Handbook of mathematics for engineers and scienteists part 136 docx

17.3. LINEAR FUNCTIONAL EQUATIONS 913
homogeneous equation with F(x) ≡ 0, the following three cases are possible (the notation
used here is in agreement with that of Paragraph 17.3.2-1):
(i) Equation (17.3.3.1) on I either has a one-parameter family of continuous solutions
or no solutions at all. If (17.3.3.1) has a continuous solution y
0
(x)onI, then the general
continuous solution on I is given by
y(x)=y
0
(x)+
a
G(x)
,
where a is an arbitrary constant and G(x) is given by (17.3.2.3).
(ii) Equation (17.3.3.1) on I has a continuous solution depending on an arbitrary func-
tion, or has no continuous solutions on I.
(iii) Equation (17.3.3.1) on I either has a continuous solution or no solutions at all.
2
◦
.Letx I, ξ I,andf (x) R
0
ξ
[I]. Suppose that g(x)andF (x) are continuous functions,
g(x) ≠ 0 for x ≠ ξ,andg(ξ)>1. Then equation (17.3.3.1) has a unique continuous solution
that can be represented by the series
y(x)=–
∞

n=0
F (f

[n]
(x))
G
n+1
(x)
,(17.3.3.2)
where the function G
n
(x)isdefined by (17.3.2.2).
17.3.3-2. Equations of special form with g(x)=const.
1
◦
. Consider the equation
y(f(x)) + y(x)=F (x), (17.3.3.3)
which is a special case of equation (17.3.3.1) with g(x)=–1.
(A) Let ξ
I, f (x) R
0
ξ
[I], and suppose that F (x) is a continuous function. If there
exists a continuous solution of equation (17.3.3.3), it can be represented by the power series
y(x)=
1
2
F (ξ)+
∞

n=0
(–1)
n


F (f
[n]
(x)) – F (ξ)

.
(B) Suppose that the assumptions of Item (A) hold and, moreover, there exist positive
constants δ, κ,andC such that
|F (x)–F (ξ)| ≤ C|x – ξ)|
κ
, x (ξ – δ, ξ + δ) ∩I,
and ξ is a strongly attractive fixed point of f(x). Then equation (17.3.3.3) has a continuous
solution on I.
2
◦
. Consider the functional equation
y(f(x)) – λy(x)=F (x), λ ≠ 0.(17.3.3.4)
Let x
I =(a, b]. Assume that on the interval I the function f(x) is continuous, strongly
increasing, and satisfies the condition a < f(x)<x,andF (x) is a function of bounded
variation.
Depending on the value of the parameter λ, the following cases are possible.
914 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS
Case |λ| > 1. There is a unique solution
y(x)=–
∞

n=0
F


f
[n]
(x)

λ
n+1
,
which coincides with (17.3.3.2) for g(x)=λ and G
n+1
(x)=λ
n
.
Case 0 < |λ| < 1.Foranyx
0
I and any test function ϕ(z) that has bounded variation
on [f (x
0
), x
0
] and satisfies the condition
ϕ(f(x
0
)) – λϕ(x
0
)=F (x
0
),
there is a single function of bounded variation for which equation (17.3.3.4) holds. This
solution has the same variation on the interval [f (x
0

), x
0
] as the test function ϕ(x).
Case λ = 1. If lim
x→a+0
F (x)=0 and there is a point x
0
I such that the series
∞

n=0
F

f
[n]
(x
0
)

is convergent, then equation (17.3.3.4) has a one-parameter family of
solutions
y(x)=C –
∞

n=0
F

f
[n]
(x)


(here, C is an arbitrary constant), for which there exists a finite limit lim
x→a+0
y(x).
Case λ =–1. See Item 1
◦
.
17.3.3-3. Abel functional equation.
The Abel functional equation has the form
y(f(x)) = y(x)+c, c ≠ 0.(17.3.3.5)
Remark. Without the loss of generality, we can take c = 1 (this can be achieved by passing to the
normalized unknown function ¯y = y/c).
1
◦
.Let¯y(x) be a solution of equation (17.3.3.5). Then the function
y(x)=¯y(x)+Θ

