EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 1, No. 2, 2008, (51-60) ISSN 1307-5543 – www.ejpam.com On the basisness inL2(0, 1) of the root functions in not strongly regular boundary value problems Khanlar R. Mamedov∗, Hamza Menken Mathematics Department, Science and Arts Faculty, Mersin University 3343, Ciftlikkoy Campus, Mersin, TURKEY Abstract. In the present article we consider the non-self adjoint Sturm-Liouville operators with periodic and anti-periodic boundary conditions which are not strongly regular. We obtain the asymptotic formulas for eigenvalues and eigenfunctions of these boundary value problems, when the potentialq(x) is a complex- valued function. Then using these asymptotic formulas, the Riesz basisness inL2(0, 1) of the root functions are proved. AMS subject classifications: 34L10, 34B24, 47E05 Key words: Riesz basis, periodic and anti-periodic boundary conditions, not strongly regular boundary conditions, eigenvalue, eigenfunction, non-self adjoint Sturm-Liouville operator, Bari’s theorem. 1. Introduction It is well known that the basisness of the root functions of a differential operator depends on regularity of boundary conditions generating the given differential operator. The basisness in the spaceL2(0, 1) of the root functions of a linear differential operator of ordern with regular (strongly regular, see. [1], p.71) boundary conditions is shown in [2, 3]. In [2, 4, 5] it is shown that the root functions of a boundary problem which is generated by not strongly regular boundary conditions may not be form a basis inL2(0, 1). In [6], one non-classical heat conduction problem in homogeneous rod has been studied. This problem is reduced to the following boundary value problem −y′′(x) = λy(x), 0 < x < 1, y(0) = 0, y′(0) = y′(1) whose boundary conditions are regular, but not strongly regular. All the eigenvalues of this prob- lem starting with the second one are double, the total number of associated functions is infinite. Nevertheless, in the paper it was established that the chosen specially system of the root functions forms an unconditional basis inL2(0, 1). ∗Corresponding author.Email addresses:hanlar@mersin.edu.tr (Kh. R. Mamedov), hmenken@mersin.edu.tr (H. Menken) http://www.ejpam.com 51 c© 2007 EJPAM All rights reserved. Khanlar R. Mamedov, Hamza Menken / Eur. J. Pure Appl. Math,1 (2008), (51-60) 52 After this work, in [7], the boundary-value problem generated by the differential equation y′′ + q(x)y = λy (1.1) and