Linearity of homogeneous order one solutions to elliptic equations in dimension three

7 Pages·0.14 MB·English

by  Qing Han

LINEARITY OF HOMOGENEOUS ORDER ONE
SOLUTIONS TO ELLIPTIC EQUATIONS IN DIMENSION
2
THREE
0
0
2
QING HAN, NIKOLAI NADIRASHVILI, AND YU YUAN

n
a
J

4
1. Introduction
2

In this note we prove that any homogeneous order one solution to non-
]
P divergence elliptic equations in R3 must be linear. Consider the general
A equations in Rn
. n
h
t (1.1) aij(x)Diju= 0,
a
m iX,j=1
[ where the coefficients satisfy

1 λI (aij(x)) λ−1I
≤ ≤
v
1 for some positive constant λ > 0, and the dimension n 3. Safonov con-
≥
4 structed homogeneous order α solutions, with α (0,1), to (1.1) in [10],
2 where he showed the unimprovability of the estim∈ates of H¨older exponent
1
for solutions to (1.1) by Krylov and himself. Thehomogeneity α with α < 1
0
2 plays an essential role in the construction. Later on, Safonov asked in [11,
0 p.49] whether one can construct nontrivial homogeneous order one solutions
h/ to (1.1). Our result indicates that it is impossible to do so in R3.
t
a Theorem 1.1. Any homogeneous order one strong solution u to (1.1) in
m
R3 with u W2,2 R3 must be a linear function.
: ∈ loc
v (cid:0) (cid:1)
i On the other hand, one does have nontrivial homogeneous order one solu-
X tions to (1.1) in R4 0 . In fact let (x,f (x)) Rn Rk be a nonparametric
r \{ } ∈ ×
minimal surface. Then each component of f satisfies (1.1), with (a (x)) be-
a ij
ing the inverse of the induced metric of the minimal surface in Rn+k (cf. [7,
p.3]). ThroughHopffibration,LawsonandOssermanconstructedaminimal
cone (x,f (x)) R4 R3, where
∈ ×
√5 1
f(x)= x2+x2 x2 x2,2x x +2x x ,2x x 2x x
2 x 1 2− 3− 4 1 3 2 4 2 3− 1 4
| |(cid:0) (cid:1)
(see [7, Theorem 7.1]). Now each component of f is a desired nontriv-
ial solution to (1.1) in R4 0 . Actually by noticing that the graph of
\{ }
Date: October29, 2001. Submitted on November12, 2001.
All authors are partially supported by NSF grants, and the first author also by Sloan
Research Fellowship.
1
2 HAN, NADIRASHVILI, AND YUAN
u = x2+x2 x2 x2 / x is a saddle surface, one can easily construct
1 2− 3− 4 | |
coeffi(cid:0)cients a (x) in R4(cid:1) 0 so that u satisfies (1.1) in R4 0 .
ij
\{ } \{ }
Theorem 1.1 gives a simple “PDE” proof of a well-known result obtained
by many authors in 1970’s, which states that any nonparametric minimal
coneofdimensionthreemustbeflat. Let(x,f (x)) R3 Rk betheminimal
cone with f (tx) = tf(x), f C∞ R3 0 , th∈en ea×ch component of f
∈ \{ }
satisfies (1.1). By Theorem 1.1, f m(cid:0)ust be l(cid:1)inear, or the minimal cone is
flat.
From Theorem 1.1 one also sees that any smooth homogeneous order two
solution in R3 0 to thefully nonlinear elliptic equation F D2u = 0must
\{ }
be a quadratic polynomial. To get this conclusion, one sim(cid:0)ply ap(cid:1)ply Theo-
rem 1.1 to the gradient u and (1.1) with a (x) = ∂F D2u . In contrast,
∇ ij ∂mij
thesecondauthorconstructedanon-quadratichomogene(cid:0)ouso(cid:1)rdertwosolu-
tiontosomeequation F D2u = 0inR12,whichprovidesacounterexample
to the regularity for fully(cid:0) non(cid:1)linear elliptic equations (see [8]).
