On Jost Solution of Diffusion Equation With Discontinuous Coeffcient

Fen Bilimleri Dergisi, Cilt 32, No. 1, (2011)

On Jost Solution of Diffusion Equation With Discontinuous

Coefficient

Yas¸ar C¸ AKMAK∗_{and Seval KARACAN} Department of Mathematics, Faculty of Science, Cumhuriyet University, SIVAS 58140, TURKEY

Received 08.04.2011; accepted 22.06.2011

Abstract. In this work, integral representation of Jost solution of a diffusion equation is obtained and the properties of representation of the kernel are studied.

AMS subject classifications: Primary 34A55, Secondary 34B24, 34L05 Key words: Diffusion equation, Integral representation, Jost solution

1. Introduction

An important role in the spectral theory of linear operators was played by the trans-formation operator. Marchenko [3] , [5] first applied the transtrans-formation operator to the solution of the inverse problem. Transformation operator was also used in the fundamental paper of Gelfand and Levitan [4] . Yurko [11] studied the inverse problem theory for Sturm Liouville operators on the half-line.

−y00_{+ q (x) y = λy, x > 0 has a unique solution satisfying the integral equation}

e (x, ρ) = exp (iρx) − 1

2iρ +∞_Z

x

(exp (iρ (x − t)) − exp (iρ (t − x))) q (t) e (t, ρ) dt. where ρ ∈©ρ : Imρ ≥ 0, ρ 6= 0, λ = ρ2ª_{. The function e (x, ρ) is called the Jost} so-lution [11] .

Huseynov [14] has considered the differential equation

−y00_{+ q (x) y = λ}2_{y, x ∈ (0, a) ∪ (a, +∞)}

with discontinous conditions y (a − 0) = αy (a + 0) , y0_{(a − 0) = α}−1_y0_{(a + 0),}

where α 6= 1, α > 0, λ is a complex valued function and satisfies the condition

xq (x) ∈ L1_{(I) . Then for all λ from the upper half-plane, the Jost solution is unique} and it is in the form

e (x, λ) = e0(x, λ) + +∞_Z

x

K (x, t) eiλt_dt.

∗_{Corresponding author. Email addresses: ycakmak@cumhuriyet.edu.tr}

Let’s consider the diffusion equation with discontinous coefficient

−y00_{+ [2λp (x) + q (x)] y = λ}2_{ρ (x) y, x ∈ (0, a) ∪ (a, +∞) = I. (1.1)} we assume that

(1 + x) q (x) , p (x) ∈ L1_{(I) , (1.2)}

p (x) ∈ BC (I) , (1.3)

where L1_{(I) is the space of integrable functions and BC (I) denotes bounded,} con-tinuous functions on I,

λ is a complex parameter and ρ (x) is a step function: ρ (x) =

½

1, 0 ≤ x ≤ a

α2_{, a < x ≤ π} , 0 < α 6= 1. (1.4) The function e (x, λ) , satisfying the equation (1.1) and condition at infinity lim

x→∞e (x, λ) e

−iλx _{= 1 is said to be Jost solution of (1.1). It is easy to show that if}

q (x) ≡ 0, then the Jost solution is the function e0(x, λ) = ( R1(x) eiλx , x > a α+_R 2(x) eiλµ +_(x) + α−_R 3(x) eiλµ −_(x) , 0 < x < a (1.5) where µ±_{(x) = ±}p_{ρ (x)x + a}³_{1 ∓}p_{ρ (x)}´_{and α}±₌1 2 µ α ± 1 α ¶ .

2. Main Result

The main result of the present paper is the following.

Theorem. If a complex-valued function q (x) fulfilled condition (1.2), then there exists the Jost solution e (x, λ) of equation (1.1) for all λ from the upper half-plane, it is unique and representable in the form

e (x, λ) = e0(x, λ) + +∞_Z

µ+_(x)

K (x, t) eiλtdt, (2.1) where, for each fixed x ∈ (0, a) ∪ (a, +∞), the kernel K (x, .) belongs to the space

L1(µ+(x) , ∞). Moreover, K(x, t) is countinous with partial derivatives

∂K(x, t) ∂x , ∂K(x, t)

