Ambarzumyan Type Theorem for a matrix quadratic Sturm-Liouville equation with energy dependent

0.09 MB·English

by  Emrah Yilmaz

AMBARZUMYAN TYPE THEOREM FOR A MATRIX QUADRATIC
STURM-LIOUVILLE EQUATION WITH ENERGY-DEPENDENT POTENTIAL
1 2 3
Emrah YILMAZ , Hikmet KEMALOGLU and Gamze DOLANBAY
1,2,3Department of Mathematics, University of Firat, 23119, Elazıg, TURKEY
[email protected]
3
1
0
2

n Abstract : Ambarzumyan’s theorem for quadratic Sturm-Liouville problem is extended to second order differential
a
J tems of dimension d 2. It is shown that if the spectrum is the same as the spectrum belonging to the zero potential,
≥
5
1 thematrix valued functions P(x) and Q(x) are both zero.

]
P
MSC 2000 : 34B24, 34A55.
S
.
h Key Words: Matrix Quadratic Sturm-LiouvilleEquation, Spectrum,Ambarzumyan’sTheorem.
t
a
m
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1 1. Introduction
v
6 From an historical viewpoint, Ambarzumyan’s paper may be thought of as the first paper in the theory
7
2 inverse spectral problems associated with Sturm-Liouville operators. Ambarzumyan proved the follo
3
1. theorem:
0
2
3 If n : n = 0,1,2,... is the spectral set of the boundary value problem
1 (cid:8) (cid:9)
:
v y′′+(λ q(x))y = 0, y′(0) = y′(π)= 0
i −
X
r 2
a then q = 0 in [0,π], where q(x) L [0,π] [1]. Ambarzumyan’s theorem was extended to the second
∈
differential systems of two dimensions in [2], to Sturm-Liouville differential systems of any dimension
[3], to the Sturm–Liouville equation (which is concerned only with Neumann boundary conditions)
general boundary conditions by imposing an additional condition on the potential function [4], and
multi-dimensional Dirac operator in [5]. In addition, some different results of Ambarzumyan’s theorem
been obtained in [6], [7], [8], [9], [10].
Ambarzumyan’s theorem was extended to the following boundary value problem by imposing to
dition for p
′′ 2
y +[2λp(x)+q(x)]y = λ y,x [0,π]
− ∈
with the homogeneous Neumann boundary conditions
′ ′
y (0) = y (π) = 0,
2 1
where λ is a spectral parameter and p W [0,π] and q W [0,π] by Koyunbakan, Lesnic and Panakho
2 2
∈ ∈
[11]. Thisproblemiscalled thequadraticpenciloftheSchro¨dingeroperator. Ifp(x)= 0theclassical Sturm-
Liouville operator is obtained. Some versions of the eigenvalue problem (1.2)-(1.3) were studied extensiv
in [12], [13], [14], [15], [16].
Before giving the main results, we mention some physical properties of the quadratic equation.
problem of describing the interactions between colliding particles is of fundamental interest in physics.
interesting in collisions of two spinles particles, and it is supposed that the s wave scattering matrix
−
the s wave binding energies are exactly known from collision experiments. For a radial static poten
−
V(E,x) and s wave, the Schro¨dinger equation is written as
−
′′
y +[E V(E,x)]y = 0,
−
where
V(E,x) = 2√Ep(x)+q(x).
2
We note that with the additional condition q(x)= p (x), the above equation reduces to the Klein-Gordon
−
s-wave equation for a particle of zero mass and energy √E [17].
2. Matrix Differential Equations
