In this section, we prove several lemmas.
Lemma 2.1 For (where is given in Theorem 1.1), we have .
Proof Suppose that for all . If , then we have
(2.1)
and
(2.2)
By (2.1)-(2.2), the Sobolev inequality, and the Hölder inequality, we get
(2.3)
and
(2.4)
where . Thus, using (2.3) and (2.4), we have
(2.5)
Hence, by (2.5) the desired conclusion yields. □
Lemma 2.2 If , then
Proof From , it is easy to see that
By the Sobolev inequality, we get
In addition,
The proof is completed. □
By Lemma 2.1, for we write and define
The following lemma shows that the minimizers on are ‘usually’ critical points for .
Lemma 2.3 For , if is a local minimizer for on , then in .
Proof If is a local minimizer for on , then is a solution of the optimization problem
Hence, by the theory of Lagrange multipliers, there exists such that
(2.6)
Thus,
(2.7)
From and Lemma 2.1, we have and . So, by (2.6)-(2.7) we get in . □
For each , we write
Then we have the following lemma.
Lemma 2.4 For each and , we have
-
(i)
there is a unique such that and ;
-
(ii)
is a continuous function for nonzero u;
-
(iii)
;
-
(iv)
if , then there is a unique such that and .
Proof (i) Fix . Let
Then we have , as , is concave and reaches its maximum at . Moreover,
(2.8)
Case I. .
There is a unique such that and . Now,
and
Thus, . In addition,
and
Hence, .
Case II. .
From (2.8) and
there exist unique and such that ,
and
Similar to the argument in Case I above, we have , , and
-
(ii)
By the uniqueness of and the external property of , we find that is continuous function of .
-
(iii)
For , let . By item (i), there is a unique such that , that is, . Since , we have , which implies
Conversely, let such that . Then . Therefore,
-
(iv)
By Case II of item (i). □
By and changes sign in Ω, we have is an open set in . Without loss of generality, we may assume that Θ is a domain in . Consider the following biharmonic equation:
(2.9)
Associated with (2.9), we consider the energy functional
and the minimization problem
where . Now we prove that problem (2.9) has a positive solution such that .
Lemma 2.5 For any , there exists a unique such that . The maximum of for is reached at , the map
is continuous and the induced continuous map defines a homeomorphism of the unit sphere of with .
Proof For any given , consider the function , . Clearly,
(2.10)
It is easy to verify that , for small and for large. Hence, is reached at a unique such that and . To prove the continuity of , assume that in . It is easy to verify that is bounded. If a subsequence of converges to , it follows from (2.10) that and then . Finally the continuous map from the unit sphere of to , , is inverse to the retraction . □
Define
where .
Lemma 2.6 is a critical value of K.
Proof From Lemma 2.5, we know that . Since for and t large, we obtain . The manifold separates into two components. The component containing the origin also contains a small ball around the origin. Moreover, for all u in this component, because , . Then each has to cross and . Since the embedding is compact (see [14]), it is easy to prove that is a critical value of K and a positive solution corresponding to c. □
With the help of Lemma 2.6, we have the following result.
Lemma 2.7 (i) For , there exists such that
-
(ii)
is coercive and bounded below on for all .
Proof (i) Let be a positive solution of problem (2.9) such that . Then
Set as defined by Lemma 2.4(iv). Hence, and
This implies
-
(ii)
For , we have . Then by the Hölder, Sobolev, and Young inequalities,
here .
Thus, is coercive on and
for all . □
Next, we will use the idea of Tarantello [16] to get the following results.
Lemma 2.8 For and any given , there exist and a differentiable functional such that , the function and
(2.11)
for all .
Proof Define as follows:
Since and by Lemma 2.1, we obtain
we can get the desired results applying the implicit function theorem at the point . □
Lemma 2.9 For and any given , there exist and a differentiable functional such that , the function and
(2.12)
for all .
Proof In view of Lemma 2.8, there exist and a differentiable functional such that , for all and we have (2.12). By use of , we have . In combination with the continuity of the functions and , we get as ϵ sufficiently small, this implies that . □