# Asymptotic Behaviour of Eigenvalues and Eigenfunctions of a Sturm-Liouville Problem with Retarded Argument

Authors: Erdoğan Şen; Jong Jin Seo; Serkan Araci
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

In the present paper, a discontinuous boundary-value problem with retarded argument at the two points of discontinuities is investigated. We obtained asymptotic formulas for the eigenvalues and eigenfunctions. This is the first work containing two discontinuities points in the theory of differential equations with retarded argument. In that special case the transmission coefficientsδ=γ=1 and retarded argument Δ≡0 in the results obtained in this work coincide with corresponding results in the classical Sturm-Liouville operator.


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## Body

## 1. Introduction

Delay differential equations arise in many areas of mathematical modelling, for example, population dynamics (taking into account the gestation times), infectious diseases (accounting for the incubation periods), physiological and pharmaceutical kinetics (modelling, for example, the body’s reaction to CO2, and so forth, in circulating blood) and chemical kinetics (such as mixing reactants), the navigational control of ships and aircraft, and more general control problems. Also, differential equations and nonlinear differential equations have been studied by many mathematicians in several ways for a long time cf. [1–20].Boundary value problems for differential equations of the second order with retarded argument were studied in [5–10, 13–16], and various physical applications of such problems can be found in [6].In the papers [13–16], the asymptotic formulas for the eigenvalues and eigenfunctions of a discontinuous boundary value problem with retarded argument and a spectral parameter in the boundary conditions were derived. In spite of their being already a long years, these subjects are still today enveloped in an aura of mystery within scientific community although they have penetrated numerous mathematical field.The asymptotic formulas for the eigenvalues and eigenfunctions of the Sturm-Liouville problem with the spectral parameter in the boundary condition were obtained in [17–20].In this paper we study the eigenvalues and eigenfunctions of a discontinuous boundary value problem with retarded argument. Namely, we consider the boundary value problem for the differential equation(1)y′′(x)+q(x)y(x-Δ(x))+λy(x)=0,
on [0,h1)∪(h1,h2)∪(h2,π], with boundary conditions
(2)y(0)cosα+y′(0)sinα=0,(3)y(π)cosβ+y′(π)sinβ=0,
and transmission conditions
(4)y(h1-0)-δy(h1+0)=0,(5)y′(h1-0)-δy′(h1+0)=0,(6)y(h2-0)-γy(h2+0)=0,(7)y′(h2-0)-γy′(h2+0)=0,
where the real-valued function q(x) is continuous in [0,h1)∪(h1,h2)∪(h2,π] and has finite limits
(8)q(h1±0)=limx→h1±0q(x),q(h2±0)=limx→h2±0q(x),
the real-valued function Δ(x)≥0 continuous in [0,h1)∪(h1,h2)∪(h2,π] and has finite limits
(9)Δ(h1±0)=limx→h1±0Δ(x),Δ(h2±0)=limx→h2±0Δ(x);
if x∈[0,h1) then x-Δ(x)≥0; if x∈(h1,h2) then x-Δ(x)≥h1; if x∈(h2,π) then x-Δ(x)≥h2; λ is a real spectral parameter; h1, h2, α, β, δ, γ≠0 are arbitrary real numbers such that 0<h1<h2<π and sinαsinβ≠0.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 [20].Letw1(x,λ) be a solution of (1) on [0,h1], satisfying the initial conditions
(10)w1(0,λ)=sinα,w1′(0,λ)=-cosα.
The conditions (10) define a unique solution of (1) on [0,h1] ([5], page 12).After defining the above solution, then we will define the solutionw2(x,λ) of (1) on [h1,h2] by means of the solution w1(x,λ) using the initial conditions
(11)w2(h1,λ)=δ-1w1(h1,λ),w2′(h1,λ)=δ-1w1′(h1,λ).
The conditions (11) define a unique solution of (1) on [h1,h2].After describing the above solution, then we will give the solutionw3(x,λ) of (1) on [h2,π] by means of the solution w2(x,λ) using the initial conditions
(12)w3(h2,λ)=γ-1w2(h2,λ),w3′(h2,λ)=γ-1w2′(h2,λ).
