The Best Constant of Sobolev Inequality Corresponding to Clamped Boundary Value Problem

Volume 2011, Article ID 875057,17pages doi:10.1155/2011/875057

Research Article

The Best Constant of Sobolev

Inequality Corresponding to

Clamped Boundary Value Problem

Kohtaro Watanabe,

_{Yoshinori Kametaka,}

_{Hiroyuki Yamagishi,}

Atsushi Nagai,

_{and Kazuo Takemura}

1_{Department of Computer Science, National Defense Academy, 1-10-20 Hashirimizu, Yokosuka}

239-8686, Japan

2_{Division of Mathematical Sciences, Graduate School of Engineering Science, Osaka University,}

1-3 Machikaneyama-cho, Toyonaka 560-8531, Japan

3_{Tokyo Metropolitan College of Industrial Technology, 1-10-40 Higashi-ooi, Shinagawa,}

Tokyo 140-0011, Japan

4_{Department of Liberal Arts and Basic Sciences, College of Industrial Technology, Nihon University,}

2-11-1 Shinei, Narashino 275-8576, Japan

Correspondence should be addressed to Kohtaro Watanabe,[email protected]

Received 14 August 2010; Accepted 10 February 2011

Academic Editor: Irena Rach ˚unkov´a

Copyrightq2011 Kohtaro Watanabe et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Green’s functionGx, y of the clamped boundary value problem for the differential operator −1M_d/dx2M _{on the interval−s, sis obtained. The best constant of corresponding Sobolev}

inequality is given by max|y|≤sGy, y. In addition, it is shown that a reverse of the Sobolev best

constant is the one which appears in the generalized Lyapunov inequality by Das and Vatsala 1975.

1. Introduction

ForM1,2,3, . . .,s >0, letHHM

0 −s, sbe a SobolevHilbertspace associated with the

inner product·,·M:

HHM u|uM∈L2−s, s, ui±s 00≤i≤M−1,

u, vM _s

−s

uMxvMxdx, u2M u, uM.

The fact that·,·Minduces the equivalent norm to the standard norm of the SobolevHilbert space ofMth order follows from Poincar´e inequality. Let us introduce the functionalSuas follows:

Su

sup_|y|≤suy 2

u2M

. 1.2

To obtain the supremum ofSi.e., the best constant of Sobolev inequality, we consider the following clamped boundary value problem:

−1Mu2Mfx −s < x < s,

ui±s 0 0≤i≤M−1.

BVPM

Concerning the uniqueness and existence of the solution toBVPM, we have the following proposition. The result is expressed by the monomialKjx:

Kjx KjM;x

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

x2M−1−j

2M−1−j!

0≤j≤2M−1,

0 2M≤j.

1.3

Proposition 1.1. For any bounded continuous functionfxon an interval−s < x < s,BVPM has a unique classical solutionuxexpressed by

ux

s

−s

Gx, yfydy −s < x < s, 1.4

where Green’s functionGx, y GM;x, y −s < x, y < sis given by

Gx, y

−1M

2

K0x−yD−1

Kij2s Ki

s−y

Kjsx 0

Kij2s Ki

sy

Kjs−x 0

1.5

−1MD−1

Kij2s Ki

sx∧y

Kj

s−x∨y 0

−s < x, y < s. 1.6

D is the determinant ofM×MmatrixKij2s 0 ≤ i, j ≤ M−1,x∧y minx, y, and

With the aid of Proposition 1.1, we obtain the following theorem. The proof of

Proposition 1.1is shown in AppendicesAandB.

Theorem 1.2. iThe supremumCM;−s, s(abbreviated asCMif there is no confusion) of the Sobolev functionalSis given by

CM;−s, s sup

u∈H, u /≡0

Su max

|y|≤sG

y, yG0,0 s

2M−1

22M−1_{2M−1{M−1!}}2 . 1.7

Concretely,

C1,−s, s s

2, C2,−s, s

s3

24, C3,−s, s

s5

640, C4,−s, s

s7

32256, . . . . 1.8

iiCM;−s, sis attained byuGx,0, that is,SGx,0 CM;−s, s.

