The critical points of
, that is,

for all
, are weak solutions of problem (1.1). So we need only to consider the existence of nontrivial critical points of
.

Denote by
, and
the generic positive constants. Denote by
the Lebesgue measure of
.

To study the existence of solutions for problem (1.1) in the first case, we additionally impose the following conditions.

(A-1)
and
,
.

(B-1) There exists a function
, such that
and
for
,
.

(C-1) there exist
such that
for
,
, where
and
.

(D-1)
, for all
.

Theorem 3.1.

Under assumptions (F) and (A-1)–(C-1), problem (1.1) admits a nontrivial solution.

Proof.

First we show that any

sequence is bounded. Let

and

, such that

and

in

. By (A-1) and (B-1),

, and

. Let

and

, then

,

,

,

and

. Let

and

. Then

By (B-1), we get

By (F), we get
, so there exist
such that
on
. Note
, so we have

By Young's inequality, for
, we get

Take
sufficiently small so that
.

Note that
, by Young's inequality again and for
, we get

Take
sufficiently small so that
.

From the above remark, we have

As
,
and
, we have
. Since
and
, by Theorem 2.6 we have

when
and
is sufficiently large. Then it is easy to see that
is bounded in
. Next we show that
possesses a convergent subsequence (still denoted by
).

Note that

Because

is bounded in

, there exists a subsequence

(still denoted by

), such that

weakly in

. By Theorem 2.11, there are compact embeddings

and

, then

in

and

. So we get

Hence
as
.

By (F), we have

and similarly for every

,

and

in

and

, we obtain

and

. Similarly,

Because
is bounded, we get
, as
. From the above remark, we conclude
, as
.

Thus
, as
. Then we get
. As in the proof of Theorem
in [6, 7], we divide
into the following two parts:

On
, we have

Then
as
.

On
, we have

so
, as
.

Thus we get
. Then
in
.

From (F) and (B-1) we have
, for all
,
. So we get
, for all
,
. for all
, take
, then
. Since
and
, there exists
such that
, and
, for
. Let
, such that
for
,
for
, and
in
. Then we have

where
. So if
is sufficiently large, we obtain
.

From (F) and (C-1), we have
, then
. So we get

Let

, then

. By Theorem 2.11,

,

and

. When

is sufficiently small,

,

and

. For any

, as

, for any

, we can find

such that

and

whenever

. Take

, then

.

is an open covering of

. As

is compact, we can pick a finite subcovering

for

from the covering

. If

we define

on

. We can use all the hyperplanes, for each of which there exists at least one hypersurface of some

lying on it, to divide

into finite open hypercubes

which mutually have no common points. It is obvious that

and for each

there exists at least one

such that

. Let

, then

and we have

If

is sufficiently small such that

we have
.

The mountain pass theorem guarantees that
has a nontrivial critical point
.

Since
is a separable and reflexive Banach space, there exist
and
such that

For
, denote
,
,
.

Theorem 3.2.

Under assumptions (F), (A-1)–(D-1), problem (1.1) admits a sequence of solutions
such that
.

Proof.

Let
. We first show that
is weakly strongly continuous. Let
weakly in
. By the compact embedding
, we have
and
a.e. on
. By the inequality
and the Vitali Theorem, we get
.

Note that

When

,

When

,

is bounded. So we get

By the compact embedding

, we get

in

. So by Theorem 2.12 we obtain

, that is,

. Hence we obtain that

is weakly strongly continuous. By Proposition

in [

8],

as

for

. For all

, there exists a positive integer

such that

for all

. Assume

for each

. Define

in the following way:

Note that

as

. Hence for

with

, we get

Note that
and
. Since the dimension of
is finite, any two norms on
are equivalent, then
. If
, it is immediate that
. If
, then
. As in the proof of Theorem 3.1 we can find hypercubes
which mutually have no common points such that
and
, where
. Then we need only to consider the case:
for every
. We have

Let

,

. Let

and

. Denote

. Let

be sufficiently large such that

. There at least exists one

such that

. We have

and
as
. Hence we obtain that
as
. Thus for each
, there exists
such that
for
. From Theorem 3.1
satisfies
condition. In view of (D-1), by Fountain Theorem (see [16]), we conclude the result.

