920 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS Remark. System (17.4.2.1), and therefore the original functional equation, may have several (even in- finitely many) solutions or no solutions at all. Example. Consider the nonlinear equation y 2 (x)=f(x)y(a – x). (17.4.2.2) Replacing x by a – x in (17.4.2.2), we get y 2 (a – x)=f(a – x)y(x). (17.4.2.3) Eliminating y(a – x) from (17.4.2.2)–(17.4.2.3), we obtain two solutions of the original equation: y(x)=[f 2 (x)f(a – x)] 1/3 and y(x) ≡ 0. 2 ◦ . 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 with the unknown function having the same arguments: F a/x, y(a/x), y(x) = 0. Eliminating y(a/x) from this 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 a system of two algebraic (or transcendental) equations: F (x, y, t)=0, F (a/x, t, y)=0, where t is a parameter. 17.4.2-2. Reciprocal (cyclic) functional equations of general form. Reciprocal functional equations have the form F x, y(x), y(ϕ(x)), y(ϕ [2] (x)), , y(ϕ [n–1] (x)) = 0.(17.4.2.4) Here we use the notation ϕ [n] (x)=ϕ(ϕ [n–1] (x)) and ϕ(x) is a cyclic function satisfying the condition ϕ [n] (x)=x.(17.4.2.5) The value n is called the order of a cyclic (reciprocal) equation. Successively replacing n times the argument x by ϕ(x) in the functional equation (17.4.2.4), we obtain the following system (the original equation coincides with the first equation of this system): F x, y 0 , y 1 , , y n–1 = 0, F ϕ(x), y 1 , y 2 , , y 0 = 0, , F ϕ [n–1] (x), y n–1 , y 0 , , y n–2 = 0, (17.4.2.6) where we have set y 0 = y(x), y 1 = y(ϕ(x)), , y n–1 = y(ϕ [n–1] (x)); condition (17.4.2.5) implies that y n = y 0 . Eliminating y 1 , y 2 , , y n–1 from the system of nonlinear algebraic (or transcendental) equations (17.4.2.6), we obtain a solution of the functional equation (17.4.2.4) in implicit form Ψ(x, y 0 )=0,wherey 0 = y(x). 17.4. NONLINEAR DIFFERENCE AND FUNCTIONAL EQUATIONS WITH A SINGLE VARIABLE 921 17.4.3. Nonlinear Functional Equations Reducible to Difference Equations 17.4.3-1. Functional equations with arguments proportional to x. Consider the equation F x, y(a 1 x), y(a 2 x), , y(a n x) = 0.(17.4.3.1) The transformation y(x)=w(z), z =lnx reduces (17.4.3.1) to the difference equation F e z , w(z + h 1 ), w(z + h 2 ), , w(z + h n ) = 0, h k =lna k . 17.4.3-2. Functional equations with powers of x as arguments. Consider the equation F x, y(x n 1 ), y(x n 2 ), , y(x n m ) = 0.(17.4.3.2) The transformation y(x)=w(z), z =lnlnx reduces (17.4.3.2) to the difference equation F e e z , w(z + h 1 ), w(z + h 2 ), , w(z + h n ) = 0, h k =lnlnn k . 17.4.3-3. Functional equations with exponential functions of x as arguments. Consider the equation F x, y(e λ 1 x ), y(e λ 2 x ), , y(e λ n x ) = 0. The transformation y(x)=w(ln z), z =lnx reduces this equation to the difference equation F e z , w(z + h 1 ), w(z + h 2 ), , w(z + h n ) = 0, h k =lnλ k . 17.4.3-4. Equations containing iterations of the unknown function. Consider the equation F x, y(x), y [2] (x), , y [n] (x) = 0, where y [2] (x)=y y(x) , , y [n] (x)=y y [n–1] (x) . We seek its solution in parametric form x = w(t), y = w(t + 1). Then the original equation is reduced to the following nth-order difference equation: F w(t), w(t + 1), w(t + 2), , w(t + n) = 0. 922 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS 17.4.4. Power Series Solution of Nonlinear Functional Equations In some situations, a solution of a functional equation can be found in the form of power series expansion. Consider the nonlinear functional equation y(x)–F x, y(ϕ(x) = 0.