Clearly,Theorem 1.2i,iiis rewritten equivalently as follows.

Corollary 1.3. Let u ∈ H, then the best constant of Sobolev inequality (corresponding to the embedding ofHintoL∞−s, s)

sup

|y|≤s _u

y

2

≤C

_s

−s

uMx2dx, 1.9

isCM;−s, s. Moreover the best constantCM;−s, sis attained byux cGx,0, wherecis an arbitrary complex number.

Next, we introduce a connection between the best constant of Sobolev- and Lyapunov-type inequalities. Let us consider the second-order differential equation

u pxu0 −s≤x≤s, 1.10

wherepx ∈ L1−s, s∩C−s, s. If the above equation has two pointss1 ands2 in−s, s

satisfyingus1 0us2, then these points are said to be conjugate. It is wellknown that if

there exists a pair of conjugate points in−s, s, then the classical Lyapunov inequality _s

−s

pxdx > 2

s, 1.11

holds, where px : maxpx,0. Various extensions and improvements for the above

result have been attempted; see, for example, Ha 1, Yang 2, and references there in. Among these extensions, Levin3and Das and Vatsala4extended the result for higher order equation

For this case, we again call two distinct pointss1and s2 conjugateif there exists a nontrivial

C2M_{−s, s∩C}M−1_{−s, ssolution of1.12satisfying}

uis1 0uis2 i0, . . . , M−1. 1.13

We point out that the constant which appears in the generalized Lyapunov inequality by Levin3and Das and Vatsala4is the reverse of the Sobolev best embedding constant.

Corollary 1.4. If there exists a pair of conjugate points on−s, swith respect to1.12, then _s

−s

pxdx > 1

CM;−s, s, 1.14

whereCM;−s, sis the best constant of the Sobolev inequality1.9.

Without introducing auxiliary equation u2M ₋₁M−1

pu 0 and the existence

result of conjugate points as2,4, we can prove this corollary directly through the Sobolev inequalitythe idea of the proof origins to Brown and Hinton5, page 5.

Proof ofCorollary 1.4.Consider

s2

s1

uMx 2dx

s2

s1

pxux2dx≤

sup s1≤x≤s2|ux|

2_s2

s1

pxdx

≤CM;s1, s2

_s2

s1

uMx 2dx

_s2

s1

pxdx.

1.15

In the second inequality, the equality holds for the function which attains the Sobolev best constant, so especially it is not a constant function. Thus, for this function, the first inequality is strict, and hence we obtain

1

CM;s1, s2 <

_s2

s1

pxdx. 1.16

Since

1

CM;−s, s ≤

1

CM;s1, s2 <

s2

−s1

pxdx≤

s

−s

pxdx, 1.17

we obtain the result.

Here, at the end of this section, we would like to mention some remarks about

4, Theorem 2.1is given by some finite series ofxandyon the other hand, the expression

ofProposition 1.1 is by the determinant. This complements the results of7–9, where the

concrete expressions of Green’s functions for the equation −1Mu2M _{f but different}

boundary conditions are given, and all of them are expressed by determinants of certain matrices asProposition 1.1.

2. Reproducing Kernel

First we enumerate the properties of Green’s functionGx, yofBVPM.Gx, yhas the following properties.

Lemma 2.1. Consider the following:

1

∂2xMG

x, y0 −_{s < x, y < s, x /}y, 2.1

2

∂ixG

x, y

x±s0

0≤i≤M−1, −s < y < s, 2.2

3

∂ixG

x, y

yx−0−∂

i xG

x, y

yx0

⎧ ⎪ ⎨ ⎪ ⎩

0 0≤i≤2M−2,

−1M i2M−1 −s < x < s,

2.3

4

∂ixG

x, y

xy0−∂

i xG

x, y

xy−0

⎧ ⎪ ⎨ ⎪ ⎩

0 0≤i≤2M−2,

−1M i2M−1−s < y < s.

