A singular initial value problem for some functional differential equations

International Journal of Stochastic Analysis, Jul 2018

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A singular initial value problem for some functional differential equations

Hindawi Publishing Corporation Journal of Applied Mathematics and Stochastic Analysis RAVI P. AGARWAL 0 DONAL O'REGAN 0 OLEKSANDR E. ZERNOV 0 0 Donal O'Regan: Department of Mathematics, National University of Ireland , Galway , Ireland E-mail address: For the initial value problem tr x (t) = at + b1x(t) + b2x(q1t) + b3tr x (q2t) + ?(t, x(t), x(q1t), x (t), x (q2t)), x(0) = 0, where r > 1, 0 < qi 1, i ? {1, 2}, we find a nonempty set of continuously differentiable solutions x : (0, ?] ? R, each of which possesses nice asymptotic properties when t ? +0. In this paper, we will consider singular initial value problems for a class of functional differential equations (FDEs). This problem has received very little attention in the literature to date; we refer the reader to [2, 3, 5, 7] where the question of solvability (in various spaces) and the number of solutions have been discussed. We remark that, even now, the asymptotic properties of solutions of FDEs are only partially understood. Our approach to this problem is essentially different from others known in the literature. In particular, we use qualitative methods [1, 7, 8] together with standard fixed point theorems [6]. Our technique leads to the existence of continuously differentiable solutions with nice asymptotic properties. where t ? (0, ?) is a real variable, x : (0, ?) ? R is a real unknown function, r, a, b1, b2, b3, q1, q2 are constants, r > 1, b1 = 0, 0 < qi 1, i ? {1, 2}, and ? : ? R is a continuous + ? t; x(t); x q1t; x (t); x q2t 1. Introduction 2. Main result Consider the initial value problem tr x (t) = at + b1x(t) + b2x q1t + b3tr x q2t function; here ? t, y1, y2, y3, y4 ? ? s, y1, y2, y3, y4 ? t, y1, y2, y3, y4 ? ? t, z1, z2, z3, z4 = t, y1, y2, y3, y4 : t ? (0, ?), y1 < ?1t, y2 < ?1q1t, y3 < ?2t1?r , y4 < ?2 q2t 1?r . In what follows, we will assume that the following conditions are satisfied: |a| < ?1 b1 + b2q1 , Further, we will assume that where c = ?a(b1 + b2q1)?1, ? : (0, ?) ? (0, +?) is a continuously differentiable function and, moreover, l1(?)|t ? s|, t, y1, y2, y3, y4 ? , s, y1, y2, y3, y4 ? , 0 < ? t, 0 < ? s, l2(t) y1 ? z1 + y2 ? z2 + tr y3 ? z3 + tr y4 ? z4 , t, y1, y2, y3, y4 ? , t, z1, z2, z3, z4 ? , where l1 : (0, ?) ? (0, +?) is a continuous nonincreasing function, l2 : (0, ?) ? (0, +?) is a continuous nondecreasing function, limt?+0 l2(t) = 0, and (2.3) (2.4) (2.5) (2.6) (2.7) (2.8) lim ?(t) = 0, t?+0 0 tl?im+0 tr?1 ?(t) ?1 = L1, tl?im+0 t? (t) ?(t) ?1 = L2, Li < +?, i ? {1, 2}. Definition 2.1. For any ? ? (0, ?), a continuously differentiable function x : (0, ?] ? R is said to be ?-solution of problem (2.1), (2.2) if (1) (t, x(t), x(q1t), x (t), x (q2t)) ? for all t ? (0, ?]; (2) x identically satisfies the differential equation (2.1) for all t ? (0, ?]