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Axioms
, Volume 9 (4) – Nov 13, 2020

/lp/multidisciplinary-digital-publishing-institute/regularization-method-for-singularly-perturbed-integro-differential-7EQ080tljf

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axioms Article Regularization Method for Singularly Perturbed Integro-Differential Equations with Rapidly Oscillating Coefﬁcients and Rapidly Changing Kernels 1, 2 Burkhan Kalimbetov * and Valeriy Safonov Department of Mathematics, Akhmed Yassawi University, B. Sattarkhanov 29, Turkestan 161200, Kazakhstan Department of Higher Mathematics, National Research University «MPEI», Krasnokazarmennaya 14, 111250 Moscow, Russia; Singsaf@yandex.ru * Correspondence: burkhan.kalimbetov@ayu.edu.kz Received: 30 September 2020; Accepted: 10 November 2020; Published: 13 November 2020 Abstract: In this paper, we consider a system with rapidly oscillating coefﬁcients, which includes an integral operator with an exponentially varying kernel. The main goal of the work is to develop an algorithm for the regularization method for such systems and to identify the inﬂuence of the integral term on the asymptotic behavior of the solution of the original problem. Keywords: singular perturbation; integro-differential equation; rapidly oscillating coefﬁcient; regularization; asymptotic convergence; resonant exhibitors MSC: 34K26; 45J05 1. Introduction In the study of various issues related to dynamic stability, with the properties of media with a periodic structure, in the study of other applied problems, one has to deal with differential equations with rapidly oscillating coefﬁcients. Equations of this kind can describe some mechanical or electrical systems that are under the inﬂuence of high-frequency external forces, automatic control systems with a linear adjustable object, etc. As an example, we can cite the principle of operation of an oscillator with a small mass and a nonlinear restoring force, in which a high-frequency periodic force with a large amplitude acts. The presence of high-frequency terms creates serious problems for their direct numerical solutions. Therefore, asymptotic methods are usually applied to such equations ﬁrst, the most famous of which are the Feshchenko–Shkil–Nikolenko splitting method [1–5] and the Lomov’s regularization method [6–8]. It should also be noted that singularly perturbed equations are the object of study by several Russian researchers, as well as other scientists (see, for example [9–22]). In this paper, the Lomov’s regularization method is generalized to previously unexplored integro-differential equations with rapidly oscillating coefﬁcients and with rapidly decreasing kernels of the form t R dz b(t) 1 m(q)dq 0 # a(t)z #g(t) cos z e K(t, s)z(s, #)ds = h(t), z(t , #) = z , t 2 [t , T] (1) 0 0 dt # where z = z(t, #), h(t), b (t) > 0, a(t) > 0, m(t) < 0, a(t) 6= m(t) (8t 2 [t , T]) , g(t) are scalar functions, z is a constant, # > 0 is a small parameter. In the case b (t) = 2g (t) , and of the absence of an integral term, such a system was considered in [6–8]. Axioms 2020, 9, 131; doi:10.3390/axioms9040131 www.mdpi.com/journal/axioms Axioms 2020, 9, 131 2 of 12 The limit operator a(t) has a spectrum l t = a(t), functions l t = ib t and ( ) ( ) ( ) 1 2 b(t) l (t) = +ib (t) are associated with the presence in Equation (1) of a rapidly oscillating cos , and the function l (t) = m(t) characterizes the rapid change in the kernel of the integral operator. We introduce the following notations: l (t) = (l (t) , ..., l (t)) , 1 4 m = m , ..., m is multi-index with non-negative components m , j = 1, 4, ( ) 1 4 j jmj = m is multi-index height m, j=1 (m, l (t)) = m l (t) . j j j=1 Assume that the following conditions are met: ¥ ¥ (1) a(t), b(t), m(t) 2 C ([t , T] , R) , g(t), h(t) 2 C ([t , T] , C) , 0 0 K(t, s) 2 C ft s t T,Cg ; (2) the relations (m, l (t)) = 0, (m, l (t)) = l (t) , j 2 f1, ..., 4g for all multi-indices m with jmj 2 or are not fulﬁlled for any t 2 [t , T] , or are fulﬁlled identically on the whole segment t 2 [t , T] . In other words, resonant multi-indices are exhausted by the following sets G = m : (m, l (t)) 0, m 2,8t 2 [t , T] , f j j g 0 0 G = m : (m, l (t)) l (t) , jmj 2,8t 2 [t , T] , j = 1, 4. j j 0 Under these conditions, we will develop an algorithm for constructing a regularized [6] asymptotic solution of the problem (1). 2. Regularization of the Problem (1) i i b(t ) + b(t ) 0 0 # # Denote by s = s # independent of the t quantities s = e , s = e , and rewrite ( ) j j 1 2 the Equation (1) in the form R R t t i 0 i 0 b (q)dq + b (q)dq g(t) dz # t # t 0 0 L z(t, #) # a(t)z # e s + e s z 1 2 dt 2 (2) m(q)dq 0 e K(t, s)z(s, #)ds = h(t), z(t , #) = z , t 2 [t , T]. 0 0 We introduce regularizing variables y (t) t = l (q)dq , j = 1, 4 (3) j j # # and instead of problem (2) we consider the problem g(t) ¶z ˜ 4 ¶z ˜ t t 2 3 L z ˜ (t, t, #) # + å l (t) l (t)z ˜ # (e s + e s ) z ˜ j 1 1 ¶t j=1 ¶t 2 (4) R t t y(s) l (q)dq # s e K(t, s)z ˜ s, , # ds = h(t), z ˜(t, t, #)j = z , t 2 [t , T] t=t ,t=0 0 t # 0 for the function z ˜ = z ˜ (t, t, #) , where it is indicated (according to (3)): t = (t , ..., t ) , y = (y , ..., y ) . 1 4 1 4 y(t) It is clear that if z ˜ = z ˜ t, t, # is the solution of the problem (4), then the function z = z ˜ t, , # is an ( ) exact solution of the problem (2), therefore, the problem (4) is an extension of the problem (2). However, (4) cannot be considered completely regularized, since the integral term y(s) l (q)dq # s Jz ˜ = e K (t, s) z ˜ s, , # ds has not been regularized in it. To regularize J, we introduce a class M , asymptotically invariant with respect to the operator Jz ˜ (see [6]; p. 62). Axioms 2020, 9, 131 3 of 12 We ﬁrst consider the space U of functions z t, t , representable by sums ( ) 4 t m (m,t) z (t, t, s) = z (t, s) + z (t, s) e + z (t, s) e , å å 0 i i=1 2jmjN (5) m ¥ z (t, s) , z (t, s) , z (t, s) 2 C ([t , T] , C) , i = 1, 4, 2 jmj N 0 i 0 where the asterisk above the sum sign indicates that in it the summation forjmj 2 occurs only over nonresonant multi-indices m = (m , ..., m ) , i.e., over m 2 / G . 