not strongly regular boundary conditions y(0)− y(1) = 0, y′(0)− y′(1) = 0 (1.2) or y(0) + y(1) = 0, y′(0) + y′(1) = 0 (1.3) was considered. Here,q(x) ∈ C(4)[0, 1] was a complex valued function satisfying the condition q(0) 6= q(1). In this paper, it was shown that the root functions of the boundary problems (1.1), (1.2) and (1.1), (1.3) formed Riesz basis inL2(0, 1). Let us present briefly the main definitions and fact which will be used in what follows. Definition 1.1. A system{ϕn}∞n=1 forms a basis in a Banach spaceX if for any elementf ∈ X there exists a unique expansion of it in the elements of the system, i.e. the series ∞∑ j=1 cjϕj convergent to f in the norm of the spaceX. Definition 1.2. [8,9] A system{ϕn}∞n=1 is called a Riesz basis of the Hilbert spaceH if there exists a bounded linear invertible operatorA such that the system{Aϕn}∞n=1 forms an orthonormal basis inH. Theorem 1.1. [8,9] If the sequence{ϕj}∞j=1 is complete in the Hilbert spaceH, there corresponds to it a complete biorthogonal sequence{ψj}∞j=1, and for anyf ∈ H one has ∞∑ j=1 |(f, ϕj)| < ∞, ∞∑ j=1 |(f, ψj)|2 <∞, then the sequence{ψj}∞j=1 forms a Riesz basis inH. We consider the boundary-value problems (1.1), (1.2) and (1.1), (1.3), whereq(x) ∈ C(4)[0, 1] is a complex-valued function. Without loss of generality, we can assume that 1∫ 0 q(x)dx = 0. In the present paper, in Section 2 we obtain the asymptotic formulas of eigenvalues and eigen- functions of the boundary problems (1.1), (1.2). In Section 3, using these asymptotic formulas and Theorem 1.1, we prove the basisness inL2(0, 1) of the root functions of the boundary problem (1.1), (1.2). In Section 4, similar results are obtained for the boundary problem (1.1), (1.3). 2. The asymptotic formulas for eigenvalues and eigenfunctions of the periodic problem First we shall prove the following lemma. Khanlar R. Mamedov, Hamza Menken / Eur. J. Pure Appl. Math,1 (2008), (51-60) 53 Lemma 2.1. All eigenvalues of the boundary-value problem (1.1), (1.2), starting from some num- ber, are simple and form two infinite sequencesλk,1, λk,2, k = N,N+1, · · ·, whereN is a positive integer and λk,1 = −(2kπ)2 − q′(1)− q′(0) + 1∫ 0 q2(t)dt (4kπ)2 +O( 1 k3 ), (2.1) λk,2 = −(2kπ)2 + q′(1)− q′(0)− 1∫ 0 q2(t)dt (4kπ)2 +O( 1 k3 ), (2.2) and the corresponding eigenfunctions are of the form yk,1(x) = sin 2kπx+O( 1 k ), (2.3) yk,2(x) = cos 2kπx+O( 1 k ). (2.4) Proof. We assume thatq(0) = q(1). The caseq(0) 6= q(1) was investigated in [7]. Consider