Existence Results for a Mixed Boundary Value Problem

In the present paper, we obtain an existence result for a class of mixed boundary value problems for second-order differential equations. A critical point theorem is used, in order to prove the existence of a precise open interval of positive eigenvalues λ, for which the considered problem admits at least one non-trivial classical solution u λ. It is proved that the norm of u λ tends to zero as λ → 0.


Introduction
The aim of this paper is to study the following mixed boundary value problem where p ∈ C 1 ([a, b]) and q ∈ C 0 ([a, b]) are positive functions, λ is a positive parameter, where a 1 , a 2 ≥ 0 and r ∈ ]1, +∞[ , and g : R → R is a Lipschitz continuous function with the Lipschitz constant L > 0, i.e., |g(t 1 ) for every t 1 , t 2 ∈ R, and g(0) = 0. Our goal here is to obtain some sufficient conditions which imply that the problem (1.1) has at least one classical solution (see Theorem 3.1).We use the variational method and a critical point theorem.
Motivated by the fact that such kind of problems are used to describe a large class of physical phenomena, many authors looked for existence and multiplicity of solutions for A. Hadjian and S. Tersian second-order ordinary differential nonlinear equations, with mixed conditions at the ends.We cite the papers [1-3, 6-10, 14].For instance, in [7], Bonanno and Tornatore established the existence of infinitely many weak solutions for the mixed boundary value problem q(x) ≥ 0, Carathéodory function and λ is a positive parameter.
We also refer the reader to the paper [11] in which, by means of an abstract critical points result of Ricceri [13], existence of at least three solutions for the following two-point boundary value problem where λ and µ are positive parameters, f : The paper is organized as follows.In Section 2 we introduce our abstract framework and we give some notations.In Section 3 we prove the main result (Theorem 3.1), while Section 4 is devoted to some consequences and remarks on the results of the paper.Here we give an application of the results (Example 4.7).

Preliminaries
In order to prove our main result, that is Theorem 3.1, we report here the result obtained in [5] (see [5,Theorem 3.1 and Remark 3.3]).Theorem 2.1.Let X be a reflexive real Banach space, let Φ, Ψ : X → R be two Gâteaux differentiable functionals such that Φ is strongly continuous, sequentially weakly lower semicontinuous and coercive in X and Ψ is sequentially weakly upper semicontinuous in X.Let I λ be the functional defined as I λ := Φ − λΨ, λ ∈ R, and for any r > inf X Φ let ϕ be the function defined as Then, for any r > inf X Φ and any λ ∈ ]0, 1/ϕ(r)[, the restriction of the functional I λ to Φ −1 (] − ∞, r[) admits a global minimum, which is a critical point (precisely a local minimum) of I λ in X.
the usual norm in X is defined by For every u, v ∈ X, we define Clearly, (2.2) defines an inner product on X whose corresponding norm is Then, it is easy to see that the norm • on X is equivalent to • X .In fact, put q(x) > 0, m := min{p 0 , q 0 } > 0, and q(x), M := max{p 1 , q 1 }.
Then, we have In the following, we will use • instead of • X on X.Note that X is a reflexive real Banach space.By standard regularity results, since f is a continuous function, It is well known that the embedding for all u ∈ X (see, e.g., [15]).Fixing r ∈ [1, +∞[ , from the Sobolev embedding theorem, there exists a positive constant c r such that and, in particular, the embedding Suppose that the Lipschitz constant L > 0 of the function g satisfies (2.5) Consider the energy functional I λ : X → R associated to (1.1) defined as follows where Lemma 2.2.Let the functional Φ be defined as above.Then we have the following estimates for every u ∈ X: Proof.Since g is Lipschitz continuous and satisfies g(0) = 0, we have and so, for every u ∈ X, and thus (2.6) follows.
On the other hand, the inequality for every u ∈ X.Therefore, we deduce (2.7).The proof is complete.
By the condition (2.5) and Lemma 2.2 we deduce that Φ is coercive.By standard arguments, we have that Φ is Gâteaux differentiable and sequentially weakly lower semicontinuous and its Gâteaux derivative is the functional Φ (u) ∈ X * , given by for every v ∈ X.Furthermore, the differential Φ : X → X * is a Lipschitzian operator.Indeed, for any u, v ∈ X, there holds Recalling that g is Lipschitz continuous and the embedding In particular, we derive that Φ is continuously differentiable.On the other hand, the fact that X is compactly embedded into C 0 ([a, b]) implies that the functional Ψ is well defined, continuously Gâteaux differentiable and with compact derivative, whose Gâteaux derivative is given by for every v ∈ X. Hence Ψ is sequentially weakly (upper) continuous (see [16,Corollary 41.9]).
Fixing the real parameter λ, Hence, the critical points of I λ are exactly the weak (classical) solutions of (1.1).
In conclusion, we cite a recent monograph by Kristály, Rȃdulescu and Varga [12] as a general reference on variational methods adopted here.

