Non-compact perturbation of the spectrum of multipliers given by a special form Текст научной статьи по специальности «Математика»

NANOSYSTEMS:

PHYSICS, CHEMISTRY, MATHEMATICS

Kucharov R.R., et al. Nanosystems: Phys. Chem. Math., 2024,15 (1), 31-36.

http://nanojournal.ifmo.ru

Original article

DOI 10.17586/2220-8054-2024-15-1-31-36

Non-compact perturbation of the spectrum of multipliers given by a special form

Ramziddin R. Kucharov1'2 "'6, Tillohon M. Tuxtamurodova2 c

1Tashkent International University of Financial Management and Technology, Tashkent, Uzbekistan 2National University of Uzbekistan, Tashkent, Uzbekistan

[email protected], [email protected],[email protected] Corresponding author: R. R. Kucharov, [email protected], [email protected]

Abstract In this paper, the change of the spectrum of multiplier H0f (x,y) = k0(x,y)f (x,y) for perturbation with non-compact partially integral operators is studied. In addition, the existence of resonance is investigated in the model H = Ho — (71T1 + Y2T2).

Keywords essential spectrum, discrete spectrum, lower bound of the essential spectrum, non-compact partial integral operator, resonance with zero energy.

For citation Kucharov R.R., Tuxtamurodova T.M. Non-compact perturbation of the spectrum of multipliers given by a special form. Nanosystems: Phys. Chem. Math., 2024,15 (1), 31-36.

1. Introduction

Self-adjoint partial integral operators appear in the theory of discrete Schrodinger operators. The study of the theory of elasticity [1], continuum mechanics [2-4], aerodynamics [5] and other problems leads to the problem of spectral analysis of the partial integral operators. In 1969, Uchiyama [6,7] obtained the first results on the finiteness of the discrete spectrum of N-particle Hamiltonians with N > 2. He found sufficient conditions for the finiteness of the number of discrete eigenvalues for the energy operators. In 1971, Zhislin [8] assumed the total charge of the system to be less than -1 and proved that the discrete spectrum of the energy operators is finite in the symmetry spaces of negative atomic ions of molecules with any mass of nucleus and infinitely heavy nuclei.

Let H be a separable Hilbert space and the operator H0 : H ^ H be self-adjoint and have only essential spectrum (ct(Ho) = aess(H0)), i.e. the operator H0 lacks the discrete spectrum (adisc(H0) = 0). Let's assume that the operator H0 is perturbed by the self-adjoint operator T, i.e. consider the operator H0 + eT, e > 0. The main questions in the theory of perturbation of spectra are as follows:

1) How is the structure of the spectrum of the operator H0 + eT related to the spectrum of the original (unperturbed) operator H0?

2) What are the properties of the spectrum as a function of e > 0?

Let H0 be a multiplier in ¿2(ft) (ft C Rm — compact): Hof (x) — u(x)f (x), where u(x) is a given real valued continuous function on ft, T : L2(ft) ^ L2(ft) is a linear self adjoint compact operator. The operator H0 + eT, e > 0, is an operator in the Friedrichs model. It is known for such an operator that aess(H0 + eT) = <t(H0) [9]. In addition, a number of methods have been developed [10-12] to investigate the existence of an eigenvalue in the discrete spectrum Vdisc(H0 + eT) and to study the finiteness (infiniteness) of the discrete spectrum adisc(H0 + eT). If the operator T is non-compact, then there is no general way to study the spectrum of the perturbed operator H0 + eT. In [13,14], a method is proposed for studying the spectrum of the operator H0 + eT : L2 (ft 1 x ft2) ^ L2 (01 x ft2) (ft 1 C Rmi, ft2 C Rm2 are nonempty compact sets), when H0 is a multiplier defined by a continuous function k0(x, y) on ft1 x ft2 and T = T + T2 is a linear bounded self adjoint operator with partial integrals in L2(ft1 x ft2), i.e. T\, T2 are partially integral operators (PIO). It should be stressed that T and T2 with a non-zero kernel are not compact. In [13] it is proved that aess (H0 + eT) = a(H0 + eT\) U a(H0 + eT2), in the case when the kernels of T and T2 are continuous functions.

