Neutron fluctuations a treatise on the physics on branching processes Imre Pã¡Zsit

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Neutron fluctuations a treatise on the physics on
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Author(s): Imre PÃƒÂ¡zsit, LÃƒÂ©nÃƒÂ¡rd PÃƒÂ¡l
ISBN(s): 9780080450643, 0080450644
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Year: 2007
Language: english

Prelims-I045064.tex 27/6/2007 9: 20 Page xi
PREFACE
Thorough descriptions of branching processes can be found in almost every book and monograph that deals
with stochastic processes [1–5]. Moreover, in the monographs by T.E. Harris [6] and B.A. Sevast’yanov [7],
nearly every problem of the theory is discussed with mathematical rigour. There are innumerable publications
available about the applications of the theory of branching processes in the different fields of natural sciences
such as physics [8], nuclear engineering [9–11], and biology [12]. With regard to the fluctuations in branching
processes concerning nuclear chain reactions, these are synonymous with zero power neutron noise, or neutron
fluctuations in zero power systems. In this respect, already in 1964, earlier than the books by Stacey [9] and
Williams [11] appeared in print, a remarkable general work, amounting to a monograph, was published by
D.R. Harris [13] on this topic.
However, it is somewhat surprising that no monograph has been published since 1974 on neutron fluctu-
ations. There appears to be a need for a self-contained monograph on the theory and principles of branching
processes that are important both for the studies of neutron noise and for the applications, and which at the
same time would treat the recent research problems of neutron noise by accounting for new developments.
The ambition to fill this gap constitutes the motivation for writing this book.
This book was thus written with two objectives in mind, and it also consists of two parts, although the
objectives and parts slightly overlap. The first objective was to present the theory and mathematical tools used
in describing branching processes which can be used to derive various distributions of the population with
multiplication. The theory is first developed for reproducing and multiplying entities in general, and then is
applied to particles and especially neutrons in particular, including the corresponding detector counts. Hence,
the text sets out by deriving the basic forward and backward forms of the master equations for the probability
distributions and their generating functions induced by a single particle.Various single and joint distributions
and their special cases are derived and discussed. Then the case of particle injection by an external source
(immigration of entities) is considered. Attention is given to the case when some entities (particles) are born
with some time delay after the branching event. Moments, covariances, correlations, extinction probabilities,
survival times and other special cases and special probabilities are discussed at depth.
All the above chapters concern an infinite homogeneous material. In Chapter 7 space dependence is
introduced. A one-dimensional case is treated as an illustration of a simple space-dependent process, in which
a number of concrete solutions can be given in closed compact form.
Whereas the first part treats concepts generally applicable to a large class of branching processes, Part II of
this book is specifically devoted to neutron fluctuations and their application to problems of reactor physics
and nuclear material management. The emphasis is on the elaboration of neutron fluctuation based methods
for the determination of the reactivity of subcritical systems with an external source. First, in Chapter 8, a
detailed derivation of the Pál–Bell equation, together with its diffusion theory approximation, is given. The
original publication of the Pál–Bell equation constituted the first theoretical foundation of the zero power
noise methods which had been suggested earlier by empirical considerations. Thereafter, Chapters 9 and 10
deal with the applications of the general theory to the derivation of the Feynman and Rossi-alpha methods.
Chapter 9 concerns the derivation of the classical formulae for traditional systems, whereas Chapter 10 reflects
the recent developments of these methods in connection with the so-called accelerator-driven systems, i.e.
subcritical cores driven with a spallation source, and/or with pulsed sources. Finally, Chapter 11 touches
upon the basic problems and methods of identifying and quantifying small samples of fissile material from
the statistics of spontaneous and induced neutrons and photons. This area of nuclear safeguards, i.e. nuclear
xi

Prelims-I045064.tex 27/6/2007 9: 20 Page xii
xii Preface
material accounting and control, is under a rapidly increasing attention due to the general increase of safety
and safeguards needs worldwide.
A special new contribution of this book to the field of neutron noise is constituted by Chapter 6, in
which the so-called zero power neutron noise, i.e. branching noise, is treated in systems with time-varying
properties. Neutron noise in systems with temporally varying properties is called‘power reactor noise’. Neutron
fluctuations in low power steady systems and high power systems with fluctuating parameters have constituted
two disjoint areas so far which were treated with different types of mathematical tools and were assumed to
be valid in non-overlapping operational areas. The results in Chapter 6 are hence the first to establish a bridge
between zero power noise and power reactor noise. Due to space limitations, theLangevin techniqueand the
theory of theparametric noiseare not discussed. The interested reader is referred to the excellent monographs
byVan Kampen [14] and Williams [11].
Since thegenerating functionsplay a decisive role in many considerations of this book, the theorems most
frequently used in the derivations are summarised in Appendix A.
This book is not primarily meant for mathematicians, rather for physicists and engineers, and notably for
those working with branching processes in practice, and in the first place for physicists being concerned with
reactor noise investigations and problems of nuclear safeguards. However, it can also be useful for researchers
in the field of biological physics and actuarial sciences.
The authors are indebted to many colleagues and friends who contributed to the realisation of this book in
one way or another and with whom they collaborated during the years. One of us (I.P.) is particularly indebted
to M.M.R.Williams, from whom he learnt immensely on neutron noise theory and with whom his first paper
on branching processes was published. He also had, during the years, a very intensive and fruitful collaboration
with several Japanese scientists, in particular withY.Yamane andY. Kitamura of Nagoya University. Chapters 9
and 10 are largely based on joint publications. Research contacts and discussions on stochastic processes and
branching processes with H. Konno of the University of Tsukuba are acknowledged with thanks. Parts of this
book were written during an inspiring visit to Nagoya and Tsukuba. The experimental results given in the
book come from the Kyoto University Critical Assembly at KURRI, and contributions from the KURRI
staff are gratefully acknowledged. The chapter on nuclear safeguards is largely due to a collaboration with Sara
A. Pozzi of ORNL, who introduced this author to the field.
Both authors are much indebted to Maria Pázsit, whose contributions by translating early versions of the
chapters of Part I from Hungarian can hardly be overestimated. She has also helped with typesetting and
editing the LaTeX version of the manuscript, as well as with proofreading. The authors acknowledge with
thanks constructive comments on the manuscript from M.M.R.Williams and H. van Dam and thank S. Croft
for reading Chapter 11 and giving many valuable comments.
Without the funding contribution of many organisations this book would not have been possible. Even if
funding specifically for this book project was not dominating, it is a must to mention that the research of one
of the authors (I.P.) was supported by the Swedish Nuclear Inspectorate (SKI), the Ringhals power plant, the
Swedish Centre for Nuclear Technology (SKC), the Adlerbert Research Foundation, the Japan Society for the
Promotion of Science (JSPS) and the Scandinavia–Japan Sasakawa Foundation.Their contribution is gratefully
acknowledged.
We had no ambition to cite all published work related to the problems treated in this monograph. The
books and papers listed in the ‘List of Publications’ represent merely some indications to guide the reader.
One has to mention the excellent review of source papers in reactor noise by Saito [15]. This review contains
practically all of the important publications until 1977 which are in strong relation with the topic of this book.
Imre Pázsit
Gothenburg
Lénárd Pál
Budapest
February 2007

Prelims-I045064.tex 27/6/2007 9: 20 Page xiii
ACKNOWLEDGEMENT
The authors are grateful to Elsevier Ltd for granting permission to reproduce the material detailed below:
•Figure 5 from the article byY. Kitamuraet al.inProgr. Nuc. Ener., 48(2006) 569.
•Figures 3 and 4 from the article by I. Pázsitet al.inAnn. Nucl. Ener., 32(2006) 896.
•Figures 4 and 5 from the article byY. Kitamuraet al.inProgr. Nucl. Ener., 48(2006) 37.
•Figures 1 and 3 from the article by A. Enqvist, I. Pázsit and S. Pozzi inNucl. Instr. Meth. A, 566(2006) 598.
xiii

Prelims-I045064.tex 27/6/2007 9: 20 Page xv
LISTOFMOSTFREQUENTLYUSEDNOTATIONS
Symbol Description
P{···} Symbol of probability
E{···} Symbol of expectation
D
2
{···} Symbol of variance
Q Intensity of a reaction
ν Number of progeny (neutrons) in one reaction
P{ν=k}=f
k Probability of{ν=k}
q(z)=
≤
∞
k=0
fkz
k
Basic generating function
E{ν}=q

(1)=q 1 Expectation of the progeny number in one reaction
E{ν(ν−1)}=q

(1)=q 2 Second factorial moment of the progeny number in one reaction
Q
a=Qf0 Total intensity of absorption
Q
b=Qf1 Intensity of renewal
Q
m=Q(1−f 0−f1) Intensity of multiplication
n(t) Number of particles at timet
P{n(t)=n|n(0)=1}=p
n(t) Probability of findingnparticles at timetin the case of one starting
particle
g(z,t)=
≤
∞
n=0
pn(t)z
n
Generating function ofp n(t)
q Number of particles produced by one injection (spallation) event
P{q=j}=h
j Probability of{q=j}; probability that there arejemitted neutrons
per spallation event
r(z)=
≤
∞
j=0
hjz
j
Generating function of probabilityh j
E{q}=r

(1)=r 1 Expectation of the particle number produced by one injection
event;expectation of the number of neutrons emitted per spallation
event
E{q(q−1)}=r

(1)=r 2 Second factorial moment of the particle number produced by
one injection event; second factorial moment of the number of
neutrons emitted per spallation event
D
ν=q2/q
2
1
,D q=r2/r
2
1
Diven factors ofνandq
s(t) Intensity of the injection process at time t
N(t) Number of particles at timetin the case of particle injection
P{N(t)=n|n(t
0)=0}=P n(t|t0) Probability of findingnparticles at timetif the particle injection
started att
0≤t
G(z,t|t
0)=
≤
∞
n=0
Pn(t|t0)z
n
Generating function ofP n(t|t0)
α=Q(q
1−1)>0 Multiplication intensity (q 1>1)
a=−α=Q(1−q
1)>0 Decay intensity (q 1<1)
m
1(t) Expectation ofn(t)
m
2(t) Second factorial moment ofn(t)
M
1(t) Expectation ofN(t)
M
2(t) Second factorial moment ofN(t)
n
a(t−u,t) Number of absorptions in the time interval [ t−u,t],u≥0
P{n
a(t−u,t)=n|n(0)=1}=p(n,t,u) Probability of absorbingnparticles in the time interval [t−u,t]in
the case of one starting particle
xv

Prelims-I045064.tex 27/6/2007 9: 20 Page xvi
xvi List of Most Frequently Used Notations
Symbol Description
N
a(t−u,t) Number of absorptions in the time interval [ t−u,t],u≥0inthe
case of particle injection
P{N
a(t−u,t)=n|n(0)=0}=P (n,t,u) Probability of absorbingnparticles in the time interval [t−u,t]in
the case of particle injection
m
(a)
1
(t,u) Expectation of the number of absorbed particles in the time
interval [t−u,t] in the case of one starting particle
m
(a)
2
(t,u) Second factorial moment of the number of absorbed particles in
the time interval [t−u,t] in the case of one starting particle
M
(a)
1
(t,u) Expectation of the number of absorbed particles in the time
interval [t−u,t] in the case of particle injection
M
(a)
2
(t,u) Second factorial moment of the number of absorbed particles in
the time interval [t−u,t] in the case of particle injection
D
2
{Na(t−u,t)} Variance ofN a(t−u,t)
{S(t)=S
},∈Z
(+)
Medium is in the stateS at timet
U Subset of the coordinate-velocity space
u={βr,βv} Phase point in the coordinate-velocity space
n(t,U) Number of neutrons in the subsetUat timet
p[t
0,u0;t,n(U)] Probability of findingnneutrons in the subsetUat timet, when
one neutron started from the phase pointu
0at timet 0≤t
m
1(t0,u0;t,U) Expectation of the number of neutrons in the subsetUat timet,
when one neutron started from the phase pointu
0at timet 0≤t
m
2(t0,u0;t,U) Second factorial moment of the number of neutrons in the subset
Uat timet, when one neutron started from the phase pointu
0at
timet
0≤t
C(t) Number of the delayed neutron precursors at timet
Z(t,t
d) Number of the detected neutrons in the time interval [t d,t]
λ
c Intensity of capture
λ
f Intensity of fission
λ
d Intensity of detection
λ Decay constant
S Source intensity
p
f(n,m) Probability of emittingnneutrons andmprecursors in one fission
g
f(x,y) Generating function ofp f(n,m)
∂g
f(x,y)/∂x| x=y=1 =ρνp Average number of prompt neutrons per fission
∂g
f(x,y)/∂y| x=y=1 =ρνd Average number of delayed neutrons per fission
ρν=ν
p+ν d Average number of neutrons per fission
β=ρν
d/ρν Effective delayed neutron fraction
ρ Reactivity
=1/ρνλ
f Prompt neutron generation time
α=(β−ρ)/ Prompt neutron decay constant used in Chapters 9 and 10
=λ
d/λf Detector efficiency
P(N,C,Z,t|t
0) Probability of findingNneutrons andCprecursors at timetin
the system driven by a source, and of countingZneutrons in the
time interval [0,t]
G(x,y,v,t|t
0) Generating function ofP(N,C,Z,t|t 0)
Z(t) Asymptotic expectation of the number of detected neutrons in the
time interval [0,t]
µ
ZZ(t,0|t 0) Modified second factorial moment

Prelims-I045064.tex 27/6/2007 9: 20 Page xvii
List of Most Frequently Used Notations xvii
Symbol Description
lim
t0→−∞µZZ(t,0|t 0)=µ ZZ(t) Asymptotic modified second factorial moment
Y(t)=µ
ZZ(t)/Z(t) Y(t) in the Feynman-alpha formula
p(n,c,z,T,t) Probability that there arenneutrons andcprecursors at timetin the
system, induced by one initial neutron att=0, and there have beenz
detector counts in the time interval [t−T,t]
P(N,C,Z,T,t) Probability that there areNneutrons andCprecursors at timetin the
system, induced by a source of intensityS, and that there have beenZ
detector counts in the time interval [t−T,t], provided that there were
no neutrons and precursors in the system at timet=0 and no neutron
counts have been registered up to timet=0
g(x,y,v,T,t) Generating function ofp(n,c,z,T,t)
G(x,y,v,T,t) Generating
function ofP(N,C,Z,T,t)
ν Total number of neutrons produced in a cascade
µ Total number of gamma photons produced in a cascade
∞ν
1 Neutron singles
∞ν
2 Neutron doubles
∞ν
3 Neutron triples
M Leakage multiplication
ϕ
r Average number of neutrons generated in a sample
M
γ Gamma multiplication per one initial neutron
∞µ
1 Gamma singles
∞µ
2 Gamma doubles
∞µ
3 Gamma triples
P(n) Number distribution of neutrons generated in a sample
F(n) Number distribution of gamma photons generated in a sample

Ch01-I045064.tex 27/6/2007 9: 23 Page 3
CHAPTER ONE
Basic Notions
Contents
1.1 Definitions 3
1.2 Equations for the Generating Functions 4
1.3 Investigation of the Generating Function Equations 11
1.4 Discrete Time Branching Processes 20
1.5 Random Tree as a Branching Process 23
1.6 Illustrative Examples 25
1.1 Deﬁnitions
First, the basic definitions will be summarised, and for the sake of easier overview, the simplest way of
treatment is chosen.
The medium, in which certain objects are capable for not only to enter reactions but also can multiply
themselves, is called amultiplying medium. Suppose that this medium is homogeneous and infinite. The medium
will often be referred to as asystemas well. For example, objects can be bacteria on a nourishing soil, or
particles suitable for chemical or nuclear chain reactions, etc. In the following, we will use the nameparticle
instead of object.
Suppose that at a certain time instantt
0,only one particle capable for multiplication exists in the multiplying
medium. Denote the number of particles at the time instancet≥t
0byn(t). It is evident thatn(t)∈Z, where
Zis the set of non-negative integers. The event which results in eitherabsorptionorrenewal,or multiplication
of the particle is calledreaction. Let τbe the interval between the time of appearance of the particle in the
multiplying medium and that of its first reaction. Suppose that the probability distribution function
P{τ>t −t
0|t0}=T(t 0,t),t 0≤t,
in whicht
0is the time instance when the particle appears in the multiplying medium, satisfies the functional
equation
T(t
0,t)=T(t 0,t
δ
)T(t
δ
,t),t 0≤t
δ
≤t.
In our considerations, letT(t
0,t) be the exponential distribution given by the equation
T(t
0,t)=e
−Q(t−t 0)
, (1.1)
whereQis theintensityof the reaction. Further, letνbe the number of new particles born in the reaction,
replacing the particle inducing the reaction, and let
P{ν=k}=f
k,k∈Z (1.2)
Neutron fluctuations © 2008 Elsevier Ltd.
ISBN-13: 978-0-08-045064-3 All rights reserved.
3

