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On calculation of eigenvalues and eigenfunctions of a Sturm-Liouville type problem with retarded argument which contains a spectral parameter in the boundary condition

Abstract

In this study, a discontinuous boundary-value problem with retarded argument which contains a spectral parameter in the boundary condition and with transmission conditions at the point of discontinuity is investigated. We obtained asymptotic formulas for the eigenvalues and eigenfunctions.

MSC (2010): 34L20; 35R10.

1 Introduction

Boundary-value problems for differential equations of the second order with retarded argument were studied in [15], and various physical applications of such problems can be found in [2].

The asymptotic formulas for the eigenvalues and eigenfunctions of boundary problem of Sturm-Liouville type for second order differential equation with retarded argument were obtained in [5].

The asymptotic formulas for the eigenvalues and eigenfunctions of Sturm-Liouville problem with the spectral parameter in the boundary condition were obtained in [6].

In the articles [79], the asymptotic formulas for the eigenvalues and eigenfunctions of discontinuous Sturm-Liouville problem with transmission conditions and with the boundary conditions which include spectral parameter were obtained.

In this article, we study the eigenvalues and eigenfunctions of discontinuous boundary-value problem with retarded argument and a spectral parameter in the boundary condition. Namely, we consider the boundary-value problem for the differential equation

p ( x ) y ( x ) + q ( x ) y ( x - Δ ( x ) ) + λ y ( x ) = 0
(1)

on 0 , π 2 π 2 , π , with boundary conditions

y ( 0 ) = 0 ,
(2)
y ( π ) + λ y ( π ) = 0 ,
(3)

and transmission conditions

γ 1 y π 2 - 0 = δ 1 y π 2 + 0 ,
(4)
γ 2 y π 2 - 0 = δ 2 y π 2 + 0 ,
(5)

where p ( x ) = p 1 2 if x 0 , π 2 and p ( x ) = p 2 2 if x π 2 , π , the real-valued function q(x) is continuous in 0 , π 2 π 2 , π and has a finite limit q ( π 2 ± 0 ) = lim x π 2 ± 0 q ( x ) , the real-valued function Δ(x) ≥ 0 continuous in 0 , π 2 π 2 , π and has a finite limit Δ ( π 2 ± 0 ) = lim x π 2 ± 0 Δ ( x ) ,x-Δ ( x ) 0, if x 0 , π 2 ; x-Δ ( x ) π 2 if x π 2 , π ; λ is a real spectral parameter; p1, p2, γ1, γ2, δ1, δ2 are arbitrary real numbers and |γ i | + |δi| ≠ 0 for i = 1, 2. Also, γ1δ2p1 = γ2δ1p2 holds.

It must be noted that some problems with transmission conditions which arise in mechanics (thermal condition problem for a thin laminated plate) were studied in [10].

Let w1(x, λ) be a solution of Equation 1 on [ 0 , π 2 ] , satisfying the initial conditions

w 1 ( 0 , λ ) = 0 , w 1 ( 0 , λ ) = - 1 .
(6)

The conditions (6) define a unique solution of Equation 1 on [ 0 , π 2 ] [2, p. 12].

After defining above solution, we shall define the solution w2 (x, λ) of Equation 1 on [ π 2 , π ] by means of the solution w1(x, λ) by the initial conditions

w 2 π 2 , λ = γ 1 δ 1 - 1 w 1 π 2 , λ , ω 2 π 2 , λ = γ 2 δ 2 - 1 ω 1 π 2 , λ .
(7)

The conditions (7) are defined as a unique solution of Equation 1 on [ π 2 , π ] .

Consequently, the function w (x, λ) is defined on 0 , π 2 π 2 , π by the equality

w ( x , λ ) = ω 1 ( x , λ ) , x 0 , π 2 , ω 2 ( x , λ ) , x π 2 , π

is a such solution of Equation 1 on 0 , π 2 π 2 , π ; which satisfies one of the boundary conditions and both transmission conditions.

