Symmetry of Solutions of Minimal Gradient Graph Equations on Punctured Space

Symmetry of solutions of minimal gradient graph equations on punctured space

Zixiao Liu, Jiguang Bao*

April 23, 2021

Abstract In this paper, we study symmetry and existence of solutions of minimal gradient graph equations on punctured space Rn \{0}, which include the Monge-Ampere` equation, inverse harmonic Hessian equation and the special Lagrangian equation. This extends the classiﬁca- tion results of Monge-Ampere` equations. Under some conditions, we also give the character- ization of the solvability on exterior Dirichlet problem in terms of their asymptotic behaviors. Keywords: minimal gradient graph equation, optimal symmetry, existence. MSC 2020: 35J60; 35B06.

1 Introduction

We consider the following fully nonlinear elliptic equations

2 n Fτ (λ(D u)) = C0 in R \{0}, (1.1)

2 2 where C0 is a constant, λ(D u) = (λ1, λ2, ··· , λn) are n eigenvalues of Hessian matrix D u, π τ ∈ [0, 2 ] and

 1 n  X  ln λi, τ = 0,  n  √ i=1  a2 + 1 n λ + a − b  X i π  ln , 0 < τ < 4 ,  2b λi + a + b arXiv:2104.10904v1 [math.AP] 22 Apr 2021  i=1  √ n  X 1 π Fτ (λ) := − 2 , τ = 4 , 1 + λi  √ i=1  2 n  a + 1 X λi + a − b  arctan , π < τ < π ,  b λ + a + b 4 2  i=1 i  n  X  arctan λ , τ = π ,  i 2 i=1

*Supported in part by National Natural Science Foundation of China (11871102 and 11631002).

1 a = cot τ, b = p|cot2 τ − 1|. Equations (1.1) origin from gradient graph (x, Du(x)) with zero n n mean curvature under various metrics equipped on R × R . In 2010, Warren [32] ﬁrst proved 2 2 n that if u ∈ C (Ω) is a solution of Fτ (λ(D u)) = C0 in Ω ⊂ R , then the volume of (x, Du(x)) π π π 1 is a maximal for τ ∈ [0, 4 ) and minimal for τ ∈ ( 4 , 2 ] among all homologous, C , space-like n n n-surfaces in (R × R , gτ ), where h π i g = sin τδ + cos τg , τ ∈ 0, , τ 0 0 2 is the linearly combined metric of standard Euclidean metric

n n X X δ0 = dxi ⊗ dxi + dyj ⊗ dyj, i=1 j=1 and the pseudo-Euclidean metric

n n X X g0 = dxi ⊗ dyi + dyj ⊗ dxj. i=1 j=1

If τ = 0, then (1.1) becomes the Monge-Ampere` equation

det D2u = enC0 .

n For the Monge-Ampere` equation on entire R , the classical theorem by Jorgens¨ [22], Calabi [11] 2 n and Pogorelov [30] states that any convex classical solution of det D u = 1 in R must be a quadratic polynomial. See Cheng-Yau [13], Caffarelli [5] and Jost-Xin [23] for different proofs n and extensions. For the Monge-Ampere` equation in exterior domain of R , there are exterior Jorgens-Calabi-Pogorelov¨ type results by Ferrer-Mart´ınez-Milan´ [15] for n = 2 and Caffarelli-Li [7], which state that any locally convex solution must be asymptotic to quadratic polynomials (for n = 2 we need additional ln-term) near infinity. For the Monge-Ampere` equation on punctured n space R \{0}, there are classification results by Jorgens¨ [22] for n = 2 and Jin-Xiong [21] that shows every locally convex function of det D2u = 1, modulo an affine transform, has to be |x| 1 R n n u(x) = 0 (r + c) dr for some c ≥ 0. For further generalization and refined asymptotics on Jorgens-Calabi-Pogorelov¨ type results we refer to [3, 8, 18, 31] and the references therein. π If τ = 2 , then (1.1) becomes the special Lagrangian equation n X 2 arctan λi D u = C0. (1.2) i=1 n For special Lagrangian equation on entire R , there are Bernstein-type results by Yuan [33, 34], which state that any classical solution of (1.2) and ( −KI, n ≤ 4, n − 2 D2u ≥ or C > π, (1.3) −( √1 + (n))I, n ≥ 5, 0 3 2 must be a quadratic polynomial, where I denote the identity matrix, K is any constant and (n) n is a small dimensional constant. For special Lagrangian equation on exterior domain of R , there

2 is an exterior Bernstein-type result by Li-Li-Yuan [25], which states that any classical solution of (1.2) on exterior domain with (1.3) must be asymptotic to quadratic polynomial (for n = 2 we need additional ln-term) near infinity. Furthermore, Chen-Shankar-Yuan [12] proved that a convex viscosity solution of (1.2) must be smooth. π Pn 1 If τ = , then (1.1) is a translated inverse harmonic Hessian equation 2 = 1, 4 i=1 λi(D u) n which is a special form of Hessian quotient equation. For Hessian quotient equation on entire R , there is a Bernstein-type result by Bao-Chen-Guan-Ji [1]. π For general τ ∈ [0, 2 ], Warren [32] proved the Bernstein-type results under suitable semi- convex conditions by the results of Jorgens¨ [22]-Calabi [11]-Pogorelov [30], Flanders [16] and Yuan [33, 34]. In our earlier work [28], we generalized the results to equation (1.1) on exterior domain and provide finer asymptotic expansions. For further generalization with perturbed right hand side we refer to [29] etc. In this paper, we focus on n ≥ 3 and prove a type of classification results as in Jorgens¨ [22] and Jin-Xiong [21]. T n T Hereinafter, we let x denote the transpose of vector x ∈ R , O denote the transpose of n×n matrix O, Sym(n) denote the set of symmetric n × n matrix, σk(λ) denote the kth elementary form of λ and DFτ (λ(A)) denote the matrix with elements being value of partial derivative of Fτ (λ(M)) w.r.t Mij variable at matrix A. A function u(x) is called generalized symmetric with respect to A ∈ Sym(n) if it relies only on the value of xT Ax. As in [6], we say u ∈ C0(Ω) is n 2 a viscosity subsolution (supersolution) of (1.1) in Ω ⊂ R if for any function ψ ∈ C (Ω) with 2 2 2 2 π π D ψ > 0, D ψ > (−a + b)I, D ψ > −I, D ψ > −(a + b)I for τ = 0, τ ∈ (0, 4 ), τ = 4 , π π τ ∈ ( 4 , 2 ) respectively and point x ∈ Ω satisfying ψ(x) = u(x) and ψ ≥ (≤)u in Ω, we have 2 Fτ (λ(D ψ(x))) ≥ (≤)C0. If u is both viscosity subsolution and supersolution of (1.1), we say it is a viscosity solution. By a direct computation (see for instance [19, 32]), if λi > −a − b, for i = 1, 2, ··· , n, then n n X λi + a − b X λi + a nπ arctan = arctan − . (1.4) λ + a + b b 4 i=1 i i=1

