Solvability of boundary value problem with p-Laplacian at resonance

Jiang Boundary Value Problems 2014, 2014:36 http://www.boundaryvalueproblems.com/content/2014/1/36 RESEARCH Open Access Solvability of boundary value problem with p-Laplacian at resonance Weihua Jiang* * Correspondence: College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei 050018, P.R. China Abstract By generalizing the extension of the continuous theorem of Ge and Ren and constructing suitable Banach spaces and operators, we investigate the existence of solutions for p-Laplacian boundary value problems at resonance. An example is given to illustrate our results. MSC: 34B15 Keywords: continuous theorem; resonance; p-Laplacian; boundary value problem 1 Introduction In this paper, we will study the boundary value problem ⎧ ⎨(ϕ (u )) (t) = f (t, u(t), u (t), u (t)), p  ⎩u() = u () = , u () =  k(t)u (t) dt, (.) and ⎧ ⎨(ϕ (u )) (t) = f (t, u(t), u (t), u (t)), p  ⎩u () = , u () =  g(t)u (t) dt,  u () =  h(t)u (t) dt, (.)    where ϕp (s) = |s|p– s, p > ,  k(t) dt = ,  g(t) dt = ,  h(t) dt = . A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a non-trivial solution. Mawhin’s continuous theorem [] is an effective tool to solve this kind of problems when the differential operator is linear, see [–] and references cited therein. But it does not work for nonlinear cases such as boundary value problems with a p-Laplacian, which attracted the attention of mathematicians in recent years [–]. Ge and Ren extended Mawhin’s continuous theorem [] and many authors used their results to solve boundary value problems with a p-Laplacian, see [, ]. In this new theorem, two projectors P and Q must be constructed. But it is difficult to give the projector Q in many boundary value problems with a p-Laplacian. In this paper, we generalize the extension of the continuous theorem and show that the p-Laplacian problem is solvable when Q is not a projector. And we will use this new theorem to discuss problems (.) and (.), respectively. In this paper, we will always suppose that ©2014 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 2014, 2014:36 http://www.boundaryvalueproblems.com/content/2014/1/36 Page 2 of 12 (H ) k(t), g(t), h(t) ∈ L [, ] are nonnegative and k = g = h = , where k :=   |k(t)| dt. (H ) f (t, u, v, w) is continuous in [, ] × R . 2 Preliminaries Definition . [] Let X and Y be two Banach spaces with 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 . Let P : X → X be a projector and  ⊂ X an open and bounded set with the origin θ ∈ . Definition . Suppose Nλ :  → Y , λ ∈ [, ] is a continuous and bounded operator. Denote N by N . Let  λ = {x ∈  ∩ dom M : Mx = Nλ x}. Nλ is said to be M-quasi-compact in  if there exists a vector subspace Y of Y satisfying dim Y = dim X and two operators Q, R with Q : Y → Y , QY = Y , being continuous, bounded, and satisfying Q(I – Q) = , R :  × [, ] → X ∩ dom M 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 let  ⊂ X be an open and bounded nonempty set. Suppose M : X ∩ dom M → Y is a quasi-linear operator and that Nλ :  → Y , λ ∈ [, ] is M-quasi-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 . Proof The proof is similar to the one of Lemma . and Theorem . in []. We can easily get the following inequalities. Lemma . For any u, v ≥ , we have () ϕp (u + v) ≤ ϕp (u) + ϕp (v),  < p ≤ . () ϕp (u + v) ≤ p– (ϕp (u) + ϕp (v)), p ≥ . In the following, we will always suppose that q satisfies /p + /q = .  Jiang Boundary Value Problems 2014, 2014:36 http://www.boundaryvalueproblems.com/content/2014/1/36 Page 3 of 12 3 The existence of a solution for problem (1.1) Let X = C  [, ] with norm u = max{u∞ , u ∞ , u ∞ }, Y = C[, ] × C[, ] with norm (y , y ) = max{y ∞ , y ∞ }, where y∞ = maxt∈[,] |y(t)|. We know that (X, ·) and (Y ,  · ) are Banach spaces. Define operators M : X ∩ dom M → Y , Nλ : X → Y as follows:   λf (t, u(t), u (t), u (t)) , Nλ u =   (ϕp (u )) (t) , Mu = T(ϕp (u )) (t)  where Ty = c, y ∈ C[, ], c satisfying     k(t)   s y(r) – c dr ds dt = , ϕq  t   (.)  ∈ C [, ], u() = u () =  . dom M = u ∈ X | ϕp u  Lemma . For y ∈ C[, ], there is only one constant c ∈ R such that Ty = c with |c| ≤ y∞ and that T : C[, ] → R is continuous. Proof For y ∈ C[, ], let  s     F(c) = k(t)  y(r) – c dr ds dt. ϕq  t Obviously, F(c) is continuous and strictly decreasing in R. Take a = mint∈[,] y(t), b = maxt∈[,] y(t). It is easy to see that F(a) ≥ , F(b) ≤ . Thus, there exists a unique constant c ∈ [a, b] such that F(c) = , i.e. there is only one constant c ∈ R such that Ty = c with |c| ≤ y∞ .  For y , y ∈ C[, ], assume Ty = c , Ty = c . By k(t) ≥ ,  k(t) dt =  and ϕq being strictly increasing, we obtain, if c – c > maxt∈[,] (y (t) – y (t)), then   =   k(t)  k(t)  s ϕq   k(t)   y (r) – c + y (r) – y (r) – (c – c ) dr  ds dt  t   <     y (r) – c dr ds dt ϕq t   =  s  s y (r) – c dr ds dt = . ϕq  t This is a contradiction. On the other hand, if c – c < mint∈[,] (y (t) – y (t)), then = k(t)  y (r) – c + y (r) – y (r) – (c – c ) dr  s y (r) – c dr ds dt = . ϕq t     k(t)   s ϕq t   >    k(t)  y (r) – c dr ds dt ϕq t   =  s       ds dt Jiang Boundary Value Problems 2014, 2014:36 http://www.boundaryvalueproblems.com/content/2014/1/36 Page 4 of 12 This is a contradiction, too. So, we have mint∈[,] (y (t) – y (t)) ≤ c – c ≤ maxt∈[,] (y (t) – y (t)), i.e. |c – c | ≤ y – y ∞ . So, T : C[, ] → R is continuous. The proof is completed.  It is clear that u ∈ dom M is a solution if and only (...truncated)