17.2. LINEAR DIFFERENCE EQUATIONS WITH A SINGLE CONTINUOUS VARIABLE 885 The next value, y m , is calculated by substituting the initial values into the right-hand side of equation (17.1.7.5) with n = 0.Wehavey m = f(0, y 0 , y 1 , , y m–1 ). Then one takes n = 1 in (17.1.7.5). We get y m+1 = f (1, y 1 , y 2 , , y m ). (17.1.7.6) Substituting the initial values y 1 , , y m–1 and the calculated value y m into this relation, we find y m+1 . Further, taking n = 2 in (17.1.7.6) and using the initial values y 2 , , y m–1 and the calculated values y m , y m+1 ,wefind y m+2 . In a similar way, we consecutively find all subsequent values y m+3 , y m+4 , The above method of solving difference equations is called the step method. 17.2. Linear Difference Equations with a Single Continuous Variable 17.2.1. First-Order Linear Difference Equations 17.2.1-1. Homogeneous linear difference equations. General properties of solutions. 1 ◦ .Afirst-order homogeneous linear difference equation has the form y(x + 1)–f(x)y(x)=0,(17.2.1.1) where f(x) is a given continuous real-valued function of real argument and y(x)isan unknown real-valued function, 0 ≤ x < ∞. Let y 1 = y 1 (x) be a nontrivial particular solution of equation (17.2.1.1), y 1 (x) 0.Then the general solution of equation (17.2.1.1) is given by y(x)=Θ(x)y 1 (x), (17.2.1.2) where Θ(x)=Θ(x + 1) is an arbitrary 1-periodic function. Example 1. The equation with constant coefficients y(x + 1)–ay(x)=0,(17.2.1.3) with a > 0, admits a particular solution y 0 (x)=a x . Therefore, the general solution of equation (17.2.1.3) has the form y(x)=Θ(x)a x ,(17.2.1.4) where Θ(x) is an arbitrary periodic function with unit period. In the special case of a = 1, the general solution of equation (17.2.1.3) is an arbitrary 1-periodic function. Remark 1. When using formula (17.2.1.2) for obtaining continuous particular solutions from a given continuous particular solution, one should consider not only continuous but also discontinuous or unbounded periodic functions Θ(x). For instance, the equation y(x + 1)+y(x)=0,(17.2.1.5) which determines antiperiodic functions of unit period, has continuous particular solutions y 1 (x)=cos(πx), y 2 (x)=sin(πx). In order to pass from the first of these particular solutions to the second with the help of (17.2.1.2), one should take the unbounded periodic function Θ(x)=tan(πx), which is undefi ned at the points x = 1 2 + n, n = 0, 1, 2, See also Remark 2. 886 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS Remark 2. The general solution of a difference equation may be represented by quite different formulas. Thus, the general solution of equation (17.2.1.5) can be represented by the following formulas: y(x)=(–1) [x] Θ(x), (17.2.1.6a) y(x)=Θ 1 (x)cos(πx)+Θ 2 (x)sin(πx), (17.2.1.6b) y(x)=Θ 3 x + 1 2 – Θ 3 x 2 , (17.2.1.6c) where Θ(x), Θ 1 (x), Θ 2 (x), and Θ 3 (x) are arbitrary periodic functions of period 1, [x] is the integer part of x. Let us show that formulas (17.2.1.6a), (17.2.1.6b), and (17.2.1.6c) are equivalent. First, we note that (17.2.1.6b) contains one redundant arbitrary function and can be written as y(x)= Θ 1 (x)cos(πx)=Θ 2 (x)sin(πx), Θ 1 (x)=Θ 1 (x)+tan(πx)Θ 2 (x), Θ 2 (x)=cot(πx)Θ 1 (x)+Θ 2 (x). Here, Θ 1 (x)andΘ 2 (x) are arbitrary 1-periodic functions, since tan(πx)andcot(πx) have period 1. Therefore, without the loss of generality, we can take Θ 1 (x) ≡ 0 in (17.2.1.6b)(orΘ 2 (x) ≡ 0). And we should consider bounded, as well as unbounded, periodic functions Θ 2 (x)(orΘ 1 (x)). Formula (17.2.1.6a) with a continuous Θ(x) yields discontinuous solutions, in general (for Θ(x) different from zero at integer points). This formula can be represented in the form (17.2.1.6b): y(x)=sin(πx)Θ 2 (x), Θ 2 (x)= (–1) [x] sin(πx) Θ(x). Here, Θ 2 (x) is an arbitrary periodic function of period 1, since the function (–1) [x] / sin(πx) has period 1. Formula (17.2.1.6c) can be written in the form (17.2.1.6a): y(x)=(–1) [x] Θ(x), Θ(x)=(–1) [x] Θ 3 x + 1 2 – Θ 3 x 2 . For the investigation of continuous (smooth) solutions it is more convenient to use formulas (17.2.1.6b) and (17.2.1.6c), where Θ 1 (x), Θ 2 (x), and Θ 3 (x) are arbitrary continuous (smooth) 1-periodic functions. 