R E S E A R C H
Open Access
On the basis property of the root functions of
some class of non-self-adjoint Sturm-Liouville
operators
Cemile Nur
*and Oktay A Veliev
*Correspondence:
Department of Mathematics, Dogus University, Acıbadem, Kadiköy, Istanbul, Turkey
Abstract
We obtain the asymptotic formulas for the eigenvalues and eigenfunctions of the Sturm-Liouville operators with some regular boundary conditions. Using these formulas, we find sufficient conditions on the potential q such that the root functions of these operators do not form a Riesz basis.
MSC: 34L05; 34L20
Keywords: asymptotic formulas; regular boundary conditions; Riesz basis
1 Introduction and preliminary facts
Let T, T, T, and Tbe the operators generated in L[, ] by the differential expression
l(y) = –y+ q(x)y ()
and the following boundary conditions:
y+ βy= , y– y= , ()
y+ βy= , y+ y= , ()
y– y= , y+ αy= ()
and
y+ y= , y+ αy= , ()
respectively, where q(x) is a complex-valued summable function on [, ], β= ± and α = ±.
In conditions (), (), (), and () if β = , β = –, α = , and α = –, respectively, then any
λ∈ C is an eigenvalue of infinite multiplicity. In () and () if β = – and α = – then they
are periodic boundary conditions; in () and () if β = and α = then they are antiperiodic boundary conditions.
These boundary conditions are regular but not strongly regular. Note that, if the bound-ary conditions are strongly regular, then the root functions form a Riesz basis (this result was proved independently in [, ] and []). In the case when an operator is associated ©2014 Nur and Veliev; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
with the regular but not strongly regular boundary conditions, the root functions gen-erally do not form even a usual basis. However, Shkalikov [, ] proved that they can be combined in pairs, so that the corresponding -dimensional subspaces form a Riesz basis of subspaces.
In the regular but not strongly regular boundary conditions, periodic and antiperiodic boundary conditions are the ones more commonly studied. Therefore, let us briefly de-scribe some historical developments related to the Riesz basis property of the root func-tions of the periodic and antiperiodic boundary value problems. First results were ob-tained by Kerimov and Mamedov []. They established that, if
q∈ C[, ], q()= q(),
then the root functions of the operator L(q) form a Riesz basis in L[, ], where L(q) de-notes the operator generated by () and the periodic boundary conditions.
The first result in terms of the Fourier coefficients of the potential q was obtained by Dernek and Veliev []. They proved that if the conditions
lim n→∞ ln|n| nqn = , () qn∼ q–n ()
hold, then the root functions of L(q) form a Riesz basis in L[, ], where qn=: (q, eiπ nx)
is the Fourier coefficient of q and everywhere, without loss of generality, it is assumed that q= . Here (·, ·) denotes the inner product in L[, ] and an∼ bnmeans that an=
O(bn) and bn= O(an) as n→ ∞. Makin [] improved this result. Using another method he
proved that the assertion on the Riesz basis property remains valid if condition () holds, but condition () is replaced by a less restrictive one: q∈ Ws
[, ],
q(k)() = q(k)(), ∀k = , , . . . , s –
holds and|qn| > cn–s–with some c > for sufficiently large n, where s is a nonnegative integer. Besides, some conditions which imply the absence of the Riesz basis property were presented in []. Shkalilov and Veliev obtained in [] more general results, which cover all results discussed above.
The other interesting results as regards periodic and antiperiodic boundary conditions were obtained in [–].
The basis properties of other some operators with regular but not strongly regular boundary conditions are studied in [–]. It was proved in [] that the system of the root functions of the operator generated by () and the boundary conditions
y() – (–)σy() + γ y() = ,
y() – (–)σy() =
forms an unconditional basis of the space L[, ], where q(x) is an arbitrary complex-valued function from the class L[, ], γ is an arbitrary nonzero complex number and σ =
, . Kerimov and Kaya [, ] investigated the basis properties of fourth order differential operators with some regular boundary conditions.
In this paper we prove that if lim
n→∞ ln|n|
nsn
= , ()
where sk= (q, sin π kx), then the large eigenvalues of the operators Tand Tare simple. Moreover, if there exists a sequence{nk} such that () holds when n is replaced by nk, then
the root functions of these operators do not form a Riesz basis. Similarly, if lim n→∞ ln|n| nsn+ = , ()
then the large eigenvalues of the operators Tand Tare simple and if there exists a se-quence{nk} such that () holds when n is replaced by nk, then the root functions of these
operators do not form a Riesz basis.