Hence replacing u(x) by  (a−b) 2 π  u(x) + 2 |x| , 0 < τ < 4 ,  1 2 π u(x) + 2 |x| , τ = 4 , (1.5)  u(x) a 2 π π  b + 2b |x| , 4 < τ < 2 , it is equivalent to consider 2 n Gτ (λ(D u)) = C0 in R \{0}, (1.6)

3 where  n  1 X  ln λi, τ = 0,  n  √ i=1  2 n  a + 1 X λi  ln , 0 < τ < π ,  2b λ + 2b 4  i=1 i  n  √ X 1 G (λ) := − 2 , τ = π , τ λ 4  i=1 i  √ n !  a2 + 1 X nπ  arctan λ − π < τ < π ,  b i 4 4 2  i=1  n  X  arctan λ , τ = π .  i 2  i=1 Our ﬁrst main result provides a type of symmetry and relationship between asymptotic at inﬁnity and value at origin.

2 n 0 n Theorem 1.1. Let u ∈ C (R \{0}) be a classical convex solution of (1.6). Then u ∈ C (R ),

2 n Gτ (λ(D u)) ≥ C0 in R (1.7)

n in viscosity sense and there exist c ∈ R, β ∈ R and A ∈ Sym(n) with Gτ (λ(A)) = C0 such that 1 u(x) ≤ xT Ax + βx + c in n. (1.8) 2 R Furthermore, u(x) − βx is symmetric in eigenvector directions of A, i.e.,

n T T u(xe) − βxe = u(x) − βx, ∀ {xe ∈ R :(O xe)i = ±(O x)i, ∀ i = 1, 2, ··· , n}, (1.9) where O is an orthogonal matrix such that OAOT is diagonal.

Remark 1.2. After a change of coordinate y := OT x, u(y) − βy is symmetric with respect to n coordinate planes. Especially, u(y) − βy is central symmetry i.e., u(y) − βy = u(−y) + βy.

Remark 1.3. The solutions of (1.1) and (1.6) have asymptotic behavior and higher order expansion at inﬁnity, see for instance earlier results by the authors [28, 29].

π π Remark 1.4. For τ ∈ 4 , 2 case, the convex assumption on u in Theorem 1.1 can be relaxed to either of the following

 D2u > −I, n ≤ 4,  n o D2u > − min 1, √1 + (n) I, n ≥ 5, 3 (1.10)  D2u > −I and √bC0 + nπ > n−2 π, n ≥ 5, a2+1 4 2

π where (n) is as in (1.3). For τ = 2 case, the convex assumption can be relaxed to (1.3). π Our second main result shows that the symmetry (1.9) is optimal for τ ∈ (0, 2 ] cases in the following sense.

4 Proposition 1.5. Let 0 < A ∈ Sym(n) satisfy Gτ (λ(A)) = C0. If there exists a generalized 1 T symmetric convex classical solution u(x) = U( 2 x Ax) of (1.6). Then for τ = 0 case,

1 (xT Ax) 2 Z 1 C0 n u = e (r + c1) n dr + c2 0 π π π π π for some c1 ≥ 0 and c2 ∈ R. For cases τ ∈ (0, 4 ), τ = 4 , τ ∈ ( 4 , 2 ) and τ = 2 , either A is a multiplication of identity matrix i.e., ! √ 2b 2C A = − 2b I,A = − 0 I, 1 − exp( √2b C ) 2n n a2+1 0 b π C0 A = tan √ C0 + I,A = tan I n a2 + 1 4 n 1 T respectively, or u is the quadratic function 2 x Ax up to some constant c.

Remark 1.6. Proposition 1.5 still holds if equation (1.6) is restricted to an annulus BR2 \ BR1 with any 0 < R1 < R2, where Br denotes the ball centered at origin with radius r. By (1.8) in Theorem 1.1, c ≥ u(0). Our final main result shows that c ≥ u(0) is also a sufficient condition for a classical solution exists. More explicitly, we prove the existence of π solution of (1.1), τ ∈ (0, 4 ) with prescribed value at origin and asymptotic behavior at infinity for all c ≥ u(0).

n Theorem 1.7. For any given c ≥ u0, β ∈ R and 0 < A ∈ Sym(n) with Gτ (λ(A)) = C0 satisfying n X min{λ1(A), ··· , λn(A)} + 2b δ := > 2, (1.11) 0 λ (A) + 2b i=1 i 0 n there exists a unique convex viscosity solution u ∈ C (R ) of  G (λ(D2u)) = C , in n \{0},  τ 0 R u(0) = u0, (1.12)  1 T 2−δ0 u(x) = 2 x Ax + βx + c + O(|x| ), as |x| → ∞,

π with τ ∈ (0, 4 ). Remark 1.8. At the end of Section 3, we prove that the solution found in Theorem 1.7 also satisﬁes (1.7) and (1.8). Condition (1.11) holds if A has two same minimal eigenvalues. For any C0 < 0, condition (1.11) is possible to fail. For instance there exists sequences such that Gτ (λ1, λ2, λ3) = C0 but 2b λ1() → − 2b and λ2(), λ3() → +∞ 1 − exp( √ 2b C ) a2+1 0 as → 0+. By a direct computation, the left hand side of (1.11) tends to 1 as → 0+. Whether (1.11) is optimal remains unknown.

