THE THIRD REGULARIZED TRACE FORMULA FOR SECOND ORDER DIFFERENTIAL EQUATIONS WITH SELF-ADJOINT NUCLEAR CLASS OPERATOR COEFFICIENTS

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1 Mathematica Aeterna, Vol.,, no. 5, 49-4 THE THIRD REGULARIZED TRACE FORMULA FOR SECOND ORDER DIFFERENTIAL EQUATIONS WITH SELF-ADJOINT NUCLEAR CLASS OPERATOR COEFFICIENTS Gamidulla I. Aslanov Institute of Mathematics and Mechanics, National Academy of Scienses of Azerbaijan Kenul G. Badalova Sumgait State University Abstract In this work we will obtain a formula for the third regularized trace of the self adjoint differential equation of second order with nuclear class operator coefficient given in a finite interval. Mathematics Subject Classification: 34L, 3E5, 34B4 Keywords: Hilbert space, self-adjoint operator, eigenvalue, spectrum, resolvent, kernel operator, regularized trace Introduction Let H be a separable Hilbert space. In the Hilbert space H = L,π;H), we consider the self-adjoint operators L and L generated by the differential expressions l y) = y x), and the boundary conditions ly) = y x)+qx)yx) ) y) =, y π) = ) Suppose that the operator function Qx) in the expression ly) satisfies the following conditions:

2 4 Gamidulla Aslanov and Kenul Badalova. For all x,π), Qx) : H H is a self-adjoint nuclear operator. Morever, Qx) hasacontinuous derivative of fourth order with respect to the norm in space σ H) in the interval,π and for x,π), Q i) x) : H H are self-adjoint operators i =,,3,4).. sup Qx) H <. x π 3. There is an orthonormal basis {ϕ n } n= of the space H such that Qx)ϕ n H <. n= 4. Qx)dx = 5.Q i ) x) = Q i) π) =, i =,) Here σ H) denotes the space of the nuclear operators from H to H. { ) } ThespectrumoftheoperatorL istheset, andeverypoint m= of this set is an eigenvalue of L with infinite multiplicity. The orthonormal eigenfunctions corresponding to the eigenvalue have the form ψ mn = π sin ) x ϕ n, n =,,... 3) In this work, we will fund a formula for the sum of the series m= n= λ 3 mn ) ) 6 3 ) 4π trq x)dx 3 6π tr Q I x) dx π 3 gx)dx+ h = 64 trq IV) π) trq IV) o) 3 8π trq II π)qπ) trq II o) + gπ) go) h, 4π where h = 5 8 i= j= β ij, β ij = π 3 n= q= s= Qx)ϕ n,ϕ q ) H cosixdx Qx)ϕ q,ϕ S ) H cosi j)xdx Qx)ϕ S,ϕ n )cosjxdx, gx) = Qx)ϕ n,ϕ q ) H Qx)ϕ q,ϕ S ) H Qx)ϕ s,ϕ n ) n= q=s= This formula is said to be the third regularized trace formula.

3 The third regularized trace formula 4 The sequences {λ mn } n= are the eigenvalues of the operator L for every m=,,...which belong to the interval ) Q, H + Q ) H. The trace formulas for the selfadjoint Sturm-Liouville differential operators have been found by Gelfand and Levitan, Dikiy 7, Gasimov and Levitan 9, Sadovnichiy and Podolskiy 4 and many others. The trace formulas of the abstract self-adjoint operators with a continuous spectrum were first analyzed by Krein, Faddeyev 8, Bayramoglu 5. On the other hand, trace formulas for the differential operators with operator coefficient have been investigated by Abdukadirov, Maksudov et al. 3, Adiguzelov and Baksi, Adiguzelov and Sezer 3, Albayrak et al. 4. Investigating the spectrum of the operator L some equalities about the resolvents Let R o λ and R λ be the resolvents of the operators L and L, respectively. Lemma.. If the operator function Qx) satis fies the condition 3 and λ ρl ), then QR λ : H H is a nuclear operator: QR λ σ H ). Proof. The eigenfunctions system {ψ mn } m,n= of the operator L is an orthonormal basis of the space H. As known from 6, to show that QR λ is a nuclear operator, it is enough to prove that the series is convergent. From 3) we obtain QR λ ψ mn m= n= H 4) m= n= QRλψ o mn o H = n= ) m= λ o mn = H = n= m= π ) ) λ π sin x Qx)ϕn H dx / π m= n= ) λ Qx)ϕn = ) H m= λ n= Qx)ϕ n H C λ m= m n= Qx)ϕ n H 5)

