Solutions of nonlinear differential equations

Transkript

1 Nonlinear Differ. Equ. Appl. 17 (21), c 29 Birkhäuser Verlag Basel/Switzerland /1/ published online December 11, 29 DOI 1.17/s Nonlinear Differential Equations and Applications NoDEA Solutions of nonlinear differential equations Nadzeya Bedziuk and Aleh Yablonski Abstract. We consider an ordinary nonlinear differential equation with generalized coefficients as an equation in differentials in the algebra of new generalized functions. The solution of such an equation is a new generalized function. In this article we formulate necessary and sufficient conditions for when the solution of the given equation in the algebra of new generalized functions is associated with an ordinary function. Moreover, a class of all possible associated functions is described. Mathematics Subject Classification (2). Primary 34A36; Secondary 46F3. Keywords. Algebra of new generalized functions, Differential equations with generalized coefficients, Functions of finite variation. 1. Introduction The dynamics of many real systems or phenomena can be described by nonlinear differential equations with generalized coefficients. Unfortunately, the theory of generalized functions allows one to only formulate such equations and it is inapplicable to solving them. Therefore different interpretations of the solution of nonlinear differential equations were proposed by many mathematicians. In general, different interpretations of the same equation lead to different solutions see, e.g., [2,6,8,14,15,18]. In order to choose an adequate interpretation of the given equation, one has to consider additional reasons which arise from the modeling of the real dynamics of the system. In this paper we consider the following nonlinear equation with generalized coefficients ẋ(t) =f(t, x(t)) L(t), (1.1) where t [a; b] R and L(t) is a derivative in the distributional sense. Generally, since L(t) is a distribution and the function f(t, x(t)) is not smooth, the product f(t, x(t)) L(t) is not well defined and the solution of Eq. (1.1) essentially depends on the interpretation.

2 25 N. Bedziuk and A. Yablonski NoDEA We investigate Eq. (1.1) by using the algebra of mnemofunctions (new generalized functions). It is worth mentioning that the first algebra of new generalized functions was proposed by Colombeau [5]. Definitions of other algebras can be found in [7,16]. The general methods of construction of such algebras were proposed by Antonevich and Radyno [1]. In this paper we interpret Eq. (1.1) as an equation in differentials in the algebra of new generalized functions from [1]. Such an approach allows us to investigate ordinary and stochastic differential equations from a common ground [11, 12]. The algebraic interpretation states that the solution of Eq. (1.1) is an element of the algebra of new generalized functions. However, a problem arises with finding conditions on coefficients of equation in differentials which allow us to associate an ordinary function with the solution of this equation. If such an ordinary function exists, we call it a solution of Eq. (1.1). The sufficient conditions under which the solution exists and the set of possible solutions in this sense were described in [17]. In this paper we prove necessary conditions for some classes of coefficients. 2. The algebra of mnemofunctions In this section we recall the definition of the algebra of mnemofunctions from [1] (see also [11, 17]). First we define an extended real line R using a construction typical for non-standard analysis. Let R = {(x n ) n=1 : x n R for all n N} be a set of real sequences. We call two sequences {x n } R and {y n } R equivalent if there is a natural number N such that x n = y n for all n>n. The set R of equivalence classes is called the extended real line and any of the classes a generalized real number. It follows easily that R R, as one may associate with any ordinary number x R a class containing a stationary sequence x n = x. The product xỹ of two generalized real numbers is defined as the class of sequences equivalent to the sequence {x n y n }, where {x n } and {y n } are the arbitrary representatives of the classes x and ỹ respectively. It is evident that R is an algebra. For any segment T =[a; b] R one can construct an extended segment T in a similar way. Consider the set of sequences of infinitely differentiable functions {f n (x)} on R. We call two sequences {f n (x)} and {g n (x)} equivalent if there is a natural number N such that f n (x) =g n (x) for all n>n and x R. The set of classes of equivalent functions is denoted by G(R) and its elements are called mnemofunctions. Similarly, one can define the space G(T) for any interval T =[a; b]. If we endow all these spaces with the natural operations of addition and multiplication they become algebras. For each distribution f we can construct a sequence f n of smooth functions such that f n converges to f (e.g., one can consider the convolution of f with some δ-sequence). This sequence defines the mnemofunction which corresponds to the distribution f. Thus, the space of distributions is a subset

3 Vol. 17 (21) Solutions of nonlinear DE 251 of the algebra of mnemofunctions. However, in this setting, the infinite set of mnemofunctions corresponds to one distribution (e.g., by taking different δ-sequences). We say that the mnemofunction f =[{f n }] is associated with a function f from some topological space if f n converges to f in this space. Let f =[{f n (x)}] and g =[{g n (x)}] be mnemofunctions. Then there is a composition defined by f g =[{f n (g n (x))}] G(R). In the same way one can define the value of the mnemofunction f at the generalized real point x =[{x n }] R as f( x) =[{f n (x n )}]. Let H denote the subset of R of the nonnegative infinitely small numbers : H = { h R : h =[{h n }], h n >, lim h n =}. For each h =[{h n }] H and f =[{f n (x)}] G(R) we define a differential d h f G(R) byd h f =[{fn (t + h n ) f n (t)}]. The construction of the differential was proposed by Lazakovich [1]. 3. Main results In this section we formulate the main results of this article. Using the introduced algebras we can now give an interpretation of Eq. (1.1). We replace the ordinary functions in Eq. (1.1) by the corresponding mnemofunctions and then write the equation in differentials in the algebra G(T) d h X( t) = f( t, X( t))d h L( t), (3.1) with initial value X [ã; h) = X, where h = [{h n }] H, ã = [{a}] T, t =[{t n }] T, X =[{Xn }], f =[{fn }], X =[{Xn}], L =[{Ln }] are elements of G(T). Moreover, f and L are associated with f and L, respectively. If X is associated with some function X, then we say that X is a solution of Eq. (1.1). It was shown in [11] that under some minor restrictions on the initial conditions there exists a unique solution of Eq. (3.1). The purpose of the present paper is to investigate when the solution X of Eq. (3.1) converges to some ordinary function and to describe all possible limits. Let L(t), t T =[a; b] be a right-continuous function of finite variation. We assume that L(t) = L(b) ift>band L(t) = L(a) ift<a. Denote the total variation of function L on the interval [u; v] T by Vu v L. Suppose that f is a Lipschitz continuous function with constants M 1 and M 2 with respect to t and x, respectively, and that it also has bounded growth, i.e., there exists a constant K such that for all x R and t T f(t, x) K(1 + x ). (3.2)