. We denote by (?, M) the set of all continuously differentiable functions u : (0, ?] ? R such that |u(t) ? ct| Mt?(t), t ? (0, ?]; here ?, M are (positive) constants, ? < ?. The main result of this paper is the following theorem. Theorem 2.2. There exist constants ?, M such that (a) if b1 > 0, then problem (2.1), (2.2) possesses an infinite set of ?-solutions x : (0, ?] ? R, each of which belongs to (?, M). Moreover, for any constant ? which satisfies the condition |? ? c?| < M??(?), there exists at least one ?-solution x? ? (?, M) such that x?(?) = ?; 0 < ? < 1miin9 ?i, where all the ?i are defined below (we choose ? small enough, M and Q large enough so that the choice of Q, M, and ? ensures the validity of all the reasoning given below). Let be the space of continuously differentiable functions x : [0, ?] ? R with the norm Let ? be such that for every element u ? , u : [0, ?] ? R satisfies the inequalities x = max t?[0,?] x(t) + x (t) . u(t) ? ctr+1 Mtr+1?(t), u (t) ? (r + 1)ctr QMt?(t), t ? (0, ?], (b) if b1 < 0, then problem (2.1), (2.2) possesses at least one ?-solution x : (0, ?] ? R which belongs to (?, M). Proof. We can choose constants Q and M (cf. (2.6)) so that 2 b1 < Q < q2r?1 b3 ?1 Next, we choose ? such that and u(0) = 0, u (0) = 0; moreover, ?? > 0, ?u ? , ?ti ? [0, ?], i ? {1, 2} : t1 ? t2 ?(?) =? u t1 ?u t2 where ?(?) = ?(8B(t?))?1; here B(t?) = l1(t?) + t??r , t? is a constant such that t? ? (0, ?), and if t ? (0, t?], then the following inequalities are satisfied: QMt?(t) ? , 33 (1 + r)|c|tr ? 33 . It is easy to see that is a closed, bounded, and convex set. Moreover, (use the Arzela? theorem). There exists a (small enough) ?1 > 0 such that if ? < ?1, then is a compact set t, u(t)t?r , u q1t q1t ?r , u (t)t?r ? ru(t)t?r?1, u q2t q2t ?r ? ru q2t q2t ?r?1 ? (2.15) for all u ? , t ? (0, ?]. (2.9) (2.10) (2.11) (2.12) ?, (2.13) (2.14) Let x = yt?r , where y : (0, ?) ? R is a new unknown function. Then we obtain the following initial value problem: y (t) = at + b1 y(t)t?r + r y(t)t?1 + b2 y q1t q1t ?r + b3q2?r y q2t ? rb3q2?r?1t?1 y q2t + ? t, y(t)t?r , y q1t q1t ?r , y (t)t?r ? r y(t)t?r?1, (2.16) y q2t q2t ?r ? r y q2t q2t ?r?1 , Now we will consider the initial value problem y (t) = at + b1 y(t)t?r + r y(t)t?1 + b2u q1t q1t ?r + b3q2?r u q2t ? rb3q2?r?1t?1u q2t + ? t, u(t)t?r , u q1t q1t ?r , u (t)t?r ? ru(t)t?r?1, u q2t q2t ?r ? ru q2t q2t ?r?1 , y(0) = 0, where u ? is an arbitrary but fixed function. Let 0 = (t, y) : t ? (0, ?], y ? R . In 0, for (2.17), conditions for the existence and uniqueness theorem [4] and conditions for the continuous dependence of the initial data theorem [4] are fulfilled. Now we denote Let the function A1 : 0 ? [0, +?) be defined by the equality ?1 = (t, y) : t ? (0, ?], y ? ctr+1 = Mtr+1?(t) , 1 = (t, y) : t ? (0, ?], y ? ctr+1 < Mtr+1?(t) , H = (t, y) : t = ?, y ? c?r+1 < M?r+1?(?) . A1(t, y) = y ? ctr+1 2 tr+1?