On Inverse Problem for Differential Operators with Deviating Argument
V. A. Yurko
Vjacheslav A. Yurko, https://orcid.org/0000-0002-4853-0102, Saratov State University, 83, Astrakhans-kaya Str., Saratov, 410012, Russia, YurkoVA@info.sgu.ru
Second-order functional differential operators with a constant delay are considered. Properties of their spectral characteristics are obtained, and a nonlinear inverse spectral problem is studied, which consists in constructing operators from their spectra. We establish the uniqueness and develop a constructive procedure for solution of the inverse problem.
Keywords: differential operators, deviating argument, inverse spectral problem.
DOI: https://doi.org/10-18500/1816-9791-2018-18-3-328-333
INTRODUCTION
We study the inverse spectral problem for Sturm - Liouville differential operators with a constant delay. Such problems often appear in natural sciences and engineering (see, for example, monograph [1] and the references therein). Inverse spectral problems consist in constructing operators with given spectral characteristics. For the classical Sturm - Liouville operators the inverse problems have been studied fairly completely; the main results can be found in [2,3]. However, differential operators with delay are essentially more difficult for investigating, since the main methods in the inverse problem theory (the transformation operator method and the method of spectral mappings [2,3]) do not work for operators with delay. Note that some particular results on the inverse problems for operators with delay were obtained in [4-7].
Consider the boundary value problems Lj(q), j = 1,2:
-y"(x) + q(x)y(x - a) = Ay(x), x G (0, n), (1)
У( 0) = y'(n) + Hj У(п) = °, (2)
where A is the spectral parameter, a G (0, n), q(x) G L(a,n) is a complex-valued function, and q(x) = 0 for x G [0,a]. We study the inverse problem of constructing the potential q(x) and the coefficients Hj from the given two spectra of the boundary value problems Lj(q). More precisely, let {^nj-}n^0, j = 1, 2 be the eigenvalues of the problems Lj(q).
Inverse problem 1. Given {^nj-}n^0, j = 1,2, construct q(x) and Hj.
We note that in the case of large delay when a ^ n/2, the characteristic functions of the problems Lj(q) depend on the potential q(x) linearly, i.e. the inverse problem becomes linear. This linear case was studied in [5,7]. For a < n/2 the characteristic functions depend on the potential nonlinearly, i.e. the inverse problem becomes nonlinear. This nonlinear case is seriously more difficult for investigating and for constructing the global solution of the inverse problem. In this paper we study namely nonlinear case. For definiteness, let a G [2n/5,n/2). The case a < 2n/5 requires separate consideration. The main results of the paper are Theorem 1 and Algorithm 1, where a global constructive procedure for solving the inverse problem is provided, and the uniqueness of the solution is proved.
© Yurko V. A, 2018
1. PROPERTIES OF SPECTRAL CHARACTERISTICS
Let S(x, A) be the solution of Eq. (1) under the initial conditions S(0, A) = 0, S'(0, A) = 1. Eigenvalues of the boundary value problem (1)-(2) coincide with the zeros of its characteristic function
Pj(A) := S'(n, A) + HjS(n, A), j = 1, 2. (3)
Lemma 1. Boundary value problem Lj has a countable set of eigenvalues {ynj}n^0f and for n ^ to:
yfhnj
1
П + 2
A0 cos(n + 1/2)a H (1
+------^-----— + — + o -
2nn nn V n
(4)
where A0 = q(t) dt.
J a
Lemma 2. The specification of the spectrum {ynj}n^0 uniquely determines pj (A) via
Ж Л
Pj (A) = nj? ■ (5)
n=0 v 1 7
Let us study the connections between the characteristic functions pj (A) and the potential q(x). Let A = p2. The function S(x, A) satisfies the integral equation
S(x, A) =
sin px P
+
sin p(x P
'd q(t)S(t
a, A) dt.
(6)
Solving (6) we get for x ^ 2a:
S(x, A)
= So(x, A) + S1 (x, A) + S2(x, A),
where
So(x, A) =
sin px
p
Si(x, A) = / —t)q(t)So(t - a, A) dt,
J a p
S2(x, A) = f -i——-------)q(t)S1 (t — a, A) dt,
2a p
Using (7) and (8), we calculate
cos p(x — a)
1
Si (x, A) =------—------ q(t) dt + — q(t) cos p(x — 2t + a) dt.