4 252 N. Bedziuk and A. Yablonski NoDEA In this paper we consider specific types of representatives of the mnemofunctions. We take the following convolutions with δ-sequence as representatives of L from Eq. (3.1) L n (t) =(L ρ n )(t) = 1/n L(t + s)ρ n (s)ds, (3.3) where ρ n C (R), ρ n, supp ρ n [; 1/n] and 1/n ρ n (s)ds =1. In the same way we set f n (t, x) =(f ρ n )(t, x) = f(t + u, x + v) ρ n (u, v)dudv, (3.4) [,1/n] 2 where ρ n C (R 2 ), ρ n, supp ρ n [, 1/n] 2, ρ [,1/n] 2 n (u, v)dudv =1. By using representatives we can rewrite Eq. (3.1) as follows: { xn (t + h n ) x n (t) =f n (t, x n (t))(l n (t + h n ) L n (t)), x n [a;a+hn)(t) =x (3.5) n(t). The solution of Eq. (3.5) is constructed inductively starting from the interval [a; a + h n ) where the initial conditions are given. Let t be an arbitrary point in T. There exists m t N and τ t [a; a + h n ) such that t = τ t + m t h n. Set t k = τ t + kh n, k =, 1,...,m t. Then the solution of Eq. (3.5) is defined explicitly as m t 1 x n (t) =x n(τ t )+ f n (t k,x n (t k ))(L n (t k+1 ) L n (t k )). (3.6) k= The solution x of Eq. (3.1) is associated with some function if and only if the sequence of the solutions x n of Eq. (3.5) converges. Therefore, we have to investigate a limiting behavior of the sequence x n. Consider the function F n :[ ;+ ] [; 1] given by F n (x) = 1/n x ρ n (s)ds. (3.7) Since ρ n (s), then F n is a non-increasing function, F n (x) 1and F n (+ ) =,F n ( ) = 1. Denote the inverse function of F n by Fn 1, i.e., : [; 1] [ ;+ ] and F 1 n Fn 1 (u) =sup{x : F n (x) =u}. (3.8) In order to describe the limits of the sequence x n, we consider the integral equation t x(t)=x + f(s, x(s))dl c (s)+ (ϕ( L(s)f(s, ),x(s ), 1) x(s )), a a<s t (3.9) where t T, L c is a continuous part of the function L, L(s) =L(s+) L(s ) is a size of the jump of the function L at the point s, and for any function z

5 Vol. 17 (21) Solutions of nonlinear DE 253 and x R, ϕ(z,x,u) denotes the solution of the integral equation ϕ(z,x,u) =x + z(ϕ(z,x,v))µ(dv), (3.1) [;u) where µ(du) is a probability measure defined on the Borel subsets of the interval [; 1]. As it was shown in [17] there exists a unique solution of Eq. (3.9) iff is a Lipschitz continuous function. Definition 3.1. We say that the function σ : [; 1] [; 1] belongs to class G if there is a system of pairwise disjoint intervals (a i ; b i ] [; 1], i I such that σ(u) = { bi, u (a i ; b i ], u, u / i I (a i; b i ]. (3.11) The following theorems describe the limits of the sequence X n. Theorem 3.2 ([3]). Let f be a Lipschitz continuous function satisfying Eq. (3.2), L is a right-continuous function of finite variation. Suppose that t T x n(τ t ) x dt and F n (Fn 1 (u) δh n ) σ(u) as n and h n for all δ (; 1) and for any continuity point u [; 1] of σ. Thenσ belongs to the class G and x n (t) x(t) dt T as n and h n, wherex n (t) is a solution of Eq. (3.5) and x(t) is a solution of the Eq. (3.9) with the measure µ generated by the function σ. Theorem 3.3. Let L be a right-continuous function of finite variation. Suppose that for each Lipschitz continuous function f satisfying Eq. (3.2), the solution of Eq. (3.5) x n converges in L 1 (T) as n, h n. If the function L is continuous, then the limit of x n is a solution of Eq. (3.9). IfL is discontinuous, then there exists a function σ G such that F n (Fn 1 (u) δh n ) σ(u) as n, h n, for all δ (, 1) and for any continuity point u [; 1] of σ, and the limit of x n is a solution of Eq. (3.9) with the measure µ generated by the function σ. 4. Auxiliary statements Suppose that for any t T and n N the partition of the interval [; 1] is given by = ξ n (t) ξ1 n (t) ξp+2(t) n = 1, where p depends on n. Consider the recurrent sequence ϕ n k (t) =ϕn k (z,x,t) fort T,k =, 1,...,p+2,n N,x R, z is a Lipschitz continuous function, { ϕ n k+1 (t) =ϕ n k (t)+z(ϕn k (t))(ξn k+1 (t) ξn k (t)) ϕ n (4.1) (t) =x.

