linear nonlinear analysis algebra and its...

Nonlinear Analysis 127 (2015) 143–156

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Nonlinear Analysis

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Linear Algebra and its Applications 466 (2015) 102–116

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Linear Algebra and its Applications

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Inverse eigenvalue problem of Jacobi matrix with mixed data

Ying Wei 1

Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, PR China

a r t i c l e i n f o a b s t r a c t

Article history:Received 16 January 2014Accepted 20 September 2014Available online 22 October 2014Submitted by Y. Wei

MSC:15A1815A57

Keywords:Jacobi matrixEigenvalueInverse problemSubmatrix

In this paper, the inverse eigenvalue problem of reconstructing a Jacobi matrix from its eigenvalues, its leading principal submatrix and part of the eigenvalues of its submatrix is considered. The necessary and sufficient conditions for the existence and uniqueness of the solution are derived. Furthermore, a numerical algorithm and some numerical examples are given.

© 2014 Published by Elsevier Inc.

E-mail address: [email protected] Tel.: +86 13914485239.

http://dx.doi.org/10.1016/j.laa.2014.09.0310024-3795/© 2014 Published by Elsevier Inc.

Over-determined problems for k-Hessian equations in ring-shapeddomainsBo Wang, Jiguang Bao∗School of Mathematical Sciences, Beijing Normal University, Laboratory of Mathematics and ComplexSystems, Ministry of Education, Beijing 100875, China

a r t i c l e i n f o

Article history:Received 5 March 2015Accepted 29 June 2015Communicated by Enzo Mitidieri

MSC:primary 35J60secondary 35J15

Keywords:Radial symmetryOver-determined problemk-Hessian equationRing-shaped domainMoving plane methodCorner lemma

a b s t r a c t

In this paper, we firstly use a variant of the moving plane method of Alexandroffto obtain radial symmetry of solutions for k-Hessian equations in annulus-typedomains, which can be regarded as a generalization of Gidas–Ni–Nirenberg result in1979. Then we consider an over-determined problem for k-Hessian equations in ring-shaped domains and prove the radial symmetry of the solutions and the domains.

© 2015 Elsevier Ltd. All rights reserved.

1. Introduction and main results

The study of over-determined problems has undergone a long history. In 1971, Serrin [18] considered anover-determined problem for the Poisson equation which arose from the theory of elasticity (see [20]), that is

∆u = n, x ∈ Ω ,u = 0, x ∈ ∂Ω ,∂u

∂ν= 1, x ∈ ∂Ω ,

(1.1)

where Ω is a smooth bounded domain in Rn (n ∈ N and n ≥ 2), ν(x) is the unit outward normal to ∂Ωat x. He proved that if there exists a solution u ∈ C2(Ω) to (1.1), then up to a translation, Ω is a unitball and u(x) = |x|

2−12 . One technique that Serrin used is the well-known moving plane method, introduced

∗ Corresponding author.E-mail addresses: [email protected] (B. Wang), [email protected] (J. Bao).

http://dx.doi.org/10.1016/j.na.2015.06.0320362-546X/© 2015 Elsevier Ltd. All rights reserved.

http://dx.doi.org/10.1016/j.na.2015.06.032http://www.sciencedirect.comhttp://www.elsevier.com/locate/nahttp://crossmark.crossref.org/dialog/?doi=10.1016/j.na.2015.06.032&domain=pdfmailto:[email protected]:[email protected]://dx.doi.org/10.1016/j.na.2015.06.032

144 B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156

by Alexandroff in 1958 for his research of surfaces with constant mean curvature (see [12]). The other isa corner lemma which can be regarded as a modification of the classical Hopf’s lemma. Just after Serrin,Weinberger [22] proved the same conclusion by using Green’s formula.

After Serrin’s contribution to over-determined problems, lots of results have been obtained to extend thisresult. In 1995, Reichel [15] considered an over-determined problem for the Poisson equation in ring-shapeddomains, that is

∆u+ f(u) = 0, a1 < u < a2, x ∈ Ω ,

u = a1,∂u

∂γ= c1, x ∈ ∂Ω1,

u = a2,∂u

∂γ= c2, x ∈ ∂Ω2,

(1.2)

where Ω1 and Ω2 are two simply connected domains with C2 boundaries in Rn and Ω1 ⊂ Ω2, Ω := Ω2\Ω1 iscalled a ring-shaped domain, γ(x) denotes the unit inward normal with respect to Ω at x ∈ ∂Ω , a1, a2, c1, c2are constants and f : [a1, a2]→ R satisfies that f = f1 + f2 with f1 Lipschitz continuous and f2 increasing.He obtained that the existence of the solution to (1.2) in C2(Ω) implies that Ω is an annulus and u isradially symmetric and decreasing in |x|. We refer to [1,2,8,10,17,16,21,23] for more results based on themoving plane method of Alexandroff.

