In this work, we show the existence of a spectral function for a singular Sturm-Liouville problem with transmission condition

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

EIGENFUNCTION EXPANSION FOR SINGULAR STURM-LIOUVILLE PROBLEMS WITH

TRANSMISSION CONDITIONS

BILENDER P. ALLAHVERDIEV, H ¨USEYIN TUNA Communicated by Adrian Constantin

Abstract. In this work, we show the existence of a spectral function for a singular Sturm-Liouville problem with transmission condition. Also we estab- lish a Parseval equality and expansion formula in eigenfunctions in terms of the spectral function.

1. Introduction

Sturm-Liouville problems are one of the important research areas of mathemat- ical physics. They arise when we apply the method of separation of variables to equations in mathematical physics. To study the problem of expanding an arbitrary function as a series of eigenfunctions, we need eigenfunction expansions theorems for which there are a lot of studies, see [13, 21].

On the other hand, the Sturm-Liouville problems with transmission conditions arise in problems of heat and mass transfer, various physical transfer problems [15], radio science [16], and geophysics [12]. Such conditions are known by various names including transmission conditions, interface conditions, jump conditions and discontinuous conditions. Regular problems were investigated in [1, 8, 9, 10, 17, 18, 19, 20, 23, 24, 25], and singular problems in [2, 3, 4, 5, 6, 7, 22]. Li et al.

[14] investigated a singular Sturm-Liouville problems with transmission conditions at finitely many interior points. They gave a definition of Weyl function for such problems in the limit circle case.

In this article, we consider singular Sturm-Liouville problems with transmission conditions. We prove the existence of a spectral function, and give a Parseval equality and an expansion formula in eigenfunctions, for such problems.

2. Main results We consider the Sturm-Liouville expression

l(y) :=− p(x)y⁰0

+q(x)y, x∈(a, c)∪(c, b),

2010Mathematics Subject Classification. 34B24, 34A37, 34L10, 47E05.

Key words and phrases. Sturm-Liouville operator; singular point; transmission condition;

spectral function; Parseval equality; eigenfunction expansion.

c

2019 Texas State University.

Submitted December 11, 2017. Published January 6, 2018.

1

whereI1:= [a, c),I2:= (c, b],−∞< a < c < b <+∞andI:=I1∪I2. We assume that the pointsa,b andc are regular for the differential expressionl. pand qare real-valued, Lebesgue measurable functions onIand 1/p, q∈L¹(Ik),k= 1,2. The pointcis regular if 1/p, q∈L¹[c−, c+] for some >0.

Let us consider the Sturm-Liouville equation

l(y) =λy, x∈I, (2.1)

with the boundary condition

y(a) cosβ+ (py⁰)(a) sinβ= 0, β∈R:= (−∞,∞), (2.2) and transmission conditions

Y(c+) =CY(c−), Y = y

py⁰

, C∈M₂(R), detC=δ >0,

(2.3) whereM2(R) denotes the the 2×2 matrices with entries fromR.

Now, we introduce the Hilbert spaceH₁=L²(I₁)+L^{· 2}(I₂) with the inner product hf, gi_H1 :=

Z c a

f⁽¹⁾g⁽¹⁾dx+γ Z b

c

f⁽²⁾g⁽²⁾dx, γ= 1 δ, where

f(x) =

(f⁽¹⁾(x), x∈I1

f⁽²⁾(x), x∈I₂, g(x) =

(g⁽¹⁾(x), x∈I1

g⁽²⁾(x), x∈I₂.

Denote by D the set of linear functions y ∈ H₁ such that y, py⁰ are locally absolutely continuous functions on I, one-sided limits y(c±),(py⁰)(c±) exist and are finite and l(y) ∈ H1. The operator L defined by Ly = l(y) is called the maximal operatorT onH1.

