A shape optimization problem for the first mixed Steklov-Dirichlet eigenvalue
AA SHAPE OPTIMIZATION PROBLEM FOR THE FIRSTMIXED STEKLOV-DIRICHLET EIGENVALUE
DONG-HWI SEO
Abstract.
We consider a shape optimization problem for the firstmixed Steklov-Dirichlet eigenvalues of domains bounded by two ballsin two-point homogeneous space. We give a geometric proof which ismotivated by Newton’s shell theorem. Introduction
Let M m be a Riemannian manifold of dimension m ≥ ⊂ M a bounded domain with Lipschitz boundary ∂ Ω. Let ∂ Ω = C ∪ C with C ∩ C = φ . A mixed Steklov-Dirichlet eigenvalue problem is to find σ ∈ R for which there exists u ∈ C ∞ (Ω) satisfying ∆ u = 0 in Ω u = 0 on C ∂u∂η = σu on C , (1)where η is the outward unit normal vector along C . When C = φ and C isconnected, the problem becomes the Steklov eigenvalue problem introducedby Steklov in 1902 [18]. We will find a domain maximizing the lowest σ in a class of subsets in M . We call this problem by a shape optimizationproblem of the first eigenvalue.The shape optimization problem of the first nonzero Steklov eigenvaluein Euclidean space has been studied since the 1950s. In 1954, Weinstockconsidered the case when M = R [21]. He showed that the disk is themaximizer among all the simply connected domains with the same bound-ary lengths. Recently, Bucur, Ferone, Nitsch, and Trombetti studied thisperimeter constraint shape optimization problem in any dimension amongall the convex sets, and showed that the ball is the maximizer [6]. Withoutthe convexity condition, Fraser and Schoen proved the ball cannot be a max-imizer even among all the smooth contractible domains of fixed boundaryvolume in R m , m ≥ R m , m ≥
2. Note that he does not need any topologicalrestriction.
Mathematics Subject Classification.
Primary 58J50, 35P15; Secondary 43A85.
Key words and phrases. shape optimization problem, two-point homogeneous space,Steklov eigenvalue. a r X i v : . [ m a t h . SP ] S e p DONG-HWI SEO
These shape optimization problems have been extended to non-Euclideanspaces as well. The first result in this direction was given by Escobar [8]who showed that the first nonzero eigenvalue is maximal for the geodesicdisk among all the simply connected domains with fixed domain area insimply connected complete surface M with constant Gaussian curvature.In 2014, Binoy and Santhanam extended this result to noncompact rank onesymmetric spaces of any dimension [3].Regarding mixed Steklov-Dirichlet eigenvalue problems, it was consid-ered by Hersch and Payne in 1968 [10]. They considered the problem (1)when Ω ⊂ R is a doubly connected region bounded by the inner and theouter boundaries, C and C , respectively. Then among all the conformallyequivalent domains with fixed perimeter of C , the annulus bounded by twoconcentric circles is the maximizer. Recently, Verma considered connectedregions in R m with m ≥ M and the closure of a set A ⊂ M by inj ( M ) and cl( A ), respectively. Theorem 1.
Let M be a two-point homogeneous space. Let B and B (cid:48) begeodesic balls of radii R , R > , respectively, such that cl( B ) ⊂ B (cid:48) and R < inj ( M ) / . Then the first mixed Steklov-Dirichlet eigenvalue of theproblem ∆ u = 0 in B (cid:48) \ cl( B ) u = 0 on ∂ B ∂u∂η = σu on ∂ B (cid:48) (2) ( η : the outward unit normal vector along ∂ B (cid:48) ) attains maximum if andonly if B and B (cid:48) are concentric. Two-point homogeneous space has similar geometric properties with Eu-clidean space. For example, for two geodesic balls B and B (cid:48) of radii R and R , respectively, satisfying cl( B ) ⊂ B (cid:48) , B (cid:48) \ cl( B ) is isometric to B (cid:48) \ cl( B ) if and only if the distance of the centers of B and B (cid:48) is equalto that of B and B (cid:48) . Furthermore, using additional angles, which are notusual Riemannian angles, there are laws of trigonometry and conditions fortriangle conditions (for example, see Proposition 1) in two-point homoge-neous space.In order to prove the theorem, we estimate the first eigenvalue by sub-stituting an appropriate test function on the Rayleigh quotient (see (3) inSection 2.1). We suggest a geometric proof to obtain the lower bound of thedenominator of the quotient (see Corollary 2). It is similar to the proof ofNewton’s shell theorem (see Remark after Proposition 3). SHAPE OPTIMIZATION PROBLEM FOR THE FIRST EIGENVALUE 3
Newton’s shell theorem is first proved by Newton [15] (see Propositio LXXTheorema XXX in Sectio XII). It is extended to constant curvature spacesby Kozlov [14] and Izmestiev and Tabachnikov [12]. We prove that it is alsoholds for two-point homogeneous spaces with some restriction (see Corollary1 and the following Remark).In Section 2, we will briefly review the variational characterization of themixed Steklov-Dirichlet eigenvalue problem (2) as well as two-point homo-geneous spaces and its trigonometry. Section 3 is devoted to the proof of themain theorem. In Section 3.1 we calculate the first mixed Steklov-Dirichleteigenfunction on the annulus. In Section 3.2, we introduce some cruciallemmas (Section 3.2.1) and prove the main theorem (in Section 3.2.2 (non-compact rank one symmetric space, noted nCROSS) and in Section 3.2.3(compact rank one symmetric space, noted CROSS)). Especially, in Section3.2.1, we give a proof of Newton’s shell theorem for a two-point homoge-neous space.