(t) ?2 and let a1 : 0 ? R be the derivative of the function A1. Using (2.17), we have a1(t, y) = 2 tr+1?(t) ?2t?r b1 ? tr?1 1 + t? (t) ?(t) ?1 y ? ctr+1 2 + y ? ctr+1 ?1(t) , (2.17) (2.18) (2.19) (2.20) (2.21) (2.22) b3 Mtr+1?(t) q1 b2 + Q qr?1 + 2 b3 ? 1 c L1 + 1 M + ?1(t) , tl?im+0 ?1(t) = 0. There exists a (small enough) ?2 > 0 (see our choice of M at the beginning of the proof) such that if ? < ?2, then ?1(t) < Mtr+1?(t) b1 , t ? (0, ?]. Since Mtr+1?(t) = |y ? ctr+1| when (t, y) ? ?1, it is easy to see that sign a1(t, y) = sign b1 when (t, y) ? ?1. (1) Let b1 > 0 and therefore a1(t, y) > 0 when (t, y) ? ?1. We prove that the integral curve of (2.17) which intersects ?1 at an arbitrary point (t0, y0) for sufficiently small |t ? t0| (t ?) lies in 1 if t < t0 and lies outside of 1 if t > t0. In fact, let P(t0, x0) be an arbitrary point belonging to ?1 and let JP : (t, yP(t)) be the integral curve of (2.17) which passes through the point P. Then A1 t0, yP t0 = M2, a1 t0, yP t0 > 0. Therefore, if t0 ? (0, ?), then there exists ? > 0 such that where But this means that (t, yP(t)) lies in 1 if t ? (t0 ? ?, t0) and (t, yP(t)) lies outside of 1 if t ? (t0, t0 + ?). If t0 = ?, then there exists ? > 0 such that A1 t, yP(t) < A1 ?, yP(?) , t ? (? ? ?, ?), yP(t) ? ctr+1 tr+1?(t) ?1 < M, t ? (? ? ?, ?), and this means that (t, yP(t)) lies in 1, t ? (? ? ?, ?). It follows from this that each of the integral curves of (2.17) which intersect H remains in 1 for all t ? (0, ?]. We consider an arbitrary point G(?, yG) ? H and let Ju : (t, yu(t)) be the integral curve of (2.17) such that yu(?) = yG. It is easy to see that yu(t) ? ctr+1 Mtr+1?(t), t ? (0, ?], yu(t) ? (r + 1)ctr 2 b1 + ?2(t) Mt?(t), tl?im+0 ?2(t) = 0. There exists a (small enough) ?3 > 0 such that if ? < ?3 (see our choice of Q at the beginning of the proof), then Now we set (b) If ti ? [t?, ?], i ? {1, 2}, and |t1 ? t2| (small enough) ?4 > 0 such that if ? < ?4, then Moreover, there exists a (small enough) ?5 > 0 such that if ? < ?5, then Therefore, ?. Moreover, since q2ti ? [0, ?], i ? {1, 2}, and we have, at the same time, |u (q2t1) ? u (q2t2)| ?. Thus, q2t1 ? q2t2 = q2 t1 ? t2 q2?(?) ?(?), yu t1 ? yu t2 8? + 58? + B t? ?(?) = 78? < ?. (2.33) (2.34) (2.35) (2.36) (2.37) (2.38) (2.39) (2.40) (2.41) (c) If t1 ? [0, t?], t2 ? [t?, ?], or vice versa, and |t1 ? t2| longing to [0, t?], t?, t2 belonging to [t?, ?], and |t? ? t2| Therefore, in view of (2.35) and (2.41), . (2) Let b1 < 0 and therefore a1(t, y) < 0 when (t, y) ? ?1. Similarly, as in the case b1 > 0, it can be proved that if an integral curve of (2.17) intersects ?1 at an arbitrary point (t0, y0), then for sufficiently small |t ? t0| (t ?), the curve indicated lies in 1 if t > t0 and it lies outside of 1 if t < t0. It follows that at least one of the integral curves of (2.17) which intersect H is defined for all t ? (0, ?] and lies in 1 if t ? (0, ?]. In fact, if an integral curve of (2.17) intersects ?1, then it has to intersect H. Let the mapping ? : ?1 ? H be defined in the following way: the point ?