CALIFORNIA STATE UNIVERSITY, NORTHRIDGE. On the Ratio of the First Two Eigenvalues of the Complex Projective Space

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1 CALIFORNIA STATE UNIVERSITY, NORTHRIDGE On the Ratio of the First Two Eigenvalues of the Complex Projective Space A thesis submitted in partial fulfillment of the requirements for the degree of Master of Science in Mathematics by Gerardo Zelaya Eufemia August 13

2 The thesis of Gerardo E. Zelaya Eufemia is approved: Dr. Werner Horn Date Dr. Terry Fuller Date Dr. Chad Sprouse, Chair Date California State University, Northridge ii

3 Dedications To my parents: Edgar Zelaya and Ligia Eufemia, whose discipline and constant encouragment have motivated me over all these years. iii

4 Acknowledgments I am grateful to my mentor, Dr. Chad Sprouse at the California State University Northridge, for his valuable guidance and his patience during this research project. iv

5 Table of Contents Signature Page Dedications Acknowledgements Abstract ii iii iv vi 1 Introduction 1 Chapter Two: Preliminaries 3.1 Rearrangements The Dirichlet Problem and the Faber Krahn s Inequality Liouville Transformation Bank s Inequality Doubly Warped Products and The Complex Projective Space Connections The Laplacian and the Level Sets The Hopf Fibration as a Riemannian Submersion The Mean Curvature On the First Eigenvalue of some modified Berger Spheres Separation of Variables On the ratio of the first two Eigenvalues of the Complex Projective Space 4.1 The PPW conjecture in a hemisphere of S n Results on CP Results on CP n v

6 Abstract On the Ratio of the First Two Eigenvalues of the Complex Projective Space by Gerardo Zelaya Eufemia Master of Science in Mathematics Working toward a possible generalization to the Payne-Polya-Weinberger conjecture for the complex projective space, we prove that for a geodesic ball in CP n of geodesic radius r, the first Dirichlet eigenvalue λ 1 (r, is such that r λ 1 (r is a decreasing function of r; while the quotient between the first two Dirichlet eigenvalues, λ /λ 1 is an increasing function of r for < r < π. We prove it following the methods of M. S. Ashbaugh and R. D. Benguria by separating variables and obtaining a family of ordinary differential equations in the radial variable, and then applying Rayleigh- Ritz methods for estimating lowest eigenvalues, basic Frobenius theory for ordinary differential equations, and Rayleigh-Schrodinger perturbation theory. vi

7 Chapter 1 Introduction In 1955 and 1956 Payne, Polya, and Weinberg conjectured that the ratio of the first two eigenvalues of the Laplacian for a bounded domain Ω R with Dirichlet boundary conditions on δω is bounded by that ratio for a disk with the same volume as Ω. i.e.: λ (Ω λ λ (1.1 Disk λ 1 In 1969, after several advances, C.J. Thomson improved the conjecture by generalizing it to a bounded domain in R n λ (Ω λ 1 (Ω λ ( jn/,1 = (1. n-dimensional ball λ 1 j n/ 1,1 Where j p,k denotes the k-th positive zero of the Bessel function J p (x. In 1991 and 199, Mark S. Ashbaugh and Rafael D. Benguria proved these conjectures. See [] and [3]. By using similar methods to those in the later paper, they established a Sharp bound for that Ratio of a Domain in a Hemisphere of S n. See [1]. Ashbaugh and Benguria achieved their results through the following tools: 1. The Rayleigh-Ritz inequality for estimating λ : Ω λ λ 1 P u 1 dv P Ω u 1 dv where Ω P u 1 dv = and P on Ω. (1.3. A Center of Mass result that guarantees the condition Ω P u 1 dv = for specific choices of functions P. 3. Rearrangement results for functions and domains. 4. General monotonicity properties of the eigenvalues and eigenfunctions in the respective spaces. 5. A Chiti s comparison argument for the eigenfunctions. See [6] In order to adapt their techniques, we divide this thesis document in three main chapters: Chapter : Preliminaries containing some groundwork knowledge used in the proof of the Payne-Polya-Weinberger conjecture. This includes other material that will be used here such as Liouville Transformation, Bank s Inequality, and Laplacians on the distance function level sets. The rest of the material could be applied for a generalization to the Complex Projective Space such as Rearrangements. 1

8 Chapter 3: Doubly Warped Products and the Complex Projective Space introduces us to the Fubini-Study metric, which cross sections the generalized Hopf Fibration as a Riemannian Submersion with Geodesic Fibres. From there, we compute both the mean curvature of metric spheres and the first eigenvalue of the Berger Spheres. These two elements will allow us to separate variables in the Dirichlet problem on a geodesic ball of radius r in CP n (where r (, π/ in order to obtain the family of ordinary differential equations in the radial variable r. Chapter 4: This chapter presents our main results in two sections: the first section proves our results in the -dimensional complex projective space, while the second section proves them in a larger dimension. We divided them in two parts as the proof in larger dimensions represent a more elaborated challenge relying on Bank s Inequality.

9 Chapter Chapter Two: Preliminaries.1 Rearrangements The symmetrization of a planar figure Γ with respect to a coplanar line L is another coplanar figure Γ with the following characteristics: i Γ is symmetric with respect to L. ii A line M perpendicular to L either intersect both Γ and Γ or none. Additionally, if it does, the length of the intersection segment between M and Γ is equal to that of the intersection segment between M and Γ. i.e., for all M L: M Γ = M Γ = Length(M Γ = Length(M Γ iii A line M perpendicular to L, intersecting both Γ and Γ, intersects Γ in a single segment bisected by L. i.e., For all M L: M Γ L M is the midpoint of M Γ Figure.1: Graph of the Symmetrization of an area Γ with respect to the line L. More generally, the symmetrization of a planar figure Γ with respect to a coplanar point P is another coplanar planar figure Γ with the following characteristics: i Γ is symmetric with respect to the point P. ii A coplanar line L passing through P either intersect both Γ and Γ or none. Additionally, if it does, the length of the intersection segment between L and Γ is equal to that of the intersection segment between L and Γ. i.e., For all P L: L Γ = L Γ = L Γ Length(L Γ = Length(L Γ 3

10 iii A coplanar line L passing through P, intersecting both Γ and Γ, intersects Γ in a single segment bisected by P. i.e., for all P L: L Γ P is the midpoing of L Γ Now, the symmetric rearrangment A of a planar figure A is in escence a symmetrization of A with respect to the origin. A is the circle centered at the origin with the same area as A. A = {(x, y R : π (x + y Area(A} Figure.: The symmetric rearrangment of the area between x 1 and the upper semicircle of (x y = 1 is the area contained in the circle x + y = (1/ With this in mind, the symmetric rearrangment of an n-dimensional measurable set E is the n-dimensional sphere E centered at the origin and with the same n- dimensional volume as E. E = {x R n : ω n x n m(e} where ω n is the volume of the unitary n-dimensional sphere. Furthermore, the symmetric decreasing rearragement of a measurable real-valued function f with finite level sets is the symmetric decreasing (from to ± measurable function f whose level set have the same measure as these of f. See f (x = 1 χ {y R n :f(y>t} (xdt We list some basic properties of the Symmetric Decreasing rearrangment: P-I A motivational property for any non-negative function is: f(x = χ {y:f(y>t} (xdt Proof. : Notice that given a t R, χ {y:f(y>t} (x = 1 is equivalent to say 4

