On the Solvability of a Boundary Value Problem for P-laplacian Differential Equations

Using barrier strip conditions, we study the existence of C 2 [0, 1]-solutions of the boundary value problem (φ p (x)) = f (t, x, x), x(0) = A, x (1) = B, where φ p (s) = s|s| p−2 , p > 2. The question of the existence of positive monotone solutions is also affected.


Introduction
This paper is devoted to the solvability of the boundary value problem (BVP) (φ p (x )) = f (t, x, x ), t ∈ [0, 1], (1.1) Here φ p (s) = s|s| p−2 , p > 2, the scalar function f (t, x, y) is defined for (t, x, y) ∈ [0, 1] × D x × D y , where the sets D x , D y ⊆ R may be bounded, and B ≥ 1. Besides, f is continuous on a suitable subset of its domain.
The solvability of various singular and nonsingular BVPs with p-Laplacian has been studied, for example, in [1-5, 7-12, 14].Conditions used in these works or do not allow the main nonlinearity to change sign, [2,11], or impose a growth restriction on it, [3,9,11], or require the existence of upper and lower solutions, [1,3,5,8,9,12]; other type conditions have been used in [7], where the main nonlinearity may changes its sign.As a rule, the obtained results guarantee the existence of positive solutions.
Another type of conditions have been used in [10] for studying the solvability of (1.1), (1.2) in the case p ∈ (1, 2).The existence of at least one positive and monotone C 2 [0, 1]-solution is established therein under the following barrier condition: H.There are constants L i , F i , i = 1, 2, and a sufficiently small σ > 0 such that where the constants m and M are, respectively, the minimum and the maximum of Let us recall, the strips are called "barrier" because they limit the values of the first derivatives of all C 2 [0, 1]-solution of (1.1), (1.2) between themselves.Recently, it was shown in [13] that conditions of form (1.3) and (1.4) guarantee C 1 [0, 1]-solutions to the φ-Laplacian equation with boundary conditions (1.2), where φ : R → R is an increasing homeomorphism and It turned out that the cases 1 < p < 2 and p > 2 require different technical approaches for the use of H for studying the solvability of (1.1), (1.2).So, in the present paper we show that H with the additional requirement guarantees the existence of at least one monotone, and positive in the case A ≥ 0, C 2 [0, 1]solution to (1.1), (1.2) with p > 2. In fact, our main result is the following.
The paper is organized as follows.In Section 2 we present preliminaries needed to formulate the Topological Transversality Theorem, which is our basic tool, and prove auxiliary results.In Section 3 we give the proof of Theorem 1.1, formulate a corollary and give an example.

Fixed point theorem, auxiliary results
Let K be a convex subset of a Banach space E and U ⊂ K be open in K. Let L ∂U (U, K) be the set of compact maps from U to K which are fixed point free on ∂U; here, as usual, U and ∂U are the closure of U and boundary of U in K.
A map F in L ∂U (U, K) is essential if every map G in L ∂U (U, K) such that G/∂U = F/∂U has a fixed point in U.It is clear, in particular, every essential map has a fixed point in U.
The following fixed point theorem due to A. Granas et al. [6].
Then H(x, λ), λ ∈ [0, 1], has at least one fixed point in U and in particular there is a x 0 ∈ U such that x 0 = F(x 0 ).
The following results is important for our consideration.It can be found also in [6].
Theorem 2.2.Let l ∈ U be fixed and F ∈ L ∂U (U, K) be the constant map F(x) = l for x ∈ U. Then F is essential.
Further, we need the following fact.
Proposition 2.3.Let the constants B and M be such that B ≥ 1 and B > M > 0. Then Proof.The inequality is evident for r = 1.For M ∈ (0, B) consider the function g In summary, we have proved that g (r) < 0 for each r ∈ [0, ∞).Then, the result follows from the fact that g(1) = 0.
Let us emphasize explicitly that we conduct the rest consideration of this section for an arbitrary fixed p > 2.
For λ ∈ [0, 1] consider the family of BVPs where For convenience set where F 1 , σ, m and M are as in H.
The next result gives a priori bounds for the C 2 [0, 1]-solutions of family (2.1 ) (as well as of (2.1)).
Proof.The proof of the bounds for x and x is the same as the corresponding part of the proof of [10, Lemma 3.1], but we will state it for completeness.So, assume on the contrary that is not true.Then, x (1) = B ≤ L 1 together with x ∈ C[0, 1] implies that is not empty.Moreover, there exists an interval [α, β] ⊂ S + with the property Then, by the fundamental theorem of calculus applied to x , (2.3) implies that there is a γ ∈ (α, β) such that x (γ) < 0.
To reach the bounds for x (t) from we obtain firstly Next, multiplying both sides of this inequality by λM ≥ 0 and λm ≤ 0, for t ∈ [0, 1] obtain respectively On the other hand, for all λ, t ∈ [0, 1], from where, keeping in mind that x (t) > 0 on [0, 1], we get which yields the required bounds for x (t).

