Open Access

One-sided and two-sided Green’s functions

  • Rubens de Figueiredo Camargo1Email author,
  • Ary Orozimbo Chiacchio2 and
  • Edmundo Capelas de Oliveira2
Boundary Value Problems20132013:45

DOI: 10.1186/1687-2770-2013-45

Received: 16 October 2012

Accepted: 14 February 2013

Published: 6 March 2013

Abstract

We discuss the one-sided Green’s function, associated with an initial value problem and the two-sided Green’s function related to a boundary value problem. We present a specific calculation associated with a differential equation with constant coefficients. For both problems, we also present the Laplace integral transform as another methodology to calculate these Green’s functions and conclude which is the most convenient one. An incursion in the so-called fractional Green’s function is also presented. As an example, we discuss the isotropic harmonic oscillator.

1 Introduction

There are several methods to discuss a second-order linear partial differential equation. Among them we mention the simplest one, the method of separation of variables, and the method of integral transforms, particularly the Laplace transform, which is in many cases most convenient [1].

On the other hand, after the method of separation of variables, we get the following general second-order linear ordinary differential equation:
a 1 ( x ) d 2 d x 2 y ( x ) + a 2 ( x ) d d x y ( x ) + a 3 ( x ) y ( x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equa_HTML.gif
on the interval a < x < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq1_HTML.gif, whose corresponding nonhomogeneous one is given by
a 1 ( x ) d 2 d x 2 y ( x ) + a 2 ( x ) d d x y ( x ) + a 3 ( x ) y ( x ) = F ( x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equb_HTML.gif
where F ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq2_HTML.gif is a forcing term. Assuming that a 1 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq3_HTML.gif is a continuously differentiable positive function on this interval and that a 2 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq4_HTML.gif and a 3 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq5_HTML.gif are continuous functions, we can write the above ordinary differential equation as follows:
d d x [ p ( x ) d d x y ( x ) ] + q ( x ) y ( x ) = f ( x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equ1_HTML.gif
(1)

the so-called self-adjoint form known also as an ordinary differential equation in the Sturm-Liouville form. In this equation, p ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq6_HTML.gif and q ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq7_HTML.gif are continuous functions that are related to the coefficients a 1 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq3_HTML.gif, a 2 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq4_HTML.gif and a 3 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq5_HTML.gif. The nonhomogeneous term, f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq8_HTML.gif, is also related to F ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq2_HTML.gif [1].

The methodology of the Laplace integral transform is adequate to discuss the one-sided Green’s functiona because the initial conditions, in general, are given in terms of the own function and the first derivative. A simple question arises when we discuss the two-sided Green’s function associated with a problem involving boundary conditions, i.e., is the Laplace transform methodology convenient to discuss this problem? The answer depends on the sort of problem we are studying, as we will see in the following sections.

On the other hand, the fractional harmonic oscillator was discussed in a series of papers by Narahari et al. [25] where they presented the dynamic of the fractional harmonic oscillator, including also the damping, and by Tofighi [6] who discusses the intrinsic damping.

In this paper we discuss Eq. (1) associated with an initial value problem and a boundary value problem. In both cases, we present two methodologies, the Laplace integral transform and the Green’s function methodology. After that we conclude which methodology is the most convenient one. We sum the paper up presenting the corresponding fractional case where we discuss the Green’s function associated with the fractional harmonic oscillator. Finally, we present our concluding remarks.

2 One-sided Green’s function

To solve the self-adjoint differential equation, we introduce the so-called one-sided Green’s function, also called influence function, a two-parameter function, denoted by G i ( x | ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq9_HTML.gif, which describes the influence of a disturbance (also known as impulse) on the value of y ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq10_HTML.gif at a point x, concentrated at the point ξ. If we know the Green’s function, as we construct below, the solution of an initial value problem composed of the self-adjoint differential equation and the initial conditions can be written as follows:
y ( x ) = a x G i ( x | ξ ) f ( ξ ) d ξ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equ2_HTML.gif
(2)

where y ( a ) = 0 = y ( a ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq11_HTML.gif.

