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HOLOMORPHIC SOLUTIONS TO FUNCTIONAL DIFFERENTIAL 
EQUATIONS 
A THESIS PRESENTED IN PARTIAL 
FULFILMENT OF THE REQUIREMENTS 
FOR THE DEGREE OF 
DOCTOR OF PHILOSOPHY 
IN 
MATHEMATICS 
AT MASSEY UNIVERSITY, PALMERSTON NORTH, 
NEW ZEALAND. 
JONATHAN CRAIG MARSHALL 
2005 
1 
Abstract 
FUnctional differential equations with an entire functional argument 9 are exam­
ined and theory regarding the presence of holomorphic solutions to these equations 
presented. There are two main problems analysed, each related to the other. The 
first is the existence of local holomorphic solutions about a point fixed under g, and 
the second is the analytic continuation of such a solution throughout the complex 
plane. The local behaviour of 9 about its fixed points determines whether holo­
morphic solutions exist about such points, whilst the global behaviour of 9 under 
iteration determines the analytic continuation of these solutions. The dynamics of 
the functional argument g, therefore, is the driving force in both problems. 
Both a local and global theory is developed for the existence of solutions, and for 
defining where such solutions are holomorphic. The case where 9 is a polynomial is 
considered in detail, although much of the theory applies equally well to the general 
case where 9 is entire. 
111 
Acknowledgements 
First and foremost, I would like to thank Assoc. Prof. Bruce van Brunt, whose 
supervision and ideas have pushed me in the right direction, and have more often 
than not highlighted the key properties needed to obtain the results presented in 
this thesis. I could not have hoped for a more suitable supervisor, and know that 
without his guidance and insight , the work would be far less complete. I 'd also 
like to thank Prof. Robert McLachlan for pointing out the similarity between my 
graphs of iterated inverse map sets and those of the Julia set, which led to the 
key realisation that the dynamics of the functional argument is the driving force in 
these equations. My thanks also go to those others who have had input into the 
work, including Prof. Graeme Wake. 
I am grateful to Massey University for supplying the scholarship monies which 
allowed me to remain a student without having to get a "real" job, and to the 
many friends, collegues, and students in and around the Mathematics Department 
for their support and for making it enjoyable to come to work. 
Lastly, I 'd like to thank my wife Keren for her love, understanding and endless 
patience and my family for all their encouragement. It's been a long time coming, 
but it 's here at last . Enjoy. 
Contents 
Abstract 
Acknowledgements 
1 Introduction 
2 First Order Functional Differential Equations 
2 . 1  The Pantograph Equation 
2 .2  Dirichlet Series Solutions 
2.3 Uniqueness . . .  
2 .4 Laplace Analysis 
2 .5 Equations with Variable Coefficients 
2 .5 . 1 The Mellin Transform . . . .  
3 Second Order Pantograph Equations 
3 . 1  Dirichlet Series Solutions 
3 .2 Laplace Analysis . . . .  
3 .2 . 1  The case where Tl i=- T2Ci for all integers k 
3.2 .2 The case where Tl = T2 ak for some positive integer k 
i 
iii 
1 
7 
7 
9 
12  
1 7  
29 
33 
39 
39 
43 
47 
50 
VI 
3.2 .3 The case where rl = r2 . . . . . . . . . . . . . . . . . . . . .  53 
4 Natural Boundaries for Functional Differential Equations 
4 . 1  
4.2 
Examples . . . . . . . . . . . . . . . . .  . 
4. 1 . 1  
4.1.2 
4 . 1 .3 
Example 1 :  y' (z) + ay(z) = by(etz) 
Example 2: y'(z) = Y(Z2) . . 
Example 3: y' (z) = y(z2 + EZ) 
Fixed and Periodic Points 
4.3 Julia Sets . . . . . . . 
4.4 The Natural Boundary 
4.5 Generalisations . . . . 
4 .5 . 1 Higher order equations 
4.5 .2 
4.5.3 
Entire functional arguments 
Equations with polynomial coefficients 
5 Holomorphic Solutions about Repelling Fixed Points 
5 . 1  The Eigenvalue Problem for Holomorphic Solutions 
5.2 Spectrum of the Pantograph Equation 
5.3 
5.4 
Non-linear Polynomial Arguments . 
Non-homogeneous Equations . . . .  
5.5 A Natural Boundary for the Pantograph Equation . 
5 .5 . 1 
5 .5 .2 
5.5.3 
The case where et is a positive integer . 
The general first order equation . . . . 
The second order pantograph equation 
59 
59 
60 
61 
62 
65 
70 
77 
81 
81 
85 
87 
95 
96 
100 
104 
106 
1 10 
1 1 1  
1 12 
1 16 
6 Holomorphic Solutions about Neutral Fixed Points 
6.1 The Linear Case 
6.2 Types Of Neutral Points 
6.3 Holomorphic Solutions A bout Siegel Points. 
7 Conclusion 
7.1 Future Work . 
A Notes on Computer Graphics 
A.l Introduction . . . . 
A.2 Distance Estimator 
A.3 Identifying Regions of the Fatou Set . 
A.3.1 Finding attractive and neutral periodic points 
A.3.2 Orbits tend to the point at infinity 
A.3.3 Orbits tend to some attracting periodic cycle. 
A.3.4 Orbits map into an attractive petal of a parabolic periodic 
point ............. . 
A.3.5 Orbits map into a Siegel disc. 
A.4 The A lgorithm ........... . 
Bibliography 
vii 
119 
120 
122 
128 
135 
137 
139 
139 
140 
141 
142 
143 
143 
144 
144 
145 
147 
List of Figures 
4. 1 Plot of no, . . .  , n20 for g(z) = z2 + 0.8z . .  64 
4.2 Plot of no, . . .  , n20 for g(z) = z2 + 0.9iz. 65 
4.3 The Julia set for the function g(z) = Z2 - 1 + 0.2i . .  71 
4.4 The Julia set for the function g(z) = Z2 + i, a dendrite. 77 
6. 1 The Julia set for g(z) = z2 + e2'71{i z with � = 2/5. 1 23 
6.2 The Julia set for g( z) = Z2 + e271'�i z with � = 1 +
2
V5 . .  1 26 
CHAPTER 1 
Introduction 
Functional differential equations are ordinary differential equations in which one or 
more of the terms has a function of the independent variable as its argument. A 
simple such equation is the Pantograph equation 
y' (x )  + by ( x )  = cy (ax) ( 1 . 1 ) 
where b, c, and a are constants such that c #- O. This equation, and its higher order 
analogues, has been studied extensively by many researchers, both on the real line 
and in the complex plane. A simple power series argument is sufficient to show 
that there exists a non-trivial solution holomorphic about x = 0 in the case where 
l a l :s; 1 ,  while no such solution exists in the case l a l > 1 unless b and c satisfy 
some compatibility condition. The major difference between these two cases is that 
in the first case, points around the origin are moved closer to the origin under 
the functional argument , whereas in the second case they are pushed further away. 
This leads to the conjecture that the local behaviour of the functional argument 
9 about the point x = 0, dictates whether local holomorphic solutions exist in a 
neighbourhood of 0 to the more general equation 
y' (x) + by ( x )  = cy ( g ( x ) ) ,  ( 1 .2) 
where 9 is some function. In this thesis, we show that this conjecture holds true 
for such equations when 9 is entire. In fact, we will show that the functional argu­
ment 9 also heavily influences the global behaviour of any solutions, with equations 
generally admitting solutions with natural boundaries determined by g .  
2 Introduction 
The work is divided into two main sections. Chapters 2 and 3 deal with the first 
and second order pantograph equations, where we look at the case where lad > 1 
in more detail. Due to the lack of solutions holomorphic at x = 0, we look instead 
for solutions defined on the positive real axis, that satisfy boundary conditions at 
x = ° and x = 00. In Chapters 4, 5, and 6 we discuss the existence of holomorphic 
solutions about fixed points of g, and the analytic continuation of these solutions. 
For the first order pantograph equation, Hall and Wake [11] examined the case 
where c = ba, and found a Dirichlet series solution via Laplace transforms. Kim [19] 
showed further that there exists a unique solution in L 1  [0 , 00 ) , for a small range 
of c. We extend this work in Chapter 2, by showing via analysis of the related 
Laplace problem that, in general, there exists an infinite number of solutions to 
such equations. Uniqueness can be obtained, however, if the decay rate of solutions 
is restricted. The solutions found exhibit an interesting symmetry in that they can 
be written in the form 
00 
y(x) = L enj(anx) (1.3) 
n=O 
for some function j, where the en 's are constant. Motivated by the fact that solu­
tions of this form have been shown for some particular second order problems (cf. 
Kim [19]), we look to characterize first order equations that admit such solutions 
in Section 2.5. The Mellin transform is used here to convert the problem into a 
functional equation, and we show how the solutions to the Mellin problem can be 
used to find the form of the solution to the original problem. 
Chapter 3 extends the work presented in Chapter 2 to second order pantograph 
equations. We now have the possibility of two linearly independent solutions, one 
for each root of the associated indicial equation that has positive real part . In fact , 
we find our two solutions are linearly independent only if the roots r1 and r2 are 
such that r1 i= akr2 for all integers k. If r1 = akr2 for some integer k, then the 
second solution reduces to a multiple of the first . This is similar to the method of 
F'robenius for ordinary differential equations. B .  van-Brunt et. al. [36] studied the 
associated Laplace equation in the case where only one root of the indicial equation 
has positive real part , and we extend this work to cover the case where both roots 
have positive real part . This leads to the discovery of a second linearly independent 
solution in the case where the roots differ by an integral power of a, and thus gives 
an analog to F'robenius' method for pantograph equations. 
3 
In Chapter 4 we look at the more general equation ( 1 .2) where 9 is a non-linear 
polynomial. We seek solutions to ( 1 .2) holomorphic about points that are fixed 
under g. Given suitable initial conditions, we obtain a unique holomorphic solution 
if the fixed point Zo is attracting. The solution can then be analytically continued by 
iterating the functional argument g .  This leads to an overview of complex dynamics 
in Section 4.3, where we present well-known results based on the work of Milnor [26] . 
We find that the Julia set J(g) (or a subset of it) forms a natural boundary for our 
solution. Much of the material from this chapter has been published in the paper 
"Natural Boundaries for Solutions to a Certain Class of Functional Differential 
Equations" [22] , with the exception of the generalisations in Section 4.5 .  
The existence of holomorphic solutions about repelling fixed points is discussed 
in the first four sections of Chapter 5 ,  where we present work from the paper 
"An Eigenvalue problem for holomorphic solutions to a certain class of functional 
differential equations" [23] . As noted above, the pantograph equation admits non­
trivial holomorphic solutions in the case where the fixed point is repelling only under 
certain arrangements of the coefficients b and c. We generalize this to polynomial 
functional arguments, and show, through a modest extension to the work of Oberg 
[28], that we can rewrite the equation as the eigenvalue problem 
.cy(Z) = AY(Z) ( 1 .4) 
where £, is a compact operator. Thus, there is a discrete set of eigenvalues A 
such that ( 1 .4) has a holomorphic solution. Holomorphic solutions to the original 
equation, therefore, occur only as special cases. We illustrate this by finding the 
spectrum and associated eigenfunctions of the second order pantograph equation. 
Equations with non-linear polynomial functional arguments are then examined. 
Using the results on complex dynamics from Chapter 4, we show that generically, 
such equations have at most one non-zero eigenvalue. Finally, we exploit the results 
from compact operator theory (in particular the Fredholm Alternative) to establish 
an existence and uniqueness result for holomorphic solutions to non-homogeneous 
problems. 
We return to the pantograph equation in Section 5 .5 ,  presenting work from the 
paper "A natural boundary for solutions to the second order pantograph equation" 
[25] . We show that, if b > 0 and a > 1 is a positive integer, the Dirichlet series 
solution developed in Chapter 2 has the imaginary axis as a natural boundary. This 
4 Introduction 
result can be generalized by using the properties of iterated functions observed in 
Chapter 4. We show that any solution singular at the origin, but holomorphic 
in an infinite strip S whose boundary contains the imaginary axis, must have the 
imaginary axis as a natural boundary. This is done by transforming the problem 
into one involving a set whose boundary is the unit circle. The functional argument 
of the transformed differential equation then has a multivalent inverse, which is used 
to show that any solutions must have a natural boundary on the unit circle. We 
thus establish the imaginary axis as a natural boundary for solutions to the original 
problem. We conclude Chapter 5 by showing how the analysis can be extended to 
higher order equations by considering the second order pantograph equation. 
Chapter 6 presents results on holomorphic solutions about neutral fixed points 
from the paper "Holomorphic solutions to pantograph type equations with neutral 
arguments" [24]. We begin with a study of the pantograph equation in the case 
where l a l = 1 .  As in the attracting case, entire solutions are obtained. The order 
of the solutions, however, depends on the arithmetic properties of a .  A specific 
example is studied, and we show that if a is a root of unity, polynomial (order 
zero) solutions are obtained, whereas if a obeys a Siegel condition, the solution is 
of order p, where p E (0, 1 )  is bounded away from zero. 
Section 6.2 introduces the different types of neutral fixed point, including Siegel 
points and Cremer points. An overview of the local behaviour about parabolic 
points is presented, as are the well-known results on the characterization of Siegel 
and Cremer points in terms of continued fraction expansions. 
Holomorphic solutions about Siegel points (which are in the Fatou set ) are then 
sought in Section 6.3. A careful study of the results in Chapter 4 on the existence 
and uniqueness of holomorphic solutions about attracting points, shows that the 
key requirement is that 9 maps small discs about the fixed point into themselves, 
i.e . ,  for each 8 sufficiently small, g ( D ( zo ;  8) ) � D ( zo ; 8) where D ( zo ;  8) is the disc 
with centre Zo and radius 8. Attracting fixed points, however, obey the strict 
inequality g ( D (zo ;  8) ) c D ( zo ;  8 ) .  The allowance for equality leaves an opening for 
the Siegel case. We thus establish the existence of a unique holomorphic solution in 
the component of the Fatou set containing zo , called a Siegel disc. The boundary 
of the Siegel disc is then shown to form a natural boundary for the solution by 
applying the results on the Julia set from Chapter 4. We conclude Chapter 6 by 
showing that for equations with non-linear functional arguments, Siegel points are, 
5 
in general, the only type of neutral fixed point that admit holomorphic solutions. 
The final chapter contains conclusions along with suggestions for future work in the 
field. In particular, we discuss the likelihood of holomorphic solutions being found 
within the attractive petals of parabolic points. 
This thesis contains a number of images of the Julia sets of polynomials, which have 
been generated using a computer program written specifically for this purpose. The 
algorithm is detailed in Appendix A, along with notes on the techniques used to 
decrease computation time, and provide sharper images. 
6 Introduction 
CHAPTER 2 
First Order Functional 
Differential Equations 
In this chapter we consider equations of the form 
y'(x ) + p(x)y(x ) = q(x )y(ax ) , 
where p and q are functions defined on the positive real axis, and a is a positive 
real constant. We seek solutions defined on the positive real axis. We start by 
considering the simplest such equation, where the functions p and q are constant. 
2 . 1  The Pantograph Equation 
Consider the functional differential equation 
y'(x ) + by(x ) = cy(ax ), (2. 1 )  
where b and c are complex constants, with c #- O and a is a positive constant . We 
note that if a = I ,  then (2. 1 )  is a first order ordinary differential equation, which 
may be solved using the usual methods. We therefore assume that a #- 1 .  Equation 
(2 . 1 )  has been studied extensively by Derfel and Iserles [5] , Kato and McLeod [18] , 
and Iserles [16] , among others. The equation arises in many applications ranging 
from current collection in an electric train [29] to cell growth [11] . Second and 
higher order analogues have been studied by Iserles [16] , Kim [19] and Cooper et 
8 First Order Functional Differential Equations 
al. [37], among others. Iserles gives further references to the pantograph equation 
and its applications. 
We first seek a non-trivial solution to (2 . 1 )  holomorphic about the point x = o. 
There are two separate cases to consider, depending on whether a < 1 or a > 1 .  
Indeed, substituting a formal solution of the form y(x ) = L�=o anxn into equation 
(2 . 1 )  yields the recurrence relation 
for n = 0, 1 ,  . . . .  We thus obtain the formal solution 
00 nn-l ( k b ) 
( ) _ """' k=O 
ca  - n yx -ao� I x. 
n=O n. 
(2 .2) 
If a < 1 ,  then it is clear that the above series converges for all x E C so that 
(2 . 1 )  has a solution holomorphic in C given by (2 .2) where the constant a o is 
determined by an initial condition at x = O. If, however, a > 1 ,  then the series 
(2.2) diverges for all x =1= 0 unless one of the terms in the product is zero. Thus, 
if c =1= :&- for all non-negative integers k, then there is no solution holomorphic at Q 
x = O. If c = � for some non-negative integer k, then the series terminates, and Q 
the solution is a polynomial. We therefore conclude that only certain equations of 
the form (2. 1 )  yield a solution holomorphic at x = 0 if a > l. In fact, Kato and 
McLeod [18] showed that, though holomorphic solutions do not generally exist for 
the initial value problem, there are nevertheless an infinite number of solutions to 
(2 . 1 )  satisfying a specific initial condition y(O) = Yo =1= O. Thus the initial value 
problem is ill-posed. 
For the remainder of this chapter, we consider a boundary value problem for the 
case where a > l . In particular, we are interested in solutions to (2. 1 )  that satisfy 
the conditions 
y(O) = Yo and lim y(x ) = O .  
x->+oo 
(2 .3) 
where Yo is a constant, possibly zero. A.J .  Hall and G.C.  Wake [11] showed that a 
cell growth model based on a cell population structure with a single size variable x ,  
where each parent cell divides evenly into a daughters all of the same size, results 
in an equation of the form (2. 1 )  with c = ba  where b > O. They found a unique 
solution satisfying the boundary conditions (2.3) with Yo = O. In their model, y 
2.2 Dirichlet Series Solutions 9 
represented the steady size distribution and so they required y to be a probability 
density function; therefore, they had the further requirement that 
1
00 
y(x )dx = 1 .  
Their solution was of the form of a Dirichlet series 
00 
y(x ) = Lane-ranx, (2 .4) 
n=O 
where r and the an's are constants. They found their unique solution by transform­
ing the equation using the Laplace transform and solving the resulting equation 
explicitly. We will use this same technique in §2.4 when we investigate uniqueness 
criteria. 
2 .2  Dirichlet Series Solutions 
Let y be a solution to (2. 1 )  of the form (2.4) , where a > 1 .  Then 
and 
00 
y'(x ) = - L ranane-ranx , 
n=O 
00 ex:> 
y (ax) = L ane-ran+lx = L an_le-ranx . 
n=O n=l 
Substituting the series into (2. 1 )  gives 
00 00 
L (b - ran)ane-ranx = L can_le-ranx. 
n=O n=l 
Now the set of functions 
is functionally independent on [0, 00); hence, we can equate the coefficients of like 
exponentials to obtain 
(b - r)ao=O, 
10 First Order Functional Differential Equations 
.and 
(2.5) 
for n = 1 , 2, . . . .  For a non-trivial solution we require ak i= 0 for some k, hence 
r = :k and b i= O. The recurrence relation (2.5) then gives 
for n = k, k + 1, . . . , where we use the convention I1�=1 (1  - Qm ) = 1 .  Hence, 
which we can re-index to give the solution 
(2.6) 
where the constant A is arbitrary. Note that r = � for any non-negative integer 
k yields the same solution, so we will assume from now on that all such Dirichlet 
series have ao i= O. 
As Q > 1 ,  it is clear that the series converges uniformly for all x such that Re( bx) � 
0, and so we require Re(b) > 0 in order to satisfy the boundary condition y(oo) = O. 
The constant A then must satisfy 
Assuming the series above is non-zero, a constant A can be found so that y(O) = Yo. 
If the above series is zero, then we must have Yo = 0 if a Dirichlet series solution is 
to exist. The constant A can then be chosen arbitrarily. 
P. Fredrickson [10] used a Dirichlet series of the form 
00 
q;(x, (J) = L ane{Janx 
n=-oo 
involving an increased range of n to examine solutions to (2. 1 )  subject to the 
2.2 Dirichlet Series Solutions 1 1  
general initial condition y (O )  = k. He showed that if Ibl � Icl and Re(b) < 0 ,  
then there is only one Dirichlet series solution. If Ibl < Icl and f3 =J. bQ-N for a 
positive integer N, then the problem has a one-parameter family of Dirichlet series 
solutions in Lp[O, 00) where max(l, :�!(I�P < p � 00 if Re(f3) < O. Another solution 
when Ibl < lel suggested by Bowen (cf. L. Fox, D . Mayers, J. Ockendon and A. 
Taylor [9] ) has the form 
00 n 00 n-l b y (x) = A L IT her - l)-l (� )nelryonx + B LIT hQ-m - l) (�teb-yo-nx, n=O m=l n=l m=O 
where A,B are real constants, , =J. a-N for any positive integer N and Re(lry) < o. 
T.S.Ferguson [7] showed an example of the case when I bl = lel and b < 0 in a paper 
on probability theory. He showed that there is a unique solution to the equation 
y'(x) = y(ax) - y (x), 
satisfying the boundary conditions 
y(O )  = 0, y (oo) = l, 
such that the solution can be expressed as 
00 
y(x) = 1 - h L ene-onx, 
n=O 
For the case b = 0 and c = 1, G.  R. Morris, A. Feldstein and E. W. Bowen [27] 
showed the existence of a solution of the form 
00 
y(x) = L (_rota-in(n-l)e-rOO-nx, 
n=-oo 
for x E IR where ro is a constant, provided Re(rox) � O .  
Finally, for the more general case when b = 0, Bowen (cf. L. Fox, D. Mayers, 
12 First Order Functional Differential Equations 
J. Ockendon and A. Taylor [9] ) suggested a solution of the form 
n=-oo 
where f3 E lR and C{3 is a constant . 
2.3  Uniqueness 
From the last section it is clear that there exist solutions to ( 2 . 1 )  satisfying ( 2 . 3 )  
when a > 1 and Re b > o. We now seek conditions under which these solutions are 
unique. We do this by adding the further restriction that yE L l  [0, (0 ) , so that 
100 y (x ) dx < 00. ( 2 . 7) 
The dirichlet series solution ( 2 . 6 ) is uniformly convergent in the half-plane { z  : 
Re(bz) 2 O } ,  and can thus be integrated term by term to show that it satisfies 
( 2 . 7) .  
Integrating equation ( 2 . 1 )  from 0 to 00 yields 
c 100 y ( O) = (b - - ) y (x ) d x .  a 0 
Hence, if c -=1= ba, we may replace the boundary conditions ( 2 . 3 )  with the normalizing 
conditions 
100 y (x ) dx = Lo , lim y (x )  = 0, 
x-++oo 
( 2 . 8 )  
where Lo is some finite constant . 
If c = ba, it is clear that any solution in L I lO ,  (0 ) must have y ( O )  = o. As noted 
earlier, this is the case studied by Hall and Wake [ 1 1 ] ,  who showed the existence of 
a unique solution when Lo= 1 .  Kim [ 1 9] showed further that we can transform the 
problem into a Fredholm equation and thus obtain uniqueness for a range of values 
of c. 
2.3 Uniqueness 
Multiplying both sides of (2. 1 ) by the integrating factor ebx gives 
and integrating the above equation from 0 to x yields 
ebXy ( x )  = Yo + - ebay ( s) d s ;  
c lax · 
a 0 
therefore, y satisfies the Volterra equation 
Let f (x) = yocbx and Ky (x) = Jooo K ( x ,  s ) y ( s) d s  where 
Then y satisfies 
{ le -bCX- ;;- )  
K ( x , s) = a 
o 
if 0 < s < ax, 
if s > ax. 
y (x) = f (x )  + cKy (x) , 
13  
(2.9) 
(2 . 10) 
which is a Fredholm equation of the second kind. Let I l f l l = Jooo I f (x) l dx .  It is 
known that if K is a bounded operator with the property that 
M< 00, 
then equation (2 .10) has a unique solution y E L dO ,  (0) for all f E L1 [0 , (0 ) and 
sufficiently small lc l . We can determine an appropriate range of c by the Contraction 
Mapping Theorem. 
Theorem 2.1  If Re(b) > 0, a > 1 and le l < a Re(b) then there exists a unique 
solution to (2 . 1 ) in L1 [0, (0) satisfying (2.8) . 
Proof: For y E L I lO ,  (0) , the condition Ic l < a Re(b) guarantees that Yo is defined 
uniquely in terms of Lo . Let T be the operator defined by 
14 First Order Functional Differential Equations 
Then, 
I ITy l l::; lyol (Xl e -Re(b)xdx+ lS {'XJ {CX
X e -Re(b)(x-;-) ly (s) l dsdx . lo ex lo lo 
Exchanging the order of integration gives 
I IT Y I I ::; � + B 
{OO ( (OO e-Re(b)(X-;-)dX) l y (s ) l ds Re(b) ex lo l � 
Cl< 
IYo l  I c l (OO = Re(b) + ex Re(b) lo ly(s)lds 
IYo l  I cl = Re(b) + ex Re(b) I l y l l· 
Thus, T maps L 1 [0, 00) into L 1 [0, 00). 
For Y1, Y2 E LdO, 00), 
I IT Y1 -TY211 = '�'l°O 1 cx
X 
e -b(x-;-) (Y1 (S ) - Y2 (S) ) dsl dx 
s: '�'l°O (hOO e -R.(bIlX-!ldx) I Yl (S) - y, (s)lds 
Ic l 
ex Re(b) 
I IY1 - Y211· 
The result thus follows by the Contraction Mapping Theorem. • 
The Dirichlet series (2.6) is in LdO, oo ), and is therefore the unique solution to 
(2. 1 )  satisfying (2.8), if I cl < ex Re(b) . We can extend the range of c further if we 
add an additional decay condition on y . Let In [O, 00) be the set of functions f 
defined on [0,00) such that Xk f E L 1[O, 00) for k = 0, 1 ,  . . .  , n, where n is a non­
negative integer. On this space of functions we can impose the following normalizing 
condition 
lim y (x )  = 0, X"'" +00 
(2. 1 1 )  
where Ln is some finite constant . Note that Ln is the n-th moment of y . We 
can establish a relationship between the moments of Y by the use of the Mellin 
transform 
M(s) = 100 xS-1y (x )dx .  (2. 12)  
In general the Mellin transform is used with s E lR; however we restrict the analysis 
2.3 Uniqueness 15 
to s E Z because we are concerned only with the moments of y. To avoid confusion, 
we use the notation of (2.11) (i .e . Ls = M(s + 1) ). Use of the Mellin Transform 
will be further examined in §2.5 where we look at determining types of equations 
that exhibit solutions displaying certain symmetries. 
Multiplying (2.1) by XS and integrating between 0 and 00 gives 
Integrating the first term by parts and changing the variable of integration from x 
to ax in the last term yields 
Now, assuming that the integrals converge (i .e. that the s-th moment of y exists) , 
xS y (x )  � 0 as x � 00, and thus 
( b  _ _ c ) L = (}(.+1 L s-l s· 
S 
(2.13) 
Clearly if Ln exists for a non-negative integer n, then L o, L1 , ... , L n-1 exist, and 
can be found recursively using ( 2 .13). In particular, 
(2.14) 
The Dirichlet series (2.6) is in Is [0, 00) for all s 2: O. Indeed, 
SA � b -banx d = � � bQ s-l -ban xd 
100 00 (c)n A 
00 
( c )n 1
00 
X  ITn (1 ) e x b  IT n (1 ) x e x o n=O m=l - am n=O m=l - am 0 
A , 
00
( c )n = � � bas 
bs  IT n (1 - am) . n=O m=l 
which converges for all s as a > 1. We can extend Theorem 2.1 as follows. 
Theorem 2.2 Let Re (b) > 0, a > 1 and n be a non- negative integer. If Ic l  < 
an+! Re (b) , there exists a unique s o lution to ( 2.1) in I n [O , 00) satisfying ( 2 .11). 
Proof: As shown above, the Dirichlet series ( 2.6) satisfies the conditions of the 
16 First Order Functional Differential Equations 
Theorem, so we need only show uniqueness. When n = 0, In [O, oo) = LdO, 00 ) , 
and the conditions ( 2 . 1 1 ) and (2.8) are equivalent so that Theorem 2 . 1 applies. 
Let n = k be a positive integer greater than 1 and suppose the Theorem holds 
for n = k - 1 .  Suppose there exist two solutions Yl and Y2 to ( 2 . 1 )  in h[O, oo) 
satisfying (2 . 1 1 ) .  Let Z l (X )  = Jxoo Yl (t ) dt and Z2 ( X ) = Jxoo Y2 ( t ) dt . We show that 
Zl == Z2· Integrating ( 2 . 1 )  from x to 00 gives 
100 c 100 -y ( x )  + b y (t ) dt = - y ( t ) dt . 
x Q ax 
Making the substitution z(x) = Jxoo y (t ) dt yields the equation 
Z'(x) + bz(x) = �z(Qx) .  Q ( 2 . 1 5 )  
Clearly Z l  and Z2 are solutions to ( 2 . 1 5 ) , so it remains to show that Zl , Z2 E 
h-dO, oo) and that Z l , Z2 satisfy the same normalizing condition. For i = 1 , 2 
we have 
The integral on the left is a monotonically increasing function of m that is bounded 
above, and therefore it converges to some finite number. Hence Zl , Z2 Eh-dO, 00) . 
