Abstract

We study the bifurcation and the exact multiplicity of solutions for a class of Neumann boundary value problem with indefinite weight. We prove that all the solutions obtained form a smooth reversed S-shaped curve by topological degree theory, Crandall-Rabinowitz bifurcation theorem, and the uniform antimaximum principle in terms of eigenvalues. Moreover, we obtain that the equation has exactly either one, two, or three solutions depending on the real parameter. The stability is obtained by the eigenvalue comparison principle.

1. Introduction

The existence and multiplicity of solutions of Neumann problems have been investigated by many authors; see, for example, [15]. It is well known that determining the exact number of solutions of semilinear equations is usually a very difficult and challenging task. The results on exact multiplicity of solutions for Neumann problems are very few in the previous literature.

In this paper, we study the bifurcation and the exact multiplicity of solutions for the Neumann problemdepending on a real parameter , where , , and are given continuous functions and and may change sign.

Llibre and Roberto [6] studied the existence and the stability of periodic solutions of Duffing equationChen and Li [7] also studied the Duffing equation (2) in a very particular case, i.e., , and constants. They obtained that (2) has exactly three T-periodic solutions. Lomtatidze et al. [8] also studied the existence of periodic solutions of Duffing type equations.

Tzeng et al. [9] also studied the global bifurcation and exact multiplicity of positive solutions of Dirichlet problem with cubic nonlinearityusing the time-map method, where are constants. Equation (3) is an autonomous system, and the time-map method has been successfully employed to solve the problem (3), but it is not applicable to study the nonautonomous system (1). There are many results on exact multiplicity of solutions for the Dirichlet problems; see [912].

In [13], under Neumann boundary value conditions, the bifurcation of solutions to a logistic equation with harvesting has been investigated using the uniform antimaximum principle and Crandall-Rabinowitz bifurcation theorem. The uniform antimaximum principle plays an important role in proving the main results. More theories and applications of antimaximum principle can be seen, for example, [14, 15]. As continuation of [13], in this paper, the bifurcation of solutions for a Neumann problem with cubic nonlinearity is investigated using the uniform antimaximum principle, Crandall-Rabinowitz bifurcation theorem, the topological degree theory, and the continuation method. More detailed results on bifurcation theory can be seen in [16, 17]. The topological degree theory and the uniform antimaximum principle play an important role in proving the main results in this paper.

2. Preliminaries

Definition 1 (see [18]). We call a solution of the equationa stable solution if the principal eigenvalue of the equationis strictly positive. The solution is unstable if the principal eigenvalue is negative.

Definition 2 (see [18]). We call a solution of the equationa nondegenerate solution if the linearized equationdoes not admit any nontrivial solutions. The solution is degenerate (singular) if the linearized equation (7) has nontrivial solutions.

Definition 3 (see [19]). A mapping ( and are topological spaces) is said to be proper if for every compact set , the set is compact in .

Lemma 4 (see [19]). Let be a real Banach space and a linear, compact map in . Suppose that and is not an eigenvalue of . Let be an open bounded and Thenwhere is the sum of the algebraic multiplicities of all the eigenvalues satisfying , and if has no eigenvalues of this kind.

Lemma 5. Consider the eigenvalue problemThen (9) has the Green functionandwhere for any , denotes a ball of radius , and .

Proof. It is easy to obtain (10) and the proof is omitted. The eigenvalue problem (9) is equivalent to the equation . The equation has nontrivial solutions if and only if is the eigenvalue of the operator . For , the general solution of equation (9) isBy Neumann boundary value condition, we haveWhen , we have , where , . Therefore, is the eigenvalue of , where , is the one-dimensional subspace of Banach space spanned by .
For , Lemma 4 implies thatThis completes the proof.

Consider the eigenvalue problemWe recall some propositions listed as follows.

Proposition 6. It is well known that the problem (15) has the eigenvalues . The first eigenvalue is real and simple, and the corresponding eigenfunction does not change sign. When , we denote by . It is obvious that is the first eigenvalue and is the second eigenvalue of the eigenvalue equation

Proposition 7. The comparison theorem of eigenvalues can be stated as follows: is strictly decreasing in the sense that implies that , where , namely, , with the strict inequality on a set of positive measure.

Proposition 8. Suppose that . Then the equationdoes not admit any nontrivial solutions.

Proof. Argue by contradiction, (17) admits nontrivial solutions. Let be a nontrivial solution of the equation (17). It follows from that and , for . It follows from (17) that is the eigenvalue, contradiction. Therefore, (17) admits only trivial solution . Equation (17) does not admit any nontrivial solutions. This completes the proof.

In order to prove our main theorem using the topological degree theory, the following lemma is essential. In the following lemma, the weight function may change sign.

Lemma 9. Suppose that satisfies the equationThenwhere , is defined in (10), for any , denotes a ball of radius .