(17.4.4.1) Suppose that the following conditions hold: 1) there exist a and b such that ϕ(a)=a, F (a, b)=b; 2) the function ϕ(x) is analytic in a neighborhood of a and |ϕ (a)| < 1; 3) the function F is analytic in a neighborhood of the point (a, b)and| ∂F ∂b (a, b)| < 1. Then the formal power series solution of equation (17.4.4.1), y(x)=b + ∞ k=1 c k (x – a) k ,(17.4.4.2) has a positive radius of convergence. The formal solution is obtained by substituting the expansions ϕ(x)=a + ∞ k=1 a k (x – a) k , F (x, y)=b + ∞ i,j=1 b ij (x – a) i (y – b) j , and (17.4.4.2) into equation (17.4.4.1). Gathering the terms with the same powers of the difference ξ =(x – a) and then equating to zero the coefficients of different powers of ξ,we obtain a triangular system of algebraic equations for the coefficients c k . See also Section T12.2, which gives exact solutions of some nonlinear difference and functional equations with one independent variable. 17.5. Functional Equations with Several Variables 17.5.1. Method of Differentiation in a Parameter 17.5.1-1. Classes of equations. Description of the method. Consider linear functional equations of the form w(x, t)=θ(x, t, a) w ϕ(x, t, a), ψ(x, t, a) ,(17.5.1.1) where x and t are independent variables, w = w(x, t) is the function to be found, θ = θ(x, t, a), ϕ = ϕ(x, t, a), ψ = ψ(x, t, a) are given functions, and a is a free parameter that can take any value (on some interval). Assume that for a particular a = a 0 ,wehave θ(x, t, a 0 )=1, ϕ(x, t, a 0 )=x, ψ(x, t, a 0 )=t,(17.5.1.2) i.e., for a = a 0 the functional equation (17.5.1.1) turns into identity. 17.5. FUNCTIONAL EQUATIONS WITH SEVERAL VARIABLES 923 Let us expand (17.5.1.1) in powers of the small parameter a in a neighborhood of a 0 , taking into account (17.5.1.2), and then divide the resulting equation by a – a 0 and pass to the limit for a → a 0 . As a result, we obtain a fi rst-order linear partial differential equation for the function w: ϕ ◦ a (x, t) ∂w ∂x + ψ ◦ a (x, t) ∂w ∂t + θ ◦ a (x, t)w = 0,(17.5.1.3) where we have used the following notation: ϕ ◦ a (x, t)= ∂ϕ ∂a a=a 0 , ψ ◦ a (x, t)= ∂ψ ∂a a=a 0 , θ ◦ a (x, t)= ∂θ ∂a a=a 0 . In order to solve equation (17.5.1.3), one should consider the corresponding system of equations for characteristics (see Subsection 13.1.1): dx ϕ ◦ a (x, t) = dt ψ ◦ a (x, t) =– dw θ ◦ a (x, t)w .(17.5.1.4) Let u 1 (x, t)=C 1 , u 2 (x, t, w)=C 2 (17.5.1.5) be independent integrals of the characteristic system (17.5.1.4). Then the general solution of equation (17.5.1.3) has the form u 2 (x, t, w)=F u 1 (x, t) ,(17.5.1.6) where F (z) is an arbitrary function. The function w should be expressed from (17.5.1.6) and substituted into the original equation (17.5.1.1) for verification [there is a possibility of redundant solutions; it is also possible that a solution of the partial differential equation (17.5.1.3) is not a solution of the functional equation (17.5.1.1); see Example 3 below]. Remark 1. Equation (17.5.1.3) can be obtained from (17.5.1.1) by differentiation in the parameter a,after which one should take a = a 0 . Remark 2. It is convenient to choose the second integral in (17.5.1.5) to be a linear function of w, i.e., u 2 (x, t, w)=ξ(x, t)w, and rewrite (17.5.1.6) as a relation resolved with respect to w. 17.5.1-2. Examples of solutions of some specific functional equations. Example 1. Self-similar solutions, which often occur in mathematical physics, may be defined as solutions invariant with respect to the scaling transformation, i.e., solutions satisfying the functional equation w(x, t)=a k w(a m x, a n t), (17.5.1.7) where k, m, n are given constants, and a > 0 is an arbitrary constant. Equation (17.5.1.7) turns into identity for a = 1. Differentiating (17.5.1.7) in a and taking a = 1,wecome to the first-order partial differential equation mx ∂w ∂x + nt ∂w ∂t + kw = 0.(17.5.1.8) The first integrals of the corresponding characteristic system of ordinary differential equations dx mx = dt nt =– dw kw can be written in the form xt –m/n = C 1 , t k/n w = C 2 (n ≠ 0). 