On the Dirichlet problem associated with bounded perturbations of positively-(p, q)- homogeneous Hamiltonian systems

The existence of solutions for the Dirichlet problem associated with bounded perturbations of positively-(p, q)-homogeneous Hamiltonian systems is considered both in nonresonant and resonant situations. To deal with the resonant case, the existence of a couple of lower and upper solutions is assumed. Both the well-ordered and the non-well-ordered cases are analysed. The proof is based on phase-plane analysis and topological degree theory.

Its proof relies on the so called shooting method, following the ideas presented in [10] (see also [3,23]), and will be given in Section 3.
If p = 2, the differential equation in (6) models an asymmetric oscillator, with τ + = π/ √ μ, τ − = π/ √ ν, and it is well known that some care on the choice of μ, ν must be taken to avoid resonance phenomena; when h is bounded, the existence of a solution depends on the position of (μ, ν) with respect to the Fučík spectrum, cf. [10]. In Fig. 1, the assumption of Theorem 1 can be visualized, when b − a = π p , taking the values (μ −1/p , ν −1/p ) in the white regions.
A different way of avoiding resonance would be the assumption of the existence of a well-ordered couple of lower/upper solutions α ≤ β, together with some Nagumo-type conditions. This method goes back to the pioneering papers [20][21][22]. Some more care is needed in the non-well-ordered case α ≤ β, Figure 1. The Fučík spectrum see [1,8,9,16,17]. We refer to the book [7] for an extensive exposition on the theory of lower and upper solutions for scalar second order differential equations.
The concept of lower and upper solution has been recently extended to planar systems in [15] for the periodic problem (see also [12]), and in [14] for Sturm-Liouville type problems, including the Dirichlet problem. We recall in Section 4 the main definitions in this case.
Here is our result for problem (2), in the well-ordered case.
The proof will be given in Section 5. We will first need some properties of positively-(p, q)-homogeneous Hamiltonian systems, which we provide in Section 2. Then, the main issue will be the construction of some guiding curves in the phase plane so to enter the framework of [14,Theorem 11].
In the sequel, we denote by C j, loc the space of C j -smooth real functions whose jth derivative is locally -Hölder continuous, with > 0. Moreover, defining the function we introduce the following order relation: for any continuous function x : if and only if We will write either u v or v u when u − v 0. Concerning the non-well-ordered case, we recall that the existence of a pair of lower/upper solutions does not guarantee the existence of a solution, since resonance phenomena can occur with respect to the higher part of the spectrum. However, at least for scalar second-order equations, it is well known that resonance can be handled with respect to the first eigenvalue, cf. [7]. This observation leads us to assume, in the non-well-ordered case, that Here is our result, in the non-well-ordered case.
Then, if (8) holds, there exists a solution (x, y) of (2) such that α x and x β.
The above theorem generalizes [14,Theorem 19]. Its proof is provided in Section 6 by the use of topological degree techniques, which require the above regularity assumptions. It would be interesting to know whether the result still holds when the functions φ, ψ are only assumed to be continuous. Notice that assumption (9) is surely verified for problems like (6), where ∂H ∂y (x, y) = |y| q−2 y.
In the final section of the paper, we extend the previous results to systems in R 2N which can be considered as weakly coupled planar systems of the above type. The different planar systems involved could have either wellordered or non-well-ordered lower and upper solutions. We are able to deal with this mixed type of situations, still obtaining an existence result, thus carrying out the investigation opened in [14]. However, for the non-well-ordered case, we need to ask the lower/upper solutions to be strict, a concept we will introduce in Section 7.