2 ◦ . Consider the equation y(x + 1)–af(x)y(x)=0 (17.2.1.7) with a parameter a.Lety 1 = y 1 (x) be a nontrivial particular solution of equation (17.2.1.7) with a = 1. Then the following results hold: For a > 0, the general solution of equation (17.2.1.7) has the form y(x)=a x Θ(x)y 1 (x), (17.2.1.8) where Θ(x) is an arbitrary periodic function with period 1. For a < 0, the general solution of equation (17.2.1.7) can be represented by any of the formulas y(x)=(–1) [x] |a| x Θ(x)y 1 (x), y(x)=|a| x Θ 1 (x)cos(πx)+Θ 2 (x)sin(πx) y 1 (x), y(x)=|a| x Θ 3 x + 1 2 – Θ 3 x 2 y 1 (x), (17.2.1.9) which generalize formulas (17.2.1.6). 3 ◦ .Lety 1 = y 1 (x) be a solution of equation (17.2.1.1). Then the equation y(x + 1)–f(x + a)y(x)=0 admits the solution y(x)=y 1 (x + a). 17.2. LINEAR DIFFERENCE EQUATIONS WITH A SINGLE CONTINUOUS VARIABLE 887 4 ◦ .Lety 1 = y 1 (x) be a positive solution of equation (17.2.1.1). Then the equation y(x + 1)–[f (x)] k y(x)=0, for any k, admits the solution y(x)=[y 1 (x)] k . 5 ◦ .Lety 1 = y 1 (x) be a solution of equation (17.2.1.1). Then the equation y(x + 1)–ϕ(x)f (x)y(x)=0, where ϕ(x)=ϕ(x + 1) is an arbitrary positive 1-periodic function, admits the solution y(x)=y 1 (x)[ϕ(x)] x+1 . 6 ◦ . A solution of the linear homogeneous difference equation y(x + 1)+f 1 (x)f 2 (x) f n (x)y(x)=0 can be represented as the product y(x)=y 1 (x)y 2 (x) y n (x), where y k (x) are solutions of the linear homogeneous difference equations y k (x + 1)+f k (x)y k (x)=0, k = 1, , n. 7 ◦ . The equation y(x + a)–f (x)y(x)=0 can be reduced to an equation of the form (17.2.1.1) with the help of the transformation z = x/a, y(x)=w(z). And we obtain w(z + 1)–f(az)w(z)=0. 17.2.1-2. Linear difference equations with rational and exponential functions. Below we give some particular solutions of some homogeneous linear difference equations with rational and exponential functions. Their general solutions can be obtained as a product of a particular solution and an arbitrary 1-periodic function; see (17.2.1.2). 1 ◦ . The general solution of the first-order homogeneous linear difference equation with constant coefficients (17.2.1.3) is determined by (17.2.1.8) and (17.2.1.9) with y 1 (x) ≡ 1. 2 ◦ . The equation y(x + 1)–xy(x)=0 admits a particular solution y(x)=Γ(x), Γ(x)= ∞ 0 t x–1 e –t dt, where Γ(x) is the gamma-function. 888 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS 3 ◦ . Consider the first-order equation with rational coefficients P n (x)y(x + 1)–Q m (x)y(x)=0, where P n (x)andQ m (x) are given polynomials of degrees n and m, respectively. Suppose that these polynomials are represented in the form P n (x)=a(x – ν 1 )(x – ν 2 ) (x – ν n ), Q m (x)=b(x – μ 1 )(x – μ 2 ) (x – μ m ), ab > 0. Direct verification shows that the function y(x)= b a x Γ(x – μ 1 )Γ(x – μ 2 ) Γ(x – μ m ) Γ(x – ν 1 )Γ(x – ν 2 ) Γ(x – ν n ) is a particular solution of the equation under consideration, where Γ(x) is the gamma- function. This solution can have polar singularities at the points x = μ k – s (k = 1, 2, ; s = 0, 1, ). 4 ◦ . The equation y(x + 1)–e λx y(x)=0 with the parameter λ admits the solution y(x)=exp 1 2 λx 2 – 1 2 λx . 