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on time or on position and time. In general, the di erential equation (14.23) must be

supplemented with boundary conditions to give an unique solution. In this section the

position of the boundary is denoted by rB and it is assumed that the function u has the

value uB at the boundary:

u(rB ) = uB : (14.25)

Let us rst nd a single solution to the di erential equation without bothering about

boundary conditions. We will follow the same treatment as in section (11.7) where in

equation (11.54) the input of a linear function was written as a superposition of delta

functions. In the same way, the source function can be written as:

Z

(r r0 )F (r0 )dV 0 :

F(r) = (14.26)

;

This expression follows from the properties of the delta function. One can interpret this

expression as an expansion of the function F(r) in delta functions because the integral

(14.26) describes a superposition of delta functions (r r0 ) centered at r = r0 each of

these delta functions is given a weight F(r0 ). We want to use a Green's function to

;

CHAPTER 14. GREEN'S FUNCTIONS, PRINCIPLES

182

construct a solution. The Green's function G(r r0 ) is the response at location r due to a

delta function source at location r0 , i.e. the Green's function satis es:

LG(r r0) = (r r0) : (14.27)

;

The response to the input (r r0 ) is given by G(r r0 ), and the source functions can

be written as a superposition of these delta functions with weight F(r0 ). This suggest that

;

a solution of the problem (14.23) is given by a superposition of Green's functions G(r r0 )

where each Green's function has the same weight factor as the delta function (r r0 ) in

;

the expansion (14.26) of F(r) in delta functions. This means that the solution of (14.23)

is given by: Z

uP (r) = G(r r0)F (r0 )dV 0 : (14.28)

Problem b: In case you worked through section (11.7) discuss the relation between this

expression and equation (11.55) for the output of a linear function.

It is crucial to understand at this point that we have used three steps to arrive at (14.28):

(i) The source function is written as a superposition of delta functions, (ii) the response of

the system to each delta function input is de ned and (iii) the solution is written as the

same superposition of Green's function as was used in the expansion of the source function

R

in delta functions:

(r r0 ) F(r) = (r r0 )F (r0 )dV 0

; $ ;

(14.29)

R G(r r0)F (r0 )dV 0

0)

+ +

G(r r uP (r) =

$

Problem c: Although this reasoning may sound plausible, we have not proven that uP (r)

in equation (14.28) actually is a solution of the di erential equation (14.23). Give

a proof that this is indeed the case by letting the operator L act on (14.28) and by

using equation (14.27) for the Green's function. Hint: the operator L acts on r while

the integration is over r0 , the operator can thus be taken inside the integral.

It should be noted that we have not solved our problem yet, because uP does not

necessarily satisfy the boundary conditions. In fact, the solution (14.28) is just one of the

many possible solutions to the problem (14.23). It is a particular solution of the inhomo-

geneous equation (14.23), and this is the reason why the subscript P is used. Equation

(14.23) is called an inhomogeneous equation because the right-hand-side is nonzero. If

the right-hand-side is zero one speaks of the homogeneous equation. This implies that a

solution u0 of the homogenous equation satis es

Lu0 = 0 : (14.30)

Problem d: In general one can add a solution of the homogeneous equation (14.30) to a

particular solution, and the result still satis es the inhomogeneous equation (14.23).

Give a proof of this statement by showing that the function u = uP +u0 is a solution

of (14.23). In other words show that the general solution of (14.23) is given by:

Z

u(r) = u0(r)+ G(r r0)F (r0 )dV 0 : (14.31)

14.5. RADIOGENIC HEATING AND THE EARTH'S TEMPERATURE 183

Problem e: The problem is that we still need to enforce the boundary conditions (14.25).

This can be achieved by requiring that the solution u0 satis es speci c boundary

conditions at rB . Insert (14.31) in the boundary conditions (14.25) and show that

the required solution u0 of the homogeneous equation must satisfy the following

boundary conditions:

Z

G(rB r0 )F (r0 )dV 0 :

u0(rB ) = uB (rB ) (14.32)

;

This is all we need to solve the problem. What we have shown is that:

the total solution (14.31) is given by the sum of the particular solution (14.28)

plus a solution of the homogeneous equation (14.30) that satis es the boundary

condition (14.32).

This construction may appear to be very complex to you. However, you should realize

that the main complexity is the treatment of the boundary condition. In many problems,

the boundary condition dictates that the function vanishes at the boundary (uB = 0) and

the Green's function also vanishes at the boundary. It follows from (14.31) that in that

case the boundary condition for the homogeneous solution is u0 (rB ) = 0. This boundary

condition is satis ed by the solution u0 (r) = 0 which implies that one can dispense with

the addition of u0 to the particular solution uP (r).

Problem f: Suppose that the boundary conditions do not prescribe the value of the

solution at the boundary but that instead of (14.25) the normal derivative of the

solution is prescribed by the boundary conditions:

@u (r ) = n ru(rB ) = wB

@n B ^ (14.33)

where n is the unit vector perpendicular to the boundary. How should the theory

^

of this section be modi ed to accommodate this boundary condition?

The theory of this section is rather abstract. In order to make the issues at stake more

explicit the theory is applied in the next section to the calculation of the temperature in

the earth.

14.5 Radiogenic heating and the earth's temperature

As an application of the use of Green's function we consider in this section the calculation of

the temperature in the earth and speci cally the e ect of the decay of radioactive elements

in the crust on the temperature in the earth. Several radioactive elements such as U235

do not t well in the lattice of mantle rocks. For this reason, these elements are expelled

from material in the earth's mantle and they accumulate in the crust. Radioactive decay

of these elements then leads to a production of heat at the place where these elements

accumulate.

As a simpli ed example of this problem we assume that the temperature T and the

radiogenic heating Q depend only on depth and that we can ignore the sphericity of the

earth. In addition, we assume that the radiogenic heating does not depend on time and

that we consider only the equilibrium temperature.

CHAPTER 14. GREEN'S FUNCTIONS, PRINCIPLES

184

Problem a: Show that these assumptions imply that the temperature is only a function

of the z-coordinate: T = T(z).

The temperature eld satis es the heat equation derived in section (8.4):

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

@t r

Problem b: Use this expression to show that for the problem of this section the temper-

ature eld satis es

d2 T = Q(z) : (14.34)

dz 2 ;

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