6 254 N. Bedziuk and A. Yablonski NoDEA For each u [; 1], t T, n =1, 2,..., define σ n (u, t) andφ n (u, t) as follows { ξ σ n n (u, t) = k (t),ξk 1 n (t) <u ξn k (t) (4.2),u= { ϕ φ n n (u, t) = k (t),ξk 1 n (t) <u ξn k (t) (4.3) x, u =. It is easy to see that φ n is a solution of the following integral equation, φ n (u, t) =x + z(φ n (s, t))σ n (ds, t). [;u) Lemma 4.1 ([17]). Suppose that there is a nonincreasing left-continuous function σ(u),u [; 1] such that σ n (u, t) σ(u) dt as n for any continuity point u of σ. Then σ belongs to the class G and φ n (u, t) φ(u) dt T T as n for any continuity point u of φ, whereφ(u) is a solution of the equation φ(u) =x + z(φ(s))dσ(s). [;u) For any fixed point t, let us denote φ n (u, t) andσ n (u, t) asφ n (u) and σ n (u), respectively. Lemma 4.2. Suppose that there exists an x R such that for each Lipschitz continuous function z : R R of bounded growth number sequence φ n (1) = ϕ n p n (z,x) converges as n. Then the sequence of measures generated by the functions σ n converges weakly. Proof. By the definition of σ n (u) it is evident that σ n (u) u. Therefore, we can represent the interval [; 1] as a union of the disjoint sets A, B, andc, where { } A = u : lim σn (u) =u, { } B = u : lim σn (u) lim σ n (u) >u, { } C = u : lim σn (u) > lim σ n (u) =u. Let us consider Lipschitz continuous functions z q,ε for q, ε Q, ε> defined as follows: 1, p x + q 1 z q,ε (p) = ε (x + q + ε p), x+ q<p x + q + ε, p > x+ q + ε.

7 Vol. 17 (21) Solutions of nonlinear DE 255 Consider an arbitrary point u B. By the definition of the set B we have that lim σ n (u) =b>u.foreachn N there exists a number m such that ξ n m 1 <u ξ n m. Then for large enough n we have σ n (u) =ξ n m (b + u)/2. In this case for each function z q,ε such that u q<q+ ε (b + u)/2, we conclude that φ n (1) = x + ξ n m. Since φ n (1) converges, σ n (u) =ξ n m converges as well. At the same time convergence of σ n (u) leads to the following equality: lim σn (u) = lim σ n (u) =b>u. (4.4) Consider an arbitrary point u C. By the definition of the set C we have that u = lim σ n (u) < lim σ n (u) =d. Aslim σ n (u) =d, there is a subsequence {n k } such that lim nk σ n k (u) =d. Suppose that there exists a subsequence {n i } such that lim ni σ ni (u) =s, where u<s<d.butifwe take the subsequence {n k } {n i } then the same arguments as in (4.4) imply that lim ni σ ni (u) =b, which leads to the contradiction with the definition of n i. Thus, we can represent the sequence {n} as a disjoint union {n } {n }, where lim n σ n (u) =d and lim n σ n (u) =u. According to Helly s selection principle [4], σ n is a weakly compact sequence, which means that from each subsequence of σ n one can select a subsequence that converges weakly. Let us show that the whole sequence σ n also converges weakly. Suppose that there exist two subsequences converging to different functions: σ n σ, σ n σ and σ σ. Using Lemma 4.1, we obtain that σ and σ belong to the class G, which implies that u = 1 is a continuity point for both σ and σ.moreover,φ n (u) φ (u) andφ n (u) φ (u) for each continuity point of σ and σ, respectively. In addition, φ and φ satisfy the integral equations φ (u)=x+ z(φ (s))dσ (s), [;u) φ (u)=x+ z(φ (s))dσ (s). [;u) (4.5) According to the conditions of the lemma φ n (1) converges, therefore φ (1) = φ (1). Let us fix an arbitrary continuity point u [; 1] of both σ and σ such that σ (u) σ (u) and suppose that σ (u) <σ (u). Since σ (u) and σ (u) are two different limits of subsequences of σ n (u), then u belongs to the set C. As it was shown above in this case, the sequence {n} is the disjoint union {n } {n }, where lim n σ n (u) =d and lim n σ n (u) =u. Consequently σ (u) =u and σ (u) =d>u. Now consider Eq. (4.5) for the function z = z q,ε, where q, ε Q and u q < q + ε < d. Denote their solutions by φ q,ε and φ q,ε, respectively. Taking wq,ε = min{1, inf{t q : σ (t) ε}} one can write φ q,ε explicitly as

