^{1}

^{2}

^{2}

^{1}

^{2}

By establishing the corresponding variational framework and using the mountain pass theorem, linking theorem, and Clark theorem in critical point theory, we give the existence of multiple solutions for a fractional difference boundary value problem with parameter. Under some suitable assumptions, we obtain some results which ensure the existence of a well precise interval of parameter for which the problem admits multiple solutions. Some examples are presented to illustrate the main results.

Variational methods for dealing with difference equations have appeared as early as 1985 in [

Recently, fractional differential and difference “operators” are found themselves in concrete applications, and hence attention has to be paid to associated fractional difference and differential equations under various boundary or side conditions. For example, a recent paper by Atici and Eloe [

In order to handle the existence problem for fractional BVPs, various methods (among which are some standard fixed-point theorems) can be used. In this paper, however, we show that variational methods can also be applied. A good reason for picking such an approach is that, in Atici and Şengül [

More specifically, in this paper, we are interested in the existence of multiple solutions for the following

By establishing the corresponding variational framework and using critical point theory, we will establish various existence results (which naturally depend on

For convenience, throughout this paper, we arrange

We first collect some basic lemmas for manipulating discrete fractional operators. These and other related results can be found in [

First, for any integer

The

Let

For

A real symmetric matrix

Let

Let

Recall that, in the finite dimensional setting, it is well known that a coercive functional satisfies the (PS) condition.

Let

Let

there are constants

there is

Then

Let

there are constants

there is a finite dimensional subspace

Let

Firstly, we establish variational framework. Let

Define a functional on

By computation,

Next, observe by Definition

We let

By Lemma

Since

Then

By direct verifications, we may find that

For convenience, we list the following assumptions.

There exists

There is a constant

There exists a constant

There is a positive constant

If (C_{1}) holds, then for all

By (C_{1}), we obtain

Combining with (

So, in view of our assumption

If

If (C_{2}) holds, then for all

Similar to the proof Theorem

By (C_{2}), there exists

Assume (C_{2}) and (C_{3}) hold. Then, for

First, we know from Theorem _{2}) and (

In view of

By (C_{3}), there exists

For _{1}) in Lemma

Since _{2}) in Lemma

Similarly, we can also choose

Assume that (C_{1}) and (C_{4}) hold and that there exists

By (C_{1}) and Theorem

By (C_{4}), for

So, for

Since

Thus, for

Similarly, for

Assume (C_{2}), (C_{3}), and (C_{5}) hold. Then, for each

By (C_{5}), functional _{3}) and (I_{4}) of Lemma

First, in view of (C_{3}), there exists

For

So for

Thus, for _{3}) of Lemma

Next if we choose _{2}) and Theorem _{4}) of Lemma

Therefore, for

In Theorem

Obviously, compared with Theorem _{5}) ensures that the problem (

Suppose (C_{1}), (C_{5}), and (C_{6}) hold. Then for every

_{5}). From (C_{1}), we obtain _{6}), there exists

For

From Theorem

In the final section, we apply the results developed in Section

Consider the following problem
_{2}) and (C_{3}). Since
_{2}) and (C_{3}) hold. Thus, by Theorem

Consider the problem
_{1}) and (C_{4}), then for _{1}) and (C_{4}) are satisfied. So, in view of Theorem

Consider the problem
_{5}) is satisfied. If we choose _{2}) and (C_{3}), then by some simple calculation, we may show that the hypotheses (C_{2}) and (C_{3}) are fulfilled. Therefore, by Theorem

Finally, consider the problem
_{1}), (C_{5}), and (C_{6}) hold. Thus, by Theorem

This Project was supported by the National Natural Science Foundation of China (11161049).