Free Boundary Minimal Surfaces in the Unit 3-Ball

Transcription

1 Free Boundary Minimal Surfaces in the Unit 3-Ball T. Zolotareva (joint work with A. Folha and F. Pacard) CMLS, Ecole polytechnique December

2 Free boundary minimal surfaces in B 3 Denition : minimal surfaces in B 3 which meet S 2 = B 3 orthogonally. Free boundary minimal surfaces = Properly embedded Topological classication : - J.C.C. Nitsche 1980 : The only simply connected free boundary minimal surface in B 3 is the equatorial disk. A. Fraser, M. Li 2012, Conjecture : The unique (up to congruences) free boundary minimal annulus in B 3 is the critical catenoid: (s, θ) 1 s cosh s (cosh s e iθ, s), s tanh s = 1 - A. Fraser, R. Schoen, 2013 : n N, there exists a genus 0 free boundary minimal surface in B 3 with n boundary components. 2 / 32

3 Free boundary minimal surfaces in B 3 Denition : minimal surfaces in B 3 which meet S 2 = B 3 orthogonally. Free boundary minimal surfaces = Properly embedded Topological classication : - J.C.C. Nitsche 1980 : The only simply connected free boundary minimal surface in B 3 is the equatorial disk. A. Fraser, M. Li 2012, Conjecture : The unique (up to congruences) free boundary minimal annulus in B 3 is the critical catenoid: (s, θ) 1 s cosh s (cosh s e iθ, s), s tanh s = 1 - A. Fraser, R. Schoen, 2013 : n N, there exists a genus 0 free boundary minimal surface in B 3 with n boundary components. 2 / 32

4 Free boundary minimal surfaces in B 3 Denition : minimal surfaces in B 3 which meet S 2 = B 3 orthogonally. Free boundary minimal surfaces = Properly embedded Topological classication : - J.C.C. Nitsche 1980 : The only simply connected free boundary minimal surface in B 3 is the equatorial disk. A. Fraser, M. Li 2012, Conjecture : The unique (up to congruences) free boundary minimal annulus in B 3 is the critical catenoid: (s, θ) 1 s cosh s (cosh s e iθ, s), s tanh s = 1 - A. Fraser, R. Schoen, 2013 : n N, there exists a genus 0 free boundary minimal surface in B 3 with n boundary components. 2 / 32

5 Free boundary minimal surfaces in B 3 Denition : minimal surfaces in B 3 which meet S 2 = B 3 orthogonally. Free boundary minimal surfaces = Properly embedded Topological classication : - J.C.C. Nitsche 1980 : The only simply connected free boundary minimal surface in B 3 is the equatorial disk. A. Fraser, M. Li 2012, Conjecture : The unique (up to congruences) free boundary minimal annulus in B 3 is the critical catenoid: (s, θ) 1 s cosh s (cosh s e iθ, s), s tanh s = 1 - A. Fraser, R. Schoen, 2013 : n N, there exists a genus 0 free boundary minimal surface in B 3 with n boundary components. 2 / 32

6 Steklov eigenproblem (M 2, g) - compact Riemannian manifold, with M u C ( M) û : { g û = 0 on M, û = u on M Dirichlet-to-Neumann operator : L g : C ( M) C ( M), L g u = û η non-negative, self-adjoint, diskrete spectrum: σ 0 < σ 1 σ 2 (Steklov eigenvalues) 3 / 32

7 Genus 0 free boundary minimal surfaces σ (0, n) := sup σ 1 (g) Length g ( M). 1. Weinstock theorem : σ (0, 1) = 2π, Σ 0 - at unit disk 2. A. Fraser, R. Schoen : For all n N a maximizing metric is achieved by a free boundary minimal surface Σ n in B 3 ( if n = 2, then σ (0, 2) = 4π/1.2 and Σ 1 is a critical catenoid) n, Σn converges on compact sets of B 3 to a double equatorial disk. (σ (0, n) converges to 4π). For large n, Σ n is approximately a pair of nearby parallel plane disks joined by n boundary bridges scaled down versions of half-catenoids. 4 / 32

