# A Sturm-Liouville Problem with a Discontinuous Coefficient and Containing an Eigenparameter in the Boundary Condition

Authors: Erdoğan Şen
Journal: Physics Research International
 (2013)

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## Abstract

We study a Sturm-Liouville operator with eigenparameter-dependent boundary conditions and transmission conditions at two interior points. We give an operator-theoretic formulation, construct fundamental solutions, investigate some properties of the eigenvalues and corresponding eigenfunctions of the discontinuous Sturm-Liouville problem and then obtain asymptotic formulas for the eigenvalues and eigenfunctions and find Green function of the discontinuous Sturm-Liouville problem.


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## Body

## 1. Introduction

In recent years, more and more researchers are interested in the discontinuous Sturm-Liouville problem for its application in physics (see [1–16]). Such problems are connected with discontinuous material properties, such as heat and mass transfer, varied assortment of physical transfer problems, vibrating string problems when the string loaded additionally with point masses, and diffraction problems. Moreover, there has been a growing interest in Sturm-Liouville problems with eigenparameter-dependent boundary conditions; that is, the eigenparameter appears not only in the differential equations but also in the boundary conditions of the problems (see [1–16] and corresponding bibliography).In this paper, following [8] we consider the boundary value problem for the differential equation
(1)τu:=-u′′+q(x)u=λω(x)u,
for x∈[-1,h1)∪(h1,h2)∪(h2,1] (i.e., x belongs to [-1,1], but the two inner points are x=h1 and x=h2), where q(x) is a real-valued function, continuous in [-1,h1), (h1,h2), and (h2,1] with the finite limits q(±h1)=limx→±h1, q(±h2)=limx→±h2; ω(x) is a discontinuous weight function such that ω(x)=ω12 for x∈[-1,h1), ω(x)=ω22 for x∈(h1,h2), and ω(x)=ω32 for x∈(h2,1], ω>0 together with the standard boundary condition at x=-1(2)L1u:=u′(-1)=0,
the spectral parameter-dependent boundary condition at x=1(3)L2u:=λ(β1′u(1)-β2′u′(1))+(β1u(1)-β2u′(1))=0,
and the four transmission conditions at the points of discontinuity x=h1 and x=h2(4)L3u:=γ1u(h1-0)-δ1u(h1+0)=0,(5)L4u:=γ2u′(h1-0)-δ2u′(h1+0)=0,(6)L5u:=γ3u(h2-0)-δ3u(h2+0)=0,(7)L6u:=γ4u′(h2-0)-δ4u′(h2+0)=0,
in the Hilbert space L2(-1,h1)⊕L2(h1,h2)⊕L2(h2,1), where λ∈ℂ is a complex spectral parameter; and all coefficients of the boundary and transmission conditions are real constants. We assume that β2′≠0 and |β1|+|β2|≠0. Moreover, we will assume that ρ:=β1′β2-β1β2′>0. Some special cases of this problem arise after application of the method of separation of variables to the diverse assortment of physical problems, heat and mass transfer problems (e.g., see [11]), vibrating string problems when the string was loaded additionally with point masses (e.g., see [11]), and a thermal conduction problem for a thin laminated plate.