For a fixed value of ξ, the one-sided Green’s function is the solution of the corresponding homogeneous initial value problem, i.e., for x > ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq12_HTML.gif we have the homogeneous ordinary differential equation
d d x [ p ( x ) d d x G i ( x | ξ ) ] + q ( x ) G i ( x | ξ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equc_HTML.gif
and the initial conditions
G i ( ξ | ξ ) = 0 and d d x G i ( x | ξ ) | x = ξ = 1 p ( ξ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equd_HTML.gif

2.1 Constant coefficients

As a particular case, we consider the problem involving an ordinary differential equation with constant coefficients, which can represent a problem related to the classical harmonic oscillator and transmission lines, for example,
d 2 d x 2 y ( x ) + 2 b d d x y ( x ) + ( a 2 + b 2 ) y ( x ) = f ( x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equ3_HTML.gif
(3)

where a and b are positive constants, and the homogeneous initial conditions are given by y ( 0 ) = 0 = y ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq13_HTML.gif.

First of all, we write the ordinary differential equation in the corresponding self-adjoint ordinary differential equation
d d x [ e 2 b x d d x y ( x ) ] + e 2 b x ( a 2 + b 2 ) y ( x ) = e 2 b x f ( x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Eque_HTML.gif

where we identify p ( x ) exp ( 2 b x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq14_HTML.gif.

Thus, the one-sided Green’s function, denoted by G i ( x | ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq15_HTML.gif, satisfies the following homogeneous ordinary differential equation for ξ fixed:
d d x [ e 2 b x d d x G i ( x | ξ ) ] + e 2 b x ( a 2 + b 2 ) G i ( x | ξ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equf_HTML.gif
and the initial conditions
G i ( ξ | ξ ) = 0 and d d x G i ( x | ξ ) | x = ξ = exp ( 2 b ξ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equg_HTML.gif
Two linearly independent solutions of the homogeneous ordinary differential equation are given by
y 1 ( x ) = e b x a sin a x and y 2 ( x ) = e b x a cos a x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equh_HTML.gif
which furnishes for the general solution (the Green’s function)
G i ( x | ξ ) = A ( ξ ) a e b x sin a x + B ( ξ ) a e b x cos a x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equi_HTML.gif
where A ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq16_HTML.gif and B ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq17_HTML.gif must be calculated by means of the initial conditions, i.e., by the following system:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equj_HTML.gif
Solving the system above, we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equk_HTML.gif
which furnish for the one-sided Green’s function, for x > ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq18_HTML.gif,
G i ( x | ξ ) = exp [ b ( x + ξ ) ] a sin [ a ( x ξ ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equ4_HTML.gif
(4)

and, as we know, satisfies the property G i ( x | ξ ) = G i ( ξ | x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq19_HTML.gif.

Finally, the solution of our initial value problem is given by
y ( x ) = 1 a 0 x e b ( x ξ ) sin [ a ( x ξ ) ] f ( ξ ) d ξ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equ5_HTML.gif
(5)

2.2 Laplace transform

Another way to calculate this one-sided Green’s function is by means of the methodology of Laplace transform. Multiplying Eq. (3) by the kernel of Laplace transform, exp ( s x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq20_HTML.gif with Re ( s ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq21_HTML.gif and integrating from zero to infinity, we have
s 2 F ( s ) s y ( 0 ) y ( 0 ) + 2 b [ s F ( s ) y ( 0 ) ] + ( a 2 + b 2 ) F ( s ) = G ( s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equl_HTML.gif
where
F ( s ) = 0 e s x y ( x ) d x and G ( s ) = 0 e s x f ( x ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equm_HTML.gif
Using the initial conditions, we obtain an algebraic equation whose solution is given by
F ( s ) = G ( s ) ( s + b ) 2 + a 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equn_HTML.gif
By means of the convolution product and the relation involving the inverse Laplace transform
L 1 [ 1 ( s + b ) 2 + a 2 ] = exp ( b x ) a sin a x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equo_HTML.gif
we get
y ( x ) = 1 a 0 x e b ( x ξ ) sin [ a ( x ξ ) ] f ( ξ ) d ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equp_HTML.gif

which is exactly Eq. (5).