5 ◦ . The equation y(x + 1)–e μx 2 +λx y(x)=0 with the parameters μ and λ admits the solution y(x)=exp 1 3 μx 3 + 1 2 (λ – μ)x 2 + 1 6 (μ – 3λ)x . 6 ◦ . The equation y(x + 1)–exp[P n (x)]y(x)=0, P n (x)= n k=1 b k x k , has a particular solution of the form y(x)=exp[Q n+1 (x)], Q n+1 (x)= n+1 k=1 c k x k , where c k can be found by the method of indefinite coefficients. 7 ◦ . The particular solutions from Items 1 ◦ – 6 ◦ allow us to obtain solutions of more intricate linear difference equations with the help of formulas from Paragraph 17.2.1-1. Example 2. Consider the equation y(x + 1)–(x + a) k y(x)=0 with the parameters a and k. As a starting point, we take the solution from Item 1 ◦ corresponding to the special case of the equation with a = 0, k = 1. Consecutive utilization of the formulas from Items 3 ◦ and 4 ◦ of Paragraph 17.2.1-1 yields the following solution of the equation under consideration: y(x)=[Γ(x + a)] k , where Γ(x) is the gamma-function. 8 ◦ . The general solution of equation (17.2.1.1) with arbitrary f (x) can be constructed with the help of formulas from Paragraph 17.2.1-3. 17.2. LINEAR DIFFERENCE EQUATIONS WITH A SINGLE CONTINUOUS VARIABLE 889 17.2.1-3. Homogeneous linear difference equations. Cauchy’s problem. 1 ◦ . Cauchy’s problem: Find a solution of equation (17.2.1.1) with the initial condition y(x)=ϕ(x)for0 ≤ x < 1,(17.2.1.10) where ϕ(x) is a given continuous function defined on the interval 0 ≤ x ≤ 1. A solution of problem (17.2.1.1), (17.2.1.10) is obtained by the step method: on the interval 1 ≤ x < 2, the solution is constructed from equation (17.2.1.1) with the initial condition (17.2.1.10) taken into account; on the interval 2 ≤ x < 3, one utilizes equation (17.2.1.1) and the solution obtained for 1 ≤ x < 2; on the interval 3 ≤ x < 4, one uses equation (17.2.1.1) and the solution obtained for 2 ≤ x < 3; etc. As a result, we get y(x)=ϕ(x)for0 ≤ x < 1, y(x)=f(x – 1)ϕ(x – 1)for 1 ≤ x < 2, y(x)=f(x – 1)f(x – 2)ϕ(x – 2)for2 ≤ x < 3, y(x)=f(x – 1)f(x – 2) f(x – n)ϕ(x – n)forn ≤ x < n + 1, (17.2.1.11) where n = 3, 4, The sequence of formulas (17.2.1.11) that determine a solution of the Cauchy problem (17.2.1.1), (17.2.1.10) can be written as a single formula y(x)=ϕ({x}) [x] k=1 f(x – k), (17.2.1.12) where [x]and{x} denote, respectively, the integer and the fractional parts of x (x = [x]+{x}), and the product over the empty set of indexes (for [x]=0) is assumed equal to unity. Solution (17.2.1.12) is continuous if it is continuous at the integer points x = 1, 2, , and this brings us to the condition ϕ(1)=f(0)ϕ(0). (17.2.1.13) Example 3. Consider the Cauchy problem for equation (17.2.1.1), where f(x)=f (x + 1) is an arbitrary (nonnegative) 1-periodic function. In the initial condition (17.2.1.10), take ϕ(x)=Θ(x)[f(x)] x+1 , where Θ(x)=Θ(x + 1) is an arbitrary 1-periodic function. It is easy to check that the continuity condition (17.2.1.13) is satisfied. Using (17.2.1.11), we obtain the solution of the problem in closed form y(x)=Θ(x)[f(x)] x+1 (0 ≤ x < ∞). (17.2.1.14) Remark. The general solution (17.2.1.4) of the equation with constant coefficients (17.2.1.3) can be obtained by substituting f (x)=a > 0 into (17.2.1.14) and making the transformation aΘ(x) → Θ(x). 2 ◦ . The general solution of the homogeneous linear difference equation (17.2.1.1) is ob- tained by replacing ϕ({x}) with Θ(x) in (17.2.1.12), where Θ(x) is an arbitrary 1-periodic function. 890 DIFFERENCE EQUATIONS AND OTHER FUNCTIONAL EQUATIONS 17.2.1-4. Nonhomogeneous linear difference equations. General solution. 