¯y(x)
c

,
where Θ(z) is an arbitrary 1-periodic function, is also a solution of equation (17.3.3.5).
2
◦
. Suppose that the following conditions hold:
(i) f (x) is strictly increasing and continuous for 0 ≤ x ≤ a;
(ii) f (0)=0 and f(x)<x for 0 < x < a;
(iii) the derivative f

(x) exists, has bounded variation on the interval 0 < x < a,and

lim
x→+0
f

(x)=1. Then, for any x, x
0
(0, a], there exists the limit
y(x) = lim
n→∞
f
[n]
(x)–f
[n]
(x
0
)
f
[n–1]
(x
0
)–f
[n]
(x
0
)
,(17.3.3.6)
which defines a monotonically increasing function satisfying the Abel equation (17.3.3.5)
with c =–1 (thisistheL
´
evy solution).

17.3. LINEAR FUNCTIONAL EQUATIONS 915
3
◦
.Letf (x)bedefined on a submodulus interval I and suppose that there is a sequence d
n
for which
lim
n→∞
f
[n+1]
(x)–f
[n]
(x)
d
n
= c, x I.
If for all x
I there exists the limit
y(x) = lim
n→∞
f
[n]
(x)–f
[n]
(x
0
)
d
n
,(17.3.3.7)

where x
0
is an arbitrary fixed point from the interval I, then the function y(x) satisfies
equation (17.3.3.5).
4
◦
. Let us describe a procedure for the construction of monotone solutions of the Abel
equation (17.3.3.5). Assume that the function f(x) is continuous, strictly increasing on the
segment [0, a], and the following conditions hold: f(0)=0 and f (x)<x ≤ a for x ≠ 0.
We define a function ϕ(y)onthesemiaxis[0, ∞)by
ϕ(y)=f
[n]

ϕ
0
({y})

, n =[y], (17.3.3.8)
where [y]and{y} are, respectively, the integer and the fractional parts of y,andϕ
0
(y)is
any continuous strictly decreasing function on [0, 1) such that ϕ
0
(0)=a, ϕ
0
(1 – 0)=f(a).
On the intervals [n, n + 1)(n = 0, 1, ), the function (17.3.3.8) takes the values
ϕ(y)=f
[n]


ϕ
0
(y – n)

, n ≤ y < n + 1,(17.3.3.9)
and therefore is continuous and strictly decreasing, being a superposition of continuous
strictly increasing functions and a continuous strictly decreasing function. At integer
points, too, the function ϕ(y) is continuous, since (17.3.3.8) and (17.3.3.9) imply that
ϕ(n – 0)=f
[n–1]

ϕ
0
(1 – 0)

= f
[n–1]

f(a)

= f
[n]
(a),
ϕ(n)=f
[n]

ϕ
0
(0)


= f
[n]
(a).
Therefore, the function ϕ(y) is also continuous and strictly decreasing on the entire semiaxis
[0, ∞). By continuity, we can define the value ϕ(∞)=0.
It follows that there exists the inverse function ϕ
–1
defined by the relations
x = ϕ(y), y = ϕ
–1
(x)(0 ≤ x ≤ a, 0 ≤ y ≤ ∞), (17.3.3.10)
and ϕ
–1
(0)=∞, ϕ
–1
(a)=0.
From (17.3.3.8), it follows that the function ϕ(y) satisfies the functional equation
ϕ(y + 1)=f

ϕ(y)

, 0 ≤ y < ∞.(17.3.3.11)
Applying ϕ
–1
to both sides of (17.3.3.11) and then passing from y to a new variable x
with the help of (17.3.3.10), we come to the Abel equation (17.3.3.5) with c = 1,where
y(x)=ϕ
–1
(x). Thus, finding a solution of the Abel equation is reduced to the inversion of
the known function (17.3.3.8).

