INITIAL-VALUE PROBLEMS FOR FIRST-ORDER DIFFERENTIAL RECURRENCE EQUATIONS WITH AUTO-CONVOLUTION

Transcription

1 Electronic Journal of Differential Equations, Vol (), No, pp 3 ISSN: URL: or ftp ejdemathtxstateedu INITIAL-VALUE PROBLEMS FOR FIRST-ORDER DIFFERENTIAL RECURRENCE EQUATIONS WITH AUTO-CONVOLUTION MIRCEA CÎRNU Abstract A differential recurrence equation consists of a sequence of differential equations, from which must be determined by recurrence a sequence of unknown functions In this article, we solve two initial-value problems for some new types of nonlinear (quadratic) first order homogeneous differential recurrence equations, namely with discrete auto-convolution and with combinatorial auto-convolution of the unknown functions In both problems, all initial values form a geometric progression, but in the second problem the first initial value is exempted and has a prescribed form Some preliminary results showing the importance of the initial conditions are obtained by reducing the differential recurrence equations to algebraic type Final results about solving the considered initial value problems, are shown by mathematical induction However, they can also be shown by changing the unknown functions, or by the generating function method So in a remark, we give a proof of the first theorem by the generating function method Different cases of first order differential recurrence equations and their solutions are presented, including those from a previous work Applications of the equations considered here will be given in subsequent articles Introduction We consider first-order differential recurrence equations of the form G n ( x n ), x n, x n,, x ), n,,,, with unknowns x, x,, x n,, complex-valued differentiable functions defined on an open interval I of real numbers, the functions G n being given For t I, is called Cauchy initial-value problem for such an equation, the determination of its solutions x n, with given initial values x n (t ), n,,, In this article we solve some initial-value problems for first order homogeneous differential recurrence equations with (discrete) auto-convolution n x n a x k x n k, t I, n,,,, () k Mathematics Subject Classification B37, 34A Key words and phrases Differential recurrence equations; discrete auto-convolution; algebraic recurrence equations c Texas State University - San Marcos Submitted September, Published January 4,

2 M CÎRNU EJDE-/ and with combinatorial auto-convolution n ( ) n y n a y k y n k, t I, n,,,, () k k where a is a given integrable function The second type of equation reduces to the first by change of unknown functions y n n!x n, n,,, Equations () and () are considered for the first time, they being different from those verified by special functions, from de Branges equation (see [8) or other types of differential recurrence equations, studied so far (see, for example, [6, 7) The discret convolution or Cauchy product of numerical sequences and its inversediscrete deconvolution (see [5 for these definitions) were formely used by the author in a series of papers [,, 3 for solving numerical difference, differential and polynomial equations In the following we denote A adt, B + x A(t ) x (t )A, t I (3) Obviously, B(t ), B x (t )a, t I (4) Algebraic recurrence equation with auto-convolution Lemma Let b n, n,,,, be a sequence of real or complex numbers Following statements are equivalent: (i) (n + )b b n n k b kb n k, n,,, ; (ii) (n )b b n n b kb n k, n, 3, ; (iii) b n b b b n, n, 3, ; (iv) The numbers b n, n,,,, are in geometric progression, namely k k b n b [ b b n b n b n, n,,, () Proof (i) (ii) and (iii) (iv) are obvious (iv) (i) If the sequence b n is given by formula (), we have n n b k b n k n b n b k b n k b k b n k b n (n + ) bn b n (n + )b b n, k n,,,, hence the sequence b n satisfies (i) in Lemma (ii) (iv) (First proof by induction) For n, from (ii) we obtain b b b For n, we suppose that b k bk, for k,,, n Then from (ii) results b k n b n b k b n k (n )b (n )b (n )b n b n b n n b k b k (n )b (n ) bn b n b n k b n k bn b n In conformity with induction axiom, formula () is true for every n,,,

3 EJDE-/ DIFFERENTIAL RECURRENCE EQUATIONS 3 (i) (iv) (Second proof by the generating function method, [9) Denoting G(z) n b nz n generating function of the numerical sequence b n, given by a formal series, item (i) in Lemma gives a differential equation with the successive forms b [ zg(z) G (z), b G (z) G(z) [ G(z) b z, b zg (z) + b G(z) G (z), G (z) G (z) G(z) b G(z) z Integrating results in ln G(z) b G(z) ln kz, hence G(z) b G(z) kz, so G(z) b kz b k kn z n, where k and k ± k are arbitrary constants It results b n b k n, n,,, For n, we have b b k, hence k b b and b n bn, b n n,,, Corollary For a a given number, the sequence b n, n, 3,, is solution of the equation if and only if n (n )b b n a b k b n k, n, 3,, () b n an b n b n, n,, (3) Proof Making the change of variables b a b, b n b n, n,,, the equation () reduces to (n ) b bn n b k bn k, n, 3, In conformity with Lemma, we have b n b n e b n e b n an b n b n, n,, Corollary 3 The sequence b n, n,,,, is solution of the equation n ( ) n (n + )b b n b k b n k, n,,,, k if and only if k b n n!bn, n,,, b n Proof Making the change of variables b n n! b n, n,,,, the considered equation reduces to (n + ) b bn n b k k bn k, n,,,, and Corollary 3 follows from Lemma Remark Other results, related to those of Lemma and its Corollary 3, were given in [4, Theorem and Corollary Lemma 3 The functions 3 First initial-value problem x n x n(t ) B n+, t I, n,,,, (3) are solutions of () if and only if their initial values x n (t ) n,,,, are in geometric progression

