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. . 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. 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. 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