## 2. Operator-Theoretic Formulation of the Problem

In this section, we introduce a special inner product in the Hilbert space(L2(-1,h1)⊕L2(h1,h2)⊕L2(h2,1))⊕ℂ and define a linear operator A in it so that the problem (1)–(5) can be interpreted as the eigenvalue problem for A. To this end, we define a new Hilbert space inner product on H:=(L2(-1,h1)⊕L2(h1,h2)⊕L2(h2,1))⊕ℂ by
(8)〈F,G〉H=ω12∫-1h1f(x)g(x)¯dx+ω22δ1δ2γ1γ2∫h1h2f(x)g(x)¯dx+ω32δ1δ2δ3δ4γ1γ2γ3γ4∫h21f(x)g(x)¯dx+δ1δ2δ3δ4ργ1γ2γ3γ4f1g1¯,
for F=(f(x)f1) and G=(g(x)g1)∈H. For convenience, we will use the notations
(9)R1(u):=β1u(1)-β2u′(1),R1′(u):=β1′u(1)-β2′u′(1).
In this Hilbert space, we construct the operator A:H→H with domain
(10)D(A)={F=(f(x)f1)∣f(x),f′(x)areabsolutelycontinuousin[1,h1]∪[h1,h2]∪[h2,1]andhavefinitelimitsf(h1±0),f(h2±0),f′(h1±0),f′(h2±0),τf∈L2(-1,h1)⊕L2(h1,h2)⊕L2(h2,1),L1f=L3f=L4f=L5f=L6f=0,f1=R1′(f)(f(x)f1)}
which acts by the rule
(11)AF=(1ω(x)[-f′′+q(x)f]-R1(f))withF=(f(x)R1′(f))∈D(A).
Thus we can pose the boundary value transmission problem (1)–(7) in H as
(12)AU=λU,U:=(u(x)R1′(u))∈D(A).
It is readily verified that the eigenvalues of A coincide with those of the problem (1)–(7).Theorem 1.
The operatorA is symmetric.Proof.
LetF=(f(x)R1′(f)) and G=(g(x)R1′(g)) be arbitrary elements of D(A). Twice integrating by parts, we find
(13)〈AF,G〉H-〈F,AG〉H=W(f,g¯;h1-0)-W(f,g¯;-1)+δ1δ2γ1γ2(W(f,g¯;h2-0)-W(f,g¯;h1+0))+δ1δ2δ3δ4γ1γ2γ3γ4(W(f,g¯;1)-W(f,g¯;h2+0))+δ1δ2δ3δ4ργ1γ2γ3γ4(R1′(f)R1(g¯)-R1(f)R1′(g¯)),
where, as usual, W(f,g;x) denotes the Wronskian of f and g; that is,
(14)W(f,g;x):=f(x)g′(x)-f′(x)g(x).
Since F,G∈D(A), the first components of these elements, that is, f and g, satisfy the boundary condition (2). From this fact, we easily see that
(15)W(f,g¯;-1)=0.
Further, as f and g also satisfy both transmission conditions, we obtain
(16)W(f,g¯;h1-0)=δ1δ2γ1γ2W(f,g¯;h1+0),W(f,g¯;h2-0)=δ1δ2δ3δ4γ1γ2γ3γ4W(f,g¯;h2+0).
Moreover, the direct calculations give
(17)R1′(f)R1(g¯)-R1(f)R1′(g¯)=-ρW(f,g¯;1).
Now, inserting (15)–(17) in (13), we have
(18)〈AF,G〉H=〈F,AG〉H(F,G∈D(A)),
and so A is symmetric.Recalling that the eigenvalues of (1)–(7) coincide with the eigenvalues of A, we have the next corollary.Corollary 2.
All eigenvalues of (1)–(7) are real.Since all eigenvalues are real, it is enough to study only the real-valued eigenfunctions. Therefore, we can now assume that all eigenfunctions of (1)–(7) are real-valued.