Acknowledgement
The author wishes to express his gratitude to Jaigyoung Choe for helpful dis-cussions. This research was partially supported by NRF-2018R1A2B6004262.2.
Background
The eigenvalue problem.
A mixed Steklov-Dirichlet eigenvalue prob-lem (1) is equivalent to the eigenvalue problem of the Dirichlet-to-Neumannoperator : L : C ∞ ( C ) → C ∞ ( C ) u (cid:55)→ ∂ ˆ u∂η , where ˆ u is the harmonic extension of u satisfying the following ∆ˆ u = 0 in Ωˆ u = 0 on C ˆ u = u on C . Then L is a positive-definite, self-adjoint operator with discrete spectrum(see for instance [1]), 0 < σ (Ω) ≤ σ (Ω) ≤ · · · → ∞ , provided that C (cid:54) = φ . We call σ k (Ω) by the k th mixed Steklov-Dirichleteigenvalue, or simply the k th eigenvalue. An eigenfunction of L correspond-ing to σ k (Ω) is called the k th mixed Steklov-Dirichlet eigenfunction, or the DONG-HWI SEO
CROSS nCROSS Isotropy representation K = R R m S m , R P n R H n ( O ( m ) , R m ) K = C · C P n C H n ( U ( n ) , R n ) K = H · H P n H H n ( Sp (1) × Sp ( n ) , R n ) K = O · O P O H ( Spin (9) , R ) Table 1.
Two-point homogeneous spaces, m ≥ , n ≥ k th eigenfunction. Then the first eigenvalue σ (Ω) is characterized varia-tionally as follows σ (Ω) = inf (cid:40) (cid:82) Ω |∇ v | dV (cid:82) C v ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) v ∈ H (Ω) \ { } and v = 0 on C (cid:41) . (3)For convenience we shall call the harmonic extension of the k th eigenfunctionby the k th mixed Steklov-Dirichlet eigenfunction or the k th eigenfunction.2.2. Two-point homogeneous spaces and triangle congruence con-ditions.
Three points in a Euclidean space determine a triangle when threepoints are not lie on a single line. In classical geometry, there are several con-gruence conditions on triangles and it is determined by lengths of sides andangles. For example, side-angle-side (SAS) congruence is given by two sidelengths and the included angle. In two-point homogeneous spaces, analogousproperties also hold with additional angles. These facts are obtained by thelaws of trigonometry. In this section, we give some information about two-point homogeneous spaces and its congruence conditions of triangles whichwill be used later. See [22],[11],[4] for more details.
Definition 1.
A connected Riemannian manifold M is called two-point ho-mogeneous space if x i , y i ∈ M, i = 1 , with dist ( x , y ) = dist ( x , y ) , thereis an isometry g of M such that g ( x ) = x and g ( y ) = y . In fact, two-point homogeneous spaces are Euclidean spaces or rank onesymmetric spaces. We will call the latter spaces by ROSSs. Further-more, compact ROSS and noncompact ROSS are denoted by CROSS andnCROSS, respectively. Then two-point homogeneous spaces with their isotropyrepresentations are classified as in the Table 1 (see [22],[11]). Here m ≥ , n ≥ m = dim R M = n · dim R K .An angle is given by two directions at a point P . It is classified by itscongruence classes which are given by the orbit space of U P M × U P M/K ,where U P M is the unit sphere in the tangent space of M at P , and K isthe isotropy subgroup of the isometry group M at P . The orbit space canbe seen by fixing the first component by the action of K . More precisely, itis equivalent to an orbit space U P M/H of an isotropy group H ⊂ K withrespect to a point in U P M . Then it can be checked that for given (cid:126)v ∈ U P M , H -invariant subspaces are R · (cid:126)v , K (cid:48) · (cid:126)v , and the subspace orthogonal to K · (cid:126)v ,where K = R , C , H , and O and K (cid:48) is the set of pure imaginary numbers in K . SHAPE OPTIMIZATION PROBLEM FOR THE FIRST EIGENVALUE 5
