Theorem 8 Assume the following conditions hold:
(M0) is a continuous function and satisfies the condition
for all , where and are positive real numbers;
(f0) satisfies the Carathéodory condition and there exist positive constants and such that
for all and , where such that . Then (P) has a weak solution.
Proof By the assumptions (M0) and (f0), we have
Therefore, by Proposition 3 and Proposition 5, it follows
By the assumption , I is coercive. Since I is weakly lower semicontinuous, I has a minimum point u in and u is a weak solution of (P). □
Theorem 9 Assume the following conditions hold:
(M1) is a continuous function and satisfies the condition
for all , where , and α real numbers such that and ;
(M2) M satisfies
for all ;
(f1) satisfies the Carathéodory condition and there exist positive constants and such that
for all and , where such that for all and ;
(f2) , uniformly for ;
(f3) There exists such that for and all ;
(AR) Ambrosetti-Rabinowitz’s condition holds, i.e., , such that
Then (P) has at least one nontrivial weak solution.
To obtain the result of Theorem 9, we need to show that Lemma 10 and Lemma 11 hold.
Lemma 10 Suppose (M1), (M2), (AR) and (f1) hold. Then I satisfies the (PS) condition.
Proof Let us assume that there exists a sequence in such that
Since , we have . Therefore,
By the above inequalities and assumptions (M1), (M2) and (AR), we get
This implies that is bounded in . Passing to a subsequence if necessary, there exists such that . Therefore, by Proposition 5, we have
By (3.2), we have . Thus
From (f1) and Proposition 1, it follows
If we consider the relations given in (3.3), we get
From (M1), we get
Since the functional (3.4) is of type (see Proposition 3.1 in ), we get in . We are done. □
Lemma 11 Suppose (M1), (AR) and (f1)-(f3) hold. Then the following statements hold:
There exist two positive real numbers γ and a such that , with ;
There exists such that , .
Proof (i) Let . Then by (M1) and Proposition 3, we have
Since , by Proposition 5 we have the continuous embeddings and , and also there are positive constants , and such that
From (f1) and (f2), we get for all and , where is small enough and . Therefore, by (M1), Proposition 3 and (3.5), (3.6), it follows
providing that . Since and , there exist two positive real numbers γ and a such that , with .
(ii) From (AR) and (f3), one easily deduces
for all and . Therefore, for and nonnegative such that , we get
(recall that and almost everywhere). On the other hand, when , from (M1) we obtain that
Since , it is obvious . Hence, for , we have
From the assumption on θ (see (AR)), we conclude as . □
Proof of Theorem 9 From Lemma 10, Lemma 11 and the fact that , I satisfies the mountain pass theorem (see [25, 26]). Therefore, I has at least one nontrivial weak solution. The proof of Theorem 9 is completed. □
In the sequel, using Krasnoselskii’s genus theory (see [25, 27]), we show the existence of infinitely many solutions of the problem (P). So, we recall some basic notations of Krasnoselskii’s genus.
Let X be a real Banach space and set
Definition 12 Let and . The genus of E is defined by
If such a mapping does not exist for any , we set . Note also that if E is a subset which consists of finitely many pairs of points, then . Moreover, from the definition, . A typical example of a set of genus k is a set which is homeomorphic to a dimensional sphere via an odd map.
Now, we will give some results of Krasnoselskii’s genus which are necessary throughout the present paper.
Theorem 13 Let and ∂ Ω be the boundary of an open, symmetric and bounded subset with . Then .
Corollary 14 .
Remark 15 If X is of an infinite dimension and separable and S is the unit sphere in X, then .
Theorem 16 Suppose that M and f satisfy the following conditions:
(M3) is a continuous function and satisfies the condition
for all , where , , δ and α are real numbers such that and ;
(f4) is a continuous function and there exist positive constants , , and such that
for all and , where such that for all ;
(f5) f is an odd function according to t, that is,
for all and .
If for all and , then the problem (P) has infinitely many solutions.
The following result obtained by Clarke in  is the main idea which we use in the proof of Theorem 16.
Theorem 17 Let be a functional satisfying the (PS) condition. Furthermore, let us suppose that:
J is bounded from below and even;
There is a compact set such that and .
Then J possesses at least k pairs of distinct critical points and their corresponding critical values are less than .
Lemma 18 Suppose (M3), (f4) and the inequality hold.
I is bounded from below;
I satisfies the (PS) condition.
Proof (i) By the assumptions (M3) and (f4), we have
By Proposition 3 and Proposition 5, we get
for large enough. Hence, I is bounded from below.
(ii) Let us assume that there exists a sequence in such that
From (3.8) we have . This fact combined with (3.7) implies that
where . Since , we obtain that is bounded in .
Hence, we may extract a subsequence and such that in . In the rest of the proof, if we consider similar relations given in (3.3) and growth conditions assumed on f and apply the same processes which we used in the proof of Lemma 10, we can see that I satisfies the (PS) condition. □
Proof of Theorem 16 Set (see [7, 25])
then we have
Now, we will show that for every . Since is a reflexive and separable Banach space, for any , we can choose a k-dimensional linear subspace of such that . As the norms on are equivalent, there exists such that with implies .
Set . By the compactness of and the condition (f4), there exists a constant such that
for all . If we consider (M3) and (f4), for and , we have
providing that . Since , we can find and such that
It is clear that , so . Finally, by Lemma 18 above, we can apply Theorem 17 to obtain that the functional I admits at least k pairs of distinct critical points, and since k is arbitrary, we obtain infinitely many critical points of I. The proof is completed. □
Theorem 19 Suppose (M3), (f4) and (f5) hold. If for all , then the problem (P) has a sequence of solutions such that .
Proof In the beginning, we will show that I is coercive. If we follow the same processes applied in the proof of Theorem 8 and consider the fact , it is easy to get the coerciveness of I. Since I is weak lower semi-continuous, I attains its minimum on , i.e., (P) has a solution. By help of coerciveness, we know that I satisfies the (PS) condition on . Moreover, from the condition (f5), I is even.
In the rest of the proof, since we develop the same arguments which we used in the proof of Theorem 16, we omit the details. Therefore, if we follow similar steps to those in (3.9) and (3.10) and consider the inequalities , we can find and such that
Obviously, , so . By Krasnoselskii’s genus, each is a critical value of I, hence there is a sequence of solutions such that . □