Consider the multiplier H0, given by the function h0(x, y), having the following form: h0(x, y) = u(x) + w(x, y) + v(x), and PIO T\, T2 with kernels identically equal to one.

Let the multiplier H0 be perturbed by a non-compact operator T = y1T1 + y2T2, where y1 > 0, y2 > 0. The purpose of the work is to apply the method from [13] to study the structure of the essential spectrum of the operator H0 — (71T1 + y2T2) and to study the existence of resonance in the model H = H0 — (y^ + y2T2).

We denote by a( ), aess( ) and <rdisc(-), respectively, the spectrum, the essential spectrum and the discrete spectrum of the self-adjoint operators.

The number

Emin(H) =inf{A : A G aess(H)}

is called the bound edge (or the lower edge) of the essential spectrum of the operator H.

© Kucharov R.R., Tuxtamurodova T.M., 2024

2. Non-compact perturbation of the essential spectrum

Let Qi = [0,1]V1 c RV1 and Q2 = [0,1]V2 C RV2 (vu V2 G N). In the space L2Q x Q2), let us consider a linear bounded self-adjoint operator H of the form

H = Ho - (Y1T1 + Y2T2), Yi > 0, Y2 > 0, (1)

where H0 is the multiplier given by the continuous real valued function ko(x, y), i.e. H0f (x, y) = ko(x, y)f (x, y), and the operators T1, T2 in the space L2(Q1 x Q2) are defined by the following formulas:

Tif (x,y) = J f (s,y)d^1(s), T2f (x,y) = J f (x, t)d^2 (t), n1 n2

where ^ (•) is the Lebesgue measure on Qj, j = 1,2.

It is known that a(H0) = [kmin, kmax] C aess(H), where kmin = min k0(x,y), kmax = max k0(x,y), and aess(H) = a(W1) U a(W2), where Wk = Ho - YkTk, k = 1, 2 (see. [13]).

Assume that k0(x, y) = u(x) + v(y), where u(x) and v(y) are real valued continuous functions on Q1 and Q2, respectively. Then the operator H (1) will be unitarily equivalent to the operator H1 ( E + E ( H2, where H1, H2 are operators in the Friedrichs model, E is the identity operator and "(" means the tensor product of operators [13]. Using the spectral properties of the tensor product of operators [15,16], it can be argued that for all positive values of the parameters Y1 and y2, the operator H has at most one eigenvalue outside the essential spectrum and Emin(H) < 0. The eigenvalue A G ffdisc(H) of the operator H is simple and A < Emin(H).

Suppose that ko(x, y) = u(x)v(y), where u(x) and v(y) are non-negative continuous functions on Q1 Q2, respectively, and 0 G Ran(u) n Ran(v). Then Emin (H) < 0 and the operator H has at most one eigenvalue below the lower edge of the essential spectrum. The eigenvalue A < Emin (H) of the operator H is simple [9,10].

Let u(x, y) is a non-negative continuous function on Q1 x Q2 and 0 G Ran(u). Assume that u(0) = v(0) = 0 and u(x, 0) = u(0, y) = 0, x G Q1, y G Q2, where the zero element in the corresponding linear space is denoted by 0. Let the multiplier in (1) be given by the function

ho(x,y) := ko(x,y) = u(x) + u(x,y) + v(y). Here, we study the spectral properties of the operator:

H = Ho - (Y1T1 + Y2T2), Y1,Y2 > 0, (2)

in the case

Ho f (x, y) = ho (x, y)f (x, y) and under the following assumptions: the following integrals exist and are finite

/ds f dt

us), J vt).

n1 Q2

For each fi G Q2, we define the self-adjoint operator H1 (fi) : L2(Q1) ^ L2(Q1) in the Friedrichs model:

H1(]3)<(x) = ho(x, fi)<(x) - Y1 j <(s)ds.

n1

Similarly, for each a G Q1, we define the operator H2(a) : L2(Q2) ^ L2(Q2) in the Friedrichs model:

H2(a)^(y) = ho(a,y)^(y) - Y2 J ^(t)dt.