Ch01-I045064.tex 27/6/2007 9: 23 Page 4
4 Imre Pázsit & Lénárd Pál
be the probability thatν=k. It is obvious thatf 0is the probability of absorption,f 1that of renewal, whilef k,
k>1 is the probability of multiplication. The quantitiesQ,f
k,k∈Zare the parameters determining thestate
of the multiplying medium.
1
The first case to be treated is the determination of the conditional probability
P{n(t)=n|n(t
0)=1}=p (n,t|1,t 0)
in the case when the process ishomogeneous in time, i.e. the probability p(n,t|1,t
0) depends only on the time
differencet−t
0. Hence, one can chooset 0=0 and accordingly write
P{n(t)=n|n(0)=1}=p (n,t|1, 0)=p
1n(t)=p n(t). (1.3)
In the sequel the notationp
n(t)=p 1n(t) will be used. Hence,p n(t) is the probability that exactlynparticles
exist in the medium at timet∈T, provided that att=0 there was only one particle in the medium. Here,T
denotes the set of the non-negative real numbers. This description is usually called theone-point model, since
the branching process in the homogeneous infinite medium is characterised by the number of particles in the
medium atone given time instant.
2
For determining the probabilityp n(t), thegenerating function
g(z,t)=E{z
n(t)
|n(0)=1}=
∞
τ
n=0
pn(t)z
n
,|z|≤1 (1.4)
will be used.
3
1.2 Equations for the Generating Functions
1.2.1 Intuitive solution
To begin with, an intuitive solution will be given which starts by progressing from backwards in the branching
processn(t), considering the mutually exclusive complete set of first events following the time instantt=0.
The following theorem will be proven.
Theorem 1.The generating function g(z,t)=
ν
∞
n=1
pn(t)z
n
,|z|≤1of the probability pn(t)with the initial condition
g(z,0)=z satisfies the backward-type equation
∂g(z,t)
∂t
=−Qg(z,t)+Qq[g(z,t)], (1.5)
in which
q(z)=E{z
ν
}=
∞
τ
k=0
fkz
k
(1.6)
is the so-called basic generating function.
Proof.The proof is based on the fact that the event{n(t)=n|n(0)=1}is the sum of the following mutually
exclusive two events.
1.The single particle in the medium at the time instantt
0=0 does not enter into a reaction until timet>0,
hence the number of particles will be exactly 1 at timet.
1
In the theory of branching processes, this process belongs to the category of the so-calledage-dependent processes.
2
The notion ‘one-point model’ is not to be mixed up with the point model of reactor theory, where the phrase ‘point’ refers to a spatial property.
3
A short summary of the characteristics of generating functions is found in Appendix A.

Ch01-I045064.tex 27/6/2007 9: 23 Page 5
BasicNotions 5
2.The single particle which exists in the medium at timet 0=0 will have a first reaction until timet>0, such
that the first reaction will take place in some subinterval (t
δ
,t
δ
+dt
δ
] of the interval (0,t], wheret
δ
runs
through every point of the interval (0,t] and every new particle born in the reaction under the remaining
timet−t
δ
will generate so many further new particles independently from each other that their number
together will be exactlynat timet>0.
Based on this, one can write that
p
n(t)=e
−Qt
δn1
+Q
≤
t
0
e
−Qt
δ
⎡
⎣f
0δn0+
∞
≥
k=1
fk
≥
n1+···+n k=n
k
→
j=1
pnj
(t−t
δ
)
⎤
⎦dt
δ
,
and this is the same as
p
n(t)=e
−Qt
δn1
+Q
≤
t
0
e
−Q(t−t
δ
)
⎡
⎣f
0δn0+
∞
≥
k=1
fk
≥
n1+···+n k=n
k
→
j=1
pnj
(t
δ
)
⎤
⎦dt
δ
.
From this one immediately obtains the integral equation
g(z,t)=e
−Qt
z+Q
≤
t
0
e
−Q(t−t
δ
)
∞
≥
k=0
fk[g(z,t
δ
)]
k
dt
δ
for the generating function of (1.4) which, by taking into account the definition in (1.6), can be written in the
following form:
g(z,t)=e
−Qt
z+Q
≤
t
0
e
−Q(t−t
δ
)
q[g(z,t
δ
)]dt
δ
. (1.7)
By derivation of this equation with respect tot, one obtains (1.5). The initial conditiong(z,0)=zfollows
immediately also from (1.7).
This equation derived for the generating functiong(z,t) of the probabilityp n(t) belongs to the family of
the so-calledbackward Kolmogorov equations.
4
Equation (1.5) can also be obtained directly by considering the probabilities of the two mutually exclusive
events (in first order indt) of having a reaction or having no reaction between 0≤t≤dt.
In many cases, one may need theexponential generating functionof the probabilityp
n(t|1) which is defined
by the infinite series
g
exp(z,t)=
∞
≥
n=0
pn(t)e
nz
,|e
z
|≤1,
that satisfies the equation
∂g
exp(z,t)
∂t
=−Qg
exp(z,t)+Qq exp[logg exp(z,t)] (1.8)
with the initial conditiong
exp(z,0)=e
z
, where
q
exp(z)=
∞
≥
k=0
fke
kz
,|e
z
|≤1.
4
According to the terminology of master equations, this equation, especially in the differential equation form, is a ‘mixed’-type equation, since the
variabletrefers to the final (terminal) time, and not the initial time on which the backward master equation operates on.

Ch01-I045064.tex 27/6/2007 9: 23 Page 6
6 Imre Pázsit & Lénárd Pál
Derive now the so-calledforward Kolmogorov equationdetermining the probabilityp n(t). In this case, the
probabilityp
n(t+∞t) will be expressed by probabilities due to an earlier time instantt. For the generating
function (1.4) the following theorem will be proved.
Theorem 2.The generating function g(z,t)with the initial condition g(0,z)=z satisfies the linear forward-type
differential equation
∂g(z,t)
∂t
=Q[q(z)−z]
∂g(z,t)
∂z
. (1.9)
Proof.Considering thatnQ∞t+o(∞t) is the probability that one reaction takes place in the medium containing
nparticles in the interval (t,t+∞t] at the time instantt>0, one can write that
p
n(t+∞t)=p n(t)(1−nQ∞t)
+Q∞t
n
≥
k=0
(n−k+1)f kpn−k+1(t)+o(∞t).
After rearranging the equation and performing the limit∞t→0, one obtains
dp
n(t)
dt
=−Qnp
n(t)+Q
n
≥
k=0
fk(n−k+1)p n−k+1(t). (1.10)
The corresponding initial condition isp
n(0)=δ n1. From this, equation (1.9) immediately follows for the
generating function
g(z,t)=
∞
≥
n=0
pn(t)z
n
,|z|≤1,
with the initial conditiong(z,0)=z.
Also in this case, it is worth quoting the equation
∂g
exp(z,t)
∂t
=Q[e
−z
q(e
z
)−1]
∂g
exp(z,t)
∂z
(1.11)
for the exponential generating function
g
exp(z,t)=
∞
≥
n=0
pn(t)e
nz
,|e
z
|≤1,
and which is appended by the initial conditiong
exp(z,0)=e
z
.
Remark.In many applications, it is practical and hence customary to separate the reactions leading to
absorbtion and multiplication. LetQ
abe the intensity of absorption, whereasQ fthat of multiplication.
5
Hence, one can writeQ=Q a+Qf. Letp f(k) denote the probability that a numberk∈Znew particles are
generated by the incoming particle which disappears in themultiplyingreaction (i.e. fission). In this case, one
obtains for the generating functiong(z,t) the integral equation
g(z,t)=e
−Qt
z+Q a
≤
t
0
e
−Q(t−t
δ
)
dt
δ
+Qf
≤
t
0
e
−Q(t−t
δ
)
qf[g(z,t
δ
)]dt
δ
, (1.12)
5
This separation is motivated on physical grounds, with absorption corresponding to capture, and multiplication to fission, including the possibility
of zero neutrons generated in fission.

Ch01-I045064.tex 27/6/2007 9: 23 Page 7
BasicNotions 7
where
q
f(z)=
∞
≥
k=0
pf(k)z
k
.
The relationship between thef
kand thep f(k) can be written as
f
k=
Q
f
Q
p
f(k)+
Q
a
Q
δ
k,0, (1.13)
which can be inverted as
p
f(0)=
Q
Qf
f0−
Q
a
Qf
andp f(k)=
Q
Qf
fk,k=1, 2,.... (1.14)
Using (1.14) in (1.12), one regains immediately the more concise equation (1.7). It can also be seen that
∞
≥
k=0
fk=
Q
f
Q
∞
≥
k=0
pf(k)+
Q
a
Q
=
Q
f
Q
+
Q
a
Q
=1,
and
∞
≥
k=0
kpf(k)=E{ν f}=
Q
Qf
∞
≥
k=0
kfk=
Q
Qf
E{ν}=
Q
Qf
q1, (1.15)
whereE{ν
f}is the expectation of the number of neutrons perfission,
6
whereasq 1≡q
⎡
(1) is the expectation
of the number of neutrons perreaction. In a similar manner one finds that
E{ν
f(νf−1)}=
Q
Qf
q2 (1.16)
withq
2≡q
⎡⎡
(1), and hence the important relationship
E{ν
f(νf−1)}
E{νf}
=
q
2
q1
(1.17)
holds. This identity will be very instrumental when transferring results and expressions from Part I to II of the
book, where the processes of absorption and fission will be separated, and the formalism will be built on the
use of the distributionp
f(k) and its moments.
1.2.2 Solution according to Kolmogorov and Dmitriev
In the following, the solution of the problem will be described by using the methods of Kolmogorov and
Dmitriev [16].
LetTdenote the set of non-negative real numbers [0,∞) andn(t) be an integer valued random process,
homogeneous in time, defined over theparameter spaceT. Let us call the set of non-negative integersZ, i.e.,
the values which may be assumed byn(t), thephase spaceofn(t). The random processn(t),t∈Tgenerates a
homogeneous Markov processif thetransition probability
P{n(t)=j|n(0)=i}=p
ij(t) (1.18)
6
In Chapters 9–11, where the probability distributionp f(k) and its factorial moments will be used, and in Section 6.4,E{ν f}will be simply denoted
as⎦ν andE{ν
f(νf−1)}as⎦ν(ν−1) .

Ch01-I045064.tex 27/6/2007 9: 23 Page 8
8 Imre Pázsit & Lénárd Pál
fulfils the following conditions:
(a)
p
ij(t)≥0,∀i,j∈Zandt∈T;
(b)
∞
≥
j=0
pij(t)=1,∀i∈Zandt∈T;
(c)
p
ij(t)=
∞
≥
k=0
pik(u)pkj(t−u),∀i,j∈Zand 0≤u≤t,u,t∈T;
(d)
p
ij(0)=δ ij=

1, ifi=j,
0, ifi=j.
Iftvaries continuously, then, in addition to condition (d) we shall also suppose that
lim
t↓0
pii(t)=1. (1.19)
From this and the conditions (a) and (b), it immediately follows that for everyi, for whichi=j, the transition
probabilityp
ij(t) converges continuously to zero ift↓0, i.e.
lim
t↓0
pij(t)=0. (1.20)
Further, from condition (c), it follows that the transition probabilitiesp
ij(t),i,j=0, 1,...are continuous at
every time instantt∈T.
Definition 1.The Markov processn(t),t∈T, defined in the phase spaceZ, is called a branching process if
p
kn(t)=
≥
n1+···+n k=n
pn1
(t)pn2
(t)···p nk
(t). (1.21)
This equation expresses the fact that thekparticles existing in the system att=0 initiate branching
processes independently from each other.
7
Letn i(t) denote the number of progeny created by theith particle
att. Obviously, the number of progeny fort>0 generated by thekparticles present in the system att=0is
expressed by the random process
n(t)=n
1(t)+n 2(t)+···+n k(t), (1.22)
in whichn
i(t),i=1,...,kare independent from each other and have the same distribution
P{n
i(t)=n|n i(0)=1}=p n(t),i=1,...,k.
It follows then that the transition probabilityp
kn(t) is simply thek-fold convolution of the transition probability
p
n(t), as expressed by (1.21). In the further considerations, the following theorem is of vital importance.
7
It has to be emphasised that this assumption is only valid in a medium whose properties do not vary in time. This question is discussed in detail
in Chapter 6.

Ch01-I045064.tex 27/6/2007 9: 23 Page 9
BasicNotions 9
Theorem 3.The generating function g(z,t)of the probability p n(t)fulfils the functional equation
g(z,t+u)=g[g(z,u),t], (1.23)
and the initial condition g(z,0)=z.
Proof.Equation (1.23) is a direct consequence of the fact thatn(t) is a branching Markov process, i.e.
p
n(t+u)=
∞
≥
k=0
pk(t)pkn(u),
and
p
kn(u)=
≥
n1+···+n k=n
pn1
(u)...p nk
(u).
Hence, in this case one has
g(z,t+u)=
∞
≥
k=0
pk(t)
∞
≥
n=0
pkn(u)z
n
=
∞
≥
k=0
pk(t)[g(z,u)]
k
=g[g(z,u),t)],
and this is exactly what was to be proven.
The initial condition, in its turn, follows from the fact thatp n(0)=δ n1, and accordinglyg(z,0)=z.By
considering the condition (1.19), one can write
lim
t→0
p1(t)=1. (1.24)
From this it follows that ift→0 theng(z,t)→z, moreover this is valid uniformly to everyzfor which
the condition|z|≤1 is satisfied. Also, it can easily be proven thatg(z,t)isuniformly continuousintfor every
t∈[0,∞), provided that|z|≤1.
It is then assumed that ift→0, the probabilitiesp
n(t),n=0, 1,...can be written in the following form:
p
1(t|1)=1+w 1t+o(t), (1.25)
and
p
n(t|1)=w nt+o(t),n=1. (1.26)
Since 0≤p
1(t)≤1 must be hold, the inequalityw 1<0 has to be fulfilled. Considering that
∞
≥
n=0
pn(t)=1,∀t∈[0,∞),
the equality
∞
≥
n=0
wn=0 (1.27)
has to be satisfied. By introducing the notations
w
n=Q[f n−δn,1],
where 0≤f
1<1, one obtains
∈
∞
n=0
fn=1. Due to this, the quantity 0≤f n≤1 can be interpreted as the
probability of the event that exactlynparticles are born during the reaction, hence its meaning is equal to
the probability defined in (1.2). Moreover, the quantityQhaving a dimension [time]
−1
is the intensity of the
reaction.
After these preparations,the basic theorem of branching processescan be stated.

Ch01-I045064.tex 27/6/2007 9: 23 Page 10
10 Imre Pázsit & Lénárd Pál
Theorem 4.Introducing the function
s(z)=
∞
≥
n=0
wnz
n
=Q
∞
≥
n=0
fnz
n
−Qz=Q[q(z)−z], (1.28)
where q(z)was defined in(1.6), for every|z|≤1the generating function g(z,t)fulfils the backward differential equation
∂g(z,t)
∂t
=s[g(z,t)]=−Qg(z,t)+Qq[g(z,t)], (1.29)
and the forward linear partial differential equation
∂g(z,t)
∂t
=s(z)
∂g(z,t)
∂z
=Q[q(z)−z]
∂g(z,t)
∂z
, (1.30)
respectively, under the initial condition g(z,0)=z.
For the proof, the following lemma is needed.
Lemma 1.If the conditions in(1.25)–(1.27) are fulfilled, then the asymptotic formula
g(z,t)=z+s(z)t+o(t)=z+Q[q(z)−z]+o(t) (1.31)
is uniformly valid for every z for which it is true that|z|≤1, for the case t↓0.
Proof.One only has to prove that the absolute value of
g(z,t)−z
t
−s(z)
converges to zero fort↓0, since the statement in (1.31) immediately follows from it.To this order, let us write
the inequality

g(z,t)−z
t
−s(z)

≤

p
1(t)−1
t
−w
1

|z|+
≥
k=1

p
k(t)
t
−w
k

|z|
k
≤

p
1(t)−1
t
−w
1

+
≥
k=1,k≤N

p
k(t)
t
−w
k

+
≥
k>N
pk(t)
t
+
≥
k>N
wk.
The last term on the right-hand side can be made arbitrarily small ifNis chosen sufficiently large. By fixing
nowNat this value and selecting a sufficiently small value fort, it is obvious that even the first and second
terms on the right-hand side can be made arbitrarily small. Further, from (1.26), it is seen that in the case
oft→0
≥
k>N
pk(t)
t
→
≥
k>N
wk,
hence we have proved that even the third term can be made arbitrarily small. By virtue of the foregoing, the lemma is proved.
Proof.Now, the generating function equations (1.29) and (1.30) can easily be derived. By using (1.23), one has
g(z,∞t+t)=g[g(z,t),∞t],
i.e.
g(z,t)=g[g(z,t−∞t),∞t].

