ) , ν ( t )) ( e ⃗ 1 , e ⃗ 2 , e ⃗ 3 ) (\vec e_1,\vec e_2,\vec e_3) ( e 1 , e 2 , e 3 ) R 3 \mathbb{R}^3 R 3 N ( t ) = γ ( t ) N(t)=\gamma(t) N ( t ) = γ ( t ) S 2 \mathbb{S}^2 S 2 γ ( t ) \gamma(t) γ ( t ) ν ( t ) \nu(t) ν ( t ) ∂ Ω \partial\Omega ∂ Ω
ν ( t ) = γ ′ ( t ) × N ( t ) = ( sin ( ϕ ( t ) ) θ ′ ( t ) + sin ( θ ( t ) ) cos ( θ ( t ) ) cos ( ϕ ( t ) ) ϕ ′ ( t ) , − cos ( ϕ ( t ) ) θ ′ ( t ) + sin ( θ ( t ) ) cos ( θ ( t ) ) sin ( ϕ ( t ) ) ϕ ′ ( t ) , − sin 2 ( θ ( t ) ) ϕ ′ ( t ) ) . (25) \nu(t)=\gamma'(t)\times N(t)=\left(\sin(\phi(t))\theta'(t)+\sin(\theta(t))\cos(\theta(t))\cos(\phi(t))\phi'(t),\right.
\left.-\cos(\phi(t))\theta'(t)+\sin(\theta(t))\cos(\theta(t))\sin(\phi(t))\phi'(t),-\sin^2(\theta(t))\phi'(t)\right). \tag*{(25)} ν ( t ) = γ ′ ( t ) × N ( t ) = ( sin ( ϕ ( t )) θ ′ ( t ) + sin ( θ ( t )) cos ( θ ( t )) cos ( ϕ ( t )) ϕ ′ ( t ) , − cos ( ϕ ( t )) θ ′ ( t ) + sin ( θ ( t )) cos ( θ ( t )) sin ( ϕ ( t )) ϕ ′ ( t ) , − sin 2 ( θ ( t )) ϕ ′ ( t ) ) . (25) Then, the geodesic curvature k Ω k_\Omega k Ω ∂ Ω \partial\Omega ∂ Ω ν \nu ν
k Ω = − ⟨ γ ′ ′ ( t ) , ν ( t ) ⟩ , (26) k_\Omega=-\langle\gamma''(t),\nu(t)\rangle, \tag*{(26)} k Ω = − ⟨ γ ′′ ( t ) , ν ( t )⟩ , (26) where the scalar product is the standard product of R 3 \mathbb{R}^3 R 3 S 2 \mathbb{S}^2 S 2
We also have
e ⃗ θ = ( cos ( θ ( t ) ) cos ( ϕ ( t ) ) , cos ( θ ( t ) ) sin ( ϕ ( t ) ) , − sin ( θ ( t ) ) ) . (27) \vec e_\theta=(\cos(\theta(t))\cos(\phi(t)),\cos(\theta(t))\sin(\phi(t)),-\sin(\theta(t))). \tag*{(27)} e θ = ( cos ( θ ( t )) cos ( ϕ ( t )) , cos ( θ ( t )) sin ( ϕ ( t )) , − sin ( θ ( t ))) . (27) Using (25) , (26) and (27) we find that
cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ = sin 2 ( θ ( t ) ) ϕ ′ ( t ) 3 + 2 cos 2 ( θ ( t ) ) θ ′ ( t ) 2 ϕ ′ ( t ) + sin 2 ( θ ( t ) ) cos 2 ( θ ( t ) ) ϕ ′ ( t ) 3 + sin ( θ ( t ) ) cos ( θ ( t ) ) ( θ ′ ( t ) ϕ ′ ′ ( t ) − ϕ ′ ( t ) θ ′ ′ ( t ) ) . (28) \begin{aligned}
\cos(\theta)k_\Omega+\sin(\theta)\langle\nu,\vec e_\theta\rangle
={}&\sin^2(\theta(t))\phi'(t)^3+2\cos^2(\theta(t))\theta'(t)^2\phi'(t)\\
&+\sin^2(\theta(t))\cos^2(\theta(t))\phi'(t)^3\\
&+\sin(\theta(t))\cos(\theta(t))\left(\theta'(t)\phi''(t)-\phi'(t)\theta''(t)\right). \tag*{(28)}
\end{aligned} cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e θ ⟩ = sin 2 ( θ ( t )) ϕ ′ ( t ) 3 + 2 cos 2 ( θ ( t )) θ ′ ( t ) 2 ϕ ′ ( t ) + sin 2 ( θ ( t )) cos 2 ( θ ( t )) ϕ ′ ( t ) 3 + sin ( θ ( t )) cos ( θ ( t )) ( θ ′ ( t ) ϕ ′′ ( t ) − ϕ ′ ( t ) θ ′′ ( t ) ) . (28) We use now that 1 = θ ′ ( t ) 2 + sin 2 ( θ ( t ) ) ϕ ′ ( t ) 2 1=\theta'(t)^2+\sin^2(\theta(t))\phi'(t)^2 1 = θ ′ ( t ) 2 + sin 2 ( θ ( t )) ϕ ′ ( t ) 2 (28) as ( θ ′ ( t ) 2 + sin 2 ( θ ( t ) ) ϕ ′ ( t ) 2 ) sin 2 ( θ ( t ) ) ϕ ′ ( t ) (\theta'(t)^2+\sin^2(\theta(t))\phi'(t)^2)\sin^2(\theta(t))\phi'(t) ( θ ′ ( t ) 2 + sin 2 ( θ ( t )) ϕ ′ ( t ) 2 ) sin 2 ( θ ( t )) ϕ ′ ( t ) (28) to finally obtain
cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ = sin 2 ( θ ( t ) ) ( ϕ ′ ( t ) ( ϕ ′ ( t ) 2 + 1 + cos 2 ( θ ( t ) ) sin 2 ( θ ( t ) ) θ ′ ( t ) 2 ) + cot ( θ ( t ) ) ( θ ′ ( t ) ϕ ′ ′ ( t ) − ϕ ′ ( t ) θ ′ ′ ( t ) ) ) . (29) \begin{aligned}
\cos(\theta)k_\Omega+\sin(\theta)\langle\nu,\vec e_\theta\rangle
={}&\sin^2(\theta(t))\left(\phi'(t)\left(\phi'(t)^2+\frac{1+\cos^2(\theta(t))}{\sin^2(\theta(t))}\theta'(t)^2\right)\right.\\
&\left.+\cot(\theta(t))\left(\theta'(t)\phi''(t)-\phi'(t)\theta''(t)\right)\right). \tag*{(29)}
\end{aligned} cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e θ ⟩ = sin 2 ( θ ( t )) ( ϕ ′ ( t ) ( ϕ ′ ( t ) 2 + sin 2 ( θ ( t )) 1 + cos 2 ( θ ( t )) θ ′ ( t ) 2 ) + cot ( θ ( t )) ( θ ′ ( t ) ϕ ′′ ( t ) − ϕ ′ ( t ) θ ′′ ( t ) ) ) . (29) Therefore from (5) it follows that (24) is always strictly positive and never vanishes, and we conclude that ⟨ K 1 , ∇ u ⟩ \langle K_1,\nabla u\rangle ⟨ K 1 , ∇ u ⟩ ∂ Ω \partial\Omega ∂ Ω
Proof of Theorem 1.3 (uniqueness of the critical point)
We will prove here Theorem 1.3 . Since the proof is analogous to that of Theorem 1.2 we shall only highlight the main computational differences. We remark that in case of H 2 \mathbb{H}^2 H 2 K 1 , K 2 K_1,K_2 K 1 , K 2
Without loss of generality we can assume that Ω \Omega Ω ∂ Ω \partial\Omega ∂ Ω ν = − ∇ u ∣ ∇ u ∣ \nu=-\frac{\nabla u}{|\nabla u|} ν = − ∣∇ u ∣ ∇ u ∣ ∇ u ∣ ≠ 0 |\nabla u|\ne0 ∣∇ u ∣ = 0 ∂ Ω \partial\Omega ∂ Ω H 2 \mathbb{H}^2 H 2 [ 0 , + ∞ ) × [ 0 , 2 π ] [0,+\infty)\times[0,2\pi] [ 0 , + ∞ ) × [ 0 , 2 π ] r r r d r 2 + sinh 2 ( r ) d ϕ 2 dr^2+\sinh^2(r)d\phi^2 d r 2 + sinh 2 ( r ) d ϕ 2
The auxiliary vector field V V V
The vector field V V V Theorem 1.3 has the same form of (3.2), namely
V = − V ~ ⊥ V=-\widetilde{V}^{\perp} V = − V ⊥ with
V ~ = ⟨ K 1 , ∇ u ⟩ ∇ ( ⟨ K 2 , ∇ u ⟩ ) − ⟨ K 2 , ∇ u ⟩ ∇ ( ⟨ K 1 , ∇ u ⟩ ) . \widetilde{V}=\langle K_1,\nabla u\rangle\nabla(\langle K_2,\nabla u\rangle)-\langle K_2,\nabla u\rangle\nabla(\langle K_1,\nabla u\rangle). V = ⟨ K 1 , ∇ u ⟩ ∇ (⟨ K 2 , ∇ u ⟩) − ⟨ K 2 , ∇ u ⟩ ∇ (⟨ K 1 , ∇ u ⟩) . where now K 1 , K 2 K_1,K_2 K 1 , K 2
Critical points of u u u
The proof that the critical points of u u u p p p
A fundamental ingredient in the proof is the fact that given K K K K 1 , K 2 K_1,K_2 K 1 , K 2 Z : = ⟨ K , ∇ u ⟩ Z:=\langle K,\nabla u\rangle Z := ⟨ K , ∇ u ⟩ ∂ Ω \partial\Omega ∂ Ω
The zeros of V V V u u u
Since at any point of H 2 \mathbb{H}^2 H 2 K 1 , K 2 K_1,K_2 K 1 , K 2 u u u V V V
The vector field V V V ⟨ V , ν ⟩ > 0 \langle V,\nu\rangle>0 ⟨ V , ν ⟩ > 0 ∂ Ω \partial\Omega ∂ Ω
As in Subsection 3.4, we see that
⟨ V , ν ⟩ ∣ ∂ Ω = − 1 ∣ ∇ u ∣ ⟨ V ~ , ∇ ⊥ u ⟩ \left.\langle V,\nu\rangle\right|_{\partial\Omega}=-\frac{1}{|\nabla u|}\langle\widetilde{V},\nabla^\perp u\rangle ⟨ V , ν ⟩ ∣ ∂ Ω = − ∣∇ u ∣ 1 ⟨ V , ∇ ⊥ u ⟩ and that
− ⟨ V ~ , ∇ ⊥ u ⟩ = ⟨ K 1 , ∇ u ⟩ D K 2 ( ∇ u , ∇ ⊥ u ) − ⟨ K 2 , ∇ u ⟩ D K 1 ( ∇ u , ∇ ⊥ u ) − ⟨ K 1 , ∇ u ⟩ D 2 u ( K 2 , ∇ ⊥ u ) + ⟨ K 2 , ∇ u ⟩ D 2 u ( K 1 , ∇ ⊥ u ) . (30) \begin{aligned}
-\langle\widetilde{V},\nabla^\perp u\rangle
={}&\langle K_1,\nabla u\rangle DK_2(\nabla u,\nabla^\perp u)-\langle K_2,\nabla u\rangle DK_1(\nabla u,\nabla^\perp u) \\
&-\langle K_1,\nabla u\rangle D^2u(K_2,\nabla^\perp u)+\langle K_2,\nabla u\rangle D^2u(K_1,\nabla^\perp u). \tag*{(30)}
\end{aligned} − ⟨ V , ∇ ⊥ u ⟩ = ⟨ K 1 , ∇ u ⟩ D K 2 ( ∇ u , ∇ ⊥ u ) − ⟨ K 2 , ∇ u ⟩ D K 1 ( ∇ u , ∇ ⊥ u ) − ⟨ K 1 , ∇ u ⟩ D 2 u ( K 2 , ∇ ⊥ u ) + ⟨ K 2 , ∇ u ⟩ D 2 u ( K 1 , ∇ ⊥ u ) . (30) We consider first (4.2) and note that the same computations of Subsection 3.4 apply. Then (4.2) can be written as
∣ ∇ u ∣ 3 A ( K 1 , K 2 ) k Ω |\nabla u|^3A(K_1,K_2)k_\Omega ∣∇ u ∣ 3 A ( K 1 , K 2 ) k Ω where k Ω k_\Omega k Ω ∂ Ω \partial\Omega ∂ Ω A ( K 1 , K 2 ) A(K_1,K_2) A ( K 1 , K 2 ) K 1 , K 2 K_1,K_2 K 1 , K 2 ( K 1 , K 2 ) (K_1,K_2) ( K 1 , K 2 ) H 2 \mathbb{H}^2 H 2
A ( K 1 , K 2 ) = 1 + x 2 + y 2 1 − x 2 − y 2 . A(K_1,K_2)=\frac{1+x^2+y^2}{1-x^2-y^2}. A ( K 1 , K 2 ) = 1 − x 2 − y 2 1 + x 2 + y 2 . Recalling that in polar coordinates based at the origin x 2 + y 2 = tanh ( r / 2 ) x^2+y^2=\tanh(r/2) x 2 + y 2 = tanh ( r /2 )
A ( K 1 , K 2 ) = 1 + x 2 + y 2 1 − x 2 − y 2 = cosh ( r ) . A(K_1,K_2)=\frac{1+x^2+y^2}{1-x^2-y^2}=\cosh(r). A ( K 1 , K 2 ) = 1 − x 2 − y 2 1 + x 2 + y 2 = cosh ( r ) . On the other hand, following Subsection 3.4, we rewrite (4.1) as
∣ ∇ u ∣ 2 ⟨ F , ∇ u ⟩ , |\nabla u|^2\langle F,\nabla u\rangle, ∣∇ u ∣ 2 ⟨ F , ∇ u ⟩ , where
F = D K 2 ( K 1 ∣ K 1 ∣ , K 1 ⊥ ∣ K 1 ∣ ) K 1 − D K 1 ( K 2 ∣ K 2 ∣ , K 2 ⊥ ∣ K 2 ∣ ) K 2 , F=DK_2\left(\frac{K_1}{|K_1|},\frac{K_1^\perp}{|K_1|}\right)K_1-DK_1\left(\frac{K_2}{|K_2|},\frac{K_2^\perp}{|K_2|}\right)K_2, F = D K 2 ( ∣ K 1 ∣ K 1 , ∣ K 1 ∣ K 1 ⊥ ) K 1 − D K 1 ( ∣ K 2 ∣ K 2 , ∣ K 2 ∣ K 2 ⊥ ) K 2 , which is a combination of the two Killing fields K 1 , K 2 K_1,K_2 K 1 , K 2 D K i DK_i D K i ( e ⃗ 1 , e ⃗ 2 ) (\vec e_1,\vec e_2) ( e 1 , e 2 ) T p H 2 T_p\mathbb{H}^2 T p H 2 p p p D K i ( K j ∣ K j ∣ , K j ⊥ ∣ K j ∣ ) = D K i ( e ⃗ 1 , e ⃗ 2 ) DK_i\left(\frac{K_j}{|K_j|},\frac{K_j^\perp}{|K_j|}\right)=DK_i(\vec e_1,\vec e_2) D K i ( ∣ K j ∣ K j , ∣ K j ∣ K j ⊥ ) = D K i ( e 1 , e 2 ) i , j ∈ { 1 , 2 } i,j\in\{1,2\} i , j ∈ { 1 , 2 } D K i DK_i D K i F = x ∂ x + y ∂ y F=x\partial_x+y\partial_y F = x ∂ x + y ∂ y x ∂ x + y ∂ y x\partial_x+y\partial_y x ∂ x + y ∂ y 2 x 2 + y 2 1 − x 2 − y 2 \frac{2\sqrt{x^2+y^2}}{1-x^2-y^2} 1 − x 2 − y 2 2 x 2 + y 2 x 2 + y 2 = tanh ( r / 2 ) x^2+y^2=\tanh(r/2) x 2 + y 2 = tanh ( r /2 ) (30) can be written as
− ∣ ∇ u ∣ 2 sinh ( r ) ⟨ ν , e ⃗ r ⟩ . -\lvert\nabla u\rvert^2\sinh(r)\langle\nu,\vec e_r\rangle. − ∣ ∇ u ∣ 2 sinh ( r ) ⟨ ν , e r ⟩ . We conclude that under the condition (8) we have that ⟨ V , ν ⟩ ∣ ∂ Ω = ∣ ∇ u ∣ 2 ( cosh ( r ) k Ω − sinh ( r ) ⟨ ν , e ⃗ r ⟩ ) > 0 \left.\langle V,\nu\rangle\right|_{\partial\Omega}=\lvert\nabla u\rvert^2\left(\cosh(r)k_\Omega-\sinh(r)\langle\nu,\vec e_r\rangle\right)>0 ⟨ V , ν ⟩ ∣ ∂ Ω = ∣ ∇ u ∣ 2 ( cosh ( r ) k Ω − sinh ( r ) ⟨ ν , e r ⟩ ) > 0
The index of any zero of V V V
The computations are exactly the same of the corresponding case of Subsection 3.5 when p p p
Application of Poincaré-Hopf and conclusion
We have proved that V V V ∇ u \nabla u ∇ u V V V
1 = χ ( Ω ) = ∑ p : V ( p ) = 0 Ind p V = ∑ p : V ( p ) = 0 1 = # { critical points of u } . 1=\chi(\Omega)=\sum_{p:V(p)=0}\operatorname{Ind}_pV=\sum_{p:V(p)=0}1=\#\{\text{critical points of }u\}. 1 = χ ( Ω ) = p : V ( p ) = 0 ∑ Ind p V = p : V ( p ) = 0 ∑ 1 = # { critical points of u } . Hence u u u
The function K 1 , u K_{1,u} K 1 , u ∂ Ω \partial\Omega ∂ Ω
We prove here that assumption (8) implies that ⟨ K 1 , ∇ u ⟩ \langle K_1,\nabla u\rangle ⟨ K 1 , ∇ u ⟩ ∂ Ω \partial\Omega ∂ Ω
First, we note that being the metric conformal to the Euclidean one, it is sufficient to prove the statement when we consider the Euclidean metric on the unit disk D D D 4 ( 1 − x 2 − y 2 ) 2 ( d x 2 + d y 2 ) \frac{4}{(1-x^2-y^2)^2}(dx^2+dy^2) ( 1 − x 2 − y 2 ) 2 4 ( d x 2 + d y 2 ) Ω \Omega Ω
We consider then an arc-length parametrization γ : [ 0 , ∣ ∂ Ω ∣ ] → D \gamma:[0,|\partial\Omega|]\to D γ : [ 0 , ∣ ∂ Ω∣ ] → D ∂ Ω \partial\Omega ∂ Ω γ ( t ) = ( x ( t ) , y ( t ) ) ∈ D \gamma(t)=(x(t),y(t))\in D γ ( t ) = ( x ( t ) , y ( t )) ∈ D ∂ Ω \partial\Omega ∂ Ω ν E = ( y ′ ( t ) , − x ′ ( t ) ) \nu_E=(y'(t),-x'(t)) ν E = ( y ′ ( t ) , − x ′ ( t )) ν E \nu_E ν E ∇ E u \nabla_Eu ∇ E u ⟨ ⋅ , ⋅ ⟩ E \langle\cdot,\cdot\rangle_E ⟨ ⋅ , ⋅ ⟩ E t ∈ [ 0 , ∣ ∂ Ω ∣ ] t\in[0,|\partial\Omega|] t ∈ [ 0 , ∣ ∂ Ω∣ ] ⟨ K 1 , ∇ E u ⟩ E = 0 \langle K_1,\nabla_Eu\rangle_E=0 ⟨ K 1 , ∇ E u ⟩ E = 0
{ x ′ ( t ) ∂ x u + y ′ ( t ) ∂ y u = 0 , 1 − x 2 ( t ) + y 2 ( t ) 2 ∂ x u − x ( t ) y ( t ) ∂ y u = 0. \begin{cases}
x'(t)\partial_xu+y'(t)\partial_yu=0,\\
\frac{1-x^2(t)+y^2(t)}{2}\partial_xu-x(t)y(t)\partial_yu=0.
