, y n–1from the system of nonlinear algebraic or transcendental equations 1, one finds the solutions of the original functional equation in implicit form: Ψ x, y0 =0, where y0= y θx.. Eq
Trang 116. F x, y(θ0 (x)), y(θ1(x)), , y(θ n–1 (x))
= 0.
Notation: θ k (x)≡θ x+ k n T
, where k =0, 1, , n –1 The functions θ(x) are assumed
to be periodic with period T , i.e., θ(x) = θ(x + T ) Furthermore, the left-hand side of the equation is assumed to satisfy the condition F (x, ) = F (x + T , ).
In the original equation, let us substitute x sequentially by x+ k n T with k =0, 1, , n–1
to obtain the following system (the original equation is given first):
F x, y0, y1, , y n–1
=0,
F x+ n1T , y1, y2, , y0
=0,
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅,
F x+ n– n1T , y n–1, y0, , y n–2
=0,
(1)
where the notation y k≡y θ k (x)
is used for brevity
Eliminating y1, y2, , y n–1from the system of nonlinear algebraic (or transcendental) equations (1), one finds the solutions of the original functional equation in implicit form:
Ψ x, y0) =0, where y0= y θ(x)
T12.3 Functional Equations in Several Independent
Variables
T12.3.1 Linear Functional Equations
T12.3.1-1 Equations involving functions with a single argument
1. f (x + y) = f (x) + f (y).
Cauchy’s equation.
Solution:
f (x) = Cx, where C is an arbitrary constant.
2. f
x + y
2
= f (x) + f (y)
Jensen’s equation.
Solution:
f (x) = C1x + C2,
where C1and C2are arbitrary constants
3. af (x) + bf (y) = f (ax + by) + c.
1◦ Solution:
f (x) =
Ax+ c
a + b –1 if a + b –1 ≠ 0,
Ax + B if a + b –1=0 and c =0,
where A and B are arbitrary constants.
2◦ If a + b –1=0and c≠ 0, then there is no solution
Trang 24. Af (a1x + b1y + c1) + Bf(a2x + b2y + c2) = Cf(a3x + b3y + c3) + D.
All continuous solutions of this equation have the form
f (x) = αx + β, where the constants α and β are determined by substituting this expression into the original
equation
5. f (x + y) + f (x – y) = 2f (x) + 2f (y).
Solution:
f (x) = Cx2,
where C is an arbitrary constant.
6. f (x + y) = f (x)e ay.
Solution:
f (x) = Ce ax,
where C is an arbitrary constant.
7. f (x + y) + f (x – y) = 2f (x) cosh y.
Solution:
f (x) = C1e x + C2e–x,
where C1and C2are arbitrary constants
8. f (x + y) + f (x – y) = 2f (x) cosh(ay) + 2f (y).
Solution:
f (x) = C[2– cosh(ax)], where C is an arbitrary constant.
9. f (x + y) + f (x – y) = 2f (x) cos y.
Solution:
f(x) = C1cos x + C2sin x, where C1and C2are arbitrary constants
10. f (x + y) + f (x – y) = 2f (x) cos(ay) + 2f (y).
Solution:
f(x) = C[2– cos(ax)], where C is an arbitrary constant.
11. f (xy) = f (x) + f (y).
Cauchy’s logarithmic equation.
Solution:
f (x) = C ln|x|,
where C is an arbitrary constant.
Trang 312. f (x n + y n) 1/n
= f (x) + f (y).
Solution:
f (x) = Cx n,
where C is an arbitrary constant.
13. f (x) + f (y) = f
x + y
1 – xy
, xy < 1.
Solution:
f (x) = C arctan x, where C is an arbitrary constant.
14. f (x) + (1 – x)f
y
1 – x
= f (y) + (1 – y)f
x
1 – y
.
A basic equation of information theory The quantities x, y, and x + y can assume values
from 0 to 1
Solution:
f (x) = C[x ln x + (1– x) ln(1– x)], where C is an arbitrary constant.
15. f (x) + (1 – x) α f
y
1 – x
= f (y) + (1 – y) α f
x
1 – y
.
Here, the quantities x, y, and x + y can assume values from 0 to 1; α≠ 0, 1, 2
Solution:
f (x) = C[x α+ (1– x) α–1],
where C is an arbitrary constant.
16. f
xy–
(1 – x2)(1 – y2 )
= f (x) + f (y), |x| ≤1, |y| ≤1.
Solution:
f (x) = C arccos x, where C is an arbitrary constant.
17. f
xy+
(x2– 1)(y2 – 1)
= f (x) + f (y), |x| ≥ 1, |y| ≥ 1.
Solution:
f(x) = C arccosh x, where C is an arbitrary constant.
18. f (x) + g(y) = h(x + y).
Pexider’s equation Here, f (x), g(y), and h(z) are the unknown functions.
Solution:
f (x) = C1x + C2, g(y) = C1y + C3, h(z) = C1z + C2+ C3,
where C1, C2, and C3are arbitrary constants
Trang 4T12.3.1-2 Equations involving functions with two arguments.
19. f (x, y) = f (y, x).
This equation may be treated as a definition of functions symmetric with respect to permu-tation of the arguments.
1◦ Solution:
f (x, y) = Φ(x, y) + Φ(y, x),
whereΦ(x, y) is an arbitrary function with two arguments.
2◦ Particular solutions may be found using the formula
f (x, y) = Ψ(ϕ(x) + ϕ(y))
by specifying the functionsΨ(z), ϕ(x).
20. f (x, y) = –f (y, x).
This equation may be treated as a definition of functions antisymmetric with respect to permutation of the arguments.
1◦ Solution:
f (x, y) = Φ(x, y) – Φ(y, x),
whereΦ(x, y) is an arbitrary function with two arguments.