Then a direction (cid:126)v is determined up to H -action by the following angularinvariants (for more details, see [11],[4]): • λ ( (cid:126)v , (cid:126)v ) = ∠ ( (cid:126)v , (cid:126)v ) ; 0 ≤ λ ≤ π , • ϕ ( (cid:126)v , (cid:126)v ) = ∠ ( (cid:126)v , K · (cid:126)v ) ; 0 ≤ ϕ ≤ π ,where ∠ ( (cid:126)v , (cid:126)v ) is the usual (Riemannian) angle and ∠ ( (cid:126)v , K · (cid:126)v ) is theangle between (cid:126)v and the subspace K · (cid:126)v . Note that when K = R , λ = ϕ or λ = π − ϕ . Then angular invariants satisfy following relations : λ ( (cid:126)v , − (cid:126)v ) = π − λ ( (cid:126)v , (cid:126)v ) , (4) ϕ ( (cid:126)v , − (cid:126)v ) = ϕ ( (cid:126)v , (cid:126)v ) . (5)Using the previous H -invariant decomposition, we can write the metricof ROSS M explicitly. Let s ( r ) and c ( r ) be functions defined as follows : s ( r ) = sin r with 0 ≤ r < π if M = S m sin r with 0 ≤ r < π if M = R P n , C P n , Q P n , O P sinh r if M is nCROSSand c ( r ) = (cid:40) cos r with 0 ≤ r < π if M = C P n , Q P n , O P cosh r if M is nCROSS . Then the metric ( ds ) is given by( ds ) = ( dr ) + ( s ( r )) ( c ( r )) g + ( s ( r )) h, (6)where ( dr ) , g, and h are written by σ with the coframe σ dual to (cid:126)v ; σ + · · · + σ k with coframes σ , . . . , σ k dual to orthonormal basis of K (cid:48) · (cid:126)v ; σ k +1 + · · · + σ m with coframes σ k +1 , . . . , σ m dual to the complementorthonormal basis of R m . Since the density function ω only depends ondistance, we may define ω as a one-variable function ω ( r ) = ( s ( r )) m − ( c ( r )) k − . Then the sectional curvature K M of M : (cid:40) ≤ K M ≤ M is CROSS − ≤ K M ≤ − M is nCROSS . (7)In particular, S m and R P n has sectional curvature 1. Then the condition0 < R < inj ( M )2 in Theorem 1 implies: < R < π if M = S m < R < π if M = R P n , C P n , H P n , O P < R otherwise . (8)Now consider a triangle ( P QR ) in M with the metric (6), which con-sists of three distinct points P, Q, R ∈ M and three connecting geodesics QR, RP, P Q . The side lengths will be denoted by p, q , and r , respectively DONG-HWI SEO and the two angular invariants λ, ϕ determined by the two tangent vectorsof geodesic rays
P Q and
P R at P will be denoted by λ ( P ) and ϕ ( P ), re-spectively. Furthermore we can denote λ ( Q ), ϕ ( Q ), λ ( R ), and ϕ ( R ) in ananalogous way. Then it is known that there are congruent conditions oftriangles. We introduce some conditions which will be used later. For moreconditions, see [4]. Proposition 1.
A triangle (PQR) in ROSS with the metric (6) is uniquelydetermined up to isometry as follows :(a) p, q, and λ ( P ) with < p, q, r < π and q < p < π if M is S m .(b) p, q, and λ ( P ) with < p, q, r < π and q < p < π if M is R P n .(c) p, q, λ ( P ) , and ϕ ( P ) with < p, q, r < π and ( p − q )(cos p − sin q cos ϕ ( P )) > if M is C P n , H P n or O P .(d) p, q, λ ( P ) , and ϕ ( P ) with < p, q, r and q < p if M is nCROSS.Proof. The proof of (a) can be found in Section VI in [19]. In fact, thecondition p < π can be replaced by p + q < π . The proof of (b) follows from(a). The proofs of (c) and (d) can be found in (ix) and (ix’) of Theorem 4and 4’ in [4]. (cid:3) Main proof
Let M be a ROSS with the metric (6). Let X and C be the centers of B and B (cid:48) , respectively. Define B to be the ball of radius R , centered at X .3.1. The first eigenfunctions.
In this section, we derive an explicit for-mula for the first mixed Steklov-Dirichlet eigenfunctions in B \ cl( B ).Using the following standard argument as in [7] and [20], we can show thatthe first eigenfunction is a function that only depends on the distance from X .Using seperation of variables, a mixed Steklov-Dirichlet eigenfunction u ( r, θ , . . . θ m − ) in B \ cl( B ) is obtained by multiplying a Laplacian eigen-function f ( θ , . . . , θ m − ) on the unit sphere S m − by an appropriate radialfunction a ( r ). Here, ( r, θ , . . . θ m − ) is the polar coordinate in T X M . SinceLaplace eigenfunctions on S m − are indeed Laplace eigenfunctions on ∂ B (see Theorem 3.1 in [7], or Corollary 5.5 in [2]) and it consists of a basis of L ( ∂ B ), our mixed Steklov-Dirichlet eigenfunctions restrict to ∂ B becomea basis of L ( ∂ B ). It implies the k th mixed Steklov-Dirichlet eigenfunctionis written by a product of a Laplacian eigenfunction and a radial function.By some computations as in Section 2.1 in [20], we can conclude that the k th mixed Steklov-Dirichlet eigenfunction is corresponding to the k th Lapla-cian eigenfunction. Since the first Laplacian eigenfunctions are constants,we obtain the following. SHAPE OPTIMIZATION PROBLEM FOR THE FIRST EIGENVALUE 7
Proposition 2.