However, much less has been done for the over-determined problem for fully nonlinear equations. To ourknowledge, one of the results in this aspect is given by Brandolini, Nitsch, Salani and Trombetti [5] in 2008.They considered over-determined problem for k-Hessian equations (1 ≤ k ≤ n), that is,

σkλ(D2u)

=n

k

, x ∈ Ω ,

u = 0, x ∈ ∂Ω ,∂u

∂ν= 1, x ∈ ∂Ω ,

(1.3)

where σkλ(D2u)

is the kth elementary symmetric function of the eigenvalues of D2u and

nk

denotes the

binomial coefficient. They applied isoperimetric inequality to prove that: if u ∈ C2(Ω) is the solution to(1.3), then up to a translation, Ω is a unit ball and u(x) = |x|

2−12 . In particular, when k = 1, (1.3) becomes

(1.1); when k = n, (1.3) is the over-determined problem for the Monge–Ampère equations. In the same year,they studied the stability of (1.3) for k = 1 in [4] and k = n in [6], respectively. For more results based onisoperimetric inequality, see, e.g. [7,9,14].

There are also many open problems about over-determined problems which have been proposed in [19].In this paper, we firstly consider the following problem

σkλ(D2u)

= f(|x|, u, |Du|), a1 < u < a2, x ∈ A,

u = a1, |x| = R1,u = a2, |x| = R2,

(1.4)

where 0 < R1 < R2, A := {x ∈ Rn : R1 < |x| < R2}, a1, a2 ∈ R and f : [R1, R2] × [a1, a2] × (0,+∞) → Rsatisfies the following conditions:

f(r, z, q) is positive and decreasing in r,f(r, z, q) = f1(r, z, q) + f2(r, z) withf1 Lipschitz continuous in z, q, and f2 decreasing in z.

(1.5)

Let Φ2k(A) = {v ∈ C2(A) : σiλD2v (x)

> 0, x ∈ A, i = 1, 2, . . . , k−1, k}. By applying the moving plane

method, we obtain the following theorem.

B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156 145

Theorem 1.1. Let u ∈ Φ2k(A) be a solution to (1.4) and f satisfy (1.5), then u is radially symmetric andincreasing in |x|.

Remark 1.2. If we take k = n and f = (1 + q2)n+22 in (1.4), then the equation is the well-known equation ofprescribed Gauss curvature (see, e.g. [11]).

Next we study an over-determined problem for k-Hessian equations in ring-shaped domains. Let usconsider σk

λ(D2u)

= f(|x|, u, |Du|), a1 < u < a2, x ∈ Ω ,

u = ai,∂u

∂γ= ci, x ∈ ∂Ωi, i = 1, 2,

(1.6)

where Ω is the ring-shaped domain defined as after (1.2), γ(x) denotes the unit inward normal with respectto Ω at x ∈ ∂Ω and a1, a2, c1, c2 are constants.

Denoting Φ2,1k (Ω) = {v ∈ C2,1(Ω) : σiλD2v (x)

> 0, x ∈ Ω , i = 1, 2, . . . , k − 1, k}, we deduce the

following result.

Theorem 1.3. Let u ∈ Φ2,1k (Ω) be a solution to (1.6) and f satisfy (1.5), then Ω is an annulus and u isradially symmetric and increasing in |x|.

At the end, we point out that if Ω1 or Ω2 is a ball, then by considering the following problem withoutthe prescribed Neumann condition on ∂Ω1 or ∂Ω2, that is

σkλ(D2u)

= f(|x|, u, |Du|), a1 < u < a2, x ∈ Ω ,

u = ai, x ∈ ∂Ωi, i = 1, 2,∂u

∂γ= ci, x ∈ ∂Ωi, if we do not know whether Ωi is a ball or not, i = 1, 2,

(1.7)

an analogous conclusion can be obtained as follows.

Corollary 1.4. Let u ∈ Φ2,1k (Ω) be a solution to (1.7) and f satisfy (1.5), then Ω is an annulus and u isradially symmetric and increasing in |x|.

2. Proof of Theorem 1.1

2.1. Notations and preliminaries

2.1.1. Hessian operatorWe first introduce the definition of kth elementary symmetric function.For a = (a1, a2, . . . , an−1, an) ∈ Rn and k ∈ {1, 2, . . . , n − 1, n}, the kth elementary symmetric function

of a is defined as

σk(a) =

1≤i1


and

Sijk (M) =∂

∂mijSk(M), i, j = 1, 2, . . . , n− 1, n, (2.2)

then Euler identity for homogeneous functions gives us

Sk(M) =1kSijk (M)m

ij ,

here and throughout the paper, we adopt the Einstein summation convention for repeated indices.Moreover, let M denote

M = m11 −m1j−mi1 mij

and D denote

D =−1 00 δij

.

Then D−1MD = M , which means M and M are similar. So we have λ(M) = λ(M). It follows thatSk(M) = Sk(M). (2.3)

For i = 2, 3, . . . , n− 1, n, from differentiating (2.3) with respect to mi1, it follows that

Si1k (M) = Si1k (M)∂ mi1∂mi1 = −Si1k (M).So we have

Si1k (M) + Si1k (M) = 0, i = 2, 3, . . . , n− 1, n. (2.4)Let Ω be an open subset of Rn and u ∈ C2(Ω). We call Sk(D2u) the k-Hessian operator of u. It is obvious

that

S1(D2u) = ∆u and Sn(D2u) = det(D2u).

