MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 17 doc

19.157 For weak damping the solution reduces to As expected, in the t + 00 limit this becomes FO sin [wot - 71 zt = - 19.1.12 Green's Function for the Helmholtz Equation in Three Di

TIME-INDEPENDENT GREENS FUNCTIONS 593

with the initial conditions z(0) = k(0) = 0, C1 and C2 in Equation (19.147) is zero, hence the solution will be written as

where we have defined

(19.156) One can easily check that s(t) satisfies the differential equation

d'z(t) d z ( t ) 2

dt2 + 2E- dt + w o z ( t ) = & - a t (19.157)

For weak damping the solution reduces to

As expected, in the t + 00 limit this becomes

FO sin [wot - 71 z(t) = -

19.1.12 Green's Function for the Helmholtz Equation in Three

Dimensions

The Helmholtz equation in three dimensions is given as

We now look for a Green's function satisfying

594 GREEN’S FUNCTIONS

Using Green’s theorem

JJJ, ( F V ~ G - G V ~ F ) d3F = JJ (PVG - GVF) iids, (19.162)

S where S is a closed surface enclosing the volume V with the outward unit normal ii, we obtain

Interchanging the primed and the unprimed variables and assuming that the Green’s function is symmetric in anticipation of the corresponding boundary conditions to be imposed later, we obtain the following remarkable formula:

Boundary conditions:

The most frequently used boundary conditions are:

i) Dirichlet boundary conditions, where G is zero on the boundary ii) Neumann boundary conditions, where the normal gradient of G on the surface is zero:

surface term in the above equation vanishes, thus giving

For any one of these cases, the Green’s function is symmetric and the

TIME-INDEPENDENT GREEN 5 FUNCTIONS 595

where H is a linear differential operator H has a complete set of orthonormal eigenfunctions, {$Ẳ)), which are determined by the eigenvalue equation

where X stands for the eigenvalues and the eigenfunctioris satisfy the homoge- neous boundary conditions given in the previous section We need a Green's function satisfying the equation

H G ( 7 , ?') = S(? - 7') (19.167) Expanding @(?) and F(?) in terms of this complete set of eigenfunctions

we write

(19.168)

@(-i-.i> = Ex axdx(?;') F(?) = Ex cx$x(?") where the expansion coefficients are

This gives the Green's function as

(19.172) This Green's function can easily be generalized to the equation

(H -A,) @(?) = F(?), (19.173) for the operator (H - A,) as

(19.174)

The normalized eigenfunctions are easily obtained as

where the eigenvalues are

klmn = - a2 + - b2 + - c2 ’ (1, m, n = positive integer) (19.177) Using these eigenfunctions [Eq (19.176)] we can now write the Green’s func- tion as

(19.178)

19.1.14

Green’s function for the Laplace operator V2 satisfies

Green’s Function for the Laplace Operator Inside a Sphere

q 2 G ( 7, ?’) = S( ?, ?’) (19.179) Using spherical polar coordinates this can be written as

T2G(?, 7’) = ~ S ( r - r ’ ) S(cm e - cos8’)6(4 - 4’) (19.180)

r ’2

TIME-INDEPENDENT GREEN’S FUNCTIONS 597

where we have used the completeness relation of the spherical harmonics For the Green’s function inside a sphere, we use the boundary conditions

We now substitute Equation (19.184) into Equation (19.181) to find the dif-

ferential equation that gl(r, r’) satisfies as

1 gl(r,r’) = -6(r r - r’)

‘2

1 d2 1 ( 1 + 1)

- - [Tgl ( T , ?-’)I - 7

r dr2

A general solution of the homogeneous equation

can be obtained by trying a solution as

~ r ‘ + c l r - ( l + l )

(19.185)

(19.186)

(19.187)

We can now construct the radial part of the Green’s function for the inside of

a sphere by finding the appropriate u and the u solutions as

r < r‘,

r > r ’

(19.188)

