LINEAR DIFFERENCE EQUATIONS SIGMUNDUR GUDMUNDSSON. [ November 2017 ]

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1 LINEAR DIFFERENCE EQUATIONS SIGMUNDUR GUDMUNDSSON [ November 2017 ] In these notes we shall by N, R and C denote the sets of natural, real and complex numbers, respectively. All the definitions and most of the results mentioned below can be formulated both for the real and for the complex numbers. Here we have chosen to present the complex version, but if the same also applies to the real case we have added the symbol [R] in the corresponding header. Definition 0.1. [R] A complex-valued sequence is a function f : N C. We denote by C the set of all complex-valued sequences i.e. C = {f f : N C}. For two elements f, g C and α C we define the addition + and the product by i. (f + g)(n) = f(n) + g(n) for all n N, ii. (α f)(n) = α f(n) for all n N. Exercise 0.2. [R] Prove that the triple (C, +, ) is a complex linear space. Remark 0.3. [R] If n N and f C we often write f n instead of f(n). Definition 0.4. [R] A linear difference of degree k with constant complex coefficients is an of the form (0.1) c k a n + c k 1 a n c 0 a n k = g n, where c 0,..., c k are complex numbers such that c 0 0 c k, g C is a sequence and a n,..., a n k unknown complex numbers. If g 0 then the is said to be homogeneous, but inhomogeneous otherwise. Definition 0.5. [R] A solution to (0.1) is a sequence f C such that c k f n + c k 1 f n c 0 f n k = g n, for all natural numbers n k. 1. The Homogeneous Case Proposition 1.1. [R] Let c 0,..., c k be complex numbers such that c 0 0 c k and V be the set of all solutions to the homogeneous linear difference i.e. V = {f C c k f n + c k 1 f n c 0 f n k = 0 for n k}. 1

2 Then V is a linear subspace of C. Proof. [R] Let f, g C be solutions to the and α, β C, then c k (αf n + βg n ) + c k 1 (αf n 1 + βg n 1 ) + + c 0 (αf n k + βg n k ) = α(c k f n + c k 1 f n c 0 f n k ) = 0. +β(c k g n + c k 1 g n c 0 g n k ) This proves that the sequence (αf + βg) is an element of V, which implies that V is a linear subspace of C. Proposition 1.2. [R] Let c 0,..., c k and α 0,..., α k 1 be complex numbers such that c 0 0 c k. Then the linear homogeneous difference (1.1) has a unique solution f C which satisfies the following initial conditions f 0 = α 0, f 1 = α 1,..., f k 1 = α k 1. Proof. [R] We have assumed that c k 0, so for each n k we have f n = (c k 1 f n c 0 f n k )/c k. The function values f 0, f 1,..., f k 1 are given by the initial conditions, so it follows from the induction principle that the sequence f C is uniquely determined. The next result gives a nice description of the solution space in the complex case. For the real case see Theorem Theorem 1.3. Let c 0,..., c k be complex numbers such that c 0 0 c k. Then the set V of all solutions to the homogeneous linear difference is a k-dimensional complex linear subspace of C. Proof. Let α denote the vector α = (α 0,..., α k 1 ) and {e 1, e 2,..., e k } be the canonical basis for C k. Then the k different sets of initial conditions α = e 1, α = e 2,..., α = e k generate k linearly independent solutions. This shows that the dimension of V is k. Definition 1.4. [R] For complex numbers c 0,..., c k such that c 0 0 c k we call p(λ) = c k λ k + c k 1 λ k c 1 λ + c 0 the characteristic polynomial of the homogeneous linear difference. 2

