Abstract
We investigate the one-dimensional prescribed mean curvature equation with concave-convex nonlinearities in the form of , , , , where is a parameter and p, q satisfy , and we obtain new exact results of positive solutions. Our methods are based on a detailed analysis of time maps.
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1 Introduction
Mean curvature equations arise in differential geometry, physics, and other applied subjects. For example, the negative solutions of prescribed mean curvature equations can describe pendent liquid drops in the equilibrium state (see [1]), or the corneal shape (see [2]). In recent years, increasing attention has been paid to the study of the prescribed mean curvature equations by different methods (see [3]–[31]).
A typical model of the prescribed mean curvature equation is
where Ω is a bounded domain in and is continuous.
It is well known that a solution u of (1.1) defines a Cartesian surface in whose mean curvature is prescribed by the right-hand side of the equation. Classical existence theorems for this problem (and in particular for the minimal surface problem) are presented in [15] with references to the original papers by Bombieri, Finn, Miranda, etc. (see e.g.[1], [14], [16] and the references therein). The existence theorems established in most of those papers are concerned with solutions of the prescribed mean curvature problem as global minimizers of the corresponding energy functionals. Some papers studied the prescribed mean curvature problems by using the sub-supersolution method (see [7], [17], [18], [28]), the time-map analysis method (see [8], [9], [13], [19], [29]–[31]) or Mawhin’s continuation theorem (see [20], [24]–[26]) and so on.
The one-dimensional version of (1.1) is
There are some papers considering the exact number of positive solutions of (1.2) in special case of f (see [6], [8], [9], [11]–[13], [19], [28]–[31]). The study derived from an open problem proposed by Ambrosetti et al. in [32], which concerned the exact number and the detailed property of solutions of the semilinear equation
where . Since then, related problems have been studied by many authors; see [33]–[36] and the references cited therein.
Generally, it is difficult to obtain the exact multiplicity results for nonlinear boundary value problems. If the operator is replaced by , then the problems will become more complicated. In [6], Habets and Omari considered the nonlinear boundary value problem of the one-dimensional prescribed mean curvature equation with , where . By using an upper and lower solution method, the authors obtained the exactness results of positive solutions. In [8], Pan and Xing derived the exact numbers of positive solutions of (1.2) for the nonlinearities (), , and ().
However, there are few articles dealing with the case involving negative exponent. We note that, in particular, Bonheure et al.[28] is the first paper where the problem with singularities has been considered. In [28], the authors considered a general class of involving a singularity, and they obtained the result that there exists a positive solution for a small parameter, and they pointed out , , (where ) as the special case. In [30] and [12], the authors studied global bifurcation diagrams and the exact multiplicity of positive solutions for the cases and (), respectively.
In [9], Li and Liu examined problem (1.2) for with or . Very recently, the authors [19] studied the exact number of solutions of problem (1.2) for with . However, to the best of our knowledge, no paper has considered the exact number of solutions of problem (1.2) with till now. In this paper, we will try to solve it.
Consider the following boundary value problem of the one-dimensional prescribed mean curvature equation:
where and is a parameter. In this paper, a positive solution is a function satisfying (1.3).
The rest of this paper is organized as follows. In Section 2, we introduce and analyze the time map which plays a key role in the paper. The main result will be stated and proved in Section 3.
2 Time maps
In this section, we will make a detailed analysis of the so-called time map of problem (1.3).
Let be a solution of problem (1.3). Then it is well known that takes its maximum at , is symmetric with respect to c, for and for . Hence problem (1.3) is equivalent to the following problem defined on :
Denote
and
Let . If is a solution of (2.1) with , then is a solution of the following problem defined on :
Since satisfies
and , we see that
Therefore,
Then
Integrating (2.3) from 0 to c, it leads to
The function is called the time map of f.
Choosing in (2.2), we see that and
Therefore, if is a solution of (2.1) with , then s satisfies (2.4) and (2.5). Conversely, for a given λ, if s satisfies (2.4) and (2.5), then (2.2) together with defines a function on which satisfies and , and then it is easy to see that is a solution of (2.1) with . So the number of solutions of (2.1) is equal to the number of s satisfying (2.4) and (2.5). This leads us to investigate the shape of the graph of .
Let . From (2.4) and (2.5) we see that T is defined on Σ by
For simplicity, we write
It follows that
The following lemmas give the properties of .
Lemma 2.1
([19])
has continuous derivatives up to the second order on Σ with respect to s and
Lemma 2.2
is strictly decreasing on Σ with respect to λ.
Proof
By a direct calculation, we have
which implies that is strictly decreasing on Σ with respect to λ. □
Lemma 2.3
([19])
is strictly decreasing onwith respect to λ, and
Lemma 2.4
-
(1)
For fixed , .