5
◦
. The Abel equation (17.3.3.5) can be reduced to the Schr
¨
odinger–Koenig equation
u(f(x)) = su(x),
with the help of the replacement u(x)=s
y(x)/c
; see equation (17.3.2.5).
916 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS
17.3.4. Linear Functional Equations Reducible to Linear Difference
Equations with Constant Coefficients
17.3.4-1. Functional equations with arguments proportional to x.
Consider the linear equation
a
m
y(b
m
x)+a
m–1
y(b
m–1
x)+···+ a
1
y(b
1
x)+a
0
y(b
0

x)=f(x). (17.3.4.1)
The transformation
y(x)=w(z), z =lnx
reduces (17.3.4.1) to a linear difference equation with constant coefficients
a
m
w(z + h
m
)+a
m–1
w(z + h
m–1
)+···+ a
1
w(z + h
1
)+a
0
w(z + h
0
)=f(e
z
), h
k
=lnb
k
.
Note that the homogeneous functional equation (17.3.4.1) with f(x)≡ 0 admits particular
solutions of power type, y(x)=Cx
β

,whereC is an arbitrary constant and β is a root of the
transcendental equation
a
m
b
β
m
+ a
m–1
b
β
m–1
+ ···+ a
1
b
β
1
+ a
0
b
β
0
= 0.
17.3.4-2. Functional equations with powers of x as arguments.
Consider the linear equation
a
m
y

x

n
m

+ a
m–1
y

x
n
m–1

+ ···+ a
1
y

x
n
1

+ a
0
y(x
n
0
)=f (x). (17.3.4.2)
The transformation
y(x)=w(z), z =lnlnx
reduces (17.3.4.2) to a linear difference equation with constant coefficients
a
m

w(z +h
m
)+a
m–1
w(z +h
m–1
)+···+a
1
w(z +h
1
)+a
0
w(z +h
0
)=f(e
e
z
), h
k
=lnlnn
k
.
Note that the homogeneous functional equation (17.3.4.2) admits particular solutions of
the form y(x)=C| ln x|
p
,whereC is an arbitrary constant and p is a root of the transcendental
equation
a
m
|n

m
|
p
+ a
m–1
|n
m–1
|
p
+ ···+ a
1
|n
1
|
p
+ a
0
|n
0
|
p
= 0.
17.3.4-3. Functional equations with exponential functions of x as arguments.
Consider the linear equation
a
m
y

e
λ

m
x

+ a
m–1
y

e
λ
m–1
x

+ ···+ a
1
y

e
λ
1
x

+ a
0
y(e
λ
0
x
)=f (x).
The transformation
y(x)=w(ln z), z =lnx

reduces this equation to a linear difference equation with constant coefficients
a
m
w(z + h
m
)+a
m–1
w(z + h
m–1
)+···+ a
1
w(z + h
1
)+a
0
w(z + h
0
)=f (e
z
), h
k
=lnλ
k
.
17.3. LINEAR FUNCTIONAL EQUATIONS 917
17.3.4-4. Equations containing iterations of the unknown function.
Consider the linear homogeneous equation
a
m
y

[m]
(x)+a
m–1
y
[m–1]
(x)+···+ a
1
y(x)+a
0
x = 0
with successive iterations of the unknown function, y
[2]
(x)=y

y(x)

, , y
[n]
(x)=
y

y
[n–1]
(x)

.
A solution of this equation is sought in the parametric form
x = w(t), y = w(t + 1). (17.3.4.3)
Then the original equation is reduced to the following mth-order linear equation with
constant coefficients (equation of the type (17.2.3.1) with integer differences):

a
m
w(t + m)+a
m–1
w(t + m – 1)+···+ a
1
w(t + 1)+a
0
w(t)=0.
Example. Consider the functional equation
y

y(x)

+ ay(x)+bx = 0.(17.3.4.4)
The transformation (17.3.4.3) reduces this equation to a difference equation of the form (17.2.2.1). Let λ
1
and
λ
2
be distinct real roots of the characteristic equation
λ
2
+ aλ + b = 0,(17.3.4.5)
λ
1
≠ λ
2
. Then the general solution of the functional equation (17.3.4.4) in parametric form can be written as
x = Θ