## 3. Asymptotic Formulas for Eigenvalues and Fundamental Solutions

Let us define fundamental solutions(19)ϕ(x,λ)={ϕ1(x,λ),x∈[-1,h1),ϕ2(x,λ),x∈(h1,h2),ϕ3(x,λ),x∈(h2,1],χ(x,λ)={χ1(x,λ),x∈[-1,h1),χ2(x,λ),x∈(h1,h2),χ3(x,λ),x∈(h2,1],
of (1) by the following procedure. We first consider the next initial value problem:
(20)-u′′+q(x)u=λω12u,x∈[-1,h1],(21)u(-1)=1,(22)u′(-1)=0.
By virtue of [1, Theorem 1.5], the problem (20)–(22) has a unique solution u=ϕ1(x,λ) which is an entire function of λ∈ℂ for each fixed x∈[-1,h1]. Similarly,
(23)-u′′+q(x)u=λω22u,x∈[h1,h2],(24)u(h1)=γ1δ1ϕ1(h1,λ),(25)u′(h1)=γ2δ2ϕ1′(h1,λ),
have a unique solution u=ϕ2(x,λ) which is an entire function of λ∈ℂ for each fixed x∈[h1,h2]. Continuing in this manner,
(26)-u′′+q(x)u=λω32u,x∈[h2,1],(27)u(h2)=γ3δ3ϕ2(h2,λ),(28)u′(h2)=γ4δ4ϕ2′(h2,λ),
have a unique solution u=ϕ3(x,λ) which is an entire function of λ∈ℂ for each fixed x∈[h2,1]. Slightly modifying the method of [1, Theorem 1.5], we can prove the initial value problem
(29)-u′′+q(x)u=λω32u,x∈[h2,1],(30)u(1)=β2′λ+β2,(31)u′(1)=β1′λ+β1,
have a unique solution u=χ3(x,λ) which is an entire function of spectral parameter λ∈ℂ for each fixed x∈[h2,1]. Similarly,
(32)-u′′+q(x)u=λω22u,x∈[h1,h2],(33)u(h2)=δ3γ3χ3(h2,λ),(34)u′(h2)=δ4γ4χ3′(h2,λ),
have a unique solution u=χ2(x,λ) which is an entire function of λ∈ℂ for each fixed x∈[h1,h2]. Continuing in this manner,
(35)-u′′+q(x)u=λω32u,x∈[-1,h1],(36)u(h1)=δ1γ1χ2(h1,λ),(37)u′(h1)=δ2γ2χ2′(h1,λ),
have a unique solution u=χ1(x,λ) which is an entire function of λ∈ℂ for each fixed x∈[-1,h1].By virtue of (21) and (22), the solution ϕ(x,λ) satisfies the first boundary condition (2). Moreover, by (24), (25), (27), and (28), ϕ(x,λ) satisfies also transmission conditions (4)–(7). Similarly, by (30), (31), (33), (34), (36), and (37) the other solution χ(x,λ) satisfies the second boundary condition (3) and transmission conditions (4)–(7). It is well known from the theory of ordinary differential equations that each of the Wronskians Δ1(λ)=W(ϕ1(x,λ),χ1(x,λ)), Δ2(λ)=W(ϕ2(x,λ),χ2(x,λ)), and Δ3(λ)=W(ϕ3(x,λ),χ3(x,λ)) is independent of x in [-1,h1], [h1,h2], and [h2,1], respectively.Lemma 3.
The equalityΔ1(λ)=(δ1δ2/γ1γ2)Δ2(λ)=(δ1δ2δ3δ4/γ1γ2γ3γ4)Δ3(λ) holds for each λ∈ℂ.Proof.
Since the above Wronskians are independent ofx, using (27), (28), (30), (31), (33), (34), (36), and (37), we find
