Volume 2011, Article ID 875057, 17 pagesdoi:10.1155/2011/875057 Research Article The Best Constant of Sobolev Inequality Corresponding to Clamped Boundary Value Problem 1 Department of C
Trang 1Volume 2011, Article ID 875057, 17 pages
doi:10.1155/2011/875057
Research Article
The Best Constant of Sobolev
Inequality Corresponding to
Clamped Boundary Value Problem
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,wata@nda.ac.jp
Received 14 August 2010; Accepted 10 February 2011
Academic Editor: Irena Rach ˚unkov´a
Copyrightq 2011 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 function Gx, y of the clamped boundary value problem for the differential operator
−1M d/dx 2Mon the interval−s, s is obtained The best constant of corresponding Sobolev
inequality is given by max|y|≤s Gy, 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
For M 1, 2, 3, , s > 0, let H H M
0 −s, s be a Sobolev Hilbert space associated with the
inner product·, · M:
H H M u | u M ∈ L2−s, s, u i ±s 0 0 ≤ i ≤ M − 1,
u, v M
s
−s u M xv M xdx, u2
M u, u M
1.1
Trang 2The fact that·, · Minduces the equivalent norm to the standard norm of the SobolevHilbert
space of Mth order follows from Poincar´e inequality Let us introduce the functional Su as
follows:
S u
sup|y|≤su
y 2
u2
M
To obtain the supremum of S i.e., the best constant of Sobolev inequality, we consider the
following clamped boundary value problem:
−1M u 2M fx −s < x < s,
u i ±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 monomial K j x:
K j x K j M; x
⎧
⎪
⎪
x 2M−1−j
2M − 1 − j
!
0≤ j ≤ 2M − 1,
2M ≤ j
.
1.3
Proposition 1.1 For any bounded continuous function fx on an interval −s < x < s, BVPM
has a unique classical solution ux expressed by
u x
s
−s G
x, y
f
y
dy −s < x < s, 1.4
where Green’s function Gx, y GM; x, y −s < x, y < s is given by
G
x, y
−1M
2
K0x − y D−1
K ij 2s K i
s − y
K j s x 0
K ij 2s K i
s y
K j s − x 0
1.5
−1M
D−1
K ij 2s K i
s x ∧ y
K j
s − x ∨ y
0
−s < x, y < s. 1.6
D is the determinant of M × M matrix K ij 2s 0 ≤ i, j ≤ M − 1, x ∧ y minx, y, and
x ∨ y maxx, y.
Trang 3With the aid of Proposition 1.1, we obtain the following theorem The proof of
Proposition 1.1is shown in AppendicesAandB
Theorem 1.2 i The supremum CM; −s, s (abbreviated as CM if there is no confusion) of the
Sobolev functional S is given by
C M; −s, s sup
u∈H, u /≡ 0S u max
|y|≤s G
y, y
22M−1 2M − 1{M − 1!}2 . 1.7
Concretely,
C 1, −s, s s
2, C 2, −s, s s3
24, C 3, −s, s s5
640, C 4, −s, s s7
32256, 1.8
ii CM; −s, s is attained by u Gx, 0, that is, SGx, 0 CM; −s, s.
Clearly,Theorem 1.2i, ii is rewritten equivalently as follows
Corollary 1.3 Let u ∈ H, then the best constant of Sobolev inequality (corresponding to the
embedding of H into L∞−s, s)
sup
|y|≤s
u
y2≤ Cs
−s
u M x2
is CM; −s, s Moreover the best constant CM; −s, s is attained by ux cGx, 0, where c is 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
where px ∈ L1−s, s ∩ C−s, s If the above equation has two points s1 and s2 in−s, s satisfying us1 0 us2, 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
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, Levin3 and Das and Vatsala 4 extended the result for higher order equation
−1M u 2M − pxu 0 −s ≤ x ≤ s. 1.12
Trang 4For this case, we again call two distinct points s1and s2 conjugate if there exists a nontrivial
C 2M −s, s ∩ C M−1 −s, s solution of 1.12 satisfying
u i s1 0 u i s2 i 0, , M − 1. 1.13
We point out that the constant which appears in the generalized Lyapunov inequality by Levin3 and Das and Vatsala 4 is the reverse of the Sobolev best embedding constant
Corollary 1.4 If there exists a pair of conjugate points on −s, s with respect to 1.12, then
s
−s pxdx > 1
where CM; −s, s is the best constant of the Sobolev inequality 1.9.