∂t for t 6= µ

−_{(x) and it satisfies the following properties}

a)+∞R µ+_(x) |K (x, t)| dt ≤ eσ(x)_{−1, σ (x) =}+∞R x [2 |p (t)| + (1 + |t|) |q (t)|] dt (2.2) b)α+_R00 2(x) + 2iλαα+R 0 2(x) − 2λα+p (x) R2(x) +2ip (x) K (x, µ+_(x))−α+_{q (x) R} 2(x)−2αdK (x, µ +_(x)) dx = 0 (2.3)

c)α−_R00 3(x) − 2iλαα−R 0 3(x) − 2λα−p (x) R3(x) −2ip (x) K (x, µ−_{(x) − 0) + 2ip (x) K (x, µ}−_{(x) + 0)} −α−_{q (x) R} 3(x)−2αdK (x, µ −_{(x) − 0)} dx +2α dK (x, µ−_{(x) + 0)} dx = 0 (2.4) d)∂ 2_{K (x, t)} ∂x2 −α 2∂2K (x, t) ∂t2 +2ip (x) ∂K (x, t) ∂t = q (x) K (x, t) (2.5) e) lim x+t→∞ ∂K (x, t) ∂x = 0, x+t→∞lim ∂K (x, t) ∂t = 0 (2.6)

Here e0(x, λ) is in the form of (1.5) which is shown above.

e (x, λ) is in the form e (x, λ) = e0(x, λ) + +∞_Z x S0(x, t, λ) [2λp (t) + q (t)] e (t, λ) dt, (2.7) where S0(x, t, λ) =                sin λ (x − t) λ , a < x < t α+sin λ (µ+(x) − t) λ + α −sin λ (µ−(x) − t) λ , x < a < t sin λα (x − t) λα , x < t < a (2.8) Proof:

Substituting (2.1) into (2.7) , we get

e0(x, λ) + +∞_R µ+_(x) K (x, t) eiλt_{dt = e} 0(x, λ) ++∞R x S0(x, t, λ) [2λp (t) + q (t)] Ã e0(t, λ) + +∞_R µ+_(t) K (t, ξ) eiλξ_dξ ! dt.

Firstly, for x > a, we arrive +∞_R x K (x, t) eiλt_{dt + (R} 1(x) − 1) eiλx= =+∞R x S0(x, t, λ) (2λp (t) + q (t)) µ R1(t) eiλt+ +∞_R t K (t, ξ) eiλξ_dξ ¶ dt

Now, if one evaluates the right-hand side of above equation, then it follows that +∞_R x S0(x, t, λ) [2λp (t) + q (t)] R1(t) eiλtdt + +∞_R x S0(x, t, λ) [2λp (t) + q (t)] µ_+∞ R t K (t, ξ) eiλξ_dξ ¶ dt

=+∞R x sin λ (x − t) 2p (t) R1(t) eiλt_{dt +}+∞R x sin λ (x − t) λ q (t) R1(t) e iλt_dt ++∞R x sin λ (x − t) 2p (t) µ_+∞ R t K (t, ξ) eiλξ_dξ ¶ dt ++∞R x sin λ (x − t) λ q (t) µ_+∞ R t K (t, ξ) eiλξ_dξ ¶ dt = −i +∞_R x p (t) R1(t) eiλxdt + i +∞_R x p (t) R1(t) eiλ(−x+2t)dt +1 2 +∞_R x q (t) R1(t) à x R −x+2t eiλu_du ! dt − i +∞_R x p (t) µ_+∞ R t K (t, ξ) eiλ(x−t+ξ)_dξ ¶ dt +i+∞R x p (t) µ_+∞ R t K (t, ξ) eiλ(−x+t+ξ)_dξ ¶ dt +1 2 +∞_R x q (t) à +∞_R t K (t, ξ) x−t+ξ_R −x+t+ξ eiλu_dudξ ! dt = −ieiλx+∞R x p (t) R1(t) dt +i 2 +∞_R x p µ x + t 2 ¶ R1 µ x + t 2 ¶ eiλt_dt −1 2 +∞_R x à +∞_R (x+t)/2 q (u) R1(u) du ! eiλt_dt −i+∞R x µ_+∞ R x p (u) K (u, t − x + u) du ¶ eiλt_dt +i+∞R x à (x+t)/2_R x p (u) K (u, t + x − u) du ! eiλt_dt +1 2 +∞_R x à +∞_R x q (u)t−x+uR t+x−u K (u, ξ) dξdu ! eiλt_dt.