Moreover, we obtain asymptotic formulas of arbitrary precision for the eigenvalues and eigenfunctions of the operators T, T, T, and T.
2 Main results
We will focus only on the operator T. The investigations of the operators T, T, and T are similar. It is well known that (see (a) and (b) on page of []) the eigenvalues of the operators T(q) consist of the sequences{λn,}, {λn,} satisfying
λn,j= (nπ )+ O
n/ ()
for j = , . From this formula one can easily obtain the following inequality: λn,j– (π k)=(n – k)π(n + k)π+ O
n> n ()
for j = , ; k= n; k = , , . . . , and n ≥ N, where N denotes a sufficiently large positive integer, that is, N .
Let us denote by Tj() the operator Tjwhen q(x) = . The eigenvalues of the operator
T() are λn= (π n)for n = , , . . . . The eigenvalue is simple and the corresponding
eigenfunction is . The eigenvalues λn= (π n)for n = , , . . . are double and the
corre-sponding eigenfunctions and associated functions are
yn(x) = cos π nx and φn(x) = β + β – x sinπ nx π n , ()
respectively. Note that for any constant c, φn(x) + cyn(x) is also an associated function
corresponding to λn, since one can easily verify that it satisfies the equation and boundary
conditions for the associated functions. It can be shown that the adjoint operator T∗() is associated with the boundary conditions
y+ βy= , y– y= .
It is easy to see that is a simple eigenvalue of T∗() and the corresponding eigenfunction is y∗(x) = x –+β . The other eigenvalues λ∗n= (π n) for n = , , . . . , are double and the
corresponding eigenfunctions and associated functions are y∗n(x) = sin π nx and φn∗(x) = x– + β cosπ nx π n () respectively. Let ϕn(x) := π n(β + ) β– φn(x) = (β + ) β– β + β – x sinπ nx () and ϕn∗(x) :=π n(β + ) β– φ ∗ n(x) = (β + ) β– x– + β cosπ nx ()
(see () and ()). The system of the root functions of T∗() can be written as{fn: n∈ Z},
where
f–n= sin π nx, ∀n > and fn= ϕn∗(x), ∀n ≥ . ()
One can easily verify that it forms a basis in L[, ] and the biorthogonal system{gn: n∈
Z} is the system of the root functions of T(), where
g–n= ϕn, ∀n > and gn= cos π nx, ∀n ≥ , ()
since (fn, gm) = δn,m.
To obtain the asymptotic formulas for the eigenvalues λn,jand the corresponding nor-malized eigenfunctions n,j(x) of T(q) we use () and the well-known relations
λN,j– (π n) ( N,j, sin π nx) = (q N,j, sin π nx) () and λN,j– (π n) N,j, ϕn∗ – γn( N,j, sin π nx) = q N,j, ϕ∗n , () where γ= π (β + ) β– ,
which can be obtained by multiplying both sides of the equality –( N,j)+ q(x) N,j= λN,j N,j
by sin π nx and ϕ∗n, respectively. It follows from () and () that ( N,j, sin π nx) = (q N,j, sin π nx) λN,j– (π n) ; N= n, () N,j, ϕn∗ =γn(q N,j, sin π nx) (λN,j– (π n)) + (q N,j, ϕ ∗ n) λN,j– (π n) ; N= n. ()
Moreover, we use the following relations: ( N,j, q sin π nx) = ∞ n= (qϕn, sin π nx)( N,j, sin π nx) + (q cos π nx, sin π nx) N,j, ϕ∗n , () N,j, qϕ∗n = ∞ n= qϕn, ϕ ∗ n ( N,j, sin π nx) + q cosπ nx, ϕn∗ N,j, ϕn∗ , () (q N,j, sin π nx)< M, () q N,j, ϕn∗< M ()
for N , where M = sup |qn|. These relations are obvious for q ∈ L(, ), since to obtain () and () we can use the decomposition of q sin π nx and qϕn∗by the basis (). For
q∈ L(, ) see Lemma of [].