(P) ? H is assigned to the point P ? ?1 if both of these points belong to a common integral curve of (2.17). We denote ?(?1) = {?(P) : P ? ?1}. The set H\?(?1) is a nonempty one (H is a closed set, but ?(?1) is a nonclosed set since ?(?1) is the image of the nonclosed set ?1). Let Ju : (t, yu(t)) be an integral curve of (2.17) such that (?, yu(?)) ? H\?(?1). Then Ju : (t, yu(t)) cannot have common points with ?1. Therefore, Ju : (t, yu(t)) is defined for all t ? (0, ?] and it lies in 1 if t ? (0, ?]. As in the case b1 > 0, we can prove that the inequalities (2.31), (2.33) are fulfilled, and we can introduce notation (2.34) and prove that condition (2.13) is fulfilled. Thus, yu ? . Now we will prove in the case b1 < 0 that if t ? +0, then all integral curves of (2.17) leave the set 1\{(0, 0)}, with one exception, Ju : (t, yu(t)). Let b1 < 0; we denote ?2(?) = (t, y) : t ? (0, ?], y ? yu(t) = ?tr+1?(t)(? ln t) , 2(?) = (t, y) : t ? (0, ?], y ? yu(t) < ?tr+1?(t)(? ln t) , where ? is a parameter, ? ? (0, 1]. Let the function A2 : 0 ? [0, +?) be defined by the equality A2(t, y) = y ? yu(t) 2 tr+1?(t)(? ln t) ?2 and let a2 : 0 ? R be the derivative of the function A2. Using (2.17), we have a2(t, y) = 2 tr+1?(t)(? ln t) ?2t?r y ? yu(t) 2 b1 ? tr?1 1 + t? (t) ?(t) ?1 + (ln t)?1 (2.43) (2.44) There exists a (small enough) ?6 > 0 such that if ? < ?6, then a2(t, y) < 0 when (t, y) ? 0, y = yu(t). In particular, a2(t, y) < 0 when (t, y) ? ?2(?) for any ? ? (0, 1]. Therefore, for any ? ? (0, 1], an integral curve of (2.17) which intersects ?2(?) in an arbitrary point (t0, y0), for sufficiently small |t ? t0| (t ?), lies in 2(?) if t > t0 and lies outside of 2(?) if t < t0 (the proof is similar to the one for ?1). Let P?(t?, y?) ? 1\{(0, 0)}, y? = yu(t?). Then there exists ?? ? (0, 1] such that P? ? ?2(??). As above, the integral curve of (2.17) J? : (t, y?(t)), which passes through P?, lies outside of 2(??) if t ? (t?, t?), where (t?, t?) is the left maximal existence interval for the solution y?. On the other hand, if (t, y) ? 1\{(0, 0)}, then y ? yu(t) y ? ctr+1 + yu(t) ? ctr+1 2Mtr+1?(t) < ??tr+1?(t)(? ln t) (2.46) if t ? (0, t??), where t?? ? (0, ?) and t?? is small enough. Thus, if (t, y) ? 1\{(0, 0)} and t ? (0, t??), then (t, y) ? 2(??). Let t? = min{t?, t??}. As above, the integral curve J? : (t, y?(t)) lies outside of 1\{(0, 0)} if t ? (t?, t?) and we have proved our statement. Now we can introduce an operator T : ? by (Tu)(t) = yu(t). We prove that T : ? is a continuous operator. Let ui ? , i ? {1, 2}, be arbitrary functions and let Tui = yi, i ? {1, 2}. If u1 = u2, then y1 = y2. Let u1 ? u2 = d, d > 0. We will investigate the behavior of the integral curves of the differential equation y (t) = at + b1 y(t)t?r + r y(t)t?1 + b2u1 q1t q1t ?r + b3q2?r u1 q2t ? rb3q2?r?1t?1u1 q2t + ? t, u1(t)t?r , u1 q1t q1t ?r , u1(t)t?r ? ru1(t)t?r?1, u1 q2t q2t ?r ? ru1 q2t q2t ?r?1 . (It is obvious that y1 : (0, ?] ? R is the solution of (2.47).) We introduce the following notations: ?3 = (t, y) : t ? (0, ?], y ? y2(t) = ?d? tr+1?