5 π Remark 1.9. The study on exterior Dirichlet problem by Li-Li [24] for τ = 4 and Li [27] for π π τ ∈ ( 4 , 2 ] also works for punctured space after minor modiﬁcation. From their proof, the results in Theorem 1.7 still holds with δ0 changed into

σn−1(λ(A)) · min{λ1(A), ··· , λn(A)} δ0 = σn(λ(A))

π for τ = 4 ,

Pn kc ( nπ + √ b C )σ (λ(A)) k=0 k 4 a2+1 0 k δ0 = Pn ξ ( nπ + √ b C , λ(A))c ( nπ + √ b C )σ (λ(A)) k=0 k 4 a2+1 0 k 4 a2+1 0 k

π π for τ ∈ ( 4 , 2 ) and Pn k=0 kck(C0)σk(λ(A)) δ0 = Pn k=0 ξk(C0, λ(A))ck(C0)σk(λ(A)) π for τ = 2 , where

j+1 c2j(C) := (−1) sin C if k = 2j, j ∈ N, ck(C) := j c2j+1(C) := (−1) cos C if k = 2j + 1, j ∈ N,

Pn 2 2 i=1 σk−1;i(λ(A))λi (A)xi n Ξk(λ(A), x) := Pn 2 ∀ x ∈ R \{0}, σk(λ(A)) i=1 λi(A)xi ¯ ξk(λ(A)) := sup Ξk(λ(A), x), ξ (λ(A)) := inf Ξk(λ(A), x), k n x∈Rn\{0} x∈R \{0} and ξ¯ (λ(A)) if c (C) > 0, ξ (C, λ(A)) := k k k ξ (λ(A)) if c (C) ≤ 0. k k For τ = 0, optimal existence of Dirichlet problem on exterior domain and punctured space have been studied in [3, 7, 21, 26] etc.

The paper is organized as follows. In the next three sections, we prove Theorem 1.1, Theorem 1.7 and Proposition 1.5 respectively.

2 Proof of Theorem 1.1

In this section, we prove Theorem 1.1 by maximum principle and removable singularity of viscosity solutions. The following knowledge on convex functions is necessary.

Theorem 2.1 (Proposition 2.1 of [21]). Let v be a locally convex function in B1 \{0} with n ≥ 2. Then u can be uniquely extended to be a convex function in B1.

n 0 Theorem 2.2 (Theorem 1.1 of [9]). Let Ω ⊂ R be a domain, x ∈ Ω and u ∈ C (Ω) be a viscosity solution of F (D2u) ≥ f(x) in Ω \{x},

6 where F ∈ C0(Sym(n)) is an elliptic operator and f ∈ C0(Ω). Suppose that u is upperconical at x, i.e., sup ((u + η)(x) − (u + η)(x) + |x − x|) > 0 (2.1) x∈Ω for any η ∈ C∞(Ω) and > 0. Then

F (D2u) ≥ f(x) in Ω in viscosity sense. Remark 2.3. If u is a convex function in Ω, then for all x ∈ Ω, u is upperconical at x. By contradiction, we suppose there exist ∈ (0, 1) and η ∈ C∞(Ω) such that

(u + η)(x) − (u + η)(x) + |x − x| ≤ 0 in Ω.

By Taylor expansion of η near x, there exists 0 < δ < 1 such that u(x) ≤ u(x) − ∇η(x) · (x − x) − |x − x|, ∀ 0 < |x − x| < δ. 2

Since u is convex, take any e ∈ ∂B1 and we have δ δ δ 2u(x) ≤ u(x + e) + u(x − e) ≤ 2u(x) − < 2u(x), 2 2 4 which is a contradiction. For more examples and discussions on condition (2.1), we refer to Remark 1.2 of [9].

2 n Let u ∈ C (R \{0}) be a classical solution of (1.6) as in Theorem 1.1 or Remark 1.4. By n the asymptotic behavior result (see for instance Theorem 1.3 of [28]), there exists c ∈ R, β ∈ R and A ∈ Sym(n) with Gτ (λ(A)) = C0 such that 2+k−n k 1 T lim sup |x| D u(x) − x Ax + βx + c < ∞ (2.2) |x|→+∞ 2 for all k ∈ N. Step 1. we prove (1.7) and (1.8). The proof is separated into ﬁve cases according to τ. If u is a classical convex solution as in Theorem 1.1, (1.7) follows immediately from Theorems 2.1 and 2.2. By (2.2), we have 1 u(x) → xT Ax + βx + c =: V (x) as |x| → ∞. 2 2 n By a direct computation, V (x) is a quadratic solution of Gτ (λ(D u)) = C0 on entire R . By maximum principle (see for instance Theorem 17.1 of [17]), 1 u(x) ≤ V (x) = xT Ax + βx + c in n. 2 R π π π For τ ∈ ( 4 , 2 ) case in Remark 1.4, the desired result follows from τ = 2 case, which will be π proved immediately. For τ = 2 case of Remark 1.4, we prove that there exists K0 ≥ 0 such that 2 D u > −K0I.