8 Main result R 3 C R, S n O(3) a subgroup of isometries generated by (z, t) ( z, t), (z, t) (z, t), (z, t) (e 2πi n z, t) Theorem (A. Folha, F. Pacard, ) For n large enough we Give an alternative proof of existence of genus 0 free boundary minimal surfaces with n boundary components (invariant under S n ), Σ n n double copie of D 2. Prove the xistence of genus 1 free boundary minmal surfaces Σ n with n boundary components (invariant under S n ), Σ n n double copie of D 2 \ {0}. 5 / 32

9 CMC surfaces by doubling constructions Ingredients: An initial compact, oriented embedded minimal surface Σ and a set of points p 1,..., p n Σ. Construct two nearby copies Σ ±ε of Σ, which converge uniformly to Σ when ε 0 : Σ ±ε = Σ ± ε Ψ N Σ, Ψ C 2 (Σ). Perform a connected sum of Σ ±ε at p 1,..., p k and then deform it to a CMC surface. 6 / 32

10 CMC surfaces by doubling constructions Question : Under what conditions is it possible to carry out a doubling construction based on given minimal surface? Neck conguration : The number, the size and the positions of the necks. Jacobi operator : J Σ = Σ + A σ 2 + Ric(N Σ, N Σ ). Σ is said to be nondegenerate if J Σ w = 0, w Σ = 0 w = 0. F. Pacard, T. Sun : If Σ is a nondegenerate minimal hypersurface in a Riemannian manifold, one can choose Ψ. s.t. H(Σ ± Ψ N Σ ) = ±1 and produce a CMC surface with H = ε by doubling Σ at any nondegenerate critical point of Ψ (neck size ε ). 7 / 32

11 CMC surfaces by doubling constructions Question : What if Σ is degenerate? Green's function method (R. Mazzeo, F. Pacard, D. Pollack) : consists to study the solutions to k J Σ Γ = c i δ pi, Γ Σ = 0, i=1 and construct Σ ±ε as a normal graphs about Σ of the functions ± ε Γ (Σ ±ε converge to Σ uniformly on compact sets when ε 0). Balancing considerations (using the rst variation formula). A. Butscher, F. Pacard : Surfaces with H = ε in S 3 can be constructed by doubling the minimal Cliord torus at the points of a square lattice that contains 2π Z 2 (neck size ε ). 8 / 32

12 Minimal surfaces by doubling constructions N. Kapouleas, S.D. Yang : Construction of minimal surfaces in S 3 by doubling the Cliord torus at the points of the square lattice 2π n Z, with n large enough. N. Kapoules : Construction of minimal surfaces in S 3 by doubling the equatorial sphere D. Wiygul : Construction of minimal surfaces in S 3 by stacking Cliord tori.! A certain relation should be satised between the number of necks and the size of the necks Construction works only for large n! 9 / 32

13 Minimal surfaces by doubling constructions The expansion of the Green's function in a neighborhood of the poles reads : ε Γ = ε c(n) + ε log r +... On the other hand, catenoid scaled by a factor ε (neck size) Xcat ε : (s, θ) R S 1 ε (cosh s cos θ, cosh s sin θ, s) can be seen as a bi-graph of the function G ε = ε log 2 ( ) ε 3 ε + ε log r + O r 2! We need the constant terms to match exactly : c(n) = log 2 ε, which gives the relation between the size and the number of the necks. 10 / 32

14 Parametrization of the unit ball We choose the following parametrization of B 3 : X : D 2 R B 3, X(z, t) = A(z, t) (z, B(z) sinh t), where B(z) = 1+ z 2 2 and A(z, t) = 1 1+B(z)(cosh t 1). t = 0 - horizontal disk D 2 {0} z = 1 - unit sphere S 2 t = t 0 - CMC leaf with H = sinh t 0 that meets S 2 orthogonally. Induced metric : X g eucl = A 2 (z, t) ( dz 2 + B 2 (z) dt 2) 11 / 32

