IS S N 1 3 0 3 –5 9 9 1
REGULARIZED TRACE OF STURM-LIOUVILLE EQUATION WITH SINGILARITY ON A BOUNDED SEGMENT
ILYAS HA¸SIMO ¼GLU
Abstract. Let 1 2 ::: n ::: and 1 2 ::: n :::
be the eigenvalues of the operatorsL0and L which are formed by di¤erential
expressions `o[y] = y00+ v2 1=4 x2 y; `[y] = y 00+v2 1=4 x2 y + q(x)y; 1/2
respectively and with the same boundary conditions Eq.(2.4) y(0) = y(1) = 0
We prove that under some conditions on q(x), the following formula for traces
1 X n=1 ( n n) = 2 q(0) + q(1) 4 with Eq.(2.4) holds.
1. Introduction
In literature there are numerous papers devoted to the calculation of regular-ized trace of scalar di¤erential operators which is the generalization of concept of matrix trace. First work in this direction belongs to I.M.Gelfand and B.M Levi-tan [1], where the formula for the sum of di¤rences of eigenvalues of two regular Sturm-Liouville operators on the [0; ], was obtained. This work has numerous con-tinuations. In [2-16] the regularized traces are calculated in several cases. In [2] the formula was obtained for the sum of di¤rences of eigenvalues of two singular self-adjoint Sturm-Liouville operators which di¤er by …nite potentials. In [4-8] by using zeta function and teta function V.A.Sadovnichiy has obtained formulae for regu-larized traces for wide class of di¤erential operators. In [7] the author considered the following operator
Ly = y00+
2 1/4
x2 y + q(x)y = 2
y
Received by the editors Dec. 20, 2011, Accepted: Feb. 15, 2012. 2000 Mathematics Subject Classi…cation. 47A10, 47A75, 34L10, 34L05. Key words and phrases. Regularized trace; Sturm-Liouville operators.
c 2 0 1 2 A n ka ra U n ive rsity
and is a su¢ ciently smooth function. In this case the series P1
n=1
( n (n +2 12)2)
is calculated.
The aim of this article is to calculate the regularized trace with the peculiarity on the bounded segment. Namely we consider the following problem.
Let L0and L be operators acting in L2(0; 1)and formed by di¤erential expressions
`o[y] = y00+v
2 1=4
x2 y
`[y] = y00+v2 1=4
x2 y + q(x)y; 1/2
respectively and with same boundary conditions Eq.(2.4)
y(0) = y(1) = 0;
where potential q(x) is a bounded function and satis…es the following conditions: a) q(x) satis…es the Hölder condion with order > 0 on the neigbour of zero, i.e. there is & > 0such that for any x 2 [0; &] the following inequality holds
jq(x) q(0)j < const: x b) 1 R 0 q(x)dx = 0
Our aim is to …nd the regularized trace of these operators.
2. Main results
Let 1 2 ::: n ::: and 1 2 ::: n ::: be the eigenvalues
of the operators L0and L, respectively. Then the following theorem holds
Theorem. Let the function q(x) satisfy conditions a) and b). Then P1 n=1 ( n n) = 2 q(0)+q(2:4) 4 It is shown in [11] that P1 n=1
( n n (q`n; `n)) = 0, where `nare eigenfunctions
of L0. If 1 P n=1 (q`n; `n) is convergent then 1 P n=1 ( n n) = 1 P n=1 (q`n; `n):
Eigenfunctions of the operator L0 have the form [2]
`n =
p
2xJ v(jnx)
Jv+1(jn) where Jv(x) is the Bessel function, j1 j2 j3 ::: are
1 X n=1 (q`n; `n) = lim N!1 N X n=1 (q`n; `n) (2.1) = lim N!1 N X n=1 1 Z 0 2xJ 2 v(jnx) J2 v+1(jn) q(x)dx (2.2) = lim N!1 1 Z 0 N X n=1 2xJ2 v(jnx) J2 v+1(jn) ! q(x)dx (2.3)
To …nd this limit we will study the asymptotical behavior of the function TN(x) = N P n=1 2xJv2(jnx) J2 v+1(jn) with and " > 0
For later use we note the following lemma Lemma. For the function TN(x);
TN(x) = AN cos(2xAN v ) 2 sin x + (ANx) x where AN = (N +v2 +14) , when 0 < x < 1; (ANx) ! 0 as N ! 1.