7 n−2 For the ﬁrst subcase of (1.3), the result is immediate. For C0 > 2 π, then for all i = 1, 2, ··· , n, n n − 1 X arctan λ (D2u) + π > arctan λ (D2u) = C . i 2 j 0 j=1

n−1 Thus we may take K0 := min{tan(C0 − 2 π), 0}. Let K v(x) := u(x) + 0 |x|2, 2 then 2 2 n D v = D u + K0I > 0 in R \{0} and v satisﬁes n X 2 n arctan(λi(D v) − K0) = C0 in R \{0}. i=1 By a direct computation, ∂ 1 arctan(λi − K0) = 2 > 0. ∂λi 1 + (λi − K0) Together with asymptotics (2.2), D2v is bounded from above. Hence the differential operator is elliptic (see for instance [10]) and the desired results follow from Theorems 2.1 and 2.2. Step 2. we prove the symmetric property (1.9) for diagonal case i.e., A = diag(a1, a2, ··· , an). We only need to prove for β = 0 case, otherwise replace u by u − βx. Similar to the proof of Theorem 1.1 in [21], we introduce the following maximum principle. Lemma 2.4. Let w be a classical solution of

n aijDijw = 0 in R \{0}, where aij(x) is a positive deﬁnite matrix. Suppose w(0) = 0 and lim w(x) = 0, |x|→∞ then w ≡ 0. n Proof. By contradiction, we may suppose there exists x0 ∈ R \{0} such that u(x0) > 0. Then by n continuity and boundary conditions, there exists x1 ∈ R \{0} such that u(x1) = maxRn\{0} u > n 1 0. Applying standard maximum principle in level set {x ∈ R \{0} : u > 2 u(x1)}, we have a 1 contradiction 2 u(x1) = u(x1). For any i = 1, 2, ··· , n, let

U(x) := u(xe), where xe = (x1, ··· , xi−1, −xi, xi+1, ··· , xn). By a direct computation, u, U satisfy

 2 2 n  Gτ (λ(D u)) = Gτ (λ(D U)) = C0, in R \{0}, u(x) = U(x), at x = 0,  1 T u(x),U(x) → 2 x Ax + c , as |x| → ∞.

8 By Newton-Leibnitz formula between u, U and applying maximum principle as in Lemma 2.4, we have u(x) = U(x) = u(xe). Hence u(x) is hyperplane symmetry with respect to all n coordinate hyperplane. Step 3. we prove (1.9) for general A. Since A is a symmetric matrix, by eigen-decomposition there exists an orthogonal matrix O such that

T A = O ΛO, where Λ = diag(a1, ··· , an). (2.3)

Let v(x) := u(O−1x), then D2v(x) = OD2u(O−1x)O−1 has the same eigenvalues as D2u(O−1x). Hence 2 n Gτ (λ(D v)) = C0 in R \{0} and 1 1 v(0) = u(0), v(x) → xT OAO−1x + c = xT Λx + c 2 2 as |x| → ∞. By the previous step, we have

n v(xe) = v(x) ∀ {xe ∈ R : xei = ±xi, ∀ i = 1, 2, ··· , n}. Thus n T T u(xe) = u(x) ∀ {xe ∈ R :(O xe)i = ±(O x)i, ∀ i = 1, 2, ··· , n} and (1.9) follows immediately.

3 Proof of Theorem 1.7

π In this section, we consider the existence result of Dirichlet problem for τ ∈ (0, 4 ) with prescribed asymptotic behavior and value at origin. By similar translation as (1.5) and eigen-decomposition as (2.3), we may assume β = 0 and consider only the following Dirichlet problem.

 2 n  Gτ (λ(D u)) = C0, in R \{0}, u(0) = u0, (3.1)  1 T u(x) → 2 x Ax + c , as |x| → ∞, where 0 < A = diag(a1, ··· , an) satisﬁes Gτ (a1, ··· , an) = C0 < 0.

We will write the vector (a1, ··· , an) simply as a when no confusion can arise. Similar to the strategy as in [2, 24, 27], we seek for generalized symmetric subsolution of form

n 1 X u(x) = U(s), where s(x) := a x2, (3.2) 2 i i i=1 and apply Perron’s method. By a direct computation,

0 0 00 Diu(x) = U (s)aixi,Diju(x) = U (s)aiδij + U (s)aiajxixj.

9 By Sylvester’s determinant theorem, we have

2 σn(λ(D u))  0     U a1 a1x1  ..  00  .   = det  .  + U (s)  .  · (a1x1, ··· , anxn) 0 U an anxn  0  U a1  ..  = det  .  0 U an   0 −1    (U a1) a1x1  00  ..   .  · 1 + U (a1x1, ··· , anxn)  .   .  0 −1 (U an) anxn n 0 n 00 0 n−1 X 2 = (U ) σn(a) + U (U ) σn(a) aixi , i=1 i.e., 2 0 n 00 0 n−1 σn(λ(D u)) = (U ) σn(a) + 2sU (U ) σn(a). (3.3) Similarly, 2 0 00 Dij(u + b|x| ) = (U ai + 2b)δij + U aiajxixj and

2 2 σn(λ(D (u + b|x| )))  0     U a1 + 2b a1x1  ..  00  .   = det  .  + U (s)  .  · (a1x1, ··· , anxn) 0 U an + 2b anxn  0  U a1 + 2b  ..  = det  .  0 U an + 2b   0 −1    (U a1 + 2b) a1x1  00  ..   .  · 1 + U (a1x1, ··· , anxn)  .   .  0 −1 (U an + 2b) anxn n n Y 0 00 X 2 2 Y 0 = (U ai + 2b) + U ai xi (U aj + 2b) i=1 i=1 j=1,··· ,n j6=i i.e.,

n n 2 2 Y 0 2b 00 X 2 Y 0 2b σn(λ(D (u + b|x| ))) = σn(a) (U + ) + U σn(a) aixi (U + ). (3.4) ai aj i=1 i=1 j=1,··· ,n j6=i

10 By (3.3) and (3.4), u(x) = U(s) is a subsolution of (1.6) as long as

0 n 00 0 n−1 (U ) σn(a) + 2sU (U ) σn(a)   ≥ c0, n n Y 0 00 X 2 Y 0 2b (U ai + 2b) + U σn(a) aixi (U + )  aj  i=1 i=1 j=1,··· ,n j6=i i.e., (U 0)n + 2sU 00(U 0)n−1   ≥ c0, (3.5) n n Y 0 2b 00 X 2 Y 0 2b (U + ) + U aixi (U + ) ai  aj  i=1 i=1 j=1,··· ,n j6=i where c = exp( √ 2b C ) ∈ (0, 1) satisﬁes 0 a2+1 0

n n Y ai Y 2b c = , i.e., 1 = c · (1 + ). (3.6) 0 a + 2b 0 a i=1 i i=1 i Hereinafter, we may assume without loss of generality that

0 < a1 ≤ a2 ≤ · · · ≤ an.

