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

Authors: Wenguo Shen
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We shall establish the global bifurcation results from the trivial solutions axis or from infinity for the Monge-Ampère equations:det(D2u)=λm(x)-uN+m(x)f1(x,-u,-u′,λ)+f2(x,-u,-u′,λ), in B, u(x)=0, on ∂B, where D2u=(∂2u/∂xi∂xj) is the Hessian matrix of u, where B is the unit open ball of RN, m∈C(B¯,[0,+∞)) is a radially symmetric weighted function and m(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.


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## Body

## 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:(1)detD2u=λmx-uN+Fx,-u,-u′,λ,inB,ux=0,on∂B,where D2u=(∂2u/∂xi∂xj) is the Hessian matrix of u, B is the unit ball of RN, m(x) is a weighted function, λ is a positive parameter, and F∈C(B¯×R+3,R+).Kutev [9] and Delano [10] treated the existence of convex radial solutions of problem (1) with m≡1, F=0 and λm(-u)N+F=λexp⁡fx,u,∇u, respectively. Caffarelli et al. [11] have investigated problem (1) in general domains of RN.In [9, 12], the authors have shown that problem (1) can reduce to the following boundary value problem:(2)u′N′=λNrN-1mr-uN+NrN-1Fr,-u,-u′,λ,r∈0,1,u′0=u1=0,where r=x with x∈B. By a solution of problem (2), we understand that it is a function which belongs to C2[0,1] and satisfies (2). It has been known that any negative solution of problem (2) is strictly convex in (0,1). Wang [13] and Hu and Wang [12] (m≡0; F=f(-u)) 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 λ1 of problem (1) with λm≡λN, F=0 via constructive proof.By global bifurcation theorem, Dai and Ma [15] and Dai [16] studied the Monge-Ampère equations (1) with λm(x)-uN+F equal λN(-uN+g(-u)) and λNmx-uN+g-u, respectively, where g:[0,+∞)→[0,+∞) satisfies lims→0+g(s)/sN=0.(H0)m(x)∈C(B¯) is radially symmetric and m(r)≥0, m(r)≢0 on any subinterval of [0,1], where r=|x| with x∈B¯.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 F=m(x)f1+f2, with f1,f2∈CB¯×R+3,R+ being radially symmetric with respect to x, and R+=[0,+∞).It is clear that the radial solutions of (1) are equivalent to the solutions of problem (2), where m satisfies (H0) and f1 and f2 satisfy the following conditions.(H1)f1∈C([0,1]×R+3) and there exist a0, a0∈(0,∞) such that (3)a0≤liminfs→0+f1r,s,y,λsN<limsups→0+f1r,s,y,λsN≤a0 uniformly for r∈[0,1], 0<y≤1, and for all λ∈R+.(H2)f1∈C([0,1]×R+3) and there exist a∞, a∞∈(0,∞) such that (4)a∞≤liminfs→+∞f1r,s,y,λsN<limsups→+∞f1r,s,y,λsN≤a∞ uniformly for r∈[0,1], y≥C for some positive constant C large enough, and for all λ∈R+.(H3)f2(r,s,y,λ)=o(sN+yN) near (s,y)=(0,0) uniformly for r∈(0,1) and λ on bounded sets.(H4)f2(r,s,y,λ)=o(sN+yN) near (s,y)=(+∞,+∞) uniformly for r∈(0,1) and λ 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 Theorems3 and 6), we shall investigate the existence of radial solutions for the following problem:(5)detD2u=γhxF-u,inB,ux=0,on∂B,where F∈C(R+,R+).It is clear that the radial solutions of (5) are equivalent to the solutions of the following problem:(6)u′N′=γNrN-1hrF-u,r∈0,1,u′0=u1=0,where γ is a positive parameter and h and F=f1+f2 with f1,f2∈C(R+,R+) satisfy the following.(A0)h(x)∈C(B¯) is radially symmetric and h(r)≥0, h(r)≢0 on any subinterval of [0,1], where r=|x| with x∈B¯.(A1)f1∈C(R+,R+) and there exist a0,a0,a∞,a∞∈(0,∞) such that (7)a0≤liminfs→0+f1ssN<limsups→0+f1ssN≤a0,a∞≤liminfs→+∞f1ssN<limsups→+∞f1ssN≤a∞.(A2)f2∈C(R+,R+) and f2(s)>0 for s∈(0,∞) and there exist f20,f2∞∈(0,∞) such that (8)f20=lims→0+f2ssN,f2∞=lims→+∞f2ssN.The rest of this paper is arranged as follows. In Section2, 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 γ which ensure existence of single or multiple strictly convex solutions for problem (6) under the assumptions of (A1)-(A2).


