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

We shall establish the global bifurcation results from the trivial solutions axis or from infinity for the Monge-Ampère equations: det(D2u) = λm(x)(−u)N+m(x)f1(x, −u, −u󸀠, λ)+f2(x, −u, −u󸀠, λ), in B, u(x) = 0, on ∂B, whereD2u = (∂2u/∂xi∂xj) is the Hessian matrix of u, where B is the unit open ball ofRN,m ∈ C(B, [0, +∞)) is a radially symmetric weighted function andm(r) := m(|x|) ̸ ≡ 0 on any subinterval of [0, 1], λ is a positive parameter, and the nonlinear term f1, f2 ∈ C(B × (R+)3,R+), but f1 is not necessarily differentiable at the origin and infinity with respect to u, where R = [0, +∞). Some applications are given to the Monge-Ampère equations and we use global bifurcation techniques to prove our main results.

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].
It is clear that the radial solutions of (1) are equivalent to the solutions of problem (2), where  satisfies (H0) and  1 and  2 satisfy the following conditions.
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  ∈ (R + , R + ).
It is clear that the radial solutions of (5) are equivalent to the solutions of the following problem: where  is a positive parameter and ℎ and  =  1 +  2 with  1 ,  2 ∈ (R + , R + ) satisfy the following.
The first main result for (10) is the following theorem.

Theorem 3. Let (H0), (H1), and (H3) hold. Let 𝐼
To prove Theorem 3, we introduce the following auxiliary approximate problem: To prove Theorem 3, the next lemma will play a key role.
Similar to (4.12) in Lemma 4.5 of [16], by some simple calculations, we have that where Similar to the proof of (4.12) in Lemma 4.5 of [16], we can show that the left-hand side of ( 14) equals 0. The Young's inequality implies that  ≥ 0. It follows that Similarly, we can also show that Similar to [17, Lemma 1], we can easily show that for  large enough.It follows from ( 16) and ( 18) that which implies  ≥  1 −  0 .Similarly, it follows from ( 17) and ( 18) that  ≤  1 −  0 .Therefore, we have that  ∈  0 1 .
Proof Theorem 3. We divide the rest of the proofs into two steps.
Step 2. We prove that C is unbounded.Suppose on the contrary that C is bounded.Using a similar method to prove [20, Theorem 1.1] with obvious changes, we can find a neighborhood O of C such that O ∩ S = 0.
In order to complete the proof of this theorem, we consider problem (11).For  > 0, it is easy to show that the nonlinear term (, V|V|  , )+(, V, ) satisfies condition (H3).Let By Lemma 1, there exists an unbounded continuum C  of S  bifurcating from ( 1 , 0) such that So there exists (11) shows that (  , V  ) is bounded in R ×  2 independently of .By the compactness of   , one can find a sequence   → 0 such that (   , V   ) converges to a solution (, V) of ( 11).So, V ∈  + .If V ∈  + , then from Lemma 2 it follows that V ≡ 0. By Lemma 4,  ∈  0 1 , which contradicts the definition of O. On the other hand, if V ∈  + , then (, V) ∈ S ∩ O which contradicts S ∩ O = 0.
From Theorem 3 and its proof, we can easily get a corollary.(10) in R × , bifurcating from  0 1 × {0}, such that
According to Rabinowitz [22], our second main result for (10) is the following theorem.

Applications
In this section, we shall investigate the existence and multiplicity of convex solutions of problem (6).With a simple transformation V = −, problem (6) can be written as By [16], in order to prove our main results, we need the following Sturm type comparison result.
The main results of this section are the following theorems.
Proof of Theorem 8.It suffices to prove that problem (25) has at least one solution V such that it is positive and strictly concave in (0, 1).
Firstly, we study the bifurcation phenomena for the following eigenvalue problem: where  > 0 is a parameter.Let  ∈ (R + , R + ) be such that It is not difficult to verify V/ is bounded in R + .From fact and it follows that lim sup where  = /2.So, we have Further, it follows from (41) that as ‖V‖ → +∞. (42) Hence, (33), (37), and (42) imply that conditions (H2) and (H4) hold.Moreover, we have that . By Theorem 6, there is a distinct unbounded component D ∞ of T ∪ ( ∞ 1 × {∞}), which satisfies the alternates of Theorem 6.Moreover, there exists a neighborhood , where T 0 denotes the closure in R ×  of the set of nontrivial solutions (, V) of (25) under conditions (H1) and (H3) with V ∈ , where Hence,  * ∈  0 1 , and it follows that D 0 = D ∞ .Next, we shall show that (2 o ) of Theorem 6 does not occur.Suppose on the contrary that (2 o ) of Theorem 6 occurs; then, we shall deduce a contradiction.Firstly, we show that D ∞ −M has a bounded projection on R. We may claim o ) of Theorem 6 occurs, which is a contradiction.
On the contrary, we suppose that It follows that In view of (A1) and (A2), we have that lim for any  ∈ (0, 1).By the Sturm comparison (Lemma 7), we get that V  has at least one zero in (0, 1) for  large enough, and this contradicts the fact that V  does not change its sign in (0, 1).
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 − 0 <  0 2 ≤ − 0 and  ∞ 2 ≤ − ∞ or  0 2 ≤ − 0 and  ∞ 2 ≤ − ∞ , (45) and (46) are impossible, and it follows that the subsets  0 1 ×  and  ∞ 1 ×  of R ×  cannot be separated by the hyperplane {1} × .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.