Lemma 1. Let w (x, λ) be a solution of Equation 1 and λ > 0. Then, the following integral equations hold:

w 1 ( x , λ ) = - p 1 s sin s p 1 x - 1 s 0 x q ( τ ) p 1 sin s p 1 ( x - τ ) w 1 ( τ - Δ ( τ ) , λ ) d τ s = λ , λ > 0 ,
(8)
w 2 ( x , λ ) = γ 1 δ 1 w 1 π 2 , λ cos s p 2 x - π 2 + γ 2 p 2 w 1 ( π 2 , λ ) s δ 2 sin s p 2 x - π 2 - 1 s π 2 x q ( τ ) p 2 sin s p 2 ( x - τ ) w 2 ( τ - Δ ( τ ) , λ ) d τ s = λ , λ > 0 .
(9)

Proof. To prove this, it is enough to substitute - s 2 p 1 2 ω 1 ( τ , λ ) - ω 1 ( τ , λ ) and - s 2 p 2 2 ω 2 ( τ , λ ) - ω 2 ( τ , λ ) instead of - q ( τ ) p 1 2 ω 1 ( τ - Δ ( τ ) , λ ) and - q ( τ ) p 2 2 ω 2 ( τ - Δ ( τ ) , λ ) in the integrals in (8) and (9), respectively, and integrate by parts twice.

Theorem 1. The problem (1)-(5) can have only simple eigenvalues.

Proof. Let λ ̃ be an eigenvalue of the problem (1)-(5) and

ũ ( x , λ ̃ ) = ũ 1 ( x , λ ̃ ) , x 0 , π 2 , ũ 2 ( x , λ ̃ ) , x π 2 , π

be a corresponding eigenfunction. Then, from (2) and (6), it follows that the determinant

W ũ 1 ( 0 , λ ̃ ) , w 1 ( 0 , λ ̃ ) = ũ 1 ( 0 , λ ̃ ) 0 ũ 1 ( 0 , λ ̃ ) - 1 = 0 ,

and by Theorem 2.2.2 in [2], the functions ũ 1 ( x , λ ̃ ) and w 1 ( x , λ ̃ ) are linearly dependent on [ 0 , π 2 ] . We can also prove that the functions ũ 2 ( x , λ ̃ ) and w 2 ( x , λ ̃ ) are linearly dependent on [ π 2 , π ] . Hence,

ũ 1 ( x , λ ̃ ) = K i w i ( x , λ ̃ ) ( i = 1 , 2 )
(10)

for some K1 ≠ 0 and K2 ≠ 0. We must show that K1 = K2. Suppose that K1K2. From the equalities (4) and (10), we have

γ 1 ũ π 2 - 0 , λ ̃ - δ 1 ũ π 2 + 0 , λ ̃ = γ 1 ũ 1 π 2 , λ ̃ - δ 1 ũ 2 π 2 , λ ̃ = γ 1 K 1 w 1 π 2 , λ ̃ - δ 1 K 2 w 2 π 2 , λ ̃ = γ 1 K 1 δ 1 γ 1 - 1 w 2 π 2 , λ ̃ - δ 1 K 2 w 2 π 2 , λ ̃ = δ 1 ( K 1 - K 2 ) w 2 π 2 , λ ̃ = 0 .

Since δ1 (K1 - K2) ≠ 0, it follows that

w 2 π 2 , λ ̃ = 0 .
(11)

By the same procedure from equality (5), we can derive that

w 2 π 2 , λ ̃ = 0 .
(12)

From the fact that w 2 ( x , λ ̃ ) is a solution of the differential equation (1) on [ π 2 , π ] and satisfies the initial conditions (11) and (12) it follows that w 1 ( x , λ ̃ ) =0 identically on [ π 2 , π ] (cf. [2, p. 12, Theorem 1.2.1]).

By using we may also find

w 1 π 2 , λ ̃ = w 1 π 2 , λ ̃ = 0 .

From the latter discussions of w 2 ( x , λ ̃ ) , it follows that w 1 ( x , λ ̃ ) =0 identically on 0 , π 2 π 2 , π . But this contradicts (6), thus completing the proof.

2 An existance theorem

The function ω(x, λ) defined in Section 1 is a nontrivial solution of Equation 1 satisfying conditions (2), (4) and (5). Putting ω(x, λ) into (3), we get the characteristic equation

F ( λ ) w ( π , λ ) + λ ω ( π , λ ) = 0 .
(13)

By Theorem 1.1, the set of eigenvalues of boundary-value problem (1)-(5) coincides with the set of real roots of Equation 13. Let q 1 = 1 p 1 0 π 2 |q ( τ ) |dτ and q 2 = 1 p 2 π 2 π q ( τ ) dτ.