Thus, in the case of x > a, one can obtain the following equations

R1(x) = 1 − i +∞_R x p (t) R1(t) dt K (x, t) = i 2p µ x + t 2 ¶ R1 µ x + t 2 ¶ −1 2 +∞_R (x+t)/2 q (u) R1(u) du −i +∞_R x p (u) K (u, t − x + u) du + i (x+t)/2_R x p (u) K (u, t + x − u) du +1 2 +∞_R x q (u)t−x+uR t+x−u K (u, ξ) dξdu. (2.9)

Similiarly, for 0 < x < a, we obtain R2(x) − 1 = −i α a R x p (t) R2(t) dt − i +∞_R a p (t) R1(t) dt R3(x) − 1 = i α a R x p (t) R3(t) dt − i +∞_R a p (t) R1(t) dt,

where K (x, t) ≡ 0 for |ξ| < αt − αa + a. Further one can get the following integral equations for the kernel K (x, t) ,

If 0 < x < a, αx − αa + a < t < −αx + αa − a ; K (x, t) = iα + 2α2p µ t + αx + αa − a 2α ¶ R2 µ t + αx + αa − a 2α ¶ −α+ 2α a R (t+αx+αa−a)/2α q (u) R2(u) du +iα − 2α2p µ −t + αx + αa + a 2α ¶ R3 µ −t + αx + αa + a 2α ¶ −α− 2α a R (−t+αx+αa+a)/2α q (u) R3(u) du − α+ 2 +∞_R a q (u) R1(u) du +iα − 2 p µ t − αx + αa + a 2 ¶ R1 µ t − αx + αa + a 2 ¶ +α− 2 (t−αx+αa+a)/2_R a q (u) R1(u) du − i α a R x

p (u) K (u, t − αx + αu) du

+i

α

(t+αx+αa−a)/2α_R

x

p (u) K (u, t + αx − αu) du − 1

2α a R x q (u)t−αx+αuR t+αx−αu K (u, ξ) dξdu −iα++∞R a

p (u) K (u, t − αx + αa − a + u) du −α

+ 2 +∞_R a q (u)t−αx+αa−a+uR t+αx−αa+a−u K (u, ξ) dξdu +iα−(t−αx+αa+a)/2R a

p (u) K (u, t − αx + αa + a − u) du

+α − 2 (t−αx+αa+a)/2_R a q (u) t−αx+αa+a−u_R t+αx−αa−a+u K (u, ξ) dξdu (2.10) if 0 < x < a, -αx + αa − a < t < −αx + αa + a; K (x, t) = iα + 2α2p µ t + αx + αa − a 2α ¶ R2 µ t + αx + αa − a 2α ¶ −α + 2α a R (t+αx+αa−a)/2α q (u) R2(u) du

+iα − 2α2p µ −t + αx + αa + a 2α ¶ R3 µ −t + αx + αa + a 2α ¶ −α − 2α a R (−t+αx+αa+a)/2α q (u) R3(u) du −α + 2 +∞_R a q (u) R1(u) du +iα − 2 p µ t − αx + αa + a 2 ¶ R1 µ t − αx + αa + a 2 ¶ +α − 2 (t−αx+αa+a)/2_R a q (u) R1(u) du − i α a R x

p (u) K (u, t − αx + αu) du

+i

α

(t+αx+αa−a)/2α_R

x

p (u) K (u, t + αx − αu) du − 1 2α a R x q (t)t−αx+αuR t+αx−αu K (u, ξ) dξdu −iα++∞R a

p (u) K (u, t − αx + αa − a + u) du −α + 2 +∞_R a q (u)t−αx+αa−a+uR t+αx−αa+a−u K (u, ξ) dξdu +iα−(t−αx+αa+a)/2R a

p (u) K (u, t − αx + αa + a − u) du

+α − 2 (t−αx+αa+a)/2_R a q (u)t−αx+αa+a−uR t+αx−αa−a+u K (u, ξ) dξdu (2.11) if 0 < x < a, −αx + αa + a < t < +∞; K (x, t) = iα + 2 p µ t + αx − αa + a 2 ¶ R1 µ t + αx − αa + a 2 ¶ −α + 2 +∞_R (t+αx−αa+a)/2 q (u) R1(u) du +iα− 2 p µ t − αx + αa + a 2 ¶ R1 µ t − αx + αa + a 2 ¶ −i α a R x

p (u) K (u, t − αx + αu) du

+i

α

a

R

x

p (u) K (u, t + αx − αu) du − 1 2α a R x q (u)t−αx+αuR t+αx−αu K (u, ξ) dξdu −iα++∞R a