8 256 N. Bedziuk and A. Yablonski NoDEA x + σ (t), if t q, x + q + ε εe (σ (t) q)/ε (1 σ (s)/ε)e σ (s)/ε, if q<t wq,ε, φ q,ε(t) = q s<t x + q + ε εe (σ (w q,ε ) q)/ε e σ (w q,ε )/ε (1 σ (s)/ε)e σ (s)/ε, if t>wq,ε. q s w q,ε Letting ε in the given formula, we obtain that wq,ε wq = min(1, inf{t q : σ (t) > }) andwq,ε wq.moreover, { x + σ φ q(1) = lim φ ε q,ε(1) = (q), if q<wq x + σ (wq)+ σ (wq), if q = wq. Similarly, one can obtain φ u(1) = lim q u φ q(1) = { x + σ (u), if u<w u x + σ (w u)+ σ (w u), if u = w u. Since u is a continuity point of σ, then σ (wu)=asu = wu and φ u(1) = x + σ (u) =x + u. Let us consider φ q,ε. Taking into account that ε d u, we obtain that φ q,ε(1) = x + σ (u) =x + d. Letting ε, and then q u we obtain that φ u (1) = x + d. Thus, it is shown that φ u(1) = x + u and φ u (1) = x + d, where u d. According to the conditions of the lemma φ q,ε(1) = φ q,ε(1). Therefore, φ u(1) = φ u (1), which leads to the contradiction. Hence, σ n (u) σ(u) for all continuity points u [; 1] of σ. Remark 4.3. Lemma 4.2 holds under weaker conditions: the sequence of measures generated by the functions σ n converges weakly if the number sequence φ n (1) = ϕ n p n (z,x) converges for a countable set of functions z q,ε of the given type, where q, ε Q. Let j be a number such that t j 1/n < t j+1. Set ξk n(t) =F n( t j+k ), t T, n N and k =, 1,...,p+ 2, where p =[1/(nh n )]. Note that ξk n(t) depends on t T, since t j+k depends on t. Denote the sequence which is defined by formula (4.3) byϕ n k (z,x,t) and let σn i (u, t) be the sequence of functions which is given by formula (4.2). The following lemma from [13] gives the necessary and sufficient conditions for σi n (u, t) to satisfy the statement of Lemma 4.1 and Lemma 4.2. Lemma 4.4 ([13]). Suppose that F n (Fn 1 (u) δh n ) σ(u) as n and h n for all δ (; 1) and all continuity points u [; 1] of σ. Then, σ n (u, t) σ(u) dt T as n and h n for all continuity points u [; 1] of σ. Conversely, if there exists such a function σ(u, t), u [; 1], t T that σ(u, t) L 1 (T) for any u [; 1], σ(u, t) is left continuous and nondecreasing

9 Vol. 17 (21) Solutions of nonlinear DE 257 on u for any t T, and for each continuous function z : [; 1] R, we have that 1 1 z(u)σ n (du, t) z(u)σ(du, t) dt, T Then σ(u, t) does not depend on t, i.e., σ(u, t) = σ(u) and F n (Fn 1 (u) δh n ) σ(u) for all continuity points u [; 1] of σ and for any δ (; 1). Proposition 4.5. Let ϕ(z,x,u) be a solution of Eq. (3.1) with real-valued function z such that z(x) z(y) K 1 x y and z(x) K 2 (1 + x ) for all x, y R. Then the following inequalities hold for all x, y R, u, v [; 1], u<v ϕ(z,x,u) ϕ(z,y,u) x y exp(k 1 ) ϕ(z,x,u) ( x + K 2 )exp(k 2 ) ϕ(z,x,u) x K 2 ( x +1)exp(K 2 ) ϕ(z,x,u) x ϕ(z,y,u)+y x y K 1 exp(k 1 ) ϕ(z,x,u) ϕ(z,x,v) K 2 (1 + ( x + K 2 )exp(k 2 ))µ([u, v)) Proof. By using the Gronwall inequality, see [9], these statements are easily shown from the definition of ϕ(z,x,u). Hereafter we denote the constant which depends only on M 2, K, T, and Va b L by C. It does not depend on n, h n and t T and its value can change from one formula to another. Proposition 4.6. Let the function f be Lipschitz continuous with constant M 2 satisfying Eq. (3.2). Then for the solutions x and x n of Eqs. (3.9) and (3.5), respectively, the following inequalities hold for all t, s T, t>sand l, n N x(t) C(1 + x ) (4.6) x n (t) C(1 + x n (τ t ) ) (4.7) x(t) x(s) C(1 + x )V t s L x n (t + lh n ) x n (t) C(1 + x n (τ t ) t+lh )V n+ 1 n t L (4.8) Proof. The proof of these inequalities is standard and uses the definitions of x and x n, the Gronwall inequality, inequality (3.2), and the Lipschitz continuity of f. 5. Proof of the main theorem In this section we prove Theorem 3.3. Let us recall the conditions of the theorem. The function L is right-continuous of finite variation. For any Lipschitz continuous function f of bounded growth, the solution of Eq. (3.5) x n converges in L 1 (T)asn, h n. Let us show that if L has no discontinuity points, then the limit of x n is a solution of the integral equation (3.9); and if L has at least one point of discontinuity, then there exists a function σ G such that F n (Fn 1 (u) δh n ) σ(u) as

10 258 N. Bedziuk and A. Yablonski NoDEA n, h n, for all δ (, 1) and for all continuity points u [, 1] of σ. Moreover, the limit of x n is a solution of the integral equation (3.9) with the measure µ generated by σ. Proposition 5.1. Under the conditions of Theorem 3.3 there exists an x R such that x n (τ t ) x dt. (5.1) T Proof. As was shown above, each point t can be represented as t = τ t + m t h n, where τ t [a, a + h n ], m t N and the solution of Eq. (3.5) can be written as follows m t 1 x n (t) =x n (τ t )+ f n (t k,x n (t k ))[L n (t k+1 ) L n (t k )]. (3.6) k= Set f, then x n (t) =x n (τ t )=x n(τ t ). According to the conditions of Theorem 3.3, x n converges in L 1 (T) and therefore x n(τ t ) converges. Taking an arbitrary function g C(T), consider the following limit I = lim b g(t)x n(τ t )dt = lim [ b a hn ] a+khn g(t)x n a k=1 a+(k 1)h n Directly from the definition of fractional part we obtain I = lim 1 [ b a hn ] ( { }) t a a + h n dt. h n g(h n s + a +(k 1)h n )h n x n(a + h n s)ds. k=1 Since the function g is bounded on T then by the Lebesgue dominated convergence theorem we have I = lim 1 b a g(t)dt x n(a + h n s)ds = b Denote x = lim 1 x n(a + h n s)ds, then a g(t)dt lim 1 x n(a + h n s)ds. I = b a g(t)x dt. Since x n(τ t ) converges in L 1 (T), then x n(τ t ) x in L 1 (T). Remark 5.2. Without loss of generality we can say that the initial conditions x n(τ t ) are bounded, i.e., there exists a constant C R such that x n(τ t ) C for any t T, n N. If the initial conditions x n(τ t ) are unbounded, we can construct new bounded initial conditions such that the solution of Eq. (3.5) with the new initial conditions converges to the same limit as the solution x n (t) ofeq.(3.5) with unbounded initial conditions. Proposition 5.3. Theorem 3.3 holds for the continuous function L.