3 Proposition 1.5. [R] Let λ 0 be a non-zero complex number. Then the sequence f : n λ n 0 is a solution to the homogeneous linear difference if and only if λ 0 is a root of the corresponding characteristic polynomial. Proof. The result follows directly from the following c k f n + c k 1 f n c 1 f n k+1 + c 0 f n k = c k λ n + c k 1 λ n c 1 λ n k+1 + c 0 λ n k = λ n k (c k λ k + + c 1 λ + c 0 ) = λ n k p(λ). The next important result is well-known but rather messy to prove in its full generality. The reader should note that we have left out the symbol [R]. Proposition 1.6. Let c 0,..., c k and α 0,..., α k 1 be complex numbers such that c 0 0 c k and λ 1,..., λ t be the different roots of the characteristic polynomial p(λ) = (c k λ k + + c 1 λ + c 0 ), with multiplicities m 1,..., m t. Then the k = m 1 + m m t sequences f 1,1 : n λ n 1 f 1,2 : n nλ n 1... f 1,m1 : n n m 1 1 λ n 1 f 2,1 : n λ n 2 f 2,2 : n nλ n 2... f 2,m2 : n n m 2 1 λ n 2. f t,1 : n λ n t f t,2 : n nλ n t... f t,mt : n n m t 1 λ n t are linearily independent solutions to the homogeneous linear difference. Exercise 1.7. Prove Proposition 1.6 for the special cases when k = 1, 2, 3. Lemma 1.8. Let the real numbers c 0, c 1,..., c k satisfying c 0 0 c k be the coefficients of the homogeneous linear difference s (1.2). Then the following conditions are equivalent i. f C is a solution to (1.2), ii. f C is a solution to (1.2), iii. Ref R and Imf R are solutions to (1.2). 3

4 Proof. The sequence f C is a solution to (1.2) if and only if c k f n + c k 1 f n c 0 f n k = 0 for all n k. By conjugating this and using the fact that the coefficients are all real we obtain the equivalent c k fn + c k 1 fn c 0 fn k = 0 which holds for all n k. But this means that the sequence f C is a solution to (1.2). From this we conclude that i. and ii. are equivalent. As a direct consequence of the definitions of the real and imaginary parts of f C we have (1.3) Ref = (f + f)/2 Imf = (f f)/2i and (1.4) f = Ref + i Imf f = Re i Imf It follows from the fact that the solution space V is complex linear and s (1.3) that i. and ii. imply iii. Similarily we can conclude from s (1.4) that iii. implies i. and ii. Example 1.9. Let α and β be real numbers such that β 0. Then the characteristic polynomial of the linear difference (1.5) a n 2αa n 1 + (α 2 + β 2 )a n 2 = 0 is given by p(λ) = λ 2 2αλ + (α 2 + β 2 ). This polynomial has the complex roots λ 1 = α + iβ and λ 2 = α iβ. According to Proposition 1.6 the sequences f : n (α + iβ) n and f : n (α iβ) n form a basis for the two dimensional solution space V. Writing α + iβ in polar form re iθ we see that the two sequences are given by f : n r n e inθ and f : n r n e inθ. Following Lemma 1.8 we see that the real-valued sequences Ref : n r n cos nθ and Imf : n r n sin nθ are linearily independent solutions to the homogeneous linear difference (1.5) with real coefficients. The next result is the real version of Theorem 1.3. Theorem Let c 0,..., c k be real numbers such that c 0 0 c k. Then the set V of all real-valued solutions to the homogeneous linear difference is a k-dimensional real linear subspace of R. Proof. The coefficients of the characteristic polynomial are all real, which means that its roots are either real or come in pairs (α + iβ, α iβ). It then follows that all possible real initial values generate a real k-dimensional solution space. 4

5 2. The Inhomogeneous Case Lemma 2.1. [R] Let c 0,..., c k be complex numbers such that c 0 0 c k and g C. If h C is a solution to the inhomogeneous c k a n + c k 1 a n c 0 a n k = g n, then the following conditions are equivalent. i. (f + h) C is a solution to the inhomogeneous, ii. f C is a solution to the corresponding homogeneous. Proof. This result follows directly from the following c k (f + h) n + c k 1 (f + h) n c 0 (f + h) n k g n = (c k f n + c k 1 f n c 0 f n k ) +(c k h n + c k 1 h n c 0 h n k ) g n = c k f n + c k 1 f n c 0 f n k. Theorem 2.2. [R] Let c 0,..., c k be complex numbers such that c 0 0 c k, g C and h C be a solution to the inhomogeneous linear difference c k a n + c k 1 a n c 0 a n k = g n. Further let V be the solution space to the corresponding homogeneous. Then the set S g of all solutions to the inhomogeneous is given by S g = h + V = {(h + f) C f V }. Proof. This result is a direct consequence of Lemma 2.1. Department of Mathematics, Faculty of Science, Lund University, Box 118, S Lund, Sweden address: Sigmundur.Gudmundsson@math.lu.se 5