Green's Function Solution of Non-Homogenous Regular Sturm

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h DOI: 10.4172/2168-9679.1000362 f

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Research Article Open Access 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

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 d dy d22 ydp() x dy d y =+=+ So px() px() 22p()x p(x) bxz() The series solution of differential equation yields an infinite series dx dx dx dx dx dx which often converges slowly. Thus it is difficult to obtain an insight Thus equation (3) can be written as into over-all behavior of the solution [1,2]. The Green’s function d dy approach would allow us to have an integral representation of the px() ++ qxy()()λ rxy =0 (4) solution instead of an infinite series. dx dx To obtain the filed y, caused by distributed source we calculate the Where q(x)=p(x)c(x) and r(x)=p(x)d(x) effect of each elementary portion of source and add (integral) them all. Equation in form (4) is known as Sturm-Liouville equation. Satisfy If G(r,r0) is the field at the observers point r caused by a unit source the boundary conditions at the source point r0, then the field at r caused by distribution f(r0) is Regular Sturm-Liouville Problem 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]. In case p(a)≠0 and p(b)≠0, p(x), q(x), r(x) are continuous, the The Green’s function is powerful tool of mathematical method Sturm-Liouville equation (4) can be expressed as which used I solving linear non-homogenous differential equation L[y]=λr(x)y (5) (ordinary and partial) [6-9]. d dy Where L=  px() + qx() (6) Preliminaries dx dx If the above equation is associated with the following boundary Sturm-Liouville problem condition Consider a linear second order differential equation α1y(a)+α2y (a)=0 d2 y dy + ++λ = (1) Ax() 2 Bx() C(x) y D(x) y 0 β1y(b)+β2y′(b)=0 (7) dx∂ x Where α +α ≠0 and β +β ≠0 Where λ is a parameter to be determined by the boundary 1 ′ 2 1 2 conditions? A(x) is positive continuous function, then by dividing The equation (4) and the boundary condition (7) are called regular every term by A(x), equation (1) can be written as Sturm -Liouville problem (RSLP). d2 y dy (a) Singular Sturm-Liouville problem +Bx() ++ C(x) yλ D(x) y = 0 (2) dx2 ∂ x Consider the equation Bx() Cx() Dx() Where b(x)= , c(x)= and d(x)= L[y]+λr(x)y=0 a

x   *Corresponding author: Abdelgabar Adam Hassana, Department of Mathematics, Px()()= exp∫ bξξ d College of Science and Arts at Tabrjal, AlJouf University, Kingdom of Saudi Arabia,  a  Tel: +966 14 624 7493; E-mail: [email protected] Multiplying equation (2) by p(x), we have Received April 05, 2017; Accepted April 25, 2017; Published May 11, 2017 d2 y dy Citation: Hassana AA (2017) Green’s Function Solution of Non-Homogenous + ++λ = Regular Sturm-Liouville Problem. J Appl Computat Math 6: 362. doi: 10.4172/2168- px() 2 pxbx()() p(x)c()() x y pxd (x) y 0 (3) dx ∂x 9679.1000362

xx bd()ζζ bd() ζζ x Copyright: © 2017 Hassana AA. This is an open-access article distributed under dp() x dd∫∫ Since = eaa= e b(ζζ )d = pxbx()() the terms of the Creative Commons Attribution License, which permits unrestricted  ∫ use, distribution, and reproduction in any medium, provided the original author and dx dx  dx a  source are credited.

J Appl Computat Math, an open access journal Volume 6 • Issue 3 • 1000362 ISSN: 2168-9679 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