## 2. Global Interval Bifurcation

LetY=C[0,1] with the norm u∞=maxr∈[0,1]|u(r)|. Let E:={u(r)∈C1(0,1)∣u′(0)=u(1)=0} with the usual norm u=max{u∞,u′∞}. Let P+={u∈E:u(r)>0,r∈(0,1)}. Set K+=R×P+ under the product topology.Now, we consider the following eigenvalue problem:(9)-v′N′=λNrN-1mrvN,inr∈0,1,u′0=u1=0.By [16, (4.2) of Section 4, p.11], the same proof as in Theorem 1.1 of [14], we can show that problem (9) possesses the first eigenvalue λ1 which is positive, simple, and unique, and the corresponding eigenfunctions are positive in (0,1) and concave on [0,1].By Sections3-4 in [16], with a simple transformation v=-u, problem (2) can be equivalently written as(10)-v′N′=λNrN-1mrvN+NrN-1Fr,v,v′,λ,r∈0,1,v′0=v1=0,where m satisfies (H0) when F∈C([0,1]×(R+)3,R+) satisfies (H2). According to Rabinowitz [21], using the same method to prove [16, Theorems 4.1 and 4.2] with obvious changes, we may get the following global bifurcation result.Lemma 1 ([16, Theorem 4.2]).
Assume that (H0) holds andF satisfies (H3). Then (λ1,0) is the unique bifurcation point of problem (10) and there exists an unbounded bifurcation continuum C⊆(K+∪{(λ1,0)}) of solutions to problem (10) emanating from (λ1,0).
Hence,F=m(r)f1+f2, where f1,f2 satisfy conditions (H1) and (H3), respectively. Let S denote the closure in R×E of the set of nontrivial solutions (λ,v) of (10) with v∈P+ 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 4.1]).
If(λ,v) is a solution of (10) under the assumptions of (H0), (H1), and (H3) and v has a double zero, then v≡0.
The first main result for (10) is the following theorem.Theorem 3.
Let (H0), (H1), and (H3) hold. LetI10=[λ1-a0,λ1-a0]. The component C of S∪(I10×{0}) contains I10×{0}, such that(i)
C⊂(K+∪(I10×{0}));(ii)
C is unbounded.
To prove Theorem3, we introduce the following auxiliary approximate problem:(11)-v′N′=λNrN-1mrvN+NrN-1mrf1r,vvϵ,v′,λ+NrN-1f2r,v,v′,λ,r∈0,1,v′0=v1=0.
To prove Theorem3, the next lemma will play a key role.Lemma 4.
Letϵn, 0<ϵn<1, be a sequence converging to 0. If there exists a sequence (λn,vn)∈K+ such that (λn,vn) is a nontrivial solution of problem (11) corresponding to ϵ=ϵn and (λn,vn) converges to (λ,0) in R×E, then λ∈I10.Proof.
Letwn=vn/vn; then, wn satisfies the problem(12)-wn′N′=λNrN-1mrwnN+NrN-1mrf1nr+NrN-1f2nr,r∈0,1,wn′0=wn1=0,(13)f1nr=f1r,vnrvnrϵ,vn′r,λvnN,f2nr=f2r,vnr,vn′r,λvnN.
Letf2¯(r,v,λ)=max0≤s≤vf2r,s,λ for all r∈(0,1) and λ on bounded sets, and then f2¯ is nondecreasing with respect to v and limv→0+f2¯(r,v,λ)/vN=0 uniformly for r∈[0,1] and λ on bounded sets. By an argument similar to that of [16, (4.7)] with obvious changes, we can show that f2n(r)→0 as v→0 uniformly for r∈(0,1) and λ on bounded sets. By (H1), we have that wn is convergent in E. Without loss of generality, we may assume that wn→w in E with w=1. Obviously, we have w∈P+.
Now, we deduce the boundedness ofλ. Let ψ∈P+ be an eigenfunction of problem (9) with F=0 corresponding to λ1.
Similar to(4.12) in Lemma 4.5 of [16], by some simple calculations, we have that(14)∫01wnN+1-ψ′NψN-wn-wn′N=A+B, where (15)A=∫01λ1-λnmr-f1nrwnN-f2nrwnNrN-1NwnN+1dr,B=-wn′N+1+N-wnψ′ψN+1+N+1wnNwn′-ψ′ψN.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 B≥0. It follows that(16)∫01λ1-λnmr-f1nrwnN-f2nrwnNrN-1NwnN+1dr≤0.Similarly, we can also show that(17)∫01λn-λ1mr+f1nrwnN+f2nrwnNrN-1NψN+1dr≤0.