15 Parametrization of the unit ball 12 / 32

16 Free boundary graphs over D 2 {0} Take w C 2 (D 2 ) and consider the image in B 3 of the vertical graph of w : z X(z, w(z)). Lemma 1. X(z, w(z)) is orthogonal to S 2 = B 3 at the boundary i r w r=1 = 0; 2. Mean curvature of X(z, w(z)) is given by H(w) = 1 ( A 2 (,w) B 2 w A 3 (,w) B div 1+B 2 w 2 ) B 2 w 2 sinh w. 13 / 32

17 Linearized mean curvature operator The linearized mean curvature operator is given by ( 1 + z 2 L gr w = (Bw) = 2 Remark: ) w The operator L gr has a kernel : { } 2x1 Ker(L gr ) = 1 + z 2, 2x z 2. This corresponds to tilting the unit disk D 2 {0} in B 3. One can eliminate the kernel by asking the function w to be invariant under a group of rotations around the coordinate axis Ox 3 (We arrange the catenoidal bridges periodically along S 1 ) 14 / 32

18 Linearized mean curvature operator The linearized mean curvature operator is given by ( 1 + z 2 L gr w = (Bw) = 2 Remark: ) w The operator L gr has a kernel : { } 2x1 Ker(L gr ) = 1 + z 2, 2x z 2. This corresponds to tilting the unit disk D 2 {0} in B 3. One can eliminate the kernel by asking the function w to be invariant under a group of rotations around the coordinate axis Ox 3 (We arrange the catenoidal bridges periodically along S 1 ) 14 / 32

19 Linearized mean curvature operator The linearized mean curvature operator is given by ( 1 + z 2 L gr w = (Bw) = 2 Remark: ) w The operator L gr has a kernel : { } 2x1 Ker(L gr ) = 1 + z 2, 2x z 2. This corresponds to tilting the unit disk D 2 {0} in B 3. One can eliminate the kernel by asking the function w to be invariant under a group of rotations around the coordinate axis Ox 3 (We arrange the catenoidal bridges periodically along S 1 ) 14 / 32

20 Green's functions Take ε, ε R + and z m := e 2πim/n, m = 1,..., n and consider the catenoids in R 3 centered at z = 0 and z = z m : (s, θ) R S 1 ( ) ε cosh s e iθ, εs [ π (s, θ) R 2, 3π 2 ] which can be seen as bi-graphs of the functions ( ) ε cosh s e iθ + z m, εs G ε,0 (z) = ( z, ± ( ε log ε2 ε log z + O ( ε 3 / z 2))) G ε,m (z) := ( z, ± (ε log ε 2 ε log z z m + O ( ε 3 / z z m 2))) in small neighborhoods of z = 0 and z = z m. 15 / 32

21 Green's function Goal : To nd the solution to the problem ( ) invariant under rotations by the angle 2π n. ( ) { (BΓ) = c0 δ 0 in D 2 r Γ = n m=1 c m δ zm on S 1 If Γ(z) = G(z n )/B(z), then G satises ( ) { G = d0 δ 0 in D 2 r G 1 n G = d 1 δ 1 on S 1 in the sense of distributions. 16 / 32

22 Green's function We construct explicitly G 0 and G 1, such that (1) { { G0 = d 0 δ 0 in D 2 G1 = 0 in D 2 r G 0 1 n G (2) 0 = 0 on S 1 r G 1 1 n G 1 = d 1 δ 1 on S 1 G 0 := log z n - satises (1) k N, H k (z) = 1 n k Re j=1 zj j k+1 H 0 (z) = log 1 z, r H k r=1 = 1 n H k 1 G 1 (z) := n 2 + k=0 H k(z) - satises (2) α, β R, Γ n (z) := 1 B(z) (α G 0(z n ) + β G 1 (z n )) satises ( ) 17 / 32

23 Green's function We construct explicitly G 0 and G 1, such that (1) { { G0 = d 0 δ 0 in D 2 G1 = 0 in D 2 r G 0 1 n G (2) 0 = 0 on S 1 r G 1 1 n G 1 = d 1 δ 1 on S 1 G 0 := log z n - satises (1) k N, H k (z) = 1 n k Re j=1 zj j k+1 H 0 (z) = log 1 z, r H k r=1 = 1 n H k 1 G 1 (z) := n 2 + k=0 H k(z) - satises (2) α, β R, Γ n (z) := 1 B(z) (α G 0(z n ) + β G 1 (z n )) satises ( ) 17 / 32

