Open Access

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

Journal of Inequalities and Applications20112011:113

DOI: 10.1186/1029-242X-2011-113

Received: 7 June 2011

Accepted: 17 November 2011

Published: 17 November 2011

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.

Keywords

differential equation with retarded argument transmission conditions asymptotics of eigenvalues and eigenfunctions

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 ) | d τ and q 2 = 1 p 2 π 2 π q ( τ ) d τ .

Lemma 2. (1) Let λ 4 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 0 x π 2 , λ < , and ω 2 s ( x , λ ) for π 2 x π , 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 ) = sin s π p 1 + p 2 2 p 1 p 2 + O ( 1 ) . Its derivative, which has the form ϕ ( s ) = sin s π p 1 + p 2 2 p 1 p 2 + s π p 1 + p 2 2 p 1 p 2 cos s π 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.

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Arts and Science, Namık Kemal University

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© Şen and Bayramov; licensee Springer. 2011

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