Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 31, pp. 1–19. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu EXISTENCE AND MULTIPLICITY FOR RADIALLY SYMMETRIC SOLUTIONS TO HAMILTON-JACOBI-BELLMAN EQUATIONS XIAOYAN LI, BIAN-XIA YANG Abstract. This article concerns the existence and multiplicity of radially symmetric nodal solutions to the nonlinear equation −M±C (D2u) = µf(u) in B, u = 0 on ∂B, whereM±C are general Hamilton-Jacobi-Bellman operators, µ is a real param- eter and B is the unit ball. By using bifurcation theory, we determine the range of parameter µ in which the above problem has one or multiple nodal solutions according to the behavior of f at 0 and ∞, and whether f satisfies the signum condition f(s)s > 0 for s 6= 0 or not. 1. Introduction and main results In scientific fields such as engineering, economical business, and mechanics, one encounters the problem of how to control systems in an optimal way [2, 6, 10]. For such systems, states are governed by the stochastic differential equation dXt dt = τ(Xt, αt)ξt + b(Xt, αt) for t ≥ 0, X0 = x ∈ RN , where ξt is the typical ‘white noise’, τ and b are matrix-valued and vector-valued functions defined on RN ×A respectively, A is a separable metric space, and αt as the control process is an stochastic process taking its values in A. Then one defines a cost function J(x, αt) = E (∫ ∞ 0 f(Xt, αt) exp ( − ∫ t 0 c(Xs, αs)ds ) dt ) , where E denotes the expectation, f(x, α) and c(x, α) are real valued functions on RN ×A, and c is a function often called the discount factor. The purpose of optimal stochastic control theory is to determine the optimal cost function (also called the value function, or the criterion) u(x) = inf {J(x, αt) αt stochastic process with values in A } . (1.1) 2010 Mathematics Subject Classification. 35B32, 35B40, 35B45, 35J60, 34C23. Key words and phrases. Radially symmetric solution; extremal operators; bifurcation; nodal solution. c©2021 Texas State University. Submitted November 29, 2020. Published April 24, 2021. 1 2 X. Y. LI, B. X. YANG EJDE-2021/31 A fundamental tool for finding u is given by the dynamic programming principle introduced by Bellman [1]. This principle indicates that u should, in some way, be the solution of the partial differential equation sup α∈A {Aαu(x)− fα(x) = 0 in RN}, (1.2) where fα(·) = f(·, α), Aα = − ∑ i,j aij(x, α)∂ij − ∑ i bi(x, α)∂i + c(x, α) and a = 1 2ττ T . Equation (1.2) is called the Hamilton-Jacobi-Bellman (HJB in short) equa- tion associated with the control problem (1.1). In some sense it is an extension of the classical first-order Hamilton-Jacobi equations occurring in Calculus of varia- tions, see P. L. Lions [6]. We refer to the book of Bensoussan and J. L. Lions [2] or the papers of P. L. Lions [7, 8, 9] for further relation between a general HJB and stochastic control. 1.1. Existing results. Quass and Allendes [14] considered the radially symmetric fully nonlinear equation involving extremal operators of Pucci type, −M±C (D2u) = f(u) in B, u = 0 on ∂B, (1.3) where B is the unit ball in RN with N ≥ 1, M±C are general HJB operators. Specifically these operators are defined as M+ C (M) = sup σ(A)∈C tr(AM), M−C (M) = inf σ(A)∈C tr(AM), where C is any subset of the cube [λ,Λ]N which is invariant with respect to permu- tations of coordinates, σ(A) is the set of eigenvalues of A, and the parameters λ,Λ satisfy 0 < λ ≤ Λ. These operators reduce to classical Pucci type operators when C = [λ,Λ]N , and to the Laplacian when λ = Λ = 1. When λ ≤ 1 N ,Λ = 1−λ(N−1) and C = {a ∈ [λ,Λ]N | ∑N i=1 ai = 1}, the operator corresponds to Pucci’s operators, see [12, 13]. Clearly, problem (1.3) is a special case of (1.2). Based on the bifurcation theory, Quass and Allendes established a multiplicity result for (1.3) and showed that the eigenvalue problem −M+ C (D2u) = µu in B, u = 0 on ∂B (1.4) has two unbounded increasing sequences µ+ k and µ−k , such that 0 < µ+ 1 < µ+ 2 < · · · < µ+ k < . . . , 0 < µ−1 < µ−2 < · · · < µ−k < . . . . Moreover, the set of radial solutions of (1.4) for µ = µ+ k is positively spanned by a function ϕ+ k , which is positive at the origin and has exactly k − 1 zeros in (0, 1), all these zeros being simple. The same holds for µ = µ−k , but considering ϕ−k is negative at the origin. Then they studied the global bifurcation phenomenon of the problem −M+ C (D2u) = µu+ f(u, µ) in B, u = 0 on ∂B, (1.5) where f is continuous, f(s, µ) = o(|s|) near s = 0, uniformly for µ ∈ R. They showed that, for each k ∈ N, k ≥ 1, there are two connected components H±k ⊂ S ± k of nontrivial solutions to (1.5), whose closures contain (µ±k , 0). Moreover, H±k are EJDE-2021/31 HAMILTON-JACOBI-BELLMAN EQUATIONS 3 unbounded