The heuristic idea of the proof of Theorem 1.1 is the following sim-
ple geometric observation. The closed saddle surface u S2 must be a
∇
point. Therefore, u is linear. More precisely, one conside(cid:0)r th(cid:1)e surface Σ
parametrized by the gradient u: S2 R3. Because of (1.1), Σ is a saddle
∇ →
surface at u(x) with D2u(x) = 0. However the supporting plane with
∇ 6
normal x touches Σ at u(x). Thus D2u(x) 0 and it follows that u is
∇ ≡
linear, see Section 2.
The ideas of gradient map and supporting plane are already in an early
paper by Alexandrov [1]. In fact, under the assumption that the homo-
geneous order one function u is analytic in R3 0 and the Hessian D2u
\{ }
is either non-definite or 0 at each point, Alexandrov showed that u must
be a linear function. Roughly speaking, if u is analytic in R3 0 , the set
(u)= x S2 D2u(x) = 0 is either isolated or the whole S2.\{Al}exandrov
S ∈ |
proved t(cid:8)hat the supporting p(cid:9)lane to Σ is unique at u(x) with x being an
∇
isolated point of (u). Since the surface Σ has supporting planes with nor-
S
mal along all the directions in R3, Alexandrov excluded case of (u) being
S
isolated and proved the result. It is interesting to note that the concept
of gradient maps and supporting planes was further employed in the later
development of the Alexandrov-Bakel’man-Pucci maximum principle.
When the coefficients a are Cα and the solution u is C2,α away from
ij
the origin, we can show that the set (u) is either isolated or the whole
S2. Coupled with Alexandrov’s argumeSnt, one sees that Theorem 1.1 holds
in the C2,α setting. After the work in C2,α case was done, we found that
Pogorelov [9] generalized the above Alexandrov’s result to C2 functions.
The argument in [9] is more involved. Since our approach in C2,α case is
interesting in its own right and short, we also include it here in Section 3.
The authors wish to thank L. Nirenberg for informing and mailing them
the inspiring reference [1] .
LINEARITY OF HOMOGENEOUS ORDER ONE SOLUTIONS 3
2. Proof of the Main Theorem
Let h(x ,x ) = u(x ,x ,1). Then the homogeneous order one function
1 2 1 2
u(x ,x ,x ) = x h x1,x2 , and the gradient u and the Hessian D2u
1 2 3 3 x3 x3 ∇
(cid:16) (cid:17)
have the following representation
(2.1) u(x ,x ,1) = (h ,h ,h x h x h ),
1 2 1 2 1 1 2 2
∇ − −
(2.2)
1 0 0 h h 0 1 0 x
11 12 1
D2u(x1,x2,1) =  0 1 0  h21 h22 0  0 1 −x2 .
−
x x 1 0 0 0 0 0 1
1 2
 − −   
From our assumption that the homogeneous order one solution u is in
W2,2 R3 , it follows that h W2,2 R2 is a strong solution to
loc ∈ loc
(cid:0) (cid:1) (cid:0) (cid:1)
2
(2.3) A (x ,x )D h = 0,
ij 1 2 ij
iX,j=1
where the coefficients A are in terms of a (x ,x ,1), x , x and satisfy
ij ij 1 2 1 2
the ellipticity condition with some λ(x ,x ).
1 2
Lemma 2.1. For any ν S2, the supporting plane with normal ν must
touch the surface Σ param∈etrized by u: S2 R3 at u(ν) or u( ν).
∇ → ∇ ∇ −
Proof. First we notice that u Cα S2 . In fact each component of
∇ ∈
(h ,h ), say h is a W1,2 R2 weak solu(cid:0)tion(cid:1)to a divergence equation
1 2 1 loc
(cid:0) (cid:1)
2
D (B D h) = 0,
i ij j
iX,j=1
where B = A /A , B = 2A /A , B = 1. It follows that h
11 11 22 12 12 22 22
C1,α S2 (cf. [5,p,284-285orTheorem12.4]). By(2.1)wesee u Cα S2∈
∇ ∈
with(cid:0)αde(cid:1)pendingonlyontheoriginalellipticityλ.Therefore,foranyν (cid:0)S2(cid:1),
∈
thesupportingplaneP with normalν must touch thesurfaceΣ somewhere
ν
(Σ is on the opposite side of ν).