3 Two-sided Green’s function

In the case that we have a two-point boundary value problem, i.e., when the boundary conditions are fixed on the extremes of the interval, we have a two-sided Green’s function, also called Green’s function, only.

We consider the second-order ordinary differential equation in the Sturm-Liouville form
d d x [ p ( x ) d d x y ( x ) ] + q ( x ) y ( x ) = f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equq_HTML.gif

for x 0 < x < x 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq22_HTML.gif and the boundary conditions y ( x 0 ) = 0 = y ( x 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq23_HTML.gif, and comparing this problem with the corresponding one-sided Green’s function, we have put f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq24_HTML.gif in the place of f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq8_HTML.gif for convenience only.

The two-sided Green’s function, denoted by G b ( x | ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq25_HTML.gif for each ξ, satisfies the following problem consisting of homogeneous ordinary differential equation
d d x [ p ( x ) d d x G b ( x | ξ ) ] + q ( x ) G b ( x | ξ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equr_HTML.gif
for x ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq26_HTML.gif and the homogeneous boundary conditions
G b ( x 0 | ξ ) = 0 = G b ( x 1 | ξ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equs_HTML.gif
This Green’s function must satisfy also the continuity
G b ( ξ + | ξ ) = G b ( ξ | ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equt_HTML.gif
and a jump discontinuity on the first derivative [1]
d d x G b ( x | ξ ) | x = ξ + d d x G b ( x | ξ ) | x = ξ = 1 p ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equu_HTML.gif
The solution y ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq10_HTML.gif can be interrelated as a displacement and be given by
y ( x ) = x 0 x 1 G b ( x | ξ ) f ( ξ ) d ξ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equv_HTML.gif

where f ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq27_HTML.gif is a force per unit length. G b ( x | ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq25_HTML.gif is the displacement at x due to a force of unit magnitude concentrated at ξ. In this case, we have G b ( x | ξ ) = G b ( ξ | x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq28_HTML.gif, the so-called reciprocity law.

3.1 Constant coefficients

As an example, we discuss the following ordinary differential equation:
d 2 d x 2 y ( x ) + 2 b d d x y ( x ) + ( a 2 + b 2 ) y ( x ) = f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equw_HTML.gif

with the homogeneous boundary conditions y ( x 0 ) = 0 = y ( x 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq29_HTML.gif.

First, we construct the corresponding Green’s function, as we have seen before. Two linearly independent solutions of the homogeneous ordinary differential equation are
y 1 ( x ) = e b x a sin a x and y 2 ( x ) = e b x a cos a x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equx_HTML.gif
Imposing that y 1 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq30_HTML.gif satisfies y 1 ( x 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq31_HTML.gif, we can write
y 1 ( x ) = A ( ξ ) a e b x sin [ a ( x x 0 ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equy_HTML.gif
On the other hand, to satisfy y 2 ( x 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq32_HTML.gif, we put
y 2 ( x ) = B ( ξ ) a e b x sin [ a ( x x 1 ) ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equz_HTML.gif

where A ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq16_HTML.gif and B ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq17_HTML.gif must be determined.

