Estimates for sums of eigenvalues of the Laplacian by

Pawel Kr¨oger∗ Mathematisches Institut Universit¨ at Erlangen-N¨ urnberg Bismarckstr. 1 1/2 91054 Erlangen Germany Abstract. The aim of this paper is to give bounds for the eigenvalues of the Laplacian on a domain in Euclidean space and on a compact Riemannian manifold. First, we consider the eigenvalue problem for the Laplacian on a bounded domain in Euclidean space under Dirichlet and Neumann boundary conditions. Our method for obtaining an upper bound for sums of eigenvalues under Dirichlet boundary conditions is closely related to the method used in [5] for the task of getting an upper bound for sums of eigenvalues under Neumann boundary conditions. On the other hand, we modify the method used in [6] for obtaining a lower bound for sums of eigenvalues under Dirichlet boundary conditions in order to get a lower bound for sums of eigenvalues under Neumann boundary conditions under the assumption that the domain under consideration is Lipschitz equivalent to a ball. Finally, we derive estimates for sums of squares of eigenvalues on a compact Riemannian manifold without boundary. 1. Introduction and statement of the results for domains in Euclidean space Let Ω be a bounded (connected) domain in Rn . Let dist(x, ∂Ω) denote the distance of a point x ∈ Ω from the boundary of Ω. We introduce the notation Ωr ≡

{x ∈ Ω | dist(x, ∂Ω) <

1 }. r

We consider the eigenvalues of −4 on Ω under Dirichlet boundary conditions increasingly ordered as follows: 0 < λ1 < λ2 ≤ λ3 ≤ ... (0)

Theorem 1. Suppose that there exists a constant CΩ such that vol(Ωr ) ≤ r. Then the following holds: k X

λj ≤ (2π)2

j=1

n+2 2 n (0) n+1 (vol(Ω)vol(B1 ))− n (k n + Cn(1) CΩ k n ) n+2

(0)

CΩ r

vol(Ω)

n−1 n

for every

(1)

1

for every k ≥ Cn(0) n and a constant Cn(1) which depends only on the dimension n. Now suppose that that there exists a bi-Lischitz map f which maps Ω onto an open ball B in Rn . Let CΩ be a common Lipschitz constant for f and f −1 . Since the spectrum of −4 on Ω under Neumann boundary conditions is discrete, the eigenvalues can be increasingly ordered as follows: 0 = µ1 < µ2 ≤ µ3 ≤ ... ∗

Research supported by the Deutsche Forschungsgemeinschaft

1

Theorem 2. Under the above assumptions on Ω the following holds: k X

µj

(2π)2

j=1

n+2 n+1 2 n 2 (vol(Ω)vol(B1 ))− n (k n − Cn(2) CΩ2n +2n k n ). n+2

(2)

for every k and a constant Cn(2) which depends only on the dimension n. Remark. The assumptions of Theorem 2 are obviously stronger than the assumptions of Theorem 1. Corollary. Suppose that eqn. (1) and eqn. (2) hold. Then we can conclude that there exists a constant Cn(3) which depends only on the dimension n such that the following holds: 2

2

2 −n

2 n

|λk − (2π)2 (vol(Ω)vol(B1 ))− n k n | |µk − (2π)2 (vol(Ω)vol(B1 ))

k |

3

(0)

Cn(3) CΩ vol(Ω)k 2n

2n2 +2n

Cn(3) CΩ

and 3 2n

vol(Ω)k .

Remarks 1. It is easy to see that the order of the second term in k on the right-hand sides of (1) and (2), resp., is correct (cf. for instance [7] ). On the other hand, the order of k in the terms on the right-hand sides of the estimates of the Corollary does not seem to be sharp. It would be a big step towards a proof of P´olya’s Conjecture (cf. [8] ) to give estimates similar to those given in the Corollary which contain two terms of correct order in k. 2. We notice that our proofs have no ”hard analysis” component. A similar requirement was one of the motivations for the paper [3] . 2. Proofs of the results from Section 1 We set χ(α) ≡ min{max{0, α} , 1} for every α ∈ R. Moreover, we set χ(r) (x) ≡ χ(rdist(x, ∂Ω)) for x ∈ Ω and χ(r) (x) ≡ 0 for x 6∈ Ω. Proof of Theorem 1. Let {φj }kj=1 be a set of orthonormal eigenfunctions for the eigenvalues λ1 , ..., λk . We introduce the function Φk on Ω × Ω by Φk (x, y) ≡

k X

φj (x)φj (y)

for x, y ∈ Ω.