The conditions (12) define a unique solution of (1) on [h2,π].Consequently, the functionw(x,λ) is defined on [0,h1)∪(h1,h2)∪(h2,π] by the equality
(13)w(x,λ)={w1(x,λ),x∈[0,h1),w2(x,λ),x∈(h1,h2),w3(x,λ),x∈(h2,π],
is a solution of (1) on [0,h1)∪(h1,h2)∪(h2,π], which satisfies one of the boundary conditions and four transmission conditions.Lemma 1.
Letw(x,λ) be a solution of (1) and λ>0. Then the following integral equations hold:
(14)w1(x,λ)=sinαcossx-cosαssinsx-1s∫0xq(τ)sins(x-τ)w1(τ-Δ(τ),λ)dτ(s=λ,λ>0),(15)w2(x,λ)=1δw1(h1,λ)coss(x-h1)+w1′(h1,λ)sδsins(x-h1)-1s∫h1xq(τ)sins(x-τ)w2(τ-Δ(τ),λ)dτ(s=λ,λ>0),(16)w3(x,λ)=1γw2(h2,λ)coss(x-h2)+w2′(h2,λ)sγsins(x-h2)-1s∫h2xq(τ)sins(x-τ)w3(τ-Δ(τ),λ)dτ(s=λ,λ>0).Proof.
To prove this lemma, it is enough to substitute-s2w1(τ,λ)-w1′′(τ,λ), -s2w2(τ,λ)-w2′′(τ,λ), and -s2w3(τ,λ)-w3′′(τ,λ) instead of  -q(τ)w1(τ-Δ(τ),λ), -q(τ)w2(τ-Δ(τ),λ), and -q(τ)w3(τ-Δ(τ),λ) in the integrals in (14), (15), and (16), respectively, and integrate by parts twice.Theorem 2.
The problem (1)–(7) can have only simple eigenvalues.Proof.
Letλ~ be an eigenvalue of the problem (1)–(7) and
(17)y~(x,λ~)={y~1(x,λ~),x∈[0,h1),y~2(x,λ~),x∈(h1,h2),y~3(x,λ~),x∈(h2,π],
be a corresponding eigenfunction. Then, from (2) and (10), it follows that the determinant
(18)W[y~1(0,λ~),w1(0,λ~)]=|y~1(0,λ~)sinαy~1′(0,λ~)-cosα|=0,
and, by Theorem  2.2.2, in [5] the functions y~1(x,λ~) and w1(x,λ~) are linearly dependent on [0,h1]. We can also prove that the functions y~2(x,λ~) and w2(x,λ~) are linearly dependent on [h1,h2] and y~3(x,λ~) and w3(x,λ~) are linearly dependent on [h2,π]. Hence
(19)y~i(x,λ~)=Kiwi(x,λ~)(i=1,2,3),
for some K1≠0,  K2≠0, and K3≠0. We must show that K1=K2 and K2=K3. Suppose that K2≠K3. From the equalities (6) and (19), we have
(20)y~(h2-0,λ~)-γy~(h2+0,λ~)=y2~(h2,λ~)-γy3~(h2,λ~)=K2w2(h2,λ~)-γK3w3(h2,λ~)=K2γw3(h2,λ~)-K3γw3(h2,λ~)=γ(K2-K3)w3(h2,λ~)=0.
Since γ(K2-K3)≠0 it follows that
(21)w3(h2,λ~)=0.
By the same procedure from equality (7) we can derive that
(22)w3′(h2,λ~)=0.
From the fact that w3(x,λ~) is a solution of the differential (1) on [h2,π] and satisfies the initial conditions (21) and (22), it follows that w3(x,λ~)=0 identically on [h2,π].
By using this method, we may also find(23)w2(h2,λ~)=w2′(h2,λ~)=0,w1(h1,λ~)=w1′(h1,λ~)=0.
From the latter discussions of w3(x,λ~), it follows that w2(x,λ~)=0 and w1(x,λ~)=0 identically on (h1,h2) and [0,h1), but this contradicts (10), thus completing the proof.