Also, for all m > 0, 
hence 
( 2 . 1 6) 
The Theorem holds for n = k - 1 by hypothesis, and so there exists a unique 
solution to (2 . 1 5 )  in h-l [O, 00) satisfying ( 2 . 1 6 ) .  Clearly Zl and Z2 satisfy these 
2.4 Laplace Analysis 17 
conditions and hence ZI Z2· Thus Yl = Y2 , and the Theorem is established by 
induction. • 
Note that the normalizing condition ( 2 . 1 1 ) may be replaced by the boundary con­
ditions ( 2 . 3 )  as long as Ln is uniquely defined in terms of Yo. Equation ( 2. 1 4 )  shows 
that this occurs as long as c =1= bak for positive integers k. We may thus rewrite 
Theorem 2 . 2  as follows 
Corollary 2.3 Let n be a non-negative integer, Re ( b )  > 0, a > 1 and c =1= bO'.k 
for k = 1 , 2 ,  . . .  , n .  If lel < an+1 Re(b) , there exists a unique so lution to ( 2 . 1 )  in 
In [0, 00) satisfying ( 2 . 3 ) . 
As shown above, the Dirichlet series solution (2 . 6 )  is  in In [O, 00) for all positive n .  
Hence, it i s  the unique solution specified in Corollary 2 . 3  for any n .  Therefore, any 
other solutions that exist must satisfy a maximal decay condition dependent on c, 
as they cannot be in In [O, 00) for 
In(�(lb) ) n >  In a - 1 .  
2 .4  Laplace Analysis 
In this section we use the Laplace transform to derive both the Dirichlet series 
solution presented earlier and more general solutions to the boundary value problem 
( 2 . 1 ) , ( 2 . 3 ) .  In particular, we seek solutions in LdO, 00 ) . 
Suppose Y E Ll [O, 00) is a solution to ( 2 . 1 )  satisfying ( 2 . 3) . Then y is continuous 
on the interval [0, 00) and hence the Laplace transform 
is defined for all p � O. Furthermore, L(p) must be holomorphic in the right half 
plane ITo = {p Ere: Re(p) > O} . Applying the Laplace transform to both sides of 
equation (2 . 1 )  and using ( 2 . 3) yields the functional equation 
c P pL(p) + bL (p) = -L( - ) + Yo, 
a 0'. ( 2 . 1 7) 
18 First Order Functional Differential Equations 
and the condition that yE L l [0,00) gives that 
lim L (p ) < 00. 
p-+o 
( 2 . 1 8 )  
We find solutions to equation ( 2 . 1 7) satisfying ( 2 . 1 8) by first constructing solutions 
holomorphic at p = 0 and then constructing the general solution. The reason 
for looking first at solutions holomorphic at the origin becomes apparent with the 
following result . 
Definition 2.4 (Exponential Decay) A function f defined on [0,00) decays ex­
ponentially if there exists positive constants A and r such that 
for all x E [0,00) . 
Theorem 2.5 Let y be any function in LdO,oo) satisfying ( 2 . 3 ) . Then y decays 
exponentially if and only if its Laplace transform is holomorphic at p = O. 
Proof: Let y be a function defined on the positive real axis, and suppose that there 
are positive constants A and r such that 
ly (x )1 < Ae-rx, 
for all x E [0, 00) . The Laplace transform L of y is given by 
For fixed x, the integrand is an entire function of p, thus expanding e -PX as a power 
series, and integrating term by term yields 
It e
-pxy (x )dX I s; � �!Ipnl t Ixny (x )ldx 
� 1 1 In An
! 
::;  I p n+l n .  r n=O 
_ 
A 00 ( Ip l )n --L: -r r n=O 
2.4 Laplace Analysis 19 
Thus, it is clear that L is holomorphic for Ip l < r. 
Suppose now that L (p) is the Laplace transform of a function y and that L is 
holomorphic at p = O. Then there exists an r > 0 such that L is holomorphic in 
the disc D(O ;  r) , hence the integral 
converges for all pE D (O; r) . In particular, it must converge for p = - � , and as y 
satisfies (2 .3), it must decay exponentially. • 
Solutions to (2 . 1) that exhibit exponential decay (e.g. the Dirichlet series solution) 
are characterised by their Laplace transforms being holomorphic at the origin. We 
have the following existence and uniqueness result . 
Lemma 2.6 There exists a solution to (2. 17) where a > 0 unique among functions 
holomorphic at p = 0) provided c =f=. bak for all positive integ ers k .  If c = bak for 
some positive integ er k then there exists a solution to (2. 1 7) holomorphic at p = 0 
only if Yo = O. This solution is unique up to an arbitrary multiplicative constant. 
Proof: Let L (p) = 2:::'=0 anpn where the an 's are constants. Substituting into (2. 1 7) 
and equating like powers of p gives 
and 
c (b - -)ao = Yo, a 
(2. 19) 
for n = 1, 2, . . . . Thus the an 's are uniquely determined if and only if c =f=. bak for 
all k = 1, 2, . . . . If c = bak for some positive integer k � 2 then ak-2 = . . . = ao = 0 
and Yo must be O. The constant ak-1 is then arbitrary. The recurrence relation 
(2. 19) then holds for n � k, thus the an 's are determined up to an arbitrary constant 
multiple. 
In either case, Cl! > 1 and (2. 19) implies 
. an li ( c )-1 1 hm -- = m -- - b = --. n-+oo an-1 n-+oo an+1 b 
20 First Order Functional Differential Equations 
The series thus has a non-zero radius of convergence; consequently, L is holomorphic 
at the origin as required. • 
Given that there exists a solution holomorphic at p = 0, we wish to continue it into 
a global solution, holomorphic in the right half plane ITo, so that it can be inverted 
to a solution of (2 . 1 ) .  We have the following continuation result. 
Lemma 2.7 The solution L to equation ( 2 . 1 7) that is holomorphic at the orig in 
can be continued to produce a function meromorphic in C .  The only singularities 
that L can have are simple poles at p = -ban for some n = 0, 1 , 2 ,  . . . .  
Proof: Evaluating equation ( 2 . 1 7) at � gives 
which, upon substitution in ( 2 . 1 7) gives 
Repeating the above steps n times yields 
( 2 . 20) 
We know there exists ab> 0 such that L is holomorphic in the disc Ip l < b. Choose 
p E D(O, b) . The right hand side of ( 2 . 20) is holomorphic in D(O, an+1b) except 
at the points {-barn }�=o where it has, at worst, simple poles. Thus L may be 
continued to a meromorphic function in D(O, an+1b) . Equation ( 2 . 2 0 )  holds for any 
n , and hence L must be meromorphic in C. • 
Since L is meromorphic in C, with possibly simple poles at p = -barn for m = 
0, 1 , 2 , . . .  , we must have 
00 
L(p ) = L :�an + g (p ), n=O p 
2.4 Laplace Analysis 21  
where 9 is an entire function and the an's are constants. In  fact , we find that 9 - 0 ,  
and the Mittag-Leffier expansion 
00 
L( ) _ "" an P - L +ban' n=O P 
( 2 . 2 1 )  
where the an's are constants suffices. We show this as follows. Substituting the 
Mittag-Leffier expansion into equation ( 2 . 1 7) gives 
and the recurrence relation 
for n = 1 , 2 ,  . . . .  Thus, 
Let 
_ 
(�)n an - ao TIn ( _ m) . m=l 1 a 
then (2 . 22 )  may be written as 
aoF(c )  = YO· 
(2 . 2 2 )  
(2 . 2 3 )  
Thus, assuming F(c ) i= 0 ,  ao can be found for any initial value Yo . The zeros of 
F, therefore, play the crucial role in determining whether a solution of this form 
exists. Setting q = � gives F in the form of a partition function 
which can be converted to a product using the Euler identity [2] . Recall that , 
for It I < 1 ,  I q l < 1 ,  the Euler identity is 
Therefore 00 c qn+l 00 c F(c ) = IT ( 1 - -b-) = IT (1 - ban)' n=O n=l 
( 2 . 2 4 )  
( 2 . 2 5 )  
22 First Order Functional Differential Equations 
provided that I b� I < 1. However, even if I b� I :2: 1 identity (2 .25) is still valid 
by analytic continuation, since F(c) is an entire function and the infinite product 
converges uniformly for all c E C. 
It is clear, then, that if c -=1= bak for all positive integers k, the Mittag-LefRer 
expansion (2 .21 )  is a solution to (2.1 7) holomorphic at p = O. In the case where 
c = bak for a positive integer k, we must have Yo = 0, and ao can be chosen 
arbitrarily. In either case, Lemma 2 .6 implies that it is the only such solution 
holomorphic at p = O. Assuming Re (b) > 0, we may invert (2 .21 )  term by term, 
giving the following result regarding solutions to (2.1 ) .  
Theorem 2 . 8  If Re (b) > 0 ,  a > 0 and c -=1= bak for all positive integ ers k, then 
there exists a unique solution to equation (2.1 ) satisfying (2.3) that decays expo­
nentially. If c = bak for a positive integ er k, then there exists a solution to (2.1 ) 
satisfying (2 .3) that decays exponentially if and only if Yo = O .  This solution is 
unique up to an arbitrary constant . 
We now drop the condition that L is  holomorphic at p = 0 and instead require 
only that the solution be holomorphic in ITo so that the transform can be inverted. 
Clearly the Mittag-LefRer solution (2 .21 )  is still valid, however there is now no 
guarantee of uniqueness. Let L denote a solution to (2 .17) satisfying (2 .18 ), Lh 
denote the Mittag-LefRer solution, and let Z = L - Lh . Then Z must satisfy 
c 
Z(p ) = 1 � � Z(� ) ,  (2.26) 
and we must have 
lim Z (p )  < 00. p-+o (2.27) 
The condition (2.27) precludes a pole at p = 0, but Z may have a branch point at 
p = O .  With this knowledge, we can find the general solution of (2 .26) . 
Let 
Z (p )  = W(p )U (p ), 
where 
(2 .28 )  
2.4 Laplace Analysis 23 
and 
In I�I + i4> 
K, = ----''-'''''-'-:--In(n ) . (2 .29) 
Here we are using pit = eltLog(P) , where Log(z) is the principal branch of log(z) and 
-1f < 4> = Arg(�) < 7f. Note that this precludes the case where E is a negative 
real number. We can however choose a suitable branch of log( z) to take care of 
this scenario. 
We have chosen W(p) to be a particular solution to (2 .26) . Indeed, 
p I pit 1+� W(�) = nit rrlO ( 1 +...E..-) = �W(p) , n=l botn 
and nit = �. Thus, substituting the expression for Z into equation (2 .26) gives the 
relation 
U (p) = U ( p ) .  n 
It is helpful at this point to classify functions with this property. 
Definition 2.9 (Log-Periodic Functions) A function u such that 
u (z) = u (qz) ,  
for all z E C where q > 1 is a positive real constant is said to be log-periodic 
with period q. 
We have the following result . 
Theorem 2 . 10 Let u (z) be a log -periodic junction, and suppose that 
lim u (z) = c, z-+o 
jor some finite constant c. Then u (z) - c for all z E C.  
Proof: Let u (z) be a log-periodic function with period q. Then 
z u (z) = u (  - ) , q 
24 First Order Functional Differential Equations 
for all z E C. Therefore 
z u (z) = u (-), qn 
for all n = 1 ,  2, . . .  , and so 
z 
u(z) = lim u (-) = u(O) = c, n--+oo qn 
for any z in C. Thus u(z) = c as required. • 
If Ic l � Iba l then Re(/'\:) � 0 so that limp-+o W(p ) does not exist . However, Z must 
satisfy (2 .27) , therefore, limp-+o U (p ) = 0 and thus U (p )  _ 0 by Theorem 2 . 10. If 
Ic l > Ibal then Re(/'\:) > 0 and so limp-+o W(p ) = 0, and thus Z satisfies (2 .27) as 
required. In summary, we have the following result on the solutions to the Laplace 
problem. 
Theorem 2.11  The general solution to (2 . 1 7) in ITo satisfying (2 . 18) 1,S of the 
form 
if le l � Iba l; and of the form 
00 
L( ) _ � 
an p 
- �p+ban ' 
� an p"U (p ) L (p ) = � p + ban + I1�=o ( 1  + &)' 
(2 .30) 
(2 .31 ) 
if Ic l > Iba l· The constants an and /'\: are defined by (2.22)) (2 .23)) and (2 .29)) and 
U is an arbitrary log -periodic function of period a) holomorphic in the ITo . 
Thus, in general, there is an infinite number of solutions defining the Laplace trans­
form if l e l > Iba l. However, if we define additional conditions at p = 0, we can 
obtain uniqueness for any value of c .  We have the following result . 
Theorem 2.12 There exists a unique solution to (2 . 1 7) in ITo satisfying (2. 18) 
such that 
(2 .32) 
exists) provided lel < Iban+l l) a > 1 and c =f:. bak for all k = 2, 3, . . .  , n . If c = bak 
for some k = 2, 3, . . .  ,n then the solution is unique up to an arbitrary constant. 
2.4 Laplace Analysis 25 
Proof: The Theorem is true for n = 0 by Theorems 2.11 and 2.8, so assume n > O. 
Let 
j(p) = pItU(p) , 
where U is a log-periodic function of period a, and 
1 
g (p) = n:=o (1 + �) 
so that Z(p) = j(p)g(p). It is clear that (2.32) exists for the Mittag Leffier solution 
(2.30), so we just need to check that limp->o Zen) (p) exists for p E rIo .  Now, 
(2.33) 
and it is clear that 9 is holomorphic at p = 0 as the infinite product is an entire 
function, non-zero in rIo as Re(b) > O. We have 
so that 
n k-l 
j(n)(p) = L IT (�- j)plt-ku(n-k) (p) , 
k=Oj=O 
n k-l 
j(n)(ap) = L alt-k IT (�- j)plt-ku(n-k) (ap) 
k=O j=O 
as u(n-k)(ap) = ak-nU(n-k) (p) . We may thus write 
where U is a log-periodic function of period a. From (2.29) we have 
ln l b�1 In(an) Re (�) = In(a) < In(a) = n, 
so for limp->o j(n)(p) to exist we require limp->o U(p) = o. Theorem 2.10 gives 
U(p) - 0 and so Z(p) _ O. The only solution, therefore, is the Mittag-Leffier 
solution (2.30), which is holomorphic at the origin, so we apply Theorem 2.8 to 
obtain the uniqueness result . • 
26 First Order Functional Differential Equations 
Although we clearly have an infinite number of solutions to the Laplace equation, 
these need not correspond to transforms of solutions to the boundary problem. We 
still need to show that the functions defined in (2.31) are invertible. Let H2(110) 
denote the Hardy space of functions g holomorphic in the right halfplane 110 = {z : 
Re (z ) > O} such that for any fixed J-L > 0 the integral 
is finite. We wish to show that (2.31) is in H2(110) in order to show that inversion 
is valid. We first require the following lower bound for the growth of polynomials. 
Lemma 2 .13 Let f (z )  = L-�=o fjzj be a polynomial of degree k, where fk =1= o. 
Then 
If (z )1 > I:k l lz lk, 
for all lzl > Ilkl max{ l fo l , IfIl,· · · , Ifk l}· 
Proof: Using the triangle inequality, we have 
I
f (z )
l
> If 1- � Jbl zk - k L Iz lk-j , J=O 
and so for Iz l > dkl max{lfol, I fI l,· · ·, Ifk l}, 
• 
Theorem 2.14 Let Re ( b) > 0, Cl > 1, and let L be a solution to (2.17) of the form 
(2.31) where the log-periodic function U is bounded in 110. Then L E H2(110). 
Proof: Let L (p ) = M (p )  + U (p )W (p ) where M is the Mittag-Leffier solution, W is 
as in (2.28), and U is log-periodic with period a and bounded in 110 (note that a 
2.4 Laplace Analysis 27  
constant function U satisfies this condition.) The series for M is uniformly conver­
gent in ITo, and each term is in H2(ITO)' hence M E H2(ITO) .  U(p) is bounded in 
ITo, so all that remains is to show that W E H2(ITO) .  
Let N be a positive integer such that N > Re(�) . Then, for pE ITo, 
eRe(� Log(p)) eRe(�) In Ipl-Im(�) Arg(p) IW(p)1 = n:=o 11 + �I n�=N+111 + �I < n:=o 11 + �I 
Now, Arg(p) > - 7r/ 2 so that 
and Lemma 2 . 1 3  gives 
for I p l sufficiently large. Hence, 
for some constant A and I p l sufficiently large. Thus W E H2(ITO) and the theorem 
holds. • 
Note that the condition that U is bounded does not preclude non-constant func­
tions. 
Lemma 2 .15  Let u be of the form 
00 2 n7ri 
u (z) = 
n�oo 
Wn exp( ln(a) Log(z)) , 
where the Wn 's are constants such that 
� Inl7r2 
n�oo 
Iwnl eXP( ln(a)) 
( 2 . 3 4 )  
28 First Order Functional Differential Equations 
converg es . Then u is log -periodic of period aJ bounded in ITa· 
Proof: Let u be of the form ( 2 . 34 ) . Then 
� 2mri u(az) = 
n�oo 
Wn e XP( ln(a) Log(az)) 
� 2mri = 
n�oo 
Wn e XP( ln(a) (Log(z) + In(a))) 
= u (z) ,  
so that u is log-pe riodic of period a.  Now, 
� 2mri lu(z) 1 �  Iwn l  exp(Re(ln(a) Log(z))) n=-oo 
� - 2mr = 
n�oo 
Iwn l exp( In(a) Im(Log(z))) 
� 2mr =  Iwn l e XP(ln(a) Arg(z)) n=-oo 
� ( I
n l7r2
) �  Iwn l  exp In(a) . n=-oo 
The series on the right converges, and thus u is bounded in ITo. • 
The significance of Theorem 2 . 1 4  is that we can invoke the Pale y-Wiener theorem 
(cf. Hoffman [14] ) to show that the re is a unique y E L2 [0 , 00 ) such that 
Such an inverse y must be a solution to ( 2 . 1 )  satisfying ( 2 . 3 ) .  
Theorem 2 . 1 2  shows that the D irichlet series solution ( 2 . 6 )  is the unique solution to 
( 2 . 1 )  satisfying ( 2 . 1 1 ) .  The above result extends this by giving an infinite number 
of solutions in L2 [0, 00) . One would expect that the condition that limp ...... o+ L(p) 
exists and is finite can be used to show that y is in L 1 [0 , 00 ) , however this require s 
the use of a Tauberian result and has not yet been  shown. Any inverse transforms 
of ( 2 . 3 1 )  that are in LdO, 00) , howe ver, must decay slower than x n1+1 as x ---+ 00.  
2.5  Equations with Variable Coefficients 
2 .5  Equations with Variable Coefficients 
29 
Note that both the Dirichlet series solution ( 2 . 6 ) , and the solutions to the Laplace 
equation (2 . 3 0 )  and (2 . 3 1 ) ,  have increasing powers of a present. Indeed, the Dirich­
let series solution takes the form 
00 
y (x) = L anyo (anx) , ( 2 . 3 5 )  
n=O 
where the an 's are constants, and yo (x) = e-bx • In this case, Yo is a solution to the 
homogeneous ordinary differential equation y' + by = O. In this section we consider 
more general equations, and derive conditions on the types of equation that admit 
solutions of this form. 
Consider the equation with variable coefficients 
y' (x) + p(x)y (x) = q (x)y (ax) , ( 2 . 3 6 )  
where p and q are functions holomorphic in the right half-plane. We wish to find 
conditions on p and q such that ( 2 . 36 )  admits solutions in Re (x) > 0 of the form 
( 2 . 35) , where Yo is a solution to the associated ordinary differential equation 
y' (x) + p(x)y (x) = O .  ( 2 . 3 7 )  
Substituting (2 . 3 5 )  into ( 2 . 36) gives 
00 00 
n=O n=l 
Now, Yo is a solution to ( 2 . 37) , so we have 
hence, 00 00 
n=O n=l 
30 First Order Functional Differential Equations 
Equating the coefficients of yo(anx) gives 
for n > O. Hence 
for n = 1 , 2, . . . .  Setting n = 1 we have 
1 ao 
p(ax) = -[P(x) - -q(x)] . a al 
Evaluating (2 .39) at ax, using n = 1 ,  gives 
al 2 q (ax) = -[p(ax) - ap(a x)] . ao 
Hence 
Let g(x) = x-�q(x) where 
/'\, 
= 
In(a) 
. 
Then 
Thus, q is of the form 
(2.38) 
(2.39) 
(2.40) 
by (2 .40) 
by (2.39) 
(2.41 ) 
(2 .42) 
where 9 is an arbitrary log-periodic function of period a. Using (2.39) with n = 1 
we have 
p(x) - p(ax) p(x) a� - aOa2 p(ax) 
- --
q(x) q (x) aOa2 q (ax) 
2 .5  Equations with Variable Coefficients 31 
Setting r (x) = ���� and using (2.41 )  gives 
Differentiating gives 
and thus r ' is of the form 
where U is an arbitrary log-periodic function of period a. Thus it is clear r can be 
written as 
r (x) = xK-1v(x) + C 
where v is log-periodic, and C is a constant of integration. The form of p is therefore 
1 p (x) = -w (x) + Cq (x) x 
for some log-periodic function w of period a. This leads us to the following result. 
Theorem 2 .16  Let q (x) = XKUl (X) and p (x) = �U2(X)+Cq (x) where ul and u2 are 
arbitrary log -periodic functions holomorphic in Re( x) > 0 with period a, and C and 
K, are constants such that C f:. 0 and K, > O. Let Yo be a solution to (2 .37) , bounded 
on [0, 00) and suppose the set of functions {yo (anx)} are functionally independent . 
Then the functional differential equation (2 .36) has a solution of the form (2.35) . 
Proof: Let y be of the form (2 .35) ,  where Yo is as in the hypotheses. We need only 
verify that the constants an may be found so that the series converges and is a 
solution to (2.36) . Substituting into equation (2.36) yields 
00 00 
n=O n=l 
As Yo is a solution to (2.37) , we have that Yb (anx) = -p (anx)yo (anx) giving 
00 00 
n=O n=l 
32 First Order Functional Differential Equations 
Now, 
which gives 
00 00 
n=O n=l 
Hence we must have 
so that 
00 1 
y(x) = ao � en I1�=o( l - a( 1+K)m) yo (a
nx) . (2 .43) 
As a > 1, r;, > 0 and Yo is bounded on [0, 00) ,  the series (2 .43) converges for all 
x E [0, 00) .  • 
In the special case where r;, is an integer, and Ul and U2 are constant, we can 
transform the equation into one with constant coefficients. Consider the functional 
differential equation 
(2.44) 
where A ,  B and C are complex constants, k is a positive integer, and a > 1 is a 
constant , along with the conditions 
lim xAy(x) = Yo 
X--->O+ 
and lim xAy(x) = O. 
X--->OO 
( 2 .45) 
where Yo is some constant. Applying the transformations z = ��; and w (z) = 
xAy(x) to (2.44) yields the equation 
C w'(z) + Bw(z) = -w(ak+lz) aA 
(2 .46) 
which is of the same form as (2. 1 ) .  The boundary conditions (2 .45)  transform to 
the conditions 
w(O) = Yo and w (oo) = 0. (2 .47) 
Thus the results of the previous section can be readily extended for solutions to 
(2.44) . For example we have the following existence and uniqueness result. 
2.5  Equations with  Variable Coefficients 33 
Theorem 2 .17  If Re (B) > 0, a > 1 ,  and IC I  < Re (B)a Re(A) +k+l then there exists 
a unique solution to (2 .44) satisfying (2 .45) in the space of functions g such that 
xAg E L dO ,  (0) . 
Proof: Let Yl (X) and Y2(X) be solutions to (2.44) satisfying (2.45) such that XAYl, 
XAY2 E LdO, (0) . Let Z = ��:, Wl (Z) = XAYl (X) and W2 (Z) = XAY2(X) . It is clear 
that Wl and W2 are solutions to (2.46) satisfying (2 .47) and that Wl, W2 E L d O ,  (0 ) . 
Theorem 2. 1 thus implies that Wl - W2 and hence Yl = Y2 . • 
In order to better understand the symmetry in powers of a that arises, we look at 
the Mellin Transform of y . 
2 .5 . 1  The Mellin Thansform 
We begin with an example. Suppose we wish to find an exponentially decaying 
solution to (2.44). Multiplying (2.44) through by xs- 1 and integrating from 0 to 00 
gives 
Integrating the first term by parts, and changing the variable of integration from x 
to ax in the last term yields, for s � 1 ,  
lim xS-ly (x) + (A - s + 1 ) 100 xS-2y (x)dx + B (OO xS+k-ly (x)dx 
X--+OO 0 lo 
c 100 = s+k (axy+k-ly (ax)d(ax), a 0 
and, as we are looking for solutions that decay exponentially, limx--+oo xS-1y (x) = O .  
Using the Mellin transform as defined in (2. 12) ,  we therefore have 
(s - A - l )M (s - 1 )  = (B - a�k ) M (s + k) , (2 .48) 
for s � 1. This is a difference equation in s, for which a solution may be found as 
follows. Let M (s ) = F(s)Q(s ) where F satisfies the difference equation 
(s - A - l)F(s - 1 ) = BF(s + k) . (2.49) 
34 First Order Functional Differential Equations 
Then, assuming that (s - A - l ) F (s - 1 ) is not identically zero, (2 .48) reduces to 
(2 .50) 
Splitting up M in this way is useful, as F has been chosen so as to be independent of 
a. In fact , F corresponds to the Mellin transform of a solution to the homogeneous 
equation (2.37) , as (2 .49) is the transform of (2.37) . We can therefore find F by 
solving the difference equation (2.49) , or by finding the Mellin transform of the 
solution to (2.37) . Either method yields 
s - A  
F(s) = (k + 1 ) k+l r ( s - A ) .  B k + 1  
Rewriting (2.50) using s in place of s - 1 gives 
Q (s) = (1 - Ba�k+ l
) 
Q (s + k + I ) ,  
thus 
Q(s) = (1 - Ba�k+ l
) (1 - Bas�(k+ l ) ) Q (s + 2(k + 1 ) ) ,  
and therefore a solution to (2 .50) is 
The difference equation (2.48) gives information about M only at values of s sep­
arated by intervals of length k + I ,  i .e. , if we know M at a particular point So , 
then equation (2.48) will yield information about the function only at the points 
So + n( k + 1 )  for integers n. We may therefore define M arbitrarily on any interval 
[so, So + k + I ) ,  and can thus write the general solution to (2.48) as 
( k + 1 ) �+� s - A 00 ( C ) M(s) = U (s) � r( k + 1 ) IT 1 - B as+n(k+ l ) . (2 .51 )  
where U is an arbitrary periodic function of period k + 1 .  Note, however, that 
this solution to (2 .48) may not correspond to the Mellin transform of a suitable 
solution to the original problem. Indeed, the addition of the arbitrary function U is 
not required when we consider that the inversion will only require knowledge of M 
2 .5  Equations with Variable Coefficients 35 
at the points s = n( k + 1) for all positive integers n. This is equivalent to knowing 
every k+1-th moment of some unknown function y and, using only this information, 
determining y completely. This is related to the well-known "Moment problem" 
(cf. Hausdorff [12] , [ 13] ) .  The moment problem consists of finding conditions on 
the sequence of numbers {JJ,d�o such that there exists a function a(t) of bounded 
variation in the interval (0, 1 )  such that 
for all non-negative integers k. A variation to this problem, involving finding a 
function given information about its Laplace transform at the origin, is considered 
by D.V. Widder (cf. Widder [38] pp. 1 00- 10 1 ) .  We exploit the relationship between 
the Mellin transform and derivatives of the Laplace transform at the origin to obtain 
the following result. 
Theorem 2 .18  Let M be the Mellin transform of an unknown func tion y that 
satisfies 
k l y (x)1 < Ae-rx , 
for constants A > 0, l' > 0 and some p ositive integ er k . If M (nk) is known for all 
positive integ ers n, then y can be found comp le te ly. 
Proof: From the definition of the Mellin transform, we have 
Setting x = f (w ) = W1/k gives 
M (nk) = 100 wn-1y (f (w ) )dw . 
We therefore know all the moments of the function Y(w) = y (f (w ) ) .  Furthermore, 
as Y satisfies 
IY (w)1 < Ae-rw , 
for some constants A > 0, l' > O. Therefore, 
1 00 (_pw)n Y(w )dw . 1 < A 100 Ip l
nwn e -rWdw < A lp l
n
, n' n' rn+1 o · o · 
36 First Order Functional Differential Equations 
so that the series 
f ( _�) n M ( (n + 1 ) k) = f 100 ( _p�) n Y(w)dw 
n=O n . n=O 0 n. 
converges absolutely for I p l  < T.  We may therefore exchange the order of summation 
and integration to obtain 
� ( -�l " M ( (n + I ) k) = f (� (_��)n) Y(w)dw 
= 100 e-pwY(w)dw = L (p ) , 
where L is the Laplace transform of Y. The Laplace transform L of y is therefore 
defined uniquely, and as y decays exponentially, is holomorphic in ITo . We may 
therefore invert L to find y [32] . • 
We may therefore drop the arbitrary periodic function U in favour of a constant 
K. We thus have M(s) = KF(s)Q(s) where a is only present in the Q term. The 
inverse of F is the solution Yo to the homogeneous equation (2 .37) ,  whereas the 
inverse of M is a solution to the functional differential equation (2 .36) .  The Q 
term, therefore, can be regarded as containing information about how Yo should be 
modified to produce a solution to (2.36) . 
