RENDICONTI DEL SEMINARIO MATEMATICO

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).