-
(2)
For fixed ,
-
(3)
Denote . Then is continuous, strictly decreasing on , and
-
(4)
Let . Then is continuous and
Furthermore, if, then; if, thenunder the condition.
-
(5)
If , , then there exists such that for fixed , .
-
(6)
If , , then there exist such that for fixed , and for fixed , .
-
(7)
Assume that . If , then for fixed , .
Proof
For fixed . It is clear that
On the other hand, by the uniformly convergent of integral a direct calculation, we obtain
If ,
If , then by the L’Hopital rule we have
So the results (1) and (2) are proved.
The results of (3) can be proved in the same way as in Lemma 3.2 of [9].
Next to prove that the result (4) holds. By the chain rule of differentiation, we have
Since , , we obtain in the case of .
If , by the fact that
we have
Let . Then
Let . Since
by , we have . It follows that .
Other parts are similar to the proof of Lemma 2.3 in [9]. So we omit them. Next we prove that the result (5) holds.
Let , then there exists such that , and for , for . For fixed , for simplicity, we denote by α. It follows from (2.8) that
Let
Then we can see that , , and
It is obvious that for . If , then , by (2.12) we have
It follows that , . Then , which implies that .
Then the result (5) is proved.
Now we prove that the result (6) holds.
By (2.6) we have
Then
It follows that
If there exists such that , then from the third formula of (2.15) and the fact that for we have for .
On the other hand, from (2.15), by a simple calculation we can see that . In fact, by the last formula of (2.15), we have
where
Then
It follows that . Then there exists such that for . This proves (6).
Note that for and , it follows from (2.8) that
The first integral in (2.16) can be estimated as
For the second integral in (2.16), we see that
Here the transformation has been used. Combining (2.16), (2.17) with (2.18) we can see if
then . This proves (7). □
Remark 2.1
It follows from the proof of (6) that if then the conclusion of (6) still holds.
Remark 2.2
From the proof of [9], one can see that the inequality
plays an important role in guaranteeing that is decreasing on λ. In (4) of Lemma 2.4, we replace (2.19) by
and the method used to prove that is decreasing on λ is completely different from that of [9].
Remark 2.3
It follows from the proof of [9] that the inequality
guarantees that . If p and q satisfy , ; , ; , , and , , we can prove that without the inequality (2.20); see, for example, the proof of (5), (6) in Lemma 2.4.
The proofs of Lemmas 2.5 and 2.6 are similar to those of [37]. So we omit them.
Lemma 2.5
For, we have
Lemma 2.6
For, we have
The proof of Lemma 2.7 is similar to Lemma 2.7 in [19]. For convenience of the reader, we prove it in the following.
Lemma 2.7
Suppose that. Let
For fixed, we have
Proof
From Lemma 2.5 and Lemma 2.6, we have . We still use the symbols such as ΔF, , and when λ is replaced by , so we have
Let
and
Then for , we have
By the fact that , we have for .
It follows that
Therefore, for we have
i.e.,
Hence
Since , we have and then (2.21) follows. □
3 Main results
In this section, we apply the Lemmas 2.1-2.7 to establish the exact number of solutions for problem (1.3). We consider the following six cases: , ; , ; , ; ; , , and , . Case , is treated in the following theorem.
Theorem 3.1
Assume that, . Then there existsuch that (1.3) has exactly one solution for, exactly two solutions for, and no solution for.
Proof
For fixed , by (2.15) and the fact that we have . Combining this with Lemma 2.7, for fixed , has only one critical point , which is a maximum point, and for , for . From and (4) of Lemma 2.4, we know that is strictly decreasing in . Selecting such that , we obtain for . Combining this with the fact that and the continuity of we see that there is only one s satisfying (2.4) for .
Choosing such that , there is only one s satisfying (2.4) for and no s satisfying (2.4) for by (3) of Lemma 2.4.
When , by the fact that and , it follows that there are two s satisfying (2.4). Theorem 3.1 is proved. □
The following theorem deals with the case , .
Theorem 3.2
Assume that, . Then there existsuch that (1.3) has exactly one solution forand no solution for.
Proof
From (2.8) we see that
then , it follows that is decreasing on s.
On the other hand, by (2) of Lemma 2.4, we know . Then for fixed , , .
From , and (4) of Lemma 2.4, we know that is strictly decreasing in . Selecting such that , we obtain for . Then there is no s satisfying (2.4) for .
If , then and there is no s satisfying (2.4) for . By the monotone property of on s we see that there is only one s satisfying (2.4) for . This completes the proof. □
Case 3: , .
Theorem 3.3
Assume that, . Then the following conclusions hold.
-
(a)
For any , (1.3) has at most two solutions.
-
(b)
There exist such that (1.3) has exactly one solution for and has no solution for .
-
(c)
If, in addition, , then there exist such that (1.3) has exactly one solution for , exactly two solutions for , and no solution for .