and (µ, u) ∈ S+ k (S−k ) implies that u possesses exactly k − 1 zeros in (0, 1), u is positive (negative) near 0. For notational simplicity, we write M±C in (1.3) to mean the two problems, one with the operator M+ C and the other with M−C . In the remaining, the situation is similar. Dai [5], by using bifurcation approach with the generalized limit theorem, studied the existence and multiplicity of nodal solutions for the special problem −M±λ,Λ(D2u) = µf(u) in B, u = 0 on ∂B, (1.6) where M±λ,Λ denote Pucci’s extremal operator. Motivated by works mentioned above, our aim is to extend the results from C = [λ,Λ]N in [5, Theorems 1.8-1.9] to the general HJB operators. To be specific, in consideration of the Rabinowitz global bifurcation theory, and according to the asymptotic behavior of the nonlinear term f at 0 and ∞, with signum condition, we focus on the existence, multiplicity and nonexistence of nodal solutions for the radially symmetric non proper equation of the type −M±C (D2u) = µf(u) in B, u = 0 on ∂B. (1.7) Moreover, we consider the global behavior of nodal solutions for (1.7) without signum condition. We shall show that the branches bifurcating from infinity and the trivial solution line are disjoint. Hence the essential role is played by the fact whether f possesses zeros in R \ {0} or not. 1.2. Statement of main results. To obtain our main results, we shall give a bifurcation theorem from infinity for problem (1.5) under the assumption that lim |s|→+∞ f(µ, s) s = 0 uniformly for µ ∈ R. (1.8) Let E = {u ∈ C[0, 1] : u′(0) = u(1) = 0} with the usual norm ‖ · ‖∞. Let S+ k denote the set of functions in E which have exactly k− 1 interior nodal (i.e. non-degenerate) zeros in (0, 1) and are positive at 0. Set S−k = −S+ k and Sk = S+ k ∪ S − k . It is clear that S+ k and S−k are disjoint and open in E. Theorem 1.1. Let condition (1.8) hold. There exists an unbounded component Dνk ⊂ ({(µνk,∞)} ∪ (R× Sνk )) of solutions to problem (1.5). Moreover, either (1) Dνk meets R = {(µ, 0) : µ ∈ R}, or (2) Dνk has an unbounded projection on R. On account of Theorem 1.1 and [14, Theorem 1.4], we shall investigate the existence and multiplicity of nodal solutions for problem (1.7). We use the following assumptions: (A1) f(s)s > 0 for any s 6= 0; (A2) there exist f0, f∞ ∈ [0,+∞] such that lim |s|→0 f(s) s = f0, lim |s|→+∞ f(s) s = f∞; 4 X. Y. LI, B. X. YANG EJDE-2021/31 (A3) there exist two constants s2 < 0 < s1 such that f(s2) = f(s1) = 0 and f(s)s > 0 for s ∈ R \ {s2, 0, s1}; (A4) there exist two constants γ1 > 0 and γ2 < 0 such that lim s→s−1 f(s) s1 − s = γ1, lim s→s+2 f(s) s− s2 = γ2; According to the asymptotic behavior of f at 0 and ∞, we have the results in Theorem 1.2. f0, f∞ ∈ (0,+∞), f0 > f∞ f0 ∈ (0,+∞), f∞ = 0 f0 = 0, f∞ ∈ (0,+∞) f0 = f∞ = 0 f0 = +∞, f∞ = 0 f0 = +∞, f∞ ∈ (0,+∞) Figure 1. Bifurcation diagrams for Theorem 1.2 Theorem 1.2. Suppose that f satisfies (A1) and (A2). (a) If f0, f∞ ∈ (0,+∞) with f0 6= f∞, then for k ∈ N, µ ∈ (min{µ ν k f0 , µνk f∞ },max{µ ν k f0 , µνk f∞ }), problem (1.7) has at least one nodal solution uνk, such that νuνk has exactly k − 1 simple zeros in (0, 1) and is positive near 0, where ν ∈ {+,−}. (b) If f0 ∈ (0,+∞) and f∞ = 0, then for any k ∈ N, µ ∈ ( µνk f0 ,+∞), problem (1.7) has at least one nodal solution uνk, such that νuνk has exactly k−1 simple zeros in (0, 1) and is positive near 0. (c) If f0 = 0 and f∞ ∈ (0,+∞), then for any k ∈ N, µ ∈ ( µνk f∞ ,+∞), problem (1.7) has at least one nodal solution uνk, such that νuνk has exactly k−1 simple zeros in (0, 1) and is positive near 0. EJDE-2021/31 HAMILTON-JACOBI-BELLMAN EQUATIONS 5 (d) If f0 = f∞ = 0, then there exists µ∗ > 0, such that for k ∈ N, µ ∈ (µ∗,+∞), problem (1.7) has at least two nodal solutions u1+ k , u2+ k . Moreover, u1+ k , u2+ k have exactly k−1 simple zeros in (0, 1) and are positive near 0; there exists µ′ > 0, such that problem (1.7) has at least two nodal solutions u1− k , u2− k for all µ ∈ (µ′,+∞), and u1− k , u2− k have exactly k − 1 simple zeros in (0, 1) and are negative near 0. Furthermore, there exists µ̃∗ > 0, such that the problem (1.7) has no nodal solution for µ ∈ (0, µ̃∗). (e) If f0 = +∞ and f∞ = 0, then for any µ ∈ (0,+∞), k ∈ N, problem (1.7) has at least one nodal solution uνk, such that νuνk has exactly k − 1 simple zeros in (0, 1) and is positive near 0. (f) If f0 = +∞ and f∞ ∈ (0,+∞), then for any µ ∈ (0, µνk f∞ ), k ∈ N, problem (1.7) has at least one nodal solution uνk, such that νuνk has exactly k−1 simple zeros in (0, 1) and is positive near 0. See illustrations in Figure 1. It is worth mentioning that the signum condition f(s)s > 0 for s 6= 0 plays an important role in the Theorem 1.2. In the following, one