Now the complete Green’s function can be written

into Equation (19.184) by substituting this result

19.1.15 Green’s Functions for the Helmholtz Equation for All

Space-Poisson and Schrdinger Equations

We now consider the operator

H o = q 2 + X (19.189)

in the continuum limit Using this operator we can write the following differ- ential equation:

Ho@(?;’) = F’(?) (19.190)

Using the Green's theorem [Eq

Equation (19.193) as

(19.162)] we can write the first term in

(19.194)

where S is a surface with an outward unit normal 2 enclosing the volume V

We now take our region of integration as a sphere of radius R and consider the limit R -+ 00 In this limit the surface term becomes

where 6 = ^er and dR = sin0dOd4 If the function *I(?) goes to zero suffi- ciently rapidly as 14 -+ 00, that is, when @(?) goes to zero faster than :, then the surface term vanishes, thus leaving us with

in Equation (19.194) Consequently, Equation (19.193) becomes

TIME-INDEPENDENT GREEN’S FUNCTIONS 5%

C a s e l : X S O :

In this case we can write X = - K ~ ; thus the denominator (k2 + K ~ ) in

* ( k ) = - - k2 + K 2 (19.198) never vanishes Taking the inverse Fourier transform of this, we write the general solution of Equation (19.190) as

where [(?) denotes the solution of the homogeneous equation

Hot(?) = (T2 - K ~ ) [ ( T ) = 0 (19.200) Defining a Green’s function G(?;‘,?:”) as

we can express the general solution of Equation (19.190) as

@(?) = [ ( F ) + /// G(?,7:”)F(?:”)d3?, (19.202) The integral in the Green’s function can be evaluated by using complex contour integral techniques Taking the k vector as

600 GREEN’S FUNCTIONS

Using Jordan’s lemma (Section 13.7) we can show that the integral over the circle in the upper half complex k-plane goes to zero as the radius goes to infinity; thus we obtain I as

In this solution if F ( 7 ’ ) goes to zero sufficiently rapidly as lr’l -+ (x, or

if F(?) is zero beyond some lr’l = ro, we see that for large r , q ( 7 ) decreases exponentially as

(19.212) This is consistent with the neglect of the surface term in our derivation

V%$(T+) = -4Tp(T+) (19.213) into an integral equation In this case X = 0; thus the solution is given

as

4(7) = -4~/// G ( 7 , ?;f’)p(?’)d3F’, (19.214)

V

TIME-INDEPENDENT GREEN'S FUNCTIONS 601

X = ( q * i & ) , E > O (19.219) Substituting this in Equation (19.197) we get

k = (q f Z E ) ,

and the solution as

The choice of the f sign is very important In the limit as /?“I -+ 00

this solution behaves as

or

e- iqr

K ( * ) + E ( ? ’ f ) + C - , (19.228) where C is a constant independent of r, but it could depend on 6 and

4 The f signs physically correspond to incoming and outgoing waves

We now look a t the solutions of the homogeneous equation:

(“2 + 9’) t ( 7 ) = 0, (19.229)

TIME-INDEPENDENT GREEN 5 FUNCTIONS 603

’ + i

which are now given as plane waves, ez

becomes

’

; thus the general solution

The constant A and the direction of the 7 vector come from the initial conditions

Example 19.10 Green’s function for the Schrodinger equation-E 2 0:

An important application of the X > 0 case is the Schriidinger equa- tion for the scattering problems, that is, for states with E 2 0 Using the Green’s function we have found [Eq (19.225)] we can write the

(19.233)

Equation (19.232) is known as the Lipmann-Schwinger equation

For bound state problems it is easier to work with the differential equa- tion version of the Schodinger equation, hence it is preferred How- ever, for the scattering problems, the Lipmann-Schwinger equation is the starting point of modern quantum mechanics Note that we have written the result free of E in anticipation that the c -+ 0 limit will not cause any problems

19.1.16 General Boundary Conditions and Applications to

Electrostatics

In the problems we have discussed so far the Green’s function and the solution were required to satisfy the same homogeneous boundary conditions (Section 19.1.12) However, in electrostatics we usually deal with cases in which we are interested in finding the potential of a charge distribution in the presence