In the second case, we additionally impose the following condition:

(A-2)
and
.

Theorem 3.3.

Under assumptions (F), (A-2), (B-1), and (C-1) there exist
such that when
, problem (1.1) admits a nontrivial solution.

Proof.

It is obvious that

. Let

be such that

. Since

, there exists

and

such that

, for all

. Thus

for all

. Let

be as defined in Theorem 3.1. By (C-1),

, and

, when

. Then for any

and

, we have

If
is sufficiently small,
.

From (F) and (C-1), we have
and
. By Theorems 2.8 and 2.11, there exist positive constants
such that
,
,
. When
is sufficiently small, we have
,
, and
. As in the proof of Theorem 3.1 we can find hypercubes
which mutually have no common points such that
,
and
, where
. Then

Since

, for all

, when

,

. Fix

such that

. Then we have

Let
. When
,
. As in the proof of Theorem
in [17], denote
, we have
and
. Let
. Applying Ekeland's variational principle to the functional
, we find
such that
,
,
, and
. Thus we get a sequence
such that
and
. It is clear that
is bounded in
. As in the proof of Theorem 3.1, we get a subsequence of
, still denoted by
, such that
in
. So
and
.

Theorem 3.4.

Under assumptions (F), (A-2), and (B-1)–(D-1), problem (1.1) has a sequence of solutions
such that
.

Proof.

First we show that any

sequence is bounded. Let

and

, such that

and

in

. By (B-1), there exist

such that

. From (F), (A-2), and (B-1)–(D-1), we have

By Young's inequality, for
, we get

Take

,

and

sufficiently small so that

,

and

, then

Therefore by Theorems 2.6 and 2.9, we get that

is bounded in

. Then as in the proof of Theorem 3.1

possesses a convergent subsequence

(still denoted by

). By Theorem 3.2, we can also have

As in the proof of Theorem 3.1 we can find hypercubes
which mutually have no common points such that
and
, where
. Since the dimension of
is finite, any two norms on
are equivalent. Then we need only to consider the cases
and
for every
. We have

Hence
as
. As in the proof of Theorem 3.2, we complete the proof.

In the third case, we additionally impose the following condition:

(A-3)
, and
,

(B-3) there exist
such that
for
.

Theorem 3.5.

Under assumptions (F), (A-3), and (B-3), problem (1.1) admits a nontrivial solution.

Proof.

By Young's inequality, for

, we get

. By (F), we have

. Thus

Take

sufficiently small so that

. From Theorem 2.11,

. If

,

is bounded. Then we need only to consider the case

. As in the proof of Theorem 3.1 we can find hypercubes

which mutually have no common points such that

and

, where

. Then we have

where
, and
,
. As in the proof of Theorem 3.2, we obtain that
is coercive, that is,
as
. Thus
has a critical point
such that
and further
is a weak solution of (1.1).

Next we show that
is nontrivial. Let
be the same as that in Theorem 3.3. By (B-3),
. Then

If
is sufficiently small,
.

In the fourth case, we additionally impose the following condition:

(A-4)
and
.

Theorem 3.6.

Under assumptions (F), (A-4), and (D-1), problem (1.1) admits a sequence of solutions
such that
.

Proof.

First we show that any
sequence is bounded. Let
and
, such that
and
in
. Denote
and
. We have
and
.

We can get

By Young's inequality, for

, we get

Take

and

sufficiently small so that

and

. Then

As in the proof of Theorem 3.5,

, when

. Thus, we conclude that

is bounded in

. Then as in the proof of Theorem 3.1

possesses a convergent subsequence

(still denoted by

). By Theorem 3.2, we can also get

As in the proof of Theorem 3.1 we can find hypercubes
which mutually have no common points such that
and
, where
. Since the dimension of
is finite, any two norms on
are equivalent. Then we need only to consider the cases
and
for every
. We have

Hence we obtain
as
. As in the proof of Theorem 3.2, we complete the proof.