Without introducing auxiliary equation u 2M −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 Hinton 5, page 5
Proof of Corollary 1.4 Consider
s2
s1
u M x 2dx
s2
s1
p xux2
dx ≤
sup
s1≤x≤s2
|ux|
2s
2
s1
pxdx
≤ CM; s1, s2
s2
s1
u M x 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
C M; s1, s2 <
s2
s1
Since
1
C M; −s, s ≤
1
C M; 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
1.12 The generalized Lyapunov inequality of the form 1.14 was firstly obtained by Levin 3 without proof; see Section 4 of Reid 6 Later, Das and Vatsala 4 obtained the same inequality1.14 by constructing Green’s function for BVPM The expression
of the Green’s function of Proposition 1.1 is different from that of 4 The expression of
Trang 54, Theorem 2.1 is given by some finite series of x and y on 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 −1M
u 2M 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 function Gx, y of BVPM Gx, y has the
following properties
Lemma 2.1 Consider the following:
1
∂ 2M x G
x, y
0−s < x, y < s, x / y, 2.1
2
∂ i x G
x, y
x±s 00≤ i ≤ M − 1, −s < y < s, 2.2
3
∂ i x G
x, y
yx−0 − ∂ i
x G
x, y
⎧
⎪
⎪
0 0 ≤ i ≤ 2M − 2,
−1M i 2M − 1 −s < x < s,
2.3
4
∂ i x G
x, y
xy0 − ∂ i
x G
x, y
⎧
⎪
⎪
0 0 ≤ i ≤ 2M − 2,
−1M i 2M − 1−s < y < s.
2.4
Proof For k 1 ≤ k ≤ 2M and −s < x, y < s, x / y, we have from 1.5
∂ k x G
x, y
−1M
2
sgn
x − yk
K kx − y
D−1
K ij 2s K i
s − y
K kj s x 0
K ij 2s K i
s y
−1k
K kj s − x 0
.
2.5
Trang 6For k 2M, noting the fact K j x 0 2M ≤ j, we have 1 Next, for 0 ≤ k ≤ M − 1 and
−s < y < s, we have from 2.5
∂ k x G
x, y
x−s
−1M
2
−1k K k
s y
D−1
K ij 2s K i
s − y
K kj0 0
K ij 2s K i
s y
−1k K kj 2s 0
.
2.6 SinceK k 0, , K kM−1 0 0, , 0, we have
−1Mk 2 ∂ k x G
x, y
x−s K k
s y
D−1
K ij 2s K i
s y
K kj 2s 0
K k
s y
D−1
K ij 2s K i
s y
0 · · · 0 −K k
s y
0.
2.7
Note that subtracting the kth row from Mth row, the second equality holds Equation
∂ k Gx, y| xs 0 is shown by the same way Hence, we have 2 For 0 ≤ k ≤ 2M − 1, we
have
∂ k x G
x, y
yx−0 − ∂ k
x G
x, y
yx0
−1M 2
1− −1k
K k0
⎧
⎪
⎪
0 0 ≤ k ≤ 2M − 2,
−1M k 2M − 1 −s < x < s,
2.8
where we used the fact K k 0 0 k / 2M−1, 1 k 2M−1 So we have 3, and 4 follows
from3
UsingLemma 2.1, we prove that the functional space H associated with inner norm
·, · Mis a reproducing kernel Hilbert space
Lemma 2.2 For any u ∈ H, one has the reproducing property
u
y
u ·, G·, yM
s
−s u M x∂ M
x G
x, y
dx
−s ≤ y ≤ s. 2.9
Proof For functions u ux and v vx Gx, y with y arbitrarily fixed in −s ≤ y ≤ s, we
have
u M v M − u−1 M
v 2M
⎛
⎝M−1
j0
−1M−1−j
u j v 2M−1−j
⎞
⎠
Trang 7Integrating this with respect to x on intervals −s < x < y and y < x < s, we have
s
−s u M xv M xdx −
s
−s u x−1 M v 2M xdx
⎡
⎣M−1
j0
−1M−1−j
u j xv 2M−1−j x
⎤
⎦xy−0 x−s xs
xy0
M−1
j0
−1M−1−j
u j sv 2M−1−j s − u j −s v 2M−1−j −s
M−1
j0
−1M−1−j u j
y
v 2M−1−j
y − 0
− v 2M−1−j
y 0
.