## 2. An Existence Theorem

The functionw(x,λ) defined in Section 1 is a nontrivial solution of (1) satisfying conditions (2) and (4)–(7). Putting w(x,λ) into (3), we get the characteristic equation
(24)F(λ)≡w(π,λ)cosβ+w′(π,λ)sinβ=0.By Theorem2 the set of eigenvalues of boundary-value problem (1)–(7) coincides with the set of real roots of (24). Let
(25)q1=∫0h1|q(τ)|dτ,q2=∫h1h2|q(τ)|dτ,q3=∫h2π|q(τ)|dτ.Lemma 3.
(1) Letλ≥4q12. Then for the solution w1(x,λ) of (14), the following inequality holds:
(26)|w1(x,λ)|≤1q14q12sin2α+cos2α,x∈[0,h1].
(2) Letλ≥max{4q12,4q22}. Then for the solution w2(x,λ) of (15), the following inequality holds:
(27)|w2(x,λ)|≤4q1|δ|4q12sin2α+cos2α,x∈[h1,h2].
(3) Letλ≥max{4q12,4q22,4q32}. Then for the solution w3(x,λ) of (16), the following inequality holds:
(28)|w3(x,λ)|≤16q1|δγ|4q12sin2α+cos2α,x∈[h2,π].Proof.
LetB1λ=max[0,h1]|w1(x,λ)|. Then from (14), it follows that, for every λ>0, the following inequality holds:
(29)B1λ≤sin2α+cos2αs2+1sB1λq1.
If s≥2q1 we get (26). Differentiating (14) with respect to x, we have
(30)w1′(x,λ)=-ssinαsinsx-cosαcossx-∫0xq(τ)coss(x-τ)w1(τ-Δ(τ),λ)dτ.
From expressions of (30) and (26), it follows that, for s≥2q1, the following inequality holds:
(31)|w1′(x,λ)|s≤1q14q12sin2α+cos2α.
Let B2λ=max[h1,h2]|w2(x,λ)|. Then from (11), (26), and (31) it follows that, for s≥2q1 and s≥2q2, the following inequality holds:
(32)B2λ≤2q1|δ|4q12sin2α+cos2α+12q2B2λq2.
Hence, if λ≥max{4q12,4q22}, it reduces to (27). Differentiating (15) with respect to, we get
(33)w2′(x,λ)=-sδw1(h1,λ)sins(x-h1)+w1′(h1,λ)δcoss(x-h1)-∫h1xq(τ)coss(x-τ)w2(τ-Δ(τ),λ)dτ.(s=λ,λ>0).
From (26) and (33), it follows that, for s≥2q1 and s≥2q2, the following inequality holds:
(34)|w2′(x,λ)|s≤4|δ|q14q12sin2α+cos2α.
Let B3λ=max[h2,π]|w3(x,λ)|. Then from (16), (27), and (34) it follows that, for s≥2q1,  s≥2q2 and s≥2q3, the following inequality holds:
(35)B3λ≤8q1|δγ|4q12sin2α+cos2α+1sB3λq3.
Hence, if λ≥max{4q12,4q22,4q32} we procure (28).Theorem 4.
The problem (1)–(7) has an infinite set of positive eigenvalues.Proof.
Differentiating (16) with respect to x, we readily see that
(36)w3′(x,λ)=-sγw2(h2,λ)sins(x-h2)+w2′(h2,λ)γcoss(x-h2)-∫h2xq(τ)coss(x-τ)w3(τ-Δ(τ),λ)dτ.(s=λ,λ>0).
