Volume 1, Number1 (2010), 137 146
EQUICONVERGENCE THEOREMS
FOR STURMLIOVILLE OPERATORS
WITH SINGULAR POTENTIALS
(RATE OF EQUICONVERGENCE IN
W 2 θNORM)
I.V. Sadovnihaya
Communiated by M.Otelbaev
Keywords and phrases: equionvergene, SturmLioville operators, singular
potential.
Mathematis Subjet Classiation: 34L10, 34L20,47E05.
Abstrat.Westudy the SturmLiouville operator
Ly = l(y) = − d 2 y
dx 2 + q(x)y
withDirihlet boundary onditions
y(0) = y(π) = 0
in the spaeL 2 [0, π]
. We assumethat the potential has the form
q(x) = u ′ (x)
, whereu ∈ W 2 θ [0, π]
with0 < θ <
1/2
. HereW 2 θ [0, π] = [L 2 , W 2 1 ] θ
is the Sobolev spae. We onsider the problem ofequionvergene in
W 2 θ [0, π]
norm of two expansions of a funtionf ∈ L 2 [0, π]
.The rst one is onstruted using the system of the eigenfuntions and assoiated
funtions of the operator
L
. The seond one is the Fourier expansion in the seriesof sines. We show that the equionvergene holds for any funtion
f
in the spaeL 2 [0, π]
.1 Introdution
In this paperwe deal with the SturmLiouville operator
Ly = l(y) = − d 2 y
dx 2 + q(x)y, (1)
with Dirihlet boundary onditions
y(0) = y(π) = 0
in the spaeL 2 [0, π]
. Weassume that the potential
q
is omplex-valued and has the formq(x) = u ′ (x)
,where
u ∈ W 2 θ [0, π]
with0 < θ < 1/2
. Here the derivative is treated in thedistributional sense, and
W 2 θ [0, π] = [L 2 , W 2 1 ] θ
is the Sobolev spae with frationalorderof smoothnessdened by interpolation.This lass of operatorswas dened in
the papers of A.M. Savhuk and A.A. Shkalikov [7℄[9℄. In partiular, itwas shown
there that
L
is bounded frombelowand has purely disretespetrum.Weonsider the problemof equionvergene in
W 2 θ [0, π]
normof two expansionof a funtion
f ∈ L 2 [0, π]
. The rst one is onstruted using the system of theeigenfuntions and assoiated funtions of the operator
L
, while the seond one isTheproblemofuniformequionvergene (i.e.inthenormofthespae
C[0, π]
)iswellstudiedinthelassialtheoryofSturmLiouvilleoperatorsforregularpotentials
q ∈ L 1 (0, π)
. Inthe monograph of V.A.Marhenko [4℄the uniform equionvergene wasproved foranyfuntionf ∈ L 2 [0, π]
.In 1991V.A. Il'in[2℄proved thisresult forany
f ∈ L 1 [0, π]
. V.A. Vinokurov and V.A. Sadovnihii [10℄ proved a theorem onthe equionvergene for SturmLiouvilleoperatorswhose potentialisthe derivative
of a funtionof bounded variation (and alsofor any
f ∈ L 1 [0, π]
).The problem of the rate of equionvergene (for regularpotentials) was studied
in the paper of A.M. Gomilko and G.V. Radzievskii [1℄. In the author's paper [5℄
results were obtained in the ase in whih
q = u ′
,u ∈ W 2 θ [0, π]
with0 < θ < 1/2
.It wasshown that forany
f ∈ L 2 [0, π]
one an estimatethe rateof equionvergene uniformlyover the ballu ∈ B θ,R = { v ∈ W 2 θ [0, π] : k v k W 2 θ 6 R }
. This result is neweven inthe lassialase
q ∈ L 2 [0, π]
.In thispaperweproveequionvergene and obtainasimilarestimateof the rate
of equionvergene inthe norm of the spae
W 2 θ [0, π]
with0 < θ < 1/2
.2 Preliminary results
Webegin with the following results about operator (1).