Lemma 3.1. Let U be a classical solution of

 n Y 2b  (U 0)n − c (U 0 + )  0 a  i=1 i  n ! 00 0 n−1 Y 0 2b (3.7)  +2(s + 1)U (U ) − c0 (U + ) = 0, in s > 0,  ai  i=2  U 0 ≥ 1,U 00 ≤ 0, in s > 0.

2 n Then u(x) = U(s) is convex and satisﬁes Gτ (λ(D u)) ≥ C0 in R \{0}. 0 Proof. By the assumptions on ai, for all U ≥ 1,   n n  n−1  n−1 X 2 Y 0 2b X 2 Y 0 2b Y 0 2b aixi (U + ) ≤ aixi (U + ) = 2s (U + ).  aj  aj aj i=1 j=1,··· ,n i=1 j=1 j=1 j6=i

We claim that the denominator part of (3.5) is positive. By U 0 ≥ 1,U 00 ≤ 0, we only need to prove U 0 + 2sU 00 > 0, (3.8)

11 then   n n Y 0 2b 00 X 2 Y 0 2b (U + ) + U aixi (U + ) ai  aj  i=1 i=1 j=1,··· ,n j6=i n n−1 Y 2b Y 2b ≥ (U 0 + ) + 2sU 00 (U 0 + ) a a i=1 i i=1 i n−1 ! Y 2b 2b = (U 0 + ) · U 0 + + 2sU 00 a a i=1 i n > 0. By (3.7), we have

n 0 n Y 0 2b (U ) − c0 (U + ) ai 00 00 i=1 2sU ≥ 2(s + 1)U = − n . Y 2b (U 0)n−1 − c (U 0 + ) 0 a i=2 i Consequently, n n ! Y 0 2b 0 Y 0 2b c0 (U + ) − U (U + ) ai ai 0 00 i=1 i=2 U + 2sU ≥ n . Y 2b (U 0)n−1 − c (U 0 + ) 0 a i=2 i Notice that for all U 0 ≥ 1, in view of (3.6),

n n ! Y 2b Y 2b (U 0)n−1 − c (U 0 + ) = (U 0)n−1 1 − c (1 + ) > 0. 0 a 0 a U 0 i=2 i i=2 i Combining the calculus above, (3.8) is proved. By (3.3) and Proposition 1.2 of [2],

n 2 0 k 00 0 k−1 X 2 σk(λ(D u)) = (U ) σk(a) + U (U ) (aixi) σk−1;i(a). (3.9) i=1

Since U 0 ≥ 1 and U 00 ≤ 0, together with (3.8) and the fact that

σk(a) = σk;i(a) + aiσk−1;i(a), for all k = 1, 2, ··· , n, we have

n 2 0k 00 0k−1 X 2 σk λ D u = U σk(a) + U U aixi (σk(a) − σk;i(a)) i=1 0 k 00 0 k−1 ≥ (U ) σk(a) + 2sU (U ) σk(a) > 0.

12 2 n It follows that D u > 0 in R \{0}. By the calculus above, (3.5) holds as long as n 0 n Y 0 00 0 n−1 (U ) σn(a) − c0 (U ai + 2b) + 2sσn(a)U (U ) i=1  n 00 X 2 Y 0 2b −c0U σn(a) aixi (U + ) ≥ 0.  aj  i=1 j=1,··· ,n j6=i

Dividing both sides by σn(a) > 0, it is equivalent to   n n 0 n Y 0 2b 00 0 n−1 00 X 2 Y 0 2b (U ) − c0 (U + ) + 2sU (U ) − c0U aixi (U + ) ≥ 0. (3.10) ai  aj  i=1 i=1 j=1,··· ,n j6=i By the assumptions, we have   n n X 2 Y 0 2b Y 0 2b aixi (U + ) ≥ 2s (U + )  aj  ai i=1 j=1,··· ,n i=2 j6=i and   n 00 0 n−1 00 X 2 Y 0 2b 2sU (U ) − c0U aixi (U + )  aj  i=1 j=1,··· ,n j6=i n ! Y 2b ≥ 2sU 00 (U 0)n−1 − c (U 0 + ) . 0 a i=2 i By a direct computation, as long as U 0 ≥ 1, n n n ! Y 2b Y ai Y ai + 2b 2b (U 0)n−1−c (U 0+ ) ≥ (U 0)n−1· 1 − ( ) ( ) = (U 0)n−1· > 0. 0 a a + 2b a a + 2b i=2 i i=1 i i=2 i 1 Hence (3.10) holds if U 0 ≥ 1,U 00 ≤ 0 and n n ! Y 2b Y 2b (U 0)n − c (U 0 + ) + 2(s + 1)U 00 (U 0)n−1 − c (U 0 + ) = 0 0 a 0 a i=1 i i=2 i Especially when U satisﬁes (3.7), u(x) = U(s) becomes a subsolution of (3.1) and this ﬁnishes the proof.