To obtain the asymptotic formulas for the eigenvalues and eigenfunctions we iterate () and () by using () and (). First let us prove the following obvious asymptotic formulas, namely (), for the eigenfunctions n,j. The expansion of n,jby the basis () can be written in the form
n,j= un,jϕn(x) + vn,jcosπ nx + hn,j(x), () where un,j= ( n,j, sin π nx), vn,j= n,j, ϕn∗ , () hn,j(x) = ∞ k= k=n ( n,j, sin π kx)ϕk(x) + n,j, ϕk∗ cosπ kx ,
and ϕn(x), ϕn∗(x) are defined in () and (), respectively. Using (), (), (), and ()
one can readily see that there exists a constant C such that suphn,j(x) ≤C k=n |λn,j– (π k)| + n |(λn,j– (π k))| = O lnn n . ()
Hence by () and () we obtain
n,j= un,jϕn(x) + vn,jcosπ nx + O lnn n . ()
Since n,jis normalized, we have = n,j = ( n,j, n,j) =|un,j| ϕn +|vn,j| cos πnx + un,jvn,j(ϕn, cos π nx) + vn,jun,j(cos π nx, ϕn) + O lnn n = |β|– Re β + |β – | |un,j|+ |vn,j| + O lnn n ,
that is, a|un,j|+ |vn,j| = + O lnn n , () where a= |β|– Re β + |β – | .
Note that a= , since |β|+ >|β| and by () we see that at least one of u
n,jand vn,jis different from zero.
Now let us iterate (). Using () in () we get λn,j– (π n) ( n,j, sin π nx) = ∞ n=
(qϕn, sin π nx)( n,j, sin π nx) + (q cos π nx, sin π nx)
n,j, ϕn∗
.
Isolating the terms in the right-hand side of this equality containing the multiplicands ( n,j, sin π nx) and ( n,j, ϕn∗) (i.e., the case n = n), using () and () for the terms ( n,j, sin π nx) and ( n,j, ϕn∗), respectively (in the case n= n), we obtain
λn,j– (π n)– (qϕn, sin π nx)
( n,j, sin π nx) – (q cos π nx, sin π nx) n,j, ϕn∗ = ∞ n= n=n
(qϕn, sin π nx)( n,j, sin π nx) + (q cos π nx, sin π nx)
n,j, ϕn∗ = n a(λn,j) q(x) n,j, sin π nx + b(λn,j) q(x) n,j, ϕn∗ , where a(λn,j) = (qϕn, sin π nx) λn,j– (π n) +γn(q cos π nx, sin π nx) (λn,j– (π n)) , b(λn,j) = (q cos π nx, sin π nx) λn,j– (π n) .
Using () and () for the terms (q n,j, sin π nx) and (q n,j, ϕn∗) of the last summation
we obtain
λn,j– (π n)– (qϕn, sin π nx)
( n,j, sin π nx) – (q cos π nx, sin π nx) n,j, ϕn∗ = n a(λn,j)(q n,j, sin π nx) + b(λn,j) q n,j, ϕ∗n = n a ∞ n= (qϕn, sin π nx)( n,j, sin π nx) + (q cos π nx, sin π nx) n,j, ϕn∗
+ n b ∞ n= qϕn, ϕ ∗ n ( n,j, sin π nx) +q cosπ nx, ϕn∗ n,j, ϕn∗ .
Now isolating the terms for n= n we get
λn,j– (π n)– (qϕn, sin π nx)
( n,j, sin π nx) – (q cos π nx, sin π nx) n,j, ϕn∗ = n a(qϕn, sin π nx) + b qϕn, ϕ∗n ( n,j, sin π nx) + n a(q cos π nx, sin π nx) + b q cosπ nx, ϕn∗ n,j, ϕ∗n(x) = n,n a(qϕn, sin π nx) + b qϕn, ϕ ∗ n ( n,j, sin π nx) + n,n a(q cos π nx, sin π nx) + b q cosπ nx, ϕn∗ n,j, ϕ∗n .
Here and below the summations are taken under the conditions ni= n and ni= , , . . . for
i= , , . . . . Introduce the notations
C=: a, M=: b, Ck+=: Ckak++ MkAk+, Mk+=: Ckbk++ MkBk+; k= , , . . . , where ak+= ak+(λn,j) = (qϕnk+, sin π nkx) λn,j– (π nk+) +γnk+(q cos π nk+x, sin π nkx) (λn,j– (π nk+)) , bk+= bk+(λn,j) = (q cos π nk+x, sin π nkx) λn,j– (π nk+) , Ak+= Ak+(λn,j) = (qϕnk+, ϕn∗k) λn,j– (π nk+) +γnk+(q cos π nk+x, ϕ ∗ nk) (λn,j– (π nk+)) , Bk+= Bk+(λn,j) = (q cos π nk+x, ϕ∗nk) λn,j– (π nk+) .