Ch01-I045064.tex 27/6/2007 9: 23 Page 11
BasicNotions 11
With the help of the lemma proved above, one obtains from these the equations
g(z,∞t+t)=g(z,t)+s[g(z,t)]∞t+o(∞t),
and
g(z,t)=g(z,t−∞t)+s[g(z,t−∞t)]∞t+o(∞t).
Sinceg(z,t) is uniformly continuous for everyt∈[0,∞)if|z |≤1, it is obvious that the above leads to
∂g(z,t)
∂t
=s[g(z,t)]=−Qg(z,t)+Qq[g(z,t)],
which agrees exactly with (1.29).
To derive (1.30), one applies equation (1.23) in an alternative way. From the relationships
g(z,t+∞t)=g[g(z,∞t),t]=g[z+s(z)∞t+o(∞t),t]
=g(z,t)+
∂g(z,t)
∂z
s(z)∞t+o(∞t),
and
g(z,t)=g[g(z,∞t),t−∞t]=g[z+s(z)∞t+o(∞t),t−∞t]
=g(z,t−∞t)+
∂g(z,t−∞t)
∂z
s(z)∞t+o(∞t),
after rearrangement and performing the limit∞t→0,the generating function equation in (1.30) is immediately
obtained.
The initial conditiong(0,z)=zis the consequence of the relation
p
n(0)=δ n1,∀n≥0
as it was pointed out before.
1.3 Investigation of the Generating Function Equations
1.3.1 Uniqueness of the solution, regular and irregular branching processes
From the theory of the differential equations, it follows thatthe solutions of the generating function equations(1.29)
and (1.30)are identical. Hence, it is sufficient to investigate only (1.29) by taking into account the initial
conditiong(z,0)=z. According to theexistence theoremof the differential equations, equation (1.29) for every
point|z|<1 has only one solutiong(z,t) which satisfies the initial conditiong(z,0)=zand equation (1.23).
However, it has to be specifically investigated under what conditions this solution satisfies also the limit
relationship
lim
z↑1
g(z,t)=g(1,t)=
∞
τ
n=0
pn(t|1)=1, (1.32)
i.e.under which conditions the solution g(z,t)can be considered a probability generating function.
For this purpose, we shall use the integral equation
g(z,t)=Q
∂
t
0
q[g(z,t−u)]e
−Qu
du+ze
−Qt
, (1.33)
which is equivalent with the differential equation (1.29) and the integral equation (1.7).

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12 Imre Pázsit & Lénárd Pál
Theorem 5.The integral equation(1.33)has a single solution g(z,t)which satisfies the inequality|g(z,t)|≤1 in
every point|z|≤1and the limit relation(1.32)if and only if
q
1=

dq(z)
dz

z=1
=
∞
≥
n=0
nfn<+∞. (1.34)
Proof.For the proof, suppose the opposite of the statement. In the first step, assume that (1.33) has two solutions
in the interval [0,t
0]. Let these beg 1(z,t) andg 2(z,t). It will be shown thatg 1(z,t) andg 2(z,t) cannot be
different in the interval [0,t
0), i.e.
g
1(z,t)=g 2(z,t),∀t∈[0,t 0].
To prove this, one has to make use of the property of the generating function that if|u|≤1 and|v|≤1 then
8
|q(u)−q(v)|≤q 1|u−v|. (1.35)
Hence, one has
|g
1(z,t)−g 2(z,t)|≤Qq 1
≤
t
0
|g1(z,t−u)−g 2(z,t−u)|e
−Qu
du.
Define the function
K(t
⎡
,t)=sup
t
⎡
≤u≤t
|z|≤1
|g1(z,u)−g 2(z,u)| (1.36)
by the use of which one obtains that
|g
1(z,t)−g 2(z,t)|≤q 1(1−e
−Qt
)K(0,t). (1.37)
Select now a value oft
0>0 such that the inequality
0<q
1(1−e
−Qt0
)<1 (1.38)
is fulfilled. It follows from (1.37) that
sup
0≤t≤t 0
|z|≤1
|g1(z,t)−g 2(z,t)|≤q 1(1−e
−Qt0
)K(0,t 0),
i.e. one can write that
K(0,t
0)≤q 1(1−e
−Qt0
)K(0,t 0) (1.39)
which, by virtue of the inequality (1.38), can only be fulfilled ifK(0,t
0)=0. This means that in every point of
the interval [0,t
0) one has
g
1(z,t)=g 2(z,t),∀|z|≤1. (1.40)
In the next step, it will be shown that the equality (1.40) is valid also in the interval [t
0,2t0]. Since (1.33) has
only one solution in the interval [0,t
0], therefore for everytwhich lies in the intervalt 0<t<2t 0, the equation
g
1(z,t)−g 2(z,t)=Q
≤
t
0
{q[g1(z,t−u)]−q[g 2(z,t−u)]}e
−Qu
du
holds. Based on this, one can write that
|g
1(z,t)−g 2(z,t)|≤Qq 1
≤
t
t
0
|g1(z,t−u)−g 2(z,t−u)|e
−Qu
du.
8
The proof of the inequality can be found in Section A.3.

Ch01-I045064.tex 27/6/2007 9: 23 Page 13
BasicNotions 13
By applying the previous procedure, one arrives at the inequality
K(t
0,2t0)≤q 1(1−e
−Qt0
)K(t0,2t0), (1.41)
in which
K(t
0,2t0)=sup
t0≤t≤2t 0
|z|≤1
|g1(z,t)−g 2(z,t)|.
Since 0<q
1(1−e
−Qt0
)<1, the relation (1.41) can only be valid if and only if the equalityK(t 0,2t0)=0, i.e.
g
1(z,t)=g 2(z,t),∀|z|≤1 (1.42)
is fulfilled in every pointtof the interval [t
0,2t0).
By continuing this procedure, it is seen that the equality (1.42) must be valid in every pointtof the interval
[0,+∞]. On the other hand, from this it follows that (1.33) has one and only one solution for every|z|≤1
in the interval 0≤t<∞.Ifz =1 then (1.33) can be written in the following form:
g(1,t)=Q
≤
t
0
q[g(1,t−u)]e
−Qu
du+e
−Qt
,0≤t<∞.
Becauseg(1,t)=1 is a solution of this equation,it is obvious that this is the only solution in the pointz=1.
Based on the foregoing, the branching processn(t)iscalled regularif
lim
z↑1
g(z,t)=g(1,t)=
∞
≥
n=0
P{n(t)=n|n(0)=1}=
∞
≥
n=0
pn(t|1)=1, (1.43)
and the condition of this is that the inequalityq
1<∞should be fulfilled. In the case whenq 1=∞, then
lim
z↑1
g(z,t)=g(1,t)=
∞
≥
n=0
P{n(t)=n|n(0)=1}=
∞
≥
n=0
pn(t|1)<1. (1.44)
This process is calledirregularor in other wordsexplosive. The notation explosive is motivated by the fact that
in this case
P{n(t)=∞|n(0)=1}=1 −g(1,t)>0,
i.e. an infinite number of progeny can be generated during a finite time interval with non-zero probability.
(In the case of a regular processP{n(t)=∞|n(0)=1}=0.) Naturally, this can only happen if there is a non-
zero probability that an infinite number of particles can be generated in a single multiplication reaction. In
reality, of course, such a process can hardly exist.
It can be shown that instead of the regularity conditionq
1<∞of the branching processes, a more general
condition can also be formulated. Define the integral
C
→=
≤
1
1−→
du
u−q(u)
, (1.45)
in which→>0.
Theorem 6.The branching processn(t)corresponding to the generating function q(z)is regular if C
→=∞, and explosive
if C
→<∞.

Ch01-I045064.tex 27/6/2007 9: 23 Page 14
14 Imre Pázsit & Lénárd Pál
Proof.Starting from (1.29) one has
Qt=
≤
z
g(z,t)
du
u−q(u)
, (1.46)
in whichzis now a real number in the interval [0, 1]. From this equation, it is seen that the right-hand side
has to be bounded for every finitet. Define the function
R(z)=
≤
z
0
du
u−q(u)
,
and letz
0be the smallest positive number for which
z
0−q(z 0)=0.
Ifz
0<z≤1 thenz−q(z)≥0, i.e.R(z) is a non-decreasing function ofzon the interval (z 0, 1]. From (1.46)
it follows that
Qt=R(z)−R[g(z,t)],∀z∈(z
0, 1],
i.e. the inequality
R(1)−R[g(t, 1)]<∞,∀t<∞ (1.47)
has to hold for every finitetat the pointz=1 as well. Since ifC
→=∞, then
lim
z↑1
R(z)=R(1)=∞,
and (1.47) can only be fulfilled if
lim
z↑1
g(z,t)=g(1,t)=1,
i.e. when the branching process is regular. If, however,C
→<∞then
lim
z↑1
R(z)=R(1)<∞,
and in this case the inequality
0<R(1)−R[g(1,t)],∀0<t<∞ (1.48)
can only be fulfilled if
lim
z↑1
g(z,t)=g(1,t)<1,
i.e. when the branching process is explosive.
The simple condition of regularityq 1<∞arises immediately from the more general condition ofC →=∞.
Namely, if
q(u)=1+q
1(u−1)+o(u−1),
then
C
→=
≤
1
1−→
du
(q1−1)(1−u)+o(1−u)
,
and from this it can be seen thatC
→is diverging ifq 1<∞.
In the case, however, when
q(u)=u−A(1−u)
α
+o[(1−u)
α
],
where 0<α<1 and 0 <A<∞thenC
→is finite, i.e.g(t,1)<1 and
lim
u↑1
q
⎡
(u)=q 1=1+αAlim
u↑1
(1−u)
−(1−α)
=∞,
which is simply the condition of explosiveness.

Ch01-I045064.tex 27/6/2007 9: 23 Page 15
BasicNotions 15
1.3.2 Moments: subcritical, criticaland supercritical systems
Start with a recollection of some well-known definitions and relations. The expectation
E{n(t)
k
}=
∞
≥
n=0
n
k
pn(t)=m
(k)
(t),∀t∈[0,∞) (1.49)
is called thekth ordinary momentof the branching processn(t), whereas the expectation
E{n(t)[n(t)−1]...[n(t)−k+1]}=
∞
≥
n=k
n(n−1)···(n−k+1)p n(t)=m k(t),∀t∈[0,∞) (1.50)
is called thekthfactorial moment. The ordinary moments can be expressed by the factorial moments, and vice
versa, factorial moments by the ordinary moments as [17]
m
(k)
(t)=
k
≥
j=1
S(k,j)m j(t) andm k(t)=
k
≥
j=1
∫(k,j)m
(j)
(t),
where∫(k,j) denotes the Stirling numbers of the first kind, whereasS(k,j) that of the second kind.
Since it may be the case thatg(z,t) is not determined for the valuesz>1, thekth factorial moment is
given by the limit
E{n(t)[n(t)−1]···[n(t)−k+1]}=m
k(t)=lim
z↑1
∂
k
g(z,t)
∂z
k
(1.51)
for every positive real numberk. According to the Abelian theorem concerning the summation of series [18],
the expression
∂
k
g(z,t)
∂z
k
=
∞
≥
n=k
n(n−1)···(n−k+1)z
n−k
pn(t)
converges to
E{n(t)[n(t)−1]···[n(t)−k+1]}=
∞
≥
n=k
n(n−1)···(n−k+1)p n(t)
for every fixedt∈[0,∞)ifz ↑1.
It the following, we will mostly need the first three ordinary or factorial moments. ThevarianceD
2
{n(t)}
of the processn(t) can be easily calculated from the factorial moments, namely
D
2
{n(t)}=m 2(t)+m 1(t)−[m 1(t)]
2
. (1.52)
For calculating the first and second factorial moments ofn(t), we will need the first and second factorial
moments of the random variableν. From thebasic generating function q(z) defined in (1.6), the first factorial
moment is given by

dq(z)
dz

z=1
=E{ν}=q
⎡
(1)=q 1, (1.53)
whereas the second is given by

d
2
q(z)
dz
2

z=1
=E{ν(ν−1)}=q
⎡⎡
(1)=q 2. (1.54)
In the following, the notationsq
1andq 2will be used.

Ch01-I045064.tex 27/6/2007 9: 23 Page 16
16 Imre Pázsit & Lénárd Pál
The first and second factorial moments ofn(t) can be determined from the integral equation (1.33). By
differentiating with respect tozonce and twice, as well as substitutingz=1, one obtains
m
1(t)=Qq 1
≤
t
0
e
−Q(t−u)
m1(u)du+e
−Qt
, (1.55)
and
m
2(t)=Qq 1
≤
t
0
e
−Q(t−u)
m2(u)du+Qq 2
≤
t
0
e
−Q(t−u)
[m1(u)]
2
du, (1.56)
respectively. Introduce the Laplace transforms
˜m
k(s)=
≤
∞
0
e
−st
mk(t)dt,k=1, 2.
It follows from (1.55) that
˜m
1(s)=
1
s+Q(1−q 1)
,
i.e.
m
1(t)=e
αt
, (1.57)
where
α=Q(q
1−1)=−a (1.58)
is thefundamental exponentcharacterising the multiplying medium. Ifα<0 then the expectation of the number
of particles generated by one particle decreases exponentially withtfrom unity to zero. Ifα=0 then the
expectation of the particle number remains unity for everyt. Finally, ifα>0 then the expectation grows
exponentially to infinity.
The solution of (1.56) can also be obtained by Laplace transform methods. For the Laplace transform one
obtains
˜m
2(s)=
Qq
2
s+a
V(s),
where
V(s)=
≤
∞
0
e
−st
[m1(t)]
2
dt=
≤
∞
0
e
−st
e
−2at
dt=
1
s+2a
,
i.e.
˜m
2(s)=
Qq
2
a

1
s+a
−
1
s+2a

,ifa=0,
and
˜m
2(s)=
Qq
2
s
2
,ifa=0.
For the second moment one obtains the expression
m
2(t)=
⎧
⎨
⎩
Q
q
2
α
e
αt
(e
αt
−1), ifα=0,
Qq
2t,i fα=0.
(1.59)
By using (1.52), the variance of the processn(t)isgivenas
D
2
{n(t)}=
⎧
⎨
⎩

Q
q
2
α
−1

e
αt

e
αt
−1

,ifα=0,
Qq
2t,i fα=0.
(1.60)

Ch01-I045064.tex 27/6/2007 9: 23 Page 17
BasicNotions 17
Definition 2.According to the formulae (1.57) and (1.60), the branching processes can be divided into three categories
depending on in which media they occur: forα<0, the process is called subcritical; ifα=0and Qq
2>0, the process is
critical; and finally ifα>0, the process is supercritical.
It is remarkable that the characteristics of the medium, and hence also that of the process, are exclusively
determined by the quantitiesq
1andq 2. From the definition it follows that ifq 1<1 then the medium is
subcritical; ifq
1=1 andq 2>0 then it is critical; and finally, ifq 1>1 then it is supercritical.
It is also worth remarking that, according to (1.57) and (1.60), in a critical system, while the expectation
of the particle number is constant, the variance grows linearly in time, and diverges asymptotically. The
implications of this fact for the operation of nuclear reactors in the critical state are sometimes discussed in the
literature [11]. We will return to this question in connection with the extinction probability.
1.3.3 Semi-invariants
In many cases, in addition to the ordinary and factorial moments of then(t), knowledge of itssemi-invariants
κ
n(t)=lim
z→0
∂
n
loggexp(z,t)
∂z
n
,n=1, 2,... (1.61)
is also needed. An important property of the semi-invariants is expressed by the theorem below.
Theorem 7.The semi-invariants of the branching processn(t)satisfy the linear differential equation system
dκ
n(t)
dt
=
n
τ
j=1

n
j−1

R
n−j+1κj(t),n=1, 2,... (1.62)
with the initial conditions
κ
n(0)=

1,if n=1,
0,if n>1.
The coefficients R
jin the equation system are given by the formula
R
j=QE{(ν−1)
j
}=Q
∞
τ
k=1
(k−1)
j
fk. (1.63)
Proof.Introduce the logarithmic generating function
φ(z,t)=logg
exp(z,t).
From equation (1.11), it follows that at every instanttand pointzwhereg
exp(z,t)=0, the equation
∂φ(z,t)
∂t
=Q[q(e
z
)e
−z
−1]
∂φ(z,t)
∂z
(1.64)
holds with the initial conditionφ(0,z)=z. Let
Q[q(e
z
)e
−z
−1]=R(z),