1
(t)λ
t
1
+ Θ
2
(t)λ
t
2
,
y = Θ
1
(t)λ
t+1
1
+ Θ
2
(t)λ
t+1
2
,
(17.3.4.6)
where Θ
1
(t)andΘ
2
(t) are arbitrary 1-periodic functions, Θ
k
(t)=Θ
k

(t + 1).
Taking Θ
1
(t)=C
1
and Θ
2
(t)=C
2
in (17.3.4.6), where C
1
and C
2
are arbitrary constants, and eliminating
the parameter t, we obtain a particular solution of equation (17.3.4.4) in implicit form:
λ
2
x – y
λ
2
– λ
1
= C
1

λ
1
x – y
C
2

(λ
1
– λ
2
)

γ
, γ =
ln λ
1
ln λ
2
.
17.3.4-5. Babbage equation and involutory functions.
1
◦
. Functions satisfying the Babbage equation
y(y(x)) = x (17.3.4.7)
are called involutory functions.
Equation (17.3.4.7) is a special case of (17.3.4.4) with a = 0, b =–1. The corresponding
characteristic equation (17.3.3.5) has the roots λ
1,2
= 1. The parametric representation of
solution (17.3.4.6) contains the complex quantity (–1)
t
= e
iπt
=cos(πt)+i sin(πt)andis,
therefore, not very convenient.
2

◦
. The following statements hold:
(i) On the interval x (a, b), there is a continuous decreasing solution of equation
(17.3.4.7) depending on an arbitrary function. This solution has the form
y(x)=

ϕ(x)forx
(a, c],
ϕ
–1
(x)forx (c, b),
where c
(a, b) is an arbitrary point, and ϕ(x) is an arbitrary continuous decreasing function
on (a, c] such that
lim
x→a+0
ϕ(x)=b, ϕ(c)=c.
(ii) The only increasing solution of equation (17.3.4.7) has the form y(x) ≡ x.
918 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS
3
◦
. Solution in parametric form:
x = Θ

t
2

, y = Θ

t + 1

2

,
where Θ(t)=Θ(t + 1) is an arbitrary periodic function with period 1.
4
◦
. Solution in parametric form (an alternative representation):
x = Θ
1
(t)+Θ
2
(t)sin(πt),
y = Θ
1
(t)–Θ
2
(t)sin(πt),
where Θ
1
(t)andΘ
2
(t) are arbitrary 1-periodic functions, Θ
k
(t)=Θ
k
(t + 1), k = 1, 2.In
this solution, the functions sin(πt) can be replaced by cos(πt).
17.4. Nonlinear Difference and Functional Equations
with a Single Variable
17.4.1. Nonlinear Difference Equations with a Single Variable

17.4.1-1. Riccati difference equation.
The Riccati difference equation has the general form
y(x)y(x + 1)=a(x)y(x + 1)+b(x)y(x)+c(x), (17.4.1.1)
where the functions a(x), b(x), c(x) satisfy the condition a(x)b(x)+c(x)
0.
1
◦
. The substitution
y(x)=
u(x + 1)
u(x)
+ a(x)
reduces the Riccati equation to the linear second-order difference equation
u(x + 2)+[a(x + 1)–b(x)]u(x + 1)–[a(x)b(x)+c(x)]u(x)=0.
2
◦
.Lety
0
(x) be a particular solution of equation (17.4.1.1). Then the substitution
z(x)=
1
y(x)–y
0
(x)
reduces equation (17.4.1.1) to the first-order linear nonhomogeneous difference equation
z(x + 1)+
[y
0
(x)–a(x)]
2

a(x)b(x)+c(x)
z(x)+
y
0
(x)–a(x)
a(x)b(x)+c(x)
= 0.
17.4.1-2. First-order nonlinear difference equations with no explicit dependence on x.
Consider the nonlinear functional equation
y(x + 1)=f

y(x)

