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is given by:

xGreen(t) = (1 F )eF1 1 ) + F + (1 F )eF2 2 ) + F (14.57)

1 ;(t;t 2 ;(t;t

1 2

; ;

for times larger than both t1 and t2 . You could verify by direct substitution that this

function is not a solution of the di erential equation (14.50). However, this process is

rather tedious and there is a simpler way to see that the function xGreen(t) violates the

di erential equation (14.50).

Problem g: To see this, show that the solution xGreen(t) has the following long-time

behavior:

lim xGreen(t) = 2 : (14.58)

t!1

This limit is at odds with the limit (14.52) that every solution of the di erential equation

(14.50) should satisfy when the forcing vanishes after a certain nite time. This proves

that xGreen(t) is not a solution of the Verhulst equation.

This implies that the Green's function technique introduced in the previous sections

cannot be used for a nonlinear equation such as the forced Verhulst equation. The reason

for this is that Green's function are based on the superposition principle by knowing the

response to a delta-function forcing and by writing a general forcing as a superposition of

delta functions one can construct a solution by making the corresponding superposition

of Green's functions, see (14.29). However, solutions of a nonlinear equation such as the

Verhulst equation do not satisfy the principle of superposition. This implies that Green's

function cannot be used e ectively to construct behavior of nonlinear systems. It is for

this reason that Green's function are in practice only used for constructing the response

of linear systems.

CHAPTER 14. GREEN'S FUNCTIONS, PRINCIPLES

190

Chapter 15

Green's functions, examples

In the previous section the basic theory of Green's function was introduced. In this

chapter a number of examples of Green's functions are introduced that are often used in

mathematical physics.

15.1 The heat equation in N-dimensions

In this section we consider once again the heat equation as introduced in section (8.4):

@T = 2T + Q (8:27) again

@t r

In this section we will construct a Green's function for this equation in N space dimensions.

The reason for this is that the analysis for N dimensions is just as easy (or di cult) as

the analysis for only one spatial dimension.

The heat equation is invariant for translations in both space and time. For this reason

the Green's function G(r t r0 t0 ) that gives the temperature at location r and time t to a

delta-function heat source at location r0 and time t0 depends only on the relative distance

r r0 and the relative time t t0.

; ;

Problem a: Show that this implies that G(r t r0 t0) = G(r r0 t t0).; ;

Since the Green's function depends only on r r0 and t t0 it su ces to construct the

; ;

simplest solution by considering the special case of a source at r0 = 0 at time t0 = 0. This

means that we will construct the Green's function G(r t) that satis es:

@G(r t) 2 G(r t) = (r) (t) : (15.1)

@t ; r

This Green's function can most easily be constructed by carrying out a spatial Fourier

transform. Using the Fourier transform (11.27) for each of the N spatial dimensions one

nds that the Green's function has the following Fourier expansion:

1 Z g(k t)eik r dN k :

G(r t) = (15.2)

(2 )N

191

CHAPTER 15. GREEN'S FUNCTIONS, EXAMPLES

192

Note that the Fourier transform is only carried out over the spatial dimensions but not over

time. This implies that g(k t) is a function of time as well. The di erential equation that

g satis es can be obtained by inserting the Fourier representation (15.2) in the di erential

equation (15.1). In doing this we also need the Fourier representation of 2 G(r t).

r

Problem b: Show by applying the Laplacian to the Fourier integral (15.1) that:

Z

2 G(r k2 g(k t)eik r dN k :

t) = (15.3)

;1

)N

r

(2

Problem c: As a last ingredient we need the Fourier representation of the delta function

in the right hand side of (15.1). This multidimensional delta function is a shorthand

notation for (r) = (x1 ) (x2 ) (xN ). Use the Fourier representation (11.31) of

the delta function to show that:

1 Z eik rdN k :

(r) = (15.4)

(2 )N

Problem d: Insert these results in the di erential equation (15.1) of the Green's function

to show that g(k t) satis es the di erential equation

@g(k t) + k2 g(k t) = (t) : (15.5)

@t

We have made considerable progress. The original equation (15.1) was a partial di er-

ential equation, whereas equation (15.5) is an ordinary di erential equation for g because

only a time-derivative is taken. In fact, you have seen this equation before when you have

read section (13.4) that dealt with the response of a particle in syrup. Equation (15.5)

is equivalent to the equation of motion (13.28) for a particle in syrup when the forcing

forcing is a delta function.

Problem e: Use the theory of section (14.3) to show that the causal solution of (15.5) is

given by:

k2 t :

g(k t) = exp (15.6)

;

This solution can be inserted in the Fourier representation (15.2) of the Green's function,

this gives:

1 Z e; k2 t+ik r dN k :

G(r t) = (15.7)

(2 )N

The Green's function can be found by solving this Fourier integral. Before we do this, let

us pause and consider the solution (15.6) for the Green's function in the wave-number-time

domain. The function g(k t) gives the coe cient of the plane wave component exp (ik r)

as a function of time. According to (15.6) ; Fourier component decays exponentially

each

in time with a characteristic decay time 1= k2 .

Problem f: Show that this implies that in the Fourier expansion (15.2) plane waves with

a smaller wavelength decay faster with time than plane waves with larger wavelength.

Explain this result physically.

15.1. THE HEAT EQUATION IN N-DIMENSIONS 193

In order to nd the Green's function, we need to solve the Fourier integral (15.7). The

integrations over the di erent components ki of the wave-number integration all have the

same form.

Problem g: Show this by giving a proof that the Green's function can be written as:

1 Z e; k1 t+ik1 x1 dk Z e; k2 t+ik2 x2 dk Z

e; kN t+ikN xN dkN

G(r t) = 2 2 2

1 2

N

(2 )

(15.8)

You will notice that the each of the integrals is of the same form, hence the Green's

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