11 Figure.3: Graph of the Symmetric Decreasing Rearrangement of a function f. Note that the level sets of the rearrangement have the same measure as these of the original function x {y : f(y > t}, which in turn is equivalent to f(x > t. Therefore, χ {y:f(y>t} (x dt = = f(x f(x = f(x χ {y:f(y>t} (x dt + 1 dt + f(x dt f(x χ {y:f(y>t} (x dt P-II By the monotonicity of the volume function, the symmetric rearrangement is order preserving: f g on E f g on E Proof. : For sets, it is clear that if A B, then A B. Now, notice that given an x E, if f(x g(x, then for all t R : {y : f(y > t} {y : g(y > t}, which in turn implies {y : f(y > t} {y : g(y > t}, and hence, χ {y:f(y>t} χ {y:g(y>t}. Thus, for all x E: f (x = 1 1 = g (x χ {y R n :f(y>t} (xdt χ {y R n :g(y>t} (xdt where we also apply monotonicity of integration. P-III The level sets of the rearrangement have the same measure as these of the 5

12 original function: m{x R : f(x > t} = m{x R : f (x > t} P-IV The Hardy-Littlewood Inequality: f g f g Notice first that given two sets, it is possible that neither one of them contains the other set. However, since their symmetric rearrangements are disks centered at the origin, then the rearrangement of the smaller set in volume is always contained in that of the larger set in volume. Thus, for any two sets A and B, Vol R n (A B Vol R n (A B. Hence, R n f g = = = = = = ( R n R n ( χ {y:f(y>t} dt R n V ol R n χ {y:g(y>s} ds χ {y:f(y>t} {y:g(y>s} dt ds ({y : f(y > t} {y : g(y > s} dt ds V ol R n ({y : f(y > t} {y : g(y > s} dt ds χ {y:f(y>t} {y:g(y>s} dt ds R ( n ( χ {y:f(y>t} dt χ {y:g(y>s} ds R n f g where the second and fifth equalities are from Fubini s Theorem and the fact that χ A χ B = χ A B, and the inequality comes from the remark above. P-V If f is an L p function, then so is f, and f p = f p Moreover, the rearrangement decreases the L p distance between functions: f g p f g p This is proved by symilar reasonings than those in the proof of Hardy-Littlewood s Inequality. P-VI Talenti s Inequality: Let f be a non-negative measurable smooth function with 6

13 compact support in R n, and let u and v the unique solutions to u = f and v = f. Then, u v in R n. P-VII From Talenti s Inequality follows the Polya-Szego Inequality: if 1 p < and f W 1,p, then f p f p (.1 For complete proofs, see [5].. The Dirichlet Problem and the Faber Krahn s Inequality The Dirichlet boundary value problem tries to find the eigenfunctions on a given bounded domain Ω which are zero on δω. That is, to find real numbers λ and realvalued functions u associated to the λ s defined on Ω which satisfy the following system: u + λu =, u δω =. It is well known that the spectrum of eigenvalues λ is an infinite discrete set of values such that: = λ < λ 1 < λ λ 3... λ n... as n Among the most celebrated results for the Dirichlet eigenvalue problem is the Fraber-Krahn s inequality. It addresses to the physical problem of finding what membrane of a given area has the lowest fundamental tone or lowest principal frequency. Baron Rayleigh conjectured that among all planar figures (respectively solids with the same area (respectively volume, the circle (resp. the sphere bounds the domain with the minimal first non-zero Dirichlet eigenvalue. Mathematicians G. Faber and Edgar Khran proved it independently through the use of symmetrization. The fundamental technique for determining the first eigenvalue is the Rayleigh quotient or Rayleigh-Ritz principle: For all Ω R n : φ dv λ 1 = inf φ L (Ω φ dv and that the corresponding eigenfunction u 1 (x is non-negative. i.e., u 1 (x. Yet, further advances showed that there is a function ϕ L (Ω such that: λ 1 = min φ L (Ω φ dv φ dv = ϕ dv ϕ dv Originally, Faber and Khran increased the integral in the numerator while fixing the value of the integral in the denominator. Here we present Polya s proof from [5]. 7

14 Theorem..1 Given a bounded domain Ω R with first Dirichlet eigenvalue λ 1 (Ω, then λ 1 (Ω λ 1 (Ω (. where Ω is the symmetric rearrangement of Ω and λ 1 (Ω is the first Dirichlet eigenvalue of Ω. Proof. : We will only prove the result for simply connected compact domains whose level sets are single closed curves. Using the Rayleigh-Ritz principle, denote u the minimizing function for: u dv u dv Hence, in order to avoid carrying complicated constant factors, we can assume that u is non-negative function, max(u = 1, and u is non-zero in the interior of Ω. For all ρ [, 1], let C ρ := u 1 (ρ Ω be the level sets of u, so C = δω and C 1 are the points where u reaches its maximum. For all ρ [, 1], let A ρ be the area inside of C ρ, so A = Area(Ω. Let L ρ be the length of the level set C ρ. Let R ρ the radius of the circle with area A ρ, i.e.: A ρ = πrρ. In this symmetric decreasing rearrangement, Ω will get replaced by a circle D of radius R, and the level sets C ρ are replaced by circumferences D ρ of radius R ρ centered at the origin. Since the gradient is the vector in the direction of the greatest rate of increase of the function, then we can denote the gradient as: u = dρ dn where dρ > and dn is the piece of the normal to the level curve C ρ between the curves C ρ and C ρ+dρ. Additionally, u dv = dρ ds dn = dρ ds dρ = u ds dρ. dn dn Hence, using the notation P (ρ = C ρ u ds, we rewrite the Dirichlet Integral as: Ω u dv = ρ 1 u ds dρ = C } ρ {{} P (ρ 1 P (ρdρ i.e.: The contribution of the ring-shaped domain between the curves C ρ and C ρ+dρ is P (ρ, and the area of this ring-shaped domain is A ρ A ρ+dρ = A ρ dρ. And another way of computing such area, dv = ds dn = ds dρ. Thus, u A ρ = A ρ = u 1 ds C ρ By Cauchy Scwarz s Inequality: ( ( ( u ds G 1 ds ds = L ρ C ρ C ρ C ρ 8

