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ISSN: 2168-9679
Journal of Applied & Computational Mathematics
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Green’s Function Solution of Non-Homogenous Regular Sturm-Liouville Problem

Abdelgabar Adam Hassana*

Department of Mathematics, College of Science and Arts at Tabrjal, AlJouf University, Kingdom of Saudi Arabia

*Corresponding Author:
Abdelgabar Adam Hassana
Department of Mathematics
College of Science and Arts at Tabrjal
AlJouf University, Kingdom of Saudi Arabia
Tel: +966 14 624 7493
E-mail: [email protected]

Received Date: April 05, 2017; Accepted Date: April 25, 2017; Published Date: May 11, 2017

Citation: Hassana AA (2017) Green’s Function Solution of Non-Homogenous Regular Sturm-Liouville Problem. J Appl Computat Math 6: 362. doi: 10.4172/2168- 9679.1000362

Copyright: © 2017 Hassana AA. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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Abstract

In this paper, we propose a new method called exp(−ϕ(ξ)) fractional expansion method to seek traveling wave solutions of the nonlinear fractional Sharma-Tasso-Olver equation. The result reveals that the method together with the new fractional ordinary differential equation is a very influential and effective tool for solving nonlinear fractional partial differential equations in mathematical physics and engineering. The obtained solutions have been articulated by the hyperbolic functions, trigonometric functions and rational functions with arbitrary constants

Introduction

The series solution of differential equation yields an infinite series which often converges slowly. Thus it is difficult to obtain an insight into over-all behavior of the solution [1,2]. The Green’s function approach would allow us to have an integral representation of the solution instead of an infinite series.

To obtain the filed y, caused by distributed source we calculate the effect of each elementary portion of source and add (integral) them all. If G(r,r0) is the field at the observers point r caused by a unit source at the source point r0, then the field at r caused by distribution f(r0) is the integral of f(r0) G(r,r0) over the whole range of r0 occupied by the course. The function G is called Green’s function [3-5].

The Green’s function is powerful tool of mathematical method which used I solving linear non-homogenous differential equation (ordinary and partial) [6-9].

Preliminaries

Sturm-Liouville problem

Consider a linear second order differential equation

Equation (1)

Where λ is a parameter to be determined by the boundary conditions? A(x) is positive continuous function, then by dividing every term by A(x), equation (1) can be written as

Equation (2)

Where Equation

Let us define integrating factor p(x) by

Equation

Multiplying equation (2) by p(x), we have

Equation (3)

Equation

Equation

Thus equation (3) can be written as

Equation (4)

Where q(x)=p(x)c(x) and r(x)=p(x)d(x)

Equation in form (4) is known as Sturm-Liouville equation. Satisfy the boundary conditions

Regular Sturm-Liouville Problem

In case p(a)≠0 and p(b)≠0, p(x), q(x), r(x) are continuous, the Sturm-Liouville equation (4) can be expressed as

L[y]=λr(x)y (5)

Where Equation (6)

If the above equation is associated with the following boundary condition

Equation (7)

Where α1+α2≠0 and β1+β2≠0

The equation (4) and the boundary condition (7) are called regular Sturm -Liouville problem (RSLP).

(a) Singular Sturm-Liouville problem

Consider the equation

L[y]+λr(x)y=0 a<x<b (8)

Where L is defined by (6), p(x) is smooth and r(x) is positive, then the Sturm- Liouville problem is called singular if one of the following situations is occurred.

(i) If p(a)=0 or p(b)= 0 or both

(ii)The interval (a,b) is infinite.

(b)Eigenvalue and eigenfunction

The Eigenvalue from equation (4) defining by a Sturm-Liouville operator can be expressed as

Equation (9)

The non-trival solutions that satisfy the equation and boundary conditions are called eigenfunctions. Therefore the eigenfunction of the Sturm-Liouville problem from complete sets of orthogonal bases for the function space is which the weight function is r(x).

The Dirac delta function

The delta function is defined as

Equation (10)

But such that the integral of δ (x−ζ) is normalized to unit

Equation (11)

In fact the first operator where Dirac used the delta function is the integration

Equation (12)

Where f(x) is a continuous function, we have to find the value of the integration (12). Since δ () is zero for xζ, the limit of integration may be change to ζε and ζ+ε, where ε is a small positive number, f(x) is continuous at xζ, it’s values within the interval (ζε,ζε) will not different much from f(ζ), approximately that:

Equation (13)

With the approximation improving as ε approaches zero.

From (11), we have

Equation (14)

From all values of ε, then by letting ε→0, we can exactly have

Equation (15)

Despite the delta function considered as fundamental role in electrical engineering and quantum mechanics, but no conventional could be found that satisfies (10) and (11), then the delta function sought to be view as the limit of the sequence of strongly peaked function δn(x)such that

Equation (16)

As

Equation (17)

(c) Some important properties of Dirac delta function

Property (1): Symmetry

δ(−x)=δ(x) (18)

Proof

Let ζ=−x, then dx=−

We can write:

Equation (19)

But, Equation (20)

Therefore, from (19) and (20), we conclude that δ(−x)=δ(x)

Property (2): Scaling

Equation (21)

Proof

Let ζ=ax, then Equation

If a>0, then

Equation

Since Equation

Therefore:

Equation

Similarly for a<0,

Equation

then

We can write:

Equation

Property (3)

Equation (22)

Proof

The argument of this function goes to zero when x = a and x=−a, wherefore

Equation

Only at the zero of the argument of the delta function that is:

Equation (23)

Near the two zeros x2a2 can be approximated as:

Equation

In the limit as ε→0 the integral (23) becomes:

Equation

Therefore: Equation

Green’s Function

The concept of Green’s function

In the case of ordinary differential equation we can express this problem as

L[y]=f (24)

Where L is a linear differential operation f (x) is known function and y(x) is desired solution. We will show that the solution y(x) is given by an integral involving that Green’s function G(x,ξ).

