Handbook of mathematics for engineers and scienteists part 210 ppt

T12.2. NONLINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1431
T12.2.1-3. Functional equations involving y(x)andy(ax).
10. y(2x) – ay
2
(x) =0.
A special case of equation T12.2.3.3.
Particular solution:
y(x)=
1
a
e
Cx
,
where C is an arbitrary constant.
11. y(2x) –2y
2
(x) + a =0.
A special case of equation T12.2.3.3.
Particular solutions with a = 1:
y(x)=0, y(x)=cos(Cx), y(x)=cosh(Cx),
where C is an arbitrary constant.
12. y(x)y(ax) = f(x).
A special case of equation T12.2.3.3.
T12.2.1-4. Functional equations involving y(x)andy(a/x).
13. y(x)y(a/x) = b
2
.
Solution:
y(x)=
b exp


Φ(x, a/x)

,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
On setting Φ(x, z)=C(ln x –lnz), one arrives at particular solutions of the form
y =
ba
–C
x
2C
,
where C is an arbitrary constant.
14. y(x)y(a/x) = f
2
(x).
The function f(x) must satisfy the condition f(x)=
f(a/x). For definiteness, we take
f(x)=f (a/x).
Solution:
y(x)=
f(x)exp

Φ(x, a/x)

,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
On setting Φ(x, z)=C(ln x –lnz), one arrives at particular solutions of the form
y =
a
–C

x
2C
f(x),
where C is an arbitrary constant.
15. y
2
(x) + Ay(x)y(a/x) + By
2
(a/x) + Cy(x) + Dy(a/x) = f(x).
A special case of equation T12.2.3.4.
Solution in parametric form (w is a parameter):
y
2
+ Ayw + Bw
2
+ Cy + Dw = f (x),
w
2
+ Ayw + By
2
+ Cw + Dy = f (a/x).
Eliminating w gives the solutions in implicit form.
1432 FUNCTIONAL EQUATIONS
T12.2.1-5. Other functional equations with quadratic nonlinearity.
16. y(x
2
) – ay
2
(x) =0.
Solution:

y(x)=
1
a
x
C
,
where C is an arbitrary constant. In addition, y(x) ≡ 0 is also a continuous solution.
17. y(x)y(x
a
) = f(x), a >0.
A special case of equation T12.2.3.8.
18. y(x)y

a – x
1+bx

= A
2
.
Solutions:
y(x)=
A exp

Φ

x,
a – x
1 + bx



,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
19. y(x)y

a – x
1+bx

= f
2
(x).
The right-hand side function must satisfy the condition f(x)=
f

a – x
1 + bx

. For definite-
ness, we take f(x)=f

a – x
1 + bx

.
Solutions:
y(x)=
f(x)exp

Φ

x,

a – x
1 + bx


,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
20. y
2
(x) + Ay(x)y

a – x
1+bx

+ By(x) = f(x).
A special case of equation T12.2.3.5.
21. y(x)y


a
2
– x
2

= b
2
, 0 ≤ x ≤ a.
Solutions:
y(x)=
b exp


Φ

x,

a
2
– x
2

,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
22. y(x)y


a
2
– x
2

= f
2
(x), 0 ≤ x ≤ a.
The right-hand side function must satisfy the condition f(x)=
f

√
a
2
– x
2


. For defi-
niteness, we take f(x)=f

√
a
2
– x
2

.
Solutions:
y(x)=
f(x)exp

Φ

x,

a
2
– x
2

,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
T12.2. NONLINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1433
23. y(sin x)y(cos x) = a
2
.

Solutions in implicit form:
y(sin x)=
a exp

Φ(sin x,cosx)

,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
24. y(sin x)y(cos x) = f
2
(x).
The right-hand side function must satisfy the condition f(x)=
f

π
2
–x

. For definiteness,
we take f(x)=f

π
2
– x

.
Solutions in implicit form:
y(sin x)=
f(x)exp


Φ(sin x,cosx)

,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
25. y(x)y

ω(x)

= b
2
, where ω

ω(x)

= x.
Solutions:
y(x)=
b exp

Φ

x, ω(x)

,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
26. y(x)y

ω(x)

= f

2
(x), where ω

ω(x)

= x.
The right-hand side function must satisfy the condition f (x)=
f

ω(x)

. For definiteness,
we take f(x)=f

ω(x)

.
Solutions:
y(x)=
f(x)exp

Φ

x, ω(x)

,
where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments.
T12.2.2. Functional Equations with Power Nonlinearity
1. y(x + a) – by
λ

(x) = f(x).
A special case of equation T12.2.3.1.
2. y
λ
(x)y(a – x) = f(x).
A special case of equation T12.2.3.2.
Solution:
y(x)=

f(x)

–
λ
1–λ
2

f(a – x)