Similar to [17, Lemma 1], we can easily show that(18)a0∫01mNrN-1wN+1dr≤∫01f1nrwnNmNrN-1wnN+1dr≤a0∫01mNrN-1wN+1drfor n large enough.
It follows from (16) and (18) that(19)∫01λ1-λ-a0mrNrN-1wN+1≤0, which implies λ≥λ1-a0.
Similarly, it follows from (17) and (18) that λ≤λ1-a0.
Therefore, we have thatλ∈I10.Proof of Theorem3.
We divide the rest of the proofs into two steps.
Step  1. Using a similar method to prove [20, Theorem 1.1] with obvious changes, we may prove that C⊂(K+∪(I10×{0})).
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=∅.
In order to complete the proof of this theorem, we consider problem (11). For ϵ>0, it is easy to show that the nonlinear term f(r,vvϵ,λ)+g(r,v,λ) satisfies condition (H3). Let (20)Sϵ=λ,v:λ,vsatisfies11,v≢0¯R×E.
By Lemma1, there exists an unbounded continuum Cϵ of Sϵ bifurcating from (λ1,0) such that (21)Cϵ⊂K+∪λ1,0. So there exists (λϵ,vϵ)∈Cϵ∩∂O for all ϵ>0. Since O is bounded in R×P+, (11) shows that (λϵ,vϵ) is bounded in R×C2 independently of ϵ. By the compactness of GN, one can find a sequence ϵn→0 such that (λϵn,vϵn) converges to a solution (λ,v) of (11). So, v∈P+¯. If v∈∂P+, then from Lemma 2 it follows that v≡0. By Lemma 4, λ∈I10, which contradicts the definition of O. On the other hand, if v∈P+, then (λ,v)∈S∩∂O which contradicts S∩∂O=∅.From Theorem3 and its proof, we can easily get a corollary.Corollary 5.
Let (H0), (H1), and (H3) hold. There exists a subcontinuumD of solutions of (10) in R×E, bifurcating from I10×{0}, such that(i)
D⊂(K+∪(I10×{0}));(ii)
D is unbounded.
We add the points{(λ,∞)∣λ∈R} to space R×E. Let T denote the closure in R×E of the set of nontrivial solutions (v,λ) of (10) under conditions (H2) and (H4) with v∈P+. Let SN denote the spectral set of problem (9). Let I¯∞=[λ¯-a∞,λ¯-a∞], where λ¯∈SN\{λ1}.
According to Rabinowitz [22], our second main result for (10) is the following theorem.Theorem 6.
Let (H0), (H2), and (H4) hold. LetI1∞=[λ1-a∞,λ1-a∞]. There exists a connected component D of T∪(I1∞×{∞}), containing I1∞×{∞}. Moreover, if Λ⊂R is an interval such that Λ∩(⋃λ¯∈SN\{λ1}(I¯∞∪I1∞))=I1∞ and M is a neighborhood of I1∞×{∞} whose projection on R lies in Λ and whose projection on E is bounded away from 0, then either(1o)  D-M is bounded in R×E in which case D-M meets R={(λ,0)∣λ∈R};(2o)  D-M is unbounded.
If(2o) occurs and D-M has a bounded projection on R, then D-M meets I¯∞×{∞}. Moreover, there exists a neighborhood N⊂M of I1∞×{∞} such that (D∩N)⊂(K+∪(I1∞×{∞})).Proof.
We use a similar method to prove [18, Theorem 2.2] with obvious changes, but we give a rough sketch of the proof for readers’ convenience. If (λ,v)∈T with v≠0, dividing (10) by v2 and setting w=v/v2 yield(22)-w′N′=λNrN-1mrwN+NrN-1Fr,v,v′,λv2N,r∈0,1,w′0=w1=0.Define(23)h~r,w,w′,λ=w2Nhr,ww2,w′w2,λ,ifw≠0,0,ifw=0.
Clearly, (22) is equivalent to(24)-w′N′=λNrN-1mrwN+NrN-1mrf1~r,w,w′,λ+NrN-1f2~r,w,w′,λ,r∈0,1,w′0=w1=0.