j=1

The function ωz(r) which is defined by ωz(r) (x) ≡ exp(i < z, x >)χ(rdist(x, ∂Ω)) for every x ∈ Ω vanishes at the boundary of Ω. The projection of ωz(r) onto the subspace of L2 (Ω) spanned by φ1 , ..., φk can be written in terms of the Fourier transform (χ(r) Φk )∧x of χ(r) Φk with respect to the x−variable: Z ωz(r) (x)Φk (x, y) dx = (2π)n/2 (χ(r) Φk )∧x (z, y). Ω

Hence, we can conclude from the Rayleigh-Ritz formula that R

λk+1 ≤ inf r

R Br Ω

R

|∇y (ωz(r) (y) − (2π)n/2 (χ(r) Φk )∧x (z, y))|2 dy dz R

Br Ω

(r)

|ωz (y) − (2π)n/2 (χ(r) Φk )∧x (z, y)|2 dy dz

2

(∗);

here Br stands for the ball with radius r and center 0 in Rn . An elementary calculation yields that Z

Z

Br

|ωz(r) (y) − (2π)n/2 (χ(r) Φk )∧x (z, y)|2 dy dz

Z

=

Z

Br

|ωz(r) (y)|2 dy dz

(2π)n

Z

Z X k

Br

n

≥ r vol(B1 )(vol(Ω) − vol(Ωr ))

Ω j=1 n

(2π)

|(χ(r) φj )∧ (z)|2 |φj (y)|2 dy dz

k Z X j=1 Br

|(χ(r) φj )∧ (z)|2 dz.

For an estimate of the expression in the nominator on the right-hand side of (*), we take into account that Z

Z

Br

< ∇y (ωz(r) (y) − (2π)n/2 (χ(r) Φk )∧x (z, y)), (χ(r) φj )∧ (z)∇y (φj (y)) > dy dz

Z

=

Z

Br

∂Ω

Z

+

(ωz(r) (y) − (2π)n/2 (χ(r) Φk )∧x (z, y)) (χ(r) φj )∧ (z)

Z

Br

∂ (φj (y)) dy dz ∂ny

(ωz(r) (y) − (2π)n/2 (χ(r) Φk )∧x (z, y)) λj (χ(r) φj )∧ (z)φj (y) dy dz

= 0 (the boundary integral vanishes because ωz(r) and the φj vanish at ∂Ω). Moreover, Z Ω

|∇y ωz(r) (y)|2 dy dz

|z|2 vol(Ω) + 2r|z|vol(Ωr ) + r2 vol(Ωr ).

Finally, Z

Z

Br

|∇y (χ(r) Φk )∧x (z, y))|2 dy dz =

k X j=1

λj

Z Br

|(χ(r) φj )∧ (z)|2 dz.

Thus, we can conclude that λk+1 ≤

1 rn+2 nvol(B1 )( n+2 vol(Ω) +

2 vol(Ωr ) n+1

rn vol(B1 )(vol(Ω) − vol(Ωr ))

Pk

(r) ∧ 2 j=1 λj Br |(χ φj ) (z)| R P (2π)n kj=1 Br |(χ(r) φj )∧ (z)|2 dz

+ n1 vol(Ωr )) − (2π)n

R

for every r such that the denominator is positive. By Parseval’s identity, Z Br

(r)

2

|(χ φj ) (z)| dz

Z

|(χ(r) φj )(x)|2 dx

≤ 1.

Since λj ≤ λk+1 for every j ≤ k, an elementary induction argument yields that λk+1

1 2 rn+2 nvol(B1 )( n+2 vol(Ω) + n+1 vol(Ωr ) + n1 vol(Ωr )) − (2π)n ≤ rn vol(B1 )(vol(Ω) − vol(Ωr )) − (2π)n k

Pk

j=1

λj

for every r such that the denominator is positive. By Federer’s co area formula, (

r1 n ) vol(Ωr1 ) ≤ vol(Ωr2 ) ≤ vol(Ωr1 ) r2

3

for every r1 , r2 with 0 < r1 ≤ r2 .

dz

We set r0 ≡ (k + 1)1/n vol(Ω)−1/n and 1/n (r0 <) r ≡ 2π(k + 1) ((vol(Ω) − vol(Ωr0 ))vol(B1 ))−1/n . We can conclude that there is a constant Cn(0) which depends only on the dimension n such that k X

λj ≤ (2π)2

j=1

n+2 2 n (max{vol(Ω) − Cn(0) vol(Ωr0 ); 0}vol(B1 ))− n k n . n+2

The assertion of the theorem follows immediately. Proof of Theorem 2. We will restrict ourselves for the proof of our estimate to the case vol(Ω) = 1 and B = B1 where B1 denotes the unit ball. Let {ψj }kj=1 be a set of orthonormal eigenfunctions for the eigenvalues µ1 , ..., µk . We introduce the function Ψk on Ω × Ω by k X

Ψk (x, y) ≡

for x, y ∈ Ω.

