Ambrosetti-Prodi-type results for a class of difference equations with nonlinearities indefinite in sign

Theorem A

[1, Theorem 6] Assume f : Z × R → R is continuous, with T-periodicity in the t variable, s ∈ R . If

Then there exists an s 0 ∈ R such that

• if s < s 0 , there is no T-periodic solution of equation (1.3),

• if s = s 0 , there is at least one T-periodic solution of equation (1.3),

• if s > s 0 , there are at least two T-periodic solutions of equation (1.3).

1. f : Z × R → R is continuous upon u ∈ R , f ( t , u ) = f ( t + T , u ) .

2. There exist a , b : [ 1 , T ] Z → R , p ∈ ( 0 , 1 ] , such that f ( t , u ) ⩾ a ( t ) ∣ u ∣ p + b ( t ) , t ∈ [ 1 , T ] Z , for all u ∈ R .

3. ∑ t = 1 T a ( t ) > 0 .

Theorem 1.1

Assume (H1)–(H3) hold, there exists s 0 ∈ R , such that

• if s < s 0 , there is no T-periodic solution of equation (1.1),

• if s = s 0 , there is at least one T-periodic solution of equation (1.1),

• if s > s 0 , there are at least two T-periodic solutions of equation (1.1).

Remark 1.2

Obersnel and Omari [15] investigated an Ambrosetti-Prodi-type result of first-order differential equation; they studied the existence and multiplicity of solutions when the parameter s exceeds a constant s 0 using normal-order upper and lower solutions and reverse-order upper and lower solutions. However, for first-order difference equations, reverse order upper and lower solutions cannot be used; in addition, lower solutions must be smaller than the upper solutions to make the method conclusive, and relevant conclusions can be found in [6]. Hence, the multiplicity of solutions when the parameter s exceeds a constant is the difficulty in this article.

Remark 1.3

In [6], Bereanu and Mawhin showed counterexamples when T ≥ 2 is odd, T > 2 is even and T = 2 , respectively. These counterexamples show that first-order difference equations have no solution when lower solutions are larger than upper solutions.

Example 1.4

First-order difference equation

We take f ( t , u ) = ( sin t + 1 / 2 ) ∣ u + 1 ∣ + cos t , f ( t + T , u ) = f ( t , u ) , and T = 2 π ; hence, (H1) holds. There exist a ( t ) = sin t + 1 / 3 , b ( t ) = cos t − 1 / 2 and p = 1 / 3 such that f ( t , u ) ⩾ a ( t ) ∣ u ∣ p + b ( t ) , t ∈ [ 1 , T ] Z , for all u ∈ R ; hence, (H2) holds. Obviously, ∑ t = 1 T a ( t ) > 0 , and hence, (H3) holds. According to Theorem 1.1, we can obtain s 0 ∈ R such that

1. if s < s 0 , there is no T -periodic solution of equation (1.5);

2. if s = s 0 , there is at least one T -periodic solution of equation (1.5);

3. if s > s 0 , there are at least two T -periodic solutions of equation (1.5).

Definition 2.1

α : [ 1 , T ] Z → R is a lower solution of problem (2.1), referring to α satisfies

β : [ 1 , T ] Z → R is an upper solution of problem (2.1), referring to β satisfies

α : [ 1 , T ] Z → R is a strict lower solution of problem (2.1), referring to α satisfies

β : [ 1 , T ] Z → R is a strict upper solution of problem (2.1), referring to β satisfies

Lemma 2.2

Problem (2.1) has a lower solution α and an upper solution β , such that α ( t ) ⩽ β ( t ) , t ∈ [ 1 , T ] Z , then problem (2.1) has at least one solution u ( t ) , such that α ( t ) ⩽ u ( t ) ⩽ β ( t ) , t ∈ [ 1 , T ] Z .

Proof

Construct auxiliary function γ : [ 1 , T ] Z × R → R by

Consider the modified problem

Using Brouwer fixed point theorem, at least one solution can be obtained for problem (2.2) in X , whose elements can be characterized by the coordinates u ( 1 ) , … , u ( T ) . Indeed, the operator L is given by

which is one to one, hence invertible, and (2.2) is equivalent to the fixed point problem

in X . It remains to show that if u ( t ) is a solution of (2.2), t ∈ [ 1 , T ] Z , then α ( t ) ⩽ u ( t ) ⩽ β ( t ) , so that u ( t ) is a solution of (2.1), t ∈ [ 1 , T ] Z . Suppose by contradiction that there exists a τ ∈ [ 1 , T ] Z , such that α ( τ ) − u ( τ ) > 0 , then

we can obtain

which contradicts with the definition of the lower solution.

