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
Trang 1homogeneous 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 y0(x) on I, then the general continuous solution on I is given by
y (x) = y0(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◦ Let x
I , ξI , and f (x)R0
ξ [I] Suppose that g(x) and F (x) are continuous functions,
g (x)≠ 0for x≠ξ , and g(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) is defined 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)R0
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
2F (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, and C 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 xI = (a, b] Assume that on the interval I the function f (x) is continuous, strongly increasing, and satisfies the condition a < f (x) < x, and F (x) is a function of bounded
variation
Depending on the value of the parameter λ, the following cases are possible.
Trang 2Case|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
Case0<|λ|<1 For any x0I and any test function ϕ(z) that has bounded variation
on [f (x0), x0] and satisfies the condition
ϕ (f (x0)) – λϕ(x0) = F (x0), 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 (x0), x0] as the test function ϕ(x).
x→a+0F (x) = 0 and there is a point x0 I such that the series
∞
n=0F f
[n] (x0)
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+0y (x).
Case λ = –1 See Item1◦.
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
,
whereΘ(z) is an arbitrary1-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 for0 ≤x≤a;
(ii) f (0) =0and f (x) < x for0< x < a;
(iii) the derivative f (x) exists, has bounded variation on the interval 0 < x < a, and
lim
x→+0f
(x) =1 Then, for any x, x0(0, a], there exists the limit
y (x) = lim
n→∞
f[n] (x) – f[n] (x0)
f[n–1 ](x0) – f[n] (x0), (17.3.3.6) which defines a monotonically increasing function satisfying the Abel equation (17.3.3.5)
with c = –1(this is the L´evy solution).
Trang 33◦ Let f (x) be defined 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)
I
If for all xI there exists the limit
y (x) = lim
n→∞
f[n] (x) – f[n] (x0)
where x0 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) =0and f (x) < x≤a for x≠ 0
We define a function ϕ(y) on the semiaxis [0,∞) by
ϕ (y) = f[n] ϕ0({y})
, n = [y], (17.3.3.8)
where [y] and{y}are, respectively, the integer and the fractional parts of y, and ϕ0(y) is
any continuous strictly decreasing function on [0,1) such that ϕ0(0) = a, ϕ0(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] ϕ0(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] ϕ0(1–0)
= f[n–1] f (a)
= f[n] (a),
ϕ (n) = f[n] ϕ0(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–1defined 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)
Trang 417.3.4 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–1y (b m–1x) +· · · + a1y (b1x ) + a0y (b0x ) = f (x). (17.3.4.1) The transformation
y (x) = w(z), z = ln x
reduces (17.3.4.1) to a linear difference equation with constant coefficients
a m w (z + h m ) + a m–1w (z + h m–1) +· · · + a1w (z + h1) + a0w (z + h0) = f (e z), h k = ln b k.
Note that the homogeneous functional equation (17.3.4.1) with f (x)≡ 0admits particular
solutions of power type, y(x) = Cx β , where C is an arbitrary constant and β is a root of the
transcendental equation
a m b β
m + a m–1b β
m–1+· · · + a1b β
1 + a0b β0 =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–1y x n m–1
+· · · + a1y x n1
+ a0y (x n0) = f (x). (17.3.4.2) The transformation
y (x) = w(z), z = ln ln x
reduces (17.3.4.2) to a linear difference equation with constant coefficients
a m w (z + h m ) + a m–1w (z + h m–1) +· · · +a1w (z + h1) + a0w (z + h0) = f (e e z), h k = ln ln n k.
Note that the homogeneous functional equation (17.3.4.2) admits particular solutions of
the form y(x) = C|ln x|p , where C 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+· · · + a1|n1|p + a
0|n0|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–1y e λ m–1x
+· · · + a1y e λ1x
+ a0y (e λ0x ) = f (x).
The transformation
y (x) = w(ln z), z = ln x
reduces this equation to a linear difference equation with constant coefficients
a m w (z + h m ) + a m–1w (z + h m–1) +· · · + a1w (z + h1) + a0w (z + h0) = f (e z), h k = ln λ k.
Trang 517.3.4-4 Equations containing iterations of the unknown function.
Consider the linear homogeneous equation
a m y[m] (x) + a m–1y[m–1 ](x) + · · · + a1y (x) + a0x=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–1w (t + m –1) +· · · + a1w (t +1) + a0w (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 λ1 and
λ2be distinct real roots of the characteristic equation
λ2+ aλ + b =0 , (17 3 4 5 )
λ1≠λ2 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+ Θ 2(t)λ t,
where Θ 1(t) andΘ 2(t) are arbitrary1 -periodic functions, Θk (t) =Θk (t +1 ).
Taking Θ 1(t) = C1 and Θ 2(t) = C2in (17.3.4.6), where C1and C2 are arbitrary constants, and eliminating
the parameter t, we obtain a particular solution of equation (17.3.4.4) in implicit form:
λ2x – y
λ2– λ1 = C1
λ1x – y
C2(λ1– λ2 )
γ , γ= ln λ1
ln λ2
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 λ1, 2= 1 The parametric representation of solution (17.3.4.6) contains the complex quantity (–1) = e iπt = cos(πt) + i sin(πt) and is,
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) for x
(a, c],
ϕ–1(x) for x
(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+0ϕ (x) = b, ϕ (c) = c.
(ii) The only increasing solution of equation (17.3.4.7) has the form y(x)≡x
Trang 63◦ Solution in parametric form:
x=Θ2t, y =Θt+21, whereΘ(t) = Θ(t +1) is an arbitrary periodic function with period 1
4◦ Solution in parametric form (an alternative representation):
x=Θ1(t) +Θ2(t) sin(πt),
y=Θ1(t) –Θ2(t) sin(πt),
whereΘ1(t) andΘ2(t) are arbitrary1-periodic functions, Θk (t) =Θk (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◦ Let y0(x) be a particular solution of equation (17.4.1.1) Then the substitution
z (x) = 1
y (x) – y0(x)
reduces equation (17.4.1.1) to the first-order linear nonhomogeneous difference equation
z (x +1) + [y0(x) – a(x)]
2
a (x)b(x) + c(x) z (x) +
y0(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) be a function defined on the semiaxis [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, and ϕ(x) is any
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,∞).
Trang 717.4.1-3 Nonlinear mth-order difference equations.
1◦ In the general case, a nonlinear difference equation has the form
F x , y(x + h0), y(x + h1), , 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 h0, h1, , h m are called differences or
deviations of the argument If all h kare 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◦ An mth-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 interval0 ≤x≤m:
y = ϕ(x) at 0 ≤x≤m (17.4.1.6)
The values of y(x) for m≤x ≤m+1are calculated by substituting the initial values (17.4.1.6) into the right-hand side of equation (17.4.1.5) for0 ≤x≤ 1 We have
y (x + m) = f x , ϕ(x), ϕ(x +1), , ϕ(x + m –1)
(17.4.1.7)
Then, replacing x by x +1in 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) for m +1 ≤x≤m+2are determined by the right-hand side of (17.4.1.8) for0 ≤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) for m +2 ≤x≤m+3, we replace x by x +1in equation (17.4.1.8) and consider this equation on the interval0 ≤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) for m +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) and y(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 formΨ x , y(x)
=0
In other words, solutions of the original functional equation y = y(x) are defined 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.