2.4

Proof. Fork 1≤k≤2Mand−s < x, y < s,_{x /}y, we have from1.5

∂kxG

x, y −1

M

2

sgnx−ykKkx−y

D−1

Kij2s Ki

s−y

Kkjsx 0

Kij2s Ki

sy

−1kKkjs−x 0

.

Fork 2M, noting the factKjx 0 2M≤j, we have1. Next, for 0≤k ≤M−1 and −s < y < s, we have from2.5

∂kxG

x, y

x−s

−1M

2

−1kKk

syD−1Kij2s Ki

s−y

Kkj0 0

Kij2s Ki

sy

−1kKkj2s 0

.

2.6

SinceKk0, . . . , KkM−10 0, . . . ,0, we have

−1Mk2 ∂kxG

x, y

x−sKk

syD−1 Kij2s Ki

sy

Kkj2s 0

Kk

syD−1

Kij2s Ki

sy

0 · · · 0 −Kk

sy

0.

2.7

Note that subtracting the kth row from Mth row, the second equality holds. Equation

∂k

xGx, y|xs 0 is shown by the same way. Hence, we have2. For 0 ≤ k ≤ 2M−1, we have

∂kxG

x, y

yx−0−∂

k xG

x, y

yx0

−1M

2

1−−1k Kk0 ⎧ ⎪ ⎨ ⎪ ⎩

0 0≤k≤2M−2,

−1M k2M−1 −s < x < s,

2.8

where we used the factKk0 0k /2M−1, 1 k2M−1. So we have3, and4follows from3.

UsingLemma 2.1, we prove that the functional spaceHassociated with inner norm

·,·Mis a reproducing kernel Hilbert space.

Lemma 2.2. For anyu∈H, one has the reproducing property

uyu·, G·, y_M

s

−s

uMx∂Mx G

x, ydx −s≤y≤s. 2.9

Proof. For functionsuuxandvvx Gx, ywithyarbitrarily fixed in−s≤y≤s, we have

uMvM−u−1Mv2M

⎛ ⎝M−1

j0

−1M−1−j ujv2M−1−j

⎞

Integrating this with respect toxon intervals−s < x < yandy < x < s, we have

_s

−s

uMxvMxdx−

_s

−s

ux−1Mv2Mxdx

⎡ ⎣M−1

j0

−1M−1−jujxv2M−1−jx

⎤ ⎦xy−0

x−s xxsy0

M−1

j0

−1M−1−j ujsv2M−1−js−uj−sv2M−1−j−s

M−1

j0

−1M−1−jujyv2M−1−jy−0−v2M−1−jy0.

2.11

Using1,2, and4inLemma 2.1, we have2.9.

3. Sobolev Inequality

In this section, we give a proof ofTheorem 1.2andCorollary 1.3.

Proof ofTheorem 1.2andCorollary 1.3. Applying Schwarz inequality to2.9, we have

_u

y2≤

_s

−s ∂M

x G

x, y2dx

_s

−s

uM_x2_{dxGy, y} s

−s

uM_x2_{dx. 3.1}

Note that the last equality holds from2.9; that is, substituting2.9,u· G·, y. Let us assume that

CM;−s, s CM max

|y|≤sG

y, yG0,0, 3.2

holdsthis will be proved in the next section. From definition ofCM, we have

sup

|y|≤s |uy|

2

≤CM

_s

−s

uMx2dx. 3.3

Substitutingux Gx,0∈Hin to the above inequality, we have

sup

|y|≤s

|Gy,0| 2

≤CM

s

−s

∂Mx Gx,0

2

Combining this and trivial inequalityCM2 G0,02≤sup_|y|≤s|Gy,0|2, we have

CM2≤

sup

|y|≤s _G

y,0

2

≤CM

_s

−s

∂Mx Gx,0

2

dx CM2. 3.5

Hence, we have

sup

|y|≤s

|Gy,0| 2

CM

s

−s

∂Mx Gx,0

2

dx, 3.6

which completes the proof ofTheorem 1.2andCorollary 1.3.

Thus, all we have to do is to prove3.2.