4 4 M CÎRNU EJDE-/ Proof If the functions x n are given by formula (3), the equation () takes successively the form (n + )x n(t )B n x k (t ) x n k (t ) B n+ a B k+ B n k+, k (n + )x n (t )x (t )a B n+ a n B n+ x k (t )x n k (t ), k n (n + )x (t )x n (t ) x k (t )x n k (t ) In conformity with Lemma, the last equality is true if and only if the values x n (t ), n,,,, are in geometric progression Theorem 3 The differentiable functions x n, n,,,, with initial values x n (t ), n,,,, in geometric progression, are solutions of () if and only if they are given by (3) Proof We suppose that the functions x n are solutions of () and have their initial values x n (t ), n,,,, are in geometric progression, hence x n (t ) x (t ) [ x (t ) n x n (t ) x (t ) x n, n,,, (3) (t ) For n, the equation () has the form x ax, hence x a x By integration, we obtain x A + C, hence x A+C, where C is an arbitrary constant For t t, it results in C +x(t)a(t) x (t ) ; therefore, x In conformity with (3), A +x(t)a(t) x (t ) k x (t ) + x (t )A(t ) x (t )A x x (t ) B (33) For n, the equation () has the form x ax x, hence By integration, we obtain x it results C x (t ), hence x x B B C B, with C an arbitrary constant For t t, x x (t ) B (34) For n, equation () has the form n x n ax x n + a x k x n k, hence, using the relation a B x (t ), obtained from (4), x n + B n B x n a x k x n k,

5 EJDE-/ DIFFERENTIAL RECURRENCE EQUATIONS 5 with general solution ( B )[ x n exp B dt [ B C n x (t ) ( B ) exp B dt n a n B B x k x n k dt x n x n k dt + C n where C n is an arbitrary constant For n, from (34) and (35), it results x [ B C B B x x (t ) dt [ B C x (t ) B x (t ) B dt [ B C + x (t ) x (t )B From (36) for t t and (3) for n, it results hence C, and (36) becomes Then x (t ) C + x (t ) x (t ) x (t ) x (t ), x, (35) (36) x (t ) x (t )B 3 (37) For n fixed and k,,,, n, we suppose that n x k x n k x k n n x k (t ) x k (t )B k+ x k (t ) x k (t )B k+ x n (t ) (t )B n+ x n k (t ) x n k (t )B n k+ (n )xn (t ) (t )B n+ and (35) becomes x n [ B C n (n )xn (t ) B x n (t ) B n dt [ x n (t ) B C n + x n (t )B n From which, for t t and (3), results x n (t ) C n + C n, and (38) becomes x n xn (t) x n x n (t) (t ) x n (38) (t ), hence x n (t ) x n, n,,, (39) (t )B n+ According to induction axiom, (39) is satisfied for any natural number n From (3) and (39), it results (3)

6 6 M CÎRNU EJDE-/ Reciprocally, if the functions x n are given by (3), with the initial values in geometric progression, then we have (39), hence x n (n + )xn (t )B x n (n + )xn (t )a n (t )B n+ (t )B n+ a x n (t ) k (t )B n+ n x k a (t ) x n k (t ) n x k (t )B k+ x n k (t )B n k+ a x k x n k ; k therefore, the functions x n satisfy () This also results by Lemma 3 Remark Theorem 3 can also be demonstrated using the generating function G(t, z) n x nz n, of the sequence of functions x n, n,,,, given by a formal series Then () is equivalent to equation t G(t, z) ag (t, z), with solution G(t, z) u(z) A, where u(z) is an arbitrary function For z, we obtain x G(t, ) u() A Let v(z) u() u(z), with v() Using geometric series, we have G(t, z) u() A v(z) u() A v(z) u() A n k v n (z) ( u() A ) n+ The choice v(z) Cz, where C is an arbitrary constant, gives C n G(t, z) ( ) n+ z n, u() A hence x n n C n ( u() A ) n+, n,,, (3) For t t and n, from (3) it results x (t ) +x (t )A(t ) x (t ), and u() A + x (t )A(t ) x (t ) For t t and n, from (3) results x (t ) ( From (3), (3) and (3) it results u() A(t ), hence u() A B x (t ) (3) C u() A(t ) ) Cx (t ), hence C x (t ) x (t ) (3) x n xn (t ) x n+ (t ) x n (t ) B n+ x n (t ) x n, n,,, (33) (t )B n+ For t t, from (33) we obtain (3), hence (33) takes the form (3) and the initial values x n (t ), n,,,, are in geometric progression This last statement also follows by Lemma 3 The reciprocal affirmation results both by direct calculation as above or by Lemma 3