Proof of Lemma. To get formula for TN(x) we express the mth term of the
sum TN(x) in the form of residue at point jmof some function of complex variable
z, which has poles at the points j1; j2; :::; jN:
Consider the following complex function. zxfJ2
v(xz) Jv 1(xz)Jv+1(xz)g
J2 v(z)
First, prove that this function has a residue 2xJv2(jmx) J2
v+1(jm) at z = jm:
If z = jm+ ; where is small, then
J (z) = J0(jm) + 1 2 2J00 (jm) + ::: and hence zJv2(z) = 2jmJ02(jm) + 3(jm) fjmJ00(jm) + J0(jm)g + :::
Using Bessel di¤erential equation it is easy to prove that the coe¢ cient of 3 on the right hand side of this expression is zero.
Therefore the residue of function g(z) zJ2 v(z) =z 2x J2 v(xz) Jv 1(xz)Jv+1(xz) zJ2 v(z) (2.4)
g0(jm) jmJ02 (jm) = 2x Jv2(jmx) J2 v+1(jm)
As a contour of integration we take the rectangle with vertex at iB; AN iB.
Here B ! 1 and jN < AN < jN +1. For AN we take the value N +v2+14 , in
the case when N is su¢ ciently large this value is between jN and jN +1:
It is easy to prove that the function (2.4) is an odd function of z, therefore the integral on the left sides of rectangle is zero. If z = u + iw, then for large jwj and for u 0 integrand will have an order O(e 2(1 x)jwj) and consequently for given
value of AN, integrals on upper and lower sides converge to zero as B ! 1; when
0 < x < 1: Thus we obtain TN(x) = 1 2 i ANZ+iB AN iB zxfJz v(xz) Jv+1(xz)Jv 1(xz)g J2 v(z) dz
When jN1+" x < 1; where 0 < " < 1=2; jxzj ! 1; the Bessel functions in the integrant can be replaced by the corresponding asymptotes with large arguments
Jv2(xz) Jv 1(xz)Jv+1(xz) = 2 xz 1 1 2xzcos(2xz v ) 1 + 0 1 (xz)2 Jv2(z) = 2 z 1 2 + sin(2z v ) 2 + cos(2z v )(2v 1)(2v + 1) 8z 1 + 0 1 z2
Then if N ! 1 TN(x) = 1 2 iBlim!1 ANZ+iB AN iB xzfJ2 v(xz) Jv+1(xz)Jv 1(xz)g J2 v(xz) dz = 1 2 iBlim!1 ANZ+iB AN iB z[1 2xz1 cos(2xz v )(1 + 0 (xz)12 dz 1 2 + sin(2z v ) 2 + cos(2z v )(2v 1)(2v + 1) 8z 1 + 0 1 z2 1 i ANZ+i1 AN i1 z [ 1 2xz1 cos(2xz v ) ] 1 + sin(2z v ) dz = 1 i ANZ+i1 AN i1 [ z 1 + sin(2z v ) 1 2x cos(2xz v ) 1 + sin(2z v )]dz = 1 +1 Z 1 (AN+ iw)dv 1 + cos(2iw) 1 2x 1 Z 1 cos(2xAN+ 2ixw ) 1 + cos(2iv) dw = AN cos(2xAN v ) 2x 1 Z 1 cosh(2xw) 1 + cosh(2w)dw = AN cos(2xAN v ) 2 sin x That is we obtain TN(x) = AN cos(2xAN v ) 2 sin x + (ANx) x ;
where AN = (N +v2 +14) and (ANx) = O( lim B+1 ANR+i1 AN i1 sin(2xz v ) xz(1+sin(2z v ))dz) when 0 < x < 1; (ANx) ! 0 as N ! 1.
