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Three Solutions Theorem for -Laplacian Problems with a Singular Weight and Its Application
We prove Amann type three solutions theorem for one dimensional p-Laplacian problems with a singular weight function. To prove this theorem, we define a strong upper and lower solutions and compute the Leray-Schauder degree on a newly established weighted solution space. As an application, we consider the combustion model and show the existence of three positive radial solutions on an exterior domain.
Let us consider the following -Laplacian problem with a sign-changing singular weight: where ,, , and may change sign. Moreover, and satisfies , for for some , where a class of weight functions is given as It is well known that .
If , then all solutions of are in and based on the Leray-Schauder degree argument on -space: Ben-Naoum and de Coster  proved three solutions theorem for . On the other hand, if , then solutions of may not be in ; for example, take , , and , then and the solution for corresponding problem of is given by which is not in . Thus if , the three solutions theorem in  can not be applied. Our main interest in this paper is to establish three solutions theorem for problem for those weights satisfying .
Main step for the proof of three solutions theorem, in general, is to compute the Leray-Schauder degree on a sector in solution space made from strong sense of upper and lower solutions. Since the sector needs to be open in the space, strong sense of ordering and the sector made from the ordering are closely related to the topology of solution space. A typical situation in application usually happens as follows.
It is comparatively easy to find a lower solution and an upper solution of satisfying , for all and . Denote , . In -analysis, that is, , we see that is nonempty and open in by providing additional conditions like and which implies a strong sense of ordering. On the other hand, in -analysis, that is, , we see so that the Leray-Schauder degree on is not even defined. Three solutions theorem with no use of such is very restrictive in application. To overcome this difficulty in our problem, we introduce a weighted solution space, say , which is finer than and also introduce a strong sense of ordering suitable to -space which makes the degree computation effective (see Section 2 in detail).
As to a question of triple multiplicity of solutions, besides the Amann type three solutions theorems, many works are done by using the variational method (see [2–5] and the references therein) and by using several extensions of the Liggett-Williams fixed point theorem and Guo-Krasnoselskii fixed point theorem, especially for positive solutions (see [6–8] and the references therein). For the problem we are concerned with in this paper, the variational setup is not possible due to lack of regularity of solutions. Three solutions theorem proved in this paper is for the case that is not only but also possibly sign-changing. By this aspect, it is new as far as the authors know. For the case , one may refer to  for a partial result about the theorem.
As an application, we study the existence of triple positive solutions for certain nonlinear -Laplacian problems with positive singular coefficient function and give an example of a combustion model defined on an exterior domain. Applying this three solutions theorem to the case having sign-changing coefficient function could be an interesting and challenging problem.
We organize this paper as follows. In Section 1, we introduce a weighted solution space , the strong upper and lower solutions of , and prove three solutions theorem for problem . In Section 3, we prove a multiplicity result for certain nonlinear -Laplacian problems by using three solutions theorem introduced in Section 2. In Section 4, we apply the result in Section 3 to a combustion model to show the existence of three positive radial solutions on exterior domain.
For , define by And also define ; then is a Banach space. We give a proof for reader's convenience.
Lemma 1. Let ; then is a Banach space with a norm .
Proof. Let be a Cauchy sequence in . Then and are Cauchy sequences in so that there exist such that and in . It is sufficient to show that on . Since for , there exists such that for all . For , we know in . This implies in . Since is arbitrary, pointwise in . Therefore, on . Since converges uniformly to on , we have Thus and on . This implies in and the proof is completed.
Then we see that for given , being a solution of implies . In fact, if and is a solution of , then for , and by using L’Hospital’s rule, we have For the cases that and (or ), by the same argument, we have
Example 2. Let us consider the following example: where , and is given by Then we see that sign-changing and every solution satisfies by using (9). We also see by calculation that can be given as and by using (10).
To establish corresponding integral operator for problem , let us first consider the problem where . We remind the reader that needs not be integrable near or . Integrating on for , we have Denoting , Since and is a fixed constant, we can see that so that we may integrate on . Using a boundary condition , we get We note that in (18) is a solution of only on the interval . Doing similar computation on the interval , we get It is known by Lemma 2.2 in  that the equation has a unique zero in for each . Therefore it is natural to paste ’s in (18) and (19) in a continuous way. Now let us define a function by Then satisfies and is a unique solution of problem .
Remark 3. We understand the number in the above as a function of defined on . That is, . It is known that maps bounded sets in into bounded sets in ([10, Lemma 3.1]). It is also known that is completely continuous on ([10, Theorem 3.4]).
As mentioned in Introduction, the regularity of solutions of problem sensitively depends on the shape of nonlinear term even if , and we are concerned with the case that problem does not have -solutions. Therefore, it is interesting to consider operator restricted on to complete three solutions theorem for those problems with no -solutions.
Define the restriction of on .
In what is to follow, we assume and we now prove the complete continuity of on the solution space . Before doing that, we give a remark useful to calculate -Laplacians.
Remark 4. If , then where
Theorem 5. is completely continuous.
