#### Abstract

We consider the equations involving the one-dimensional -Laplacian   , and where and . We show the existence of sign-changing solutions under the assumptions and . We also show that has exactly one solution having specified nodal properties for for some . Our main results are based on quadrature method.

#### 1. Introduction and Main Results

Existence and multiplicity of positive solutions of nonlinear second order boundary value problem and its generalized forms have been extensively studied via the fixed point theorem in cones, bifurcation theory, quadrature method, and fixed index theorem in the past four decades; see Erbe and Wang [1], Henderson and Wang [2], Laetsch [3], Fink et al. [4], Ma and Thompson [5, 6], and the references therein.

Existence and multiplicity of positive solutions of the corresponding one-dimensional -Laplacian have also been studied by several authors; see Lee and Sim [7], Wang [8], Kong and Wang [9], Aranda and Godoy [10], Bouguima and Lakmeche [11], and de Coster [12] for references along this line.

Recently, Lee and Sim [7] consider the existence and multiplicity of positive solutions of (2), (3) under the assumptions They proved the following.

Theorem A (see [7, Theorem 3.14]). Assume (4) hold. Then, there exist such that (2), (3) have at least one positive solution for and no positive solution for .

Of course, natural question is as follows. What would happen if we allow that ?

It is the purpose of this paper to study sign-changing solutions of (2), (3) under the assumptions and or The main tool is the quadrature method.

We will make the following assumptions: (H0) for ; (H1);  (H2) and .

Let , where . The main results of this paper are the following.

Theorem 1. Let (H0), (H1), and (H2) hold. Assume that satisfies Then, for , (2), (3) have two solutions and for each : has zeros in and is positive near , and has zeros in and is negative near . Moreover, there exists a constant , such that for each the above solution is unique.

Theorem 2. Let (H0), (H1), and (H2) hold. Assume that . Then, for , (2), (3) have two solutions and for each : has zeros in and is positive near , and has zeros in and is negative near . Further, there exists a constant independent of , such that for each the above solution is unique.

Theorem 3. Let (H0) and (H1) hold. Assume that . Then, for , there exists a constant small independent of , such that for each (2), (3) have two solutions and : and have zeros in and are positive near 0; and problems (2), (3) have two solutions and , where and have zeros in and are negative near .

Remark 4. For , the existence of positive and sign-changing solutions has been extensively studied by many authors [16], but they did not give any information about the uniqueness of nodal solutions.

Remark 5. It is worth noticing that Lee and Sim [7] studied the nonautonomous cases (2), (3) and obtained the existence of positive solutions with . They gave no information about the sign-changing solutions. In Theorem 1, we show the existence of solutions having specified nodal properties.

Remark 6. Very little is known in the available literature even in the special case . We establish uniqueness results in this paper; see Theorems 1 and 2.

Remark 7. Let us consider the problem where . Obviously, satisfies (H0) and (H1). Since it is easy to see that (H2) is fulfilled. Thus, Theorem 1 implies that, for , (6) have two solutions and for each : has zeros in and is positive near , and has zeros in and is negative near . Moreover, there exists a constant , such that for each the above solution is unique.

For other results dealing with -Laplacian operators and the bifurcation behavior of solutions, see [1324] and the references therein.

The rest of the paper is arranged as follows. In Section 2, we state and prove some preliminary results. Finally, in Section 3, we give the proofs of Theorems 1, 2, and 3.

#### 2. Quadrature Method and Preliminaries

Let for and .

Lemma 8. If is any solution of (2), (3) and is such that , then .

Proof. Since is autonomous, both and satisfy the initial value problem By Reichel and Walter [14, Theorem 2] and [14, (iii) and (v) in the case () of Theorem 4], (8) has a unique solution defined on . Therefore, .

Now, we divide the discussion into two cases.

Case 1 (.). In this case, we attempt to find a solution of (2), (3) with zeros in (0, 1) and and a solution of (2), (3) with zeros in (0, 1) and .
Obviously, if is a sign-changing solution with zeros in and , then, thanks to Lemma 8 and the fact that (2) is autonomous, we only need to study on the intervals and .

