Propagators & Greens Functions

8/3/2019 Propagators & Greens Functions

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Appendix C

Propagators and GreensFunctions

C.1 Introduction

The technique of solution for partial differential equations by means of inter-grals over propagators and Greens functions is the most powerful and usefulone available. Its exibility and generalizability make it useful in almost allapplications of eld theory. The basis for its application is Greens integraltheorems. The second theorem

V 2 2 dV = V n n dS (C.1)which was originally developed by Green for the solution to problems of ow and electricity and magnetism [Green 1828]. The theorem is easilydeveloped from Gausss theorem. In this section, we will apply an offshootof this technique to the simplest case of a eld theory in (0 , 1) dimensions;that is an ordinary differential equations of functions of one variable. Thissimple example will allow us to set some of the terminology and provide astarting point to the analysis of the n-spatial dimensional cases which willbe analyzed in the subsequent sections.

The problem at hand for the (0 , 1)1

eld is the ordinary differential1 This is a standard notation for the size of a space-time. Our space-time being (3 , 1),

three Euclidean dimensions and one time. In this case, three coordinate intervals whosedifference contributes to the metric with the same sign and one with the opposite sign,ds 2 = dx 2 + dy 2 + dz 2 c2 dt 2 .

177


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178 APPENDIX C. PROPAGATORS AND GREENS FUNCTIONS

equation

d2

dt 2 = f (t) (C.2)

with boundary conditions. This example is not as trivial as it appears butis one whose solution we all know. It is the case of a time dependent forcef (t) acting on a unit mass. It is also the Poisson equation in one spatialdimension, a subject of the next section.

This problem is a special case of the more general case of the Sturm-Liouville system. Here the differential is

d2

dx 2+ q(x)

ddx

+ r (x) (x) L (x) (C.3)

and the homogeneous Sturm-Liouville equation is L = 0 and the inhomo-geneous equation is L = f (x). Consider two solutions to the homogeneousSturm-Liouville equation, 1(x) and 2(x). The Wronskian of these twosolutions is dened as

W (1, 2) 1(x)d2dx

(x) d1dx

(x)2(x) W (x). (C.4)The solutions 1(x) and 2(x) are linearly dependent if W (x) = 0. If thetwo solutions are linearly independent W (x) = const e R dxq (x ) . Thegeneral form of the solution for the inhomogeneous Sturm-Liouville equationis (x) = c(x)+ p(x) where c(x) solves the homogeneous Sturm-Liouvilleequation and has two adjustable parameters and is called the complementarysolution and p(x) is any solution of the inhomogeneous equation and iscalled the particular solution.

We dene a Greens function as the solution to

LG (x|x ) = (x x ) (C.5)where the boundary conditions are chosen to suit the problem. The partic-ular solution is then

p(x) =

dx G (x

|x )f (x ) (C.6)

The initial value problem (IVP) is the case in which the value of (x1) = and ddx (x1) = are given. Let 1(x) be a solution of the homogeneousSturm-Liouville equation satisfying the boundary conditions and 2(x) be


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C.1. INTRODUCTION 179

any solution of the homogeneous Sturm-Liouville equation linearly indepen-

dent of 1(x). Then c(x) = 1(x) and the Greens function isG (x|x ) = (x x ){

2(x )1(x) 1(x )2(x)}W (x )

. (C.7)

That this form is a solution of Equation C.5 is easily seen by integrating thatequation in an interval around x . The only term which is non-zero is dGdx =(x x )2. This jump in the rst derivative is a general observation whenusing Greens function techniques for second order systems. Differentiationof our construction, Equation C.7, obviously satises this condition.

Lets apply this general result to our particular example with initialvalue boundary conditions, (0) = and ddx (0) = . The homogeneouscase of our differential equation, Equation C.2, has solutions of the form(t) = At + B . Thus our 1(t) = c(t) = t + . The solution 2(t) canbe a constant, . The Wronskian is which is a non-zero constant andconsistent with the two solutions linear independence. The Greens functionis G (t|t ) = (tt ) ( t t) and thus the solution is (t) = t +