For arbitrary functionsy, z∈ D, we have Green’s formula Z b

a

l(y)zdx− Z b

a

yl(z)dx= [y, z]_c−−[y, z]a+ [y, z]b−[y, z]c+, (2.4) where [y, z]_x=y(x)(pz⁰)(x)−(py⁰)(x)z(x) (x∈I). Denote by

φ(x, λ) =

(φ⁽¹⁾(x, λ), x∈I₁ φ⁽²⁾(x, λ), x∈I2

the solution of (2.1) satisfying the initial conditions

φ(a, λ) = sinβ, (pφ⁰)(a, λ) =−cosβ, (2.5) and transmission conditions

Φ(c+, λ) =CΦ(c−, λ), Φ(x, λ) :=

φ(x, , λ) (pφ⁰)(x, , λ)

, C∈M2(R), detC=δ >0.

(2.6) Now, to problem (2.1)-(2.3) we add the boundary condition

(py⁰)(b) sinα+y(b) cosα= 0, α∈R. (2.7) Then, (2.1)-(2.3), (2.7) is a regular problem for a Sturm-Liouville equation with transmission conditions.

In [8, 9, 10, 23, 24] the authors proved that the regular self-adjoint boundary- value problem (2.1)-(2.3), (2.7) with transmission conditions has a compact resol- vent, so it has a purely discrete spectrum.

Letλ_m,b (m∈N:={1,2, . . .}) denote the eigenvalues of this problem and φ_m,b(x) =

(φ⁽¹⁾_m,b(x), x∈I1

φ⁽²⁾_m,b(x), x∈I₂, φm,b(x) :=φ(x, λm,b)

the corresponding real-valued eigenfunctions which satisfy conditions (2.2), (2.3), (2.7). Iff ∈H1 is real-valued function with

f(x) =

(f⁽¹⁾(x), x∈I1

f⁽²⁾(x), x∈I₂, and

α²_m,b= Z c

a

(φ⁽¹⁾_m,b(x))²dx+γ Z b

c

(φ⁽²⁾_m,b(x))²dx, then

kfk²_H

1 = Z c

a

f⁽¹⁾(x)2

dx+γ Z b

c

f⁽²⁾(x)2

dx

=

∞

X

m=1

1 α²_m,b

nZ c a

f⁽¹⁾(x)φ⁽¹⁾_m,b(x)dx+γ Z b

c

f⁽²⁾(x)φ⁽²⁾_m,b(x)dxo² .

(2.8)

which is called the Parseval equality [9, 10].

Now, let us define a nondecreasing step function onR,

%b(λ) =







−P

λ<λ_m,b<0 1

α²_m,b, forλ <0 P

0≤λm,b<λ 1

α²_m,b forλ≥0. (2.9)

Then (2.8) can be written as Z c

a

(f⁽¹⁾(x))²dx+γ Z b

c

(f⁽²⁾(x))²dx= Z ∞

−∞

F²(λ)d%b(λ), (2.10) where

F(λ) = Z c

a

f⁽¹⁾(x)φ⁽¹⁾(x, λ)dx+γ Z b

c

f⁽²⁾(x)φ⁽²⁾(x, λ)dx.

We will show that the Parseval equality for problem (2.1)-(2.3), (2.7) can be ob- tained from (2.10) by lettingb→ ∞.

A function f defined on an interval [a, b] is said to be of bounded variation if there is a constantC >0 such that

n

X

k=1

|f(x_k)−f(x_k−1)| ≤C for every partitiona=x₀< x₁<· · ·< x_n=bof [a, b].

Letf be a function of bounded variation. Then, by the total variation off on [a, b], denoted by

V^b_a(f) := sup

n

X

k=1

|f(xk)−f(x_k−1)|,

where the supremum is taken over all (finite) partitions of the interval [a, b] (see [11]).

Lemma 2.1. For any positive N, there is a positive constant Υ = Υ(N)indepen- dent of b such that

V^N_−N(%b(λ)) = X

−N≤λ_m,b<N

1

α_m,b² =%b(N)−%b(−N)<Υ. (2.11) Proof. Let sinβ 6= 0. Sinceφ(x, λ) is continuous in domain−N ≤λ≤N,a≤x≤ c, by the condition φ(a, λ) = sinβ, there is a small positive number k such that, for|λ| ≤N,

1 k²

Z k a

φ⁽¹⁾(x, λ)dx2

>1

2sin²β. (2.12)

Let us define

fk(x) =

(1/k, a≤x < k 0, x≥k.