Let r X : M → [0 , ∞ ) be the distance function from X . Let a : [ R , ∞ ) → R be a function defined by a ( r ) = (cid:90) rR ω ( t ) dt. Then the first mixed Steklov-Dirichlet eigenfunction in B \ cl( B ) is a ◦ r X up to constant.Proof. By the argument in the paragraph, the first eigenfunction can bewritten by a ◦ r X , where a : [ R , ∞ ) → R is a real-valued function. Then, the harmonicity ofthe eigenfunction implies0 = ∆ a ( r ) = a (cid:48)(cid:48) ( r ) + ω (cid:48) ( r ) ω ( r ) a (cid:48) ( r ) = 1 ω ( r ) ( a (cid:48) ( r ) ω ( r )) (cid:48) . Here, we used r instead of r X for simplicity of notation. With the fact that a ( R ) = 0 from the boundary condition, we obtain the formula of a ( r ) upto constant. (cid:3) Crucial lemmas and the proof for nCROSS.
We begin with twodefinitions.
Definition 2.
For given X ∈ B (cid:48) , a vector-valued function (cid:126)v X : M \ { X } → U X M is defined by P ∈ M \ { X } and (cid:126)v X ( P ) ∈ U X M such that (cid:126)v X ( P ) isthe unit tangent vector of the geodesic ray XP at X . For a given parametrization of M around X , we can identify T X M with R m . Then we can give the following definition. Definition 3.
For given X ∈ B (cid:48) and a parametrization of M around X , amap π X : S m − ∼ = U X M → ∂ B (cid:48) is defined by π X ( v ) = exp X ([0 , ∞ ) · v ) ∩ ∂ B (cid:48) ,i.e. π X ( v ) is the point of ∂ B (cid:48) in the geodesic emanating from X in v direction. Note that π X has the inverse map. Thus, for any P ∈ ∂ B (cid:48) , we canfind P s ∈ S m − such that P = π X ( P s ). Furthermore, let C s ∈ S m − suchthat the geodesic ray exp X ([0 , ∞ ) · C s ) passes through C . Then we candefine − P s and ¯ P s in S m − such that they are the symmetric points of P s with respect to X and the line passing through X and C s , respectively. Inaddition, − ¯ P s can be defined as the symmetric point of ¯ P s with respect to X . Now we denote exp X ( − P s ), exp X ( ¯ P s ), and exp X ( − ¯ P s ) by − P , ¯ P , and − ¯ P , respectively. Figure 1 explains the situation. DONG-HWI SEO
CX P = π X ( P s )¯ P − P − ¯ P P s − ¯ P s − P s ¯ P s C s Figure 1.
Description of P, ¯ P , − P , and − ¯ P . The dottedcircle and the bigger circle represent ∂ B and ∂ B (cid:48) , respec-tively.3.2.1. Properties of angles and distances.
In this section, we prove the lem-mas which are essential in the proof of the main proposition in the nextsection. We prove a lemma about the “symmetric properties” of angles anddistances. In addition, we obtain a lemma which is motivated from theconcept of solid angle. As a corollary, we introduce Newton’s shell theoremwith an infinitesimally thin “shell” in ROSS. We begin with a lemma, whichare useful for the lemmas below.
Lemma 1.
A triangle ( P QR ) in ROSS M with the metric (6) satisfies :(a) If M = S m , < p, q, r < π , and p ≤ q < π , then λ ( P ) < π .(b) If M = R P n , C P n , H P n , or O P , < p, q, r < π , and p ≤ q < π ,then λ ( P ) < π .(c) If M is nCROSS, < p, q, r , and p ≤ q , then λ ( P ) < π .Proof. (a) Suppose λ ( P ) ≥ π . Using the law of cosines of sphericaltriangles (see p. 179 in [13]),cos p = cos q cos r + sin q sin r cos P < cos q cos r. Combining the previous inequality with cos p, cos q >
0, we obtaincos r > p < cos q . It implies p > q , contradiction to ourassumption.(b) Suppose λ ( P ) ≥ π . Since M has sectional curvature K M ≤ p ≤ cos 2 q cos 2 r + sin 2 q sin 2 r cos P < cos 2 q cos 2 r. Then by an analogous argument in (a), we obtain a contradiction.(c) Suppose λ ( P ) ≥ π . Since M has sectional curvature K M ≤ − SHAPE OPTIMIZATION PROBLEM FOR THE FIRST EIGENVALUE 9 [13]).cosh p ≥ cosh q cosh r − sinh q sinh r cos P > cosh q. Thus p > q , which contradicts to our assumption. (cid:3)
For P ∈ ∂ B (cid:48) , consider a triangle ( P XC ) in cl( B (cid:48) ) defined in the begin-ning of Section 3, which consists of the center X of B , the center C of B (cid:48) , P , and godesics connecting two of them. Then the next lemma explainsrelations of distances from X to P, ¯ P , − P , and − ¯ P and relations of anglesat those points.(a) (b) (c) Figure 2.