\end{cases} { x ′ ( t ) ∂ x u + y ′ ( t ) ∂ y u = 0 , 2 1 − x 2 ( t ) + y 2 ( t ) ∂ x u − x ( t ) y ( t ) ∂ y u = 0. Since ∇ u ≠ 0 \nabla u\ne0 ∇ u = 0 ∂ Ω \partial\Omega ∂ Ω t ∈ [ 0 , ∣ ∂ Ω ∣ ] t\in[0,|\partial\Omega|] t ∈ [ 0 , ∣ ∂ Ω∣ ] t t t
F ( t ) : = x ′ ( t ) x ( t ) y ( t ) + y ′ ( t ) ( 1 − x 2 ( t ) + y 2 ( t ) 2 ) = 0. F(t):=x'(t)x(t)y(t)+y'(t)\left(\frac{1-x^2(t)+y^2(t)}{2}\right)=0. F ( t ) := x ′ ( t ) x ( t ) y ( t ) + y ′ ( t ) ( 2 1 − x 2 ( t ) + y 2 ( t ) ) = 0. From now on we will drop the dependence on t t t F F F [ 0 , ∣ ∂ Ω ∣ ] [0,|\partial\Omega|] [ 0 , ∣ ∂ Ω∣ ] K 1 K_1 K 1 ∂ Ω \partial\Omega ∂ Ω
First, we note that since Ω \Omega Ω x y ′ − x ′ y ≥ δ ≥ 0 xy'-x'y\ge\delta\ge0 x y ′ − x ′ y ≥ δ ≥ 0 δ \delta δ y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0 x ( 0 ) > 0 x(0)>0 x ( 0 ) > 0 t ∗ t^* t ∗ y ( t ∗ ) = 0 y(t^*)=0 y ( t ∗ ) = 0 x ( t ∗ ) < 0 x(t^*)<0 x ( t ∗ ) < 0 y > 0 y>0 y > 0 ( 0 , t ∗ ) (0,t^*) ( 0 , t ∗ ) y ′ ( 0 ) > 0 y'(0)>0 y ′ ( 0 ) > 0 y ′ ( t ∗ ) < 0 y'(t^*)<0 y ′ ( t ∗ ) < 0 F ( 0 ) > 0 F(0)>0 F ( 0 ) > 0 F ( t ∗ ) < 0 F(t^*) < 0 F ( t ∗ ) < 0 F F F ( 0 , t ∗ ) (0,t^*) ( 0 , t ∗ ) F ′ F' F ′
F ′ = y + x y ′ ′ + y ′ ′ ( 1 − x 2 + y 2 2 ) . F' = y + xy'' + y''\left(\frac{1-x^2+y^2}{2}\right). F ′ = y + x y ′′ + y ′′ ( 2 1 − x 2 + y 2 ) . Let now t ∈ ( 0 , t ∗ ) t \in(0,t^*) t ∈ ( 0 , t ∗ ) F ( t ) = 0 F(t)=0 F ( t ) = 0 x ′ ( t ) ≠ 0 x'(t) \ne0 x ′ ( t ) = 0 y ′ ( t ) = 0 y'(t)=0 y ′ ( t ) = 0 t t t x ′ ( t ) < 0 x'(t)<0 x ′ ( t ) < 0 x ( t ) y ′ ( t ) > 0 x(t)y'(t)>0 x ( t ) y ′ ( t ) > 0 F ( t ) ≠ 0 F(t) \ne0 F ( t ) = 0 F ′ F' F ′ x ′ x' x ′ x ′ x y = − y ′ ( 1 − x 2 + y 2 2 ) x'xy=-y'\left(\frac{1-x^2+y^2}{2}\right) x ′ x y = − y ′ ( 2 1 − x 2 + y 2 ) t t t
x ′ ( t ) F ′ ( t ) = x ′ y + x ′ x y x ′ ′ + x ′ y ′ ′ ( 1 − x 2 + y 2 2 ) = ( 1 − x 2 + y 2 2 ) ( x ′ y ′ ′ − y ′ x ′ ′ ) + x ′ y = 1 − x 2 + y 2 1 + x 2 + y 2 ( 1 + x 2 + y 2 2 ( x ′ y ′ ′ − y ′ x ′ ′ ) + ( 1 + x 2 + y 2 1 − x 2 + y 2 ) x ′ y ) = 1 − x 2 + y 2 1 + x 2 + y 2 ( 1 + x 2 + y 2 2 ( x ′ y ′ ′ − y ′ x ′ ′ ) − x y ′ + x ′ y ) = 1 − x 2 + y 2 1 + x 2 + y 2 ( 1 + x 2 + y 2 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E ) . (31) \begin{aligned}
x'(t)F'(t)
&=x'y+x'xyx''+x'y''\left(\frac{1-x^2+y^2}{2}\right) \\
&=\left(\frac{1-x^2+y^2}{2}\right)(x'y''-y'x'')+x'y \\
&=\frac{1-x^2+y^2}{1+x^2+y^2}\left(\frac{1+x^2+y^2}{2}(x'y''-y'x'')+\left(\frac{1+x^2+y^2}{1-x^2+y^2}\right)x'y\right) \\
&=\frac{1-x^2+y^2}{1+x^2+y^2}\left(\frac{1+x^2+y^2}{2}(x'y''-y'x'')-xy'+x'y\right) \\
&=\frac{1-x^2+y^2}{1+x^2+y^2}\left(\frac{1+x^2+y^2}{2}k_{\Omega_E}-\langle\nu_E,(x,y)\rangle_E\right). \tag*{(31)}
\end{aligned} x ′ ( t ) F ′ ( t ) = x ′ y + x ′ x y x ′′ + x ′ y ′′ ( 2 1 − x 2 + y 2 ) = ( 2 1 − x 2 + y 2 ) ( x ′ y ′′ − y ′ x ′′ ) + x ′ y = 1 + x 2 + y 2 1 − x 2 + y 2 ( 2 1 + x 2 + y 2 ( x ′ y ′′ − y ′ x ′′ ) + ( 1 − x 2 + y 2 1 + x 2 + y 2 ) x ′ y ) = 1 + x 2 + y 2 1 − x 2 + y 2 ( 2 1 + x 2 + y 2 ( x ′ y ′′ − y ′ x ′′ ) − x y ′ + x ′ y ) = 1 + x 2 + y 2 1 − x 2 + y 2 ( 2 1 + x 2 + y 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E ) . (31) Here we have denoted by k Ω E k_{\Omega_E} k Ω E ∂ Ω \partial\Omega ∂ Ω F ( t ) = x ′ x y + y ′ ( 1 − x 2 + y 2 2 ) = 0 F(t)=x'xy+y'\left(\frac{1-x^2+y^2}{2}\right)=0 F ( t ) = x ′ x y + y ′ ( 2 1 − x 2 + y 2 ) = 0 (31) never vanishes in ( 0 , t ∗ ) (0,t^*) ( 0 , t ∗ ) F ′ F' F ′ ( 0 , t ∗ ) (0,t^*) ( 0 , t ∗ ) F F F ( t ∗ , ∣ ∂ Ω ∣ ) (t^*,|\partial\Omega|) ( t ∗ , ∣ ∂ Ω∣ )
In order to conclude, we must prove that condition (1.8) implies that the last term of (31) is positive an never vanishes on ( 0 , t ∗ ) (0,t^*) ( 0 , t ∗ ) k Ω k_\Omega k Ω ∂ Ω \partial\Omega ∂ Ω k Ω E k_{\Omega_E} k Ω E ∂ Ω \partial\Omega ∂ Ω
k Ω = 1 − x 2 − y 2 2 k Ω E + ⟨ ν E , ( x , y ) ⟩ E . k_\Omega=\frac{1-x^2-y^2}{2}k_{\Omega_E}+\langle\nu_E,(x,y)\rangle_E. k Ω = 2 1 − x 2 − y 2 k Ω E + ⟨ ν E , ( x , y ) ⟩ E . Hence, we can write
cosh ( r ) k Ω + sinh ( r ) ⟨ ν , e ⃗ r ⟩ = 1 + x 2 + y 2 1 − x 2 − y 2 k Ω + 2 1 − x 2 − y 2 ⟨ ν E , ( x , y ) ⟩ E = 1 + x 2 + y 2 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E . (32) \begin{aligned}
\cosh(r)k_\Omega+\sinh(r)\langle\nu,\vec e_r\rangle
&=\frac{1+x^2+y^2}{1-x^2-y^2}k_\Omega+\frac{2}{1-x^2-y^2}\langle\nu_E,(x,y)\rangle_E \\
&=\frac{1+x^2+y^2}{2}k_{\Omega_E}-\langle\nu_E,(x,y)\rangle_E. \tag*{(32)}
\end{aligned} cosh ( r ) k Ω + sinh ( r ) ⟨ ν , e r ⟩ = 1 − x 2 − y 2 1 + x 2 + y 2 k Ω + 1 − x 2 − y 2 2 ⟨ ν E , ( x , y ) ⟩ E = 2 1 + x 2 + y 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E . (32) The proof of the claim is now completed.
Optimality of the conditions (1.5) and (1.8) in Theorems 1.2 and 1.3
In this section, we focus on the case of the torsion problem
{ − Δ u = 1 on Ω u = 0 on ∂ Ω . (33) \begin{cases}
-\Delta u=1 & \text{on }\Omega\\
u=0 & \text{on }\partial\Omega. \tag*{(33)}
\end{cases} { − Δ u = 1 u = 0 on Ω on ∂ Ω. (33) Our goal is to construct examples of solutions to (33) in certain geodesically star-shaped domains of M = S 2 M=\mathbb{S}^2 M = S 2 H 2 \mathbb{H}^2 H 2 Ω \Omega Ω
The examples we construct demonstrate that the conditions (1.5) and (1.8) in Theorems 1.2 and 1.3, which guarantee the uniqueness of the critical point of u u u
This type of example first appeared in [14 ] and [9 ] , where the authors showed the optimality of the convexity assumption on Ω \Omega Ω [6 ] when Ω ⊂ R 2 \Omega\subset\mathbb{R}^{2} Ω ⊂ R 2 [14 ] was later generalized in [10 ] to the case of a general d d d M M M f ( u ) f(u) f ( u )
The resulting example is a smooth domain which collapses to a point but does not fully inherit the intrinsic features of the manifold M M M
Here, we aim to produce more constructive examples of solutions to (33) , which retain all the desirable properties of those in [14 ] and [10 ] , by working directly in S 2 \mathbb{S}^{2} S 2 H 2 \mathbb{H}^{2} H 2 [2 ] for the Euclidean case: a small perturbation of a convex domain does not necessarily produces additional critical points.
Through all this section we will consider coordinates ( x , y ) ∈ R 2 (x,y) \in\mathbb{R}^{2} ( x , y ) ∈ R 2 S 2 \mathbb{S}^{2} S 2 H 2 \mathbb{H}^{2} H 2 S 2 \mathbb{S}^{2} S 2 ( x , y ) ∈ R 2 (x,y) \in\mathbb{R}^{2} ( x , y ) ∈ R 2 H 2 \mathbb{H}^{2} H 2 ( x , y ) ∈ D ⊂ R 2 (x,y) \in D \subset\mathbb{R}^{2} ( x , y ) ∈ D ⊂ R 2 D D D
Theorem 5.1. For any integer n ≥ 2 n \ge2 n ≥ 2 Ω b \Omega_b Ω b S 2 \mathbb{S}^{2} S 2 H 2 \mathbb{H}^{2} H 2 u b u_b u b (33) in Ω b \Omega_b Ω b b b b
a) the solution u b u_b u b n n n Ω b \Omega_b Ω b
b) we have Ω b ‾ → [ − x ‾ η , x ‾ η ] × { 0 } \overline{\Omega_b} \to[-\overline{x}_{\eta},\overline{x}_{\eta}] \times\{0\} Ω b → [ − x η , x η ] × { 0 } b → 0 b \to0 b → 0 x ‾ η ∈ ( 0 , 1 ) \overline{x}_{\eta} \in(0,1) x η ∈ ( 0 , 1 )
c) Ω b \Omega_b Ω b S 2 \mathbb{S}^{2} S 2
d) for b b b
∂ Ω b = { ( x , y ) : − x ‾ η ≤ x ≤ x ‾ η , y = ± b 2 ( 1 − η f ( x ) ) ( 1 + x 2 ) ( 1 + o ( 1 ) ) } ⊂ S 2 , ∂ Ω b = { ( x , y ) : − x ‾ η ≤ x ≤ x ‾ η , y = ± b 2 ( 1 − η f ( x ) ) ( 1 − x 2 ) ( 1 + o ( 1 ) ) } ⊂ H 2 , (34) \begin{aligned}
\partial\Omega_b&=\left\{(x,y):-\overline{x}_{\eta}\le x\le\overline{x}_{\eta},\ y=\pm\sqrt{\frac{b}{2}(1-\eta f(x))(1+x^{2})(1+o(1))}\right\}\subset\mathbb{S}^{2},\\
\partial\Omega_b&=\left\{(x,y):-\overline{x}_{\eta}\le x\le\overline{x}_{\eta},\ y=\pm\sqrt{\frac{b}{2}(1-\eta f(x))(1-x^{2})(1+o(1))}\right\}\subset\mathbb{H}^{2}, \tag*{(34)}
\end{aligned} ∂ Ω b ∂ Ω b = { ( x , y ) : − x η ≤ x ≤ x η , y = ± 2 b ( 1 − η f ( x )) ( 1 + x 2 ) ( 1 + o ( 1 )) } ⊂ S 2 , = { ( x , y ) : − x η ≤ x ≤ x η , y = ± 2 b ( 1 − η f ( x )) ( 1 − x 2 ) ( 1 + o ( 1 )) } ⊂ H 2 , (34) where η > 0 \eta>0 η > 0 f f f 2 n 2n 2 n
e) (sharpness of (5) and (8) ). We have that
lim b → 0 ( cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ ) ≥ 0 on ∂ Ω b , ⊂ S 2 , (35) \lim_{b\to0}\left(\cos(\theta)k_{\Omega_b}+\sin(\theta)\langle\nu,\vec{e}_{\theta}\rangle\right)\ge0\quad\text{on }\partial\Omega_b,\ \subset\mathbb{S}^{2}, \tag*{(35)} b → 0 lim ( cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e θ ⟩ ) ≥ 0 on ∂ Ω b , ⊂ S 2 , (35) lim b → 0 ( cosh ( r ) k Ω b − sinh ( r ) ⟨ ν , e ⃗ r ⟩ ) ≥ 0 on ∂ Ω b ⊂ H 2 , (36) \lim_{b\to0}\left(\cosh(r)k_{\Omega_b}-\sinh(r)\langle\nu,\vec{e}_{r}\rangle\right)\ge0\quad\text{on }\partial\Omega_b\subset\mathbb{H}^{2}, \tag*{(36)} b → 0 lim ( cosh ( r ) k Ω b − sinh ( r ) ⟨ ν , e r ⟩ ) ≥ 0 on ∂ Ω b ⊂ H 2 , (36) where θ = dist ( ⋅ , 0 ) \theta=\operatorname{dist}(\cdot,0) θ = dist ( ⋅ , 0 ) r = dist ( ⋅ , 0 ) r=\operatorname{dist}(\cdot,0) r = dist ( ⋅ , 0 ) H 2 \mathbb{H}^{2} H 2
The two examples are built in a similar way: in both cases we start with a solution of the torsion problem (33) in a strip around a geodesic (i.e., the set of points at distance at most c > 0 c>0 c > 0 b v bv b v b b b v v v 2 n 2n 2 n
It is interesting to note that, taking a geodesic γ \gamma γ S 2 \mathbb{S}^{2} S 2 H 2 \mathbb{H}^{2} H 2 P ∈ γ P\in\gamma P ∈ γ S = { x : dist ( x , γ ) < c } S=\{x:\operatorname{dist}(x,\gamma)<c\} S = { x : dist ( x , γ ) < c } c > 0 c>0 c > 0 r = dist ( ⋅ , P ) r=\operatorname{dist}(\cdot,P) r = dist ( ⋅ , P ) (5) and (8) are identically zero on ∂ S \partial S ∂ S
Proof of Theorem 5.1
Proof of Theorem 5.1. We will consider the following two models:
( R 2 , 4 ( 1 + x 2 + y 2 ) 2 ( d x 2 + d y 2 ) ) \left(\mathbb{R}^2,\frac{4}{(1+x^2+y^2)^2}(dx^2+dy^2)\right) ( R 2 , ( 1 + x 2 + y 2 ) 2 4 ( d x 2 + d y 2 ) ) for S 2 S^2 S 2
( D , 4 ( 1 − x 2 − y 2 ) 2 ( d x 2 + d y 2 ) ) \left(D,\frac{4}{(1-x^2-y^2)^2}(dx^2+dy^2)\right) ( D , ( 1 − x 2 − y 2 ) 2 4 ( d x 2 + d y 2 ) ) for H 2 \mathbb{H}^2 H 2 D = { ( x , y ) : x 2 + y 2 < 1 } D=\{(x,y):x^2+y^2<1\} D = {( x , y ) : x 2 + y 2 < 1 } R 2 \mathbb{R}^2 R 2 D D D R 2 ∖ D \mathbb{R}^2\setminus D R 2 ∖ D
Let b > 0 b>0 b > 0 γ : = { ( x , y ) : y = 0 } \gamma:=\{(x,y):y=0\} γ := {( x , y ) : y = 0 }
S b : = { p ∈ M : dist ( p , γ ) < d M ( tanh ( b / 2 ) ) } . (37) S_b:=\{p\in M:\operatorname{dist}(p,\gamma)<d_M(\sqrt{\tanh(b/2)})\}. \tag*{(37)} S b := { p ∈ M : dist ( p , γ ) < d M ( tanh ( b /2 ) )} . (37) Here dist \operatorname{dist} dist d M : [ 0 , + ∞ ) → [ 0 , + ∞ ) d_M:[0,+\infty)\to[0,+\infty) d M : [ 0 , + ∞ ) → [ 0 , + ∞ )
d M ( ⋅ ) = { 2 arctan ( ⋅ ) , when M = S 2 , 2 arctanh ( ⋅ ) , when M = H 2 . d_M(\cdot)=
\begin{cases}
2\arctan(\cdot), & \text{when }M=S^2,\\
2\operatorname{arctanh}(\cdot), & \text{when }M=\mathbb{H}^2.
\end{cases} d M ( ⋅ ) = { 2 arctan ( ⋅ ) , 2 arctanh ( ⋅ ) , when M = S 2 , when M = H 2 . In other words, S b S_b S b d M ( tanh ( b / 2 ) ) d_M(\sqrt{\tanh(b/2)}) d M ( tanh ( b /2 ) ) γ \gamma γ γ \gamma γ ( − 1 , 1 ) × { 0 } (-1,1)\times\{0\} ( − 1 , 1 ) × { 0 } Figure 1 . In particular, in the spherical case, S b S_b S b Figure 2 .
Figure 1. The strip S b S_b S b S 2 S^2 S 2 H 2 \mathbb{H}^2 H 2
Figure 2. In the stereographic projection (from the south pole) of the sphere, the origin corresponds to the north pole, the x x x
A more explicit way of describing S b S_b S b
S b = { { ( x , y ) ∈ R 2 : 4 y 2 < ( 1 + x 2 + y 2 ) 2 ( 1 − e − 2 b ) } , M = S 2 , { ( x , y ) ∈ D : 4 y 2 < ( 1 − x 2 − y 2 ) 2 ( e 2 b − 1 ) } , M = H 2 . (38) S_b=
\begin{cases}
\{(x,y)\in\mathbb{R}^2:4y^2<(1+x^2+y^2)^2(1-e^{-2b})\}, & M=S^2,\\
\{(x,y)\in D:4y^2<(1-x^2-y^2)^2(e^{2b}-1)\}, & M=\mathbb{H}^2.
\end{cases} \tag*{(38)} S b = { {( x , y ) ∈ R 2 : 4 y 2 < ( 1 + x 2 + y 2 ) 2 ( 1 − e − 2 b )} , {( x , y ) ∈ D : 4 y 2 < ( 1 − x 2 − y 2 ) 2 ( e 2 b − 1 )} , M = S 2 , M = H 2 . (38) In the stereographic projection of S 2 S^2 S 2 H 2 \mathbb{H}^2 H 2 Figure 1 .