Now, introduce sets
and, in case that H holds, Introduce also the map (2.4) , is well defined and continuous.
Proof.Clearly, because of (2.4), Integrating this inequality from 1 to t, t ∈ [0, 1), we get from where it follows and By (1.5) and Proposition 2.3, we have and then, 0 Further, introduce the sets and the map Φ p : K → C 1 + [0, 1] defined by Φ p x = φ p (x ).
Lemma 2.6.The map Φ p is well defined and continuous.

It is well known that the inverse function of φ
Using it, we introduce the map But, for y ∈ C 1 + [0, 1] we have y(t) > 0 on [0, 1] and so Lemma 2.7.The map Φ q : C 1 + [0, 1] → K is well defined, the inverse map of Φ p and continuous.
Proof.For each fixed y ∈ C 1 + [0, 1] we get a unique x(t) = (Φ q y)(t) = t 0 (y(s)) 1 p−1 ds + A. In fact, to establish the veracity of the first two assertions, we have to show that x ∈ K or, what is the same, to show that x is a unique C 2 [0, 1]-solution to the BVP The last follows immediately from x (t) = (y(t)) 1 p−1 = B and x(0) = A. Now, the continuity of y (t) and y(t) > 0 on [0, 1] imply that exists and is continuous on [0, 1].Thus, x(t) is a solution to (2.6) and is in To complete the proof we just have to observe that the continuity of Φ q follows from the continuity of y 1/(p−1) (t) on [0, 1].

Proof of main result
Proof of Theorem 1.1.We will prove the assertion for an arbitrary fixed p > 2. Introduce the set and consider the homotopy defined by H λ (x) := Φ q Λ λ j, where j : U → C 1 [0, 1] is the embedding jx = x.To show that all assumptions of Theorem 2.1 are fulfilled observe firstly that U is an open subset of K, and K is a convex subset of the Banach space C 2 [0, 1].For the fixed points of H λ , λ ∈ [0, 1], we have which is the operator form of the family Thus, the fixed points of H λ coincide with the C 2 [0, 1]-solutions of (3.1).But, it is obvious that each C 2 [0, 1]-solution of (3.1) is a C 2 [0, 1]-solution of (2.1).So, all conclusions of Lemma 2.4 are valid in particular and for the C 2 [0, 1]-solutions of (3.1) which allow us to conclude that the C 2 [0, 1]-solutions of (3.1) do not belong to ∂U and so the homotopy is fixed point free on ∂U.On the other hand, it is well known that j is completely continuous, that is, it maps each bounded set to a compact one.Thus, j(U) is a compact set.Besides, it is clear that j(U) ⊂ V.Then, according to Lemma 2.5, Λ λ (j(U)) ⊆ C 1 + [0, 1] is compact.Finally, the set Φ q (Λ λ (j(U)) ⊂ K is compact, by Lemma 2.7.So, the homotopy is compact.Now, since for x ∈ U we have Λ 0 j(x) = φ p (B) = B p−1 , the map H 0 maps each x ∈ U to the unique solution l = Bt + A ∈ K to the BVP x = B, t ∈ (0, 1),  √ x + 1 + 100 , t ∈ (0, 1), where p > 2 is fixed.