1 4 i i=0 Note that in (5) the degree N of the polynomial z (t, t, s) to exponentials e depends on the element z. The elements of the space U depend on bounded in # > 0 constants s = s (#) and 1 1 s = s (#), which do not affect the development of the algorithm described below, therefore in the 2 2 notation of element (5) of this space U we omit the dependence on s = (s , s ) for brevity. We show 1 2 that the class M = Uj is asymptotically invariant with respect to the operator J. t=y(t)/# The image of the operator J on the element (5) of the space U has the form: Z Z R R 4 R t t 1 s t t 1 1 l (q)dq l (q)dq l (q)dq i # t 4 4 # s # s 0 Jz t, t = e K t, s z s ds + e K t, s z s e ds+ ( ) ( ) ( ) ( ) ( ) 0 i t t 0 0 i=1 Z R R s t 1 (m,l(q))dq l (q)dq m 4 # t # s 0 + e K (t, s) z (s) e ds = 2jmjN Z R Z t t t t 1 l (q)dq l (q)dq 4 # t # s 0 = e K (t, s) z (s) ds + e K (t, s) z (s) ds+ 0 4 t t 0 0 R Z R t t s 1 1 l (q)dq (l (q)l (q))dq 4 i 4 # # t t 0 0 + e K (t, s) z (s) e ds+ å i i=1,i6=4 R Z R t s 1 t 1 l (q)dq (me ,l(q))dq 4 m 4 # t # t 0 0 + e K (t, s) z (s) e ds. 2jmjN Integrating in parts, we have Z R Z R s s t 1 t 1 K (t, s) z (s) l (q)dq l (q)dq 4 4 # t # t 0 0 J (t, #) = K (t, s) z (s) e ds = # de = 0 0 l (s) t t 0 0 4 R Z R s=t s s 1 t 1 K (t, s) z (s) l (q)dq ¶ K (t, s) z (s) l (q)dq 0 0 4 4 # t # t 0 0 = # e # e ds = l (s) ¶s l (s) 4 4 s=t 0 R Z R t t s 1 1 K (t, t) z (t) K (t, t ) z (t ) ¶ K (t, s) z (s) l (q)dq l (q)dq 0 0 0 0 0 4 4 # t # t 0 0 = # e # e ds. l (t) l (t ) ¶s l (s) 4 4 0 0 4 Continuing this process further, we obtained the decomposition l (q)dq ¥ n n+1 n 4 n # t J (t, #) = (1) # I (K (t, s) z (s)) e I (K (t, s) z (s)) , 0 0 0 n = 0 0 0 s=t s=t 1 1 ¶ 0 n n1 I = , I = I (n 1) . 0 0 0 l (s) l (s) ¶s 4 4 Next, apply the same operation to the integrals: R R t s 1 1 l (q)dq (l (q)l (q))dq 4 i 4 # t # t 0 0 J (t, #) = e K (t, s) z (s) e ds = 4,i i R R t s 1 1 l (q)dq (l (q)l (q))dq t K(t,s)z (s) 4 i 4 # t i # t 0 0 = #e de = t l (s)l (s) i 4 Axioms 2020, 9, 131 4 of 12 R R s=t t s 1 1 K (t, s) z (s) l (q)dq (l (q)l (q))dq 4 i 4 # t # t 0 0 = #e e l (s) l (s) i 4 s=t Z R t s ¶ K (t, s) z (s) (l (q)l (q))dq i i 4 # t # e ds = ¶s l s l s t ( ) ( ) i 4 R R t t 1 1 l (q)dq l (q)dq K(t,t)z (t) K(t,t )z (t ) i 4 i # t 0 i 0 # t 0 0 = # e e l (t)l (t) l (t )l (t ) i 4 i 0 4 0 R R t s 1 R 1 l (q)dq t K t,s z s (l (q)l (q))dq 4 ( ) ( ) i 4 # t ¶ i # t 0 0 #e e ds = t ¶s l (s)l (s) 0 i 4 R R t t 1 1 l (q)dq l (q)dq ¥ i 4 n n+1 n # t n # t 0 0 = (1) # I (K (t, s) z (s)) e I (K (t, s) z (s)) e , i i n=0 i s=t i s=t 1 1 ¶ 0 n n1 I = , I = I , n 1, i = 1, 3 . i i l (s) l (s) l (s) l (s) ¶s i 4 i 4 Denote bay e = (0, 0, 0, 1). Then R Z R t t s 1 1 l (q)dq (me ,l(q))dq 4 m 4 # t # t 0 0 J (t, #) = e K (t, s) z (s) e ds = R Z R t m s 1 t 1 K (t, s) z (s) l (q)dq (me ,l(q))dq 4 4 # t # t 0 0 = #e de = (m e , l (s)) 0 4 R R s=t t s 1 1 K (t, s) z (s) l (q)dq (me ,l(q))dq 4 4 # # t t 0 0 = #e [ e (m e , l (s)) s=t t s ¶ K (t, s) z (s) (me ,l(q))dq e ds] = ¶s (m e , l (s)) t 4 ¥ t (m,l(q))dq n n+1 n m # t = (1) # [ I (K (t, s) z (s)) e 4,m s=t n=0 l (q)dq n m 4 # t I (K (t, s) z (s)) e ], 4,m s=t 1 1 ¶ 0 n n1 I = , I = I , n 1, 4,m 4,m 4,m (m e , l (s)) (m e , l (s)) ¶s 4 4 2 jmj N . Here it is taken into account that (m e , l (s)) 6= 0, since by the deﬁnition of the space U multi-indices m 2 / G . The image of the operator J on the space U element (5) is represented as a series R Z R t t 1 t 1 l (q)dq l (q)dq 4 n n+1 n 4 # t # t 0 0 Jz (t, t) = e K (t, s) z (s) ds + (1) # ( I (K (t, s) z (s))) e 4 å 0 0 s=t n=0 i 4 ¥ l (q)dq n n n+1 n i # t ( I (K (t, s) z (s))) + (1) # ( I (K (t, s) z (s))) e 0 å å i 0 s=t i s=t i=1,i6=4 n=0 l (q)dq n # ( I (K (t, s) z (s))) e + i s=t ¥ t (m,l(q))dq n n+1 n m # t + (1) # I (K (t, s) z (s)) e å å 4,m s=t n=0 2jmjN l (q)dq n m # I (K (t, s) z (s)) e , t = y (t) /#. 