Let's put M1(fi) = max ho(x, fi), M2(a) = max ho(a, y).

XEQ1 ye^2

By Weyl's theorem [1] on the compact perturbation of the essential spectrum, we have aess(Hk (£)) = [0,Mk (£)], £ G

Qj, j = k, j, k = 1, 2.

Lemma 2.1. [18] The number A G R\[0, M1(fi)] is the eigenvalue of the operator H1(fi)if and only if A1(fi; Y1, A) = 0, where

ds

A^fi; Y1,A) = 1 - Y1 J

Let's define the function h1 (fi) on Q2 by the formula

ho(s,fi) - A" n1

ho(s,fi )

The function h1(fi) is continuous on the set of Q2 and h1(fi) > 0, fi G Q2.

h1(fi) = J

n1

ds

We define [20] non-positive and continuous functions n1(y) on ft2 and n2(x) on ft1 using the following equalities

n1(y) = inf (H1(y)y, y), y G ft2, n2(x) = inf (H2(x)^,^), x G Q1.

Let's put nmin = min n,(£), ^max = max n(£), j = k, j, k = 1, 2, hmax = max h0(x,y).

Proposition 2.1. The following conditions hold for the operators W1 and W2

a) a(W1) = [nmin,nmax] U a(Ho);

b) a(W2) = [n2min,n2max] U a(Ho).

Proof. a) In [13], the equality a(W1) = a(H0) U 01 is proven, where

01 = {A G p(Ho) : A^£o; A,y) = 0 for some £0 G ft2}.

Let n1(^0) < 0 for some £0 G ft2. Then, due to the minimax principle, solution A0 (£0) of the equation A1(^0; 71,A) = 0, is defined using continuous function n1(^0). i.e. A0(£0) = n1 (£0). Therefore, n1(^0) G 01. If n^£) < 0 for all £ G ft2, then A(£) = n1(£) G 01,01 = [n^n^] and0(^1) = 0(Ho)U01 = [0, hmax]U[nmin, n^]. Ifn^o) = 0 for some £0 G ft2, then = 0. Hence, we obtain 0(W1) = 0(Ho) U 01 = [0, hmax] U [n^, n^] = [n^, h^]. The equality 0(^2) = [0, h^] U [nfn nrx is proved similarly.

Proposition 2.2. If 71 < h-1(0), then 0(H1(£)) = 0ess№(£)) for all £ G ft2-Proof. Since

ho(x, y) = u(x) + w(x, y) + v(y) > u(x), x G Q1, y G Q2,

then

H1(£) > H1(0), £ G ft2. (3)

However, Emin(H1(0)) = 0 and

. , „ ,s , ds

A

71, A) = 1 — 71J

n1

u(s) — A

The function A1(A) = A1(0; 71, A) on (—to, 0) is strictly decreasing, while lim A1 (A) = 1 and lim A1(A) =

1 — 71h1(0) > 0. Hence, one obtains that A1(A) = A1(0; 71, A) > 0 for A G (—to, 0). Then, according to Lemma 2.1, 0disc(H1W) = 0, i.e. 0(H1(0)) = [0,M1(0)]. It follows from (3) that

inf (H1(£)y,y) > inf (H^)y,y) = 0, £ G ft2.

Ilvll=1 n^y=1

However, 0 G 0(H1(£)), £ G ft2 and consequently inf (H1(£)y, y) = Emin(H1(£)) = 0, £ G ft2. Hence, due to

llvll=1

the minimax principle [1], it follows that 0disc(H1(£)) = 0, for all £ G ft2. Now we define the function h2 (a) on ft1 by the following formula

h2(a) f dt J ho(a, t)'

Obviously, the function h2(a) is continuous in ft1 and h2 (a) > 0, a G ft1.

Just as in proposition 2.2, it is proved that if 72 < h—1(0),then 0(H2(a)) = 0ess(H2(a)) for all £ G ft2. Hence, due to Proposition 2.2, the following theorem is proved:

Theorem 2.1. a) if 71 < h—1 (0), then 0(^1) = 0(Ho) = [0, h^]; b) if Y2 < h—1(0), then 0(^2) = 0(Ho) = [0,hmax].