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Where L is defined by (6), p(x) is smooth and r(x) is positive, then n  δ = the Sturm- Liouville problem is called singular if one of the following n ( x) 22  π 1+ nx situations is occurred. ( )   n −nx22  (i) If p(a)=0 or p(b)= 0 or both δn ( xe) =  (17) π  (ii)The interval (a,b) is infinite. sin2 (nx)  δ =  n ( x) 2 etc (b)Eigenvalue and eigenfunction nxπ  The Eigenvalue from equation (4) defining by a Sturm-Liouville (c) Some important properties of Dirac delta function operator can be expressed as Property (1): Symmetry −1 d dy λ = px( ) + qx( ) (9) δ(−x)=δ(x) (18) r( x)  dx dx The non-trival solutions that satisfy the equation and boundary Proof conditions are called eigenfunctions. Therefore the eigenfunction of Let ζ=−x, then dx=−dζ the Sturm-Liouville problem from complete sets of orthogonal bases for the function space is which the weight function is r(x). We can write: +∞ −∞ +∞ The Dirac delta function ∫∫∫fxxdxfdfdf( )δ(−=) (ζδζ) ( ) ζ ≈( ζδζ) ( ) ζ =(0) (19) −∞ +∞ −∞ The delta function is defined as +∞ 0 x ≠ξ But, f( x)δ ( x) dx= f (0) (20) (10) ∫ δξ( x −=)  −∞ ∞=x ξ Therefore, from (19) and (20), we conclude that δ(−x)=δ(x) But such that the integral of δ (x−ζ) is normalized to unit +∞ Property (2): Scaling ∫ δξ( x−=) dx 1 (11) 1 −∞ δδ(ax)= ( x) (21) a In fact the first operator where Dirac used the delta function is the Proof integration 1 Let ζ=ax, then dx= dζ +∞ a ∫ f( x)δξ( x− ) dx (12) If a>0, then −∞ +∞ −∞ ζζ1 1+∞ 10  Where f(x) is a continuous function, we have to find the value of f( x)δ( ax) dx= f δζ( ) d ζ= f δζ( ) d ζ= f  f (0) ∫∫a a a ∫ a aa  the integration (12). Since δ (xζ) is zero for x≠ζ, the limit of integration −∞ +∞ −∞ may be change to ζ−ε and ζ+ε, where ε is a small positive number, f(x) +∞ +∞ 11 1 is continuous at x−ζ, it’s values within the interval (ζ−ε,ζ−ε) will not Since ∫∫f(ζδ) ( x) dx= f( x)δ( x) dx=  f (0) different much from f(ζ), approximately that: −∞ aa−∞  a +∞ ξξ+∞ +∞ Therefore: f xδξ x−= dx f x δξ x −≈ dx f ξδξ x − dx (13) ∫∫∫( ) ( ) ( ) ( ) ( ) ( ) 1 −∞ ξξ−∞ −∞ δδ(ax)= ( x) a With the approximation improving as ε approaches zero. Similarly for a<0, 1 From (11), we have δδ(ax)= ( x) −a +∞ ξ +∞ then I=∫∫δξ( x −=) dx δξ( x −=) dx 1 (14) −∞ ξ −∞ We can write: From all values of ε, then by letting ε→0, we can exactly have 1 δδ(ax)= ( x) +∞ a ∫ f( x)δξ( x−=) dx f ( ζ) (15) −∞ Property (3)

22 1 Despite the delta function considered as fundamental role in δ(x− a) =  δδ( xa +) +( xa −) (22) electrical engineering and quantum mechanics, but no conventional 2a could be found that satisfies (10) and (11), then the delta function Proof sought to be view as the limit of the sequence of strongly peaked The argument of this function goes to zero when x = a and x=−a, function δn(x)such that wherefore +∞ +∞ 22 δδ( xx)= lim n ( ) (16) fζδ x− a dx = f x δ x + a + δ x − a dx n→∞ ∫∫( ) ( ) ( )( ) ( ) −∞ −∞ As Only at the zero of the argument of the delta function that is:

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+∞ −ae + ae+ 22 22 22 αGa( ) += αζ G′( a,0) ∫∫f(ζδ) ( x−= a) dx f( x)δ( x −+ a) dx ∫ f( x)δ( x − a) dx (23) 12 (31) −∞ −ae − ae− β12Gb( )+= βζ G′( b,0) Near the two zeros x2−a2 can be approximated as: Now consider the region a≤x<ζ.

22 (−2,axa)( +) x →− a ( x− a) =−( xaxa)( +=)  Let y1(x) be a nontrivial solution at x=a, i.e (+2,axa)( −) x →+ a α y (a)+α y (a)=0 (32) In the limit as ε→0 the integral (23) becomes: 1 1 2 +∞ −ae + ae+ Then α y (a,ζ)+α G (a,ζ)=0 (33) 22 1 1 ′ 2 ∫∫f(ζδ) ( x− a) dx = f(x)δ −( 2 a)( x + a)  dx + ∫ f(x)δ ( 2 a)( x − a)  dx −∞ −ae − ae− The wronskian of y1′ and G must vanish at x=a or 11ae++ae = f(x)δδ( x) ( x + a) + f(x)δ( x− a) dx ∫∫ y1(a)G (a,ζ)+y 1(a)G(a,ζ)=0 22aaae−−ae +∞ 1 So G(x,′ζ)= u1y1′(x) for a≤x<ζ (34) =∫ fx( ) δδ[ xa−+] ( xadx −) −∞ 2a Where u1 is an arbitrary constant. Similarly if the nontrivial