Lemma 3.2. For any c ≥ 0, there exists a solution of (3.7) with U(0) = 0, U 0(0) ≥ 1 and U(s) = s + c + o(1) as s → ∞, (3.11) if a1, ··· , an satisﬁes n X a1 + 2b > 2. (3.12) a + 2b i=1 i

13 Proof. Step 1. existence of initial value problem  g(ψ(s))  ψ0(s) = , in s ∈ (0, ∞), s + 1 (3.13)  ψ(0) = α ≥ 1, where n Y 2b ψn − c (ψ + ) 0 a i=1 i g(ψ) := − n !. Y 2b 2 ψn−1 − c (ψ + ) 0 a i=2 i By a direct computation and (3.6), for any ψ > 1, Qn 2b n n c0 i=1(ψ + a ) Y 2b Y 2b i = c (1 + ) < c (1 + ) = 1, (3.14) ψn 0 a ψ 0 a i=1 i i=1 i for any ψ ≥ 1, Qn 2b n n c0 i=2(ψ + a ) Y 2b Y 2b i = c (1 + ) < c (1 + ) = 1 (3.15) ψn−1 0 a ψ 0 a i=2 i i=1 i and g(1) = 1. Thus for all ψ > 1, g(ψ) < 0 and ψ = 1 is the unique root of g(ψ) = 0 in ψ ≥ 1. If α = 1, then ψ ≡ 1 is the only solution by the uniqueness theorem of ordinary differential equations. If α > 1, by the local existence result of ODE, ψ exists in a neighbourhood of origin. Since ψ = 1 is the only root of g(ψ) = 0 in the range of ψ ≥ 1 and g(ψ) < 0 as long as ψ > 1, ψ(s) is monotone decreasing with respect to s. Thus ψ is bounded from above by α and hence by the extension theorem, ψ exists on entire s ∈ [0, ∞). Since g(1) = 0, by the uniqueness theorem of ODE, ψ(s) > 1 for all s ∈ [0, ∞). Hence ψ(s) admits a limit ψ∞ at inﬁnity. We claim that the limit ψ∞ must be 1, i.e., ψ(s) → 1 as s → +∞. (3.16)

By contradiction, if ψ∞ > 1, then by equation in (3.13), there exists > 0 and C = C(ψ∞, ) > 0 such that g(ψ(s)) ψ0(s) = < −, ∀ s > C. s + 1 Integral over s ∈ (C, ∞) and it contradicts to the fact that ψ converge to ψ∞ at infinity. This finishes the proof of (3.16). Step 2. refined asymptotics at infinity. By a direct computation and applying (3.14), (3.15),   n n !−1 0 1 X Y 2b Y 2b g (1) = − n − c0 (1 + ) · 1 − c0 (1 + ) . 2  aj  ai i=1 j=1,··· ,n i=2 j6=i Furthermore by (3.6),   n n n n X Y 2b X Y aj Y aj + 2b X ai c0 (1 + ) =  ( ) = aj  aj + 2b aj  ai + 2b i=1 j=1,··· ,n i=1 j=1 j=1,··· ,n i=1 j6=i j6=i

14 and n !−1 n n !−1 Y 2b Y ai Y ai + 2b a1 + 2b 1 − c (1 + ) = 1 − = . 0 a a + 2b a 2b i=2 i i=1 i i=2 i Thus n n 1 a1 + 2b X 2b 1 X a1 + 2b δ0 g0(1) = − = − = − < −1 2 2b a + 2b 2 a + 2b 2 i=1 i i=1 i and ψ − 1 satisﬁes g(ψ) − g(1) (ψ − 1)0 = and ψ − 1 = o(1) s + 1 as s → ∞. By standard ODE analysis, we prove

0 ψ(s) = 1 + O(sg (1)) as s → ∞. (3.17)

In fact, t := ln(s + 1) ∈ (0, +∞), ϕ(t) := ψ(s(t)) − 1 satisﬁes ϕ0(t) = ψ0(s(t)) · et = g(ϕ + 1) − g(1) =: h(ϕ) in (0, +∞). with h(0) = 0, h0(0) = g0(1) and ϕ(t) → 0 as t → +∞. Since g0(1) < 0 and h(ϕ) = g0(1)ϕ + O(ϕ2) as ϕ → 0, by the asymptotic stability of ODE (see for instance Theorem 1.1 of Chap.13 in [14] or Theorem 2.16 of [4] for more detailed analysis), we have

0 ϕ(t) = O(eg (1)t) as t → +∞, which is exactly (3.17). Step 3. construction of solution of second order ODE (3.7). By taking Z s U(s) := u0 + ψ(r)dr, s ≥ 0, 0 we have U 0(s) = ψ(s) ≥ 1 and thus U becomes a solution of (3.7). Step 4. computation on the asymptotics at inﬁnity. By (3.17), for any α ≥ 1, there exist a solution of (3.13) (denoted by ψ(s, α)) and a constant µ(α) ≥ 0 such that

g0(1)+1 U(s) = u0 + s + µ(α) + O(s ) as s → ∞.

More explicitly, by Z s U(s) − s − u0 = (ψ(r, α) − 1)dr, 0 sending s → +∞ we have Z +∞ µ(α) = (ψ(s, α) − 1)ds. (3.18) 0

15 By the theorem of the differentiability of the solution with respect to the initial value, we can differentiate (3.13) with respect to α and obtain

( 0 ∂ψ(s, α) v0 = g (ψ(s,α)) · v, in s ∈ (0, ∞), v(s) := satisﬁes s+1 ∂α v(0) = 1, which is solved by Z s g0(ψ(r, α)) v(s) = exp dr > 0. 0 r + 1 Hence ψ(s, α) is monotone increasing w.r.t. α. Now we claim that for any ﬁxed s > 0, lim ψ(s, α) = +∞. (3.19) α→+∞ ∞ By contradiction, we suppose there exist s1 > 0, M > 1 and {αk}k=1 such that

1 < αk → +∞, as k → ∞ and ψ(s1, αk) ≤ M for all k ∈ N. By a direct computation, there exists C > 0 independent of β such that 0 = g(1) < −g(ψ) < Cψ, ∀ ψ > 1. By (3.13), we have dψ dψ ds ≤ = − . Cψ −g(ψ) s + 1