Multiplying (2) throughout by , we obtain and integrating we have If and , then and . Substituting and in (10), we get and . Hence, is such that Thus, we have Integrating (12) and (13) on and , respectively, we obtain Hence, substituting in (14) and in (15), we have Multiplying (17) by and adding to (16), we can see that and satisfy In fact, the following result holds.

Lemma 9. Given , if there exists such that , then (2), (3) have a sign-changing solution with interior zeros satisfying . Further, is a continuous function in and it is also differentiable with the derivative given by where .

The proof of the above theorem follows by carefully extending the arguments used in [15, Theorem 2.2] for second order differential equation to the case of one-dimensional -Laplacian.

Using the same argument, with obvious changes, we may deduce the following.

If is a sign-changing solution with zeros in and , the corresponding is

Case 2 (). In this case, if is a sign-changing solution with zeros in and , the corresponding is Similarly, we may get the same function as above when is a sign-changing solution with zeros in with .

#### 3. The Proofs of the Main Results

Proof of Theorem 1. First, we consider .
It follows from the quadrature method that a solution with zeros in (0,1) exists if for there exists such that . To prove this, we will show that . We achieve this by proving (A),  (B).
Proof of (A). Recall that
First let us consider
To this end, we have from (H1) that, for any , there exists , such that
If , it follows from (23) and (24) that we have that where It follows from the fact that is sufficiently large and (25) that
Next, we know that as (). We consider To this end, we have from (H1) that, for any , there exists , such that
If , it follows from (28) and (29) that we have that where It follows from the fact that is sufficiently large and (30) that Therefore, from (27) and (32), we have that .
Proof of (B). Recall that
First, let us consider
Since , then, for any , there exists such that Thus, if , the second part of (35) implies that Similarly, from the first part of (35), we have that It follows from (36), (37) and the fact is arbitrary that In fact, as (); we consider From , then, for any , there exists such that Thus, if , the second part of (40) implies that Similarly, from the first part of (40), we have that It follows from (41), (42) and the fact that is arbitrary that we have that
Therefore, from (40) and (43), we have that By analysing defined in (20) instead of in the proof of the above, we have the same result. Thus, we have shown that there are two solutions with interior zeros, which are negative near 0 and positive near 0 for , respectively.
Now, in order to achieve the existence of , we will first establish that for large enough. In fact, First, we consider where . From the first part of (H2), it follows that if is large enough.
Next, let us consider where . From the second part of (H2), we have that for and large enough. In fact, as . Consequently, we get that for large enough.
Finally, if , this clearly follows by analysing defined in (21) instead of in the proof of the case .

Proof of Theorem 2. First, we consider .
It follows from the quadrature method that a solution with interior zeros exists if for there exists such that . To prove this, we will show that . We achieve this by proving  (A1), (B1). The proof of (A1) is the same as the proof of (A) of Theorem 1, so we omit it here; we are only to prove (B1). Recall that
First let us consider
From , then, for any , there exists such that Thus, if , from (50), we have that
Next, in fact, as   ; we consider
Since , then, for any , there exists such that Thus, if , from (54), we have that From the fact that is small and combining (52) and (55), we get that By analyzing defined in (20) instead of in the proof of the above, we have the same result. The proof of is similar to the proof of Theorem 1. We omit it here.
Finally, if , then it clearly follows by analyzing defined in (21) instead of in the proof of the case .

Proof of Theorem 3. First, we consider .
It follows from the quadrature method that a solution with interior zeros exists if for there exists such that . We prove this by proving  (A2), (B2).The proof of (A2) is the same as the proof of (A) of Theorem 1, so we omit it here; we are only to prove (B2).
Recall that
First, consider
Since , then, for any large , there exists such that Thus, if , from (58) and (59), we have that
Next, in fact, as (); we consider
Since , then, for any large , there exists such that
Thus, if , from (62), we have that From the fact that is arbitrary and large and combining (60) and (63), we get that By analyzing defined in (20) instead of in the proof of the above, we have the same result.
Finally, if , then it clearly follows by analyzing defined in (20) instead of in the proof of the case .

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work is supported by the NSFC (no. 11361054), SRFDP (no. 20126203110004), and Gansu Provincial National Science Foundation of China (no. 1208RJZA258).