Without loss of generality,we assume ν = (1,0,0).
Claim: P touches Σ at u(x ,x ,0) with (x ,x ,0) S2.
(1,0,0) 1 2 1 2
∇ ∈
SupposeP touchesΣat u(x ,x ,x )withx = 0,sayx > 0.Then
(1,0,0) 1 2 3 3 3
∇ 6
u (x ,x ,1) = u x1,x2,1 = h x1, x2 would achieve its maximum at
1 1 2 1 x3 x3 1 x3 x3
(cid:16) (cid:17) (cid:16) (cid:17)
x1, x2 . Because of (2.3), h satisfies the strong maximum principle (cf.
x3 x3 1
(cid:16) (cid:17)
[5,Theorem 8.19]). We then have a contradiction unless h const.. In the
1
≡
latter case, u (x ,x ,x ) const. for x > 0, our claim holds.
1 1 2 3 3
≡
Similarly, by applying the above argument to u(x ,1,x ), we see that
1 3
P also touches Σ at u(x ,0,x ) with (x ,0,x ) S2. Therefore,
(1,0,0) 1 3 1 3
∇ ∈
P must touch Σ at u(1,0,0) or u( 1,0,0).
(1,0,0)
∇ ∇ −
4 HAN, NADIRASHVILI, AND YUAN
We now present the proof of Theorem 1.1.
Proof. By our assumption on u, u W2,2 S2 . If D2u = 0 almost every-
∈
where on S2, then u is already linear. Othe(cid:0)rwi(cid:1)se, we pick a Lebesgue point
x∗ S2 for D2u with D2u(x∗) = 0, say x∗ = (1,0,0). We may also assume
∗ ∈ 6
x is a Lebesgue point for u and (a (x)). By Lemma 2.1, the supporting
ij
∇
planesP(1,0,0) andP(−1,0,0) touches Σ at u(1,0,0) or u( 1,0,0). Ifboth
∇ ∇ −
planes touch Σ at the same point, then we see that u c. Consequently
3
≡
the homogeneous order one function v(x ,x ) = u cx satisfies (1.1) and
1 2 3
−
it follows that v is linear, or u is also linear.
So we are left with the case that P(1,0,0) and P(−1,0,0) touch Σ at different
point. We may assume P touches Σ at u(1,0,0). It means that
(1,0,0)
∇
(2.4) u (x ,x ,1) = h x h x h h(0) near (0,0).
3 1 2 1 1 2 2
− − ≤
By (2.1) and (2.2), (0,0) is a Lebesgue point for D2h and h. It follows
∇
that (cf. [4, Appendix C])
2
1
h(x ,x ) = h(0)+h (0)x +h (0)x + h (0)x x +0 x 2 .
1 2 1 1 2 2 ij i j
2 | |
iX,j=1 (cid:16) (cid:17)
Because of (2.3) and D2h(0,0) = 0, we may assume h (0) = a, h (0) =
11 12
6
0, h (0) = b with a,b > 0. A simple computation yields u(0,x ,1) =
22 2
−
h(0)+ 1bx2+o x2 which contradicts (2.4). This completes the proof.
2 2 2
(cid:0) (cid:1)
3. Another Proof in the C2,α Case
In this section, we present yet another proof of Theorem 1.1 in the
C2,α case. Suppose u : R3 0 R is a C2,α homogeneous order one
\ { } →
function u = rg(θ ,θ ), where (r,θ ,θ ) is the spherical coordinates with
1 2 1 2
x = rcosθ cosθ , x = rcosθ sinθ , x = rsinθ . Then the Hessian
1 2 1 2 2 1 3 2
D2u= ∂2u has the following representation
(cid:16)∂xi∂xj(cid:17)
(3.1) D2u(x) = 1R(θ ,θ )HR′(θ ,θ )
1 2 1 2
r
where
0 0 0
H =  0 cos12θ2∂∂2θ1g2 −tanθ2∂∂θg2 +g ∂θ∂12∂gθ2 + csoisn2θθ22∂∂θg1 
∂2g
 0 ∗ ∂θ22 +g 
and R(θ ,θ ) is the rotation from (∂x ∂x ,∂x ) to ∂r, 1∂θ ,1∂θ . We
1 2 1, 2 3 r 1 r 2
see that D2u= 0 if and only if H = 0. We define the s(cid:0)ingular set S(u(cid:1)) as
S(u) = x S2 D2u(x) = 0 .