4 4 Gamidulla Aslanov and Kenul Badalova Here C λ is a positive constant related only to λ. By virtue of condition 3 from 5) we obtain QRλ o ψo n H <. m= n= Teorem.. If the operator function Qx) satisfies conditions -3, then the spectrum of the operator L is a subset of the union of the disjoint intervals Ω m = ) Q, H )+ Q H, m =,,3,... and the following conditions are satisfied: a) Each point of the spectrum of the operator L which is different from ) in Ωm is as isolated eigenvalue which has finite multiplicity. b) ) may be an eigenvalue of the operator L which has finite or infinite multiplicity. The equality n lim λ mn = ) holds. Here {λmn } n= are the eigenvalues, belonging to the interval Ω m of the operator L, and each eigenvalue has been repeated according to multiplicity. Proof. The resolvent R λ of the operator L satisfies the equation If λ R m= R o λ R λqr o λ = R λ 6) ) Q H, ) + Q H, then we have λ ) > Q H,m =,,..., 7) For the self-adjoint operator R o λ = L λi), R o λ H = max m holds. From here and 7), we obtain And so, we get λ ) R o λ H < Q H. QR o λ H Q H R o λ <. Thus, AB) = R o λ BQRo λ isacontractionoperatorfromlh )tolh ). Here LH ) is the linear bounded operator space from H to H. According to this, AR λ )R λ, that is, equation 6) has a unique solution R λ LH ). Thus,

5 The third regularized trace formula 43 every point λ / Ω m is the regular point of the self-adjoint operator L. So, m= the spectrum of operator L is σl) Um=Ω m. From formula 6) and lemma., for every λ ρl) ρl ), R λ Rλ o belongs to σ H ), that is, R λ Rλ o is a nuclear operator. In this case, as it was proved in Kato,page44, the continuous parts of the spectra of operators L and L coincide. According to this, and since the spectrum of operator { L is continuous, the continuous part ) } of the spectra operator L is the set. This also means that m= assertions a), b) and c) of Theorem. are satisfied. In the case when conditions b) and c) are satisfied, it can be proved that the series λ mn ), m =,,...) n= areabsolutely convergent. Ontheotherhand, if weconsider R λ Rλ o σ H ) for every λ ρl), then we get trr λ Rλ o ) = m= n= λ mn λ ). λ If we multiply both sides of this equality by λ3 and integrate this equality πi over the circle λ = bp = p ) +p, p ), then we find πi λ3 trr λ Rλ )dλ = πi λ =b P λ 3 p m= + πi λ3 n= m=p+ λ mn λ dλ ) λ n= For m p and p by condition, we can write λ mn λ ) λ Q ) λ H mn ) + Q < H p +p = b P. ) There fore we get Moreover, for m > p, λ mn λ mn < b p, m p, p, n =,,... 9) ) Q H p+ ) Q > H p +p = b p ) dλ 8) Thus we get λ mn > b p ; m > p; n =,,... )

6 44 Gamidulla Aslanov and Kenul Badalova Using 9), ), from 8) we have πi λ =b P λ 3 trr λ Rλ)dλ o = p n= m= πi + m=p+ n= πi λ 3 dλ λ λ ) πi λ =b 3 dλ P λ λ mn + λ λ =b 3 dλ P λ λ ) πi λ =b 3 dλ P λ λ mn = ) = p m= ) n= 6 λ 3 mn Moreover from the formula R λ = R o λ R λ QR o λ, we obtain the following equality R λ Rλ o N = ) j ) Rλ o QR j λ + ) N+ R λ QRλ o )N+ ) j= where N is any natural number. From ) and ), we have p ) 6 n= m= λ 3 mn = N ) j j= + )N+ πi λ =b P λ 3 tr R λ QR o λ )N+ dλ πi λ =b P λ 3 trr o λqr o λ j dλ+ 3) Let M pj = )j+ πi M PN = )j+ πi Then from 3), 4) and 5) ve have p λ 3 mn m= n= λ 3 tr Rλ o QRo λ )j dλ 4) λ =b P λ 3 tr RλQR o λ) o N+ dλ 5) λ =b P ) 6 = N M Pj +M PN 6) j= Since the operator function QRλ o in the domain C { } is analytic m= with respect to the norm in space σ H ), one can show for M Pj the following formula is true M Pj = 3 )j λ trqr πij λ o dλ 7) λ =b P

