Equidistribution of Neumann data mass on triangles

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by  Hans Christianson

EQUIDISTRIBUTION OF NEUMANN DATA MASS ON
TRIANGLES
HANS CHRISTIANSON
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Abstract. In this paper we study the behaviour of the Neumann data of Dirich-
1
let eigenfunctions on triangles. We prove that the L2 norm of the (semi-classical)
0
2 Neumann data on each side is equal to the length of the side divided by the area of
the triangle. The novel feature of this result is that it is not an asymptotic, but an
n
exact formula. The proof is by simple integrations by parts.
a
J

0
1
1. Introduction
]
P
Given a compact surface or manifold with boundary, it is an interesting question
A
to consider restrictions of eigenfunctions to hypersurfaces; either the Dirichlet data or
.
h Neumann data (or both, the Cauchy data) can be considered. Perhaps the simplest
t
a question is to consider boundary values. That is, if we consider Dirichlet (respectively
m Neumann) eigenfunctions, we may try to study the Neumann (respectively Dirichlet)
[ data on the boundary.

In this short note, we consider one of the simplest planar domains, a planar triangle
1
v T. Our main result is that the L2 mass of the semi-classical Neumann data on each
3 sideofT equalsthelengthofthesidedividedbytheareaofT. Itshouldbeemphasized
9
that these formulae are equalities, not asymptotics or estimates.
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2
Theorem1. LetT beaplanartrianglewithsidesA,B,C, oflengthsa,b,crespectively.
0
. Consider the (semi-classical) Dirichlet eigenfunction problem:
1
0 (cid:40)
(−h2∆−1)u = 0, in T,
7 (1.1)
1 u| = 0,
∂T
:
v
and assume the eigenfunctions are normalized (cid:107)u(cid:107) = 1.
i L2(T)
X
Then the (semi-classical) Neumann data on the boundary satisfies
r
a (cid:90) a
|h∂ u|2dS = ,
ν
Area(T)
A
(cid:90)
b
|h∂ u|2dS = ,
ν
Area(T)
B
and
(cid:90)
c
|h∂ u|2dS = ,
ν
Area(T)
C
where h∂ is the semi-classical normal derivative on ∂T, dS is the arclength measure,
ν
and Area(T) is the area of the triangle T.
1
2 H.CHRISTIANSON
We pause to note briefly that the the semiclassical parameter h takes discrete values
as h → 0 (reciprocals of eigenvalues).
Remark 1.1. We are calling this “equidistribution” of Neumann mass since it says
that the Neumann data has the same mass to length ratio. Of course it does not say
anything about local equidistribution to subsets of the sides.
To the author’s knowledge, no exact formula such as this exists in the previous
literature, except in cases where explicit formulae for the eigenfunctions are known.
Eveninthesecases, theformulaetypicallydependonh. AstatementsuchasTheorem
1isfalseingeneralforotherplanarpolygons. SeeSection3fortheexampleofasquare.
In order to better understand these formulae, in subsequent works, the author will
study non-Euclidean triangles, and higher dimensional problems, as well as explore
weaker lower bounds and interior lower bounds in some simple polygons.
1.1. History. Previous results on restrictions primarily focused on upper bounds.
In the paper of Burq-G´erard-Tzvetkov [BGT07], restrictions of the Dirichlet data to
arbitraryhypersurfaceswereconsidered. Anupperboundofthenorm(squared)ofthe
restrictions of O(h−1/2) was proved, and shown to be sharp. Of course this shows that
there are some eigenfunctions with a known lower bound on the norms of restrictions.
In the author’s paper with Hassell-Toth [CHT13], an upper bound of O(1) was proved
for(semi-classical)Neumanndatarestrictedtoarbitraryhypersurfaces,andalsoshown
to be sharp. Again, this gives a lower and upper bound for some eigenfunctions.
In the case of quantum ergodic eigenfunctions, a little more is known. In the pa-
pers of G´erard-Leichtnam [GL93] and Hassell-Zelditch [HZ04], the Neumann (respec-
tively Dirichlet) boundary data of Dirichlet (respectively Neumann) quantum ergodic
eigenfunctions is studied, and shown to have an asymptotic formula for a density one
subsequence. That means that there is a lower bound, and explicit local asymptotic
formula in this special case, at least for most of the eigenfunctions. Similar state-
ments were proved for interior hypersurfaces by Toth-Zelditch [TZ12,TZ13]. Again,
potentially a sparse subsequence may behave differently. In the the author’s paper
with Toth-Zelditch [CTZ13], an asymptotic formula for the whole weighted Cauchy
data is proved for the entire sequence of quantum ergodic eigenfunctions, however it
is impossible to separate the behaviour of the Dirichlet versus Neumann data.