Using these notations and repeating this iteration k times we get λn,j– (π n)– (qϕn, sin π nx) – Ak(λn,j) ( n,j, sin π nx) =(q cos π nx, sin π nx) + Bk(λn,j) n,j, ϕn∗(x) + Rk, () where Ak(λn,j) = k m= αm(λn,j), Bk(λn,j) = k m= βm(λn,j), αk(λn,j) = n,...,nk Ck(qϕn, sin π nkx) + Mk qϕn, ϕn∗k ,
βk(λn,j) = n,...,nk Ck(q cos π nx, sin π nkx) + Mk q cosπ nx, ϕn∗k , Rk= n,...,nk+ Ck+(q n,j, sin π nk+x) + Mk+ q n,j, ϕ∗nk+ .
It follows from (), (), (), and () that
αk(λ) = O ln|n| n k , βk(λ) = O ln|n| n k , Rk(λ) = O ln|n| n k+ ()
for λ = λn,jand for all λ∈ U(n), where U(n) = {λ : |λ – (πn)| ≤ n}. Therefore letting k tend to infinity, we obtain
λn,j– (π n)– Qn– A(λn,j) un,j= Pn+ B(λn,j) vn,j, where
Pn= (q cos π nx, sin π nx), Qn= (qϕn, sin π nx), ()
A(λ) = ∞ m= αm(λ), B(λ) = ∞ m= βm(λ) and by () we have A(λ) = O ln|n| n , B(λ) = O ln|n| n ()
for λ = λn,jand for all λ∈ U(n).
Thus iterating () we obtain (). Now iterating () instead of (), using () and (), and arguing as in the previous iteration, we get
λn,j– (π n)– P∗n– Ak(λn,j) vn,j= γn+ Q∗n+ Bk(λn,j) un,j+ Rk, () where P∗n=q cosπ nx, ϕn∗, Q∗n=qϕn, ϕn∗ , () Ak(λn,j) = k m= αm(λn,j), Bk(λn,j) = k m= βm (λn,j), αk(λn,j) = n,...,nk Ck(q cos π nx, sin π nkx) + Mk q cosπ nx, ϕn∗k , βk(λn,j) = n,...,nk Ck(qϕn, sin π nkx) + Mk qϕn, ϕ∗nk , Rk= n,...,nk+ Ck+(q n,j, sin π nk+x) + Mk+ q n,j, ϕ∗nk+ ,
Ck+= Ckak++ MkAk+, M k+= Ckbk++ MkBk+; k= , , . . . , C= A(λn,j) = (qϕn, ϕn∗) λn,j– (π n) +γn(q cos π nx, ϕ ∗ n) (λn,j– (π n)) , M= B(λn,j) = (q cos π nx, ϕ∗n) λn,j– (π n) .
Similar to () one can verify that
αk(λ) = O ln|n| n k , βk(λ) = O ln|n| n k , Rk(λ) = O ln|n| n k+ ()
for λ = λn,jand for all λ∈ U(n). Now letting k tend to infinity in (), we obtain λn,j– (π n)– P∗n– A(λn,j) vn,j= γn+ Q∗n+ B(λn,j) un,j, where A(λn,j) = ∞ m= αm(λn,j), B(λn,j) = ∞ m= βm(λn,j) and by () we have A(λ) = O ln|n| n , B(λ) = O ln|n| n ()
for λ = λn,jand for all λ∈ U(n).
