Journal of Function Spaces

Volume 2018, Article ID 9269458, 7 pages

https://doi.org/10.1155/2018/9269458

## Global Bifurcation from Intervals for the Monge-Ampère Equations and Its Applications

Department of Basic Courses, Lanzhou Institute of Technology, Lanzhou 730050, China

Correspondence should be addressed to Wenguo Shen; moc.361@963gwnehs

Received 8 August 2017; Accepted 20 November 2017; Published 9 January 2018

Academic Editor: Vijay Gupta

Copyright © 2018 Wenguo Shen. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We shall establish the global bifurcation results from the trivial solutions axis or from infinity for the Monge-Ampère equations: , in , , on , where is the Hessian matrix of , where is the unit open ball of , is a radially symmetric weighted function and on any subinterval of , is a positive parameter, and the nonlinear term , but is not necessarily differentiable at the origin and infinity with respect to , where . Some applications are given to the Monge-Ampère equations and we use global bifurcation techniques to prove our main results.

#### 1. Introduction

The Monge-Ampère equations are a type of important fully nonlinear elliptic equations [1–3]. Existence and regularity results of the Monge-Ampère equations can be found in [4–8] and the references therein.

We first consider the following real Monge-Ampère equations:where is the Hessian matrix of , is the unit ball of , is a weighted function, is a positive parameter, and .

Kutev [9] and Delano [10] treated the existence of convex radial solutions of problem (1) with , and , respectively. Caffarelli et al. [11] have investigated problem (1) in general domains of .

In [9, 12], the authors have shown that problem (1) can reduce to the following boundary value problem:where with . By a solution of problem (2), we understand that it is a function which belongs to and satisfies (2). It has been known that any negative solution of problem (2) is strictly convex in . Wang [13] and Hu and Wang [12] (; ) also considered the existence of strictly convex solutions for problem (2) by using fixed index theorem. Lions [14] proved the existence of the first eigenvalue of problem (1) with , via constructive proof.

By global bifurcation theorem, Dai and Ma [15] and Dai [16] studied the Monge-Ampère equations (1) with equal and , respectively, where satisfies .

(H0) is radially symmetric and , on any subinterval of , where with .

However, the nonlinearities of the above papers are differentiable at the origin. In 1977, Berestycki [17] established an interval bifurcation theorem for the problems involving nondifferentiable nonlinearity. The main difficulties when dealing with this problem lie in the bifurcation results of [15, 16] which cannot be applied directly to obtain our results. Recently, Ma and Dai [18] and Dai and Ma [19, 20] also considered similar problems to [17].

Motivated by the above papers, we shall consider problem (1), where , with being radially symmetric with respect to , and .

It is clear that the radial solutions of (1) are equivalent to the solutions of problem (2), where satisfies (H0) and and satisfy the following conditions.

(H1) and there exist , such that uniformly for , , and for all .

(H2) and there exist , such that uniformly for , for some positive constant large enough, and for all .

(H3) near uniformly for and on bounded sets.

(H4) near uniformly for and on bounded sets.

Under assumptions (H1) and (H3), we shall establish interval bifurcation of (2) from the trivial solutions axis by Rabinowitz [21]. Moreover, by the global bifurcation from infinity of Rabinowitz [22], we shall also establish a result involving global bifurcation of (2) from infinity with conditions (H2) and (H4).

Following the above theory (see Theorems 3 and 6), we shall investigate the existence of radial solutions for the following problem:where .

It is clear that the radial solutions of (5) are equivalent to the solutions of the following problem:where is a positive parameter and and with satisfy the following.

(A0) is radially symmetric and , on any subinterval of , where with .

(A1) and there exist such that

(A2) and for and there exist such that

The rest of this paper is arranged as follows. In Section 2, we establish the bifurcation results which bifurcate from the trivial solutions axis and from infinity for problem (2), respectively. In Section 3, on the basis of the interval bifurcation result (Theorems 3 and 6), we give the intervals for the parameter which ensure existence of single or multiple strictly convex solutions for problem (6) under the assumptions of (A1)-(A2).

#### 2. Global Interval Bifurcation

Let with the norm . Let with the usual norm . Let . Set under the product topology.

Now, we consider the following eigenvalue problem:By [16, of Section , p.11], the same proof as in Theorem of [14], we can show that problem (9) possesses the first eigenvalue which is positive, simple, and unique, and the corresponding eigenfunctions are positive in and concave on .