Using the continuity at x = ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq33_HTML.gif and the jump discontinuity of the first derivative, we obtain a system of two algebraic equations involving A ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq16_HTML.gif and B ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq17_HTML.gif whose solution is
A ( ξ ) = sin [ a ( x 1 ξ ) ] sin [ a ( x 1 x 0 ) ] and B ( ξ ) = sin [ a ( ξ x 0 ) ] sin [ a ( x 1 x 0 ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equaa_HTML.gif
Thus, the Green’s function is given by the following expression:
G b ( x | ξ ) = exp [ b ( x + ξ ) ] a sin [ a ( x 1 x 0 ) ] { sin [ a ( x x 0 ) ] sin [ a ( x 1 ξ ) ] , x 0 < x < ξ , sin [ a ( ξ x 0 ) ] sin [ a ( x 1 x ) ] , ξ < x < x 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equab_HTML.gif
Using this Green’s function, the solution of the boundary value problem can be written as follows:
y ( x ) = x 0 x 1 G b ( x | ξ ) f ( ξ ) d ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equac_HTML.gif

3.2 Laplace transform

As we have already said, the Laplace transform converts the differential equation with constant coefficients into an algebraic equation whose solution is
F ( s ) = s y ( 0 ) + y ( 0 ) ( s + b ) 2 + a 2 G ( s ) ( s + b ) 2 + a 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equad_HTML.gif

where F ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq34_HTML.gif and G ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq35_HTML.gif are the corresponding Laplace transforms of y ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq10_HTML.gif and f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq8_HTML.gif, respectively.

Using the inverse Laplace transform and the convolution product, we get
y ( x ) = y ( 0 ) e b x cos a x a + y ( 0 ) e b x sin a x a 1 a 0 x e b ( x ξ ) sin [ a ( x ξ ) ] f ( ξ ) d ξ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equae_HTML.gif
Thus, substituting the boundary conditions, we obtain two algebraic equations, a system involving y ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq36_HTML.gif and y ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq37_HTML.gif. Solving this algebraic system, we obtain
y ( x ) = 1 / a sin [ a ( x 1 x 0 ) ] x 0 x e b ( x + t ) sin [ a ( x 1 x ) ] sin [ a ( x 0 t ) ] f ( t ) d t 1 / a sin [ a ( x 1 x 0 ) ] x x 1 e b ( x + t ) sin [ a ( x 1 t ) ] sin [ a ( x 0 x ) ] f ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equaf_HTML.gif
which can be rewritten as follows:
y ( x ) = x 0 x 1 G b ( x | t ) f ( t ) d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equag_HTML.gif
where the Green’s function is given by
G b ( t | x ) = exp [ b ( x + t ) ] a sin [ a ( x 1 x 0 ) ] { sin [ a ( x 1 x ) ] sin [ a ( t x 0 ) ] , x 0 < t < x , sin [ a ( x 1 t ) ] sin [ a ( x x 0 ) ] , x < t < x 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equah_HTML.gif

which is the same expression as that obtained in Section 3.1.

At this point we conclude that for a problem involving initial conditions, the Laplace integral transform is more convenient since y ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq36_HTML.gif and y ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq37_HTML.gif are known. On the other hand, i.e., for a problem involving boundary conditions, the Sturm-Liouville, as opposed to the Laplace integral transform, is more convenient in the sense that the calculation is much more simple.

4 Fractional Green’s function

Fractional calculus is one of the most accurate tools to refine the description of natural phenomena. The usual way to use this tool is to replace the integer-order derivatives of the partial differential equation that describes one specific phenomenon by a derivative of non-integer order. For many expected reasons, the solution of a fractional partial differential equation used to be much more complicated than the solution of the corresponding integer-order partial differential equation.

On the other hand, many important results and generalizations were obtained using this procedure in several areas such as fluid flow, diffuse transport, electrical networks, probability, biomathematics and others [712]. Here, as a generalization to the integer case, we present a calculation associated with the so-called fractional one-sided and two-sided Green’s function relative to the fractional differential equation with constant coefficients, i.e., we obtain the fractional Green’s function for the fractional differential equation whose coefficients are constants. We discuss the problem by means of the Laplace integral transform, and as an application, we present explicitly the Green’s function associated with the fractional harmonic oscillator.