Integrate in ψ ∈ (M, αk) for large k and using ψ(1, αk) = αk, we have

Z αk dψ Z ψ(1,αk) dψ Z 1 ds s + 1 ≤ = − = ln 1 < ∞. M Cψ ψ(s1,αk) −g(ψ) s1 s + 1 2 Sending k → +∞ and we have 1 α s + 1 ln k < ln 1 < +∞, C M 2 which becomes a contradiction. Consequently we prove that µ(α) → +∞ as α → +∞. (3.20) By contradiction, if µ(α) is uniformly bounded by some M for all α ≥ 1, then by (3.18) we have Z +∞ 2M ≥ (ψ(s, α) − 1)ds. 0

Fix s = 1 in (3.19), then there exists sufﬁciently large α0 such that ψ(1, α0) > 3M + 1. Since ψ(s, α) ≥ 1 is monotone decreasing with respect to s, we have ψ(s, α0) > 3M + 1 for all s ∈ (0, 1). This leads to a contradiction and proves (3.20). On the other hand, when α = 1, we have Z s U(s) = u0 + 1dr = s + u0, 0 which gives µ(1) = 0. By the continuity of solution w.r.t α, such a solution of (3.7) with U(0) = 0 and asymptotics (3.11) exists for all c ≥ 0.

16 By standard Perron’s method, we have the following existence result.

Theorem 3.3. For any given c ≥ u0 and 0 < A ∈ Sym(n) satisfying Gτ (λ(A)) = C0 and (1.11), 0 n there exists a unique convex viscosity solution u ∈ C (R ) of (3.1).

Proof. By Lemmas 3.1, 3.2 and the computation at the start part of this section, for any c ≥ u0 1 T we have a generalized symmetric subsolution u(x) := U( 2 x Ax) with 1 u(0) = u and u(x) = xT Ax + c + O(|x|2−δ0 ) 0 2

2 as |x| → ∞. By a direct computation, u is convex near origin and satisﬁes Gτ (λ(D u)) ≥ C0 in n R \{0}. By Theorem 2.2, 2 n Gτ (λ(D u)) ≥ C0 in R . On the other hand, 1 u(x) := xT Ax + c 2 2 n is a smooth convex solution of Gτ (λ(D u)) = C0 in R . From the proof of Lemma 3.2, we have n u ≤ u in R . In fact by (3.20), Z s 1 T u(x) − x Ax + c = U − s − u0 − µα = (ψ(r, α) − 1)dr < 0, 2 +∞ where α ≥ 1 is the unique constant such that c = u0 + µα. Let

2 n u(x) := sup{v(x): Gτ (λ(D v)) ≥ C0, v(0) = u0 and u ≤ v ≤ u, in R \{0}}.

0 n 1 T By a direct computation, u ∈ C (R ) is convex, u(0) = u0 and u(x) → 2 x Ax + c at inﬁnity. By Perron’s method (see for instance [2, 20]), u(x) deﬁned above is a viscosity solution of (3.1). The uniqueness follows similarly by maximum principle as in Lemma 2.4.

Eventually, we ﬁnish this section by proving that the solution found satisﬁes (1.7) and (1.8). 2 n 2 In fact, by equation Gτ (λ(D u)) = C0 in R \{0} and convexity D u > 0, Theorem 2.2 implies n u is a subsolution on entire R . From the proof of Theorem 3.3, we have 1 u(x) ≤ u(x) = xT Ax + c in n. 2 R Hence (1.8) follows immediately.

4 Proof of Proposition 1.5

In this section, we prove Proposition 1.5 following the line of Bao-Li-Li [2]. Let U, s(x) be as in (3.2). For τ = 0 case, the result follows from the classiﬁcation results by Jin-Xiong [21]. π For τ ∈ (0, 4 ) case, we only need to prove the following result for Gτ operator and desired result follows by eigen-decomposition (2.3).

17 Proposition 4.1. For any 0 < s1 < s2 < ∞, if there exists a generalized symmetric classical convex solution u(x) = U(s) of

1 G (λ(D2u)) = C in x : xT Ax ∈ (s , s ) , (4.1) τ 0 2 1 2 where A = diag(a1, ··· , an) > 0 satisﬁes Gτ (λ(A)) = C0. Then either 1 a = a = ··· = a or u = xT Ax + c (4.2) 1 2 n 2 for some c ∈ R.

q 2s Proof. For any s ∈ (s1, s2) and i0 = 1, ··· , n, take x = (0, ··· , 0, , 0, ··· , 0). By (3.3) ai0 and (3.4), u(x) is a classical solution of (4.1) if

0 n 00 0 n−1 (U ) σn(a) + 2sU (U ) n = c0. Y 2b Y 2b (U 0 + ) + 2sU 00 (U 0 + ) ai ai i=1 i=1,··· ,n i6=i0 Consequently, n Y 2b (U 0)n + 2sU 00(U 0)n−1 − c (U 0 + ) 0 a i=1 i Y 2b (4.3) = c 2sU 00 (U 0 + ). 0 a i=1··· ,n i i6=i0 00 1 T If U ≡ 0 in (s1, s2), then u = 2 x Ax + c immediately. 00 If U (s) 6= 0 at a point s ∈ (s1, s2), since the left hand side of (4.3) is independent of choice of i0, we have Y 2b 2sU 00(s) (U 0(s) + ) = C(s) ∀ i = 1, 2, ··· , n. a 0 i=1··· ,n i0 i6=i0 Thus in this case, a1 = a2 = ··· = an.

π For τ = 4 case, similarly by eigen-decomposition, we only need to prove the following result.

Proposition 4.2. For any 0 < s1 < s2 < ∞, if there exists a generalized symmetric classical convex solution u(x) = U(s) of

1 G (λ(D2u)) = C in x : xT Ax ∈ (s , s ) , (4.4) τ 0 2 1 2 where A = diag(a1, ··· , an) > 0 satisﬁes Gτ (λ(A)) = C0. Then (4.2) holds for some c ∈ R.