(38)Δ1(λ)=ϕ1(h1,λ)χ1′(h1,λ)-ϕ1′(h1,λ)χ1(h1,λ)=(δ1γ1ϕ2(h1,λ))(δ2γ2χ2′(h1,λ))-(δ2γ2ϕ2′(h1,λ))(δ1γ1χ2(h1,λ))=δ1δ2γ1γ2Δ2(λ)=(δ1δ3γ1γ3ϕ3(h2,λ))(δ2δ4γ2γ4χ3′(h2,λ))-(δ2δ4γ2γ4ϕ3′(h2,λ))(δ1δ3γ1γ3χ3(h2,λ))=δ1δ2δ3δ4γ1γ2γ3γ4Δ3(λ).Corollary 4.
The zeros ofΔ1(λ), Δ2(λ), and Δ3(λ) coincide.In view of Lemma3, we denote Δ1(λ), (δ1δ2/γ1γ2)Δ2(λ), and (δ1δ2δ3δ4/γ1γ2γ3γ4)Δ3(λ) by Δ(λ). Recalling the definitions of ϕi(x,λ) and χi(x,λ), we infer the next corollary.Corollary 5.
The functionΔ(λ) is an entire function.Theorem 6.
The eigenvalues of (1)–(7) coincide with the zeros of Δ(λ).Proof.
LetΔ(λ0)=0. Then, W(ϕ1(x,λ0),χ1(x,λ0))=0 for all x∈[-1,h1]. Consequently, the functions ϕ1(x,λ0) and χ1(x,λ0) are linearly dependent; that is, χ1(x,λ0)=kϕ1(x,λ0), x∈[-1,h1], for some k≠0. By (21) and (22), from this equality, we have
(39)χ′(-1,λ0)=kϕ1′(-1,λ0)=0,
and so χ(x,λ0) satisfies the first boundary condition (2). Recalling that the solution χ(x,λ0) also satisfies the other boundary condition (3) and transmission conditions (4)–(7), we conclude that χ(x,λ0) is an eigenfunction of (1)–(7); that is, λ0 is an eigenvalue. Thus, each zero of Δ(λ) is an eigenvalue. Now, let λ0 be an eigenvalue, and let u0(x) be an eigenfunction with this eigenvalue. Suppose that Δ(λ0)≠0, whence W(ϕ1(x,λ0),χ1(x,λ0))≠0, W(ϕ2(x,λ0),χ2(x,λ0))≠0, and W(ϕ3(x,λ0),χ3(x,λ0))≠0. From this, by virtue of the well-known properties of Wronskians, it follows that each of the pairs ϕ1(x,λ0), χ1(x,λ0); ϕ2(x,λ0), χ2(x,λ0) and ϕ3(x,λ0), χ3(x,λ0) is linearly independent. Therefore, the solution u0(x) of (1) may be represented as
(40)u0(x)={c1ϕ1(x,λ0)+c2χ1(x,λ0),x∈[-1,h1),c3ϕ2(x,λ0)+c4χ2(x,λ0),x∈(h1,h2),c5ϕ3(x,λ0)+c6χ3(x,λ0),x∈(h2,1],
where at least one of the coefficients ci(i=1,6¯) is not zero. Considering the true equalities
(41)Lυ(u0(x))=0,υ=1,6¯
as the homogenous system of linear equations in the variables ci(i=1,6¯) and taking (24), (25), (27), (28), (33), (34), (36), and (37) into account, we see that the determinant of this system is equal to -((δ1δ2δ3δ4)2/γ1γ2γ3γ4)Δ4(λ0), and so it does not vanish by assumption. Consequently, the system (41) has the only trivial solution ci=0(i=1,6¯). We thus get at a contradiction, which completes the proof.Theorem 7.
Letλ=μ2 and Imμ=t. Then, the following asymptotic equalities hold as |λ|→∞:
(42)ϕ1(k)(x,λ)=dkdxkcos[μω1(x+1)]+O(1|μ|1-kexp(|t|ω1(x+1))),(43)ϕ2(k)(x,λ)=γ1δ1dkdxkcos[μ(ω2x+ω1h1+ω1)]+O(1|μ|1-kexp(|t|(ω2x+ω1h1+ω1))),(44)ϕ3(k)(x,λ)=γ1γ3δ1δ3dkdxkcos[μ(ω3x+ω2h2+ω1)]+O(1|μ|1-kexp(|t|(ω3x+ω2h2+ω1))),
for k=0 and k=1. Moreover, each of these asymptotic equalities holds uniformly for x.Proof.