4.1 Fractional one-sided Green’s function

Let a, b and c be real constants. We present the solution of the fractional differential equation
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equ6_HTML.gif
(6)

where 1 < α 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq38_HTML.gif and 0 < β 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq39_HTML.gif and the fractional derivatives are taken in the Caputo sense [13]. We also consider y ( 0 ) = 0 = y ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq13_HTML.gif as the initial conditions. In the case where α = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq40_HTML.gif and β = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq41_HTML.gif, we recover the results associated with the integer case, and taking a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq42_HTML.gif and b 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq43_HTML.gif, we recover the equation associated with the fractional relaxor-oscillator as discussed in [5].

Introducing the Laplace transform and using the initial conditions, we obtain an algebraic equation whose solution can be written as follows:
L [ y ( x ) ] = F ( s ) a s α + b s β + c , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equai_HTML.gif
where F ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq34_HTML.gif is the Laplace transform of the f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq8_HTML.gif. This expression can be manipulated, using the geometric series, to obtain
L [ y ( x ) ] = F ( s ) a k = 0 ( b a ) k s β k ( s α + c / a ) k + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equaj_HTML.gif

which is valid for | s λ c / a | < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq44_HTML.gif.

Using the Laplace transform of the generalized Mittag-Leffler function and its corresponding inverse [13],
L [ t μ 1 E ν , μ ρ ( λ t ν ) ] = s ν ρ μ ( s ν λ ) ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equak_HTML.gif
with Re ( s ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq21_HTML.gif, Re ( μ ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq45_HTML.gif, λ C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq46_HTML.gif and | λ s ν | < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq47_HTML.gif, we have
y ( x ) = 1 a k = 0 ( b a ) k 0 x f ( ξ ) ( x ξ ) ( α β ) k + α 1 E α , ( α β ) k + α k + 1 [ c a ( x ξ ) α ] d ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equal_HTML.gif

which is the solution of the fractional differential equation.

Thus, the one-sided fractional Green’s function can be written as follows:
G ( x | ξ ) = 1 a ( x ξ ) α 1 E α , α [ c a ( x ξ ) α ] + 1 a k = 1 ( b a ) k ( x ξ ) ( α β ) k + α 1 E α , ( α β ) k + α k + 1 [ c a ( x ξ ) α ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equ7_HTML.gif
(7)
Taking b = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq48_HTML.gif, a = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq49_HTML.gif and c = ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq50_HTML.gif in Eq. (7), we get
G ( x | ξ ) = ( x ξ ) α 1 E α , α [ ω 2 ( x ξ ) α ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equam_HTML.gif
which is the fractional one-sided Green’s function associated with the fractional harmonic oscillator. The one-sided Green’s function associated with the classical harmonic oscillator is recovered by introducing α = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq40_HTML.gif in the last equation, i.e.,
G ( x | ξ ) = ( x ξ ) E 2 , 2 [ ω 2 ( x ξ ) 2 ] , x > ξ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equan_HTML.gif
which can also be written as follows:
G ( x | ξ ) = 1 ω sin [ ω ( x ξ ) ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equao_HTML.gif

which is the same expression as that obtained in Eq. (4) in the case b = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq48_HTML.gif.

To conclude this section, we plot, in Figure 1, a graphic t = x ξ × t α 1 E α , α ( t α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq51_HTML.gif for particular values of the parameter α to compare with the sine function. This graphic is plotted using the program (MatLab R2009a). For reader interested in the Mittag-Leffler function, we suggest a recent nice book [14] and the paper [15] particularly regarding the asymptotic algebraic behavior of this function.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Fig1_HTML.jpg
Figure 1

Graphics for f ( t ) = t α 1 E α , α ( t α ) × t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq52_HTML.gif with t x ξ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq53_HTML.gif .

4.2 Fractional two-sided Green’s function

Let a, b and c be real constants. We present the solution of the fractional differential equation, Eq. (6), with the homogeneous boundary conditions, y ( x 0 ) = 0 = y ( x 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq29_HTML.gif.

