Untitled - Cdm.unimo.it

Eigenvalue Analysis 161
We note that constant polynomials are eigenfunctions of problem (8.2.13), corresponding
to the eigenvalue µ. Very little theory has been developed. We only present the following
result.
Theorem 8.2.4 - Let α = β = 0. Then, for any n ≥ 2, if γ = 1
2n(n + 1), the
eigenvalues of problem (8.2.13) are real and strictly positive.
Proof - Noting that γ ˜w (n)
0
(8.2.14)
= γ ˜w(n)
n = 1 (see (3.5.6)), we get
1
|p
−1
′ n,m| 2 n
dx + µ
i=0
= [p ′ n,m¯pn,m](1) − [p ′ n,m¯pn,m](−1) −
= γ ˜w (n)
n [p ′ n,m¯pn,m](1) − γ ˜w (n)
0 [p ′ n,m¯pn,m](−1) +
The proof follows easily.
= λn,m
n
i=0
[|pn,m|(η (n)
i )] 2 ˜w (n)
i
1
p
−1
′′ n
n,m¯pn,m dx + µ [|pn,m|(η
i=0
(n)
i )] 2 ˜w (n)
i
n
i=0
[(−p ′′ n,m + µpn,m)¯pn,m](η (n)
i ) ˜w (n)
i
[|pn,m|(η (n)
i )] 2 ˜w (n)
i , 0 ≤ m ≤ n.
Real or complex eigenvalues, with a relatively small imaginary part, are obtained for
many other values of the parameters α and β, both for problems (8.2.12) and (8.2.13).
In the latter case, γ is required to be positive and larger than a constant γn, which
is in general proportional to n 2 . To our knowledge, currently, there are no theoretical
results to support these conjectures.
It is worthwhile to mention that an incorrect specification of the sign of the equa-
tion, when imposing boundary conditions (see for instance (7.4.13)), can result in a
matrix with at least one negative eigenvalue. For example, replacing the second equa-
tion in (8.2.12) by p ′ n,m(η (n)
0 ) = λn,mpn,m(η (n)
0 ), we still obtain real eigenvalues, but
one of them is negative.