Elementary properties of positively-(p, q)-homogeneous
Hamiltonian systems For the autonomous system (3) the origin (0, 0) is an isochronous center, all solutions having minimal period τ > 0. We denote by S(t) = (S 1 (t), S 2 (t)) the periodic solution such that S 1 (0) = 0 and S 2 (0) > 0, with H(S 1 (t), S 2 (t)) = 1 for every t. Then, the periodic solutions of system (3) having energy H(x, y) = E can be written as On the Dirichlet problem associated Page 5 of 32 66 for some σ ∈ R. We will use the notations For any function u = (x, y) : [a, b] → R 2 \ {(0, 0)}, we can introduce the generalized polar coordinates where For every D > 0, one has that Condition (1) can be rewritten as for every λ > 0 and (x, y) = (0, 0). Then, We can write the generalized Euler formula We can rewrite (12) and (13) as for every μ > 0. We will need the following property. Hence, we can find δ 0 > 0 and c 0 > 0 such that If we set then, for every x ∈ [−L, L] we get, using (16) and (17), So, for any x ∈ [−L, L] and μ > 1, from (16) we obtain The conclusion follows.
The proof of Theorem 1 is now completed.

Remarks.
Other types of boundary conditions can be considered, leading to similar results. The case of a Neumann-type problem, with boundary conditions y(a) = 0 = y(b), is nothing but the previous Dirichlet-type problem by a simple switch in the variables x y. The mixed problem can be considered, as well. Going back to the solution S = (S 1 , S 2 ) of the unperturbed system (3), we can find the positive values τ 1 , τ 2 , τ 3 , τ 4 such that leading to the following.

Theorem 5.
Assume that there is an integer n ≥ 0 such that one of the following alternatives hold

Then, problem (27) has a solution.
Similarly, we can consider the boundary conditions y(a) = 0 = x(b), as well, thus obtaining the corresponding existence result. Finally, Carathéodory conditions on φ and ψ could be assumed. We avoid entering into details, for briefness.

The definition of lower and upper solutions
Let us consider the boundary value problem where f, g : [a, b]×R 2 → R are continuous functions, and recall the definitions introduced in [14].
We say that (α, β) is a well-ordered pair of lower/upper solutions of problem (28), if α and β are respectively a lower and an upper solution for problem (28) and they satisfy On the other hand, if the above inequality does not hold, we say that the pair (α, β) is non-well-ordered.

Proof of Theorem 2
Define This can be obtained as an immediate consequence of the next two lemmas.

Lemma 8.
For any L > 0 and y 0 > 0 we can define two continuously differ- and such that (29)  Proof. Since both φ and ψ are bounded, let us consider K > 0 as in (18). From Lemma 4 we can find y 1 ≥ y 0 such that Since H is C 1 , we can find a positive constant c 1 such that Now, since 2 − q < 1, we can find M > 1 sufficiently large so to have We define γ + 1,2 : [−L, L] → R as then, since from (36) and (33) the denominator in (37) is positive, we get both for every t ∈ [a, b] and x ∈ [−L, L] and the lemma will be proved. Hence, we need to prove the validity of (37). Using equality (15) and the estimate (34), we have that Moreover, using equality (16) and recalling (33), we have that From (38) and (39), for every x ∈ [−L, L] and v ∈ [y 1 , y 1 + 2L] .
Since, using (36), for all Hence, recalling (33) and (36), we have where the last estimate is given by (35). We have thus proved (37), and so the proof of the lemma is completed.

Lemma 9.
For any L > 0 and y 0 > 0 we can define two continuously differ- The proof of this lemma is analogous to the previous one, so we omit it, for briefness.
Let us now conclude the proof of Theorem 2. Choosing we can apply Lemmas 8 and 9 to get the existence of the needed curves γ ± i . Moreover, for every t ∈ [a, b], . This is an immediate consequence of Lemma 4, choosing a larger value of y 0 , if necessary. Then, [14,Theorem 11] applies, thus completing the proof of Theorem 2.