18 q 2s Proof. For any s ∈ (s1, s2) and i0 = 1, ··· , n, take x = (0, ··· , 0, , 0, ··· , 0). As a ai0 necessary condition for such solution exists, C0 < 0. Since u(x) is a classical solution of (4.4), we have 2 2 c0 · σn(λ(D u)) = σn−1(λ(D u)), √ 2 where c0 = − 2 C0 > 0. By (3.3) and Proposition 1.2 of [2],

n 0 n 00 0 n−1 X 2 c0 · (U ) σn(a) + c0 · U (U ) σn(a) aixi i=1 n 0 n−1 00 0 n−2 X 2 = (U ) σn−1(a) + U (U ) (aixi) σn−2;i(a), i=1 i.e.,

0 n 00 0 n−1 0 n−1 00 0 n−2 c0(U ) σn(a) + 2c0sU (U ) σn(a) − (U ) σn−1(a) = 2sU (U ) ai0 σn−2;i0 (a). (4.5)

00 0 σn−1(a) 1 T If U ≡ 0 in (s1, s2), then U = = 1 and u = x Ax + c is a quadratic solution. c0σn(a) 2 00 If U (s) 6≡ 0 at a point s ∈ (s1, s2), since the left hand side of (4.5) is independent of choice of i0, we have 00 0 n−2 2sU (U ) ai0 σn−2;i0 (a) = C(s) ∀ i0 = 1, 2, ··· , n. By the continuity of solution, we may assume without loss of generality that U 0(s) 6= 0, otherwise there exists s0 close to s such that U 00(s0) 6= 0 and U 0(s0) 6= 0. Together with the fact that

σn−1(a) = σn−1;i(a) + aiσn−2;i(a) ∀ i = 1, 2, ··· , n,

we have σn−1;i0 (a) is independent of the choice of i0. Thus in this case, all ai0 are the same. π π π For τ ∈ ( 4 , 2 ), the equations are very similar to τ = 2 case and only need to prove for the latter case for simplicity.

Proposition 4.3. For any 0 < s1 < s2 < ∞, if there exists a generalized symmetric classical convex solution u(x) = U(s) of

n X 1 arctan λ (D2u) = C in x : xT Ax ∈ (s , s ) , (4.6) i 0 2 1 2 i=1 Pn where A = diag(a1, ··· , an) > 0 satisﬁes i=1 arctan ai = C0. Then (4.2) holds up to some constant c.

Proof. By the algebraic form of (4.6) (see for instance [10]), λ(D2u) satisﬁes

X k 2 X k 2 cos C0 (−1) σ2k+1 λ D u − sin C0 (−1) σ2k λ D u = 0. 0≤2k+1≤n 0≤2k≤n

19 q 2s For any s ∈ (s1, s2) and i0 = 1, ··· , n, take x = (0, ··· , 0, , 0, ··· , 0). Since u(x) is a ai0 classical solution of (4.6), by (3.3) and (3.9) we have   X k 0 2k+1 00 0 2k cos C0  (−1) (U ) σ2k+1(a) + 2sU (U ) ai0 σ2k;i0 (a)  0≤2k+1≤n   X k 0 2k 00 0 2k−1 = sin C0  (−1) (U ) σ2k(a) + 2sU (U ) ai0 σ2k−1;i0 (a)  , 0≤2k≤n i.e.,

 00 T 2sp1U   00 0 ai0 n−2  2sp2U U  X  00 0 2   ai0 σ1;i0 (a)  2sp U 00(U 0)ka σ (a) =  2sp3U (U )  ·  = G(s), (4.7) k i0 k;i0    .   .   .  k=0  .  00 0 n−2 ai0 σn−2;i0 (a) 2spn−2U (U ) where k k p = (−1)k cos C + − π k 0 2 2 for k = 0, 1, ··· , n − 1 and   X k 0 2k+1 G(s) = − cos C0  (−1) (U ) σ2k+1(a) 0≤2k+1≤n   X k 0 2k + sin C0  (−1) (U ) σ2k(a) 0≤2k≤n 00 0 n−1 −2spn−1U (U ) σn(a).

Since sin C0 = 0 and cos C0 = 0 cannot hold at the same time, we may only consider the non-zero part of (4.7) and the following proof still works. Hereinafter, we only prove for sin C0 ·cos C0 6= 0 case. 00 0 1 T If U ≡ 0 in (s1, s2), then U ≡ 1 and u = 2 x Ax + c is a quadratic polynomial. 00 If U (s) 6≡ 0 at a point s ∈ (s1, s2), then by the continuity of solution, we may assume 00 0 without loss of generality that U (s) 6= 0 and U (s) 6= 0 in (s3, s4) ⊂ (s1, s2). Pick n − 1 points such that s3 < r1 < r2 < ··· < rn−1 < s4, ri 6= 0 and we prove that the matrix formed by coefﬁcients of (4.7) is invertible, i.e.,

 00 00 0 n−2  2r1p1U (r1) ··· 2r1pn−2U (r1)(U (r1)) 00 00 0 n−2  2r2p1U (r2) ··· 2r2pn−2U (r2)(U (r2))  det   6= 0. (4.8)  . .. .   . . .  00 00 0 n−2 2rnp1U (rn−1) ··· 2rn−1pn−2U (rn−1)(U (rn−1))

20 By (4.7) and (4.8), (ai0 , ··· , ai0 σn−1;i0 (a)) is a constant vector independent of i0. Thus in this case, all ai0 are the same. 00 0 To prove (4.8), since sin C0 · cos C0 6= 0, 0 < r1 < ··· < rn−1, U (s) 6= 0 and U (s) 6= 0, by the determinate of Vandermonde form matrix, we have

 0 0 n−2  1 U (r1) ··· (U (r1)) 0 0 n−2  1 U (r2) ··· (U (r2))  Y det   = (U 0(r ) − U 0(r )) 6= 0.  . . .. .  i j  . . . .  i>j 0 0 n−2 i,j=1,2,··· ,n−1 1 U (rn−1) ··· (U (rn−1))

Hence (4.8) is proved and the result follows immediately.

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Z.Liu & J. Bao School of Mathematical Sciences, Beijing Normal University Laboratory of Mathematics and Complex Systems, Ministry of Education Beijing 100875, China Email: [email protected] Email: [email protected]