Let Mo ( s )  = KF(s) , (3 = ak+l and let 
for n = 1 , 2, . . .  , so that M (s) = limn-too Mn (s ) . We investigate the form of the 
solution Y to (2.44) by examining the inverses of Mn in turn. 
Clearly Mo inverts to the solution Yo of the homogeneous ODE (2.37) . Define Yn(x ) 
as the inverse of Mn (s) for n > O. For any function h where the integrals are finite, 
100 1 100 x S-1h (ax )dx = ----; (ax )S-lh (ax )d(ax ), 
o a 0 
(2.52) 
2.5 Equations with Variable Coefficients 37 
so that ;. Mn-1 (s) inverts to Yn-l (ax) ;  hence, 
C 
Yn(x) = Yn-l (X) - B{3n Yn- 1
(ax) .  (2 .53) 
Solving this recursion for the first few values of n yields 
Yl (X) = Yo(x) - � (�) Yo(ax) 
Y2 (X) = Yo(x) - � (� + ;2) Yo(ax) + (�) (;3) Yo(a2x) 
C ( l 1 1 ) (C)
2
( 1 1 1 ) 2 Y3 (X) = Yo(x) - B {3 + {32 + {33 Yo (ax) + B {33 + {34 + {35 Yo (a x) 
_ (�) 
3 
U.) yo(a3x) .  
While it is not immediately obvious what the exact expression for Yn will be, it is 
clear that it will be a linear combination of the terms Yo(akx) for k = 0 , 1 ,  . . .  n.  
Thus, as expected, the inverse Mellin transform of M is a series in yo(anx) .  We can 
find the series exactly by rewriting the product Q(s) as a sum, and then inverting 
term by term. 
Setting c = B�' and {3 as before, we can write 
using the Euler Identity. Thus, 
where 
00 1 
M(s) = K � an-F(s) , � ans n=O 
_ 
(� )n 
an - rr�=l ( 1  - {3m) '  
and K is an arbitrary constant. Replacing a in equation (2.52) with an and invert­
ing term by term yields the inverse transform of M (and thus a solution to ( 2 . 44 ) )  
as 
38 First Order Functional Differential Equations 
While the analysis above applies to one specific example, the principle holds for 
all functional differential equations. Assuming we can solve the associated Mellin 
equation, and that the solution found can be split into a product Q(s)F(s) where 
F(s) is independent of a,  then Q will give an indication as to how to alter the 
inverse transform of F in order to obtain a solution to the original equation. 
We end this section with a necessary condition for equation (2 .36) to yield solutions 
of the form (2.35) . 
Theorem 2.19 There exists a solution to (2.36) of the form (2 .35 ) , where Yo is an 
exponentially decaying solution to the homog eneous equation (2.37) , only if there 
exists a solution M to the associated Mellin equation that can be written in the form 
M (s) = F(s)Q(s), 
where F is independent of a,  and Q is holomorphic in a-s . 
Proof: Suppose y is a solution to (2.36) of the form (2 .35), where Yo is an exponen­
tially decaying solution to (2.37) . Then y must also decay exponentially, so that 
the Mellin transform M of y is defined for all s � 1 .  We therefore have 
As Yo decays exponentially, we may integrate the series term by term so that 
00 
= L ana-SnF(s) . 
n=O 
where F is the Mellin transform of yo, which is necessarily independent of a .  Thus, 
M has the form required. • 
Consider the equation 
CHAPTER 3 
Second Order 
Pantograph Equations 
y"(x) + ay' (x) + by (x) = ,Xy (exx) , (3. 1 )  
where a , b, ,x ,  and ex are real constants such that ,x #- 0 and ex > 1 .  As in the 
first order case, ex > 1 implies that the only solutions holomorphic at the origin are 
polynomials. This will be shown in detail in Chapter 5 .  For now, we concentrate 
on finding solutions defined on some halfplane, and for definiteness, we require 
solutions to be defined on the positive real axis. 
3 . 1  Dirichlet Series Solutions 
We expect solutions similar to those derived for the first order case, and therefore 
look for solutions of the form 
00 
y (x) = L ene-a
nrx , (3.2) 
n=O 
where r and the coefficients en are to be determined. Substituting the above series 
into equation (3. 1 )  leads to the recurrence relation 
(3.3) 
40 Second Order Pantograph Equations 
for n = 1 , 2 ,  . . .  , and the indicial equation 
r2 - ar + b = O. (3 .4) 
A formal solution of (3. 1 )  is thus given by 
(3.5) 
where Co is an arbitrary constant, and r is a solution to (3.4) . This series converges 
uniformly in the halfplane Re( rx) � 0, and thus it is clear that the nature of the 
solution is determined by the roots of the indicial equation (3.4) . This technique 
is analogous to the Frobenius method of finding solutions to ordinary differential 
equations, and as we shall see, there are a number of cases to consider. 
The first complication encountered is the presence of two roots to the indicial 
equation (3 .4) . This allows the possibility of two linearly independent solutions to 
our functional differential equation. There are three cases to consider: 
1 .  No solutions of the form (3.2) exist (no roots to (3.4) such that Re(r) > 0) . 
2. One solution of the form (3.2) exists (one root to (3.4) such that Re(r) > 0) . 
3. Two solutions of the form (3.2) exist (two roots to (3.4) such that Re(r) > 0) . 
The first case leads to a divergent Dirichlet series and is consequently of little 
interest . The second case is satisfied if a > 0 and b < 0, and we can find similar 
results for existence and uniqueness (or lack thereof) as those found in Chapter 2 
to the first order problem. Kim presents second order analogues to Theorem 2 . 1  
and Theorem 2 .2  ( [ 19] pp. 48,49) and Laplace analysis similar to  that done in 
Section 2.4 is covered by van-Brunt et . al. [36] where analogues to Lemma 2.6 and 
Theorem 2. 1 1  are presented. 
We briefly examine the third case, where the possibility of two Dirichlet senes 
solutions exists. If we have 2 roots to the real quadratic (3.4) ,  both with positive 
real part, then it is clear that both must be real and positive, or they are complex 
conjugates. 
3.1  Dirichlet Series Solutions 41 
Suppose there exists distinct positive real roots TI and T2 , where TI > T2 ,  to the 
indicial equation (3.4) . Then two convergent Dirichlet series solutions Y I (x) = 
L:�=o ene-anTlx and Y2 ( X )  = L:�=o dne-anT2x corresponding to TI and T2 can be 
constructed. However, we must be careful with the treatment of Y2 when TI = anT2 
for some n E N since 
and hence d':l has a singularity. This is analogous to the case where roots differ 
by an integer in the Frobenius method. Note that we have no problem in the 
construction of YI as 
2n 2 n b ( n ) (  n ) a T I - aa T I  + = a TI - TI a TI - T2 
= ( anTI - T I ) (a2n T2 - T2 ) 
= T I T2 ( an - 1 ) (a2n - 1 ) > O. 
Indeed, as in the Frobenius case, Y2 is just a constant multiple of YI .  Suppose that 
TI = akT2 for some k E N . The recurrence relation for the dn 's gives 
and thus dk- I = O. We can then follow the recurrence relation back and find that 
dn = 0 for 0 ::; n ::; k - 1 .  Thus, 
00 
Y2 (X) = L dne-anT2x 
n=k 
00 
- ""' d -an-kTlX - � ne , 
n=k 
00 
- ""' d e-anrlx - � n+k , 
n=O 
and so the constants dn+k must be identical up to a multiplicative constant to 
the constants en. Thus Y2 = AYI and only one linearly independent solution ex­
ists. With the method of Frobenius, a second linearly independent solution can be 
found to the ordinary equation by multiplying the first solution by an appropriate 
logarithmic function. An analogous method for the pantograph equation will be 
42 Second Order Pantograph Equations 
shown in the next section on Laplace analysis. 
In the case where the roots differ by a non-integer power of ex,  we get linear inde­
pendence. Indeed, for large x, the behaviour of the solutions is dominated by the 
first term in the series, i .e . ,  
Yl (X) '" coe-TIX and Y2 (X) '" doe -r2x 
Y� (x) '" -coTl e-r1x and y; (x) '" -doT2e-r2x 
as x --t 00. Therefore the Wronskian of Yl and Y2 becomes 
hence, Yl (x )Y� (x) - y� (x )Y2 (x) is not identically zero for all x, and hence Yl and Y2 
are linearly independent . Thus, there exists an infinite number of solutions of the 
form y(x) = Yl (X) + Y2 (X) since Co and do are arbitrary constants. 
Lastly, we consider the case where Tl and T2 are complex conjugates. The solutions 
Yl and Y2 are then complex valued functions. We let Tl = U + iv where u = � and 
v = J4b2-a2 , and thus T2 = u - iv . The constants en and dn are also complex, and 
we find that dn = Aen for some constant A. Indeed, the recurrance relation for en 
gives 
and so the dn and en follow the same recurrence relation, and are thus identical up 
to a multiplicative constant . We thus find that 
00 
Y2 = A L cne-anrlx 
n=O 00 
= A L -en-e--a-:::n-rl-x 
n=O 
The real and imaginary parts of Yl (or Y2) provide two linearly independent real 
3.2 Laplace Analysis 43 
solutions. Let en = CIn + i C2n . These solutions can then be expressed as 
00 
WI (X )  = Re (I:: ( c In + i C2n ) e -an (U+iV)X ) 
n=O 
00 
= I:: ( C In cos(an v x )  + C2n sin ( anv x ) ) e -an UX 
n=O 
and 
00 
W2 (X )  = Im(I:: (cIn + i C2n ) e -an (u+iv)X ) 
n=O 
00 
= I:: (C2n COs ( an v x )  - c In sin ( an v x ) ) e -an ux . 
n=O 
A study of the long term behaviour of these functions indicates that they are linearly 
independent. 
3 .2  Laplace Analysis 
As with the first order case, transformation of the equation using the Laplace 
transform is useful in further tracking both the Dirichlet series solutions and other 
possible solution forms. Laplace analysis of equations of the form ( 3 . 1 )  has been 
studied by van-Brunt et . al. [36] , although they consider only the case where there 
is a single root of the indicial equation ( 3 . 4 )  with Re(r) > 0, and where solutions 
were subject to the initial conditions y (O) = 0 and y' (O) = � - b. We look at the 
case where both roots to the indicial equation ( 3 . 4 )  have Re(r) > 0 and derive a 
further solution form in the special case where the roots of ( 3 . 4) differ by an integer 
power of a .  
Consider equation ( 3 . 1 )  along with the initial conditions 
y(O) = Yo , y' (O) = YI . ( 3 . 6 )  
Applying the Laplace transform to both sides of equation ( 3 . 1 ) , we get the func­
tional relation 
L(p) = �
L(;) + (p + a)yo + YI 
(p + rl ) (p + r2 ) ( 3 . 7) 
44 Second Order Pantograph Equations 
where rl and r2 are the roots of the indicial equation ( 3 . 4 ) . Equation ( 3 . 7) is an 
example of a class of equations known as 'non-homogeneous q-difference equations' 
with q = � < 1 .  These equations have the form 
J(p )Y (qp ) + g (p )Y(p ) + h (p ) = O. 
Non-homogeneous q-difference equations have been studied by Adams [ 1 ]  in addi­
tion to other researchers for the cases where the coefficients are holomorphic at 
p = O. In particular, Adams shows that there are meromorphic solutions to equa­
tions such as ( 3 . 7) that are holomorphic at p = O. Indeed, a simple power series 
substitution as was done with the first order problem in Chapter 2 is sufficient to 
show this . We have the following result . 
Lemma 3 .1  There exists a unique holomorphic solution to ( 3 . 7 ) among Junctions 
holomorphic at p = 0, provided A =I anb Jor n = 1 , 2 ,  . . . . 
Proof: Substituting a formal power series expression L (p )  = 2::=0 Cnpn into (3 .7) 
and equating like powers gives 
and the recurrence relation 
A 
(b - -)co = YI + ayo, a 
A (b - 2 )CI = Yo - aco, 
a 
A 
(b - an+! )Cn = -aCn-1 - Cn-2, ( 3 . 8 )  
for n = 2 , 3, . . . . It is clear that as long as A =I :n for n = 1 , 2 ,  . . .  , these relations 
uniquely define the Cn 's, so we just need to check that the series converges. We do 
this by majorizing the series. Let Co = Ico l  and Cl = l e l l  and define the rest of the 
Cn's by the recurrence relation 
where b = minnEN{ I A  - :n I } .  Clearly the series F(p ) = 2::=0 Cnpn majorizes our 
formal power series solution. It is also clear that F(p ) is the solution of the equation 
bF(p ) - CIP - Co = l a lp (F(p ) - Co ) + p 2 F(p ), 
3.2 Laplace Analysis 
which gives 
F (p) = ( CI �- Co l a l )p + Co . 
b - I a lp - p2 
45 
Since A =1= blan for all n E  N, we have b > 0;  thus, F is holomorphic at p = O .  Now, 
the power series for F majorizes our formal power series, so that L must also be 
holomorphic at p = o. • 
Note that in the case where A = ban for some positive integer n, Cn-I is not 
uniquely defined, and can be chosen arbitrarily. This however forces a compatibility 
condition between the initial conditions Yo and YI , and the constants a, b and a,  
that must be met. Assuming that this compatibility condition is met, the solution 
is still holomorphic at p = 0 and is unique up to the multiplicative constant Cn-I .  
The solution holomorphic at the origin can be continued to a solution meromorphic 
in the complex plane as shown by the following Lemma. 
Lemma 3.2  The solution L to equation ( 3 . 7) that is holomorphic at the origin can 
be continued to a function meromorphic in C .  If L has any singularities, then they 
must be simple poles at p = -anrI or P = - an r2 for some n = 0, 1 , 2, . . .  / unless 
rI = akr2 for some integer k .  In this case L may have double poles at p = - an r I . 
Proof: Evaluating equation ( 3 . 7) at pia gives 
which, upon substitution in ( 3 . 7) gives 
Repeating the above steps n times yields 
We know there exists a 8 > 0 such that L is holomorphic in the disc Ipl < 8 .  Choose 
46 Second Order Pantograph Equations 
any complex number p and choose n so large that 
l a�+1 1 < 8. 
If p =j:. -rIam and p =j:. -r2am for all m = 0 , 1 , 2 , . . .  , n , then ( 3 . 9) implies that L 
must be holomorphic at p. If p = -amrl or P = -amr2 for some m = 0 , 1 , 2 ,  . . .  , n ,  
then p can, at worst , be a pole of order 1 ,  assuming rl =j:. akr2 for all integers k .  If 
rl = akr2 for an integer k then we obtain at worst poles of order 2. We may choose 
p to be any point in the complex plane, hence L must be meromorphic in C. • 
Since L is meromorphic in C,  with possible poles at p = amrl and p = amr2 for 
m = 0, 1 , 2 , . . .  , we anticipate a solution in the form of a Mittag-Leffier expansion 
( 3 . 1 0 )  
in the case where rl =j:. akr2 for all integers k ,  and of the form 
( 3 . 1 1 )  
when r l = akr2 for some integer k. 
We may assume without loss of generality that I r l l � I r2 1 .  Thus, there are 3 cases 
to consider: 
2. rl = akr2 for some positive integer k ,  
We treat each of these cases in a separate section in order to simplify the notation. 
3.2 Laplace Analysis 
3.2 .1  The case where Tl i- T2ak for all integers k 
47 
Substituting the expression ( 3 . 1 0 )  into equation ( 3 . 7) and equating principal parts 
yields the recurrance relations 
and 
ACn- l  
dn = 
Adn- l 
( rl - r2cxn ) (r2 - r2 cxn ) , 
for n = 1 , 2, . . . . The coefficients Co and do must satisfy the relations 
and 
( 3 . 1 2 )  
( 3 . 1 3 ) 
It is clear from the recurrence relations for the Cn and dn that the series will converge 
to a meromorphic function in re with the correct poles, and so all that is left to do 
is verify that there exist a Co and a do so that ( 3 . 1 0 )  is a solution to ( 3 . 7) .  
Adding equations ( 3 . 1 2 ) and ( 3 . 1 3 )  yields the relation 
which when combined with the recurrence relations for Cn and dn gives 
00 00 
L Cn + L dn = Yo · 
n =O n=O 
Similarly, subtracting ( 3 . 1 2 )  and ( 3 . 1 3 ) yields the further relation 
00 
( 3 . 1 4 )  
(rl - r2 ) ( cO - d o )  = (rl + r2 - 2a) yo + 2Y l  - L { cn (2rl cxn - a) + dn (2r2 CXn - an , 
n= l 
which can be simplified with the use of ( 3 . 1 4 )  and the fact that rl + r2 = a to the 
48 
expression 
Let 
and 
Then, 
and 
00 
Second Order Pantograph Equations 
00 00 
Tl L anen + Tl L Qndn = -Yl · 
00 
n=O n=O 
00 
L en = COF(A) , L dn = doG()") ,  
n=O n=O 
00 
( 3.15) 
L Qnen = coF(a).. ) ,  L andn = doG(a).. ) .  
n=O n=O 
The relations ( 3.14) and ( 3.15) can therefore be reduced to the conditions 
COF(A) + dOG(A) = Yo , ( 3.1 6 ) 
and 
( 3.17 ) 
For a fixed value of A, equations ( 3.16 ) and ( 3.17 )  form a system of linear equations 
for the unknown constants Co and do. The constants are determined uniquely 
provided 
( 3.18) 
To check that ( 3 . 18) is satisfied for a given value of ).. , we use the inherent recurrence 
in the terms of both F and G to establish functional identities that reduce ( 3 . 18) 
to something more tractable. Indeed, we see that 
= )"F()") . 
3.2 Laplace Analysis 49 
so by replacing A with � and using the relationships b = TlT2 and a = Tl + T2 , we 
have the functional identity 
(b - �) A 
TlF(aA) - aF(A) = - Q F( - ) .  Tl a 
Similarly, we can establish the identity 
(b - �) A 
T2G(aA) - aG(A) = - Q G( - ) .  T2 a 
Multiplying the first identity by G(A) and subtracting the second identity multiplied 
by F(A) gives 
Applying this relation n times yields 
As n ---t 00 we have that F( Q>'n ) ---t F(O) = 1 and G(Q>;") ---t G(O) = 1 ;  hence, 
00 A 
TlF(aA)G(A) - T2F(A)G (aA) = (Tl - T2) IT ( 1  - amb ) '  
m=l 
( 3 . 1 9 )  
Since Tl - T2 =I- 0, condition ( 3 . 1 8 )  is satisfied for all A E C unless A = amb for some 
positive integer m. 
In summary, provided that A =I- amb for any positive integer m, Lemma 3 . 1  implies 
that the Mittag-Leffier expansion ( 3 . 1 0 ) is the only solution to equation ( 3 . 7) that 
is holomorphic at the origin. In the case where A = amb for some positive integer 
rn, the coefficients Co and do are not determined uniquely, and furthermore the 
compatibility condition 
( 3 . 2 0 )  
must be met. Assuming that Yo and Yl satisfy ( 3 . 20 ) ,  the solution is unique up to 
an arbitrary constant. 
50 Second Order Pantograph Equations 
3.2 .2  The case where rl  = r2ci for some positive integer k 
The second case is similar to case 1 ,  although there is a coupling introduced between 
the constants en and dn that makes life a little more difficult . 
Substituting the expression ( 3 . 1 1 ) into equation ( 3 . 7) and equating principal parts 
yields the recurrance relations 
and 
>..dn-1 + (2T2an - a)cn-k 
(Tl - T2an ) (T2 - T2an ) 
0 < n < k ,  
n > k ,  
where Co = Adk_ 1 • The coefficients do and dk must satisfy the relations T2-Tl 
and 
( 3 . 2 1  ) 
( 3 . 2 2 )  
It is clear that the Mittag-Leffier expansion converges to a meromorphic function 
in C with the correct poles, but we need to verify that there exist constants do and 
dk such that ( 3 . 1 1 )  is a solution to ( 3 . 7) .  
Subtracting equations ( 3 . 2 1 )  and ( 3 . 2 2 ) , and using the recurrence relations for en 
and dn gives 
00 
L dn = Yo· 
n=O 
Similarly, adding ( 3 . 2 1 ) and (3.22) gives 
00 00 
2 L Cn - 2:)2T2an - a)dn = ayo + 2Yl , 
n=O n=O 
( 3 . 2 3 )  
3.2  Laplace Analysis 
which can be simplified with the use of ( 3 . 23 )  to give 
00 00 
L en - r2 L andn = YI . 
n=O n=O 
Solving the recurrence relations for the en and dn , we find 
and 
Thus, 
where 
and 
en = 
TIn ( m ) (  )
co , 
m=l rl - rIa r2 - rIam 
>.n 
TI�=I (rl - r2am ) (r2 - r2 am) 
do 
n 
IT (rl - r2am) (r2 - r2am ) 
m=k+ 1 
00 
L dn = dk F(>.) + doG(>') , 
n=O 
n 
51  
( 3 . 24 )  
n < k 
n > k .  
IT ( r l  - r2am) (r2 - r2am) 
m=1 
Similarly, we find 
and 
00 
m=1 mfk 
L andn = dkak F(aA) + doG(aA ) ,  
n=O 
00 
L en = doCo ( >. ) F(a>.), 
n=O 
52 Second Order Pantograph Equations 
where 
Ak 
CO(A) = k . (T2 - Tl )  TIm=l (Tl - T2an) (T2 - T2an) 
We therefore reduce the relations ( 3 . 2 3 )  and ( 3 . 24) to 
dkF()..) + doG()") = Yo , 
dkT1F(aA) + do (T2G(a)..) - CO(A)F(aA)) = -Yl · 
( 3 . 2 5 )  
( 3 . 26 )  
For a fixed value of A, equations ( 3 . 2 5 )  and ( 3 . 26) form a system of linear equations 
for the unknown constants do and dk , where the constants are determined uniquely 
provided 
TIF(aA)G(A) - T2F()..)G(aA) + CO(A)F(A)F(aA) =I- O. ( 3 . 27) 
In the same way as we did for the case where Tl =I- T2ak , we use the inherent 
recurrence in the expressions for F and G to establish the functional identities 
and 
(b - �) A 
TIF(aA) - aF(A) = - a F( -) ,  Tl a 
(b - �)  ( A 1 A ).. ) 
T2G(aA) - CO(A)F(aA) - aG()") = - a G( -)  + -Co(  -)F( -) . T2 a Tl a a 
Multiplying the first by G(A) and subtracting the second multiplied by F(A) yields 
TIF(aA)G(A) - T2F(A)G(aA) + Co(A)F(A)F(a)..) = 
).. ( ).. A A A ) 
( 1  - -) TIF(A)G( - )  - T2F( -)G(A) + Co( - )F(  - )F(A) , ab a a a a ( 3 . 28) 
and applying relation ( 3 . 2 8) n times gives 
Now, G(;n ) - G(O) = 1 , F(;n )  - F(O) = 1 ,  and CoC\n ) - Co(O) = 0 as n - 00;  
3.2  Laplace Analysis 53 
hence, 
00 .\ T1 F(Q.\ ) G ( .\) - T2F ( .\ ) G(Q.\) + Co ( '\) F ('\) F(Q.\) = (Tl - T2) IT ( 1 - Qmb ) ' ( 3. 29) 
m= l  
Since T l  - T2 #- 0, condition (3. 27) i s  satisfied for all .\ E C unless .\ = Qmb for some 
positive integer rn. 
Thus, provided that .\ #- Qmb for any positive integer rn, the Mittag-Leffier expan­
sion (3 . 1 1 )  is a valid solution to (3. 7 )  that is holomorphic at p = O. If .\ = Qmb for 
some positive integer rn, the coefficients do and dk are not determined uniquely, and 
furthermore the compatibility condition ( 3 . 20) must be met. Assuming that Yo and 
Yl satisfy (3 . 20), the solution is still valid, and will be unique up to an arbitrary 
constant . 
3.2 .3  The case where Tl  = T2 
In this last case, we have a repeated root to the indicial equation. Much of the 
analysis of the case where Tl = T2Qk holds for this case as well, so we will present 
just the highlights. Once again, we may substitute the series (3 . 1 1 )  with T2 replaced 
by Tl into (3 . 7) and equate principal parts. The recurrence relations are 
and 
d _ .\dn- 1 + 2 ( T1 Q
n - Tl ) Cn-k n - (Tl - T1 Qn)2 
. 
The relations obtained for Co and do yield the equations 
and 
00 
L dn = Yo , 
n=O 
00 00 
L Cn - Tl L Qndn = Yl · 
n=O n=O 
(3.30) 
(3 .3 1 )  
Solving the recurrence relations for Cn and dn , equations (3 .30) , (3. 3 1 )  can be  recast 
54 Second Order Pantograph Equations 
in the form 
where 
and 
do F ( A )  + COG(A) = Yo ,  
dOT1 F(aA) + cO (r1 G(aA) - F(aA)) = - Y1 . 
An � 2am 
00 � (Tl - Tl am) 2 
G(A) = L -----::::;_=1 ___ _ 
n= l II (T1 - Tl am? 
m= l  
( 3 . 3 2 )  
( 3 . 3 3 )  
For a fixed value of A , equations (3 . 32) and ( 3 . 33) form a system of linear equations 
for the unknown constants Co and do . They determine the constants uniquely, 
provided 
TI F(aA)G (A) - TIF(A) G(aA) + F(A)F(aA) =1= O. ( 3 . 34) 
Using the recurrence relations established so far, we obtain 
for any n > O. As n --+ 00, FC;n ) --+ F ( O) = 1 and G ( ;n ) --+ G (O) = 1 ;  hence, 
00 A TI F(aA) G(A) - TI F(A)G(aA) + F ( A ) F(aA) = II ( 1 - am b ) '  ( 3 . 3 5) 
m= l  
Thus, the condition ( 3 . 34) is satisfied for all A E re unless A = amb for some positive 
integer m. 
In all 3 cases, the Mittag-Leffier solution is a valid solution to (3 . 7) that is holomor­
phic at p = O. Lemma 3 . 1  implies that it is the only such solution if A =1= bam for all 
positive integers m. It is clear that the inverse Laplace transform Y of our Mittag­
Leffier solution L decays exponentially, as it transforms to a series of exponentials. 
In fact we can show that y is the only solution to ( 3 . 1 )  satisfying ( 3 . 6 )  that decays 
3.2  Laplace Analysis 55  
exponentially. Suppose we have two solutions Y1 , Y2 to ( 3 . 1 )  satisfying ( 3 . 6 )  that 
decay exponentially. Then their Laplace transforms L1 ,  L2 would be solutions to 
( 3 . 7 )  holomorphic at p = O. If A =1= ba.rn for all positive integers m, Lemma 3 . 1 
implies that L1 L2 ; hence, Yl - Y2 · We therefore have the following result . 
Theorem 3.3 If A =1= barn for all positive integers m, there exists a unique solution 
Y to ( 3 . 1 )  satisfying ( 3 . 6 )  that decays exponentially. 
Furthermore, we note that in the case where A = barn for some non-negative integer 
m, if Yo and Y1 satisfy ( 3 . 20 ) , then we have an exponentially decaying solution to 
( 3 . 1 )  unique up to a multiplicative constant. In fact, we know the form of the 
solution completely. We thus have the following analog of the Frobenius method 
for ordinary differential equations. 
Theorem 3.4 (Frobenius Method for the Pantograph Equation) Let r1 and r2 
be roots of the indicial equation ( 3 . 4 )  such that Re(r1 ) � Re(r2) > O. Then the 
solutions to ( 3 . 1 )  that decay exponentially take the form 
where Y1 and Y2 are linearly independent solutions to ( 3 . 1 ) .  The solution Y1 is of 
the form 
00 
Y1 (X) = L Cne -TIOnX , 
n=O 
and the solution Y2 is one of the following forms. 
00 
Y2 (X) = L dn e-T2onx 
n=O 
2. If r2 = r1 am for some non-negative integer m, 
00 00 
Y2(X) = L Cnanxe-TIOnX + L dne-T2onx . 
n=O n=O 
56 Second Order Pantograph Equations 
As in §2.4, the uniqueness of the solution to (3 .7) depends crucially on the re­
quirement that L be holomorphic at p = o. We can drop this condition to obtain 
other solutions to (3 .7) and therefore non-Dirichlet series solutions to (3 . 1 ) .  Indeed, 
all we require is that the solution to (3 .7) be holomorphic in the right half plane 
IIo = {z : Re( z) > O }  so that the transform can be inverted. 
Let L(p) = Lh (p) + Z(p) where Lh is the Mittag-Leffier solution that we know is 
holomorphic at the origin. Then Z must satisfy 
(3 .36) 
The function Z is therefore meromorphic in CC and using an argument similar to 
that used in Lemma 3 .2 , we see that 'at worst' Z has simple poles at p = -riak 
and p = -r2ci for non-negative integers k, unless ri = r2am for some non-negative 
integer m where we may get double poles. In fact , (3 .36) can be solved explicitly 
for Z. 
