Proof of Theorem 1.1
Arguing by contradiction, let us assume that problem has solutions as stated in Theorem 1.1. Recall that is written as
with orthogonal to each and their derivatives with respect to and , where denotes the k th component of (see (2.2) and (2.3)). For simplicity, we will write , , and . From Proposition 3.2, for each , with , . We have
Furthermore, an easy computation shows that
On the other hand, following the proof of Proposition 3.3, we have, for each ,
We distinguish many cases depending on the set
and we will prove that all these cases cannot occur.
We remark that if we derive and as .
Furthermore, the behavior of depends on the set Ϝ. In fact we have, assuming that ,
First we start by proving the following crucial lemmas.
Remark 4.1 Ordering the ’s: , adding , and using (4.2), it is easy to derive a contradiction if we have .
Lemma 4.2 Let . Then there exists a positive constant such that
Proof The proof will be by contradiction.
Proof of (i). Assume that . In this case, we have
which implies that . Using Remark 4.1, we derive a contradiction. In the same way, we prove that . Hence the proof of Claim (i) is completed.
Proof of (ii). Assume that . By Claim (i), we have . Four cases may occur.
Case 1. or . Using (4.5), implies that
By Claim (i) and , we obtain . By Remark 4.1, this case cannot occur.
Case 2. , , and . In this case, it is easy to obtain . Using Remark 4.1, we derive a contradiction.
Case 3. , , , and . In this case, we see that is bounded and . Hence, we derive that , which implies that for . Thus
Then by Remark 4.1, we get a contradiction.
Case 4. , , and . In this case, it is easy to get .
Using the formula , we deduce that , which implies that . Hence by Remark 4.1, we derive a contradiction and Claim (ii) is thereby completed.
Proof of (iii). Without loss of generality, we can assume that . First, as in the proof of Claim (i), we get . Now assume that , which implies
Two cases may occur.
Case 1. or . Using , we obtain
and we derive a contradiction from .
Case 2. and . Let such that . Using Claim (ii) and the fact that , we derive that , which implies that , , and is bounded. Using (4.3) for , we get
Since is bounded and , we derive that
which implies that
By Remark 4.1, we get a contradiction. □
There exists a positive constant
Proof Without loss of generality, we can assume that .
Proof of (i). Assume that . First we claim that . In fact, arguing by contradiction we assume that , we get , , and . Hence, . From , we obtain
Let be the outward normal vector at . Since , , and are of the same order, we have (see  and )
Using , we get , which implies that . From , we derive a contradiction. Hence our claim is proved.
Thus there exists a positive constant c so that . Now, since we have assumed that , Lemma 4.2 implies that . Finally, using Remark 4.1, we get a contradiction and the proof of Claim (i) follows.
Proof of (ii). Assume that . Note that Claim (i) and imply that (4.9) holds.
Now, following the proof of (i), we obtain a contradiction. □
We turn now to the proof of Theorem 1.1. By the previous lemmas, we know that and are of the same order, and , for where c is a positive constant.
Hence, implies that (4.9) holds. Furthermore, for implies that
We denote by the eigenvector associated to whose norm is 1. We point out that we can choose so that all their components are positive (see  and ).
Let , , and . From (4.11), we have
The scalar product of (4.12) by gives
Since the components of are positive and , are of the same order, there exists a positive constant c, such that . Hence, we get
We deduce from (4.3) and (4.11) that
Observe that Λ may be written in the form
Using (4.15), we get
Since for and , the matrix is bounded.
Furthermore, we have , which implies that
Let us consider the equality
and derivative it with respect to ; we obtain
The scalar product with gives
Using (4.18), we obtain
Hence, we derive a contradiction from (4.14), (4.20), and the fact that 0 is a regular value of ρ. Thus the proof of our theorem follows.
Proof of Theorem 1.2
Arguing by contradiction, let us assume that problem has solutions as stated in Theorem 1.2. From Section 2, these solutions have to satisfy (2.2) and (2.3).
As in the proof of Proposition 3.2, we have, for each ,
Observe that, if , we have is bounded (by the assumption) which implies that
where c is a positive constant. Using (4.21), easy computations show that
Thus, using (4.22), can be written as
and for ,
The proof will depend on the value of l which is defined in the theorem.
Case 1. . From the definition of l we get . Now using (4.1) and , we derive that
Now, using (4.23) and , we derive the estimate of and by induction we get
Finally, using (4.22), (4.23), (4.24), and we obtain
which gives a contradiction.
Case 2. . Using (4.1), an easy computation implies that
Then from , , (4.1), (4.25), and the fact that and (since ), we obtain
Now using and (4.26) we get (4.23) and as before, (4.24) is satisfied. Hence we also derive a contradiction from .
Case 3. . Recall that in this case we have assumed that . This implies that
Hence, using , the definition of l and (4.1) we obtain the first part of (4.23). The second part follows from and the first one. Finally, as before we derive a contradiction from .
Hence, our theorem is proved.