The kth Hessian operators are uniformly elliptic if restricted to the class of k-convex functions

{v ∈ C2(Ω) : SiD2v (x)

> 0, x ∈ Ω , i = 1, 2, . . . , k − 1, k}.

2.1.2. Moving plane methodBefore applying the moving plane method to prove Theorem 1.1, we would like to introduce some

notations. Let λ ∈ (0, R2),

Tλ = {x ∈ Rn : x1 = λ} the hyperplane,xλ = (2λ− x1, x′) the reflection of x about Tλ,Bλ = {xλ : x ∈ B} the reflection of a set B about Tλ,Σ (λ) = {(x1, x′) ∈ A : x1 > λ} the right hand cap,

Σ (λ) = Σ (λ)\BR1(0)λ

the reduced right hand cap.

Alessandrini [2] and Willms, Gladwell, Siegel [23] make similar use of reduced right hand caps. The twolemmas below will be used later and the proofs can be found in [15].

Lemma 2.1. For any component Z of Σ (λ), the following holds:

E := ∂Z ∩ {x : |x| = R2, x1 > λ} ≠ ∅.

In particular, there is a sequence {x(k)}+∞k=1 ⊂ E such that x(k)1 ↘ λ as k → +∞.


Lemma 2.2. Suppose D is a bounded domain in Rn with C2 boundary and u ∈ C2(D). Let x ∈ ∂D andη ∈ Rn satisfying γ(x) · η ̸= 0. If there exists a ball Bρ(x) such that u = constant, ∂u∂η ≤ 0 or

∂u∂η ≥ 0 on

∂D ∩Bρ(x) and ∂u∂η (x) = 0, then Du(x) = 0 and∂2u∂η2 (x) = ∆u(x) (γ(x) · η)

2.

Proposition 2.3. Suppose that u ∈ Φ2k(A). For x ∈ ∂A, let η ∈ Rn and γ(x) · η > 0. Then there exists a ballBρ(x) such that ∂u∂η < 0 in Bρ(x) ∩A if |x| = R2 and

∂u∂η > 0 in Bρ(x) ∩A if |x| = R1.

Proof. For x ∈ ∂A and |x| = R2, we have ∆u > 0 in A since u ∈ Φ2k(A). It follows that ∂u∂η (x) < 0 by Hopf’slemma and u < a2 in A. Then by the continuity of ∂u∂η at x, there exists a ball Bρ(x) such that

∂u∂η < 0 in

Bρ(x) ∩A. So the conclusion can be obtained due to the continuity of ∂u∂η at x.

For x ∈ ∂A and |x| = R1, by the definition of ∂u∂η (x) and u > a1 in A, we have

∂u

∂η(x) = lim

t→0+

u(x+ tη)− u(x)t

≥ 0.

If ∂u∂η (x) > 0, the continuity of∂u∂η at x yields the conclusion.

If ∂u∂η (x) = 0, then by Lemma 2.2 and u ∈ Φ2k(A), we have

∂2u

∂η2(x) = ∆u(x)(γ(x) · η)2 > 0.

By the continuity of ∂2u∂η2 at x, there exists a ball Bρ(x) such that

∂2u∂η2 > 0 in Bρ(x) ∩ A. Taking ρ1 smaller

than ρ and for any y ∈ Bρ1(x) ∩ A, we can find a point y0 where |y0| = R1 and ∂u∂η (y0) ≥ 0 such thaty − y0 = t0η(t0 > 0). Next we define g(t) := ∂u∂η (y − tη), 0 ≤ t ≤ t0. Then by Newton–Leibniz formula,

∂u

∂η(y0)−

∂u

∂η(y) = g(t0)− g(0) =

t00g′(t)dt = −

t00

∂2u

∂η2(y − tη)dt < 0.

So ∂u∂η (y) >∂u∂η (y0) ≥ 0. Therefore, we have

∂u∂η > 0 in Bρ1(x) ∩A. �

2.2. Proof of Theorem 1.1

Proof of Theorem 1.1. Since (1.4) is rotationally invariant, without loss of generality, it suffices to provethat

u(x1, x′) ≥ u(−x1, x′), ∀x = (x1, x′) ∈ A, x1 > 0. (2.5)

For λ ∈ (0, R2), we define the comparison function as

w(x, λ) := v(x, λ)− u(x) := u(xλ)− u(x), x ∈ Σ (λ).

In order to prove (2.5), we only need to get the following two properties of w(x, λ) for any λ ∈ (0, R2):

(i) w(x, λ) ≤ 0 in Σ (λ);

(ii) ∂w∂x1

(x, λ) < 0 on A ∩ Tλ.

Indeed, by taking λ→ 0 in (i), we can get (2.5). Since ∂w∂x1 (x, λ) = −2∂u∂x1

(x) < 0, the monotonicity can beobtained.

Our proof of (i) and (ii) will be divided into three steps.


Step 1. (Initial step) In this step, we will prove that there exists ϵ > 0 such that for any λ ∈ (R2− ϵ, R2),(i) and (ii) hold.