Remark 2.3

Assume that α is a strict lower solution of (2.1), β is the strict upper solution of (2.1), then the problem (2.1) admits at least one solution u such that α < u < β . Define the open set Ω α , β = { u ∣ u ∈ X , α < u < β } and the open ball B ρ with the radius of ρ . The mapping Φ : R × X → R is defined by Φ ( s , u ( t ) ) = Δ u ( t − 1 ) − f ( t , u ( t ) ) + s , t ∈ [ 1 , T ] Z . If ρ is large enough, using the additivity-excision property of Brouwer degree, we have

Proof of Theorem 1.1

Step 1. We verify that for every s ∈ R , there is ξ 0 ∈ R , such that, for all ξ ⩽ ξ 0 , any solution u of the Cauchy problem

is a strict lower solution of the T -periodic problem

Hence, by (H2), u is a strict lower solution of problem (2.1).

We consider the case p ∈ ( 0 , 1 ) and prove the following claim first.

Claim For any m ∈ R , there is ξ m ⩽ m such that, for every ξ ⩽ ξ m , any solution u of (3.1) satisfies max t ∈ [ 1 , T ] Z u ( t ) < m .

Assume, by contradiction, that there exists m 0 ∈ R such that, for every n ∈ Z − , with n < − ∣ m 0 ∣ , there is a solution u n of problem (3.1) satisfying u n ( 0 ) ⩽ n and max t ∈ [ 1 , T ] Z u n ( t ) ⩾ m 0 . Let s n , t n ∈ [ 1 , T ] Z be such that s n + 1 < t n on [ s n , t n ] Z , [ s n , t n ] Z ≔ { s n , s n + 1 , … , t n − 1 , t n } , n ⩽ u n ( t ) ⩽ m 0 , t ∈ [ s n , t n ] Z , u n ( s n ) = n and u n ( t n ) = m 0 , then

For fixed s , we obtain a contradiction if n → − ∞ ; thus, our claim is proved.

In the case of p ∈ ( 0 , 1 ) , suppose that there is a sequence ( ξ n ) n ∈ R , with lim n → − ∞ ξ n = − ∞ and the solution ( u n ) n of problem (3.1) with ξ = ξ n , for any n ∈ Z − , satisfies u n ( T ) ⩽ u n ( 0 ) . By the claim above, we can assume that max t ∈ [ 1 , T ] Z u n ( t ) ⩽ n . Thus,

We obtain the contradiction 0 ⩾ ∑ t = 1 T a ( t ) > 0 when n → − ∞ . Hence, we have u ( T ) > u ( 0 ) , and u is a solution of (3.1).

The validity of step 1 when p = 1 can be verified by a direct inspection is obtained as follows:

where C is an arbitrary constant, choose

Then, we have u ( 0 ) < u ( T ) and u ( t ) is a solution of (3.1).

Step 2. We show that there exists s ∗ such that, for all s > s ∗ , equation (1.1) has at least one T -periodic solution. Indeed, it is easily verified that there exists s ∗ ∈ R such that, for all s > s ∗ , the constant β ∈ R , sup t ∈ [ 1 , T ] Z f ( t , β ) < + ∞ , β is a strict upper solution of problem (2.1). Furthermore, by the results proved in Step 1, problem (2.1) admits one strict lower solution α 1 satisfying α 1 ( t ) ⩽ β for all t ∈ [ 1 , T ] Z . Therefore, equation (1.1) has at least one T -periodic solution u 1 , satisfying α 1 ( t ) ⩽ u 1 ( t ) ⩽ β for all t ∈ [ 1 , T ] Z , u 1 ≠ α 1 , β .