Letus reallthat the operator
L
is bounded frombelow andhas purely disretespetrum.Let
{ λ n } ∞ 1
bethe sequene of alleigenvaluesof the operatorL
.Weshallenumerate theeigenvalues insuh away that
| λ 1 | 6 | λ 2 | 6 . . .
, andweassume thateah eigenvalue is repeated as many times as its algebrai multipliity. By
{ y n } ∞ 1
we denote the system of the eigenfuntions and assoiated funtions. We assume
that the funtion
y n
orresponds to the eigenvalueλ n
andk y n k L 2 = 1
ify n
is aneigenfuntion.
Statement 1 (A.M. Savhuk, A.A. Shkalikov). Let
u ∈ L 2 [0, π]
. Then the system{ y n } ∞ 1
of theeigenfuntionsandassoiatedfuntionsof theoperatorL
forms aRieszbasis in the spae
L 2 [0, π]
.By the above there exists the biorthogonal system
{ w n } ∞ 1
, i.e.(y n , w m ) = δ nm
where
(f, g) = R π
0
f (x)g(x)dx
. For more detailssee [5℄.Let
l θ 2 =
x = { x n } ∞ n=1 :
X ∞ n=1
| x n | 2 n 2θ
! 1/2
= k{ x n }k l 2 θ < ∞
.
Wereall that for any
u ∈ W 2 θ [0, π]
with0 < θ < 1/2 C 1 k{ u n }k l 2 θ 6 k u k W 2 θ 6 C 2 k{ u n }k l θ 2
where
u n = r 2
π (u(x), sin nx)
andC 1 , C 2 > 0
are independent ofu
.Statement 2 (A.M. Savhuk, A.A. Shkalikov). Let
R > 0
,0 < θ < 1/2
, andu ∈ B θ,R
. Then there exists a natural number 2N = N θ,R
suh that for alln > N
the eigenvalues
λ n
of the operatorL
are simple,y n (x) = q
2
π sin nx + ϕ n (x), w n (x) = q
2
π sin nx + ψ n (x), y n ′ (x) = n q
2
π cos nx + η n (x)
+ u(x) q
2
π sin nx + ϕ n (x)
, (2)
where the funtions
ϕ n
,ψ n
andη n
are suh that the sequene{ γ n } ∞ n=N = {k ϕ n (x) k C + k ψ n (x) k C + k η n (x) k C } ∞ n=N ∈ l 2 θ
and its norm in this spae is boundedby a quantity depending only on
θ
andR
. Moreover forn > N ψ n (x) = ψ n,0 (x) + ψ n,1 (x),
where
ψ n,0 (x) = α n sin nx + β n x cos nx − r 2
π Z x
0
u(t) sin n(x − 2t)dt, (3)
andthenumbers
α n
,β n
andthefuntionsψ n,1
aresuhthatthenormofthesequene{| α n | + | β n |} ∞ n=N
in the spael 2 θ
is bounded by a quantity depending only onθ
andR
, and the sequene{k ψ n,1 (x) k C } ∞ n=N
belongs to the spael 1 τ
for anyτ < 2θ
.Statement 3. Let
R > 0
,0 < θ < 1/2
andu ∈ B θ,R
. Then there exists a naturalnumber
N = N θ,R
suh that for alln > N
the operatorP n (u) : L 2 [0, π] → W 2 1 [0, π], P n (u) : f 7→
X n k=1
(f, w k )y k ,
is ontinuous on
B θ,R
, i.e. for anyu 0 ∈ B θ,R
k P n (u) − P n (u 0 ) k L 2 → W 2 1 → 0
ask u − u 0 k W 2 θ → 0, u ∈ B θ,R .
This statement follows from the results of the paper [8℄ (Theorem 1.9) and
lassial theorems (see [3, Theorems IV.2.23 and IV.3.16℄). For detailed proof see
[6℄.
