Handbook of mathematics for engineers and scienteists part 211 potx

1438 FUNCTIONAL EQUATIONS 16. F  x, y(θ 0 (x)), y(θ 1 (x)), , y(θ n–1 (x))  =0. Notation: θ k (x) ≡ θ  x + k n T  ,wherek = 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 theoriginal 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, y 0 , y 1 , , y n–1  = 0, F  x + 1 n T , y 1 , y 2 , , y 0  = 0, ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅, F  x + n–1 n T , y n–1 , y 0 , , y n–2  = 0, (1) where the notation y k ≡ y  θ k (x)  is used for brevity. Eliminating y 1 , y 2 , , y n–1 from the system of nonlinear algebraic (or transcendental) equations (1), one finds the solutions of the original functional equation in implicit form: Ψ  x, y 0 )=0,wherey 0 = 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) 2 . Jensen’s equation. Solution: f(x)=C 1 x + C 2 , where C 1 and C 2 are 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 ◦ .Ifa + b – 1 = 0 and c ≠ 0, then there is no solution. T12.3. FUNCTIONAL EQUATIONS IN SEVERAL INDEPENDENT VARIABLES 1439 4. Af(a 1 x + b 1 y + c 1 ) + Bf(a 2 x + b 2 y + c 2 ) = Cf(a 3 x + b 3 y + c 3 ) + 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)=Cx 2 , 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)coshy. Solution: f(x)=C 1 e x + C 2 e –x , where C 1 and C 2 are 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)cosy. Solution: f(x)=C 1 cos x + C 2 sin x, where C 1 and C 2 are 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. 1440 FUNCTIONAL EQUATIONS 12. 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,andx + 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,andx + 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–x 2 )(1–y 2 )  = f (x) + f(y), |x| ≤ 1, |y| ≤ 1. Solution: f(x)=C arccos x, where C is an arbitrary constant. 17. f  xy +  (x 2 –1)(y 2 –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)=C 1 x + C 2 , g(y)=C 1 y + C 3 , h(z)=C 1 z + C 2 + C 3 , where C 1 , C 2 ,andC 3 are arbitrary constants. T12.3. FUNCTIONAL EQUATIONS IN SEVERAL INDEPENDENT VARIABLES 1441 T12.3.1-2. 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), where Φ(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 functions Ψ(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), where Φ(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)andΨ(z). 21. f(x, y) = f(x + ak 1 , y + ak 2 ). Traveling-wave equation. Here, a is an arbitrary number and k 1 and k 2 are some constants. Solution: f(x, y)=Φ(k 2 x – k 1 y), where Φ(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), where Φ(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)andβ is a fixed number called the order of homogeneity. Solution: f(x, y)=x β Φ(y/x), where Φ(x) is an arbitrary function. 1442 FUNCTIONAL EQUATIONS 24. f(ax, a β y) = f(x, y). Here, a is an arbitrary number (a ≠ 0)andβ is some constant. Solution: f(x, y)=Φ  yx –β  , where Φ(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)andβ and γ are some constants. Solution: f(x, y)=x γ Φ  yx –β  , where Φ(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)andn is some constant. Solution: f(x, y)=y –n Φ(x + y), where Φ(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)andn and m are some constants. Solution: f(x, y)=e –ny Φ  xe –my  , where Φ(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), where Φ(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 y 2 if x > 0, y ≠ 0; f(x, y)=ϕ(xy)ln –x y 2 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. T12.3. FUNCTIONAL EQUATIONS IN SEVERAL INDEPENDENT VARIABLES 1443 T12.3.2. 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) ≡ 0 is 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) ≡ 0 is 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 x 2 /2 , where C is an arbitrary constant. In addition, the function f (x) ≡ 0 is also a solution. 1444 FUNCTIONAL EQUATIONS 8. 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) ≡ 0 is also a solution. 9. f(x + y)f(x – y) = f 2 (x). Lobachevsky’s equation. Solution: f(x)=C 1 exp(C 2 x), where C 1 and C 2 are arbitrary constants. 10. f(x + y + a)f(x – y + a) = f 2 (x) + f 2 (y) –1. Solutions: f(x)= 1, f(x)= cos nπx a , where n = 1, 2, For trigonometric solutions, a must be nonzero. 11. f(x + y + a)f(x – y + a) = f 2 (x) – f 2 (y). Solutions: f(x)=0, f(x)=C sin 2πx a , f(x)=C sin (2n + 1)πx a , 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) ≡ 0 is 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. . for brevity. Eliminating y 1 , y 2 , , y n–1 from the system of nonlinear algebraic (or transcendental) equations (1), one finds the solutions of the original functional equation in implicit form: Ψ  x,. C 2 , where C 1 and C 2 are 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. c 2 ) = Cf(a 3 x + b 3 y + c 3 ) + 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.