0 dt (t

t )( t t)f (t ) = t + t

0 dt (t t)f (t ). Lets check that this solutiondoes satisfy Equation C.2 and the IVP boundary conditions. Obviously,(0) = . Since ddt (t) = +

t0 dt f (t ),

ddt (0) = . Also

d 2 dt 2 (t) = f (t) as

required.For boundary value problems, the unknown function or its rst deriva-

tive is known at two points, for example (x1) = and (x2) = , and theproblem is it nd the form of the solution in the interval. This problem is theone dimensional version of the problem that Green originally addressed forthe two dimensional case. First consider the case of homogeneous boundaryconditions. These can be Dirichlet (DBC), (x1) = (x2) = 0, or Neumann(NBC), ddx (x1) =

ddx (x2) = 0, or mixed. The procedure is similar to the

previous analysis. For the case of DBC, choose 1(x) to satisfy the ho-mogeneous Sturm-Liouville equation with boundary condition 1(x1) = 0.Choose 2(x) to satisfy the homogeneous Sturm-Liouville equation withboundary condition 2(x2) = 0. Thus, 1(x) and 2(x) will be linearlyindependent. The unique solution to the inhomogeneous Sturm-Liouvilleproblem is the particular solution, Equation C.6, with

G (x|x ) = {(x

x)2(x )1(x) + (x

x )1(x )2(x)

}W (x ) . (C.8)The solution for the case of homogeneous NBC or mixed the solution con-struction is similar.

2 See Appendix B for the rules for integration of generalized functions.


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Applying this to our example differential equation, Equation C.2, in

the interval between t = 0 and t = 1, with mixed homogeneous bound-ary conditions, 1(t) = t and 2(t) = and the Wronskian is .The Greens function is (t t)t + (t t )t and the solution for theinhomogeneous equation with homogeneous mixed boundary conditions is(t) = c(t) + p(t) = t

1t dt f (t ) +

t0 dt t f (t ) since the only solution

to the homogeneous Sturm-Liouville equation with homogeneous boundaryconditions is the trivial function c(t) = 0. Again, it is an easy exercise toshow that (0) = 0. Since ddt (t) =

1t dt f (t ) tf (t) + tf (t) =

1t dt f (t ),

ddt (1) = 0. If the BVP had inhomogeneous boundary conditions, the c(t)would have provided them. For example, if the boundary conditions were(0) = and ddt = , then c(t) = t + .

It is worthwhile to introduce some of the usual language that is as-sociated with the use of Greens integral techniques. From the deningequation for the Greens function, Equation C.5, or the particular solution,Equation C.6, it is easy to see why the Greens function is often called theinuence function. G (t|t ) is the response at time t of the system to a unitsource at t . Note also that from Equation C.5 that G (t|t ) = G (t |t). Thissymmetry of the Greens function is called reciprocity. A unit source att produces and effect G (t|t ) at t and a unit source at t produces an ef-fect G (t |t) at t . There is also a very popular notation used in the pre-sentation of Greens functions. Consider our example Greens function,(t

t)t + (t

t )t . By dening a simple function of the two variables t

and t , t> Max (t, t ) = t t > tt t < t , the Greens function for this examplecan be written simply as G (t|t ) = t> . There is a corresponding form usedto select the other variable, t< Min (t, t ) =

t t < tt t > t

and our simple

Greens function could as well be written G (t|t ) = t< or G (t|t ) = t> . Themost usual form for the Greens function is G (t|t ) = g(t> )h (t< ).Note that one essential ingredient in the construction of the Greens func-

tion is the identication of the two functions 1(t) and 2(t) that satisfy thehomogeneous Sturm-Liouville equation with with homogeneous boundary

conditions. We require that these solutions be linearly independent, theWronskian be non-zero. Although this does not happen in the case of oursimple example, Equation C.2, it is possible with more general forms of theSturm-Liouville equation to have solutions at the two boundaries that arenot be linearly independant. If the Sturm-Liouville equation with homoge-


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C.1. INTRODUCTION 181

neous BV has zero eigenvalue 3, a single non-trivial solution can be found

that ts the boundary conditions.