Taking x = (R2, 0, . . . , 0) and η = (−1, 0, . . . , 0), by Proposition 2.3, there exists a ball Bρ(x) such that∂u∂x1> 0 in Bρ(x) ∩ A. Hence there exists ϵ > 0 such that for any λ ∈ (R2 − ϵ, R2), Σ (λ) ∪ Σ (λ)λ ⊂ Bρ(x)

and then ∂u∂x1 > 0 in Σ (λ). If λ ∈ (R2−ϵ2 , R2) and x ∈ Σ (λ), [x, x

λ] ⊂ Σ (R2− ϵ). Notice that 2λ− 2x1 < 0.Let g(t) = u(x1 + t, x′), 2λ− 2x1 ≤ t ≤ 0, then

u(xλ)− u(x) = g(2λ− 2x1)− g(0) = 2λ−2x1

0g′(t)dt =

2λ−2x10

∂u

∂x1(x1 + t, x′)dt < 0,

i.e.

w(x, λ) < 0 in Σ (λ);∂w

∂x1(x, λ) = −2 ∂u

∂x1(x) < 0 on A ∩ Tλ,

for λ ∈ (R2 − ϵ2 , R2).

Step 2. By Step 1, there exists λ > 0 such that (i) and (ii) hold. In this step, we want to strengthen theinequality in (i) and (ii) for such λ.

By (2.3), it is easy to see that v(x, λ) := u(xλ) satisfies σkλ(D2v)

= f(|xλ|, v, |Dv|) in Σ (λ). Then by

using (1.5), w ≤ 0 in Σ (λ) in Step 1 and |xλ| ≤ |x| for λ ≥ 0, we have

aij(x)Dijw = f(|xλ|, v, |Dv|)− f(|x|, u, |Du|)≥ f(|x|, v, |Dv|)− f(|x|, u, |Du|)≥ f1(|x|, v, |Dv|)− f1(|x|, u, |Du|)= bi(x)Diw + c(x)w,

where

aij(x) = 1

0SijksD2u(x) + (1− s)D2v(x)

ds,

bi(x) = f1(x, u, u1, . . . , ui−1, vi, . . . , vn)− f1(x, u, u1, . . . , ui, vi+1, . . . , vn)vi − ui

,

c(x) = f1(|x|, v, |Dv|)− f1(|x|, u, |Dv|)v − u

.

Then

aij(x)Dijw − bi(x)Diw − c+(x)w ≥ aij(x)Dijw − bi(x)Diw − c(x)w ≥ 0, x ∈ Σ (λ).

By strong maximum principle, we have either w < 0 or w ≡ 0 on a component of Σ (λ). Suppose that thelatter case holds. Then by Lemma 2.1, there is a sequence {x(k)}+∞k=1 ⊂ ∂Z with |x(k)| = R2 and x

(k)1 ↘ λ

as k → +∞. We can deduce that u(x(k)) = a2 and u(x(k),λ) = a2. So |x(k),λ| = R2 and λ = 0, which is acontradiction with λ > 0.

Step 3. (Continuation step) Due to Step 1, we can define

µ = inf{α > 0 : (i) holds for λ ∈ (α,R2)}.

In this step, we will show that µ = 0.


Suppose µ > 0, then there exist sequences {λk}+∞k=1 and {x(k)}+∞k=1 such that λk ↗ µ and w(x(k), λk) > 0.

Next we choose x(k) such that w(x(k), λk) attains its positive maximum over Σ (λk) at x(k). For x ∈ ∂Σ (λk),

w(x, λk) =

0, x ∈ ∂Σ (λk) ∩ Tλk ,u(xλk)− a2, x ∈ ∂Σ (λk) ∩ {x : |x| = R2},a1 − u(x), x ∈ ∂Σ (λk) ∩ {x : |xλk | = R1},

then w(x, λk) ≤ 0 on ∂Σ (λk). So x(k) ∈ Σ (λk) and Dw(x(k), λk) = 0.

Since {x(k)}+∞k=1 is bounded, there exists a subsequence converging to x ∈ Σ (µ) with Dw(x, µ) = 0 andw(x, µ) ≥ 0. For (i) holds at λ = µ, we have w(x, µ) = 0. By Step 2, x ∈ ∂Σ (µ). Also owing to the strongmaximum principle, x cannot lie on the smooth part of ∂Σ (µ). Hence x ∈ (∂A∩Tµ)∪ (∂BR1 (0)

µ ∩∂A\Tµ).Since w(x, µ) = a− b < 0 on ∂BR1 (0)

µ ∩ ∂A\Tµ, it follows that x ∈ ∂A ∩ Tµ.

By Proposition 2.3 with x = x, there exists a ball Bρ(x) such that ∂u∂x1 > 0 in Bρ(x) ∩ A. For k bigenough, we have x(k), x(k),λk ∈ Bρ(x). Then

u(x(k))− u(x(k),λk) = x(k)1

2λk−x(k)1

∂u

∂x1(t, x(k)

′)dt > 0,

i.e. w(x(k), λk) < 0 which contradicts with w(x(k), λk) > 0. �

3. Proof of Theorem 1.3

3.1. Notations and preliminaries

Before applying moving plane method to prove Theorem 1.3, we also need to introduce some notations.Let

Tλ = {x ∈ Rn : x1 = λ} the hyperplane,xλ = (2λ− x1, x′) the reflection of x about Tλ,Σ1(λ) = {(x1, x′) ∈ Ω1 : x1 > λ} the inner cap,Σ2(λ) = {(x1, x′) ∈ Ω2 : x1 > λ} the outer cap,Γi(λ) = {(x1, x′) ∈ ∂Ωi : x1 > λ} the right hand boundary, i = 1, 2,Mi = sup{x1 : (x1, x′) ∈ Ωi} the x1-extent of Ωi, i = 1, 2.