Illustration of Lemma 2. The circles in (a),(b),(c)represent ∂ B (cid:48) . Lemma 2.
Let λ X : ∂ B (cid:48) → [0 , π ] be an angle function with respect to X that assigns to each P ∈ ∂ B (cid:48) an angle λ ( P ) of the triangle (PXC). Define r X as in the Proposition 2. Then, λ X and r X satisfy the following.(a) ≤ λ X ( P ) < π .(b) λ X ( P ) = λ X ( ¯ P ) , r X ( P ) = r X ( ¯ P ) for all P ∈ ∂ B (cid:48) .(c) λ X ( P ) = λ X ( − P ) , r X ( P ) ≥ r X ( − P ) for all P ∈ ∂ B (cid:48) satisfying ∠ ( (cid:126)v X ( P ) , (cid:126)v X ( C )) ≤ π . The equality holds if and only if ∠ ( (cid:126)v X ( P ) , (cid:126)v X ( C )) = π .Proof. We will prove this lemma when M = C P n , H P n or O P . Then wehave R < inj ( M )2 = π .(a) Note that R < π and | CX | < | CP | = R . Then the statementfollows from Lemma 1.(b) Consider two triangles ( P XC ) and ( ¯
P XC ). By the constructions of P and ¯ P , λ ( X ) of ( P XC ) and ( ¯
P XC ) are identical. The same holdsfor ϕ ( X ). Note that the two triangles have the common edge XC and | CP | = | C ¯ P | = R . From the fact that | CX | < | CP | = R < π we have sin | CX | < cos | CP | . Therefore by Proposition 1, ( P XC )and ( ¯
P XC ) are congruent. Then our statement follows. (c) Using the fact that B (cid:48) is convex (see p. 148 in [16]), we can define apoint R ∈ B (cid:48) in the complete geodesic containing X and P such thatthe geodesic meets CR perpendicularly. Then under the conditionon P , we claim that | P R | ≤ |
P X | . It is equivalent to showing that λ ( XR, XC ) ≤ π . If X = R , λ ( XR, XC ) = π . Otherwise, we have | RC | < | XC | < π . Then by Lemma 1 for ( XCR ), our claim follows.On the other hand, two triangles (
P RC ) and ( − P RC ) are congruentby (4),(5), and Proposition 1 as in the proof of (b). Thus we obtainthat λ X ( P ) = λ X ( − P ) and | P R | = | − P R | , which imply the desiredconclusion.A slight change in the proof shows it also holds if M is S m , R P n or nCROSSs. (cid:3) Now we will give another lemma that explains an “infinitesimal area of ∂ B from X ” can be calculated by λ X and r X . X A π − X ( A ) R Figure 3.
Description of R in the proof of Lemma 3. Thedotted circle and the bigger circle represent ∂ B and ∂ B (cid:48) ,respectively. Lemma 3.
Let µ be the Lebesgue measure on S m − and consider the push-forward π X µ on ∂ B (cid:48) . Then for a measurable set A ⊂ ∂ B (cid:48) , we have π X µ ( A ) = µ ( π − X ( A )) = (cid:90) A cos λ X ω ( r X ) dS (cid:48) , where S (cid:48) is the induced measure on ∂ B (cid:48) from the metric of M . Equivalently, dS (cid:48) = ω ( r X )cos λ X dπ X µ. Proof.
It is clear that S (cid:48) and π X µ are σ -finite and π X µ (cid:28) S (cid:48) that isto say ( π X ) µ is absolutely continuous with respect to S (cid:48) . Furthermore, S (cid:48) (cid:28) π X µ . By the Radon-Nikodym theorem, there are functions f and SHAPE OPTIMIZATION PROBLEM FOR THE FIRST EIGENVALUE 11 f on ∂ B (cid:48) such that π X µ ( A ) = (cid:90) A f dS (cid:48) and dS (cid:48) ( A ) = (cid:90) A f dπ X µ. Consider a vector field F on M \ { X } defined by F ( Y ) = (cid:18) ω ( r X ) ∂∂r (cid:19) ( Y ) , where ∂∂r ( Y ) is the vector in T Y M obtained by the parallel transport of theunit tangent vector (cid:126)v X ( Y ) along XY . Thendiv( F ) = 1 ω ( r X ) ∂∂r (cid:18) ω ( r X ) · ω ( r X ) (cid:19) = 0 . Now consider a region R that is the region of the solid cone from X over ageodesic ball B ⊂ ∂ B (cid:48) bounded by ∂ B and ∂ B (cid:48) . Equivalently, R = { exp X ( t · (cid:126)v X ( Y )) | Y ∈ B , R ≤ t ≤ r X ( Y ) } . Let
R ∩ ∂ B = B . Then applying the divergence theorem to F on R , wehave 0 = (cid:90) R div F = (cid:90) B cos λ X ω ( r X ) dS (cid:48) − (cid:90) B ω ( R ) dS , where S is the measure on ∂ B induced by the metric of M . Combining itwith the fact that (cid:90) B ω ( R ) dS = µ ( π − X ( A )) , the first statement is proved for B . Then by Theorem 4.7 in [17], the firststatement is proved. Since cos λ X (cid:54) = 0 from Lemma 2, the second argumentfollows. (cid:3) The following corollary is not necessary for the proof of the main theorem.