ψj (x)ψj (y)

j=1

We introduce a function F on Rn by Z

F (z) ≡

|

Z Ω

for every z ∈ Rn .

exp(i < z, x >)χ(r) (x)Ψk (x, y) dx|2 dy

We will restrict ourselves to the case r ≥ 1. (Our definition of F is motivated by the definition of R ˆ the function f (z) ≡ |Φ(z, y)|2 dy from p. 313 below in [6] and by the ”duality” of the Neumann boundary value problem under consideration with the situation which was considered in Theorem 1. We have put the word duality in quotation marks because we need here somewhat stronger assumptions and are hence unable to give a precise meaning to that duality.) R Since the function y 7→ Ω exp(i < z, x >)χ(r) (x)Ψk (x, y) dx coincides with the orthogonal projection of exp(i < z, x >)χ(r) (x) on the span of ψ1 , ..., ψk , we can conclude that (recall that by assumption vol(Ω) = 1): F (z) ≤ || exp(i < z, x >)χ(r) (x)||22 ≤ 1 for every z ∈ Rn . Moreover, by Parseval’s identity, −n

(2π)

Z

R

n F (z) dz

k X

=

||χ(r) ψj ||22

j=1

k − exp(µk /r2 )

Z

pN (1/r2 ; x, x) dx,

Ωr

where pN (t; x, x) denotes the Neumann heat kernel on Ω. Finally, −n

(2π)

Z

R

2

n

|z| F (z) dz

=

k Z X j=1 Ω

=

k Z X j=1 Ω

|∇(χ(r) ψj )|2 dx |ψj ∇χ(r) |2 + < ∇ψj , ∇((χ(r) )2 ψj ) > dx

2

2

r exp(µk /r ) 4

Z Ωr

N

2

p (1/r ; x, x) dx +

k X j=1

µj .

It is known that

2

µk ≤ Cn(4) CΩ2n+2 k n for a constant Cn(4) which depends only onR n (cf. [1] , Introduction). We aim to give an upper bound for Ωr pN (1/r2 ; x, x) dx. Our argument is closely related to standard arguments used for the proof of the continuity of the paths of a Brownian motion process (cf. [2] ). We set t ≡ r−2 . Let E x denote the expectation operator for a Brownian motion starting t denote the density of the distribution of at x ∈ Ω and reflected at the boundary of Ω and let dX dm the Brownian particle at time t with respect to the Lebesgue measure. Thus, we can write the above as integral as a sum of two expectation densities as follows: Z

pN (1/r2 ; x, x) dx =

Z

Ωr

Ωr

+

dXt (x) ; Xs ∈ Ωr for every s ∈ (0, t)] dx dm dXt E x[ (x) ; ∃s ∈ (0, t) such that Xs 6∈ Ωr ] dx. dm

E x[ Z Ωr

The first integral on the right-hand side of the last equation can be estimated above by the heat kernel pN Ωr (t; x, x) with Neumann boundary condition on both components of the boundary of Ωr r (the corresponding eigenvalues of Ωr will be denoted by µΩ j ): Z

E x[

Ωr

Z dXt pN (x) ; Xs ∈ Ωr for every s ∈ (0, t)] dx ≤ Ωr (t; x, x) dx dm Ωr

=

∞ X

r exp(−tµΩ j )

k=1 n(n+1) n−1

≤ Cn(5) CΩ

r

for a constant Cn(5) which depends only on n (for the last inequality cf. also [1], Theorem2). Let Ak for an integer k ≥ 0 denote the event Ak ≡ {There exists an integer j ∈ [0, 2k ] with Xs 6∈ ∂Ω for every s ∈ [

j j+1 t, k t]}. 2k 2

The second of the above integrals can be represented as a sum of integrals: R Ωr

=

dXt (x) ; ∃s ∈ (0, t) such that Xs 6∈ Ωr ] dx dm ∞ Z X dXt (x) ; (∃s ∈ (0, t) such that Xs 6∈ Ωr ) and (Ak \ Ak−1 )] dx. E x[ dm k=1 Ωr

E x[

The event {(∃s ∈ (0, t) such that Xs ∈ 6 Ωr ) and (Ak \ Ak−1 )} implies the existence of a pair of j i t, 2i+1 t) and J = ( 2k+1 t, 2j+1 disjoint intervals I = ( 2k+1 k+1 k+1 t) such that sup |Xs1 − Xs2 | ≥

s1 ,s2 ∈I

1 −1 r 2

and

Xs 6∈ ∂Ω for each s ∈ J.