Asymptotic formulas forϕ1(x,λ) and ϕ2(x,λ) are found in [9, Lemma 1.7] and [8, Theorem 3.2], respectively. But the formulas for ϕ3(x,λ) need individual considerations, since this solution is defined by the initial condition with some special nonstandard form. The initial-value problem (26)–(28) can be transformed into the equivalent integral equation
(45)u(x)=γ3δ3ϕ2(h2,λ)cosμω3x+γ4μω3δ4ϕ2′(h2,λ)sinμω3x+ω3μ∫h2xsin[μω3(x-y)]q(y)u(y)dy.
Inserting (43) in (45), we have
(46)ϕ3(x,λ)=γ1γ3δ1δ3cos[μ(ω3x+ω2h2+ω1)]+ω3μ∫h2xsin[μω3(x-y)]q(y)ϕ3(y,λ)dy+O(1|μ|exp(|t|(ω3x+ω2h2+ω1))).
Multiplying this by exp(-|t|(ω3x+ω2h2+ω1)) and denoting F(x,λ)=exp(-|t|(ω3x+ω2h2+ω1))ϕ3(x,λ), we have the next “asymptotic integral equation”:
(47)F(x,λ)=γ1γ3δ1δ3exp(-|t|(ω3x+ω2h2+ω1))×cos[μ(ω3x+ω2h2+ω1)]+ω3μ∫h2xsin[μω3(x-y)]×exp(-|t|ω3(x-y))q(y)F(y,λ)dy+O(1μ).
Putting M(λ)=maxx∈[h2,1]|F(x,λ)|, from the last equation, we derive that
(48)M(λ)≤M0(|γ1γ3δ1δ3|+1μ),
for some M0>0. Consequently, M(λ)=O(1) as |λ|→∞, and so ϕ3(x,λ) = O(exp(|t|(ω3x+ω2h2+ω1))) as |λ|→∞. Inserting the integral term of (46) yields (44) for k=0. The case k=1 of (44) follows at once on differentiating (43) and making the same procedure as in the case k=0.Theorem 8.
Letλ=μ2, μ=σ+it. Then, the following asymptotic formula holds for the eigenvalues of the boundary value transmission problem (1)–(7):
(49)μn=π(n-1)ω3+ω2h2+ω1+O(1n).Proof.
Puttingx=1 in Δ3(λ)=ϕ3(x,λ)χ3′(x,λ)-ϕ3′(x,λ)χ3(x,λ) and inserting χ3(1,λ)=β2′λ+β2, χ3′(1,λ)=β1′λ+β1, we have the following representation for Δ3(λ):
(50)Δ3(λ)=(β1′λ+β1)ϕ3(1,λ)-(β2′λ+β2)ϕ3′(1,λ).
Putting x=1 in (44) and inserting the result in (50), we derive now that
(51)Δ3(λ)=δ2δ4γ2γ4ω3β2′μ3sin[μ(ω3+ω2h2+ω1)]+O(|μ|2exp(2|t|(ω+ω2h2+ω1))).
By applying the well-known Rouché theorem which asserts that if f(z) and g(z) are analytic inside and on a closed contour Γ and |g(z)|<|f(z)| on Γ, then f(z) and f(z)+g(z) have the same number of zeros inside Γ provided that the zeros are counted with multiplicity on a sufficiently large contour, it follows that Δ3(λ) has the same number of zeros inside the contour as the leading term in (51). Hence, if λ0<λ1<λ2… are the zeros of Δ3(λ) and μn2=λn, we have
(52)π(n-1)ω3+ω2h2+ω1+δn,
for sufficiently large n, where |δn|<π/4(ω3+ω2h2+ω1) for sufficiently large n. By putting in (51), we have δn=O(1/n), and the proof is completed.Theorem 9.
The following asymptotic formula holds for the eigenfunctions(53)ϕλn(x)={ϕ1(x,λn),x∈[-1,h1),ϕ2(x,λn),x∈(h1,h2),ϕ3(x,λn),x∈(h2,1],
