20 C. Bereanu - J. MawhinTHEOREM 2. If (10) has a lower solution α = (α 1 ,...,α n ) and an uppersolution β = (β 1 ,...,β n ) such that α m ≤ β m (1 ≤ m ≤ n), then (10) has a solutionx = (x 1 ,..., x n ) such that α m ≤ x m ≤ β m (1 ≤ m ≤ n). Moreover, if α and β arestrict, then α m < x m < β m (1 ≤ m ≤ n − 1).Proof. I. A modified problem.Let γ m : R → R (1 ≤ m ≤ n − 1) be the continuous functions defined by⎧⎨ β m if x > β m(12)γ m (x) = x if α m ≤ x ≤ β m⎩α m if x < α m .We consider the modified problem(13) Dx m − x m + f m ◦ γ m (x m ) + γ m (x m ) = 0 (1 ≤ m ≤ n − 1), x 1 = x n ,and show that if x = (x 1 ,..., x n ) is a solution of (13) then α m ≤ x m ≤ β m (1 ≤m ≤ n), and hence x is a solution of (10). Suppose by contradiction that there is some1 ≤ i ≤ n such that α i − x i > 0 so that α m − x m = max 1≤ j≤n (α j − x j ) > 0. If1 ≤ m ≤ n − 1, thenwhich givesα m+1 − x m+1 ≤ α m − x m ,Dα m ≤ Dx m = x m − α m − f m (α m ) ≤ x m − α m + Dα m < Dα m ,a contradiction. Now the condition α 1 ≥ α n shows that the maximum is reached atm = n only if it is reached also at m = 1, a case already excluded. Analogously wecan show that x m ≤ β m (1 ≤ m ≤ n). We remark that if α,β are strict, the samereasoning gives α m < x m < β m (1 ≤ m ≤ n − 1).II. Solution of the modified problem.We use Brouwer degree to study the zeros of the continuous mapping G : R n−1 →R n−1 defined by(14) G m (x) = (Lx) m − x m + f m ◦ γ m (x m ) + γ m (x m ) (1 ≤ m ≤ n − 1).By Lemma 1, L − I : R n−1 → R n−1 is invertible. On the other hand the mappingwith components f m ◦ γ m + γ m (1 ≤ m ≤ n − 1) is bounded on R n−1 . Consequently,Theorem 1 implies the existence of R > 0 such that, for all ρ > R, one has(15)|d B [G, B(ρ), 0]| = 1,and, in particular, G has a zero ˜x ∈ B(ρ). Hence, x = (˜x, x 1 ) is a solution of (13),which means that α m ≤ x m ≤ β m (1 ≤ m ≤ n) and x is a solution of (10). Moreoverif α,β are strict, then α m < x m < β m (1 ≤ m ≤ n − 1).
Periodic solutions of difference equations 21Suppose now that α (resp. β) is a strict lower (resp. upper) solution of (10).Define the open set(16) αβ = {(x 1 ,..., x n−1 ) ∈ R n−1 : α m < x m < β m (1 ≤ m ≤ n − 1)},and the continuous mapping H : R n−1 → R n−1 by(17)H m (x) = (Lx) m + f m (x m ) (1 ≤ m ≤ n − 1).COROLLARY 1. Assume that the conditions of Theorem 2 hold with strict lowerand upper solutions. Then(18)|d B [H, αβ , 0]| = 1,with αβ defined in (16).Proof. If ρ is large enough, then, using the additivity-excision property of Brouwerdegree [7], we have|d B [G, αβ , 0]| = |d B [G, B(ρ), 0]| = 1.On the other hand, H is equal to G on αβ , and then|d B [G, αβ , 0]| = |d B [H, αβ , 0]|.A simple but useful consequence of Theorem 2, goes as follows.COROLLARY 2. Assume that there exists numbers α ≤ β such thatf m (α) ≥ 0 ≥ f m (β) (1 ≤ m ≤ n − 1).Then problem (10) has at least one solution with α ≤ x m ≤ β (1 ≤ m ≤ n − 1).Proof. Just observe that (α,...,α) is a lower solution and (β,...,β) an upper solutionfor (10).COROLLARY 3. For each p > 0, a m > 0 and b m ∈ R (1 ≤ m ≤ n − 1) theproblemDx m − a m |x m | p−1 x m = b m (1 ≤ m ≤ n − 1), x 1 = x nhas at least one solution.() 1/p|bProof. If R ≥ max m |1≤m≤n−1 a m , then (−R,...,−R) is a lower solution and(R,..., R) an upper solution.