924 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS Therefore, the general solution of the partial differential equation (17.5.1.8) has the form w(x, t)=t –k/n F (z), z = xt –m/n ,(17.5.1.9) where F (z) is an arbitrary function. Direct verification shows that expression (17.5.1.9) is a solution of the functional equation (17.5.1.7). Example 2. Consider the functional equation w(x, t)=a k w(a m x, t + β lna), (17.5.1.10) where k, m, β are given constants, a > 0 is an arbitrary constant. Equation (17.5.1.10) turns into identity for a = 1. Differentiating (17.5.1.10) in a and taking a = 1,we come to the first-order partial differential equation mx ∂w ∂x + β ∂w ∂t + kw = 0. (17.5.1.11) The corresponding characteristic system of ordinary differential equations dx mx = dt β =– dw kw admits the first integrals x exp(–mt/β)=C 1 , w exp(kt/β)=C 2 . Therefore, the general solution of the partial differential equation (17.5.1.11) has the form w(x, t)=exp(–kt/β)F (z), z = x exp(–mt/β), (17.5.1.12) where F (z) is an arbitrary function. Direct verification shows that (17.5.1.12) is a solution of the functional equation (17.5.1.10). Example 3. Now consider the functional equation w(x, t)=a k w x +(1 – a)t, a n t , (17.5.1.13) where a > 0 is arbitrary and n is a constant. Equation (17.5.1.13) turns into identity for a = 1. Differentiating (17.5.1.13) in a and taking a = 1,we come to the first-order partial differential equation –t ∂w ∂x + nt ∂w ∂t + kw = 0. (17.5.1.14) The corresponding characteristic system – dx t = dt nt =– dw kw has the first integrals t + nx = C 1 , wt k/n = C 2 . Therefore, the general solution of the partial differential equation (17.5.1.14) has the form w(x, t)=t –k/n F (nx + t), (17.5.1.15) where F (z) is an arbitrary function. Substituting (17.5.1.15) into the original equation (17.5.1.13) and dividing the result by t –k/n , we obtain F (nx + t)=F (nx + σt), σ =(1 – a)n + a n . (17.5.1.16) Hence, for F (z) ≠ const we have σ = 1 or (1 – a)n + a n = 1. (17.5.1.17) Since (17.5.1.16) must hold for all a > 0, it follows that (17.5.1.17), too, must hold for all a > 0. This can take place only if n = 1. In this case, the solution of equation (17.5.1.13) is given by the following formula [see (17.5.1.15) for n = 1]: w(x, t)=t –k F (x + t), (17.5.1.18) where F (z) is an arbitrary function. If n ≠ 1, then equation (17.5.1.13) admits only a degenerate solution w(x, t)=Ct –k/n ,whereC is an arbitrary constant [the degenerate solution corresponds to F = const in (17.5.1.16)]. 17.5. FUNCTIONAL EQUATIONS WITH SEVERAL VARIABLES 925 17.5.2. Method of Differentiation in Independent Variables 17.5.2-1. Preliminary remarks. 1 ◦ . In some situations, differentiation in independent variables can be used to eliminate some arguments of the functional equation under consideration and reduce it to an ordinary differential equation (see Example 1 below). The solution obtained in this way should be then inserted into the original equation in order to get rid of redundant integration constants, which may appear due to the differentiation. 2 ◦ . In some situations, differentiation in independent variables should be combined with the multiplication (division) of the equation and the results of its differentiation by suitable functions. Sometimes it is useful to take logarithm of the equation or the results of its transformation (see Example 2 below). 3 ◦ . In some situations, differentiation of a functional equation in independent variables allows us to eliminate some arguments and reduce the equation to a simpler functional equation whose solution is known (see Subsection 17.5.5). 17.5.2-2. Examples of solutions of some specific functional equations. Example 1. Consider the Pexider equation f(x)+g(y)=h(x + y), (17.5.2.1) where f(x), g(y), h(z) are the functions to be found. Differentiating the functional equation (17.5.2.1) in x and y, we come to the ordinary differential equation h zz (z)=0,wherez = x + y. Its solution is the linear function h(z)=az + b.(17.5.2.2) Substituting this expression into (17.5.2.1), we obtain f(x)+g(y)=ax + ay + b. Separating the variables, we find the functions f and g: f(x)=ax + b + c, g(y)=ay – c. (17.5.2.3) Thus, the solution of the Pexider equation (17.5.2.1) is given by the formulas (17.5.2.2), (17.5.2.3), where a, b, c are arbitrary constants. Example 2. Consider the nonlinear functional equation f(x + y)=f(x)+f(y)+af(x)f ( y), a ≠ 0,(17.5.2.4) which occurs in the theory of probability with a =–1. Differentiating both sides of this equation in x and y,weget f zz (z)=af x (x)f y (y), (17.5.2.5) where z = x + y. Taking the logarithm of both sides of equation (17.5.2.5) and differentiating the resulting relation in x and y, we come to the ordinary differential equation [ln f zz (z)] zz = 0.