15 By the Isoperimetric Inequality, which is also valid for a system of mutually exclusive curves: L ρ 4πA ρ Thus, for all ρ > : P (ρ 4πA ρ A ρ In this setting, we want a function f such that f( = and the function v defined for all ρ [, 1] and for all z D ρ : v(z = f(ρ that satisfies: u dv = v dv Ω D while Ω u dv v dv D Hence, since P (ρ and A ρ are integrable, for all ρ > define f(ρ := ρ [ ] 1/ P (t 1/ 4πAρ dt A ρ Therefore, f(ρ goes to as ρ goes to, making v satisfy the boundary condition v( = f( =. Notice that we can rewrite the area element as A (ρdρ = πr ρ dr ρ. Hence, ( ( dv dρ v = = [f (ρ ] = [f (ρ ] 4π Rρ dr ρ dr ρ [A ρ] = [f (ρ ] 4πA ρ [A ρ] = P (ρ A ρ Since the area between the ricrcles of radii R ρ and R ρ+dρ is the same as that between C ρ and C ρ+dρ, which in turn is A ρ dρ, then D v dv = 1 P (ρdρ = Ω u dv i.e.: The Dirichlet s Integral (the numerator integral is unchanged by this rearrangement. However, ρ [ f(ρ = P (t 4πA ] 1/ ρ ρ dt dt = ρ A ρ Hence, D v dv = 1 ρ A ρ dρ 1 f(ρ A ρ dρ = Ω u dv Note that a much shorter proof of the same result can be directly achieved by the 9

16 symmetric decreasing rearrangement using the Rayleigh-Ritz principle and Polya- Szego s Inequality (.1: λ 1 (Ω = inf φ dv φ L (ω Ω = ϕ dv Ω ϕ dv Ω inf φ dv φ L (ω Ω = λ 1 (Ω where ϕ is non-negative normalization of the minimizing function for the eigenvalue of Ω..3 Liouville Transformation Theorem.3.1 (Liouville Transformation If y satisfies the equation y py + qy = λy, then z = e p y satisfies the equation z + Iz = λz where: I = q + p + p 4 (.3 Proof. First notice that z = e p y implies that: Also, z = p e p y + e p y z = p p p p p p e y + 4 e y + e y + p p e y + e p y ( = e p y + pe p p y + + p e p y 4 1

17 Therefore, z + (q + p + p 4 ( z = e p y pe p p y + (q + p + p e p y 4 = e p ( y py + qy = e p (λy = λe p y = λz p 4 e p y.4 Bank s Inequality Theorem.4.1 (Bank s Inequality Assume a(x is a positive and increasing (nonconstant function on [, R]. Assume b 1(x b is an increasing function on [, R]. Then, (x R a(xb 1(xdx R b 1(xdx > R a(xb (xdx R b (xdx (.4 Proof. For this proof, we will use the following notation for the L norm: f = ( R 1/. f (x dx First we claim that b ˆb1 = 1 ( and ˆb b R = b 1 (xdx1/ ( must R b (xdx1/ have exactly one crossing point in (, R. First, if they have no crossing point, then either ˆb 1 > ˆb or the other way around. Thus, ˆb 1 > ˆb or the other way around. However, since ˆb 1 = 1 = ˆb, then this is a contradiction, and hence they have at least one crossing point. Additionally, since b 1 b is an increasing function in [, R], then ˆb 1 ( < b ˆ (. On the other hand, if they have more than one crossing point, say at least two crossing points, say < α < β < R, then b 1 (y = b (y b 1 b for y either equal to α or β, which implies that b 1 (y b (y = b 1 b 11

18 Since the right hand side is a constant, then b 1 (α b (α = b 1(β b (β Since b 1 b is increasing, then this is a contradiction. Now, assume α is the only crossing point, and hence ˆb 1 (R > ˆb 1 (R, then R a b R 1 dx R b 1 dx a b dx R b dx = = = R R α a ˆb 1 dx R a(ˆb 1 ˆb dx a(ˆb 1 ˆb dx + α a ˆb dx R α a(ˆb 1 ˆb dx R > a(α ˆb 1 ˆb dx + a(α ˆb 1 ˆb dx α [ α R ] = a(α ˆb 1 ˆb dx + ˆb 1 ˆb dx = a(α = a(α R R ˆb 1 ˆb dx R ˆb 1 dx a(α ˆb dx α = a(α ˆb 1 a(α ˆb = 1

19 Chapter 3 Doubly Warped Products and The Complex Projective Space 3.1 Connections Connections are rules for taking coordinate-invariant directional derivatives of vector fields on Manifolds. For a vector bundle π : E M over the manifold M, let ɛ(e be the smooth sections of E, then a Connection is a function with the following properties: : T (M ɛ(e ɛ(e (X, Y X Y Con-i It is linear over C (M in X: For all f, g C (M: fx1 +gx Y = f X1 Y + g X Y Con-ii It is linear over R in Y : For all a, b R: X ay 1 + by = a X Y 1 + b X Y Con-iii It is a Product Rule: For all f C (M: Moreover, some derived properties are: X (fy = f X Y + (XfY Con-iv For all covector field ω and a vector field Y : X ω, Y = X ω, Y + ω, X Y Con-iiv For any (k, h-tensor F T k h (M, vector fields Y i (1 i k, and 1-forms ω j (1 j h: X F (ω 1,..., ω h, Y 1,..., Y k = X [ F (ω 1,..., ω h, Y 1,..., Y k ] h (ω 1,..., X ω j,..., ω h, Y 1,..., Y k j=1 k (ω 1,..., ω h, Y 1,..., X Y i,..., Y k In a Riemannian Manifold there is a unique connection, which seems so desirable for computations, called the Riemannian Connection or the Levi-Civita connection of the metric g. It has two key properties: i=1 13

20 RCon-i Compatibility with the metric g: For all vector fields X, Y, Z: X Y, Z = X Y, Z + Y, X Z RCon-ii Symmetric: the torsion tensor of the connection τ : T (M T (M T (M such that τ(x, Y = X Y + Y X [X, Y ] vanishes identically. i.e.: X Y + Y X [X, Y ] 3. The Laplacian and the Level Sets Definition 3..1 (Riemannian Submersion A Riemannian Submersion is a map ϕ : (M, g (N, h such that for each p M, Dϕ : ker (Dϕ T ϕ(p N is a linear isometry. Definition 3.. A function f : U I R is a distance function if and only if f is a Riemannian submersion. For a fixed distance function where U (M, g is an open subset of a Riemannian manifold, we will denote f as r, a notation that comes from our warped product metric dr See [1]. The level sets of this distance function U r = f 1 (r are smooth hypersurfaces in U, g r is the induced metric on U r, and r is the Riemannian connection on (U r, g r. If X and Y are vector fields tangent to the distance function level sets U r : r XY = tan( X Y Therefore, X Y = r X Y II(X, Y r. = X Y g( X Y, r r = X Y + g( X r, Y r = X Y + II(X, Y r Proposition 3..3 (The Laplacian on the Level Sets of a Manifold f = f r + H(M f r + r f Proof. On the other hand, the Laplacian of f on an orthonormal frame {e i } N i=1 is given by n n f = g( ei f, e i = g( r f, r + g( ei f, e i i=1 i= 14