From (2.10), (2.11) and (2.12), we obtain Z k

a

f_k²(x)dx= k−a k² =

Z ∞

−∞

1 k

Z k a

φ⁽¹⁾(x, λ)dx2

d%_b(λ)

≥ Z N

−N

1 k

Z k a

φ⁽¹⁾(x, λ)dx2

d%b(λ)

> 1 2sin²β

Z N

−N

d%b(λ)

= 1

2sin²β{%_b(N)−%_b(−N)}, which proves the inequality (2.11).

If sinβ= 0, then we define the function fk(x) =

(1/k², a≤x < k 0, x≥k.

So, we obtain (2.11) by applying the Parseval equality.

Now, we recall the following two well-known Helly’s theorems.

Theorem 2.2([11]). Let (w_n)_n∈Nbe a uniformly bounded sequence of real nonde- creasing function on a finite interval a≤λ≤b. Then there exists a subsequence (wn_k)_k∈Nand a nondecreasing function wsuch that

k→∞lim wn_k(λ) =w(λ), a≤λ≤b.

Theorem 2.3 ([11]). Assume (w_n)_n∈N is a real, uniformly bounded, sequence of nondecreasing function on a finite intervala≤λ≤b, and suppose

n→∞lim w_n(λ) =w(λ), a≤λ≤b.

If f is any continuous function ona≤λ≤b, then

n→∞lim Z b

a

f(λ)dwn(λ) = Z b

a

f(λ)dw(λ).

We introduce the Hilbert spaceH :=L²(I1)+^· L²(I3), (I1:= [a, c),I3:= (c,∞)) with the inner product

hf, giH:=

Z c a

f⁽¹⁾g⁽¹⁾dx+γ Z ∞

c

f⁽²⁾g⁽²⁾dx, where

f(x) =

(f⁽¹⁾(x), x∈I1

f⁽²⁾(x), x∈I3, g(x) =

(g⁽¹⁾(x), x∈I1

g⁽²⁾(x), x∈I3.

Let %be any nondecreasing function on −∞< λ < ∞. Denote by L²_%(R) the Hilbert space of all functionsf :R→Rwhich are measurable with respect to the Lebesque-Stieltjes measure defined by%and such that

Z ∞

−∞

f²(λ)d%(λ)<∞, with the inner product

(f, g)% :=

Z ∞

−∞

f(λ)g(λ)d%(λ).

The main result of this article reads as follows.

Theorem 2.4. For the Sturm-Liouville problem (2.1)-(2.3), there exists a nonde- creasing function%(λ)on−∞< λ <∞ with the following properties:

(i) If

f(x) =

(f⁽¹⁾(x), x∈I1

f⁽²⁾(x), x∈I3

is real valued function and f ∈H, then there exist a functionF ∈L²_%(R) such that

n→∞lim Z ∞

−∞

n F(λ)−

Z c a

f⁽¹⁾(x)φ⁽¹⁾(x, λ)dx−γ Z n

c

f⁽²⁾(x)φ⁽²⁾(x, λ)o2

d%(λ) = 0, (2.13) and the Parseval equality

kfk²_H = Z c

a

f⁽¹⁾(x)² dx+γ

Z ∞ c

(f⁽²⁾(x))²dx= Z ∞

−∞

F²(λ)d%(λ). (2.14) (ii) The integral R∞

−∞F(λ)φ(x, λ)d%(λ)converge tof inH; that is,

n→∞lim nZ c

a

[f⁽¹⁾(x)− Z n

−n

F(λ)φ⁽¹⁾(x, λ)d%(λ)]²dx +γ

Z ∞ c

[f⁽²⁾(x)− Z n

−n

F(λ)φ⁽²⁾(x, λ)d%(λ)]²dxo

= 0.