Green’s function for ordinary differential equation

Here we consider non-homogenous ordinary differential equation

L[y]=f (25)

Where L is an ordinary linear differential operator that can be represented bySturm-Liouville operator, i.e.

Equation (26)

And the Sturm-Liouville type is gives by

Equation (27)

Where λ is a parameter. Now consider the linear non homogenous ordinary differential equation of the form

Equation (28)

With the boundary condition

Equation (29)

where the constant are such that α1+α20 and 1+β2≠0 if λ=0 then equation (27) and equation (28) are identical in the interval and r(x) are real and positive in that interval.

Now we are seeking to determine the Green’s function G for the equation satisfies the following

Equation (30)

With the boundary condition

Equation (31)

Now consider the region ax<ζ.

Let y1(x) be a nontrivial solution at x=a, i.e

α1y1(a)+α2y′(a)=0 (32)

Then α1y1(a,ζ)+α2G′(a,ζ)=0 (33)

The wronskian of y1 and G must vanish at x=a or

y1(a)G′(a,ζ)+y1(a)G(a,ζ)=0

So G(x,ζ)= u1y1(x) for ax<ζ (34)

Where u1 is an arbitrary constant. Similarly if the nontrivial solution y2(x) satisfies the homogeneous equation and the condition at x=b, then

G(x,ζ)= u2y2(x) for ζx<b (35)

Now by integrating equation (29) from ζε to ζ+ε we obtain

Equation (36)

Since G(x,ζ) and q(x) are continuous at x=ζ then we have

Equation (37)

The continuity condition of G and the Jump discontinuity of G′ at x=ζ from equation (33) , (34) and equation (36) imply

Equation (38)

we can solve equation (37) for u1 and u2 provided the wronskian y1 and y2 doesn’t varnish at x=ζ or

Equation (39)

Equation

The system of equation (37) has the solution

Equation (40)

Where w(ζ) is the wronskian of y1(x) and y2(x) at x=ζ

Therefore

Equation (41)

Equation (42)

Now from (42) the solution (27) can be expressed as

Equation (43)

Equation (44)

Some properties of Green’s function:

The following properties followed Green’s function

Property (i)

G(x,ζ) is exit because both p(x)≠0, w(x)≠0

Proof

From equation (32) and (33) we obtain

Equation (45)

And from equation (35) at x=ζ we have

Equation (46)

Substituting from (44) into (45) we obtain

Equation (47)

Let Equation

Equation (48)

Equation (49)

but u1 and u2 are arbitrary constant

∴ both p(x)≠ 0 and w(x)≠ 0

Property (ii)

G(x,ζ) satisfies the homogenous equation except at x=ζ

Proof

From equation (37) δ(xζ)=0 except at x=ζ

Equation (50)

where p(x)≠0 , p′(x),q(x) are continuous on [a,b]

Property (iii)

G(x,ζ) is continuous at x=ζ

Proof

Equation

Therefore G(x,ζ) is continuous at x=ζ

Property (iv)

The first and second derivatives are continuous for all

xζ in ax, ζb

Proof

Differentiation equation (32) with respect to x

Equation (51)

But G(x,ζ) is continuous everywhere, there we have

G(x, x+)=G(x, x) so that

Equation

Differentiation once more gives

Equation (52)

The second and fourth terms on the right side will not cancel in this case to the contrary

Equation (53)

We note that the term Equation denotes a differentiation of G(x,ζ)with respect to x using the x>ζ, at ζx. Thus

Equation

For: dG(x, x+) we use the x<ζ then

Equation

Property (v)

G(x,ζ) is symmetric in x and ζ

Poof

Equation

Equation

Problem(1)

Find the solution of the following problem by construction the Green’s function

Equation (54)

Subject to the boundary conditions

y(0)+ y(L) =0 (55)

With k≠0

Solution

Let G(x,ζ) be the Green’s function of the problem, then

Equation (56)

With G(0,ζ)+G(L,ζ)=0 (57)

The general solution to the homogenous equation is given by

y(x)=c1coskx+c2sinkx (58)

Now applying to the above solution

y(0)=0

y(0)=c1coskx+ c2(0)=0⇒c1=0

y1(x)=c2sinkx (59)

and y(L)=0⇒0=c1cosL+c2sin kL

Equation

Equation (60)

The Wronskian is given by

Equation (61)

The Green’s function is given by

Equation (62)

The solution is given by

Equation (63)

Problem (2)

Solve the problem by construction the Green’s function

Equation (64)

with the boundary condition

y(0)+y(L)=0 where a≠0 (65)

Solution

Let G(x,ζ) be the Green’s function of the problem, then

Equation (66)

With the boundary condition G(0,ζ) +G(L,ζ)=0

The general solution to the homogenous equation is given by

Equation (67)

Applying to the above solution

then, and y(0)=0 then c2=0, and

Equation (68)

Applying the boundary condition, then y(L)=0 Equation, andEquation (69)

The Wronskian is given by

Equation (70)

The Green’s function is given by

Equation

Equation

Where we chose c1=1, then

Equation (71)

The solution is given by

Equation

Equation (72)

References

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