4,m s=t 0 Axioms 2020, 9, 131 5 of 12 It is easy to show (see, for example, [23], pp. 291–294) that this series converges asymptotically for # ! +0 (uniformly in t 2 [t , T]). This means that the class M is asymptotically invariant (for # ! +0) with respect to the operator J. Let as introduce the operators R : U ! U, acting on each element z (t, t) 2 U of the form (5) according to the law: R z (t, t) = e K (t, s) z (s) ds, (6 ) 0 0 0 t 0 R z (t, t) = I (K (t, s) z (s)) e I (K (t, s) z (s)) + 1 0 0 0 0 s=t s=t 0 t 0 t i 4 + I K t, s z s e I K t, s z s e + 6 ( ( ) ( )) ( ( ) ( )) ( ) i i 1 å i i s=t s=t i=1 0 m (m,t) 0 m t + I (K (t, s) z (s)) e I (K (t, s) z (s)) e , å 4,m 4,m s=t s=t 2jmjN h i n t n R z (t, t) = ( I (K (t, s) z (s))) e ( I (K (t, s) z (s))) + n+1 0 0 0 s=t 0 s=t 3 h i n n t n t i 4 + (1) ( I (K (t, s) z (s))) e ( I (K (t, s) z (s))) e + (6 ) å i i n+1 i i s=t s=t i=1 h i n m (m,t) n m t + I (K (t, s) z (s)) e I (K (t, s) z (s)) e , n 1. 4,m 4,m s=t s=t 2jmjN Let now z ˜ (t, t, #) be an arbitrary continuous function in (t, t) 2 [t , T] t : Ret 0, j = 1, 4 0 j with the asymptotic expansion z ˜ (t, t, #) = # z (t, t) , z (t, t) 2 U, (7) å k k k=0 converging as # ! +0 (uniformly in t, t 2 t , T t : Ret 0, j = 1, 4 ). Then the image ( ) [ ] 0 j Jz ˜ (t, t, #) of this function is expanded in the asymptotic series ¥ ¥ r k r Jz ˜ t, t, # = # Jz t, t = # R z t, t j . ( ) ( ) ( ) rs s å k å å t=y(t)/# k=0 r=0 s=0 This equality is the basis for introducing the extension of the operator J on the series type (7): ¥ ¥ r de f k r ˜ ˜ Jz ˜ t, t, # J # z t, t = # R z t, t . ( ) ( ) ( ) rs s å k å å k=0 r=0 s=0 Although the operator J is formally deﬁned, its usefulness is obvious, since in practice they usually construct the N-th approximation of the asymptotic solution of problem (2), in which only the N-th partial sums of the series (7) will take part, which do not have a formal but true meaning. Now we can write down a problem that is completely regularized with respect to the original problem (2): g(t) ¶z ˜ 4 ¶z ˜ t t 2 3 ˜ L z ˜ (t, t, #) # + l (t) l (t)z ˜ # (e s + e s ) z ˜ Jz ˜ = h(t), j 1 1 2 j=1 2 ¶t ¶t (8) z ˜(t, t, #)j = z , t 2 [t , T]. t=t ,t=0 0 0 Axioms 2020, 9, 131 6 of 12 3. Iterative Problems and Their Solvability in the Space U Substituting series (7) into (8) and equating the coefﬁcients for the same powers #, we obtain the following iterative problems: ¶z 0 0 L z (t, t) l (t) l (t)z R z = h (t) , z (t , 0) = z ; (9 ) 0 1 0 0 0 0 0 0 å j ¶t j=1 ¶z g(t) t t 2 3 L z (t, t) = + (e s + e s ) z + R z , z (t , 0) = 0; (9 ) 1 1 2 0 1 0 1 0 1 ¶t 2 ¶z g(t) t t 2 3 L z (t, t) = + (e s + e s ) z + R z + R z , z (t , 0) = 0; (9 ) 2 1 2 1 1 1 2 0 0 0 2 ¶t 2 ¶z g(t) k1 t t 2 3 L z (t, t) = + (e s + e s ) z + R z + ... + R z , z (t , 0) = 0, k 1. (9 ) 1 2 0 1 0 k k1 k k1 k k ¶t 2 Each of the iterative problems can be written as ¶z L z (t, t) l (t) l (t)z R z = H (t, t) , z (t , 0) = z , (10) 1 0 0 å j ¶t j=1 t m (m,t) where H (t, t) = H (t) + H (t) e + H (t) e is the known function of the space å å 0 i i=1 2jmjN U, z is the known number of complex the space C, and the operator R has the form (see (6 )) 0 0 0 1 de f t m (m,t) t @ i A 4 R z R z (t) + z (t) e + z (t) e = e K (t, s) z (s) ds. 0 0 0 å i å 4 i=1 2jmjN We introduce the scalar product (for each t 2 [t , T]) in the space U : t m (m,t) < z, w >< z (t) + z (t) e + z (t) e , å i å i=1 2jmjN de f t m (m,t) w (t) + w (t) e + w (t) e > = 0 å i å i=1 2jmjN de f m m = z t , w t + z t , w t + z t , w t , ( ( ) ( )) ( ( ) ( )) ( ( ) ( )) 0 0 å i i å i=1 2jmjmin(N ,N ) z w where ( , ) we denote the ordinary scalar product in the complex space C: (u, v) = u v ¯. We prove the following statement. Theorem 1. Suppose that conditions (1) and (2) are satisﬁed and the right-hand side H (t, t) = H (t) + 4 t m (m,t) + H (t) e + H (t) e of the Equation (10) belongs to the space U. Then for the å å i=1 2jmjN solvability of the Equation (10) in U it is necessary and sufﬁcient that the identities < H (t, t) , e > 0, 8t 2 [t , T] (11) hold true. Axioms 2020, 9, 131 7 of 12 Proof. We will determine the solution of the Equation (10) in the form of an element (5) of the space U: t m m,t ( ) z (t, t) = z (t) + z (t) e + z (t) e . (12) å i å i=1 2jmjN Substituting (12) into the Equation (10), we have t m (m,t) l t z t + l t l t z t e + m, l t l t z t e ( ) ( ) [ ( ) ( )] ( ) [( ( )) ( )] ( ) 1 0 å i 1 i å 1 i=1 2jmjN t t m (m,t) e K (t, s) z (s) ds = H (t) + H (t) e + H (t) e . 4 0 å i å i=1 2jmjN Equating here separately the free terms and coefﬁcients at the same exponents, we obtained the following equations: l (t) z (t) = H (t) , (13 ) 1 0 0 0 [l (t) l (t)] z (t) = H (t) , i = 1, 3; (13 ) i i i i l t l t z t K t, s z s ds = H t ; 13 [ ( ) ( )] ( ) ( ) ( ) ( ) ( ) 4 1 4 4 4 4 m m [(m, l (t)) l (t)] z (t) = H (t) , m 2 / G , 2 m N . (13 ) j j 1 1 H m Since the function l (t) 6= 0 8t 2 [t , T], the Equation (13 ) has a unique solution 0 0 z (t) = l (t) H (t). Since l (t) l (t) 6= 0 8t 2 [t , T], then the Equation (13 ) can be written as 0 0 4 1 0 4 1 1 z (t) = [l (t) l (t)] K (t, s) z (s) ds [l (t) l (t)] H (t) . (14) 4 4 1 4 4 1 4 Due to the smoothness of the kernel l t l t K t, s and heterogeneity [ ( ) ( )] ( ) 4 1 [l (t) l (t)] H (t), this Volterra integral equation has a unique solution 4 1 4 z t 2 C t , T , C . The Equations 13 and 13 also have unique solutions ( ) ([ ] ) ( ) ( ) 4 0 2 3 z (t) = [l (t) l (t)] H (t) 2 C ([t , T] , C) , i = 2, 3, i i 1 i 0 since l (t) 6= l (t) , i = 2, 3. The Equation (13 ) is solvable in the space C ([t , T] , C) if and only i 1 1 0 if identities ( H (t) , e ) 0 8t 2 [t , T] hold. It is easy to see that this identity coincides with 1 0 identity (11). Further, since (m, l (t)) 6= l (t) , 2 jmj N (8m 2 / G ), then the Equation (13 ) has a 1 H 1 m unique solution m m ¥ z (t) = [(m, l (t)) I A (t)] H (t) 2 C ([t , T] , C) , 2 jmj N . 0 H Thus, condition (11) is necessary and sufﬁcient for the solvability of the Equation (10) in the space U. The Theorem 1 is proved. Remark 1. If identity (11) holds, then under conditions (1) and (2), the Equation (10) has the following solution in the space U : 4 t m (m,t) t i 1 z (t, t) = z (t) + å z (t) e + å z (t) e z (t) + a (t) e + 0 0 1 i=1 2jmjN (15) t t t m (m,t) 2 3 4 +h (t)e + h (t)e + z t e + P t e , ( ) å ( ) 21 31 4 2jmjN H Axioms 2020, 9, 131 8 of 12 where a t 2 C t , T , C are arbitrary function, z t = l (t) H (t), z t is the solution of the ( ) ([ ] ) ( ) ( ) 1 0 0 0 4 integral Equation (14), and introduced notations