3 Main theorem
Theorem. Let
R > 0
,0 < θ < 1/2
.Consideroperator(1)atinginthespaeL 2 [0, π]
with the homogeneous Dirihlet boundary onditions. Suppose that the omplex-
valued potential
q(x) = u ′ (x)
, whereu(x) ∈ B θ,R
.Let
{ y n (x) } ∞ n=1
be the system of the eigenfuntions and assoiated funtions of the operatorL
and{ w n (x) } ∞ n=1
be the biorthogonal system.2
Here andin thesequelwhen we write
N θ,R
,C θ,R
etwemean that these quantities dependonlyon
θ
andR
.For an arbitrary funtion
f ∈ L 2 [0, π]
denotec n := (f(x), w n (x)), c n,0 := p
2/π(f (x), sin nx).
Then there exist a natural number
M = M θ,R
and a positive numberC = C θ,R
suhthat for all
m > M
and for allf ∈ L 2 [0, π]
X m n=1
c n y n (x) − X m n=1
r 2
π c n,0 sin nx W 2 θ
6 C
s X
n>m 1−θ
| c n,0 | 2 + k f k L 2
m θ(1 − θ)
. (4)
Proof.
Step 1 (operators
B m,N
andB m
).For given integers
m
,m > N
, we introdue the operatorB m,N : L 2 [0, π] → W 2 θ [0, π]
dened byB m,N f (x) :=
X m n=N
c n y n (x) − X m n=N
r 2
π c n,0 sin nx
and denote
B m,1 =: B m
. Then for any funtionf ∈ L 2 [0, π]
the following equalityholds:
B m,N f (x) = X m n=N
r 2
π (f (t), ψ n (t)) sin nx+
+ X m n=N
r 2
π (f (t), sin nt)ϕ n (x) + X m n=N
(f(t), ψ n (t))ϕ n (x). (5)
Let
N = N θ,R
be the maximal of the positive integers dened in Statements 2and 3.
Step 2 (estimation of the norm of the operator
B m,N
).There existsapositive number
C θ,R
suhthatforallnaturalm
satisfyingm > N k B m,N (u) k L 2 → W 2 θ 6 C θ,R . (6)
Letusestimateeahtermoftheright-handsidein(5)separately.Byasymptoti
formulae (2)we have:
X m n=N
(f (t), ψ n (t)) sin nx W 2 θ
6 C 1 X m n=N
| (f (t), ψ n (t)) | 2 n 2θ
! 1/2
6
6 C 1
X m n=N
k f k 2 L 2 k ψ n k 2 L 2 n 2θ
! 1/2
6 C 2 k f k L 2 ,
where
C 1 , C 2 > 0
are independent off
. Consider the seond and the third terms.We willshow that
{k ϕ n k W 2 θ } ∞ n=N ∈ l 2
provided that0 < θ < 1/2
. Asymptotis (2)implythe estimate
k ϕ n k W 2 θ 6 C 3 k ϕ n k 1 L − 2 θ k ϕ n k θ W 2 1 6 C 4 k ϕ n k 1 L − 2 θ k ϕ ′ n k θ L 2 6
6 C 4 k ϕ n k 1 L − 2 θ
k η n k θ L 2 n θ + r 2
π
! θ
k u k θ L 2 + k uϕ n k θ L 2
6 C 5 k ϕ n k 1 L − 2 θ k η n k θ L 2 n θ + 1
=
= C 5 k ϕ n k L 2 n θ 1 − θ
( k η n k L 2 n θ ) θ + (n θ − 1 ) θ
, (7)
where
C 3 , C 4 , C 5 > 0
are independent ofn
. Here we used the inequalityk f k W 2 1 6
√ π 2 + 1 k f ′ k L 2
fullledforallf ∈ W 2 1 [0, π]
satisfyingf(0) = 0
.NextweuseHolder'sinequality
X m n=N
(a 1 n − θ b θ n ) 2 = X m n=N
(a 2 n ) 1 − θ (b 2 n ) θ 6 X m n=N
a 2 n
! 1 − θ X m n=N
b 2 n
! θ
andnote thatthe sequenes
{k ϕ n k L 2 n θ } ∞ n=N
,{k η n k L 2 n θ } ∞ n=N
,and{ n θ − 1 } ∞ n=1
belongto
l 2
spae for any0 < θ < 1/2
. Therefore (7) impliesthe inequalityX m
n=N
k ϕ n k 2 W 2 θ 6 C 6 , (8)