2. Proof of Theorem 1
Assume the sides A,B,C are listed in clockwise orientation. We assume that A is
the shortest side, followed by B and C with respective lengths a (cid:54) b (cid:54) c.
We use rectangular coordinates (x,y) in the plane, and orient our triangle so that
the corner between B and C is at the origin (0,0). We further assume that the side A
is parallel to the y axis.
Webreakouranalysisintothetwocasesofacutetriangles(includingrighttriangles)
and obtuse. See Figures 1 and 2 for a picture of the setup.
2.1. Acute triangles. Let (cid:96) be the segment on the x axis beginning at (0,0) and
perpendicular to the side A. Of course (cid:96) can be computed in terms of the sides, but
its value is not necessary for this computation, other than to note that the area of T
NEUMANN DATA ON TRIANGLES 3
C
A
a
2
(cid:96)
a
1
B
Figure 1. Setup for acute (and right) triangles
A
a
2
C
B
a
1
(cid:96)
Figure 2. Setup for obtuse triangles
is a(cid:96)/2. Write A = A ∪A , where A is the part of A under the x axis and A is the
1 2 1 2
part above. Let a ,a denote the respective sidelengths.
1 2
We can parametrize B and C with respect to x.
(cid:110) a (cid:111)
C = (x,y) ∈ R2 : y = 2x, 0 (cid:54) x (cid:54) (cid:96) ,
(cid:96)
and
(cid:110) a (cid:111)
B = (x,y) ∈ R2 : y = − 1x, 0 (cid:54) x (cid:54) (cid:96) .
(cid:96)
Then the arclength parameters are
((cid:96)2+a2)1/2 c
γ = (1+(a /(cid:96))2)1/2 = 2 = ,
C 2
(cid:96) (cid:96)
4 H.CHRISTIANSON
and
((cid:96)2+a2)1/2 b
γ = (1+(a /(cid:96))2)1/2 = 1 = ,
B 1
(cid:96) (cid:96)
and the unit tangent vectors are
(cid:18) (cid:19)
(cid:16) a (cid:17) (cid:96) a
τ = 1, 2 γ−1 = , 2
C (cid:96) C c c
and
(cid:18) (cid:19)
(cid:16) a (cid:17) (cid:96) a
τ = 1,− 1 γ−1 = ,− 1 .
B (cid:96) B b b
From this we have the outward unit normal vectors
(cid:18) (cid:19)
a (cid:96)
2
ν = − ,
C
c c
and
(cid:18) (cid:19)
a (cid:96)
1
ν = − ,− .
B
b b
Of course the outward normal to A is ν = (1,0).
A
We are assuming Dirichlet boundary conditions, which implies that the tangential
derivatives of u vanish on ∂T. That is,
∂ u = 0
y
on A, and
(cid:96) a
2
τ ·∇u = ∂ u+ ∂ u = 0
C x y
c c
on C. Similarly,
(cid:96) a
1
τ ·∇u = ∂ u− ∂ u = 0
B x y
b b
along B. Rearranging, we have
a
2
h∂ u = − h∂ u
x y
(cid:96)
on C and
a
1
h∂ u = h∂ u
x y
(cid:96)
on B.
Making the substitutions, along C we have
h∂ u = ν ·h∇u
νC C
a (cid:96)
2
= − h∂ u+ h∂ u
x y
c c
(cid:18)a2 (cid:96)(cid:19)
= 2h∂ u+ h∂ u
x y
c(cid:96) c
(cid:18)a2+(cid:96)2(cid:19)
= 2 h∂ u
y
c(cid:96)
c
= h∂ u.
y
(cid:96)
NEUMANN DATA ON TRIANGLES 5
Hence
(cid:96)
h∂ u = h∂ u
y c νC
on C. Substituting again, we have
a a
2 2
h∂ u = − h∂ u = − h∂ u
x (cid:96) y c νC
along C. Similarly, along B, we have
(cid:96)
h∂ u = − h∂ u
y b νB
and
a
1
h∂ u = − h∂ u.
x b νB
We now consider the vector field
X = (x+m)∂ +(y+n)∂ ,
x y
where m,n are parameters independent of x and y. Since m∂ commutes with −h2∆
x
as well as n∂ , the usual computation yields
y
[−h2∆−1,X] = −2h2∆.
