Handbook of mathematics for engineers and scienteists part 136 docx

17.3. LINEAR FUNCTIONAL EQUATIONS 913 homogeneous equation with F(x) ≡ 0, the following three cases are possible (the notation used here is in agreement with that of Paragraph 17.3.2-1): (i) Equation (17.3.3.1) on I either has a one-parameter family of continuous solutions or no solutions at all. If (17.3.3.1) has a continuous solution y 0 (x)onI, then the general continuous solution on I is given by y(x)=y 0 (x)+ a G(x) , where a is an arbitrary constant and G(x) is given by (17.3.2.3). (ii) Equation (17.3.3.1) on I has a continuous solution depending on an arbitrary func- tion, or has no continuous solutions on I. (iii) Equation (17.3.3.1) on I either has a continuous solution or no solutions at all. 2 ◦ .Letx I, ξ I,andf (x) R 0 ξ [I]. Suppose that g(x)andF (x) are continuous functions, g(x) ≠ 0 for x ≠ ξ,andg(ξ)>1. Then equation (17.3.3.1) has a unique continuous solution that can be represented by the series y(x)=– ∞  n=0 F (f [n] (x)) G n+1 (x) ,(17.3.3.2) where the function G n (x)isdefined by (17.3.2.2). 17.3.3-2. Equations of special form with g(x)=const. 1 ◦ . Consider the equation y(f(x)) + y(x)=F (x), (17.3.3.3) which is a special case of equation (17.3.3.1) with g(x)=–1. (A) Let ξ I, f (x) R 0 ξ [I], and suppose that F (x) is a continuous function. If there exists a continuous solution of equation (17.3.3.3), it can be represented by the power series y(x)= 1 2 F (ξ)+ ∞  n=0 (–1) n  F (f [n] (x)) – F (ξ)  . (B) Suppose that the assumptions of Item (A) hold and, moreover, there exist positive constants δ, κ,andC such that |F (x)–F (ξ)| ≤ C|x – ξ)| κ , x (ξ – δ, ξ + δ) ∩I, and ξ is a strongly attractive fixed point of f(x). Then equation (17.3.3.3) has a continuous solution on I. 2 ◦ . Consider the functional equation y(f(x)) – λy(x)=F (x), λ ≠ 0.(17.3.3.4) Let x I =(a, b]. Assume that on the interval I the function f(x) is continuous, strongly increasing, and satisfies the condition a < f(x)<x,andF (x) is a function of bounded variation. Depending on the value of the parameter λ, the following cases are possible. 914 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS Case |λ| > 1. There is a unique solution y(x)=– ∞  n=0 F  f [n] (x)  λ n+1 , which coincides with (17.3.3.2) for g(x)=λ and G n+1 (x)=λ n . Case 0 < |λ| < 1.Foranyx 0 I and any test function ϕ(z) that has bounded variation on [f (x 0 ), x 0 ] and satisfies the condition ϕ(f(x 0 )) – λϕ(x 0 )=F (x 0 ), there is a single function of bounded variation for which equation (17.3.3.4) holds. This solution has the same variation on the interval [f (x 0 ), x 0 ] as the test function ϕ(x). Case λ = 1. If lim x→a+0 F (x)=0 and there is a point x 0 I such that the series ∞  n=0 F  f [n] (x 0 )  is convergent, then equation (17.3.3.4) has a one-parameter family of solutions y(x)=C – ∞  n=0 F  f [n] (x)  (here, C is an arbitrary constant), for which there exists a finite limit lim x→a+0 y(x). Case λ =–1. See Item 1 ◦ . 17.3.3-3. Abel functional equation. The Abel functional equation has the form y(f(x)) = y(x)+c, c ≠ 0.(17.3.3.5) Remark. Without the loss of generality, we can take c = 1 (this can be achieved by passing to the normalized unknown function ¯y = y/c). 1 ◦ .Let¯y(x) be a solution of equation (17.3.3.5). Then the function y(x)=¯y(x)+Θ  ¯y(x) c  , where Θ(z) is an arbitrary 1-periodic function, is also a solution of equation (17.3.3.5). 2 ◦ . Suppose that the following conditions hold: (i) f (x) is strictly increasing and continuous for 0 ≤ x ≤ a; (ii) f (0)=0 and f(x)<x for 0 < x < a; (iii) the derivative f  (x) exists, has bounded variation on the interval 0 < x < a,and lim x→+0 f  (x)=1. Then, for any x, x 0 (0, a], there exists the limit y(x) = lim n→∞ f [n] (x)–f [n] (x 0 ) f [n–1] (x 0 )–f [n] (x 0 ) ,(17.3.3.6) which defines a monotonically increasing function satisfying the Abel equation (17.3.3.5) with c =–1 (thisistheL ´ evy solution). 