To get the main results of this paper we use the following system of equations, obtained above, with respect to un,jand vn,j,
λn,j– (π n)– Qn– A(λn,j) un,j= Pn+ B(λn,j) vn,j, () λn,j– (π n)– P∗n– A(λn,j) vn,j= γn+ Q∗n+ B(λn,j) un,j, () where Qn= – (β + ) β– xq(x) dx +(β + ) β– xq(x), cos π nx – β β– q(x), cos π nx () = –(β + ) β– xq(x) dx + o(), () P∗n=(β + ) β– xq(x) dx +(β + ) β– xq(x), cos π nx– β– q(x), cos π nx () =(β + ) β– xq(x) dx + o(), ()
Pn= (q, sin π nx) = o(), () Q∗n= β+ β– q(x) β + β – x x– + β sinπ nx dx = o() ()
(see () and ()). Note that (), () with (), () give λn,j– (π n)– Qn+ O ln|n| n un,j= Pn+ O ln|n| n vn,j, () λn,j– (π n)– Pn∗+ O ln|n| n vn,j= γn+ Q∗n+ O ln|n| n un,j. () Introduce the notations
cn= (q, cos π nx), sn= (q, sin π nx), cn,= (xq, cos π nx), sn,= (xq, sin π nx), () cn,= xq, cos π nx, sn,= xq, sin π nx. Then, by ()-() and () we have
Qn= – (β + ) β– xq(x) dx +(β + ) β– cn,– β β– cn, () P∗n=(β + ) β– xq(x) dx +(β + ) β– cn,– β– cn, () Pn= sn, () Q∗n= – β+ β– sn,+ β+ β– sn,– β (β – )sn. ()
Theorem The following statements hold:
(a) Any eigenfunction n,jof Tcorresponding to the eigenvalue λn,jdefined in()
satisfies n,j=
√
cos π nx + On–/. ()
Moreover, there exists N such that for all n > N the geometric multiplicity of the
eigenvalue λn,jis .
(b) A complex number λ∈ U(n), where U(n) is defined in (), is an eigenvalue of Tif
and only if it is a root of the equation
λ– (π n)– Qn– A(λ) λ– (π n)– Pn∗– A(λ) –Pn+ B(λ) γn+ Q∗n+ B(λ) = . ()
Moreover, λ∈ U(n) is a double eigenvalue of Tif and only if it is a double root of().
Proof (a) By () the left-hand side of () is O(n/), which implies that u
n,j= O(n–/). Therefore from () we obtain (). Now suppose that there are two linearly independent
eigenfunctions corresponding to λn,j. Then there exists an eigenfunction satisfying
n,j= √
sin π nx + o(), which contradicts ().
(b) First we prove that the large eigenvalues λn,jare the roots of (). It follows from (), (), and () that vn,j= . If un,j= then multiplying () and () side by side and then canceling vn,jun,jwe obtain (). If un,j= then by () and () we have Pn+ B(λn,j) = and λn,j– (π n)– Pn∗– A(λn,j) = , which means that () holds. Thus in any case λn,jis a root of ().
Now we prove that the roots of () lying in U(n) are the eigenvalues of T. Let F(λ) be the left-hand side of (), which can be written as
F(λ) =λ– (π n)–Qn+ A(λ) + Pn∗+ A(λ) λ– (π n) +Qn+ A(λ) P∗n+ A(λ)–Pn+ B(λ) γn+ Q∗n+ B(λ) () and G(λ) =λ– (π n).
Using () and (), one can easily verify that the inequality F(λ) – G(λ)<G(λ)
holds for all λ from the boundary of U(n). Since the function G(λ) has two roots in the set U(n), by the Rouche theorem we find that F(λ) has two roots in the same set. Thus T has two eigenvalues (counting with multiplicities) lying in U(n) that are the roots of (). On the other hand, () has preciously two roots (counting with multiplicities) in U(n). Therefore λ∈ U(n) is an eigenvalue of Tif and only if () holds.
If λ∈ U(n) is a double eigenvalue of T then it has no other eigenvalues in U(n) and hence () has no other roots. This implies that λ is a double root of (). By the same way one can prove that if λ is a double root of () then it is a double eigenvalue of T.