It is obvious that(λ,0) is always the solution of (24). By simple computation, we can show that assumptions (H2) and (H4) imply that f1~, f2~ satisfy (H1) and (H3). Now, applying Theorem 3 to problem (24), we have that the component C of S∪(I10×{0}), containing I10×{0}, is unbounded and lies in K+∪(I10×{0}). Under the inversion w→w/w2=v, C→D satisfying problem (10). By an argument similar to that of [18, Theorem 2.3], we can prove the existence of a neighborhood N⊂M of I1∞×{∞} such that (D∩N)⊂(K+∪(I1∞×{∞})).


## 3. Applications

In this section, we shall investigate the existence and multiplicity of convex solutions of problem (6). With a simple transformation v=-u, problem (6) can be written as(25)-v′N′=γNrN-1hrFv,r∈0,1,v′0=v1=0.By [16], in order to prove our main results, we need the following Sturm type comparison result.Lemma 7 (see [16, Lemma 4.6]).
Letbi(r)∈C(0,1),i=1,2, such that b2(r)≥b1(r) for r∈(0,1) and the inequality is strict on some subset of positive measure in (0,1). Also, let v1 and v2 be solutions of the following differential equations:(26)-v′N′=birvN,r∈0,1,i=1,2,v′0=v1=0,respectively. If v1≠0 in (0,1), then v2 has at least one zero in (0,1).
The main results of this section are the following theorems.Theorem 8.
Let (A0), (A1), and (A2) hold. Iff20>-a0 and f2∞>-a∞, for either(27)λ1f20+a0<γ<λ1f2∞+a∞or(28)λ1f2∞+a∞<γ<λ1f20+a0,then problem (6) has at least one solution u such that it is negative and strictly convex in (0,1).Theorem 9.
Let (A0), (A1), and (A2) hold. Iff20>-a0 and -a∞≤f2∞≤-a∞, for(29)λ1f20+a0<γ<λ1f2∞+a∞, then problem (6) has at least one solution u such that it is negative and strictly convex in (0,1).Theorem 10.
Let (A0), (A1), and (A2) hold. Ifg0>-a0 and f2∞≤-a∞, for(30)γ>λ1f20+a0, then problem (6) has at least one solution u such that it is negative and strictly convex in (0,1).Theorem 11.
Let (A0), (A1), and (A2) hold. If-a0<f20≤a0 and f2∞>-a∞, for(31)λ1f2∞+a∞<γ<λ1f20+a0, then problem (6) has at least one solution u such that it is negative and strictly convex in (0,1).Theorem 12.
Let (A0), (A1), and (A2) hold. If-a0≤g0 and f2∞>-a∞, for(32)γ>λ1f2∞+a∞, then problem (6) has at least one solution u such that it is negative and strictly convex in (0,1).Remark 13.
Note that iff≡0, then Theorem 8 is equivalent to Theorem 4.1 of [15] or Theorem 5.1 of [16]. In this sense, Theorem 8 is also a generalization of Theorem 4.1 of [15] or Theorem 5.1 of [16].Proof of Theorem8.
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:(33)-v′N′=λγNrN-1hrf2v+γNrN-1hrf1v,r∈0,1,v′0=v1=0,where λ>0 is a parameter.
Letζ∈C(R+,R+) be such that(34)f2s=f20sN+ζs with lims→0+ζ(s)/sN=0. Let ζ¯(v)=max0≤|s|≤v|ζ(s)|; then, ζ¯(v) is nondecreasing and(35)lims→0+ζ¯ssN=0.
Further, it follows from (35) that(36)ζvvN≤ζ¯vvN≤ζ¯r∞vN≤ζ¯rvNasv⟶0. Hence, (33), (34), and (36) imply that conditions (H1) and (H3) hold. Moreover, we have that I10=[λ1/γf20-a0/f20,λ1/γf20-a0/f20].
By Theorem3, there is a distinct unbounded component D0 of S∪(I10×{0}), containing I10×{0} and lying in (K+∪(I10×{0})).
Letξ∈C(R+,R+) be such that(37)f2s=f2∞sN+ξs with lims→+∞ξ(s)/sN=0. Let v¯=max0≤|s|≤v|ξ(s)|, and then ξ¯ is nondecreasing.
Define(38)v~=maxv/2≤s≤vξs. Then, we can see that(39)lims→+∞ξ~ssN=0,v¯≤v2¯+v~.It is not difficult to verify that v¯/s is bounded in R+. From this fact and (37), it follows that(40)limsups→+∞ξ¯vvN≤limsups→+∞ξ¯v/2vN=limsups→+∞ξ¯vvN, where t=u/2. So, we have(41)lims→+∞ξ¯ssN=0.
Further, it follows from (41) that(42)ξvvN≤ξ¯vvN≤ξ¯v∞vN≤ξ¯vvNasv⟶+∞.