21 On the first term, we know that: where r r = since r = r g( r f, r = r g( f, g( f, r r = r f g( f, r = f r = f r is unitary. Additionally, since every e = e i is tangent to the distance function level sets U r, then g( e f, e = e g ( f, e g ( f, e e = e g ( f, e g ( f, r ee II(e, e r = e g ( f, e g ( f, r ee + II(e, eg ( f, r = e g ( f, e [e g ( f, e g ( r e f, e] + II(e, e f r = g ( r e f, e + II(e, e f r Therefore, combining all of these results, we obtain that: f = g( r f, r + n g( ei f, e i i= = f n r + II(e i, e i f n r + g( r e i f, e i i= i= = f r + H(M f r + r f (3.1 where H(M is the mean curvature of the Manifold, and r is the Laplacian in the distance function level sets. 3.3 The Hopf Fibration as a Riemannian Submersion First, let us review a couple of basic definitions. Definition (Rotationally Symmetric Metrics A rotationally symmetric metric is a metric of the type dt + ϕ (tds n 1 defined on I S n 1, where ds n 1 is the canonical metric on S n 1 (1. 15

22 Finally, Definition 3.3. (Doubly Warped Product A Doubly Warped Product is a metric of the type dt + ϕ (tds p + ψ (tds q defined on I S p S q, where ds p and ds q are the canonical metrics on S p (1 and S q (1 respectively. A general Riemannian submersion leading to the generalized Hopf Fibration can be written as: ((, π S n+1 S 1, dt + ϕ (tds n+1 + ψ (tdθ ( (, π Sn+1 S 1 S 1, dt + ϕ (tg + ϕ (t ψ (t ϕ (t + ψ (t h (3. Where ds n+1 = g + h, h corresponds to the metric along the fiber and g is the orthogonal component. For the generalized Hopf fibration π : S n+3 CP n+1, we let ϕ = sin(t, ψ(t = cos(t, and adopt the Fubini-Study metric on CP n+1. Definition The following is the Fubini-Study metric on the complex projective space CP n+1, which makes the generalized Hopf fibration a Riemannian submersion: ( dr + ϕ ψ (r (r g + ϕ (r + ψ (r h = dr + sin (r ( g + cos (rh ( The Mean Curvature The Mean Curvature, a local extrinsic measure of curvature, is given by the trace of the second fundamental form. Thus, the second fundamental form of the Hopf Fibration with the Fubini-Study metric can be computed making use of a fundamental Equation of Riemannian Geometry: First, the Fubini-Study metric on CP n is: r (g ij = (S k i (g kj (3.4 g = dr + ϕ (rds n + ϕ (rψ (r ϕ (r + ψ (r dθ Thus, g ij = if i j 1 if i = j = 1 ϕ (r if 1 < i = j < n ϕ (rψ (r ϕ (r+ψ (r if i = j = n 16

23 This implies that Now, Therefore, r [g ij ] = S j i = if i j if i = j = 1 ϕ(rϕ (r if 1 < i = j < n ϕ(rψ(r[ψ3 (rϕ (r+ϕ 3 (rψ (r] [ϕ (r+ψ (r] if i = j = n if i j 1 if i = j = 1 ϕ (r ϕ(r if 1 < i = j < n [ψ 3 (rϕ (r+ϕ 3 (rψ (r] ϕ(rψ(r[ϕ (r+ψ (r] if i = j = n H = (n 1 ϕ (r ϕ(r + [ψ3 (rϕ (r + ϕ 3 (rψ (r] ϕ(rψ(r[ϕ (r + ψ (r] For the generalized Hopf fibration, ϕ(r = sin(r and ψ(r = cos(r, so we have: H = (n 1 cot(r tan(r 3.5 On the First Eigenvalue of some modified Berger Spheres We note that the generalized Hopf Fibration π : (S n+3, g (CP n+1, j is a Riemannian submersion with totally geodesic fibres since at each point x of S n+3 the restriction of T x π to the horizontal space H x S n+3 (i.e.: the space orthogonal for g to the kernel V x S n+3, also called the vertical space is an isometry from ( H x S n+3, g HxS n+3 to ( T π(x CP n+1, j π(m. Thus, we can split the Laplacian in S n+3 into two operators called the vertical Laplacian and the horizontal Laplacian as follows: is the second- Definition (Vertical Laplacian The vertical Laplacian v order differential operator defined on a C function f on S n+3 by v [f](x = Fx [f Fx ](m (3.5 Where F x = π 1 π(x is the fibre of π through x and Fx the metric induced by S n+3 on F x. is the Laplace operator of Definition 3.5. (Horizontal Laplacian The difference operator h := v is called the horizontal Laplacian. In [7], we note that in a Riemannian submersion with totally geodesic fibres, the Laplacian operators, v and h commute with each other. Additionally, the Hilbert space L (S n+3 admits Hilbert basis consisting of simultaneous eigenfunction for and v. Moreover, the metric g on S n+3 has a canonical variation associated with the submersion: 17

24 Definition (Canonical Variation Metric For each positive real number t, let g t be the unique Riemannian metric on M such that: (i g t VmM H mm =, (ii g t VmM = t g VmM, (iii g t HmM = g H mm Moreover, the Riemannian submersion π : S n+3 CP n+1 still has totally geodesic fibres. Proposition If we denote S n+3 t Laplacian by t, then: the Riemannian manifold (S n+3, g t and its t = t v + h = t + (1 t h (3.6 Proof. For a proof see [7], where they use moving frames (X i p i=1 and (U j n p j=1, which are orthonormal basis of H x S n+3 and V S n+3 around x respectively. Thus, the moving frames (X i p i=1 and (t 1 U j n p j=1 are orthonormal basis for g t of H x S n+3 and V x S n+3 respectively. Therefore, n p ( t v = t Uj u j D Uj U j = t v (3.7 j=1 And, t h = h (3.8 Hence, as a generalization to their corollary 6.3 on page 193, for t 6 1/ the first nonzero eigenvalue of t is µ 1 (t = n + t (3.9 In this way, the first eigenvalue for Berger Spheres, components of the complex projective space with the Fubini-Study metric defined in (3.3, is λ 1 = sin (r ( (n + cos (r ( Separation of Variables In the Dirichlet problem with boundary conditions on CP n, assume the eigenfunction can be separated by u(r, θ = ρ(rg r (θ, where r is the distance variable, ρ and g r are 18

25 the separated factors of u. Now, using (3.1 and (3.1, we obtain ( CP nu = r H(CP n r BS n 1 (r u u r H(CP n u r BS n 1 (r (ρ(r g r (θ u r H(CP n u r ρ(r BS n 1 (r (g r (θ u r H(CP n u r + λ 1 ( BS n 1 u (3.11 where we denote BS n to the n-dimensional Berger Sphere or the distance function n-level set. 19