We define the sets D0 C ft2 and D1 C ft2:

do = {£ G ft2 : Y1 < h—1(£)}, D1 = ft2 \ do.

Lemma 2.2. a) If D0 = ft2 (i.e. D1 = 0), then n1(t) = 0;

b) if Do = 0 D1 = 0, then < = 0;

c)if Do = 0, then < < 0.

Proof. Obviously, for every fixed £ G ft2 and 71 > 0, the function A1(A) = A1(£; 71, A) is strictly decreasing on

(—to, 0) and

lim A1(A) = 1 and lim A1(A) = 1 — 71 h1(£).

A^ — TO A^0—

a) Let D0 = ft2. Then 1 — 71h1(£) > 0 for all £ G ft2. Due to monotonicity of the function A1(A) for (—to, 0) we have A1(£; 71, A) > 0 for all A G (—to, 0) for each £ G ft2. Hence, by virtue of Lemma 2.1, we obtain 0(H1(£)) = 0ess(H1(£)), £ G ft2. Then, by the minimax principle n1(t) = 0 for all t G ft2.

b) Let V0 = 0. Then there exists a point P0 e V0 c Q2, such that,

lim A1P; Y1,A) = 1 - YiMPo) > 0,

i.e. A1(p0; y1, A) > 0 on 0). Hence, due to the lemma 2.1, we obtain that a(H1(P0)) = aess(H1(P0)). Therefore, we have n (P0) = 0. Since n (t) < 0, t e Q2, we have nmax = n (P0) = 0. If V1 = 0, then there exists P1 eVj c Q2 such that

lim A1(p1; Y1,A) = 1 - Y^M < 0.

Hence, the equation A1(P1; y1 ,A) =0 on 0) has unique solution A0 < 0. By Lemma 2.1, the number A0 is an eigenvalue of the operator H1 (P1). Hence, following the minimax principle, we obtain that n1(p1) = A0 < 0, i.e.

nmin < MP1) < 0.

c) Let V0 = 0. Then V1 = Q2 and therefore for each P e Q2, we have

lim A1(p; Y1,A) = 1 - Y1^(P) < 0.

Due to the monotonicity of the function A1(P; y1,A) on 0) there is a negative number A = A(P) such that A1(P; y1,A(P)) = 0, i.e. the number A(P) is the eigenvalue of the operator H1 (P). Then, by the minimax principle, we obtain that n1(P) = A(P), p e Q2 .It follows from the continuity of the function n1(t) on Q2 that nmax < 0.

We prove that n™ < nmax. Let's assume the opposite: let n™ = nmax. Then the solutions A0, A0 < 0, and A1,

A1 < 0 of the equations A1(0; y1,A) = 0 and A1(P; y1,A) = 0, p e Q2, P = 0 coincide, i.e.

A1(d; Y1,A0) = A1(P; yua))=0.

Therefore, we obtain

f h,0(s, P) - u(s)

J (u(s) - A0)(h0(s,P) - A0)

Q,

ds = 0. (4)

However, h0(s, P) - u(s) > 0, s e Q2 and the function

h0(s, P) - u(s)

F1(s,P)

_max

(u(s) - A0)(h0(s,P) - A0)

is non-negative continuous on Q2 and distinct from a constant. Hence, in accordance with the property of the Lebesgue integral, we obtain that J F1(s, P)ds > 0. This contradicts equality (4). Therefore, ^m™ = n2m

Q2

We put:

hfn = min h3 (£) and hmmax = max h3 (£), j = 1, 2, k = 1, 2, j = k. 3 ten* 3Vsy 3 ten* 3Vsy' ^ ^ r

Lemma 2.2 implies the proof the theorem

Theorem 2.2.a)if Y1 > (hfn)-1, then nmx < 0;

b) if (h^)-1 <Y1 < (htn)-1, then nfn < 0 nmax = 0;

c) if Y1 < (Kax)-1, then n (t) = 0.