considers the global behavior of nodal solutions for (1.7) without signum condition. We shall show that the branches bifurcating from infinity and the trivial solution line are disjoint. Concretely, it has the following interesting results. Theorem 1.3. Let (A2)–(A4) hold. If f0, f∞ ∈ (0,+∞) with f0 6= f∞, then problem (1.7) has at least one nodal solution uνk for µ ∈ ( min {µ ν k f0 , µνk f∞ },max {µ ν k f0 , µνk f∞ } ) , k ∈ N, such that νuνk has exactly k− 1 simple zeros in (0, 1) and is positive near 0; (1.7) has at least four nodal solutions u+ k,0, u + k,∞, u − k,0 and u−k,∞ for µ ∈ ( max { µ+ k f0 , µ−k f0 , µ+ k f∞ , µ−k f∞ },+∞ ) , such that they have exactly k − 1 simple zeros in (0, 1), u+ k,0 and u+ k,∞ are positive near 0, u−k,0 and u−k,∞ are negative near 0. Moreover, it derives ‖u+ k,0‖ → s−1 and ‖u−k,0‖ → (−s2)− as µ→ +∞. See illustrations in Figure 2. Figure 2. Bifurcation diagrams of Theorem 1.3. 6 X. Y. LI, B. X. YANG EJDE-2021/31 The results obtained above are also valid for problem (1.7) if we replace M+ C by M−C , so to simplify our presentation, we only consider the operator M+ C . For simplicity, we only consider the case when f does not depend on x, and f is asymp- totically linear at 0 and ∞. In fact, Theorem 1.3 is still valid for the case of f depending on x or f satisfying other asymptotic behaviors with obvious changes. The conclusions of Theorem 1.3 are not only significate in theory, but also mean- ingful in economics. For example, the conclusion of Theorem 1.3 with ν = + means that if the reaction function f which can denote the investment strategy is linear near 0 and ∞, and the diffusion coefficient d := 1/µ which can denote the rate of investment belongs to some interval (α, β) with 0 < α < β < +∞, then there at least exists one optimal cost function. This article is arranged as follows. In Section 2, we recall some preliminary results and give the proof of Theorem 1.1. In Section 3, according to the differ- ent asymptotic behaviors of f at 0 and ∞, we prove Theorem 1.2 and derive the existence, nonexistence and multiplicity of nodal solutions for problem (1.7) with signum condition. In Section 4, we give the proof of Theorem 1.3, which considers the global behavior of nodal solutions for (1.7) without signum condition, and we shall show that the branches bifurcating from infinity and the trivial solution line are disjoint. 2. Preliminary results and Proof of Theorem 1.1 We start this section by studying the operator acting on radial functions, details can be seen in [14, Section 3]. We define the operator M+ C acting on C2 radially symmetric functions as M+ C (D2u) = sup (a1,a2)∈C̃ ( a1u ′′ + (N − 1)a2u ′ r ) , where C̃ := {(a1, 1 N−1 ∑N i=2 ai) ∈ R2 : (a1, a2, . . . , an) ∈ C}. In the rest of this article we write C for C̃ to simplify the notation. To describe the set C in a more convenient way. To avoid trivialization, we make an additional assumption. (A5) The set C ⊂ R2 + is compact, convex and its projection onto the y−axis is not a singleton. Assuming (A5) we exclude the case when the projection of C onto the y-axis is a singleton, which is equivalent to C = {(a1, a2)}. This particular case can be analyzed as the radial Laplacian. Observe that C is a symmetric set. Under assumption (A5), we can describe ∂C by means of two functions. Let 0 < θmin < θmax be defined as θmin = min{θ : (x, θ) ∈ C} and θmax = max{θ : (x, θ) ∈ C}, and define the functions S, S̃ : [θmin, θmax]→ R+ as S(θ) = min{x : (x, θ) ∈ C}, S̃(θ) = max{x : (x, θ) ∈ C}. With these definitions we see that S is convex, S̃ is concave and C = {(x, θ) : θ ∈ [θmin, θmax], S(θ) ≤ x ≤ S̃(θ)}. Being S convex, it has one-sided derivatives S′−(θ) and S′+(θ), consequently it is locally Lipschitz continuous in (θmin, θmax). The sub-differential of S is then defined as ∂S(θ) = [S′−(θ), S′+(θ)] for θ ∈ (θmin, θmax). The cases θ = θmin and θ = θmax EJDE-2021/31 HAMILTON-JACOBI-BELLMAN EQUATIONS 7 are special. At θmax we have two possibilities, either S′−(θmax) exists, and then we define ∂S(θmax) = [S′−(θmax),+∞), or lim t→0− S(θmax + t)− S(θmax) t = +∞. An analogous situation occurs at θmin. We observe that with these definitions, for every Q ∈ R there is at least one solution θ ∈ [θmin, θmax], such that ∂S(θ)θ − S(θ) 3 Q. In each of this maximal intervals [θ−i , θ + i ], with θ−i < θ+ i where the function S is affine, we may write S(θ) = diθ −Qi, ∀θ ∈ [θ−i , θ + i ], for numbers di and Qi. We define the function d : R → R as d(Q) ∈ ∂S(θ) such that d(Q)θ − S(θ) = Q. All the above hold for S̃ with natural modification since S̃ is concave and ∂S̃ is the super-differential of S̃. We consider Θ : R→ R as Θ(Qi) = θ+ i in each interval where S or