For simplicity, A denotes entry of matrix A at the i th row and j th column and I is a
ij d
− −
identity matrix and 0 is a d d zero matrix.
d
×
We are interested in the eigenvalue problem
′′ 2
φ +[2λP(x)+Q(x)]φ= λ φ
−
′ ′
Aφ(0)+Bφ(0) = Cφ(π)+Dφ(π) = 0,
where P(x) = diag[p1(x),p2(x),...,pd(x)] and Q(x) are d d real symmetric matrix-valued functions,
×
those n n matrices A,B,C and D satisfy the following conditions
×
∗
DC : Self-Adjoint
∗
BA = 0
rank[A,B]= rank[C,D]= n.
To study (2.1)-(2.2), we introduce the following matrix differential equation
′′ 2
Y +[2λP(x)+Q(x)]Y = λ Y,x [0,π]
− ∈
′
Y(0,λ) = I ,Y (0,λ) = 0
d d
whereλisaspectralparameter,Y(x) = [yk(x)],k = 1,disacolumnvector,P(x) = diag[p1(x),p2(x),...,
2 1
and Q(x) are d d real symmetric matrix-valued functions, P W [0,π] and Q W [0,π], where W
2 2
× ∈ ∈
(k = 1,2)denotesasetwhoseelementisak thordercontinuously differentiablefunctioninL2[0,π],
−
and µ = σ +it C. Then, λ is an eigenvalue of (2.1)-(2.2), if the matrix which is called charachteristic
∈
function
′
W(µ) = CY(π,µ)+DY (π,µ)
is singular.
In order to describe W(µ) explicitly, we must know how to express the solution Y(x,µ). The solution
Y(x,µ) of (2.6)-(2.7) can be expressed as
x x
Y(x,µ) = cos[λI x α(x)]+ A(x,t)cos(λt)dt+ B(x,t)sin(λt)dt
d
− Z Z
0 0
where A(x,t) and B(x,t) are symmetric matrix-valued functions whose entries have continuous
derivatives up to order two respect to t and x, and it can described by the following lemma.
Lemma 2.1. [18] Let A and B be as in (2.8). Then, A and B satisfy following conditions
2 2
∂ A(x,t) ∂B(x,t) ∂ A(x,t)
2P(x) Q(x)A(x,t) =
∂x2 − ∂t − ∂t2
2 2
∂ B(x,t) ∂A(x,t) ∂ B(x,t)
+2P(x) Q(x)B(x,t) =
∂x2 ∂t − ∂t2
∂A(x,t)
A(0,0) = h , B(x,0) = 0 , = 0 ,
d d d
∂t (cid:12)
(cid:12)t=0
(cid:12)
x
(cid:12)
with α(x) = P(t)dt. Moreover, there holds
Z
0
x
2[cosα(x)A(x,x)+sinα(x)B(x,x)] = 2h+ T1(t)dt
Z
0
and
x
2[sinα(x)A(x,x) cosα(x)B(x,x)] = P(x) P(0)+ T2(t)dt
− − Z
0
where
2
T1(x) = P (x)+cosα(x)Q(x)cosα(x)+sinα(x)Q(x)sinα(x)
and
T2(x) = sinα(x)Q(x)cosα(x) cosα(x)Q(x)sinα(x).
−
Lemma 2.2. [18] The eigenvalues of the operator L(P,Q;h,H) are
π
α
j
λ = n+ ,n = 0, 1, 2, 3,...,j = 1,d and α = p (x)dx.
n j j
π ± ± ± Z
0
3. Main Results
In this section, some uniqueness theorems are given for the equation (2.6) with the Robin boundary
conditions. It is shown that an explicit formula of eigenvalues can determine the functions Q(x) and
be zero both. Results are some generalizations of [11].
Consider the a second matrix quadratic initial value problem
∼
′′ 2
Y + 2λP(x)+Q(x) Y = λ Y,x [0,π]
− (cid:20) (cid:21) ∈
′
Y(0,λ) = I ,Y (0,λ) = 0
d d
∼
where Q has the same properties of Q.
∼ ∼
The problems (2.6)-(2.7) and (3.1)-(3.2) will be denoted by L(P,Q;h,H) and L(P,Q;h,H) and
∼ ∼
trums of these problems wil be denoted by σ(P,Q) and σ(P,Q), respectively.
π
∼ ∼ ∼
Theorem 3. 1. Suppose that σ(P,Q) = σ(P,Q) and α(π) = 0, then Q(x) Q(x) dx = 0
Z (cid:20) − (cid:21)
0
everywhere on [0,π].
∼ ∼
Proof: Since σ(P,Q) = σ(P,Q), it follows that λ σ(P,Q) are large eigenvalues. Then we can
n
∈
from (2.8) that
′