Proof
By Lemma 2.7, for , has at most one maximum point . If , then for . In this case there is at most one s satisfying (2.4) for some . If , then for and for . Then there are at most two s satisfying (2.4) for some . It follows that (1.3) has at most two solutions. This proves (a).
Let such that . By (3) of Lemma 2.4, we see that there is no s satisfying (2.4) for .
On the other hand, by , there exists such that for . Combining this with the proof of (a) we see that there is only one s satisfying (2.4) for . This gives (b).
Considering (c), from , , and (4) of Lemma 2.4, we know that is strictly decreasing in . Selecting such that , we obtain for . Then there is only one s satisfying (2.4) for .
By (7) of Lemma 2.4, for fixed , . Combining this with Lemma 2.7, has only one critical point , which is a maximum point, and for , for .
Choosing such that , there is one s satisfying (2.4) for and no s satisfying (2.4) for by (3) of Lemma 2.4.
When , by the fact that and , it follows that there are two s satisfying (2.4). Theorem 3.3 is proved. □
Case 4: .
Theorem 3.4
Assume that. Then there existssuch that (1.3) has exactly one solution for, and no solution for.
Proof
From Lemma 2.4 we know for fixed . By (2.14) and the fact that we have . Then for fixed , . Combing this with (4) of Lemma 2.4, there exists such that and for , for . Then (1.3) has exactly one solution for , and no solution for . □
Case 5: , .
Theorem 3.5
Assume that, . Then the following conclusions hold.
-
(a)
For any , (1.3) has at most two solutions.
-
(b)
There exist such that (1.3) has exactly one solution for and has no solution for .
-
(c)
Furthermore, suppose that holds. Let be defined in the same way as in (5) of Lemma 2.4.
-
(i)
If is such that , then there exist such that (1.3) has exactly one solution for , exactly two solutions for , and no solution for .
-
(ii)
If is such that .
Case 1. If, then there existsuch that (1.3) has exactly one solution for, exactly two solutions for, and no solution for.
Case 2. If, then there existsuch that (1.3) has exactly one solution for, exactly two solutions for, and no solution for.
Case 3. If, then there existsuch that (1.3) has exactly one solution for, exactly two solutions for, and no solution for.
Proof
The proof of (a), (b) is similar to that of Theorem 3.3.
Considering (c)(i), from and (4) of Lemma 2.4, we know that is strictly decreasing in . Selecting such that , we obtain for . Then there is only one s satisfying (2.4) for .
By (5) of Lemma 2.4, for fixed , . Combining this with Lemma 2.7, has only one critical point , which is a maximum point, and for , for .
If , choosing such that , there is one s satisfying (2.4) for and no s satisfying (2.4) for by (3) of Lemma 2.4. If , then let .
When , by the fact that and , it follows that there are two s satisfying (2.4). The proof of (c)(i) is complete.
Next, turning to (c)(ii), since , there exists such that and there is no s satisfying (2.4) for .
By the fact , if , then we have ; if , then we have ; if , then we have . The proof of the other conclusions follows by a similar method to (c)(i). Then the result (c)(ii) follows. □
Remark
If we assume that , , then we have similar results to those of Theorem 3.5. It is worth to point that is different from that of Theorem 3.5 in this case. Then we have Theorem 3.6.
Case 6: , .
Theorem 3.6
Assume that, . Then the following conclusions hold.
-
(a)
For any , (1.3) has at most two solutions.
-
(b)
There exist such that (1.3) has exactly one solution for and has no solution for .
-
(c)
Furthermore, suppose that holds. Let be defined in the same way as in (6) of Lemma 2.4.
-
(i)
If is such that , then there exist such that (1.3) has exactly one solution for , exactly two solutions for , and no solution for .
-
(ii)
If is such that .
Case 1. If, then there existsuch that (1.3) has exactly one solution for, exactly two solutions for, and no solution for.
Case 2. If, then there existsuch that (1.3) has exactly one solution for, exactly two solutions for, and no solution for.
Case 3. If, then there existsuch that (1.3) has exactly one solution for, exactly two solutions for, and no solution for.
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Acknowledgements
This work is sponsored by the project NSFC (11301178, 11171032), the Fundamental Research Funds for the Central Universities (2014MS58) and the improving project of graduate education of Beijing Information Science and Technology University (YJT201416). The authors are grateful to the anonymous referees for their constructive comments and suggestions, which have greatly improved this paper.
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MF completed the main study and carried out the results of this article. XZ checked the proofs and verified the calculation. All the authors read and approved the final manuscript.
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Feng, M., Zhang, X. Time-map analysis to establish the exact number of positive solutions of one-dimensional prescribed mean curvature equations. Bound Value Probl 2014, 193 (2014). https://doi.org/10.1186/s13661-014-0193-4
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DOI: https://doi.org/10.1186/s13661-014-0193-4