, 0 ≤ x < ∞ ,(17.4.1.2)
where f(y) is a continuous function.
Let y(x)beafunctiondefinedonthesemiaxis[0, ∞)by
y(x)=f
[n]

ϕ({x})

, n =[x], (17.4.1.3)
where [x]and{x} are, respectively, the integer and the fractional parts of x,andϕ(x)isany
continuous function on [0, 1) such that ϕ(0)=a, ϕ(1 – 0)=f (a), a is an arbitrary constant.
Direct verification shows that the function (17.4.1.3) satisfies equation (17.4.1.2) and is
continuous on the semiaxis [0, ∞).
17.4. NONLINEAR DIFFERENCE AND FUNCTIONAL EQUATIONS WITH A SINGLE VARIABLE 919
17.4.1-3. Nonlinear mth-order difference equations.
1
◦

. In the general case, a nonlinear difference equation has the form
F

x, y(x + h
0
), y(x + h
1
), , y(x + h
m
)

= 0.(17.4.1.4)
This equation contains the unknown function with different values of its argument differing
from one another by constants. The constants h
0
, h
1
, , h
m
are called differences or
deviations of the argument. If all h
k
are integers, then (17.4.1.4) is called an equation with
integer differences. If h
k
= k (k = 0, 1, , m) and the equation explicitly involves y(x)
and y(x + m), then it is called a difference equation of order m.
2
◦
.Anmth-order difference equation resolved with respect to the leading term y(x + m)

has the form
y(x + m)=f

x, y(x), y(x + 1), , y(x + m – 1)

.(17.4.1.5)
The Cauchy problem for this equation consists of finding its solution with a given initial
distribution of the unknown function on the interval 0 ≤ x ≤ m:
y = ϕ(x)at0 ≤ x ≤ m.(17.4.1.6)
The values of y(x)form ≤ x ≤ m + 1 are calculated by substituting the initial values
(17.4.1.6) into the right-hand side of equation (17.4.1.5) for 0 ≤ x ≤ 1.Wehave
y(x + m)=f

x, ϕ(x), ϕ(x + 1), , ϕ(x + m – 1)

.(17.4.1.7)
Then, replacing x by x + 1 in equation (17.4.1.5), we obtain
y(x + m + 1)=f

x + 1, y(x + 1), y(x + 2), , y(x + m)

.(17.4.1.8)
The values of y(x)form + 1 ≤ x ≤ m +2 are determined by the right-hand side of (17.4.1.8)
for 0 ≤ x
≤ 1, the quantities y(x+1), , y(x +m–1) are determined by the initial condition
(17.4.1.6), and y(x + m) is taken from (17.4.1.7).
In order to find y(x)form + 2 ≤ x ≤ m + 3, we replace x by x + 1 in equation (17.4.1.8)
and consider this equation on the interval 0 ≤ x ≤ 1. Using the initial conditions for y(x + 2),
, y(x + m – 1) and the quantities y(x + m), y(x + m + 1) found on the previous steps, we
find y(x + m + 2).

In a similar way, one can find y(
x)form + 3 ≤ x ≤ m + 4, etc. This method for solving
difference equations is called the step method.
17.4.2. Reciprocal (Cyclic) Functional Equations
17.4.2-1. Reciprocal functional equations depending on y(x)andy(a – x).
1
◦
. Consider a reciprocal equation of the form
F

x, y(x), y(a – x)

= 0.
Replacing x by a – x, we obtain a similar equation containing the unknown function
with the same arguments:
F

a – x, y(a – x), y(x)

= 0.
Eliminating y(a – x) from this equation and the original equation, we come to the usual
algebraic (or transcendental) equation of the form Ψ

x, y(x)

= 0.
In other words, solutions of the original functional equation y = y(x)aredefined in a
parametric manner by means of two algebraic (or transcendental) equations:
F (x, y,t)=0, F (a – x, t, y)=0,(17.
4.2.1)

where t is a parameter.