Ch01-I045064.tex 27/6/2007 9: 23 Page 18
18 Imre Pázsit & Lénárd Pál
and notice that
R(z)=Q

∞
≥
k=0
fke
(k−1)z
−1

=Q
⎡
⎣
∞
≥
k=0
fk
∞
≥
j=0
(k−1)
j
z
j
j!
−1
⎤ ⎦
=Q
∞
≥
j=1
E{(ν−1)
j
}
z
j
j!
=
∞
≥
j=1
Rj
z
j
j!
,
whereR
jis identical with (1.63). If the semi-invariantsκ k(t),k=1, 2,...exist then one can write that
φ(z,t)=
∞
≥
k=1
κk(t)
z
k
k!
. (1.65)
Substitute now the power series ofφ(z,t) andR(z) with respect tozinto equation (1.64). One obtains that
∞
≥
n=1
dκn(t)
dt
z
n
n!
=
∞
≥
j=1
Rj
z
j
j!
∞
≥
k=1
κk(t)
z
k−1
(k−1)!
.
The coefficient ofz
n
on the right-hand side is equal to
R
nκ1(t)
1
n!0!
+R
n−1κ2(t)
1
(n−1)!1!
+···
R
n−i+1κi(t)
1
(n−i+1)!(i−1)!
+···+R
1κn(t)
1
(n−1)!1!
=
1
n!
n
≥
i=1

n
n−i+1

R
n−i+1κi(t).
Hence
dκ
n(t)
dt
=
n
≥
i=1

n
n−i+1

R
n−i+1κi(t),
and this is identical with equation (1.62).
Determine now the semi-invariantsκ 1(t) andκ 2(t). First of all, one notices that
κ
1(t)=E{n(t)}=m 1(t)
and
κ
2(t)=D
2
{n(t)}=m 2(t)+m 1(t)[1−m 1(t)].
Based on (1.62), one obtains
dκ
1(t)
dt
=R
1κ1and
dκ
2(t)
dt
=2R
1κ2+R2κ1(t).
The initial conditions areκ
1(0)=1 andκ 2(0)=0. From (1.63) one has
R
1=QE{ν−1}=Q (q 1−1)=α,
R
2=QE{(ν−1)
2
}=Q [E{ν(ν−1)}−E{ν −1}]=α

Qq
2
α
−1

,
and accordingly,
κ
1(t)=e
R1t
=e
αt
.

Ch01-I045064.tex 27/6/2007 9: 23 Page 19
BasicNotions 19
Further,
κ
2(t)=
⎧
⎨
⎩
[Q
q
2α
−1]e
αt
(e
αt
−1), ifα=0,
Qq
2t,i fα=0.
Performing the Laplace transform of equation (1.62), and using the notation
˜κ
n(s)=
∂
∞
0
e
−st
κn(t)dt=˜κ n,
one can write that
(s−nR
1)˜κn=
n
τ
k=1

n
k

R
k˜κn−k+1 +δn1, (1.66)
i.e.
(s−R
1)˜κ1=1
−R
2˜κ1+(s−R 2)˜κ2=0
−R
3˜κ1−

3
2

R 2˜κ2+(s−3R 1)˜κ3=0
.
.
.
−R
n˜κ1−

n
n−1

R n−1˜κ2+···+(s−nR 1)˜κn=0.
From this, the following solution is obtained:
˜κ
n=
1
Dn

s−R
1 00 ···1
−R
2 s−2R 2 0 ···0
.
.
.
.
.
.
−R
n−

n
n−1

R
n−1−

n
n−2

R
n−2···0

in which
D
n=
n
→
k=1
(s−kR 1).
As an illustration, the Laplace-transforms of the first three semi-invariants are given as follows:
˜κ
1(s)=
1
s−R 1
,
˜κ
2(s)=
1
(s−R 1)(s−2R 1)

s−R
11
−R
20

=
R
2
(s−R 1)(s−R 2)
,
and
˜κ
3(s)=
1
(s−R 1)(s−2R 1)(s−3R 1)

s−R
1 01
−R
2s−2R 10
−R
3−3R20

=
3R
2
2
+R3(s−2R 1)
(s−R 1)(s−2R 1)(s−3R 1)
.

Ch01-I045064.tex 27/6/2007 9: 23 Page 20
20 Imre Pázsit & Lénárd Pál
Investigate now the dependence of thenth semi-invariant ontin the case when the medium is critical,
i.e. whenR
1=α=0 andR 2=Qq2>0. It is easy to confirm that in this case whenn=1, 2 then
˜κ
1(s)=
1
s
and˜κ
2(s)=R 2
1
s
2
, (1.67)
whereas ifn>2 then
˜κ
n(s)=
n
→
k=3

k
2

(Qq
2)
n−1
1
s
n

1+˜
n−2(s)

, (1.68)
where the function˜
k(s)isakth order polynomial ofs. From this, it obviously follows that in the critical state,
ifn=1, 2 then
κ
1(t)=1 andκ 2(t)=Qq 2t,
whereas ifn>2 then
κ
n(t)=
n
→
k=3

k
2

(Qq
2)
n−1
t
n−1
(n−1)!
[1+
n−2(1/t)],
where
k(1/t)isakth order polynomial of 1/t . It is worth noting that thet-dependence of the semi-invariants
of the branching processn(t) in the critical state is dominantly determined by the second factorial momentq
2.
1.4 Discrete Time Branching Processes
It is well-known that F. Galton and H.W.Watson were the first to deal with branching processes in the 1870s,
in order to determine the probability of the extinction of families. The number of articles and monographs
on the discrete time branching processes named after them is exceedingly large. An excellent survey on the
Galton–Watson processes is given in the by now classic monograph by T.E. Harris [6]. In this book, however,
discrete time branching processes are not dealt with. Only some elementary questions are discussed here that
are necessary, among others, for the modelling of branching processes.
Divide the interval [0,t) intoTequal and mutually non-overlapping subintervals∞t. The concept of the
reaction will be defined as before with the associated number distribution, such thatf
k,k=0, 1, 2,...is the
probability thatkparticles are born in a reaction. Obviously,f
0is the probability of the absorption,f 1is that of
renewal, andf
k,k=2, 3,...is that of the actual multiplication. Suppose that in every subinterval∞tat most
one reaction can occur. Moreover, letWbe the probability of the occurrence of the reaction, while 1−Wis
the probability of the non-occurrence of the reaction.
Letn(j),j=0, 1,...,Tdenote the number of particles in the multiplying medium at thejth discrete time
point, i.e. in the subinterval [(j−1)∞t,j∞t]. Determine the probability
P{n(j)=n|n(0)=1}=p
n(j),j=0, 1,...,T (1.69)
of the event that exactlynparticles are present in the multiplying system at thejth discrete time instant,
provided that there was just one particle present at the 0th time instant. Then from obvious considerations one
can write down the backward equation forj>1as
p
n(j)=(1−W)p n(j−1)
+Wf
0δn0+W
∞
≥
k=1
fk
≥
n1+···+n k=n
k
→
i=1
pni
(j−1). (1.70)
By introducing the generating functions
g(z,j)=
∞
≥
n=0
pn(j)z
n
(1.71)

Ch01-I045064.tex 27/6/2007 9: 23 Page 21
BasicNotions 21
and
q(z)=
∞
≥
k=0
fkz
k
, (1.72)
after some elementary considerations one obtains the equation
g(z,j)=(1−W)g(z,j−1)+Wq[g(z,j−1)], (1.73)
and sincep
n(0)=δ n1, one has
g(z,0)=z.
Further, obviously if
∞
≥
n=0
pn(j)=1 and
∞
≥
k=0
fk=1, theng(1,j)=1.
With adequate rigour, based on the fundamental relation (1.23), one can discuss thediscrete time homogeneous
branching processes, the so-called Galton–Watson processes. Let us callone generation of particlesthe particles that
are born from one particle under unit time. Denote their number withn(1) and introduce the notations
P{n(1)=n|n(0)}=p
n(1)=p n,
and
g(z,1)=
∞
≥
n=0
pnz
n
=g(z),
respectively. From (1.23), by selectingt=j−1 andu=1, one obtains
g(z,j)=g[g(z, 1),j−1]=g[g(z),j−1]. (1.74)
One notes that
g(z,0)=z,g(z,1)=g(z),g(z,2)=g(g(z)),...,
i.e.g(z,j) is equal to thejthiterationofg(z,1)=g(z). Accordingly,
g(z,j)=g(g(...g(z)...))=g
j(z), (1.75)
whereg
j(z) denotes thejth iteration ofg(z).
Equation (1.73) can also be obtained by an iteration process starting fromj=0. Then
p
n(1)=(1−W)δ n1+W
∞
≥
k=0
fkδnk,
and hence
g(z,1)=g(z)=(1−W)z+Wq[z].
By continuing, one can write that
g(z,2)=g(g(z, 1), 1)=(1−W)g(z,1)+Wq[g(z, 1)],
g(z,3)=g(g(z, 1), 2)=g(g(g(z, 1), 1), 1)=g(g(z, 2), 1)
=(1−W)g(z,2)+Wq[g(z, 2)],
and so on. This leads to the conclusion that
g(z,j)=(1−W)g(z,j−1)+Wq[g(z,j−1)],

Ch01-I045064.tex 27/6/2007 9: 23 Page 22
22 Imre Pázsit & Lénárd Pál
and this is the same as equation (1.73). The expectationm 1(j)=E{n(j)|n(0)=1}can be calculated from (1.73)
by the relation

dg(z,j)
dz

z=1
=m1(j).
One obtains that
m
1(j)=[1−W(1−q 1)]m1(j−1),
and sincem
1(0)=1,
m
1(j)=[1−W(1−q 1)]
j
, (1.76)
whereq
1=E{ν}.
9
It is seen that for fixed∞t
lim
j→∞
m1(j)=
⎧
⎨
⎩
0, ifq
1<1,
1, ifq
1=1,
∞,ifq
1>1.
Accordingly, one can state that the discrete time processξ(j),j=0, 1,...is subcritical ifq
1<1, critical ifq 1=1
and supercritical ifq
1>1.
The second factorial moment
m
2(j)=E{n(j)[n(j)−1]|n(0)=1}
can be calculated by using the relation

d
2
g(z,j)
dz
2

z=1
=m2(j).
From (1.73) one obtains
m
2(j)=[1−W(1−q 1)]m2(j−1)+Wq 2[m1(j−1)]
2
.
By introducing the notations
1−W(1−q
1)=aandWq 2=b,
and by taking into consideration (1.76), from the previous equation the recursive expression
m
2(j)=am 2(j−1)+ba
2(j−1)
is obtained, which by using the generating function
γ(s)=
∞
≥
j=0
m2(j)s
j
is simplified to
γ(s)=asγ(s)+bs
1
1−a
2
s
,a
2
s<1.
9
Letj∞t=tand introduce the notationWj=Q jt, whereW=W(∞t). If, for a fixedt
lim
j→∞
Qj=lim
∞t→0
W
∞t
=Q,
then
lim
∞t→0
m1(j)=lim
j→∞

1−
Q
j(1−q 1)t
j

j
=e
−Q(1−q 1)t
,
and this agrees exactly with the expectation (1.57) of the continuous time parameter process.

Ch01-I045064.tex 27/6/2007 9: 23 Page 23
BasicNotions 23
After some elementary steps, one obtains the formula
γ(s)=b
s
1−a

1
1−as
−a
1
1−a
2
s

.
From this it immediately follows that
m
2(j)=ba
j
1−a
j
1−a
=Wq
2m1(j)
1−m
1(j)
1−m 1(1)
,ifq
1=1. (1.77)
Ifq
1→1 thenm 1(j)→1, thus
lim
q1→1
1−m 1(j)
1−m 1(1)
=limq1→1
1−[1−W(1−q 1)]
j
1−[1−W(1−q 1)]
=j,
hence in a critical system
m
2(j)=Wq 2j. (1.78)
For the variance of the processn(j), one obtains
D
2
{n(j)|n(0)=1}=
⎧
⎨
⎩

1+
q
2
1−q 1

m
1(j)[1−m 1(j)], ifq 1=1,
Wq
2j,i fq 1=1.
(1.79)
1.5 Random Tree as a Branching Process
It has long been well known that to every Galton–Watson process one can order a graph called atree
which displays thedevelopment of the populationformed by the successors of a single particle entering the
multiplying medium att=0 by mutually independent reactions at discrete time instantst=1, 2,.... A wealth
of outstanding work has been published on these Galton–Watson trees. These will, however, not be discussed
here, rather the interested reader is referred to the excellent publication of Bollobás [19].
For the illustration of branching processes,however,a brief discussion will be given of the process describing
the development of the random trees in the case whenthe period between the consecutive branching points is
a continuous random variable with exponential distribution.Suppose that the tree consists ofactiveandinactive
branching points, callednodes.
10
The nodes are connected with branches. The development of the tree can be
described as follows: at the momentt=0, the tree consists of only one node (root) which becomes inactive
by the random timeτand creates new active nodesk=0, 1, 2,...with probabilitiesf
kat the ends of branches
of the same length. The same happens with these mutually independent nodes as with the root node and this
continues as long as an active node is generated.
The first step of the random development of the tree is illustrated in Fig. 1.1, while a possible form of the
development until a given timetis shown in Fig. 1.2. The horizontal dotted lines denote the random time
instants where the new nodes appear after being created by the node that became inactive. It is obvious that
the active nodes are always the end points of the tree.
This random tree corresponds to a branching process in which the number of particles attis equal to the
number of active nodes of the tree, whereas the number of the particles absorbed untilt, equals the number of
the inactive nodes. Denote the number of active nodes withn
a(t) and that of the inactive nodes withn i(t)at
t. Determine now the generating function
g
(a,i)
(za,zi,t)=
∞
τ
na=0
∞
τ
ni=0
p
(a,i)
(na,ni,t|1, 0)z
na
a
z
ni
i
(1.80)
10
Results of a more detailed investigation of the problem can be found in the works [20–23].