Lemma 2. (1) Let λ4 q 1 2 . Then, for the solution w1(x, λ) of Equation 8, the following inequality holds:

w 1 ( x , λ ) p 1 q 1 , x 0 , π 2 .
(14)
  1. (2)

    Let λ max 4 q 1 2 , 4 q 2 2 . Then, for the solution w2 (x, λ) of Equation 9, the following inequality holds:

    w 2 ( x , λ ) 2 p 1 q 1 γ 1 δ 1 + p 2 γ 2 p 1 δ 2 , x π 2 , π .
    (15)

Proof. Let B 1 λ = max 0 , π 2 w 1 ( x , λ ) . Then, from (8), it follows that, for every λ > 0, the following inequality holds:

B 1 λ p 1 s + 1 s B 1 λ q 1 .

If s ≥ 2q1, we get (14). Differentiating (8) with respect to x, we have

w 1 ( x , λ ) = - cos s p 1 x - 1 p 1 2 0 x q ( τ ) cos s p 1 ( x - τ ) w 1 ( τ - Δ ( τ ) , λ ) d τ .
(16)

From (16) and (14), it follows that, for s ≥ 2q1, the following inequality holds:

w 1 ( x , λ ) s 2 p 1 2 + 1 + 1 .

Hence,

w 1 ( x , λ ) s 1 q 1 .
(17)

Let B 2 λ = max π 2 , π w 2 ( x , λ ) . Then, from (9), (14) and (17), it follows that, for s ≥ 2q1, the following inequalities holds:

B 2 λ p 1 q 1 γ 1 δ 1 + p 2 γ 2 δ 2 1 q 1 + 1 2 q 2 B 2 λ q 2 , B 2 λ 2 p 1 q 1 γ 1 δ 1 + p 2 γ 2 p 1 δ 2 .

Hence, if λ max 4 q 1 2 , 4 q 2 2 , we get (15).

Theorem 2. The problem (1)-(5) has an infinite set of positive eigenvalues.

Proof. Differentiating (9) with respect to x, we get

w 2 ( x , λ ) = - s γ 1 p 2 δ 1 w 1 π 2 , λ sin s p 2 x - π 2 + γ 2 w 1 ( π 2 , λ ) δ 2 cos s p 2 x - π 2 - 1 p 2 2 π 2 x q ( τ ) cos s p 2 ( x - τ ) w 2 ( τ - Δ ( τ ) , λ ) d τ .
(18)

From (8), (9), (13), (16) and (18), we get

- s γ 1 p 2 δ 1 - p 1 s sin s π 2 p 1 - 1 s p 1 0 π 2 q ( τ ) sin s p 1 π 2 - τ ω 1 ( τ - Δ ( τ ) , λ ) d τ × sin s π 2 p 2 + γ 2 δ 2 - cos s π 2 p 1 - 1 p 1 2 0 π 2 q ( τ ) cos s p 1 π 2 - τ ω 1 ( τ - Δ ( τ ) , λ ) d τ × cos s π 2 p 2 - 1 p 2 2 π 2 π q ( τ ) cos s p 2 ( π - τ ) ω 2 ( τ - Δ ( τ ) , λ ) d τ + λ γ 1 δ 1 - p 1 s sin s π 2 p 1 - 1 s p 1 0 π 2 q ( τ ) sin s p 1 π 2 - τ ω 1 ( τ - Δ ( τ ) , λ ) d τ × cos s π 2 p 2 + γ 2 p 2 δ 2 s - cos s π 2 p 1 - 1 p 1 2 0 π 2 q ( τ ) cos s p 1 π 2 - τ ω 1 ( τ - Δ ( τ ) , λ ) d τ × sin s π 2 p 2 - 1 s p 2 π 2 π q ( τ ) sin s p 2 ( π - τ ) ω 2 ( τ - Δ ( τ ) , λ ) d τ = 0 .
(19)

Let λ be sufficiently large. Then, by (14) and (15), Equation 19 may be rewritten in the form

s sin s π p 1 + p 2 2 p 1 p 2 + O ( 1 ) = 0 .
(20)

Obviously, for large s, Equation 20 has an infinite set of roots. Thus, the theorem is proved.