1
1–λ
2
.
3. y
2n+1
(x) + y
2n+1
(a – x) = b, n =1, 2,
The change of variable w(x)=y
2n+1
(x) leads to a linear equation of the form T12.1.1.22:

w(x)+w(a – x)=b.
1434 FUNCTIONAL EQUATIONS
4. y(ax) = by
k
(x).
Solution for x > 0, a > 0, b > 0,andk > 0 (a ≠ 1):
y(x)=b
1
1–k
exp

x
ln k
ln a
Θ

ln x
ln a


,
where Θ(x)=Θ(x + 1) is an arbitrary periodic function with period 1.
5. y
λ
(x)y(a/x) = f(x).
A special case of equation T12.2.3.4.
Solution:
y(x)=

f(x)


–
λ
1–λ
2

f(a/x)

1
1–λ
2
.
6. y
λ
(x)y

a – x
1+bx

= f(x).
A special case of equation T12.2.3.5.
7. y
λ
(x)y

ax – β
x + b

= f(x), β = a
2

+ ab + b
2
.
A special case of equation T12.2.3.13.
8. y
λ
(x)y

bx + β
a – x

= f(x), β = a
2
+ ab + b
2
.
A special case of equation T12.2.3.13.
9. y
λ
(x)y(x
a
) = f(x).
A special case of equation T12.2.3.8.
10. y
λ
(x)y


a
2

– x
2

= f(x).
A special case of equation T12.2.3.9.
11. y
λ
(sin x)y(cos x) = f(x).
A special case of equation T12.2.3.10.
T12.2.3. Nonlinear Functional Equation of General Form
1. F

x, y(x), y(x + a)

=0.
We assume that a > 0. Let us solve the equation for y(x + a) to obtain
y(x + a)=f

x, y(x)

.(1)
1
◦
. First, let us assume that the equation is defined on a discrete set of points x = x
0
+ ak
with integer k. Given an initial value y(x
0
), one can make use of (1) to find sequentially
y(x

0
+ a), y(x
0
+ 2a), etc.
Solving the original equation for y(x) yields
y(x)=g

x, y(x + a)

.(2)
On setting x = x
0
– a here, one can find y(x
0
– a) and then likewise y(x
0
– 2a)etc.
Thus, given initial data, one can use the equation to find y(x) at all points x
0
+ak,where
k = 0, 1, 2,
T12.2. NONLINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1435
2
◦
. Now assume that x in the equation can vary continuously. Also assume that y(x)isa
continuous function defined arbitrarily on the semi-interval [0, a). Setting x = 0 in (1), one
finds y(a). Then, given y(x)on[0, a], one can use (1) to find y(x)onx
[a, 2a], then on
x [2a, 3a],andsoon.
Remark. The case of a < 0 can be reduced, using the change of variable z = x + a, to an equation of the

form F

z + b, y(z + b), y(z)

= 0 with b =–a > 0, which was already considered above.
2. F

x, y(x), y(a – x)

=0.
Substituting x with a – x, one obtains F

a – x, y(a – x), y(x)

= 0. Now, eliminating
y(a – x) from this equation and the original one, one arrives at an ordinary algebraic (or
transcendental) equation of the form Ψ

x, y(x)

= 0.
Toputitdifferently, thesolutiony =y(x) of the original functional equation is determined
parametrically with the system of two algebraic (or transcendental) equations
F (x, y, t)=0, F (a – x, t, y)=0,
where t is a parameter.
3. F


x, y(x), y(ax)


=0, a >0.
The transformation z =lnx, w(z)=y(x) leads to an equation of the form T12.2.3.1:
F

e
z
, w(z), w(z + b)

= 0, b =lna.
4. F

x, y(x), y(a/x)

=0.
Substituting x with a/x yields F

a/x, y(a/x), y(x)

= 0. Eliminating y(a/x) from this
equation and the original one, we arrive at an ordinary algebraic (or transcendental) equation
of the form Ψ

x, y(x)

= 0.
Toputitdifferently, thesolutiony =y(x) of the original functional equation is determined
parametrically with the system of two algebraic (or transcendental) equations
F (x, y, t)=0, F (a/x, t, y)=
0,
where t is the parameter.

5. F

x, y(x), y

a – x
1+bx


=0.
Substituting x with
a – x
1 + bx
yields
F

a – x
1 + bx
, y

a – x
1 + bx

, y(x)

= 0.
Eliminating y

a – x
1 + bx


from this equation and the original one, we arrive at an ordinary
algebraic (or transcendental) equation of the form Ψ

x, y(x)

= 0.
In other words, the solution y = y(x) of the original functional equation is determined
parametrically with the system of two algebraic (or transcendental) equations
F (x, y, t)=0, F

a – x
1 + bx
, t, y

= 0,
where t is the parameter.
1436 FUNCTIONAL EQUATIONS
6. F

x, y(x), y

ax – β
x + b


=0, β = a
2
+ ab + b
2
.