Ch01-I045064.tex 27/6/2007 9: 23 Page 24
24 Imre Pázsit & Lénárd Pál
Root f
0 f
1 f
2 f
3
...
...
Figure 1.1The first step of the random development of the tree. The active root node becomes inactive and
creates 0, 1, 2, 3,...new active nodes with probabilitiesf
0,f1,f2,f3,.... Active nodes are marked by light circles,
while inactive nodes by dark ones.
Root
Living tree
Figure 1.2A possible realisation of the tree development. The active nodes capable for further development are
marked by light circles. The inactive nodes that do not take part in the development any longer are marked by
dark circles. The horizontal dotted lines denote the random time instants where the new nodes appear after being
created by the node that became inactive (i.e. the branching of the previously active node).
of the probability
P{n
a(t)=n a,ni(t)=n i|na(0)=1,n i(0)=0}=p
(a,i)
(na,ni,t|1, 0).
By considering the mutually exclusive two first events that can occur after the momentt=0, one arrives at
∂g
(a,i)
(za,zi,t)
∂t
=−Qg
(a,i)
(za,zi,t)+Qz iq[g
(a,i)
(za,zi,t)] (1.81)
with the initial conditiong
(a,i)
(za,zi,0)=z a. It is immediately seen that the generating function
g
(a,i)
(za=z,z i=1,t)=g
(a)
(z,t)
satisfies equation (1.5), i.e.g
(a)
(z,t)=g(z,t). From (1.81), one can immediately write down the probabilities
and variances of the numbers of the active and inactive nodes. Without going into details, using the previous

Ch01-I045064.tex 27/6/2007 9: 23 Page 25
BasicNotions 25
notations, one obtains
m
(a)
1
(t)=e
αt
andm
(i)
1
(t)=
Q
α
(e
αt
−1),
ifq
1=1. Whenq 1=1 then
m
(a)
1
(t)=1 andm
(i)
1
(t)=Qt.
The variances are given by the expressions
D
2
{na(t)}=
Q
α
E{(ν−1)
2
}(e
αt
−1)e
αt
and
D
2
{ni(t)}=

Qα

3
[D
2
{ν}+E{( ν−1)
2
}e
αt
](e
αt
−1)−2

Q
α

2
D
2
{ν}Qte
αt
ifq1=1. Ifq 1=1, then
D
2
{na(t)}=q 2QtandD
2
{ni(t)}=Qt +
1
3
q
2(Qt)
3
.
The variance of the number of inactive nodes for the caseq
1<1 converges to the limitD
2
{ν}/(1−q 1)
3
if
t→∞.This means that in the case of branching processes in a subcritical medium, the variance of the number
of absorbed particles converges to a finite value if the duration of the process converges to infinity.
It is worth calculating also the covariance functions ofn
a(t) andn i(t),
Cov{n
a(t)ni(t)}=E{n a(t)ni(t)}−E{n a(t)}E{n i(t)}.
After some elementary operations, forq
1=1 one obtains
Cov{n
a(t)ni(t)}=

1+
D
2
{ν}
(q1−1)
2

e
αt
(e
αt
−1)−
D
2
{ν}
q1−1
Qte
αt
, (1.82)
and forq
1=1
Cov{n
a(t)ni(t)}=
1
2
q
2(Qt)
2
. (1.83)
For the case ofq
1=1, in a critical medium, the correlation function shows a peculiar behaviour. One obtains
Cov{n
a(t)ni(t)}
D{na(t)}D{n i(t)}
=
√
3
2

1+
3
q2(Qt)
2

−1/2
, (1.84)
and from this it follows that in a critical medium, a sufficient long time after the start of the process, the correl-
ation between the numbers of the active and non-active particles is
√
3/2, i.e. it is constant, independently
from any parameter influencing the process. More detailed calculations can be found in [20–23].
The investigation of the random trees of continuous time parameter has enriched the theory of branching
processes with many valuable results; however, their full description lies somewhat outside the basic subject of
this monograph.
1.6 Illustrative Examples
In the following, both in the illustrative examples and for the exactly solvable problems, it is assumed that
the generating functionq(z) of the medium is known. The use of the quadratic expression
q(z)=f
0+f1z+f2z
2
=1+q 1(z−1)+
1
2
q
2(z−1)
2
,f0+f1+f2=1 (1.85)

Ch01-I045064.tex 27/6/2007 9: 23 Page 26
26 Imre Pázsit & Lénárd Pál
0.5 1 1.5 2 2.5
0.5
1
1.5
2
2.5
Permitted region
q
1
q
2
Figure 1.3The permitted values ofq 1andq 2if the generating functionq(z) of the random variableνis a quadratic
function ofz.
is advantageous because it represents the effects of the generating functionq(z), which is unknown but possesses
three finite first factorial moments, sufficiently well. Besides, physically it describes a process in which at most
two particles can be born in a reaction (collision), which is a good model of atomic collision cascades with
recoil production. From (1.85) it follows that theallowable values of q
1and q2are contained in a single, specific
domain of the plane (q
1,q2).
This domain is illustrated in Fig. 1.3. In the forthcoming, the generating function (1.85) will be called a
quadratic generating function. The probabilities f
i,i=0, 1, 2 can be expressed by the momentsq 1andq 2as
f
0=1−q 1+
1
2
q
2,
f
1=q1−q2,
f
2=
1
2
q
2.
The roots of the equation
q(z)−z=f
2z
2
−(1−f 1)z+f 0=0.
will also be needed. These are obtained as
z
i=
⎧
⎪
⎨
⎪
⎩
1, ifi=1,
1+
2(1−q
1)
q2
,ifi=2.
(1.86)
1.6.1 Regular processes
Yule–Furry process
One of the simplest regular processes is the Yule–Furry process, well-known from the literature. In this case
f
0=f1=0 andf 2=1, i.e.q(z)=z
2
, and henceq 1=q2=2. In other words, in this specific medium the particle
is doubled in every reaction. Let us write down both the backward equation
∂g(z,t)
∂t
=−Qg(z,t)[1−g(z,t)] (1.87)
and the forward equation
∂g(z,t)
∂t
=Qz(z−1)
∂g(z,t)
∂z
(1.88)

Ch01-I045064.tex 27/6/2007 9: 23 Page 27
BasicNotions 27
for the generating function. We require also the conditions
g(z,0)=zandg(1,t)=1. (1.89)
As mentioned earlier, it follows from the theory of differential equations that the solution of these two equations
is the same function
g(z,t)=
ze
−Qt
1−z(1−e
−Qt
)
. (1.90)
However, for better insight, this result will be derived explicitly.
Backward equationOne notes that
dg(z,t)
g(z,t)[g(z,t)−1]
=Qdt,
from which the equality
Ce
−Qt
=
g(z,t)
g(z,t)−1
follows. By taking into account the initial conditiong(0,z)=z, one obtains
C=
z
z−1
,
and from this, with some basic algebra, one arrives at (1.90) immediately. Forward equationThe characteristic equation of the homogeneous, linear, first order partial differential
equation (1.88) is:
Qdt+
dz
z(z−1)
=0
whose integral
ψ(z,t)=Qt+log

1−
1
z

is at the same time thebasic integralof the partial differential equation. It is known that any continuous and
differentiable functionH(u), in whichu=ψ(z,t), is also an integral of (1.88), i.e.
g(z,t)=H

Qt+log

1−
1
z

.
Accounting for the initial conditiong(0,z)=z, the functional equation
H

log

1−
1
z

=z
is obtained. By introducing the notationv=log

1−
1
z

, it is seen that
H(v)=
1
1−e
v
.
Hence
g(z,t)=H[ψ(z,t)]=
1
1−e
ψ(z,t)
=
1
1−(1−
1
z
)e
Qt
.
By multiplying the numerator and denominator withze
−Qt
, the solution (1.90) is obtained.

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28 Imre Pázsit & Lénárd Pál
Quadratic process
In the following, theregular processwill be dealt with, defined by the quadratic generating function which will
be called a quadratic process. By taking (1.85) into account, the backward equation (1.29) takes the following
form:
Qdt=
dg
f0+(f1−1)g+f 2g
2
. (1.91)
The roots of the denominator
g
1,2=
1−f
1±

(1−f 1)
2
−4f0f2
2f2
on the right-hand side can be expressed by the quantitiesq 1=f1+2f2andq 2=2f2. One finds that
g
k=

1, ifk=1,
1+2(1−q
1)/q2,ifk=2.
(1.92)
By utilising this, one arrives at
Q(1−q
1)dt=

1
g−1−2(1−q 1)/q2
−
1
g−1

dg,
and further to
g(z,t)=1−
2C(z)(1−q
1)/q2
e
Q(1−q 1)t
−C(z)
.
HereC(z) is the integration constant which, from the initial conditiong(z,0)=zcan be determined from
the equation
z=1−
2C(z)(1−q
1)/q2
1−C(z)
.
This yields
C(z)=
1−z
1−z+2(1−q 1)/q2
,
and hence finally one obtains
g(z,t)=1−
1−z
e
Q(1−q 1)t
+(1−z)
q2
2(1−q 1)
(e
Q(1−q 1)t
−1)
, (1.93)
ifq
1=1. For the case whenq 1=1,i.e. if the medium is critical,from equation (1.93) by applying the L’Hospital
rule one obtains
g(z,t)=1−
1−z
1+(1−z)q 2Qt/2
. (1.94)
It is worth to demonstrate another procedure for the determination of the generating functiong(z,t). By using
the expression of the generating functionq(z)
q(z)=1+q
1(z−1)+
1
2
q
2(z−1)
2
,
one obtains from (1.29)
dg
dt
=−Q(1−q
1)(g−1)+
1
2
Qq
2(g−1)
2
.

Ch01-I045064.tex 27/6/2007 9: 23 Page 29
BasicNotions 29
By introducing the function
h(z,t)=
1
1−g(z,t)
for which one has
h(z,0)=
1
1−z
,
one can immediately write
dh
dt
=
1
(1−g)
2
dg
dt
=Q(1−q
1)
1
1−g
+
1
2
Qq
2
=Q(1−q 1)h+
1
2
Qq
2
.
The solution forq
1=1 by using the formula forh(0,z) is equal to
h(z,t)=
1
1−z
e
Q(1−q 1)t
+
q
2
2(1−q 1)
(e
Q(1−q 1)t
−1).
From this it follows that
g(z,t)=1−
1−z
e
Q(1−q 1)t
+(1−z)
q2
2(1−q 1)
(e
Q(1−q 1)t
−1)
,
and this agrees with the formula (1.93).
The expression for the caseq
1=1 can be obtained from this by a simple limit procedure. Figure 1.4 shows
the shape of the surface determined by the generating functiong(z,t) in a subcritical medium (q
1=0.95) with
parametersq
2=0.5 andQ=0.4.
11
By expanding the right-hand side of (1.93) and (1.94) into a power series with respect tozand introducing
the notation
U(t)=
1−e
−Q(1−q 1)t
Q(1−q 1)
, (1.95)
0
10
20
30
40
50
t
0
0.25
0.5
0.75
1
z
0.6
0.7
0.8
0.9
1
g (t, z)
0
10
20
30
40
q
2
≥ 0.5
q
1
≥ 0.95
Q ≥ 0.4
Figure 1.4The generating functiong(z,t) in a subcritical medium.
11
Notations of the dimensions of parameters will be omitted both here and in the following. Units for the figures will be chosen such that they
show the essential characteristics of the phenomena.

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30 Imre Pázsit & Lénárd Pál
Table 1.1Values of probabilities
f0 f1 f2 q1 q2
0.30 0.45 0.25 0.95 0.5
0.25 0.50 0.25 1.00 0.5
0.20 0.55 0.25 1.05 0.5
0 10 20 30 40 50
Time (t )
0
0.2
0.4
0.6
0.8
1
Extinction probability

q
1
≥ 0.95
Q ≥ 1
q
1
≥ 1.05q
2
≥ 0.5
q
1
≥ 1

Figure 1.5Dependence of the extinction probabilityp 0(t) on timetin subcritical, critical, and supercritical
systems.
one obtains the probabilities
p
0(t)=
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
1−
e
−Q(1−q 1)t
1+Qq 2U(t)/2
,ifq
1=1,
1−
1
1+Qq 2t/2
,if q
1=1,
(1.96)
and
p
n(t)=
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
e
−Q(1−q 1)t
[Qq2U(t)/2]
n−1
[1+Qq 2U(t)/2]
n+1
,ifq 1=1,
(Qq
2t/2)
n−1
(1+Qq 2t/2)
n+1
,i fq 1=1,
(1.97)
n=1, 2,....
The values of the probabilitiesf
0,f1,f2that were used in calculations are listed in Table 1.1.
The quantityp
0(t) is called theextinction probabilityandR(t)=1−p 0(t), thesurvival probability. The extinc-
tion probability converges to 1 in both subcritical and critical media fort→∞. In a supercritical medium
one obtains that
lim
t→∞
p0(t)=1−2
q
1−1
q2
. (1.98)
According to this, the survival probabilityR(t) converges to a value larger than zero only in a supercritical
medium ift→∞. Figure 1.5 displays the time-dependence of the extinction probability in the case of
subcritical, critical and supercritical processes. In Fig. 1.6, thet-dependence of the probabilitiesp
1(t) andp 2(t)
is shown also for three different processes.
Calculate now the time instantt
maxin a critical system (q 1=1) for which the probabilityp n(t) is a maximum.
From
dp
n(t)
dt
=Q(2/q
2)
2
(Qt)
n−2
(2/q2+Qt)
n+2
[2(n−1)/q 2−2Qt]=0

Ch01-I045064.tex 27/6/2007 9: 23 Page 31
BasicNotions 31
0 10 20 30 40 50
Time (t ) Time (t )
0
0.2
0.4
0.6
0.8
1
Probability ( p
1
(t))
Probability ( p
2
(t))
0 10 20 30 40 50
0
0.05
0.1
0.15

Q ≥ 1
q
2
≥ 0.5
Q ≥ 1
q
2
≥ 0.5

q
1
≥ 0.95
q
1 ≥ 1.05
q
1
≥ 1

q
1
≥ 0.95
q
1 ≥ 1.05
q
1
≥ 1
Figure 1.6Dependence of probabilitiesp 1(t) andp 2(t) on timetin subcritical, critical, and supercritical systems.
it follows that
t
max=
n−1
Qq2
andp n(tmax)=

2
n+1

2
n−1
n+1

n−1
,
wheren=1, 2,....
Remark.It is worthwhile to put into context the fact that the extinction probability converges to 1 in
critical systems, while the expectation remains constant. As it was seen in Section 1.3.2, at the same time, the
variance diverges linearly with time. These facts together are consistent. The time-dependence of the variance
shows that the branching process is not stationary, and hence not ergodic either. This means that the ensemble
average is not equal to the time average, and the realisations of processes in several identical systems have
different asymptotic behaviour. An illustrative explanation is that, if we take an increasing number of systems
and look at the asymptotic properties, in most systems except a small fraction of them, the population will
die out asymptotically, while in the remaining systems it will diverge. Increasing the number of systems, the
number of the systems in which the population diverges remains finite, whereas the number in which the
population dies out, goes to infinity. It is this, asymptotically negligible measure (in terms of the number of
systems considered) of divergence that guarantees that the expectation can remain constant while the extinction
probability converges to unity. Concerning the implications for the operation of for example nuclear reactors,
the above facts have little relevance. Partly,both the divergence of the variance and the certainty of extinction are
only asymptotic properties; and partly, due to the fact that the process is not ergodic, the asymptotic behaviour
of the moments of the population does not say anything about the long-term behaviour of the system. An
illustration of the different behaviour of the different realisations in a critical systems is shown in Fig. 1.9.
1.6.2 Explosive process
As a second example, a simple explosive process is chosen, for which it is known that its generating function
is not a real probability generating function, sinceg(t,1)<1ift>0. Let the function
q(z)=1−
√
1−z
define this simple process for which it is easily seen that
q
1=∞.
Naturally, thisq(z) satisfies also the requirement for the explosive processes, arising from the general condition
in (1.42), according to which the integral
C
→=
≤
1
1−→
dz
z−q(z)
has to be finite for every 0<→<1. One obtains that
C
→=−2log

1−
√
→

<∞,∀0<→<1.