1 ◦ . Consider a first-order nonhomogeneous linear difference equation y(x + 1)–f(x)y(x)=g(x), (17.2.1.15) where f(x)andg(x) are given continuous functions, y(x) is the sought function, 0 ≤ x < ∞. The general solution of the nonhomogeneous equation (17.2.1.15) can be represented as the sum y(x)=u(x)+y(x), where the first term u(x) is the general solution of the corresponding homogeneous equation (with g ≡ 0), and the second term y(x) is a particular solution of equation (17.2.1.15). A formula for the general solution of equation (17.2.1.15) is given in Paragraph 17.2.1-7, Item 2 ◦ . 2 ◦ .Letg(x)=g(x + 1)bea1-periodic function and let y 1 (x) be a solution of equation (17.2.1.15) in the special case of g(x) ≡ 1. Then the function y(x)=g(x)y 1 (x)(17.2.1.16) is a solution of equation (17.2.1.15). Example 4. Consider the difference equation, which is a special case of equation (17.2.1.15) with f(x) ≡ a > 0: y(x + 1)–ay(x)=g(x), (17.2.1.17) where g(x)=g(x + 1)isagiven1-periodic function. Equation (17.2.1.17) with g(x) ≡ 1 admits the particular solution y 1 (x)= x if a = 1, 1 1 – a if a ≠ 1. The corresponding particular solution of equation (17.2.1.17) is found with the help of (17.2.1.16), and the general solution has the form y(x)= Θ(x)+xg(x)ifa = 1, Θ(x)a x + 1 1 – a g(x)ifa ≠ 1, where Θ(x) is an arbitrary 1-periodic function. 3 ◦ . Consider the difference equation which is a special case of (17.2.1.15) with f(x) ≡ 1: y(x + 1)–y(x)=g(x). (17.2.1.18) Let x (a, ∞) with an arbitrary a. Suppose that the function g(x) is monotone, strictly convex (or strictly concave), and satisfies the condition lim x→∞ g(x + 1)–g(x) = 0, and let x 0 (a, ∞) be an arbitrary fixed point. Then for every y 0 , there exists exactly one function y(x) (monotone and strictly convex/concave) satisfying equation (17.2.1.18), together with the condition y(x 0 )=y 0 . This solution is given by the formulas y(x)=y 0 +(x – x 0 )g(x 0 )– ∞ n=0 g(x + n)–g(x 0 + n)–(x – x 0 ) g(x 0 + n + 1)–g(x 0 + n) . 17.2. LINEAR DIFFERENCE EQUATIONS WITH A SINGLE CONTINUOUS VARIABLE 891 4 ◦ . The functional equation y(x + 1)+y(x)=g(x), after the transformation y(x)=u x + 1 2 – u x 2 , ξ = x 2 , is reduced to an equation of the form (17.2.1.18): u(ξ + 1)–u(ξ)=g(2ξ). 17.2.1-5. Nonhomogeneous linear equations with right-hand sides of special form. 1 ◦ . The equation y(x + 1)–y(x)= n k=0 a k x k with a polynomial right-hand side admits the particular solution y(x)= n k=0 a k k + 1 B k+1 (x), where B k (x)areBernoulli polynomials. The Bernoulli polynomials are defined with the help of the generating function te xt e t – 1 ≡ ∞ n=0 B n (x) t n n! (|t| < 2π). The first six Bernoulli polynomials have the form B 0 (x)=1, B 1 (x)=x – 1 2 , B 2 (x)=x 2 – x + 1 6 , B 3 (x)=x 3 – 3 2 x 2 + 1 2 x, B 4 (x)=x 4 – 2x 3 + x 2 – 1 30 , B 5 (x)=x 5 – 5 2 x 4 + 5 3 x 3 – 1 6 x. See also Subsection 18.18.1. 2 ◦ . For the equation with polynomial right-hand side y(x + 1)–ay(x)= n k=0 b k x k , a ≠ 1,(17.2.1.19) a particular solution is sought by the method of indefinite coefficients in the form of a polynomial of degree n. A particular solution of equation (17.2.1.19) may also be defined by the formula y(x)= n k=0 b k d k dλ k e λx e λ – a λ=0 . . general solution of the corresponding homogeneous equation (with g ≡ 0), and the second term y(x) is a particular solution of equation (17.2.1.15). A formula for the general solution of equation. some particular solutions of some homogeneous linear difference equations with rational and exponential functions. Their general solutions can be obtained as a product of a particular solution and. to the special case of the equation with a = 0, k = 1. Consecutive utilization of the formulas from Items 3 ◦ and 4 ◦ of Paragraph 17.2.1-1 yields the following solution of the equation under