2.11
Using1, 2, and 4 inLemma 2.1, we have2.9
3 Sobolev Inequality
In this section, we give a proof ofTheorem 1.2andCorollary 1.3
Proof of Theorem 1.2 and Corollary 1.3 Applying Schwarz inequality to2.9, we have
u
y2≤
s
−s
∂ M
x, y2
dx
s
−s
u M x2
dx G
y, y s
−s
u M x2
Note that the last equality holds from2.9; that is, substituting 2.9, u· G·, y Let us
assume that
C M; −s, s CM max
|y|≤s G
y, y
holdsthis will be proved in the next section From definition of CM, we have
sup
|y|≤s |uy
|
2
≤ CM
s
−s
u M x2
Substituting ux Gx, 0 ∈ H in to the above inequality, we have
sup
|y|≤s |Gy, 0
|
2
≤ CM
s
−s
∂ M
x G x, 02
dx CM2
Trang 8Combining this and trivial inequalityCM2 G0, 02≤ sup|y|≤s |Gy, 0|2, we have
CM2≤
sup
|y|≤s
G
y, 02≤ CMs
−s
∂ M
x G x, 02
dx CM2. 3.5
Hence, we have
sup
|y|≤s |Gy, 0
|
2
CM
s
−s
∂ M
x G x, 02
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 for M 1, 2, 3
G 1; x, x
s2− x2
2s , G 2; x, x
s2− x23
24s3 , G 2; x, x
s2− x25
650s5 . 4.1
Thus, we can expect that Gx, x takes the form GM; x, x const K0M; 1xK0M; 1−x.
Precisely, we have the following proposition
Proposition 4.1 Consider
G x, x −1 M
D−1
K ij 2s K i s − x
K j s x 0
2M − 1
M − 1
1
K02s K0s xK0s − x
2M − 1
M − 1
1
K02s
s2− x22M−1
{2M − 1!}2.
4.2
Hence,
C M; −s, s sup
|x|≤s G x, x G0, 0 −1 M
D−1
K ij 2s K i s
K j s 0
s 2M−1
22M−1 2M − 1!
2M − 1
M − 1
22M−1 2M − 1M − 1!2,
4.3
where i, j satisfy 0 ≤ i, j ≤ M − 1.
Trang 9To prove this proposition, we prepare the following two lemmas.
Lemma 4.2 Let ux c1Gx, x, where
c−11 −1M
22M − 1
2M − 1
D−1
1
K ij 2s 0
.
0
1 0 · · · 0 0
(i, j satisfy 0 ≤ i, j ≤ M − 1), then it holds that
− u 22M−1 1 −s < x < s, 4.5
u 2M−1 s −
2M − 1
M − 1
Lemma 4.3 Let ux c2K0s xK0s − x −s < x < s, where c−1
2 22M−1
2M−1
, then it holds that4.6 and u 2M−1 s −K02sc2.