3 Asymptotic formulas for eigenvalues and eigenfunctions

Now, we begin to study asymptotic properties of eigenvalues and eigenfunctions. In the following, we shall assume that s is sufficiently large. From (8) and (14), we get

ω 1 ( x , λ ) = O ( 1 ) on 0 , π 2 .
(21)

From (9) and (15), we get

ω 2 ( x , λ ) = O ( 1 ) on π 2 , π .
(22)

The existence and continuity of the derivatives ω 1 s ( x , λ ) for 0x π 2 , λ <, and ω 2 s ( x , λ ) for π 2 xπ, λ <, follows from Theorem 1.4.1 in [?].

ω 1 s ( x , λ ) = O ( 1 ) , x 0 , π 2 and ω 2 s ( x , λ ) = O ( 1 ) , x π 2 , π .
(23)

Theorem 3. Let n be a natural number. For each sufficiently large n, there is exactly one eigenvalue of the problem (1)-(5) near p 1 2 p 2 2 ( p 1 + p 2 ) 2 ( 2 n + 1 ) 2 .

Proof. We consider the expression which is denoted by O(1) in Equation 20. If formulas (21)-(23) are taken into consideration, it can be shown by differentiation with respect to s that for large s this expression has bounded derivative. It is obvious that for large s the roots of Equation 20 are situated close to entire numbers. We shall show that, for large n, only one root (20) lies near to each 4 n 2 p 1 2 p 2 2 ( p 1 + p 2 ) 2 . We consider the function ϕ ( s ) =sinsπ p 1 + p 2 2 p 1 p 2 +O ( 1 ) . Its derivative, which has the form ϕ ( s ) =sinsπ p 1 + p 2 2 p 1 p 2 +sπ p 1 + p 2 2 p 1 p 2 cossπ p 1 + p 2 2 p 1 p 2 +O ( 1 ) , does not vanish for s close to n for sufficiently large n. Thus, our assertion follows by Rolle's Theorem.

Let n be sufficiently large. In what follows, we shall denote by λ n = s n 2 the eigenvalue of the problem (1)-(5) situated near 4 n 2 p 1 2 p 2 2 ( p 1 + p 2 ) 2 . We set s n = 2 n p 1 p 2 p 1 + p 2 + δ n . From (20), it follows that δ n =O 1 n . Consequently

s n = 2 n p 1 p 2 p 1 + p 2 + O 1 n .
(24)

The formula (24) makes it possible to obtain asymptotic expressions for eigenfunction of the problem (1)-(5). From (8), (16) and (21), we get

ω 1 ( x , λ ) = O 1 s ,
(25)
ω 1 ( x , λ ) = O ( 1 ) .
(26)

From (9), (22), (25) and (26), we get

ω 2 ( x , λ ) = O 1 s .
(27)

By putting (24) in (25) and (27), we derive that

u 1 n = w 1 ( x , λ n ) = O 1 n , u 2 n = w 2 ( x , λ n ) = O 1 n .

Hence, the eigenfunctions u n (x) have the following asymptotic representation:

u n ( x ) = O 1 n for x 0 , π 2 π 2 , π .

Under some additional conditions, the more exact asymptotic formulas which depend upon the retardation may be obtained. Let us assume that the following conditions are fulfilled:

  1. (a)

    The derivatives q'(x) and Δ(x) exist and are bounded in 0 , π 2 π 2 , π and have finite limits q ( π 2 ± 0 ) = lim x π 2 ± 0 q ( x ) and Δ ( π 2 ± 0 ) = lim x π 2 ± 0 Δ ( x ) , respectively.

  2. (b)

    Δ'(x) ≤ 1 in 0 , π 2 π 2 , π , Δ(0) = 0 and lim x π 2 + 0 Δ ( x ) =0.

