1424 FUNCTIONAL EQUATIONS 2 ◦ . On the interval x (a, b), there exists a decreasing solution of the original equation involving an arbitrary function: y(x)= ϕ(x)forx (a, c], ϕ –1 (x)forx (c, b), where c is an arbitrary point belonging to the interval (a, b), and ϕ(x) is an arbitrary continuous decreasing function on (a, c] such that lim x→a+0 ϕ(x)=b, ϕ(c)=c. 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: 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 periodic functions with period 1. 5 ◦ . The original functional equation has a single increasing solution: y(x)=x. 6 ◦ . Particular solutions of the equation may be represented in implicit form using the algebraic (or transcendental) equation Φ(x, y)=0, where Φ(x, y)=Φ(y, x) is some symmetric function with two arguments. 8. y y(x) + ay(x) + bx =0. 1 ◦ . General solution in parametric form: 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 , where λ 1 and λ 2 are roots of the quadratic equation λ 2 + aλ + b = 0 and Θ 1 = Θ 1 (t)andΘ 2 = Θ 2 (t) are arbitrary periodic functions with period 1. 2 ◦ .ForΘ 1 (t)=C 1 = const and Θ 2 (t)=C 2 = const, we have a particular solution in implicit form λ 2 x – y(x) λ 2 – λ 1 = C 1 λ 1 x – y(x) C 2 (λ 1 – λ 2 ) k , k = ln λ 1 ln λ 2 . 9. y y(y(x)) – x =0. A special case of equation T12.1.2.21. 1 ◦ . Particular solutions: y 1 (x)=– C 2 C + x , y 2 (x)=C – C 2 x , y 3 (x)=C 1 – (C 1 + C 2 ) 2 C 2 + x , where C, C 1 , C 2 are arbitrary constants. 2 ◦ . Solution in parametric form: x = Θ t 3 , y = Θ t + 1 3 , where Θ(t)=Θ(t + 1) is an arbitrary periodic function with period 1. T12.1. LINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1425 T12.1.2-3. Equations involving unknown function with three different arguments. 10. Ay(ax) + By(bx) + y(x) =0. This functional equation has particular solutions of the form y(x)=Cx β ,whereC is an arbitrary constant and β is a root of the transcendental equation Aa β + Bb β + 1 = 0. 11. Ay(x a ) + By(x b ) + y(x) =0. This functional equation has 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|a| p + B|b| p + 1 = 0. 12. Ay(x) + By ax – β x + b + Cy bx + β a – x = f(x), β = a 2 + ab + b 2 . Let us substitute x in the equation first by ax – β x + b and then by bx + β a – x to obtain two more equations. So we get the system (the original equation is given first) Ay(x)+By(u)+Cy(w)=f (x), Ay(u)+By(w)+Cy(x)=f (u), Ay(w)+By(x)+Cy(u)=f(w), (1) where u = ax – β x + b and w = bx + β a – x . Eliminating y(u)andy(w) from the system of linear algebraic equations (1) yields the solution of the original functional equation. 13. f 1 (x)y(x) + f 2 (x)y ax – β x + b + f 3 (x)y bx + β a – x = g(x), β = a 2 + ab + b 2 . Let us substitute x in the equation first by ax – β x + b and then by bx + β a – x to obtain two more equations. So we get the system (the original equation is given first) f 1 (x)y(x)+f 2 (x)y(u)+f 3 (x)y(w)=g(x), f 1 (u)y(u)+f 2 (u)y(w)+f 3 (u)y(x)=g(u), f 1 (w)y(w)+f 2 (w)y(x)+f 3 (w)y(u)=g(w), (1) where u = ax – β x + b , w = bx + β a – x . Eliminating y(u)andy(w) from the system of linear algebraic equations (1) yields the solution y = y(x) of the original functional equation. T12.1.2-4. Higher-order linear difference equations. 14. y n+m + a m–1 y n+m–1 + ···+ a 1 y n+1 + a 0 y n =0. A homogeneous mth-order linear difference equation defined on a discrete set of points x = 0, 1, 2, Notation adopted: y n = y(n). 1426 FUNCTIONAL EQUATIONS Let λ 1 , λ 2 , , λ m be roots of the characteristic equation P (λ) ≡ λ m + a m–1 λ m–1 + ···+ a 1 λ + a 0 = 0.(1) If all the roots of the characteristic equation (1) are distinct, then the general solution of the original difference equation has the form y n = m–1 i=0 y i m–i–1 j=0 a i+j+1 m k=1 λ n+1 k P (λ k ) ,(2) where the prime denotes a derivative. Formula (2) involves the initial values y 0 , y 1 , , y m . They can be set arbitrarily. In the case of complex conjugate roots, one should separate the real and imaginary parts in solution (2). 