(t) 1?? , 3 = (t, y) : t ? (0, ?], y ? y2(t) < ?d? tr+1?(t) 1?? , where ? = (r + 2)?1, ? = 3|b1|?1(2M)1??(|b2| + 1) + 1. Let the function A3 : 0 ? [0, +?) be defined by the equality A3(t, y) = y ? y2(t) 2 tr+1?(t) ?2(1??) and let a3 : 0 ? R be the derivative of the function A3. Using (2.47), we have a3(t, y) = 2 t1+r ?(t) ?2(1??)t?r b1 + ? 1 + ?(1 + r) ? (1 ? ?)t? (t) ?(t) ?1 tr?1 y ? y2(t) 2 + y ? y2(t) ?3(t) , and there exists a (small enough) ?7 > 0 such that if ? < ?7, then ?3(t) (2M)1?? b2 + 1 d? t1+r ?(t) 1?? since u1(t) ? u2(t) u1 qit ? u2 qit u1(t) ? u2(t) (1) Let b1 > 0 and therefore a3(t, y) > 0 when (t, y) ? ?3. Then an integral curve of (2.47) which intersects ?3 in an arbitrary point (t0, y0), for sufficiently small |t ? t0| (t ?), lies in 3 if t < t0 and lies outside of 3 if t > t0. (The proof is similar to the one for ?1 in the case b1 > 0.) In addition, y1(?) = y2(?) = yG. Thus, if t decreases from t = ? to t = 0, then the integral curve of (2.47), J : (t, y1(t)), lies in 3 for all t ? (0, ?] because this integral curve cannot intersect ?3. We have y1(t) ? y2(t) y1(t) ? y2(t) ?d? tr+1?(t) 1??, t ? (0, ?], b1 + rtr?1 ? + (2M)1?? b2 + 1 t(1+r)(1??)?r ?(t) 1??d?. Here 1 ? ? > 0, (1 + r)(1 ? ?) ? r = (r + 2)?1 > 0. Thus, there exists a (small enough) ?9 > 0 such that if ? < ?9, then y1(t) ? y2(t) + y1(t) ? y2(t) d?, t ? (0, ?]. (2) Let b1 < 0 and therefore a3(t, y) < 0 when (t, y) ? ?3. Then an integral curve of (2.47) which intersects ?3 in an arbitrary point (t0, y0), for sufficiently small |t ? t0| (t ?), lies in 3 if t > t0 and lies outside of 3 if t < t0. (The proof is similar to the one for ?1 in the case b1 < 0.) In addition, y1(t) ? y2(t) y1(t) ? ctr+1 + y2(t) ? ctr+1 2Mtr+1?(t) < ?d? tr+1?(t) 1?? (2.56) when t ? (0, t(d)], where t(d) ? (0, ?) is small enough. Therefore, if t ? (0, t(d)], then the integral curve of (2.47), J : (t, y1(t)), lies in 3. As above, if t increases from t = t(d) to t = ?, then the integral curve of (2.47), J : (t, y1(t)), cannot intersect ?3 and therefore this curve lies in 3 for all t ? (0, ?]. Further, we can get (2.54) and (2.55) as in the case b1 > 0. Thus, if u1 ? u2 = d < ?(?) = (?/2)1/?, then max t?[0,?] y1(t) ? y2(t) + y1(t) ? y2(t) The reasoning given above does not depend on the choice of ? > 0 and ui ? , i ? {1, 2}. Therefore T : ? is the continuous operator. To complete the proof of the theorem, it suffices to apply the Schauder fixed point theorem to the operator T : . ? Ravi P. Agarwal: Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901, USA E-mail address: Oleksandr E. 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Ravi P. Agarwal, Donal O'Regan, Oleksandr E. Zernov. A singular initial value problem for some functional differential equations, International Journal of Stochastic Analysis, DOI: 10.1155/S1048953304405012