A special case of equation T12.2.3.13 below.
7. F

x, y(x), y

bx + β
a – x


=0, β = a
2
+ ab + b
2
.
A special case of equation T12.2.3.13 below.
8. F

x, y(x), y(x
a
)

=0.
The transformation ξ =lnx, u(ξ)=y(x) leads to an equation of the form 3 above:
F

e
ξ
, u(ξ), u(aξ)

= 0.

9. F

x, y(x), y


a
2
– x
2

=0, 0 ≤ x ≤ a.
Substituting x with
√
a
2
– x
2
yields
F


a
2
– x
2
, y


a
2

– x
2

, y(x)

= 0.
Eliminating y

√
a
2
– x
2

from this equation and the original one, we arrive at an ordinary
algebraic (or transcendental) equation of the form Ψ

x, y(x)

= 0.
In other words, the solution y = y(x) of the original functional equation is determined
parametrically with the system of two algebraic (or transcendental) equations
F (x, y, t)=0, F


a
2
– x
2
, t, y


= 0,
where t is the parameter.
10. F

x, y(sin x), y(cos x)

=0.
Substituting x with
π
2
–x yields F

π
2
–x, y(cos x), y(sin x)

= 0. Eliminating y(cos x) from
this equation and the original one, we arrive at an ordinary algebraic (or transcendental)
equation of the form Ψ

x, y(sin x)

= 0 for y(sin x).
11. F

x, y(x), y

ω(x)


=0, where ω

ω(x)

= x.
Substituting x with ω(x) yields F

ω(x), y

ω(x)

, y(x)

= 0. Eliminating y

ω(x)

from
this equation and the original one, we arrive at an ordinary algebraic (or transcendental)
equation of the form Ψ

x, y(x)

= 0.
Thus, the solution y = y(x) of the original functional equation is determined parametri-
cally with the system of two algebraic (or transcendental) equations
F (x, y, t)=0, F

ω(x), t, y


= 0,
where t is the parameter.
12. F

x, y(x), y(x +1), y(x +2)

=0.
A second-order nonlinear difference equation of general form. A special case of equation
T12.2.3.14.
T12.2. NONLINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1437
13. F

x, y(x), y

ax – β
x + b

, y

bx + β
a – x


=0, β = a
2
+ ab + b
2
.
Let us substitute x first with
ax – β

x + b
and then with
bx + β
a – x
to obtain two more equations.
As a result, we get the following system (the original equation is given first):
F

x, y(x), y(u), y(w)

= 0,
F

u, y(u), y(w), y(x)

= 0,
F

w, y(w), y(x), y(u)

= 0.
(1)
The arguments u and w are expressed in terms of x as
u =
ax – β
x + b
, w =
bx + β
a – x
.

Eliminating y(u)andy(w) from the system of algebraic (transcendental) equations (1),
we arrive at the solutions y = y(x) of the original functional equation.
14. F

x, y(x), y(x +1), , y(x + n)

=0.
An nth-order nonlinear difference equation of general form.
Let us solve the equation for y(x + n) to obtain
y(x + n)=f

x, y(x), y(x + 1), , y(x + n – 1)

.(1)
1
◦
. Let us assume that the equation is defined on a discrete set of points x = x
0
+ k with
integer k. Given initial values y(x
0
), y(x
0
+ 1), , y(x
0
+ n – 1), one can make use of (1)
to find sequentially y(x
0
+ n), y(x
0

+ n + 1), etc.
Solving the original equation for y(x)gives
y(x)=g

x, y(x + 1), y(x + 2), , y(x + n)

.(2)
On setting x = x
0
– 1 here, one can find y(x
0
– 1), then likewise y(x
0
– 2)etc.
Thus, given initial data, one can use the equation to find y(x) at all points x
0
+ k,where
k = 0, 1, 2,
2
◦
. Now assume that x in the equation can vary continuously. Also assume that y(x)isa
continuous function defined arbitrarily on the semi-interval [0, n). Setting x = 0 in (1), one
finds y(n). Then, given y(x)on[0, n], one can use (1) to find y(x)onx
[n, n + 1], then
on x [n + 1, n + 2],andsoon.
15. F

x, y(x), y
[2]
(x), , y

[n]
(x)

=0.
Notation: y
[2]
(x)=y

y(x)

, , y
[n]
(x)=y

y
[n–1]
(x)

.
Solutions are sought in the parametric form
x = w(t), y = w(t + 1). (1)
Then the original equation is reduced to an nth-order difference equation (see the previous
equation):
F

w(t), w(t + 1), w(t + 2), , w(t + n)

= 0.(2)
The general solution of equation (2) has the structure
x = w(t)=ϕ(t; C

1
, , C
n
),
y = w(t + 1)=ϕ(t + 1; C
1
, , C
n
),
where C
1
= C
1
(t), , C
n
= C
n
(t) are arbitrary periodic functions with period 1, that is,
C
k
(t)=C
k
(t + 1), k = 1, 2, , n.