Positive solutions for anisotropic singular (p,q)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{(p,q)}$$\end{document}-equations

We consider a nonlinear elliptic Dirichlet problem driven by the anisotropic (p, q)-Laplacian and with a reaction which is nonparametric and has the combined effects of a singular and of a superlinear terms. Using variational tools together with truncation and comparison techniques, we show that the problem has at least two positive smooth solutions.


Introduction
Let Ω ⊆ R N be a bounded domain with a C 2 -boundary ∂Ω. In this paper, we study the following anisotropic singular (p, q)-equation (double phase problem) Given r ∈ C(Ω) with 1 < min Ω r, by Δ r(z) we denote the r(z)-Laplace differential operator defined by Δ r(z) = div |Du| r(z)−2 Du for all u ∈ W 1,r(z) 0 (Ω).
In Problem (1.1), we have the sum of two such operators (double phase problem). In the reaction (right-hand side of (1.1)), we have the competing effects of two different terms of different nature. One is the singular term u −η (z) , and the other term is a Carathéodory perturbation f (z, x) (that is, for all x ∈ R, z → f (z, x) is measurable and for a.a. z ∈ Ω, x → f (z, x) is continuous) which exhibits (p + − 1)superlinear growth as x → +∞ (here p + = max Ω p). We point out that problem (1.1) is nonparametric.
Our aim is to prove the existence and the multiplicity of positive solutions for problem (1.1).

Preliminaries-auxiliary results and hypotheses
Let C 0,1 (Ω) denote the space of Lipschitz continuous functions. If r ∈ C 0,1 (Ω), we set r − = min Ω r and r + = max Ω r. We introduce the sets We identify two elements in M (Ω) if they differ only on a set of zero Lebesgue measure.
We have that A comprehensive analysis of variable exponent Lebesgue and Sobolev spaces can be found in the book of Diening-Harjulehto-Hästo-Ruzicka [7].
Let A r(z) : W 1,r(z) 0 (Ω) → W −1,r (z) (Ω) be the nonlinear operator defined by The following proposition summarizes the main properties of this operator (see Gasiński  In addition to the variable exponent spaces, we will also use the Banach space C 1 0 (Ω) = {u ∈ C 1 (Ω) : u| ∂Ω = 0}. This is an ordered Banach space with positive cone C + = {u ∈ C 1 0 (Ω) : u(z) ≥ 0 for all z ∈ Ω}. This cone has a nonempty interior given by with n(·) being the outward unit normal on ∂Ω.
When X is a Banach space and ϕ ∈ C 1 (X, R), we set Also, we say that ϕ(·) satisfies the C-condition, if the following property holds: This is a compactness-type condition on the functional ϕ(·). In most cases of interest, the ambient space X is infinite dimensional and so it is not locally compact. So, the burden of compactness is passed on the functional ϕ(·). Using the C-condition, one can prove a deformation theorem from which follow the minimax theorems of critical point theory (see Papageorgiou-Rȃdulescu-Repovš [30,Chapter 5]).
uniformly for a.a. z ∈ Ω; (iv) there exist τ ∈ C(Ω), δ > 0 and ϑ > 0 such that (v) there exists ξ ϑ > 0 such that for a.a. z ∈ Ω, the function Remarks. Since we aim to find positive solutions and all the above hypotheses concern the positive semiaxis R + = [0, +∞), without any loss of generality, we may assume that However, this superlinearity is not expressed using the well-known Ambrosetti-Rabinowitz condition (the AR-condition for short, see Ambrosetti-Rabinowitz [1]). Instead, we employ hypothesis H 1 (iii) which is less restrictive and incorporates in our framework (p + − 1)-superlinear nonlinearities with "slower" growth near +∞. For example, consider the following function Then, this function satisfies hypotheses H 1 , but fails to satisfy the AR-condition. On account of hypotheses H 1 (i),(iv), we have We introduce the following truncation of the right-hand side of (2.2): with ϑ > 0 as in hypothesis H 1 (iv). Evidently, this is a Carathéodory function. Using k(·, ·) as the source term, we consider the following auxiliary Dirichlet problem: Proof. First we prove the existence of a positive solution. So, let K(z, x) = x 0 k(z, s) ds and consider the (Ω). Let u ∈ int C + and choose t ∈ (0, 1) small such that tu(z) ≤ ϑ for all z ∈ Ω. Then, using (2.3) we have Since 1 < τ − < q + < p + < r + , by choosing t ∈ (0, 1) even smaller if necessary, we have (Ω). We obtain Next in (2.6), we choose h = (u − ϑ) + ∈ W 1,p(z) 0 (Ω). Then, ≤ 0 (see hypothesis H 1 (iv)), We have proved that u ∈ [0, ϑ], u = 0.
We conclude that Next we show the uniqueness of this positive solution.

Positive solutions
In this section, we prove a multiplicity theorem for the positive solutions of problem (1.1).
To produce the first positive solution of (1.1), we use (2.7) and (2.9) to define the following truncation of the reaction in problem (1.1): This is a Carathéodory function. We set E(z, x) = x 0 e(z, s) ds and introduce the functional ψ : (Ω). From (2.9) it follows that ψ ∈ C 1 (W 1,p(z) 0 (Ω)) (see also Proposition 3]). Using this functional, we can now produce the first positive solution of (1.1).
To produce a second positive solution for problem (1.1), we introduce the following truncation of the reaction: This is a Carathéodory function. We set L(z, x) = x 0 l(z, s) ds and consider the functional ϕ : As before, on account of (2.9), we have that ϕ ∈ C 1 (W Proof. From the proof of Proposition 3.1, we know that u 0 ∈ int C + is a minimizer of ψ(·) and u 0 (z) < ϑ for all z ∈ Ω, From (3.8) and (3.9), it follows that (Ω) − minimizer of ϕ (see Fan [8], Gasiński-Papageorgiou [12] and Tan-Fang [42]).