Let 
Z(p) = U(p)W(p) , (3 .37) 
where 
and 
_ 
In I � I  + icf;>  
K - ln a 
. 
As in Chapter 2, we use the principle branch of log(z) , and -7r < cf;> = arg(;b ) < 7r. 
Note that W(p) so defined is a solution to (3.36) and, upon substituting (3.37) into 
equation (3.36) , we see that 
U(p) = U(E.. ) ,  
a 
so that U is log-periodic with period a. We have thus found all possible solutions 
to (3.7) . 
Theorem 3.5 The general solution to (3.7) is of the form 
(3 .38) 
3.2 Laplace Analysis 57  
if r l =I- r2ak for all integers k ;  and of the form 
(3 . 3 9 )  
if rl = r2 ak for some non-negative integer k .  Here, t h e  constants Cn, dn and K, a re  
as defined earlier and U i s  a log-periodic function of period a .  
We therefore have an infinite number of  solutions to  the Laplace equation (3 . 7 ) ,  but 
this does not imply that each solution corresponds to a solution to ( 3 . 1 )  satisfying 
( 3 . 6) . We need to first show that the solutions of Theorem 3 . 5  can be inverted. As in 
the first order case, however, we can show that these solutions must be in the Hardy 
space H2(IIo ) ,  and hence by the Paley-Weiner theorem (cf. Hoffman [ 1 4] )  they are 
invertible to some function y E  L2 [0, 00) that is a solution to ( 3 . 1 )  satisfying (3 . 6 ) .  
Thus, the second order pantograph equation has, in general, an infinite number of 
solutions defined on the positive real axis. 
In order to obtain unique solutions to ( 3 . 1 ) ,  we must specify a further decay con­
dition on y such as requiring the solution be in In [O , 00) . We have the following 
result. 
Theorem 3.6  If 1 >' 1 < an H l b l and A =I- ak b for k = 0 , 1 ,  . . .  , n ,  then there exists a 
unique solution to (3 . 1 )  in In [0 , 00 ) satisfying ( 3 . 6 ) . 
Proof: Let y be any solution in In [0, 00 ) to ( 3 . 1 )  satisfying (3 . 6) , and let L (p) be 
it's Laplace transform. Then L is a solution to ( 3 . 7 ) ,  and 
so that 
is finite. By Theorem 3 . 38 ,  L can be written in the form 
L (p )  = L h (p) + Z(p ) . 
where Lh and Z are as defined earlier. Now, limp-+o Lhn) (p) is clearly finite, so that 
58 Second Order Pantograph Equations 
limp-+o z(n) (p) must be finite. Let 
j (p ) = p"'U(p) , 
and 
so that Z(p) = j (p )g (p) . Clearly, g(p) holomorphic at p = 0 as the infinite product 
is entire and non-zero for Ip l < T2 , so that limp-+o g ( k) (p) exists for all k. As was 
done in the proof of Theorem 2 . 1 2 ,  we can show that limp-+o j ( k) (p ) exists for 
k = 0 ,  1 ,  . . .  , n - 1 ,  and that we may write 
where u is some log-periodic function of period a. Now, in order for limp-+o z (n) (p) 
to be finite, we require limp-+o j (n) (p ) to be finite, and as 
In I A I Re(/'\.) = ab < n, 
ln a 
we therefore must have limp-+o u(p) = O. Theorem 2 . 1 0  then implies that u _ 0 ,  
hence Z - 0 ,  and L consists solely of the Mittag-Leffier solution. Thus, L is 
holomorphic at p = 0 ,  and hence must be unique by Theorem 3 . 1 .  • 
Clearly, the unique solution in Theorem 3 . 6  is the Dirichlet series solution. The 
additional information that Theorem 3.6 provides is that, for a given value of A, the 
decay rate of any non-Dirichlet series solutions to ( 3 . 1 )  must be limited. Indeed, if 
I A I < an+ l l b l , then any non-Dirichlet series solution must decay slower than xn1+1 
as x ---t 00.  
CHAPTER 4 
Natural Boundaries for 
Functional Differential 
Equations 
In this chapter we consider functional differential equations of the form 
y'(z) + ay (z) = by ( g ( z ) ) ,  (4. 1 )  
where a and b are complex constants, and g (z) is a non-constant entire function, 
along with an initial condition 
y(zo) = Yo · (4.2 ) 
Note that this problem is not well posed if g (zo) =J. zo , because the initial condition 
does not determine y ' ( zo )  since y ( g (zo ) )  is unknown. We thus limit ourselves to 
the case where Zo is fixed under g. Much of the material presented in this chapter 
(excluding §4.5) has been published in [22] . We start by considering some examples. 
4. 1 Examples 
In this section we look at two simple examples that can be solved using power series 
expressions, and then look at a more complicated example in a somewhat ad-hoc 
manner in order to obtain an understanding of the requirements for the existence 
of holomorphic solutions. 
60 Natural Boundaries for Functional Differential Equations 
4. 1 . 1  Example 1 
We start with the first order pantograph equation as discussed in Chapter 2 .  Con­
sider the initial value problem 
y' ( z )  + ay(z) = by (az) , (4 . 3 )  
y(O) = yo, (4 . 4 )  
where a ,  b and a are complex constants, and Yo =1= O .  If a solution y is holomorphic 
at the origin, then y can be represented as a power series 
00 
(4 . 5 )  
with a non-zero radius of convergence. Substituting this expression into equation 
(4 . 3 )  gives us 
ban - a 
Cn+ l = n + 1 Cn · (4 . 6) 
If l a l  < 1 ,  then ( 4 . 6 )  shows that either the series terminates in the case where 
b = a/ ak for some non-negative integer k, or I �l I - 0 as n - 00, so that the 
series converges for all z. In either case, we obtain the entire solution 
( ) _ 
� rr��lo (baj - a) n y Z - Yo L , z . 
n=O n.  
( 4 . 7) 
If l a l > 1 ,  then it is clear that the same formal power series solution would be 
obtained; however, the series would diverge unless the constants a and b were 
related such that one of the terms in the product is zero, thus terminating the 
series. We elucidate further on this case in Chapter 5 .  
The difference between these two cases is clear. The case where l a l < 1 has a 
retarded functional argument . That is, all points are moved closer to the origin 
under the mapping 9 (i .e . ,  I g ( z) 1 < I z l ) · For the second case, however, the functional 
argument is advanced: all points are moved further from the origin under 9 (i .e . ,  
I g ( z ) 1 > I z l ) · This behaviour of the origin attracting, or repelling neighbouring 
points will be characterised in the next section, and is crucial to the underlying 
behaviour of solutions to functional differential equations. 
4.1  Examples 61  
A third case occurs when l a d  = 1 .  Here, the functional argument i s  neutral, 
neither advanced nor retarded. In this case, it is clear that the formal power series 
( 4 . 7) we have established converges for all z. This, however, is a special case. The 
rotations Q Z  where I Q I = 1 are the only entire functions for which I g (z) 1 = I z l  
for all z .  We consider the more general case for neutral functional arguments in 
Chapter 6. 
The solution to the retarded version of the pantograph equation is entire, and one 
may wonder whether or not this is always the case. The functional argument Q Z ,  
however, is special among entire functions in that it is the only function that has 
the origin fixed, and always remains retarded (i .e . I g (z) 1 < I z l  for all z E C) .  
Liouville's theorem precludes any other entire function having this property other 
than multiples of QZ. Indeed, let J ( z )  be an entire function that fixes the origin ,  
and is  retarded for all z ,  (i .e. I J (z ) 1 < I z l  for all z ) . Then fSz) is an entire 
function, bounded above by 1 .  It is therefore a constant with modulus less than 1 
by Liouville's theorem, and hence J ( z )  = (3z for some 1 (3 1 < 1 .  
In contrast to the above example, we now present one whose solutions have a natural 
boundary. 
4.1 .2 Example 2 
Consider the equation 
(4 . 8 )  
along with initial condition ( 4 . 4 ) . It is clear that the functional argument g ( z) = z2 
is retarded for all I z l  < 1 ,  but is advanced for all I z l  > 1 .  We begin by seeking a 
solution of the form (4. 5 ) .  Upon substitution into the differential equation, we find 
that en = 0 unless n is of the form n = 2 k - 1  for some non-negative integer k. The 
solution is 
(4 . 9 )  
and it is clear that the gaps between the non-zero terms of the series are growing 
of order 2k as k --+ 00. Indeed, there exists a positive number v such that the 
sequence of non-zero terms {nk} = { 2k - I }  satisfies nk+ 1 > ( 1 + v)nk for all k 2: 1 .  
The Hadamard gap theorem [33] can thus be used to assert that the circle I z l  = 1 
is a natural boundary for (4 . 9 ) . 
62 Natural Boundaries for Functional Differential Equations 
It is evident that if we were to replace the functional argument Z2 in this example 
with any g ( z )  = zj , where j � 3, then the natural boundary at I z l  = 1 would 
persist . If we "unfold" these polynomials to more general j-th order polynomials 
that preserve the retarded character around the origin, then it is less clear, but 
certainly expected that a natural boundary will be present. The geometric form of 
this boundary, however, is not obvious. In this regard it is illuminating to unfold 
the z2 in the above example, and approach it from a geometric perspective. 
The natural boundary that occured above coincided with the boundary of the region 
in which Z2 remained retarded. It is intuitive to think that this property will hold 
in general, however it quickly becomes clear that a better definition of the idea of 
a function being retarded is needed. 
4.1 .3 Example 3 
Consider the example 
y' ( z )  = Y ( Z2 + fZ) (4 . 1 0) 
where f is some complex number. Then, using our present definition for a retarded 
function, the functional argument g ( z )  = Z2 + fZ would be retarded in the set such 
that 
which reduces to 
D ( - f ; l ) = { z  E re :  I Z + f l < I } .  
This set must include the origin, hence I t l  < 1 .  Is this a good place to start looking 
for a possible natural boundary? Certainly, the function is retarded in this set; 
however, points that start in the set may have images outside of the set. Suppose 
that a solution y is holomorphic in some set O. Then y' is also holomorphic in 0, 
and the differential equation then implies that y must then be holomorphic in g (O ) . 
It is clear, therefore, that the maximal set 0 for which the solution y is holomorphic 
must satisfy 
g (O) � O. (4. 1 1 ) 
The region D ( - f ;  1 ) does not satisfy this property. For instance, consider the point 
on the boundary of D ( -f;  1 ) furthest from the origin, fa = - f ( 1;1 + 1 ) .  Then 
4.1 Examples 63 
Ig(Eo ) 1  = I Eo l  and arg g(Eo) = 2 arg Eo, hence g(EO) is outside of D(-E; 1 ) ,  except for 
the special case where E is real and negative. By continuity, there exist points in 
D( -E;  1 ) arbitrarily close to EO which also get mapped outside of D ( -E, 1 ) . Thus, 
D ( -E;  1) does not satisfy (4 . 1 1 ) .  For the special case where E is real and negative, 
other suitable example points can be found, such as -E + i .  
If we look instead at the largest disc centred at the origin contained in D ( -E; 1 ) we 
find that we have a region that satisfies ( 4 . 1 1 ) .  Indeed, let no = D (O ;  1 - I t l ) ,  and 
let z E no . Then 
Ig ( z ) 1 = I z l l z  + El < ( 1 - I E I ) ( l  - I E I  + I E I )  < 1 - I t \ ,  
and hence no satisfies ( 4 . 1 1 ) .  This set, however, need not be  maximal. We can 
extend this set by using the differential equation. Suppose y is holomorphic in no . 
Choose any z such that g( z )  E Do . Then the right hand side of the differential 
equation is holomorphic at z , which implies the left hand side must also be holo­
morphic at z. Thus, we can expand our region of holomorphicity to the larger set 
n1 = { z  : g(z) E no} .  Repeating the procedure we find we can generate a sequence 
of regions, each larger than the previous one, that our solution must be holomorphic 
in. We can therefore build up an approximation to the boundary of the maximal 
region in which the solution to (4 . 1 0 ) is holomorphic by successively plotting the 
image of the circle I z l  = 1 - I E I  under inverses of g. 
The boundaries of the successive sets 
for the function g (z) = Z2 + 0 . 8z up to k = 20 are given in figure 4 . 1 .  Evidently the 
shape of the sets nk increases in complexity as k increases, though the sequence of 
sets appears to be converging. 
In figure 4 .2 the same sets are plotted for the function g ( z )  = z2 + 0 . 9i z ,  and exhibit 
a similar behaviour, although now the sets are disconnected. The initial set no is 
the interior of the circle in the top centre of the figure. The second set n1 extends 
no slightly, and also adds the interior of the circle in the bottom centre. The next 
set , n2 further expands these two circles and adds 2 new ones in the upper left and 
lower right parts of the figure. This continues, with each set consisting of more 
64 Natural Boundaries for Functional Differential Equations 
Figure 4 . 1 :  Plot of 00, . . .  , 020 for g(z) = z2 + O.8z. 
and more disconnected components, whilst also enlarging each of the components 
in turn. In fact , from k = 7 onwards, the sets "pull-in" and reconnect some of 
the disconnected elements, with rh containing the figure eight shaped region in the 
center of the plot which encloses the first two components of 01 through 06 . It 
appears as if the convergence observed in the previous case is also occuring in this 
case, although it is at a slower rate. In both of these plots, however, it is clear that 
the resulting sets have a very different shape than the simple circle in Example 2 .  
The preceding arguments are informal and gloss over many of the details; however, 
they suggest that the region in which any solution to a functional differential equa­
tion is holomorphic is governed by how points surrounding the initial point map 
under successive applications of 9 and its inverse. In the next section we introduce 
the definitions and machinery required to formally prove the above arguments. 
4.2  Fixed and Periodic Points 65 
Figure 4.2: Plot of Do, . . .  , D20 for g(z) = z2 + O.9iz. 
4.2  Fixed and Periodic Points 
The above discussion indicates that a precise definition of the region in which a 
function is retarded or advanced is required. Such a definition is linked to the 
behaviour of a function 9 about its fixed points. 
Definition 4.1  (Fixed Point) A fixed point of g (z) is a point Zo such that g ( zo )  = 
zo ° 
Fixed points can be characterized into 3 types. 
Definition 4.2 A fixed point Zo of a function 9 is 
1 .  attracting if I g ( z )  - zo l < I z  - zo l for a l l  z i n  some neighbourhood of Z0 0 
2. repelling if I g (z ) - zo l > I z - zo l for all z in some neighbourhood of Z00 
66 Natural Boundaries for Functional Differential Equations 
If Zo is neither attracting n o r  repe lling, we say that Zo is neutral . 
The definition of the derivative of 9 at Zo gives us the following method of classifi­
cation, which is easier to use. 
Theorem 4.3 A fixed point Zo of an analytic function g (z )  is 
1. attracting if and o n ly if I g' (zo ) I < 1 .  
2. repelling if and only if I g' (zo) 1 > 1 .  
3. neutral if and only if I g' (zo ) 1 = 1 .  
We observed in the last section that compositions of g - 1 with itself are used to 
define the region in which a solution is likely to be holomorphic. Let 
denote the n-th composition of 9 with itself. If Zo is a fixed point of g, then it 
will also be a fixed point of gn for all n .  It turns out that the fixed points of 
compositions of 9 are also useful. 
Definition 4.4 (Periodic Point) A fixed point Zo of gn ( z )  is a periodic point 
of g ( z ) . The period of Zo is the smallest n for which gn ( zo ) = zo o The points 
zo , g ( zo) , g2 ( ZO ) ,  . . .  , gn- 1 ( zo ) are all periodic points of 9 and are known as a peri­
odic orbit . 
Periodic points of a function 9 can be classified as attracting, repelling or neutral 
in a similar way as was done for fixed points. 
Theorem 4.5 Let Zo be a periodic point of the function g ( z )  with period n .  Then 
Zo zs 
1 .  attracting if and only if I (gn ) ' (zo ) I  < 1 .  
2. repelling if and only if I (gn ) '  ( zo ) I > 1 .  
4.2 Fixed and Periodic Points 
3. neutral if and only if I (gn ) ' (  zo ) I = 1 . 
67 
Definition 4 .6  (Basin of Attraction) Let Zo b e  a n  attrac ting fixed point. The basin 
of attraction of Zo for 9 i s  
Azo = { z :  lim gn ( z )  = zo } . n->oo 
We can extend this definition to include periodic points by altering the definition 
to Azo = {z : limn->oo gmn ( z )  = zo } ,  where m is the period of zoo 
The set Azo need not be connected. The definition of an attracting fixed point, 
however, indicates that there is some neighbourhood U of Zo that is in Azo '  so that 
there is a non-empty connected open subset of Azo that contains zoo We define 
the immediate bas in of attraction A�o as the component of Azo containing 
the point zo o We are now in a position to show that there exist solutions to (4 . 1 )  
holomorphic in A�o . Let H ( O )  denote the set o f  functions holomorphic on the open 
set O .  
Theorem 4.7 Let Zo b e  a n  attracting fixed point of the non-constant entire function 
g .  Then there exists a solution to the initial-value problem (4. 1 ) , ( 4 . 2 )  that is unique 
among functions holomorphic at Zo . 
Proof: The differential equation ( 4 . 1 )  can be written in the form 
y'(z) = H ( y ( z) , y ( g ( z ) ) ) ,  
where H ( y ( z ) , y (g ( z) ) )  = by ( g ( z ) )  - ay ( z ) . We establish the existence of a local 
solution by showing that the associated integral equation 
y ( z) = 1: H ( y (�) ,  y ( g (� ) ) ) d� + Yo (4 . 1 2 )  
has a unique solution in H ( D (zo, 8 ) )  for some 8 > O .  We do this by showing that 
the integral operator is a contraction mapping for small 8. 
Let HOO ( D (zo ;  8 ) )  denote the Banach space of functions in H ( D (zo; 8 ) )  equipped 
68 Natural Boundaries for Functional Differential Equations 
with the norm 1 1 . 1 1  defined by 
I l y l l  = sup l y ( z) l ,  
ZED(zo;8) 
for all y E 'HOO ( D (zo; 8 ) ) , and let T be the operator defined by 
Ty = 1: H ( y (� ) ,  y (g (�) ) ) d� + Yo · 
The function H is holomorphic in both variables, and since 9 E 'H ( D ( zo ,  8) )  for any 
8 >  0 we have that T maps functions in 'HOO ( D (zo ;  8) ) into 'H (X) ( D ( zo ;  8) ) .  Now, for 
any u , v E 'H(X) ( D ( zo; 8) ) ,  
I I Tu - Tv l l = 1 1 : H (u (� ) , u (g (�) ) ) d� - 1: H (v (� ) , v ( g (� ) ) ) d� 1 1 
::; I z - Zo I I I H (u (�) , u ( g (�) ) )  - H (v (�) ,  v (g (� ) ) ) 1 1  
= I z  - zo l ( �up I b (u (g (z) ) - v (g ( z ) ) )  - a (u ( z )  - V ( Z ) ) I) 
zED(zo;8) 
::; 8 ( I a l �up l u (z) - v (z) 1 + I b l �up l u ( g (z) ) - V ( g ( Z ) ) I) . 
zED(zo;8) zED(zo;8) 
By hypothesis, Zo is an attracting fixed point of g, and hence for 8 sufficiently small 
g ( D (zo; 8) )  � D (zo ;  8) . 
Thus, for small 8 > 0 ,  
and hence 
sup l u ( g ( z ) ) - v ( g ( z ) ) I ::; sup l u ( z )  - v ( z ) l ,  
ZED(zo;8) ZED(zo;8) 
I I Tu - Tv l l  ::; 8 ( l a l  + I b l ) l l u - v i i ·  
(4. 13) 
We may choose 8 small enough so that 8 ( l a l + I b j ) < 1 ,  and hence the operator 
T is a contraction and we conclude that the integral equation (4. 12) has a unique 
solution in 'HOO ( D (zo; 8 ) ) .  Therefore, the initial value problem (4. 1 ) , (4 .2) has a 
unique solution in 'H ( D ( zo ;  8 ) )  for small 8. • 
4.2 Fixed and Periodic Points 69 
We remark here that the above proof relies only on the condition that H is holo­
morphic in both its arguments, and that the fixed point Zo is attracting. It can be 
readily extended to higher order functional differential equations by simply rewrit­
ing the higher order equation as a first order system, and then adjusting the Banach 
space accordingly. It also does not rely on the polynomial character of g. Indeed, 
all we require is that 9 E 'H. ( D ( zo ;  8 ) ) .  We address these issues among others in 
§4.5. 
The local solution of Theorem 4 .7 can be analytically continued using the functional 
differential equation (4. 1 )  in a similar way as was done in the previous section. 
Theorem 4 .8  Let Zo be an attracting fixed point of the non- constant entire function 
g .  Then there exists a unique solution to the initial value pro blem (4. 1 ) , (4.2) that 
is holomorphic in the set A�o . 
Proof: Theorem 4.7 shows that there is a 8 > 0 and a unique y E 'H. ( D (zo ;  8 ) )  
such that y i s  a solution to the initial value problem for all z E D ( zo ; 8) , and 
D ( zo ;  8 ) )  � A�o . Suppose there exists a w E A�o at which y is not holomorphic. 
Since y is not holomorphic at w, the function y' + ay is also not holomorphic at w ,  
for if it was, y would be a solution to the equation y' + ay = f where f is some 
holomorphic function, which yields a holomorphic solution for y. Thus, y' + ay is 
not holomorphic at w ,  and equation (4. 1 )  shows that y is not holomorphic at g (w ) .  
The same argument can now be applied to  g (  w )  to show that y i s  not holomorphic 
at g2 ( W )  also. A simple inductive argument shows that y is not holomorphic at 
gn ( w )  for all n E N. But w E A�o ' and therefore there is an N > 0 such that 
I gn ( w )  - zo l < 8 for all n > N. Hence, there must be an n E N such that 
gn ( w )  E D ( zo ; 8 )  and y is not holomorphic at gn ( w ) .  This contradicts the fact that 
y E 'H. ( D ( zo ;  8 ) )  and hence y must be holomorphic for all w E A�o . Now, A�o is 
connected, and contains D ( zo ; 8) , therefore the local solution holomorphic at Zo can 
be analytically continued to A�o . • 
Of interest here, is the arguments in the proof above apply equally well for any 
w E Azo ' not just those in the component A�o . This means that the local solution 
defines not only a unique function in 'H. ( A�o ) '  but it also defines unique functions 
holomorphic in other components of Azo . 
70 Natural Boundaries for Functional Differential Equations 
We will show that the boundary of the immediate basin of attraction forms a natural 
boundary for our solution. To accomplish this, we need a number of results from 
Complex Dynamics, specifically results on Julia sets. 
4.3 J ulia Sets 
In this section we review some relevant facts regarding Julia sets .  Many of the 
proofs in this section are based on those given by Milnor [26] , where a more detailed 
account can be found. 
We start with the definition of a normal family of functions. 
Definition 4 .9  A collection F of functions from a set D � re to a set E � re is 
normal if every infinite s equence of maps from F contains a subsequence which 
converges locally uniformly to some limit function g : D --+ E, or diverges uniformly 
to 00. 
We can now state the following definition for the Julia set . 
Definition 4 .10  For a given function 9 = g ( z ) ,  we define the Julia set of g as 
J ( g )  = { z  : { gn} is not a normal family in any neighbourhood of z} . 
The above definition shows us that the Julia set must be closed, as any neighbour­
hood of any point on the boundary of the Julia set must include points in the Julia 
set , upon which { gn } is not a normal family. The complement , re " J is called 
the Fatou set ,  or stable set . This definition of the Julia and Fatou sets is not 
convenient for determining the sets computationally for a particular function. We 
can, however, quickly determine some of the elements and properties of the Fatou 
and J ulia sets as the following results show. 
Theorem 4.11  (Milnor p. 39) The Julia set and Fatou set of a function 9 are 
invariant under g .  
4.3 J ulia Sets 71 
Proof: For any open set U E C, some sequence of iterates { gnj }  converges uni­
formly on compact subsets of U (or diverges uniformly to 00) if and only if the 
corresponding sequence { gnj + 1 } converges uniformly (or diverges uniformly to 00 ) 
on compact subsets of the open set g- I (U) . • 
This has two immediate consequences. The first is that the Julia set of a function 
must present a large amount of self-similarity. Indeed, if we have g (Zl ) = Z2 for 
any Zl E J(g) where g'(Zl ) =J. 0, then there is a neighbourhood NI of ZI that maps 
isomorphic ally under 9 to a neighbourhood N2 of Z2 , and NI n J(g) maps precisely 
onto N2nJ(g) . Thus, any shape observed in the Julia set is generally seen infinitely 
many times. See figure 4 . 3  for example. 
The second implication, is that we can use an inductive argument to conclude that 
the Julia set of gk coincides with the Julia set of 9 for any positive integer k. This 
allows us to quickly find where the attracting and repelling periodic orbits must lie. 
Figure 4.3: The Julia set for the function g(z) = z2 - 1 + O.2i. 
Theorem 4.12  (Milnor p. 43) Every attracting periodic orbit i s  contained i n  the 
Fatou set of g .  In fact, the entire basin of attraction for any attracting orbit is in 
the Fatou set. Every repel ling periodic point is contained in the Julia set. 
Proof: Consider a fixed point Zo with g'(zo ) = a. Then if J a J  > 1 ,  the first derivative 
of gk at Zo is ak which diverges as k -+ 00,  hence no sequence of iterates of 9 can 
72 Natural Boundaries for Functional Differential Equations 
converge uniformly near zo, and so Zo must be in the Julia set . On the other hand, 
if IQ I < 1 ,  we have an attracting fixed point , then there exists a constant c such that 
IQ I < c < 1 and I g (z) - zo l :::; c l z  - zo l  for all z sufficiently close to zo o Thus, for any 
positive integer k ,  we have that Igk (z) - zo l  :::; ck l z  - zo l  for all z sufficiently close 
to Zo; hence, the successive iterates of 9 restricted to a small neighbourhood of Zo 
converge uniformly to the constant function h(z) = zo o This is extended readily to 
the basin of attraction of Zo by noting that for any z E Azo there exists an integer 
N > 0 such that IgN (z) - zo l < c lz - zo l  where c is as before. Thus, the sequence of 
iterates {gN+k} will converge uniformly to the constant function h = Zo as before. 
These statements regarding fixed points generalize immediately to periodic points 
by making use of Theorem 4 . 1 1 ,  since periodic points are simply fixed points of 
some iterate gm . • 
The case of neutral periodic points is not so clear - indeed it turns out that some 
are in the Julia set, and some are in the Fatou set . The simplest case is as follows. 
Definition 4.13 Let Zo be a neutral periodic point of period rn, and let Q = 
(gm)' (zo) . If Q is a root of unity, and no iterate of 9 is the identity map, then 
Zo is called parabolic. 
Theorem 4 .14 (Milnor p. 44) Every parabolic periodic point belongs to the Julia 
set. 
Before we prove this result ,  let us first note that we may always consider any 
periodic point to be at the origin (or any other point that makes the analysis more 
convenient ) by using a change of coordinates. Indeed, let Zo be any periodic point 
of g, and consider the transformation w = z - Zo and h (w) = g (w + zo) - zo o Then 
h(O) = g (zo) - Zo and h' (O) = g'(zo ) .  Also, 
h2(O) = h(h(O)) 
= h(g(zo) - zo)) - Zo 
= g (  (g(zo) - zo) + zo) - Zo 
= g2 (ZO ) - zo 0 
4.3 Julia Sets 73 
and so ( h2 ) ' ( 0) = (g2 ) ' ( ZO ) .  A simple inductive argument then shows that hk ( O )  = 
gk ( zO ) - Zo and (hk ) ' (O) = ( gk ) ' (zo ) for all positive integers k. Thus, h ( w )  will 
have a periodic point at the origin with the same properties as zo 0 We can perform 
whatever analysis is necessary with h ,  and then transform the results back to the 
original coordinate system. 
Proof: Following the above argument, we may assume the periodic point is at the 
origin. Since the periodic point is parabolic, there is an iterate gm that can be 
written using a Taylor series expansion as gm ( z )  = z + aqzq + aq+ l Zq+ 1 + . . . , where 
q � 0, and aq =1= O. It thus follows that any iterate gmk corresponds to a power 
series of the form z + kaq zq + . . . .  Thus, the q-th derivative of gmk at 0 is equal 
to q ! kaq which diverges to infinity as k � 00. Hence, no subsequence { gmki }  can 
converge uniformly as kj � 00.  • 
Parabolic points are of course just one type of neutral point. Other types of neutral 
fixed points will be considered in Chapter 6, where we look for local holomorphic 
solutions about neutral fixed points. 
The following result provides a characterisation of the Julia set in terms of excep­
tional points. 
Definition 4.15 (Exceptional Points) An exceptional point w E re  fo r a func­
tion f is a point such that f ( z) =I- w for all z in some open s e t  U � re .  
Theorem 4.16  (Montel) Let { fn } b e  a family of functions i n  U where U c re is 
an open set. Suppose there exist a , b  E re such that fn ( z )  =1= a and fn ( z )  =1= b for all 
n E  N and all z E U .  Then { fn } is a normal family in U.  
A proof of this result can be found in  [26] and is often used a s  a stepping stone to 
Picard's theorem [4] . 
Corollary 4 .17 Let z E J ( g )  and U be any neighbourhood of z .  The set G defined 
as 
00 
omits at most one point in re. 