(17.5.2.6) Integrating (17.5.2.6) in z twice, we get f zz (z)=C 1 exp(C 2 z), (17.5.2.7) where C 1 and C 2 are arbitrary constants. Substituting (17.5.2.7) into (17.5.2.5), we obtain the equation C 1 exp[C 2 (x + y)] = af x (x)f y (y), which admits separation of variables. Integration yields f(x)=A exp(C 2 x)+B, A = 1 C 2 C 1 a .(17.5.2.8) Substituting (17.5.2.8) into the original equation (17.5.2.4), we find the values of the constants: A =–B = 1/a and C 2 = β is an arbitrary constant. As a result, we obtain the desired solution f(x)= 1 a e βx – 1 . 926 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS 17.5.3. Method of Substituting Particular Values of Independent Arguments In order to solve functional equations with several independent variables, one often uses the method of substituting particular values of independent arguments. This procedure is aimed at reducing the problem for a functional equation with several independent variables to a problem for a system of algebraic (transcendental) equations with a single independent variable. Basic ideas of this method can be demonstrated by the example of a fairly general functional equation of the form Φ f(x), f(x + y), f(x – y), x, y = 0.(17.5.3.1) 1 ◦ .Takingy = 0 in this equation, we get Φ f(x), f(x), f(x), x, 0 = 0.(17.5.3.2) If the left-hand side of this relation does not vanish identically for all f(x), then this equation can be resolved with respect to f(x). Then the function f(x) obtained in this way should be inserted into the original equation (17.5.3.1) and one should find conditions under which this function is its solution. Now, assume that the left-hand side of (17.5.3.2) is identically equal to zero for all f(x). 2 ◦ . In equation (17.5.3.1), we consecutively take x = 0, y = t; x = t, y = 2t; x = t, y =–2t. We obtain a system of algebraic (transcendental) equations Φ a, f(t), f(–t), 0, t = 0, Φ f(t), f(3t), f(–t), t, 2t = 0, Φ f(t), f(–t), f(3t), t,–2t = 0 (17.5.3.3) for the unknown quantities f (t), f(–t), f(3t), where a = f(0). Resolving system (17.5.3.3) with respect to f (t) [or with respect to f(–t)orf (3x)], we obtain an admissible solution that should be inserted into the original equation for verification. 3 ◦ . In some cases the following trick can be used. In equation (17.5.3.1), one consecutively takes x = 0, y = t; x = t + a, y = a; x = a, y = t + a,(17.5.3.4) where a is a free parameter. We obtain the system of equations Φ f(0), f(t), f(–t), 0, t = 0, Φ f(t + a), f(t + 2a), f(t), t + a, a = 0, Φ f(a), f(t + 2a), f(–t), a, t + a = 0. (17.5.3.5) Letting f (0)=C 1 and f(a)=C 2 ,whereC 1 and C 2 are arbitrary constants, we eliminate f(t)andf(t + 2a) from system (17.5.3.5) (it is assumed that this is possible). As a result, we come to the reciprocal equation Ψ f(t + a), f(–t), t, a, C 1 , C 2 = 0.(17.5.3.6) In order to find a solution of equation (17.5.3.6), we replace t with –t – a.Weget Ψ f(–t), f(t + a), –t – a, a, C 1 , C 2 = 0.(17.5.3.7) Further, eliminating f(t + a) from equations (17.5.3.6)–(17.5.3.7), we come to the algebraic (transcendental) equation for f(–t). . us expand (17.5.1.1) in powers of the small parameter a in a neighborhood of a 0 , taking into account (17.5.1.2), and then divide the resulting equation by a – a 0 and pass to the limit for a. multiplication (division) of the equation and the results of its differentiation by suitable functions. Sometimes it is useful to take logarithm of the equation or the results of its transformation (see. Method of Substituting Particular Values of Independent Arguments In order to solve functional equations with several independent variables, one often uses the method of substituting particular