26 Chapter 4 On the ratio of the first two Eigenvalues of the Complex Projective Space 4.1 The PPW conjecture in a hemisphere of S n In [1], Ashbaugh and Benguria proved the Payne-Polya-Weinberger conjecture in a hemisphere of S n by proving four main results: Theorem Let Ω be contained in a hemisphere of S n and let B λ1 denote the geodesic ball in S n having the same eigenvalue λ 1 as Ω (i.e., λ 1 (Ω = λ 1 (B λ1. Then λ (Ω λ (B λ1 with equality if and only if Ω is itself a geodesic ball in S n. Theorem 4.1. The first Dirichlet eigenvalue for a geodesic ball in S n of geodesic radius r, i.e.: λ 1 (r is such that r λ 1 (r is an decreasing function of r for r (, π]. Theorem The quotient between the first two Dirichlet eigenvalues for a geodesic ball in S n, of geodesic radius r, λ /λ 1, is an increasing function of r for r (, π ]. Theorem Let Ω be contained in a hemisphere of S n. Then λ λ 1 (Ω λ λ 1 (Ω with equality if and only if Ω is itself a geodesic ball in S n. In this paper, we will reproduce similar result than those in theorems 4.1. and but for CP n, which are necessary results in proving the generalized Payne-Polya- Weinberger conjecture for the n-dimensional complex projective space, a similar result than that in theorem Figure 4.1: Graph of the radius squared times the first eigenvalue as a function of the radius of a geodesic ball in a hemisphere of S 4. Assisted by COMSOL Multiphysics 4.3b

27 Figure 4.: Graph of the first two nonzero eigenvalues as functions of the radius of a geodesic ball in S 4. Assisted by COMSOL Multiphysics 4.3b Figure 4.3: Graph of the ratio of the first two nonzero eigenvalues as a function of the radius of a geodesic ball in S 4. Assisted by COMSOL Multiphysics 4.3b 4. Results on CP In the Dirichlet problem on a geodesic sphere of radius r in the complex projective space CP where r (, π 4, one can separate variables by the decomposition stated in (3.1 to obtain a family of differential equations of the form: y H ( CP y + λ r ny = λy (4.1 where λ r m is the m-th eigenvalue for the r-berger s sphere or the distance function r-level set, and the boundary contitions are y( is finite and y(r =. Particularly, we are interested in the lowest eigenvalues of the m = and m = 1 cases of this equation. Hence, by (3.1, one knows that λ = and λ r 1 = sin (r [ + cos (r] = 3 csc (r sec (r. We will assume λ > by considering the Rayleigh quotients that define them. Hence, we will be discussing the equations: y [3 cot(r tan(r] y = λy (4. y [3 cot(r tan(r] y + [ 3 csc (r + sec (r ] y = λy (4.3 Now, we want to establish some algebraic properties of the function solutions to (4. and (4.3. 1

28 Proposition 4..1 Notice that if u satisfies (4. then satisfies (4.3. Proof. Indeed if u satisfies (4. then Hence, by differentiating both sides: u 1 = u (4.4 u [3 cot(r tan(r]u = λu (4.5 u [3 cot(r tan(r]u + [3 csc (r + sec (r]u = λu (4.6 Since u 1 = u, we obtained that u (n 1 = u (n+1. Thus, by replacing in the above equation and multiplying by 1, we obtain: u 1 [3 cot(r tan(r] u 1 + [ 3 csc (r + sec (r ] u 1 = λu 1 (4.7 Proposition 4.. If u 1 satisfies (4.3 then satisfies (4.. Proof. Let u be defined as above. Then, Hence, u = u 1 + [3 cot(r tan(r] u 1 (4.8 u = u 1 + [3 cot(r tan(r]u 1 [3 csc (r + sec (r]u 1 = λu 1 (4.9 Thus, by replacing in (4., we obtain: u = λu 1 (4.1 u [3 cot(r tan(r]u = λu 1 + [3 cot(r tan(r]λu 1 = λu (4.11 Combining (4.4 and (4.8

29 u 1 + [3 cot(r tan(r]u 1 = ( u + [3 cot(r tan(r]( u = u [3 cot(r tan(r]u = λu (4.1 Proposition 4..3 The above equation can be rewritten as: [ sin(r3 cos(r u 1 ] = λ sin(r3 cos(r u (4.13 Proof. [ ] sin 3 (r cos(r u 1 = 3 sin (r cos (ru 1 sin 4 (ru 1 + sin 3 (r cos(ru 1 = sin(r 3 cos(r [(3 cot(r tan(r u 1 + u 1] = λ sin(r 3 cos(r u (4.14 Now, by redefining the left hand side of (4.1 as operators h m applied to y, then it is clear that they are increasing operators on m. Thus, the first eigenvalue λ 1 of our Dirichlet problem is λ 1 (h while λ must be either λ 1 (h 1 or λ (h. Lemma 4..4 The first eigenvalue of the Dirichlet Laplacian on a geodesic ball of geodesic radius r (, π 4 in CP is the first eigenvalue of (4. while the second eigenvalue of the Dirichlet Laplacian is the first eigenvalue of (4.3. Proof. By Rolle s Theorem, between any two zeros of u there is a zero of u and hence of u 1, by (4.4. Similarly, between two zeros of sin(r 3 cos(r u 1 there is a zero of its derivative, which by (4.13 is λ sin(r 3 cos(r u, and hence of u. Thus for fixed λ > the zeros of u and λ sin(r 3 cos(r u on (, π interlace. i.e.: λ 1 (h < λ 1 (h 1 < λ (h. Lemma 4..5 For r 1 (, π 4, the first eigenfunction of (4.4, i.e.: y1 (r u (r, λ 1 is strictly decreasing on (, r 1. Remark: By convention, we assume y 1 > on [, r 1. Proof. Notice that by equations (4.4 and (4.13, u (r, λ 1 satisfies: [ sin 3 (r cos(r u ] = λ1 sin 3 (r cos(ru > (4.15 in [, r 1, which implies that sin 3 (r cos(r u is decreasing in [, r 1. Hence, sin 3 (r cos(r u < [ sin 3 (r cos(r u ] r= =, which proves the lemma. 3

30 Now, we will apply Liouville Transformation to (4.. Since Then, p = 3 cot(r tan(r (4.16 p = 3 csc (r sec (r = 3 csc (r 1 sec (r (4.17 Also, p 4 = 9 cot (r + tan (r 6 4 = 9 4 cot (r tan (r 3 = 9 csc (r + sec (r 16 4 = 9 4 csc (r sec (r 4 (4.18 Thus, I = }{{} q ( + 3 csc (r 1 sec (r }{{} p = 3 4 csc (r 1 4 sec (r 4 4 csc (r sec (r 4 }{{} ( 9 + p 4 = 3 4 cot (r 1 4 tan (r 7 (4.19 Therefore, (4. becomes: z + ( 3 4 cot (r 1 4 tan (r 7 z = λz (4. where any solution v relates to a solution u of (4. through the relation v = sin 3/ (r cos 1/ (ru. Furthermore, the one-dimensional Schrodinger operator associated with this equation is: H (r 1 = d dr + ( 3 4 cot (r 1 4 tan (r 7 (4.1 4

