-
Nonlinear Analysis 127 (2015) 143–156
Contents lists available at ScienceDirect
Nonlinear Analysis
www.elsevier.com/locate/na
Linear Algebra and its Applications 466 (2015) 102–116
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Linear Algebra and its Applications
www.elsevier.com/locate/laa
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
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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.
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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
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146 B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156
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 → +∞.
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B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156 147
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.
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148 B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156
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.
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B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156 149
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
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150 B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156
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′),
-
B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156 151
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).
-
152 B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156
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
-
B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156 153
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
0Si1ksD2u(x) + (1− s)D2v(x)
ds+
10Si1ksD2v(x) + (1− s)D2u(x)
ds,
= 1
0Si1ksD2u(x) + (1− s)D2v(x)
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).
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154 B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156
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)
-
B. Wang, J. Bao / Nonlinear Analysis 127 (2015) 143–156 155
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