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Earthâ€™s surface

T=0 z=0

Q(z)=Q 0

Crust

z=H

Q(z)=0

Mantle

T=T 0 z=D

Figure 14.3: De nition of the geometric variables and boundary conditions for the tem-

perature in the earth.

This equation can be solved when the boundary conditions are speci ed. The thickness of

the crust is denoted by H, see gure (14.3). The temperature is assumed to vanish at the

earth's surface. In addition, it is assumed that at a xed depth D the temperature has a

xed value T0 . This implies that the boundary conditions are:

T(z = 0) = 0 T(Z = D) = T0 : (14.35)

In this section we will solve the di erential equation (14.34) with the boundary con-

ditions (14.35) using the Green's function technique described in the previous section.

Analogously to expression (14.28) we will rst determine a particular solution TP of the

di erential equation (14.34) and worry about the boundary conditions later. The Green's

function G(z z 0 ) to be used is the temperature at depth z due to delta function heating

at depth z 0 :

d2 G(z z 0 ) = (z z0 ) : (14.36)

dz 2 ;

Problem c: Use the theory of the previous section that the following function satis es

the heat equation (14.34):

1 Z D G(z z 0 )Q(z 0 )dz 0 :

TP (z) = (14.37)

;

0

14.5. RADIOGENIC HEATING AND THE EARTH'S TEMPERATURE 185

Before further progress can be made it is necessary to nd the Green's function, i.e. to

solve the di erential equation (14.36). In order to do this the boundary conditions for the

Green's function need to be speci ed. In this example we will use a Green's function that

vanishes at the endpoints of the depth interval:

G(z = 0 z 0 ) = G(Z = D z0 ) = 0 : (14.38)

Problem d: Use (14.36) to show that for z = z0 the Green's function satis es the di er-

ential equation d2 G(z z 0 )=dz 2 = 0 and use this to show that the Green's function

6

that satis es the boundary conditions (14.38) must be of the form

(

for z < z 0

z

G(z z 0 ) = (14.39)

(z D) for z > z 0

;

with and constants that need to be determined.

Problem e: Since there are two unknown constants, two conditions are needed. The rst

condition is that the Green's function is continuous for z = z 0 . Use the theory of sec-

tion (14.3) and the di erential equation (14.36) to show that the second requirement

is:

dG(z z0 ) z=z +" = 1

0

lim (14.40)

dz z=z ;"

"#0 0

i.e. that the rst derivative makes a unit jump at the point of excitation.

Problem f: Apply these two conditions to the solution (14.39) to determine the constants

and and show that the Green's function is given by:

(

for z < z 0

D;z0 z

G(z z 0 ) = D (14.41)

;

for z > z 0

z0 (D ; z)

D

;

In this notation the two regions z < z 0 and z > z 0 are separated. Note, however, that the

solution in the two regions has a highly symmetric form. In the literature you will nd

that a solution such as (14.41) is often rewritten by de ning z> to be the maximum of z

and z 0 and z< to be the minimum of z and z 0 :

z> max(z z0 ) (14.42)

z< min(z z 0 )

Problem g: Show that in this notation the Green's function (14.41) can be written as:

G(z z 0 ) = D D z> z< : (14.43)

;

;

As a particular heating function we will assume that the heating Q is only nonzero in

the crust. This is a rst-order description of the radiogenic heating in the shallow layers

in the earth. The reason for this is that many of the radiogenic elements such as U235

t much better in the crystal lattice of crustal material than in mantle material. For

simplicity we will assume that the radiogenic heating is constant in the crust:

(

Q(z) = Q0 for 0 < z < H (14.44)

for H < z < D

0

CHAPTER 14. GREEN'S FUNCTIONS, PRINCIPLES

186

Problem h: Show that the particular solution (14.37) for this heating function is given

by

8 Q H2 n ; z 2 z o

<0 H for 0 < z < H

TP (z) = : Q0H 2 ; H + H 2 D

2 (14.45)

; ;

z for H < z < D

1D

2 ;

Problem i: Show that this particular solution satis es the boundary conditions

TP (z = 0) = TP (z = D) = 0 : (14.46)

Problem j: This means that this solution does not satisfy the boundary conditions

(14.35) of our problem. Use the theory of section (14.4) to derive that to this partic-

ular solution we must add a solution T0 of the homogenous equation d2 T0 =dz 2 = 0

that satis es the boundary conditions T0 (z = 0) = 0 and T0 (z = D) = T0 .

Problem k: Show that the solution to this equation is given by T0(z) = T0 z=D and that

the nal solution is given by

8 z Q H2 n ; z 2 z o

< T0 D + 0 H for 0 < z < H

T (z) = : z Q2H 2 ; H + H 2 D (14.47)

; ;

z

T0 D + 0 1 D for H < z < D

2 ;

Problem l: Verify explicitly that this solution satis es the di erential equation (14.34)

with the boundary conditions (14.35).

As shown in expression (8.25) of section (8.4) the conductive heat- ow is given by

J= Since the problem is one-dimensional the heat ow is given by

; rT.

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