of (1)–(7):(54)
ϕ
λ
n
(
x
)
=
{
cos
[
ω
1
π
(
n
-
1
)
(
x
+
1
)
ω
2
+
ω
1
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+
O
(
1
n
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∈
[
-
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1
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,
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1
δ
1
cos
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x
+
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1
h
1
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1
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(
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ω
2
+
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1
h
1
+
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(
1
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x
∈
(
h
1
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h
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1
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3
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1
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3
cos
[
(
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x
+
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2
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1
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2
+
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(
1
n
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,
x
∈
(
h
2
,
1
]
.Proof.
Inserting (44) in the integral term of (51), we easily see that
(55)∫h2xsin[μω3(x-y)]q(y)ϕ3(y,λ)dy=O(exp(|t|(ω3x+ω2h2+ω1))).
Inserting in (42) yields
(56)ϕ3(x,λ)=γ1γ3δ1δ3cos[μ(ω3x+ω2h2+ω1)]+O(1|μ|exp|t|(ω3x+ω2h2+ω1)).
We already know that all eigenvalues are real. Furthermore, putting λ=-H, H>0 in (51), we infer that ω(-H)→∞ as H→+∞, and so ω(-H)≠0 for sufficiently large R>0. Consequently, the set of eigenvalues is bounded below. Letting λn=μn in (56), we now obtain
(57)ϕ3(x,λn)=γ1γ3δ1δ3cos[μn(ω3x+ω2h2+ω1)]+O(1μn),
since tn=lmμn for sufficiently large n. After some calculation, we easily see that
(58)cos[μn(ω3x+ω2h2+ω1)]=cos[(ω3x+ω2h2+ω1)π(n-1)ω3+ω2h2+ω1]+O(1n).
Consequently,
(59)ϕ3(x,λn)=γ1γ3δ1δ3cos[(ω3x+ω2h2+ω1)π(n-1)ω3+ω2h2+ω1]+O(1n).
In a similar method, we can deduce that
(60)ϕ2(x,λn)=γ1δ1cos[(ω2x+ω1h1+ω1)π(n-1)ω2+ω1h1+ω1]+O(1n),ϕ1(x,λn)=cos[ω1π(n-1)(x+1)ω2+ω1]+O(1n).
Thus is the proof of the theorem, completed.