We note that the function % is called a spectral function for boundary value problem (2.1)-(2.3).

Proof. Assume that

fξ(x) =

(f_ξ⁽¹⁾(x), x∈[a, c) f_ξ⁽²⁾(x), x∈(c, ξ]

satisfies the following conditions.

(1) f_ξ(x) vanishes outside the set [a, c)∪(c, ξ] withξ < b;

(2) (pf_ξ⁰)(x) has a continuous derivative.

(3) fξ(x) satisfy the conditions (2.2)-(2.3).

When we apply the Parseval equality (2.10) tofξ(x), we obtain Z c

a

(f_ξ⁽¹⁾(x))²dx+γ Z ξ

c

(f_ξ⁽²⁾(x))²dx= Z ∞

−∞

F_ξ²(λ)d%(λ), (2.15) where

F_ξ(λ) = Z c

a

f_ξ⁽¹⁾(x)φ⁽¹⁾(x, λ)dx+γ Z ξ

c

f_ξ⁽²⁾(x)φ⁽²⁾(x, λ)dx. (2.16) Sinceφ(x, λ) satisfies (2.1), we see that

φ(x, λ) = 1 λ

−(pφ⁰)⁰(x, λ) +q(x)φ(x, λ) .

By (2.16), we obtain F_ξ(λ) = 1 λ

Z c a

f_ξ⁽¹⁾(x)

− pφ⁽¹⁾⁰0

(x, λ) +q(x)φ⁽¹⁾(x, λ) dx

+ 1 λ

Z ξ c

f_ξ⁽²⁾(x)

− pφ⁽²⁾⁰⁰

(x, λ) +q(x)φ⁽²⁾(x, λ) dx.

Sincefξ(x) vanishes in a neighborhood of the pointbandfξ(x) andφ(x, λ) satisfy the boundary conditions (2.2), (2.3) we obtain

F_ξ(λ) = 1 λ

Z c a

φ⁽¹⁾(x, λ)

− pf_ξ⁽¹⁾⁰0

(x) +q(x)f_ξ⁽¹⁾(x) dx

+ 1 λ

Z b c

φ⁽²⁾(x, λ)

− pf_ξ⁽²⁾⁰0

(x) +q(x)f_ξ⁽²⁾(x) dx,

using integration by parts.

For any finiteN >0, using (2.10), we have Z

|λ|>N

F_ξ²(λ)d%b(λ)

≤ 1 N²

Z

|λ|>N

nZ c a

φ⁽¹⁾(x, λ)

− pf_ξ⁽¹⁾⁰⁰

(x) +q(x)f_ξ⁽¹⁾(x) dx

+γ Z b

c

φ⁽²⁾(x, λ)

− pf_ξ⁽²⁾⁰0

(x) +q(x)f_ξ⁽²⁾(x) dxo2

d%b(λ)

≤ 1 N²

Z ∞

−∞

nZ c a

φ⁽¹⁾(x, λ)

− pf_ξ⁽¹⁾⁰0

(x) +q(x)f_ξ⁽¹⁾(x) dx

+γ Z b

c

φ⁽²⁾(x, λ)

− pf_ξ⁽²⁾⁰0

(x) +q(x)f_ξ⁽²⁾(x) dxo²

d%b(λ)

= 1 N²

Z c a

− pf_ξ⁽¹⁾⁰0

(x) +q(x)f_ξ⁽¹⁾(x)² dx

+ 1 N²γ

Z ξ c

−pf_ξ⁽²⁾⁰0

(x) +q(x)f_ξ⁽²⁾(x)2

dx.

From (2.15), we see that

Z c

a

(f_ξ⁽¹⁾(x))²dx+γ Z ξ

c

(f_ξ⁽²⁾(x))²dx− Z N

−N

F_ξ²(λ)d%_b(λ)

< 1 N²

Z c a

− pf_ξ⁽¹⁾⁰⁰0

(x) +q(x)f_ξ⁽¹⁾(x)² dx

+ 1 N²γ

Z ξ c

− pf_ξ⁽²⁾⁰0

(x) +q(x)f_ξ⁽²⁾(x)2

dx.