Proof. Equation (18) is equivalent to the equationIn (11), let and ; we haveNext we calculate . Let . Now we prove that for and . Argue by contradiction and assume that ; that is, for , the equationhas solution satisfying on .
Case 1. When does not change sign:
(i) For , choose ; that is, for , , by Proposition 7, we have . It follows that (22) has only trivial solution . Contradiction, therefore, we have for , . Using the homotopy invariance we getTherefore, for , that is, ,(ii) For , choose , that is, for , , we have and . By Proposition 8, it follows that (22) has only trivial solution , contradiction. Therefore, we have for , . Using the homotopy invariance, we getTherefore, for , , we haveCase 2. When changes sign, let , where , .
(i) For , consider the following equation:where . Obviously,By Proposition 7 and (28), we haveWhen , the solution of (27) isLet ; since , we haveThe solution of (27) isLet . It is easy to prove that for , . Therefore, using the homotopy invariance, we get(ii) For , , by Proposition 7, ; similarly, letSince , we haveLet . It is easy to prove that for , . Therefore, using the homotopy invariance we getThis completes the proof.

Definition 10 (see [20]). Let be open set in Banach space . Suppose is completely continuous operator, . Let be an isolated fixed point of in ; that is, there exists , such that , and is only fixed point of in . Define

Lemma 11 (see [20]). Let be open set in Banach space . Suppose is completely continuous operator; has no fixed point on . Suppose that there are finite isolated fixed points . Then

Lemma 12 (see [20]). Let be open set in Banach space . Suppose is completely continuous operator, , . Suppose that is Fréchet differentiable at and is not the eigenvalue of derived operator . Then is an isolated fixed point of andwhere is the sum of the algebraic multiplicities of all the eigenvalues in .

Now we state three known results, which plays a key role in proving the main results in this paper. We do not provide their proof which can be seen in [21].

Lemma 13. Suppose that , and such thatThen(1)the possible solution of (18) is either or for each ;(2) cannot both admit nontrivial solutions if .

Lemma 14. Let on and satisfyIf is a solution of the nonhomogeneous differential equationthen the following statements hold:(1)either or for all ;(2)maximum principle: , if ;(3)uniform antimaximum principle: , if .

For the reader’s convenience, we denote the continuous function in (1). The assumptions on function are listed as follows:(f1) is locally differentiable with respect to the second variable with ;(f2), uniformly in

Lemma 15. Let be an order Banach space and an order set. Assume that the given continuous function satisfies (f1). Then(1)the solutions of (1) are totally ordered;(2)(1) cannot admit three distinct solutions in if is strictly increasing or strictly decreasing in

For the convenience, we recast (1) in the operator form

Lemma 16 (see [16]). Let and be Banach spaces. Assume that is continuously differentiable on satisfying the following three conditions:(1) for some , and the null space ;(2)the Fredholm index of is zero, ;(3). Then there is a continuously differentiable curve through ; that is, there exists such thatand all solutions of in a neighborhood of belong to the curve,

3. Main Results

Lemma 17. Assume (f2) holds. Then for every fixed , (1) has at least a solution andfor large enough, where and is defined in (10).

Proof. It follows from (f2) that there exists large enough such that and . Therefore, and are subsolution and supersolution of (1), respectively. Therefore, there exists at least one solution of (1) between and . Next, we show that all the solutions of (1) are between and for large enough. Argue by contradiction, suppose that there exists such that attains its maximum value and . We haveIt follows from and thatFrom the above there exists large enough such that , contradiction. Similar, suppose that there exists such that attains its minimum value and ; we will obtain contradiction. Therefore, there exists large enough such that for all the solutions . (1) is equivalent to the following equation:Next we calculate . Let . The equation is equivalent to the following equation:It is evident that and are subsolution and supersolution of (49), respectively. Therefore, all the solutions of (49) must be between and ; that is, the equation has no solution on . Therefore, for , . By the homotopy invariance properties of the topological degree we haveThis completes the proof.

Lemma 18. Suppose that , , and for all . Then (1) has exactly three solutions , for , where is the unique positive stable solution and is the unique negative stable solution.

Proof. For the mapping defined in (43), let , and we haveNext we prove that any solution of is nondegenerate.
First, we have the Fréchet derivative of (51)It is obvious that is the solution of , and therefore, when , in (52). It follows from that is nondegenerate.
Next, we prove that any nontrivial solution of is nondegenerate. is equivalent to the following equation:where and is defined in (10). The equationis equivalent to , where denotes the derivative operator of . When , , by Lemma 9, Definition 10, and Lemma 12, there exists , and we haveIf is a nontrivial solution of , it follows from Lemma 13 that for all . Therefore,Let and . It follows from the above hypothesis that is a nontrivial solution of . By the second conclusion of Lemma 13, we have that does not have nontrivial solution, which implies that is a nondegenerate solution.
Finally, we prove the positive solution is unique and stable. We denote by the nontrivial solution of (51). It is obvious that is also the nontrivial solution of (51). It follows from (56) and the comparison of eigenvalues that . By Lemma 9, Definition 10, and Lemma 12, we haveLet be the number of nontrivial solutions of . Hence, is proper. Since is a regular value of , must be finite. According to Lemma 17, we have for sufficiently large . By Lemmas 9 and 11 and the index formula, we have thatIt is obvious that . Let for . Therefore, has exactly three solutions , and . Since , we have that the positive solution is unique and stable. The negative solution is also stable. This completes the proof.