H (t) H (t) 2 3 m 1 m h (t) , h (t) , P t m, l t l t H t . ( ) [( ( )) ( )] ( ) 21 31 1 l (t) l (t) l (t) l (t) 2 1 3 1 4. The Remainder Term Theorem Along with problem (10), we consider the equation ¶z g (t) t t 2 3 L w (t, t) = + (e s + e s ) z + R z + Q (t, t) , (16) 1 1 ¶t 2 where z = z (t, t) is the solution (15) of Equation (10), Q (t, t) 2 U is the known function of the space U (this form will have problems (9 ) after calculating the solution of the problem (9 ) in U). The right k+1 k side of this equation: ¶z g (t) t t 2 3 G (t, t) + (e s + e s ) z + R z + Q (t, t) = 1 2 1 ¶t 2 2 3 t m (m,t) 4 5 = z (t) + z (t) e + z (t) e + å i å ¶t i=1 2jmjN 2 3 g (t) t t t m (m,t) 2 3 4 i 5 + (e s + e s ) z (t) + z (t) e + z (t) e + R z + Q (t, t) , 1 2 0 å i å 1 i=1 2jmjN may not belong to the space U, if z = z (t, t) 2 U. Indeed, taking into account the form (15) of function z = z (t, t) 2 U, we consider in G (t, t) , for example, the terms g t ( ) t t t m (m,t) 2 3 i 5 Z (t, t) (e s + e s ) z (t) + z (t) e + z (t) e = 1 2 0 å i å i=1 2 m N j j g (t) g (t) t t t +t t +t 2 3 i 2 i 3 = z (t) (e s + e s ) + z (t) e s + e s + 0 1 2 å i 1 2 2 2 i=1 g (t) t t m (m,t) 2 3 + (e s + e s ) P (t) e . 1 2 å 2jmjN Function Z (t, t) 2 / U, since it contains resonant exponentials t +t (m,t) t +(m,t) t +(m,t) 2 3 2 3 e = e j , e (m + 1 = m ) , e (m + 1 = m ) , and, therefore, 2 3 3 2 m=(0,1,1,0) the right-hand side G (t, t) = Z (t, t) of the Equation (16) also does not belong to the U. Then, according to the well-known theory (see [6], p. 234), we need to embed^: G (t, t) ! G (t, t) the right-hand side G (t, t) of the Equation (16) into the space U. This operation is defined as follows. m (m,t) Let the function G (t, t) = w (t) e contain resonant exponentials, i.e., G (t, t), it has jmj=0 the form 4 4 N N j j t m m ,t m (m,t) ( ) G (t, t) = w (t) + w (t) e + w (t) e + w (t) e . 0 å i å å å j j j i=1 j=0 m =2:m 2G jmj=2,m6=m ,j=0,4 j j j Then 4 4 N N t m t m (m,t) ˆ i G (t, t) = w (t) + w (t) e + w (t) e + w (t) e . 0 å i å å å j j j i=1 j=0 jmj=2,m6=m ,j=0,4 jm j=2: m 2G j Axioms 2020, 9, 131 9 of 12 Therefore, the embedding operation acts only on the resonant exponentials and replaces them with a unit or exponents e of the ﬁrst dimension according to the rule: ^ ^ (m,t) 0 (m,t) t e j = e = 1, e j = e , j = 1, 4. m2G m2G 0 j Therefore, the right-hand sides of iterative problems (9 ) (if they solve sequentially) may not belong to the space U. Then, according to [6] (p. 234), the right-hand sides of these problems must be embedded in U according to the above rule. As a result, we obtained the following problems: ¶z Lz t, t l t A(t)z R z = h t , z t , 0 = z ; (9 ) ( ) ( ) ( ) ( ) 0 j 0 0 0 0 0 0 ¶t j=1 ¶z g(t) t t 2 3 Lz (t, t) = + (e s + e s ) z + R z , z (t , 0) = 0; (9 ) 1 1 2 0 1 0 1 0 1 ¶t 2 ¶z g(t) 1 t t 2 3 Lz t, t = + e s + e s z + R z + R z , z t , 0 = 0; (9 ) ( ) ( ) ( ) 2 1 2 1 1 1 2 0 0 0 2 ¶t 2 h i ¶z g(t) k1 t t 2 3 Lz (t, t) = + (e s + e s ) z + R z + ... + R z , k 1 2 k1 k 0 1 k1 ¶t (9 ) z (t , 0) = 0, k 1 k 0 (images of linear operators and R do not need to be embedded in the space U, since these operators ¶t act from U to U). Such a replacement will not affect the construction of an asymptotic solution to y(t) the original problem (1) (or its equivalent problem (2)), since the narrowing t = of the series of problems 9 will coincide with the series of problems (9 ) (see [6], pp. 234–235). k k It is easy to show that applying Theorem 1 to iterative problems 9 , we can ﬁnd their solutions uniquely in the space U. As a result, we can construct series (7) with coefﬁcients z (t, t) 2 U. As in [23] (pp. 303–308), we proved the following statement. Theorem 2. Suppose that conditions (1)–(2) are satisﬁed for the Equation (2). Then, when # 2 (0, # ](# > 0 0 0 is sufﬁciently small) the Equation (2) has a unique solution z(t, #) 2 C ([t , T],C); at the same time there is the estimate N+1 jjz(t, #) z (t)jj c # , 8N = 0, 1, 2, . . . , #N N C[t ,T] y(t) where z (t) is the narrowing (for t = ) N-th partial sum of the series (7) (with coefﬁcients z (t, t) 2 U #N k satisfying the iterative problems (9 )), and the constant c > 0 does not depend # on # 2 (0, # ]. k N 0 5. Construction of the Solution of the First Iteration Problem in the Space U Using Theorem 1, we will tried to ﬁnd a solution to the ﬁrst iterative problem 9 . Since the right-hand side h (t) of the equation 9 satisﬁes condition (11), this equation has (according to (15)) a solution in the space U in the form (0) (0) z (t, t) = z (t) + a (t) e , (17) 0 1 (0) (0) h(t) where a (t) 2 C ([t , T] , C) are arbitrary function, z (t) = . Subordinating (17) to the 1 0 l (t) initial condition z (t , 0) = z , we have 0 0 (0) (0) (0) 0 0 1 z (t ) + a (t ) = z , a (t ) = z + l (t ) h (t ) . 0 0 0 0 0 0 1 1 1 Axioms 2020, 9, 131 10 of 12 (0) To fully calculate the function a (t), we pass to the next iterative problem 9 . Substituting the solution (17) of the equation 9 , into it, we arrived at the following equation: (0) K (t, t) z (t) d d (0) (0) t 0 t 1 4 L z (t, t) = z (t) a (t) e + e 0 1 dt dt l (t) (0) K (t, t ) z (t ) g(t) 0 0 (0) (0) 0 t t t 2 3 1 + e s + e s z t + a t e + (18) ( ) ( ) ( ) 1 2 0 1 l (t ) 2 4 0 (0) (0) K (t, t ) a (t ) K (t, t) a (t) 0 0 + e , l t l t ( ) ( ) 1 1 0 (here we used the expression (6 ) for R z (t, t) and took into account that when z (t, t) = z (t, t) in 1 1 0 t t 1 4 the sum (6 ) only terms with e and remain e ). Let us calculate g(t) (0) (0) t t t 2 3 1 M = (e s + e s ) z (t) + a (t) e = 1 2 h i (0) (0) (0) (0) t +t t +t t t 2 1 3 1 2 3 = g (t) s a (t) e + s a (t) e + s z (t) e + s z (t) e . 1 2 1 2 1 1 0 0 Let us analyze the exponents of the second dimension included here for their resonance: R R t t 1 0 1 0 (ib (q)+a(q))dq (+ib (q)+a(q))dq t +t # t +t # t t 2 1 3 1 0 0 e j = e , e j = e , t=y(t)/# t=y(t)/# 2 2 0, 0, 6 6 a, a, 6 6 6 6 0 0 0 0 ib + a = 6 ib , , Æ; +ib + a = 6 ib , , Æ. 6 6 0 0 4 4 +ib , +ib , m, m, t +t t +t 2 1 3 1 Thus, exponents e ang e are not resonant. Then, for solvability the Equation (18) it is necessary and sufﬁcient that the condition (0) d K (t, t) a (t) (0) a (t) + = 0 dt l (t) (0) is satisﬁed. Attaching the initial condition a (t ) = z + l (t ) h (t ) , to this equation, we found 0 0 0 1 1 uniquely the function K (s, s) 0 0 ( ) ( ) a (t) = a (t ) exp ds , 1 1 l (s) 0 1 and therefore, we uniquely calculate the solution (17) of the problem 9 in the space U. In this case, the leading term of the asymptotics of the solution to the problem (2) has the form Z R t t K (s, s) l (q)dq (0) (0) # t z (t) = z (t) + a (t ) exp ds e , #0 0 0 1 l (s) 0 1 (0) (0) 0 1 where a (t ) = z + A (t ) h (t ) , z (t) = l (t ) h (t) . 