In this section with Δ \Delta Δ S 2 S^2 S 2 H 2 \mathbb{H}^2 H 2 Δ R 2 \Delta_{\mathbb{R}^2} Δ R 2
Euclidean metric, namely Δ R 2 = ∂ x x 2 + ∂ y y 2 \Delta_{\mathbb{R}^2}=\partial_{xx}^2+\partial_{yy}^2 Δ R 2 = ∂ xx 2 + ∂ y y 2 ψ b \psi_b ψ b (33) in S b S_b S b ψ b \psi_b ψ b
{ − Δ ψ b = 1 on S b , ψ = 0 on ∂ S b (39) \begin{cases}
-\Delta\psi_b=1 & \text{on } S_b,\\
\psi=0 & \text{on } \partial S_b \tag*{(39)}
\end{cases} { − Δ ψ b = 1 ψ = 0 on S b , on ∂ S b (39) which is equivalent to
{ − Δ R 2 ψ b = 4 ( 1 + x 2 + y 2 ) 2 when S b ⊂ S 2 , u = 0 on ∂ S b (40) \begin{cases}
-\Delta_{\mathbb{R}^2}\psi_b=\dfrac{4}{(1+x^2+y^2)^2} & \text{when } S_b\subset S^2,\\
u=0 & \text{on } \partial S_b \tag*{(40)}
\end{cases} ⎩ ⎨ ⎧ − Δ R 2 ψ b = ( 1 + x 2 + y 2 ) 2 4 u = 0 when S b ⊂ S 2 , on ∂ S b (40) and
{ − Δ R 2 ψ b = 4 ( 1 − x 2 − y 2 ) 2 when S b ⊂ H 2 , u = 0 on ∂ S b . (41) \begin{cases}
-\Delta_{\mathbb{R}^2}\psi_b=\dfrac{4}{(1-x^2-y^2)^2} & \text{when } S_b\subset\mathbb{H}^2,\\
u=0 & \text{on } \partial S_b. \tag*{(41)}
\end{cases} ⎩ ⎨ ⎧ − Δ R 2 ψ b = ( 1 − x 2 − y 2 ) 2 4 u = 0 when S b ⊂ H 2 , on ∂ S b . (41) It is easily seen that
ψ b ( x , y ) = { 1 2 log ( 1 − 4 y 2 ( 1 + x 2 + y 2 ) 2 ) + b , when M = S 2 , − 1 2 log ( 1 + 4 y 2 ( 1 − x 2 − y 2 ) 2 ) + b , when M = H 2 , (42) \psi_b(x,y)=
\begin{cases}
\dfrac{1}{2}\log\left(1-\dfrac{4y^2}{(1+x^2+y^2)^2}\right)+b, & \text{when } M=S^2,\\
-\dfrac{1}{2}\log\left(1+\dfrac{4y^2}{(1-x^2-y^2)^2}\right)+b, & \text{when } M=\mathbb{H}^2,
\end{cases} \tag*{(42)} ψ b ( x , y ) = ⎩ ⎨ ⎧ 2 1 log ( 1 − ( 1 + x 2 + y 2 ) 2 4 y 2 ) + b , − 2 1 log ( 1 + ( 1 − x 2 − y 2 ) 2 4 y 2 ) + b , when M = S 2 , when M = H 2 , (42) is the unique solution to (39) in the strip S b S_b S b
Moreover ∇ ψ b ( x , y ) = 0 ⟺ y = 0 ⟺ ( x , y ) ∈ γ \nabla\psi_b(x,y)=0\Longleftrightarrow y=0\Longleftrightarrow(x,y)\in\gamma ∇ ψ b ( x , y ) = 0 ⟺ y = 0 ⟺ ( x , y ) ∈ γ
To obtain a function which verifies the properties of Theorem 5.1 in Ω b \Omega_b Ω b ψ b \psi_b ψ b 2 n 2n 2 n n > 1 n>1 n > 1 − η v ( x , y ) -\eta v(x,y) − η v ( x , y ) η > 0 \eta>0 η > 0
u b = ψ b − b η v u_b=\psi_b-b\eta v u b = ψ b − b η v still verifies the torsion equation − Δ u b = 1 -\Delta u_b=1 − Δ u b = 1 Ω b \Omega_b Ω b Figure 3 and Figure 4 . The advantage of working with metrics that are conformal to the Euclidean one, is that we can look for the functions v v v
Figure 3. The unperturbed domain S b S_b S b Ω b \Omega_b Ω b
Figure 4. The unperturbed domain S b S_b S b Ω b \Omega_b Ω b
Next we introduce the function v v v n > 1 n>1 n > 1 n n n 0 < a 1 < a 2 < ⋯ < a n < 1 0<a_1<a_2<\cdots<a_n<1 0 < a 1 < a 2 < ⋯ < a n < 1
f ( x ) = ∏ i = 1 n ( x 2 − a i 2 ) (43) f(x)=\prod_{i=1}^{n}(x^2-a_i^2) \tag*{(43)} f ( x ) = i = 1 ∏ n ( x 2 − a i 2 ) (43) so that f ( x ) f(x) f ( x )
{ f ( 1 ) > sup x ∈ ( − a n , a n ) [ f ( x ) − x ( 1 + x 2 ) 2 ( 1 − x 2 ) f ′ ( x ) ] when M = S 2 , f ( 1 ) > sup x ∈ ( − a n , a n ) [ f ( x ) − x ( 1 − x 2 ) 2 ( 1 + x 2 ) f ′ ( x ) ] when M = H 2 . (44) \begin{cases}
f(1)>\sup_{x\in(-a_n,a_n)}\left[f(x)-\dfrac{x(1+x^2)}{2(1-x^2)}f'(x)\right] & \text{when }M=\mathbb{S}^2,\\
f(1)>\sup_{x\in(-a_n,a_n)}\left[f(x)-\dfrac{x(1-x^2)}{2(1+x^2)}f'(x)\right] & \text{when }M=\mathbb{H}^2. \tag*{(44)}
\end{cases} ⎩ ⎨ ⎧ f ( 1 ) > sup x ∈ ( − a n , a n ) [ f ( x ) − 2 ( 1 − x 2 ) x ( 1 + x 2 ) f ′ ( x ) ] f ( 1 ) > sup x ∈ ( − a n , a n ) [ f ( x ) − 2 ( 1 + x 2 ) x ( 1 − x 2 ) f ′ ( x ) ] when M = S 2 , when M = H 2 . (44) Let us make some remarks on the properties of f f f f ( x ) f(x) f ( x ) x x x 2 n 2n 2 n ± a i \pm a_i ± a i i = 1 , … , n i=1,\ldots,n i = 1 , … , n n n n x 1 , … , x n ∈ ( − a n , a n ) x_1,\ldots,x_n\in(-a_n,a_n) x 1 , … , x n ∈ ( − a n , a n ) n − 1 n-1 n − 1 x ~ 1 , … , x ~ n − 1 ∈ ( − a n , a n ) \widetilde{x}_1,\ldots,\widetilde{x}_{n-1}\in(-a_n,a_n) x 1 , … , x n − 1 ∈ ( − a n , a n )
Moreover f f f ∣ x ∣ ≥ a n |x|\geq a_n ∣ x ∣ ≥ a n c ∈ ( sup x ∈ ( − a n , a n ) f ( x ) , f ( 1 ) ) c\in\left(\sup_{x\in(-a_n,a_n)}f(x),f(1)\right) c ∈ ( sup x ∈ ( − a n , a n ) f ( x ) , f ( 1 ) ) x ˉ c ∈ ( 0 , 1 ) \bar{x}_c\in(0,1) x ˉ c ∈ ( 0 , 1 ) f ( x ˉ c ) = c f(\bar{x}_c)=c f ( x ˉ c ) = c
x f ′ ( x ) ≥ 0 for any a n ≤ ∣ x ∣ . (45) xf'(x)\geq0 \qquad\text{for any }a_n\leq|x|. \tag*{(45)} x f ′ ( x ) ≥ 0 for any a n ≤ ∣ x ∣. (45) Now we define
v ( x , y ) : = Re ( f ( z ) ) = Re ( ∏ i = 1 n ( z 2 − a i 2 ) ) (46) v(x,y):=\operatorname{Re}(f(z))=\operatorname{Re}\left(\prod_{i=1}^{n}(z^2-a_i^2)\right) \tag*{(46)} v ( x , y ) := Re ( f ( z )) = Re ( i = 1 ∏ n ( z 2 − a i 2 ) ) (46) where f ( z ) f(z) f ( z ) f ( x ) f(x) f ( x ) v ( x , y ) v(x,y) v ( x , y ) 2 n 2n 2 n x 2 x^2 x 2 y 2 y^2 y 2
We take
η = 1 2 [ 1 f ( 1 ) + 1 sup x ∈ ( − a n , a n ) [ f ( x ) − x ( 1 + x 2 ) 2 ( 1 − x 2 ) f ′ ( x ) ] ] (47) \eta=\frac{1}{2}\left[\frac{1}{f(1)}+\frac{1}{\sup_{x\in(-a_n,a_n)}\left[f(x)-\frac{x(1+x^2)}{2(1-x^2)}f'(x)\right]}\right] \tag*{(47)} η = 2 1 f ( 1 ) 1 + sup x ∈ ( − a n , a n ) [ f ( x ) − 2 ( 1 − x 2 ) x ( 1 + x 2 ) f ′ ( x ) ] 1 (47) in the spherical case and
η = 1 2 [ 1 f ( 1 ) + 1 sup x ∈ ( − a n , a n ) [ f ( x ) − x ( 1 − x 2 ) 2 ( 1 + x 2 ) f ′ ( x ) ] ] (48) \eta=\frac{1}{2}\left[\frac{1}{f(1)}+\frac{1}{\sup_{x\in(-a_n,a_n)}\left[f(x)-\frac{x(1-x^2)}{2(1+x^2)}f'(x)\right]}\right] \tag*{(48)} η = 2 1 f ( 1 ) 1 + sup x ∈ ( − a n , a n ) [ f ( x ) − 2 ( 1 + x 2 ) x ( 1 − x 2 ) f ′ ( x ) ] 1 (48) in the hyperbolic case (alternatively we can choose η ∈ ( 1 f ( 1 ) , 1 sup x ∈ ( − a n , a n ) [ f ( x ) − x ( 1 + x 2 ) 2 ( 1 − x 2 ) f ′ ( x ) ] ) \eta\in\left(\frac{1}{f(1)},\frac{1}{\sup_{x\in(-a_n,a_n)}\left[f(x)-\frac{x(1+x^2)}{2(1-x^2)}f'(x)\right]}\right) η ∈ ( f ( 1 ) 1 , s u p x ∈ ( − a n , a n ) [ f ( x ) − 2 ( 1 − x 2 ) x ( 1 + x 2 ) f ′ ( x ) ] 1 ) S 2 \mathbb{S}^2 S 2 H 2 \mathbb{H}^2 H 2
Of course, due to (44) 1 η ∈ ( sup x ∈ ( − a n , a n ) [ f ( x ) − x ( 1 + x 2 ) 2 ( 1 − x 2 ) f ′ ( x ) ] , f ( 1 ) ) \frac{1}{\eta}\in\left(\sup_{x\in(-a_n,a_n)}\left[f(x)-\frac{x(1+x^2)}{2(1-x^2)}f'(x)\right],f(1)\right) η 1 ∈ ( sup x ∈ ( − a n , a n ) [ f ( x ) − 2 ( 1 − x 2 ) x ( 1 + x 2 ) f ′ ( x ) ] , f ( 1 ) )
u b ( x , y ) = ψ b ( x , y ) − b η v ( x , y ) u_b(x,y)=\psi_b(x,y)-b\eta v(x,y) u b ( x , y ) = ψ b ( x , y ) − b η v ( x , y ) where ψ b \psi_b ψ b (42) , namely
u b ( x , y ) = { 1 2 log ( 1 − 4 y 2 ( 1 + x 2 + y 2 ) 2 ) + b ( 1 − η v ( x , y ) ) , when M = S 2 , − 1 2 log ( 1 + 4 y 2 ( 1 − x 2 − y 2 ) 2 ) + b ( 1 − η v ( x , y ) ) , when M = H 2 , (49) u_b(x,y)=
\begin{cases}
\frac{1}{2}\log\left(1-\frac{4y^2}{(1+x^2+y^2)^2}\right)+b(1-\eta v(x,y)),&\text{when }M=\mathbb{S}^2,\\
-\frac{1}{2}\log\left(1+\frac{4y^2}{(1-x^2-y^2)^2}\right)+b(1-\eta v(x,y)),&\text{when }M=\mathbb{H}^2,
\end{cases} \tag*{(49)} u b ( x , y ) = ⎩ ⎨ ⎧ 2 1 log ( 1 − ( 1 + x 2 + y 2 ) 2 4 y 2 ) + b ( 1 − η v ( x , y )) , − 2 1 log ( 1 + ( 1 − x 2 − y 2 ) 2 4 y 2 ) + b ( 1 − η v ( x , y )) , when M = S 2 , when M = H 2 , (49) and we set
Ω b : = connected component of { ( x , y ) : u b > 0 } containing ( 0 , 0 ) . (50) \Omega_b:=\text{connected component of }\{(x,y):u_b>0\}\text{ containing }(0,0). \tag*{(50)} Ω b := connected component of {( x , y ) : u b > 0 } containing ( 0 , 0 ) . (50) In the following, we will always use the symbol Ω b \Omega_b Ω b u b u_b u b (40) and (41) in Ω b \Omega_b Ω b b b b Ω b \Omega_b Ω b
We start by showing some properties of the domain Ω b \Omega_b Ω b
Lemma 5.2. Let u b u_b u b (49) and Ω b \Omega_b Ω b (50) . The set Ω b \Omega_b Ω b { ( x , 0 ) : − x ˉ η < x < x ˉ η } \{(x,0):-\bar{x}_\eta<x<\bar{x}_\eta\} {( x , 0 ) : − x ˉ η < x < x ˉ η } x ˉ η ∈ ( a n , 1 ) \bar{x}_\eta\in(a_n,1) x ˉ η ∈ ( a n , 1 ) f ( x ) = 1 η f(x)=\frac{1}{\eta} f ( x ) = η 1 ( ± x ˉ η , 0 ) ∈ ∂ Ω b (\pm\bar{x}_\eta,0)\in\partial\Omega_b ( ± x ˉ η , 0 ) ∈ ∂ Ω b
b b b x ˉ η \bar{x}_{\eta} x ˉ η ∂ Ω b ∩ { y = 0 } \partial\Omega_b \cap\{y = 0\} ∂ Ω b ∩ { y = 0 } b b b Ω b \Omega_b Ω b ( − x ˉ η , x ˉ η ) × ( − δ , δ ) (-\bar{x}_{\eta},\bar{x}_{\eta}) \times(-\delta,\delta) ( − x ˉ η , x ˉ η ) × ( − δ , δ ) δ ∈ ( 0 , 1 ) \delta\in(0,1) δ ∈ ( 0 , 1 )
Ω b ‾ → { ( x , 0 ) : − x ˉ η ≤ x ≤ x ˉ η } as b → 0. (51) \overline{\Omega_b} \to\{(x,0) : -\bar{x}_{\eta} \le x \le\bar{x}_{\eta}\} \text{ as } b \to0. \tag*{(51)} Ω b → {( x , 0 ) : − x ˉ η ≤ x ≤ x ˉ η } as b → 0. (51) Proof. First we compute u b ( 0 , 0 ) = b − b η v ( 0 , 0 ) = b ( 1 − η f ( 0 ) ) > 0 u_b(0,0) = b - b\eta v(0,0) = b(1-\eta f(0)) > 0 u b ( 0 , 0 ) = b − b η v ( 0 , 0 ) = b ( 1 − η f ( 0 )) > 0 f ( 0 ) ≤ sup x ∈ ( − a n , a n ) f ( x ) < 1 η f(0) \le\sup_{x \in(-a_n,a_n)} f(x) < \frac{1}{\eta} f ( 0 ) ≤ sup x ∈ ( − a n , a n ) f ( x ) < η 1 η \eta η f ( x ) f(x) f ( x ) Ω b \Omega_b Ω b b b b
Now we compute u b u_b u b x x x
u b ( x , 0 ) = b ( 1 − η v ( x , 0 ) ) = b ( 1 − η f ( x ) ) . (52) u_b(x,0) = b(1-\eta v(x,0)) = b(1-\eta f(x)). \tag*{(52)} u b ( x , 0 ) = b ( 1 − η v ( x , 0 )) = b ( 1 − η f ( x )) . (52) Then (47) , the definition of x ˉ η \bar{x}_{\eta} x ˉ η f ( x ) f(x) f ( x ) f ( x ˉ η ) = 1 η f(\bar{x}_{\eta}) = \frac{1}{\eta} f ( x ˉ η ) = η 1 f ( x ) < 1 η f(x) < \frac{1}{\eta} f ( x ) < η 1 ∣ x ∣ < x ˉ η |x| < \bar{x}_{\eta} ∣ x ∣ < x ˉ η f ( x ) > 1 η f(x) > \frac{1}{\eta} f ( x ) > η 1 ∣ x ∣ > x ˉ η |x| > \bar{x}_{\eta} ∣ x ∣ > x ˉ η { ( x , 0 ) : − x ˉ η < x < x ˉ η } ⊂ Ω b \{(x,0) : -\bar{x}_{\eta} < x < \bar{x}_{\eta}\} \subset\Omega_b {( x , 0 ) : − x ˉ η < x < x ˉ η } ⊂ Ω b b b b ( ± x ˉ η , 0 ) (\pm\bar{x}_{\eta},0) ( ± x ˉ η , 0 ) ∂ Ω b ∩ { y = 0 } \partial\Omega_b \cap\{y = 0\} ∂ Ω b ∩ { y = 0 } b b b Ω b ⊂ D \Omega_b \subset D Ω b ⊂ D u b ( x , y ) u_b(x,y) u b ( x , y ) x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1 y = 0 y=0 y = 0
u b ( ± 1 , 0 ) = b ( 1 − η v ( ± 1 , 0 ) ) = b ( 1 − η f ( 1 ) ) < 0 u_b(\pm1,0) = b(1-\eta v(\pm1,0)) = b(1-\eta f(1)) < 0 u b ( ± 1 , 0 ) = b ( 1 − η v ( ± 1 , 0 )) = b ( 1 − η f ( 1 )) < 0 by the definition of η \eta η v ( x , y ) v(x,y) v ( x , y ) δ > 0 \delta> 0 δ > 0 v ( ± 1 − y 2 , y ) < 1 η v(\pm\sqrt{1-y^2},y) < \frac{1}{\eta} v ( ± 1 − y 2 , y ) < η 1 ∣ y ∣ < δ |y| < \delta ∣ y ∣ < δ u b ( ± 1 − y 2 , y ) < 0 u_b(\pm\sqrt{1-y^2},y) < 0 u b ( ± 1 − y 2 , y ) < 0 ∣ y ∣ < δ |y| < \delta ∣ y ∣ < δ ∣ y ∣ ≥ δ |y| \ge\delta ∣ y ∣ ≥ δ
u b ( ± 1 − y 2 , y ) = 1 2 log ( 1 − 2 y 2 ) + b ( 1 − η v ( ± 1 − y 2 , y ) ) ≤ 1 2 log ( 1 − 2 δ 2 ) + b ( 1 + η sup ∣ y ∣ ≤ 1 v ( ± 1 − y 2 , y ) ) < 0 \begin{aligned}
u_b(\pm\sqrt{1-y^2},y)
&= \frac{1}{2}\log(1-2y^2)+b(1-\eta v(\pm\sqrt{1-y^2},y)) \\
&\le\frac{1}{2}\log(1-2\delta^2)+b\left(1+\eta\sup_{|y|\le1}v(\pm\sqrt{1-y^2},y)\right)<0
\end{aligned} u b ( ± 1 − y 2 , y ) = 2 1 log ( 1 − 2 y 2 ) + b ( 1 − η v ( ± 1 − y 2 , y )) ≤ 2 1 log ( 1 − 2 δ 2 ) + b ( 1 + η ∣ y ∣ ≤ 1 sup v ( ± 1 − y 2 , y ) ) < 0 if b b b
b < − 1 2 log ( 1 − 2 δ 2 ) 1 + η sup ∣ y ∣ ≤ 1 v ( ± 1 − y 2 , y ) . b < \frac{-\frac{1}{2}\log(1-2\delta^2)}{1+\eta\sup_{|y|\le1}v(\pm\sqrt{1-y^2},y)}. b < 1 + η sup ∣ y ∣ ≤ 1 v ( ± 1 − y 2 , y ) − 2 1 log ( 1 − 2 δ 2 ) . This proves that u b ( x , y ) u_b(x,y) u b ( x , y ) b b b Ω b ⊂ D \Omega_b \subset D Ω b ⊂ D
Now we prove that Ω b ⊂ { ( x , y ) ∈ D : − x ˉ η < x < x ˉ η } \Omega_b \subset\{(x,y) \in D : -\bar{x}_{\eta} < x < \bar{x}_{\eta}\} Ω b ⊂ {( x , y ) ∈ D : − x ˉ η < x < x ˉ η } b k b_k b k b k → 0 b_k \to0 b k → 0 k → ∞ k \to\infty k → ∞ y k ∈ ( − 1 , 1 ) y_k \in(-1,1) y k ∈ ( − 1 , 1 ) u b k ( x ˉ η , y k ) > 0 u_{b_k}(\bar{x}_{\eta},y_k)>0 u b k ( x ˉ η , y k ) > 0 ∣ b ( 1 − η v ( x ˉ η , y k ) ) ∣ ≤ C b k → 0 |b(1-\eta v(\bar{x}_{\eta},y_k))| \le Cb_k \to0 ∣ b ( 1 − η v ( x ˉ η , y k )) ∣ ≤ C b k → 0 1 2 log ( 1 − 4 y 2 ( 1 + x ˉ η 2 + y 2 ) 2 ) < 0 \frac{1}{2}\log\left(1-\frac{4y^2}{(1+\bar{x}_{\eta}^2+y^2)^2}\right)<0 2 1 log ( 1 − ( 1 + x ˉ η 2 + y 2 ) 2 4 y 2 ) < 0 ∣ y ∣ > 0 |y|>0 ∣ y ∣ > 0 y k → 0 y_k \to0 y k → 0 k → ∞ k \to\infty k → ∞