Ch01-I045064.tex 27/6/2007 9: 23 Page 32
32 Imre Pázsit & Lénárd Pál
0
5
10
15t
0
0.25
0.5
0.75
Q τ 0.4
1
z
0
0.25
g(t, z)
0.5
0.75
1
Figure 1.7Generating function of an explosive process.
The generating functiong(z,t) satisfies the following equation:
dg
dt
=Q(1−g)−Q

1−g, (1.99)
from which, by taking into account the initial conditiong(z,0)=z, one can write down the solution in the
form
∂
g(z,t)
z
dx
1−x−
√
1−x
=Qt.
Considering that
∂
dx
1−x−
√
1−x
=−2log(1−
√
1−x)+const.,
one arrives at
g(z,t)=1−[1−e
−Qt/2
(1−
√
1−z)]
2
, (1.100)
and further to
g(1,t)=1−(1−e
−Qt/2
)
2
<1, ift>0.
Figure 1.7 shows the surface defined by the generating function of equation (1.100). For illustration, the
probabilities are listed as
p
0(t)=0,
p
1(t)=e
−Qt/2
,
p
2(t)=
1
4
e
−Qt/2
(1−e
−Qt/2
).
It is obvious that
P{n(t)=∞|n(0)=1}=1 −g(1,t)=(1−e
−Qt/2
)
2
(1.101)
is the probability thatunder a finite period of time t, an infinite number of progeny is generated.
1.6.3 Modelling of branching processes
The modelling is based on equation (1.70). For the sake of simplicity, suppose that a particle generates a reaction
under the time∞twith the probabilityW, in which the particle is either absorbed, renewed or gets converted
into two particles. Let the probabilities of these events bef
0,f1andf2.
The first step is the random generation of the numbers 0, 1, 2 with a frequency corresponding to the
probabilitiesf
0,f1,f2.The next step produces the realisations of the number of particlesn(t) at the discrete time
instantst=1, 2,...,T, supposing that the number of particles wasratt=0. Hence, with the help of a simple

Ch01-I045064.tex 27/6/2007 9: 23 Page 33
BasicNotions 33
0 10 20 30 40 50
Time steps
10
20
30
40
50
60
Number of particles
W τ 0.4, q
2
τ 0.5
q
1
τ 0.95
Figure 1.8Three particular realisations illustrating the evolution of the number of particles in a subcritical
medium.
0 10 20 30 40 50
Time steps
20
40
60
80
100
120
Number of particles
W τ 0.4, q
2
τ 0.5
q
1
τ 1
Figure 1.9Three particular realisations illustrating the evolution of the number of particles in a critical medium.
0 10 20 30 40 50
Time steps
40
60
80
100
120
140
160
Number of particles
W τ 0.4, q
2
τ 0.5
q
1
τ 0.5
Figure 1.10Three realisations illustrating the formation of the number of particles in a supercritical medium.
program, one can describe any possible history of therparticles present att=0 in the multiplying medium,
with fixed probabilitiesW,f
0,f1,f2.
In Fig. 1.8, one can see three realisations of the discrete processn(t) defined by the probabilities
W=0.4,f
0=0.3,f 1=0.45,f 2=0.25, of asubcritical medium, starting from the initial condition n(0)=50,
att=1, 2,..., 50. For the sake of illustration, the discrete points are connected by linear sections.
Figures 1.9 and 1.10 each show three realisations of the discrete processn(t) defined by the probabilities
W=0.4,f
0=0.25,f 1=0.5,f 2=0.25 andW=0.4,f 0=0.2,f 1=0.55,f 2=0.25 of acriticaland asupercritical
medium, respectively. The curves in the figures illustrate well the fact that the characteristic behaviour in the
subcritical,critical and supercritical media develops through significant fluctuations. Moreover,even realisations

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34 Imre Pázsit & Lénárd Pál
01020304050
Time steps Time steps
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Expectation
C
S
W τ 0.4, q
2
τ 0.5
q
1
τ 0.95
N τ 10
4
0 20 40 60 80 100
0
0.5
1
1.5
2
2.5
3
Variance
q
2 τ 0.5
q
1 τ 0.95
W τ 0.4N τ 10
4
C
S
Figure 1.11Estimation of the time-dependence of the expectation and the variance in a subcritical medium.
0 20 40 6080 100
Time steps Time steps
0.94
0.96
0.98
1
1.02
1.04
1.06
1.08
Expectation
W τ 0.4, q
2
τ 0.5
W τ 0.4, q
2
τ 0.5
q
1
τ 1
q
1
τ 1
N τ 10
4
N τ 10
4
0 10 20 30 40 50
0
2
4
6
8
10
Variance
C
S
C
S
Figure 1.12Estimation of the time-dependence of the expectation and the variance in a critical medium.
can occur that do not show the typical behaviour for the given medium at all. It can happen, for example, that
in a supercritical medium, one of the realisations leads to the extinction of the particles for some finitet, i.e.
the staten(t)=0,t>0 is realised.
It is interesting to demonstrate the estimation of the time-dependence of the expectationE{n(t)|n(0)=1}
and the varianceD
2
{n(t)|n(0)=1}by using a relatively not very large number of realisations of the process.The
estimated values are compared to those calculated directly from the formula (1.76), and from (1.79), respectively,
but there is no strive to achieve high precision since the goal is only to illustrate the random behaviour of the
process.
Figure 1.11 shows the time-dependence of the expectation and the variance, respectively, estimated by
N=10 000 realisations of the processn(t) in a subcritical medium, starting fromn(0)=1. The symbol∗
corresponds to the estimated values, while the points of the continuous curves to the values calculated from
the corresponding formula. For a critical process, Fig. 1.12 displays the time-dependence of the expectation
and the variance. One can see that the estimated expectations (S ) hardly show more than 2% fluctuation around
the exact values (C ). The time-dependence of the estimated and the exact variances, on the other hand, are
nearly identical.
As seen in Fig. 1.13, in a supercritical medium, the time-dependence of the estimated expectations and the
variance (S ) agrees outstandingly well with the exact time-dependence (C). We note that in both the critical
and the supercritical cases, altogether onlyN=10 000 realisations were used for the calculations.
The realisations can be used for the calculation of the probabilitiesp
n(t),t=0, 1,.... As an example,
determine the time-dependence of the extinction probabilityp
0(t). In connection with the direct calculation
ofp
0(t), it is worth noting thatp 0(t)=g(t, 0), hence if
q(z)=f
0+f1z+f2z
2
,

Ch01-I045064.tex 27/6/2007 9: 23 Page 35
BasicNotions 35
0 10 20 30 40 50
Time steps Time steps
1
1.25
1.5
1.75
2
2.25
2.5
Expectation
C
S
0 10 20 30 40 50
0
10
20
30
40
Variance
C
S
W ≥ 0.4, q
2
≥ 0.5
W ≥ 0.4
q
1
≥ 1.05 q 1
≥ 1.05
q
2 ≥ 0.5
N ≥ 10
4
N ≥ 10
4
Figure 1.13Estimation of the time-dependence of the expectation and the variance in a supercritical medium.
0 10 20 30 40 50
Time steps
0.2
0.4
0.6
0.8
1
Extinction probability
calc
calc
calc
q
2
≥ 0.5
W ≥ 0.4
N ≥ 10
4
q
1
≥ 0.95
q
1
≥ 1.05
q
1
≥ 1
Figure 1.14Estimation of the time-dependence of the extinction probability in subcritical, critical and supercritical
systems.
then from
g(z,t)=(1−W)g(z,t−1)+Wq[g(z,t−1)]
one obtains the formula
p
0(t)=Wf 0+[1−W(1−f 1)]p0(t−1)+Wf 2[p0(t−1)]
2
.
Define the functionh(x)=h
0+h1x+h 2x
2
, in which
h
0=Wf0,h1=1−W(1−f 1) andh 2=Wf2.
By using this function, one can write thatp
0(t)=h[p 0(t−1)], by noticing thatp 0(0)=0. This formula was
used for the direct calculation of the probabilitiesp
0(t). It can easily be shown that ift→∞, then the value
of the extinction probability is given by the smallest positive root of the equationh(x)=x. It can be proven
that this root equals unity in both subcritical and critical media, while in a supercritical medium it is less than
1 and is equal to the value of
lim
t→∞
p0(t)=p 0(∞)=1−2
q
1−1
q2
.
It can be seen in Fig. 1.14 that the curvep
0(t) corresponding toq 1=1.05 tends to the value 0.8ift →∞.

Ch02-I045064.tex 27/6/2007 9: 34 Page 36
CHAPTER TWO
Generalisation of the Problem
Contents
2.1 Joint Distribution of Particle Numbers at Different Time Instants 36
2.2 Branching Process with Two Particle Types 39
2.3 Extinction and Survival Probability 41
2.1 Joint Distribution of Particle Numbers at Different Time Instants
Lett1=tandt 2=t+ube two time instants whereu≥0. Letn(t 1) andn(t 2) denote the number of
particles present in a multiplying system at the time instantst
1andt2, respectively. Then,
P{n(t
2)=n 2,n(t1)=n 1|n(0)=1}=p 2(n1,n2,t1,t2) (2.1)
is the probability that in the multiplying system,n
1particles are present att 1andn 2particles att 2≥t1, provided
that one particle existed in the system att=0. It is customary to call this description thetwo-point model.We
will now prove the following important theorem.
Theorem 8.In the case of a homogeneous process, the generating function
g
2(z1,z2,t1,t2)=
∞
≥
n1=0
∞
≥
n2=0
p2(n1,n2,t1,t2)z
n1
1
z
n2
2
(2.2)
satisfies the functional equation
g
2(z1,z2,t1,t2)=g[z 1g(z2,t2−t1),t1], (2.3)
in which g(z,t)is the solution of(1.29)or its equivalent(1.30).
Proof.Since one has
P{n(t
2)=n 2,n(t1)=n 1|n(0)=1}
=P{n(t
2)=n 2|n(t1)=n 1}P{n(t 1)=n 1|n(0)=1},
Neutron fluctuations © 2008 Elsevier Ltd.
ISBN-13: 978-0-08-045064-3 All rights reserved.
36

Ch02-I045064.tex 27/6/2007 9: 34 Page 37
Generalisation of the Problem 37
and since each of then 1particles found in the system at timet 1will start a branching process independently
from the others, one can write that
∞
≥
n1=0
∞
≥
n2=0
P{n(t 2)=n 2,n(t1)=n 1|n(0)=1}z
n2
2
z
n1
1
=
∞
≥
n1=0
∞
≥
n2=0
P{n(t 2)=n 2|n(t1)=n 1}z
n2
2
×P{n(t 1)=n 1|n(0)=1}z
n1
1
=
∞
≥
n1=0
[z1g(z2,t2−t1)]
n1
P{n(t 1)=n 1|n(0)=1}.
This proves the theorem, as expressed by equation (2.3).
As a generalisation of the case, let us now determine the generating function of the probability distribution
P{n(t
j)=n j,j=1,...,k|n(0)=1}=p k(n1,...,n k,t1,...,t k).
Introduce the notations
t
1=tandt j−tj−1=uj−1,j=2,...,k.
By using these, the relation (2.3) can be written in the form
g
2(z1,z2,t,u1)=g[z 1g(z2,u1),t].
Defining the operation
ˆG
jzj=zjg(zj+1,uj), (2.4)
one notes that the expressiong[z
1g(z2,u1),t] can be considered as the transform of the function
g
1(z1,t)=g(z 1,t) that results from the operation
ˆG
1z1=z1g(z2,u1)
applied directly to the variablez
1. By using (2.4), one can write that
ˆG
j+1
ˆG
jzj=zjg(ˆGj+1zj+1,uj)=z jg[zj+1g(zj+2,uj+1),uj)],
by the virtue of which one has
g
3(z1,z2,z3,t,u1,u2)=g(ˆG 2
ˆG
1z1,t)=g{z 1g[z2g(z3,u2),u1,t]}.
By induction, one arrives at
g
j(z1,...,z j,t,u1,...,u j−1)=g(ˆG j−1···ˆG 1z1,t), (2.5)
whereg
1(z1,t)=g(z 1,t) andj=2,...,k. Expression (2.5) is the generating function of thej-point model.
2.1.1 Autocorrelation function of the particle number
In many cases, one needs the autocorrelation function
E{[n(t)−m
1(t)][n(t+u)−m 1(t+u)]}=R n,n(t+u,t)
which will now be calculated. Actually, it would be more correct to call the functionR
n,n(t+u,t) the
autocovariance function. However, whenever it does not lead to confusion, following common practice,R
n,nwill
be referred to as the autocorrelation.

Ch02-I045064.tex 27/6/2007 9: 34 Page 38
38 Imre Pázsit & Lénárd Pál
First, determine the expectation
E{n(t)n(t+u)}=
∞
∂
2
g2(z1,z2,t,u)
∂z1∂z2

z1=z2=1
.
Introducing the notation
s=s(z
1,z2)=z 1g(z2,u)
and considering that
∂g(s,t)
∂z1
=
dg
ds
∂s
∂z1
, as well as
∂g(s,t)
∂z2
=
dg
ds
∂s
∂z2
,
one can write down the relation
∂
2
g2
∂z1∂z2
=
d
2
g
ds
2
∂s
∂z1
∂s
∂z2
+
dg
ds
∂
2
s
∂z1∂z2
.
From this it follows that
E{n(t)n(t+u)}=[m
2(t)+m 1(t)]m1(u). (2.6)
After a short calculation, one obtains
E{[n(t)−m
1(t)][n(t+u)−m 1(t+u)]}=R n,n(t+u,t)=D
2
{n(t)}e
αu
, (2.7)
in which, according to (1.60)
D
2
{n(t)}=
→
[Q
q2
α
−1]e
αt
(e
αt
−1), ifα=0,
Qq
2t,i fα=0.
It is seen thatD
2
{n(0)}=0, and if
α=−a<0, then lim
t→∞
D
2
{n(t)}=0,
hence
lim
t→∞
Rn,n(t+u,t)=0.
The normalised covariance function
C
n,n(t+u,t)=
R
n,n(t+u,t)
D{n(t)}D{n(t+u)}
=
D{n(t)}
D{n(t+u)}
e
αu
(2.8)
is called the (traditional)autocorrelation functionofn(t). In a critical system, i.e. whenα=0, one has
C
n,n(t+u,t)=
δ
t
t+u
,
and further thatn(t) andn(t+u)−n(t)areuncorrelatedfor every finitetandu. This follows from (2.7), since
forα=0
E{[n(t)−m
1(t)][n(t+u)−m 1(t+u)]}−E{[ n(t)−m 1(t)]
2
}=0,
i.e.
E{[n(t)−m
1(t)][n(t+u)−n(t)]}=0.
This observation implies that in a subcritical system near to criticality, thevariation in the number of particles
(increase or decrease) under a time periodufollowing a time instantt is not correlated with the number of particles
at t. This statement may have important consequences regarding the characteristics of the fluctuations in
quasi-critical systems.

Ch02-I045064.tex 27/6/2007 9: 34 Page 39
Generalisation of the Problem 39
2.2 Branching Process with Two Particle Types
Suppose that in a multiplying system, two different particle types, which can be converted into each other,
develop a branching process. Again, it is assumed that a particle of any type found in the multiplying system at
any given time can induce a reaction and generate a number of progeny independently from its own past and
from the history of the other particles present.
Denote the two particle types asT
1andT 2, respectively. Letn 1(t) be the number of particles of type
T
1andn 2(t) that of the particles of typeT 2in the multiplying system att≥0, respectively. Further, let
Q
it+o(t),i=1, 2 be the probability that during the time periodt→0, a particle of typeT i(i=1, 2)
will induce a reaction. As a result of the reaction,νparticles of typeT
1andµparticles of typeT 2will be
generated.
1
Let
P{ν=k,µ=l|T
i}=f
(i)
k,l
(2.9)
denote the probability that in a reaction induced by one particle of the typeT
i, the number of generated
particles of typesT
1andT 2will bekandl, respectively. The normalisation condition reads as
∞
≥
k=0
∞
≥
l=0
f
(i)
k,l
=1,i=1, 2.
Define the probabilities
P{n
1(t)=n 1,n2(t)=n 2|n1(0)=1,n 2(0)=0}=p
(1)
(n1,n2,t) (2.10)
and
P{n
1(t)=n 1,n2(t)=n 2|n1(0)=0,n 2(0)=1}=p
(2)
(n1,n2,t). (2.11)
Obviously,p
(i)
(n1,n2,t) is the probability that at timet≥0 there will ben 1particles of typeT 1andn 2particles
of typeT
2in the system, provided that att=0 there was one particle of typeT iand no particle of the opposite
type in the system. Introduce the generating functions
g
(i)
(z1,z2,t)=
∞
≥
n1=0
∞
≥
n2=0
p
(i)
(n1,n2,t)z
n1
1
z
n2
2
i=1, 2 (2.12)
and
q
(i)
(z1,z2)=
∞
≥
k=0
∞
≥
l=0
f
(i)
k,l
z
k
1
z
l
2
,i=1, 2. (2.13)
Theg
(i)
andq
(i)
fulfil the conditions
g
(i)
(1, 1,t)=1,q
(i)
(1, 1)=1
and
g
(i)
(z1,z2,0)=z i,i=1, 2. (2.14)
The backward Kolmogorov equation will now be derived for the probabilities
p
(1)
(n1,n2,t) andp
(2)
(n1,n2,t).
1
Naturallyνandµare random variables.