For i = 1, 2, it is known from [3] that if λ is a little less than Mi, Σi(λ)λ ⊂ Ωi and the x1-direction isexternal at each point of Γi(λ). It is also known from [13] that Ωi contains Σi(λ)λ until one of the followingtwo events (critical positions) occurs:

(1) ∂Σi(λ)λ becomes internally tangent to ∂Ωi at P ̸∈ Tλ;(2) Tλ reaches a position where it is orthogonal to ∂Ωi at some point Q ∈ Tλ ∩ Γi(λ).

We denote this critical value of λ by mi and let m = max{m1,m2}.From the above argument, we can see that the admissible range for λ is (m,M2). For λ ∈ (m,M2), we

can define the reduced right hand caps as

Σ (λ) = Σ2(λ)\Ω1λ.

During the proof below, we will discuss two critical cases respectively, and a key ingredient in the proofis a corner lemma. For readers’ convenience, we will state it below and the proof can be found in [18].

Let D∗ be a bounded domain with C2 boundary in Rn and let T be a hyperplane containing the unitouter normal to ∂D∗ at some point Q. D denotes the portion of D∗ lying on some particular side of T . We


assume bi and c are bounded functions in D, i = 1, 2, . . . , n− 1, n, and there exist three positive constantsK1, K2 and K such that

K1|ξ|2 ≤ âij(x)ξiξj ≤ K2|ξ|2, (3.1)

and

|âij (x) ξiηj | ≤ K (|ξ · η|+ |ξ|d (x)) , (3.2)

where x ∈ D, ξ = (ξ1, ξ2, . . . , ξn−1, ξn) is an arbitrary real vector, η = (η1, η2, . . . , ηn−1, ηn) is the unitnormal to the hyperplane T and d(x) is the distance from x to T .

Lemma 3.1. Suppose w ∈ C2(D) satisfies

âij(x) ∂2w

∂xi∂xj+ bi(x) ∂w

∂xi− c+(x)w ≥ 0, x ∈ D,

and w ≤ 0 in D and w(Q) = 0. Let s be any direction at Q which enters D non-tangentially. Then at Q,we have either

∂w

∂s< 0, or ∂

2w

∂s2< 0,

unless w ≡ 0 in D.

Remark 3.2. From the above lemma, we can see that w ̸≡ 0 implies that Dw(Q) ̸= 0 or D2w(Q) ̸= 0.

At the end of this subsection, we will give a proposition which is useful in the proof of Theorem 1.3.

Proposition 3.3. Suppose that u ∈ Φ2k(A). For x ∈ ∂Ω , let η ∈ Rn and γ(x) · η > 0. Then there exists a ballBρ(x) such that ∂u∂η > 0 in Bρ(x) ∩ Ω if x ∈ Ω1 and

∂u∂η < 0 in Bρ(x) ∩ Ω if x ∈ Ω2.

Remark 3.4. The proof of this proposition is similar with Proposition 2.3.

3.2. Proof of Theorem 1.3

Proof of Theorem 1.3. By the rotation invariance of (1.6), we only need to prove the symmetry of u and Ωin x1-direction.

For λ ∈ (m,M2), we define the comparison function as

w(x, λ) := v(x, λ)− u(x) := u(xλ)− u(x), x ∈ Σ (λ).

We will divide our proof into seven steps.

Step 1. (Initial step) In this step, we will prove that there exists ϵ > 0 such that for any λ ∈ (M2−ϵ,M2):

(i) w(x, λ) ≤ 0 in Σ (λ);

(ii) ∂w∂x1

(x, λ) < 0 on Ω ∩ Tλ.

Let

K = {x = (x1, x′) ∈ ∂Ω2 : x1 =M2}.

For any x ∈ K, by Proposition 3.3, there exists a ball Bρ(x) such that ∂u∂x1 > 0 in Bρ(x) ∩ Ω . By thecompactness of K, we can find a radius σ independent on x such that ∂u∂x1 > 0 in Bσ(x) ∩ Ω for all x ∈ K.LetKσ :=

x∈K Bσ(x)∩Ω . Hence there exists ϵ > 0 such that for any λ ∈ (M2−ϵ,M2), Σ (λ) ∪ Σ (λ)λ ⊂ Kσ,

and then ∂u∂x1 > 0 in Σ (λ). If λ ∈ (M2−ϵ2 ,M2) and x ∈ Σ (λ), [x, x

λ] ⊂ Σ (M2− ϵ). Let g(t) = u(x1 + t, x′),


2λ− 2x1 ≤ t ≤ 0, then

u(xλ)− u(x) = g(2λ− 2x1)− g(0) = 2λ−2x1

0g′(t)dt =

2λ−2x10

∂u

∂x1(x1 + t, x′)dt < 0,

i.e.

w(x, λ) < 0 in Σ (λ),∂w

∂x1(x, λ) = −2 ∂u

∂x1(x) < 0 on Ω ∩ T (λ),

for λ ∈ (R2 − ϵ2 , R2).