+iα+(t+αx−αa+a)/2R

a

p (u) K (u, t + αx − αa + a − u) du −α + 2 +∞_R a q (u)t−αx+αa−a+uR t+αx−αa+a−u K (u, ξ) dξdu −iα−+∞R a

p (u) K (u, t + αx − αa − a + u) du

+iα−(t−αx+αa+a)/2R a

p (u) K (u, t − αx + αa + a − u) du α− 2 (t−αx+αa+a)/2_R a q (u)t−αx+αa+a−uR t+αx−αa−a+u K (u, ξ) dξdu. (2.12)

Let’s shown by the method of successive approximations that, equations of (2.10) , (2.11) and (2.12) have solution K (x, .) ∈ L1(µ+_{(x) , ∞) satisfying property of (2.2).}

In the case of x > a K0(x, t) = i 2p µ x + t 2 ¶ R1 µ x + t 2 ¶ −1 2 +∞_R (x+t)/2 q (u) R1(u) du K1(x, t) = −i +∞_R x p (u) K0(u, t − x + u) du +i (x+t)/2_R x p (u) K0(u, t + x − u) du + 1 2 +∞_R x q (u) t−x+u_R t+x−u K0(u, ξ) dξdu Kn(x, t) = −i +∞_R x p (u) Kn−1(u, t − x + u) du +i (x+t)/2_R x p (u) Kn−1(u, t + x − u) du +1 2 +∞_R x q (u)t−x+uR t+x−u Kn−1(u, ξ) dξdu K (x, t) = ∞ X n=0 Kn(x, t) (2.13)

Let us show by induction that +∞_Z αx−αa+a |Kn(x, t)| dt ≤σ n+1_(x) (n + 1)!, where, +∞_Z x [2 |p (t)| + (1 + |t|) |q (t)|] dt = σ (x) . (2.14) Indeed,

+∞_R x |K0(x, t)| dt ≤ 1 2 +∞_R x ¯ ¯ ¯ ¯p µ x + t 2 ¶¯_¯ ¯ ¯ ¯ ¯ ¯ ¯R1 µ x + t 2 ¶¯_¯ ¯ ¯ dt +1 2 +∞_R x +∞_R (x+t)/2

|q (u)| |R1(u)| dudt. We obtain +∞_R x |K0(x, t)| dt ≤ +∞_R x [2 |p (t)| + (1 + |t|) |q (t)|] dt. Thus +∞_Z x |K0(x, t)| dt ≤ σ (x) . (2.15) For n = 1 +∞_R x |K1(x, t)| dt ≤ +∞_R x +∞_R x

|p (u)| |K0(u, t − x + u)| dudt ++∞R

x

(x+t)/2_R

x

|p (u)| |K0(u, t + x − u)| dudt +1 2 +∞_R x +∞_R x |q (u)|t−x+uR t+x−u |K0(u, ξ)| dξdudt. ≤+∞R x [2 |p (t)| + (1 + |t|) |q (t)|] σ (t) dt. We get +∞_Z x |K1(x, t)| dt ≤ σ 2_(x) 2! . (2.16) Suppose that +∞_Z x |Kn−1(x, t)| dt ≤σ n_(x) n! (2.17)

is valid for n − 1. Let’s show that +∞_Z x |Kn(x, t)| dt ≤ σn+1_(x) (n + 1)!. (2.18) +∞_R x |Kn(x, t)| dt ≤ +∞_R x +∞_R x

|p (u)| |Kn−1(u, t − x + u)| dudt

+ +∞_R

x

(x+t)/2_R

x

|p (u)| |Kn−1(u, t + x − u)| dudt+1

2 +∞_R x +∞_R x |q (u)| t−x+u_R t+x−u |Kn−1(u, ξ)| dξdudt ≤ +∞_R x [2 |p (t)| + (1 + |t|) |q (t)|]σ n_(t) n! dt.