17.3. LINEAR FUNCTIONAL EQUATIONS 915 3 ◦ .Letf (x)bedefined on a submodulus interval I and suppose that there is a sequence d n for which lim n→∞ f [n+1] (x)–f [n] (x) d n = c, x I. If for all x I there exists the limit y(x) = lim n→∞ f [n] (x)–f [n] (x 0 ) d n ,(17.3.3.7) where x 0 is an arbitrary fixed point from the interval I, then the function y(x) satisfies equation (17.3.3.5). 4 ◦ . Let us describe a procedure for the construction of monotone solutions of the Abel equation (17.3.3.5). Assume that the function f(x) is continuous, strictly increasing on the segment [0, a], and the following conditions hold: f(0)=0 and f (x)<x ≤ a for x ≠ 0. We define a function ϕ(y)onthesemiaxis[0, ∞)by ϕ(y)=f [n]  ϕ 0 ({y})  , n =[y], (17.3.3.8) where [y]and{y} are, respectively, the integer and the fractional parts of y,andϕ 0 (y)is any continuous strictly decreasing function on [0, 1) such that ϕ 0 (0)=a, ϕ 0 (1 – 0)=f(a). On the intervals [n, n + 1)(n = 0, 1, ), the function (17.3.3.8) takes the values ϕ(y)=f [n]  ϕ 0 (y – n)  , n ≤ y < n + 1,(17.3.3.9) and therefore is continuous and strictly decreasing, being a superposition of continuous strictly increasing functions and a continuous strictly decreasing function. At integer points, too, the function ϕ(y) is continuous, since (17.3.3.8) and (17.3.3.9) imply that ϕ(n – 0)=f [n–1]  ϕ 0 (1 – 0)  = f [n–1]  f(a)  = f [n] (a), ϕ(n)=f [n]  ϕ 0 (0)  = f [n] (a). Therefore, the function ϕ(y) is also continuous and strictly decreasing on the entire semiaxis [0, ∞). By continuity, we can define the value ϕ(∞)=0. It follows that there exists the inverse function ϕ –1 defined by the relations x = ϕ(y), y = ϕ –1 (x)(0 ≤ x ≤ a, 0 ≤ y ≤ ∞), (17.3.3.10) and ϕ –1 (0)=∞, ϕ –1 (a)=0. From (17.3.3.8), it follows that the function ϕ(y) satisfies the functional equation ϕ(y + 1)=f  ϕ(y)  , 0 ≤ y < ∞.(17.3.3.11) Applying ϕ –1 to both sides of (17.3.3.11) and then passing from y to a new variable x with the help of (17.3.3.10), we come to the Abel equation (17.3.3.5) with c = 1,where y(x)=ϕ –1 (x). Thus, finding a solution of the Abel equation is reduced to the inversion of the known function (17.3.3.8). 5 ◦ . The Abel equation (17.3.3.5) can be reduced to the Schr ¨ odinger–Koenig equation u(f(x)) = su(x), with the help of the replacement u(x)=s y(x)/c ; see equation (17.3.2.5). 916 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS 17.3.4. Linear Functional Equations Reducible to Linear Difference Equations with Constant Coefficients 17.3.4-1. Functional equations with arguments proportional to x. Consider the linear equation a m y(b m x)+a m–1 y(b m–1 x)+···+ a 1 y(b 1 x)+a 0 y(b 0 x)=f(x). (17.3.4.1) The transformation y(x)=w(z), z =lnx reduces (17.3.4.1) to a linear difference equation with constant coefficients a m w(z + h m )+a m–1 w(z + h m–1 )+···+ a 1 w(z + h 1 )+a 0 w(z + h 0 )=f(e z ), h k =lnb k . Note that the homogeneous functional equation (17.3.4.1) with f(x)≡ 0 admits particular solutions of power type, y(x)=Cx β ,whereC is an arbitrary constant and β is a root of the transcendental equation a m b β m + a m–1 b β m–1 + ···+ a 1 b β 1 + a 0 b β 0 = 0. 17.3.4-2. Functional equations with powers of x as arguments. Consider the linear equation a m y  x n m  + a m–1 y  x n m–1  + ···+ a 1 y  x n 1  + a 0 y(x n 0 )=f (x). (17.3.4.2) The transformation y(x)=w(z), z =lnlnx reduces (17.3.4.2) to a linear difference equation with constant coefficients a m w(z +h m )+a m–1 w(z +h m–1 )+···+a 1 w(z +h 1 )+a 0 w(z +h 0 )=f(e e z ), h k =lnlnn k . Note that the homogeneous functional equation (17.3.4.2) admits 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 m |n m | p + a m–1 |n m–1 | p + ···+ a 1 |n 1 | p + a 0 |n 0 | p = 0. 