31 Symilarly, in (4.3 I = 3 csc (r + sec (r }{{} q 3 csc (r 1 sec (r }{{} p ( + = 15 4 csc (r sec (r 4 = 15 4 cot (r tan (r + 1 Therefore, (4.3 becomes: z + ( csc (r sec (r 4 }{{} p 4 (4. ( 15 4 cot (r tan (r + 1 z = λz (4.3 where any solution v 1 relates to a solution u 1 of (4.3 through the relation v 1 = sin 3/ (r cos 1/ (ru 1. Furthermore, the one-dimensional Schrodinger operator associated with this equation is: H 1 (r 1 = d dr + ( 15 4 cot (r tan (r + 1 And the results in (4.1 and (4.4 can be rewritten as: (4.4 H i (r 1 = d dr + ( a i cot (r + b i tan (r + c i (4.5 where (a, b, c = ( 3, 1, and (a1, b 1, c 1 = ( 15, 3, We now use perturbation theory to study how λ 1 and λ vary with r 1. On a fixed interval (, r 1, the eigenvalue problem H m (cr 1 v = λv on (, cr 1 can be rescaled to: [ 1c d dr + ( ] a i cot (cr + b i tan (cr + c i v = λv for r (, r 1 (4.6 or equivalently, [ d dr + ( a i c cot (cr + b i c tan (r + c i c ] v = c λv for r (, r 1 (4.7 The one-dimensional Schrodinger operator on L (, r 1 associated with these equations are: H m (c = d dr + ( a i c cot (cr + b i c tan (r + c i c (4.8 Clearly, λ k ( H m (c = c λ k (H m (c. In particular, λ 1 (cr 1 = c λ 1 ( H (c and λ (cr 1 5

32 = c λ 1 ( H 1 (c. Additionally, notice that: λ j(r 1 = d λ j (r 1 = 1 dr 1 r 1 dλ j (cr 1 dc c=1 (4.9 Since H m (c is an analytic family in c for c approximately 1, we can apply regular Rayleigh Schrodinger perturbation theory. Hence, V i (r, c = H i (c H i (r 1 = a i ( c cot (cr cot (r + b i ( c tan (cr tan (r + c i (c 1 (4.3 Since V m (r, c is analytic in c for c approximately 1 we can compute the derivatives of the eigenvalues of H m (c using the first order perturbation formula: dλ 1 ( H m (c [ Vm = c (r, c ] c=1 v m dr dc r1 (4.31 v c=1 m dr where the function v m (r denote first eigenfunctions of H m (r 1 = H m (1. Thus, And V i (r, c c These functions can be rewritten as: ( = a i cot (r r cot(r csc (r c=1 +b i ( tan (r + r tan(r sec (r +c i (4.3 cot (r r cot(r csc (r = csc (r 1 r cot(r csc (r = csc (r [1 r cot(r] 1 (4.33 tan (r + r tan(r sec (r = sec (r 1 + r tan(r sec (r = sec (r [1 + r tan(r] 1 (4.34 Thus, let s define L(x = cot(r r csc (r, M(x = L (x = csc (r [1 r cot(r], N(x = tan(r + r sec (r, and P (x = N (x = sec (r [1 + r tan(r]. They satisfy the relations, L(r cot(r = M(r 1 and N(r tan(r = P (r 1. 6

33 Notice that L is non-positive, decreasing, and concave; M is positive, increasing, and convex; N is non-negative, increasing, and convex; and P is positive, increasing, and convex. Hence, V i (r, c c = a i (M(r 1 + b i (P (r 1 + c i c=1 = a i M(r + b i P (r + (c i a i b i (4.35 Thus, V i (r, c c = d i M(r + e i P (r + f i (4.36 c=1 where (d, e, f = ( 3, 1, 8 and (d 1, e 1, f 1 = ( 15, 3, 8. Turning back our attention to our eigenvalues, let λ 1 (r = λ 1 [H (r] and λ (r = λ 1 [H 1 (r]. Second, λ 1 (cr = c λ 1 [ H (c] and λ (cr = c λ 1 [ H 1 (c]. Third, λ 1 (cr = λ 1 [ H (c] = c λ 1 (cr and λ (cr = λ 1 [ H 1 (c] = c λ (cr. Theorem 4..6 The first Dirichlet eigenvalue for a geodesic ball in CP of geodesic radius r 1, i.e.: λ 1 (r 1 is such that r1 λ 1 (r 1 is an decreasing function of r 1 for r 1 (, π 4. 7

34 Figure 4.4: Graph of the radius squared times the first eigenvalue as a function of the radius of a geodesic ball in CP. Assisted by COMSOL Multiphysics 4.3b Proof. 1 r 1 d dr 1 ( r 1 λ 1 (r 1 = d dc λ 1 (c = dλ 1( H dc = = [ 3 (M(r 1 1(P (r 1 7] v dr v dr [ 3 M(r 1 P (r 8] v dr v dr (4.37 However, P (r is a positive, increasing function for r (, π ; and M(r is a positive, increasing function between M( = 1 and M(π/ = 1. Hence, 3 3 M(x 1 P (x 8 < (4.38 Which in turn implies that the integrand in the numerator of dλ 1( H is negative, see dc (4.37, and hence, dλ 1 ( H < (4.39 dc Now, we proceed to prove the main result: Theorem 4..7 For the Dirichlet Laplacian on a geodesic ball of geodesic radius r 1 (, π 4 in CP, λ (r 1 increases with r λ 1 (r

35 Figure 4.5: Graph of the first two eigenvalues as functions of the radius of a geodesic ball in CP. Assisted by COMSOL Multiphysics 4.3b Figure 4.6: Graph of the ratio of the first two eigenvalues as a function of the radius of a geodesic ball in CP. Assisted by COMSOL Multiphysics 4.3b Proof. First notice that [ ] d λ (r 1 = 1 [ ( ] d λ (cr 1 dr 1 λ 1 (r 1 r 1 dc λ 1 (cr 1 c=1 = 1 ( ( ] dλ (cr 1 [λ 1 (r 1 dλ1 (cr 1 r 1 dc λ (r 1 1 c=1 dc c=1 λ 1 (r 1 ( ( = 1 dλ (cr 1 dλ1 (cr dc 1 dc c=1 c=1 λ (r 1 r 1 λ (r 1 λ 1 (r 1 λ 1 (r 1 ( ( = 1 dλ (cr 1 dc λ 1 (r 1 dλ1 (cr 1 dc λ (r 1 c=1 c=1 λ (r 1 r 1 λ 1 (r 1 λ (r 1 λ 1 (r 1 = 1 dλ (cr 1 dc dλ 1(cr 1 dc c=1 c=1 1 dλ 1 (cr 1 dc [λ (r 1 λ 1 (r 1 ] c=1 r 1 λ 1 (r r 1 λ 1 (r 1 (4.4 Hence, since λ (r 1 > λ 1 (r 1 >, and dλ 1(cr 1 dλ dc = r 1 (r 1 dr ] c=1 1 > reduces to proving that d dr [ λ (r λ 1 (r <, then proving that d dc (λ (cr c=1 d dc (λ 1(cr c=1 > (4.41 9