u b k ( x ˉ η , y k ) = u b k ( x ˉ η , 0 ) + y k ∂ u b k ∂ y ( x ˉ η , 0 ) + 1 2 y k 2 ∂ 2 u b k ∂ y 2 ( x ˉ η , 0 ) + o ( y k 2 ) u_{b_k}(\bar{x}_{\eta},y_k)=u_{b_k}(\bar{x}_{\eta},0)+y_k\frac{\partial u_{b_k}}{\partial y}(\bar{x}_{\eta},0)+\frac{1}{2}y_k^2\frac{\partial^2u_{b_k}}{\partial y^2}(\bar{x}_{\eta},0)+o(y_k^2) u b k ( x ˉ η , y k ) = u b k ( x ˉ η , 0 ) + y k ∂ y ∂ u b k ( x ˉ η , 0 ) + 2 1 y k 2 ∂ y 2 ∂ 2 u b k ( x ˉ η , 0 ) + o ( y k 2 ) as k → ∞ k \to\infty k → ∞ u b k ( x ˉ η , 0 ) = 0 u_{b_k}(\bar{x}_{\eta},0)=0 u b k ( x ˉ η , 0 ) = 0 x ˉ η \bar{x}_{\eta} x ˉ η ∂ u b k ∂ y ( x ˉ η , 0 ) = 0 \frac{\partial u_{b_k}}{\partial y}(\bar{x}_{\eta},0)=0 ∂ y ∂ u b k ( x ˉ η , 0 ) = 0 u b ( x , y ) u_b(x,y) u b ( x , y ) y 2 y^2 y 2 ∂ 2 u b k ∂ y 2 ( x ˉ η , 0 ) = − 4 ( 1 + x ˉ η 2 ) 2 − b k η ∂ 2 v ∂ y 2 ( x ˉ η , 0 ) \frac{\partial^2u_{b_k}}{\partial y^2}(\bar{x}_{\eta},0)=-\frac{4}{(1+\bar{x}_{\eta}^2)^2}-b_k\eta\frac{\partial^2v}{\partial y^2}(\bar{x}_{\eta},0) ∂ y 2 ∂ 2 u b k ( x ˉ η , 0 ) = − ( 1 + x ˉ η 2 ) 2 4 − b k η ∂ y 2 ∂ 2 v ( x ˉ η , 0 ) v ( x , y ) v(x,y) v ( x , y ) ∣ ∂ 2 v ∂ y 2 ( x ˉ η , 0 ) ∣ ≤ C \left|\frac{\partial^2v}{\partial y^2}(\bar{x}_{\eta},0)\right|\le C ∂ y 2 ∂ 2 v ( x ˉ η , 0 ) ≤ C ∂ 2 u b k ∂ y 2 ( x ˉ η , 0 ) < − 2 1 + x ˉ η 2 < 0 \frac{\partial^2u_{b_k}}{\partial y^2}(\bar{x}_{\eta},0)<-\frac{2}{1+\bar{x}_{\eta}^2}<0 ∂ y 2 ∂ 2 u b k ( x ˉ η , 0 ) < − 1 + x ˉ η 2 2 < 0 b < b ˉ b<\bar{b} b < b ˉ b ˉ \bar{b} b ˉ x ˉ η \bar{x}_{\eta} x ˉ η η ∂ 2 v ∂ y 2 ( x ˉ η , 0 ) \eta\frac{\partial^2v}{\partial y^2}(\bar{x}_{\eta},0) η ∂ y 2 ∂ 2 v ( x ˉ η , 0 ) k k k b k < b ˉ b_k<\bar{b} b k < b ˉ
u b k ( x ˉ η , y k ) < y k 2 ( − 2 1 + x ˉ η 2 + o ( y k ) ) < 0 u_{b_k}(\bar{x}_{\eta},y_k)<y_k^2\left(-\frac{2}{1+\bar{x}_{\eta}^2}+o(y_k)\right)<0 u b k ( x ˉ η , y k ) < y k 2 ( − 1 + x ˉ η 2 2 + o ( y k ) ) < 0 and gives a contradiction with u b k ( x ˉ η , y k ) > 0 u_{b_k}(\bar{x}_{\eta},y_k)>0 u b k ( x ˉ η , y k ) > 0 Ω b ⊂ { ( x , y ) ∈ D : − x ˉ η < x < x ˉ η } \Omega_b \subset\{(x,y) \in D : -\bar{x}_{\eta}<x<\bar{x}_{\eta}\} Ω b ⊂ {( x , y ) ∈ D : − x ˉ η < x < x ˉ η }
Finally (51) is an easy consequence of the fact that ∣ v ( x , y ) ∣ ≤ C \lvert v(x,y)\rvert\le C ∣ v ( x , y )∣ ≤ C Ω b \Omega_b Ω b
u b ( x , y ) → { 1 2 log ( 1 − 4 y 2 ( 1 + x 2 + y 2 ) 2 ) ≤ 0 , when M = S 2 , − 1 2 log ( 1 + 4 y 2 ( 1 − x 2 − y 2 ) 2 ) ≤ 0 , when M = H 2 , u_b(x,y) \to\begin{cases}
\frac{1}{2}\log\left(1-\frac{4y^2}{(1+x^2+y^2)^2}\right)\le0, & \text{when } M=\mathbb{S}^2,\\
-\frac{1}{2}\log\left(1+\frac{4y^2}{(1-x^2-y^2)^2}\right)\le0, & \text{when } M=\mathbb{H}^2,
\end{cases} u b ( x , y ) → ⎩ ⎨ ⎧ 2 1 log ( 1 − ( 1 + x 2 + y 2 ) 2 4 y 2 ) ≤ 0 , − 2 1 log ( 1 + ( 1 − x 2 − y 2 ) 2 4 y 2 ) ≤ 0 , when M = S 2 , when M = H 2 , and the convergence is uniform on compact subsets of M M M
Next lemma will be used several times. It claims that ∂ Ω b \partial\Omega_b ∂ Ω b
Lemma 5.3. Let ( x b , y b ) ∈ ∂ Ω b (x_b,y_b)\in\partial\Omega_b ( x b , y b ) ∈ ∂ Ω b b → 0 b\to0 b → 0 x b → x ˉ ∈ [ − x ˉ η , x ˉ η ] x_b\to\bar{x}\in[-\bar{x}_\eta,\bar{x}_\eta] x b → x ˉ ∈ [ − x ˉ η , x ˉ η ]
y b = { ± 1 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + x ˉ 2 ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ 2 , y b 2 b ∣ x b − x ˉ ∣ ) ) , when M = S 2 , ± 1 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 − x ˉ 2 ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ 2 , y b 2 b ∣ x b − x ˉ ∣ ) ) , when M = H 2 . y_b=\begin{cases}
\pm\sqrt{\frac{1}{2}\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)}(1+\bar{x}^2)\left(1+O\left(y_b^2,\lvert x_b-\bar{x}\rvert^2,\frac{y_b^2}{b}\lvert x_b-\bar{x}\rvert\right)\right), & \text{when } M=\mathbb{S}^2,\\
\pm\sqrt{\frac{1}{2}\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)}(1-\bar{x}^2)\left(1+O\left(y_b^2,\lvert x_b-\bar{x}\rvert^2,\frac{y_b^2}{b}\lvert x_b-\bar{x}\rvert\right)\right), & \text{when } M=\mathbb{H}^2.
\end{cases} y b = ⎩ ⎨ ⎧ ± 2 1 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + x ˉ 2 ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ 2 , b y b 2 ∣ x b − x ˉ ∣ ) ) , ± 2 1 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 − x ˉ 2 ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ 2 , b y b 2 ∣ x b − x ˉ ∣ ) ) , when M = S 2 , when M = H 2 . Proof. We prove the result in the case M = S 2 M=\mathbb{S}^2 M = S 2 ( x b , y b ) ∈ ∂ Ω b (x_b,y_b)\in\partial\Omega_b ( x b , y b ) ∈ ∂ Ω b
1 2 log ( 1 − 4 y b 2 ( 1 + x b 2 + y b 2 ) 2 ) = b ( η v ( x b , y b ) − 1 ) . (53) \frac{1}{2}\log\left(1-\frac{4y_b^2}{(1+x_b^2+y_b^2)^2}\right)=b\left(\eta v(x_b,y_b)-1\right). \tag*{(53)} 2 1 log ( 1 − ( 1 + x b 2 + y b 2 ) 2 4 y b 2 ) = b ( η v ( x b , y b ) − 1 ) . (53) Since ∣ x b ∣ , ∣ y b ∣ ≤ C \lvert x_b\rvert,\lvert y_b\rvert\le C ∣ x b ∣ , ∣ y b ∣ ≤ C (53) is bounded by C b Cb C b b → 0 b\to0 b → 0 x b → x ˉ ∈ [ − x ˉ η , x ˉ η ] x_b\to\bar{x}\in[-\bar{x}_\eta,\bar{x}_\eta] x b → x ˉ ∈ [ − x ˉ η , x ˉ η ] y b → 0 y_b\to0 y b → 0 (53) and we get
η v ( x b , y b ) − 1 = η v ( x ˉ , 0 ) − 1 + η ∂ v ∂ x ( x ˉ , 0 ) ( x b − x ˉ ) + η 1 2 ∂ 2 v ∂ y 2 ( x ˉ , 0 ) y b 2 + O ( ∣ ( x b − x ˉ ) ∣ 2 ) + o ( y b 2 ) \eta v(x_b,y_b)-1=\eta v(\bar{x},0)-1+\eta\frac{\partial v}{\partial x}(\bar{x},0)(x_b-\bar{x})+\eta\frac{1}{2}\frac{\partial^2v}{\partial y^2}(\bar{x},0)y_b^2+O(\lvert(x_b-\bar{x})\rvert^2)+o(y_b^2) η v ( x b , y b ) − 1 = η v ( x ˉ , 0 ) − 1 + η ∂ x ∂ v ( x ˉ , 0 ) ( x b − x ˉ ) + η 2 1 ∂ y 2 ∂ 2 v ( x ˉ , 0 ) y b 2 + O (∣( x b − x ˉ ) ∣ 2 ) + o ( y b 2 ) (where we used that ∂ v ∂ y ( x ˉ , 0 ) = 0 \frac{\partial v}{\partial y}(\bar{x},0)=0 ∂ y ∂ v ( x ˉ , 0 ) = 0 ∂ 2 v ∂ x ∂ y ( x ˉ , 0 ) = 0 \frac{\partial^2v}{\partial x\partial y}(\bar{x},0)=0 ∂ x ∂ y ∂ 2 v ( x ˉ , 0 ) = 0
η v ( x b , y b ) − 1 = η f ( x ˉ ) − 1 + η f ′ ( x ˉ ) ( x b − x ˉ ) + O ( y b 2 , ∣ ( x b − x ˉ ) ∣ 2 ) \eta v(x_b,y_b)-1=\eta f(\bar{x})-1+\eta f'(\bar{x})(x_b-\bar{x})+O(y_b^2,\lvert(x_b-\bar{x})\rvert^2) η v ( x b , y b ) − 1 = η f ( x ˉ ) − 1 + η f ′ ( x ˉ ) ( x b − x ˉ ) + O ( y b 2 , ∣( x b − x ˉ ) ∣ 2 ) and
1 2 log ( 1 − 4 y b 2 ( 1 + x b 2 + y b 2 ) 2 ) = − 2 y b 2 ( 1 + x ˉ 2 ) 2 + O ( y b 4 , y b 2 b ∣ x b − x ˉ ∣ ) . \frac{1}{2}\log\left(1-\frac{4y_b^2}{(1+x_b^2+y_b^2)^2}\right)=-\frac{2y_b^2}{(1+\bar{x}^2)^2}+O\left(y_b^4,\frac{y_b^2}{b}\lvert x_b-\bar{x}\rvert\right). 2 1 log ( 1 − ( 1 + x b 2 + y b 2 ) 2 4 y b 2 ) = − ( 1 + x ˉ 2 ) 2 2 y b 2 + O ( y b 4 , b y b 2 ∣ x b − x ˉ ∣ ) . Then (53) becomes
2 y b 2 ( 1 + x ˉ 2 ) 2 + O ( y b 4 , y b 2 b ∣ x b − x ˉ ∣ ) = b [ 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) + O ( y b 2 , ( x b − x ˉ ) 2 ) ] . \frac{2y_b^2}{(1+\bar{x}^2)^2}+O\left(y_b^4,\frac{y_b^2}{b}\lvert x_b-\bar{x}\rvert\right)=b\left[1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})+O\left(y_b^2,(x_b-\bar{x})^2\right)\right]. ( 1 + x ˉ 2 ) 2 2 y b 2 + O ( y b 4 , b y b 2 ∣ x b − x ˉ ∣ ) = b [ 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) + O ( y b 2 , ( x b − x ˉ ) 2 ) ] . This implies that the ratio y b 2 b \frac{y_b^2}{b} b y b 2
y b 2 b = 1 2 ( 1 + x ˉ 2 ) 2 [ 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ] ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ 2 , y b 2 b ∣ x b − x ˉ ∣ ) ) (54) \frac{y_b^2}{b}=\frac{1}{2}(1+\bar{x}^2)^2\left[1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right]\left(1+O\left(y_b^2,\lvert x_b-\bar{x}\rvert^2,\frac{y_b^2}{b}\lvert x_b-\bar{x}\rvert\right)\right) \tag*{(54)} b y b 2 = 2 1 ( 1 + x ˉ 2 ) 2 [ 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ] ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ 2 , b y b 2 ∣ x b − x ˉ ∣ ) ) (54) and (5.21) follows.
Remark 5.4. As consequence of the previous lemma, when x ˉ ≠ x ˉ η \bar{x}\ne\bar{x}_\eta x ˉ = x ˉ η y b 2 ∼ b y_b^2\sim b y b 2 ∼ b 1 − η f ( x ˉ ) ≠ 0 1-\eta f(\bar{x})\ne0 1 − η f ( x ˉ ) = 0 x ˉ = x ˉ η \bar{x}=\bar{x}_\eta x ˉ = x ˉ η y b 2 ∼ b ( x b − x ˉ η ) y_b^2\sim b(x_b-\bar{x}_\eta) y b 2 ∼ b ( x b − x ˉ η ) 1 − η f ( x ˉ η ) = 0 1-\eta f(\bar{x}_\eta)=0 1 − η f ( x ˉ η ) = 0 f ′ ( x ˉ η ) ≠ 0 f'(\bar{x}_\eta)\ne0 f ′ ( x ˉ η ) = 0
Next aim is to prove that the superlevel set
ω b : = { ( x , y ) ∈ Ω b : u b ( x , y ) > b } (55) \omega_b:=\{(x,y)\in\Omega_b:u_b(x,y)>b\} \tag*{(55)} ω b := {( x , y ) ∈ Ω b : u b ( x , y ) > b } (55) has n n n
Lemma 5.5. Let x 1 , … , x n x_1,\ldots,x_n x 1 , … , x n f ( x ) f(x) f ( x ) ( − a n , a n ) (-a_n,a_n) ( − a n , a n ) b > 0 b>0 b > 0 ( x i , 0 ) ∈ ω b (x_i,0)\in\omega_b ( x i , 0 ) ∈ ω b i = 1 , … , n i=1,\ldots,n i = 1 , … , n
Proof. We compute
u b ( x i , 0 ) = b − η b f ( x i ) > b (56) u_b(x_i,0)=b-\eta b f(x_i)>b \tag*{(56)} u b ( x i , 0 ) = b − η b f ( x i ) > b (56) since f ( x i ) < 0 f(x_i)<0 f ( x i ) < 0 i = 1 , … , n i=1,\ldots,n i = 1 , … , n f f f
Next we have
Lemma 5.6. Let x ~ 1 , … , x ~ n − 1 \tilde{x}_1,\ldots,\tilde{x}_{n-1} x ~ 1 , … , x ~ n − 1 f ( x ) f(x) f ( x ) ( − a n , a n ) (-a_n,a_n) ( − a n , a n ) b 1 > 0 b_1>0 b 1 > 0 ω b \omega_b ω b x = x ~ i x=\tilde{x}_i x = x ~ i i = 1 , … , n − 1 i=1,\ldots,n-1 i = 1 , … , n − 1 0 < b < b 1 0<b<b_1 0 < b < b 1
Proof. We compute
u b ( x ~ i , y ) ≤ b ( 1 − η v ( x ~ i , y ) ) . u_b(\tilde{x}_i,y)\leq b(1-\eta v(\tilde{x}_i,y)). u b ( x ~ i , y ) ≤ b ( 1 − η v ( x ~ i , y )) . By Lemma 5.3 if ( x ~ i , y ) ∈ Ω b (\tilde{x}_i,y)\in\Omega_b ( x ~ i , y ) ∈ Ω b y → 0 y\to0 y → 0 b → 0 b\to0 b → 0 v ( x ~ i , y ) = v ( x ~ i , 0 ) + O ( ∣ y ∣ 2 ) = f ( x ~ i ) + O ( b ) v(\tilde{x}_i,y)=v(\tilde{x}_i,0)+O(|y|^2)=f(\tilde{x}_i)+O(b) v ( x ~ i , y ) = v ( x ~ i , 0 ) + O ( ∣ y ∣ 2 ) = f ( x ~ i ) + O ( b ) v v v y y y y b = O ( b ) y_b=O(\sqrt{b}) y b = O ( b )
u b ( x ~ i , y ) ≤ b ( 1 − η f ( x ~ i ) ) ( 1 + o ( 1 ) ) < b ( 1 − η 2 f ( x ~ i ) ) < 0 u_b(\tilde{x}_i,y)\leq b(1-\eta f(\tilde{x}_i))(1+o(1))<b\left(1-\frac{\eta}{2}f(\tilde{x}_i)\right)<0 u b ( x ~ i , y ) ≤ b ( 1 − η f ( x ~ i )) ( 1 + o ( 1 )) < b ( 1 − 2 η f ( x ~ i ) ) < 0 if b b b η \eta η □ \square □
Corollary 5.7. The set ω b \omega_b ω b n n n b b b
Proof. By Lemma 5.5 we have that the n n n ( x i , 0 ) (x_i,0) ( x i , 0 ) i = 1 , … , n ) i=1,\ldots,n) i = 1 , … , n ) ω b \omega_b ω b x = x ~ i x=\tilde{x}_i x = x ~ i ω b \omega_b ω b i = 1 , … , n − 1 i=1,\ldots,n-1 i = 1 , … , n − 1 □ \square □
In next proposition we prove the the domain Ω b \Omega_b Ω b b b b Ω b \Omega_b Ω b Ω b \Omega_b Ω b
Proposition 5.8. The set Ω b \Omega_b Ω b b b b
Proof. We want to show that
⟨ ν E , ( x , y ) ⟩ E ≥ α b > 0 (57) \langle\nu_E,(x,y)\rangle_E\geq\alpha_b>0 \tag*{(57)} ⟨ ν E , ( x , y ) ⟩ E ≥ α b > 0 (57) for some α b > 0 \alpha_b>0 α b > 0 ( x , y ) ∈ ∂ Ω b (x,y)\in\partial\Omega_b ( x , y ) ∈ ∂ Ω b ⟨ ⋅ , ⋅ ⟩ E \langle\cdot,\cdot\rangle_E ⟨ ⋅ , ⋅ ⟩ E R 2 \mathbb{R}^2 R 2 ν E ( x , y ) \nu_E(x,y) ν E ( x , y ) ∂ Ω b \partial\Omega_b ∂ Ω b ( x , y ) (x,y) ( x , y ) b b b (16) we will show that
⟨ ∇ u b ∣ ∇ u b ∣ ( x , y ) , ( x , y ) ⟩ E ≤ − α b < 0 \left\langle\frac{\nabla u_b}{|\nabla u_b|}(x,y),(x,y)\right\rangle_E\leq-\alpha_b<0 ⟨ ∣∇ u b ∣ ∇ u b ( x , y ) , ( x , y ) ⟩ E ≤ − α b < 0 for every ( x , y ) ∈ ∂ Ω b (x,y)\in\partial\Omega_b ( x , y ) ∈ ∂ Ω b b b b Ω b \Omega_b Ω b
Let us consider the spherical case first. Some computations give that
∂ u b ( x , y ) ∂ x = 8 x y 2 ( 1 + x 2 + y 2 ) [ ( 1 + x 2 + y 2 ) 2 − 4 y 2 ] − b η ∂ v ( x , y ) ∂ x \frac{\partial u_b(x,y)}{\partial x}