4. Diagonal Value of Green’s Function

In this section, we consider the diagonal value of Green’s function, that is,Gx, x. From

Proposition 1.1, we have forM1,2,3

G1;x, x

s2_−x2

2s , G2;x, x

s2_−x23

24s3 , G2;x, x

s2_−x25

650s5 . 4.1

Thus, we can expect thatGx, xtakes the formGM;x, x const. K0M; 1xK0M; 1−x.

Precisely, we have the following proposition.

Proposition 4.1. Consider

Gx, x −1MD−1

Kij2s Kis−x

Kjsx 0

2M−1

M−1

1

K02sK0sxK0s−x

2M−1

M−1

1

K02s

s2_−x22M−1

{2M−1!}2.

4.2

Hence,

CM;−s, s sup

|x|≤s

Gx, x G0,0 −1MD−1

Kij2s Kis

Kjs 0

s2M−1

22M−1_2M−1!

2M−1

M−1

s2M−1

22M−1_2M−1M−1!2,

4.3

To prove this proposition, we prepare the following two lemmas.

Lemma 4.2. Letux c1Gx, x, where

c−₁1 −1M

22M−1 2M−1

D−1

1

Kij2s 0

.. .

0

1 0 · · · 0 0

, 4.4

(i, jsatisfy0≤i, j≤M−1), then it holds that

−u22M−11 −s < x < s, 4.5

ui±s 0 0≤i≤2M−2, 4.6

u2M−1s −

2M−1

M−1

c1. 4.7

Lemma 4.3. Letux c2K0sxK0s−x −s < x < s, wherec2−1

22M−1

2M−1 , then it holds

that4.6andu2M−1_{s −K02sc2.}

Proof ofProposition 4.1. From Lemmas 4.2 and 4.3, ux c1Gx, x and ux c2K0s

xK0s−xsatisfy BVP2M−1 in case offx 1−s < x < s. So we have

c1Gx, x c2K0sxK0s−x −s < x < s, 4.8

2M−1

M−1

c1 K02sc2. 4.9

Inserting4.9into4.8, we haveProposition 4.1.

Proof ofLemma 4.2.Let

ux c1Gx, x c1−1MD−1vx, vx

Kij2s Kis−x

Kjsx 0

, 4.10

then differentiatingvxktimes we have

vkx

k

l0

−1l

k l

wk,lx, wk,lx

Kij2s Klis−x

Kk−ljsx 0

At first, fork22M−1, we have

v22M−1x

22M−1

l0

−1l

22M−1

l

w22M−1,lx

2M−2

l0

−1l

22M−1

l

w22M−1,lx−

22M−1 2M−1

w22M−1,2M−1x

22M−1

l2M

−1l

22M−1

l

w22M−1,lx.

4.12

The first term vanishes because

K22M−1−ljsx K2M2M−2−ljsx 0 0≤l≤2M−2. 4.13

The third term also vanishes because

Klis−x 0 2M≤l≤22M−1. 4.14

Thus, we have

v22M−1x −

22M−1 2M−1

w22M−1,2M−1x,

w22M−1,2M−1x

Kij2s K2M−1is−x

K2M−1jsx 0

1

Kij2s 0 .. .

0

1 0 · · · 0 0

.

4.15

Hence, we have

−u22M−1x −c1−1MD−1 v22M−1x 1, 4.16

by which we obtain4.5. Next, for 0≤k≤M−1, we have

vks

k

l0

−1l

k l

wk,ls, wk,ls

Kij2s Kli0

Kk−lj2s 0

Since 0≤li≤2M−2, we havewk,ls 0. Thus, we havevks 00 ≤k≤M−1. For

M≤k≤2M−2, we have

vks

M−1

l0

−1l

k l

wk,ls k

lM

−1l

k l

wk,ls. 4.18

The first term vanishes becauseKli0 00 ≤l ≤ M−1. Next, we show that the second term also vanishes. Let

wk,ls

Kj2s 0

..

. ...

K2M−2−lj2s 0

K2M−1−lj2s 1

K2M−lj2s 0 ..

. ...