36 which is equivalent to prove that: [ 15 M(r + 3P (r 8] v 1 dr v 1 dr > [ 3 M(r 1 P (r 8] v dr v dr (4.4 However, notice that for r 1 > : 3M(r 1P (r 3 < ; and 15M(r+ 3P (r 5 >. Therefore, our task reduces to prove: [ 15 M(r + 3P (r 5 r1 v 1 dr which is certainly true. ] v 1 dr 11 > [ 3 M(r 1P (r 3 r1 v dr ] v dr 13 ( Results on CP n. In the Dirichlet problem on a geodesic sphere of radius r in the complex projective space CP n where r (, π 4, one can separate variables by the decomposition stated in (3.1 to obtain a family of differential equations of the form: y H(CP n y + λ r ny = λy (4.44 where λ r m is the m-th eigenvalue function for the r-berger s sphere or distance function r-level set, and the boundary contitions are y( is finite and y(r =. Particularly, we are interested in the lowest eigenvalues of the m = and m = 1 cases of this equation. Hence, by (3.1, one knows that λ = and λ r 1 = (n 1 csc (r sec (r. We will assume λ > by considering the Rayleigh quotients that define them. Hence, we will be discussing the equations: y [(n 1 cot(r tan(r] y = λy (4.45 y [(n 1 cot(r tan(r] y + [(n 1 csc (r + sec (r] y = λy (4.46 }{{}}{{} H(CP λ r 1 Now, we want to establish some algebraic properties of the function solutions to (4.45 and (4.46. Proposition Notice that if u satisfies (4.45 then satisfies (4.46. u 1 = u (4.47 Proof. The proof follows the same lines as that of Proposition

37 Proposition 4.3. If u 1 satisfies (4.46 then u = u 1 + [(n 1 cot(r tan(r]u 1 (4.48 satisfies (4.45. Proof. The proof follows the same lines as that of Proposition 4... Combining (4.47 and (4.48 u 1 + [(n 1 cot(r tan(r]u 1 = ( u + [(n 1 cot(r tan(r]( u = u [(n 1 cot(r tan(r]u = λu (4.49 Proposition The above equation can be rewritten as: [ sin n 1 (r cos(ru 1 ] = λ sin n 1 (r cos(ru (4.5 Proof. The proof follows the same lines as that of Proposition Now, by redefining the left hand side of (4.44 as operators h m applied to y, then it is clear that they are increasing operators on m. Thus, the first eigenvalue λ 1 of our Dirichlet problem is λ 1 (h while λ must be either λ 1 (h 1 or λ (h. Lemma The first eigenvalue of the Dirichlet Laplacian on a ball of radius r 1 (, π 4 in CP n is the first eigenvalue of (4.45 while the second eigenvalue of the Dirichlet Laplacian is the first eigenvalue of (4.46. Proof. By Rolle s Theorem, between any two zeros of u there is a zero of u and hence of u 1, by (4.47. Similarly, between two zeros of sin n 1 (r cos(ru 1 there is a zero of its derivative, which by (4.5 is λ sin n 1 (r cos(ru, and hence of u. Thus for fixed λ > the zeros of u and λ sin n 1 (r cos(ru on (, π 4 interlace. i.e.: λ 1 (h < λ 1 (h 1 < λ (h. Lemma For r 1 (, π 4, the first eigenfunction of (4.47, i.e.: y1 (r u (r, λ 1 is strictly decreasing on (, r 1. Remark: By convention, we assume y 1 > on [, r 1. Proof. Notice that by equations (4.47 and (4.5, u (r, λ 1 satisfies: [ sin n 1 (r cos(ru ] = λ1 sin n 1 (r cos(ru > (

38 in [, r, which implies that sin n 1 (r cos(ru is decreasing in [, r. Hence, which concludes our proof. sin n 1 (r cos(ru < [ sin n 1 (r cos(ru ] r= = (4.5 Now, we will apply Liouville Transformation to (4.45. ( I = }{{} + n 1 csc (r 1 ( (n 1 sec (r + csc (r sec (r n q }{{}}{{} p p 4 [ = (n 1 1 ] csc (r sec (r n [ = (n 1 1 ] cot (r tan (r 4n 1 (4.53 Therefore, (4.45 becomes: [( z + (n 1 1 cot (r tan (r 4n 1 ] z = λz (4.54 where any solution v relates to a solution u of (4.45 through the relation v = sin (n 1/ (r cos 1/ (ru. Furthermore, the one-dimensional Schrodinger operator associated with this equation is: H (r 1 = d dr + [( (n cot (r 1 4 tan (r 4n 1 ] (4.55 Symilarly, in (4.46 I 1 = = [ n 1 ] csc (r sec (r n [ n 1 ] cot (r tan (r + 1 (4.56 Therefore, (4.46 becomes: [( z + n 1 cot (r tan (r + 1 ] z = λz (4.57 where any solution v 1 relates to a solution u 1 of (4.46 through the relation v 1 = sin (n 1/ (r cos 1/ (ru 1. Furthermore, the one-dimensional Schrodinger operator as- 3

39 sociated with this equation is: H 1 (r 1 = d dr + [( n 1 cot (r tan (r + 1 ] (4.58 And the results of (4.55 and (4.58 can be rewritten: H i (r 1 = d dr + ( a i cot (r + b i tan (r + c i (4.59 where (a, b, c = ( (n 1 1, 1, 4n and (a1, b 1, c 1 = ( n 1, 3, We now use perturbation theory to study how λ 1 and λ vary with r 1. On a fixed interval (, r 1, the eigenvalue problem H m (cr 1 v = λv on (, cr 1 can be rescaled to: [ 1c d dr + ( ] a i cot (cr + b i tan (cr + c i v = λv for r (, r 1 (4.6 or equivalently, [ d dr + ( a i c cot (cr + b i c tan (r + c i c ] v = c λv for r (, r 1 (4.61 The one-dimensional Schrodinger operator on L (, r 1 associated with these equations are: H m (c = d dr + ( a i c cot (cr + b i c tan (r + c i c (4.6 Clearly, λ k ( H m (c = c λ k (H m (c. In particular, λ 1 (cr 1 = c λ 1 ( H (c and λ (cr 1 = c λ 1 ( H 1 (c. Additionally, notice that: λ j(r 1 = d λ j (r 1 = 1 dλ j (cr 1 dr 1 r 1 dc (4.63 c=1 Since H m (c is an analytic family in c for c approximately 1, we can apply regular Rayleigh Schrodinger perturbation theory. Hence, V i (r, c = H i (c H i (r 1 = a i ( c cot (cr cot (r + b i ( c tan (cr tan (r +c i (c 1 (4.64 Since V m (r, c is analytic in c for c approximately 1 we can compute the derivatives of the eigenvalues of H m (c using the first order perturbation formula: dλ 1 ( H m (c [ Vm = c (r, c ] c=1 v m dr dc r1 (4.65 v c=1 m dr 33