Proof. Let be a bounded subset of . Then for any sequence , we need to show the relative compactness of with respect to -norm. We know by Remark 3 that is completely continuous on so that there exists and a subsequence of , say again such that in . To complete the proof, we need to show the following.
and there is a subsequence of such that as in and is continuous on .
Claim 1. is uniform bounded in .
Since is bounded in , there exists such that and , for all . We also know by Remark 3 that there is such that , for all . For , by using Remark 4, we get where . From the fact that and the definition of , we see thus we have for . For , by similar calculation, we get the same upper bound of as in (27) and by taking limt and in (27), we have the same upper bound of and as in (27). This proves that is bounded in .
Claim 2. is equicontinuous on .
If , then since , for all and , we get for all . This implies that is equicontinuous in and by Arzela-Ascoli theorem, there exist a subsequence of and such that converges uniformly to on as . Thus using Lebesgue Dominated Convergence theorem, we obtain uniformly on . This implies that is equicontinuous in .
On the other hand, for the case of , suppose that is not equicontinuous on . Then there exists such that we may choose a subsequence of and sequences , satisfying As for sequences and , it is easy to see that . We show that or . Suppose it is not true so let . Then taking satisfying , we see that and uniformly on . By the same argument of the above case of , we can prove that is equicontinuous on . Thus there is sufficiently large such that and this contradicts with (30). Now we consider the case . The argument for the case is similar. It is easy to see that this case implies . Since , , for all and we get Now we want to calculate .
Since is bounded and , we have On the other hand, We show .
Indeed, is close to for sufficiently large , since and . Therefore, without loss of generality, we may assume that so that . Thus using the fact in , we have Next we show Indeed, using the fact for sufficiently large , we get If , then we can easily verify . Since , we see that the limit is . On the other hand, if . We note that , for given . By using L’Hospital’s rule, we get the conclusion.
Therefore we get By the same argument, also has the same limit and this contradicts with (30). Consequently, is equicontinuous in . By Arzela-Ascoli theorem, there exists a subsequence of and such that
Claim 3. and in .
It is enough to show that . Since for , let ; then and for , uniformly in and thus on . Since is arbitrary, on and from (39), we have Therefore on . By similar argument near , we get on and thus .
Claim 4. is continuous on .
Assume that in . By compactness of , there is a subsequence of and such that in . It is suffice to see that , for all . Since is continuous in and in , we have in . Thus and by standard limit argument, we see that . This completes the proof.
Now we define a strong sense of ordering in .
Definition 6. For , one says that if and only if(i) for all ,(ii)either or ,(iii)either or .
Definition 8. One says that is a strict lower solution of if and only if is a lower solution of and satisfies where is a solution of such that , .
We say that is a strict upper solution of if and only if is an upper solution of and satisfies where is a solution of such that , .
Theorem 9. Assume that there exist a strict lower solution and a strict upper solution of such that . Then problem has at least one solution such that . Moreover, for large enough, the Leray-Schauder degree can be computed as where .
Proof. Consider the modified problem where is defined by Then it is well known that if is a solution of , then and thus is solution of . Define by . Then is bounded and thus there exists such that for all . By the homotopy invariance property of degree, we have where . Thus has a solution and has a solution satisfying . Since and are strict lower and upper solutions, respectively, by the definition of strict lower and upper solution, we have . Moreover, by using the fact that on , (44) and excision property, we conclude that there exists large enough such that
Theorem 10 (three solutions theorem). Assume that there exist a lower solution , an upper solution , a strict lower solution , and a strict upper solution of such that and there exists with . Then problem has at least three solutions , and such that
Proof. Consider the modified problem, where is defined by For any , and are strict lower solution and strict upper solution of . In fact, if is a solution of , then we have . By Theorem 5, there is a sufficient large : where is defined by and Then by excision and additive property, we have Thus there are three solutions of , , , and . Since all solutions of satisfy , they are solutions of and the proof is done.
In this section, we prove the existence of triple positive solutions for a problem of the form where and . Let us assume with ,, , is nondecreasing.
The existence of two positive solutions for problem was proved in  under a stronger condition on such as for some . Since , we can easily see that any solution of problem is in but not in so that with the aid of three solutions theorem in Section 2, we prove the following theorem.
Proof. For , , it is trivial that is a lower solution of . Let . Then Thus is an upper solution of . To show that is a strict upper solution of , assume that is a solution of such that . We first show that , for all . Suppose it is not true, then there exist and with in such that and . Integrate from to , by (55) and monotonicity of , we have the following contradiction: Since , it suffices to show that . The inequality can be proved similarly. For the case that , we know and exist. Since we know that there exists such that . Indeed, otherwise, for all ; then by integrating this from to 1, we have the contradiction Integrating (57) from to 1, we have and thus and For the case that , From the monotonicity of and choice of with help of L’Hospital’s rule, we have Thus we proved that is a strict upper solution of . Now, since , we may choose satisfying ; then since , we may choose such that . Let be the solution of where when and , for . We note that and let us show that for . It is clear that for . We note that from the symmetry of . For