Step 2. In this step, we will show that if w ≡ 0 in Z, which is a component of Σ (λ), then Ω = Z ∪ Zλ.

Now we can divide ∂Z into three disjoint parts

Z1 = {x = (x1, x′) ∈ ∂Z : x1 = λ, x ∈ Ω},Z2 = {x = (x1, x′) ∈ ∂Z : x1 ≥ λ, x ∈ ∂Ω2},Z3 = {x = (x1, x′) ∈ ∂Z : x1 ≥ λ, x ∈ ∂Ωλ1 }.

By the assumption that w ≡ 0 in Z, we have u = a2 on Zλ2 and u = a1 on Z3. This implies that Zλ2 ⊂ ∂Ω2and Z3 ⊂ ∂Ω1 since a1 < u < a2 in Ω . So ∂Z\Tλ ⊂ ∂Ω and ∂Zλ\Tλ ⊂ ∂Ω .

We define

X = Z ∪ Zλ ∪ (∂Z ∩ Ω) ∪ (∂Zλ ∩ Ω).

It is easy to see that

∂X ⊂ (∂Z ∪ ∂Zλ)\(∂Z ∩ Ω)\(∂Zλ ∩ Ω) ⊂ ∂Ω .

If X is also open, then since Ω is connected and X ⊂ Ω , we have Ω = X.

Thus it remains to prove that X is open. It is obvious that Z ∪ Zλ is a subset of the interior of X.

(a) For x ∈ ∂Z ∩ Ω , we have x ∈ Tλ ∩ Ω since ∂Z\Tλ ⊂ ∂Ω . By the openness of Ω , there exists a ballBρ(x) ⊂ Ω and we define

B> = Bρ(x) ∩ {x1 > λ},B< = Bρ(x) ∩ {x1 < λ},B= = Bρ(x) ∩ {x1 = λ}.

Clearly Z ∩ B> ̸= ∅. If we also have Zc ∩ B> ̸= ∅, then ∂Z ∩ B> ̸= ∅ which is impossible since∂Z ∩ B> ⊂ (∂Ω ∪ Tλ) ∩ B> = ∅. Hence B> ⊂ Z, B< ⊂ Zλ and B= ⊂ Ω ∩ (Z ∩ Tλ) ⊂ Ω ∩ ∂Z. Theseresult in Bρ(x) ⊂ X.

(b) For x ∈ ∂Zλ∩Ω , we have x ∈ Tλ since ∂Zλ\Tλ ⊂ ∂Ω . Then there exists a sequence {x(k)}+∞k=1 ⊂ Zλ∩Ωconverging to x as k → +∞, and {x(k),λ}+∞k=1 ⊂ Z also converges to x = xλ. So x ∈ Z∩Tλ∩Ω ⊂ ∂Z∩Ω .Then by following the argument in (a) we can get x lies in the interior of X.

Step 3. From Step 1, there exists λ > m such that (i) and (ii) hold. In this step, we want to strengthenthe inequality in (i) and (ii) for such λ.

Since u is the solution to (1.6), v satisfies

σkλ(D2v)

= f(|xλ|, v, |Dv|), x ∈ Σ (λ).


It follows that

âij(x)Dijw = f(|xλ|, v, |Dv|)− f(|x|, u, |Du|)≥ f(|x|, v, |Dv|)− f(|x|, u, |Du|)≥ f1(|x|, v, |Dv|)− f1(|x|, u, |Du|)= bi(x)Diw + c(x)w,

where

âij(x) = 1

0SijksD2u(x) + (1− s)D2v(x)

ds,

bi(x) = f1(x, u, u1, . . . , ui−1, vi, . . . , vn)− f1(x, u, u1, . . . , ui, vi+1, . . . , vn)vi − ui

,

c(x) = f1(|x|, v, |Dv|)− f1(|x|, u, |Dv|)v − u

.

Then we have

âij(x)Dijw − bi(x)Diw − c+(x)w ≥ aij(x)Dijw − bi(x)Diw − c(x)w ≥ 0, x ∈ Σ (λ).

It follows from u, v ∈ C2,1Σ (Tλ)

are k-convex that

âij(x)

is positive definite. By strong maximum

principle, we have either w < 0 or w ≡ 0 on a component of Σ (λ). Suppose that the latter case holds. Thenby Step 2, Ω is symmetric about x1 = λ which is impossible for λ > m.

Step 4. (Continuation step) Due to Step 1, we can define

µ = inf{α > m : (i) holds for λ ∈ (α,M2)}.

In this step, we will show that µ = m.