17.3.4-3. Functional equations with exponential functions of x as arguments. Consider the linear equation a m y  e λ m x  + a m–1 y  e λ m–1 x  + ···+ a 1 y  e λ 1 x  + a 0 y(e λ 0 x )=f (x). The transformation y(x)=w(ln z), z =lnx reduces this equation to a linear difference equation with constant coefficients a m w(z + h m )+a m–1 w(z + h m–1 )+···+ a 1 w(z + h 1 )+a 0 w(z + h 0 )=f (e z ), h k =lnλ k . 17.3. LINEAR FUNCTIONAL EQUATIONS 917 17.3.4-4. Equations containing iterations of the unknown function. Consider the linear homogeneous equation a m y [m] (x)+a m–1 y [m–1] (x)+···+ a 1 y(x)+a 0 x = 0 with successive iterations of the unknown function, y [2] (x)=y  y(x)  , , y [n] (x)= y  y [n–1] (x)  . A solution of this equation is sought in the parametric form x = w(t), y = w(t + 1). (17.3.4.3) Then the original equation is reduced to the following mth-order linear equation with constant coefficients (equation of the type (17.2.3.1) with integer differences): a m w(t + m)+a m–1 w(t + m – 1)+···+ a 1 w(t + 1)+a 0 w(t)=0. Example. Consider the functional equation y  y(x)  + ay(x)+bx = 0.(17.3.4.4) The transformation (17.3.4.3) reduces this equation to a difference equation of the form (17.2.2.1). Let λ 1 and λ 2 be distinct real roots of the characteristic equation λ 2 + aλ + b = 0,(17.3.4.5) λ 1 ≠ λ 2 . Then the general solution of the functional equation (17.3.4.4) in parametric form can be written as 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 , (17.3.4.6) where Θ 1 (t)andΘ 2 (t) are arbitrary 1-periodic functions, Θ k (t)=Θ k (t + 1). Taking Θ 1 (t)=C 1 and Θ 2 (t)=C 2 in (17.3.4.6), where C 1 and C 2 are arbitrary constants, and eliminating the parameter t, we obtain a particular solution of equation (17.3.4.4) in implicit form: λ 2 x – y λ 2 – λ 1 = C 1  λ 1 x – y C 2 (λ 1 – λ 2 )  γ , γ = ln λ 1 ln λ 2 . 17.3.4-5. Babbage equation and involutory functions. 1 ◦ . Functions satisfying the Babbage equation y(y(x)) = x (17.3.4.7) are called involutory functions. Equation (17.3.4.7) is a special case of (17.3.4.4) with a = 0, b =–1. The corresponding characteristic equation (17.3.3.5) has the roots λ 1,2 = 1. The parametric representation of solution (17.3.4.6) contains the complex quantity (–1) t = e iπt =cos(πt)+i sin(πt)andis, therefore, not very convenient. 2 ◦ . The following statements hold: (i) On the interval x (a, b), there is a continuous decreasing solution of equation (17.3.4.7) depending on an arbitrary function. This solution has the form y(x)=  ϕ(x)forx (a, c], ϕ –1 (x)forx (c, b), where c (a, b) is an arbitrary point, and ϕ(x) is an arbitrary continuous decreasing function on (a, c] such that lim x→a+0 ϕ(x)=b, ϕ(c)=c. (ii) The only increasing solution of equation (17.3.4.7) has the form y(x) ≡ x. 918 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS 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 (an alternative representation): 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 1-periodic functions, Θ k (t)=Θ k (t + 1), k = 1, 2.In this solution, the functions sin(πt) can be replaced by cos(πt). 17.4. Nonlinear Difference and Functional Equations with a Single Variable 17.4.1. Nonlinear Difference Equations with a Single Variable 17.4.1-1. Riccati difference equation. The Riccati difference equation has the general form y(x)y(x + 1)=a(x)y(x + 1)+b(x)y(x)+c(x), (17.4.1.1) where 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) reduces the Riccati equation 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 equation (17.4.1.1). Then the substitution z(x)= 1 y(x)–y 0 (x) reduces equation (17.4.1.1) to the first-order linear nonhomogeneous 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. 