11 Vol. 17 (21) Solutions of nonlinear DE 259 Proof. Let y be a solution of the integral equation (3.9). Since L is continuous, then L c = L and Eq. (3.9) can be represented as follows y(t) =x + t a f(s, y(s))dl c (s), where x is a limit of the initial conditions x n(τ t ). According to Proposition 5.1, the value x exists and it is unique. We will show that b a x n(t) y(t) dt as. Then, due to the uniqueness of the limit, x will coincide with y. Taking into account the explicit forms of x n and y, inequalities (4.6), (4.7), and also properties of f n and f with the help of standard methods one can show that x n (t) y(t) x n (τ t ) x m t 1 + f n (t k,x n (t k ))[L c n(t k+1 ) L c n(t k )] k= x n (τ t ) x + C/n + C(1 + x n (τ t ) + x ) m t 1 + C x n (t k ) y(t k ) [L c (t k+1 ) L c (t k )]. k= t a f(s, y(s))dl c (s) sup u v h n+1/n V v u L c Applying the Gronwall inequality to the above expression we obtain that [ x x n (t) y(t) n (τ t ) x + C/n + C(1 + x n (τ t ) + x ) sup Vu v L c] exp(cva b L c ). u v h n+1/n Taking the integral over T and considering the limit as n we obtain that b a x n (t) y(t) dt. Thus Theorem 3.3 holds for the continuous function L. Now let us show that Theorem 3.3 holds for the discontinuous function L. By the conditions of Theorem 3.3, the solution of Eq. (3.5) converges for any Lipschitz continuous function f of bounded growth. But, of course, for two different functions f the solutions of Eq. (3.5) differ. In order to emphasize that x n substantially depends on the function f, we denote it as x n = x n (f), and its value at point t as x n (t) =x n (f,t). Thus, we need to show that if for any Lipschitz continuous function f of bounded growth the function sequence x n (f) converges in L 1 (T), then F n (Fn 1 (u) δh n ) converges for all δ (, 1) and almost all (in general, for all except for no more than countable number of points) u [, 1].

12 26 N. Bedziuk and A. Yablonski NoDEA For the arbitrary fixed Lipschitz continuous function f, using Remark 5.2 it can be shown that inequality (4.7) takes the form x n (t) C(1+C ). Introduce the function f(t, C(1 + C )), if x>c(1 + C ), f(t, x) = f(t, x), if x C(1 + C ), f(t, C(1 + C )), if x< C(1 + C ). As f coincides with f for all x n (f,t), it means that the solution x n (f) of Eq. (3.5) will always coincide with the solution x n ( f). Therefore, hereafter we can regard f as a bounded function with respect to x. Combining this condition with inequality (Eq. 3.2) we obtain that f(t, x) K(1 + C(1 + C )). (5.2) For any t T define t k = t k (t) =τ t + kh n, k =, 1, 2,... as above. Pick any point T. Letq N be a number such that t q 1 (t) < t q (t). (5.3) Note that t q (t + h n )=t q (t), i.e., t q (t) is a periodic function with period h n. Proposition 5.4. For any Lipschitz continuous function f of bounded growth and for each T there exists x + () R such that lim b x n (t q (t)) x + () dt =. (5.4) Proof. For each δ (; 1) we introduce the following function 1, if t + δ/2, g(δ,, t) = 2(δ + t)/δ, if + δ/2 <t + δ,, if t>+ δ. Set f δ (t, x) =f(t, x)g(δ,, t), then { f δ f(t, x), if t + δ/2, (t, x) =, if t + δ. Moreover, the function f δ is Lipschitz continuous with respect to x and it satisfies (Eq. 3.2) with the same constants as f does. Using the argument preceding inequality (5.2), we claim that this inequality holds for the function f. Then function f δ is also Lipschitz continuous with respect to t. Let x δ n = x n (f δ )andx n = x n (f) be solutions of Eq. (3.5) with functions f δ and f, respectively. Note that x δ n(t) =x n (t) ift<+ δ/2 1/n and x δ n(t q (t)) = x n (t q (t)) for large enough n and t T. Moreover,x δ n(t) =c δ n(t) ift>+ δ where c δ n is some h n -periodic function. Now x δ n converges to x δ in L 1 (T), therefore the sequence of functions c δ n also converges in L 1 ([ + δ, b]). Denote the limit of c δ n by c δ. The same arguments as in Proposition 5.1 imply that the function c δ does not depend on t, i.e., it is a constant. Let us show that the number sequence c δ converges as δ. Consider arbitrary numbers δ 1,δ 2 (, 1) and corresponding functions f δ1, f δ2. Note that fn δ1 (t, x) =fn δ2 (t, x) only if t ( + min{δ 1,δ 2 }/2 1/n, + max{δ 1,δ 2 }].