Suppose that i < j. The probability of the event sups1 ,s2 ∈I |Xs1 − Xs2 | ≥ 12 r−1 for a given i can be estimated above by Cn(6) exp(−Cn(6) 2k ) for a constant Cn(6) . The maximum of the density of the expectation at time 2j+1 k+1 t of a Brownian motion starting at j time 2k+1 t which does not hit the boundary ∂Ω for s ∈ J can be estimated above by (2πt)−n/2 2n(k+1) . Because of the symmetry of a reflected Brownian motion with respect to the time variable, we can deal with the case j > i in a similar way. 5

Thus, we can conclude that dXt (x) ; ∃s ∈ (0, t) such that Xs 6∈ Ωr ] dx dm

Ωr

E x[

vol(Ωr )

k=1 (7) n−1 n−1 Cn CΩ r

R

∞ X

n

2k+1 (2k+1 − 1) Cn(6) exp(−Cn(6) 2k ) (2πt)− 2 2n(k+1)

for a constant Cn(7) (a suitable upper bound for vol(Ωr ) can be easily obtained if we cover ∂B1 by balls of radius 2(rCΩ )−1 ). 1/2 We set r ≡ µk . We have finally arrived at Z

R

n

F (z) dz

(n)

and Z

R

2 n |z| F (z) dz

2 +n

(2π)n k − C7 CΩ2n

(2π)n

k X

k

n−1 n

2 +2n

µj + Cn(8) Ω2n

k

n+1 n

j=1

for a constant Cn(8) which depends only on n. By [6] , Lemma 1, we obtain that: k X j=1

µj

(2π)2

n+2 n+1 2 n 2 (n) vol(B1 )− n (k n − C8 CΩ2n +2n k n ). n+2

Remark. It is easy to see that similar estimates hold true for domains Ω which are not simply connected provided that there exists a bi-Lipschitz map from Ω onto a suitable domain with smooth boundary. Proof of the Corollary. We aim to prove the upper bound for λk from the first part of the assertion. The lower bound can be shown in exactly the same manner. Suppose that vol(Ω)vol(B1 ) = (2π)n . We have by (1) and [5], Theorem 1 that λk

≤ ≤

k+l 1X λj l k+1 n+2 n+1 n+2 1 n n (0) { ((k + l) n + Cn(1) CΩ (k + l) n ) − k n }. l n+2 n+2

For |l − k 1−1/2n | ≤ 1/2 we can conclude from Taylor’s formula that there exists a constant Cn(3) such that 2 (0) 3 λk ≤ k n + Cn(3) CΩ k 2n . The second part of the assertion follows in a similar way from (2) and [5] , Corollary 1. 3. The case of a compact Riemannian manifold Let M be a compact Riemannian manifold without boundary. We denote the (increasingly ordered) eigenvalues of the Laplace-Beltrami operator −4M on M by 0 = λ1 < λ2 ≤ λ3 ≤ .... We consider a system {φj }∞ j=1 of orthonormal eigenfunctions for the eigenvalues λ1 , λ2 , λ3 , ... We set again Pk Φk (x, y) ≡ j=1 φj (x)φj (y) for every x, y ∈ Ω and every k ∈ N. Let i(M ) denote the injectivity radius and let |K| denote an upper bound for the modulus of the sectional curvature of M . 6

First, we aim to deal with the problem of obtaining upper bounds for the eigenvalues λk . Let χ be a smooth nonnegative function on R with χ(λ) = 0 for λ ≤ 1/2 and λ ≥ 1 and R1 n vol(B1 ) 1/2 χ(λ) dλn = 1 . We introduce the test functions ωy(ρ,s) on M for every ρ, s ∈ R with ρ ≥ ρ0 ≡ max{2/i(M ); |K|1/2 } and y ∈ M by ωy(ρ,s) (expy u) ≡ exp(is|u|)χ(ρ|u|) for every u ∈ Rn ; here expy stands for the exponential map from Ty M in M . The orthogonal projection of a function ω in L2 (M ) onto the subspace generated by φ1 , ..., φk R will be denoted by Pk (ω). By definition, Pk (ω)(x) = M Φk (x, y)ω(y) dy for every x ∈ M . We can conclude from the Rayleigh-Ritz formula that for every r the following holds: Rr R