where
C 6 > 0
is independent ofm
andN
. ThenX m n=N
(f (t), sin nt)ϕ n (x) W 2 θ
6 X m n=N
| f n |k ϕ n k W 2 θ 6 C 7 k f k L 2 , (9)
where
f n = (f (t), sin nt)
andC 7 = √ C 6
.For the lastterm of the right-handside in(5)wehave:
X m n=N
(f (t), ψ n (t))ϕ n (x) W 2 θ
6 X m n=N
k f k L 2 k ψ n k L 2 k ϕ n k W 2 θ 6
6 k f k L 2
X m n=N
k ψ n k 2 L 2
! 1/2 X m n=N
k ϕ n k 2 L 2
! 1/2
6 C 8 k f k L 2 , (10)
where
C 8 > 0
is independent ofm
andN
, beause the onditions of our theoremimplythat
{k ψ n k L 2 } ∞ n=N ∈ l 2
.Weproved inequality (6). Step 2is ompleted.
Step 3 (estimation of the norm of the operator
B m
).There existsapositive number
C θ,R
suhthatfor allnaturalm
satisfyingm > N k B m (u) k L 2 → W 2 θ 6 C θ,R . (11)
Obviously the operator
B m (u)
an berepresented asthe sum:B m (u) = B N − 1 (u) + B m,N (u) = P N − 1 (u) − P N − 1 (0) + B m,N (u),
where
P n (u)
is the operator dened in statement 3. Henek B m (u) k L 2 → W 2 θ 6 2 sup
v ∈ B θ,R
k P N − 1 (v) k L 2 → W 2 θ + k B m,N (u) k L 2 → W 2 θ 6 6 2 sup
v ∈ B θ,R
k P N − 1 (v) k L 2 → W 2 1 + k B m,N (u) k L 2 → W 2 θ .
Let
0 < θ 1 < θ
, sayθ 1 = θ 2
. Bystatement3 thefuntionk P N − 1 (v ) k L 2 → W 2 1 : B θ 1 ,R → R
is ontinuous onB θ 1 ,R
. It is known that the ballB θ,R
is a ompat subset ofB θ 1 ,R
. Therefore the funtionk P N − 1 (v ) k L 2 → W 2 1
is bounded onB θ,R
by a onstantdepending only on
θ
andR
. This, together with inequality (6), imply (11). Step 3is ompleted.
Step 4 (proof of the equionvergene).
Let
f ∈ L 2 [0, π]
. Thenm lim →∞ k B m f k W 2 θ = 0. (12)
First let us hek that the system
{ y k } ∞ 1
of the eigenfuntions and assoiated funtions is minimal in the spaeW 2 θ [0, π]
. Assume the onverse. Theny n ∈ span { y k } k 6 =n
(here the losure is taken subjet toW 2 θ [0, π]
norm) for some naturalnumber
n
. Thereforey n ∈ span { y k } k 6 =n
, where the losure is taken subjet toL 2 [0, π]
norm. It means that the system{ y k } ∞ 1
is not minimal inthe spaeL 2 [0, π]
ontradition with statement 1. Next, we proved (see (8)) that the system
{ y k } ∞ 1
is lose to the orthogonal basis{ p
2/π sin kx } ∞ 1
in the spaeW 2 θ [0, π]
. Letus now onsider the orthonormal basis
nq 2
π (1 + k 2 ) − θ/2 sin kx o ∞ 1
and the system
{ (1 + k 2 ) − θ/2 y k } ∞ 1
inW 2 θ [0, π]
. We see that these two systems are also lose in thespae
W 2 θ [0, π]
.Itanbeeasilyproved thatthe system{ (1 + k 2 ) − θ/2 y k } ∞ 1
isminimalinthe spae
W 2 θ [0, π]
.Hene, this system formsthe Baribasis inthe spaeW 2 θ [0, π]
and onsequently the system
{ y k } ∞ 1
is totalin this spae.Let usonsider the image ofthe funtion
y k
underthe mapB m
:(B m y k )(x) = X m n=1
(y k (x), w n (x))y n (x) − 2 π
X m n=1
(y k (x), sin nx) sin nx.
