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dT :

J= (14.48)

dz

;

Problem m: Compute the heat- ow at the top of the layer (z = 0) and at the bottom

(z = D). Assuming that T0 and Q0 are both positive, does the heat- ow at these

locations increase or decrease because of the radiogenic heating Q0 ? Give a physical

interpretation of this result. Use this result also to explain why people who often

feel cold like to use an electric blanket while sleeping.

The derivation of this section used a Green's function that satis ed the boundary

conditions (14.38) rather than the boundary conditions (14.35) of the temperature eld.

However, there is no particular reason why one should use these boundary conditions. To

wit, one might think one could avoid the step of adding a solution T0 (z) of the homogeneous

˜

equation by using a Green's function G that satis es the di erential equation (14.39) and

the same boundary conditions as the temperature eld:

G(z = 0 z 0 ) = 0 G(z = D z 0 ) = T0 :

˜ ˜ (14.49)

Problem n: Go through the same steps as you did earlier in this section by constructing

the Green's function G(z z 0 ), computing the corresponding particular solution TP (z),

˜ ˜

verifying whether the boundary conditions (14.35) are satis ed by this particular

solution and if necessary adding a solution of the homogeneous equation in order to

satisfy the boundary conditions. Show that this again leads to the solution (14.47).

14.6. NONLINEAR SYSTEMS AND GREEN'S FUNCTIONS 187

Problem o: If you carried out the previous problem you will have discovered that the

trick to use a Green's function that satis ed the boundary condition at z = D did

not lead to a particular solution that satis ed the same boundary condition at that

point. Why did that trick not work?

The lesson from the last problems is that usually one needs to add to solution of the ho-

mogeneous equation to a particular solution in order to satisfy the boundary conditions.

However, suppose that the boundary conditions of the temperature eld would be homoge-

neous as well (T = (z = 0) = T(z = D) = 0). In that case the particular solution (14.45)

that was constructed using a Green's function that satis es the same homogeneous bound-

ary conditions (14.38) satis es the boundary conditions of the full problem. This implies

that it only pays o to use a Green's function that satis es the boundary conditions of

the full problem when these boundary conditions are homogeneous, i.e. when the function

itself vanishes (T = 0) or when the normal gradient of the function vanishes (@T=@n = 0)

or when a linear combination of these quantities vanishes (aT + b@T=@n = 0). In all other

cases one cannot avoid adding a solution of the homogeneous equation in order to satisfy

the boundary conditions and the most e cient procedure is usually to use the Green's

function that can most easily be computed.

14.6 Nonlinear systems and Green's functions

Up to this point, Green's function were applied to linear systems. The de nition of a linear

system was introduced in section (11.7). Suppose that a forcing F1 leads to a response

x1 and that a forcing F2 leads to a response x2 . A system is linear when the response

to the linear combination c1 F1 + c2 F2 (with c1 and c2 constants) leads to the response

c1 x1 + c2 x2.

Problem a: Show that this de nition implies that the response to the input times a

constant is given by the response that is multiplied by the same constant. In other

words show that for a linear system an input that is twice as large leads to a response

that is twice as large.

Problem b: Show that the de nition of linearity given above implies that the response

to the sum of two forcing functions is the sum of the responses to the individual

forcing functions.

This last property re ects that a linear system satis es the superposition principle which

states that for a linear system one can superpose the response to a sum of forcing functions.

Not every system is linear, and we will exploit here to what extent Green's functions

are useful for nonlinear systems. As an example we will consider the Verhulst equation:

x = x x2 + F(t) :

_ (14.50)

;

This equation has been used in mathematical biology to describe the growth of a popu-

lation. Suppose that only the term x was present in the right hand side. In that case

the solution would be given by x(t) = C exp (t). This means that the rst term on the

right hand side accounts for the exponential population growth that is due to the fact

CHAPTER 14. GREEN'S FUNCTIONS, PRINCIPLES

188

that the number of o spring is proportional to the size of the population. However, a

population cannot grow inde nitely, when a population is too large limited resources re-

strict the growth, this is accounted for by the 2 term in the right hand side. The term

;x

F(t) accounts for external in uences on the population. For example, a mass-extinction

could be described by a strongly negative forcing function F(t). We will consider rst the

solution for the case that F(t) = 0. Since the population size is positive we consider only

positive solutions x(t).

Problem c: Show that for the case F(t) = 0 the change of variable y = 1=x leads to the

linear equation y = 1 y. Solve this equation and show that the general solution of

_ ;

(14.50) (with F(t) = 0) is given by:

1

x(t) = Ae;t + 1 (14.51)

with A an integration constant.

Problem d: Use this solution to show that any solution of the unforced equation goes to

1 for in nite times:

lim x(t) = 1 : (14.52)

t!1

In other words, the population of the unforced Verhulst equation always converges to the

same population size. Note that when the forcing vanishes after a nite time, the solution

after that time must satisfy (14.51) which implies that the long-time limit is then also

given by (14.52).

Now, consider the response to a delta function excitation at time t0 with strength F0 .

The associated response g(t t0 ) thus satis es

g g + g2 = F0 (t t0) :

_ (14.53)

; ;

Since this function is the impulse response of the system the notation g is used in order to

bring out the resemblance with the Green's functions used earlier. We will consider only

causal solution, i.e. we require that g(t t0 ) vanishes for t < t0 : g(t t0 ) = 0 for t < t0 . For

t > t0 the solution satis es the Verhulst equation without forcing, hence the general form

is given by (14.51). The only remaining task is to nd the integration constant A. This

constant follows by a treatment similar to the analysis of section (14.3).

Problem e: Integrate (14.53) over t from t0 " to t0 + ", take the limit " 0 and show

; #

that this leads to the following requirement for the discontinuity in g:

lim g(t t0 )]tt0 +" = F0 : (14.54)

0 ;"

"#0

Problem f: Use this condition to show that the constant A in the solution (14.51) is

given by A = F10 1 exp t0 and that the solution is given by:

;

(

for t < t0

0

g(t t0 ) = (14.55)

F0 for t > t0

(1;F0 )e +F0

;(t;t0 )

14.6. NONLINEAR SYSTEMS AND GREEN'S FUNCTIONS 189

At this point you should be suspicious for interpreting g(t t0 ) as a Green's function. An

important property of linear systems is that the response is proportional to the forcing.

However, the solution g(t t0 ) in (14.55) is not proportional to the strength F0 of the

forcing.

Let us now check if we can use the superposition principle. Suppose the forcing function

is the superposition of a delta-function forcing F1 at t = t1 and a delta-function forcing

F2 at t = t2 :

F(t) = F1 (t t1 ) + F2 (t t2 ) : (14.56)

; ;

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