Proof of Theorem 3
As a first step, we need the following two lemmas, where the ordering relation defined in Introduction is used.
Let us introduce a C ∞ -smooth cut-off function χ : R → [0, 1] such that Vol. 24 (2022) On the Dirichlet problem associated Page 13 of 32 66 and, for every d ≥ 1, consider the modified problem where Notice that (2) and (41)  Proof. Assume by contradiction that there exist a diverging sequence (d n ) n and some solutions u n = (x n , y n ) of (41), with d = d n , such that α / x n , x n / β and N p (u n ) > n. We introduce the functions v n = x n N p (u n ) and w n = y n N p (u n ) p−1 .
In particular, N p (v n , w n ) = 1. Notice that, from (11) and (10), we have v n ∞ ≤ S and w n ∞ ≤ S .
By (12) and (13), we see that (v n , w n ) solves where Then, by (42), since (v n , w n ) solves (43), it is bounded in C 1 × C 1 . By a standard compactness argument, a subsequence converges to some (v, w) in Hence, we have either (v(t), w(t)) = S(t), or (v(t), w(t)) = S(t + τ + ). More precisely, using (8), the first case is possible if and only if b − a = τ + , the second one if and only if b−a = τ − . In the first case, since v n C 1 -converges to S 1 0, we have that v n 1 2 S 1 when n is sufficiently large, so that, recalling Lemma 12, we get the contradiction In the second case, since v n C 1 -converges to S 1 = S 1 ( · + τ + ) 0, we have that v n 1 2 S 1 when n is sufficiently large, thus providing the contradiction The proof is thus completed.
We now fix D > 1 as in Lemma 13, assuming also From Lemma 13, if u = (x, y) is any solution of (41) with d ≥ D, satisfying both α / x and x / β, then u is a solution of (2), too.

Lemma 14.
The functions α and β are a lower and an upper solution of (41), respectively, provided that d is chosen large enough.
Proof. From Lemma 4 and the boundedness of φ, we deduce the existence of a constant K > 0 such that, for every d ≥ 1, we have both φ d (t, α(t), y) ≥ − K when y > 0, and φ d (t, α(t), y) ≤ K when y < 0, for every t ∈ [a, b]. Moreover, using again Lemma 4, we can take d large enough so that, if y > d p/q S then On the other hand, if y ∈ [−d p/q S, d p/q S], then (α(t), y) belongs to the region where the problem has not been modified and the desired estimates hold. The inequality for y α (t) still holds since, for every t ∈ [a, b], also the point (α(t), y α (t)) belongs to the region where the problem has not been modified.
Vol. 24 (2022) On the Dirichlet problem associated Page 15 of 32 66 Our aim now is to construct a lower solution α and an upper solution β such that α β and α β. To do so, let us consider, for every d > 1, the slowed autonomous system It is isochronous, all solutions being periodic with minimal period τ d d−1 > τ.
Proof. If we parametrize the solutions (x, y) with the energy E = d−1 d H(x, y), we see that they cross the line L = {(x, y) ∈ R 2 : x = ξ} if and only if E is greater than a well determined value E ξ > 0, and assumption (9) ensures that there are exactly two crossing points (ξ,ȳ + ξ ) and (ξ, , the solution (x, y) we are looking for cannot follow the path to the left of L, hence, if it exists, it has to be x ξ (t) > ξ for every t ∈ ]a, b[ . Let us denote by T ξ (E) the time needed to go from (ξ, y + ξ ) to (ξ, y − ξ ). We have thus defined a function T ξ : ]E ξ , +∞[ → R which is continuous, positive and strictly increasing.
there is a unique value E ξ > E ξ for which T ξ (E ξ ) = b − a, and this value of the energy determines the solution we are looking for.
We have thus proved the first part of the lemma; the proof of the second part is similar.
We are now ready to define the lower and upper solutions α and β. [−2dS, 2dS], since ξ > 2dS, using assumption (9), we have Moreover, since equality holds. Finally, x ξ (a) ≥ 0 and x ξ (b) ≥ 0, thus proving that x ξ is an upper solution. Analogously one proves that x −ξ is a lower solution. Now, since from (44) we have ξ > 2dS > max{ α ∞ , β ∞ }, one has thus ending the proof of the lemma.

Letā = min α andb = max β. By Lemma 4, we can find
We now need to introduce the guiding curves.