Therst term ofthe right-handsideinthe lastequalityisequalto0for
m < k
andis equalto
y k (x)
form > k
.The seondone is the partialsum of the Fourier seriesfor the funtion
y k
. Reallthat the funtiony k ∈ W 2 1
. This implies that its Fourierseries onverges to
y k
inW 2 1
norm. Consequently this series onverges toy k
in thespae
W 2 θ
. This yields that (12) holds for any funtionf ∈ span { y k }
. To onludethe proof of the Step 4, it remains to apply ompleteness of the system
{ y k }
andinequality(11).
Nowletusproveinequality(4).Let
g k (x) = P k n=1
c n y n (x)
,wherek > N
.Itislearthat for any funtion
f ∈ L 2 [0, π]
and any natural numberm
k B m f k W 2 θ 6 k B m (f − g k ) k W 2 θ + k B m g k k W 2 θ . (13)
Step 5 (estimation of the norm of
B m (f − g k )
in the spaeW 2 θ [0, π]
).There exists a positive number
C θ,R
suh that for all naturalk
,m
satisfyingk, m > N
and for allf ∈ L 2 [0, π]
k B m (f − g k ) k W 2 θ 6 C θ,R
X ∞ n=k+1
| c n,0 | 2
! 1/2
+ k f k L 2
k θ
. (14)
By asymptotiformula(2) wehave:
k f(x) − g k (x) k L 2 =
X ∞ n=k+1
c n y n (x) 6
6
X ∞ n=k+1
2
π (f (x), sin nx) sin nx L 2
+
X ∞ n=k+1
r 2
π (f (x), ψ n (x)) sin nx L 2
+
+
X ∞ n=k+1
r 2
π (f (x), sin nx)ϕ n (x) L 2
+
X ∞ n=k+1
(f (x), ψ n (x))ϕ n (x) L 2
6
6
X ∞
n=k+1
| c n,0 | 2
! 1/2
+
X ∞ n=k+1
| (f(x), ψ n (x)) | 2
! 1/2
×
×
1 + X ∞
n=k+1
k ϕ n k 2 L 2
! 1/2
6
6
X ∞
n=k+1
| c n,0 | 2
! 1/2
+ k f k L 2
k θ
1 + C θ,R
k θ
. (15)
Now inequality (14) follows from (15) and (11). Step 5is ompleted.
Next we estimate the seond term in (13). For a given natural
m
,m > k
, weintrodue the operator
S m : W 2 1 [0, π] → W 2 θ [0, π]
dened byS m h(x) = 2/π
X ∞ n=m+1
(h(t), sin nt) sin nx .
Notethat
B m g k (x) = g k (x) − 2 π
X m n=1
(g k (t), sin nt) sin nx = S m g k (x) .
(The expression
S m g k (x)
is well dened, sine all eigenfuntions and assoiated funtionsof the operatorL
belong to the spaeW 2 1 [0, π]
.) Thereforek B m g k k W 2 θ 6 k S m (g k − g N ) k W 2 θ + k S m g N k W 2 θ . (16)
Step 6 (estimation of the norm of
S m (g k − g N )
in the spaeW 2 θ [0, π]
).There existsa positive number
C θ,R
suh thatforallnaturalk, m
satisfyingm >
k > N
and for allf ∈ L 2 [0, π]
k S m (g k − g N ) k W 2 θ 6 C θ,R k f k L 2 m θ − 1 k 1 − θ . (17)
First let usreall that
y n (x) = p
2/π sin nx + ϕ n (x)
(see (2)) and represent theleft-hand side of inequality (17) as follows:
k S m (g k (x) − g N (x)) k W 2 θ =
X k n=N +1
c n S m y n (x) W 2 θ
=
X k n=N +1
c n S m ϕ n (x) W 2 θ
.