Then using eigenfunction equation (1.1), we have
(cid:90) (cid:90)
([−h2∆−1,X]u)u¯dV = −2 (h2∆u)u¯dV
T T
(cid:90)
= 2|u|2dV
T
= 2,
since u is normalized.
On the other hand, again using the eigenfunction equation (1.1) again, we have
(cid:90)
([−h2∆−1,X]u)u¯dV
T
(cid:90) (cid:90)
= ((−h2∆−1)Xu)u¯dV − (X(−h2∆−1)u)u¯dV
T T
(cid:90)
= ((−h2∆−1)Xu)u¯dV.
T
Integrating by parts and using the eigenfunction equation and Dirichlet boundary
conditions, we have
(cid:90)
((−h2∆−1)Xu)u¯dV
T
(cid:90)
= (Xu)((−h2∆−1))u¯dV
T
(cid:90) (cid:90)
− (h∂ hXu)u¯dS + (hXu)(h∂ u¯)dS
ν ν
∂T ∂T
(cid:90)
= (hXu)(h∂ u¯)dS,
ν
∂T
6 H.CHRISTIANSON
Hence we have computed:
(cid:90)
2 = (hXu)(h∂ u¯)dS.
ν
∂T
Let us break up the analysis into the three different sides. In order to simplify
notation somewhat, set
(cid:90)
I = |h∂ u|2dS,
A ν
A
and similarly for B and C. Notice we have left the surface measure dS alone, even
though we could write it explicitly in terms of the arclength parameters computed
above. This is not necessary for the analysis.
Returning now to our computations of the normal derivatives, we have
(cid:90)
(hXu)(h∂ u¯)dS
ν
A
(cid:90)
= (((x+m)h∂ +(y+n)h∂ )u)(h∂ u¯)dS
x y ν
A
= ((cid:96)+m)I ,
A
since x = (cid:96) on A.
Continuing, using that along C, we have y = (a /(cid:96))x:
2
(cid:90)
(hXu)(h∂ u¯)dS
ν
C
(cid:90)
= (((x+m)h∂ +(y+n)h∂ )u)(h∂ u¯)dS
x y ν
C
(cid:90) (cid:16) (cid:16)a (cid:17) (cid:17)
2
= ((x+m)h∂ + x+n h∂ )u (h∂ u¯)dS
x y ν
(cid:96)
C
(cid:90) (cid:18)(cid:18) (cid:16) a (cid:17) (cid:16)a (cid:17)(cid:18)(cid:96)(cid:19)(cid:19) (cid:19)
2 2
= (x+m) − + x+n h∂ u (h∂ u¯)dS
c (cid:96) c νC νC
C
(cid:90) (cid:18)(cid:18) (cid:19) (cid:19)
a (cid:96)
2
= − m+ n h∂ u (h∂ u¯)dS
c c νC νC
C
(cid:18) (cid:19)
a (cid:96)
2
= − m+ n I .
C
c c
NEUMANN DATA ON TRIANGLES 7
Similarly, along B, we have y = −(a /(cid:96))x:
1
(cid:90)
(hXu)(h∂ u¯)dS
ν
B
(cid:90)
= (((x+m)h∂ +(y+n)h∂ )u)(h∂ u¯)dS
x y ν
B
(cid:90) (cid:16) (cid:16) a (cid:17) (cid:17)
1
= ((x+m)h∂ + − x+n h∂ )u (h∂ u¯)dS
x y ν
(cid:96)
B
(cid:90) (cid:18)(cid:18) (cid:16) a (cid:17) (cid:16) a (cid:17)(cid:18) (cid:96)(cid:19)(cid:19) (cid:19)
1 1
= (x+m) − + − x+n − h∂ u (h∂ u¯)dS
b (cid:96) b νB νB
B
(cid:90) (cid:18)(cid:18) (cid:19) (cid:19)
a (cid:96)
1
= − m− n h∂ u (h∂ u¯)dS
b b νB νB
B
(cid:18) (cid:19)
a (cid:96)
1
= − m− n I .
B
b b
Summing up, we have:
(cid:18) (cid:19) (cid:18) (cid:19)
a (cid:96) a (cid:96)
1 2
2 = ((cid:96)+m)I + − m− n I + − m+ n I . (2.1)
A B C
b b c c
First, set m = n = 0. Then we get 2 = (cid:96)I , or
A
2
I =
A
(cid:96)
a
=
a(cid:96)/2
a
= .