13 Vol. 17 (21) Solutions of nonlinear DE 261 Let r and r be numbers such that t r 1 + min{δ 1,δ 2 }/2 1/n < t r t r + max{δ 1,δ 2 } <t r +1. Then from equality (3.6) weget x δ 1 n (t) x δ2 n (t) mt 1 = = k= [f δ1 r [f δ1 k=r n (t k,x δ1 n (t k )) fn δ2 (t k,x δ2 n (t k ))][L n (t k+1 ) L n (t k )] n (t k,x δ1 n (t k )) fn δ2 (t k,x δ2 n (t k ))][L n (t k+1 ) L n (t k )]. Using inequality (4.8) with the help of standard methods it can be shown that x δ 1 n (t) x δ2 n (t) C(1 + xn (t r ) )V +max{δ1,δ2}+hn 1/n+min{δ L 1,δ 2}/2 n C(1 + x n (τ t ) )V +max{δ 1,δ 2}+h n+1/n 1/n+min{δ 1,δ 2}/2 L. Taking the integral over the interval [ + max{δ 1,δ 2 },b] and letting n we obtain that c δ 1 c δ2 (b max{δ1,δ 2 }) = lim b +max{δ 1,δ 2} x δ 1 n (t) x δ2 n (t) dt b lim C (1 + x n(τ t ) )dt V +max{δ1,δ2}+hn+1/n 1/n+min{δ 1,δ 2}/2 L +max{δ 1,δ 2} C(1 + x +max{δ )V 1,δ 2} +min{δ L. 1,δ 2}/2 Since L is a right-continuous function, the above inequality implies that c δ is a Cauchy sequence. Thus, there exists lim δ c δ = x + (). Now let us recall that c δ n is h n -periodic function and x δ n(t q (t)) = x n (t q (t)) for large enough n. Then x n (t q (t)) c δ n(t q (t)) = x δ n(t q (t)) c δ n(t). Since for t>+ δ we have that f δ (t, x) =andx δ n(t) =c δ n(t), then using inequality (4.7) we deduce that x δ n(t q (t)) c δ n(t) = x δ n(t q (t)) x δ n(t) C(1 + x δ n (t q (t)) +δ+h )V n+1/n L C(1 + x n (τ t ) +δ+h )V n+1/n L. Thus, we obtain the following inequality x n (t q (t)) x + () x δ n(t q (t)) c δ n(t) + c δ n(t) x + () C(1 + x n (τ t ) +δ+h )V n+1/n L + c δ n(t) x + ().

14 262 N. Bedziuk and A. Yablonski NoDEA Taking the integral over the interval [ + δ, b] and letting n we have that lim b + lim +δ b x n (t q (t)) x + () dt C(1 + x )V +δ L +δ c δ n(t) c δ dt + c δ x + () (b δ) = C(1 + x )V +δ L + c δ x + () (b δ). Letting δ we obtain the required equality b lim x n (t q (t)) x + () dt =. Now for t T and m N let t j (t, m) =τ t + jh n be a point such that t j (t, m) < 1 m t j+1(t, m). (5.5) Notice that t j (t + h n,m)=t j (t, m) for all t T. Proposition 5.5. For any Lipschitz continuous function f of bounded growth and for each T there exists an x () such that b lim x n (t j (t, n)) x () dt =. (5.6) Proof. Applying equality (5.4) at the point 1 m we have that b ( lim x n (t j+1 (t, m)) x + 1 ) dt =. (5.7) m Now we show that the sequence x + ( 1 m ) converges as m.itis sufficient to prove that x + ( 1 m ) x+ ( 1 k ) asm, k. Let us introduce k, m N such that k<mand δ< 1 mk.letfδ,m (t, x) = f(t, x)g(δ, 1 m,t) and let xδ,m n the function f δ,m. Due to the special form of δ we have that x δ,m n = x n (f δ,m ) be a solution of Eq. (3.5) with (t) =x δ,k n (t) if t< 1/k 1/n. Moreover, as it was shown above, there exists a number c δ ( 1 m ) such that It is easy to see that ( cδ 1 k lim b n (t) c δ ( 1 m ) dt =. xδ,m ) c δ ( 1 m ) = 1 b lim b x δ,m n (t) x δ,k n (t) dt.

15 Vol. 17 (21) Solutions of nonlinear DE 263 Equality (3.6) implies that x δ,m n (t) x δ,k n (t) mt 1 = [fn δ,m i= (t i,x δ,m n [L n (t i+1 ) L n (t i )]. (t i )) fn δ,k (t i,x δ,k n (t i ))] Since δ< 1 δ,m mk, we have that fn = fn δ,k if t ( 1 k 1 n ; 1 m + 1 mk ). Denote the first point that belongs to this interval by t r and the last one by t r. Then using inequality (4.8) and the same technique as previously we have that x δ,m n (t) x δ,k n (t) r = [fn δ,m i=r (t i,x δ,m n C(1 + x n(τ t ) k 1 )V (t i )) fn δ,k (t i,x δ,k mk +hn+ 1 n 1 k 1 n L. n (t i ))][L n (t i+1 ) L n (t i )] After taking the integral over [,b] and letting n we obtain that ( cδ 1 ) ( c δ 1 ) = 1 k m b lim b x δ,m n C(1 + x )V k 1 mk L. 1 k (t) x δ,k n (t) dt Recall that c δ (u) x + (u) asδ. Therefore, the given inequality implies that ( x+ 1 m ) x + ( 1 k ) C(1 + x )V k 1 mk L. 1 k Thus, ( x + m) ( ) 1 x + 1 k asm, k. Therefore, the sequence x ( ) + 1 k converges to the limit x (). From equality (5.7) it now follows that lim m lim b x n (t j+1 (t, m)) x () dt =. We will now calculate lim b x n(t j (t, n)) x () dt. It is evident that for each m and large enough n we have for all t T that t j+1 (t, m) <t j (t, n).