24 Green's function We construct explicitly G 0 and G 1, such that (1) { { G0 = d 0 δ 0 in D 2 G1 = 0 in D 2 r G 0 1 n G (2) 0 = 0 on S 1 r G 1 1 n G 1 = d 1 δ 1 on S 1 G 0 := log z n - satises (1) k N, H k (z) = 1 n k Re j=1 zj j k+1 H 0 (z) = log 1 z, r H k r=1 = 1 n H k 1 G 1 (z) := n 2 + k=0 H k(z) - satises (2) α, β R, Γ n (z) := 1 B(z) (α G 0(z n ) + β G 1 (z n )) satises ( ) 17 / 32

25 Green's function We construct explicitly G 0 and G 1, such that (1) { { G0 = d 0 δ 0 in D 2 G1 = 0 in D 2 r G 0 1 n G (2) 0 = 0 on S 1 r G 1 1 n G 1 = d 1 δ 1 on S 1 G 0 := log z n - satises (1) k N, H k (z) = 1 n k Re j=1 zj j k+1 H 0 (z) = log 1 z, r H k r=1 = 1 n H k 1 G 1 (z) := n 2 + k=0 H k(z) - satises (2) α, β R, Γ n (z) := 1 B(z) (α G 0(z n ) + β G 1 (z n )) satises ( ) 17 / 32

26 Neck congurations c 0 (n) + 2n α log z + O(α z 2 ), as z 0 Γ n (z) = c 1 (n) + β log z z m + O(β z z m ), as z z m, where c 0 (n) c 1 (n) n. We obtain : ε ε, n log ε, α ε n, β ε. and do the "gluing" in the regions z ε 1/2, z z m ε 2/3, m = 1,..., n 18 / 32

27 Catenoidal bridges orthogonal to S 2 C := {ζ C : Re(ζ) 0}. Consider the dieomorphisms λ m (ζ) : C D 2, λ m (ζ) = e 2πim/n 1 + ζ 1 ζ and Λ m (ζ, t) : C R D 2 R, Λ m (z, t) = (λ m (z), 2t) 19 / 32

28 Approximate solution A n A n = connected sum of the graph of the Green's function Γ n with n half-catenoidal bridges and 1 catenoidal neck. Orthogonality to S 2 at the boundary: The half-catenoid X cat is foliated by horizontal leafs orthogonal to C R at the boundary. The restriction X Λ m to the horizontal planes is conformal. The surface parametrized by X Λ m X cat is foliated by spherical-cap leafs orthogonal to S 2 at the boundary. 20 / 32

29 Perturbation argument Dene a vector eld Ξ transverse to A n (smoothly interpolating between the normal to the catenoids and the vertical vector eld), s.t. if ξ t is the associated ow : ξ (A n, 0) = A n and ξ t = Ξ(ξ(, t)) then A n,t := ξ t (A n ) meets S 2 orthogonally along A n,t. 21 / 32

30 Mean curvature of the perturbed surface Take w C 2 (A n ) and put A n (w) = {ξ(p, w(p)), p A n }. The Taylor expansion of the mean curvature of A n (w) satises H(A n (w)) = H(A n ) + L n w + Q n (w, w, 2 w), L n - the linearized mean curvature operator, Q n - smooth nonlinear function, s.t. Q(0, 0, 0) = DQ(0, 0, 0) = 0. If w satises the Neumann boundary condition : w η = 0 on A n A n (w) meets S 2 orthogonally at the boundary. 22 / 32

31 Minimal surface equation Our goal is to solve the equation H(A n (w)) = 0 or L n w = ( H(A n ) + Q n (w, w, 2 w) ) If H(A n ) n 0 in a suitable topology and if L n were invertible with its inverse bounded uniformly in n we could apply Banach xed point theorem in a ball of an appropriate Banach space (the radius of the ball being dened by H(A n ) ). 23 / 32