{ ∈ | }
LINEARITY OF HOMOGENEOUS ORDER ONE SOLUTIONS 5
Lemma 3.1. Let u be as in Theorem 1.1. Then one of the following holds:
(1) S(u) is empty;
(2) S(u) consists of finitely many points;
(3) S(u) = S2.
Proof. Suppose S(u) is not empty. For any p S(u), we prove that either
p is isolated or S(u) contains a neighborhood∈of p on S2. Without loss of
generality, we assume p = (1,0,0), a (p) = δ . Note that S(u) = S(u+
ij ij
ax +bx +cx ),wemay also assume u(p)= 0,then u(p) = p u(p) = 0.
1 2 3
∇ ·∇
Correspondingly, we have
∂2g ∂g ∂g
(0,0) = 0, , (0,0) = 0, and g(0,0) = 0.
(cid:18)∂θ ∂θ (cid:19) (cid:18)∂θ ∂θ (cid:19)
i j 1 2
Also g satisfies
2 ∂2g 2 ∂g
A (θ ,θ ) + B (θ ,θ ) +C(θ ,θ )g = 0,
ij 1 2 i 1 2 1 2
∂θ ∂θ ∂θ
iX,j=1 i j Xi=1 i
where the Cα coefficients (A ) satisfy the ellipticity condition with the
ij
same constant λ, A (0,0) = δ , B and C are Cα functions near (0,0).
ij ij i
Suppose g vanishes up to infinity order at (0,0), then by the Carleman
unique continuation (cf. [3, p.124]), g 0 in a neighborhood of (0,0). It
follows that D2u = 0 on S2, since u is C≡2. Suppose g vanishes up to order
k 1at (0,0).By our assumptionk 3.We applythe resultof [2, Theorem
− ≥
1] to obtain
g = P (θ ,θ )+R(θ ,θ ),
1 2 1 2
wherethehomogeneousorderkpolynomialP satisfies 2 A (0,0) ∂2P =
i,j=1 ij ∂θi∂θj
0, or P = 0, and the remainder satisfies P
△
R O θ k+α , R O θ k−1+α , D2R O θ k−2+α
∼ | | ∇ ∼ | | ∼ | |
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
with α (0,1). It is a simple fact that (θ ,θ ) D2P = 0 = (0,0) and
1 2
∈ | { }
consequently (θ ,θ ) D2H = 0 = (0(cid:8),0) . Hence p is (cid:9)an isolated zero
1 2
| { }
pointofD2u.T(cid:8)hereforeS(u)cons(cid:9)istsoffinitelymanypointsinthiscase.
We consider the surface Σ parametrized by u : S2 R3. From (3.1),
∇ →
it follows that the Hessian always has one zero eigenvalue. Let λ (x) and
1
λ (x) be the other two eigenvalues of D2u(x). Because of equation (1.1),
2
λ (x)λ (x) < 0 for all x S2 S(u).
1 2
∈ \
Lemma 3.2. For any x S2 S(u), the surface Σ is C2,α at u(x) with a
∈ \ ∇
normal vector givenbyxandthe two principle curvatures givenby 1/λ (x)
1
−
and 1/λ (x).
2
−
Proof. We may assume x= p = (0,0,1) S2 S(u). Then locally at u(p),
∈ \ ∇
Σ can be represented by
(3.2) F(x ,x )= u(x ,x , 1 x2 x2).