16 264 N. Bedziuk and A. Yablonski NoDEA Equation (3.6) yields b lim x n (t j (t, n)) x () dt lim m lim b + lim m lim lim m lim b b x n (t j (t, n)) x n (t j+1 (t, m)) dt x n (t j+1 (t, m)) x () dt C(1+ x n (τ t ) )V t j(t,n)+ 1 n t j+1(t,m) Ldt lim C(1+ x )V L=, m 1 m since L is a right-continuous function. Thus, b lim x n (t j (t, n)) x () dt =. Proposition 5.6. Let T be a discontinuity point of L, then ϕ n p+2( L()f(, ),x (),t) x + () in L 1 (T) as n, (5.8) where x + () and x () are defined as in Propositions 5.4 and 5.5, p =[1/(nh n )], and the function ϕ n k (z,x,t) is defined by Eq. (4.1) with ξn k (t) = F n ( t j+k (t)). Proof. Recall that t q = t q (t) andt j = t j (t, n) are defined by the inequalities (5.3) and (5.5), respectively. As p =[1/(nh n )], then p+1 x n (t j+p+2 ) x n (t j )= f n (t j+i,x n (t j+i ))(L n (t j+i+1 ) L n (t j+i )). i= It is easy to see that t j+p+2 is equal to t q or t q+1. Therefore, equalities (5.4) and (5.6) imply that as n, p+1 I n = f n (t j+i,x n (t j+i ))(L n (t j+i+1 ) L n (t j+i )) x + () x () inl 1 (T). i= (5.9) Now let us represent the function L in the following form L = L + L d, where { L d, if t<, (t) = L(), if t.

17 Vol. 17 (21) Solutions of nonlinear DE 265 Then, we have that L d n(t j+i+1 ) L d n(t j+i )= = tj+i+1 t j+i+1 1/n 1/n (L d (t j+i+1 + s) L d (t j+i + s))ρ n (s)ds (L d ( u) L d ( h n u))ρ n ( t j+i+1 u)du = J. From definition of L d one can see that L d ( u) L d ( h n u) =if u [ h n ; ). Then since supp ρ n [; 1/n] we have that J = h n (L d ( u) L d ( h n u))ρ n ( t j+i+1 u)du tj+i = L() ρ n ( t j+i+1 u)du = L() ρ n (s)ds h n t j+i+1 = L()[F n ( t j+i+1 ) F n ( t j+i )] = L()[ξi+1(t) n ξi n (t)]. Applying the methods described in Lemma 5.6 from [17] and using an explicit form of f n, L n,andϕ n p+2 one can show that p+1 f n (t j+i,x n (t j+i ))(L n (t j+i+1 ) L n (t j+i )) i= (ϕ n p+2( L()f(, ),x n (t j ),t) x n (t j )) C(1 + x n(τ t ) )(1/n + h n + V 1/n L + V +1/n+hn L). Therefore, formula (5.9) yields ϕ n p+2( L()f(, ),x n (t j ),t) x n (t j ) x + () x () inl 1 (T) asn. Using Proposition 4.5 we obtain that ϕ n p+2 ( L()f(, ),x n (t j ),t) x n (t j ) ϕ n p+2( L()f(, ),x (),t)+x () C xn (t j ) x (). Since from (5.6) we have that x n (t j ) x () inl 1 (T), then ϕ n p+2( L()f(, ),x (),t) x + () inl 1 (T) asn.

18 266 N. Bedziuk and A. Yablonski NoDEA Proposition 5.7. The sequence of functions ϕ n p+2( L()f(, ),x,t) converges in L 1 (T). Proof. Consider the following set of functions, if t δ, w(δ,, t) = (t + δ )/δ, if δ<t, 1, if t>. Introduce the function g δ (t, x) =f(t, x)w(δ,, t). As in Proposition 5.4, weuse the argument preceding inequality (5.2) and claim that this inequality holds for the function f. Then the function g δ (t, x) is Lipschitz continuous with respect to both x and t. Denote the solution x n (g δ )ofeq.(3.5) byx δ n.asit was shown in Proposition 5.5, x δ n(t j ) x (,g δ ). Moreover, using equality (3.6) one can show that x δ n(t j ) x n(τ t ) C(1 + x n(τ t ) )V tj+1/n δ L. Taking the integral over T and letting n we get x (,g δ ) x C(1 + x )V δ L. Thus, x (,g δ ) converges to x as δ. Let us show that x + (,g δ ) also converges. Since L()f(, ) = L() g δ (, ), formula (5.8) implies that ϕ n p+2( L()f(, ),x (,g δ ),t) x + (,g δ ). Using the definition of ϕ n we obtain the following inequality ϕ n p+2( L()f(, ),x (,g δ1 ),t) ϕ n p+2( L()f(, ),x (,g δ2 ),t) C x (,g δ1 ) x (,g δ2 ). Therefore, x + (,g δ1 ) x + (,g δ2 ) C x (,g δ1 ) x (,g δ2 ). (5.1) Since the sequence x (,g δ ) converges, then from (5.1) we obtain that the sequence x + (,g δ ) also converges. Denote its limit by x + (,f ). The definition of ϕ n implies ϕ n p+2( L()f(, ),x,t) x + (,f ) ϕ n p+2( L()f(, ),x,t) ϕ n p+2( L()f(, ),x (,g δ ),t) + ϕ n p+2( L()f(, ),x (,g δ ),t) x + (,g δ ) + x + (,g δ ) x + (,f ) C x (,g δ ) x + ϕ n p+2( L()f(, ),x (,g δ ),t) x + (,g δ ) + x + (,g δ ) x + (,f ). Taking the integral over the interval [; b] and letting n we have that lim b ϕ n p+2( L()f(, ),x,t) x + (,f ) dt C x (,g δ ) x + C x + (,g δ ) x + (,f ). Letting δ we obtain that ϕ n p+2( L()f(, ),x,t) converges in L 1 (T).