32 Weighted Banach spaces Weight function : γ(x) = x n Π m=1 2πim x e n, x A n. For ν R, w L ν (A n ) i γ ν w <. Then, we nd H(A n ) L ν 2 c ε 5 3 ν c e n( 5 3 ν) and take ν (0, 1). Moreover we have : L n : L ν (A n ) L ν 2(A n ). In the same manner we dene Hölder weighted spaces C k,α ν. 24 / 32

33 Properties of the linearized operator As n, In the graph region : L n L gr = (B ), In the catenoidal regions : L n J cat = 1 ε 2 cosh 2 s ( ) s 2 + θ cosh 2 s (Jacobi operator about the catenoid ) We should study the corresponding operators in noncompact domains D 2 \ {0, z 1,..., z n }, R S 1, and R [ π 2, 3π ] 2. Problem : The catenoid is degenerate there are small eigenvalues of the operator L n (eigenvalues that tend to 0 as fast as n ). 25 / 32

34 Linear analysis on the half-catenoid Study the homogeneous problem { 1 ε 2 cosh 2 s θ w = 0 ( 2 s + 2 θ + 2 cosh 2 s Fourier series w = j Z w j (s) e iθj. ) w = 0 R [ π 2, 3π ] 2, R { π 2, 3π } 2. j 2, maximum principle w j = 0; j = 1 (rotations + horizontal translations), Neumann boundary condition + symmetry w(s, θ) = w(s, 2π θ) w ±1 = 0; j = 0 (dilatation + vertical translation), symmetry w(s, θ) = w( s, θ) + suitable function space w 0 = / 32

35 Linear analysis on the half-catenoid Lemma For ν < 1 and for all f (ε cosh s) ν 2 L (R [ π 2, 3π ] 2 ) there exist v cat (ε cosh s) ν L (R [ π 2, 3π ] 2 ) and ccat R, s.t. w cat = v cat + c cat satises { ( ) 1 2 ε 2 cosh 2 s s + θ w cosh 2 s cat = f θ w cat = 0 R [ π 2, 3π ] 2, R { π 2, 3π } 2, and (ε cosh s) ν v cat L + c cat C (ε cosh s) ν+2 f L. Deciency space : D cat = span{1}. 27 / 32

36 Linear analysis in the punctured disk We are interested in the problem { (Bw) = f in D 2 \ {0}, ( ) r w = 0 on S 1 \ {z 1,..., z n }. Suppose that f and w are invariant under rotations by the angle 2π n no bounded kernel. Put w(z) = W (z n )/B(z), then ( ) is equivalent to { W = F in D 2 \ {0}, ( ) r W 1 n W = 0 on S1 \ {1}. 28 / 32

37 Linear analysis in the punctured disk Fix a cut-o function χ C (D 2 ), which is identically equal to 1 in a neighborhood of z = 1 and to 0 in a neighborhood of z = 0. Weight function : γ(z) = z z 1. Lemma Assume that ν (0, 1), then n large enough and for all F, such that F L ν 2 (D2 ), there exists a unique function V gr and unique constants c 0 and c 1, such that W gr = V gr + c 0 n + c 1 χ is a solution to ( ) and V gr L ν + c 0 + c 1 C F L ν 2 Deciency space : D gr = span{n, χ}. 29 / 32

38 Gluing the parametrices together We need to solve L n w = f, for f L ν 2(A n ). We divide A n in 2 symmetric parts, extend f to and solve the problem in R S 1, R [ π 2, 3π ] 2 and D 2 \ {0, z 1,..., z n }, restrict the solutions to A n, and glue the restricted solutions together to produce : w n : L n w n f f. (We need to treat the terms decaying at innity and deciency terms separately). 30 / 32

39 Conclusion So, we have M n : f w n, L n M n Id = R n, R n 1 Finally, L 1 n := M n (Id R n ) 1. Remark: L 1 n n - explodes when n. Conclusion: L 1 n L 1 Q n - a contraction mapping in a ball of radius n H(A n ) of Cν 2,α (A n ). Banach xed point theorem is applied. 31 / 32

40 Thank you for your attention! 32 / 32