Suppose µ > m, then there exist sequences {λk}+∞k=1 and {x(k)}+∞k=1 such that λk ↗ µ, x(k) ∈ Σ (λk) and

w(x(k), λk) > 0. Next we choose x(k) such that w(x(k), λk) attains its positive maximum over Σ (λk) at x(k).For x ∈ ∂Σ (λk),

w(x, λk) =

0, x ∈ ∂Σ (λk) ∩ Tλk ,u(xλk)− a2, x ∈ ∂Σ (λk) ∩ ∂Ω2,a1 − u(x), x ∈ ∂Σ (λk) ∩ ∂Ω1,

then w(x, λk) ≤ 0 on ∂Σ (λk). So x(k) ∈ Σ (λk) and Dw(x(k), λk) = 0.

Since {x(k)}+∞k=1 is bounded, then there exists a subsequence converging to x ∈ Σ (µ) with Dw(x, µ) = 0and w(x, µ) ≥ 0. For (i) holds at λ = µ, we have w(x, µ) = 0. By Step 3, x ∈ ∂Σ (µ). Also owing to the strongmaximum principle, x cannot lie on the smooth part of ∂Σ (µ). Hence x ∈ (∂Ω ∩ Tµ) ∪ (∂Ωµ1 ∩ (∂Ω\Tµ)).Since on the latter set w(x, µ) = a1 − a2 < 0, it follows that x ∈ ∂Ω ∩ Tµ.

By Proposition 3.3 with x = x, there exists a ball Bρ(x) such that ∂u∂x1 > 0 in Bρ(x) ∩ Ω . For k bigenough, we have x(k), x(k),λk ∈ Bρ(x). Then

u(x(k))− u(x(k),λk) = x(k)1

2λk−x(k)1

∂u

∂x1(t, x(k)

′)dt > 0,

i.e. w(x(k), λk) < 0 in contradiction with w(x(k), λk) > 0. This shows µ = m and finishes the continuationstep.

Up to now, we have obtained that w(x,m) ≤ 0 in Σ (m). Then by the strong maximum principle, for anycomponent Z of Σ (m), we have either w < 0 or w ≡ 0 in Z. Step 2 shows that the latter case implies the


symmetry of Ω . Therefore, we only need to show that for the two critical cases mentioned in (1) and (2) ofSection 3.1, w < 0 in Z is impossible.

Step 5. Let us consider the first critical case, that is, ∂Σ (m)m becomes internally tangent to ∂Ωi atP ̸∈ Tm. Without loss of generality we may assume that i = 2. Since P ∈ ∂Ωi ∩ ∂Σ (m)m, we haveu(Pm) = v(Pm,m). Thus w(Pm,m) = v(Pm,m)− u(Pm) = 0 and

∂w

∂γ(Pm,m) = ∂v

∂γ(Pm,m)− ∂u

∂γ(Pm) = 0.

However, by Hopf’s Lemma, we have

∂w

∂γ(Pm,m) < 0,

which is a contradiction. Hence w < 0 in Z is impossible for the first critical case.

Step 6. Let us consider the second critical case, that is, Tm reaches a position where it is orthogonal to∂Ωi at some point Q ∈ Tm.

We shall use Lemma 3.1 to make a contradiction. Since u, v ∈ C2,1Σ (m)

are k-convex, it is obvious

that (3.1) in Lemma 3.1 holds in our case. Now let us verify the condition (3.2) in Lemma 3.1.

If x ∈ ∂Σ (m) ∩ Tm, by the definition of v, we can get

sD2v(x) + (1− s)D2u(x) = sD2u(x) + (1− s)D2v(x), s ∈ [0, 1]. (3.3)

Then by (3.3) and (2.4), for i = 2, 3, . . . , n− 1, n and x ∈ ∂Σ (m) ∩ Tm, we have

2âi1(x) = 2 1

0Si1ksD2u(x) + (1− s)D2v(x)

ds

= 1


ds+

10Si1ksD2v(x) + (1− s)D2u(x)

ds,

= 1


ds+

10Si1k

sD2u(x) + (1− s)D2v(x)

ds,

= 0.

By u, v ∈ C2,1Σ (m)

and the definition of âi1, we can get âi1(x) is Lipschitz continuous on Σ (m). So

there exists L > 0, for x = (x1, x′) ∈ Σ (m) and x0 = (m,x′) ∈ ∂Σ (m) ∩ Tm, such that

|âi1(x)| = |âi1(x)− âi1(x0)| ≤ L|x− x0| = Ld(x),

where d(x) is the distance from x to Tm. Now the unit outer normal of Tm is η = (1, 0, . . . , 0, 0), so for anarbitrary vector ξ = (ξ1, ξ2, . . . , ξn−1, ξn), we have

ni,j=1âij(x)ξiηj

=ni=1âi1(x)ξi

,≤n−1i=1|âi1(x)| |ξi|+ |â11(x)| |ξ1|

≤ (n− 1)Ld(x)|ξ|+K2|ξ · η|≤ K (|ξ|d (x) + |ξ · η|) .

This completes the proof of condition (3.2).


Since w < 0 in Z and w(Q) = 0, from Lemma 3.1, at Q we obtain that

∂w

∂s< 0 or ∂

2w

∂s2< 0,

which contradicts with the fact Dw(Q) = 0 and D2w(Q) = 0 that will be obtained in Step 7. Hence w < 0in Z is also impossible for the second critical case.