=
\frac{8xy^2}{(1+x^2+y^2)\left[(1+x^2+y^2)^2-4y^2\right]}
-b\eta\frac{\partial v(x,y)}{\partial x} ∂ x ∂ u b ( x , y ) = ( 1 + x 2 + y 2 ) [ ( 1 + x 2 + y 2 ) 2 − 4 y 2 ] 8 x y 2 − b η ∂ x ∂ v ( x , y ) ∂ u b ( x , y ) ∂ y = − 4 y ( 1 + x 2 − y 2 ) ( 1 + x 2 + y 2 ) [ ( 1 + x 2 + y 2 ) 2 − 4 y 2 ] − b η ∂ v ( x , y ) ∂ y . \frac{\partial u_b(x,y)}{\partial y}
=
\frac{-4y(1+x^2-y^2)}{(1+x^2+y^2)\left[(1+x^2+y^2)^2-4y^2\right]}
-b\eta\frac{\partial v(x,y)}{\partial y}. ∂ y ∂ u b ( x , y ) = ( 1 + x 2 + y 2 ) [ ( 1 + x 2 + y 2 ) 2 − 4 y 2 ] − 4 y ( 1 + x 2 − y 2 ) − b η ∂ y ∂ v ( x , y ) . Let ( x b , y b ) (x_b,y_b) ( x b , y b ) ∂ Ω b \partial\Omega_b ∂ Ω b y b → 0 y_b\to0 y b → 0 b → 0 b\to0 b → 0 Lemma 5.3 ) and we assume that
x b → x ˉ ∈ [ − x ˉ η , x ˉ η ] . x_b\to\bar{x}\in[-\bar{x}_\eta,\bar{x}_\eta]. x b → x ˉ ∈ [ − x ˉ η , x ˉ η ] . Since Ω b \Omega_b Ω b x = 0 x=0 x = 0 y = 0 y=0 y = 0 x ≥ 0 x\geq0 x ≥ 0 y ≥ 0 y\geq0 y ≥ 0
∂ v ( x b , y b ) ∂ x = ∂ v ( x ˉ , 0 ) ∂ x + ∂ 2 v ( x ˉ , 0 ) ∂ x 2 ( x b − x ˉ ) + ∂ 2 v ( x ˉ , 0 ) ∂ x ∂ y y b + O ( y b 2 , ( x b − x ˉ ) 2 ) = f ′ ( x ˉ ) + O ( y b 2 , ∣ x b − x ˉ ∣ ) \begin{aligned}
\frac{\partial v(x_b,y_b)}{\partial x}
&=\frac{\partial v(\bar{x},0)}{\partial x}
+\frac{\partial^2v(\bar{x},0)}{\partial x^2}(x_b-\bar{x})
+\frac{\partial^2v(\bar{x},0)}{\partial x\partial y}y_b
+O\left(y_b^2,(x_b-\bar{x})^2\right)\\
&=f'(\bar{x})+O\left(y_b^2,|x_b-\bar{x}|\right)
\end{aligned} ∂ x ∂ v ( x b , y b ) = ∂ x ∂ v ( x ˉ , 0 ) + ∂ x 2 ∂ 2 v ( x ˉ , 0 ) ( x b − x ˉ ) + ∂ x ∂ y ∂ 2 v ( x ˉ , 0 ) y b + O ( y b 2 , ( x b − x ˉ ) 2 ) = f ′ ( x ˉ ) + O ( y b 2 , ∣ x b − x ˉ ∣ ) and
∂ v ( x b , y b ) ∂ y = ∂ v ( x ˉ , 0 ) ∂ y + ∂ 2 v ( x ˉ , 0 ) ∂ x ∂ y ( x b − x ˉ ) + ∂ 2 v ( x ˉ , 0 ) ∂ y 2 y b + O ( y b 2 , ( x b − x ˉ ) 2 ) = ∂ 2 v ( x ˉ , 0 ) ∂ y 2 y b + O ( y b 2 , ( x b − x ˉ ) 2 ) \begin{aligned}
\frac{\partial v(x_b,y_b)}{\partial y}=\frac{\partial v(\bar{x},0)}{\partial y}+\frac{\partial^2v(\bar{x},0)}{\partial x\partial y}(x_b-\bar{x})+\frac{\partial^2v(\bar{x},0)}{\partial y^2}y_b+O\left(y_b^2,(x_b-\bar{x})^2\right) \\
&=\frac{\partial^2v(\bar{x},0)}{\partial y^2}y_b+O\left(y_b^2,(x_b-\bar{x})^2\right)
\end{aligned} ∂ y ∂ v ( x b , y b ) = ∂ y ∂ v ( x ˉ , 0 ) + ∂ x ∂ y ∂ 2 v ( x ˉ , 0 ) ( x b − x ˉ ) + ∂ y 2 ∂ 2 v ( x ˉ , 0 ) y b + O ( y b 2 , ( x b − x ˉ ) 2 ) = ∂ y 2 ∂ 2 v ( x ˉ , 0 ) y b + O ( y b 2 , ( x b − x ˉ ) 2 ) by the symmetry of v ( x , y ) v(x,y) v ( x , y )
8 x b y b 2 ( 1 + x b 2 + y b 2 ) [ ( 1 + x b 2 + y b 2 ) 2 − 4 y b 2 ] = 8 x ˉ y b 2 ( 1 + x ˉ 2 ) 3 ( 1 + O ( ∣ x b − x ˉ ∣ , y b 2 ) ) \frac{8x_by_b^2}{(1+x_b^2+y_b^2)\left[(1+x_b^2+y_b^2)^2-4y_b^2\right]}=\frac{8\bar{x}y_b^2}{(1+\bar{x}^2)^3}\left(1+O\left(|x_b-\bar{x}|,y_b^2\right)\right) ( 1 + x b 2 + y b 2 ) [ ( 1 + x b 2 + y b 2 ) 2 − 4 y b 2 ] 8 x b y b 2 = ( 1 + x ˉ 2 ) 3 8 x ˉ y b 2 ( 1 + O ( ∣ x b − x ˉ ∣ , y b 2 ) ) and
− 4 y b ( 1 + x b 2 − y b 2 ) ( 1 + x b 2 + y b 2 ) [ ( 1 + x b 2 + y b 2 ) 2 − 4 y b 2 ] = − 4 y b ( 1 + x ˉ 2 ) 2 ( 1 + O ( ∣ x b − x ˉ ∣ , y b 2 ) ) . \frac{-4y_b(1+x_b^2-y_b^2)}{(1+x_b^2+y_b^2)\left[(1+x_b^2+y_b^2)^2-4y_b^2\right]}=-\frac{4y_b}{(1+\bar{x}^2)^2}\left(1+O\left(|x_b-\bar{x}|,y_b^2\right)\right). ( 1 + x b 2 + y b 2 ) [ ( 1 + x b 2 + y b 2 ) 2 − 4 y b 2 ] − 4 y b ( 1 + x b 2 − y b 2 ) = − ( 1 + x ˉ 2 ) 2 4 y b ( 1 + O ( ∣ x b − x ˉ ∣ , y b 2 ) ) . Using (54) we get the expansion of the first derivative of u b u_b u b ( x b , y b ) (x_b,y_b) ( x b , y b )
∂ u b ( x b , y b ) ∂ x = b ( 4 x ˉ 1 + x ˉ 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ) ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) (58) \frac{\partial u_b(x_b,y_b)}{\partial x}=b\left(\frac{4\bar{x}}{1+\bar{x}^2}(1-\eta f(\bar{x})-\eta f'(\bar{x}))\right)\left(1+O(y_b^2,|x_b-\bar{x}|)\right) \tag*{(58)} ∂ x ∂ u b ( x b , y b ) = b ( 1 + x ˉ 2 4 x ˉ ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ )) ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) (58) and
∂ u b ( x b , y b ) ∂ y = [ − 4 b 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 + x ˉ 2 − b η ∂ 2 v ( x ˉ , 0 ) ∂ y 2 y b ⏟ = O ( b y b ) ] ( 1 + O ( ∣ x b − x ˉ ∣ , y b 2 ) ) = − 4 b 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 + x ˉ 2 ( 1 + O ( ∣ x b − x ˉ ∣ , b y b ) ) . (59) \begin{aligned}
\frac{\partial u_b(x_b,y_b)}{\partial y}
&=\left[-4\frac{\sqrt{\frac{b}{2}(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x}))}}{1+\bar{x}^2}-b\eta\underbrace{\frac{\partial^2v(\bar{x},0)}{\partial y^2}y_b}_{=O(by_b)}\right]\left(1+O(|x_b-\bar{x}|,y_b^2)\right)\\
&=-4\frac{\sqrt{\frac{b}{2}(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x}))}}{1+\bar{x}^2}\left(1+O\left(|x_b-\bar{x}|,\sqrt{by_b}\right)\right). \tag*{(59)}
\end{aligned} ∂ y ∂ u b ( x b , y b ) = − 4 1 + x ˉ 2 2 b ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ )) − b η = O ( b y b ) ∂ y 2 ∂ 2 v ( x ˉ , 0 ) y b ( 1 + O ( ∣ x b − x ˉ ∣ , y b 2 ) ) = − 4 1 + x ˉ 2 2 b ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ )) ( 1 + O ( ∣ x b − x ˉ ∣ , b y b ) ) . (59) So we have
∣ ∇ u b ( x b , y b ) ∣ 2 = b ( 8 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ( 1 + x ˉ 2 ) 2 + b ( 4 x ˉ ( 1 + x ˉ 2 ) 2 ( 1 − η f ( x ˉ ) ) − η f ′ ( x ˉ ) ) 2 ) ( 1 + o ( 1 ) ) . (60) \begin{aligned}
|\nabla u_b(x_b,y_b)|^2
&=b\left(8\frac{1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})}{(1+\bar{x}^2)^2}
+b\left(\frac{4\bar{x}}{(1+\bar{x}^2)^2}(1-\eta f(\bar{x}))-\eta f'(\bar{x})\right)^2\right)(1+o(1)). \tag*{(60)}
\end{aligned} ∣∇ u b ( x b , y b ) ∣ 2 = b ( 8 ( 1 + x ˉ 2 ) 2 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) + b ( ( 1 + x ˉ 2 ) 2 4 x ˉ ( 1 − η f ( x ˉ )) − η f ′ ( x ˉ ) ) 2 ) ( 1 + o ( 1 )) . (60) Now if x ˉ ≠ x ˉ η \bar{x}\ne\bar{x}_{\eta} x ˉ = x ˉ η 1 − η f ( x ˉ ) ≠ 0 1-\eta f(\bar{x})\ne0 1 − η f ( x ˉ ) = 0 (60) gives
∣ ∇ u b ( x b , y b ) ∣ = 2 2 b 1 − η f ( x ˉ ) 1 + x ˉ 2 ( 1 + o ( 1 ) ) . (61) |\nabla u_b(x_b,y_b)|=2\sqrt{2}\sqrt{b}\frac{\sqrt{1-\eta f(\bar{x})}}{1+\bar{x}^2}(1+o(1)). \tag*{(61)} ∣∇ u b ( x b , y b ) ∣ = 2 2 b 1 + x ˉ 2 1 − η f ( x ˉ ) ( 1 + o ( 1 )) . (61) On the other hand if x ˉ = x ˉ η \bar{x}=\bar{x}_{\eta} x ˉ = x ˉ η 1 − η f ( x ˉ η ) = 0 1-\eta f(\bar{x}_{\eta})=0 1 − η f ( x ˉ η ) = 0 f ′ ( x ˉ η ) ≠ 0 f'(\bar{x}_{\eta})\ne0 f ′ ( x ˉ η ) = 0 (60) becomes
∣ ∇ u b ( x b , y b ) ∣ = { 2 2 b η f ′ ( x ˉ η ) ( x ˉ η − x b ) 1 + x ˉ η 2 ( 1 + o ( 1 ) ) when ∣ x ˉ η − x b ∣ b → ∞ , b 2 2 η f ′ ( x ˉ η ) ( γ + 2 x ˉ η 2 η f ′ ( x ˉ η ) ) 1 + x ˉ η 2 ( 1 + o ( 1 ) ) when ∣ x ˉ η − x b ∣ b → γ > 0 , b 4 x ˉ η 1 + x ˉ η 2 η f ′ ( x ˉ η ) ( 1 + o ( 1 ) ) when ∣ x ˉ η − x b ∣ b → 0. (62) |\nabla u_b(x_b,y_b)|=
\begin{cases}
2\sqrt{2}\sqrt{b}\dfrac{\sqrt{\eta f'(\bar{x}_{\eta})(\bar{x}_{\eta}-x_b)}}{1+\bar{x}_{\eta}^2}(1+o(1)) & \text{when }\dfrac{|\bar{x}_{\eta}-x_b|}{b}\to\infty,\\
b2\sqrt{2}\dfrac{\sqrt{\eta f'(\bar{x}_{\eta})(\gamma+2\bar{x}_{\eta}^2\eta f'(\bar{x}_{\eta}))}}{1+\bar{x}_{\eta}^2}(1+o(1)) & \text{when }\dfrac{|\bar{x}_{\eta}-x_b|}{b}\to\gamma>0,\\
b\dfrac{4\bar{x}_{\eta}}{1+\bar{x}_{\eta}^2}\eta f'(\bar{x}_{\eta})(1+o(1)) & \text{when }\dfrac{|\bar{x}_{\eta}-x_b|}{b}\to0.
\end{cases} \tag*{(62)} ∣∇ u b ( x b , y b ) ∣ = ⎩ ⎨ ⎧ 2 2 b 1 + x ˉ η 2 η f ′ ( x ˉ η ) ( x ˉ η − x b ) ( 1 + o ( 1 )) b 2 2 1 + x ˉ η 2 η f ′ ( x ˉ η ) ( γ + 2 x ˉ η 2 η f ′ ( x ˉ η )) ( 1 + o ( 1 )) b 1 + x ˉ η 2 4 x ˉ η η f ′ ( x ˉ η ) ( 1 + o ( 1 )) when b ∣ x ˉ η − x b ∣ → ∞ , when b ∣ x ˉ η − x b ∣ → γ > 0 , when b ∣ x ˉ η − x b ∣ → 0. (62) Let us compute
⟨ ∇ u b ( x , y ) ∣ ∇ u b ( x , y ) ∣ , ( x , y ) ⟩ E = b ∣ ∇ u b ( x , y ) ∣ ( 4 x ˉ 2 1 + x ˉ 2 ( 1 − η f ( x ˉ ) ) − η x ˉ f ′ ( x ˉ ) − 2 ( 1 − η f ( x ˉ ) ) ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) = b ∣ ∇ u b ( x , y ) ∣ 2 ( 1 − η f ( x ˉ ) ) ( x ˉ 2 − 1 ) − η ( 1 + x ˉ 2 ) x ˉ f ′ ( x ˉ ) 1 + x ˉ 2 ( 1 + o ( 1 ) ) . (63) \begin{aligned}
\left\langle\frac{\nabla u_b(x,y)}{|\nabla u_b(x,y)|},(x,y)\right\rangle_E
&=\frac{b}{|\nabla u_b(x,y)|}\left(\frac{4\bar{x}^2}{1+\bar{x}^2}(1-\eta f(\bar{x}))-\eta\bar{x}f'(\bar{x})-2(1-\eta f(\bar{x}))\right)\left(1+O(y_b^2,|x_b-\bar{x}|)\right)\\
&=\frac{b}{|\nabla u_b(x,y)|}\frac{2(1-\eta f(\bar{x}))(\bar{x}^2-1)-\eta(1+\bar{x}^2)\bar{x}f'(\bar{x})}{1+\bar{x}^2}(1+o(1)). \tag*{(63)}
\end{aligned} ⟨ ∣∇ u b ( x , y ) ∣ ∇ u b ( x , y ) , ( x , y ) ⟩ E = ∣∇ u b ( x , y ) ∣ b ( 1 + x ˉ 2 4 x ˉ 2 ( 1 − η f ( x ˉ )) − η x ˉ f ′ ( x ˉ ) − 2 ( 1 − η f ( x ˉ )) ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) = ∣∇ u b ( x , y ) ∣ b 1 + x ˉ 2 2 ( 1 − η f ( x ˉ )) ( x ˉ 2 − 1 ) − η ( 1 + x ˉ 2 ) x ˉ f ′ ( x ˉ ) ( 1 + o ( 1 )) . (63) so that the proof of the starshapedness of Ω b \Omega_b Ω b b b b
− 2 ( 1 − η f ( x ˉ ) ) ( 1 − x ˉ 2 ) − η x ˉ f ′ ( x ˉ ) ( 1 + x ˉ 2 ) ≤ − α < 0 (64) -2(1-\eta f(\bar{x}))(1-\bar{x}^{2})-\eta\bar{x}f'(\bar{x})(1+\bar{x}^{2})\leq-\alpha<0 \tag*{(64)} − 2 ( 1 − η f ( x ˉ )) ( 1 − x ˉ 2 ) − η x ˉ f ′ ( x ˉ ) ( 1 + x ˉ 2 ) ≤ − α < 0 (64) for every x ˉ ∈ [ 0 , x ˉ η ] \bar{x}\in[0,\bar{x}_{\eta}] x ˉ ∈ [ 0 , x ˉ η ] α \alpha α
We consider two different cases: in the first one we will prove (64) for 0 ≤ x < a n 0\leq x<a_n 0 ≤ x < a n x ≥ a n x\geq a_n x ≥ a n
Case 0 ≤ x ˉ ≤ a n 0\leq\bar{x}\leq a_n 0 ≤ x ˉ ≤ a n
By (47) we know that
1 η > sup ( − a n , a n ) [ f ( x ) − x ( 1 + x 2 ) 2 ( 1 − x 2 ) f ′ ( x ) ] . \frac{1}{\eta}>\sup_{(-a_n,a_n)}\left[f(x)-\frac{x(1+x^2)}{2(1-x^2)}f'(x)\right]. η 1 > ( − a n , a n ) sup [ f ( x ) − 2 ( 1 − x 2 ) x ( 1 + x 2 ) f ′ ( x ) ] . it is then easy to see that
sup ( 0 , a n ) [ − 2 ( 1 − x 2 ) + η ( 2 f ( x ) ( 1 − x 2 ) + x f ′ ( x ) ( 1 + x 2 ) ) ] < 0. \sup_{(0,a_n)}\left[-2(1-x^2)+\eta\left(2f(x)(1-x^2)+xf'(x)(1+x^2)\right)\right]<0. ( 0 , a n ) sup [ − 2 ( 1 − x 2 ) + η ( 2 f ( x ) ( 1 − x 2 ) + x f ′ ( x ) ( 1 + x 2 ) ) ] < 0. This proves (64) when x ˉ ∈ [ 0 , a n ] \bar{x}\in[0,a_n] x ˉ ∈ [ 0 , a n ]
Case a n ≤ x ˉ ≤ x ˉ η a_n\leq\bar{x}\leq\bar{x}_{\eta} a n ≤ x ˉ ≤ x ˉ η
We already know that f ( x ) ≤ f ( x n ) = 1 η f(x)\leq f(x_n)=\frac{1}{\eta} f ( x ) ≤ f ( x n ) = η 1 1 − η f ( x ) ≥ 0 1-\eta f(x)\geq0 1 − η f ( x ) ≥ 0 x ∈ [ a n , x ˉ η ] x\in[a_n,\bar{x}_{\eta}] x ∈ [ a n , x ˉ η ] f ′ ( x ) > 0 f'(x)>0 f ′ ( x ) > 0 x > a n x>a_n x > a n
sup ( a n , x ˉ η ) − 2 ( 1 − η f ( x ) ) ( 1 − x 2 ) − η x f ′ ( x ) ( 1 + x 2 ) ≤ sup ( a n , x ˉ η ) − η x f ′ ( x ) ( 1 + x 2 ) < 0. \sup_{(a_n,\bar{x}_{\eta})}-2(1-\eta f(x))(1-x^2)-\eta xf'(x)(1+x^2)\leq\sup_{(a_n,\bar{x}_{\eta})}-\eta xf'(x)(1+x^2)<0. ( a n , x ˉ η ) sup − 2 ( 1 − η f ( x )) ( 1 − x 2 ) − η x f ′ ( x ) ( 1 + x 2 ) ≤ ( a n , x ˉ η ) sup − η x f ′ ( x ) ( 1 + x 2 ) < 0. and this proves (64) . So (57) follows by (62) , (63) and (64) . Although it is not essential in the proof, we write the expansion of α b \alpha_b α b
α b = { α b 2 2 1 − η f ( x ˉ ) when x b → x ˉ ≠ x ˉ η α b 2 2 η f ′ ( x ˉ η ) ( x ˉ η − x b ) when x b → x ˉ η , ∣ x ˉ η − x b ∣ b → ∞ α 2 2 η f ′ ( x ˉ η ) ( γ + 2 x ˉ η 2 η f ′ ′ ( x ˉ η ) ) when ∣ x ˉ η − x b ∣ b → γ > 0 α 4 x ˉ η η f ′ ( x ˉ η ) when ∣ x ˉ η − x b ∣ b → 0. (65) \alpha_b=
\begin{cases}
\frac{\alpha\sqrt{b}}{2\sqrt{2}\sqrt{1-\eta f(\bar{x})}} & \text{when }x_b\to\bar{x}\neq\bar{x}_{\eta}\\[6pt]
\frac{\alpha\sqrt{b}}{2\sqrt{2}\sqrt{\eta f'(\bar{x}_{\eta})(\bar{x}_{\eta}-x_b)}} & \text{when }x_b\to\bar{x}_{\eta},\ \frac{|\bar{x}_{\eta}-x_b|}{b}\to\infty\\[6pt]
\frac{\alpha}{2\sqrt{2}\sqrt{\eta f'(\bar{x}_{\eta})(\gamma+2\bar{x}_{\eta}^{2}\eta f''(\bar{x}_{\eta}))}} & \text{when }\frac{|\bar{x}_{\eta}-x_b|}{b}\to\gamma>0\\[6pt]
\frac{\alpha}{4\bar{x}_{\eta}\eta f'(\bar{x}_{\eta})} & \text{when }\frac{|\bar{x}_{\eta}-x_b|}{b}\to0.