15. y n+m + a m–1 y n+m–1 + ···+ a 1 y n+1 + a 0 y n = f n . A nonhomogeneous mth-order linear difference equation defined on a discrete set of points x = 0, 1, 2, Notation adopted: y n = y(n). The general solution of the difference equation has the form y(x)=Y (x)+¯y(x), where Y (x) is the general solution of the corresponding homogeneous equation (with f n ≡ 0)and ¯y(x) is any particular solution of the nonhomogeneous equation. Let λ 1 , λ 2 , , λ m be roots of the characteristic equation P (λ) ≡ λ m + a m–1 λ m–1 + ···+ a 1 λ + a 0 = 0.(1) If all the roots of the characteristic equation (1) are distinct, then the general solution of the original difference equation has the form y n = m–1 i=0 y i m–i–1 j=0 a i+j+1 m k=1 λ n+1 k P (λ k ) + n ν=m f n–ν m k=1 λ ν–1 k P (λ k ) ,(2) where the prime denotes a derivative. Formula (2) involves the initial values y 0 , y 1 , , y m . They can be set arbitrarily. In the case of complex conjugate roots, one should separate the real and imaginary parts in solution (2). 16. y(x + n) + a n–1 y(x + n –1) + ···+ a 1 y(x +1) + a 0 y(x) =0. A homogeneous nth-order constant-coefficient linear difference equation. Let us write out the characteristic equation: λ n + a n–1 λ n–1 + ···+ a 1 λ + a 0 = 0.(1) Consider the following cases. 1 ◦ . All roots λ 1 , λ 2 , , λ n of equation (1) are real and distinct. Then the general solution of the original finite-difference equation has the form y(x)=Θ 1 (x)λ x 1 + Θ 2 (x)λ x 2 + ···+ Θ n (x)λ x n ,(2) where Θ 1 (x), Θ 2 (x), , Θ n (x) are arbitrary periodic functions with period 1, which means that Θ k (x)=Θ k (x + 1), k = 1, 2, , n. For Θ k (x) ≡ C k , formula (2) gives a particular solution y(x)=C 1 λ x 1 + C 2 λ x 2 + ···+ C n λ x n , where C 1 , C 2 , , C n are arbitrary constants. T12.1. LINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1427 2 ◦ .Therearem equal real roots, λ 1 = λ 2 = ··· = λ m (m ≤ n), the other roots real and distinct. In this case, the general solution of the functional equation is expressed as y = Θ 1 (x)+xΘ 2 (x)+···+x m–1 Θ m (x) λ x 1 +Θ m+1 (x)λ x m+1 +Θ m+2 (x)λ x m+2 +···+Θ n (x)λ x n . 3 ◦ .Therearem equal complex conjugate roots, λ = ρ(cos β i sin β)(2m ≤ n), the other roots real and distinct. Then the original functional equation has a solution corresponding to Θ n (x) ≡ const k : y = ρ x cos(βx)(A 1 + A 2 x + ···+ A m x m–1 ) + ρ x sin(βx)(B 1 + B 2 x + ···+ B m x m–1 ) + C m+1 λ x m+1 + C m+2 λ x m+2 + ···+ C n λ x n , where A 1 , , A m , B 1 , , B m , C 2m+1 , , C n are arbitrary constants. 17. y(x + n) + a n–1 y(x + n –1) + ···+ a 1 y(x +1) + a 0 y(x) = f(x). A nonhomogeneous nth-order constant-coefficient linear difference equation. 1 ◦ . Solution: y(x)=Y (x)+¯y(x), where Y (x) is the general solution of the corresponding homogeneous equation Y (x + n)+a n–1 Y (x + n – 1)+···+ a 1 Y (x + 1)+a 0 Y (x)=0 (see the previous equation), and ¯y(x) is any particular solution of the nonhomogeneous equation. 2 ◦ .Forf(x)= n k=0 A k x n , the nonhomogeneous equation has a particular solution of the form ¯y(x)= n k=0 B k x n ; the constants B k are found by the method of undetermined coefficients. 3 ◦ .Forf(x)= n k=1 A k exp(λ k x), the nonhomogeneous equation has a particular solution of the form ¯y(x)= n k=1 B k exp(λ k x); the constants B k are found by the method of undetermined coefficients. 4 ◦ .Forf(x)= n k=1 A k cos(λ k x), the nonhomogeneous equation has a particular solution of the form ¯y(x)= n k=1 B k cos(λ k x)+ n k=1 D k sin(λ k x); the constants B k and D k are found by the method of undetermined coefficients. 5 ◦ .Forf(x)= n k=1 A k sin(λ k x), the nonhomogeneous equation has a particular solution of the form ¯y(x)= n k=1 B k cos(λ k x)+ n k=1 D k sin(λ k x); the constants B k and D k are found by the method of undetermined coefficients. 18. y(x + b n ) + a n–1 y(x + b n–1 ) + ···+ a 1 y(x + b 1 ) + a 0 y(x) =0. There are particular solutions of the form y(x)=λ x k ,whereλ k are roots of the transcendental (or algebraic) equation λ b n + a n–1 λ b n–1 + ···+ a 1 λ b 1 + a 0 = 0. 1428 FUNCTIONAL EQUATIONS T12.1.2-5. Equations involving unknown function with many different arguments. 19. y(a n x) + b n–1 y(a n–1 x) + ···+ b 1 y(a 1 x) + b 0 y(x) =0. This functional equation has particular solutions of the form y = Cx β ,whereC is an arbitrary constant and β is a root of the transcendental equation a β n + b n–1 a β n–1 + ···+ b 1 a β 1 + b 0 = 0. 