Extremum problems for eigenvalues of discrete Laplace operators

10 REN GUO
Then 0 = ∂H
∂θ 1
and 0 = ∂H
∂θ 2
imply that
∑ m
∑ m
(a 3 ...a m + a 2 a 3 ...â i ...a m )(1 + a 2 1) = (a 3 ...a m + a 1 a 3 ...â i ...a m )(1 + a 2 2).
i=3
Since a 1 + a 2 > 0, it is equivalent to
(a 1 − a 2 )(a 1 + a 2 )(a 3 ...a m + a 1a 2 − 1
a 1 + a 2
The third factor is
cot θ 3 ...cot θ m + cot(θ 1 + θ 2 )
m
∑
i=3
i=3
a 3 ...â i ...a m ) = 0.
m∑
cot θ 3 ...ĉot θ i ...cot θ m
which is written as ∑ m
i=1 ã1...̂ã i ...ã m−1 , where ã 1 = cot(θ 1 + θ 2 ),ã i = cot θ i+1 for
i = 2,...,m−1. This expression corresponds to a cyclic (m−1)-gon. By assumption
of the induction, ∑ m
i=1 ã1...̂ã i ...ã m−1 > 0.
Hence the only possibility is a 1 = a 2 . By similar argument, we show that a i = a j
for any i,j. Hence the function ∑ m
i=1 a 1...â i ...a m has the unique critical point such
that θ i = π m
for any i = 1,...,m.
Next, we claim that ∑ m
i=1 a 1...â i ...a m > 0. Without loss of generality, we assume
that a 1 > 0,a 2 > 0,...,a m−1 > 0. Now
m∑
a 1 ...â i ...a m
i=1
i=3
m−2
∑
= a 1 a 2 ...a m−2 (a m−1 + a m ) + a 1 ...â i ...a m−2 (a m−1 a m )
i=1
m−2
∑
m−2
∑
= a 1 a 2 ...a m−2 (a m−1 + a m ) + a 1 ...â i ...a m−2 (a m−1 a m − 1) + a 1 ...â i ...a m−2
i=1
m−2
∑
m−2
a m−1 a m − 1 ∑
= (a m−1 + a m )(a 1 a 2 ...a m−2 + a 1 ...â i ...a m−2 ) + a 1 ...â i ...a m−2 .
a m−1 + a m
Let ã m−1 = am−1am−1
a m−1+a m
i=1
= cot(θ m−1 + θ m ). Then
m−2
∑ a m−1 a m − 1
a 1 a 2 ...a m−2 + a 1 ...â i ...a m−2
a m−1 + a m
i=1
m−2
∑
= a 1 a 2 ...a m−2 + a 1 ...â i ...a m−2 ã m−1 .
i=1
i=1
i=1
Consider an cyclic (m −1)-gon with angles θ 1 ,...,θ m−2 ,θ m−1 +θ m . By the assumption
of induction,
Therefore ∑ m
i=1 a 1...â i ...a m > 0.
m−2
∑
a 1 a 2 ...a m−2 + a 1 ...â i ...a m−2 ã m−1 > 0.
i=1