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The aim of this paper is to establish the existence of solutions for singular double-phase problems depending on one parameter. This work improves and complements the existing ones in the literature. There seems to be no results on the existence of solutions for singular double-phase problems.

The study of various mathematical problems involving the double-phase operator has become very attractive in recent decades. The existence and multiplicity of solutions of double-phase Dirichlet problems has been studied by several authors (see, e.g., [

But up to now, to the best of our knowledge, no paper discussing the existence of solutions for singular double-phase problems via critical point theory can be found in the existing literature. In order to fill in this gap, we study double-phase problems from a more extensive viewpoint. More precisely, we are going to prove that problem

This paper is concerned with the existence of solutions to the following singular double-phase problem:

Problems of the above type arise for instance in nonlinear elasticity. The main reasons are to describe the behavior of Lavrentiev’s phenomenon; we refer to [

In the entire paper, we suppose the following assumptions:

there exists positive measurable subsets

there exists

there exists

The following function satisfies hypotheses

We are now in the position to state our main results. Firstly, problem

Assume that

Moreover, we also show that problem

there exists

there exists a positive measurable subset

The following functions satisfy hypotheses

Assume that

The rest of this paper is organized as follows. In Section

In order to discuss problem

Denote by

The Musielak-Orlicz space

([

Set

For

If

If

([

If

Assume that

By the above Proposition, there exists

In order to discuss the problem

We know that (see [

For any

Now, we are ready to prove Theorem

To complete the proof of the main result, we need to consider the following three steps.

We first show that for every

Let

Thus, we deduce that for any

By virtue of assumption

Since

We show that there exists

Let

Hence, for any

Since

The proof of Step

We show that there exists

Let

Recall that

Now, using Hölder’s inequality, we get that, as

Analogously,

Hence, by (

Moreover, using assumptions

The above information and Hölder’s inequality imply

Again, by Proposition

Thus, using the fact that

Hence, for every

So, we complete Step

Therefore, combining the above Steps

Now, we are ready to prove Theorem

To complete the proof of the main result, we need to consider the following three steps.

We first show that for every

Firstly, due to condition

Again, using the condition

Since

We show that there exists

Let

Hence, for any

Since

The proof of Step

We show that there exists

Let

Thus, as the proof of Step

The proof of Step

Therefore, combining the above Steps

Data sharing is not applicable to this article as no new data were created or analyzed in this study.

The authors declare that they have no competing interests.

The authors declare that the study was realized in collaboration with equal responsibility. All authors read and approved the final manuscript.

This work is supported by the National Key Research and Development Program of China (No. 2018YFC0310500), the Fundamental Research Funds for the Central Universities (No. 3072020CF2401), the Natural Science Foundation of Inner Mongolia (No. 2017MS0116), the National Natural Science Foundation of China (No. 11201095), the Postdoctoral Research Startup Foundation of Heilongjiang (No. LBH-Q14044), and the Science Research Funds for Overseas Returned Chinese Scholars of Heilongjiang Province (No. LC201502).