S̃ are affine functions. Now, we can easily show that the radially symmetric solutions of problem (1.7) are also solutions to u′′(r) + (Nd − 1) u′(r) r + µf(u) θ = 0 in (0, 1), u′(0) = 0, u(1) = 0, (2.1) where θ = Θ( u (N−1)u′ ) when u′ 6= 0 and Nd = {S(θ) θ (N − 1), if u′ < 0, S̃(θ) θ (N − 1), if u′ > 0. When u′ = 0, then θ := θmin if u > 0 and θ := θmax if u < 0. Notice that the functions θ(r) and Nd(r) are measurable functions, having discontinuities whenever r is so that u(r)r (N−1)u′(r) = Qi and S, S̃ are affine functions. Moreover, both θ(r) and Nd(r) are bounded and bounded away from 0. Similar to [14], we obtain that problem (2.1) is equivalent to the problem − ( ρu(r)u′(r) )′ = µρ̃uf(u(r)) in (0, 1), u′(0) = 0, u(1) = 0, (2.2) where ρu(r) := exp ( ∫ r 0 Nd(τ)−1 τ dτ ) denotes the integral factor of the equation, ρ̃u(r) := ρu(r) θ , ρu and θ are characterized by the optimal condition. For arriving to the results in Theorem 1.2, we need the following topological lemma, see[11]. Lemma 2.1. Let X be a Banach space and let Cn be a family of closed connected subsets of X. Assume that (i) there exist zn ∈ Cn, n = 1, 2, . . . , and z∗ ∈ X, such that zn → z∗; (ii) rn = sup{‖x‖X : x ∈ Cn} = +∞; (iii) for every R > 0, (∪+∞ n=1Cn) ∩ BR is a relatively compact set of X, where BR = {x ∈ X : ‖x‖X ≤ R}. Then there exists an unbounded component C of D = lim supn→+∞ Cn and z∗ ∈ C. 8 X. Y. LI, B. X. YANG EJDE-2021/31 Now we recall the following compactness results for the Pucci’s extremal opera- tor, see [4, Proposition 2.1]. Lemma 2.2. Let {Fn}n>0 be a sequence of uniformly elliptic concave (or convex) operators with ellipticity constants λ and Λ, such that Fn → F is uniformly in compact sets of Sn×Ω (Sn is the set of symmetric matrices). In addition, suppose that un ∈ C(Ω̄) ∩W 2,N loc (Ω) satisfies Fn(D2un, x) = 0in Ω, un = 0 on ∂Ω and that un converges uniformly to u. Then, u ∈ C(Ω̄) is a solution to F (D2u, x) = 0 in Ω, u = 0 on ∂Ω. Before giving the strong maximum principle and a version of the Hopf’s boundary lemma, let us recall the notion of viscosity sub-solution and super-solution for extremal operators of Pucci type. Lemma 2.3. Given γ ≥ 0, a radially symmetric continuous function u : B → R is a viscosity super-solution (sub-solution) of −M+ C (D2u) + γu = 0 in B, (2.3) when the following condition holds: If x0 ∈ (0, 1], φ ∈ C2(0, 1), such that u− φ has a local minimum (maximum) at x0, and ϕ′(x0) 6= 0, then − sup (a1,a2)∈C ( a1ϕ ′′(x0) + (N − 1)a2ϕ ′(x0) |x0| ) ≤ γu(x0), ( − sup (a1,a2)∈C ( a1ϕ ′′(x0) + (N − 1)a2ϕ ′(x0) |x0| ) ≥ γu(x0) ) . We say that u is a viscosity super-solution (sub-solution), if u satisfies − sup (a1,a2)∈C ( a1u ′′ + (N − 1)a2u ′ r ) + γu ≥ (≤)0 in the viscosity sense. We say that u is a viscosity solution of (2.3) when it is simultaneously a viscosity sub-solution and a super-solution. Now, we give the strong maximum principle and a version of the Hopf’s boundary lemma. Lemma 2.4. For γ ≥ 0, if u ∈ C2(B) ∩ C(B̄) satisfies −M+ C (D2u) + γu ≥ 0 in B, u ≥ 0 on ∂B, then either u ≡ 0 or u > 0 in B. Moreover, lim sup x→x0 u(x0)− u(x) |x− x0| < 0, where x0 ∈ ∂B and the limit is non-tangential; that is, taken over the set of x for which the angle between x − x0 and the outer normal at x0 is less than π 2 − δ for some fixed δ > 0. EJDE-2021/31 HAMILTON-JACOBI-BELLMAN EQUATIONS 9 Proof. We start by claiming that u ≥ 0 in B. (2.4) Otherwise, suppose there exists x0 ∈ B, such that u(x0) < 0. Let ε ∈ (0,−u(x0)), then for x ∈ B(x0, ε), the function uε(x) = u(x) − ε 2 |x − x0|2 also has a strictly negative minimum which is achieved at xε ∈ B. Indeed if it is achieved on the boundary xε ∈ ∂B, then uε(xε) = u(xε)− ε 2 |xε − x0|2 ≥ u(xε)− ε 2 ≥ u(xε) + 1 2 u(x0) = 1 2 u(x0) > u(x0) = uε(x0). This is impossible. At the point xε, one has D2u(xε) ≥ εI, where I is the identity matrix. Therefore, 0 ≥ γu(xε) ≥M+ C (D2u(xε)) ≥ 1 2 λεN > 0, which is a contradiction. So (2.4) holds. On the other hand, let M and S be symmetric matrices such that S ≥ 0 (i.e. nonnegative definite), and let ā ∈ C such that M+ C (M + S) = sup a∈C N∑ i=1 aiλi(M + S) = N∑ i=1 āiλi(M + S). Then we have M+ C (M + S)− N∑ i=1 āiλi(M) = N∑ i=1 āi(λi(M + S)− λi(M)) ≤ Λ tr(S), from which it follows that M+ C (M + S)−M+ C (M) ≤ Λ tr(S). Proceeding in a similar form we obtain M+ C (M + S)−M+ C (M) ≥ λ tr(S). SoM+ C (M) ≥ λ tr(M+)−Λ tr(M−) := H(M) with S = M−, where M = M+−M− is a minimal decomposition of M into the difference of two nonnegative matrices. Hence it is sufficient to prove the conclusions when u is a super solution of H(D2u)− γu = 0. On the contrary, Suppose that u is not identically equal to zero and that there exists x0 inside B on which u(x0) = 0, we can find x1 ∈ B and R > 0, such that B(x1, 3R/2) ⊂ B, and u > 0 in B(x1, R) with |x1 − x0| = R. So u1 = inf |x−x1|=R/2 u > 0. Let us recall that if ϕ(ρ) = e−kρ, the eigenvalues of D2ϕ are ϕ′′(ρ) with multi- plicity 1 and ϕ′/ρ with multiplicity N − 1. 