20. y x a n + b n–1 y x a n–1 + ···+ b 1 y x a 1 + b 0 y(x) =0. This functional equation has 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 n | p + b n–1 |a n–1 | p + ···+ b 1 |a 1 | p + b 0 = 0. 21. y [n] (x) + a n–1 y [n–1] (x) + ···+ a 1 y(x) + a 0 x =0. Notation used: y [2] (x)=y y(x) , , y [n] (x)=y y [n–1] (x) . 1 ◦ . Solutions are sought in the parametric form x = w(t), y = w(t + 1). With it, the original equation is reduced to an nth-order linear finite-difference equation (see equation 16 above): w(t + n)+a n–1 w(t + n – 1)+···+ a 1 w(t + 1)+a 0 w(t)=0. 2 ◦ . In the special case a n–1 = = a 1 = 0 and a 0 =–1, we have the following solution in parametric form: x = Θ t n , y = Θ t + 1 n , where Θ(t)=Θ(t + 1) is an arbitrary periodic function with period 1. T12.2. Nonlinear Functional Equations in One Independent Variable T12.2.1. Functional Equations with Quadratic Nonlinearity T12.2.1-1. Difference equations. 1. y n y n+1 = a n y n+1 + b n y n + c n . Riccati difference equation. Here, n = 0, 1, and the constants a n , b n , c n satisfy the condition a n b n + c n ≠ 0. 1 ◦ . The substitution y n = u n+1 u n + a n leads to the linear second-order difference equation of the form T12.1.2.1: u n+2 +(a n+1 – b n )u n+1 –(a n b n + c n )u n = 0. T12.2. NONLINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1429 2 ◦ .Lety ∗ n be a particular solution of the Riccati difference equation. Then the substitution z n = 1 y n – y ∗ n , n = 0, 1, reduces this equation to the nonhomogeneous first-order linear difference equation z n+1 + (y ∗ n – a n ) 2 a n b n + c n z n + y ∗ n – a n a n b n + c n = 0. About this equation, see Paragraph 17.1.1-2. 2. y(x +1) – ay 2 (x) = f(x). A special case of equation T12.2.3.1. 3. y(x)y(x +1) + a[y(x +1) – y(x)] =0. Solution: y(x)= a x + Θ(x) , where Θ(x)=Θ(x + 1) is an arbitrary periodic function with period 1. 4. y(x)y(x +1) = a(x)y(x +1) + b(x)y(x) + c(x). Riccati difference equation. Here, 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) leads 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 Riccati difference equation. Then the substitution z(x)= 1 y(x)–y 0 (x) reduces this equation to the nonhomogeneous first-order linear 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. T12.2.1-2. Functional equations involving y(x)andy(a – x). 5. y(x)y(a – x) = b 2 . Solutions: 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(x – z), we arrive at particular solutions of the form y(x)= be C(2x–a) , where C is an arbitrary constant. 1430 FUNCTIONAL EQUATIONS 6. y(x)y(a – x) =–b 2 . Discontinuous solutions: y(x)= ⎧ ⎨ ⎩ bϕ(x)ifx ≥ a/2, – b ϕ(a – x) if x < a/2, where ϕ(x) is an arbitrary function. There are no continuous solutions. 7. y(x)y(a – x) = f 2 (x). The function f(x) must satisfy the condition f(x)= f(a – x). 1 ◦ . The change of variable y(x)=f(x)u(x) leads to one of the equations of the form T12.2.1.5 or T12.2.1.6. 2 ◦ .Forf(x)=f(a – x), there are solutions of the form 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(x – z), we arrive at particular solutions y(x)= e C(2x–a) f(x), where C is an arbitrary constant. 8. y 2 (x) + y 2 (a – x) = b 2 . 1 ◦ . Solutions: y(x)= 1 2 b 2 + Φ(x, a – x), where Φ(x, z)=–Φ(z, x) is any antisymmetric function of two arguments. 2 ◦ . Particular solutions: y 1,2 (x)= b √ 2 , y 3,4 (x)= b sin πx 2a , y 5,6 (x)= b cos πx 2a . 9. y 2 (x) + Ay(x)y(a – x) + By 2 (a – x) + Cy(x) + Dy(a – x) = f (x). A special case of equation T12.2.3.2. 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. . f(x)= f(a – x). 1 ◦ . The change of variable y(x)=f(x)u(x) leads to one of the equations of the form T12.2.1.5 or T12.2.1.6. 2 ◦ .Forf(x)=f(a – x), there are solutions of the form y(x)= f(x)exp Φ(x,. + By(bx) + y(x) =0. This functional equation has particular solutions of the form y(x)=Cx β ,whereC is an arbitrary constant and β is a root of the transcendental equation Aa β + Bb β + 1 = 0. 11 By(x b ) + y(x) =0. This functional equation has 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|a| p + B|b| p +