A similar theorem is true for the function n2(x).

Corollary 2.1. If Y1 < (h^)-1 Y2 < (h^)-1, then aess(H) = aH). Proof. For the essential spectrum of the operator H, the equality holds (see. [13])

aess(H) = a(W1) U a(W2),

where Wk = H0 - Yk Tk, k = 1, 2. Hence, by Theorem 2.2 and Proposition 2.1, we obtain

aess(H )= a(H0) = [0,hmax].

-/Ti0)

Corollary 2.2 Let in (1) Y1 = h-1 (0) and Y2 = h21(0). Then aess(H) = a(H0).

Proof. Consider PIO V, defined by the equality

V = H0 - (h-1 (0)T1 + h-1 (0)T2).

For y1 = h-1 (0), one has

/do

—,—;— = 0.

h0(s,0) - A

Qi

From the monotonicity of the function A1 (0; h-1(0), A) on (-<x, 0) we obtain that A1(0; h-1 (0), A) > 0 on (-<x, 0), i.e.

a(H0 - h-1(0)T1) = a(H0). Similarly, it is shown that a(W2) = a(H0). Hence, aess(V) = a(W1) U aW) = aH).

3. Zero-energy resonance of the operator H

It is said that, the operator Hi(0) (operator H2(0)) )) has a resonance with zero energy [19] if the number 1 is the

eigenvalue of the integral operator H1 : L2(Q1) ^ L2(^1) (H2 : L2(^2) ^ L2(Q2)), where

f y(s)ds f

Hiy(x) = 7w —-—, H2^(y) = Y2

J u(s) 7 v(t)

fil fil

and at least one corresponding eigenfunction y0(x) (eigenfunction ^0(y)) satisfies the condition y0(0) = 0 (^0(0) = 0).

Theorem 3.1. Let 71 = h-1(0). Then;

a) operator H1(0) has a resonance with zero energy;

b) for all ft G ft = 0 operator H1(£) has no negative eigenvalue..

Proof. a) Let y0(x) = 1. Then V1 y0 = 71h1(0) = y0(x), i.e. the equation V1y = y has a solution y0 from C(Q1) and y0(0) = 0.

b) If 71 = h-1 (0), then the condition of Proposition 2.2 is satisfied, and therefore a(H1 (ft)) = aess (H1 (ft)) for all ft G i.e. there is no negative eigenvalue for the operators H1 (ft), ft G .

Example. Let = = [0,1] and

We have

u(x) = v(x) = x1/2, w(x, y) = — cos 2— cos "2y^ .

ds u(s)

dt

= 2.

ïmm

h1(1) and h

max

h1(0)

2. It is obvious that

The function hi(x) strictly decreases on [0,1], and hence, hm

—— / L2 (ft1), i.e. ——— G L1(ft1) \ L2(^1). Hence, for 71 = - the operator H1(0) has a resonance with zero energy u(x) u(x) 2

and for all ft G (0,1] operator H1(ft) has no negative eigenvalue. 4. Conclusion

Our main goals are the description of the essential spectrum of the operator H and studying its spectral properties. This work differs from the work of other scientists because we choose the special form of the multiplier H0 and the non-compactness of the partial integral operators T1 and T2 takes place. To summarize, we applied the method of [13] for the description of the essential spectrum. Additionally, we mainly used the minimax principle from [9] to prove the theorems and found the exact description of the essential spectrum proved by conditioning the parameters 71 and 72.

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Submitted 30 December 2023; revised 13 January 2024; accepted 15 January 2024

Information about the authors:

RamziddinR. Kucharov - Tashkent International University of Financial Management and Technology; National University of Uzbekistan, Mathematics, Tashkent, 4, 100174, Uzbekistan; ORCID 0000-0002-0728-9340; [email protected]; [email protected]

TillohonM. Tuxtamurodova - National University of Uzbekistan, Mathematics, Tashkent, 4, 100174, Uzbekistan; ORCID 0009-0002-1721-880X; [email protected]

Conflict of interest: the authors declare no conflict of interest.