G(?,?‘) and Equation (19.234) with ẳ), and then subtract and integrate the result over V to write

Using the fact that for homogeneous boundary conditions the Green’s function

is symmetric we interchange 7’ and ? :

TIME-INDEPENDENT GREEN'S FUNCTIONS 605

where 6 is the outward unit normal to the surface S bounding the volume

V If we impcse the same homogeneous boundary conditions on @( ?") and G(T+,-F"), the surface term vanishes and we reach the conclusions of Section

constant

Similarly, if we fix the value of the normal derivative q@(?;t) ii on the surface, then we use a Green's function with a normal derivative vanishing on the surface Now the solution becomes

@(?) =//I P(?"')G(T+,T')d3?' -

V

(19.242)

606 GREEN 5 FUNCTIONS

19.2 TIME-DEPENDENT GREEN’S FUNCTIONS

19.2.1 Green’s Functions with First-Order Time Dependence

We now consider differential equations, which could be written as

(19.243) where T is a timelike variable, and H is a linear differential operator indepen- dent of T , which also has a complete set of orthonormal eigenfunctions In applications we frequently encounter differential equations of this type For example, the heat transfer equation is given as

c aT(7,t)

V2T(?.,t) = -

where T ( 7 , t ) is the temperature, c is the specific heat per unit volume, and

5 is conductivity Comparing this with Equation (19.243) we see that

Another example for the first-order timedependent equations is the Schrodinger equation:

TIME-DEPENDENT GREEN’S FUNCTIONS 607

where p(r, t ) is the density (or concentration) and u is the diffusion coefficient

Because {q5m} is a set of linearly independent functions, this equation cannot

he satisfied unless all the coefficients of 4m vanish simultaneously, that is,

Because the eigenfunctions satisfy the orthogonality relation

and the completeness relation

Substituting these Am(0) functions back into Equation (19.254) we obtain

Rearranging this expression as

TIME-DEPENDENT GREEN’S FUNCTIONS 609

Because Gl(?,?’, T ) does not satisfy the basic equation for Green’s func- tions, that is,

of the differential equation (19.243), that is,

From Equation (19.267) it is seen that, given the solution a t (-?’,T’) as

@(?’,T’), we can find the solution a t a later time, Q(?,T > T ’ ) , by using

G l ( ? , ? ’ , ~ , ~ ’ ) It is for this reason that Gl(?;t,?;”,~,i’) is also called the

propagator In quantum field theory and perturbation calculations, p r o p

agator interpretation of GI is very useful in the interpretation of Feynman diagrams

19.2.3 Compounding Propagators

Given a solution at TO, let us propagate it first to 7 1 > TO and then to 72 > TI

as (from now on we use j d 3 T instead of sljd3’?;’ )

610 GREEN'S FUNCTlONS

(19.271) Using the definition of propagators [Eq (19.268)] we can also write this as

(19.272)

Using the orthogonality relation

Equation (19.272) becomes

(19.273) (19.274) Using this in Equation (19.271) we obtain the propagator, G1(7'),7')", T ~ , T o ) , that takes us from TO to 7 2 in a single step in terms of the propagators, that take us from 70 to 7 1 and from 7 1 to 7 2 , respectively, as

(19.275)

19.2.4 Propagator for the Diffusion Equation with Periodic Boundary

Conditions

As an important example of the first-order time-dependent equations, we now

consider the diffusion or heat transfer equations, which are both in the form

(19.276)

TIME-DEPENDENT GREEN’S FUNCTIONS 611

To simplify the problem we consider only one dimension with

and use the periodic boundary conditions:

(19.277) Because the H operator for this problem is

612 GREENS FUNCTlONS

This gives us the propagator as

Completing the square:

r-r’ 2

(19.289)

1 lim ~ l ( z , z’, r ) = lim -e-*

r-0 T+O &

= I ( z , z’), which is one of the definitions of the Dirac-delta function; hence

Plotting Equation (19.288) we see that it is a Gaussian (Fig 19.8)

