Solvability for p-Laplacian boundary value problem at resonance on the half-line

Jiang Boundary Value Problems 2013, 2013:207 http://www.boundaryvalueproblems.com/content/2013/1/207 RESEARCH Open Access Solvability for p-Laplacian boundary value problem at resonance on the half-line Weihua Jiang* * Correspondence: College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei 050018, P.R. China Abstract The existence of solutions for p-Laplacian boundary value problem at resonance on the half-line is investigated. Our analysis relies on constructing the suitable Banach space, defining appropriate operators and using the extension of Mawhin’s continuation theorem. An example is given to illustrate our main result. MSC: 70K30; 34B10; 34B15 Keywords: p-Laplacian; resonance; half-line; multi-point boundary value problem; continuation theorem 1 Introduction A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a non-trivial solution. Resonance problems can be expressed as an abstract equation Lx = Nx, where L is a noninvertible operator. When L is linear, Mawhin’s continuation theorem [] is an effective tool in finding solutions for these problems, see [–] and references cited therein. But it does not work when L is nonlinear, for instance, p-Laplacian operator. In order to solve this problem, Ge and Ren [] proved a continuation theorem for the abstract equation Lx = Nx when L is a noninvertible nonlinear operator and used it to study the existence of solutions for the boundary value problems with a p-Laplacian: ⎧ ⎨(ϕ (u )) + f (t, u) = ,  < t < , p ⎩u() =  = G(u(η), u()), where ϕp (s) = |s|p– s, p > ,  < η < . ϕp (s) is nonlinear when p = . As far as the boundary value problems on unbounded domain are concerned, there are many excellent results, see [–] and references cited therein. To the best of our knowledge, there are few papers that study the p-Laplacian boundary value problem at resonance on the half-line. In this paper, we investigate the existence of solutions for the boundary value problem ⎧ ⎨(ϕ (u )) + f (t, u, u ) = , p ⎩u() = ,  < t < +∞,  ϕp (u (+∞)) = ni= αi ϕp (u (ξi )),  where αi > , i = , , . . . , n, (.) n i= αi = . © 2013 Jiang; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Jiang Boundary Value Problems 2013, 2013:207 http://www.boundaryvalueproblems.com/content/2013/1/207 Page 2 of 10 In order to obtain our main results, we always suppose that the following conditions hold.  (H )  < ξ < ξ < · · · < ξn < +∞, αi > , ni= αi = . (H ) f : [, +∞) × R → R is continuous, f (t, , ) = , t ∈ (, ∞) and for any r > , there exists a nonnegative function hr (t) ∈ L [, +∞) such that   f (t, x, y) ≤ hr (t), a.e. t ∈ [, +∞), x, y ∈ R, |x| ≤ r, |y| ≤ r. +t 2 Preliminaries For convenience, we introduce some notations and a theorem. For more details, see []. Definition . [] Let X and Y be two Banach spaces with the norms  · X ,  · Y , respectively. A continuous operator M : X ∩ dom M → Y is said to be quasi-linear if (i) Im M := M(X ∩ dom M) is a closed subset of Y , (ii) Ker M := {x ∈ X ∩ dom M : Mx = } is linearly homeomorphic to Rn , n < ∞, where dom M denote the domain of the operator M. Let X = Ker M and X be the complement space of X in X, then