Put
The main result in this paper is the following.inf x∈B F(x, t) t 2 = +∞, and lim inf Then, there exists a positive number λ given by , such that, for every λ ∈ ]0, λ [ , problem (1.1) admits at least one non-trivial classical solution u λ ∈ X.Moreover, lim and the function λ → I λ (u λ ) is negative and strictly decreasing in ]0, λ [.
Proof.We prove the result for the case q 0 < p 0 /(b − a) 2 .The proof for the case q 0 ≥ p 0 /(b − a) 2 is similar.Fix λ ∈ ]0, λ [ .Our aim is to apply Theorem 2.1 with the Sobolev space X and the functionals Φ and Ψ introduced in Section 2. As given in Section 2, Φ and Ψ satisfy the regularity assumptions of Theorem 2.1.Clearly, inf u∈X Φ(u) = 0. Owing to (f 1 ), one has that for any (x, ξ) ∈ [a, b] × R. Since 0 < λ < λ , there exists γ > 0 such that Now, set r ∈ ]0, +∞[ and consider the function Taking into account (3.1) it follows that Then, due to (2.7), we get for every u ∈ X such that Φ(u) < ρ.Now, from (2.4) and by using (3.3), one has In particular, we deduce that (3.4) At this point, observe that In conclusion, bearing in mind (3.2), the above inequality together with (3.4) yields In other words, and, in particular, u λ is a global minimum of the restriction of Now, we have to show that for any λ ∈ ]0, λ [ the solution u λ is not the trivial zero function.If f (•, 0) = 0, then it easily follows that u λ ≡ 0 in X, since the trivial function does not solve problem (1.1).
Let us consider the case when f (•, 0) = 0 and let us fix λ ∈ ]0, λ [ .We will prove that the function u λ cannot be trivial in X.To this end, let us show that lim sup For this, due to (f 2 ), we can fix a sequence {ξ n } ⊂ R + converging to zero and two constants σ and κ (with σ > 0) such that lim Finally, fix M > 0 and consider a real positive number η with Then, there is ν ∈ N such that ξ n < σ and for every n > ν.Now, for every n > ν, by the properties of the function v (that is, 0 ≤ ξ n v(x) < σ for n sufficiently large), one has Since M could be taken arbitrarily large, it follows that from which (3.5) clearly follows.Hence, there exists a sequence {w n } ⊂ X strongly converging to zero, such that, for every n sufficiently large, Since u λ is a global minimum of the restriction of so that u λ is not trivial in X.Moreover, from (3.6) we easily see that the map is negative.Now, we claim that lim Indeed, bearing in mind that Φ is a coercive functional and that As a consequence, using the growth condition (f 1 ) together with the property (2.4), it follows that for every λ ∈ ]0, λ γ[ .Hence, from (2.3), (3.7) and (3.8), it follows that for every λ ∈ ]0, λ γ[ .Letting λ → 0 + , we get lim λ→0 + u λ = 0, as claimed.Finally, we have to show that the map λ → I λ (u λ ) is strictly decreasing in ]0, λ [ .For this, we observe that for any u ∈ X, one has Now, let us fix 0 < λ 1 < λ 2 < λ γ and let u λ i be the global minimum of the functional Clearly, the negativity of the map λ → I λ (u λ ) in ]0, λ [ together with (3.9) and the positivity of λ imply that thanks to 0 < λ 1 < λ 2 and Φ ≥ 0 by Lemma 2.2.Then, by (3.9)-(3.11)and again by the fact that 0 < λ 1 < λ 2 , we get that so that the map λ → I λ (u λ ) is strictly decreasing in ]0, λ [ , which completes the proof.
Remark 3.2.Theorem 3.1 can be also obtained applying Theorem 2.3 of [4], which directly ensures that the local minimum is non-zero.