By means of the Laplace transform, we can write
L [ y ( x ) ] = F ( s ) + s α 1 y ( 0 ) + s α 2 y ( 0 ) a s α + b s β + c , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equap_HTML.gif
where F ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq34_HTML.gif is the Laplace transform of the f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq8_HTML.gif. This algebraic equation can be manipulated as follows:
L [ y ( x ) ] = F ( s ) a k = 0 ( b a ) k s β k ( s α + c / a ) k + 1 + y ( 0 ) a k = 0 ( b a ) k s β k + α 1 ( s α + c / a ) k + 1 + y ( 0 ) a k = 0 ( b a ) k s β k + α 2 ( s α + c / a ) k + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equaq_HTML.gif
From the inverse Laplace transform and the convolution theorem, we get
y ( x ) = 1 a k = 0 ( b a ) k 0 x f ( x ξ ) ξ ( α β ) k + α 1 E α , ( α β ) k + α k + 1 ( c a ξ α ) d ξ + y ( 0 ) a k = 0 ( b a ) k L 1 [ s β k + α 1 ( s α + c / a ) k + 1 ] + y ( 0 ) a k = 0 ( b a ) k L 1 [ s β k + α 2 ( s α + c / a ) k + 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equar_HTML.gif
which can be rewritten in the following way:
y ( x ) = 1 a k = 0 ( b a ) k 0 x f ( ξ ) ( x ξ ) ( α β ) k + α 1 E α , ( α β ) k + α k + 1 [ c a ( x ξ ) α ] d ξ + y ( 0 ) a k = 0 ( b a ) k x ( α β ) k E α , ( α β ) k + 1 k + 1 ( c a x α ) + y ( 0 ) a k = 0 ( b a ) k x ( α β ) k + 1 E α , ( α β ) k + 2 k + 1 ( c a x α ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_Equas_HTML.gif

As we have already said in Section 3, to explicitly calculate the solution, we must substitute the homogeneous boundary conditions in the last equation and determine the y ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq36_HTML.gif and y ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq37_HTML.gif. Finally, we can get the respective fractional two-sided Green’s function. We have shown this is a hard calculation. To obtain the results associated with the fractional harmonic oscillator, we introduce b = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq48_HTML.gif, a = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq49_HTML.gif and c = ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq54_HTML.gif in the last equation. Remember that to obtain the respective Green’s function, we substitute f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-45/MediaObjects/13661_2012_Article_297_IEq8_HTML.gif to the corresponding delta function.

5 Concluding remarks

We have presented and discussed the so-called one-sided and two-sided Green’s function to study, respectively, an initial value problem and a boundary value problem. Besides that, we studied the same problems by means of the Laplace transform methodology in order to conclude which methodology was most accurate for each problem. We also obtained the fractional generalization of the one-sided and two-sided Green’s function in terms of the generalized Mittag-Leffler function.

We conclude that for the initial value problem, the Laplace integral transform methodology is more convenient; on the other hand, for the boundary value problem, the two-sided Green’s function provides a much more simple calculation. It is important to note that in the present manuscript we did not consider the problem involving physical dimensions. This problem has been discussed in a recent paper by Inizan [16].

A natural continuation of this work would be to study the problems involving partial differential equations with non-constant coefficients and their fractional versions, which could provide a better description of the phenomena related to those equations. Study in this direction is upcoming.

Endnote

a A recent historical review about George Green can be found in ref. [17].

Declarations

Acknowledgements

We are grateful to Prof. J. Vaz Jr., Dr. J. Emílio Maiorino and Dr. E. Contharteze Grigoletto for several useful discussions. Besides that, we are thankful to the referees for several important suggestions which improved this article a lot.

Authors’ Affiliations

(1)
Departamento de Matemática, Faculdade de Ciências
(2)
Departamento de Matemática, Imecc, Unicamp

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© Camargo et al.; licensee Springer. 2013

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