Area(T)
Now we observe that the left hand side of (2.1) is independent of m and n so we can
differentiatewithrespecttomandntogettwonewequations. Thatis, differentiating
both sides of (2.1) with respect to m yields
a a
1 2
0 = I − I − I ,
A B C
b c
and plugging in the value of I , we have
A
a a 2
1 2
I + I = . (2.2)
B C
b c (cid:96)
Now differentiating (2.1) with respect to n, we have
(cid:96) (cid:96)
0 = − I + I ,
B C
b c
so that
b
I = I .
B C
c
8 H.CHRISTIANSON
Plugging in to (2.2), we have
(cid:18) (cid:18) (cid:19) (cid:19)
2 (cid:16)a (cid:17) b a
1 2
= + I
C
(cid:96) b c c
(cid:16)a a (cid:17)
1 2
= + I
C
c c
a
= I .
C
c
Hence
2c c
I = = ,
C
a(cid:96) Area(T)
and back substituting,
b b
I = I = .
B C
c Area(T)
This proves the theorem for acute and right triangles.
2.2. Obtuse triangles. The proof is nearly the same, with several sign changes. Us-
ing the setup in Figure 2, we have
(cid:26) (cid:27)
(a+a )
C = (x,y) : 0 (cid:54) x (cid:54) (cid:96), and y = 1 ,
(cid:96)x
and
(cid:110) a (cid:111)
B = (x,y) : 0 (cid:54) x (cid:54) (cid:96), and y = 1x .
(cid:96)
Similar computations as above lead to the following: along C,
(cid:96)
h∂ u = h∂ u,
y c νC
and
a+a
1
h∂ u = − h∂ .
x c νC
Along B we have
(cid:96)
h∂ u = − h∂ u
y b νB
and
a
1
h∂ u = h∂ u.
x b νB
The same commutator computation holds, and similar substitutions as in the acute
case yield the equation
(cid:18) (cid:19) (cid:18) (cid:19)
a (cid:96) a+a (cid:96)
1 1
2 = ((cid:96)+m)I + m− n I + − m+ n I . (2.3)
A B C
b b c c
Again first setting m = n = 0, we get
2 a
I = = .
A
(cid:96) Area(T)
Differentiating with respect to m and n we get the two equations
a a+a
1 1
0 = I + I − I , (2.4)
A B C
b c
NEUMANN DATA ON TRIANGLES 9
and
(cid:96) (cid:96)
0 = − I + I ,
B C
b c
so that again
b
I = I .
B C
c
Now substituting into (2.4), we have
(cid:18) (cid:18) (cid:19) (cid:19)
2 (cid:16)a (cid:17) b a+a
1 1
= − + I ,
C
(cid:96) b c c
or
2c c
I = = .
C
a(cid:96) Area(T)
Back substituting once again, we have also
b
I = .
B
Area(T)
This completes the proof of the obtuse triangle case, and hence proves the theorem.
3. Other polygons
Theorem 1 is false in general for other planar polygons. One can see from the
computationsabovethatitisstraightforwardtocomeupwith3independentequations
relating the Neumann data on the sides. The proof suggests that only three equations
are possible in general. Of course we do not have a proof of that. However, even for
convex polygons the theorem is false in general.
Consider the square Ω = [0,2π]2. The Dirichlet eigenfunctions are given by the
Fourier basis. Let us examine some specific choices. That is, for integers j,k let
u (x,y) = (π)−1sin(jx)sin(ky).
jk
The u vanish on ∂Ω, and are normalized. We have
jk
−∆u = (j2+k2)u
jk jk
as usual. Rescaling to a semi-classical equation, we take h = (j2+k2)−1/2 to get
−h2∆u = u .
jk jk
On x = 0 and x = 2π repectively, we have
h∂ u | = π−1(−j)hsin(ky)
ν jk x=0
and
h∂ u | = π−1jhsin(ky).
ν jk x=2π
Hence the norm of the Neumann data along either x = 0 or x = 2π is
(cid:90) 2π
|h∂ u | |2dy = π−1h2j2.
ν jk x=0,2π
0
If k (cid:29) j, we can make h2j2 as small as we like, so in fact we can say only
(cid:90) 2π
|h∂ u | |2dy (cid:62) ch2
ν jk x=0,2π
0
10 H.CHRISTIANSON
for some c > 0. A rough conjecture is that a lower bound of h2 holds for any polygon
in the plane. The author plans to revisit this question in subsequence papers.
References
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(H. Christianson) Department of Mathematics, University of North Carolina.
E-mail address: [email protected]