Hence +∞_Z x |Kn(x, t)| dt ≤ σ n+1_(x) (n + 1)!

is valid. It follows by virtue of last equation in the case of x > a, series, (2.13) convergents in L1(x, +∞) for each fixed x, and then

+∞_R x |K (x, t)| dt ≤ P∞ n=0 +∞_R x |Kn(x, t)| dt ≤ 1 − 1 + σ (x) +σ2(x) 2! + ... + σn+1_(x) (n + 1)! + ... = eσ(x)_{−1. (2.19)}

Now, we consider the case of 0 < x < a.

Then, put K (x, t) ≡ 0 for t > −αx + αa − a in the case of αx − αa + a < t <

−αx + αa − a −αx+αa−a_R αx−αa+a |K0(x, t)| dt ≤ α+ 2α2 −αx+αa−a_R αx−αa+a ¯ ¯ ¯ ¯p µ t + αx + αa − a 2α ¶¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯R2 µ t + αx + αa − a 2α ¶¯ ¯ ¯ ¯ dt +α + 2α −αx+αa−a_R αx−αa+a à a R (t+αx+αa−a)/2α |q (u)| |R2(u)| du ! dt +α− 2α2 −αx+αa−a_R αx−αa+a ¯ ¯ ¯ ¯p µ −t + αx + αa + a 2α ¶¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯R3 µ −t + αx + αa + a 2α ¶¯ ¯ ¯ ¯ dt +α − 2α −αx+αa−a_R αx−αa+a à a R (−t+αx+αa+a)/2α |q (u)| |R3(u)| du ! dt +α+ 2 −αx+αa−a_R αx−αa+a µ_+∞ R a |q (u)| |R1(u)| du ¶ dt +α − 2 −αx+αa−a_R αx−αa+a ¯ ¯ ¯ ¯p µ t − αx + αa + a 2 ¶¯_¯ ¯ ¯ ¯ ¯ ¯ ¯R1 µ t − αx + αa + a 2 ¶¯_¯ ¯ ¯ dt +α− 2 −αx+αa−a_R αx−αa+a à (t−αx+αa+a)/2_R a |q (u)| |R1(u)| du ! dt ≤ α + α +∞_R x |p (t)| dt + α++∞R x (1 + |t|) |q (t)| dt −α − α +∞_R x |p (t)| dt +α−+∞R x (1 + |t|) |q (t)| dt + αα++∞R x (1 + |t|) |q (t)| dt +α−+∞R x |p (t)| dt + 2αα−+∞R x (1 + |t|) |q (t)| dt

< α++∞R x [2 |p (t)| + (1 + |t|) |q (t)|] dt +α−+∞R x [2 |p (t)| + (1 + |t|) |q (t)|] dt +α (α+_{+ 2α}−₎+∞R x (1 + |t|) |q (t)| dt + α−+∞R x |p (t)| dt < α+_{σ (x) + α}−_{σ (x) + 2α}+∞R x (1 + |t|) |q (t)| dt + 4α+∞R x |p (t)| dt < (α+_{+ α}−_{+ 2α) σ (x) = c1σ (x) . (2.20)} Hence −αx+αa−a_Z αx−αa+a |K0(x, t)| dt ≤ c1σ (x) .

Put K (x, t) ≡ 0 for t < −αx+αa−a in the case of −αx+αa−a < t < −αx+αa+a to get −αx+αa+a_R −αx+αa−a |K0(x, t)| dt ≤ α + 2α2 −αx+αa+a_R −αx+αa−a ¯ ¯ ¯ ¯p µ t + αx + αa − a 2α ¶¯_¯ ¯ ¯ ¯ ¯ ¯ ¯R2 µ t + αx + αa − a 2α ¶¯_¯ ¯ ¯ dt +α+ 2α −αx+αa+a_R −αx+αa−a à a R (t+αx+αa−a)/2α |q (u)| |R2(u)| du ! dt +α − 2α2 −αx+αa+a_R −αx+αa−a ¯ ¯ ¯ ¯p µ −t + αx + αa + a 2α ¶¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯R3 µ −t + αx + αa + a 2α ¶¯ ¯ ¯ ¯ dt +α − 2α −αx+αa+a_R −αx+αa−a à a R (−t+αx+αa+a)/2α |q (u)| |R3(u)| du ! dt +α + 2 −αx+αa+a_R −αx+αa−a µ_+∞ R a |q (u)| |R1(u)| du ¶ dt +α− 2 −αx+αa+a_R −αx+αa−a ¯ ¯ ¯ ¯p µ t − αx + αa + a 2 ¶¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯R1 µ t − αx + αa + a 2 ¶¯ ¯ ¯ ¯ dt +α − 2 −αx+αa+a_R −αx+αa−a à (t−αx+αa+a)/2_R a |q (u)| |R1(u)| du ! dt < α + α +∞_R x |p (t)| dt + α++∞R x (1 + |t|) |q (t)| dt +α − α +∞_R x |p (t)| dt + α−+∞R x (1 + |t|) |q (t)| dt +α++∞R x (1 + |t|) |q (t)| dt + α−+∞R x |p (t)| dt