7 The third regularized trace formula 45 3 The formula of the regularized trace of the operator L In this section we will find a formula for the regularized trace of the operator L According to formula 7) M Pj = 3 λ trqr πi λ o dλ. λ =b P Sincetheeigenfunctions{ψ o mn } m=,n= ontheoperatorl isanorthonormal basis of the space H, then M Pj = 3 πi λ trqrλ odλ = 3 πi λ =b P λ n= m= QRλ oψo mn,ψo mn ) H dλ = = 3 p m= n= πi λ =b P λ dλ λ ) ) o mn,ψo mn ) H = = 3 p m= ) n= 4 o mn,ψmn o ) = = 3 p n= 4 m= ) π Qx)ϕ n,ϕ n )sin ) xdx = = 3 π p m= ) n= 4 Qx)ϕ n,ϕ n ) H dx 3 π p m= ) n= 4π Qx)ϕ n,ϕ n )cos)xdx. If we take into account the inequality 8) q n=qx)ϕ n,ϕ n ) H n= Qx)ϕ n,ϕ n ) H Qx) σ H), q n=qx)ϕ n,ϕ n )cos)x n= Qx)ϕ n,ϕ n ) H Qx) σ H) and condition then by Leveque s theorem we obtain Qx) σ H) dx < Qx)ϕ n,ϕ n ) H dx = Qx)ϕ n,ϕ n ) H dx = trqx)dx 9) n= n= Qx)ϕ n,ϕ n ) H dxcos)xdx = trqx) cos)xdx n= )

8 46 Gamidulla Aslanov and Kenul Badalova From 8), 9) and ) we have M P = 3 π m= + 3 π ) 4 trqx) cos)xdx+ m= trqx) dx ) ) Then after the application of integration on parts in the first integral of equality ) for four times we achieve M P = 3 6π + 3 π trq IV) x) cos)xdx+ m= m= Let us calculate M P. According to formula 7) trqx) dx ) ) M P = 3 λ tr QRλ o 3 4πi ) dλ = λ QR o λ =b P 4πi λ ) ψmn o,ψo mn dλ λ =b P m= n= 3) Moreover QRλ o ψo mn = λ mn ) o = QRλ o ψo mn = ) ) λ QRλ o o mn,ψrq o ψrq = i= q= ) λ i= q= From 3) and 4) we get M P = 3 ) 4πi mn,ψ o rq m= n= r= q= = 3 ) 4πi mn,ψ o rq m= n= r=q= + 3 ) 4πi mn,ψ o rq m= n= r=p+q= r ) λ o mn rq),ψo o mn 4) λ λ λ λ dλ ) λ r λ dλ ) λ r λ dλ ) λ r ) + ) + ) =

9 The third regularized trace formula ) 4πi mn,ψ o rq m=p+ n=r= q= + 3 ) 4πi mn,ψ o rq m=p+n= r=p+q= = 3 +3 m= n= r= q= m= n= r=p+q= λ λ λ dλ ) λ r λ dλ ) λ r ) + r ) mn,ψmn) o + Here we use that in the case m,r < p ) 4 ) r ) mn,ψ o rq ) ) + ) = 5) λ dλ λ ) λ ) =, 6) r In the case m = r λ dλ λ ) =. If in the expression m= n= r=q= r ) ) mn,ψo mn to after the index m with r and n with q, then we obtain m= n= r=q= r ) mn mn),ψo = hence 3 = m= n= r=q= ) ) rq,ψo mn M P = 3 ) mn m= n= ) ) r ) ) mn,ψrq o m= n=r=p+q= r ) 7) 8)

10 48 Gamidulla Aslanov and Kenul Badalova Mark α P = m= r=p+ ) r r We present α P in the following form ) ) ) mn,ψ o rq ) 9) α P = α P α P +α P3, 3) where α P = π P α P = π α P3 = π P m= r=p+ m= r=p+ m= r=p+ ) r r ) ) r r r r From 8)-33) we obtain M P = 3 m= n= ) π ) ) ) ) Re Qx)ϕ n,ϕ q ) H cosr m)xdx 3) Qx)ϕ n,ϕ q ) H cosr m)xdx Qx)ϕ n,ϕ q ) H cosr m)xdx 3) ) ) ) o ) π ) Qx)ϕ n,ϕ q ) H cosr +)xdx mn 3 n= q= Separately evaluating α P,α P and α P3, we obtain M P = m= 33) α P α P +α P3 ) 34) 3 ) trq x)dx+ 3 tr Q I x) dx+ 4π 6π + 3 tr Q II x)qx)+ Q I x) cos)xdx+o p ), 35) π where Op ) < const p. Researching in detail M P3, after very complicated calculation we obtain. M P3 = P π gx)dx gx)cosl )xdx π l= 5 P 6 p+) β ij +o), 36) i= j=