Hence, (33), (37), and (42) imply that conditions (H2) and (H4) hold. Moreover, we have that I1∞=[λ1/γf2∞-a∞/f2∞,λ1/γf2∞-a∞/f2∞].
By Theorem6, there is a distinct unbounded component D∞ of T∪(I1∞×{∞}), which satisfies the alternates of Theorem 6. Moreover, there exists a neighborhood N⊂M of I1∞×{∞} such that (D∞∩N)⊂(K+∪(I1∞×{∞})).
We claim thatD0=D∞. It suffices to show that D∞ meets some point (λ∗,0) of R. In fact, if this occurs, we can show that λ∗∈I10. Suppose on the contrary that λ∗∉I10; hence, λ∗∈I¯0. So, (D∞∩N)⊂D∞⊂D¯0⊂((R×P¯)∪(I¯0×{0})), noting (D∞∩N)∩(R×{0})=∅, which contradicts (D∞∩N)⊂(K+∪(I∞×{∞})), where T¯0 denotes the closure in R×E of the set of nontrivial solutions (λ,v) of (25) under conditions (H1) and (H3) with v∈P¯, where P¯={v∣(λ¯,v)∈(SN\{λ1})×E}, I¯0=[λ¯/γf20-a0/f20,λ¯/γf20-a0/f20], where λ¯∈SN\{λ1}. D¯0 is a connected component of T¯0∪(I¯0×{0}), containing I¯0×{0}.
Hence,λ∗∈I10, and it follows that D0=D∞.
Next, we shall show that(2o) of Theorem 6 does not occur. Suppose on the contrary that (2o) 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 D∞⊂K+. If (D∞-(D∞∩N))⊄K+, then there exists (μ,v)∈(D∞-(D∞∩N))∩(R×∂P+). Since v∈∂P+, by Lemma 4, v≡0; that is, (1o) of Theorem 6 occurs, which is a contradiction.
On the contrary, we suppose that(μn,vn)∈D∞-M such that (43)limn→∞μn=∞. It follows that(44)-vn′N′=μnNγNrN-1hrf2vn+γNrN-1hrf1vn,r∈0,1,vn′0=vn1=0.In view of (A1) and (A2), we have that limn→∞(μnNNrN-1a(r)f2(vn)/vnN+NrN-1a(r)f1(vn)/vnN)=∞ for any r∈(0,1). By the Sturm comparison (Lemma 7), we get that vn has at least one zero in (0,1) for n large enough, and this contradicts the fact that vn does not change its sign in (0,1).
In the following, we show that the case ofD∞-M meeting I¯∞×{∞} is impossible. Assume on the contrary that D∞-M meets I¯∞×{∞}. So, there exists a neighborhood N~⊂M~ of I¯∞ such that (D∞-M)∩(N~\(I¯∞×{∞}))⊂(R×P¯), where M~ is a neighborhood of I¯∞×{∞} which satisfies the assumptions of Theorem 6, which contradicts D∞⊂K+, where P¯={v∣(λ¯,v)∈(SN\{λ1})×E} and I¯∞=[λ¯/γf2∞-a∞/f2∞,λ¯/γf2∞-a∞/f2∞], where λ¯∈SN\{λ1}.
For simplicity, we writeD=D0=D∞. It is clear that any solution of (33) of the form (1,v) yields a solution v of (25).
By (27), we obtain(45)λ1Nγf20-a0f20<1<λ1Nγf2∞-a∞f2∞. By (28), we have(46)λ1γf2∞-a∞f2∞<1<λ1γf20-a0f20.
FromI10=[λ1/γf20-a0/f20,λ1/γf20-a0/f20] and I1∞=[λ1/γf2∞-a∞/f2∞,λ1/γf2∞-a∞/f2∞], it follows that the subsets I10×E and I1∞×E of R×E can be separated by the hyperplane {1}×E. Furthermore, we have D crossing the hyperplane {1}×E in R×E.Proof of Theorems9 and 10.
The proof is similar to that of Theorem8. In this case, it follows that (45) holds and (46) is impossible.Proof of Theorems11 and 12.
The proof is similar to that of Theorem8. In this case, it follows that (46) holds and (45) is impossible.Remark 14.
Note that if-a0<f20≤-a0 and f2∞≤-a∞ or f20≤-a0 and f2∞≤-a∞, (45) and (46) are impossible, and it follows that the subsets I10×E and I1∞×E of R×E cannot be separated by the hyperplane {1}×E. In this case, we cannot give a suitable interval of γ in which there exist nodal solutions for (25). It would be interesting to have more information about this case.



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