Corollary 1.
We have (cid:90) ∂ B (cid:48) (cid:126)v X ω ( r X ) dS (cid:48) = 0 . Proof.
Using the previous lemma, the left hand side is equal to (cid:90) S m − (cid:18) (cid:126)v X cos λ X ◦ π X (cid:19) dµ. (9)By Lemma 2, we have (cid:18) (cid:126)v X cos λ X (cid:19) ◦ π X ( P s ) + (cid:18) (cid:126)v X cos λ X (cid:19) ◦ π X ( − P s ) = 0for P s ∈ S m − . Then this relation gives the desired result. (cid:3) Remark.
Note that if M = R , then ω ( r ) = r . Furthermore (cid:126)v X ( π X ( p )) isthe unit vector from X to P = π X ( p ) at X . Thus the equation becomesNewton’s shell theorem, which implies that the net gravitational force of aspherical shell acting on any object inside is zero.3.2.2. The proof for nCROSS.
In this section, we prove the main theoremfor nCROSS. We use the fact that the first mixed Steklov-Dirichlet eigen-function, a ◦ r X , of the annulus B \ B is a test function in both of thevariational characterizations of σ ( B (cid:48) \ B ) and σ ( B \ B ). Substitutingthe test function into the two Rayleigh quotients, we compare the two de-nominators and the two numerators in the following two propositions.Define a map (cid:90) ∂ B (cid:48) ( a ◦ r ( · ) ) dS (cid:48) : B (cid:48) → R that assigns to X ∈ B (cid:48) (cid:90) ∂ B (cid:48) ( a ◦ r X ) dS (cid:48) In the following proposition, we show that the function has a minimum valueat C by analyzing the gradient of the function at each X ∈ B (cid:48) , ∇ (cid:32)(cid:90) ∂ B (cid:48) ( a ◦ r ( · ) ) dS (cid:48) (cid:33) ( X ) ∈ T X M. Proposition 3.
We have ∇ (cid:32)(cid:90) ∂ B (cid:48) ( a ◦ r ( · ) ) dS (cid:48) (cid:33) ( X ) = (cid:40) − g ( X ) · (cid:126)v X ( C ) if X (cid:54) = C, if X = C, where g : B (cid:48) \ { C } → R + is a positive function. Furthermore, (cid:90) ∂ B (cid:48) ( a ◦ r C ) dS (cid:48) ≤ (cid:90) ∂ B (cid:48) ( a ◦ r X ) dS (cid:48) , and equality holds if and only if X = C .Proof. The gradient is calculated at X ∈ B (cid:48) , so it does not affect on theintegration region ∂ B (cid:48) . Then for P ∈ ∂ B (cid:48) , ∇ ( a ◦ r ( · ) ( P )) ( X ) ∈ T X M .Thus −∇ (cid:32)(cid:90) ∂ B (cid:48) ( a ◦ r ( · ) ) ( P ) dS (cid:48) ( P ) (cid:33) ( X ) = (cid:90) ∂ B (cid:48) a ◦ r X ) ω ( r X ) · ( −∇ r ( · ) ( P )( X )) dS (cid:48) ( P ) . With −∇ ( r ( · ) ( P ))( X ) = (cid:126)v X ( P ) and Lemma 3, the previous equation is equalto (cid:90) S m − (cid:18) a ◦ r X ) · (cid:126)v X cos λ X (cid:19) ◦ π X dµ. (10) SHAPE OPTIMIZATION PROBLEM FOR THE FIRST EIGENVALUE 13
X P ¯ P − P − ¯ P Figure 4.
Pictorial explanation of calculation of (10). Eachthick arrows represents integrand of (10) at P, ¯ P , − P , and − ¯ P .If X = C , the integral has value 0. Otherwise, we consider the integrandat P s ∈ { v |(cid:104) v, c (cid:105) ≥ } ⊂ S m − , ¯ P s , − P s , and − ¯ P s . Note that the conditionfor P s is equivalent to ∠ ( (cid:126)v X ( P ) , (cid:126)v X ( C )) ≤ π . Then using Lemma 2, (cid:18)(cid:18) a ◦ r X ) · (cid:126)v X cos λ X (cid:19) ( P ) + (cid:18) a ◦ r X ) · (cid:126)v X cos λ X (cid:19) ( ¯ P ) (cid:19) + (cid:18)(cid:18) a ◦ r X ) · (cid:126)v X cos λ X (cid:19) ( − P ) + (cid:18) a ◦ r X ) · (cid:126)v X cos λ X (cid:19) ( − ¯ P ) (cid:19) =2( a ◦ r X )( P ) · (cid:104) (cid:126)v X ( P ) , (cid:126)v X ( C ) (cid:105) cos λ X · (cid:126)v X ( C )+2( a ◦ r X )( − P ) · (cid:104) (cid:126)v X ( − P ) , (cid:126)v X ( C ) (cid:105) cos λ X · (cid:126)v X ( C )=4 (( a ◦ r X )( P ) − ( a ◦ r X )( − P )) · (cid:104) (cid:126)v X ( P ) , (cid:126)v X ( C ) (cid:105) cos λ X · (cid:126)v X ( C ) . Furthermore, Lemma 2 implies ( a ◦ r X )( P ) − ( a ◦ r X )( − P ) > ∠ ( (cid:126)v X ( P ) , (cid:126)v X ( C )) = π/
2. Thus our integration has a form g ( X ) · (cid:126)v X ( C )for some positive function g . Note that we actually proved that the gradi-ent of the function has the opposite direction from X to C . It implies ourdesired inequality. (cid:3) Remark. • In the proof, the function g only depends on the distancebetween X and C . • The proof is similar to the proof of Corollary 1 if we compare (9)and (10). The difference between the two proofs is the fact that a isan increasing function. Corollary 2.