1 2 ∇ 1 2 q − 1− 2
6 HAN, NADIRASHVILI, AND YUAN
By differentiating the identity x u(x) = u(x) twice with respect to x ,x
i j
·∇
for i,j = 1,2, we obtain
(3.3) u (p)= 0, u (p) = u (p).
3i 3ij ij
−
By differentiating (3.2) and making use of (3.3), we have
F (0) = (u (p),u (p),0)
i i1 i2
and
F(3)(0) = u (p)
ij − ij
where F(3) denotes the third component of the vector F . Then we get
ij ij
F (0) F (0) = (0,0,(u u u2 )(p)).
1 × 2 11 22− 12
It is easy to see that (u u u2 )(p) = λ λ (p) = 0. Hence (0,0,1) is a
normal vector for Σ at 11u(2p2).−We12see that 1 u2 = G6 −1 near (0,0,1), where
G−1 is the inverse of th∇e Gauss map of Σ. T∇hus Σ is C2,α nearby. We also
get the first and second fundamental forms of Σ at u(p) as follows
∇
I = (u2 +u2 )dx2+2u (u +u )dx dx +(u2 +u2 )dx2,
11 12 1 12 11 22 1 2 21 22 2
II = u dx2 2u dx dx u dx2,
− 11 1− 12 1 2− 22 2
where all u are evaluated at p. Therefore the two principle curvatures are
ij
1/λ and 1/λ .
1 2
− −
Remark 3.3. In computing the principle curvatures of Σ in terms of the
eigenvalues of D2u, we differentiate u three times. By approximation, the
conclusion still holds for u C2,α.
∈
Now we are ready for another proof of Theorem 1.1 in the C2,α case.
Proof. We prove that S(u) = S2 by excluding the case (1) and (2) in Lemma
3.1. WeassumeS(u)consistsofatmostfinitelymanypoints. First, thegra-
dient surface Σ has supporting planes with normals along all the directions
in R3. However, the saddle points in u(x) D2u(x) = 0 cannot support
∇ | 6
any supporting planes. Hence there a(cid:8)re at most finitely m(cid:9)any points on Σ
with the supporting planes.
Claim.(Alexandrov)Foranyx S(u),thesupportingplanetoΣat u(x )
0 0
∈ ∇
can only have normal direction x .
0
ThereforethereareatmostfinitelymanysupportingplanestoΣ. Thisisa
contradiction. Nowweprovetheclaim. Supposethereisanothersupporting
plane P to Σ at u(x ) with normal direction x = x . Take a small
1 0 1 0
neighborhood U of∇x on S2 such that D2u(x) = 0 for6any x = x U and
0 0
∗ ∗ 6 6 ∈
U S = φ, with S being a great circle through x and x . This can be
1 1
∩ −
donesince x is an isolated point in (u). Now lift P along the x direction
0 1 1
S
to P so that P u(U) = C is a smooth close curve on P. We can take a
∩∇ ∗ ∗ ∗
point on C, say u(x ) with x U and x = x , such that the normal to
0
∇ ∗ ∈ 6
the plane curve C at u(x ) is along the intersection P and another plane
∗ ∇
through S . Then we see that the normal of the surface Σ at the regular
LINEARITY OF HOMOGENEOUS ORDER ONE SOLUTIONS 7
∗ ∗ ∗
point u(x ) must be on S . On the other hand, the normal at u(x ) is
∗ ∇ ∇
x U. This contradiction completes the proof of the Claim.
∈
By excluding the case (1) and (2) in Lemma 3.1, we are left with the case
(3) (u) = S2. That is D2u 0, and hence u is linear. This finishes the
S ≡
proof of Theorem 1.1 in the C2,α case.
References
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DepartmentofMathematics,NotreDameUniversity,NotreDame,IN46556
E-mail address: [email protected]
Department of Mathematics, University of Chicago, 5734 S. University
Ave., Chicago, IL 60637
E-mail address: [email protected]
Department of Mathematics, University of Chicago, 5734 S. University
Ave., Chicago, IL 60637, and University of Washington, Seattle, WA 98195
E-mail address: [email protected]