Y (π,λ ) = (λ I P(π))sin[λ I π α(π)]+A(π,π)cos(λ π)+B(π,π)sin(λ π)
n n d n d n n
− − −
x x
+ A (π,t)cos(λ t)dt+ B (π,t)sin(λ t)dt
x n x n
Z Z
0 0
and similarly for the problem (3.1)-(3.2), we can write
∼′ ∼ ∼
Y (π,λ ) = (λ I P(π))sin[λ π α(π)]+A(π,π)cos(λ π)+B(π,π)sin(λ π)
n n d n n n
− − −
x x
∼ ∼
+ A (π,t)cos(λ t)dt+ B (π,t)sin(λ t)dt.
x n x n
Z Z
0 0
∼′
′
By substracting, Y (π,λ ) and Y (π,λ ), we get
n n
∼ ∼
0 = A(π,π) A(π,π) cos(λ π)+ B(π,π) B(π,π) sin(λ π)
n n
(cid:20) − (cid:21) (cid:20) − (cid:21)
x x
∼ ∼
+ A (π,t) A (π,t) cos(λ t)dt+ B (π,t) B (π,t) sin(λ t)dt.
x x n x x n
Z (cid:20) − (cid:21) Z (cid:20) − (cid:21)
0 0
∼
By using Riemann-Lebesque lemma and for λ in Lemma 2.2., we obtain A(π,π) = A(π,
n
→ ∞
the other hand by Lemma 2.1., we know the following equalities,
d
2
2 [cosα(x)A(x,x)+sinα(x)B(x,x)] = P (x)+Q(x)
dx
d ∼ ∼ ∼
2
2 cosα(x)A(x,x)+sinα(x)B(x,x) = P (x)+Q(x).
dx (cid:20) (cid:21)
After substracting (3.3),(3.4) and integrating, we get
π
∼ ∼ ∼
Q(x) Q(x) dx = 2 A(π,π) A(π,π) cosα(π)+ B(π,π) B(π,π) sinα(π)
Z (cid:20) − (cid:21) (cid:26)(cid:20) − (cid:21) (cid:20) − (cid:21) (cid:27)
0
and
π
∼
Q(x) Q(x) dx = 0
Z (cid:20) − (cid:21)
0
∼
where α(0) = α(π)= 0 and A(π,π)= A(π,π). This completes the proof.
Theorem 3. 2. Let P(x) = diag[p1(x),p2(x),...,pd(x)] and Q(x) are two d d real symmetric matrix-
×
valued functions, and α(π) = 0. If 0 m :j = 1,2,... is a subset of the spectrum of the follo
j
{ } ∪ { }
d dimensional second order differential system
−
′′ 2 ′ ′
φ +[2λP(x)+Q(x)]φ = λ φ, φ(0) = φ(π)= 0
−
where 0 is the first eigenvalue of (3.5), m is a strictly ascending infinite sequence of positive integers,
j
0 and m are multiplicity of n, then P(x) = Q(x) = 0.
j
Proof: Suppose for the (3.5) Neumann problem, there are infinitely many eigenvalues of the form
m are positive integers, j = 1,2,... and each m is of multiplicity n. For such a case, Let A = C =
j j
B = D = I in (2.1). Then, we get
d
′
Y (π,m ) = 0.
j
On the other hand, by (2.8), we have
′
Y (x,λ ) = (λ I P(x))sin[λ x α(x)]+A(x,x)cos(λ x)+B(x,x)sin(λ x)
n n d n n n
− − −
x x
+ A (x,t)cos(λ t)dt+ B (x,t)sin(λ t)dt.
x n x n
Z Z
0 0
Equations (3.6) and (3.7) imply
π π
A(π,π)cos(m π)+ A (π,t)cos(λ t)dt+ B (π,t)sin(λ t)dt = 0.
j x n x n
Z Z
0 0
We have from (3.8) and Riemann Lebesque lemma that A(π,π)= 0. Then, by integration of
d
2
Q(x)+P (x) = 2 [cosα(x)A(x,x)+sinα(x)B(x,x)],
dx
we get
x
1
2
cosα(x)A(x,x)+sinα(x)B(x,x) = Q(t)+P (t) dt.
2 Z
0 (cid:2) (cid:3)
Since α(π) = 0, we have
π
1
2
0 = A(π,π) = Q(t)+P (t) dt
2 Z
0 (cid:2) (cid:3)
and
π π
2
Q(x)dx = P (x)dx.
Z −Z
0 0
By using the reality of 0 being the ground state of the eigenvalue problem (3.5), we may find n linearly
−
independentconstant vectors correspondingtothe sameeigenvalue 0by thevariational principleand
them by ϕ , j = 1,2,...,d. Since they should the following equation
j
′′ 2
ϕj +[2λ0P(x)+Q(x)]ϕj = λ0ϕj.
−
Then, we obtain
Q(x)ϕ = 0, 0 x π.
j
≤ ≤
Thus Q(x) = 0. If we consider (3.9) and diagonally of P, we get P(x) = 0. This completes the pro
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