Lemma 17.
There are four continuously differentiable functions γ ± i : [ā,b] → R, with i = 1, 2, satisfying and such that Proof. It follows the lines of the proofs of Lemmas 8 and 9.
Let us now introduce our functional setting for the problem (41). Set I = [a, b], denote by C 0, (I) the space of -Hölder continuous functions and by C 1, (I) the space of functions having derivative belonging to C 0, (I). Moreover, define Vol. 24 (2022) On the Dirichlet problem associated Page 17 of 32 66 We consider the linear operator L : C 1, 0 (I) × C 1, (I) → C 0, (I) × C 0, (I) L(x, y) = (x − y , y ) , and the Nemitskii operator y(t)) .

Lemma 18. The operator L is invertible with continuous inverse, and problem (41) is equivalent to
Proof. It is rather standard, using the fact that C 1, (I) is compactly imbedded in C 1 (I).
Let us define the sets where d LS denotes the Leray-Schauder degree. By excision, this definition does not depend on the choice of ρ.
Our aim is to prove that, if there are no solutions of (41) on ∂V j , then V j is admissible and This fact will be proved in Lemma 22. In the following we denote by (ϕ, η) any of the three pairs ( α, β) , (α, β) , ( α, β) .
and consider the problem which is equivalent to Lu = N ϕ,η u, with the appropriate Nemitskii operator.
Notice that the functions f ϕ,η and g ϕ,η are bounded. Moreover, if (x, y) is a solution of (51) satisfying ϕ ≤ x ≤ η and −Y < y < Y , then it is a solution of (41), too.
Since −Y < y α (t) < Y for every t ∈ [a, b] from the choice (48), the inequality for y α (t) still holds, since (α(t), y α (t)) belongs to the region where the problem has not been modified. All the other cases can be treated similarly.

Lemma 20. Every solution
Proof. Let us define the following regions As in [12] and [14], one can verify that, if u = (x, y) is a solution of η (t, x, y) , Vol. 24 (2022) On the Dirichlet problem associated Page 19 of 32 66 Figure 2. A section at a fixed time of the regions introduced in Lemmas 20 and 21, describing the dynamics of (51) then, for any t 0 ∈ [a, b], Moreover, for any t 0 ∈ [a, b], if x(t 0 ) < ϕ(t 0 ) and y(t 0 ) = y ϕ (t 0 ) , then there exists δ > 0 such that Similarly, if, for any t 0 ∈ [a, b], x(t 0 ) > η(t 0 ) and y(t 0 ) = y η (t 0 ) , then there exists δ > 0 such that Indeed, by contradiction, let u = (x, y) be a solution of (51) such that , then t 0 ∈ ]a, b[ and by the above considerations it cannot be that (t 0 , u(t 0 )) ∈ A NW ∪ A SW . Hence, y(t 0 ) = y ϕ (t 0 ), and there exists δ > 0 such that (t, u(t)) ∈ A NW for t ∈ ]t 0 − δ, t 0 [ , which leads to a contradiction. The case x(t 0 ) > η(t 0 ) leads to a similar contradiction, as well.
We now prove that y(t) < γ + (x(t)) for every t ∈ [a, b]. Assume by contradiction that there is t 0 ∈ ]a, b[ such that y(t 0 ) ≥ γ + (x(t 0 )). We distinguish two possibilities. First, if x(t 0 ) ≥ 0, then the solution remains above γ + 1 for all t ∈ ]t 0 , b]; hence, y(t) > M and x (t) > 0 for all t ∈ ]t 0 , b], leading to But then the solution remains below γ + 2 for all t ∈ [t 1 , b], in contradiction with the assumption.