\end{cases} \tag*{(65)} α b = ⎩ ⎨ ⎧ 2 2 1 − η f ( x ˉ ) α b 2 2 η f ′ ( x ˉ η ) ( x ˉ η − x b ) α b 2 2 η f ′ ( x ˉ η ) ( γ + 2 x ˉ η 2 η f ′′ ( x ˉ η )) α 4 x ˉ η η f ′ ( x ˉ η ) α when x b → x ˉ = x ˉ η when x b → x ˉ η , b ∣ x ˉ η − x b ∣ → ∞ when b ∣ x ˉ η − x b ∣ → γ > 0 when b ∣ x ˉ η − x b ∣ → 0. (65) In the hyperbolic case, reasoning as in (58) and (59) , we have
∂ u b ( x b , y b ) ∂ x = b ( − 4 x ˉ 1 − x ˉ 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ) ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) \frac{\partial u_b(x_b,y_b)}{\partial x}
=b\left(-\frac{4\bar{x}}{1-\bar{x}^{2}}(1-\eta f(\bar{x})-\eta f'(\bar{x}))\right)(1+O(y_b^{2},|x_b-\bar{x}|)) ∂ x ∂ u b ( x b , y b ) = b ( − 1 − x ˉ 2 4 x ˉ ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ )) ) ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ )) and
∂ u b ( x b , y b ) ∂ y = [ − 4 b 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 − x ˉ 2 − b η ∂ 2 v ( x ˉ , 0 ) ∂ y 2 ⏟ = O ( b y b ) y b ] ( 1 + O ( ∣ x b − x ˉ ∣ , y b 2 ) ) . \frac{\partial u_b(x_b,y_b)}{\partial y}
=\left[-4\frac{\sqrt{\frac{b}{2}(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x}))}}{1-\bar{x}^{2}}-b\eta\underbrace{\frac{\partial^{2}v(\bar{x},0)}{\partial y^{2}}}_{=O(by_b)}y_b\right](1+O(|x_b-\bar{x}|,y_b^{2})). ∂ y ∂ u b ( x b , y b ) = − 4 1 − x ˉ 2 2 b ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ )) − b η = O ( b y b ) ∂ y 2 ∂ 2 v ( x ˉ , 0 ) y b ( 1 + O ( ∣ x b − x ˉ ∣ , y b 2 )) . This gives the expansion of ∣ ∇ u b ( x b , y b ) ∣ |\nabla u_b(x_b,y_b)| ∣∇ u b ( x b , y b ) ∣
∣ ∇ u b ( x b , y b ) ∣ = { b 2 2 2 1 − η f ( x ˉ ) 1 − x ˉ 2 ( 1 + o ( 1 ) ) when x b → x ˉ ≠ x η b 2 2 2 η f ′ ( x ˉ η ) ( x ˉ η − x b ) 1 − x ˉ η 2 ( 1 + o ( 1 ) ) when ∣ x ˉ η − x b ∣ b → ∞ b 2 2 η f ′ ( x ˉ η ) ( γ + 2 x ˉ η 2 η f ′ ′ ( x ˉ η ) ) 1 − x ˉ η 2 ( 1 + o ( 1 ) ) when ∣ x ˉ η − x b ∣ b → γ > 0 b 4 x ˉ η 1 − x ˉ η 2 η f ′ ( x ˉ η ) ( 1 + o ( 1 ) ) when ∣ x ˉ η − x b ∣ b → 0. (66) |\nabla u_b(x_b,y_b)|=
\begin{cases}
\sqrt{b_2}2\sqrt{2}\sqrt{\frac{1-\eta f(\bar{x})}{1-\bar{x}^{2}}}(1+o(1)) & \text{when }x_b\to\bar{x}\neq x_{\eta}\\[6pt]
\sqrt{b_2}2\sqrt{2}\sqrt{\frac{\eta f'(\bar{x}_{\eta})(\bar{x}_{\eta}-x_b)}{1-\bar{x}_{\eta}^{2}}}(1+o(1)) & \text{when }\frac{|\bar{x}_{\eta}-x_b|}{b}\to\infty\\[6pt]
b_2\sqrt{2}\sqrt{\frac{\eta f'(\bar{x}_{\eta})(\gamma+2\bar{x}_{\eta}^{2}\eta f''(\bar{x}_{\eta}))}{1-\bar{x}_{\eta}^{2}}}(1+o(1)) & \text{when }\frac{|\bar{x}_{\eta}-x_b|}{b}\to\gamma>0\\[6pt]
b\frac{4\bar{x}_{\eta}}{1-\bar{x}_{\eta}^{2}}\eta f'(\bar{x}_{\eta})(1+o(1)) & \text{when }\frac{|\bar{x}_{\eta}-x_b|}{b}\to0.
\end{cases} \tag*{(66)} ∣∇ u b ( x b , y b ) ∣ = ⎩ ⎨ ⎧ b 2 2 2 1 − x ˉ 2 1 − η f ( x ˉ ) ( 1 + o ( 1 )) b 2 2 2 1 − x ˉ η 2 η f ′ ( x ˉ η ) ( x ˉ η − x b ) ( 1 + o ( 1 )) b 2 2 1 − x ˉ η 2 η f ′ ( x ˉ η ) ( γ + 2 x ˉ η 2 η f ′′ ( x ˉ η )) ( 1 + o ( 1 )) b 1 − x ˉ η 2 4 x ˉ η η f ′ ( x ˉ η ) ( 1 + o ( 1 )) when x b → x ˉ = x η when b ∣ x ˉ η − x b ∣ → ∞ when b ∣ x ˉ η − x b ∣ → γ > 0 when b ∣ x ˉ η − x b ∣ → 0. (66) Then
⟨ ∇ u b ( x , y ) , ( x , y ) ⟩ E = b − 2 ( 1 − η f ( x ˉ ) ) ( x ˉ 2 + 1 ) − η ( 1 − x ˉ 2 ) x ˉ f ′ ( x ˉ ) 1 − x ˉ 2 + o ( b ) (67) \langle\nabla u_b(x,y),(x,y)\rangle_E
=b\frac{-2(1-\eta f(\bar{x}))(\bar{x}^{2}+1)-\eta(1-\bar{x}^{2})\bar{x}f'(\bar{x})}{1-\bar{x}^{2}}+o(b) \tag*{(67)} ⟨ ∇ u b ( x , y ) , ( x , y ) ⟩ E = b 1 − x ˉ 2 − 2 ( 1 − η f ( x ˉ )) ( x ˉ 2 + 1 ) − η ( 1 − x ˉ 2 ) x ˉ f ′ ( x ˉ ) + o ( b ) (67) and letting − α = sup ( 0 , a n ) − 2 ( 1 − η f ( x ˉ ) ) ( x ˉ 2 + 1 ) − η ( 1 − x ˉ 2 ) x ˉ f ′ ( x ˉ ) < 0 -\alpha=\sup(0,a_n)-2(1-\eta f(\bar{x}))(\bar{x}^2+1)-\eta(1-\bar{x}^2)\bar{x}f'(\bar{x})<0 − α = sup ( 0 , a n ) − 2 ( 1 − η f ( x ˉ )) ( x ˉ 2 + 1 ) − η ( 1 − x ˉ 2 ) x ˉ f ′ ( x ˉ ) < 0 α ˉ > 0 \bar{\alpha}>0 α ˉ > 0
⟨ ν E , ( x , y ) ⟩ E > α ˉ b + o ( b ) if x b → x ˉ ≠ x η (68) \langle\nu_E,(x,y)\rangle_E>\bar{\alpha}\sqrt{b}+o(\sqrt{b})\quad\text{if }x_b\to\bar{x}\ne x_\eta \tag*{(68)} ⟨ ν E , ( x , y ) ⟩ E > α ˉ b + o ( b ) if x b → x ˉ = x η (68) and, using (65)
⟨ ν E , ( x , y ) ⟩ E = α b ≥ 0 if x b → x η . (69) \langle\nu_E,(x,y)\rangle_E=\alpha_b\geq0\quad\text{if }x_b\to x_\eta. \tag*{(69)} ⟨ ν E , ( x , y ) ⟩ E = α b ≥ 0 if x b → x η . (69) We conclude this section showing that (35) and (36) hold. It is convenient to look at the stereographic projection of a star-shaped domain with respect to the south pole − P -P − P (5) into a condition for the image domain in R 2 \mathbb{R}^2 R 2
Note that since Ω \Omega Ω P P P − P ∉ Ω -P\notin\Omega − P ∈ / Ω Ψ : S 2 → R 2 \Psi:S^2\to\mathbb{R}^2 Ψ : S 2 → R 2 − P -P − P ( x , y , z ) ∈ S 2 ⊂ R 3 (x,y,z)\in S^2\subset\mathbb{R}^3 ( x , y , z ) ∈ S 2 ⊂ R 3 ( X , Y ) = ( x 1 + z , y 1 + z ) (X,Y)=\left(\frac{x}{1+z},\frac{y}{1+z}\right) ( X , Y ) = ( 1 + z x , 1 + z y ) R 2 \mathbb{R}^2 R 2 Ω E : = Ψ ( Ω ) ⊂ R 2 \Omega_E:=\Psi(\Omega)\subset\mathbb{R}^2 Ω E := Ψ ( Ω ) ⊂ R 2 Ω \Omega Ω P P P Ω E \Omega_E Ω E 0 0 0
Let k Ω E k_{\Omega_E} k Ω E ∂ Ω E \partial\Omega_E ∂ Ω E ν E \nu_E ν E S 2 S^2 S 2 4 ( 1 + X 2 + Y 2 ) 2 ( d X 2 + d Y 2 ) \frac{4}{(1+X^2+Y^2)^2}(dX^2+dY^2) ( 1 + X 2 + Y 2 ) 2 4 ( d X 2 + d Y 2 )
k Ω = 1 + X 2 + Y 2 2 k Ω E − ⟨ ν E , ( X , Y ) ⟩ E , k_{\Omega}=\frac{1+X^2+Y^2}{2}k_{\Omega_E}-\langle\nu_E,(X,Y)\rangle_E, k Ω = 2 1 + X 2 + Y 2 k Ω E − ⟨ ν E , ( X , Y ) ⟩ E , where ⟨ ⋅ , ⋅ ⟩ E \langle\cdot,\cdot\rangle_E ⟨ ⋅ , ⋅ ⟩ E R 2 \mathbb{R}^2 R 2
cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ = 1 − X 2 − Y 2 2 k Ω E + ⟨ ν E , ( X , Y ) ⟩ E . (70) \cos(\theta)k_{\Omega}+\sin(\theta)\langle\nu,\vec e_\theta\rangle=\frac{1-X^2-Y^2}{2}k_{\Omega_E}+\langle\nu_E,(X,Y)\rangle_E. \tag*{(70)} cos ( θ ) k Ω + sin ( θ ) ⟨ ν , e θ ⟩ = 2 1 − X 2 − Y 2 k Ω E + ⟨ ν E , ( X , Y ) ⟩ E . (70) Lemma 5.9. For b b b
lim b → 0 ( cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ ) ≥ 0 on ∂ Ω b , ⊂ S 2 , lim b → 0 ( cosh ( r ) k Ω b − sinh ( r ) ⟨ ν , e ⃗ r ⟩ ) ≥ 0 on ∂ Ω b ⊂ H 2 . (71) \begin{aligned}
\lim_{b\to0}\bigl(\cos(\theta)k_{\Omega_b}+\sin(\theta)\langle\nu,\vec e_\theta\rangle\bigr)&\geq0 &&\text{on }\partial\Omega_b,\subset S^2,\\
\lim_{b\to0}\bigl(\cosh(r)k_{\Omega_b}-\sinh(r)\langle\nu,\vec e_r\rangle\bigr)&\geq0 &&\text{on }\partial\Omega_b\subset H^2. \tag*{(71)}
\end{aligned} b → 0 lim ( cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e θ ⟩ ) b → 0 lim ( cosh ( r ) k Ω b − sinh ( r ) ⟨ ν , e r ⟩ ) ≥ 0 ≥ 0 on ∂ Ω b , ⊂ S 2 , on ∂ Ω b ⊂ H 2 . (71) where θ = dist ( ⋅ , 0 ) \theta=\operatorname{dist}(\cdot,0) θ = dist ( ⋅ , 0 ) r = dist ( ⋅ , 0 ) r=\operatorname{dist}(\cdot,0) r = dist ( ⋅ , 0 ) H 2 H^2 H 2
Proof. In the proof of Proposition 5.8 we have shown that ⟨ ν E , ( x , y ) ⟩ E ≥ α b > 0 \langle\nu_E,(x,y)\rangle_E\geq\alpha_b>0 ⟨ ν E , ( x , y ) ⟩ E ≥ α b > 0 (57) . So it will be enough to show that lim b → 0 cos ( θ ) k Ω b ≥ 0 \lim_{b\to0}\cos(\theta)k_{\Omega_b}\geq0 lim b → 0 cos ( θ ) k Ω b ≥ 0 ( x , y ) ∈ ∂ Ω b (x,y)\in\partial\Omega_b ( x , y ) ∈ ∂ Ω b
Case 1, Ω b ⊂ S 2 \Omega_b\subset S^2 Ω b ⊂ S 2
Using stereographic coordinates we have,
cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ = 1 − x 2 − y 2 2 k Ω b , E + ⟨ ν E , ( x , y ) ⟩ E (72) \cos(\theta)k_{\Omega_b}+\sin(\theta)\langle\nu,\vec e_\theta\rangle=\frac{1-x^2-y^2}{2}k_{\Omega_b,E}+\langle\nu_E,(x,y)\rangle_E \tag*{(72)} cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e θ ⟩ = 2 1 − x 2 − y 2 k Ω b , E + ⟨ ν E , ( x , y ) ⟩ E (72) where k Ω b , E k_{\Omega_b,E} k Ω b , E ∂ Ω b \partial\Omega_b ∂ Ω b ⟨ ν E , ( x , y ) ⟩ E \langle\nu_E,(x,y)\rangle_E ⟨ ν E , ( x , y ) ⟩ E Proposition 5.8 , see (57) with α b \alpha_b α b (65) . Recalling that
k Ω b , E = − ∂ 2 u b ∂ x 2 ( ∂ u b ∂ y ) 2 − 2 ∂ 2 u b ∂ x ∂ y ∂ u b ∂ x ∂ u b ∂ y + ∂ 2 u b ∂ y 2 ( ∂ u b ∂ x ) 2 ∣ ∇ u b ∣ 3 , k_{\Omega_b,E}=-\frac{\dfrac{\partial^2u_b}{\partial x^2}\left(\dfrac{\partial u_b}{\partial y}\right)^2-2\dfrac{\partial^2u_b}{\partial x\partial y}\dfrac{\partial u_b}{\partial x}\dfrac{\partial u_b}{\partial y}+\dfrac{\partial^2u_b}{\partial y^2}\left(\dfrac{\partial u_b}{\partial x}\right)^2}{\lvert\nabla u_b\rvert^3}, k Ω b , E = − ∣ ∇ u b ∣ 3 ∂ x 2 ∂ 2 u b ( ∂ y ∂ u b ) 2 − 2 ∂ x ∂ y ∂ 2 u b ∂ x ∂ u b ∂ y ∂ u b + ∂ y 2 ∂ 2 u b ( ∂ x ∂ u b ) 2 , we expand the second derivatives of u b u_b u b ( x b , y b ) ∈ Ω b (x_b,y_b)\in\Omega_b ( x b , y b ) ∈ Ω b x b → x ˉ ∈ [ − x ˉ η , − x ˉ η ] x_b\to\bar{x}\in[-\bar{x}_\eta,-\bar{x}_\eta] x b → x ˉ ∈ [ − x ˉ η , − x ˉ η ] b → 0 b\to0 b → 0
of x ≥ 0 x \ge0 x ≥ 0 y ≥ 0 y \ge0 y ≥ 0 y b → 0 y_b \to0 y b → 0 b → 0 b \to0 b → 0
∂ 2 ψ b ∂ x 2 ( x b , y b ) = 8 y b 2 1 − 5 x ˉ 2 ( 1 + x ˉ 2 ) 4 ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) = 4 b ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 − 5 x ˉ 2 ( 1 + x ˉ 2 ) 2 ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) \begin{aligned}
\frac{\partial^2\psi_b}{\partial x^2}(x_b,y_b)=8y_b^2\frac{1-5\bar{x}^2}{(1+\bar{x}^2)^4}\left(1+O(y_b^2,|x_b-\bar{x}|)\right) \\
&=4b\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)\frac{1-5\bar{x}^2}{(1+\bar{x}^2)^2}\left(1+O(y_b^2,|x_b-\bar{x}|)\right)
\end{aligned} ∂ x 2 ∂ 2 ψ b ( x b , y b ) = 8 y b 2 ( 1 + x ˉ 2 ) 4 1 − 5 x ˉ 2 ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) = 4 b ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + x ˉ 2 ) 2 1 − 5 x ˉ 2 ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) ∂ 2 ψ b ∂ x ∂ y ( x b , y b ) = 16 y b x ˉ ( 1 + x ˉ 2 ) 3 ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) = 16 b 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) x ˉ ( 1 + x ˉ 2 ) 2 ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) \begin{aligned}
\frac{\partial^2\psi_b}{\partial x\partial y}(x_b,y_b)=16y_b\frac{\bar{x}}{(1+\bar{x}^2)^3}\left(1+O(y_b^2,|x_b-\bar{x}|)\right) \\
&=16\sqrt{\frac{b}{2}}\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)\frac{\bar{x}}{(1+\bar{x}^2)^2}\left(1+O(y_b^2,|x_b-\bar{x}|)\right)
\end{aligned} ∂ x ∂ y ∂ 2 ψ b ( x b , y b ) = 16 y b ( 1 + x ˉ 2 ) 3 x ˉ ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) = 16 2 b ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + x ˉ 2 ) 2 x ˉ ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) ∂ 2 ψ b ∂ y 2 ( x b , y b ) = − 4 ( 1 + x ˉ 2 ) 2 ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) . \frac{\partial^2\psi_b}{\partial y^2}(x_b,y_b)=-\frac{4}{(1+\bar{x}^2)^2}\left(1+O(y_b^2,|x_b-\bar{x}|)\right). ∂ y 2 ∂ 2 ψ b ( x b , y b ) = − ( 1 + x ˉ 2 ) 2 4 ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) . Moreover, by the symmetry of v ( x , y ) v(x,y) v ( x , y )
∂ 2 v ( x b , y b ) ∂ x 2 = ∂ 2 v ( x ˉ , 0 ) ∂ x 2 + O ( y b 2 , ∣ x b − x ˉ ∣ ) = f ′ ′ ( x ˉ ) + O ( y b 2 , ∣ x b − x ˉ ∣ ) \frac{\partial^2v(x_b,y_b)}{\partial x^2}=\frac{\partial^2v(\bar{x},0)}{\partial x^2}+O(y_b^2,|x_b-\bar{x}|)=f''(\bar{x})+O(y_b^2,|x_b-\bar{x}|) ∂ x 2 ∂ 2 v ( x b , y b ) = ∂ x 2 ∂ 2 v ( x ˉ , 0 ) + O ( y b 2 , ∣ x b − x ˉ ∣ ) = f ′′ ( x ˉ ) + O ( y b 2 , ∣ x b − x ˉ ∣ ) ∂ 2 v ( x b , y b ) ∂ x ∂ y = ∂ 2 v ( x ˉ , 0 ) ∂ x ∂ y + O ( y b , ∣ x b − x ˉ ∣ 2 ) = O ( y b , ∣ x b − x ˉ ∣ 2 ) \frac{\partial^2v(x_b,y_b)}{\partial x\partial y}=\frac{\partial^2v(\bar{x},0)}{\partial x\partial y}+O(y_b,|x_b-\bar{x}|^2)=O(y_b,|x_b-\bar{x}|^2) ∂ x ∂ y ∂ 2 v ( x b , y b ) = ∂ x ∂ y ∂ 2 v ( x ˉ , 0 ) + O ( y b , ∣ x b − x ˉ ∣ 2 ) = O ( y b , ∣ x b − x ˉ ∣ 2 ) ∂ 2 v ( x b , y b ) ∂ y 2 = ∂ 2 v ( x ˉ , 0 ) ∂ y 2 + O ( y b 2 , ∣ x b − x ˉ ∣ ) . \frac{\partial^2v(x_b,y_b)}{\partial y^2}=\frac{\partial^2v(\bar{x},0)}{\partial y^2}+O(y_b^2,|x_b-\bar{x}|). ∂ y 2 ∂ 2 v ( x b , y b ) = ∂ y 2 ∂ 2 v ( x ˉ , 0 ) + O ( y b 2 , ∣ x b − x ˉ ∣ ) . Then
∂ 2 u b ∂ x 2 ( x b , y b ) = b [ 4 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 − 5 x ˉ 2 ( 1 + x ˉ 2 ) 2 − η f ′ ′ ( x ˉ ) ] ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) \frac{\partial^2u_b}{\partial x^2}(x_b,y_b)=b\left[4\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)\frac{1-5\bar{x}^2}{(1+\bar{x}^2)^2}-\eta f''(\bar{x})\right]\left(1+O(y_b^2,|x_b-\bar{x}|)\right) ∂ x 2 ∂ 2 u b ( x b , y b ) = b [ 4 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + x ˉ 2 ) 2 1 − 5 x ˉ 2 − η f ′′ ( x ˉ ) ] ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) ∂ 2 u b ∂ x ∂ y ( x b , y b ) = b [ 16 1 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) x ˉ ( 1 + x ˉ 2 ) 2 ] ( 1 + O ( b y b , ∣ x b − x ˉ ∣ ) ) \frac{\partial^2u_b}{\partial x\partial y}(x_b,y_b)=\sqrt{b}\left[16\sqrt{\frac{1}{2}\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)}\frac{\bar{x}}{(1+\bar{x}^2)^2}\right]\left(1+O(\sqrt{b}y_b,|x_b-\bar{x}|)\right) ∂ x ∂ y ∂ 2 u b ( x b , y b ) = b [ 16 2 1 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + x ˉ 2 ) 2 x ˉ ] ( 1 + O ( b y b , ∣ x b − x ˉ ∣ ) ) ∂ 2 u b ∂ y 2 ( x b , y b ) = − 4 ( 1 + x ˉ 2 ) 2 ( 1 + O ( b , ∣ x b − x ˉ ∣ ) ) . \frac{\partial^2u_b}{\partial y^2}(x_b,y_b)=-\frac{4}{(1+\bar{x}^2)^2}\left(1+O(b,|x_b-\bar{x}|)\right). ∂ y 2 ∂ 2 u b ( x b , y b ) = − ( 1 + x ˉ 2 ) 2 4 ( 1 + O ( b , ∣ x b − x ˉ ∣ ) ) . Then we have, using (58) and (59)
