26 C. Bereanu - J. MawhinTHEOREM 3. If the functions f m (1 ≤ m ≤ n − 1) satisfy (29), there existss 1 ∈ R such that (28) has zero, at least one or at least two solutions according tos < s 1 , s = s 1 , s > s 1 .Proof. LetS j = {s ∈ R : (28) has at least j solutions } ( j ≥ 1).(a) S 1 ̸= ∅.Take s ∗ > max 1≤m≤n−1 f m (0) and use (29) to find R− ∗ < 0 such thatmin f m(R− ∗ ) > 1≤m≤n−1 s∗ .Then α with α j = R ∗ − < 0 (1 ≤ j ≤ n) is a strict lower solution and β with β j = 0(1 ≤ j ≤ n) is a strict upper solution for (28) with s = s ∗ . Hence, using Theorem 2,s ∗ ∈ S 1 .(b) If˜s ∈ S 1 and s >˜s then s ∈ S 1 .Let ˜x = (˜x 1 ,...,˜x n ) be a solution of (28) with s =˜s, and let s >˜s. Then ˜x is a strictupper solution for (28). Take now R − < min 1≤m≤n ˜x m such that min 1≤m≤n−1 f m (R − )> s It follows that α with α j = R − (1 ≤ j ≤ n) is a strict lower solution for (28), andhence, using Theorem 2, s ∈ S 1 .(c) s 1 = inf S 1 is finite and S 1 ⊃ ]s 1 ,∞[.Let s ∈ R and suppose that (28) has a solution (x 1 ,..., x n ). Then (33) holds, fromwhere we deduce that s ≥ c, with c ∈ R given in (31). To obtain the second part ofclaim (c) S 1 ⊃]s 1 ,∞[ we apply (b).(d) S 2 ⊃ ]s 1 ,∞[.We reformulate (28) to apply Brouwer degree theory. Consider the continuous mappingG : R × R n−1 → R n−1 defined byG m (s, x) = (Lx) m + f m (x m ) − s (1 ≤ m ≤ n − 1).Then (x 1 ,..., x n−1 , x 1 ) is a solution of (28) if and only if (x 1 ,..., x n−1 ) ∈ R n−1 isa zero of G(s,·). Let s 3 < s 1 < s 2 . Using Lemma 3 we find ρ > 0 such that eachpossible zero of G(s,·) with s ∈ [s 3 , s 2 ] is such that max 1≤m≤n−1 |x m | < ρ. Consequently,d B [G(s,·), B(ρ), 0] is well defined and does not depend upon s ∈ [s 3 , s 2 ].However, using (c), we see that G(s 3 , x) ̸= 0 for all x ∈ R n−1 . This implies thatd B [G(s 3 ,·), B(ρ), 0] = 0, so that d B [G(s 2 ,·), B(ρ), 0] = 0 and, by excision property,d B [G(s 2 ,·), B(ρ ′ ), 0] = 0 if ρ ′ > ρ. Let s ∈ ]s 1 , s 2 [ and ̂x = (̂x 1 ,...,̂x n ) be a solutionof (28) (using (c)). Then ̂x is a strict upper solution of (28) with s = s 2 . LetR < min 1≤ j≤n ̂x j be such that min 1≤m≤n−1 f m (R) > s 2 . Then (R,..., R) ∈ R n is astrict lower solution of (28) with s = s 2 . Consequently, using Corollary 1, (28) withs = s 2 has a solution in R̂x and|d B [G(s 2 ,·), R̂x , 0]| = 1.Taking ρ ′ sufficiently large, we deduce from the additivity property of Brouwer degree
Periodic solutions of difference equations 27that|d B [G(s 2 ,·), B(ρ ′ ) \ R̂x , 0]| = |d B [G(s 2 ,·), B(ρ ′ ), 0] − d B [G(s 2 ,·), R̂x , 0]|= |d B [G(s 2 ,·), R̂x , 0]| = 1,and (28) with s = s 2 has a second solution in B(ρ ′ ) \ R̂x .(e) s 1 ∈ S 1 .Taking a decreasing sequence (σ k ) k∈N in ]s 1 ,∞[ converging to s 1 , a correspondingsequence (x k 1 ,..., xk n ) of solutions of (28) with s = σ k and using Lemma 3, we obtaina subsequence (x j k1 ,..., x j kn ) which converges to a solution (x 1 ,..., x n ) of (28) withs = s 1 .COROLLARY 4. If p > 0, a m > 0 and b m ∈ R (1 ≤ m ≤ n − 1), there existss 1 ∈ R such that the periodic problemDx m + a m |x m | p = s + b m (1 ≤ m ≤ n − 1), x 1 = x nhas no solution if s < s 1 , at least one solution if s = s 1 and at least two solutions ifs > s 1 .Similar arguments allow to prove the following result.THEOREM 4. If the functions f m satisfy condition(34)f m (x) → −∞ as |x| → ∞ (1 ≤ m ≤ n − 1).then there is s 1 ∈ R such that (28) has zero, at least one or at least two solutionsaccording to s > s 1 , s = s 1 or s < s 1 .COROLLARY 5. If p > 0, a m > 0 and b m ∈ R (1 ≤ m ≤ n − 1), there existss 1 ∈ R such that the periodic problemDx m − a m |x m | p = s + b m (1 ≤ m ≤ n − 1), x 1 = x nhas no solution if s > s 1 , at least one solution if s = s 1 and at least two solutions ifs < s 1 .8. One-side bounded nonlinearitiesThe nonlinearity in Ambrosetti-Prodi type problems is bounded from below and coerciveor bounded from above and anticoercive. In this section, we consider nonlinearitieswhich are bounded from below or above but have different limits at +∞ and −∞.Let n ≥ 2 be an integer and f m : R → R continuous functions (1 ≤ m ≤ n−1).Consider the problem(35)Dx m + f m (x m ) = 0 (1 ≤ m ≤ n − 1), x 1 = x n .