## 4. Asymptotic Formulas for Normalized Eigenfunctions

We will first obtain the norms of the eigenelements(61)Φn:=(ϕλn(x)R1′(ϕλn)).
It is evident that the two-component vectors
(62)Φn:=(ϕλn(x)R1′(ϕλn)),n=0,1,2,…,
are the eigenelements of A with eigenvalue λn. For n≠m,
(63)〈Φn,Φm〉H=0,n,m=0,1,2,…
since A is symmetric. Putting
(64)ψn:=ϕλn(x)∥Φn∥H,
we see easily that the eigenelements
(65)Ψn=(ψλn(x)R1′(ψλn)),n,m=0,1,2,…
are orthonormal. That is, AΨn=λnΨn and 〈Ψn,Ψm〉H=δnm, where δnm is the Kronecker delta.Lemma 10.
The following asymptotic equality holds:(66)R1′(ϕλn)=O(1n).Proof.
It follows that from the equalitiesΔ3(λn)=0 that
(67)λnR1′(ϕ3λn)+R1(ϕ3λn)=0.
From (44), we obtain
(68)R1(ϕ3λn)=β1ϕ3λn(1)-β2ϕ3λn′(1)=β1O(1)-β2O(|sn|).
Applying Theorem 8, we now have
(69)R1(ϕ3λn)=O(n).
Inserting (69) in (67) and using Theorem 8, we find
(70)R1′(ϕ3λn)=-1λnR1(ϕ3λn)=O(1n).
The proof is completed.Theorem 11.
LetΦn be defined as in (61). Then, the following asymptotic formula holds for the norms ∥Φn∥H of the eigenelements Φn:(71)∥Φn∥H=ω12δ1δ3γ2γ4+ω22δ2δ3γ1γ4+ω32δ2δ4γ1γ32δ1δ3γ2γ4+O(1n).Proof.
By (54), we have
(72)∫-1h1(ϕλn(x))2dx=12+O(1n),(73)∫h1h2(ϕλn(x))2dx=γ122δ12+O(1n),(74)∫h21(ϕλn(x))2dx=γ122δ12γ32δ32+O(1n).
Using (66), (72), (73), and (74), we obtain
(75)∥Φn∥H2=ω12∫-1h1(ϕλn(x))2dx+ω22δ1δ2γ1γ2∫h1h2(ϕλn(x))2dx+ω32δ1δ2δ3δ4γ1γ2γ3γ4∫h21(ϕλn(x))2dx+δ1δ2δ3δ4ργ1γ2γ3γ4(R1′(ϕλn))2=ω12[12+O(1n)]+ω22δ1δ2γ1γ2[γ122δ12+O(1n)]+ω32δ1δ2δ3δ4γ1γ2γ3γ4[γ122δ12γ32δ32+O(1n)]+δ1δ2δ3δ4ργ1γ2γ3γ4O(1n2)=(ω12δ1δ3γ2γ4+ω22δ2δ3γ1γ4+ω32δ2δ4γ1γ3)2δ1δ3γ2γ4+O(1n).
Consequently,
(76)∥Φn∥H=ω12δ1δ3γ2γ4+ω22δ2δ3γ1γ4+ω32δ2δ4γ1γ32δ1δ3γ2γ4+O(1n),
which proves (71).Theorem 12.
The first components of the normalized eigenelements (65) have the following asymptotic representation as n→∞:(77)
ψ
n
(
x
)
=
{
2
δ
1
δ
3
γ
2
γ
4
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1
2
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3
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2
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4
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γ
1
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(
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1
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n
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1
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γ
1
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2
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1
δ
3
γ
2
γ
4
ω
1
2
δ
1
δ
3
γ
2
γ
4
+
ω
2
2
δ
2
δ
3
γ
1
γ
4
+
ω
3
2
δ
2
δ
4
γ
1
γ
3
cos
(
π
(
n
-
1
)
(
ω
2
x
+
ω
1
h
1
+
ω
1
)
ω
2
+
ω
1
h
1
+
ω
1
)
+
O
(
1
n
)
,
x
∈
(
h
1
,
h
2
)
,
γ
1
δ
1
γ
3
δ
3
2
δ
1
δ
3
γ
2
γ
4
ω
1
2
δ
1
δ
3
γ
2
γ
4
+
ω
2
2
δ
2
δ
3
γ
1
γ
4
+
ω
3
2
δ
2
δ
4
γ
1
γ
3
cos
(
π
(
n
-
1
)
(
ω
3
x
+
ω
2
h
2
+
ω
1
)
ω
3
+
ω
2
h
2
+
ω
1
)
+
O
(
1
n
)
,
x
∈
(
h
2
,
1
]
.Proof.
From (71), it follows that
(78)1∥Φn∥H=2δ1δ3γ2γ4ω12δ1δ3γ2γ4+ω22δ2δ3γ1γ4+ω32δ2δ4γ1γ3+O(1n).
Putting (54) and (78) in (64), we find the needed asymptotic formula (77).