10 X. Y. LI, B. X. YANG EJDE-2021/31 Then we take k > 0 such that k2 > 2(N − 1)Λ Rλ k + γ. If k is as above, let m be chosen such that m(e−kR/2 − e−kR) = u1 and define v(x) = m(e−kρ − e−kR) with ρ = |x − x1|. As we discussed above, u is a non negative super solution of the operator M+ C (D2u) − γu, so u is a super solution of H(D2u)−γu = 0 sinceM+ C (D2u) ≥ H(D2u). It is not difficult to check H(D2v)− γv > 0 in the annulus with the above choice k, which means that v is a strict subsolution of H(D2v)− γv = 0 in the annulus. Furthermore v = u1 ≤ u on |x− x1| = R 2 , v < 0 ≤ u on |x− x1| = 3R 2 , that is u ≥ v everywhere on the boundary of the annulus. In fact u ≥ v everywhere in the annulus, since we can use the comparison principle [3, Theorem 2.9] with F := H(D2u) − γu. Therefore, u(x0) ≥ v(x0). Actually, u(x0) = v(x0) = 0 since |x1 − x0| = R. On the other hand, based on the fact that u is a super solution of H(D2u)− γu = 0, one has H(D2v(x0))− γv(x0) = H(D2u(x0))− γu(x0) ≤ 0, which clearly contradicts the definition of v. So u cannot be zero inside B. One derives either u ≡ 0 or u > 0 in B. Now, we shall give the Hopf’s property by using the same construction. Suppos- ing that there is x0 ∈ ∂B on which u(x0) = 0, we can find x1 ∈ B and R > 0, such that B(x1, R) ⊂ B with |x1 − x0| = R. So u1 = inf |x−x1|=R 2 u > 0, and we replace the previous annulus with R/2 ≤ |x− x1| = ρ ≤ R, with the comparison principle again. Since v = 0 on |x− x1| = R,Dv 6= 0 in B and v ≤ u on the other boundary of the annulus, one arrives at u(x) ≥ m(e−kρ − e−kR) everywhere in the annulus. Then taking x = x0 − hω and letting h > 0 go to zero, where ω is the outward pointing normal to ∂Ω, it arrives u(x0)− u(x) h ≤ me−kR − e−kR+kh h → −mke−kR. � Remark 2.5. Similar to the discussion in Lemma 2.4, we have: For γ ≥ 0, if u ∈ C2(B) ∩ C(B̄) satisfies M−C (D2u)− γu ≤ 0 in B, u ≥ 0 on ∂B, then either u ≡ 0 or u > 0 in B and lim sup x→x0 u(x0)− u(x) |x− x0| < 0, where x0 ∈ ∂B and the limit is non-tangential. After giving the following important result, we study the nonlinear bifurcation problem and give the proof of Theorem 1.1. EJDE-2021/31 HAMILTON-JACOBI-BELLMAN EQUATIONS 11 Proposition 2.6. If (µ̄, 0) is a bifurcation point of problem (1.5), then µ̄ is an eigenvalue of (1.4). Proof. Since (µ̄, 0) is a nonlinear bifurcation point, there is a sequence {(µm, um)}m∈N of nontrivial solutions of problem (1.5), such that µm → µ̄ and um → 0 uniformly in B. Let vm = um/‖um‖, then vm satisfies −M+ C (D2vm) = µmvm + f(µm, um) ‖um‖ in B, vm = 0 on ∂B. (2.5) So, the right-hand side of the equation is bounded owing to f(s, µ) = o(|s|) near s = 0. By the compactness of (−M+ C )−1, see [14, Page 5], we can extract a subsequence, such that vm → v̄ as m→ +∞ and ‖v̄‖ = 1. Clearly, v̄ satisfies −M+ C (D2v̄) = µ̄v̄ in B, v̄ = 0 on ∂B, namely, µ̄ is an eigenvalue of problem (1.4). � Proof of Theorem 1.1. If (µ, u) with u 6≡ 0 is a solution pair of problem (1.5), dividing equation (1.5) by ‖u‖2 and setting w = u ‖u‖2 , it yields −M+ C (D2w) = µw + f(µ, u) ‖u‖2 in B, w = 0 on ∂B. (2.6) We define f̃(µ,w) = { ‖w‖2f(µ, w ‖w‖2 ), if w 6= 0, 0, if w = 0. Clearly, (2.6) is equivalent to −M+ C (D2w) = µw + f̃(µ,w) in B, w = 0 on ∂B. (2.7) It is easy to see that (µ, 0) is always the solution of the problem (2.7). Let f̂(µ, u) = max0≤|s|≤u |f(µ, s)| for any µ ∈ R. Then f̂ is nondecreasing with respect to u. We define f̄(µ, u) = max u/2≤|s|≤u |f(µ, s)| for any µ ∈ R. Then we arrive at f̂(µ, u) ≤ f̂(µ, u 2 ) + f̄(µ, u), (2.8) lim u→+∞ f̄(µ, u) u = 0 uniformly for µ ∈ R, (2.9) since lim|s|→+∞ f(µ,s) s = 0 uniformly for µ ∈ R. It is not difficult to verify that, for any given ρ > 0, f̂(µ,s) s is positive and bounded for s ∈ [ρ,+∞). This fact and (2.8), (2.9) imply lim sup u→+∞ f̂(µ, u) u ≤ lim sup u→+∞ f̂(µ, 2u) u = lim sup t→+∞ 2 f̂(µ, t) t 12 X. Y. LI, B. X. YANG EJDE-2021/31 uniformly for µ ∈ R, where t = 2u. So it has lim u→+∞ f̂(µ, u) u = 0 (2.10) uniformly for µ ∈ R. Furthermore, it follows from (2.10) that f(µ, u) ‖u‖ ≤ f̂(µ, |u|) ‖u‖ ≤ f̂(µ, ‖u‖) ‖u‖ → 0 as ‖u‖ → +∞ (2.11) uniformly for µ ∈ R. By a direct computation, one can see that (2.11) implies lim ‖w‖→0 f̃(µ,w) ‖w‖ = 0 uniformly for µ ∈ R. Applying [14, Theorem 1.4] to problem (2.7), we derive that the component Cνk of problem (2.7) containing (µνk, 0) is unbounded and lies in (R × Sνk ) ∪ (µνk, 0), under the inversion w → w ‖w‖2 = u and Cνk → Dνk . It is not difficult to check that Dνk emanates