40 where the function v m (r denote first eigenfunctions of H m (r 1 = H m (1. Thus, V i (r, c c ( = a i cot (r r cot(r csc (r c=1 +b i ( tan (r + r tan(r sec (r +c i (4.66 As we did before, we will make use of the functions L(r = cot(r r csc (r, which is non-positive, decreasing, and concave; M(r = L (r = csc (r [1 r cot(r], which is positive, increasing, and convex; N(r = tan(r + r sec (r, which is non-negative, increasing, and convex; and P (r = N (r = sec (r [1 + r tan(r], which is positive, increasing, and convex. Also, they satisfy the relations, L(r cot(r = M(r 1 and N(r tan(r = P (r 1. Hence, V i (r, c c = a i (M(r 1 + b i (P (r 1 + c i c=1 = a i M(r + b i P (r + (c i a i b i (4.67 Thus, where (d, e, f = V i (r, c c = d i M(r + e i P (r + f i (4.68 c=1 ( ( 4(n 1 1, 1, n 4n and (d 1, e 1, f 1 = 1,. 3, n Turning back our attention to our eigenvalues, let λ 1 (r = λ 1 [H (r] and λ (r = λ 1 [H 1 (r]. Second, λ 1 (cr = c λ 1 [ H (c] and λ (cr = c λ 1 [ H 1 (c]. Third, λ 1 (cr = λ 1 [ H (c] = c λ 1 (cr and λ (cr = λ 1 [ H 1 (c] = c λ (cr. Theorem The first Dirichlet eigenvalue for a geodesic ball in CP n of geodesic radius r 1, i.e.: λ 1 (r 1 is such that r 1λ 1 (r 1 is an decreasing function of r 1 for r 1 (, π 4. Proof. 1 r 1 d dr 1 ( r 1 λ 1 (r 1 = d dc λ 1 (c = dλ 1( H dc = ( 4(n 1 1 M(r 1P (r n v dr v dr (

41 However, P (r is a positive, increasing function for r (, π 4 ; and M(r is a positive, increasing function between M( = 1 and M(π/ = 1. Hence, 3 ( 4(n 1 1 M(r 1 P (r n < (4.7 Which in turn implies that the integrand in the numerator of dλ 1( H is negative, see dc (4.69, and hence, dλ 1 ( H < (4.71 dc Now, we proceed to prove the main result: Theorem For the Dirichlet Laplacian on a geodesic ball of geodesic radius r 1 (, π 4 in CP n, λ (r 1 increases with r λ 1 (r 1 1. ] Proof. As we saw before, proving that > reduces to proving that which is equivalent to prove that: ( r1 4n 1M(r + 3P (r n v1 dr v 1 dr Which in turn is equivalent to [ d λ (r 1 dr 1 λ 1 (r 1 d dc (λ (cr 1 c=1 d dc (λ 1(cr 1 c=1 > (4.7 ( 4n 1 M(r + 3 P (r v 1 dr v 1 dr By the properties of M and P, 4n 1 > > ( 4(n 1 1 ( 4n 1 M(r 1P (r n v dr v dr (4.73 M(r + 3P (r v dr v dr ((4n M(r + P (r v dr v dr (4.74 M(r + 3 P (r 4n 1 M( + 3 P ( = 4n = 3 n > (

42 Moreover, (4n M(r + P (r (4n M( + P ( = 4n + 3 = 4n > (4.76 Thus,(4.74 follows by Bank s inequality, see [1], Theorem.4.1, and proposition (see below. Proposition Let p(r = u (r,λ 1 = u 1(r,λ 1 u (r,λ 1 u (r,λ 1. Then p(r is positive and strictly increasing on (, r 1 for r 1 (, π 4. Moreover, p( = and p(r as r r 1. Proof. For this proof, let A = (n 1 cot(r tan(r and B = (n 1 csc (r + sec (r, ( so A = B. Then recall (4.47: u 1 = u, (4.48: λu = u 1 + Au 1, and (4.5: sin 3 (r cos(r u 1 = λ sin n 1 (r cos(r u. Now, since p = u 1 /u, then u p (r = u 1u u 1 u σ, then we claim that [ sin n 1 (r cos(rσ(r ] = sin n 1 (r cos(r [ (n 1 csc (r + sec (r ] u u 1 (4.77 Indeed, first notice that Also, σ = u 1u u u 1 = [λu Au 1 ]u + u 1 = λu + u 1 Au u 1 σ = u 1u u 1 u = [λu Au 1 A u 1 ]u + u 1 u 1 = λu u 1 A[λu Au 1 ]u + Bu u 1 + u 1 [λu Au 1 ] = λu u 1 Aλu + A u u 1 + Bu u 1 + λu u 1 Au 1 = A[λu + u 1 Au u 1 ] + Bu u 1 = Aσ + Bu u 1 (

43 Finally, [ sin n 1 (r cos(rσ(r ] = (n 1 sin n (r cos (rσ(r sin n (rσ(r + sin n 1 (r cos(rσ (r = sin n 1 (r cos(raσ(r + sin n 1 (r cos(r [ Aσ(r + Bu u 1 ] = sin n 1 (r cos(r [Aσ(r Aσ + Bu u 1 ] = sin n 1 (r cos(rbu u 1 Since sin n 1 (r cos(r > and B(r > for r [, π 4, then [sin n 1 (r cos(rσ(r] >, which means that σ (r >. Thus, σ(r σ( =, so p (r, which in turn implies that p(r is increasing. The fact that p( = and that p(r as r r1 follows from the initial conditions imposed in our family of differential equations. See (4.44. Proposition If u(r is a solution to the equation u (r [(n 1 cot(r tan(r] u (r = λu(r (4.79 then p(r = u (r u(r satisfies the Ricatti equation p p + [(n 1 cot(r tan(r] p = λ (4.8 Proof. Again, let A = (n 1 cot(r tan(r and B = (n 1 csc (r + sec (r, so A = B. Since p = u u, then p p = u u = Au + λu u = Ap + λ (4.81 Lemma Assume u satisfies u o Au = λ u and u 1 satisfies u Bu 1 = λ 1 u 1 where A = B and λ 1 > λ. Let g = u 1 u, p = u u q = f g g. Then 1 Au 1 + (as above, and ( f q = q f A + p q + [B (λ λ 1 ] f (4.8 f 37

44 Proof. First of all notice that: g g = ( ( u 1 u u 1 u u u u 1 Thus, Hence, q = = = = u 1 u 1 u u (4.83 q = ( u ( 1 u u f + 1 u f u 1 u u 1 u ( [ u (u 1 u ( ] f + u 1 + q f u 1 u u u 1 f ( Au 1 + Bu 1 λ u 1 Au λ 1 u u 1 u + Bf (λ λ 1 f + pq q f ( f = q f A ( u 1 u f (4.84 u 1 u f + [ (u ( ] u 1 + q f f u u 1 (4.85 Proposition Let g(r = y (r r r 1 π. 4 = u 1(r,λ(r 1 y 1 (r u (r,λ 1 (r 1. Then g(r is increasing for Proof. Using the notation above for A(r and B(r, by lemma if q(r = g (r g(r, then q (r = q(r ( A(r + p(r q(r + [B(r (λ λ 1 ] (4.86 Now, assume q is negative somewhere in [, r 1 ]. By the boundary conditions q( > and q is positive to the left of r 1, then by continuity, there are two points < α < β < r 1 such that q(α = q(β = and q (α and q (β. However, when q(r =, then q (r = [ (n 1 csc (r + sec (r (λ λ 1 ] (4.87 Since λ λ 1 is constant, then the right hand side is strictly decreasing, which contradict the fact that q (α q (β. Hence, q(r q( > and q is positive in [, r 1. Therefore, g is increasing by the definition of q. 38

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