Because the area under a Gaussian is constant, that is,

(19.291) the total amount of the diffusing material is conserved Using GI (z, d, T )

and given the initial concentration 8(z’,O), we can find the concentration at subsequent times as

Note that our solution satisfies the relation

(19.292)

@(Z,T)dZ = [ 8(z’,O)dz’ ( 19.293)

TIME-DEPENDENT GREEN’S FUNCTIONS 613

a Green’s function which allows us to express the solution as

O( 7 , ~ ) = Q ~ ( ? , T ) + 1 G ( 7 , ?’”,7,7’)F(T+’,T’)d3?;tlCE71, (19.295)

where @ ~ ( T , T ) represents the solution of the homogeneous part of Equation (19.294) We have seen that the propagator GI(?”, ?’,T,T’) satisfies the equation

614 GREEN 5 FUNCTIONS

differential equation (19.297) Considering that GI (?, ?’,T,T’) satisfies the relation

lim G I ( 7, 7’,r,r’) = 63( f - ?’), (19.298)

we can expect to satisfy Equation (19.297) by introducing a discontinuity at

T = T’ Let us start with

( H + -$) G(?, ?j,T,+) = G~ (7, ? / , T , ~ / ) s ( 7-r’) (19.303) Because the Dirac-delta function is zero except at T=T’, we only need the value of G1 at r = T ’ , which is equal to S3( ? - 7’); thus we can write Equation (19.303) as

H + - G(?, ?“,T,T’) = J3(?;’ - ?’)a( 7-70 (19.304)

From here we see that the Green’s function for Equation (19.294) is

G ( 7 , T”,T,T’) = GI(?, ?’,T,T’)-S( T-7’) (19.305)

TIME-DEPENDENT GREEN’S FUNCTIONS 615

and the general solution of Equation (19.294) is now written as

19.2.7 Green’s Function for the Schrdinger Equation for Free Particles

To write the Green’s function for the Schrdinger equation for a free particle,

we can use the similarity between the Schrdinger and the diffusion equations Making the replacement 7 + - in Equation (19.288) gives us the propagator for a free particle as

iiit 2m

Now the solution of the Schrodinger equation

2m 8x2 @(z,t) = ih-Q(z,t), at

(19.307)

(19.308) with the initial condition @ ( d , O ) , can he written as

616 GREEN5 FUNCTIONS

19.2.9 Second-Order Time-Dependent Green’s Functions

Most of the frequently encountered time-dependent equations with second- order time dependence can be written as

[ + $1 * ( 7 , 7 ) = 0, (19.313)

where r is a timelike variable and H is a linear differential operator inde- pendent of T We again assume that H has a complete set of orthonormal eigenfunctions satisfying

and

@(7,0) = Z C J X n [ a n - b n ] & ( T )

n

(19.321)

TIME-DEPENDEN T GREEN’S FUNCTIONS 617

Using the orthogonality relation [Eq (19.256)] of &(?) we obtain two rela- tions between a, and b, as

[a, + b,] = 1 (b3;i;)’)@(?;f’,0)d3?;f’ (19.322) and

and

(19.329)

618 GREENS FUNCTIONS

Among these functions G2 acts on @(?.’,O) and G 2 acts on 6(?’,0) They both satisfy homogeneous equation

Thus 9 ( 7 , r ) is a solution of the differential equation (19.313) Note G2 and

G2 are related to each other by

[,,+GI { G,(7, 7 , r )

G2(?, ?,T) = ZGz(7, ?.’,r) (19.331) Hence we can obtain G 2 ( 7 , T”,r) from e2(?., ?,T) by differentiation with respect to 7 Using Equation (19.328) and the completeness relation we can write

d r

G 2 ( 7 , 7 ’ , 0 ) = x(bn(?)4i(7’) = b3(? - 7’) (19.332)

n Using the completeness relation (19.257) in Equation (19.326), one can easily check that @(?, T ) satisfies the initial conditions -

For an arbitrary initial time T ’ , ly(?”,r), Gz and G2 are written as