Ch02-I045064.tex 27/6/2007 9: 34 Page 40
40 Imre Pázsit & Lénárd Pál
To be able to follow closely the evolution of the process, the intuitive method which was already used in
Section 1.2.1 will be followed. Then one can write
p
(1)
(n1,n2,t)=e
−Q1t
δn11δn20+Q1
θ
t
0
e
−Q1(t−t
δ
)
[f
(1)
0,0
δn10δn20
+S
(1)
(n1,n2,t
δ
)+S
(2)
(n1,n2,t
δ
)+S
(1,2)
(n1,n2,t
δ
)]dt
δ
,
where
S
(1)
(n1,n2,t
δ
)=
∞
∂
k=1
f
(1)
k,0
∂
a1+···+a k=n1
∂
b1+···+b k=n2
k
φ
u=1
p
(1)
(au,bu,t
δ
),
S
(2)
(n1,n2,t
δ
)=
∞
∂
l=1
f
(1)
0,l
∂
a1+···+a l=n1
∂
b1+···+b l=n2
l
φ
v=1
p
(1)
(av,bv,t
δ
),
and
S
(1,2)
(n1,n2,t
δ
)=
∞
∂
k=1
∞
∂
l=1
f
(1)
k,l
∂
a
(1)
1
+···+a
(1)
k
+a
(2)
1
+···+a
(2)
l
=n1
∂
b
(1)
1
+···+b
(1)
k
+b
(2)
1
+···+b
(2)
l
=n2
×
k
φ
u=1
l
φ
v=1
p
(1)
(a
(1)
u
,b
(1)
u
,t
δ
)p
(2)
(a
(2)
v
,b
(2)
v
,t
δ
).
Based on this, for the generating functiong
(1)
(z1,z2,t), one obtains the following equation:
g
(1)
(z1,z2,t)=e
−Q1t
z1+Q1
θ
t
0
e
−Q1(t−t
δ
)
∞
∂
k=0
∞
∂
l=0
f
(1)
k,l
[g
(1)
(z1,z2,t
δ
)]
k
[g
(2)
(z1,z2,t
δ
)]
l
dt
δ
,
which – by considering the definition (2.13) – can be written in the following form:
g
(1)
(z1,z2,t)=e
−Q1t
z1+Q1
θ
t
0
e
−Q1(t−t
δ
)
q
(1)
[g
(1)
(z1,z2,t
δ
),g
(2)
(z1,z2,t
δ
)]dt
δ
. (2.15)
In a completely analogous way, one can derive the generating function equation
g
(2)
(z1,z2,t)=e
−Q2t
z2+Q2
θ
t
0
e
−Q2(t−t
δ
)
q
(2)
[g
(1)
(z1,z2,t
δ
),g
(2)
(z1,z2,t
δ
)]dt
δ
. (2.16)
Differentiating with respect tot, from these equations one arrives at
∂g
(i)
∂t
=Q
i[q
(i)
(g
(1)
,g
(2)
)−g
(i)
],i=1, 2. (2.17)
By introducing the notations
s
(i)
(z1,z2)=Q i[q
(i)
(z1,z2)−z i],i=1, 2, (2.18)
the basic equations can be written in a rather simple form as
∂g
(i)
∂t
=s
(i)
(g
(1)
,g
(2)
),i=1, 2, (2.19)
together with the initial conditionsg
(i)
(z1,z2,0)=z i,i=1, 2.

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se on auringonsäde. — Naisen nimen tulee olla miellyttävä, suloinen,
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siunauksia." —… Niin, tuo viisas on oikeassa; todellakin, Maria, Sofia,
Esmeralda… — Kirous! aina tuo ajatus!
Ja hän sulki kirjan kiivaasti.
Hän pyyhkäisi kädellä otsaansa ikään kuin karkottaakseen
ajatuksen, joka häntä vaivasi. Sitten hän otti pöydältä naulan ja

pienen vasaran, jonka varteen oli maalattu omituisia kabbalistisia
kirjaimia.
— Viime aikoina, virkkoi hän katkerasti hymyillen, — eivät mitkään
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—… Yksi ainoa kurja ajatus, jatkoi pappi, riittää siis tekemään
ihmisen heikoksi ja hupsuksi! Oh! miten Claude Pernelle nauraisikaan
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jonka Zéchiélé lausui iskiessään vasaralla naulaa.
— Pötyä! ajatteli Jehan.
— Katsotaan, koetetaan, jatkoi arkkidiakoni vilkkaasti. Jos
onnistun, näen sinisen kipinän välähtävän naulan päästä. —

Emen-hetan! Emen-hetan! — Se ei ole se. — Sigeani! Sigeani! —
Avatkoon tämä naula haudan sille, jonka nimi on Febus!… — Kirous.
Aina, yhä, ikuisesti sama ajatus!
Ja hän paiskasi vasaran raivostuen kädestään. Sitten hän painautui
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hänen kouristuksenomaisesti pusertuneen nyrkkinsä eräällä kirjalla.
Äkkiä nousi dom Claude, tarttui harppiin ja piirsi ääneti seinään
isoilla kirjaimilla seuraavan kreikkalaisen sanan:
'AN'AGKH.
— Veljeni on hullu, virkkoi Jehan itsekseen; — olisihan ollut paljon
yksinkertaisempaa kirjoittaa Fatum [Kohtalo]. Eihän kuka tahansa ole
velvollinen taitamaan kreikkaa.
Arkkidiakoni istuutui jälleen tuoliinsa ja nojasi otsansa molempia
käsiään vasten kuin sairas, jonka pää on raskas ja polttava.
Ylioppilas katseli hämmästyneenä veljeään. Hän, jonka sydän aina
oli avoinna, joka ei nähnyt muuta lakia maailmassa kuin hyvän
luonnonlain, joka antoi intohimojensa virrata taipumustensa teitä ja
jolla suurten mielenliikutusten järvi aina oli kuivillaan, hän kun joka
aamu valmisti siihen uusia viemäreitä, hän ei tiennyt, miten
raivokkaasti tuo inhimillisten intohimojen järvi saattaa käydä ja
kuohua, kun siltä kokonaan kielletään purkautuminen, miten se
paisuu, nousee, tulvii yli reunojensa, miten se kalvaa sydäntä, miten
se purkautuu sisäisinä huokauksina ja äänettöminä kouristuksina,
kunnes se murtaa sulkunsa ja puhkaisee alustansa. Claude Frollon
ankara ja jäinen ulkomuoto, tuo jäykän ja saavuttamattoman hyveen
kylmä kuori oli aina johtanut Jehanin harhaan. Tuo iloinen ylioppilas

ei ollut koskaan uneksinutkaan, että tuon otsan, tuon Etnan jäätikön
alla virtailisi kiehuva, raivoisa, syvä laava.
Emme tiedä, heräsivätkö nämä ajatukset hänessä äkkiä, mutta
niin kevytmielinen kuin hän olikin, ymmärsi hän kuitenkin, että oli
nähnyt sellaista, mitä hänen ei olisi pitänyt nähdä, että oli yllättänyt
vanhemman veljensä hänen salaisimmasta sieluntilastaan, ja ettei
Clauden pitänyt sitä huomata. Nähtyään, että arkkidiakoni oli
vaipunut hiljaiseen mietiskelyyn, veti hän päänsä hyvin varovasti
takaisin ja kolisteli hieman jaloillaan oven ulkopuolella, aivan kuin
joku tulija olisi halunnut ilmoittaa tulostaan dom Claudelle.
— Sisään! huusi arkkidiakoni huoneestaan, olen teitä odottanut.
Jätin tahallani avaimen oveen. Käykää sisään, mestari Jacques.
Ylioppilas astui rohkeasti sisälle. Arkkidiakoni, josta veljen
tapaaminen tuntui kiusalliselta tällaisessa paikassa, hätkähti
nojatuolissaan.
— Mitä! sinäkö, Jehan?
— No J:llä alkava joka tapauksessa, vastasi ylioppilas kasvot
punakkoina, julkeina ja iloisina.
Dom Clauden kasvot olivat käyneet jälleen ankariksi.
— Mitä sinä täällä teet?
— Hyvä veli, vastasi ylioppilas koettaen näyttää siivolta,
surkuteltavalta ja häveliäältä ja pyöritellen viattoman näköisenä
lakkia käsissään, tulin pyytämään…
— Mitä?

— Hieman moraalista tukea, jota kipeästi kaipaan. Jehan ei
uskaltanut lisätä ääneensä: — Ja vähän rahaa, jota vielä kipeämmin
kaipaan. Tämä toivomus jäi sanomatta.
— Hyvä herra, virkkoi arkkidiakoni kylmästi, — olen hyvin
tyytymätön sinuun.
— Valitettavasti! huokasi ylioppilas.
Dom Claude kääntyi nojatuoleineen ja tarkasti Jehania tuikeasti.
— Olipa hyvä, että tapasin sinut.
Alku oli pahaenteinen. Jehan valmistautui saamaan ankaran iskun.
— Jehan, joka päivä saan kuulla valituksia sinusta. Mikä tappelu se
taas on ollut, jossa sinä olet piessyt erästä pientä varakreiviä, Albert
de Ramonchampia?…
— Oh! virkkoi Jehan, — vähäpätöinen juttu! hävytön hovipoika,
joka huvikseen antoi hevosensa roiskutella kuraa ylioppilaitten
päälle!
— Kuka on tuo Mahiet Fargel, jatkoi arkkidiakoni, — jonka takin
olet repinyt? Tunicam dechiraverunt, sanotaan valituksessa.
— Hoh! vaivainen Montaigu-viitta! eikö siinä ole niin!
— Valituksessa sanotaan tunicam eikä cappettam. Osaatko
latinaa?
Jehan ei vastannut.

— Niin, jatkoi pappi päätään pudistaen, siitä nähdään, miten
opintojen ja tietojen laita nykyään on. Latinaa tuskin ymmärretään,
Syyriasta ei tiedetä mitään ja kreikkaa halveksitaan niin, ettei pidetä
minään, vaikka oppineimmatkin hyppäävät kreikkalaisen sanan yli
lukematta sitä, ja että sanotaan: Graecum est, non legitur.
[Kreikkaa, en ymmärrä.]
Ylioppilas kohotti päättävästi katseensa.
— Herra veljeni, tahdotko, että käännän paikalla ranskaksi tuon
kreikkalaisen sanan, joka on tuossa seinässä?
— Minkä sanan?
— 'ANA'GKH.
Heikko puna levisi arkkidiakonin karvaille poskipäille aivan kuin
savupilvi, joka ilmaisee tulivuoren salaista kuohuntaa. Ylioppilas
tuskin huomasikaan sitä.
— No, Jehan, sammalsi vanhempi veli ilmeisen vaivalloisesti, mitä
se sana merkitsee?
— Kohtalo.
Dom Claude kävi kalpeaksi, ja ylioppilas jatkoi huolettomasti: — Ja
tuo sana, jonka on sen alle piirtänyt sama käsi, Anagneia, merkitsee
saastainen. Siitä näkee, että kreikkaa osataan.
Arkkidiakoni oli vaiti. Tämä kreikankuulustelu oli herättänyt
hänessä mietteitä. Pieni Jehan, jolla oli kaikki hemmoitellun lapsen
metkut, luuli hetkeä sopivaksi pyyntönsä esittämiseen. Hän aloitti
mitä mielistelevimmällä äänellä.

— Hyvä veljeni, oletko minuun niin suuttunut, että voit julmistua
avoimessa taistelussa vaihdetuista mitättömistä tönäisyistä ja
puskuista joittenkin pojannulikkani kanssa, quibusdam marmosetis?
— Näet, että osaan latinani, hyvä veli Claude.
Mutta tämä hymyilevä mielistely ei nyt tehonnutkaan ankaraan
isoon veljeen. Kerberos ei välittänyt hunajakakusta. Arkkidiakonin
otsalta ei ainoakaan kurttu tasoittunut.
— Mitä asiaa sinulla oli tänne? kysäsi hän kuivasti.
— No niin, asiaan! vastasi Jehan rohkeasti. — Tarvitsen rahaa.
Tämä avoin selitys sai arkkidiakonin ilmeen muuttumaan
opettavaksi ja isälliseksi.
— Tiedät hyvin, herra Jehan, että Tirechappen läänityksemme
kaksikymmentäyksi taloa tuottavat vuokraa kaikkiaan
kolmekymmentäyhdeksän livreä, yksitoista souta ja kuusi pariisilaista
denieriä. Se on puolet enemmän kuin Paclet-veljesten aikana, mutta
se ei ole paljon.
— Tarvitsen rahaa, virkkoi Jehan stoalaisen rauhallisena.
— Tiedäthän, että kaupungin hallitus on määrännyt
kaksikymentäyksi taloamme suorittamaan täyden veron
piispanistuimelle, ja että voimme ostaa takaisin verovapauden
maksamalla arvoisalle piispalle kaksi markkaa kultarahassa kuuden
pariisilaisen livren mukaan markalta. Tätä rahasummaa en
kuitenkaan ole voinut hankkia. Sinä tiedät sen.
— Tiedän, että tarvitsen rahaa, toisti Jehan kolmannen kerran.

— Ja mitä sillä aiot tehdä?
Tämä kysymys sytytti taas toivonkipinän Jehanin silmiin. Hän
aloitti uudelleen kissamaisen mielistelevästi:
— Rakas veljeni Claude, minä vakuutan, etten käänny puoleesi
missään huonossa tarkoituksessa. Ei ole aikomukseni keikailla
kapakoissa rahoillasi eikä ratsastaa Pariisin katuja pitkin kullalla
kirjailluin loimin ja lakeija jäljessäni, cum meo laquasio. Ei, veljeni,
vaan päinvastoin hyväntekeväisyyttä varten.
— Mitä hyväntekeväisyyttä? kysyi Claude hieman yllättyneenä.
— Kaksi ystävääni aikoo ostaa kapalon erään haudriette-laitoksen
köyhän lesken lapselle. Se on laupeuden työ. Se maksaa kolme
floriinia, ja minäkin tahtoisin antaa roponi.
— Minkä nimisiä ovat ystäväsi?
— Pierre Hirmunyrkki ja Baptiste Hanhihotka.
— Hm! virkkoi arkkidiakoni, omituisia nimiä, jotka sopivat
hyväntekeväisyyteen yhtä huonosti kuin pommitykki pääalttarille.
Jehan olikin sangen onnettomasti valinnut kahden ystävänsä
nimet. Hän huomasi sen liian myöhään.
— Entä mikä kapalo se sellainen on, joka maksaa kolme floriinia?
jatkoi hellittämätön Claude, ja lisäksi haudriettesisarelle? — Milloin
haudriette-leskille on tullut tavaksi saada kapalolapsia?
— No niin! minä tarvitsen rahaa mennäkseni illalla katsomaan
Isabeau la Thierryeä Rakkaudenlehtoon!

— Kurja irstailija! huudahti pappi.
— Anagneia, virkkoi Jehan.
Tämä sana, jonka ylioppilas ehkä ilkeyksissään lainasi kammion
seinältä, teki pappiin omituisen vaikutuksen. Hän puraisi huultaan ja
punastui vihasta.
— Mene tiehesi, hän sanoi Jehanille. — Minä odotan erästä
vierasta.
Ylioppilas teki vielä yrityksen.
— Veli Claude, anna minulle edes pieni pariisilainen, että saan
syödä.
— Miten pitkälle olet lukenut Gratianuksen asetuksia? kysyi dom
Claude.
— Olen kadottanut muistiinpanoni.
— Miten pitkällä olet latinassa? — Horatiukseni on varastettu.
— Miten pitkällä olet Aristoteleessa?
— Hyvä veli! eikö joku kirkkoisä ole sanonut, että kaikkien aikojen
kerettiläiset ovat kätkeneet erehdyksensä Aristoteleen metafysiikan
viidakoihin? Hiiteen koko Aristoteles! en tahdo repiä uskontoani
riekaleiksi hänen metafysiikkaansa.
— Nuori mies, virkkoi arkkidiakoni, kun kuningas viimeksi saapui
tänne, oli hänen seurueessaan eräs aatelismies nimeltä Philippe de
Comines, jonka hevosen loimessa oli lause, jota neuvoisin sinua

miettimään: Qui non laborat non manducet.[Joka ei työtä tee, ei
hänen syömänkään pidä.]
Ylioppilas oli hetken vaiti, sormi korvalla, silmät maahan luotuina
ja äkäisen näköisenä. Äkkiä hän kääntyi Clauden puoleen vilkkaana
kuin västäräkki.
— Hyvä veli, sinä siis kieltäydyt antamasta minulle yhtä
pariisilaista souta, jotta voisin ostaa leipää leipurista?
— Qui non laborat non manducet.
Näiden arkkidiakonin jyrkkien sanojen jälkeen Jehan peitti
kasvonsa käsillään kuin nyyhkivä nainen ja huudahti epätoivoisella
äänellä.
— Otototototoi!
— Mitä se on olevinaan, herraseni? kysyi Claude hämmästyneenä
tästä huudahduksesta.
— Mitäkö! vastasi ylioppilas kiinnittäen Claudeen julkeat silmänsä,
joita hän oli hieronut sormillaan saadakseen ne itkettyneen
punaisiksi; — se on kreikkaa! Se on eräs Aiskhyloksen anapesti, joka
ilmaisee suurinta tuskaa.
Ja sitten hän purskahti niin veitikkamaiseen ja rajuun nauruun,
että arkkidiakonikin hymähti. Olihan se oikeastaan Clauden vika;
miksi hän oli niin hemmotellut poikaa?
— Oh! rakas veljeni Claude, uudisti Jehan tämän hymyn
rohkaisemana, katsohan minun risaisia kenkiäni. Onko maailmassa

sen traagillisempia koturneja kuin saappaat, joiden pohjasta kieli
pistää esille?
Arkkidiakoni oli jälleen yhtä ankara kuin ennen.
— Lähetän sinulle uudet saappaat, mutta en rahaa.
— Vain pieni vaivainen pariisilainen, hyvä veli, jatkoi Jehan
rukoilevasti. Luen Gratianusta sydämeni pohjasta, uskon lujasti
Jumalaan ja minusta tulee oikea tiedon ja siveyden Pytagoras. Mutta
pieni pariisilainen, armosta! Tahdotko, että minut ahmaisee nälkä
kitaansa, joka ammottaa edessäni mustempana ja löyhkäävämpänä
kuin Tartarus tai jonkun munkin nenä?
Dom Claude ravisti kurttuista päätään:
— Qui non laborat…
Jehan keskeytti hänet.
— No niin, piru vieköön! huudahti hän. — Eläköön ilo! Juon itseni
humalaan, tappelen, isken sarkat lattiaan ja menen tyttöjen luo!
Ja hän heitti lakkinsa seinään ja läpsäytti sormiaan ikään kuin
kastanjetteja.
Arkkidiakoni katseli häntä synkkänä.
— Jehan, sinulla ei ole sielua.
— Siinä tapauksessa minulta puuttuu Epikuroksen mukaan jotakin,
en tiedä mitä, joka on tehty jostakin, jolla ei ole nimeä.
— Jehan, sinun täytyy vakavasti ajatella parannusta.