Really, if we put z = AN + iw, the integral
ANZ+iB AN iB
sin(2z v )
2ANsin(2xAN v ) 1 Z 0 cosh(2xw) x(A2 N+ w2)(1 + cosh(2w)) dw +2 cos(2xAN v ) 1 Z 0 w sinh(2xw)dw x(A2 N+ w2)(1 + cosh(2w))
When jW1+" x < 1, absolute value of (2.5) doesn’t exceed
1 xAN 1 Z 0 cosh(2xv) 1 + cosh(2v)dv + 2 xA2 N 1 Z 0 v sinh(2xv) 1 + cosh(2v)dv
This shows, that (ANx) ! 0 as N ! 1:
If x = 1 lim B!1 ANZ+iB AN iB sin(2z v ) z(1 + sin(2z v ))dz = limB!1 B Z B cosh(2w)dw (AN + iw)(1 + cosh(2w)) = lim B!1 2 42AN B Z 0 cosh(2w) (A2 N+ w2)(1 + cosh(2w)) dw i B Z B w cosh(2w)dw (A2 N+ w2)(1 + cosh(2w)) 3 5 = lim B!12AN B Z 0 cosh(2w) (A2 N+ w2)(1 + cosh(2w)) dw :
Absolute value of this integral doesn’t exceed 2AN 1 R 0 dw A2 N+w2 = .
We obtain that (ANx) ! 0 as N ! 1 when jN1+"< x < 1 and j (ANx)j < c
when x = 1:
Proof of Theorem. Now using the asymtotes of the function TN(x) = N P n=1 2xJv2(jnx) J2 v+1(jn); we get
1 P n=1 ( n n) = lim N!1 N P n=1 1 R 0 2xJ2 v(jnx) J2 v+1(jn)q(x)dx = limN!1 1 R 0 TN(x)q(x)dx = lim N!1 1 R 0 TN(x)(q(x) q(0))dx + q(0)N = lim N!1 ANR1+" 0 TN(x)(q(x) q(0))dx + 1 R AN1+" AN cos(2ANx ) 2 sin x (ANx) x (q(x) q(0)dx + q(0)N ] = lim N!1 "A 1+" NR 0 TN(x)(q(x) q(0))dx A 1+" NR 0 (AN cos(2ANx ) 2 sin x )(q(x) q(0))dx + 1 R 0 (AN cos(2ANx ) 2 sin x )(q(x) q(0))dx + q(0)N + 1 R AN1+" (ANx) x (q(x) q(0))dx 3 5 = lim N!1 2 6 4 AZN1+" 0 TN(x)(q(x) q(0))dx+ when 0 < " < +1:
Using the inequality jpxJv(x)j < const; for any x [17], it is easy to prove that
TN(x) const N (3)
Take into account (3) and condition a), one can easily see that
lim N!1 ANR1+" 0 TN(x)(q(x) q(0)dx lim N!1 ANR1+" 0 TN(x) jq(x) q(0)j dx lim N!1const ARN 0 N x dx = const lim N!1N A 1 +"( +1) N = 0 for 0 < " < +1 .
Moreover if is easy to prove that
lim N!1 AZN1+" 0 AN cos(2ANx v 2 sin x (q(x) q(0))dx = 0 (2.6)
lim N!1 A 1+" N (ANx) x (q(x) q(0)dx = 0 (2.7)
Taking into account (3) –(2.7) and condition b), we obtain lim N!1 1 R 0 AN cos(2ANx v ) 2 sin x (q(x) q(0))dx + q(0)N = lim N!1 AN 1 R 0 q(x)dx ANq(0) + N q(0) 1 4 1 R 0 cos(2N +1 2) cos2x cos(v (x 1)) sin2x (q(x) q(0))dx = v 2+ 1 4 q(0) q(1) q(0) 4 = 2vq(0)+q(1) 4 That is, 1 X n=1 ( n n) = 2vq(0) + q(1) 4 :
This completes the proof of the theorem. ÖZET: 1 2 ::: n ::: ve 1 2 ::: n :::say¬lar¬ `o[y] = y00+v 2 1=4 x2 y; `[y] = y00+ v2 1=4 x2 y+q(x)y; 1/2
diferansiyel ifadeleri ve y(0) = y(1) = 0 s¬n¬r ¸sartlar¬ile tan¬mlan-m¬¸s s¬ras¬yla L0 ve L1operatörlerinin özde¼gerleri olsun. Makalede
q(x) fonkisyonu baz¬¸sartlar¬sa¼glad¬¼g¬nda
1
P
n=1
( n n) = 2 q(0)+q(1)4 iz formülü ¬spatlanm¬¸st¬r.
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Current address : Ilyas Ha¸simo¼glu; Karabuk University, Karabuk, TURKEY and Institute of Mathematics, Academy of Sciences of Azerbaijan, Baku, Azerbaijan
E-mail address : ilyas_hashimov@yahoo.com