By Sections - in [16], with a simple transformation , problem (2) can be equivalently written aswhere satisfies (H0) when satisfies (H2). According to Rabinowitz [21], using the same method to prove [16, Theorems and ] with obvious changes, we may get the following global bifurcation result.

Lemma 1 ([16, Theorem ]). *Assume that (H0) holds and satisfies (H3). Then is the unique bifurcation point of problem (10) and there exists an unbounded bifurcation continuum of solutions to problem (10) emanating from .**Hence, , where satisfy conditions (H1) and (H3), respectively. Let denote the closure in of the set of nontrivial solutions of (10) with under the assumptions of (H0), (H1), and (H3). By an argument similar to that of [16, Lemma 4.1] with obvious changes, we can show that the following existence and uniqueness theorem is valid for problem (10).*

Lemma 2 ([16, Lemma ]). *If is a solution of (10) under the assumptions of (H0), (H1), and (H3) and has a double zero, then **The first main result for (10) is the following theorem.*

Theorem 3. *Let (H0), (H1), and (H3) hold. Let . The component of contains , such that*(i)*;*(ii)* is unbounded.**To prove Theorem 3, we introduce the following auxiliary approximate problem:**To prove Theorem 3, the next lemma will play a key role.*

*Lemma 4. Let , , be a sequence converging to . If there exists a sequence such that is a nontrivial solution of problem (11) corresponding to and converges to in , then .*

*Proof. *Let ; then, satisfies the problemLet for all and on bounded sets, and then is nondecreasing with respect to and uniformly for and on bounded sets. By an argument similar to that of [16, (4.7)] with obvious changes, we can show that as uniformly for and on bounded sets. By (H1), we have that is convergent in . Without loss of generality, we may assume that in with . Obviously, we have .

Now, we deduce the boundedness of . Let be an eigenfunction of problem (9) with corresponding to .

Similar to in Lemma of [16], by some simple calculations, we have that where Similar to the proof of in Lemma of [16], we can show that the left-hand side of (14) equals . The Young’s inequality implies that . It follows thatSimilarly, we can also show thatSimilar to [17, Lemma ], we can easily show thatfor large enough.

It follows from (16) and (18) that which implies .

Similarly, it follows from (17) and (18) that .

Therefore, we have that .

*Proof of Theorem 3. *We divide the rest of the proofs into two steps.*Step 1*. Using a similar method to prove [20, Theorem ] with obvious changes, we may prove that .*Step 2*. We prove that is unbounded. Suppose on the contrary that is bounded. Using a similar method to prove [20, Theorem ] with obvious changes, we can find a neighborhood of such that .

In order to complete the proof of this theorem, we consider problem (11). For , it is easy to show that the nonlinear term satisfies condition (H3). Let By Lemma 1, there exists an unbounded continuum of bifurcating from such that So there exists for all . Since is bounded in , (11) shows that is bounded in independently of . By the compactness of , one can find a sequence such that converges to a solution of (11). So, . If , then from Lemma 2 it follows that . By Lemma 4, , which contradicts the definition of . On the other hand, if , then which contradicts .

*From Theorem 3 and its proof, we can easily get a corollary.*

*Corollary 5. Let (H0), (H1), and (H3) hold. There exists a subcontinuum of solutions of (10) in , bifurcating from , such that(i);(ii) is unbounded.We add the points to space . Let denote the closure in of the set of nontrivial solutions of (10) under conditions (H2) and (H4) with . Let denote the spectral set of problem (9). Let , where .*

According to Rabinowitz [22], our second main result for (10) is the following theorem.

*Theorem 6. Let (H0), (H2), and (H4) hold. Let . There exists a connected component of , containing . Moreover, if is an interval such that and is a neighborhood of whose projection on lies in and whose projection on is bounded away from , then either is bounded in in which case meets ; is unbounded.If occurs and has a bounded projection on , then meets . Moreover, there exists a neighborhood of such that .*

*Proof. *We use a similar method to prove [18, Theorem ] with obvious changes, but we give a rough sketch of the proof for readers’ convenience. If with , dividing (10) by and setting yieldDefineClearly, (22) is equivalent toIt is obvious that is always the solution of (24). By simple computation, we can show that assumptions (H2) and (H4) imply that , satisfy (H1) and (H3). Now, applying Theorem 3 to problem (24), we have that the component of , containing , is unbounded and lies in . Under the inversion , satisfying problem (10). By an argument similar to that of [18, Theorem ], we can prove the existence of a neighborhood of such that .