We have (cid:90) ∂ B ( a ◦ r X ) dS ≤ (cid:90) ∂ B (cid:48) ( a ◦ r X ) dS (cid:48) , where S is the measure on ∂ B induced from the metric of M . The equalityholds if and only if B (cid:48) = B .Proof. Note that B is a ball of radius R , centered at X . Therefore wehave (cid:90) ∂ B ( a ◦ r X ) dS = (cid:90) ∂ B (cid:48) ( a ◦ r C ) dS (cid:48) . Then Proposition 3 implies the statement. (cid:3)
In the following proposition, ( ∇ ( a ◦ r X ))( Z ) for Z ∈ M \ { X } is thegradient of a ◦ r X ( · ) : M \ { X } → R at Z . Proposition 4.
We have (cid:90) B (cid:48) \ cl( B ) |∇ ( a ◦ r X ) | dV (cid:48) ≤ (cid:90) B \ cl( B ) |∇ ( a ◦ r X ) | dV, where V and V (cid:48) are measures on B and B (cid:48) induced from the metric of M ,respectively, and equality holds if and only if B (cid:48) = B .Proof. Note that |∇ ( a ◦ r X ( · )) | = |∇ a | ◦ r X ( · ) and it is easy to check that |∇ a | ( r ) = | a (cid:48) ( r ) | = ω ( r ) is a decreasing function since we only consider when M is nCROSS. Then (cid:90) B \ cl( B ) |∇ ( a ◦ r X ) | dV − (cid:90) B (cid:48) \ cl( B ) |∇ ( a ◦ r X ) | dV (cid:48) = (cid:90) B \ B (cid:48) |∇ ( a ◦ r X ) | dV − (cid:90) B (cid:48) \ B |∇ ( a ◦ r X ) | dV (cid:48) ≥ (cid:90) B \ B (cid:48) |∇ a ( R ) | dV − (cid:90) B (cid:48) \ B |∇ a ( R ) | dV (cid:48) = 0 . To satisfy the equality, | B (cid:48) \ B | = | B \ B (cid:48) | = 0, or B (cid:48) = B . (cid:3) Remark.
We used only the fact that ω ( r ) is a concave function in [0 , R ).Thus the proof also applies when M is CROSS and R < inj ( M )4 .Now we have the following proof of the main theorem when M is anCROSS. Proof of Theorem 1 for nCROSS.
Note that u ◦ r X = 0 on ∂ B . By thevariational characterization of σ ( B (cid:48) \ cl( B )), σ ( B (cid:48) \ cl( B )) ≤ (cid:82) B (cid:48) \ cl( B ) |∇ ( a ◦ r X ) | dV (cid:48) (cid:82) ∂ B (cid:48) ( a ◦ r X ) dS (cid:48) . SHAPE OPTIMIZATION PROBLEM FOR THE FIRST EIGENVALUE 15
By Corollary 2 and Proposition 4, we have σ ( B (cid:48) \ cl( B )) ≤ (cid:82) B \ cl( B ) |∇ ( a ◦ r X ) | dV (cid:82) ∂ B ( a ◦ r X ) dS . Since we have shown that a ◦ r X is the first mixed Steklov-Dirichlet eigen-function on the annulus B \ cl( B ) in Proposition 2, the right hand side is σ ( B \ cl( B )). It is the desired inequality. In addition, the equality condi-tion is followed from the equality conditions in Corollary 2 and Proposition4. (cid:3) Remark.
The method of the proof carries over to Euclidean space R m .3.2.3. The proof for CROSS.