∂ 2 u b ∂ x 2 ( ∂ u b ∂ y ) 2 − 2 ∂ 2 u b ∂ x ∂ y ∂ u b ∂ x ∂ u b ∂ y + ∂ 2 u b ∂ y 2 ( ∂ u b ∂ x ) 2 = 8 b 2 [ 4 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 − 5 x ˉ 2 ( 1 + x ˉ 2 ) 2 − η f ′ ′ ( x ˉ ) ] ⋅ 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ( 1 + x ˉ 2 ) 2 ( 1 + o ( 1 ) ) + 8 b 2 [ 16 1 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) x ˉ ( 1 + x ˉ 2 ) 2 ] ⋅ ( 4 x ˉ ( 1 − η f ( x ˉ ) ) 1 + x ˉ 2 − η f ′ ( x ˉ ) ) ⋅ 1 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 + x ˉ 2 ( 1 + o ( 1 ) ) − b 2 4 ( 1 + x ˉ 2 ) 2 ( 4 x ˉ ( 1 − η f ( x ˉ ) ) 1 + x ˉ 2 − η f ′ ( x ˉ ) ) 2 ( 1 + o ( 1 ) ) . (73) \begin{aligned}
\frac{\partial^2u_b}{\partial x^2}\left(\frac{\partial u_b}{\partial y}\right)^2
-2\frac{\partial^2u_b}{\partial x\partial y}\frac{\partial u_b}{\partial x}\frac{\partial u_b}{\partial y}
+\frac{\partial^2u_b}{\partial y^2}\left(\frac{\partial u_b}{\partial x}\right)^2
={}&8b^2\left[4\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)\frac{1-5\bar{x}^2}{(1+\bar{x}^2)^2}-\eta f''(\bar{x})\right]\\
&\qquad\cdot\frac{1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})}{(1+\bar{x}^2)^2}(1+o(1))\\
&+8b^2\left[16\sqrt{\frac{1}{2}\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)}\frac{\bar{x}}{(1+\bar{x}^2)^2}\right]\\
&\qquad\cdot\left(\frac{4\bar{x}(1-\eta f(\bar{x}))}{1+\bar{x}^2}-\eta f'(\bar{x})\right)\\
&\qquad\cdot\frac{\sqrt{\frac{1}{2}\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)}}{1+\bar{x}^2}(1+o(1))\\
&-b^2\frac{4}{(1+\bar{x}^2)^2}\left(\frac{4\bar{x}(1-\eta f(\bar{x}))}{1+\bar{x}^2}-\eta f'(\bar{x})\right)^2(1+o(1)). \tag*{(73)}
\end{aligned} ∂ x 2 ∂ 2 u b ( ∂ y ∂ u b ) 2 − 2 ∂ x ∂ y ∂ 2 u b ∂ x ∂ u b ∂ y ∂ u b + ∂ y 2 ∂ 2 u b ( ∂ x ∂ u b ) 2 = 8 b 2 [ 4 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + x ˉ 2 ) 2 1 − 5 x ˉ 2 − η f ′′ ( x ˉ ) ] ⋅ ( 1 + x ˉ 2 ) 2 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ( 1 + o ( 1 )) + 8 b 2 [ 16 2 1 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + x ˉ 2 ) 2 x ˉ ] ⋅ ( 1 + x ˉ 2 4 x ˉ ( 1 − η f ( x ˉ )) − η f ′ ( x ˉ ) ) ⋅ 1 + x ˉ 2 2 1 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + o ( 1 )) − b 2 ( 1 + x ˉ 2 ) 2 4 ( 1 + x ˉ 2 4 x ˉ ( 1 − η f ( x ˉ )) − η f ′ ( x ˉ ) ) 2 ( 1 + o ( 1 )) . (73) So when x ˉ ≠ x ˉ η \bar{x}\ne\bar{x}_\eta x ˉ = x ˉ η 1 − η f ( x ˉ ) ≠ 0 1-\eta f(\bar{x})\ne0 1 − η f ( x ˉ ) = 0
∣ ∇ u b ( x b , y b ) ∣ = 2 2 b 1 − η f ( x ˉ ) 1 + x ˉ 2 ( 1 + o ( 1 ) ) |\nabla u_b(x_b,y_b)|=2\sqrt{2}\sqrt{b}\frac{\sqrt{1-\eta f(\bar{x})}}{1+\bar{x}^2}(1+o(1)) ∣∇ u b ( x b , y b ) ∣ = 2 2 b 1 + x ˉ 2 1 − η f ( x ˉ ) ( 1 + o ( 1 )) so that
k Ω b , E = − b 2 G ( x ˉ ) b 2 / 3 ( 2 2 1 − η f ( x ˉ ) 1 + x ˉ 2 ) 3 ( 1 + o ( 1 ) ) = O ( b ) (74) k_{\Omega_b,E}=-b^2\frac{G(\bar{x})}{b^{2/3}\left(2\sqrt{2}\frac{\sqrt{1-\eta f(\bar{x})}}{1+\bar{x}^2}\right)^3}(1+o(1))=O(\sqrt{b}) \tag*{(74)} k Ω b , E = − b 2 b 2/3 ( 2 2 1 + x ˉ 2 1 − η f ( x ˉ ) ) 3 G ( x ˉ ) ( 1 + o ( 1 )) = O ( b ) (74) where
G ( x ˉ ) = [ 32 ( 1 − η f ( x ˉ ) ) 1 − 5 x ˉ 2 ( 1 + x ˉ 2 ) 2 − 8 η f ′ ′ ( x ˉ ) ] 1 − η f ( x ˉ ) ( 1 + x ˉ 2 ) 2 + 64 ( 1 − η f ( x ˉ ) ) x ˉ ( 1 + x ˉ 2 ) 3 ( 4 x ˉ ( 1 − η f ( x ˉ ) ) 1 + x ˉ 2 − η f ′ ( x ˉ ) ) − 4 ( 1 + x ˉ 2 ) 2 ( 4 x ˉ 1 + x ˉ 2 ( 1 − η f ( x ˉ ) ) − η f ′ ( x ˉ ) ) 2 . (75) \begin{aligned}
G(\bar{x})={}&\left[32\left(1-\eta f(\bar{x})\right)\frac{1-5\bar{x}^{2}}{\left(1+\bar{x}^{2}\right)^{2}}-8\eta f''(\bar{x})\right]\frac{1-\eta f(\bar{x})}{\left(1+\bar{x}^{2}\right)^{2}}\\
&+64\left(1-\eta f(\bar{x})\right)\frac{\bar{x}}{\left(1+\bar{x}^{2}\right)^{3}}\left(\frac{4\bar{x}\left(1-\eta f(\bar{x})\right)}{1+\bar{x}^{2}}-\eta f'(\bar{x})\right)\\
&-\frac{4}{\left(1+\bar{x}^{2}\right)^{2}}\left(\frac{4\bar{x}}{1+\bar{x}^{2}}\left(1-\eta f(\bar{x})\right)-\eta f'(\bar{x})\right)^{2}. \tag*{(75)}
\end{aligned} G ( x ˉ ) = [ 32 ( 1 − η f ( x ˉ ) ) ( 1 + x ˉ 2 ) 2 1 − 5 x ˉ 2 − 8 η f ′′ ( x ˉ ) ] ( 1 + x ˉ 2 ) 2 1 − η f ( x ˉ ) + 64 ( 1 − η f ( x ˉ ) ) ( 1 + x ˉ 2 ) 3 x ˉ ( 1 + x ˉ 2 4 x ˉ ( 1 − η f ( x ˉ ) ) − η f ′ ( x ˉ ) ) − ( 1 + x ˉ 2 ) 2 4 ( 1 + x ˉ 2 4 x ˉ ( 1 − η f ( x ˉ ) ) − η f ′ ( x ˉ ) ) 2 . (75) So we get
1 − x 2 − y 2 2 k Ω b , E + ⟨ ν E , ( x , y ) ⟩ E = O ( b ) , (76) \frac{1-x^{2}-y^{2}}{2}k_{\Omega_b,E}+\langle\nu_E,(x,y)\rangle_E=O(\sqrt{b}), \tag*{(76)} 2 1 − x 2 − y 2 k Ω b , E + ⟨ ν E , ( x , y ) ⟩ E = O ( b ) , (76) and so by (72) we get (71) .
When x ˉ = x ˉ η \bar{x}=\bar{x}_{\eta} x ˉ = x ˉ η 1 − η f ( x ˉ η ) = 0 1-\eta f(\bar{x}_{\eta})=0 1 − η f ( x ˉ η ) = 0 f ′ ( x ˉ η ) ≠ 0 f'(\bar{x}_{\eta})\ne0 f ′ ( x ˉ η ) = 0 (62) holds. So (73) becomes
∂ 2 u b ∂ x 2 ( ∂ u b ∂ y ) 2 − 2 ∂ 2 u b ∂ x ∂ y ∂ u b ∂ x ∂ u b ∂ y + ∂ 2 u b ∂ y 2 ( ∂ u b ∂ x ) 2 = 8 b 2 [ 4 η f ′ ( x ˉ ) ( x ˉ − x b ) 1 − 5 x ˉ 2 ( 1 + x ˉ 2 ) 2 − η f ′ ′ ( x ˉ ) ] × η f ′ ( x ˉ ) ( x ˉ − x b ) ( 1 + x ˉ 2 ) 2 ( 1 + o ( 1 ) ) − 64 b 2 x ˉ η 2 f ′ ( x ˉ ) 2 ( x ˉ − x b ) ( 1 + x ˉ 2 ) 3 ( 1 + o ( 1 ) ) − 4 b 2 η 2 ( f ′ ( x ˉ ) ) 2 ( 1 + x ˉ 2 ) 2 ( 1 + o ( 1 ) ) − 4 b 2 η 2 ( f ′ ( x ˉ ) ) 2 ( 1 + x ˉ 2 ) 2 ( 1 + o ( 1 ) ) < 0. (77) \begin{aligned}
\frac{\partial^{2}u_b}{\partial x^{2}}\left(\frac{\partial u_b}{\partial y}\right)^{2}
-2\frac{\partial^{2}u_b}{\partial x\partial y}\frac{\partial u_b}{\partial x}\frac{\partial u_b}{\partial y}
+\frac{\partial^{2}u_b}{\partial y^{2}}\left(\frac{\partial u_b}{\partial x}\right)^{2}
={}&8b^{2}\left[4\eta f'(\bar{x})(\bar{x}-x_b)\frac{1-5\bar{x}^{2}}{(1+\bar{x}^{2})^{2}}-\eta f''(\bar{x})\right]\\
&\mathrel{}\times\frac{\eta f'(\bar{x})(\bar{x}-x_b)}{(1+\bar{x}^{2})^{2}}(1+o(1))\\
&-64b^{2}\bar{x}\eta^{2}f'(\bar{x})^{2}\frac{(\bar{x}-x_b)}{(1+\bar{x}^{2})^{3}}(1+o(1))\\
&-4b^{2}\frac{\eta^{2}(f'(\bar{x}))^{2}}{(1+\bar{x}^{2})^{2}}(1+o(1))\\
&-4b^{2}\frac{\eta^{2}(f'(\bar{x}))^{2}}{(1+\bar{x}^{2})^{2}}(1+o(1))<0. \tag*{(77)}
\end{aligned} ∂ x 2 ∂ 2 u b ( ∂ y ∂ u b ) 2 − 2 ∂ x ∂ y ∂ 2 u b ∂ x ∂ u b ∂ y ∂ u b + ∂ y 2 ∂ 2 u b ( ∂ x ∂ u b ) 2 = 8 b 2 [ 4 η f ′ ( x ˉ ) ( x ˉ − x b ) ( 1 + x ˉ 2 ) 2 1 − 5 x ˉ 2 − η f ′′ ( x ˉ ) ] × ( 1 + x ˉ 2 ) 2 η f ′ ( x ˉ ) ( x ˉ − x b ) ( 1 + o ( 1 )) − 64 b 2 x ˉ η 2 f ′ ( x ˉ ) 2 ( 1 + x ˉ 2 ) 3 ( x ˉ − x b ) ( 1 + o ( 1 )) − 4 b 2 ( 1 + x ˉ 2 ) 2 η 2 ( f ′ ( x ˉ ) ) 2 ( 1 + o ( 1 )) − 4 b 2 ( 1 + x ˉ 2 ) 2 η 2 ( f ′ ( x ˉ ) ) 2 ( 1 + o ( 1 )) < 0. (77) This gives that k Ω b , E > 0 k_{\Omega_b,E}>0 k Ω b , E > 0 x ˉ − x b → 0 \bar{x}-x_b\to0 x ˉ − x b → 0 (57) we get
1 − x 2 − y 2 2 k Ω b , E + ⟨ ν E , ( x , y ) ⟩ E > 0 (78) \frac{1-x^{2}-y^{2}}{2}k_{\Omega_b,E}+\langle\nu_E,(x,y)\rangle_E>0 \tag*{(78)} 2 1 − x 2 − y 2 k Ω b , E + ⟨ ν E , ( x , y ) ⟩ E > 0 (78) which proves (71) when ∂ Ω b ⊂ S 2 \partial\Omega_b\subset\mathbb{S}^{2} ∂ Ω b ⊂ S 2
Case 2, Ω b ⊂ H 2 \Omega_b\subset\mathbb{H}^{2} Ω b ⊂ H 2
Using Euclidian coordinates we have
cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ = 1 + x 2 + y 2 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E . (79) \cos(\theta)k_{\Omega_b}+\sin(\theta)\langle\nu,\vec{e}_{\theta}\rangle
=\frac{1+x^{2}+y^{2}}{2}k_{\Omega_E}-\langle\nu_E,(x,y)\rangle_E. \tag*{(79)} cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e θ ⟩ = 2 1 + x 2 + y 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E . (79) Here the computations are very similar but the contribution of ⟨ ν E , ( x , y ) ⟩ E \langle\nu_E,(x,y)\rangle_E ⟨ ν E , ( x , y ) ⟩ E
∂ 2 u b ∂ x 2 ( x b , y b ) = − b [ 4 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 + 5 x ˉ 2 ( 1 − x ˉ 2 ) 4 + η f ′ ′ ( x ˉ ) ] ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) \frac{\partial^{2}u_b}{\partial x^{2}}(x_b,y_b)
=-b\left[4\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)\frac{1+5\bar{x}^{2}}{(1-\bar{x}^{2})^{4}}+\eta f''(\bar{x})\right]\left(1+O(y_b^{2},|x_b-\bar{x}|)\right) ∂ x 2 ∂ 2 u b ( x b , y b ) = − b [ 4 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 − x ˉ 2 ) 4 1 + 5 x ˉ 2 + η f ′′ ( x ˉ ) ] ( 1 + O ( y b 2 , ∣ x b − x ˉ ∣ ) ) ∂ 2 u b ∂ x ∂ y ( x b , y b ) = − b [ 16 1 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) x ˉ ( 1 − x ˉ 2 ) 2 ] ( 1 + O ( b y b , ∣ x b − x ˉ ∣ ) ) \frac{\partial^{2}u_b}{\partial x\partial y}(x_b,y_b)
=-\sqrt{b}\left[16\sqrt{\frac{1}{2}\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)\frac{\bar{x}}{(1-\bar{x}^{2})^{2}}}\right]\left(1+O(\sqrt{b}y_b,|x_b-\bar{x}|)\right) ∂ x ∂ y ∂ 2 u b ( x b , y b ) = − b [ 16 2 1 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 − x ˉ 2 ) 2 x ˉ ] ( 1 + O ( b y b , ∣ x b − x ˉ ∣ ) ) ∂ 2 u b ∂ y 2 ( x b , y b ) = − 4 ( 1 − x ˉ 2 ) 2 ( 1 + O ( b , ∣ x b − x ˉ ∣ ) ) . \frac{\partial^{2}u_b}{\partial y^{2}}(x_b,y_b)
=-\frac{4}{(1-\bar{x}^{2})^{2}}\left(1+O(b,|x_b-\bar{x}|)\right). ∂ y 2 ∂ 2 u b ( x b , y b ) = − ( 1 − x ˉ 2 ) 2 4 ( 1 + O ( b , ∣ x b − x ˉ ∣ ) ) . and then
∂ 2 u b ∂ x 2 ( ∂ u b ∂ y ) 2 − 2 ∂ 2 u b ∂ x ∂ y ∂ u b ∂ x ∂ u b ∂ y + ∂ 2 u b ∂ y 2 ( ∂ u b ∂ x ) 2 = \frac{\partial^2u_b}{\partial x^2}\left(\frac{\partial u_b}{\partial y}\right)^2-2\frac{\partial^2u_b}{\partial x\partial y}\frac{\partial u_b}{\partial x}\frac{\partial u_b}{\partial y}+\frac{\partial^2u_b}{\partial y^2}\left(\frac{\partial u_b}{\partial x}\right)^2 = ∂ x 2 ∂ 2 u b ( ∂ y ∂ u b ) 2 − 2 ∂ x ∂ y ∂ 2 u b ∂ x ∂ u b ∂ y ∂ u b + ∂ y 2 ∂ 2 u b ( ∂ x ∂ u b ) 2 = − 8 b 2 [ 4 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 + 5 x ˉ 2 ( 1 − x ˉ 2 ) 4 + η f ′ ′ ( x ˉ ) ] 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ( 1 − x ˉ 2 ) 2 ( 1 + o ( 1 ) ) + 8 b 2 [ 16 1 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) x ˉ ( 1 − x ˉ 2 ) 2 ] ( 4 x ˉ ( 1 − η f ( x ˉ ) ) 1 − x ˉ 2 + η f ′ ( x ˉ ) ) \begin{aligned}
-8b^2\left[4\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)\frac{1+5\bar{x}^2}{(1-\bar{x}^2)^4}+\eta f''(\bar{x})\right]\frac{1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})}{(1-\bar{x}^2)^2}(1+o(1)) \\
&+8b^2\left[16\sqrt{\frac{1}{2}\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)}\frac{\bar{x}}{(1-\bar{x}^2)^2}\right]\left(\frac{4\bar{x}(1-\eta f(\bar{x}))}{1-\bar{x}^2}+\eta f'(\bar{x})\right)
\end{aligned} − 8 b 2 [ 4 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 − x ˉ 2 ) 4 1 + 5 x ˉ 2 + η f ′′ ( x ˉ ) ] ( 1 − x ˉ 2 ) 2 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ( 1 + o ( 1 )) + 8 b 2 [ 16 2 1 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 − x ˉ 2 ) 2 x ˉ ] ( 1 − x ˉ 2 4 x ˉ ( 1 − η f ( x ˉ )) + η f ′ ( x ˉ ) ) 1 2 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) 1 − x ˉ 2 ( 1 + o ( 1 ) ) − b 2 4 ( 1 − x ˉ 2 ) 2 ( 4 x ˉ ( 1 − η f ( x ˉ ) ) 1 − x ˉ 2 + η f ′ ( x ˉ ) ) 2 ( 1 + o ( 1 ) ) . (80) \frac{\sqrt{\frac{1}{2}\left(1-\eta f(\bar{x})-\eta f'(\bar{x})(x_b-\bar{x})\right)}}{1-\bar{x}^2}(1+o(1))-b^2\frac{4}{(1-\bar{x}^2)^2}\left(\frac{4\bar{x}(1-\eta f(\bar{x}))}{1-\bar{x}^2}+\eta f'(\bar{x})\right)^2(1+o(1)). \tag*{(80)} 1 − x ˉ 2 2 1 ( 1 − η f ( x ˉ ) − η f ′ ( x ˉ ) ( x b − x ˉ ) ) ( 1 + o ( 1 )) − b 2 ( 1 − x ˉ 2 ) 2 4 ( 1 − x ˉ 2 4 x ˉ ( 1 − η f ( x ˉ )) + η f ′ ( x ˉ ) ) 2 ( 1 + o ( 1 )) . (80) As in the previous case, when x ˉ ≠ x ˉ η \bar{x}\ne\bar{x}_\eta x ˉ = x ˉ η 1 − η f ( x ˉ ) ≠ 0 1-\eta f(\bar{x})\ne0 1 − η f ( x ˉ ) = 0
∣ ∇ u b ( x b , y b ) ∣ = 2 2 b 1 − η f ( x ˉ ) 1 − x ˉ 2 ( 1 + o ( 1 ) ) \lvert\nabla u_b(x_b,y_b)\rvert=2\sqrt{2}\sqrt{b}\frac{\sqrt{1-\eta f(\bar{x})}}{\sqrt{1-\bar{x}^2}}(1+o(1)) ∣ ∇ u b ( x b , y b )∣ = 2 2 b 1 − x ˉ 2 1 − η f ( x ˉ ) ( 1 + o ( 1 )) so that again k Ω b , E = O ( b 1 2 ) → 0 k_{\Omega_b,E}=O(b^{\frac{1}{2}})\to0 k Ω b , E = O ( b 2 1 ) → 0 (66) and (67) we deduce that ⟨ ν E , ( x , y ) ⟩ E = O ( b 1 2 ) \langle\nu_E,(x,y)\rangle_E=O(b^{\frac{1}{2}}) ⟨ ν E , ( x , y ) ⟩ E = O ( b 2 1 )
cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ = 1 + x 2 + y 2 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E = O ( b 1 2 ) (81) \cos(\theta)k_{\Omega_b}+\sin(\theta)\langle\nu,\vec{e}_\theta\rangle=\frac{1+x^2+y^2}{2}k_{\Omega_E}-\langle\nu_E,(x,y)\rangle_E=O(b^{\frac{1}{2}}) \tag*{(81)} cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e θ ⟩ = 2 1 + x 2 + y 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E = O ( b 2 1 ) (81) and then (71) holds.