*3. Applications*

*3. Applications*

*In this section, we shall investigate the existence and multiplicity of convex solutions of problem (6). With a simple transformation , problem (6) can be written as*

*By [16], in order to prove our main results, we need the following Sturm type comparison result.*

*Lemma 7 (see [16, Lemma ]). Let , such that for and the inequality is strict on some subset of positive measure in . Also, let and be solutions of the following differential equations:respectively. If in , then has at least one zero in .The main results of this section are the following theorems.*

*Theorem 8. Let (A0), (A1), and (A2) hold. If and , for eitherorthen problem (6) has at least one solution such that it is negative and strictly convex in .*

*Theorem 9. Let (A0), (A1), and (A2) hold. If and , for then problem (6) has at least one solution such that it is negative and strictly convex in .*

*Theorem 10. Let (A0), (A1), and (A2) hold. If and , for then problem (6) has at least one solution such that it is negative and strictly convex in .*

*Theorem 11. Let (A0), (A1), and (A2) hold. If and , for then problem (6) has at least one solution such that it is negative and strictly convex in .*

*Theorem 12. Let (A0), (A1), and (A2) hold. If and , for then problem (6) has at least one solution such that it is negative and strictly convex in .*

*Remark 13. *Note that if , then Theorem 8 is equivalent to Theorem of [15] or Theorem of [16]. In this sense, Theorem 8 is also a generalization of Theorem of [15] or Theorem of [16].

*Proof of Theorem 8. *It suffices to prove that problem (25) has at least one solution such that it is positive and strictly concave in .

Firstly, we study the bifurcation phenomena for the following eigenvalue problem:where is a parameter.

Let be such that with Let ; then, is nondecreasing andFurther, it follows from (35) that Hence, (33), (34), and (36) imply that conditions (H1) and (H3) hold. Moreover, we have that .

By Theorem 3, there is a distinct unbounded component of , containing and lying in .

Let be such that with Let , and then is nondecreasing.

Define Then, we can see thatIt is not difficult to verify that is bounded in . From this fact and (37), it follows that where . So, we haveFurther, it follows from (41) thatHence, (33), (37), and (42) imply that conditions (H2) and (H4) hold. Moreover, we have that .

By Theorem 6, there is a distinct unbounded component of , which satisfies the alternates of Theorem 6. Moreover, there exists a neighborhood of such that .

We claim that . It suffices to show that meets some point of . In fact, if this occurs, we can show that Suppose on the contrary that ; hence, . So, , noting , which contradicts , where denotes the closure in of the set of nontrivial solutions of (25) under conditions (H1) and (H3) with , where , , where . is a connected component of , containing .

Hence, , and it follows that .

Next, we shall show that of Theorem 6 does not occur. Suppose on the contrary that of Theorem 6 occurs; then, we shall deduce a contradiction. Firstly, we show that has a bounded projection on . We may claim . If , then there exists . Since , by Lemma 4, ; that is, of Theorem 6 occurs, which is a contradiction.

On the contrary, we suppose that such that It follows thatIn view of (A1) and (A2), we have that for any . By the Sturm comparison (Lemma 7), we get that has at least one zero in for large enough, and this contradicts the fact that does not change its sign in .

In the following, we show that the case of meeting is impossible. Assume on the contrary that meets . So, there exists a neighborhood of such that , where is a neighborhood of which satisfies the assumptions of Theorem 6, which contradicts , where and , where .

For simplicity, we write . It is clear that any solution of (33) of the form yields a solution of (25).

By (27), we obtain By (28), we haveFrom and , it follows that the subsets and of can be separated by the hyperplane . Furthermore, we have crossing the hyperplane in .

*Proof of Theorems 9 and 10. *The proof is similar to that of Theorem 8. In this case, it follows that (45) holds and (46) is impossible.

*Proof of Theorems 11 and 12. *The proof is similar to that of Theorem 8. In this case, it follows that (46) holds and (45) is impossible.

*Remark 14. *Note that if and or and , (45) and (46) are impossible, and it follows that the subsets and of cannot be separated by the hyperplane . In this case, we cannot give a suitable interval of in which there exist nodal solutions for (25). It would be interesting to have more information about this case.

* Conflicts of Interest*

*Conflicts of Interest*

*The author declares that there are no conflicts of interest regarding the publication of this paper.*

*Acknowledgments*

*Acknowledgments*

*The author was supported by the NSFC (no. 11561038) and Gansu Provincial National Science Foundation of China (no. 145RJZA087).*

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