2p a 2p a
Denote Ak(A) := S(n, A), k = 0,1. In view of (3), one has
Pj(A):=Ai(A) + HjAo(A), j = 1, 2,
ж
x
(7)
(8) (9)
(10)
(11)
and consequently,
Aq (Л)
1
Hi - H
Taking (7), (9) and (10) into account, we obtain
Aq (Л) = SinPn _ Aq C0S a) + 1
(Pi (Л) _ P2 (Л)) ■
(12)
P
2p2 . 2р2 у q(t) cosp(2t - п - a) dt + S2(п, Л),
л /лч , sinр(п — a) 1 Г , ^// xx
А1(Л) = cosрп + Aq------2----- + 2“ q(t) sinp(2t - п - a) dt + S2(п,Л).
2р 2p J a
Denote
aq(p) := 2р2 (Aq(Л) - sinPn + AqC0sP(^ V р 2p
sin р(п - a)N
AJ (р) := 2p ( Ai (Л) - cos рп - Aq
2P
(13)
(14)
Then
Aq(р) = / q(t) cosp(2t - п - a) dt + 50(р), J a
pn
Aq(p) = / q(t) sinp(2t - п - a) dt + Aj(p),
a
(15)
(16)
where £0(р) = 2р2S2(п,Л), С(р) = 2pS2(п,Л). Using (9) and (10), we infer
1 p (n-2a)
2р^о(р) = -A sin р(п - 2a) + - / Q(?) sin p? d?,
2 J — (n—2a)
1 p (n — 2a)
2p5j(р) = -A cos р(п - 2a) - - / Q(?) cos p£ d£,
2 J— (n—2a)
(17)
(18)
where
/*п pt—a
A = / q(t)dt / q(s) ds,
2a a
Q(?) = Qi(?/2 + п/2 + a) - QJ(C/2 + п/2) - Q3(?/2 + п/2),
px—a pn pn
Qi(x) = q(x) / q(s) ds, Q2(x) = q(x) / q(s) ds, Q3(x) = q(s)q(s - x) ds.
J a J x+a x+a
For simplicity we assume that q(x) G AC[a,п]■ The general case requires small technical modifications. Denote qi(x) := q'(x). Taking (15)-(18) into account, we get
4pAQ (р) = Bi sin р(п - a) - 2 A sin р(п - 2a)-
/(п—a) r (n — 2a)
qo(?) sin p£d£ + / Q(£) sin p£d£,
(n—a) J — (n—2a)
4pAQ (р) = B2 cos р(п - a) - 2A cos р(п - 2a)+
(19)
+
(п—а)
qo(£) cos pC dC
-(n-a)
(n—2a)
— (n—2 a)
Q(C) cos pC dC,
(20)
where Bi = 2(q(a) + q(n)), B2 = 2(q(a) - q(n)), qo(C) = qi(C/2 + n/2 + a/2). Denote
d0 (p) = 4pA* (p) — B1 sin p(n — a) + 2A sin p(n — 2a), (21)
dl (p) = 4pA* (p) — B2 cos p(n — a) + 2 A cos p(n — 2a). (22)
It follows from (19)-(20) and (21)-(22) that
/(n—a) p(n—a)
R(C) sin pCdC, di(p) = R(C)cos pCdC, (23)
- (n—a) J— (n—a)
where R(C) = q0(C) — Q(C), and Q(C) = 0 outside the interval (—(n — 2a), (n — 2a)). In
particular, this yields
qi(x) = R(2x — n — a)+Qi(x+a/2) — Q2(x — a/2) — Qa(x — a/2), x £ (3a/2,n — a/2). (24) Denote by {An}n^i the zeros of the entire function A0(A). Then
/— Ao cos na /1\ ,^r4
^=n++0 n • (25)
2. SOLUTION OF THE INVERSE PROBLEM
In this section we present our main results: a constructive procedure for solving the inverse problem and the corresponding uniqueness theorem. The solution of Inverse problem 1 can be found by the following algorithm.