Let us consider () in detail. By () we have
F(λ) = . ()
If we substitute t =: λ – (π n)in (), then it becomes
t–Qn+ A(λ) + P∗n+ A(λ) t+Qn+ A(λ) P∗n+ A(λ) –Pn+ B(λ) γn+ Q∗n+ B(λ) = . ()
The solutions of () are
t,=
(Qn+ P∗n+ A + A)±
√
(λ)
where (λ) =Qn+ P∗n+ A + A – (Qn+ A) P∗n+ A+ (Pn+ B) γn+ Q∗n+ B , which can be written in the form
(λ) =Qn– Pn∗+ A – A + (Pn+ B) γn+ Q∗n+ B () and, as we shall see below,√ (λ) can be defined as analytic function on U(n). Clearly the eigenvalue λn,jis a root either of the equation
λ= (π n)+ Qn+ P∗n+ A + A – (λ) () or of the equation λ= (π n)+ Qn+ P∗n+ A + A + (λ) . ()
Now let us examine (λ) and√ (λ) in detail. If () holds then one can readily see from (), (), ()-(), and () that
(λ) = γnsn
+ o() ()
for λ∈ U(n). By () there exists an appropriate choice of branch of√ (λ) (depending on n) which is analytic on U(n). Taking into account (), (), (), (), and (), we see that () and () have the form
λ= (π n)– √ γ √ nsn + o(), () λ= (π n)+ √ γ √ nsn + o(). ()
Theorem If() holds, then the large eigenvalues λn,jare simple and satisfy the following
asymptotic formulas: λn,j= (π n)+ (–)j √ γ √ nsn + o() ()
for j= , . Moreover, if there exists a sequence{nk} such that () holds when n is replaced
by nk, then the root functions of Tdo not form a Riesz basis.
Proof To prove that the large eigenvalues λn,jare simple let us show that one of the eigen-values, say λn,satisfies () for j = and the other λn,satisfies () for j = . Let us prove that each of () and () has a unique root in U(n) by proving that
(π n)+ Qn+ Pn∗+ A + A ± (λ)
Ksuch that
A(λ) – A(μ)< K|λ – μ|, A(λ) – A(μ)< K|λ – μ|, ()
(λ) – (μ)< K|λ – μ| ()
for λ, μ∈ U(n), where K+ K+ K< . The proof of () is similar to the proof of () of the paper [].
Now let us prove (). By () and () we have
(λ)–= o()
for λ∈ U(n). On the other hand arguing as in the proof of () of the paper [] we get
d
dλ (λ) = O().
Hence for the large values of n we have
d dλ (λ) = d dλ (λ) √ (λ)= o()
for λ∈ U(n). Thus by the fixed point theorem, each of () and () has a unique root λ and λin U(n) respectively. Clearly by () and (), we have λ= λwhich implies that () has two simple roots in U(n). Therefore by Theorem (b), λand λare the eigenvalues of Tlying in U(n), that is, they are λn,and λn,, which proves the simplicity of the large eigenvalues and the validity of ().
If there exists a sequence{nk} such that () holds when n is replaced by nk, then by
Theorem (a)
( nk,, nk,) = + O
n–/k .
Now it follows from the theorems of [, ] (see also Lemma of []) that the root functions
of Tdo not form a Riesz basis.
Now let us consider the operators T, T, and T. First we consider the operator T. It is well known that (see (a) and (b) on page of []) the eigenvalues of the operators T(q) consist of the sequences{λn,,}, {λn,,} satisfying () when λn,jis replaced by λn,j,. The eigenvalues, eigenfunctions and associated functions of T() are
λn,= (π n); n= , , , . . . , y,(x) = x – α + α, yn,(x) = sin π nx; n= , , . . . , φn,(x) = x– α + α cosπ nx π n ; n= , , . . . ,
respectively. The biorthogonal systems analogous to () and () are cosπ nx,( + α) – α + α– x sinπ nx ∞ n= , ()
sinπ nx,( + α) – α x– α + α cosπ nx ∞ n= , () respectively.
Analogous formulas to () and () are λN,j,– (π n) ( N,j,, cos π nx) = (q N,j,, cos π nx), () λN,j,– (π n) N,j,, ϕn∗, – γn( N,j,, cos π nx) = q(x) N,j,, ϕn∗, , () respectively, where γ= π ( + α) – α .
Instead of ()-() using ()-() and arguing as in the proofs of Theorem and The-orem we obtain the following results for T.
Theorem If() holds, then the large eigenvalues λn,j,are simple and satisfy the following
asymptotic formulas: λn,j,= (π n)+ (–)j √ γ √ nsn + o()
for j= , . The eigenfunctions n,j,corresponding to λn,j,obey
n,j,= √
sin π nx + On–/.
Moreover, if there exists a sequence{nk} such that () holds when n is replaced by nk, then
the root functions of Tdo not form a Riesz basis.