Lemma 19. Suppose that . For some , (1) has a unique positive solution with , where is the unique positive solution of (1) for .

Proof. By Lemma 18, for (1) has the unique positive solution , which provided a subsolution of (1) for . There exists large enough such that . Thus, is a supersolution of (1). Therefore, we prove that there exists a positive solution of (1) for such that .
Next we will prove that the positive solution is unique. Assume, by contradiction that (1) has another positive solution for . Let ; then satisfies the following equation:Clearly, By Lemma 13, is the solution of (59), which is a contradiction. Therefore, the solution obtained above is only positive solution. For , satisfies (1) andwhere . Again since and , it follows from Lemma 14 that . Consider the linearization associated with (1)where Thus, Therefore, by Lemma 13 we have is nondegenerate. This completes the proof.

Lemma 20. For , (1) has a unique negative solution with , where is the unique negative solution of (1) for .

Proof. The proof is similar to the proof of Lemma 19.

Theorem 21. Assume that the first eigenvalue and for all . Suppose that . Then all the solutions of (1) are of one sign and lie on a unique reversed S-shaped solution curve, which is symmetric with respect to the origin. More precisely, there exists , such that(i)For , (1) has no positive solution and has a unique negative solution which is stable.(ii)For , (1) has exactly two solutions. Moreover, when , the negative solution is stable and the positive solution is degenerate. When , the positive solution is stable and the negative solution is degenerate.(iii)For , (1) has exactly three ordered solutions at the same and the middle solution is unstable and the remaining two are stable. Moreover, when , the maximal solution is positive and the other two are negative. When , the minimal solution is negative and the other two are positive.(iv)For , (1) has no negative solution and has a unique positive solution which is stable.

Proof. It follows from Lemma 19 that (1) has a unique nondegenerate positive solution for some . The solution curve can be continued a little bit such that remains positive for increasing when . By Lemma 18, (1) has a unique nondegenerate positive solution for . The positive solution curve can pass through and and can be continued further for increasing until the linearized equation (61) admits the nontrivial solutions. We claim that the curve of positive solutions cannot be continued for . Next we will prove the existence of . Since , let be the first eigenfunction of the problemMultiplying (1) by and subtracting from (62) multiplied by , after that integrating over , we obtainApplying the mean-value theorem for (63), there exists such thatFor all , the only root of is negative for . Let andand it follows from (64) that for . Therefore, there exists a such that (1) has no positive solution for .
By applying Lemma 14, we obtain that is still a positive solution of (1) for . We denote the degenerate solution . At , we verify that Lemma 16 can be applied here. It follows from (43) that . In fact, is simple and the principal eigenvalue of (5), and the first eigenfunction ; therefore, and the null space . Therefore, the Fredholm index of is zero. Conditions (1) and (2) of Lemma 16 are satisfied. Next we verify Condition (3) of Lemma 16. Suppose on the contrary that ; namely, there is a continuous function satisfyingConsider the linearized equation of (1) at Since is singular at , that is, (67) has a nontrivial solution such that . Multiplying (66) by , subtracting from (67) multiplied by and integrating by parts on , we havea contradiction, since both and are positive. Therefore, Condition (3) of Lemma 16 is satisfied. Near , the solutions ofform a curveDifferentiating (69) twice in , setting and , and evaluating at , we haveMultiplying the linearized equation (67) by , subtracting from (71) multiplied by , and integrating by parts over , we obtainTherefore, is a fold point of to the left. It follows from the above that the curve of positive solutions cannot be continued to the right indefinitely for all . Hence, the positive solution curve will make a left turn at . Near the critical point , by the Crandall-Rabinowitz bifurcation theorem, there are two branches of positive solutions denoted by the upper branch and the lower branch with . It follows from Lemmas 14, 15, and 18 that the fold point is unique, and the upper branch is monotone decreasing for all and lower branch is monotone increasing for all . Therefore, the lower branch curve is monotone increasing and continues to the left without any turnings. Rewriting (1) in the following form:since , it follows from Lemma 14 that or . This shows that the solution changes its sign only at . Since the nonlinearity is an odd function in , it follows from the symmetry that if is a solution of (1), so is . Thus, the component of solutions we constructed above forms a smooth reversed S-shaped curve with exactly two turning points and .
It follows from Lemma 18 that the positive solution is stable. For , the upper branch remains stable until it reaches the degenerate solution . Next we prove the lower branch is unstable. Let be any one solution of lower branch such that . Since , we have . Therefore, the lower branch is unstable. Since the solution set is symmetric with respect to the origin, the stability of negative solutions is obtained by using the property of symmetry. The upper branch of the negative solutions is unstable and the lower branch of negative solutions is stable. Therefore, all solutions of (1) lie on a unique reversed S-shaped solution curve. This completes the proof.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This work was completed with the support of Tian Yuan Special Funds of the National Science Foundation of China (no. 11626182).