X = X ⊕ X . On the other hand, suppose that Y is a subspace of Y , and that Y is the complement of Y in Y , i.e., Y = Y ⊕ Y . Let P : X → X and Q : Y → Y be two projectors and  ⊂ X an open and bounded set with the origin θ ∈ . Definition . [] Suppose that Nλ :  → Y , λ ∈ [, ] is a continuous operator. Denote N by N . Let λ = {x ∈  : Mx = Nλ x}. Nλ is said to be M-compact in  if there exist a vector subspace Y of Y satisfying dim Y = dim X and an operator R :  × [, ] → X being continuous and compact such that for λ ∈ [, ], (a) (I – Q)Nλ () ⊂ Im M ⊂ (I – Q)Y , (b) QNλ x = θ , λ ∈ (, ) ⇔ QNx = θ , (c) R(·, ) is the zero operator and R(·, λ)| λ = (I – P)| λ , (d) M[P + R(·, λ)] = (I – Q)Nλ . Theorem . [] Let X and Y be two Banach spaces with the norms  · X ,  · Y , respectively, and  ⊂ X an open and bounded nonempty set. Suppose that M : X ∩ dom M → Y is a quasi-linear operator and Nλ :  → Y , λ ∈ [, ] M-compact. In addition, if the following conditions hold: (C ) Mx = Nλ x, ∀x ∈ ∂ ∩ dom M, λ ∈ (, ), (C ) deg{JQN,  ∩ Ker M, } = , then the abstract equation Mx = Nx has at least one solution in dom M ∩ , where N = N , J : Im Q → Ker M is a homeomorphism with J(θ ) = θ . Jiang Boundary Value Problems 2013, 2013:207 http://www.boundaryvalueproblems.com/content/2013/1/207 Page 3 of 10 3 Main result Let X = {u|u ∈ C  [, +∞), u() = , supt∈[,+∞) |u(t)| < +∞, limt→+∞ u (t) exists} with norm +t u ∞ , u ∞ }, where u∞ = supt∈[,+∞) |u(t)|. Y = L [, +∞) with norm u = max{ +t  +∞ y =  |y(t)| dt. Then (X,  · ) and (Y ,  ·  ) are Banach spaces. Define operators M : X ∩ dom M → Y and Nλ : X → Y as follows:   Mu = ϕp u   Nλ u = –λf t, u, u , , λ ∈ [, ], t ∈ [, +∞), where      dom M = u ∈ X ϕp u ∈ AC[, +∞), ϕp u  ϕp u (+∞) = n  αi ϕp u (ξi )  ∈ L [, +∞), . i= Then the boundary value problem (.) is equivalent to Mu = Nu. Obviously, n   αi Im M = yy ∈ Y , Ker M = {at|a ∈ R}, i= +∞ y(s) ds =  . ξi It is clear that Ker M is linearly homeomorphic to R, and Im M ⊂ Y is closed. So, M is a quasi-linear operator. Define P : X → X , Q : Y → Y as  +∞ i= αi ξi y(s) ds –t n e , –ξi i= αi e n  (Pu)(t) = u (+∞)t, (Qy)(t) = where X = Ker M, Y = Im Q = {be–t |b ∈ R}. We can easily obtain that P : X → X , Q : Y → Y are projectors. Set X = X ⊕ X , Y = Y ⊕ Y . Define an operator R : X × [, ] → X :  +∞ n     i= αi ξi f (r, u(r), u (r)) dr –s  n R(u, λ)(t) = ds ϕq λ f s, u(s), u (s) – e –ξi  τ i= αi e   + ϕp u (+∞) dτ – u (+∞)t, t  +∞ where p + q = , ϕq = ϕp– . By (H ) and (H ), we get that R : X × [, ] → X is continuous. |u ∈ V } and {u (t)|u ∈ V } are both equicontinuLemma . [] V ⊂ X is compact if { u(t) +t ous on any compact intervals of [, +∞) and equiconvergent at infinity. Lemma . R : X × [, ] → X is compact. Proof Let  ⊂ X be nonempty and bounded. There exists a constant r >  such that u ≤ r, u ∈ . It follows from (H ) that there exists a nonnegative function hr (t) ∈ L [, +∞) Jiang Boundary Value Problems 2013, 2013:207 http://www.boundaryvalueproblems.com/content/2013/1/207 such that    f t, u(t), u (t)  ≤ hr (t), a.e. t ∈ [, +∞), u ∈ . For any T > , t , t ∈ [, T], u ∈ , λ ∈ [, ], we have    R(u, λ)(t ) R(u, λ)(t )    –  +t  + t    +∞ n   +∞    t    i= αi ξi f (r, u(r), u (...truncated)