KM−1j2s 0

Kk−lj2s 0

M≤l≤k≤2M−2. 4.19

Since 0≤k−l ≤2M−2−l, two rows, including the last row, coincide, and hence we have

wk,ls 0. Thus, we havevks 0M≤k≤2M−2. So we have obtaineduks 00≤

k≤2M−2. By the same argument, we haveuk_{−s 00≤k≤2M−2. Hence, we have}

4.6. Finally, we will show4.7. Fork2M−1, notingKli0 00≤l≤M−1, we have

v2M−1s

2M−1

lM

−1l

2M−1

l

w2M−1,ls, 4.20

where

w2M−1,ls

Kij2s Kli0

K2M−1−lj2s 0

Kj2s 0

..

. ...

K2M−2−lj2s 0

K2M−1−lj2s 1

K2M−lj2s 0 ..

. ...

KM−1j2s 0

K2M−1−lj2s 0

Kj2s 0

..

. ...

K2M−2−lj2s 0

K2M−1−lj2s 1

K2M−lj2s 0 ..

. ...

KM−1j2s 0

Thus, we obtainw2M−1,ls −DM≤l≤2M−1. Hence we have

v2M−11

2M−1

lM

−1l

2M−1

l

w2M−1,ls −D

2M−1

lM

−1l

2M−1

l

−D

2M−2

lM

−1l

2M−2

l−1

2M−2

l

D −1M1D

2M−1

M−1

,

4.22

that is,

u2M−1s c1−1MD−1v2M−1s −

2M−1

M−1

c1. 4.23

This completes the proof ofLemma 4.2.

Proof ofLemma 4.3.Let

ux c2K0sxK0s−x c2

2M−1!2

s2−x2 2M−1. 4.24

Differentiatingux ktimes, we have

ukx c2

k

l0

−1l

k l

Kk−lsxKls−x. 4.25

Fork22M−1, notingK22M−1−lsx 0 0≤l≤2M−2,K2M−1sx K2M−1s−x 1,

andKls−x 0 2M≤l≤22M−1, we have

−u22M−1x c2

22M−1 2M−1

1. 4.26

Thus, we have4.5. If 0≤k≤2M−2, then we have

uks c2

k

l0

−1l

k l

Kk−l2sKl0 0. 4.27

Sinceuk_{−x −1}k

uk_{x, we haveu}k_{−s 00≤k≤2M−2. Hence, we have4.6. If}

k2M−1, then we have

u2M−1s c2 2M−1

l0

−1l

2M−1

l

K2M−1−l2sKl0 −c2K02s. 4.28

Appendices

A. Deduction of

1.5

In this section,1.5in Proposition 1.1 is deduced. Suppose thatBVPM has a classical solutionux. Introducing the following notations:

u t_u

0, . . . , u2M−1, ui ui 0≤i≤2M−1, e t₀

, . . . ,0,1 2M×1 matrix,

N

⎛ ⎜ ⎜ ⎜ ⎝

0 1 0 . ..

. .. 1 0

⎞ ⎟ ⎟ ⎟ ⎠

2M×2Mnilpotent matrix,

A.1

BVPMis rewritten as

u_Nue−1M

fx −s < x < s,

ui±s 0 0≤i≤M−1.

A.2

Let the fundamental solutionExbe expressed asEx expNx KxK0−1, where

Kx Kij

x, K0 ⎛ ⎝ _{· · ·} 1

1

⎞

⎠_K0−1_{, A.3}

theni, jsatisfy 0≤i, j ≤2M−1.Exsatisfies the initial value problemE NE,E0 I.Iis a unit matrix. SolvingA.2, we have

ux Exsu−s

_x

−s

Ex−ye−1Mfydy,

ux Ex−sus−

s

x

Ex−ye−1Mfydy,

A.4

or equivalently, for 0≤i≤2M−1, we have

uix

2M−1

j0

Kijxsu2M−1−j−s _x

−s

−1MKi

x−yfydy,

uix

2M−1

j0

Kijx−su2M−1−js− _s

x

−1MKi

x−yfydy.

Employing the boundary conditionsA.2, we have

uix M−1

j0

Kijxsu2M−1−j−s _x

−s

−1MKi

x−yfydy,

uix M−1

j0

Kijx−su2M−1−js− _s

x

−1MKi

x−yfydy.