With the helps of (14), (15), (16), (24), (30), and (36), we have the following:
(37)[1γ{-1s∫h1h2q(τ)sins(h2-τ)w2(τ-Δ(τ),λ)dτ1δ(∫0h1q(τ)sinssinαcossh1-cosαssinsh1-1s∫0h1q(τ)sins(h1-τ)w1(τ-Δ(τ),λ)dτsinαcossh1-cosαssinsh1-1s∫0h1q(τ)sins)×coss(h2-h1)-1sδ(∫0h1q(τ)coss(h1-τ)w1(τ-Δ(τ),λ)dτssinαsinsh1+cosαcossh1+∫0h1q(τ)coss(h1-τ)w1(τ-Δ(τ),λ)dτ)×sins(h2-h1)-1s∫h1h2q(τ)sins(h2-τ)w2(τ-Δ(τ),λ)dτ}×coss(π-h2)+1sγ{-sδ(-1s∫0h1q(τ)sinαcossh1-cosαssinsh1-1s∫0h1q(τ)sins(h1-τ)×w1(τ-Δ(τ),λ)dτsinαcossh1-cosαssinsh1-1s∫0h1q(τ))×sins(h2-h1)+1δ(∫0h1q(τ)coss(h1-τ)w1(τ-Δ(τ),λ)dτ-ssinαsinsh1-cosαcossh1-∫0h1q(τ)coss(h1-τ)×w1(τ-Δ(τ),λ)dτ∫0h1q(τ)coss(h1-τ))×coss(h2-h1)-∫h1h2q(τ)coss(h2-τ)w2(τ-Δ(τ),λ)dτ1δ(sinαcossh1-cosαssinsh1-1s∫0h1q(τ)sins}×sins(π-h2)-1s∫h2πq(τ)sins(π-τ)w3(τ-Δ(τ),λ)dτ]cosβ+[-sγ{1δ(∫0h1q(τ)sins(h1-τ)w1(τ-Δ(τ),λ)dτsinαcossh1-cosαssinsh1-1s∫0h1q(τ)sins(h1-τ)×w1(τ-Δ(τ),λ)dτ∫0h1q(τ)sins(h1-τ))×coss(h2-h1)-1sδ(∫0h1q(τ)ssinαsinsh1+cosαcossh1+∫0h1q(τ)coss(h1-τ)×w1(τ-Δ(τ),λ)dτ∫0h1q(τ))×sins(h2-h1)-1s∫h1h2q(τ)sins(h2-τ)w2(τ-Δ(τ),λ)1δ(∫0h1q(τ)sins(h1-τ)w1(τ-Δ(τ),λ)dτsinαcossh1-cosαssinsh1}×sins(π-h2)+1γ{-sδ×(sinαcossh1-cosαssinsh1∫0h1q(τ)sins(h1-τ)-1s∫0h1q(τ)sins(h1-τ)×w1(τ-Δ(τ),λ)dτcosαs∫0h1q(τ)sins(h1-τ))×sins(h2-h1)+1δ(∫0h1q(τ)coss(h1-τ)w1(τ-Δ(τ),λ)dτ-ssinαsinsh1-cosαcossh1-∫0h1q(τ)coss(h1-τ)×w1(τ-Δ(τ),λ)dτ∫0h1q(τ)coss(h1-τ))×coss(h2-h1)-∫h1h2q(τ)coss(h2-τ)w2(τ-Δ(τ),λ)dτ(sinαcossh1-cosαssinsh1}×coss(π-h2)-∫h2πq(τ)coss(π-τ)w3(τ-Δ(τ),λ)dτ-sγ]sinβ=0.
Let λ be sufficiently big. Then, by (26), (27), and (28), (37) may be rewritten in the following form:
(38)-ssinαsinβγδ{cossh2sins(π-h2)+sinsh2coss(π-h2)}+O(1)=0,(39)ssinsπ+O(1)=0.
Obviously, for big s (39) has an infinite set of roots. Thus, the proof of the theorem is completed.


## 3. Asymptotic Formulas for Eigenvalues and Eigenfunctions

Now we begin to study asymptotic properties of eigenvalues and eigenfunctions. In the following we will assume that is sufficiently big. From (14) and (26), we obtain
(40)w1(x,λ)=O(1)on[0,h1].
By (15) and (27), this leads to
(41)w2(x,λ)=O(1)on[h1,h2].
By (16) and (28), this leads to
(42)w3(x,λ)=O(1)on[h2,π].
The existence and continuity of the derivatives w1s′(x,λ) for 0≤x≤h1, |λ|<∞, w2s′(x,λ) for h1≤x≤h2, |λ|<∞ and w3s′(x,λ) for h2≤x≤π, |λ|<∞ follows from Theorem  1.4.1 in [5].Lemma 5.
The following holds true:(43)w1s′(x,λ)=O(1),x∈[0,h1],(44)w2s′(x,λ)=O(1),x∈[h1,h2],(45)w3s′(x,λ)=O(1),x∈[h2,π].Proof.