17.4.1-2. First-order nonlinear difference equations with no explicit dependence on x. Consider the nonlinear functional equation y(x + 1)=f  y(x)  , 0 ≤ x < ∞ ,(17.4.1.2) where f(y) is a continuous function. Let y(x)beafunctiondefinedonthesemiaxis[0, ∞)by y(x)=f [n]  ϕ({x})  , n =[x], (17.4.1.3) where [x]and{x} are, respectively, the integer and the fractional parts of x,andϕ(x)isany continuous function on [0, 1) such that ϕ(0)=a, ϕ(1 – 0)=f (a), a is an arbitrary constant. Direct verification shows that the function (17.4.1.3) satisfies equation (17.4.1.2) and is continuous on the semiaxis [0, ∞). 17.4. NONLINEAR DIFFERENCE AND FUNCTIONAL EQUATIONS WITH A SINGLE VARIABLE 919 17.4.1-3. Nonlinear mth-order difference equations. 1 ◦ . In the general case, a nonlinear difference equation has the form F  x, y(x + h 0 ), y(x + h 1 ), , y(x + h m )  = 0.(17.4.1.4) This equation contains the unknown function with different values of its argument differing from one another by constants. The constants h 0 , h 1 , , h m are called differences or deviations of the argument. If all h k are integers, then (17.4.1.4) is called an equation with integer differences. If h k = k (k = 0, 1, , m) and the equation explicitly involves y(x) and y(x + m), then it is called a difference equation of order m. 2 ◦ .Anmth-order difference equation resolved with respect to the leading term y(x + m) has the form y(x + m)=f  x, y(x), y(x + 1), , y(x + m – 1)  .(17.4.1.5) The Cauchy problem for this equation consists of finding its solution with a given initial distribution of the unknown function on the interval 0 ≤ x ≤ m: y = ϕ(x)at0 ≤ x ≤ m.(17.4.1.6) The values of y(x)form ≤ x ≤ m + 1 are calculated by substituting the initial values (17.4.1.6) into the right-hand side of equation (17.4.1.5) for 0 ≤ x ≤ 1.Wehave y(x + m)=f  x, ϕ(x), ϕ(x + 1), , ϕ(x + m – 1)  .(17.4.1.7) Then, replacing x by x + 1 in equation (17.4.1.5), we obtain y(x + m + 1)=f  x + 1, y(x + 1), y(x + 2), , y(x + m)  .(17.4.1.8) The values of y(x)form + 1 ≤ x ≤ m +2 are determined by the right-hand side of (17.4.1.8) for 0 ≤ x ≤ 1, the quantities y(x+1), , y(x +m–1) are determined by the initial condition (17.4.1.6), and y(x + m) is taken from (17.4.1.7). In order to find y(x)form + 2 ≤ x ≤ m + 3, we replace x by x + 1 in equation (17.4.1.8) and consider this equation on the interval 0 ≤ x ≤ 1. Using the initial conditions for y(x + 2), , y(x + m – 1) and the quantities y(x + m), y(x + m + 1) found on the previous steps, we find y(x + m + 2). In a similar way, one can find y( x)form + 3 ≤ x ≤ m + 4, etc. This method for solving difference equations is called the step method. 17.4.2. Reciprocal (Cyclic) Functional Equations 17.4.2-1. Reciprocal functional equations depending on y(x)andy(a – x). 1 ◦ . Consider a reciprocal equation of the form F  x, y(x), y(a – x)  = 0. Replacing x by a – x, we obtain a similar equation containing the unknown function with the same arguments: F  a – x, y(a – x), y(x)  = 0. Eliminating y(a – x) from this equation and the original equation, we come to the usual algebraic (or transcendental) equation of the form Ψ  x, y(x)  = 0. In other words, solutions of the original functional equation y = y(x)aredefined in a parametric manner by means of two algebraic (or transcendental) equations: F (x, y,t)=0, F (a – x, t, y)=0,(17. 4.2.1) where t is a parameter. . Θ 1 (t)=C 1 and Θ 2 (t)=C 2 in (17.3.4.6), where C 1 and C 2 are arbitrary constants, and eliminating the parameter t, we obtain a particular solution of equation (17.3.4.4) in implicit form: λ 2 x. values of y(x)form + 1 ≤ x ≤ m +2 are determined by the right-hand side of (17.4.1.8) for 0 ≤ x ≤ 1, the quantities y(x+1), , y(x +m–1) are determined by the initial condition (17.4.1.6), and y(x. – n)  , n ≤ y < n + 1,(17.3.3.9) and therefore is continuous and strictly decreasing, being a superposition of continuous strictly increasing functions and a continuous strictly decreasing