+αα−+∞R x (1 + |t|) |q (t)| dt < α++∞R x [2 |p (t)| + (1 + |t|) |q (t)|] dt +α−+∞R x [2 |p (t)| + (1 + |t|) |q (t)|] dt + (αα−_{+ α}+₎+∞R x (1 + |t|) |q (t)| dt + α−+∞R x |p (t)| dt < α+_{σ (x) + α}−_{σ (x) + αα}++∞R x [2 |p (t)| + (1 + |t|) |q (t)|] dt = (α+_{+ α}−_{+ αα}+_{) σ (x) = c2σ (x) . (2.21)} Thus −αx+αa+a_Z −αx+αa−a |K0(x, t)| dt ≤ c2σ (x) is obtained. For −αx + αa + a < t < +∞, +∞_R −αx+αa+a |K0(x, t)| dt ≤α + 2 +∞_R −αx+αa+a ¯ ¯ ¯ ¯p µ t + αx − αa + a 2 ¶¯_¯ ¯ ¯ ¯ ¯ ¯ ¯R1 µ t + αx − αa + a 2 ¶¯_¯ ¯ ¯ dt +α+ 2 +∞_R −αx+αa+a à +∞_R (t+αx−αa+a)/2 |q (u)| |R1(u)| du ! dt +α− 2 +∞_R −αx+αa+a ¯ ¯ ¯ ¯p µ t − αx + αa + a 2 ¶¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯R1 µ t − αx + αa + a 2 ¶¯ ¯ ¯ ¯ dt < α++∞R x |p (t)| dt + α++∞R x (1 + |t|) |q (t)| dt + α++∞R x |p (t)| dt = α++∞R x [2 |p (t)| + (1 + |t|) |q (t)|] dt = α+_{σ (x) . (2.22)} So +∞_Z −αx+αa+a |K0(x, t)| dt ≤ α+σ (x) . Then, it follows from (2.20) , (2.21) and (2.22) for 0 < x < a

+∞_Z

αx−αa+a

|K0(x, t)| dt ≤ Cσ (x) , (2.23) where C = c1+ c2+ α+_{, c}

Now, for n = 1, if one evaluate integral +∞R αx−αa+a |K1(x, t)| dt then +∞_R αx−αa+a |K1(x, t)| dt ≤ 1 α +∞_R αx−αa+a µ_a R x

|p (u)| |K0(u, t − αx + αu)| du ¶ dt +1 α +∞_R αx−αa+a µ_a R x

|p (u)| |K0(u, t + αx − αu)| du ¶ dt + 1 2α +∞_R αx−αa+a à a R x |q (u)|t−αx+αuR t+αx−αu |K0(u, ξ)| dξdu ! dt +α+ +∞R αx−αa+a µ_+∞ R a

|p (u)| |K0(u, t − αx + αa − a + u)| du ¶ dt +α+ +∞R αx−αa+a à (t+αx−αa+a)/2_R a

|p (u)| |K0(u, t + αx − αa + a − u)| du ! dt +α + 2 +∞_R αx−αa+a à +∞_R a |q (u)| t−αx+αa−a+u_R t+αx−αa+a−u |K0(u, ξ)| dξdu ! dt +α− +∞R αx−αa+a µ_+∞ R a

|p (u)| |K0(u, t + αx − αa − a + u)| du ¶ dt +α− +∞R αx−αa+a à (t−αx+αa+a)/2_R a

|p (u)| |K0(u, t − αx + αa + a − u)| du ! dt +α − 2 +∞_R αx−αa+a à (t−αx+αa+a)/2_R a |q (u)|t−αx+αa+a−uR t+αx−αa−a+u |K0(u, ξ)| dξdu ! dt < CRa x [2 |p (t)| + (1 + |t|) |q (t)|] σ (t) dt +C+∞R a [2 |p (t)| + (1 + |t|) |q (t)|] σ (t) dt. (2.24) It follows from (2.24) +∞_Z αx−αa+a |K1(x, t)| dt ≤ Cσ 2_(x) 2! . (2.25) Assume that +∞_Z αx−αa+a |Kn−1(x, t)| dt ≤ Cσ n_(x) n! (2.26)