from ((R× Sνk ) ∪ (µνk × {∞})). Next, we show that there exists a neighborhood N ⊂ U of (µνk×{∞}) such that Dνk ∩N ⊂ ((R× Sνk ) ∪ (µνk × {∞})) for ν = + and − . We only prove the case of ν = +, since the proof of the other case is similar. It is easy to see that the inversion w → w ‖w‖2 = u turns (µ+ k × {0}) into (µ+ k × {∞}). Let M be a bounded neighborhood of (µ+ k × {0}). Then C+ k ∩ (M \ (µ+ k × {0})) ⊂ R × S+ k . By the inversion w → w ‖w‖2 = u, C+ k ∩ (M \ (µ+ k × {0})) is translated to a deleted neighborhood N 0 of (µ+ k × {∞}). Clearly, (µ,w) ∈ ( C+ k ∩ (M\ (µ+ k × {0})) ) implies that there exists a constant C0 such that 0 < ‖w‖ ≤ C0. It follows that (µ, u) ∈ N 0, which implies 1 C0 ≤ ‖u‖ < ∞. Consequently, we obtain D+ k ∩ N ⊂ ( (R× S+ k ) ∪ (µ+ k × {∞}) ) by taking N := N 0 ∪ (µ+ k × {∞}). � At last, we give a Sturm type comparison theorem. Lemma 2.7 ([14, Lemma 3.5]). Let a, b ∈ L∞(0, 1) with a ≥ b in (0, 1). Assume that u, v ∈ C2[0, 1] \ {0}, u′(0) = v′(0) = 0, and respectively satisfy − (ρu(r)u′(r)) ′ = ρ̃u(r)a(r)u(r) a.e. (0, 1), − (ρv(r)v ′(r)) ′ = ρ̃v(r)b(r)v(r) a.e. (0, 1), where ρu(r) denote the integral factor of the equation, ρ̃u(r) := ρu(r) θ , ρu and θ are characterized by the optimal condition. Then (i) If v has a zero in (0, 1), then u also has a zero. The first zero of u is less than or equal to the first zero of v. (ii) If (r0, r1) ⊆ [0, 1], v(r0) = v(r1) = 0, u(r) 6≡ 0, for r ∈ (r0, r1), and a ≥ b in some subset of (r0, r1), then u has at least one zero in (r0, r1). 3. Proof of Theorem 1.2 Based on [14, Theorem 1.4] and Theorem 1.1, we give the proof of Theorem 1.2. EJDE-2021/31 HAMILTON-JACOBI-BELLMAN EQUATIONS 13 Proof of Theorem 1.2. (a) Let ξ ∈ C(R,R) be such that f(s) = f0s+ ξ(s) with lim |s|→0 ξ(s) s = 0 and lim |s|→+∞ ξ(s) s = f∞ − f0. By [14, Theorem 1.4], we have that there is an unbounded continua Cνk , emanating from ( µνk f0 , 0), such that Cνk ⊂ ({(µ ν k f0 , 0)} ∪ (R× Sνk )), where ν ∈ {+,−}. To complete the proof, it will be sufficient to show that Cνk connects ( µνk f0 , 0) to ( µνk f∞ ,+∞). Let (µn, un) ∈ Cνk with un 6≡ 0 satisfying µn + ‖un‖ → +∞. We note that µn > 0 for all n ∈ N, since 0 is the only solution of the problem (1.7) for µ = 0 and Cνk ∩ ({0} × E) = ∅. We divide the remainder of the proof into two steps. Step 1: One shows if there exists a constant M > 0 such that µn ⊂ (0,M ] for sufficiently large n ∈ N, then Cνk connects ( µνk f0 , 0) to ( µνk f∞ ,+∞). In this case it follows that ‖un‖ → +∞. Let ζ ∈ C(R,R) be such that f(s) = f∞s+ ζ(s) with lim |s|→+∞ ζ(s) s = 0 and lim |s|→0 ζ(s) s = f0 − f∞. (3.1) We divide both sides of the equation −M+ C (D2un) = µnf∞un(x) + µnζ(un(x)) in B, un = 0 on ∂B by ‖un‖ and set un = un/‖un‖. Similar to the argument for (2.11), we obtain limn→+∞ ζ(un)/‖un‖ = 0 as n→ +∞. Then one derives −M+ C (D2un) = µnf∞un(x) in B, un = 0 on ∂B. By the compactness of (−M+ C )−1, see [14, Page 5], we can extract a subsequence such that ūm → ū as m→ +∞ and ‖ū‖ = 1. Clearly, ū satisfies −M+ C (D2ū) = µ̄f∞ū in B, ū = 0 on ∂B, where µ = limn→+∞ µn. Clearly, ū ∈ Sνk since ūm ∈ Sνk . Thus, µ̄f∞ = µνk, i.e., µ̄ = µνk f∞ . Therefore, Cνk connects ( µνk f0 , 0) to ( µνk f∞ ,+∞). Step 2: We show that there exists a constant M such that µn ∈ (0,M ] for suffi- ciently large n ∈ N. On the contrary, suppose that limn→+∞ µn = +∞. One notes that −M+ C (D2un(x)) = µnf̃n(x)un(x), x ∈ B, where f̃n(x) = { f(un(x)) un(x) , if un(x) 6= 0, f0, if un(x) = 0. 14 X. Y. LI, B. X. YANG EJDE-2021/31 As for (2.2), one derives that (µn, un) satisfies − ( ρun(r)u′n(r) )′ = µnρ̃un f̃n(r)un(r) in (0, 1), u′n(0) = 0, un(1) = 0, where ρun(r) := exp ( ∫ r 0 Nd(τ)−1 τ dτ ) denote the integral factor of the equation ρ̃un(r) := ρun (r) θ , ρun and θ are characterized by the optimal condition. The signum condition (A1) implies that there exists a positive constant % such that f̃n ≥ % for r ∈ [0, 1]. Thus, one has that µnf̃n > µk where µk is the k−th eigenvalue of the problem −M+ C (D2v) = µv(x) in B, v = 0 on ∂B. i.e. (µk, v) satisfies − (ρv(r)v ′(r)) ′ = µkρ̃v(r)v(r) in (0, 1), v′(0) = 0, v(1) = 0. By [14, Theorem 1.2], we know that µk is positive, simple and the corresponding eigenfunction v has exactly k − 1 simple zeros in (0, 1). By Lemma 2.7, we obtain that un has at least k zeros in (0, 1) for n large enough, and this contradicts the fact that un has exactly k − 1 zeros in (0, 1). Consequently, µn ≤M for some constant M > 0 and sufficiently large n ∈ N. (b) In view of (a), we only need to show that Cνk connects ( µνk f0 , 0) to (+∞,+∞). At the beginning, we prove that Cνk is unbounded in the direction of µ. On the contrary, suppose that there exists µM be a blow up point of parament µ and