74 Natural Boundaries for Functional Differential Equations 
Proof: Suppose G omits two points, a and b. Then { gn } is a family of functions in 
the open set U such that gn (z) =I- a and gn ( z )  =I- b for all n E N and all z E U. By 
Theorem 4. 16 ,  { gn } is a normal family, contradicting the hypothesis that z E J ( g )  . 
• 
This result will be used later to show that the Julia set forms a natural boundary for 
our functional differential equation. In the special case where g ( z )  is a polynomial, 
we can refine this result as follows. 
Theorem 4.18 Let g (z) be a non-constant polynomial. Suppose that there is a 
point W E J ( g )  and a neighbourhood U of W such that 
00 
U gn ( u )  = C "  { a }  
n= l 
for some a E C .  Then g ( z )  = a + b(z - a ) k for some b E C "  { O }  and some integer 
k ? 2 .  
Proof: We first show that 9 must be of degree at least 2 .  Suppose 9 i s  linear. 
Then we may write g ( z )  = e(z - d) + d for constants e =I- 0 and d .  Now, g n ( z )  = 
en (z - d) + d, and thus 9 has a single periodic point at z = d. This point must be 
repelling, as J ( g )  is non-empty; hence, l e l ? 1 .  Now, any neighbourhood U of d 
contains a disc D ( d ;  €) for some € > O. But gn ( D ( O ;  E) )  is a disc of radius I c l n € ,  thus 
00 
for all neighbourhoods U of d. Thus 9 is must be at least degree 2 .  
By the fundamental Theorem of Algebra, there exists a point e E C such that 
g (e) = a. This pre-image of a cannot be in any of the sets gn ( u )  as a is not . The 
only possibility is that e = a, and thus a is a fixed point of z. Furthermore, a has 
no pre-images other than itself; therefore, the only solution to g ( z )  - a = 0 is a ,  
hence 9 is of the form required. • 
Polynomials of this form (i.e. with exceptional values) will be referred to as cyclic 
polynomials. 
4.3 Julia Sets 75 
Definition 4.19 (Cyclic Polynomials) A polynomial of the form 
g ( z) = a + b ( z - a)k , 
where a and b =f. 0 are complex constants, and k 2: 2 tS an integer is a cyclic 
polynomial . 
We can characterize the fixed points of cyclic polynomials as follows. 
Theorem 4.20 A cyclic polynomial has only one attracting fixed point - the ex­
ceptional point. A l l other fixed points are repel ling. 
Proof: Let 9 be a cyclic polynomial with exceptional point a, so that 
g ( z ) = a + b( z - a)k 
for constants b =f. 0, k > 2 .  Clearly z = a is an attracting fixed point of g, as 
g (a ) = a and g ' ( a) = o. The other fixed points of 9 are given by the solutions 
Zl , ·  . .  , Zk-l  of 
b ( z - a)k-l = 1 .  
These points must be repelling, as g'( z ) = kb(z - a )k- l , and hence g ' ( zm ) = k for 
m = 1 , 2 ,  . . .  , k  - 1 . The only attracting fixed point of 9 is therefore the exceptional 
point a. • 
The above result shows that ,  for polynomials, the exceptional point must be in the 
Fatou set. For polynomials then, the iterated images of any neighbourhood of any 
point in the Julia set covers the Julia set . Indeed, the same result holds for general 
functions, although the exceptional point may be in the Julia set. In either case, 
we can establish the following result . 
Theorem 4.21 (Milnor p .  47) Let Zo be any point in J ( g ) . Then the pre-images 
of Zo are dense in J (g) , unless Zo is the exceptional point. 
Proof: Choose any point Zl E J (g ) .  If Zo is not the exceptional point, then there 
exists points z arbitrarily close to Zl whose forward orbits contain Zo by Corollary 
4. 1 7. Thus, we can find a pre-image of Zo arbitrarily close to any point in J(g ) ,  and 
so the preimages of Zo are dense in J (g ) . • 
76 Natural Boundaries for Functional D ifferential Equations 
This result gives an efficient method of computing pictures of the Julia set for 
polynomials - we simply start with any repelling point , and compute its pre-images 
until we have enough points to get a good plot of the Julia set. Note that it is 
most useful for quadratics, because the number of points generated grows as dn , 
where d is the degree of the polynomial, and n is the length of the orbits we are 
generating. This growth of the number of points is the only real disadvantage with 
this technique, as many iterations may be required in order to get close to certain 
points in the Julia set . 
The next theorem shows that the boundary of the immediate basin of attraction is 
contained in the J ulia set. 
Theorem 4.22 (Milnor p. 46) If A E C is the basin of attraction for some at­
tracting periodic orbit, then the boundary BA = A " A is equal to the entire Julia 
set. 
Proof: If N is any neighbourhood of a point of the Julia set , then (4. 16) implies 
that some gn (N) intersects A , hence N itself intersects A by the definition of an 
attracting basin. Thus, J c A. But J is disjoint from A , so it follows that J c BA. 
On the other hand, if N is any neighbourhood of a point of BA then any limit 
of iterates gn evaluated on N must have a jump discontinuity between A and BA, 
since the points in A tend to the periodic orbit , whereas those on BA do not . Hence 
BA c J, and so BA = J. • 
Note here that we must make a distinction between BA and the union of the bound­
aries of the connected components of A. Often the latter set is smaller than the 
former. For example, the quadratic function g(z) = z2+i has no attracting periodic 
points in C, and thus no basins, yet it 's Julia set is non-empty (See figure 4 .4) .  
The last two results in this section concern only polynomials, and are presented 
here without proof. These results show simply that, for polynomials, the Julia set 
is non-empty and that the number of attracting periodic points is finite and limited 
by the degree of the polynomial. The reader is referred to Milnor ( [26] pp. 78-79, 
136) for the details. 
Theorem 4.23 Every po lynomial of degree at least two must have either a repelling 
fixed point, or a parabolic fixed point, or both. 
4.4 The Natural Boundary 77 
Figure 4 .4: The Julia set for the function g(z) = z2 + i ,  a dendrite. 
Theorem 4.24 For a polynomial of degree d, the number of attracting periodic 
points is at most d - 1 .  
4.4 The Natural Boundary 
We are now in a position to prove that the boundary of the immediate basin of 
attraction forms a natural boundary for the solution to our functional differential 
equation. 
We first note that the immediate basin of attraction A�o of our fixed point Zo , has as 
its boundary a subset of the Julia set J(g) by Theorem 4.22. Thus, if our solution 
is to be analytically continued outside of A�o ' it must be holomorphic at some point 
Z E J ( g ) . 
Lemma 4.25 Let Zo be an attracting fixed point of the non- constant entire func­
tion g ,  and let y be a solution to the differentia l  equation (4. 1 )  holomorphic at Zo . 
78 Natural Boundaries for Functional Differential Equations 
Suppose that y is holomorphic at a point w E J ( g ) . The n  y is holom orphic in 
C " { go }  where go is the exceptional point of g .  
Proof: If 9 is linear, with attracting point zo, it may be written in  the form g ( z )  = 
a ( z  - zo ) + zo, where l a l  < 1 .  Thus, gn ( z )  - Zo = an ( z  - zo ) ,  and hence Azo = C 
Theorem 4 . 1 2  thus gives J ( g )  = 0,  so we need only prove the result in the case 
where 9 is non-linear. 
Suppose that y is holomorphic at w E J (g ) .  Then there is a neighbourhood U 
of w such that y E 1t ( U) . Now, y is holomorphic in U and hence the function 
y' + ay is also holomorphic in U .  Since y is a solution to equation ( 4 . 1 ) ,  y must be 
holomorphic in g ( U ) ,  i .e. y E 1t (g ( U) ) .  We apply the same argument to g ( U ) to 
deduce that y E 1t ( g2 ( U ) ) ,  and by induction y E  1t ( gn ( u ) ) for all n E  N. Therefore, 
y E  1t ( G) ,  where 
00 
Corollary 4 . 1 7  then gives that y is either entire , or is holomorphic at all but the 
exceptional point of g .  • 
In the case of polynomials, we can eliminate the possibility of a singularity at the 
exceptional point, and can then establish the existence of the natural boundary. 
Corollary 4.26 Let Zo be an attracting fixed point of the non- constant po lynomial 
g, and let y be a solution to the differential equation ( 4 . 1 ) holomorphic at Zo . Sup­
pose that y is holomorphic at a point w E J (g ) . Then y is an entire function. 
Proof: Let G be as in the proof of Lemma 4 . 2 5  and let w be any point in J ( g ) . 
As 9 is a polynomial we can apply Theorem 4 . 1 8 ,  so the set G is either the entire 
complex plane, or G = C - { a }  for some constant a, and 9 is a cyclic polynomial. If 
9 is a cyclic polynomial, then it has a as its sole attracting fixed point by Theorem 
4 . 2 0 .  By hypothesis Zo is an attracting fixed point of g, thus a = zo , and so y is 
holomorphic at a. Hence, y is entire. • 
Theorem 4.27 Le t Yo i= 0, a i= b, b i= 0, and let Zo be an attracting fixed point for 
the non-linear polynomial g .  Then the solution to equation ( 4 . 1 ) satisfying ( 4 . 2 ) 
has aA�o as a natural boundary. 
4.4 The Natural Boundary 79 
Proof: Theorem 4.8 shows that the solution y to the initial-value problem is holo­
morphic in A�o . We are therefore required to show that y is singular at every point 
on the boundary oA�o . 
Suppose that y is holomorphic at some point w E oA�o . Since oA�o � J(g) , 
Corollary 4.26 implies that y must be entire. We show by examining the growth of 
y that this is impossible. 
Let CR = oD(O; R) and let M(R) = SUPZED(O;R) l y (z) l .  Since y is entire, the 
maximum modulus principle implies that 
M(R) = sup l y (z) 1 
zECR 
for all R > O. The conditions a =I b and Yo =I 0 preclude y from being a constant, 
and in fact y cannot be linear, since equation (4. 1 )  would imply that 9 is linear, 
contrary to our hypothesis. Thus, M(RI) < M(R2) for any RI < R2. 
Let z E CR. The Cauchy integral formula implies that for any non-negative integer 
k, 
(k) - k! 1 y (E )  y ( z )  27ri 'Y (E _ z)k+ I dE , 
where I is any simple closed contour orientated anticlockwise that encloses z. Thus, 
for any 8 > 0, we may use the contour of the form 10 = {E : lE - z l  = 8} which 
yields the inequality 
Now, 10 is contained entirely in CR+8 , thus the maximum modulus principle implies 
that 
(4. 14) 
For any z E CR, inequality (4. 14) with 8 = 1 and the differential equation (4. 1 )  
yield 
ly (g (z) ) 1  :s; ( 1 + 
I
I�I
I
) M(R + 8) . (4. 15) 
Now since 9 is a polynomial with degree at least 2, there is an RI > 0 and an a > 0 
such that 
Ig (z) 1 > aR2 (4. 16) 
for all z E CR with R > RI . For such a choice of R, the curve r = {g(z) : z E CR} is 
80 Natural Boundaries for Functional Differential Equations 
closed, and encloses the curve CaR2 . The maximum modulus principle thus implies 
that there is a Zl E C R such that 
Combining this with inequality (4.15) yields 
M(aR2) < (1 + II�I) M(R + 8 ) .  
We now show that this gives a contradiction. Let 
h (z) = 
y ( z ) - y ( O ) . 
z 
(4. 1 7) 
(4. 18) 
(4 . 19) 
Since y is entire and non-constant, h must be entire, and so Mh(R) = SUPzECR I h ( z) 1 
where Mh(R) = SUPzED(O;R) I h (z) 1 by the maximum modulus principle. Taking the 
modulus of (4. 19) and using the triangle inequality yields the relations 
M (R) < M(R) + l y ( O ) 1 h - R 
M(R) � RMh(R) + l y ( O) I . 
Combining these with inequality (4. 18) yields 
For R sufficiently large, we have 
4.5 Generalisations 81  
and 
For such R, we therefore have that 
But y is not a linear function, and hence h cannot be a constant , so we have a 
contradiction to the maximum modulus principle. We conclude that y is not an 
entire function, and hence that y is not holomorphic at w.  Since our choice of 
w E aA�o was arbitrary, we see that y is not holomorphic at any point of aA�o ' and 
hence that aA�o is a natural boundary. • 
An interesting feature of solutions to (4. 1 )  is that the natural boundary does not 
depend on the choice of the constants a and b (other than requiring a =I b) : it 
is determined entirely by the functional argument g .  In the following section, 
we see how this independence from the differential operator allows some simple 
generalisations to be made. Note also that if a = b, then the initial value problem 
admits the constant solution y = Yo regardless of g.  By Theorem 4.7 this solution 
is unique for the attracting fixed point Yo ; hence, there is no natural boundary in 
this case. 
4.5  Generalisations 
In the last sections, we proved a number of results regarding the solutions to simple 
first order functional differential equations with constant coefficients, and a polyno­
mial functional argument. We show in this section how some of these results may 
be extended to more general equations. 
4.5 . 1  Higher order equations 
Let us first investigate how we may extend our results on first order equations to 
n-th order equations with constant coefficients. 
82 Natural Boundaries for Functional Differential Equations 
Consider the functional differential equation 
Ly = by (g ( z ) )  ( 4.20) 
where b -=1= 0 is a constant, g ( z) is a non-constant polynomial, and L is the linear 
differential operator 
where ao , al , " "  an- l are complex constants. We seek solutions to (4 .20) satisfying 
the initial conditions 
( ) _ ' ( ) _ , (n - l ) ( ) _ (n- l ) Y Zo - Yo , Y Zo - Yo, . . .  , Y Zo - Yo , 
where Zo is a fixed point of g. 
(4.22) 
We start by modifying the proof to Theorem 4.7 to obtain a local analyticity result 
for these higher order equations. 
Let Yl = y , Y2 = y' , . . . , Yn = y(n- l ) . The differential equation (4.20) can then be 
the written as 
y' ( z) = H ( y ( z ) , y (g (z) ) ) , 
where y = (Y l , Y2 , " . , Yn f and 
H (y (z) , y (g(z) ) )  = 
Y2 ( Z )  
Y3 ( Z )  
Yn ( z )  
byl (g ( Z ) ) - L�= l ak - 1 Yk ( Z )  
The associated integral equation is then 
y ( z )  = 1: H (y (e) , y ( g (�) ) ) de + Yo , 
h ( '  (n- l ) )T w ere Yo = Yo , Yo , . . . Yo . 
(4.23 ) 
(4.24) 
Let HOC ( D  ( zo ;  8) )  denote the Banach space of functions in the n-fold cartesian 
product H ( D ( zo ;  8) )  x H ( D ( zo ; 8 ) ) x . . .  x H ( D (zo ;  8) )  equipped with the norm 1 1 . 1 1 
4.5 Generalisations 83 
defined by 
n 
I l y l l  = L �up I Yk ( Z ) 1 , 
k= l  ZED(zo;8) 
for all y = (Y l , . . .  , Yn f E 1{OO ( D (zo ;  (5) ) ,  and let T be the operator defined by 
Ty = 1� H ( y (� ) ,  y ( g (� ) ) ) d� + Yo · 
For any 15 > 0 the function H is holomorphic in Yl , . . .  , Yn and Yl (g ) , and since 
9 E 1{ ( D ( zo ,  (5) ) we have that T maps functions in 1{OO ( D (zo; (5) ) into 7-tX) ( D ( zo ;  (5) ) . 
Now, for any u , v E 1{OO ( D ( zo ;  (5) ) ,  
I I Tu - Tv l l  = 1 1 � H ( u (� ) ,  u (g (�) ) ) d� - 1� H (v (� ) ,  v ( g (�) ) ) d� 1 1 
::; I z  - zo I I I H ( u (�) ,  u ( g (� ) ) )  - H ( v (�) , v ( g (�) ) ) 1 1  
= I z  - zo l (t �up I Uk ( Z )  - vk (z) 1 
k=2 zED(zo;8) 
+ �up I b(Ul ( g (Z) ) - Vl (g ( Z) ) )  - t ak - l (uk ( Z )  - Vk (Z) ) I) 
zED(zo;8) k= l 
::; 15 (t(1 + l ak- l l ) �up I Uk ( Z )  - vk ( z ) 1 
k=2 ZED(zo ;8) 
+ I ao l �up I U l (Z ) - vl (z) 1 + I b l �up I u ( g ( z ) ) - V ( g ( Z ) ) I) . 
zED(zo;8) zED(zo;8) 
If Zo is an attracting fixed point of g, then for sufficiently small 15 
g ( D ( zo ;  (5) ) � D (zo ; (5) . 
Thus, for small 15 > 0, the maximum modulus principle implies that 
sup I Ul ( g ( Z ) )  - vl ( g ( z) ) I ::; sup I U l (Z) - vl (z) l .  
ZED(zo;8) ZED(zo;8) 
Setting M = max { l + l a l l ,  1 + l a2 1 , . . .  , l + l an - l l , l ao l + l b l } gives, for 15 > 0 sufficiently 
small and z E D ( zo ;  8) , 
I I Tu - Tv l l  ::; I5M l l u - v i i . 
For 15 sufficiently small, 8M < 1 ,  and hence the operator T is a contraction and we 
84 Natural Boundaries for Functional Differential Equations 
conclude that the integral equation (4.24) has a unique solution in 1{= (D (zo ; 8) ) .  
We can thus generalize Theorem 4.7 to the following. 
Theorem 4.28 Let Zo be any attracting fixed point of the non-constant entire func­
tion g .  Then there exists a local solution to (4.20) satisfying (4.22) , unique among 
functions holomorphic at zo o 
Given that the n-th order differential operator L in (4 .21)  maps a holomorphic 
function y to a holomorphic function, the proof for Theorem 4.8 can be readily 
extended. 
Theorem 4.29 Let Zo be an attracting fixed point of the non-constant entire func­
tion g .  Then there exists a unique solution to the initial value problem (4 .20) , (4 .22) 
that is holomorphic in the set A�o ' 
In order to extend Theorem 4.27 to handle higher order operators, we need only 
show that we can establish an equivalent inequality to (4. 15) .  Indeed, for z E CR, 
the Cauchy integral formula and maximum modulus principle yield 
hence, 
I L I (
n !  lan-ll (n - I) ! � I I) M(R � ) . y ::; 8n + 8n-1 + . . .  + 8 + a o + u ,
I ( ( ) ) 1 
1 
(
n !  lan-l l (n - I ) !  la l l  I I) ( ) y g z ::; 1bf 8n + 8n-1 + . . .  + T + ao M R + 8 . 
For R sufficiently large, there exists a 80 such that 
Replacing inequality (4. 15) with 
in the proof of Theorem 4.27, and all references to a and 8 with ao and 80 respec­
tively, the result follows mutatis mutandis. 
4.5  Generalisations 85  
Theorem 4.30 Let Yo =I- 0 ,  a o  =I- b ,  and let Zo b e  an attracting fixed point for the 
non-linear polynomial g .  Then the solution t o  the equation ( 4 . 2 0 )  satisfying ( 4 . 2 2 )  
has aA�o a s  a natural boundary. 
4.5 .2  Entire functional arguments 
We now look to extend Theorem 4 . 27 and 4.30 to the case where 9 is a non-constanst 
entire function. The main issue is that 9 may have an exceptional point at which 
Y may not be holomorphic. 
In the polynomial case, if 9 had an exceptional point, it was the only attracting 
point of g. However, no such claims can be made in the non-polynomial case. For 
example, the entire funtion f (z) = Azez has z = 0 as an exceptional point (for z = 0 
is the only solution to f (z )  = 0 ) ,  yet 1' (0) = A .  Thus the exceptional point need 
not be in the Fatou set, and no conclusion can be immediately drawn on whether 
y is holomorphic there. 
We can conclude that if y is a solution to ( 4 . 2 0 ) , 9 is an entire function, and y is 
holomorphic at some point on the Julia set J ( g ) ,  then y must be either an entire 
function, or holomorphic at all points in C with at most one exception. In fact, we 
can alter the proof for Theorem 4 . 27 and Theorem 4 . 3 0  to eliminate the possibility 
of non-constant entire solutions. 
Theorem 4.31 Let y be any solution to (4 . 20 )  where 9 is a non- linear entire func­
tion with a fixed point Zo . If y is holomorphic at some point W E J(g ) ,  then one of 
the fo llowing must hold: 
1 .  Y is a constant. 
2. Y is holomorphic in C "  {go }  and is singular at go , where go is the exceptional 
point of g .  
Proof: Let y be a solution of  (4 . 2 0 )  holomorphic at some point W E J (g ) .  Lemma 
4 . 25 shows that either y is entire, or is holomorphic throughout C "  {go} where go 
is the exceptional point of g. All we must show, therefore, is that y cannot be a 
non-constant entire function. 
86 Natural Boundaries for Functional Differential Equations 
Assume that y is a non-constant entire function. As was done in Theorem 4.27, we 
can establish the inequality 
\ y ( g ( z) ) \  � ( 1 + ab
o ) M ( R + 6o ) ,  
for all z E C R for R sufficiently large. We are required to establish a lower bound 
on \ y ( g ( z ) ) \ for some point z E CR to replace (4. 1 7) .  In order to establish this lower 
bound, we need to be able to have g ( O )  = o. 
Suppose that g ( O) =I 0, i.e. the origin is not a fixed point of g .  Let w = z - Zo , 
h ( w )  = g ( w  + zo) - Zo , and f ( w )  = y ( w + zo ) .  Then (4.20) transforms to 
f (n) ( w )  + an_ d (n-
l ) ( w )  + . . .  + a d' (w) + ao f ( w )  = bf ( h ( w ) ) ,  ( 4.25) 
and h ( O )  = O .  Thus f is a non-constant entire solution to (4 .25) . We may therefore 
assume that g ( O )  = o. 
For entire functions f and 9 such that g ( O) = 0, P6lya [30] showed that there is a 
constant c, 0 < c < 1 such that 
( 4.26) 
for all R > O. Now, 9 is non-linear, so there exists a constant {3 > 0 such that 
Mg ( R )  > {3R2 for all R sufficiently large by Liouville 's Theorem. Hence, for R 
large, there exists a Zl E C R such that 
Setting a = c� we have established the inequality (4 . 1 7) , and the rest of the proof 
of Theorem 4.27 then shows a contradiction to the maximum modulus principle. 
Thus, y must be a constant function. • 
The first case in Theorem 4.31 can be ruled out by placing the same conditions on 
Yo ,  a and b as were used in Theorem 4 .27. Assuming 9 has no exceptional point 
(or the exceptional point is in the immediate basin of attraction) , we can therefore 
conclude that y cannot be analytically continued across the Julia set. Thus, aA�o 
must be a natural boundary. We sum up the results as follows. 
4.5 Generalisations 87 
Theorem 4.32 Let Yo =1= 0 and a =1= b ,  a n d  let y b e  any solution to the initial value 
problem (4.20) , (4.22) where Zo is an a ttracting fixed point of the n on-linear entire 
function g ( z ) . Then y is holomorphic in the entire complex plane with the exception 
of a single point, or y is ho lomorphic within the immediate basin of attraction A�o ' 
and has aA�o as a natural boundary. 
If 9 has an exceptional point that is outside of the basin of attraction, then we can 
not conclude that y can be analytically continued across the Julia set - the Julia set 
may still form a natural boundary for y .  The likelihood of (4.20) having a solution 
that is holomorphic at some point in the Julia set (in particular at a repelling fixed 
point) will be further examined in Chapter 5.  
4.5.3 Equations with polynomial coefficients 
In this last section, we examine the possibility of replacing the constants ak , 
k = 0, . . .  , n  - 1 and b in the functional differential equation (4.20) with functions 
holomorphic at the initial point Zoo 
Consider the equation 
y (n) ( z )  + an _ l ( z )y (n-
l ) (z) + . . .  + a l (z ) y' ( z )  + ao ( z ) y ( z )  = b ( z) y (g ( z ) ) ,  (4.27) 
where the functions ak (z) , k = 0, . . .  , n  - 1 , b ( z )  and g ( z )  are entire. 
Let H (y (z) , y ( g (z ) ) )  be defined as in the proof of Theorem 4.28, but with the ak 's 
and b being replaced by functions of z , and let T be as before. Then, 
I I Tu - Tv l l  ::; I z  - zo l (t �up I Uk ( Z )  - vk ( z ) 1 
k=2 ZED(zo;8) 
+ �up I b(z) (Ul (g (Z) ) - Vl ( g ( Z ) ) )  - t ak - l (Z ) (Uk (Z) - Vk (Z) ) I) 
ZED(zo;8) k= l 
::; 8 (t(1 + Ak - 1 (8) ) �up I Uk ( Z )  - vk ( z ) 1 
k=2 zED(zo;8) 
+ A o (8 )  �up I U l ( Z )  - v l ( z ) 1 + B (8 )  �up l u (g ( z ) )  - V ( g (Z) ) I) 
ZED(zo;8) ZED(zo;8) 
88 Natural Boundaries for Functional Differential Equations 
where B (8) = SUPzED(zo ;o) I b ( z ) 1 and Ak ( 8 )  = SUPzED(zo ;o) l ad z) 1 for k = 0, . . .  n - 1 .  
Hence, for 8 sufficiently small and z E D ( zo ;  8 ) ,  
I ITu - Tv l l  � 8 M l lu - v i i , 
where M = max{l + A l (8) , 1 + A2 (8) , . . .  , 1 + An- 1 (8) , A ( 8 )  + B ( 8 ) } .  The functions 
ak and b are all holomorphic in D ( zo ; 8) , hence Ak and B are monotone increasing 
functions of 8 by the maximum modulus principle. We can thus find a 8 sufficiently 
small such that 8M < 1 as before. 
Note also that the proof of Theorem 4.8 can be easily extended to this case as it 
relies only on the fact that y is holomorphic at some point W if and only if the 
differential operator is holomorphic at w. This certainly holds for the differential 
operator Ly = yn (z) + an_ l ( Z ) y(n-
l) ( z )  + . . .  + a l ( z ) y' ( z )  + ao ( z ) y ( z ) . We thus have 
the following result . 
Theorem 4.33 Let Zo be an attracting fixed point of the non-constant entire func­
tion g .  Then there exists a unique solution to (4.27) satisfying (4 .22) that is holo­
morphic in A�o ' 
There are two key difficulties that we face when attempting to prove that aA�o 
must be a natural boundary for our solution. The first problem is that the proof 
of Lemma 4.26 breaks down when we hit the zeros of b(z) . We can however work 
around this problem as follows. Suppose y is the solution from Theorem 4.33 and 
that y is holomorphic at some point W E J(g) . Then there is a neighbourhood U 
of W such that y E 7-l ( U) . Let Wl be any point in g ( U ) .  There then exists a Wo in 
U such that Wl = g(wo ) .  As y is holomorphic at Wo , the linear differential operator 
( 4.28) 
is also holomorphic at Wo . Now, y is a solution to (4.27) ; hence, either y is holo­
morphic at Wl , or Wo is a zero of b(z) and y has a pole at W l ' Suppose that y has a 
pole at W l . Then by following the same argument, y must have singularities (they 
need not be poles; indeed, they can not possibly all be poles) at all preimages of Wl , 
with the possible exceptions of those preimages where b has a zero that matches the 
singularity of y. The solution y must also be singular at all the images of Wl , and 
4.5  Generalisations 89 
all of their respective preimages as well (once again, with the possible exception of 
points where b has a matching zero) . For any point ill E J ( g  ) , there exists points 
arbitrarily close that are preimages of Wl by Theorem 4.21 , hence ill is a limit point 
of singularities, and thus must be a singularity of y .  Hence y must be singular at 
every point in the Julia set, which gives us a contradiction to our hypothesis that y 
was holomorphic at w. Therefore y is holomorphic at Wl ,  and as Wl is an arbitrary 
point in g ( U ) ,  y E 1-£ ( g ( U ) ) .  
Repeating the above arguments with g ( U) in the place of U shows that y E 
1-£ (g2 ( U ) ) ,  and an inductive argument then shows that y E 1-£(C) , where 
00 
We have thus established the following results. 
Lemma 4.34 Let Zo be an attracting fixed point of the non-constant entire function 
g ,  and let y be the s o lution to the differential equation (4 .27) satisfying (4.22) . 
Suppose that y is holomorphic at a point W E J(g) . Then y is holomorphic in 
C "- go where go is the exceptional point of g .  
Corollary 4.35 Let Zo be an attracting fixed point of the polynomial g ,  and let y 
be the s o lution to the differential equation (4.27) satisfying (4.22) . Suppose that y 
is holomorphic at a point W E J (g) . Then y is an entire function. 
The second problem that arises when trying to obtain a natural boundary for the 
solution to an equation with variable coefficients is that the growth of the differential 
operator is dependent on the growth of the coefficients ak ( z ) .  This growth must 
then be offset with the growth of y ( g ( z ) )  and b(z) . In general this can be difficult 
to track, but if we restrict ourselves to polynomials for the coefficients ak and b 
and the functional argument 9 we can make progress. We start with a lemma that 
concerns the growth of entire functions. 
Lemma 4.36 Let f be an entire function, 6, " c ,  positive constants, and C, N 
positive integers. If f s atisfies the inequality 
(4.29) 
90 Natural Boundaries for Functional Differential Equations 
for all R sufficiently large, then f is a polynomial of degree at most r d'!-l l .  