Step 3. We prove that the set of the parameters s for which equation (1.1) has at least one T -periodic solution is bounded from below. Define the set

We prove there exists s 0 ∈ R , such that s 0 = inf Ψ . Assume, by contradiction, that inf Ψ = − ∞ . Then, there exists a sequence ( s n ) n ∈ R with lim n → + ∞ s n = − ∞ , and a sequence ( u n ) n of T -periodic solutions of equation (1.1) with s = s n . We claim that lim n → + ∞ ‖ u n ‖ = + ∞ , otherwise, we would obtain

There would exist a function φ : [ 1 , T ] Z → R , such that

which is a contradiction. Moreover, by (H2) we have

Thus, we obtain

Let n → + ∞ , and using (H3) yields the contradiction 0 ⩾ ∑ t = 1 T a ( t ) > 0 .

Step 4. We show the existence of at least one T -periodic solution of equation (1.1) for s = s 0 . Let ( s n ) n be a sequence in Ψ converging to s 0 and let ( u n ) n be the corresponding sequence of T -periodic solutions of equation (1.1) with s = s n . Let us verify that there is R > 0 , such that ‖ u n ‖ ⩽ R for all n ∈ N . Indeed, otherwise, we can find a subsequence of ( u n ) n , we still denote by ( u n ) n , such that lim n → + ∞ ( u n ) n = + ∞ . Arguing similarly as in the proof of Step 3, thus easily leading to a contradiction as above. Therefore, ( u n ) n is bounded in X ; according to Weierstrass concentration theorem, we can obtain lim n → + ∞ u n ( t ) = u 0 ( t ) , t ∈ [ 1 , T ] Z . Besides, lim n → + ∞ f ( t , u n ( t ) ) = f ( t , u 0 ( t ) ) , t ∈ [ 1 , T ] Z , and when n is large enough, ∣ f ( t , u n ( t ) ) ∣ ⩽ φ ( t ) . Sequence ( f ( ⋅ , u n ) − s n ) n , i.e., ( Δ u n ) n , convergence to f ( ⋅ , u 0 ) − s 0 in X , with Δ u 0 ( t − 1 ) = f ( t , u 0 ( t ) ) − s 0 , u 0 ( T ) = u 0 ( 0 ) , t ∈ [ 1 , T ] Z , u 0 is a T -periodic solution of equation (1.1) for s = s 0 .

Step 5. We show that for all s > s 0 , equation (1.1) has at least two T -periodic solutions.

Claim For any constant c ∈ R , there exists ρ > 0 , such that, for all s ⩽ c , all possible periodic solutions u of equation (1.1) belong to open ball B ρ .

For every s ⩽ c , we have

We need to show there exists a constant c 1 , such that

By (H2), we can obtain f ( t , u ( t ) ) ⩾ a ( t ) ∣ u ( t ) ∣ p + b ( t ) , then

Thus,

Hence, all possible solutions of problem (2.1) belong to open ball B ρ .

Using the Brouwer degree theory, obviously, u ( t ) is a solution of problem (2.1) if and only if u ( t ) is a zero of Φ ( s , ⋅ ) , t ∈ [ 1 , T ] Z . Let s 2 < s 0 < s 1 , according to the claim above, we can find the corresponding ρ such that, for all s ∈ [ s 2 , s 1 ] , every possible zero points u of Φ ( s , ⋅ ) satisfy u ∈ B ρ . Consequently, the Brouwer degree deg [ Φ ( s , ⋅ ) , B ρ , 0 ] is well defined and does not depend upon s . Using the conclusion of step 3, for u ∈ X , u − Φ ( s 2 , ⋅ ) ≠ 0 . This implies that deg [ Φ ( s 2 , ⋅ ) , B ρ , 0 ] = 0 , so that deg [ Φ ( s 1 , ⋅ ) , B ρ , 0 ] = 0 . By excision property, deg [ Φ ( s 1 , ⋅ ) , B ρ ′ , 0 ] = 0 if ρ ′ > ρ .

Let u ˆ be a solution of (2.1) with s ∈ ( s 0 , s 1 ) , then u ˆ is a strict upper solution of problem (2.1) with s = s 1 . From Step 1, α 1 is a strict lower solution of problem (2.1). Consequently, using Remark 2.3, (2.1) with s = s 1 has a solution in Ω α 1 , u ˆ , and

Taking ρ ′ sufficiently large, we deduce from the additivity property of Brouwer degree that

When s = s 1 , (2.1) has the second solution in B ρ ′ \ Ω α 1 , u ˆ .□