Additional results and comments
In this section we give some consequences, remarks and an example.
Remark 4.1.By direct computation, it follows that the parameter λ in Theorem 3.1 can be expressed as where .
Remark 4.3.We observe that if f (x, 0) = 0 for a.e.x ∈ [a, b], Theorem 3.1 is a bifurcation result.Indeed, in this setting, it follows that the trivial solution solves problem (1.1) for every parameter λ.Hence, λ = 0 is a bifurcation point for problem (1.1), in the sense that the pair (0, 0) belongs to the closure of the set in the space X × R. Indeed, by Theorem 3.1 we have that Hence, there exist two sequences {u j } j∈N in X and {λ j } j∈N in R + (here u j := u λ j ) such that λ j → 0 + and u j → 0, as j → +∞.
where Then, there exists a positive number λ given by , such that, for every λ ∈ ]0, λ [, problem (1.1) admits at least one nontrivial classical solution u λ ∈ X.Moreover, lim and the function λ → I λ (u λ ) is negative and strictly decreasing in ]0, λ [ .
We state an example on the following special case of our results.
Theorem 4.5.Let f : R → R be a continuous function such that f (0) = 0, and for some 0 ≤ s < +∞.Then, there exists λ > 0 such that, for every λ ∈ ]0, λ [, the following autonomous mixed problem admits at least one nontrivial classical solution u λ ∈ X.Moreover, and the mapping Then, condition (4.Remark 4.8.We point out that the energy functional I λ associated to problem (4.3) is unbounded from below.Indeed, fix u ∈ X \ {0} and let τ ∈ R. We have as τ → +∞, bearing in mind that h < 2 < l.Hence, since the functional I λ is not coercive, the classical direct method result cannot be applied to the case treated in Example 4.7.

Theorem 3 . 1 .
Let f : [a, b] × R → R be a continuous function satisfying condition (f 1 ).In addition, if f (x, 0) = 0 for a.e.x ∈ [a, b], assume also that (f 2 ) there exist a non-empty open set D ⊆ ]a, b[ and a set B ⊆ D of positive Lebesgue measure such that lim sup t→0 + ξ) ≥ κξ 2 , for every ξ ∈ [0, σ].Now, fix a set C ⊂ B of positive measure and a function v

1 . 4 . 4 .
1 are different, thanks to the fact that the map ]0, λ [ λ → I λ (u λ ) is strictly decreasing.The next result is an immediate consequence of Remark 4.Corollary Let f : [a, b] × R → R be a continuous function with f (x, 0) = 0 for a.e.x ∈ [a, b], satisfying the following subcritical growth condition

Remark 4 . 9 .
We note that, applying Theorem 2.1, we have the relevant result of Theorem 3.1 for the following mixed boundary value problem with a complete equation−( pu ) + nu + qu = λ f (x, u) + g(u) in ]a, b[ , u(a) = u (b) = 0,(4.4)wherep ∈ C 1 ([a, b]) and q, n ∈ C 0 ([a, b]) such that p and q are positive functions, λ is a positive parameter, f :[a, b] × R → R is a continuous function such that | f (x, t)| ≤ e R(x) a 1 + a 2 |t| r−1 , a.e.x ∈ [a, b], t ∈ R,where a 1 , a 2 ≥ 0 and r ∈ ]1, +∞[ and R is a primitive of n/ p, while g : R → R is a Lipschitz continuous function with the Lipschitz constant L > 0 satisfyingL < max min x∈[a,b] e −R(x) q(x), min x∈[a,b] e −R(x) p(x) (b − a) 2 ,and g(0) = 0.In fact, since the solutions of problem (4.4) are solutions of the problem−(e −R pu ) + e −R qu = (λ f (x, u) + g(u)) e −R in ]a, b[ , u(a) = u (b) = 0,we can state and prove a result for problem (4.4) similar to Theorem 3.1.