is valid for n − 1.Let’s show that +∞_Z

αx−αa+a

|Kn(x, t)| dt ≤ Cσ n+1_(x)

is valid for n. +∞_R αx−αa+a |Kn(x, t)| dt ≤ 1 α +∞_R αx−αa+a µ_a R x

|p (u)| |Kn−1(u, t − αx + αu)| du

¶ dt +1 α +∞_R αx−αa+a µ_a R x

|p (u)| |Kn−1(u, t + αx − αu)| du

¶ dt + 1 2α +∞_R αx−αa+a à a R x |q (u)|t−αx+αuR t+αx−αu |Kn−1(u, ξ)| dξdu ! dt +α+ +∞R αx−αa+a µ_+∞ R a

|p (u)| |Kn−1(u, t − αx + αa − a + u)| du

¶ dt +α+ +∞R αx−αa+a à (t+αx−αa+a)/2_R a

|p (u)| |Kn−1(u, t + αx − αa + a − u)| du

! dt +α+ 2 +∞_R αx−αa+a à +∞_R a |q (u)|t−αx+αa−a+uR t+αx−αa+a−u |Kn−1(u, ξ)| dξdu ! dt +α− +∞R αx−αa+a µ_+∞ R a

|p (u)| |Kn−1(u, t + αx − αa − a + u)| du

¶ dt +α− +∞R αx−αa+a à (t−αx+αa+a)/2_R a

|p (u)| |Kn−1(u, t − αx + αa + a − u)| du

! dt +α − 2 +∞_R αx−αa+a à (t−αx+αa+a)/2_R a |q (u)| t−αx+αa+a−u_R t+αx−αa−a+u |Kn−1(u, ξ)| dξdu ! dt < C n! a R x [2 |p (t)| + (1 + |t|) |q (t)|] σn_{(t) dt} +C n! +∞_R a [2 |p (t)| + (1 + |t|) |q (t)|] σn_{(t) dt.} Thus +∞_Z αx−αa+a |Kn(x, t)| dt ≤ Cσ n+1_(x) (n + 1)! (2.28)

Remark: For simplicity, α was taken as α > 1. Other case can be treated in the same way.

Finally, let investigate correlation between the kernel function K (x, t) and co-efficients p (x) , q (x) and ρ (x) in the equation

−y00_{+ [2λp (x) + q (x)] y = λ}2_{ρ (x) y.} For 0 < x < a, e (x, λ) = α+_R 2(x) eiλµ +_(x) + α−_R 3(x) eiλµ −_(x) + +∞_R µ+_(x) K (x, t) eiλt_dt.

with µ+_{(x) < µ}−_{(x) < +∞. If we put e (x, λ) and e}00_{(x, λ) instead of y ve y}00 _in (1.1) , we get; α+_R00 2(x) + 2iλαα+R20 (x) − 2λα+p (x) R2(x) + 2ip (x) K (x, µ+(x)) −α+_{q (x) R} 2(x) − 2αdK (x, µ +_(x)) dx = 0 α−_R00 3(x) − 2iλαα−R03(x) − 2λα−p (x) R3(x) − 2ip (x) K (x, µ−(x) − 0) +2ip (x) K (x, µ−_{(x) + 0) − α}−_{q (x) R} 3(x) −2αdK (x, µ −_{(x) − 0)} dx + 2α dK (x, µ−_{(x) + 0)} dx = 0 ∂2_{K (x, t)} ∂x2 − α2 ∂2_{K (x, t)} ∂t2 + 2ip (x) ∂K (x, t) ∂t = q (x) K (x, t) lim x+t→∞ ∂K (x, t) ∂x = 0, x+t→∞lim ∂K (x, t) ∂t = 0

In the case x > a, since µ+_{(x) = x and α = 1, then the same properties of the} kernel function.can be easily obtained.

Hence, the proof is completed.

Acknowledgments

This work is supported by the Scientific Research Project Fund of Cumhuriyet Uni-versity under the project number F-287.

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