λ2k+1

(ρ,s) − Pk (ωy(ρ,s) ))(x)|2 dx dy vol(B1n ) dsn M M |4M (ωy Rr R R (ρ,s) (ρ,s) − Pk (ωy ))(x)|2 dx dy vol(B1n ) dsn 0 M M |(ωy Rr R P (ρ,s) 2 ||L2 (M ) − kj=1 λ2j | < φj , ωy(ρ,s) >L2 (M ) |2 0 M ||4M ωy

R

0

≤ =

(ρ,s) 2 ||L2 (M )

Rr R 0

||ωy

M

(ρ,s)

Pk

j=1 | < φj , ωy

dy vol(B1n ) dsn

>L2 (M ) |2 dy vol(B1n ) dsn

.

There exists a constant Cn(9) depending only on the dimension n and the function χ (recall that χ was a suitably normalized test function on R) such that ||4M ωy(ρ,s) (x)||22

=

Z M

|Re (exp(is|u|)4M ωy(ρ,s) )(x)|2 + |Im (exp(is|u|)4M ωy(ρ,s) )(x)|2 dx

≤ (|s|2 + Cn(9) ρ2 )2 + (Cn(9) |s|ρ)2 . We can conclude from Bishop’s Comparison Theorem that for an appropriate Cn(10) the following holds: Z

r

Z

0

M

||4M ωy(ρ,s) ||2L2 (M )

dyvol(B1n ) dsn

(n) ((1+Cn(10) |K|ρ−2 )nvol(B1 )vol(M )(

n+2 rn+4 (10) r +Cn ρ2 ) n+4 n+2

for every r with r ≥ ρ. In a similar way we obtain that Z 0

r

Z

(n)

M

||ωy(ρ,s) ||2L2 (M ) dy |s|n−1 ds ≥ (1 + Cn(10) |K|ρ−2 )−1 nvol(B1 )vol(M )

rn . n

We set ρ ≡ (ρ0 r)1/2 for every r with r ≥ ρ0 . Thus, we have arrived at (n)

λ2k+1

n+4

(1 + 2Cn(10) ρr0 )2 nvol(B1 )vol(M ) rn+4 − n (10) ρ0 −1 (n) ) nvol(B1 )vol(M ) rn r

(1 + Cn

Pk

j=1

λ2j

Rr R 0

Rr R j=1 0 M

Pk

M

| < φj , ωy(ρ,s) >L2 (M ) |2 dy vol(B1n ) dsn (ρ,s)

| < φj , ωy

>L2 (M ) |2 dy vol(B1n ) dsn

.

By Parseval’s formula and some calculations which are similar to the proof of Bishop’s Comparison Theorem, Z 0

r

Z M

| < φj , ωy(ρ,s) >L2 (M ) |2 dy vol(B1n ) dsn ≤ (2π)n (1 + Cn(10) |K|ρ−2 ) ρ0 ≤ (2π)n (1 + Cn(10) )2 . r

Z M

||χ(ρdist(y, .))φj ||22 dy

A similar induction argument as in the proof of Theorem 1 leads to k X j=1

λ2j ≤ (2π)2

n+4 2 n k −1/n (vol(M )vol(B1 ))− n k n (1 + Cn(11) ρ0 ) for every k n+4 vol(M )

7

(3)

and an appropriate constant Cn(11) which depends only on n. Recall that by definition ρ0 ≡ max{2/i(M ); |K|1/2 }. The converse inequality can be shown in a similar way as Theorem 2 if we study the function (ρ) F on [0, ∞) which is given by F (ρ) (s) ≡

Z M

Z M

|Pk ωy(ρ,s) (x)|2 dx dy.

The only problem is to estimate the integral M |4M (χ(ρdist(., y))φj )(x)|2 dx since no information on ∇x φ seems to be available. We can easily overcome that difficulty and remove the mixed integral which would lead to a term of order |s|3 ρ if we take into account that R

Z M

χφj < ∇x χ , ∇x φj > dx

=

1 Z ∇x · (χ∇x χ)φ2j dx. 2 M

Thus, we can conclude that k X

λ2j ≥ (2π)2

j=1

n+4 2 k −1/n n (vol(M )vol(B1 ))− n k n (1 − Cn(12) ρ0 ) for every k n+4 vol(M )

(4).