Notethat 3
X k n=N+1
| c n | 2 6 X ∞ n=N+1
| c n | 2 6
6 2 2 π
X ∞ n=N+1
| (f (x), sin nx) | 2 + X ∞ n=N+1
| (f (x), ψ n (x)) | 2
!
6 C θ,R (1) k f k 2 L 2 ,
Therefore
X k n=N +1
c n S m ϕ n (x) W 2 θ
6
X k n=N+1
| c n | 2
! 1/2 k X
n=N +1
k S m ϕ n (x) k 2 W 2 θ
! 1/2 6
6 q
C θ,R (1) k f k L 2
X k n=N +1
k S m ϕ n (x) k 2 W 2 θ
! 1/2
.
Furthermore
k S m ϕ n (x) k W 2 θ 6 C θ,R (2)
X ∞ j=m+1
(j 2θ + 1) | (ϕ n (x), sin jx) | 2
! 1/2
=
= C θ,R (2)
X ∞ j=m+1
j 2θ + 1
j 2 | (ϕ ′ n (x), cos jx) | 2
! 1/2
6 C θ,R (2) m θ − 1 k ϕ n (x) k W 2 1 . (18)
Here we applied integration by parts and boundary onditions
ϕ n (0) = ϕ n (π) = 0
. By asymptoti formulae (2) we obtainϕ ′ n (x) = nη n (x) + u(x)y n (x)
, where{k η n (x) k L 2 + n θ − 1 } ∈ l θ 2
. Consequentlyk ϕ ′ n (x) k L 2 6 (n k η n (x) k L 2 + k u(x)y n (x) k L 2 ) = n 1 − θ τ n ,
3
Here
C θ,R (1)
andin thesequelC θ,R (j)
withj = 2, 3...
aresomepositivequantitiesdependingonlyon
θ
andR
.where
k{ τ n }k l 2 6 C θ,R (3)
.Therefore,k S m ϕ n (x) k W 2 θ 6 C θ,R (4) m θ − 1 n 1 − θ τ n .
This impliesthat the rst term in(16) satises the followinginequality:
k S m (g k − g N ) k W 2 θ 6 C θ,R (5) k f k L 2
X k n=1
m 2θ − 2 n 2 − 2θ τ n 2
! 1/2
6 C θ,R (6) k f k L 2 m θ − 1 k 1 − θ .
Step 6 isompleted.
Step 7 (estimation of the norm of
S m g N
in the spaeW 2 θ [0, π]
).Finallyweonsidertheseondtermin(16).Weshallestimatethenorm
k S m g N k C
inthe same way as in(18):
k S m g N (x) k C 6 C θ,R (7) m θ − 1 k g N ′ (x) k L 2 6 C θ,R (7) m θ − 1 k g N k W 2 1 .
Sine
k P N (u) k L 2 → W 2 1 6 C θ,R
(see Step 3) andg N = P N f
we havek g N k W 2 1 6 C θ,R k f k L 2
.Therefore,k S m g N k C 6 C θ,R (8) k f k L 2 m θ − 1 . (19)
Hene inequalities(16), (17) and (19) imply that
k B m g k k C 6 C θ,R (9) k f k L 2 m θ − 1 k 1 − θ . (20)
Let us put
k = [m 1 − θ ] + 1
for any naturalnumberm > N 2
. Inequalities (13), (14)and (20) implyinequality(4).
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IrinaSadovnihaya
FaultyofComputationalMathematisand Cybernetis
M.V.LomonosovMosowStateUniversity
Leninskiegory
119991Mosow,Russia
E-mail:ivsadyandex.ru
Reeived:14.10.2009