µM < +∞. Then there exists a sequence nodal solutions {(µn, un)} ∈ Cνk , such that limn→+∞ µn = λM and limn→+∞ ‖un‖ = +∞. Let vn = un ‖un‖ . Then vn should be the solutions of problem −M+ C (D2vn) = µn f(un) ‖un‖ in B. Similar to the argument for (2.11), one obtains limn→+∞ f(un)/‖un‖ = 0. By the compactness of (−M+ C )−1, see [14, Page 5], we have that for a subsequence vn → v0 as n → +∞ and v0 ≡ 0. This contradicts ‖v0‖ = 1. Thus Cνk is unbounded in the direction of µ. Next, we show that Cνk is unbounded in the direction of E. Suppose that Cνk is bounded in the direction of E. Thus there exist (µn, un) ∈ Cνk and a positive constant M , such that µn → +∞ as n → +∞ and ‖un‖ ≤ M for any n ∈ N. Therefore, we can conclude that there exists a constant δ > 0, such that f(un) un ≥ δ. Similar to part (a) and by using Strum comparison lemma, Lemma 2.7, we can arrive that un has at least k simple zeros, which contradicts (µn, un) ∈ Cνk . So Cνk is unbounded in the direction of E. Therefore, conclusion (b) follows. EJDE-2021/31 HAMILTON-JACOBI-BELLMAN EQUATIONS 15 (c) If (µ, u) is a nontrivial solution of problem (1.7), dividing problem (1.7) by ‖u‖2 and setting v = u/‖u‖2, we obtain −M+ C (D2v) = µ f(u(x)) ‖u‖2 , in B, v = 0, on ∂B. (3.2) We define f̃(v) = { ‖v‖2f(v/‖v‖2), if v 6= 0, 0, if v = 0. The problem (3.2) is equivalent to −M+ C (D2v) = µf̃(v), in B, v = 0, on ∂B. (3.3) Clearly, (µ, 0) is always the solution of problem (3.3). By simple computation, we can show that (f̃)0 = f∞ and (f̃)∞ = f0. Now, applying (b) and the inversion v → v/‖v‖2 = u, we achieve the conclusions. (d) We define fn(x) =  x/n, x ∈ [−1/n, 1/n],( f( 2 n )− 1/n2 ) (nx− 2) + f( 2 n ), x ∈ (1/n, 2/n), − ( f(− 2 n ) + 1/n2 ) (nx+ 2) + f(− 2 n ), x ∈ (−2/n,−1/n), f(x), x ∈ (−∞,−2/n] ∪ [2/n,+∞). Then fn(x) ∈ C(R,R). One considers the auxiliary problem −M+ C (D2u) = µfn(u(x)), in Ω, u = 0, on ∂Ω. (3.4) It is not difficult to check that limn→+∞ fn(x) = f(x), (fn)0 = 1 n and (fn)∞ = f∞ = 0. For any fixed n ∈ N, it follows from (b) that there exists a sequence unbounded continua Cνk,n of solutions to problem (3.4) emanating from (nµνk, 0) and connecting to (+∞,+∞). Let C̃νk = lim supn→+∞ Cνk,n. For any (µ, u) ∈ C̃νk , the definition of limit superior shows that there exists a sequence (µn, un) ∈ Cνk,n, such that (µn, un) → (µ, u) as n→ +∞. Clearly, one has −M+ C (D2un) = µnf n(un(x)). One applies Lemma 2.2, then it arrives that u satisfies −M+ C (D2u) = µf(u(x)), i.e., u is a solution of (1.7). By Lemma 2.1, there exists an unbounded component Cνk of C̃νk of solutions to problem (1.7), such that (+∞, 0) ∈ Cνk and (+∞,+∞) ∈ Cνk . So there exists µ∗ > 0 such that for µ ∈ (µ∗,+∞), problem (1.7) has at least two nodal solutions u1,+ k and u2,+ k . Moreover, u1,+ k and u2,+ k have exactly k − 1 simple zeros in (0, 1) and are positive near 0; and there exists µ′ > 0 such that for µ ∈ (µ′,+∞), problem (1.7) has at least two nodal solutions u1,− k and u2,− k ; moreover, u1,− k and u2,− k have exactly k − 1 simple zeros in (0, 1) and are negative near 0. Next, one shows that there exists µ̃∗ > 0 such that problem (1.7) has no nodal solution for any µ ∈ (0, µ̃∗). On the contrary, suppose that there exists a sequence {(µn, un)} ∈ Cνk such that limn→+∞ µn = 0. On the other hand, f0 = f∞ = 0 16 X. Y. LI, B. X. YANG EJDE-2021/31 implies that there exists a positive constant M such that f(s) s ≤M for any s 6= 0. Let vn = un ‖un‖ . Then, one has vn = (−M+ C )−1 (µnf(un(x)) ‖un‖ ) . Let f̂(u) = max0≤|s|≤u |f(s)|, then f̂ is nondecreasing with respect to u. Then we arrive at lim u→0 f̂(u) u = 0. (3.5) Furthermore, it follows from (3.5) that f(u) ‖u‖ ≤ f̂(|u|) ‖u‖ ≤ f̂(‖u‖) ‖u‖ → 0 as ‖u‖ → 0. (3.6) So limn→+∞ µnf(un)/‖un‖ = 0. By the compactness of (−M+ C )−1 again, see [14, Page 5], it receives that for some convenient subsequence vn → v0 as n → +∞. Letting n → +∞, one has v0 ≡ 0. This contradicts ‖v0‖ = 1. So the conclusions follow. (e) We define the cut-off function of f as fn(x) =  nx, x ∈ [−1/n, 1/n], n(f( 2 n )− 1)(x− 1 n ) + 1, x ∈ (1/n, 2/n), −n(f(− 2 n ) + 1)(x+ 1 n )− 1, x ∈ (−2/n,−1/n), f(x), x ∈ (−∞,−2/n] ∪ [2/n,+∞). (3.7) Then fn ∈ C(R,R). One considers the auxiliary problem −M+ C (D2u) = µfn(u(x)), in B, u = 0, on ∂B. (3.8) It is easy to see that limn→+∞ fn(x) = f(x), (fn)0 = n and (fn)∞ = f∞ = 0. (b) implies that there exists a sequence unbounded continua Cνk,n of solutions to problem (3.8) emanating from ( µνk n , 0) and connecting to (+∞,+∞). With the help of Lemma 2.1, we shall finish the proof. Taking zn = ( µνk n ,+∞) and z∗ = (0, 0), it receives that zn → z∗. So Lemma 2.1 (i) is satisfied. (ii) and (iii) are obvious. So there exists an unbounded component Cνk of lim supn→+∞ Cνk,n, such that (0, 0) ∈ Cνk and (+∞,+∞) ∈ Cνk . This completes the proof. (f) We define the function fn(x) as in (3.7) and consider the auxiliary problem (3.8) again, but in this case (fn)0 = n and (fn)∞ = f∞ ∈ (0,+∞). In view of (a), one derives that there exists a sequence unbounded continua Cνk,n of solutions to problem (3.8) emanating from ( µνk n , 0) and connecting to ( µνk f∞ ,+∞). On account of Lemma 2.1 again, we can easily obtain that there exists an un- bounded component Cνk of lim supn→+∞ Cνk,n, such that (0, 0) ∈ Cνk and ( µνk f∞ ,+∞). This completes the proof. � EJDE-2021/31 HAMILTON-JACOBI-BELLMAN EQUATIONS 17 4. Proof of Theorem 1.3 Based on the Theorem 1.2, we give the proof of Theorem 1.3. Proof of Theorem 1.3. The argument of Theorem 1.2(a) implies that there is an unbounded continua Cνk , emanating from ( µνk f0 , 0), such that it satisfies Cνk ⊂ ({(µ ν k f0 , 0)} ∪ (R+ × Sνk )), where ν ∈ {+,−}. Let η ∈ C(R,R) be such that f(u) = f∞u+ η(u) with lim |s|→+∞ η(s) s = 0. Let us consider −M+ C (D2u) = µf∞u+ µη(u), in B, u = 0, on ∂B (4.1) as a bifurcation problem from infinity. Applying Theorem 1.1 to (4.1), it shows that there exists an unbounded continua Dνk of solutions of (4.1), emanating from ( µνk f∞ ,+∞). Next, we show that the components Cνk and Dνk are disjoint under the assumption (A3), that is, one wants to obtain for any (µ, u) ∈ C+ k ∪ C − k , s2 < u(r) < s1 for all r ∈ [0, 1]; and for any (µ, u) ∈ D+ k ∪ D − k , max{u(r)|r ∈ [0, 1]} > s1 or min{u(r)|r ∈ [0, 1]} < s2. On the contrary, suppose that there exists (µ, u) ∈ C+ k ∪ C − k ∪ D + k ∪ D − k , such that either max{u(r)|r ∈ [0, 1]} = s1 or min{u(r)|r ∈ [0, 1]} = s2. We only discuss the case of max{u(r)|r ∈ [0, 1]} = s1. The discussion for the other case min{u(r)|r ∈ [0, 1]} = s2 is closely similar, so we omit it here. In this case, there exists j ∈ {0, 1, . . . , k − 1} such that max{u(r) : r ∈ [0, 1]} = s1 and 0 ≤ u(r) ≤ s1 for all r ∈ [τj , τj+1], where [τj , τj+1] ⊂ [0, 1]. We claim that there exists 0 < m < +∞ such that f(s) ≤ m(s1 − s) for any s ∈ [0, s1]. With the aid of (A3), it is easy to see that the claim is true for the cases s = 0 and s = s1. For any ε ∈ (0, γ1), it follows from (A4) that there exists δ > 0 such that f(s) < (γ1 + ε)(s1 − s) for any s ∈ (s1 − δ, s1). From (A3), one arrives at max s∈[0,s1−δ] f(s) s1 − s := ρ > 0. So the claim is verified by choosing m = max{ρ, γ1 + ε}. Now, we consider an equivalent problem of (1.7) as follows, −M+ C (D2(s1 − u)) + µm(s1 − u) = µm(s1 − u)− µf(u), |x| ∈ [τj , τj+1], s1 − u > 0, |x| = τj , τj+1. It is obvious that f(s) ≤ m(s1 − s) for any s ∈ [0, s1] implies −M+ C (D2(s1 − u)) + µm(s1 − u) ≥ 0, |x| ∈ [τj , τj+1], s1 − u > 0, |x| = τj , τj+1. (4.2) 18 X. Y. LI, B. X. YANG EJDE-2021/31 Let v = s1 − u, then (4.2) is equivalent to M−C (D2v)− µmv ≤ 0, |x| ∈ [τj , τj+1], v = s1, |x| = τj , τj+1. (4.3) The strong maximum principle (Remark 2.5) implies that, if v satisfies (4.3), then it has v > 0 in [τj , τj+1]; that is s1 > u(r) on [τj , τj+1]. This is a contradiction. So for any (µ, u) ∈ C+ k ∪ C − k , s2 < u(r) < s1, r ∈ [0, 1]; for (µ, u) ∈ D+ k ∪ D − k , max{u(r)|r ∈ [0, 1]} > s1 or min{u(r)|r ∈ [0, 1]} < s2. Therefore, ( µνk f0 ,+∞) ⊆ Proj(Cνk ) and Dνk has an unbounded projection on R. Immediately, from the global structures of Cνk and Dνk , one obtain that problem (1.7) has at least one nodal solution uνk for any µ ∈ (min {µ ν k f0 , µνk f∞ },max {µ ν k f0 , µνk f∞ }), k ∈ N, such that νuνk has exactly k − 1 simple zeros in (0, 1) and is positive near 0; (1.7) at least has four nodal solutions u+ k,0, u + k,∞, u − k,0 and u−k,∞ for any µ ∈ (max {µ + k f0 , µ−k f0 , µ+ k f∞ , µ−k f∞ },+∞), such that they have exactly k − 1 simple zeros in (0, 1), u+ k,0 and u+ k,∞ are positive near 0, u−k,0 and u−k,∞ are negative near 0. Finally, we show ‖u+ k,0‖ → s−1 and ‖u−k,0‖ → (−s2)− as µ → +∞. Here, we only prove the case of ν = +. Because the proof of ν = − is similar. Suppose, by contradiction, that there exists η ∈ (0, s1), such that ‖u+ k,0‖ ≤ η. The assumption (A3) implies that there exists a positive constant δ > 0, such that f(u+ k,0) u+ k,0 ≥ δ. 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Xiaoyan Li Department of Mathematics and Statistics, Northwest Normal University, Lanzhou, Gansu 730000, China Email address: 17242502@qq.com Bian-Xia Yang (corresponding author) College of Science, Northwest A&F University, Yangling, Shaanxi 712100, China Email address: yanglina7765309@163.com 1. Introduction and main results 1.1. Existing results 1.2. Statement of main results 2. Preliminary results and Proof of Theorem 1.1 3. Proof of Theorem 1.2 4. Proof of Theorem 1.3 Acknowledgments References