## 5. Green's Function

LetA:H→H be defined by (10) and (11), and let λ not be an eigenvalue of A. For finding the resolvent operator R(λ,A)=(λI-A)-1, consider the operator equation
(79)(λI-A)U=F
for F=(f(x)f1)∈H. This operator equation is equivalent to the inhomogeneous differential equation
(80)u′′+(λω(x)-q(x))u=f(x)forx∈[-1,h1)∪(h1,h2)∪(h2,1]
subject to the inhomogeneous boundary conditions
(81)u′(-1)=0,λ(β1′u(1)-β2′u′(1))+(β1u(1)-β2u′(1))=f1
and the homogeneous transmission conditions
(82)γ1u(h1-0)-δ1u(h1+0)=0,γ2u′(h1-0)-δ2u′(h1+0)=0,γ3u(h2-0)-δ3u(h2+0)=0,γ4u′(h2-0)-δ4u′(h2+0)=0.
By applying the same techniques as in [8], we can prove that the problem (80)–(82) has a unique solution u(x,λ) which can be represented as(83)u(x,λ)={χ1(x,λ)Δ1(λ)ω12∫-1xϕ1(y,λ)f(y)dy+ϕ1(x,λ)Δ1(λ)(ω12∫xh1χ1(y,λ)f(y)dy+δ1δ2γ1γ2ω22∫h1h2χ2(y,λ)f(y)dy+δ1δ2δ3δ4γ1γ2γ3γ4ω32∫h21χ3(y,λ)f(y)dy+δ1δ2δ3δ4γ1γ2γ3γ4f1),x∈[-1,h1),χ2(x,λ)Δ2(λ)(γ1γ2δ1δ2ω12∫-1h1ϕ1(y,λ)f(y)dy+ω22∫h1xϕ2(y,λ)f(y)dy)×ϕ2(x,λ)Δ2(λ)(∫h21χ3(y,λ)f(y)dy+δ3δ4γ3γ4f1ω22∫xh2χ2(y,λ)f(y)dy+δ3δ4γ3γ4ω32∫h21χ3(y,λ)f(y)dy+δ3δ4γ3γ4f1),x∈(h1,h2),χ3(x,λ)Δ3(λ)(γ1γ2γ3γ4δ1δ2δ3δ4ω12∫-1h1ϕ1(y,λ)f(y)dy+γ3γ4δ3δ4ω22∫h1h2ϕ2(y,λ)f(y)dy+ω32∫h2xϕ3(y,λ)f(y)dy)+ϕ3(x,λ)Δ3(λ)(ω32∫x1χ3(y,λ)f(y)dy+f1),x∈(h2,1].Putting
(84)G(x,y,λ)={χ(x,λ)ϕ(y,λ)ω(λ)for-1≤y≤x≤1,ϕ(x,λ)χ(y,λ)ω(λ)for-1≤x≤y≤1,
where x≠0 and y≠0, we reduce (73) to
(85)u(x,λ)=ω12∫-1h1G(x,y,λ)f(y)dy+ω22δ1δ2γ1γ2∫h1h2G(x,y,λ)f(y)dy+ω32δ1δ2δ3δ4γ1γ2γ3γ4∫h21G(x,y,λ)f(y)dy+f1ϕ(x,λ)ω(λ).
On the other hand, by applying R1′ to Green's function with respect to the variable y and recalling that χ(x,λ) satisfies the initial conditions (30) and (31), we have
(86)R1′(G(x,·,λ))=β1′G(x,1,λ)-β2′∂G(x,1,λ)∂y=γ1γ2γ3γ4δ1δ2δ3δ4ϕ(x,λ)ω(λ)(β1′χ(1,λ)-β2′χ′(1,λ))=γ1γ2γ3γ4δ1δ2δ3δ4ϕ(x,λ)ω(λ)×(β1′(β2′λ+β2)-β2′(β1′λ+β1))=ργ1γ2γ3γ4δ1δ2δ3δ4ϕ(x,λ)ω(λ).
Inserting in (85) yields
(87)u(x,λ)=ω12∫-1h1G(x,y,λ)f(y)dy+ω22δ1δ2γ1γ2∫h1h2G(x,y,λ)f(y)dy+ω32δ1δ2δ3δ4γ1γ2γ3γ4∫h21G(x,y,λ)f(y)dy+f1δ1δ2δ3δ4ργ1γ2γ3γ4R1′(G(x,·,λ)).
We now introduce
(88)Gx,λ=(G(x,·,λ)R1′(G(x,·,λ))),
which we will call Green’s element of (80)–(82). The last formula (87) then takes the form
(89)u(x,λ)=〈Gx,λ,F¯〉,
where by F¯ we mean F¯=(f(x)¯f1¯).



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*Source: 159243-2013-09-01.xml*