19 Vol. 17 (21) Solutions of nonlinear DE 267 Proof of the theorem. Proposition 5.3 states that Theorem 3.3 holds for the continuous function L. For the discontinuous function L, let us introduce the following notation, ϕ n p+2( L()f(, ),x,t)=φ n (1,t), where ϕ n p+2 and ξk n(t) are defined in Proposition 5.6. Then the function σn (s, t) is generated by ξk n(t) accordingto(4.2) andφn is a solution of the corresponding integral equation φ n (u, t) =x + L()f(,φ n (s, t))σ n (ds, t). [;u) Proposition 5.7 implies that there exists a number x R such that for any Lipschitz continuous function f of bounded growth and for any discontinuity point of L, the sequence φ n (1,t) converges in L 1 (T). Since for each f the sequence φ n (1,t) converges in L 1 (T), there exists a subsequence n k, depending on f, such that φ n k (1,t) converges almost everywhere. Fix t T in such a way that φ n k (1,t) converges for a countable number of functions f = z q,ε, q, ε Q defined in Lemma 4.2. Denote φ n k (1) = φ n k (1,t) and σ n k (u) =σ n k (u, t). Thus, we showed that there exists a sequence n k,for which the sequence φ n k (1) converges for any functions z q,ε.fromremark4.3 we deduce that Lemma 4.2 holds. Lemma 4.2 in turn implies that the sequence of functions σ n k converges to σ G for any continuity point u of σ. This means that the conditions of Lemma 4.4 also hold, so we obtain that F nk (Fn 1 k (u) δh nk ) σ(u), (5.11) as n k, h nk for all δ (, 1) and for each continuity point u of σ. It is easy to see that for each subsequence n one can select a subsequence n k such that convergence (5.11) holds. Therefore, (5.11) holds for the whole sequence, i.e., F n (Fn 1 (u) δh n ) σ(u) (5.12) as n, h n, for all δ (, 1) and for all continuity points u [, 1] of σ. Formula (5.12) means that the conditions of Theorem 3.2 hold. Applying Theorem 3.2 we obtain that the limit of the solutions of Eq. (3.5) is a solution of the integral equation (3.9) with measure µ generated by σ, where σ G. The proof is complete. 6. Remarks Corollary 6.1. Let the δ-sequence ρ n be of the simplest type, which means that ρ n (t) =nρ(nt), whereρ C (R), ρ, supp ρ [; 1] and 1 ρ(s)ds =1. Then the sequence F n (Fn 1 (u) δh n ) converges at any continuity point of the limit function σ for any δ (; 1) and the limit does not depend on δ if and only if either 1/n = o(h n ) or h n = o(1/n).

20 268 N. Bedziuk and A. Yablonski NoDEA Moreover, suppose that f is a Lipschitz continuous function of bounded growth and t T x n(τ t ) x dt as n and h n. Then the solution x n of Eq. (3.5) converges to x(t) in L 1 (T) if and only if 1. 1/n = o(h n ), which implies that the measure µ, generated by σ, gives the mass 1 to the point. Consequently Eq. (3.1) has the following solution { x, u =, ϕ(z,x,u) = x + z(x), u (; 1]. Thus x(t) is a solution of Eq. (3.9) that can be represented in the following form x(t) =x + f(s, x(s ))dl(s). (a,t] 2. h n = o(1/n), which means that the measure µ, generated by σ, is equal to the Lebesgue measure. Hence, Eq. (3.1) can be written as follows ϕ(z,x,u) = z(ϕ(z,x,u)), u ϕ(z,x,) = x. Thus, x(t) is a solution of Eq. (3.9) with the function ϕ defined above. 7. Conclusion We demonstrated that the algebra of mnemofunctions enables us to define the solution of the differential equation with generalized coefficients (Eq. 1.1). Considering specific types of mnemofunctions (which are restricted to be a convolution of corresponding generalized coefficients with a δ-sequence) allowed us to give necessary and sufficient conditions under which the solution is an ordinary function satisfying the integral equation (3.9). It is worth mentioning that each δ-sequence define a measure µ, which is involved in the formulation of Eq. (3.9). In particular, existing interpretations of the solution of Eq. (1.1) from [2,6,8,14,15,18] can be obtained by using different δ-sequences which define appropriate measures µ. Acknowledgements We thank professor N.V. Lazakovich for his encouragement and interest, Dr. Ingemar Eriksson for the attentive reading, and anonymous reviewers for their constructive comments, which helped us improve the manuscript. References [1] Antonevich, A.B., Radyno, Y.V.: On a general method for constructing algebras of generalized functions. Sov. Math. Dokl. 43(3), (1991)

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22 27 N. Bedziuk and A. Yablonski NoDEA N. Bedziuk, A. Yablonski Department of Functional Analysis, Belarusian State University, Nezavisimosti 4, 225 Minsk, Belarus A. Yablonski Received: 19 December 28. Accepted: 25 November 29.