By differentiating (16) with respect to s, we get, by (43) and (44) the following:
(46)w3s′(x,λ)=-1s∫h2xq(τ)sins(x-τ)w3s′(τ-Δ(τ),λ)+Z(x,λ),(|Z(x,λ)|≤Z0).
Let Dλ=max[h2,π]|w3s′(x,λ)|. Then the existence of Dλ follows from continuity of derivation for x∈[h2,π]. From (46)
(47)Dλ≤1sq3Dλ+Z0.
Now let s≥2q3. Then Dλ≤2Z0 and the validity of the asymptotic formula (45) follows. Formulas (43) and (44) may be proved analogically.Theorem 6.
Letn be a natural number. For each sufficiently big n there is exactly one eigenvalue of the problem (1)–(7) near n2.Proof.
We consider the expression which is denoted byO(1) in (39). If formulas (40)–(45) are taken into consideration, it can be shown by differentiation with respect to s that for big s this expression has bounded derivative. We will show that, for big  n, only one root (39) lies near to each n. We consider the function ϕ(s)=ssinsπ+O(1). Its derivative, which has the form ϕ′(s)=sinsπ+sπcossπ+O(1), does not vanish for s close to n for sufficiently big n. Thus our assertion follows by Rolle’s Theorem.Letn be sufficiently big. In what follows we will denote by λn=sn2 the eigenvalue of the problem (1)–(7) situated near n2. We set sn=n+δn. Then from (39) it follows that δn=O(1/n). Consequently
(48)sn=n+O(1n),(49)λn=n2+O(1).
Formula (48) makes it possible to obtain asymptotic expressions for eigenfunction of the problem (1)–(7). From (14), (30), and (40), we get
(50)w1(x,λ)=sinαcossx+O(1s),(51)w1′(x,λ)=-ssinαsinsx+O(1).
From expressions of (15), (31), (39), and (41), we easily see that
(52)w2(x,λ)=sinαδcossx+O(1s),w3(x,λ)=sinαδγcossx+O(1s).
By substituting (48) into (50), (52), we find that
(53)u1n=w1(x,λn)=sinαcosnx+O(1n),u2n=w2(x,λn)=sinαδcosnx+O(1n),u3n=w3(x,λn)=sinαδγcosnx+O(1n).
Hence the eigenfunctions un(x) have the following asymptotic representation:
(54)un(x)={sinαcosnx+O(1n),forx∈[0,h1),sinαδcosnx+O(1n),forx∈(h1,h2),sinαδγcosnx+O(1n),forx∈(h2,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.(a) The derivativesq′(x) and Δ′′(x) exist and are bounded in [0,h1)∪(h1,h2)∪(h2,π] and have finite limits q′(h1±0)=limx→h1±0q′(x), q′(h2±0)=limx→h2±0q′(x) and Δ′′(h1±0)=limx→h1±0Δ′′(x),  Δ′′(h2±0)=limx→h2±0Δ′′(x), respectively.(b)Δ′(x)≤1 in [0,h1)∪(h1,h2)∪(h2,π], Δ(0)=0, limx→h1+0Δ(x)=0, and limx→h2+0Δ(x)=0.It is easy to see that, using (b)(55)x-Δ(x)≥0,x∈[0,h1),x-Δ(x)≥h1,x∈(h1,h2),x-Δ(x)≥h2,x∈(h2,π]
are obtained.By (50), (52), and (55), we have
(56)w1(τ-Δ(τ),λ)=sinαcoss(τ-Δ(τ))+O(1s),(57)w2(τ-Δ(τ),λ)=sinαδcoss(τ-Δ(τ))+O(1s),(58)w3(τ-Δ(τ),λ)=sinαδγcoss(τ-Δ(τ))+O(1s)
on [0,h1),  (h1,h2) and (h2,π], respectively.Under the conditions (a) and (b) the following formulas:(59)∫0x‍q(τ)coss(2τ-Δ(τ))dτ=O(1s),∫0x‍q(τ)sins(2τ-Δ(τ))dτ=O(1s)
can be proved by the same technique in Lemma  3.3.3 in [5].Putting the expressions (56), (57), and (58) into (37), and then using (59), after long operations we have
(60)cossπsin(α-β)-ssinαsinβsinsπδγ-sinαsinβ2δγ∫0π‍q(τ)coss(π-Δ(τ))dτ+O(1s)=0.