Proof: Let fo = f ,  eo = c, and 
h ( z )  = 
fo (z) - fo ( O )  . 
z 
Then !I is also an entire function, and by the triangle inequality we have the 
relations 
M (R) < MJo (R) + I fo (O) 1 h - R 
MJo (R) :S RMh (R) + I fo (O) I · 
Combining these with inequality (4.29) yields 
M ( RG) < MJobR
G) + I fo ( O ) 1 
h , - ,RG 
< CORNMJo (R + 8) + I fo ( O ) 1 
- ,RG 
< CORN(R + 8)Mh (R + 8) + (CORN + l ) l fo ( O ) 1 - ,RG 
< { coRN (R + 8) ( 1  + eoRN) lfo (O) I } M (R 8) - ,RG + ,RGMh (R + 8) h + . 
For R sufficiently large, there is a Cl > 0 such that 
thus 
Repeating the process inductively using 
yields further constants Ck such that for R sufficiently large, 
Eventually, we reach the value of k where N - k ( G  - 1 )  < O. For this value of k ,  
4.5 Generalisations 91  
and sufficiently large R,  we have the inequalities 
and 
Hence, by the maximum modulus theorem, fk must be a constant function (possibly 
identically zero) , and therefore f must be a polynomial of at most degree k. • 
Theorem 4.37 Let Zo be an attracting fixed point for the non- linear polynomi a l  
g .  Then the solution t o  t h e  equation (4 . 27) satisfying (4.22) either has aA�o a s  a 
natural boundary, or is a polynomial. 
Proof: Let y be the solution to (4.27) satisfying (4.22) , and let L be the differential 
operator defined in (4.28) .  Then the Cauchy integral formula, and the maximum 
modulus theorem give 
(n !  (n - I ) !  1 ) sup I L y l :::; J:n + Man_l (R) 8n-1 + . . .  + Mal (R) �  + Mao (R) My (R + 8) ZECR U U 
(4.30) 
where 0 < 8 < R. Let the degree of the polynomial ak ( z )  be Ak for k = 0, 1 ,  . . .  , n -
1 ,  and the degree of b(z) be B . Then there exists positive constants RI , (3, and ak 
such that 
(4.3 1 )  
for k = 0, 1 ,  . . .  , n - 1 and 
(4.32) 
for all R > RI by Liouville's Theorem. Now, a similar argument as used in the 
proof of Theorem 4.27, shows that there exists an R2 > 0 and , > 0 such that 
for all z E eR with R > R2, where G is the degree of g ( z ) . Thus, for such R, the 
curve r = { g ( z )  : z E eR} completely encloses the curve C-yRO , and the maximum 
modulus principle implies that there is a ZI E eR such that 
92 Natural Boundaries for Functional Differential Equations 
Combining this with inequalities (4.30) ,  (4.31)  and (4.32) yields 
where N = max:{Ao ,  AI , " . An-I ,  E} - E .  For R sufficiently large, there exists a 80 
such that 
thus, 
My(!RG) < (1 + � ) RN My (R + 80 ) , 
and so by Lemma 4.36, y must be a polynomial of degree at most r d'!..l l .  
If y is not a polynomial, then we have a contradiction and thus y is not holomorphic 
at any point on aA�o ' and hence aA�o is a natural boundary for y. • 
Note that, although polynomial coefficients allow for polynomial solutions, these 
solutions need not exist. Indeed, both the degrees of the polynomials ak , b and 
9 and their coefficients put constraints on the solution in addition to the initial 
conditions (4.22) .  For example, consider the functional differential equation 
( 4.33) 
along with the initial condition 
y (O) = Co, 
where the ak , bk l k = 0, 1 and Co are complex constants. From the proof of Theorem 
4.37 we note that any polynomial solution to (4.33) must have degree at most 1 .  
Let y (z) = CIZ + Co where Cl is a constant. Then 
4.5  Generalisations 
and thus we have the relationships 
aOC l + a l Co - bl Co = 0, 
aoCo - boCo + Cl = o. 
93 
(4.34) 
( 4.35) 
(4.36) 
(4.37) 
Suppose that Cl = O .  Then from (4.36) and (4.37) we obtain that bo = ao and 
bl = a l in order to get a non-trivial constant solution. If Cl =I- 0, then (4.34) and 
(4.35) yield that bo = a l and b1 = 0, and upon substituting these into (4.36) and 
(4.37) we have the further requirement that either ao = - 1 , or 
Thus, relatively few examples of (4.33) actually yield entire solutions - the rest all 
have natural boundaries on D(O; 1 )  by Theorem 4 .37. 
94 Natural Boundaries for Functional Differential Equations 
CHAPTER 5 
Holomorphic Solutions 
about Repelling Fixed 
Points 
In this chapter we consider solutions to functional differential equations holomor­
phic about a repelling fixed point . Consider the functional differential equation 
y" ( z )  + p ( z ) y' (z) + q ( z ) y ( z )  = Ar (z ) y (g ( z ) )  (5. 1 )  
where p, q ,  r and 9 are functions holomorphic at zo , a repelling fixed point of g ,  A 
is a constant , and r (zo) #- o. We wish to find non-trivial holomorphic solutions to 
(5. 1 )  about the point zoo The paper [23] contains much of what is presented in this 
chapter, excluding §5.5. 
From our experience with the first order pantograph equation in Chapter 2 ,  we 
would expect that this more general equation will only admit holomorphic solutions 
for specific configurations of the functions p ,  q and r and the constant A. Most of 
the literature concerning functional differential equations deals with the case where 
9 is linear ( [ 1 1 ] ,  [19] , [5] , [18] , [37] ) where little mention is made of the case where 
the solution is holomorphic at a repelling point . A notable exception is Oberg [28] 
who studied the first order equation 
y' ( z )  = A q ( Z ) y ( g ( z) ) ,  (5.2) 
where q and 9 are holomorphic at zo , and Z o  is a repelling fixed point of g.  Oberg's 
analysis shows that equation (5.2) has non-trivial solutions holomorphic at Zo only 
96 Holomorphic Solutions about Repelling Fixed Points 
for certain values of A. He also showed that the values of A that yielded a holomor­
phic solution formed a discrete set. 
The key to Oberg's results is that equation (5.2) can be replaced by one that involves 
a compact operator. Once the reformulation is done, we are left with an eigenvalue 
problem for analyticity, where the operator is compact . We can therefore use the 
formidable array of results from the spectral theory of compact operators to gain 
more understanding into the likelihood of holomorphic solutions. Oberg limits his 
use of the spectral theory to the discrete nature of the spectrum, and the finite 
dimensional nature of the corresponding eigenspaces. His analysis, however, can be 
readily adapted to more general equations such as equation (5. 1 ) .  The requirement 
of analyticity allows use of the Cauchy Integral formula, and this coupled with the 
repelling nature of the fixed point provides the mechanism whereby the problem 
can be reposed as an eigenvalue problem with a compact operator. 
5 . 1  The Eigenvalue Problem for Holomorphic So­
lutions 
We start by transforming (5. 1 )  to an eigenvalue problem. Let .c denote the operator 
defined by 
.cy ( z )  = a2 (z)y" ( z )  + a l ( z ) y ' ( z )  + ao ( z ) y ( z )  
where a o ,  a I ,  and a2 are functions holomorphic at Zo with a2 ( zO ) #- O. Setting 
and 
q ( z )  ao ( z )  = r ( z ) · 
(5 .3) 
we have that ao , a l and a2 are holomorphic at Zo , and a2 ( zO ) #- 0, so we may write 
(5 .1 ) in the form 
.cy ( z )  = Ay ( g ( Z) ) .  (5 .4) 
We thus consider the problem of determining the values of A E re such that equation 
(5.4) has a non-trivial holomorphic solution at zo0 The set of all such A, i.e. the 
spectrum, will be denoted by S. 
The character of S depends crucially on the nature of the fixed point . From Theo­
rem 4.33 we see that if Zo is attracting, then S = re. However, in the repelling case, 
5 .1  The Eigenvalue Problem for Holomorphic Solutions 97 
we will first show that the spectrum must be discrete, before using the results of 
Chapter 4 to conclude that in many cases, the spectrum consists of just one point. 
We start by altering the analysis of Oberg to show that the spectrum is discrete, 
and that >. = 0 is the only possible limit point of S. 
Since Z o  is a repelling fixed point, g' ( zo )  i= 0, and 9 has a unique inverse function 
h holomorphic at g ( zo )  = zo o Moreover, since I g' ( zo ) 1 > 1 we have I h' (zo) 1 < 1 ;  
therefore, Zo is an attracting fixed point for h .  Let .ch denote the operator defined 
by 
.ch y ( z )  = a2 ( h (z) ) y" ( h (z) ) + a l ( h ( z ) ) y' ( h ( z ) ) + ao ( h ( z ) ) y ( h ( z ) ) .  
Equation (5.4) can be replaced by the equation 
(5.5) 
and, since y is holomorphic at Zo if and only if y a h  is holomorphic at zo, the problem 
of finding holomorphic solutions to equation (5.4) at Zo is equivalent to that of 
finding holomorphic solutions to equation (5.5) at zoo This seemingly innocuous 
step has the effect of transforming the original problem into an eigenvalue problem 
with a compact operator. 
Theorem 5 . 1  Let 1{oo ( D (zo ;  p) ) denote the Banach space of functions holomorphic 
in D (zo; p) and continuous in D ( zo ;  p )  equipped with the norm 
I I f l l oo = sup I f ( z) l · 
Izl=p 
There is a p > 0 such that .ch is a compact operator from 'HOC! ( D (zo ; p) ) into 
'Hoo ( D ( zo ;  p) ) . 
Proof: Since ao , a l , a2 are holomorphic at zo , and I h' (zo) 1 < 1 ,  there are numbers 
R >  p > 0 such that ao , a l , a2 , h E  'H ( D ( zo ;  R)) and 
W c  D (zo; p) , 
where W denotes the closure of the set W = { h (z) : z E D (zo; Rn. For such a 
choice of R and p, it is evident that .ch maps 'HOO ( D ( zo ; p) ) into 'HOO ( D ( zo ; p) ) .  
98 Holomorphic Solutions about Repelling Fixed Points 
We use the Arzela-Ascoli Theorem [21 ] to show that .ch is compact. Let F = { in } 
be a bounded family of functions in '}-{OO(D(zo ; p) ) .  Then there exists a number 
M such that I l in l l oo � M for all n.  Moreover, there exists a number A such that 
I l ak l l oo � A,  for k = 0, 1 , 2. Now, for any in E F and z E D ( zo ;  p) we have that 
h(z) = w E W and 
2 
l .ch in (z) 1 � L l ak (w) l l i�k) ( w ) 1 
k=O 
< A � I� r in (�) d� 1 � 27ri J1 E. I =R (� - W ) (k+ l ) 
{ 2 1 I } < ARM (R _ p)3 + (R _ p)2 + R _ p ; 
hence, { Lh i n } is uniformly bounded in '}-{OO ( D  (zo ; p) ) .  
We now show that Lh is equicontinuous in '}-{OO(D(zo ; p) ) .  Let Zl , Z2 E D(zo ; R) . 
Then for any in E F we have W l , W2 E W and 
Since am E H(D(zo; p) for m = 0, 1 , 2 , there are numbers 'Ym such that 
5 .1  The Eigenvalue Problem for Holomorphic Solutions 
for m = 0, 1 , 2 .  We thus have 
where 
In a similar manner, we can find numbers Ko and K 1 such that 
and 
Indeed, the constants Km can be found from 
for m = 0, 1 , 2. 
m!RM { mRm-lA 
} Km = (R _ p)m (R _ p)m + Im , 
Since h E H(D(zo ; p) ) there is a number Ih such that 
99 
Thus, there is a constant K = Ih max{Ko, K1 , K2} such that for all Zl , Z2 E D(zo; R) 
and hence .ch is equicontinuous. The Arzeli-Ascoli Theorem therefore implies that 
.ch is compact . • 
A basic property of a compact operator is that its spectrum is discrete; moreover, 
the associated eigenspace must be finite dimensional. Theorem 5 . 1  thus immedi­
ately yields the following result. 
Corollary 5.2 The set S has n o  limit point except perhaps at >. = 0. The eigen­
space associated with any A E S is finite dimensional. 
100 Holomorphic Solutions about Repelling Fixed Points 
5 .2  Spectrum of the Pantograph Equation 
In Chapter 2 ,  we found that in the repelling case of the first order pantograph 
equation, the only solutions holomorphic at the origin were polynomials and they 
arose when >. = b / an. In the first order case then, 
which has a limit point at >. = ° as allowed by Corollary 5 . 2 .  
In this section, we study the spectrum associated with the second order pantograph 
equation and establish that the spectrum is the same as for the first order result, 
although to do so requires a more complicated method. Consider the functional 
differential equation 
y" ( Z )  + ay' ( z )  + by ( z )  = >. y ( a z )  ( 5 . 6 )  
where a ,  b ,  >. and a are constants with lal > 1 .  Substituting a formal power series 
y ( z) = L�=o Cn zn into ( 5 . 6 )  yields the recurrence relation 
(>.an - b)Cn - a Cn+ l ( n + 1 )  Cn+2 = ( n + 2) ( n + 1 ) . 
Unlike the first order case, it is not immediately clear from the above recurrence 
relation that the power series would diverge for z =J. 0. The coefficients Cn look as 
if they would grow large, though it is hard to draw anything conclusive from the 
above expression due to the difference present in the numerator. We instead attack 
the problem using some of the techniques from Chapter 4 .  
The Julia set of the functional argument az when lal > 1 consists entirely of the 
origin by Theorem 4 . 2 1 ,  as the only point fixed by az is z = 0, and this point is 
repelling and has no other pre-images. Thus, by the arguments used in the proof 
of Lemma 4 . 2 6 ,  any solution to ( 5 . 6 )  holomorphic at the origin must be entire. We 
start by proving the following lemma. 
Lemma 5.3 Let f be an entire function, and let M ( R) = sUPlzl=R I f ( z ) 1 for all 
R > 0. Suppose there exis ts constants f3 > 1 and fJ > 0, and a non-negative integer 
n such that 
M(f3R) ::; f3n M ( R  + fJ) , ( 5 . 7) 
5.2 Spectrum of the Pantograph Equation 101 
for all R sufficiently large. Then f i s  a polynomial of degree d ::; n .  
Proof: Polya & Szego [31 ]  show that , for any fixed number J1" 0 < J1, < 1 ,  i f  f i s  a 
transcendental entire function then 
lim 
M
(J1,
R
) = 0 
R-+oo M(R) 
. 
Given inequality ( 5 . 7 ) , there is a number " 1 < , < f3 such that 
( 5 . 8 )  
for all R sufficiently large. Note that we can choose , arbitrarily close to 1 by 
increasing R, as we require only that , > 1 + 8/ R, and 8 does not depend on R. 
Therefore 
M(p,p) > � > 0 
M(p) - f3n ' ( 5 . 9 )  
for all p large, where 1 /  f3 < J1, = , /  j3 < 1 .  Hence, relation ( 5 . 8 )  cannot be satisfied, 
and therefore f is a polynomial. Suppose that f has degree d. Then there exists 
constants T/1 , T/2 such that T/IRd < M(R) < T/2Rd for R large by Liouville's Theorem, 
where the ratio � tends to 1 as R increases. The relation (5.9) then yields 
Hence, 
T/2 d 1 -'V > -­
T/1 I - j3n-d ' 
and as j3 > 1 and both !l1. and , can be made arbitrarily close to 1 ,  it follows that T/l 
d ::; n .  • 
We are now in a position to prove that any solution holomorphic at the origin must 
be a polynomial. 
Theorem 5 .4 Suppos e  that A =1= 0 and y is a non- trivial solution to (5.6) that is 
holomorphic at z = o. Th en A E D (O ;  I b l )  and y is a polynomial. 
Proof: As y is holomorphic at a point on the Julia set, it is entire. Using the same 
102 Holomorphic Solutions about Repelling Fixed Points 
argument as in the proof for Theorem 4.27, we can quickly establish the inequality 
M ( l a I R) � I�I (:2 + 
I�I + I b l) M ( R  + 8 ) , ( 5 . 1 0) 
for constants 8, R such that 0 < 8 < R .  
Suppose that I A I  > I b l . Then for 8 and R sufficiently large we have 
Fix 8 and choose R so large that l a i R >  R + 8 .  Then the above inequality implies 
M( l a I R) < M ( R  + 8) , 
which contradicts the maximum modulus principle. Hence, I A I � I b l . 
Suppose that A E D (O ;  I b l ) .  Then there is a non-negative integer n such that 
Inequalities 5 . 1 0  and 5 . 1 1  imply 
an+ 1 ( 2 l a l ) M ( l a I R) � lbI 82 + T + I b l M ( R  + 8 ) ,  
and hence for 8 and R sufficiently large 
M ( l a I R) � l a i n+ ! M ( R  + 8 ) .  
Lemma 5 . 3  thus implies that y is a polynomial of degree d � n + 1 .  • 
( 5 . 1 1 )  
Note that if b = 0 ,  the above analysis shows that the set of non-zero eigenvalues is 
empty. For b =1= 0 we have the following. 
Corollary 5 .5  Let b =1= O. The spectrum associated with ( 5 . 6 )  is S = { bjak }k�o . 
For any non-negative integer k, the eigenfunction associated with Ak = bj ak is a 
polynomial of degree k that is determined uniquely up to a multiplicative constant. 
5 .2  Spectrum of the Pantograph Equation 103 
Proof: Theorem 5 . 4  shows that any non-trivial solution to equation ( 5 . 6 )  holomor­
phic at z = 0 must be a polynomial. Therefore, there exists a non-negative integer 
n, and n + 1 constants Cn, . . .  , Co with Cn i= 0 such that 
( ) n + n - l  y Z = CnZ Cn- I Z + . . . + C I Z + Co. ( 5 . 1 2 )  
Substituting this expression into the functional differentional equation ( 5 .6) and 
equating coefficients of zn gives 
( 5 . 1 3 )  
and hence 
We thus conclude that if >. E S then >. = b/an for some non-negative integer n.  
Suppose that >. = b/an for some non-negative integer n.  Equation ( 5 . 1 3) shows us 
that Cn can be chosen arbitrarily, and equating coefficients of zn- l gives 
an 
Cn-l = b ( l /a _ 1 )  Cn , 
and the general recurrence relation 
(k + 2 ) ( k  + 1 ) ck+2 + a( k  + l ) Ck + 1 + b ( l  - Ci-n ) Ck = 0 ,  
for k = 0, 1 ,  . . .  , n - 2 .  Given a value of Cn, we can thus determine the other 
coefficients uniquely. Hence, for each non-negative integer n the choice >. = b/an 
corresponds to a polynomial solution to ( 5 . 6 ) ,  and therefore >. E S. Moreover, the 
polynomial solution is determined uniquely up to the choice of the multiplicative 
constant Cn . • 
It is clear that the arguments used to show that the second order pantograph equa­
tion has spectrum S = { b/ an } �=o will apply equally well to higher order pantograph 
equations. All pantograph equations, therefore, have no solutions holomorphic at 
Z = 0 except the polynomial solutions that arise when >. E S. 
104 Holomorphic Solutions about Repelling Fixed Points 
5 .3  Non-linear Polynomial Arguments 
Having studied the spectrum of the pantograph equation, it seems natural to in­
vestigate the existence of solutions about repelling fixed points of equations with 
more general functional arguments. Consider the functional differential equation 
y" (z) + ay' (z) + by ( z) = Ay (g(Z ) ) , ( 5 . 1 4 )  
where a and b are complex constants, and 9 is a polynomial of degree d 2: 2 .  
One might expect that the discrete spectrum is much like that of the pantograph 
equation, albeit more complicated to determine, but it turns out that, in general, 
the spectrum is simpler. In fact we show that there is precisely one non-zero 
eigenvalue unless 9 is a cyclic polynomial. 
If we drop the condition in Lemma 4 . 2 6  that y be holomorphic at an attracting point 
zo , replacing it with the condition that 9 be non-cyclic, we obtain the following. 
Corollary 5.6 Let 9 be a non- cyclic polynomial, and let Zo be a repe l ling fixed 
point of g .  If y is a s o lution to ( 5 . 1 4 )  that is holomorphic at Zo then y is an entire 
function. 
Theorem 4 . 3 1 ,  however, completely eliminates the possibility of non-constant entire 
solutions, so any solution to ( 5 . 1 4 )  holomorphic at a repelling fixed point of 9 must 
be a constant. Substituting a constant solution into our equation quickly yields the 
spectrum as follows. 
Theorem 5.7 Let 9 be a non- cyclic polynomial of degree at least 2, and let Zo be a 
repelling fixed point for g .  There exists a solution to ( 5 . 1 4) holomorphic at Zo and 
not i dentically 0, if and only if A = b. In this case, y is a constant function. 
Theorem 5.7 shows that if 9 is non-cyclic, and if b #- 0 then the only non-zero 
eigenvalue is A = b. If b = 0, then the set of non-zero eigenvalues is empty. 
If 9 is cyclic, the character of the spectrum can be somewhat different. Consider, 
for example, the equation 
( 5 . 1 5 )  
5.3 Non-linear Polynomial Arguments 105 
where n ;::: 2 .  The functional argument zn is cyclic, and has a repelling fixed point 
at z = 1 .  It also has an exceptional point at z = O. If a solution were to be 
holomorphic at z = 1 ,  then, it must either be a constant , or be holomorphic in 
the entire plane excluding the origin by Theorem 4 . 3 1 .  The only constant solution 
available is the trivial solution, so any eigenfunctions, if they exist, will have an 
isolated singularity at z = O. Let us first try a function of the form 
1 
Yo ( z )  = Co - , zm ( 5 . 1 6 )  
where Co is a constant. Substituting ( 5 . 1 6 )  into ( 5 . 1 5 )  we find that m must satisfy 
2 m = -­n - 1 ' 
( 5 . 1 7) 
and ), = m ( m  + 1 ) .  Thus we have at least one non-zero eigenvalue and its corre­
sponding eigenfunction which is holomorphic throughout the plane except at the 
origin. Unlike the case where 9 is non-cyclic, however, we can find other solu­
tions holomorphic at z = 1 for different choices of ), =F O. In fact, ( 5 . 1 5 )  can be 
transformed into a pantograph equation. Let 
1 y ( z) = - Y ( z ) , zm 
where m satisfies ( 5 . 1 7) .  Then ( 5 . 1 5 )  transforms to 
z2 y" (z) - 2mzY' ( z) + m ( m  + l ) Y ( z )  = ),Y ( zn ) .  
Let z = eW and Y ( z) = P ( w ) .  Then the above equation transforms to 
P" ( w )  - ( 2 m  + l ) P' ( w )  + m(m + l ) P ( w )  = ), P (nw ) , ( 5 . 1 8) 
which is a second order pantograph equation of the form ( 5 . 6 )  with ex = n > 1 .  The 
repelling fixed point Zo = 1 for zn is mapped to the repelling fixed point Wo = 0 for 
n w ,  and we thus know that ( 5 . 1 8 )  has eigenvalues of the form 
), = 
m ( m  + 1 )  
k ' n 
for non-negative integers k .  We can thus deduce that ( 5 . 1 5 )  also has these eigen-
106 Holomorphic Solutions about Repelling Fixed Points 
values, and that the corresponding eigenfunctions are of the form 
where Lk (z) is a polynomial in log z of degree k. 
5 .4 Non-homogeneous Equations 
In this section, we address the existence of holomorphic solutions about repelling 
fixed points to non-homogeneous functional differential equations. Consider the 
equation 
£y (z) - Ay ( g ( Z ) )  = f ( z ) ,  (5 . 19) 
where £ is defined as in §5. 1 ,  f and 9 are given functions holomorphic at zo , and 
Zo is a repelling fixed point of g. 
It was shown in §5 .1  that the homogeneous equation (5.4) could be replaced by 
(5.5) which involves a compact operator. Evidently, (5. 19) can be replaced by 
£hY (Z) - AY ( Z )  = F ( z ) , (5 .20) 
where 
F(z) = f (h(z) ) . 
The value of this reformulation is that extensions of the Fredholm type theorems 
are available for compact operators (cf. [ 15] ,  [20] ) .  For example, the following result 
is an immediate consequence of the compactness of £ h .  
Theorem 5.8 (Existence and Uniqueness) Let Zo b e  a repelling fixed point for 9 
and suppose A =I- O .  Then (5. 19) has a s o lution y that is holomorphic at Zo for any 
given function f holomorphic at Zo if and only if the only solution to (5.4) that 
is holomorphic at Zo is the trivial solution. In this case the s o lution to (5 . 19) is 
umque. 
Indeed, if A is not in the spectrum of £h , then the solution y can be represented by 
5.4 Non-homogeneous Equations 107 
the N eumann series 
( 5 . 2 1 )  
where, for n 2: 2 ,  
Theorem 5 . 8  contrasts sharply with the analogous problem for the case where Zo is 
an attracting fixed point . In this case, it can be shown through a simple alteration 
of the proof of Theorem 4 . 28 that holomorphic solutions to ( 5 . 1 9 )  exist for any 
A E C and any choice of f that is holomorphic at zoo Given A and f ,  however, 
there is no uniqueness because it is always possible to prescribe the initial values 
y ( zo) and y' ( zo ) .  It is thus remarkable that the repelling fixed point case yields a 
unique solution without any initial data.  
In § 5 . 2  we determined the spectrum for the second order pantograph equation. 
Since this spectrum is known explicitly, we can use the above theorem to generate 
a result specific to the pantograph equation. 
Corollary 5.9 Let A and b be n on-zero numbers such that A #- b/an for any non­
negative integer n .  For any given function f holomorphic at z = 0 there exists a 
unique s o lution y to the equation 
y" (z) + ay' ( z )  + by ( z) - Ay (az) = f ( z) , 
that is holomorphic at z = o. Here, a and a are complex constants such that 
1 0. 1 > 1 .  
As a simple example, consider the equation 
y" ( z )  - y ( z )  + ay (az )  = z ,  ( 5 . 22) 
which corresponds to the choice of A = -a and f ( z) = z .  This value of A is not in 
the spectrum, and hence the above corollary asserts that there is a unique solution 
to ( 5 . 2 2 )  holomorphic at O. We may find the solution either by inspection or by use 
of the Neumann series. Using the latter approach by way of illustration, we have 
108 Holomorphic Solutions about Repelling Fixed Points 
F(z) = z/a and 
Lh F  = F" (z/a) - F ( z /a ) .  
We can quickly generate the needed terms for the N eumann series as 
and in general 
Hence 
y ( z) = � (F ( Z) - � Lh F ( z )  + � L� F ( z )  + . . .  ) a a a 
= 
� ( � + Z
3 
+ z
5 
+ . . .  ) a 0 0 0 
z 
02 - 1 ' 
In contrast , consider the equation 
y" (Z) - y ( z) + y ( o z )  = 1 ,  ( 5 . 2 3 )  
which corresponds to the choice A = -1 and J (z) = 1 .  This value of A is in the 
spectrum and the corresponding eigenfunction is a constant. Suppose that there is 
a holomorphic solution y to equation ( 5 . 23 ) .  Then y can be represented as a power 
senes 00 
with non-zero radius of convergence. Substituting the power series into equation 
( 5 . 2 3 )  gives 
and, for all k 2: 2 ,  
nk-1 ( 1 _ 2m ) m=l 0 C2k = ( 2 k ) ! . 
5.4 Non-homogeneous Equations 
Consequently, 
li 2n+2 2 I C z2n+2 1 1 1 - a2n l m = z ---+ 00 n-->oo C2nZ2n ( 2n + 2) (2n + 1 )  1 1 
109 
for all z =1= 0 and therefore the power series must diverge. There is no solution to 
( 5.23 ) holomorphic at z 
= O .  
Before we end our discussion of non-homogeneous functional differential equations, 
we note that Theorem 5.8 is part of a more general result for compact operators, 
the Fredholm A lternative , that addresses the existence problem for solutions to 
the non-homogeneous equation when A is an eigenvalue. A version of the Fred­
holm A lternative is available for compact operators in general Banach spaces (cf. 
Kreyszig [ 20] ) ,  but for our case the result would involve a Banach space adjoint 
and the dual space of Hoo. It is more tractable to first recast the problem in some 
Hilbert space. We are interested primarily in local existence results as opposed to 
the behaviour of solutions on some disc boundary, and with this in mind we can 
recast the problem to one that involves a compact operator on the Hardy space 
H2 (D(zo; p)) for a suitable choice of p > O. Recall that H2(D(zo; p)) is the set of 
all functions f E H(D(zo ;  p) ) such that 
and that H2 (D(zo ; p) )  is a Hilbert space equipped with the inner product ( . , . ) 
defined by 
1 1 1 (1, g) = -2 f(z)g(z) -dz, 7r Iz l=p z 
for f, g E H2 (D(zo; p) ) .  The dual of H2(D(zo ; p) )  is H2 (D(zo; p) ) .  Let I>.. 