22 1 solution y2(x) satisfies the homogeneous equation and the condition Therefore: δ(x− a) == δδ[ xa −+] ( xa −) 2a at x=b, then

Green’s Function G(x,ζ)= u2y2(x) for ζ≤xdifferential operator that can be , 1  (38) uy11(ζ )−= u 2 y 2  represented bySturm-Liouville operator, i.e. p(ζ ) d dy (26) we can solve equation (37) for u and u provided the wronskian y and L=  px( ) + qx( ) 1 2 1 dx dx y2 doesn’t varnish at x=ζ or And the Sturm-Liouville type is gives by y1(ζ)y 2(ζ)y 1(ζ)≠0 (39) d dy px++ qxyλ rxy =− f x (27) yy(ζζ) ( ) ( ) ( ) ( ) ( ) i.e., ′ ′ 11 dx dx w(ζ )= ,,≠0 yy12(ζζ) ( ) Where λ is a parameter. Now consider the linear non homogenous ordinary differential equation of the form The system of equation (37) has the solution d dy (28) −y , (ζ )  px( ) + qxy( ) =− f( xa) ≤≤ x b = 2 dx dx u1  pw(ζζ) ( )   (40) With the boundary condition and  ,  ,  αα12ya( ) += y( a) 0 −y (ζ )  (29) u = 1  , 2 ζζ ββ12yb( )+= y( b) 0  pw( ) ( ) where the constant are such that α +α 0 and +β ≠0 if λ=0 then equation 1 2 1 2 Where w(ζ) is the wronskian of y1(x) and y2(x) at x=ζ (27) and equation (28) are identical in the interval and r(x) are real and positive in that interval. Therefore yx12( <>) yx( ) Now we are seeking to determine the Green’s function G for the Gx( ,ζ )= (41) pw(ζζ) ( ) equation satisfies the following d dG  y( xy) (ζ )  px( ) + qxy( ) =−−δζ( x ) (30)  12, xb<≤ζ dx dx  pw(ζζ) ( ) (42) Gx( ,ζ ) =  With the boundary condition yx12( <>) yx( ) − ,ax<≤ζ  pw(ζζ) ( ) 

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Now from (42) the solution (27) can be expressed as Proof b y(x)= ∫ Gx( ,ζ) f ( ζζ ) d (43) Differentiation equation (32) with respect to x − a x dG2 ( x,ζ ) dG( x, x )  xb fd(ζζ) + d(ζ)  b ∫ So y(x)= Gx( ,ζ) f () ζζ d+ Gx( , ζ) f () ζζ d (44) d a dx dx  ∫∫ Gx( ,ζ) f( ζζ) d( )=  (51) aa ∫ b 2 + dx a dG( x,ζ ) dG( x, x )  Some properties of Green’s function: +−∫ fd(ζζ) d(ζ)  x dx dx  The following properties followed Green’s function But G(x,ζ) is continuous everywhere, there we have Property (i) G(x, x+)=G(x, x−) so that G(x,ζ) is exit because both p(x)≠0, w(x)≠0 d b xbdG( x,,ζζ) dG( x ) ∫Gxf( ,ζ) ( ζ) d( ζ)= ∫∫fd(ζζ) + fd(ζζ) Proof dx a axdx dx From equation (32) and (33) we obtain Differentiation once more gives uy( x) u = 12 (45) x dG2 ( x,ζ ) dG( x, x- )  1 fd(ζζ) + d ζ yx1 ( ) 2 b ∫ 2 d dx dx  ζ ζζ= a 2 Gx( , ) f( ) d( )  And from equation (35) at x=ζ we have ∫ b 2 + (52) dx a dG( x,ζ ) dG( x, x )  +−fdζζ d ζ ,, 1 ∫ 2 ( )  uy( x) −=− uy( x) (46) b dx dx  11 22 px( ) The second and fourth terms on the right side will not cancel in this Substituting from (44) into (45) we obtain case to the contrary yx1 ( ) -+ u12 y( xy) 1′′( x) − y 1( xy) 2( x) =− (47) dG( x,, x) dG( x x ) 1 px( ) −= (53) dx dx p(ζ ) Let wx( ) = y( xy) ,,( x) − y( xy) ( x) 12 21 dG( x, x- ) We note that the term denotes a differentiation of yx12( ) yx( ) i.e. wx= (48) dx ( ) ,, yx12( ) yx( ) G(x,ζ)with respect to x using the x>ζ, at ζ→x. Thus dG x, x− ,,ζζ −yx( ) ( ) y11( xy) ( ) y 11( xy) ( ) u =− 2 = lim = 1  dxζ → X p(ζζ) w( ) p( x) w( ζ) pxwx( ) ( )  ζ →∞  and  (49) For: dG(x, x+) we use the x<ζ then  −yx( ) + u =− 1  dG( x, x ) y,,( xy) (ζζ) y( xy) ( ) 2 pxwx( ) ( ) = lim 12= 12.  ζ → X dxζ →∞ p(ζζ) w( ) p( x) w( ζ) but u and u are arbitrary constant 1 2 Property (v) both p(x)≠ 0 and w(x)≠ 0 G(x,ζ) is symmetric in x and ζ Property (ii) ∴ Poof G(x,ζ) satisfies the homogenous equation except at x=ζ −y, ( xy) (ζ )  12 Proof  pw(ζζ) ( ) Gx,ζζ= a≤< x Let ( )  , From equation (37) δ(x−ζ)=0 except at x=ζ −y21( xy) (ζ )  p′′( xG) ( x,ζζ) += qxGx( ) ( ,0) (50)  pw(ζζ) ( ) where p(x)≠0 , p (x),q(x) are continuous on [a,b] y, ( xy) (ζ ) limGx( + ,ζ )= 21 x+ξ pw(ζζ) ( ) Property (iii) ′ G(x,ζ) is continuous at x=ζ y, ( xy) (ζ ) and limGx( + ,ζ )= 21 Proof x+ξ pw(ζζ) ( )