0 0 0 0 1 1 Example. Consider a model problem 2(ts) dz t + t # = z # cos z e t s z(s, #)ds + h(t), z(t , #) = z , t 2 [t , T](t 0), (19) 0 0 0 dt # 0 Axioms 2020, 9, 131 11 of 12 were a(t) 1, m(t) 2, b(t) t + t, K(t, s) t s. The main term of the asymptotic solution of this problem has the form 3 3 t t t t 0 0 z (t) = h(t) + [z h(t )] exp [ ]ex p[ ]. (20) #0 0 3 # For # ! +0 the function z (t) tends to the solution of the degenerate equation z + h(t) = 0 #0 uniformly on any interval [t + d, T](0 < d T t ) and at the point t = t takes on the value 0 0 0 z (t ) = z . It is seen from (20) that the leading term of the asymptotics of the solution to problem (19) #0 0 t +t does not depend on cos and spectral value m(t) 2, but depends on the kernel K(t, s) t s . y(t) Further calculations show that already the asymptotic solution z (t) = z (t) + #z t, of the ﬁrst #1 #0 1 order will depend on both m(t) 2, and the frequency b (t) = 2t + 1 of the rapidly oscillating cosine. 6. Conclusions The function z (t) shows that when passing from a differential equation of type (1) (K(t, s) 0) #0 to an integro-differential one (K(t, s) 6= 0 ), the main term of the asymptotic is inﬂuenced by the kernel K(t, s) of the integral operator. However, the main term of the asymptotics is not affected by the spectral values of the integral operator m(t) and rapidly oscillating coefﬁcients. Their effects are detected when constructing the next approximation z (t). #1 Author Contributions: All authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript. Funding: This work was supported by grant No. AP05133858 of the Ministry of Education and Science of the Republic of Kazakhstan. Conﬂicts of Interest: The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, or in the decision to publish the results. References 1. Feschenko, S.F.; Shkil, N.I.; Nikolenko, L.D. Asymptotic Methods in the Theory of Linear Differential Equations; Naukova Dumka: Kiev, Ukraine, 1966. 2. Shkil, N.I. Asymptotic Methods in Differential Equations; Naukova Dumka: Kiev, Ukraine, 1971. 3. Daletsky, Y.L.; Krein, S.G. On differential equations in Hilbert space. Ukr. Math. J. 1950, 2, 71–91. 4. Daletsky, Y.L. The asymptotic method for some differential equations with oscillating coefﬁcients. DAN USSR 1962, 143, 1026–1029. 5. Daletsky, Y.L.; Krein M.G. 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A problem with inverse time for a singularly perturbed integro-differential equation with diagonal degeneration of the kernel of high order. Izv. Math. 2016, 80, 285–298. [CrossRef] 22. Bobodzhanov, A.A.; Kalimbetov, B.T.; Safonov, V.F. Integro-differential problem about parametric amplification and its asymptotical integration. Int. J. Appl. Math. 2020, 33, 331–353. [CrossRef] 23. Safonov, V.F.; Bobodzhanov, A.A. Course of Higher Mathematics. Singularly Perturbed Equations and the Regularization Method: Textbook; Publishing House of MPEI: Moscow, Russia, 2012. Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional afﬁliations. 2020 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

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**Published: ** Nov 13, 2020

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