H2(D(zo; p) ) ---+ H2(D(zo;  p)) be the operator defined by 
7).Y(z) = .chY(z) - AY(Z) , 
and let 7;..* denote the (Hilbert space) adjoint of 7).. Since .ch : H2 (D(zo;  p) ) ---+ 
H2(D(zo;  p) ) is compact we have the following version of the Fredholm A lternative 
for functional differential equations of the form ( 5.19) 
Theorem 5.10 (Fredholm A lternative) Suppose that A =1= O .  Then one of the two 
following alternatives holds: 
110  Holomorphic Solutions about Repelling Fixed Points 
1. The homog eneous equation �y = 0 has only the trivial solution. In whic h 
case the non-homog eneous equation ( 5 . 1 9 )  has exactly one solution for each 
f E 1{2 ( D (zo ;  p) ) . 
2. The homog eneous equation �y = 0 has a non-trivial solution. In whic h case , 
for f E 1{2 ( D (zo ;  p) ) ,  the non-homog eneous equation ( 5 . 1 9 )  has a solution if 
and only if (F, K) = 0 for every solution K to the adjoint equation �* K = O .  
5 .5  A Natural Boundary for the Pantograph 
Equation 
As a final application of the theory we developed in Chapter 4 ,  we return to the 
pantograph equation, and show that if a solution is not holomorphic at the origin, 
but is holomorphic in some strip whose boundary includes a line through the origin, 
then that line forms a natural boundary for the solution. Note that the line in 
question does not correspond to the Julia set of the functional argument, which is 
{ O } ,  and thus the techniques used in Chapter 4 can not be applied directly. We can,  
however, use the properties of iterated functions that we have thus far observed to 
obtain a natural boundary nevertheless. The material below may also be found 
in [25] . We begin with the first order case as an example. 
Recall that the first order pantograph equation studied in Chapter 2 ,  
y ' (z) + by(z) = Ay ( a z )  ( 5 . 24 )  
where b, A and a are real constants with b > 0 and a > 1 ,  has Dirichlet series 
solutions of the form 
00 
y (z) = L ane-bOnz, ( 5 . 25 )  
n=O 
where the an 's are constant . We showed in Section 2 . 2  that solutions of this form 
are holomorphic in the right half plane ITa = {z E re :  Re(z) > O }  and, as they are 
clearly not polynomials, must have a singularity at z = O .  
However, we also saw in Section 2 . 4  that ( 5 . 2 4 )  exhibited non-uniqueness if I A I  
i s  large enough, and so this indicates that other non-Dirichlet series solutions are 
present . These too must be singular at z = O .  
5.5  A Natural Boundary for the Pantograph Equation 111  
In the special case where b = 0 ,  Fredrickson [ 1 0] showed that the associated Dirichlet 
series solution to equation ( 5 . 2 4 )  defines a function that has the imaginary axis as a 
natural boundary. His method of proof, however, is crucially dependent on b being 
zero, which led him to comment "The hypothesis b = 0 in this theorem seems quite 
unnatural (although it is essential in the proof) . Are any of the solutions we have 
generated continuable beyond the half plane given?" 
5 .5 . 1  The case where a is a positive integer 
We start by considering the case where a is a positive integer. We showed in 
Chapter 2, that equation ( 5 . 24) has non-trivial solutions of the form ( 5 . 2 5 )  where 
ao =1= 0 and 
( 5 . 26 )  
for n � 1.  
Consider the transformation defined by 
w = e-bz .  ( 5 . 27) 
This transformation maps the half plane IIo to the unit disc D (O; 1 ) ,  and the imag­
inary axis to the circle BD (O; 1 ) .  Under this transformation, the Dirichlet series 
( 5 . 2 5 )  becomes a series of the form 
00 
h ( w )  = L anw an . ( 5 . 28) 
n=O 
Now, a is an integer, so ( 5 . 2 8 )  is a power series in w .  Taking the modulus of a ratio 
of consecutive terms, and using ( 5 . 26) gives 
which converges to 0 as n --+ 00 if I w l  < 1 ,  yet diverges if I w l  > 1 .  Consequently, 
the power series defines a function h holomorphic in D (O; 1 )  that has at least one 
singularity on BD (O; 1 ) .  
Note that the power series ( 5 . 2 8 )  has large gaps between the non-zero terms. Indeed, 
112 Holomorphic Solutions about Repelling Fixed Points 
if we write h in the standard form 
00 
h ( w )  = I>jWj , 
j=O 
we have that Cj = 0 except when j belongs to the sequence {j d = { ak } .  Since 
Cl > 1 ,  there exists a number 11 such that jk + l > ( 1  + lI)jk for all k 2:: 1 .  We can 
thus appeal to the Hadamard gap theorem [33] to assert that o D ( O ;  1 )  is a natural 
boundary for h .  
If we assume that the Dirichlet senes solution y can be analytically continued 
across the imaginary axis, then it follows that h must be analytically continued 
across o D ( O ; 1 ) ,  which gives a contradiction. We have thus shown the following 
result . 
Theorem 5 .11  Let a >  1 be an integer, and suppose that y is a non- trivial Dirich­
let series solution of the form ( 5 . 25 )  with coefficients satisfying ( 5 . 26) . Then the 
imaginary axis is a natural boundary for y .  
ote that the conditions on Cl and y are somewhat restrictive and are certainly 
"unnatural" to use F'redrickson's term. Unfortunately, when Cl is not an integer, 
the transformed series ( 5 . 28) is not a power series, and so we cannot appeal to 
Hadamard's gap theorem. Furthermore, there is no obvious way in which we could 
extend the above analysis to non-Dirichlet series solutions. An additional compli­
cation is the existence of polynomial solutions to ( 5 . 2 4 )  for certain configurations of 
the constants b, C and Cl which clearly do not have the imaginary axis as a natural 
boundary. In order to rule the polynomial solutions out, we show in the following 
section that the condition that the solutions are singular at z = 0 is all that is 
needed to imply the existence of a natural boundary. 
5 .5 .2  The general first order equation 
Whether or not a function f has the imaginary axis as a natural boundary is only 
meaningful if f is holomorphic in some open set n e e  the boundary of which 
includes the imaginary axis. Let Y be the set of functions f such that: 
5.5  A Natural Boundary for the Pantograph Equation 113  
1 .  there exists an open set n � no such that ano � an and f E 1t ( n )  ; 
2. f has a singularity at z = 0 ;  and 
3. f satisfies equation ( 5 .24) for all z E n . 
Our goal is to show that the imaginary axis is a natural boundary for any function 
y E T . In order to do this, we use the transformation ( 5 . 27) that arose naturally 
from the Dirichlet series solutions to transform the problem to one concerning 
the unit circle aD(O; 1 ) .  This yields a new functional differential equation whose 
functional argument has a multivalent inverse. We can exploit the multivalent 
nature of this inverse to show that the unit circle forms a natural boundary for 
the transformed problem. A simple argument by contradiction then shows that the 
original problem has the imaginary axis as a natural boundary. 
Suppose that Y E T . Then there is a set n such that ano � an and f E 1t ( n) . Let 
W = {w E re :  w = e - bz , z  E n } .  Under the transformation ( 5 . 2 7 ) ,  if h ( w )  = y ( z ) ,  
then y' (z) = - bw h' ( w )  and y (o:z) = h (wO ) ;  thus, ( 5 . 24 )  transforms to the Cauchy­
Euler type functional differential equation 
( 5 . 29 )  
Let 
h(w) = y ( -� log(w )) ( 5 . 30 )  
for w E W.  Without loss of generality we can assume that n does not contain the 
point at infinity, so that W does not contain w = O. For any choice of w E W,  
then, there i s  a branch of log(w) such that h i s  holomorphic at w.  Moreover, h is 
a solution to ( 5 . 29 )  for any Y E T . We show that aD(O; 1) is a natural boundary 
for h, i.e. , there is no branch of log( w ) such that h can be analytically continued 
to values in aD(O; 1 ) .  
I n  the case where 0: is a positive integer, then the functional argument g (w) = WO 
in equation ( 5 . 29 )  is an entire function, so we can apply the results of Chapter 4 to 
obtain the natural boundary. Indeed, h is a solution to ( 5 . 29 )  and has a singularity 
at w = 1 ,  as y is singular at z = O. Using the arguments used to prove Theorem 
4 . 8 ,  we conclude that every pre-image of w = 1 is also a singularity for h, and 
1 14 Holomorphic Solutions about Repelling Fixed Points 
seeing as 1 E J ( g ) , these singularities are dense in the Julia set J ( g )  by Theorem 
4 . 2 1 .  Now, J ( g )  = 8D (0 ;  1 )  and hence 8 D ( 0 ;  1 )  is a natural boundary for h. 
The above analysis is only applicable in the case where a is an integer, as otherwise 
the functional argument 9 is not an entire function, and so we cannot apply the 
results of Chapter 4. In this more general case, however, we can use a similar argu­
ment by considering the pre-images of 1 .  We start by showing that the preimages 
of 1 under the the functional argument wO< are dense on 8D ( 0 ;  1 ) .  
Lemma 5 . 12  For all non-negative integers k ,  n let 
where a > 1 .  The set <I> defined by 
is dense on the interval [0 , 1] . 
k 
CPn,k = an ' (5 . 3 1 )  
Proof: Choose any two distinct points Po , P I  E [0 , 1] . Without loss of  generality we 
may assume that Po < PI . Since a > 1 we have that a-n --+ ° as n --+ 00, and 
hence there is a positive integer N such that 
Now the points 0, a - N ,  2 a - N ,  . . .  , laN J a -N are equally spaced on the interval [0 , 1 ] ,  
with a maximum distance a- N between consecutive points. Hence, there exists a 
k E Z· , k ::; aN such that 
k 
Po < N < PI · a 
We may choose Po and PI to be anywhere in the interval [0, 1 ] , and arbitrarily close 
together, hence the set <I> is dense on [0, 1 ] . • 
Now, let Wn,k = e2IjJn, k 7ri , where CPn, k are as in ( 5 . 3 1 ) , and let r = { wn,d .  Then 
WO,k = 1 for all k, and we have the relation 
5.5  A Natural Boundary for the Pantograph Equation 115  
for all k .  By  induction, each element of  r is a pre-image of 1 under the functional 
argument g ( w )  = WCX . Lemma 5 . 1 2  shows that the set <I> is dense on [0, 1] and 
therefore r must be dense on o D ( O ;  1 ) .  As all points in r are pre-images of 1 ,  the 
pre-images of 1 are dense on o D ( O ;  1 ) .  We can now prove the following result . 
Lemma 5.13  Suppose that h is a s o lution to ( 5 . 29 )  holomorphic in some set � � 
D (O ; 1 )  such that oD (O; 1 )  � o� . Suppose further that h has a singularity at w = 1 .  
Then oD ( O ;  1 )  is a natural boun dary for h .  
Proof: By hypothesis, h is singular at W = 1 and, as WO,k = 1 for all k ,  it i s  clear 
that h has a singularity at the points WO,k for all k. Suppose that h has singularities 
at all the points Wn,k for all k, where n is some non-negative integer. Then h' + �h 
also has singularities at these points. Since 
for all n, k, and since h is a solution to (5 . 29) , we conclude that h is also singular 
at Wn+1,k for all k. Induction then shows that h is singular at all points in r, and 
as the points in r are dense on o D ( O ;  1 ) ,  we have established the natural boundary .
• 
We are now in a position to prove the main result. 
Theorem 5 .14 If Y E Y, then the imaginary axis is a natural boundary for y .  
Proof: Suppose that y E Y,  and that there exists an analytic continuation Y of y 
across the imaginary axis. Therefore there is a point Zo with Re(zo ) = 0 such that 
Y is holomorphic at zo 0 Let h be a function defined by equation (5 . 30 ) .  Then h is 
a solution to ( 5 . 29) and for some branch of log W the function 
1 H(w) = Y (  - b  log w) 
is holomorphic in some open set 6. containing the point Wo = e -bzo E oD ( O ;  1 ) .  
Since y E Y, property ( 1 )  is satisfied, and hence there is an open set � � w � 
D ( O ;  1 )  such that 9 E 1{ ( � )  and o D ( O ;  1 )  � o � .  Since H(w) = h ( w )  for all w E � 
116 Holomorphic Solutions about Repelling Fixed Points 
and W n fj. i- 0 we have that H must be an analytic continuation of h across 
oD (O; 1 ) .  
Now, h must have a singularity at w = 1 ,  for otherwise y (z) = h(w) would imply 
that y is not singular at z = o. But Lemma 5 . 13  implies that o D ( O ;  1 )  is a natural 
boundary for h, and this contradicts the existence of the analytic continuation H . 
We thus conclude that for y E Y there does not exist an analytic continuation 
across the imaginary axis, and thus the imaginary axis is a natural boundary. • 
This result is readily extended to second and higher order pantograph type equa­
tions, as suitable transformations of the form (5.27) always result in a Cauchy-Euler 
type equation of the form (5.29), and much of the above analysis can be applied 
with only minor changes needed. We briefly discuss the second order case in order 
to show the necessary modifications. 
5 .5 .3  The second order pantograph equation 
Consider the equation 
y"(z) + ay' (z) + by (z) = )"y(etz) ,  (5.32) 
where a, b ,  ).. and et are real constants such that ).. =1= 0 and et > 1 .  In this case, the 
natural boundary will depend on the roots to the indicial equation 
r2 - ar + b = O. (5.33) 
As was shown in Chapter 3, there are two linearly independent Dirichlet series 
solutions to (5.32) corresponding to the two solutions rl and r2 to the indicial 
equation (5.33) . These solutions are holomorphic in the halfplane Re(rjx) > 0 for 
j = 1 , 2 . It is clear then, that if rl and r2 are not real and positive, the natural 
boundary will no longer be the imaginary axis ,  but rather the line Re(rjx) = O. i.e. 
Each solution may have a different natural boundary, depending on which root of 
the indicial equation that solution corresponds to. We modify our definition of the 
set Y as follows. 
Let r be a root of the indicial equation (5.33) ,  II (r) = {z  E re :  Re(rz) > O } ,  and 
define the set Y (r) as the set of functions f such that: 
5 . 5  A Natural Boundary for the Pantograph Equation 1 1 7 
1 .  there exists an open set 0 � II (r )  such that oII ( r )  � 00 and f E 'H ( O ) ;  
2 .  f has a singularity at z = 0 ;  and 
3 .  f satisfies equation ( 5 . 32 )  for all z E O .  
The transformation 
( 5 . 3 4 )  
then takes the set II ( r )  to the unit disc D(O;  1 ) ,  mapping the boundary Re (rz) = 0 
to oD(O; 1 )  and the repelling point z = 0 to w = 1 .  Setting h (w )  = y ( z ) ,  we can 
transform equation ( 5 . 3 2 )  to 
" ( )  
r - a ' ( )  b ( )  C ( at ) h w + --h w + 2"2 h w = 2"2 h w . rw r w r w ( 5 . 35 )  
For any y E Y (r ) ,  we can construct the function 
which will be (for a suitable branch of log z) a solution to ( 5 . 3 5 )  holomorphic in the 
set W = { w  E re :  w = e - rz , z E O} ,  where oD(O; 1 )  � ow. Furthermore, h must 
be singular at w = 1 .  
The differential operator does not affect the proof of Lemma 5. 1 3  so we can conclude 
that oD(O; 1 )  forms a natural boundary for h. The same arguments as used in 
Theorem 5 . 1 4  then give us the following result . 
Theorem 5 . 1 5  Let r be a root of the indicial equation ( 5 . 3 3 ) ,  and let y E Y (r ) . 
Then the line Re(rx) = 0 is a natural boun dary for y .  
Thus, a solution to a pantograph equation corresponding to a particular root of 
the associated indicial equation will have a line through the origin as a natural 
boundary, unless the solution is holomorphic at the origin .  If a particular solution 
corresponds to more than one root of the indicial equation (such as a sum of two 
linearly independent solutions) ,  then that solution will be holomorphic in the inter­
section of the halfplanes associated with each root. The boundary of the resultant 
wedge will then be the natural boundary. As an example, consider the equation 
Y" (Z) + 2y' ( z )  + 2y ( z )  = 3y (2z) . 
1 18 Holomorphic Solutions about Repelling Fixed Points 
The associated indicial equation is r2 - 2r + 2 = 0 which has roots rl = 1 + i and 
T2 = 1 - i. A Dirichlet series solution is given by 
where 
00 
Yl (Z) = L ene -2"r1z , 
n=O 
and 
00 
Y2 (Z) = L dne -2"r2z , 
n=O 
for constants en and dn . The function Yl is holomorphic in the halfplane Re(rl z) � 0 
and is singular at z = 0 , so by Theorem 5 . 1 5  has a natural boundary on the line 
{z : Re(z) = Im(z) } .  Similarly, Y2 is holomorphic in the halfplane Re(r2z) � 0 and 
thus has a natural boundary on the line {z : Re(z) = - Im(z) } .  Therefore, y must 
be holomorphic only where these two halfplanes intersect , i .e .  in the wedge 
1T 1T 
A = {z : -- < arg z < - } .  4 4 
The boundary of this wedge will be a natural boundary of y. 
Finally, we note that we cannot conclude that every solution to the pantograph 
equation that is singular at z = 0 must have a natural boundary along a line passing 
through the origin. For that we would be required to show that all solutions are 
holomorphic in some strip that includes a line through the origin as part of its 
boundary. However, the analysis used to discover the non-Dirichlet series solutions 
in §3.2, for instance, showed only that they must be in L2 [0 , 00 ) , and be singular at 
z = O. An interesting open problem, then, is to show that these general solutions 
must also have natural boundaries. 
CHAPTER 6 
Holomorphic Solutions 
about Neutral Fixed 
Points 
In Chapter 4 we discussed the existence of solutions to functional differential equa­
tions about attracting fixed points and found that there is always a holomorphic so­
lution available. In Chapter 5 we looked at the corresponding situation for repelling 
fixed points and found that the existence of holomorphic solutions is equivalent to 
an eigenvalue problem involving a compact operator - only certain equations yield 
holomorphic solutions. In this chapter we consider the case where the fixed point 
is neutral, presenting material from [24] . 
Recall that a neutral fixed point Zo of a function 9 is a point such that g ( zo) = Zo 
and I g' ( zo )  I = 1 .  Consider the functional differential equation 
y' ( z) + by (z) = >..y (g ( z ) ) ,  ( 6 . 1 )  
where b and >" are complex constants, and 9 is an entire function. We seek solutions 
to (6 . 1 )  holomorphic about a neutral fixed point Zo of g .  
There are different types of neutral points depending on whether the point is in 
the Julia set. Parabolic points, for instance, were discussed in §4.3 and were found 
to be in the Julia set. Theorem 4.31 then shows that solutions holomorphic at a 
parabolic fixed point are either constant functions, or are holomorphic everywhere 
except the exceptional point of g .  Constant functions require a specific balancing 
in the differential operator, and if 9 has no exceptional point , then holomorphic 
120 Holomorphic Solutions about Neutral Fixed Points 
solutions to (6. 1 )  do not exist at parabolic fixed points. We will show, however, that 
other types of neutral fixed points, admit local holomorphic solutions regardless of 
the differential operator involved, and that it is the arithmetic properties of g' (zo) 
that determine the type of neutral point . 
6 . 1  The Linear Case 
We start by examining the case where 9 is a linear function of the form g(z) = az 
where la l = 1 .  This is the pantograph equation with neutral argument. We study 
non-trivial solutions to (6. 1 )  that are holomorphic at z = O .  
Suppose that y is  a non-trivial solution to (6. 1 )  that is  holomorphic at z = O .  Then 
y can be represented as a power series of the form 
00 
that has a non-zero radius of convergence. Substituting this power senes into 
equation (6. 1 )  and equating coefficients of zn gives 
n 
Cn = 
Co II(Aak-l - b) 
n! k=l 
for n � 1 .  As l a l = 1 ,  we have 
so that the solution is entire. We also have 
(6.2) 
(6 .3) 
(6 .4) 
and thus have both upper and lower bounds on the coefficients en with which we 
can find the order of the entire function. 
The order p of an entire function represented by a power series is given by 
n lnn  p = lim sup ( ) . n-+oo In _1_ lenl  
(6.5) 
6.1 The Linear Case 
If I A I  =f I b l , then inequality (6.3) gives 
and so 
In C�
I
) � In( n! )  - In i co l - n ln I I A I  + I b l l , 
li 
n lnn  
p < .m ----,----,-----,---- -.,,---- --- n-+oo In( n ! ) - In I co l - n ln ( I A I + I b l )  
= 1 ;  
121  
hence, the maximum order for the entire solution i s  1 .  A similar calculation using 
the lower bound (6.4) gives, if I b l =f l A ! ,  
I 
n ln n  
p > im --------------,-­- n-+oo In( n ! ) - In l eo l - n ln I I A I - I b l l 
= 1 ,  
thus the order o f  the entire solution is 1 .  I f  I A I = I b l then we have a somewhat 
different scenario. We still have the upper bound (6.3) and so the solution is entire 
and is at most of order 1 ,  however the lower bound is now zero, so we cannot 
determine the actual order of the solution. If A = b / ak for some non-negative 
integer k, then it is clear that the solution is a polynomial, and thus order o. The 
interesting case occurs when I A I  = I b l but A =f b/ak for all non-negative integers k. 
In this case, the order of the entire solution depends on the arithmetic properties of 
a ,  in particular on whether or not a is a root of unity. To illustrate this comment, 
consider the example 
y' (z) + y ( z) = a y (az) . 
For this equation, the coefficients en are given by 
n Co IT k en = I" ( a  - 1 ) .  n .  k= l 
( 6 . 6 )  
If a is a root of unity, then a j  = 1 for some positive integer j, and consequently 
the entire solution y must be a polynomial. Suppose that a is not a root of unity. 
Evidently y cannot be a polynomial, but the order of y depends on "how close" a 
approximates a root of unity. From the discussion above, the order of the solution 
122 Holomorphic Solutions about Neutral Fixed Points 
Y is no greater than 1 .  Suppose that a is such that 
- In j an - 1 j  = O(ln n) ,  as n --+ 00. (6.7) 
This condition corresponds to Siegel's condition for a stable neutral fixed point 
( [34] , p. 608) which we will discuss further in the next section. In this case, there 
is an L > 0 such that 
- ln jan - l j < L ln n, 
for n sufficiently large; consequently, 
for n large. Therefore, 
In C�j ) ::; In(n!) + nL In n - In Co , 
I 
n ln n  
p > im -,....--,------,....----,-- n-+oo In(n! ) + nL In n - In j Co j 
1 
l + L  
For such a ,  the order p of y must satisfy 
1 -- < p < l .  
l + L  - -
One suspects that there may be other conditions (Cremer type relations) for non 
roots of unity that would limit the order of the function from above. At any rate, the 
order of the entire function for this example is linked to the arithmetic properties 
of a. 
6.2  Types Of Neutral Points 
There are different types of neutral points, some of which are in the Fatou set , 
and others of which are in the Julia set. We know, for instance, that all parabolic 
neutral points are in the Julia set. Here we introduce the other types of neutral 
fixed point .  Let Zo be a neutral fixed point of g , and set a = g' ( zo) so that j a j  = l .  
We call a the multiplier of zoo We start with the case where a i s  a root of unity 
(i .e. it is a rational rotation) . 
6.2  Types Of Neutral Points 123 
Suppose an = 1 for some positive integer n .  If gn ( z )  =1= z then we have a parabolic 
fixed point , which we know is in the Julia set . Figure 6 . 1  shows the Julia set of 
the function g ( z) = z2 + e27r�iz with � = 2/5, which has a neutral periodic point at 
z = 0 of period 5, which is the centre of the 5 "petals" in the top left of the image. 
Figure 6.1 : The Julia set for g(z) = z2 + e27r�iz with � = 2/5. 
The behaviour of points close to a parabolic fixed point Z o  gives us a little insight as 
to why they fall in the Julia set. If Zo is a parabolic fixed point of g ,  then it can be 
shown that in any neighbourhood of Zo there exist both points that are attracted 
to Zo under g, and points that are repelled from Zo under g .  Specifically, we have 
the following well-known result (cf. Milnor [26] p. 105) . 
Theorem 6 .1  (Leau-Fatou Flower Theorem) Let Zo be a parabolic fixed point of g 
such that g ' ( zo )  = a, where an = 1 for some positive integer n, where n is chosen 
to be as small as possible. Then there exist n attracting petals PI , P2 , . . . , Pn such 
that any orbit under gn containing points in Pk will converge to Zo from within 
Pk . Similarly, there exist n repel ling petals P{, P� , . . .  , P� such that any orbit under 
the inverse of gn containing points in P� will converge to Zo from within P� . The 
union of these 2n petals along with the point Zo form a neighbourhood of zo , and 
124 Holomorphic Solutions about Neutral Fixed Points 
furthermore, they are alternately arranged cyclically at Zo such that any repelling 
petal intersects o n ly the two attractive petals o n  either side. 
Figure 6 . 1  clearly shows the attractive petals about the parabolic point . The re­
pelling petals overlap the attractive petals and are each centered on one of the 
"spokes" of the Julia set emanating from the parabolic point . As there are attrac­
tive petals about zo , we may define the bas in of attraction of Zo as the set of 
all points that eventually map into an attractive petal of zo o Note, however, that 
we cannot generate local holomorphic solutions within this attractive basin, as the 
presence of repelling petals (and thus repelling periodic points arbitrarily close to 
zo) forces any local holomorphic solution to be continuable throughout the plane, 
from which we obtain only constant solutions, unless 9 has an exceptional point . 
The presence of repelling points means there is no neighbourhood U of Zo such that 
g (U )  � U, which as we will see, is the crucial property that allows (and guarantees) 
holomorphic solutions at zoo 
If an = 1 for some positive integer n and gn ( z )  = z, we no longer have a parabolic 
point. However, we find that we have already dealt with this case in the previous 
section - 9 must be a rotation. We start by showing that 9 is linear. 
Assume 9 is non-linear. As 9 is entire, there exists constants j3 > 0 and R > 0 such 
that 
for all I z l  > R by Liouville's Theorem. For R sufficiently large, then 
and in general, 
I g2 ( Z ) 1 > j33 I z l \ 
I l ( z ) 1 > j37 I z I 8 , 
for all I z l  > R. But gn (z) = z for all z, so we have a contradiction. Thus, 9 must 
be linear. 
Let g (z )  = az + b  where a and b are complex constants. Substituting this expression 
6.2  Types Of Neutral Points 
into gn(z) = z and equating the coefficients of Zl and zO gives 
an - 1 = 0 
b(an-1 + an-2 + . . .  + a1 + 1 )  = 0 ;  
125  
hence, a must be  a root of  unity. If a = 1 ,  then b = 0 and g(z) = z ,  which is a 
trivial rotation. If a -=/:  1 then b can be any complex number. Setting 
then gives 9 in the form 
b 
Zo = --1 - a ' 
9 = a(z - zo) + Zo, 
which is a rotation about the point Zo through an angle of arg(a) . We can (if 
necessary) change the problem so that Zo = 0 by using a simple translation. Thus, 
the functional differential equation is of the form analysed in the previous section, 
and will always have a solution holomorphic about zo0 In fact the solution will be 
entire. 
We now consider the case where Q is not a root of unity. In this case we can 
characterize our neutral points into a further two categories: Siegel points, and 
Cremer points. 
Definition 6.2 (Siegel and Cremer Points) Let Zo be a neutral fixed point of 9 
such that Q = g'(zo) is an irrational rotation. Suppose that there exists a simply 
connected neighbourhood U of Zo such that g(U) � U. Then Zo is a Siegel point . 
If no such neighbourhood exists, Zo is a Cremer point . 
It is clear from Definition 4 . 10 of the J ulia set that Siegel points are in the Fatou set, 
whereas Cremer points are in the Julia set. We would therefore expect holomorphic 
solutions about Siegel points to exist as they do about attracting fixed points. 
Likewise, we would expect the local theory developed for repelling points to be 
applicable in some way to Cremer fixed points. Figure 6 . 2  contains the Julia set 
for the quadratic g(z) = z2 + e27rEi where � is the Golden Ratio � = l+2J5 . The 
Siegel point at z = 0 is in the centre of the large white region in the lower left . The 
large white region containing the Siegel point is the region mapped into itself by 
126 Holomorphic Solutions about Neutral Fixed Points 
g. There is no image of a Cremer point available to the best of my knowledge (See 
Appendix 1 ) .  
Figure 6.2 :  The Julia set for g(z) = z2 + e27l'�iz with � = 1+2.;5 .  
Before we examine the theory regarding holomorphic solutions to functional differ­
ential equations about Siegel and Cremer points, we first note some results charac­
terizing such points. We begin with some definitions. 
Definition 6.3 (Diophantine Numbers) An irrational number � is said to be Dio­
phantine of order K, where K, > 0 if there exists an E > 0 so that 
for every rational number E .  q 
Definition 6.4 (Continued Fractions) For any irrational number � E (0 , 1 )  we 
define the continued fraction expansion of � as 
1 
� = ------
1 
a l + -----
1 
6.2  Types Of Neutral Points 127 
where the an are uniquely defined positive integers. The an may be computed in­
ductively by the formula 
The rational number 
Pn 
qn 
1 
1 
al +-------
is called the n-th convergent to �.  
1 
+ -
an-l 
The first characterisation we provide is due to Siegel [34] . 
Theorem 6 .5  (Siegel's Theorem) If the angle � is Diophantine of any order, then 
any neutral fixed point with multiplier a = e2nEi is a Siegel point. 
A much sharper version of Theorem 6.5 that makes use of the denominators of the 
convergents to � was proved in 1972 by Bryuno. 