Using (b), we have

x - Δ ( x ) 0 for x 0 , π 2 and x - Δ ( x ) π 2 for x π 2 , π .
(28)

From (25), (27) and (28), we have

w 1 ( τ - Δ ( τ ) , λ ) = O 1 s ,
(29)
w 2 ( τ - Δ ( τ ) , λ ) = O 1 s .
(30)

Under the conditions (a) and (b), the following formulas

0 π 2 q ( τ ) sin s p 1 π 2 - τ d τ = O 1 s , 0 π 2 q ( τ ) cos s p 1 π 2 - τ d τ = O 1 s
(31)

can be proved by the same technique in Lemma 3.3.3 in [?]. Putting these expressions into (19), we have

0 = γ 1 p 1 p 2 δ 1 sin s π 2 p 1 sin s π 2 p 2 - γ 2 δ 2 cos s π 2 p 2 - s p 1 sin s π 2 p 1 cos 2 π 2 p 2 - s γ 2 p 2 δ 2 cos s π 2 p 1 sin s π 2 p 2 + O 1 s ,

and using γ1δ2p1 = γ2δ1p2 we get

0 = γ 2 δ 2 cos s π p 1 + p 2 2 p 1 p 2 - s p 1 sin s π p 1 + p 2 2 p 1 p 2 + O 1 s .

Dividing by s and using s n = 2 n p 1 p 2 p 1 + p 2 + δ n , we have

sin n π + π ( p 1 + p 2 ) δ n 2 p 1 p 2 = O 1 n 2 .

Hence,

δ n = O 1 n 2 ,

and finally

s n = 2 n p 1 p 2 p 1 + p 2 + O 1 n 2 .
(32)

Thus, we have proven the following theorem.

Theorem 4. If conditions (a) and (b) are satisfied, then the positive eigenvalues λ n = s n 2 of the problem (1)-(5) have the (32) asymptotic representation for n → ∞.

We now may obtain a sharper asymptotic formula for the eigenfunctions. From (8) and (29),

w 1 ( x , λ ) = - p 1 s sin s p 1 x + O 1 s 2 .
(33)

Replacing s by s n and using (32), we have

u 1 n ( x ) = p 1 + p 2 2 p 2 n sin 2 p 2 n p 1 + p 2 x + O 1 n 2 .
(34)

From (16) and (29), we have

w 1 ( x , λ ) s = - cos s p 1 x s + O 1 s 2 , x 0 , π 2 .
(35)

From (9), (30), (31), (33) and (35), we have

w 2 ( x , λ ) = { γ 1 p 1 sin s π 2 p 1 s δ 1 + O ( 1 s 2 ) } cos 2 p 2 ( x π 2 ) { γ 2 p 2 cos s π 2 p 1 s δ 2 + O ( 1 s 2 ) } sin s p 2 ( x π 2 ) + O ( 1 s 2 ) , w 2 ( x , λ ) = γ 2 p 2 s δ 2 sin s ( π ( p 2 p 1 2 p 1 p 2 + x 2 p 2 ) + O ( 1 s 2 ) .

Now, replacing s by s n and using (32), we have

u 2 n ( x ) = - γ 2 ( p 1 + p 2 ) 2 n p 1 δ 2 sin n π ( p 2 - p 1 ) p 1 + p 2 + p 1 x p 1 + p 2 + O 1 n 2 .
(36)

Thus, we have proven the following theorem.

Theorem 5. If conditions (a) and (b) are satisfied, then the eigenfunctions u n (x) of the problem (1)-(5) have the following asymptotic representation for n → ∞:

u n ( x ) = u 1 n ( x ) for x 0 , π 2 , u 2 n ( x ) for x π 2 , π ,

where u1n(x) and u2n(x) defined as in (34) and (36), respectively.

4 Conclusion

In this study, first, we obtain asymptotic formulas for eigenvalues and eigenfunctions for discontinuous boundary-value problem with retarded argument which contains a spectral parameter in the boundary condition. Then, under additional conditions (a) and (b) the more exact asymptotic formulas, which depend upon the retardation obtained.

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Correspondence to Erdoğan Şen.

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5 Competing interests

The authors declare that they have no completing interests.

6 Authors' contributions

Establishment of the problem belongs to AB (advisor). ES obtained the asymptotic formulas for eigenvalues and eigenfunctions. All authors read and approved the final manuscript.

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Şen, E., Bayramov, A. On calculation of eigenvalues and eigenfunctions of a Sturm-Liouville type problem with retarded argument which contains a spectral parameter in the boundary condition. J Inequal Appl 2011, 113 (2011). https://doi.org/10.1186/1029-242X-2011-113

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