— Kas vain, huudahti ylioppilas katsellen vuoroin veljeään, vuoroin
liedellä olevia tislaimia, — täällä on kaikki sarvipäitä, niin ajatukset
kuin pullotkin.
— Jehan, sinä olet luisuvalla pinnalla. Tiedätkö minne joudut?
— Kapakkaan, vastasi Jehan.
— Kapakka vie kaakinpuuhun.
— Onhan se yhtä hyvä lyhty kuin jokin muukin, ja sen avulla olisi
Diogenes ehkä löytänyt ihmisen.
— Kaakinpuu vie hirteen.
— Hirsipuu on vaaka, jonka toisessa päässä on ihminen, toisessa
koko maailma. On ihanaa olla ihminen.
— Hirsipuu vie helvettiin.
— Se on suuri nuotio.
— Jehan, Jehan, sinun päiväsi päättyvät huonosti.
— Alku on ollut hyvä.
Tällöin kuului askeleita portaista.
— Hiljaa! sanoi arkkidiakoni pannen sormen huulilleen, siellä on
mestari Jacques. Kuule, Jehan, lisäsi hän hiljaa, muista, ettet
koskaan puhu kenellekään siitä, mitä olet täällä nähnyt ja kuullut.
Kätkeydy nopeasti tuonne uunin alle, äläkä hiisku mitään.

Ylioppilas ryömi uunin alle. Siellä hänen päähänsä pälkähti
hedelmällinen ajatus.
— Kesken kaiken, veli Claude, floriini siitä, etten hiisku mitään.
— Hiljaa! saat sen.
— Anna se heti.
— Tuossa on! virkkoi arkkidiakoni heittäen hänelle suuttuneena
kukkaronsa. Jehan kätkeytyi uunin alle, ja ovi aukeni.
V. Kaksi mustakaapua
Tulija oli mustapukuinen ja synkännäköinen mies. Ensin pistikin
ystävämme Jehanin silmiin (joka tietenkin oli asettunut loukkoonsa
niin, että saattoi mielensä mukaan nähdä ja kuulla kaikki) tulijan
puku ja kasvojen syvä surullisuus. Hänen kasvoissaan oli kuitenkin
jonkinlainen lempeä ilme, mutta se oli kissan tai tuomarin lempeyttä,
imelää lempeyttä. Hän oli ylen harmaa ja kurttuinen, noin
kuusikymmenvuotias ja räpytteli silmiään. Hänellä oli valkeat
kulmakarvat, riippuva alahuuli ja suuret kädet. Kun Jehan näki, ettei
tulija ollutkaan sen kummempi, toisin sanoen varmaankin joku
lääkäri tai lakimies, ja että miehen nenä oli hyvin etäällä suusta,
mikä oli tyhmyyden merkki, hän vetäytyi loukkoonsa toivottomana,
tietämättä miten pitkän ajan saisi viettää tuossa epämukavassa
asennossa ja huonossa seurassa.
Arkkidiakoni ei ollut edes noussut seisomaan miehen saapuessa.
Hän oli viitannut häntä istuutumaan oven luona olevalle penkille ja

hetken hiljaisuuden jälkeen, jolloin hän oli näyttänyt jatkavan
mietteitään, hän oli virkkanut hieman suojelijan omaisella
äänensävyllä:
— Hyvää päivää, mestari Jacques.
— Terve, mestari! oli mustapukuinen vastannut.
Siinä sävyerossa, millä toisaalta mestari Jaques ja toisaalta
varsinaisesti mestari lausuttiin, oli sama ero kuin sanoilla
monseigneur ja monsieur, domine ja domne. Siinä oli selvästi oppilas
opettajan edessä.
— No, virkkoi arkkidiakoni uuden tauon jälkeen, jota mestari
Jacques karttoi katkaisemasta, oletteko onnistunut?
— Ah, mestari, vastasi toinen surullisesti hymyillen, — minä
puhallan yhtenään. Tuhkaa niin paljon kuin haluan. Mutta ei
kultajyvääkään.
Dom Claude liikahti kärsimättömästi.
— En tarkoittanut sitä, mestari Jacques Charmolue, vaan
oikeusjuttuanne velhoa vastaan. Hänen nimensähän on Marc
Cenaine, tiliviraston mestari? Tunnustaako hän noituutensa? Onko
kidutus onnistunut?
— Ehei, niin hyvin eivät asiat ole, vastasi mestari Jacques jälleen
surullisesti hymyillen. Se mies on kova kuin piikivi. Meidän täytyy
keittää häntä Marché-aux-Pourceaux'lla, ennen kuin hän mitään
sanoo. Emme ole säästäneet keinoja saadaksemme esille totuuden.
Hän onkin jo aivan riepuna. Me käytämme kaikkia Pyhän
Johanneksen yrttejä, kuten vanha komediankirjoittaja Plautus sanoo:

Advorsum stimidos, laminas, crucesque, compedesque,
Nervos, catenas, carceres, numellas, pedicas, botas.
[Piikkejä, pihtejä, kidutusta, jalkarautoja, hihnoja, köysiä, koppia,
siteitä, jalkapuuta, kahleita vastaan.]
Mikään ei auta. Se on kauhea mies. En tiedä mitä tehdä.
— Ettekö ole löytänyt mitään uutta hänen asunnostaan?
— Olen, vastasi mestari Jacques pistäen käden laukkuunsa, tämän
pergamentin. Siinä on sanoja, joita me emme lainkaan ymmärrä.
Rikosasianajaja herra Philippe Lheulier osaa kuitenkin vähän
hepreaa, jota hän oppi Brysselissä Kanterstenkadun juutalaisten
oikeusjutussa.
Ja mestari Jacques kääri auki pergamenttikäärön.
— Näyttäkääpä, sanoi arkkidiakoni ja huudahti silmäiltyään sitä: —
Pelkkää noituutta, mestari Jacques! Emen-hetan! on velhojen huuto
tullessaan hornaan. Per ipsum, et cum ipso, et in ipso! on kaava,
jolla piru jälleen suljetaan helvettiin. Hax, pax, max! on lääke. Kaava
hullujen koirien puremaa vastaan. Mestari Jacques, te olette
kuninkaallinen syyttäjä kirkollisasioissa; tämä pergamentti on
inhottava.
— Panemme miehen uudelleen kidutukseen. Tässä on vielä jotain,
lisäsi mestari Jacques kopeloiden uudelleen laukkuaan, Marc
Cenainen luota löydettyä.
Se oli samantapainen astia, joita oli dom Clauden uunilla.
— Ahaa! virkkoi arkkidiakoni, alkemistinen sulatin.

— Tunnustan, sanoi mestari Jacques hymyillen hämillään, että
koetin sitä uunilla, mutta en onnistunut paremmin kuin
omallanikaan.
Arkkidiakoni tutki astiaa.
— Mitä hän on piirtänyt sulattimeensa? Och! Och! sana, joka
karkoittaa kirput. Tuo Marc Cenaine on typerys! Uskonpa todellakin,
että ette tällä kultaa tehneet! Se kelpaa enintään kesällä
makuukammioonne, siinä kaikki!
— Koska on puhe erehdyksistä, virkkoi kuninkaallinen syyttäjä, —
niin minä tarkastelin tänne tullessani portaalia tuolla alhaalla. Onko
teidän korkea-arvoisuutenne aivan varma, että luonnon tutkimisen
alku on esitettynä Hôtel-Dieun puolella, ja että niistä seitsemästä
alastomasta henkilökuvasta, jotka ovat Pyhän Neitsyen jalkojen
juuressa, se, jolla on siivet jaloissa, on Merkurius?
— Olen, vastasi pappi. Niin on kirjoittanut Augustin Nypho, tuo
italialainen tohtori, jolla oli palveluksessaan parrakas paholainen,
joka opetti hänelle kaikki. Muuten menemme alas ja minä selitän
teille siellä kaiken.
— Kiitos, mestarini, sanoi Charmolue kumartaen maahan asti. —
Mutta olinpa unohtaa! milloin haluatte minun vangituttavan tuon
pienen noitatytön?
— Kenen noitatytön?
— Tuon mustalaistytön, jonka hyvin tunnette, joka päivisin tanssii
tuomiokirkon torilla, huolimatta kirkollisviraston kiellosta. Hänellä on
noiduttu vuohi, jolla on pirun sarvet, joka lukee ja kirjoittaa, joka

laskee kuin itse Picatrix. Pelkästään senkin takia jo koko
mustalaisjoukko ansaitsisi hirsipuun. Syytös on valmis. Kaikki käy
nopeasti! Kaunis olento, totta tosiaan tuo tanssijatar! Mitä
kauneimmat mustat silmät! Oikeat egyptiläiset jalokivet. Milloin
käymme käsiksi asiaan?
Arkkidiakoni oli kalvennut.
— Kyllä ilmoitan teille, sammalsi hän tuskin kuuluvalla äänellä.
Sitten hän lisäsi ponnistaen: — Onhan teillä Marc Cenaine.
— Olkaa huoleti, virkkoi Charmolue hymyillen. Annan venyttää
hänet uudelleen nahkavuoteelle. Kyllä hän on piru miehekseen. Hän
saa itse Pierrat Torteruenkin uupumaan, miehen, jolla kuitenkin on
tukevammat kourat kuin minulla. Kuten tuo kunnon Plautus sanoo:
Nudus vinctus, centum pondo, es quando pendes per pedes. [Olet
alasti sidottu, sadan naulan painoinen, kun sinut ripustetaan jaloista
riippumaan.] Puomikidutus on paras, mitä meillä on. Sitä hän saa
maistaa.
Dom Claude näytti vaipuneen synkkään mietiskelyyn. Hän kääntyi
Charmoluen puoleen:
— Mestari Pierrat — mestari Jacques tarkoitin, onhan teillä Marc
Cenaine!
— Niin on, dom Claude. Mies parka! hän saa kärsiä kuin Mummol.
Mutta miten saattaakaan ajatella hornaan menoa! tiliviraston
talousmestari, jonka täytyi tuntea Kaarle Suuren määräys Stryga vel
masca! — Mitä tyttöön tulee — Esmeraldaan, kuten häntä nimitetään
— odotan käskyänne. — No niin! Kun menemme portaalin ohi,

selitätte kai minulle, mitä merkitsee puutarhuri, joka on maalattu
kirkon sisäänkäytävän seinälle. Eikö se ole kylväjä? — Mutta mitä te
oikein ajattelette, mestari?
Dom Claude oli syventynyt omiin mietteisiinsä eikä kuunnellut.
Charmolue, joka seurasi hänen katseensa suuntaa, näki, että se
jäykästi tuijotti ikkunassa olevaa hämähäkinverkkoa. Joku
ajattelematon kärpänen, joka pyrki maaliskuun aurinkoon, oli juuri
yrittänyt tuon verkon läpi ja tarttunut siihen kiinni. Verkkonsa
värähtelyn tunnettuaan liikahti suuri hämähäkki äkkiä verkon
keskellä ja syöksyi yhdellä hyppäyksellä kärpäsen kimppuun, jonka
se tuntosarvillaan taittoi kaksinkerroin, samalla kun sen kärsä tavoitti
kärpäsen päätä.
— Kärpäs-raukka! sanoi kuninkaallinen syyttäjä kirkollisasioissa ja
nosti kätensä auttaakseen sitä.
Arkkidiakoni, joka äkkiä heräsi kuin unesta, tarttui
kouristuksenomaisen rajusti hänen käteensä.
— Mestari Jacques, hän huusi, antakaa kohtalon täyttyä! Syyttäjä
kääntyi hämmästyneenä. Hänestä tuntui siltä kuin rautapihdit olisivat
tarttuneet hänen käsivarteensa. Papin katse oli tuijottava, villi ja
liekehtivä ja viipyi liikahtamatta tuossa kärpäsen ja hämähäkin
muodostamassa hirveässä ryhmässä.
— Niinpä niin! jatkoi pappi äänellä, joka tuntui tulevan hänen
sisimmästään, — siinä on kaiken symboli. Se lentää, se on iloinen,
äsken syntynyt; se etsii kevättä, ilmaa, vapautta; niinpä niin, mutta
jos se lentää päin kohtalokasta ruusustoa, hämähäkki hyökkää sen
kimppuun, kauhea hämähäkki! Tanssijatar raukka! tuomittu
kärpäsraukka! Mestari Jacques, antakaa olla! se on kohtalo! — Niin,

Claude, sinä olet hämähäkki. Claude, sinä olet myös kärpänen! —
Sinä lensit kohden tietoa, valoa, aurinkoa, sinä halusit vain päästä
ikuisen totuuden suuren päivän avariin ilmapiireihin; mutta
syöksyessäsi toiseen maailmaan, valon, älyn ja tiedon maailmaan
johtavaa häikäisevää aukkoa kohden, sinä sokea kärpänen, sinä
hupsu oppinut, et nähnyt hienoa hämähäkinverkkoa, jonka kohtalo
oli levittänyt valon ja sinun välillesi, sinä syöksyit sitä vasten suin
päin, kurja mielipuoli, ja nyt rimpuilet sinä murskatuin päin ja
rikkeimin siivin kohtalon rautaisten tuntosarvien välissä! — Mestari
Jacques! mestari Jacques! antakaa hämähäkin olla!
— Enhän minä lainkaan koske siihen, sanoi Charmolue ja katseli
häntä ihmeissään. Mutta laskekaa irti käteni, mestari, armahtakaa!
puristattehan kuin pihdeillä minua.
Arkkidiakoni ei kuullut hänen sanojaan.
— Oh! sinä hupsu! hän jatkoi hellittämättä tuijottaen ikkunaa.
Vaikka voisitkin kärpässiivilläsi murtautua verkon läpi, luuletko
sittenkään pääseväsi valoon! Voi! sen takanahan on tuo lasiruutu,
tuo läpinäkyvä este, tuo kristallimuuri, joka on kovempaa kuin malmi
ja joka erottaa kaikki filosofiat totuudesta. Miten luulet pääseväsi sen
läpi? Oi tietämisen turhuutta! Miten monta viisasta tuleekaan kaukaa
liihotellen murskatakseen otsansa sitä vastaan! Miten monta
filosofista järjestelmää lentääkään sikin sokin suristen tätä ikuista
lasiruutua vasten!
Hän vaikeni. Hänen ajatuksensa olivat huomaamatta siirtyneet
tieteeseen hänen omasta itsestään ja hän näytti rauhoittuneen.
Jacques Charmolue herätti hänet kokonaan todellisuuteen
kysymyksellään: — Entä, mestarini, milloin autatte minua
kullanteossa? Haluaisin onnistua.

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Neutron fluctuations a treatise on the physics on branching processes Imre Pã¡Zsit

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