Theorem 6.6 (Theorem of Bryuno) Let qn be the denominator of the n-th conver­
gent to � .  If 
� 
_
lo-=-g (:...::....qn-,-+....:...l ) � < 00, 
n=O qn 
then any neutral fixed point with multiplier a = e2nEi is a Siegel point. 
(6.8) 
Yoccoz, in 1987 showed that this is the best possible result, by completely analysing 
the quadratic case. 
Theorem 6.7 (Theorem of Yoccoz) Conversely, if the sum (6.8) diverges, then the 
quadratic map f (z) = Z2 + az with a = e2nEi has a Cremer point at z = O .  
We end this section with a sufficient condition for Cremer points. 
Theorem 6 .8  (Cremer's Theorem) Suppose g is a polynomial of degree d with a 
neutral fixed point Zo with multiplier a .  If the dLth root of laLII is unbounded as 
q - 00, then Zo is a Gremer point. 
128 Holomorphic Solutions about Neutral Fixed Points 
6 .3  Holomorphic Solutions About Siegel Points 
In  this section, we address the task of finding a holomorphic solution to  (6 . 1 )  about 
a Siegel point zoo The theory developed for attracting fixed points can be applied, 
with suitable alterations, to the Siegel point case. 
Note that the proof of the local existence and uniqueness result , Theorem 4.7, uses 
the condition that Zo is an attracting fixed point only to deduce that 
g (D(zo ; 8)) � D(zo ;  8) , (6.9) 
for all 8 > 0 sufficiently small. In fact, attracting fixed points obey the strict 
inequality g(D(zo ; 8)) c D(zo ;  8) by definition. The allowance for equality leaves 
an opening that enables a similar result to be obtained for the Siegel case. We 
also note that the discussion preceding Theorem 4.33 indicates that we can replace 
the constants a and b in (6. 1 )  with functions a(z) , b(z) holomorphic at Zo and still 
obtain a unique holomorphic solution. We can thus relax the conditions of Theorem 
4 .7  as follows. 
Corollary 6.9 Let p, q and 9 be functions holomorphic at a fixed point Zo of g ,  
and suppose that there is a 81 > 0 such that 9 satisfies (6.9) for all 8 < 81 . Then 
there exists a solution to 
y'(z) + p(z)y(z) = q(z)y (g (z) ) ,  (6 . 10) 
satisfying the initial condition (4.2) unique among functions holomorphic at zo o 
Corollary 6.9 alone is not enough to show that there exists a holomorphic solution 
to (6. 1 )  about a Siegel point ,  as the neighbourhood U in the definition of a Siegel 
point need not contain a disc that satisfies property (6.9) . We can, however, get 
around this difficulty by transforming equation (6. 1 )  to one in which the functional 
argument satisfies (6.9) . 
Let Zo be a Siegel point of g ,  and let U be a simply connected open neigbourhood 
of Zo such that g(U) � U. Let cjJ be a conformal map of the unit disc D(O; 1 )  onto 
U which takes the origin to zoo Let G = cjJ-1 0 g o cjJ be the map of D(O; 1 )  into itself 
induced by g. Now, G(O) = 0, and so by Schwarz' lemma, either I G (w ) 1 < Iw l  for 
6.3 Holomorphic Solutions About Siegel Points 129 
all W E D(O; 1 )  or else G(w) = 13w where 1 13 1  = 1 .  Clearly G'(O) = g' (O) and so 
we must have G(w) = aw, so that G satisfies relation (6.9) . We use the map cp to 
recast equation (6. 1 )  into one where the functional argument is G. 
Let z = cp(w) ,  and Y(w) = y (z) . Then (6 . 1 )  is transformed to 
Y'(w) + bcp' (w)Y(w) - 'x</>' (w)Y(aw) = O. (6. 1 1 )  
We now show that this transformed equation has a unique holomorphic solution 
that can be analytically continued throughout the unit disc. 
Lemma 6.10 For any Yo E C, there exists a unique solution Y E H(D(O; 1 )) to 
equation (6. 1 1 ) that satisfies Y(O) = Yo . 
Proof: Since </> E 1i(D(O; 1 ) )  and aw satisfies (6.9) for any 8 > 0,  the conditions of 
Corollary 6.9 are satisfied. Hence, for any Yo E re there is an E > 0 and a function 
Y E 1i(D(O; E) )  such that Y is a solution to equation (6. 1 1 ) and satisfies Y(O) = Yo. 
Moreover, Y is unique. 
Corollary 6.9 guarantees a unique solution to the initial-value problem, but the 
result is local in character and it remains to show that the holomorphic solution 
can be analytically continued throughout the disc D(O; 1 ) . If equation (6. 1 1) were 
an ordinary differential equation we could appeal to well-known results concerning 
the analytic continuation of solutions to discs of radius at least that of the radius of 
convergence of cp (cf. [3] ) .  Nonetheless, we can mimic the proof of this result for the 
neutral case and employ the method of majorants to deduce that Y E H(D(O; 1 ) ) . 
Specifically, we have that Y can be represented in the form 
00 
Y(w) = L Cnwn , 
n=O 
for w E D(O; E) , and that </>'(w) can be represented in the form 
00 
</>' (w) = L knwn , 
n=O 
130 Holomorphic Solutions about Neutral Fixed Points 
for w E D(O;  1 ) .  Substituting these power series into equation (6. 1 1 )  gives 
For any r, 0 < r < 1 there is a number AT such that 
for all n. Let w E D(O; r) ,  q = \ A \  + \ b l , 
and define the function H as 
00 
= qAT f= ( � ) n 
n=O 
qAT -
1 - :!!:'. 
. 
r 
By construction, the differential equation 
P' (w) + H(w)P(w) = 0 (6. 12) 
majorizes equation (6. 1 1 )  under the condition 
P(O) = \ Co \ ,  (6. 13) 
so that if P is represented by the power series 
00 
P(w) = L bnwn , 
n=O 
then bn > \On \ for all n. Now, equation (6. 12)  with the initial condition (6. 13)  has 
the unique holomorphic solution 
( 
W ) -rqAr 
P(w) = 1 - -;: \Co \ . 
6.3  Holomorphic Solutions About Siegel Points 131 
Since P E  H(D(O; r ) )  and r can be chosen arbitrarily close to 1 ,  we conclude that 
Y E H(D(O; 1 ) ) .  • 
The set U is simply a neighbourhood of Zo that is mapped into itself under g. We 
now find the largest such set for a particular Zoo Let 0. be the component of the 
Fatou set of 9 containing the Siegel point Zoo It is clear then that g (n) � 0., for 
by Theorem 4. 1 1  the Fatou set F(g) is invariant under g ,  and thus components of 
F(g) must map to components of F(g) under g .  The point Zo is fixed under g ,  and 
since 9 is an entire function, 9 must map the set 0. into itself. Thus, we can set 
U = 0. ,  and therefore 0. is conformally equivalent to the unit disc D(O; 1 )  under cP. 
The set 0. is called the Siegel disc associated with the Siegel point Zo o We then 
have the following result . 
Theorem 6 .11  Let Zo be a Siegel point for g, and let 0. be the Siegel disc associated 
with Zo . Then there exists a unique solution to equation (6. 1 )  satisfying the initial 
condition (4.2) holomorphic in n.  
Proof: We have a unique solution Y to (6. 1 1 )  satisfying Y(O) = Yo, where Yo is 
as in (4 .2) ,  holomorphic in D(O; 1 )  by Lemma 6. 10. Since cP defined above was a 
conformal map, y = Y 0 cP-1 will be a function holomorphic in 0. which satisfies 
(6. 1 )  and has y (zo) = Yo . Thus, there exists a solution to (6. 1 )  satisfying (4.2) 
holomorphic in H(O). • 
The local theory about Siegel points is identical to that about attracting fixed points 
- there exists a local holomorphic solution which can be analytically continued 
up to a component of the Fatou set . In the attracting case, the boundary of 
this component formed a natural boundary for our solution. This was due to the 
boundary of Fatou components being in the Julia set , which must also hold for 
Siege I discs. We can thus claim a natural boundary in the majority of cases as 
follows. 
Theorem 6 .12  Let Yo =1= 0, b =1= A, and let y be any solution to the initial value 
problem (6. 1 )  satisfying (4.2) where Zo is a Siegel point for the non-linear entire 
function g .  Then one of the following must hold. 
132 Holomorphic Solutions about Neutral Fixed Points 
1. Y is holomorphic in C " {go} where go is the exceptional point of g .  
2. y is holomorphic in the Siegel disc 0 containing zo, and has 80 as a natural 
boundary. 
Proof: We have a solution y to (6 . 1 )  satisfying (4 .2) which is holomorphic in the 
Siegel disc 0 by Theorem 6. 1 1 .  If there is no point w E 80 at which y is holo­
morphic, then 80 forms a natural boundary for y ,  and so y satisfies the second 
option. Suppose there is a w E 80 such that y is holomorphic at w. Then Theorem 
4.31 can be applied to show that either y is a constant , or y E H(C " {go} ) .  The 
conditions Yo =1= ° and b =1= A, however, rule out constant solutions; therefore, y 
satisfies the first option. • 
If the class of equations is restricted to those where 9 is a polynomial, then only 
the second option in Theorem 6.12 is available. The key result is a Corollary of 
Theorem 4.31 , which holds for all neutral fixed points, not just Siegel points, of 
non-linear polynomials. 
Corollary 6.13 Let y be a solution to (6. 1 )  where g (z) is a non-linear polynomial 
with a neutral fixed point Zo . If y is holomorphic at some point W E J (g ) ,  then y is 
a constant. 
Proof: Cyclic polynomials have no neutral fixed points by Theorem 4 .20, thus 9 
must be non-cyclic. It therefore has no exceptional point. Theorem 4 .31  can then 
be applied to show that y is a constant. • 
For non-linear polynomial functional arguments with Siegel points, non-constant 
solutions have a natural boundary. 
Corollary 6.14 Let Yo =1= 0, b =1= A, and let Zo be a Siegel point for the non-linear 
polynomial g .  Then the solution y to equation (6. 1 )  satisfying (4.2) is holomorphic 
in 0, and has 80 as a natural boundary. 
In fact, once we rule out constant solutions, we can then rule out the existence of 
holomorphic solutions about Cremer points and parabolic points altogether. 
6.3 Holomorphic Solutions About Siege I Points 133 
Corollary 6 .15  Let Yo =1= 0, b =1= A, and let Zo be a neutral fixed point of the non­
linear polynomial g .  Then there is a solution to (6. 1 )  satisfying (4.2) holomorphic 
at Zo if and only if Zo is a Siegel point. 
Proof: The neutral fixed point Zo must be either a parabolic point, a Siegel point, 
or a Cremer point. Clearly if Zo is a Siegel point, the Corollary holds. If Zo is a 
parabolic point or Cremer point, then Zo E J(g) , and so any solution holomorphic 
at Zo is a constant by Corollary 6. 13 .  The conditions Yo =1= 0 and b =1= A preclude 
constant solutions; consequently, no solutions holomorphic at Zo can exist unless Zo 
is a Siegel point. • 
t£l 
CHAPTER 7 
Conclusion 
Functional differential equations with an entire functional argument have been ex­
amined and theory regarding the presence of holomorphic solutions to these equa­
tions presented. The main results come in Chapters 4 through 6 where two separate 
problems are analysed,  each one related to the other. The first is the existence of 
local holomorphic solutions, and the second is the analytic continuation of such a 
local solution throughout the complex plane. 
Both problems depend predominantly on the behaviour of g ,  the functional ar­
gument . The local behaviour of 9 about its fixed points determines whether a 
holomorphic solution exists about that point . Similarly, the analytic continuation 
of such a solution occurs through iteration of g. The dynamics of 9 therefore is the 
driving force in both problems. 
If the initial fixed point is attracting, then a unique local holomorphic solution 
exists given appropriate initial conditions, and the dynamics of 9 dictate that the 
solution can be analytically continued into the immediate basin of attraction of 
the fixed point . A similar result can be found if the initial point is a Siegel point . 
In this case, the solution can be analytically continued throughout the Siegel disc. 
In both these cases, however, the boundary of this region of analyticity is part of 
the J ulia set J (g) , which is the closure of the set of repelling periodic points of 
g .  Therefore, in order for solutions to be analytically continued beyond these sets,  
they must be holomorphic about some repelling periodic point of g.  
Repelling fixed points are more difficult to analyse at the local level. Transforming 
the problem into one involving the inverse of 9 and considering it as an eigenvalue 
problem, however, shows that any solutions that are locally holomorphic are rare, 
136 Conclusion 
corresponding to eigenfunctions of a compact operator. Furthermore, if 9 is an 
entire map, then we can analytically continue any such solutions throughout the 
entire complex plane, with the possible exception of a single point . Thus, either 
these solutions are entire, or have a singularity at the exceptional point . Entire 
solutions, however, are at the mercy of the maximum modulus principle, and from 
this we can conclude that their existence is even less likely, as the form of equation 
severely limits the growth rate (often down to constant functions) . 
In the general case, ruling out the occurrence of solutions with a single singularity 
in the complex plane is difficult . If we restrict the functional argument 9 to be a 
polynomial, however, we can make more progress. Indeed, the only polynomials 
that have exceptional points are the cyclic polynomials. We can thus conclude 
that , functional differential equations with non-cyclic polynomial arguments, have 
no holomorphic solutions about repelling fixed points. FUrthermore, any polyno­
mial with a neutral fixed point cannot be cyclic, hence no functional differential 
equation with a polynomial functional argument can have non-entire solutions that 
are holomorphic about a Cremer or parabolic fixed point . These results, once fed 
back in to the problem regarding the analytic continuation of solutions about at­
tracting fixed points, imply that the boundary of the immediate basin of attraction 
(or Siegel disc in the case of a Siegel point) forms a natural boundary for the 
solution. 
Additionally, in Chapters 2 and 3 we considered solutions away from the fixed point , 
by examining the simplest functional differential equation with a repelling fixed 
point, the advanced pantograph equation. Through studying both the equation on 
the real line and it 's Laplace transformed analogue, we have shown that the well 
known Dirichlet series solutions are the only ones that decay exponentially along 
the positive real axis. Furthermore, we have established the existence of an infinite 
number of other solutions in L2 [0, 00) .  A decay condition is therefore necessary 
to obtain uniqueness, the required condition being dependent on the coefficients of 
the particular equation being analysed. 
7.1  Future Work 137 
7. 1 Future Work 
Of interest is how the Laplace and Mellin transforms are useful in gaining insight 
into a given equation - particularly pantograph type equations. The Laplace trans­
form often allows an advanced equation to be rewritten as a retarded equation in 
the Laplace variable, whereas the Mellin transform reduces pantograph type equa­
tions with polynomial coefficients to difference equations. Both lead to simpler 
equations in many circumstances. A useful further research project would be the 
analysing of the Mellin transformed equation and its application in determining 
possible solution forms for other pantograph type equations with polynomial coef­
ficients. The symmetries present in the Dirichlet series solutions in Chapter 2, for 
instance, may be applicable to a broader class of equations, and other symmetry 
forms may also be found. 
Another potential area for new results is in the uniqueness problem for the panto­
graph equation. The uniqueness results of Chapter 2 (Theorems 2 .2 ,  2 .8, and 2 . 12 ) ,  
limit available solutions to  the Dirichlet series solution. However, the Laplace anal­
ysis of §2.4 indicates that if we relax the conditions of these theorems, there are 
an infinite number of solutions available, though they must decay slowly along the 
positive real axis. There is an outstanding question as to whether such solutions are 
in LdO, (0 ) . The condition that limp-+o L(p) exists and is finite suggests that this 
may be true, but has not been shown. This leads to questions regarding the natural 
space for solutions to this problem. For instance, it is clear that the Dirichlet series 
solutions are in the Hardy space H 2 ( IIo ) .  A suitable choice of the solution space 
may be enough to guarantee a unique solution. 
Further work on the neutral case also looks promising. In particular, parabolic 
fixed points are known to be in the Julia set (and thus do not generally admit local 
holomorphic solutions) , but it is likely that holomorphic solutions may be found in 
a set n whose boundary contains the parabolic point . Indeed, Theorem 6 . 1  implies 
there exists a small wedge A whose tip is the parabolic point such that gn (A )  � g (A )  
for some positive integer n. With this knowledge, it  may be possible to establish 
the existence of a solution within such a wedge via a contraction mapping. Such 
a solution could then be analytically continuable throughout the attracting petal 
containing the wedge in the same manner as was done for attracting fixed points. 
Lastly, we note that there is a relatively simple extension to be made using periodic 
138 Conclusion 
rather than fixed points for the initial point Zo o Indeed, if Zo was periodic of period 
2, then we would need only know y (zo) and y (g(zo))  in order to determine all the 
derivatives of y at Zo and g(zo )  using the differential equation directly. Thus, one 
can find local holomorphic solutions about attractive periodic points. Much of the 
theory of Chapter 4 would then be readily extendible to the case where Zo was 
periodic. 
A.1  Introd uction 
ApPENDIX A 
Notes on Computer 
Graphics 
This section presents a basic overview of the techniques used to produce the images 
featured in this thesis. The methods presented apply only to polynomials. We 
divide the area we wish to investigate into a grid of pixels, and for each pixel, 
we choose a point inside (usually the midpoint of the pixel) . We then study the 
forward images of this point to determine whether it is in the Julia set or the 
Fatou set , colouring the pixel accordingly. For instance, if the forward images grow 
without bound in magnitude, then it cannot be in the Julia set as the Julia set for 
polynomials is bounded, and forward images of points in the Julia set are in the 
Julia set . Similarly, if the forward images tend towards some fixed orbit , then the 
point must have been in a basin of attraction, so once again is not in the Julia set. 
There are two problems with this approach. Firstly, the chance that the midpoint 
of a particular pixel is in the Julia set is remote. The Julia set is a closed set 
and often consists of very thin sections (The Julia set for 9 = Z2 , for instance, is 
the circle I z l  = 1 )  and thus it can pass through a pixel while not containing the 
midpoint. Thus, pixels would be characterized as being in the Fatou set, when 
they contain points in the Julia set . This problem is tackled by using a distance 
estimator to estimate how far a given point is away from any point on the Julia 
set. We detail such an estimate in section A.2. 
The second problem is that we often require a great number (i .e . millions) of 
140 Notes on Computer Graphics 
iterations in order to determine if a point is in the Fatou set - particularly when 
we are dealing with neutral fixed points. Consider, for instance, a point within a 
distance of 10-3 of the parabolic fixed point z = 0 of the function g(z) = z + z3 . In 
one iteration, we see that it will have moved no more than 1 0 - 9 ,  as Ig(z) - z l  = Iz 13 . 
It will thus take at least 1 06 iterations for the image of the point to move more 
than 10-3 . This problem is overcome by identifying known regions of the Fatou 
set , so that we may terminate the iteration whenever an orbit falls into one of the 
regions. This is detailed in section A.3. 
Finally, we note that we have not presented a figure of a Cremer point , and there 
is no way to generate one using a computer as far as I am aware. The problem is, 
of course, that a Cremer point Zo must have g'(zo )  = e27r�i where c; is an irrational 
number that is "close" to a rational number. This is close to impossible to generate 
using a computer, as all the usual computer representations of numbers are floating 
point, and therefore rationals. Thus, it is extremely difficult to create a number 
using a computer that is "close enough" to a rational, yet isn't actually rational. 
A.2 Distance Estimator 
U se of a distance estimator is crucial as the J ulia set often contains very thin sections 
or filaments that may pass through many pixels while missing their midpoints. 
Distance estimators are built by identifying a conformal isomorphism between a 
region U of the complex plane (or the Riemann sphere if we are considering the 
attractive basin at infinity) and the unit disc, and then using this isomorphism to 
determine an appropriate distance estimator. For instance, Milnor ( [26] ' Appendix 
H)  details a distance estimator for an attractive basin U of a superattracting fixed 
point at the origin. (i .e . ,  one in which g(z) = azn + b1 Zn+ l + . . .  for some n > 1 ) .  
It is shown that the distance from the boundary o f  U to a point z is given (up to 
a factor of 2) by 
I sinh(G(z) ) I  
I IG' (z) 1 1 
where, for any orbit Z o  � Zl � . . . in U, 
G( ) - l' 
log I Zk l Zo lm k '  k-+oo n 
A.3 Identifying Regions of the Fatou Set 141 
and 
I I G' ( zo ) 1 I  = lim I dzk /dzo l . k-+oo nk l zk l 
In both cases, the successive terms can be computed in an inductive manner. If Zo 
is very close to the boundary of U, then G will be small, and hence the distance 
estimator is very close to the ratio I G ( zo ) I / I I G' ( zo ) 1 I which saves the computation 
of the sinh. It also has the additional benefit that the ratio of successive terms 
for G ( zo )  and I I G' (zo ) 1 1 may be used as an approximate for the distance estimator, 
allowing additional simplification of the formula which reduces computation time. 
Indeed, we have 
I G ( zo) 1 '" I Zk l log I Zk l 
I I G' (zo) 11 '" I dzk /  dzo l 
Therefore, we need only retain I Zk l and I dzk / dzo l from the previous iteration. At 
the next iteration, these can then be updated by 
I
dZk + l 1 = I g' ( Zk ) l l dZk l , dzo dzo 
I Zk+ l l = I g ( Zk ) l ·  
Thus, the distance estimator requires only the additional computation of the deriva­
tive of g at the previous orbit point . 
Similar distance estimates can be found for other regions of the Fatou set, although, 
for polynomials, we find that just using a distance estimator on the attractive basin 
at infinity is usually enough to give a sharp picture of J (g ) .  
A.3  Identifying Regions of the Fatou Set 
For polynomials, the orbits of points in the complex plane can behave in the fol­
lowing ways: 
1 .  Orbits tend to the point at infinity. 
2. Orbits tend to some attracting periodic orbit . 
3 .  Orbits map into an attractive petal of a parabolic point. 
4. Orbits map into a Siegel disc. 
142 Notes on Computer Graphics 
5. Orbits remain in the Julia set. 
Note that the last case occurs if and only if one of the previous cases doesn't occur. 
We examine each of the first 4 cases above in turn, but we first note that we must 
be able to find the attractive and neutral periodic points of a polynomial 9 before 
we can identify such regions. 
A.3. 1 Finding attractive and neutral periodic points 
Given a polynomial g ,  we must find the periodic points of 9 that are not in J(g) . 
Theorem 4.24 gives a limit on the number of attracting periodic points that 9 may 
have, though we do not know the period of such points. Thus, we would require 
to find the roots of gk for k = 0 , 1 ,  . . .  until we had the required number of points. 
Note that this is an expensive operation, as the degree of gk is dk where d is the 
degree of g. Note also that Theorem 4.24 does not give how many attracting 
periodic points 9 has, it gives only an upper bound on the number of such points. 
Additionally, we have no information about how many neutral periodic points 9 
may have. Fortunately, this is remedied by examining the orbits of the critical 
points of g, which we will refer to as critical orbits . We make use of the following 
result , presented without proof. The reader is referred to Milnor ( [26] ' pp. 78, 146) 
for the details. 
Theorem A.1  Every attractive basin of a polynomial 9 contains some point in a 
critical orbit of g .  Similarly, every Siegel disc, or cycle of Siegel discs contains 
points from a critical orbit in its boundary. 
We find the critical points of 9 (by using a numerical polynomial solver if neces­
sary) , and then examine each of their orbits. From the orbits we can identify the 
periodicity of any periodic points, and their approximate location. We may then 
examine the appropriate iterate of g, and find the fixed point required, along with 
its type by using a numerical polynomial solver. 
Once we have identified the attractive and neutral fixed points of g, we can now 
identify regions of F(g) as detailed in the following sections. 
A.3 Identifying Regions of the Fatou Set 
A.3.2  Orbits tend to  the point at infinity 
143 
Such orbits can be identified readily, as I Zn l � 00 as n � 00. Indeed, all we need 
to find is the escape radius r such that I g ( z ) 1 > I z l for all I z l > T. We may 
use Lemma 2 . 1 3  to determine r for a given polynomial. If any of the iterates of a 
point z has modulus greater than r ,  we know immediately that z must be in the 
attractive basin of infinity. We then use the distance estimator from section A . 2  to 
check that our pixel does not contain points in J(g) away from the midpoint . 
A.3.3 Orbits tend to some attracting periodic cycle 
Identifying points within an attractive basin is accomplished by establishing a 
neighbourhood U of the periodic points such that 9 maps points in U to points 
in U. Any orbits that end up in the set U, then, must be in the basin of attraction 
of the periodic cycle. We may establish such a neighbourhood U as follows. Sup­
pose that we have an attractive periodic point Zo of a polynomial g. Then some 
iterate gk of 9 may be written in the form 
n k 2 � ' �  9 ( z) = Zo + gl (Z - zo) + ( z  - zo) � gj ( z - zo)J , 
j =2 
where n is the degree of gk . Thus 
and so for any z E D ( zo ; 1)  we have 
Now I g1 1 < 1 as Zo is an attractive fixed point of gk , so that 
144 Notes on Computer Graphics 
is positive. Choosing b < min { T, I } ,  then gives 
\g (z) - zo \ � \z - zo \ ,  
for all z E U = D(zo, b) . 
A.3.4 Orbits map into an attractive petal of a parabolic 
periodic point 
This case is similar to the previous case, although now the convergence is much 
slower, and we need to be able to differentiate between the different attractive 
petals of the parabolic periodic point zo o The Julia set will necessarily be very 
thin near parabolic points, so some pixels may contain points from two or more 
different attractive petals. We can determine such pixels by examining the orbits of 
the corners of each pixel, and examining the angle by which each orbit approaches 
zo o Points that have corners whose orbits approach Zo from different directions must 
also contain points in the Julia set , whereas points with corners whose orbits all 
approach Zo from the same direction must be in an attractive petaL Note that we 
may still require a great number of iterations to get sufficiently close to a parabolic 
point so as to be able to reliably compute the "approaching angle" . Fortunately, 
modern computers are fast enough to perform the necessary computations in a 
reasonable time frame. For instance, figure 6. 1 consists of approximately 6 million 
pixels and took around 12  hours to render on a 2GHz Athlon processor. Note that 
almost all that time was spent on points in a small neighbourhood of the parabolic 
point at z = O. Relatively few iterations are required for orbits to map into the 
disc D(O; 0 . 1 ) , for instance, but it takes many more iterations from that stage to 
distinguish each attractive petal. 
A.3.5  Orbits map into a Siegel disc 
In this last case, the problem that arises is that points in the Siegel disc do not 
move closer to the Siegel point Zo under iteration, as 9 is conformally equivalent 
to an irrational rotation. We are therefore required to know the Siegel disc in its 
entirety. Fortunately, this is a simple task, as not only does the boundary of the 
AA The Algorithm 145 
Siegel disc U contain points from a critical orbit by Theorem A.l, but these points 
are in fact dense on the boundary au (cf. Milnor [26] , pp 128) . We can therefore 
compute au to the required accuracy using the orbits of the appropriate critical 
point . It is then a trivial matter to determine if a particular orbit lies within this 
boundary or not. 
A.4 The Algorithm 
The algorithm used, therefore, is quite simple. The input data consists of a poly­
nomial g ,  a region R of the plane on which we wish to draw the Julia set J (g ) ,  and 
the resolution (number of pixels) we require. 
We begin by finding the critical points of 9 and, by examining their orbits, we obtain 
the periodicity and approximate locations of the attractive and neutral periodic 
points of g. These points are then found precisely using a numerical solver on the 
appropriate iterate of g .  Known regions F1 , F2 , . . . of the Fatou set can then be 
identified as detailed in the previous section. 
The region R is divided up into a pixel grid so as to give the required resolution, 
and each pixel's midpoint is iterated under g .  At each iteration (or after a given 
number of iterations) ,  we test to see whether the current point z in the orbit 
is contained within one of the regions Fk . If z E Fk for some k, we optionally 
use the distance estimator to check that the initial point was sufficiently far from 
the J ulia set , before assigning the pixel an appropriate colour. If z fJ. Fk for all 
k, we loop back and perform more iterations. Note that a maximum number of 
iterations must be specified, as we may have midpoints that are in the Julia set, 
whose orbits will never lie in one of the terminating sets Fk . If the maximum 
number of iterations is exceeded for a particular pixel, we issue a warning notice, 
apply the distance estimator for the attractive basin at infinity, and colour the 
pixel accordingly. Once we have coloured all our pixels, the image of the Julia set 
is displayed, and optionally saved to a permanent file. 
146 Notes on Computer Graphics 
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advanced function, 60 
attracting fixed point, 65, 66 
attracting periodic point, 66 
basin of attraction, 67, 124 
continued fraction expansion, 126 
convergents, 127 
Cremer point, 125 
critical orbits, 142 
cyclic polynomial, 75 
decays exponentially, 18  
Diophantine number, 126 
distance estimator, 140 
escape radius, 143 
Euler identity, 21 
exceptional point, 73 
Fatou set, 70 
fixed point, 65 
immediate basin of attraction, 67 
Julia set, 70 
log-periodic functions, 23 
Mellin transform, 14 
multiplier, 122 
Index 
neutral fixed point, 66 
neutral function, 61 
neutral periodic point, 67 
normal family, 70 
parabolic point, 72 
periodic orbit , 66 
periodic point, 66 
repelling fixed point, 65, 66 
repelling periodic point, 66 
retarded function, 60 
Siegel disc, 131  
Siegel point, 125