11 The third regularized trace formula 49 where gx) = Qx)ϕ n,ϕ q ) H Qx)ϕ q,ϕ s ) H Qx)ϕ s,ϕ n ) 37) n=q= s= β ij = π 3 n= q=s= Qx)ϕ n,ϕ q ) H cosixdx Qx)ϕ q,ϕ s ) H cosi j)xdx Qx)ϕ s,ϕ n ) H cosjxdx 38) Thus, as λ = bp = p +, there is such an stable c, where the following 4 inequality is true: QR o λ σ H ) < C, R o λ < Cp, R λ < Cp 39) From 4), 4) and condition we obtain M pj j λ trqrλ o j dλ b p QR o j λ =b p λ ) j dλ σ H ) b p j λ =b p QRλ) o σ H ) QRλ) o j Cb dλ p j λ =b p Q j R λ j dλ < < Cj b p j p j dλ < C b 3 p p j < C p 7 j here C and C are stable figures. From the last inequality we obtain with j = 4,5,6,7 we can also prove that From 4) and 4) we obtain lim M pj =, j 8 4) p lim M pj =. 4) p lim M pj =, j 4. 4) p From 5) and 4) we obtain M PN λ3 tr R λ QRλ o)n+ dλ b 3 p R λ QRλ o)n+ dλ σ H ) b 3 p R λ QR o λ )N+ σ H ) dλ Cb 3 p p λ =b P QR o λ H ) N QR o λ σ H ) dλ C 3 b 4 p p N < C 4 p 7 N

12 4 Gamidulla Aslanov and Kenul Badalova here C 3 and C 4 are stable figure. From the last inequality we obtain lim M PN =, N 8 43) p The main result of this article is given by the following theorem. Theorem 3.. If the operator function Qx) satisfies conditions.-5., then λ 3mn ) 6 ) m= n= 3 4π 3 tr Q I x) ) dx 6π π ) trq x)dx gx)dx+h = 3 trq IV) π) trq IV) o) 64 3 trq II π)qπ) trq II o) Qo) + 8π 4π gπ) go) h, 44) here h = 5 i= j= β 8 ij, β ij figure are defined by means of equality 38), but the function gx) is defined from the equality 36). The series on the left side of this equality is called the third regularized trace of the operator L. Proof. From formulas 6), ), 36), 4) and 43), we obtain. λ 3 mn m= n= ) 6 = 3 6π 3 ) + m= 4π trq IV) x)cos)xdx+ m= trq x)dx+ 3 tr Q I x) ) dx+ 6π + 3 tr Q II x)qx)+ Q I x) cos)xdx+ π + P π gx)dx gx)cosl )xdx π l= 5P 8 i= j= β ij 5 6 β ij +o) 45) i= j= From the Fourier theory it is known that if the function f x) is continuous on the segment,π, then the following equality is true: π n= n= f x)cos)xdx = f o) f π). 46) 4

13 The third regularized trace formula 4 From 45), 46) and condition 5. we obtain ) λ 3 mn ) 6 3 trq x)dx 3 4π 6π m= n= π gx)dx+ 5 8 β ij i= j= = 3 64 trq IV) π) trq IV) o) tr Q I x) dx 3 trq II π)qπ) trqπ)qo) + 5 gπ) go) β ij. 8π 4π 6 i= j= References E.Abdukadirov. Regularized trace formula for the Dirak systems. Vest. Mosk. un-ta, issue math. mech. 967), No4, 7-4. E.E.Adiguzelov, O.Baksi. On the regularized trace of the differential operator equation given in a finite interval. J.Eng. Natur. Sei. Sigma 4), E.E.Adiguzelov, Y.Sezer. The second regularized trace of a self adjoint differential operator coefficient. Mathematical and Computer Modeling, ), vol.53, I.Albayrak, M.Bairamoglu, E.E.Adiguzelov. The second regularized trace formula for the Sturm-Liouville problem with spectral parameter in boundary condition. Methods Funet. Anal. Tomology, ), vol.6, No3, M.Bayramoglu. The trace formula for the abstract Sturm-Liouville equation with continuous spectrum. Azerb.SSR, Inst. Fiz., Baku, Preprint 6 986), I.S.Cohberg, M.G.Krein. Introduction to the theory of linear non-self adjoint operators. in: Translation of Mathematical Monographs, vol.8. AMS. Providence, RI, L.A.Dikiy. About a formula of Gelfand-Levitan. Usp. Met. Nauk. 8953), L.D.Faddeyev. On expression for trace of difference of the singular operators of Sturm-Liouville type, Dokl. SSSR 5 957), M.G.Gasimov, B.M.Levitan. About the sum of difference of eigenvalues the operators Sturm-Liouville. Dokl. SSSR, 5, No6 963), -5.

14 4 Gamidulla Aslanov and Kenul Badalova I.M.Gelfand, B.M.Levitan. On a formula for eigenvalues of a differential operator of second order, Dokl. SSSR ). T.Kato. Perturbation theory for linear operators, Springer-Verlag, Berlin, 98. M.G.Krein. The trace formula in the perturbation theory, Matem ), F.G.Maksudov, M.Bayramoglu, E.E.Adiguzelov. On regularized trace of Sturm-Liouville operator on a finite interval with the unbounded operator coefficient. Dokl. Akad. Nauk SSSR, 3 984), V.A.Sadovnichiy and V.E.Podolskiy. Trace of operators, Usp. Math. Nauk. 6 6), Received: June,