Lemma 22.
If there are no solutions of (41) on ∂V j , then V j is admissible and Proof. For any sufficiently large ρ > 0, denoting by B ρ the open ball in X with radius ρ, centered at the origin, we claim that Indeed, let us show that there is a ρ > 0 such that, for every λ ∈ [0, 1], every solution of Lu = λN ϕ,η u satisfies u C 1 < ρ. By contradiction, if this is not true, there exist a sequence (λ n ) n in [0, 1] and some solutions u n = (x n , y n ) of Lu n = λ n N ϕ,η u n such that u n C 1 → ∞. Let v n = x n / u n C 1 and w n = y n / u n C 1 . By a standard argument it can be seen that, for a subsequence, λ n →λ ∈ [0, 1], while (v n , w n ) → (v,w) in X. Moreover,v =w,w =λv, so thatv =λv, and sincev(a) = 0 =v(b), this impliesv ≡ 0, hence alsō w ≡ 0, a contradiction. By homotopy invariance, the degree is then equal to 1. Fix ρ > 0 as above. Since there are no solutions of (41) on ∂V(ϕ, η), we also have that there are no solutions of (51) on ∂V(ϕ, η). Hence, V(ϕ, η) is admissible and, by excision, V(ϕ, η), the result follows. Proof. We recall that in Lemma 16 we provided the definition of the functions α = x −ξ and β = x ξ with the choice ξ > 2dS. Let u = (x, y) be a solution belonging to the closure of V 1 . Then α(t) ≤ x(t) ≤ β(t), for every t ∈ I. Assume by contradiction that x β. Since x(a) = 0 < ξ = β(a) and Hence, as ξ > 2dS, both (x(t), y(t)) and (x ξ (t), y ξ (t)) solve the autonomous system (45) in a neighborhood of t 0 . Since , by assumption (9) it has to be that y(t 0 ) = y ξ (t 0 ). Since autonomous planar Hamiltonian systems have the uniqueness property for Cauchy problems when the initial value is not an equilibrium (cf. [19,Theorem 1]), then the two solutions (x(t), y(t)) and (x ξ (t), y ξ (t)) coincide, as long as they remain in [ξ, +∞[ ×R, leading to a contradiction. Hence, x β. Similarly, one proves that α x. So, there are no solutions of (41) on ∂V 1 . Now, if there is a solution u = (x, y) of (41) on ∂V 2 , then x ≤ β and x β. Since α ≤ β, there is a t 0 such that x(t 0 ) ≤ β(t 0 ) < α(t 0 ), implying that α x. So, u is the solution of (28) we are looking for. A similar argument shows that if u = (x, y) is a solution of (41) on ∂V 3 , then u is the solution of (28) we are looking for.
Finally, if there are no solutions of (41) on ∂V 2 ∪ ∂V 3 , then Then, there is a solution of (41) in V 1 \V 2 ∪ V 3 , and this is the solution of (28) we are looking for. The proof of Theorem 3 is thus completed.