In this section we modify the proof of Propo-sition 4 to show that the inequality in this proposition also holds when M isCROSS and R < inj ( M )2 . Then using the same argument in Section 3.2.2,we can show that the main theorem holds in this situation. B r ( X ) denotes the ball of radius r , centered at X and d := r X ( C ) denotesthe distance between X and C . Then the difference between the two sidesof the inequality in Proposition 4 becomes (cid:90) B \ cl( B ) (cid:18) ω ( r ) (cid:19) dV − (cid:90) B (cid:48) \ cl( B ) (cid:18) ω ( r ) (cid:19) dV = (cid:90) B \ B (cid:48) (cid:18) ω ( r ) (cid:19) dV − (cid:90) B (cid:48) \ B (cid:18) ω ( r ) (cid:19) dV = (cid:90) R R − d (cid:90) π − X (( B \ B (cid:48) ) ∩ ∂B r ( X )) ω ( r ) dµdr − (cid:90) R + dR (cid:90) π − X (( B (cid:48) \ B ) ∩ ∂B r ( X )) ω ( r ) dµdr = (cid:90) d (cid:32)(cid:90) π − X (( B \ B (cid:48) ) ∩ ∂B R − s ( X )) ω ( R − s ) − (cid:90) π − X (( B (cid:48) \ B ) ∩ ∂B R s ( X )) ω ( R + s ) (cid:33) dµds. The last equality is obtained by substituting r and r by R − s and R + s for s < d , respectively. Then the integral becomes nonnegative providedthat the following two lemmas hold. Lemma 4.
We have | π − X (( B (cid:48) \ B ) ∩ ∂B R + s ( X )) | ≤ | π − X (( B \ B (cid:48) ) ∩ ∂B R − s ( X )) | for s < R . X C B B (cid:48) R + s R − s Figure 5.
The left and right thick arcs represent π − X (( B \ B (cid:48) ) ∩ ∂B R − s ( X )) and π − X (( B (cid:48) \ B ) ∩ ∂B R + s ( X )), respec-tively. In addition, the dotted circle is ∂ B and we have | XC | = d . Proof.
Consider S ∈ ( B \ B (cid:48) ) ∩ ∂B R − s ( X ). Then the triangle ( SXC ) hasside lengths | XC | = d, | XS | = R − s, | CS | ≥ R . Consider the space form S mκ of constant curvature κ , where κ ∈ R + is aconstant such that a geodesic ball of radius R is a hemisphere in S mκ . Thenwe have π √ κ = R , so κ is bigger than the sectional curvature of M . Now consider a triangle( S κ X κ C κ ) with the same side lengths as ( SXC ) in S mκ . Then by the trianglecomparison theorem (see p. 197 in [13]), ∠ SXC ≤ ∠ S κ X κ C κ . Then it implies the following inequality. | π − X (( B \ B (cid:48) ) ∩ ∂B R − s ( X )) | = |{ π − X ( S ) || XS | = R − s, | CS | ≥ R }|≥|{ S κ || X κ S κ | = R − s, | C κ S κ | ≥ R }| × s κ ( R − s )= |{ S κ | S κ ∈ (( B ) κ \ ( B (cid:48) ) κ ) ∩ ∂ ( B R − s ) κ ( X κ ) }| × s κ ( R − s ) , (11)where ( B ) κ and ( B (cid:48) ) κ are geodesic balls of radius R in S mκ , centered at X κ and C κ , respectively, and s κ ( r ) = sin √ κr √ κ . SHAPE OPTIMIZATION PROBLEM FOR THE FIRST EIGENVALUE 17
By a similar argument, we obtain | π − X (( B (cid:48) \ B ) ∩ ∂B R + s ( X )) |≤|{ S (cid:48) κ | S (cid:48) κ ∈ (( B (cid:48) ) κ \ ( B ) κ ) ∩ ∂ ( B R + s ) κ ( X κ ) }| × s κ ( R + s )(12)Since s κ ( R − s ) = s κ ( R + s ) , and the set { S κ | S κ ∈ (( B ) κ \ ( B (cid:48) ) κ ) ∩ ∂ ( B R − s ) κ ( X κ ) } is the image of the antipodal map in S mκ of { S (cid:48) κ | S (cid:48) κ ∈ (( B (cid:48) ) κ \ ( B ) κ ) ∩ ∂ ( B R + s ) κ ( X κ ) } , the right hand sides of (11) and (12) are equal. Thus our desired inequalityis obtained. (cid:3) Lemma 5.
We have ω ( R − s ) < ω ( R + s ) for < s < R .Proof. We begin with M = R P n , C P n , H P n , O P , which are CROSS exceptfor S m . Then s < R < π . We have two observations of the density function ω ( t ) = (sin t ) m − (cos t ) k − : ω (cid:48) ( t ) > t < arctan (cid:113) m − k − ,ω (cid:48) ( t ) < t > arctan (cid:113) m − k − . and ω ( t ) ≤ ω ( π − t )for t < π . The second observation follows from ω ( π − t ) − ω ( t ) = (cos t ) m − (sin t ) k − − (sin t ) m − (cos t ) k − = (sin t ) k − (cos t ) k − ((cos t ) m − k − (sin t ) m − k ) > . Therefore if R + s < arctan (cid:114) m − k − , the first observation implies ω ( R − s ) < ω ( R + s ) . Otherwise, the two observations give ω ( R − s ) < ω ( π − ( R − s )) < ω ( R + s ) . Therefore the proof for CROSS follows except for S m . The same proof alsoworks for S m if we replace π and π by π and π , respectively. (cid:3) References
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SHAPE OPTIMIZATION PROBLEM FOR THE FIRST EIGENVALUE 19
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