limGx( ,ζζ) = y2 ( ) x→ζ Problem(1)

limGx( ,ζζ) = y1 ( ) Find the solution of the following problem by construction the x→ζ Green’s function Therefore G(x,ζ) is continuous at x=ζ y +k2y=−f(x), 0

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With k≠0 1 Gx'' −=−−δζ( ) (66) Solution a With the boundary condition G(0,ζ) +G(L,ζ)=0 Let G(x,ζ) be the Green’s function of the problem, then The general solution to the homogenous equation is given by G +k2G=−δ(xζ) (56) 1 2 (67) With G(0,ζ)+G(L,ζ)=0 (57) Y( x) =++ cx12 c ″ 2a The general solution to the homogenous equation is given by Applying to the above solution y(x)=c coskx+c sinkx (58) 1 2 then, and y(0)=0 then c2=0, and Now applying to the above solution x2 y( x) = + cx (68) 112a y(0)=0 L2 ∴ c= + cL y(0)=c coskx+ c (0)=0 c =0 Applying the boundary condition, then y(L)=0 21, and 1 2 1 2a c 2 22 (69) y (x)=c sinkx (59) yx21( )=−−+( x L) cx( − L) 1 2 ⇒ 2a

and∴ y(L)=0 0=c1cosL+c2sin kL The Wronskian is given by c2 sin kL xL   ∴=−c1 ⇒ (70) wx( )=++ L c11  c  cos L aa 2  cC22 y2 ( x)=(coskx sin kL −=− sin kx cos kL) sink(Lx ) (60) The Green’s function is given by cosk L CoskL  , The Wronskian is given by y12( xy) (ζ )   pw(ζζ) ( ) C2 Gx( ,0ζζ) = ≤

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References 5. Tintang K (2007) Mathematical methods for Engineers and Scientists 2.Verlag Berlin Heidelbarg. 1. Gwaiz MA Al (2008) Sturm-Liouville theory& its Application. Verlag London limited. 6. Neta B (2002) Partial differential equations, lecture notes, Naval postgraduate School. California. 2. Blakledge GE, Yardley JP (2000) Analytic methods for Partial differential Equations. Sprenger London Limited. 7. Pinchover Y, Rubinstein J (2005) An introduction to Partial differential equations. Cambridge University. 3. Tintang K (2007) Mathematical methods for Engineers and Scientists 3. 8. Roach GF (1970) Green’s function introductory theory with Application. New 4. Gerald BH (1992) Fourier analysis and Application. Wodsworth. Inc. Belmon York Toronto Melbourne. California. 9. Addison W (1984) Boundary Value problem, Monterey California.

J Appl Computat Math, an open access journal Volume 6 • Issue 3 • 1000362 ISSN: 2168-9679