Step 7. We shall show that Dw(Q) = 0 and D2w(Q) = 0. By the definition of w, it is obvious that at Q

∂w

∂xl= 0,

∂w

∂x1= −2 ∂u

∂x1,

∂2w

∂xk∂xl= ∂

2w

∂x21= 0,

∂2w

∂x1∂xl= −2 ∂

2u

∂x1∂xl,

for k, l = 2, 3, . . . , n− 1, n. So we only need to prove that at Q

∂w

∂x1= 0,

∂2w

∂x1∂xl= 0, l = 2, 3, . . . , n− 1, n.

Since ∂Ω ∈ C2,1, we consider a rectangular coordinate frame with origin at Q, the xn axis being directedalong the inward normal to ∂Ω at Q and the x1 axis being normal to Tm. In this frame we can represent∂Ω locally by the equation

xn = φ(x1, x2, . . . , xn−2, xn−1), φ ∈ C2,1, φ(0′) = 0, D′φ(0′) = 0,

where D′ denotes ( ∂∂x1 ,∂∂x2, . . . , ∂∂xn−2 ,

∂∂xn−1

). Since u ∈ C2,1(Ω), the Dirichlet boundary condition u = bon ∂Ω0 can be expressed as a twice differentiable identity

u(x1, x2, . . . , xn−1, φ) ≡ b. (3.4)

Differentiating (3.4) with respect to xk, k = 1, 2, . . . , n− 2, n− 1, we obtain that

∂u

∂xk+ ∂u∂xn

∂φ

∂xk= 0. (3.5)

By D′φ(0′) = 0, we find that for k = 1, 2, . . . , n− 2, n− 1,

∂u

∂xk(0) = 0,

in particular,

∂w

∂x1(0) = −2 ∂u

∂x1(0) = 0.

Similarly, the Neumann boundary condition ∂u∂γ = c0 on ∂Ω0 can also be written locally as an identity

n−1k=1

∂u

∂xk

∂φ

∂xk− ∂u∂xn≡ c0

1 +

n−1k=1

∂φ

∂xk

2 12. (3.6)


Next differentiating (3.6) with respect to x1 and evaluating at 0, we can getn−1k=1

∂u

∂xk(0) ∂

2φ

∂xk∂x1(0′)− ∂

2u

∂x1∂xn(0) = 0,

i.e.∂2w

∂x1∂xn(0) = −2 ∂

2u

∂x1∂xn(0) = 0.

Now it remains to prove that∂2w

∂x1∂xl(0) = 0, l = 2, 3, . . . , n− 2, n− 1.

We do this by using the following Taylor expansion of w at 0. For ξ ∈ Σ (Tm), and ξ → 0, we have

w(ξ) = w(0) +nl=1

∂w

∂xl(0)ξl +

12

nk,l=1

∂2w

∂xk∂xl(0)ξlξk + o(|ξ|2),

=n−1l=2

∂2w

∂x1∂xl(0)ξlξ1 + o(|ξ|2).

Fix l ∈ {2, 3, . . . , n − 2, n − 1}, define ξ(δ) = δ(1, 0, . . . , 0,±1, 0, . . . , 0,−1) for δ > 0, where ±1 is the lthcomponent of ξ and when ∂

2w∂x1∂xl

(0) ≤ 0 minus sign is taken, when ∂2w

∂x1∂xl(0) > 0 plus sign is taken. Now we

choose δ sufficiently small, such that ξ(δ) ∈ Σ (Tm). It follows from the Taylor expansion that

w(ξ(δ)) = δ2 ∂2w∂x1∂xl (0)

+ o(δ2), δ → 0 + .Since w < 0 in Σ (Tm), we see that it forces ∂

2w∂x1∂xl

(0) = 0. So far we have proved that Dw(0) = 0 andD2w(0) = 0. �

4. Proof of Corollary 1.4

Proof of Corollary 1.4. Without loss of generality, we may assume that Ω1 is a ball. Then we can follow theargument from Step 1 to Step 4 in Section 3.2.

If m = m2, then we can finish our proof by following from Step 5 to Step 7 in Section 3.2.

Ifm = m1, that is Tm becomes the critical position of a ball. Then we can choose the −x1 direction insteadof the x1 direction as the approach direction of the hyperplane so that the hyperplane cannot become thecritical position of the ball. Thus we can suppose that the hyperplane sits in the critical position of Ω2 andwe can finish the proof by following from Step 5 to Step 7 in Section 3.2. �

Acknowledgments

The first author is supported in part by the scholarship from China Scholarship Council under the GrantCSC No. 201406040131. The research of the second author is partially supported by Beijing Municipal Com-mission of Education for the Supervisor of Excellent Doctoral Dissertation (20131002701). All authors werepartially supported by NNSF (11371060) and the Fundamental Research Funds for the Central Universities.

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Over-determined problems for k -Hessian equations in ring-shaped domainsIntroduction and main resultsProof of Theorem 1.1Notations and preliminariesHessian operatorMoving plane method

Proof of Theorem 1.1

Proof of Theorem 1.3Notations and preliminariesProof of Theorem 1.3

Proof of Corollary 1.4AcknowledgmentsReferences

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