the equation (1.3) or y′′ + q(x)y + µ2y = 0, (2.5) whereµ = √ −λ and √ reiϕ/2 for−π < ϕ ≤ π. From [1,10], it is well known that the eigenvalues of the boundary problem (1.1), (1.2) are asymptotically located in pairs, i.e. λk,1 = λk,2 +O(k1/2) = −(2kπ)2 { 1 + ξ0 k +O ( 1 k3/2 )} , (k = N,N + 1, · · ·). It follows from the last relation that µk,1 = √ −λk,1 = 2kπ { 1 + ξ0 2k +O ( 1 k3/2 )} , (k = N,N + 1, · · ·) µk,2 = √ −λk,2 = 2kπ { 1 + ξ0 2k +O ( 1 k3/2 )} , (k = N,N + 1, · · ·). Hence, there exists a positive numberco such that|=(µk,1)| ≤ co and|=(µk,2)| ≤ co. Thus, the relation µk,1, µk,2 ∈ Q = {µ : <(µ) ≥ 0, |=(µ)| ≤ co} holds for allk = N,N + 1, · · ·. It is easy to verify thatQ ⊂ S0 − ico ≡ T , whereS0 ={ µ : 0 ≤ argµ ≤ π 2 } . From [1,10], it is well known that in a regionT of the complex planeµ the equation (2.5) has two linear independent solutionsϕ1(x, µ), ϕ2(x, µ) satisfying the relations ϕj(x, µ) = eµωjx { 6∑ m=0 um(x) (2ωjµ)m +O( 1 µ7 ) } , (j = 1, 2), Khanlar R. Mamedov, Hamza Menken / Eur. J. Pure Appl. Math,1 (2008), (51-60) 54 ϕ′j(x, µ) = µωje µωjx { u0(x) + 6∑ m=1 um(x) + 2u′m−1(x) (2ωjµ)m +O( 1 µ7 ) } , (j = 1, 2), where ω1 = −ω2 = i, u0(x) ≡ 1, um(x) = − x∫ 0 l (um−1(t)) dt, m = 1, 2, 3, 4, 5, 6. It follows that ϕj(0, µ) = 1 +O( 1 µ7 ), ϕj(1, µ) = eµωj 1− 1 (2ωjµ)3 [q′(1)− q′(0) + 1∫ 0 q2(t)dt] + 1 (2ωjµ)4 [q′′(1)− q′′(0) + 5 2 q2(1)− 3 2 q2(0)− q(0)q(1)]− 1 (2ωjµ)5 [q′′′(1)− q′′′(0) + 7q(1)q′(1) −5q(0)q′(0)− q(0)q′(1)− q(1)q′(0) + (q(1)− q(0)) 1∫ 0 q2(t)dt +2 1∫ 0 q3(t)dt− 1∫ 0 q′ 2 (t)dt] + 1 (2ωjµ)6 [q(4)(1)− q(4)(0) + 9q(1)q′′(1) −7q(0)q′′(0)− q(0)q′′(1)− q(1)q′′(0) + 11 2 q′2(1)− 9 2 q′2(0)− q′(0)q′(1) + 15 2 q3(1)− 7 2 q3(0)− 5 2 q(0)q2(1)− 3 2 q(1)q2(0) +(q′(1)− q′(0)) 1∫ 0 q2(t)dt+ 1 2 ( 1∫ 0 q2(t)dt)2] +O( 1 µ7 )  , ϕ′j(0, µ) = µωj { 1− 2q(0) (2ωjµ)2 + 2q′(0) (2ωjµ)3 − 1 (2ωjµ)4 [2q′′(0) + 2q2(0)] + 1 (2ωjµ)5 [2q ′′′ (0) + 8q(0)q′(0)]− 1 (2ωjµ)6 [2q(4)(0) +10q′2(0) + 12q(0)q′′(0) + 4q3(0)] +O( 1 µ7 ) } , ϕ′j(1, µ) = µωje µωj 1− q(0) + q(1) (2ωjµ)2 + 1 (2ωjµ)3 [q′(1) + q′(0)− 1∫ 0 q2(t)dt] − 1 (2ωjµ)4 [q′′(1) + q′′(0) + 3 2 q2(1) + 3 2 q2(0)− q(0)q(1)] Khanlar R. Mamedov, Hamza Menken / Eur. J. Pure Appl. Math,1 (2008), (51-60) 55 + 1 (2ωjµ)5 [q′′′(1) + q′′′(0) + 5q(1)q′(1) + 5q(0)q′(0)− q(0)q′(1) −q(1)q′(0) + (q(1) + q(0)) 1∫ 0 q2(t)dt− 2 1∫ 0 q3(t)dt+ 1∫ 0 q′2(t)dt] − 1 (2ωjµ)6 [q(4)(1) + q(4)(0) + 13 2 q′2(1) + 9 2 q′2(0)− q′(0)q′(1) +7q(1)q′′(1) + 7q(0)q′′(0)− q(0)q′′(1)− q(1)q′′(0) + 7 2 q3(1) + 7 2 q3(0)− 3 2 