We notice that the right-hand side of the last equation is only positive if k > Cn(12) ρn0 vol(M ) (≥ ) ). 2n Cn(12) vol(M i(M )n Remark. Corresponding results for the eigenvalues of the Hodge-de Rham Laplacian can be shown in exactly the same way (cf. also [4]). References [1] E. B. DAVIES, Spectral properties of compact manifolds and changes of metric, Amer. J. Math. 112, 15–39 (1990). [2] E. B. DYNKIN, Fondations of the theory of Markov processes. Oxford: Pergamon 1960 [3] V. GUILLEMIN, A new proof of Weyl’s formula on the asymptotic distribution of eigenvalues, Adv. in Math. 55, 131–160 (1985). [4] H. HESS AND R. SCHRADER AND D. A. UHLENBROCK, Kato’s inequality and the spectral distribution of Laplacians on compact Riemannian manifolds, J. Differ. Geom. 15, 27–38 (1980). ¨ [5] P. KROGER, Upper bounds for the Neumann eigenvalues on a bounded domain in Euclidean space. J. Funct. Anal. 106, 353–357 (1992). [6] P. LI AND S. T. YAU, On the Schr¨odinger equation and the eigenvalue problem, Commun. Math. Phys. 88 (1983), 309–318. [7] H. P. MCKEAN, JR. AND I. M. SINGER, Curvature and the eigenvalues of the Laplacian. J. Diff. Geom. 1, 43–69 (1967) ´ [8] G. POLYA, On the eigenvalues of vibrating membranes, Proc. London Math. Soc. (3) 11, (1961), 419–433.

8

## Estimates for sums of eigenvalues of the Laplacian

Now suppose that that there exists a bi-Lischitz map f which maps Î© onto an open ball B in Rn. Let CÎ© be a .... (Our definition of F is motivated by the definition of.

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working [15], bioinformatics [16], and even commercial usage [17,18]. Recently, ..... In the AT&T database 1, there are ten different images of each of 40 distinct.

rigorous bounding of position error estimates for aircraft ...
1. Segmentation Methods for Visual Tracking of Deep. Ocean Jellyfish Using a ... opportunity such a platform offers to extend the science of marine ecology. ... Tools for performance evaluation of computer vision algorithms have arisen ... evaluation

ENTROPY ESTIMATES FOR A FAMILY OF EXPANDING ... - CiteSeerX
tangent bundle of M is transitive. Here the projective bundle is the bundle whose fiber at m â M is the projective space of TmM and the flag bundle the bundle.

Estimates on the Distribution of the Condition Number ...
Jan 31, 2006 - Hausdorff measure of its intersection with the unit disk (cf. Section 2 for ... The probability space of input data is the projective algebraic variety Î£nâ1. As we. 3 .... However, it seems to be a hard result to prove this optimali

Applications of Weil's Theorem on Character Sums ...
Sep 7, 2007 - We want to define a block transitive STS(v) generated by {0,1,Î±} under the action of maps x â¦â Ïx + m, where Ï,m ... 6-sparse Steiner triple systems. We were interested in 6-sparse STS(v)s, systems which avoid these configuration

Nonnegative Polynomials and Sums of Squares Why ...
Jan 24, 2011 - Why do bad things happen to good polynomials? Grigoriy .... There exist constants c1(d) and c2(d), dependent on the degree d only, such that.

Nonnegative Polynomials and Sums of Squares Why ...
Mar 17, 2011 - We can compute instead the best sos lower bound: Î³. â. = max f âÎ³ is sos ... For small n and 2d find good bounds on degree of q and explicit.

Challenging the popular wisdom. New estimates of ...
consumption and changes in overall economic activity taking into account addi ... Energy Information Administration, International Energy Agency, World Bank ... Romania, Singapore, Slovak R., Spain, Sri Lanka, Sweden, Switzerland, Tanzania, Tunisia .

Aerial survey abundance estimates of the loggerhead sea turtle ...
study illustrates the ability of aerial surveys to provide such data in this area. KEY WORDS: .... A simple power analysis was performed to explore the 80% power ...

Use of livestock quality estimates for improved ... - Roberto Rossi
Use of livestock quality estimates for improved product allocation planning to .... animals farmers will deliver (ai), and transport costs of animals to processors ...

Boundary estimates for solutions of non-homogeneous boundary ...
values of solutions to the non-homogeneous boundary value problem in terms of the norm of the non-homogeneity. In addition the eigenparameter dependence ...