Hence
(61)tansπ=1s(sin(α-β)sinαsinβ-12∫0π‍q(τ)coss(π-Δ(τ))dτsin(α-β)sinαsinβ)+O(1s2).
Again, if we take sn=n+δn, then from (48)
(62)tan((n+δn)π)=tanδnπ=1n(sin(α-β)sinαsinβ-12∫0π‍q(τ)coss(π-Δ(τ))dτsin(α-β)sinαsinβ)+O(1n2).
Hence for big n,
(63)δn=1nπ(sin(α-β)sinαsinβ-12∫0π‍q(τ)coss(π-Δ(τ))dτ)+O(1n2)
and finally
(64)sn=n+1nπ(sin(α-β)sinαsinβ-12∫0π‍q(τ)coss(π-Δ(τ))dτ)+O(1n2).
Thus, we have proven the following theorem.Theorem 7.
If conditions (a) and (b) are satisfied, then the eigenvaluesλn=sn2 of the problem (1)–(7) have the (64) asymptotic formula for n→∞.Now, we may obtain sharper asymptotic formulas for the eigenfunctions. From (14), (56), and (59), we have
(65)w1(x,λ)=sinαcossx[1+12s∫0x‍q(τ)sinsΔ(τ)dτ]-sinsxs[cosα+sinα2∫0x‍q(τ)cossΔ(τ)dτ]+O(1s2),x∈[0,h1).
Now, replacing s by sn and using (64), we have
(66)u1n(x)=w1(x,λn)=sinα{cosnx[1+12n∫0x‍q(τ)sin(nΔ(τ))dτ]-sinnxnπ[(sin(α-β)sinαsinβ-12∫0π‍q(τ)cos(n(π-Δ(τ)))dτsin(α-β)sinαsinβ)x+(cotα+12∫0x‍q(τ)cos(nΔ(τ))dτ)π(sin(α-β)sinαsinβ]}+O(1n2).
From (15), (57), (59), and (65) we have
(67)w2(x,λ)=sinαcossxδ[1+12s∫0x‍q(τ)sinsΔ(τ)dτ]-sinsxsδ[cosα+sinα2∫0x‍q(τ)cossΔ(τ)dτ]+O(1s2),x∈(h1,h2).
Now, replacing s by sn and using (64), we have
(68)u2n(x)=w2(x,λn)=sinαδ{cosnx[1+12n∫0x‍q(τ)sin(nΔ(τ))dτ]-sinnxnπ[(sin(α-β)sinαsinβ-12∫0π‍q(τ)cos(n(π-Δ(τ)))dτsin(α-β)sinαsinβ)x+(cotα+12∫0x‍q(τ)cos(nΔ(τ))dτ)πsin(α-β)sinαsinβ]sin(α-β)sinαsinβ}+O(1n2).From (16), (58), (59), (65), and (67) and after long operations, we have
(69)w3(x,λ)=sinαcossxδγ[1+12s∫0x‍q(τ)sinsΔ(τ)dτ]-sinsxsδγ[cosα+sinα2∫0x‍q(τ)cossΔ(τ)dτ]+O(1s2),x∈(h2,π].
Now replacing s by sn and using (64), we have
(70)u3n(x)=w3(x,λn)=sinαδγ{cosnx[1+12n∫0x‍q(τ)sin(nΔ(τ))dτ]sin(α-β)sinαsinβ-sinnxnπ[(sin(α-β)sinαsinβ-12∫0π‍q(τ)cos(n(π-Δ(τ)))dτsin(α-β)sinαsinβ)xsin(α-β)sinαsinβsin(α-β)sinαsinβ+(cotα+12∫0x‍q(τ)cos(nΔ(τ))dτ)π]}+O(1n2).Thus, we have proven the following theorem.Theorem 8.
If conditions (a) and (b) are satisfied, then the eigenfunctionsun(x) of the problem (1)–(7) have the following asymptotic formula for n→∞:
(71)un(x)={u1n(x),x∈[0,h1),u2n(x),x∈(h1,h2),u3n(x),x∈(h2,π],
where u1n(x),u2n(x), and u3n(x) are determined as in (49), (68), and (70), respectively.



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