Higher dimensional systems
Let us now introduce a higher dimensional version of problem (2). We will write x = (x 1 , . . . , x N ) ∈ R N , y = (y 1 , . . . , y N ) ∈ R N , and assume that the continuously differentiable function H : R 2N → R is of the type Moreover, for every n ∈ {1, . . . , N}, we assume that there exist p n > 1 and q n > 1, with (1/p n ) + (1/q n ) = 1, such that for every λ > 0 and (u, v) = (0, 0). We consider the problem We now adapt the definition of lower/upper solutions given in [14, Definition 31] to the higher dimensional problem and there exist two C 1 -functions y α , y β : [a, b] → R N such that, for every t ∈ [a, b], x, y ∈ R N with x ∈ α(t) , β(t) , and n ∈ {1, . . . , N}, one has f n (t, x, y) < α n (t) when x n = α n (t) and y n < y α n (t) , f n (t, x, y) > α n (t) when x n = α n (t) and y n > y α n (t) , f n (t, x, y) < β n (t) when x n = β n (t) and y n < y β n (t) , f n (t, x, y) > β n (t) when x n = β n (t) and y n > y β n (t) , Vol. 24 (2022) On the Dirichlet problem associated Page 23 of 32 66 and (y α n ) (t) ≥ g n (t, x, y) when x n = α n (t) and y n = y α n (t) , (y β n ) (t) ≤ g n (t, x, y) when x n = β n (t) and y n = y β n (t) .
Here is our result in the well-ordered case. For the non-well-ordered case, we need to introduce the notion of strict lower and upper solutions. To this aim we will follow the ideas developed in [11], and distinguish the components which are well ordered from the others. can be decomposed as x = (x J , x K ) where x J = (x j ) j∈J ∈ R #J and x K = (x k ) k∈K ∈ R #K . Similarly, we can decompose every function F : D → R N as In the following definition, we will use the relation introduced in (7). Definition 27. Let (α, β) be a pair of lower/upper solutions of (54) related to the partition (J , K). We will say that this pair of lower/upper solutions is strict with respect to the jth component, with j ∈ J , if α j β j and, for every solution (x, y) of (54), We will say that this pair of lower/upper solutions is strict with respect to the kth component, with k ∈ K if, for every solution (x, y) of (54), To prove the existence of a solution of (53), once a pair of lower/upper solutions (α, β) is given, we need to ask the strictness property with respect to the non-well-ordered components α k , β k . Moreover, we will need to ask more regularity on the functions φ and ψ: we will ask them to be locally -Hölder continuous for a certain > 0. Here is our result.
Theorem 28. Let > 0 and assume H : R 2N → R to be as in (52), with components H n being C 1, loc -smooth positively-(p n , q n )-homogeneous positivedefinite functions, for some p n > 1 and q n > 1 with (1/p n ) + (1/q n ) = 1. Let (α, β) be a pair of lower/upper solutions of (53) related to the partition (J , K) of {1, . . . , N} which is strict with respect to the kth component, for every k ∈ K. Moreover, assume that for every k ∈ K, ∂H k ∂y (x 0 , ·) is a strictly increasing function, for every x 0 ∈ R.
Let φ, ψ be uniformly bounded C 0, loc -smooth functions. If b − a ≤ min{τ + k , τ − k : k ∈ K} , then (53) has a solution (x, y) with the following properties: As in Section 2, for every n ∈ {1, . . . , N}, let S n (t) = (S 1,n (t), S 2,n (t)) be the periodic solution of the planar autonomous system such that S 1,n (0) = 0 and S 2,n (0) > 0, with H n (S 1,n (t), S 2,n (t)) = 1 for every t. We set For each n ∈ {1, . . . , N}, we will mainly follow the procedure developed in Section 5 if n ∈ J , and the one in Section 6 if n ∈ K, so that the couple of variables (x n , y n ) will overshadow the remaining ones, which will essentially act as parameters. We first need to modify problem (53) both in the kvariables, following the lines of the proof of Theorem 3, and in the j-variables, to apply a topological degree argument, as in the proof of [11,Theorem 10].
We need now to introduce the functional setting. We denote by f d and g d the functions  Fix any (ϕ μ , η μ ) ∈ Ξ. From (S1) and (S5), we see that there are no solutions of (61) on ∂V μ . Then, arguing as in the proof of Lemma 22, the set V μ is admissible and Then, we can follow the procedure in [ k ) α k x k β k , and there are t 1 k , t 2 k ∈ ]a, b[ such that x k (t 1 k ) < α k (t 1 k ) and x k (t 2 k ) > β k (t 2 k ) ; (A γ n ) γ − n (x n (t)) < y n (t) < γ + n (x n (t)), for every t ∈ [a, b] .  So, there exists a solution (x, y) of (61) belonging to Ω (4,4,...,4,4) . This solution satisfies (A 4 k ), for every k ∈ K, hence α k / x k and x k / β k , for every k ∈ K. Then, from (S2) we conclude that N p k (u k ) < D, so that (x, y) is indeed a solution of the original problem (53), since the differential equation defining the two problems coincide in the set {u ∈ R 2N : N p k (u k ) < D}. Moreover, from (62), we have α J (t) x J (t) β J (t) for every t ∈ [a, b]. The proof is thus completed.