(2.17)

By Lemma 2.1, the set{%b(λ)}is bounded. Using Theorems 2.2 and 2.3, we can find a sequence{bk} (b_k → ∞) such that the functions %_bk(λ) converge to a monotone function%(λ). Passing to the limit with respect to{b_k}in (2.17), we obtain

Z c a

(f_ξ⁽¹⁾(x))²dx+γ Z ξ

c

(f_ξ⁽²⁾(x))²dx− Z N

−N

F_ξ²(λ)d%(λ)

< 1 N²

Z c a

− pf_ξ⁽¹⁾⁰⁰

(x) +q(x)f_ξ⁽¹⁾(x)² dx

+ 1 N²γ

Z ξ c

− pf_ξ⁽²⁾⁰0

(x) +q(x)f_ξ⁽²⁾(x)² dx.

Hence, lettingN → ∞, we obtain Z c

a

(f_ξ⁽¹⁾(x))²dx+γ Z ξ

c

(f_ξ⁽²⁾(x))²dx= Z ∞

−∞

F_ξ²(λ)d%(λ).

Now, let f be an arbitrary real valued function on H. It is known that there exists a sequence of functions{fξ(x)}satisfying the condition 1-3 and such that

ξ→∞lim nZ c

a

(f⁽¹⁾(x)−f_ξ⁽¹⁾(x))²dx+γ Z ξ

c

(f⁽²⁾(x)−f_ξ⁽²⁾(x))²dxo

= 0.

Let

Fξ(λ) = Z ∞

a

f_ξ⁽¹⁾(x)φ⁽¹⁾(x, λ)dx+γ Z ∞

a

f_ξ⁽²⁾(x)φ⁽²⁾(x, λ)dx.

Then, we have Z c

a

(f_ξ⁽¹⁾(x))²dx+γ Z ξ

c

(f_ξ⁽²⁾(x))²dx= Z ∞

−∞

F_ξ²(λ)d%(λ).

Since

Z c a

(f_ξ⁽¹⁾

1 (x)−f_ξ⁽¹⁾

2 (x))²dx+ Z ∞

c

(f_ξ⁽²⁾

1 (x)−f_ξ⁽²⁾

2 (x))²dx→0 asξ1, ξ2→ ∞, we have

Z ∞

−∞

(Fξ₁(λ)−Fξ₂(λ))²d%(λ)

= Z c

a

(f_ξ⁽¹⁾

1 (x)−f_ξ⁽¹⁾

2 (x))²dx+ Z ∞

c

(f_ξ⁽²⁾

1 (x)−f_ξ⁽²⁾

2 (x))²dx→0 asξ1, ξ2→ ∞. Consequently, there is a limit functionF which satisfies

Z c a

f⁽¹⁾(x)² dx+γ

Z ∞ c

f⁽²⁾(x)² dx=

Z ∞

−∞

F²(λ)d%(λ), by the completeness of the spaceL²_%(R).

Our next goal is to show that

Kξ(λ) = Z c

a

f⁽¹⁾(x)φ⁽¹⁾(x, λ)dx+γ Z ξ

c

f⁽²⁾(x)φ⁽²⁾(x, λ)dx→F

as ξ→ ∞, in the metric of spaceL²_%(R). Let gbe another real-valued function in H. By a similar arguments, letG(λ) be defined by g. It is clear that

Z c a

f⁽¹⁾(x)−g⁽¹⁾(x)² dx+γ

Z ∞ c

f⁽²⁾(x)−g⁽²⁾(x)² d

= Z ∞

−∞

{F(λ)−G(λ)}²d%(λ).

Set

g(x) =

(f(x), x∈[a, c)∪[c, ξ]

0, x∈(ξ,∞).