C.1.1 Diversion on the Sturm-Liouville System

The Sturm-Liouville system was introduced above in Equation C.3. This isan important tool of analysis and a review of its structure is required. Con-sider the general bilinear form B [(x), (x)] a (x) ddx (x) ddx (x)+ b(x) ddx (x)(x)+c(x)(x) ddx (x) + d(x)(x)(x) for the two functions (x) and (x) denedon the interval x0 to x1. Integrating B over the interval and removing allthe derivatives of (x) by integration by parts, we obtain

x 1

x 0B [, ] =

x 1

x 0(x)L0[(x)]dx + a (x)

d

dx(x) + c(x)(x) (x) x 1x 0

where L 0[(x)] = ddx a (x)ddx (x) b(x) ddx (x) + c(x) ddx (x) + dcdx (x)(x)

d(x)(x). Now integrating B over the interval and removing all the deriva-tives of (x) by integration by parts, we obtain

x 1

x 0B [, ] =

x 1

x 0(x)M 0[(x)]dx + a (x)

ddx

(x) + c(x) (x) (x) |x 1x 0

where M 0[ (x)] = ddx a (x)ddx (x) + b(x)

ddx (x) +

dbdx (x)(x) c(x) ddx (x)

d(x) (x). Thus

x1

x 0(x)L 0[(x)]dx x

1

x 0(x)M 0[(x)]dx =

a (x)ddx

(x) ddx

(x)

+ ( c(x) b(x)) (x) (x)] x 1x 0 .The two differential operators L 0 and M 0 are dened by each other, given L 0we know M 0 and visa versa, and they are said to be adjoints of one another.If L 0(x) = M 0(x) for all functions in the interval, L0 or M 0 are said tobe self adjoint.

shinola3 A homogeneous Strum-Liouville problem has an eigenvalue if there exists solutions to

the new differential equation L (x ) = n (x ) with the same boundary conditions. Obvi-ously, if the Sturm-Liouville eigenproblem has a zero eigenvalue, 0 = 0, the associatedeigenfunction satises the homogeneous Strum-Liouville equation and the homogeneousboundary conditions.


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C.2 Potential Problems

The original application of the Greens theorem was to potential problemsin electromagnetism. The generic example is the Poisson equation,

2( x) = ( x)

0(C.9)

with either ( x) given on the enclosing surface bounding the volume understudy or x ( x) n with n the unit normal given on the bounding surface,see Section ?? . Following the pattern of Section C.1, examples of the rstcase are called Dirichlet boundary conditions (DBC) and the second areNeumann boundary conditions (NBC).

This equation in a one dimension case, either (0 , 1) or (1, 0), is the ex-ample problem that was discussed in the previous section, Section C.1. It isalso relevant to higher number of spatial dimensions if there is symmetry inall but one dimension. The spatial symmetry implies that the charge distri-bution and potential function are independent of all but one of the spatialvariables and thus the form of the Poisson equation becomes an ordinarydifferential equation and takes the form of a Sturm-Liouville problem. Forexample, the Poisson equation in three spatial dimensions with symmetryin the x y directions is d

2 dz 2 (z) =

(z )0

. In the case with angular symmetry,

the Poisson equation is 1r 2d

dr r2 d

dr (r ) =d2 dr 2 (r ) +

2r

ddr (r ) =

(r )0

. In bothof these cases, the differential operator is of the Sturm-Liouville type.

shinolaIn the case of more than one dimension, the equation is elliptic, in fact

in standard form, see Section 2.2.2. A direct application of Greens SecondIntegral Theorem, Equation C.1, and the dening equation for the construc-tion of the Greens function,

2G ( x | x ) = ( x x ), (C.10)yields

( x) =

V dn x G x

| x

( x )

0

+ S V dn 1 x G x| x x ( x ) n |S

G

x( x| x ) n ( x ) |S . (C.11)


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C.2. POTENTIAL PROBLEMS 183

This result provides a construction for the solution given boundary condi-

tions that are either Dirichlet which has ( x) |S known, or Neumann withx ( x) n |S known, or mixed with the understanding that the Greens func-tion boundary conditions are homogeneous of the same type. For example,with Dirichlet boundary conditions given for ( x), the appropriate bound-ary conditions for the Greens function are homogeneous Dirichlet boundaryconditions, G ( x| x ) = 0 . In this fashion, the unknown surface term is forcedto vanish. Thus for the Dirichlet problem the solution is

( x) = V dn x GD x| x ( x )

0 S V dn 1 x G D x ( x| x )n ( x ) |S (C.12)where G D ( x| x ) = 0 for x on the bounding surface. This solution to theDirichlet problem is unique. Consider two solutions. Due to the linear-ity of the Poisson equation, the difference of two solutions is a solution,

2( x) = 21( x) 22( x) = (x)

0+ ( x)

0= 0 except that the difference

has homogeneous boundary conditions. A general result for harmonic func-tions 4 is that there is no local minimum or maximum in the volume insidethe bounding surface. Thus with homogeneous boundary conditions theonly solution to the homogeneous Poisson equation is the trivial function,( x) = 0 and the Greens theorem construction for the Dirichlet problem isunique.