q(0)q2(1)− 3 2 q(1)q2(0) + (q′(1) + q′(0)) 1∫ 0 q2(t)dt −1 2 ( 1∫ 0 q2(t)dt)2] +O( 1 µ7 )  . Let us substitute all these expressions into the characteristic determinant ∆(µ) = ∣∣∣∣ U1(ϕ1) U1(ϕ2) U2(ϕ1) U2(ϕ2) ∣∣∣∣ , whereU1(y) = y(1)− y(0), U2(y) = y′(1)− y′(0). By elementary transformations, we obtain the relation (iµ)−1∆(µ) = e2iµ 1− 2q(0) (2iµ)2 − 1 (2iµ)3 1∫ 0 q2(t)dt− 1 (2iµ)4 [2q′′(0)− 1 2 q2(1) + 3 2 q2(0) + q(0)q(1)]− 1 (2iµ)5 [q(1)q′(1)− q(0)q′(1)− q(0)q′(0) +q(1)q′(0)− 2q(0) 1∫ 0 q2(t)dt+ 2 1∫ 0 q3(t)dt− 1∫ 0 q′2(t)dt] − 1 (2iµ)6 [2q(4)(0) + 1 2 q′2(1) + 21 2 q′2(0)− q′(0)q′(1)− q(1)q′′(1) +11q(0)q′′(0) + q(0)q′′(1) + q(1)q′′(0)− 2q3(1) + 3q3(0) +3q(0)q2(1)− ( 1∫ 0 q2(t)dt)2] +O( 1 µ7 )  Khanlar R. Mamedov, Hamza Menken / Eur. J. Pure Appl. Math,1 (2008), (51-60) 56 −2eiµ { 1− 2q(0) (2iµ)2 − 1 (2iµ)4 [2q′′(0) + 2q2(0)]− 1 (2iµ)6 [2q(4)(0) +12q(0)q′′(0) + 10q′2(0) + 4q3(0)] +O( 1 µ7 ) } + 1− 2q(0) (2iµ)2 + 1 (2iµ)3 1∫ 0 q2(t)dt− 1 (2iµ)4 [2q′′(0)− 1 2 q2(1) + 3 2 q2(0) + q(0)q(1)] + 1 (2iµ)5 [q(1)q′(1)− q(0)q′(0)− q(0)q′(1) +q(1)q′(0)− 2q(0) 1∫ 0 q2(t)d+ 2 1∫ 0 q3(t)dt− 1∫ 0 q′2(t)dt] − 1 (2iµ)6 [2q(4)(0) + 1 2 q′2(1) + 21 2 q′2(0)− q′(0)q′(1)− q(1)q′′(1) +11q(0)q′′(0) + q(0)q′′(1) + q(1)q′′(0)− 2q3(1) + 3q3(0) +3q(0)q2(1)− ( 1∫ 0 q2(t)dt)2] +O( 1 µ7 )  , (2.6) for µ ∈ T sufficiently large in absolute value. Let b(µ) be the coefficient ofe2iµ in (2.6). Using the expansion 1 1− x = 1 + x+ x2 + x3 +O(x4), x→ 0, it can be easily seen that the relation b−1(µ) = 1 + 2q(0) (2iµ)2 + 1 (2iµ)3 1∫ 0 q2(t)dt+ 1 (2ωjµ)4 [2q′′(0)− 1 2 q2(1) + 11 2 q2(0) +q(0)q(1)] + 1 (2iµ)5 [q(1)q′(1)− q(0)q′(0)− q(0)q′(1) + q(1)q′(0) +2q(0) 1∫ 0 q2(t)dt+ 2 1∫ 0 q3(t)dt− 1∫ 0 q′2(t)dt] + 1 (2iµ)6 [2q(4)(0) + 1 2 q′2(1) + 21 2 q′2(0)− q′(0)q′(1)− q(1)q′′(1) + 19q(0)q′′(0) +q(0)q′′(1) + q(1)q′′(0)− 2q3(1) + 17q3(0) + q(0)q2(1) +4q(1)q2(0) + 1 2 ( 1∫ 0 q2(t)dt)2] +O( 1 µ7 )  (2.7) Khanlar R. Mamedov, Hamza Menken / Eur. J. Pure Appl. Math,1 (2008), (51-60) 57 holds forµ ∈ T sufficiently large in absolute value. Thus, forµ ∈ T sufficiently large in absolute value, the equation∆(µ) = 0 is equivalent to the equation (iµ)−1b−1(µ)∆(µ)eiµ = 0. (2.8) Using (2.6), (2.7) and the relationsq(1) = q(0) andq′(1) 6= q′(0), from the equation (2.8), we obtain two equations µk,1 = 2kπ + q′(1)− q′(0) + 1∫ 0 q2(t)dt (4kπ)3 +O( 1 k4 ), (2.9) µk,2 = 2kπ − q′(1)− q′(0)− 1∫ 0 q2(t)dt (4kπ)3 +O( 1 k4 ). (2.10) By Rouche’s theorem, we have asymptotic expressions for the rootsµk,1 andµk,2, k = N,N+ 1, · · · , of the equations (2.9) and (2.10), respectively, whereN is a positive integer µk,1 = 2kπ + q′(1)− q′(0) (4kπ)3 +O( 1 k4 ), (2.11) µk,2 = 2kπ − q′(1)− q′(0) (4kπ)3 +O( 1 k4 ). (2.12) Note thatµk,1 andµk,2 are simple roots of the equations (2.9) and (2.10), respectively. From the relations (2.11), (2.12) and the relationsλk,1 = − µ2 k,1, λk,2 = − µ2 k,2, we obtain the formula (2.1) and observe that these eigenvalues are simple. Let us calculateU2(ϕ1(x, µk,1)) andU2(ϕ2(x, µk,1)). Since eiµk,1 − 1 = q′(1)− q′(0) + 1∫ 0 q2(t)dt (2iµk,1)3 +O( 1 µ4 k,1 ), we have U2(ϕ1(x, µk,1)) = ϕ′1(1, µk,1)− ϕ′1(0, µk,1) = iµk,1e iµk,1 [1− q(1) + q(0) (2iµk,1)2 + q′(1) + q′(0)− 1∫ 0 q2(t)dt (2iµk,1)3 +O( 1 µ4 k,1 )] −iµk,1[1− 2q(0) (2iµk,1)2 + 2q′(0) (2iµk,1)3 +O( 1 µ4 k,1 )] = q′(1)− q′(0) (2iµk,1)2 +O( 1 µ3 k,1 ). (2.13) Khanlar R. Mamedov, Hamza Menken / Eur. J. Pure Appl. Math,1 (2008), (51-60) 58 In a similar way, we obtain U2(ϕ2(x, µk,1)) = q′(1)− q′(0) (2iµk,1)2 +O( 1 µ3 k,1 ). Without loss of generality, we can assume thatq′(1)− q′(0) 6= 0. SinceU2(ϕj(x, µk,1)) 6= 0, j = 1, 2, andq′(1)−q′(0) 6= 0, we seek the eigenfunctionyk,1(x) corresponding to the eigenvalue λk,1 in the form yk,1(x) = (2iµk,1)2 2i [q′(1)− q′(0)] ∣∣∣∣ ϕ1(x, µk,1) ϕ2(x, µk,1) U2(ϕ1(x, µk,1)) U2(ϕ2(x, µk,1)) ∣∣∣∣ . (2.14) From the equalities ϕj(x, µk,1) = eµk,1ωjx [ 1 + u1(x) (2wjµk,1) + u2(x) (2wjµk,1)2 +O( 1 µ3 k,1 ) ] , j = 1, 2 and the formulas (2.13), (2.14) we obtain yk,1(x) = sinµk,1x+O( 1 µk,1 ). Therefore, the eigenfunctionyk,1(x) satisfies the asymptotic formula (2.5). In a similar way, sinceU1(ϕj(x, µk,1)) 6= 0, j = 1, 2, andq′(1)− q′(0) 6= 0, we can seek the eigenfunctionsyk,2(x) corresponding to the eigenvaluesλk,2 in the form yk,2(x) = − (2iµk,2)3 4 [q′(1)− q′(0)] ∣∣∣∣ ϕ1(x, µk,2) ϕ2(x, µk,2) U1(ϕ1(x, µk,2)) U1(ϕ2(x, µk,2)) ∣∣∣∣ . Thus, we obtain yk,2(x) = cosµk,2x+O( 1 µk,2 ). This completes the proof of the lemma. 3. The Riesz basisness inL2(0, 1) of the root functions for the periodic problem Theorem 3.1. The root functions of the boundary problem (1.1), (1.2) form a Riesz basis in L2(0, 1). Proof. The system of the root functions of the boundary problem (1.1), (1.2) is complete and minimal inL2(0, 1). The minimality of this system follows from the fact that this system has a biorthogonal system consisting of the root functions of the adjoint operator l∗(v) = v′′ + q(x)v, v(1) = v(0), v′(1) = v′(0). Khanlar R. Mamedov, Hamza Menken / Eur. J. Pure Appl. Math,1 (2008), (51-60) 59 For anyf ∈ L2(0, 1), with a direct computation we have that ∞∑ n=N |(f, yk,1)|2 < ∞, ∞∑ n=N |(f, yk,2)|2 <∞. On the other hand, the eigenfunctions of the adjoint operator have of the form υk,1(x) = 2 sin 2kπx+O( 1 k ), (3.1) υk,2(x) = 2 cos 2kπx+O( 1 k ), (3.2) and the inequalities ∞∑ n=N |(f, υk,1)|2 <∞ and ∞∑ n=N |(f, υk,2)|2 <∞ hold. According to Theorem 1.1, the root functions of the boundary problem (1.1), (1.2) form a Riesz basis inL2(0, 1). This completes the proof. 4. The basisness inL2(0, 1) of the root functions for the anti-periodic boundary-value problem Similarly, the following results are obtained for the boundary problem (1.1), (1.3). Lemma 4.1. All eigenvalues of the boundary value problem (1.1), (1.3), starting from some num- ber, are simple and form two infinite sequenceλk,1, λk,2, k = N,N+1, · · · , whereN is a positive integer and λk,1 = − [(2k + 1)π]2 + q′(1)− q′(0)− 1∫ 0 q2(x)dx [2(2k + 1)π]2 +O( 1 k3 ), (4.1) λk,2 = − [(2k + 1)π]2 − q′(1)− q′(0) + 1∫ 0 q2(x)dx [2(2k + 1)π]2 +O( 1 k3 ), (4.2) and the corresponding eigenfunctions are of the form yk,1(x) = sin(2k + 1)πx+O( 1 k ), (4.3) yk,2(x) = cos(2k + 1)πx+O( 1 k ). (4.4) Proof. In the anti-periodic case, in a similar way to the proof Lemma 2.1, we have the relations eiµ + 1 = q′(1)− q′(0)− 1∫ 0 q2(x)dx (2iµ)3 +O( 1 µ4 ), eiµ + 1 = − q′(1)− q′(0)− 1∫ 0 q2(x)dx (2iµ)3 +O( 1 µ4 ). REFERENCES 60 From these relations we can obtain (4.1) and (4.2). Again in a similar way to the proof Lemma 2.1, we obtain U1(ϕ1(x, µk,1)) = ϕ1(1, µk,1) + ϕ1(0, µk,1) = 2[q′(1)− q′(0)] (2iµk,1)3 +O( 1 µ4 k,1 ), U1(ϕ2(x, µk,1)) = ϕ2(1, µk,1) + ϕ2(0, µk,1) = −2[q′(1)− q′(0)] (2iµk,1)3 +O( 1 µ4 k,1 ). SinceU1(ϕj(x, µk,1)) 6= 0,j = 1, 2, we can seek the eigenfunctionyk,1(x) corresponding to the eigenvalueλk,1 in the form yk,1(x) = − (2iµk,1)3 [q′(1)− q′(0)] ∣∣∣∣ ϕ1(x, µk,1) ϕ2(x, µk,1) U1(ϕ1(x, µk,1)) U1(ϕ2(x, µk,1)) ∣∣∣∣ . Hence, we have yk,1(x) = sin(2k + 1)πx+O( 1 k ), i.e., the formula (4.3) satisfies. In similar way we can obtain the formula (4.4). Theorem 4.1. The root functions of the boundary problem (1.1), (1.3) form a Riesz basis in L2(0, 1). References [1] M. A. Naimark, Linear Differential Operators, Part I, Frederick Ungar Pub. Co., New York, 1967. [2] G. M. Kesel’man, On the unconditional convergence of expansions in the eigenfunc- tions of some differential operators,Izv. Vyssh. Uchebn. Zaved. Mat. [Soviet Math. (Iz. 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