Proof of Proposition 4.1 From Lemmas 4.2 and 4.3, ux c1Gx, x and ux c2K0s
xK0s − x satisfy BVP2M − 1 in case of fx 1−s < x < s So we have
c1G x, x c2K0s xK0s − x −s < x < s, 4.8
2M − 1
M − 1
Inserting4.9 into 4.8, we haveProposition 4.1
Proof of Lemma 4.2 Let
u x c1G x, x c1−1M D−1v x, vx
K ij 2s K i s − x
K j s x 0
, 4.10
then differentiating vx k times we have
v k x k
l0
−1l
k l
w k,l x, w k,l x
K ij 2s K li s − x
Trang 10
At first, for k 22M − 1, we have
v 22M−1 x 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−ljs x K 2M2M−2−lj s x 0 0 ≤ l ≤ 2M − 2. 4.13 The third term also vanishes because
K li s − x 0 2M ≤ l ≤ 22M − 1. 4.14 Thus, we have
v 22M−1 x −
22M − 1
2M − 1
w22M−1,2M−1x,
w22M−1,2M−1x
K ij 2s K 2M−1i s − x
K 2M−1j s x 0
1
K ij 2s 0
0
1 0 · · · 0 0
.
4.15
Hence, we have
− u 22M−1 x − c1−1M D−1v 22M−1 x 1, 4.16
by which we obtain4.5 Next, for 0 ≤ k ≤ M − 1, we have
v k s k
l0
−1l
k l
w k,l s, w k,l s
K ij 2s K li0
Trang 11
Since 0≤ l i ≤ 2M − 2, we have w k,l s 0 Thus, we have v k s 0 0 ≤ k ≤ M − 1 For
M ≤ k ≤ 2M − 2, we have
v k s M−1
l0
−1l
k l
w k,l s k
lM
−1l
k l
The first term vanishes because K li 0 0 0 ≤ l ≤ M − 1 Next, we show that the second
term also vanishes Let
w k,l s
K j 2s 0
.
K 2M−2−lj 2s 0
K 2M−1−lj 2s 1
K 2M−lj 2s 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
w k,l s 0 Thus, we have v k s 0 M ≤ k ≤ 2M −2 So we have obtained u k s 0 0 ≤
k ≤ 2M − 2 By the same argument, we have u k −s 0 0 ≤ k ≤ 2M − 2 Hence, we have
4.6 Finally, we will show 4.7 For k 2M − 1, noting K li 0 0 0 ≤ l ≤ M − 1, we have
v 2M−1 s 2M−1
lM
−1l
2M − 1
l
where
w 2M−1,l s
K ij 2s K li0
K 2M−1−lj 2s 0
K j 2s 0
.
K 2M−2−lj 2s 0
K 2M−1−lj 2s 1
K 2M−lj 2s 0
.
K 2M−1−lj 2s 0
K j 2s 0
K 2M−2−lj 2s 0
K 2M−1−lj 2s 1
K 2M−lj 2s 0
−D.
4.21
Trang 12Thus, we obtain w 2M−1,l s −D M ≤ l ≤ 2M − 1 Hence we have
v 2M−11 2M−1
lM
−1l
2M − 1
l
w 2M−1,l s − D 2M−1
lM
−1l
2M − 1
l
−D 2M−2
lM
−1l
2M − 2
l − 1
2M − 2
l
D −1 M1 D
2M − 1
M − 1
,
4.22
that is,
u 2M−1 s c1−1M
D−1v 2M−1 s −
2M − 1
M − 1
This completes the proof ofLemma 4.2
Proof of Lemma 4.3 Let
u x c2K0s xK0s − x c2
2M − 1!2
s2− x2 2M−1
Differentiating ux k times, we have
u k x c2
k
l0
−1l
k l
K k−l s xK l s − x. 4.25
For k 22M−1, noting K22M−1−lsx 0 0 ≤ l ≤ 2M−2, K 2M−1 sx K 2M−1 s−x 1, and K l s − x 0 2M ≤ l ≤ 22M − 1, we have
− u 22M−1 x c2
22M − 1
2M − 1
Thus, we have4.5 If 0 ≤ k ≤ 2M − 2, then we have
u k s c2
k
l0
−1l
k l
K k−l 2sK l 0 0. 4.27
Since u k −x −1 k
u k x, we have u k −s 0 0 ≤ k ≤ 2M − 2 Hence, we have 4.6 If
k 2M − 1, then we have
u 2M−1 s c2
2M−1
l0
−1l
2M − 1
l
K 2M−1−l 2sK l 0 − c2K02s. 4.28
This provesLemma 4.3