2◦ Particular solutions may be found using the formulas
f (x, y) = ϕ(x) – ϕ(y),
f (x, y) = (x – y) Ψ(ϕ(x) + ϕ(y)),
by specifying the functions ϕ(x) and Ψ(z).
21. f (x, y) = f (x + ak1, y + ak2).
Traveling-wave equation Here, a is an arbitrary number and k1and k2are some constants Solution:
f(x, y) = Φ(k2x – k1y),
whereΦ(z) is an arbitrary function.
22. f (ax, ay) = f (x, y).
Here, a is an arbitrary number (a≠ 0)
Solution:
f (x, y) = Φ(y/x),
whereΦ(z) is an arbitrary function.
23. f (ax, ay) = a β f (x, y).
Equation of a homogeneous function Here, a is an arbitrary number (a≠ 0) and β is a fixed
number called the order of homogeneity
Solution:
f (x, y) = x β Φ(y/x),
whereΦ(x) is an arbitrary function.
Trang 524. f (ax, a β y) = f (x, y).
Here, a is an arbitrary number (a≠ 0) and β is some constant.
Solution:
f (x, y) =Φ yx–β
, whereΦ(x) is an arbitrary function.
25. f (ax, a β y) = a γ f (x, y).
Equation of self-similar solutions Here, a is an arbitrary number (a≠ 0) and β and γ are
some constants
Solution:
f (x, y) = x γΦ yx–β
, whereΦ(x) is an arbitrary function.
26. f (x, y) = a n f x + (1 – a)y, ay
.
Here, a is an arbitrary number (a >0) and n is some constant.
Solution:
f(x, y) = y–n Φ(x + y),
whereΦ(x) is an arbitrary function.
27. f (x, y) = a n f (a m x, y + ln a).
Here, a is an arbitrary number (a >0) and n and m are some constants.
Solution:
f (x, y) = e–nyΦ xe–my
, whereΦ(x) is an arbitrary function.
28. f (x, y) + f (y, z) = f (x, z).
Cantor’s first equation.
Solution:
f (x, y) = Φ(x) – Φ(y),
whereΦ(x) is an arbitrary function.
29. f (x + y, z) + f (y + z, x) + f (z + x, y) = 0.
Solution:
f(x, y) = (x –2y)ϕ(x + y), where ϕ(x) is an arbitrary function.
30. f (xy, z) + f (yz, x) + f (zx, y) = 0.
Solution:
f (x, y) = ϕ(xy) ln x
y2 if x >0, y≠ 0;
f (x, y) = ϕ(xy) ln –x
y2 if x <0, y≠ 0;
f (x, y) = A + B ln|x| if x≠ 0, y=0;
f (x, y) = A + B ln|y| if x =0, y≠ 0,
where ϕ(x) is an arbitrary function and A and B are arbitrary constants.
Trang 6T12.3.2 Nonlinear Functional Equations
T12.3.2-1 Equations involving one unknown function with a single argument
1. f (x + y) = f (x)f (y).
Cauchy’s exponential equation.
Solution:
f (x) = e Cx,
where C is an arbitrary constant In addition, the function f (x)≡ 0is also a solution
2. f (x + y) = af (x)f (y).
Solution:
f (x) = 1
a e
Cx,
where C is an arbitrary constant In addition, the function f (x)≡ 0is also a solution
3. f x + y
2
=
f (x)f (y).
Solution:
f (x) = Ca x,
where a and C are arbitrary positive constants.
4. f x + y
n
=
f (x)f (y)1/n
.
Solution:
f (x) = a x,
where a is an arbitrary positive constant.
5. f (y + x) + f (y – x) = 2f (x)f (y).
D’Alembert’s equation.
Solutions:
f (x) = cos(Cx), f (x) = cosh(Cx), f (x)≡ 0,
where C is an arbitrary constant.
6. f (y + x) + f (y – x) = af (x)f (y).
Solutions:
f (x) = 2
a cos(Cx), f (x) = 2
a cosh(Cx), f (x)≡ 0,
where C is an arbitrary constant.
7. f (x + y) = a xy f (x)f (y).
Solution:
f (x) = e Cx a x2/ ,
where C is an arbitrary constant In addition, the function f (x)≡ 0is also a solution
Trang 78. f (x + y) = f (x) + f (y) – af (x)f (y), a ≠0.
For a =1, it is an equation of probability theory.
Solution:
f (x) = 1
a 1– e–βx
,
where β is an arbitrary constant In addition, the function f (x)≡ 0is also a solution
9. f (x + y)f (x – y) = f2(x).
Lobachevsky’s equation.
Solution:
f (x) = C1exp(C2x), where C1and C2are arbitrary constants
10. f (x + y + a)f (x – y + a) = f2(x) + f2(y) – 1.
Solutions:
f (x) = 1, f (x) = cosnπx
a ,
where n =1, 2, For trigonometric solutions, a must be nonzero.
11. f (x + y + a)f (x – y + a) = f2(x) – f2(y).
Solutions:
f (x) =0, f(x) = C sin 2πx
a , f(x) = C sin (2n+1)πx
where C is an arbitrary constant and n =0, 1,2, For trigonometric solutions, a must
be nonzero
12. (x – y)f(x)f(y) = xf(x) – yf(y).
Solutions:
f(x)≡ 1, f (x) = C
x + C, where C is an arbitrary constant.
13. f (xy) = af (x)f (y).
Cauchy’s power equation (for a =1)
Solution:
f (x) = 1
a|x|C,
where C is an arbitrary constant In addition, the function f (x)≡ 0is also a solution
14. f (xy) = [f (x)] y.
Solution:
f (x) = e Cx,
where C is an arbitrary constant In addition, the function f (x)≡ 0 is also a solution for
y>0