A.6

In particular, ifi0, then we have

u0x

M−1

j0

Kjxsu2M−1−j−s x

−s

−1MK0

x−yfydy,

u0x

M−1

j0

Kjx−su2M−1−js− _s

x

−1MK0

x−yfydy.

A.7

On the other hand, using the boundary conditionsA.2again, we have

0uis M−1

j0

Kij2su2M−1−j−s _s

−s

−1MKi

s−yfydy,

0ui−s M−1

j0

Kij−2su2M−1−js− _s

−s

−1MKi

−s−yfydy.

A.8

Solving the above linear system of equations with respect to u2M−1−i−s,

u2M−1−is 0≤i≤M−1, we have

u2M−1−i−s − _s

−s

−1MKij −1

2sKi

s−yfydy,

u2M−1−is _s

−s

−1MKij ₋1₋

2sKi

−s−yfydy.

A.9

SubstitutingA.9intoA.7, we have

u0x −

s

−s

−1MKj

xsKij −1

2sKi

s−yfydy

_x

−s

−1MK0x−yf

ydy,

u0x

_s

−s

−1MKj

x−sKij ₋1₋

2sKi

−s−yfydy

s

x

−1MK0x−yf

ydy.

Taking an average of the above two expressions and notingux u0x, we obtain1.4,

where Green’s functionGx, yis given by

Gx, y −1

M

2

K0x−y−

Kj

xsKij −1

2sKi

s−y

Kj

x−sKij ₋1₋

2sKi

−s−y.

A.11

Using propertiesKi−x −1i1Kix, we have

Kj

x−s −Kj

s−x−1iδij ,

Kij

−2s −1ij1Kij 2s −

−1iδij Kij

2s−1iδij ,

Ki

−s−y−1i1Ki sy

−−1iδij Ki

sy,

A.12

where δij is Kronecker’s delta defined by δij 1i j, 0 i /j. Inserting these three relations intoA.11, we have

Gx, y −1

M

2

K0x−y−

Kj

sxKij −1₂

sKi

s−y

−Kj

s−xKij ₋1

2sKi

sy.

A.13

Applying the relation

t_{a A}−1_{b −}

A b t_{a 0}

|A| , A.14

whereAis anyN×Nregular matrix andaandbare anyN×1 matrices, we have1.5.

B. Deduction of

1.6

To prove1.6, we show

K0

x−y−D−1

Kij2s Ki

s−y

Kjsx 0

−

Kij2s Ki

sy

Kjs−x 0

−s < x, y < s.

Letx≥y. IfB.1holds, substituting it to1.5, replacingxwithx∨y,ywithx∧y, then we obtain1.6. The casex≤ yis shown in a similar way. Lety−s ≤ y ≤ sbe fixed, and let

ux K0x−y. Thenusatisfies

u2M0 −s < x < s,

ui−s −1i1Ki

sy, uis Ki

s−y 0≤i≤M−1.

B.2

On the other hand, let

vx −D−1

Kij2s Ki

s−y

Kjsx 0

−

Kij2s Ki

sy

Kjs−x 0

. B.3

Differentiatingv ktimes with respect tox, we have

vkx −D−1 Kij2s Ki

s−y

Kkjsx 0

−−1k

Kij2s Ki

sy

Kkjs−x 0

. B.4

Fork2M, noticingKkjsx Kkjs−x 0, we havev2Mx 0. For 0≤k≤M−1, we have

vk−s −D−1

Kij2s Ki

s−y

Kkj0 0

−−1k

Kij2s Ki

sy

Kkj2s 0

−1kD−1

Kij2s Ki

sy

0 · · · 0 −Kk

sy

−1k1Kk

sy,

B.5

where we usedKkj0 0. Similarly, for 0 ≤ k ≤ M−1, we havevks Kks−y. So

vxsatisfies

v2M 0 −s < x < s,

vi−s −1i1Ki

sy, vis Ki

s−y 0≤i≤M−1.

B.6

which is the same equation asB.2. Hence, we havevx ux.

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