Then Z ∞

−∞

{F(λ)−Kξ(λ)}²d%(λ) =γ Z ∞

ξ

(f⁽²⁾(x))²dx→0, as ξ→ ∞,

which proves thatK_ξ converges toF in L²_%(R) asξ→ ∞. This proves (i).

Now, we prove (ii). Suppose that the real valued functionsf, g∈H, andF(λ) andG(λ) are their Fourier transforms (see (2.13)). ThenF∓Gare transforms of f∓g. Consequently, by (2.14), we have

Z c a

f⁽¹⁾(x) +g⁽¹⁾(x)² dx+γ

Z ∞ c

f⁽²⁾(x) +g⁽²⁾(x)² dx

= Z ∞

−∞

[F(λ) +G(λ)]²d%(λ), Z c

a

f⁽¹⁾(x)−g⁽¹⁾(x)2

dx+γ Z ∞

c

f⁽²⁾(x)−g⁽²⁾(x)2

dx

= Z ∞

−∞

[F(λ)−G(λ)]²d%(λ).

Subtracting the second relation from the first, we obtain Z c

a

f⁽¹⁾(x)g⁽¹⁾(x)dx+γ Z ∞

c

f⁽²⁾(x)g⁽²⁾(x)dx= Z ∞

−∞

F(λ)G(λ)d%(λ) (2.18) which is called the generalized Parseval equality. Set

f_τ^(j)(x) = Z τ

−τ

F(λ)φ^(j)(x, λ)d%(λ), j= 1,2,

where F is the function defined in (2.13). Let g ∈ H be a real valued function which equals zero outside the set [a, c)∪(c, µ]. Thus, we obtain

Z c a

f_τ⁽¹⁾(x)g⁽¹⁾(x)dx+γ Z µ

c

f_τ⁽²⁾(x)g⁽²⁾(x)dx

= Z c

a

nZ τ

−τ

F(λ)φ⁽¹⁾(x, λ)d%(λ)o

g⁽¹⁾(x)dx +γ

Z µ c

nZ τ

−τ

F(λ)φ⁽²⁾(x, λ)d%(λ)o

g⁽²⁾(x)dx

= Z τ

−τ

F(λ)nZ c a

φ⁽¹⁾(x, λ)g⁽¹⁾(x)dx+γ Z µ

c

φ⁽²⁾(x, λ)g⁽²⁾(x)dxo d%(λ)

= Z τ

−τ

F(λ)G(λ)d%(λ).

(2.19)

Subtracting (2.18) and (2.19), we have Z c

a

f⁽¹⁾(x)−f_τ⁽¹⁾(x)

g⁽¹⁾(x)dx+γ Z ∞

c

f⁽²⁾(x)−f_τ⁽²⁾(x)

g⁽²⁾(x)dx

= Z

|λ|>τ

F(λ)G(λ)d%(λ).

Using Cauchy-Schwarz inequality, we obtain Z c

a

(f⁽¹⁾(x)−f_τ⁽¹⁾(x))g⁽¹⁾(x)dx+γ Z ∞

c

(f⁽²⁾(x)−f_τ⁽²⁾(x))g⁽²⁾(x)dx2

≤ Z

|λ|>τ

F²(λ)d%(λ) Z

|λ|>τ

G²(λ)d%(λ).

We apply this inequality to the function g(x) =

(fτ(x)−f(x), x∈[a, c)∪(c, µ]

0, x∈(µ,∞),

we obtain Z c

a

f⁽¹⁾(x)−f_τ⁽¹⁾(x)2

dx+γ Z ∞

c

f⁽²⁾(x)−f_τ⁽²⁾(x)2

dx

≤ Z

|λ|>τ

F²(λ)d%(λ).

Lettingτ → ∞yields the desired result, since the right-hand side does not depend

onµ.

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Bilender P. Allahverdiev

Department of Mathematics, S¨uleyman Demirel University, 32260 Isparta, Turkey E-mail address:[email protected]

H¨useyin Tuna

Department of Mathematics, Mehmet Akif Ersoy University, 15030 Burdur, Turkey E-mail address:[email protected]