Now let us consider the operator T. It is well known that (see (a) and (b) on page of []) the eigenvalues of the operators T(q) consist of the sequences{λn,,}, {λn,,} satisfying
λn,j,= (nπ + π )+ O
n/ ()
for j = , . The eigenvalues, eigenfunctions, and associated functions of T() are (π + π n), yn,(x) = cos(n + )π x, φn,(x) = β β– – x sin(n + )π x (n + )π
for n = , , , . . . , respectively. The biorthogonal systems analogous to () and () are sin(n + )π x,(β – ) β+ x+ β– cos(n + )π x ∞ n= , () cos(n + )π x,(β – ) β+ β β– – x sin(n + )π x ∞ n= , () respectively.
Analogous formulas to () and () are λN,j,– (n + )π N,j,, sin(n + )π x =q N,j,, sin(n + )π x , () λN,j,– (n + )π N,j,, ϕ∗n, – (n + )γ N,j,, sin(n + )π x =q N,j,, ϕ∗n, , () respectively, where γ= π (β – ) β+ .
Instead of ()-() using ()-() and arguing as in the proofs of Theorem and Theo-rem we obtain the following results for T.
Theorem If() holds, then the large eigenvalues λn,j,are simple and satisfy the following
asymptotic formulas: λn,j,= (n + )π+ (–)j √ γ (n + )sn+ + o()
for j= , . The eigenfunctions n,j,corresponding to λn,j,obey
n,j,= √
cos(n + )π x + On–/.
Moreover, if there exists a sequence{nk} such that () holds when n is replaced by nk, then
the root functions of Tdo not form a Riesz basis.
Lastly we consider the operator T. It is well known that (see (a) and (b) on page of []) the eigenvalues of the operators T(q) consist of the sequences{λn,,}, {λn,,} satisfying () when λn,j,is replaced by λn,j,. The eigenvalues, eigenfunctions, and asso-ciated functions of T() are
λn,= (π + π n), yn,(x) = sin(n + )π x, φn,(x) = α – α+ x cos(n + )π x (n + )π
for n = , , , . . . , respectively. The biorthogonal systems analogous to () and () are cos(n + )π x,( – α) + α – α– x sin(n + )π x ∞ n= , () sin(n + )π x,( – α) + α α – α + x cos(n + )π x ∞ n= , () respectively.
Analogous formulas to () and () are λN,j,– (π + π n) N,j,, cos(n + )π x =q(x) N,j,, cos(n + )π x , ()
λN,j,– (n + )π N,j,, ϕn∗, – (n + )γ N,j,, cos(n + )π x =q N,j,, ϕn∗, , () respectively, where γ= π ( – α) + α .
Instead of ()-() using ()-() and arguing as in the proofs of Theorem and Theo-rem we obtain the following results for T.
Theorem If() holds, then the large eigenvalues λn,j,are simple and satisfy the following
asymptotic formulas: λn,j,= (n + )π+ (–)j √ γ (n + )sn+ + o()
for j= , . The eigenfunctions n,j,corresponding to λn,j,obey
n,j,= √
sin(n + )π x + On–/.
Moreover, if there exists a sequence{nk} such that () holds when n is replaced by nk, then
the root functions of Tdo not form a Riesz basis. Now suppose that
xq(x) dx= . () If sn+ B = o n , ()
where B is defined by (), then one can readily see from (), (), (), and ()-() that there exists a positive constant K such that
(λ)> K
for λ∈ U(n) and for the large values of n. Therefore arguing as in the proof of Theorem , we obtain the following.
Theorem Suppose that() holds. If () holds, then the large eigenvalues of the
oper-ator T are simple. Moreover, if there exists a sequence{nk} such that () holds when n
is replaced by nk, then the root functions of T do not form a Riesz basis. Similar results
continue to hold for the operators T, T, and T.
Remark Since the eigenvalues λn,and λn,are the fixed points of () and () respec-tively, using the fixed point iteration one can determine these eigenvalues with arbitrary
precision. Moreover, using these better approximations of the eigenvalues, one can also determine the better approximations for the eigenfunctions of the operator T. Similar results can be obtained for the operators T, T, and T.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript. Received: 2 January 2014 Accepted: 28 February 2014 Published: 15 March 2014
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doi:10.1186/1687-2770-2014-57
Cite this article as: Nur and Veliev: On the basis property of the root functions of some class of non-self-adjoint Sturm-Liouville operators. Boundary Value Problems 2014 2014:57.