Algorithm 1. Let the spectra {pnj-}n^0, j = 1, 2 be given.
1. Construct pj(A), j = 0,1, via (5).
2. Find Hi — H2, using (4):
Hi — H2 = n lim (д/ДПТ — л/ДПг).
3. Construct A0(A), using (12), and calculate An.
4. Find A0 from (25).
5. Calculate Hi and H2, using (4).
6. Find Ai(A) from (11).
7. Construct A*(A), j = 0,1, according to (13) and (14).
8. Calculate A, Bi and B2, using (19) and (12), and find
q (a) = (Bi + B2 )/4, q(n) = (Bi — B2 )/4.
9. Construct d0(p) and di(p) by (21) and (22).
10. Calculate R(C), using (23).
11. Find q0(C) for C £ (—(n — a), —(n — 2a)) U (n — 2a,n — a): q0(C) = R(C).
12. Calculate qi(x) = q0(2x — n — a) for x £ (a, 3a/2) U (n — a/2, n).
13. Find
q (x) = q(a) +
'X
qi(t) dt,
x G (a, 3a/2),
q(x) = q(n)
■n
qi(t) dt, x G (n — a/2,n).
14. Using (24) and knowledge of q(x) for x G (a, 3a/2) U (n — a/2,n), construct qi(x) for x G (3a/2, n — a/2):
rx-a/2
qi (x) = R(2x — n — a) + q(x + a/2) q(s) ds — q(
a
rn
— q(s)q(s — x + a/2) ds.
J x+a/2
a/2) q(s) ds—
x+a/2
15. Calculate q(x) for x G (3a/2, n — a/2). Thus, the following theorem is proved.
Theorem 1. The specification of two spectra {pnj}n^0, j = 1, 2 uniquely determines the potential q(x) and the coefficients Hi , H2. The solution of Inverse problem 1 can be found by Algorithm 1.
Acknowledgements: This work was supported by the Ministry of Education and Science of the Russian Federation (project nos. 1.1660.2017/4.6) and by the Russian Foundation for Basic Research (projects nos. 16-01-00015, 17-51-53180).
References
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Cite this article as:
Yurko V. A. On Inverse Problem for Differential Operators with Deviating Argument. Izv. Saratov Univ. (N.S.), Ser. Math. Mech. Inform., 2018, vol. 18, iss. 3, pp. 328-333. DOI: https://doi.org/10.18500/1816-9791-2018-18-3-328-333
УДК 517.984
ОБ ОБРАТНОЙ ЗАДАЧЕ ДЛЯ ДИФФЕРЕНЦИАЛЬНЫХ ОПЕРАТОРОВ С ОТКЛОНЯЮЩИМСЯ АРГУМЕНТОМ
В. А. Юрко
Юрко Вячеслав Анатольевич, доктор физико-математических наук, заведующий кафедрой математической физики и вычислительной математики, Саратовский национальный исследовательский государственный университет имени Н. Г. Чернышевского, Россия, 410012, Саратов, Астраханская,
83, YurkoVA@info.sgu.ru
Рассматриваются функционально-дифференциальные операторы второго порядка с постоянным запаздыванием. Установлены свойства их спектральных характеристик и исследуется нелинейная обратная спектральная задача, которая состоит в построении операторов по их спектрам. Доказана единственность решения обратной задачи и указана конструктивная процедура ее решения.
Ключевые слова: дифференциальные операторы, отклоняющийся аргумент, обратная спектральная задача.
Благодарности. Работа выполнена при финансовой поддержке Минобрнауки РФ (проект № 1.1660.2017/4.6) и РФФИ (проекты № 16-01-00015, № 17-51-53180).
Образец для цитирования:
Yurko V. A. On Inverse Problem for Differential Operators with Deviating Argument [Юрко В. А. Об обратной задаче для дифференциальных операторов с отклоняющимся аргументом] // Изв. Сарат. ун-та. Нов. сер. Сер. Математика. Механика. Информатика. 2018. Т. 18, вып. 3. С. 328-333. DOI: https://doi.org/10.18500/1816-9791-2018-18-3-328-333