When x ˉ = x ˉ η \bar{x}=\bar{x}_\eta x ˉ = x ˉ η 1 − η f ( x ˉ η ) = 0 1-\eta f(\bar{x}_\eta)=0 1 − η f ( x ˉ η ) = 0 f ′ ( x ˉ η ) ≠ 0 f'(\bar{x}_\eta)\ne0 f ′ ( x ˉ η ) = 0
∂ 2 u b ∂ x 2 ( ∂ u b ∂ y ) 2 − 2 ∂ 2 u b ∂ x ∂ y ∂ u b ∂ x ∂ u b ∂ y + ∂ 2 u b ∂ y 2 ( ∂ u b ∂ x ) 2 = − 4 b 2 η 2 f ′ ( x ˉ ) 2 ( 1 − x ˉ 2 ) 2 ( 1 + o ( 1 ) ) (82) \frac{\partial^2u_b}{\partial x^2}\left(\frac{\partial u_b}{\partial y}\right)^2-2\frac{\partial^2u_b}{\partial x\partial y}\frac{\partial u_b}{\partial x}\frac{\partial u_b}{\partial y}+\frac{\partial^2u_b}{\partial y^2}\left(\frac{\partial u_b}{\partial x}\right)^2=-4b^2\frac{\eta^2f'(\bar{x})^2}{(1-\bar{x}^2)^2}(1+o(1)) \tag*{(82)} ∂ x 2 ∂ 2 u b ( ∂ y ∂ u b ) 2 − 2 ∂ x ∂ y ∂ 2 u b ∂ x ∂ u b ∂ y ∂ u b + ∂ y 2 ∂ 2 u b ( ∂ x ∂ u b ) 2 = − 4 b 2 ( 1 − x ˉ 2 ) 2 η 2 f ′ ( x ˉ ) 2 ( 1 + o ( 1 )) (82) On the other hand by (66) we have for some positive constants C 1 , C 2 C_1,C_2 C 1 , C 2
k Ω b , E = { ( C 1 + o ( 1 ) ) b ∣ x b − x ˉ η ∣ 3 2 when ∣ x ˉ η − x b ∣ b → ∞ C 2 + o ( 1 ) b when ∣ x ˉ η − x b ∣ b → γ ≥ 0 (83) k_{\Omega_b,E}=
\begin{cases}
(C_1+o(1))\dfrac{\sqrt{b}}{\lvert x_b-\bar{x}_\eta\rvert^{\frac{3}{2}}} & \text{when }\dfrac{\lvert\bar{x}_\eta-x_b\rvert}{b}\to\infty\\
\dfrac{C_2+o(1)}{b} & \text{when }\dfrac{\lvert\bar{x}_\eta-x_b\rvert}{b}\to\gamma\ge0
\end{cases} \tag*{(83)} k Ω b , E = ⎩ ⎨ ⎧ ( C 1 + o ( 1 )) ∣ x b − x ˉ η ∣ 2 3 b b C 2 + o ( 1 ) when b ∣ x ˉ η − x b ∣ → ∞ when b ∣ x ˉ η − x b ∣ → γ ≥ 0 (83) and again since
⟨ ν E , ( x , y ) ⟩ E = { O ( b ∣ x b − x ˉ η ∣ ) when ∣ x ˉ η − x b ∣ b → ∞ O ( 1 ) when ∣ x ˉ η − x b ∣ b → γ ≥ 0 \langle\nu_E,(x,y)\rangle_E=
\begin{cases}
O\left(\sqrt{\dfrac{b}{\lvert x_b-\bar{x}_\eta\rvert}}\right) & \text{when }\dfrac{\lvert\bar{x}_\eta-x_b\rvert}{b}\to\infty\\
O(1) & \text{when }\dfrac{\lvert\bar{x}_\eta-x_b\rvert}{b}\to\gamma\ge0
\end{cases} ⟨ ν E , ( x , y ) ⟩ E = ⎩ ⎨ ⎧ O ( ∣ x b − x ˉ η ∣ b ) O ( 1 ) when b ∣ x ˉ η − x b ∣ → ∞ when b ∣ x ˉ η − x b ∣ → γ ≥ 0 this gives
cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e ⃗ θ ⟩ = 1 + x 2 + y 2 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E = { > 0 when ∣ x ˉ η − x b ∣ b → ∞ + ∞ when ∣ x ˉ η − x b ∣ b → γ ≥ 0 (84) \cos(\theta)k_{\Omega_b}+\sin(\theta)\langle\nu,\vec{e}_\theta\rangle=\frac{1+x^2+y^2}{2}k_{\Omega_E}-\langle\nu_E,(x,y)\rangle_E=
\begin{cases}
>0 & \text{when }\dfrac{\lvert\bar{x}_\eta-x_b\rvert}{b}\to\infty\\
+\infty& \text{when }\dfrac{\lvert\bar{x}_\eta-x_b\rvert}{b}\to\gamma\ge0
\end{cases} \tag*{(84)} cos ( θ ) k Ω b + sin ( θ ) ⟨ ν , e θ ⟩ = 2 1 + x 2 + y 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E = ⎩ ⎨ ⎧ > 0 + ∞ when b ∣ x ˉ η − x b ∣ → ∞ when b ∣ x ˉ η − x b ∣ → γ ≥ 0 (84) which proves (71) for H 2 \mathbb{H}^2 H 2
We are now ready to conclude the proof of Theorem 5.1.
Proof of Theorem 5.1. Property a) follows by Corollary 5.7 because the existence of n n n ω b \omega_b ω b (55) , implies the existence of at least n n n u b u_b u b Lemma 5.2 , see (51) , c) in Proposition 5.8 . Property d) follows from in Lemma 5.3 with x ˉ ≠ x ˉ η \bar{x}\ne\bar{x}_\eta x ˉ = x ˉ η Lemma 5.9 .
Then the condition (8) becomes
1 + x 2 + y 2 2 κ Ω b , E − ⟨ ν E , ( x , y ) ⟩ E = ( 1 + x ˉ 2 ) ( 1 − x ˉ 2 ) 3 b 2 2 1 − η f ( x ˉ ) ( − 16 ( 1 − η f ( x ˉ ) ) 2 ( 1 − x ˉ 2 ) 3 + 1 2 G ˉ ( x ˉ ) − 8 η x ˉ f ′ ( x ˉ ) ( 1 − η f ( x ˉ ) ) 2 ( 1 + x ˉ 2 ) ( 1 − x ˉ 2 ) 2 ) ( 1 + o ( 1 ) ) . \frac{1+x^{2}+y^{2}}{2}\kappa_{\Omega_b,E}-\langle\nu_E,(x,y)\rangle_E=
\frac{(1+\bar{x}^{2})(1-\bar{x}^{2})^{3}\sqrt{b}}{2\sqrt{2}\sqrt{1-\eta f(\bar{x})}}\left(-16\frac{(1-\eta f(\bar{x}))^{2}}{(1-\bar{x}^{2})^{3}}+\frac{1}{2}\bar{G}(\bar{x})-8\eta\bar{x}f'(\bar{x})\frac{(1-\eta f(\bar{x}))^{2}}{(1+\bar{x}^{2})(1-\bar{x}^{2})^{2}}\right)(1+o(1)). 2 1 + x 2 + y 2 κ Ω b , E − ⟨ ν E , ( x , y ) ⟩ E = 2 2 1 − η f ( x ˉ ) ( 1 + x ˉ 2 ) ( 1 − x ˉ 2 ) 3 b ( − 16 ( 1 − x ˉ 2 ) 3 ( 1 − η f ( x ˉ ) ) 2 + 2 1 G ˉ ( x ˉ ) − 8 η x ˉ f ′ ( x ˉ ) ( 1 + x ˉ 2 ) ( 1 − x ˉ 2 ) 2 ( 1 − η f ( x ˉ ) ) 2 ) ( 1 + o ( 1 )) . Then the condition (8) is not satisfied when H ˉ ( x ˉ ) < 0 \bar{H}(\bar{x})<0 H ˉ ( x ˉ ) < 0 x ~ i \tilde{x}_i x ~ i f ( x ) f(x) f ( x )
H ˉ ( x ~ i ) = 8 ( 1 − η f ( x ~ i ) ) 2 ( 1 − x ~ i 2 ) 4 ( − 3 x ~ i 2 ) + 4 η f ′ ′ ( x ~ i ) 1 − η f ( x ~ i ) ( 1 − x ~ i 2 ) 2 < 0. \bar{H}(\tilde{x}_i)=8\frac{(1-\eta f(\tilde{x}_i))^{2}}{(1-\tilde{x}_i^{2})^{4}}\left(-3\tilde{x}_i^{2}\right)+4\eta f''(\tilde{x}_i)\frac{1-\eta f(\tilde{x}_i)}{(1-\tilde{x}_i^{2})^{2}}<0. H ˉ ( x ~ i ) = 8 ( 1 − x ~ i 2 ) 4 ( 1 − η f ( x ~ i ) ) 2 ( − 3 x ~ i 2 ) + 4 η f ′′ ( x ~ i ) ( 1 − x ~ i 2 ) 2 1 − η f ( x ~ i ) < 0. Some examples
Convex spherical domains
Convex spherical domains. Every smooth convex spherical domain Ω \Omega Ω P ∈ Ω P\in\Omega P ∈ Ω Ω \Omega Ω P P P P P P (5) is satisfied. This extends the result of [16 ] , where Theorem 1.2 was proved only for uniformly convex domains with diameter smaller than π 2 \frac{\pi}{2} 2 π
Complements of convex spherical domains
Complements of convex spherical domains. Note that also the complements of spherical convex sets are uniformly star-shaped with respect to some points and satisfy (5) , hence Theorem 1.2 applies. For example geodesic disks of radius greater than π 2 \frac{\pi}{2} 2 π S 2 S^{2} S 2
Other examples of non-convex, uniformly star-shaped spherical domains satisfying (1.5) obtained via stereographic projection
In order to obtain other examples of spherical domains for which Theorem 1.2 applies, it is convenient to look at the stereographic projection used in the previous section.
We use (5.39) to produce examples of uniformly star-shaped, non-convex domains on S 2 S^{2} S 2 (5) holds. To do so, it is sufficient to find uniformly star-shaped domains of R 2 \mathbb{R}^{2} R 2
A first example is the following. Let Ω p : = Ψ − 1 ( Q p ) \Omega_p:=\Psi^{-1}(Q_p) Ω p := Ψ − 1 ( Q p ) Q p = { ( X , Y ) ∈ R 2 : ∣ X ∣ p + ∣ Y ∣ p < 2 − p } Q_p=\{(X,Y)\in\mathbb{R}^{2}: |X|^p+|Y|^p<2^{-p}\} Q p = {( X , Y ) ∈ R 2 : ∣ X ∣ p + ∣ Y ∣ p < 2 − p } p ∈ N p\in\mathbb{N} p ∈ N Q p Q_p Q p 0 0 0 Ω p \Omega_p Ω p (5) , and if p p p Q p Q_p Q p R 2 \mathbb{R}^{2} R 2
Analogously, one can produce examples of uniformly star-shaped domains of S 2 S^{2} S 2 (5) starting from uniformly star-shaped domains D D D R 2 \mathbb{R}^{2} R 2 not convex in R 2 \mathbb{R}^{2} R 2 Ψ − 1 ( D ) \Psi^{-1}(D) Ψ − 1 ( D ) S 2 S^{2} S 2 D a , b , k ⊂ R 2 D_{a,b,k} \subset\mathbb{R}^{2} D a , b , k ⊂ R 2
∂ D a , b , k = { ( X , Y ) = ( a + b sin ( 2 π k s ) ) ( cos ( s ) , sin ( s ) ) : s ∈ ( 0 , 2 π ) } , (85) \partial D_{a,b,k}=\{(X,Y)=(a+b\sin(2\pi ks))(\cos(s),\sin(s)):s\in(0,2\pi)\}, \tag*{(85)} ∂ D a , b , k = {( X , Y ) = ( a + b sin ( 2 π k s )) ( cos ( s ) , sin ( s )) : s ∈ ( 0 , 2 π )} , (85) where a > 0 a>0 a > 0 b ∈ R b\in\mathbb{R} b ∈ R k ∈ N k\in\mathbb{N} k ∈ N Ψ − 1 ( Q p ) \Psi^{-1}(Q_p) Ψ − 1 ( Q p ) Ψ − 1 ( D a , b , k ) \Psi^{-1}(D_{a,b,k}) Ψ − 1 ( D a , b , k ) Figure 5 .
Figure 5. Some examples of non-convex domains with a unique critical point.
On the location of the critical point for spherical convex sets
If Ω \Omega Ω P 0 ∈ Ω P_0\in\Omega P 0 ∈ Ω Ω \Omega Ω P 0 P_0 P 0 G Ω G_{\Omega} G Ω P 0 P_0 P 0 P ∈ Ω P\in\Omega P ∈ Ω P P P P ∈ G Ω P\in G_{\Omega} P ∈ G Ω ω = Ω ∩ ⋂ P ∈ G Ω B ( P , π / 2 ) \omega=\Omega\cap\bigcap_{P\in G_{\Omega}}B(P,\pi/2) ω = Ω ∩ ⋂ P ∈ G Ω B ( P , π /2 ) ω \omega ω Figure 6 for a graphic representation. We also remark that using this information in [16 ] would allow to drop the hypothesis on the diameter for uniformly convex, spherical domains. In fact, if the diameter of Ω \Omega Ω π 2 \frac{\pi}{2} 2 π ω = Ω \omega=\Omega ω = Ω Ω \Omega Ω Theorem 1.2 provides a stronger statement.
Figure 6. The critical point of the convex set Ω \Omega Ω
Non horoconvex domains
We show here that the class of hyperbolic domains to which Theorem 1.3 applies is larger than horoconvex domains.
Consider in the Poincaré disk model the ellipse Ω a , b : = { x 2 a 2 + y 2 b 2 < 1 } \Omega_{a,b}:=\left\{ \frac{x^2}{a^2}+\frac{y^2}{b^2}<1\right\} Ω a , b := { a 2 x 2 + b 2 y 2 < 1 } 0 < b ≤ a < 1 0<b\le a<1 0 < b ≤ a < 1 b < a 2 b<a^2 b < a 2 Ω a , b \Omega_{a,b} Ω a , b ( 0 , b ) (0,b) ( 0 , b ) b a 2 ≤ 1 \frac{b}{a^2}\le1 a 2 b ≤ 1 Ω a , b \Omega_{a,b} Ω a , b (8) is satisfied for suitable choices of a , b a,b a , b b < a 2 b<a^2 b < a 2 (32) )
1 + x 2 + y 2 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E > 0. \frac{1+x^2+y^2}{2}k_{\Omega_E}-\langle\nu_E,(x,y)\rangle_E>0. 2 1 + x 2 + y 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E > 0. Writing x ( ϕ ) = a cos ( ϕ ) x(\phi)=a\cos(\phi) x ( ϕ ) = a cos ( ϕ ) y ( ϕ ) = b sin ( ϕ ) y(\phi)=b\sin(\phi) y ( ϕ ) = b sin ( ϕ ) ϕ ∈ [ 0 , 2 π ] \phi\in[0,2\pi] ϕ ∈ [ 0 , 2 π ]
1 + x 2 + y 2 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E = a b ( 1 − 2 a 2 + b 2 + 3 ( a 2 − b 2 ) cos 2 ( ϕ ) ) 2 ( a 2 sin 2 ( ϕ ) + b 2 cos 2 ( ϕ ) ) 3 / 2 . (86) \frac{1+x^2+y^2}{2}k_{\Omega_E}-\langle\nu_E,(x,y)\rangle_E=\frac{ab(1-2a^2+b^2+3(a^2-b^2)\cos^2(\phi))}{2(a^2\sin^2(\phi)+b^2\cos^2(\phi))^{3/2}}. \tag*{(86)} 2 1 + x 2 + y 2 k Ω E − ⟨ ν E , ( x , y ) ⟩ E = 2 ( a 2 sin 2 ( ϕ ) + b 2 cos 2 ( ϕ ) ) 3/2 ab ( 1 − 2 a 2 + b 2 + 3 ( a 2 − b 2 ) cos 2 ( ϕ )) . (86) A sufficient condition ensuring that the right-hand side of (86) is positive and Ω a , b \Omega_{a,b} Ω a , b ( a , b ) ∈ ( 0 , 1 ) 2 (a,b)\in(0,1)^2 ( a , b ) ∈ ( 0 , 1 ) 2 2 a 2 − 1 < b < a 2 \sqrt{2a^2-1}<b<a^2 2 a 2 − 1 < b < a 2 (8) which are not horoconvex are shown in Figure 7 .
Figure 7. Some examples of non-horoconvex domains with a unique critical point.
Dipartimento di Scienze CFNM, Università degli Studi di Sassari, Via Vienna 2 - 07100 Sassari, Italy, e-mail: fgladiali@uniss.it.
Dipartimento di Scienze di Base e Applicate per l’Ingegneria, Università di Roma “La Sapienza”, Via Scarpa 12 - 00161 Roma, Italy, e-mail: massimo.grossi@uniroma1.it.
Dipartimento di Scienze di Base e Applicate per l’Ingegneria, Università di Roma “La Sapienza”, Via Scarpa 12 - 00161 Roma, Italy, e-mail: luigi.provenzano@uniroma1.it.