There is a similar construction for the Neumann problem but with addedcomplications due to the boundary conditions,

( x) = V dn x GN x| x

( x)

0

+ S V dn 1 x G N ( x| x ) x ( x ) n |S + |S , (C.13)with GN ( x| x ) where |S

R S dn 1 x (x)| S

R S dn 1 x. This extra term for the Neu-

mann case is to x the non-uniqeness of the solution. In the Neumanncase, as above, if there were two solutions, the difference between the solu-tions ( x) = 1( x) 2( x) would be a solution of the homogeneous Poissonproblem, 2 ( x) = 0, with homogeneous Neumann boundary conditions,x n |S = 0. This equation has the solution ( x) = constant. In additionfor the Neumann case, Poissons equation requires

V dn x ( x)0 = V dn x 2( x) = S V dn 1 x x n |S (C.14)4 Harmonic functions are functions that satisfy the homogeneous Poisson equation,

2 (x) = 0 .


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by Gausss theorem. This constraint must be satised for any well posed

Neumann problem.shinola

C.3 Wave Equation

For a scalar eld, ( x, t ), the wave equation is

2 ( x, t ) 1c2

t

2

2 ( x, t ) = ( x, t ) (C.15)

We seek solutions of this problem for the eld within some volume of space

for all times. This is a hyperbolic differential equation and the solutionis propagated from an initial time surface, see Section 2.2.2. The initialconditions are ( x, t 0) and t ( x, t 0) throughout the volume of interest. Inaddition, there is an elliptic differential system included in the spatial part.This then requires the solution machinery of Section C.1 and will require theinformation from a surface with the boundary conditions either Dirichlet orNeumann or mixed on the closed surface bounding our volume for all t t0.The source term, ( x, t ), given everywhere for all t . A special importantcase exists in the problem of unbounded systems. These will be discussedseparately.

Instead of developing a Greens function solution directly, we constructa related object, the propagator. The propagator satises

2K x, t | x , t = 0lim

tt +K x, t | x , t = 0

limtt +

K t

x, t | x , t = c23 x x (C.16)

with homogeneous boundary conditions of the Dirichlet type , K ( x, t | x , t ) =0, or Neumann type, n K x ( x, t | x , t ) = 0, for x on the surface boundingthe volume.If the volume is stationary, Equations C.16 are time translation invariant

and thus the propagator is a function of t t only. The propagator satisesthe symmetriesK x, t | x , t = K x, x : t t

K x, x : t t = K x , x : t t . (C.17)


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C.3. WAVE EQUATION 185

The Greens function satises

2G x, t | x , t = t t 3 x x (C.18)with homogeneous boundary conditions and G ( x, t | x , t ) = 0 for t < t .The propagator and the Greens function are related by

G x, t | x , t = t t K x, t | x , t (C.19)The solution for the eld is

( x, t ) = t +

t 0dt V d3 x G x, t | x , t x , t

+ t+

t 0dt S V d2 x G x, t | x , t x n x , t

t +t 0 dt S V d2 x Gx n x, t | x , t x , t+ V d3 x 1c2 G x, t | x , t t x , t 0+

t V d3 x G x, t | x , t 0 x , t 0 (C.20)

There is an eigenfunction expansion for the propagator with 2i ( x) = i i ( x) plus homogeneous boundary conditions.

K x, t | x , t = c21V

t t +i

i ( x) i xsin c i (t t )

c i(C.21)

where the rst term is present only if there is a 0 = 0 eigenvalue and theindicates that the sum omits the zero eigenvalue.In an unbounded space, the propagator in n dimensions is

K (n )0 R x x , t t = c (2 ) n dn k sin( kcR )k ei k R (C.22)The Greens functions are

G (3)0 R x x , t t = ( ) Rc

4R(C.23)

G (2)0

R, =

(

)

1

2

Rc 2 R

2

c212 (C.24)

G (1)0 x x , = ( )c2

|x x |

c(C.25)

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