7 EJDE-/ DIFFERENTIAL RECURRENCE EQUATIONS 7 Corollary 33 The differentiable functions y n, n,,,, whose initial values are y n (t ) n!yn (t ) y n, n,,,, (t ) are solutions of the differential recurrence equation with combinatorial autoconvolution () if and only if they are given by Lemma 4 If A(t ) and then the functions y n y n(t ) B n+, t I, n,,, 4 Second initial-value problem x (t ) x (t ) x (t ), (4) A(t ) x x (t ) B, x n x n(t )A n A n (t )B n+, t I, n,, (4) are solutions of () if and only if the initial values x n (t ), n,,, are in geometric progression Proof If the functions x n are given by formula (4), then () is obviously satisfied for n and n For n, 3,, it takes successively the form x n (t ) (n )A n A B n+ (n + )B n B A n A n (t ) B n+ ax (t )x n (t )A n A n (t )B n+ n + a x k (t )A k x n k (t )A n k A k (t )B k+ A n k (t )B n k+, x n (t ) (n )A n ab n+ + (n + )x (t )B n aa n A n (t ) B n+ ax (t )x n (t )A n A n (t )B n+ + aan n A n (t )B n+ x k (t )x n k (t ), x n (t )[ (n )B + (n + )x (t )A A(t ) x (t )x n (t )A A(t ) x k (t )x n k (t ), n + (n ) x n(t )[ B + x (t )A n x k (t )x n k (t ) A(t ) From (4) and (4) it results B + x (t )A + x (t )A(t ) x (t )A(t ), t I; (43) x (t )

8 8 M CÎRNU EJDE-/ therefore, () is still equivalent to the relations (n ) x n(t ) x n (t )A(t ) x k (t )x n k (t ), A(t ) x (t ) (n )x (t )x n (t ) x (t )x (t ) x (t ) n x k (t )x n k (t ) By Corollary, for b n x n (t ), n,,, and a x(t)x(t) x (t), the last equality is equivalent to the relation x n (t ) xn (t )x n (t ) x n (t ) x n (t ) x n (t ) x n (t ) [ (t ) x x (t ) n, (t ) x (t ) where n,, ; hence with the fact that the initial values x n (t ), n,,, are in geometric progression Theorem 4 If A(t ), while x (t ) is given by (4), then the differentiable functions x n, n,,,, with initial values x n (t ), n,,, in geometric progression, are solutions of () if and only if they are given by (4) Proof We suppose that the functions x n, n,,,, are solutions of () and the initial values x n (t ), n,,, are in geometric progression, hence [ x (t ) n x n x n (t ) x (t ) (t ) x (t ), n,, (44) (t ) As was shown in the proof of Theorem 3, the functions x, x, x n for n, and x are given by (33), (34), (35) and (36) From (36) and (3) we have x C x (t )B + x (t ) x (t )B 3 C x (t ) [ + x (t )A(t ) C x (t )A + x (45) (t ) x (t )B 3 We take x C (t ) x (t ) [ + x (t )A(t ) x (t ) x (t )A(t ) (46) the above equality resulting from (43), which in turn resulted from (4) From (45) and (46) it results x C x (t )A B 3 x (t )A A(t )B 3 (47) From (35), for n 3, (34) and (47), it results x 3 [ B C 3 B B x x dt x (t ) [ B C 3 x (t )x (t ) AB x (t )A(t ) B 3 dt (48) Using (3) and (4), we obtain A a B x ; hence (t ) [ A A B AB B B [B + x (t )AB x (t )B (49)

9 EJDE-/ DIFFERENTIAL RECURRENCE EQUATIONS 9 Using again (43), formula (49) takes the form from which we obtain From (48) and (4), it results x 3 B [ A x (t )A(t )B B x (t )x (t )B, B B x (t )x (t ) [ A (4) x (t )A(t ) B B [ C 3 + x (t ) x (t )A (t ) [ C 3 + A B x (t )A x (t )A (t )B From which for t t and (44) for n 3, it results x 3 (t ) C 3 + x (t ) x (t ) x (t ) x (t ), hence C 3 and formula (4) becomes x 3 For n 3 fixed and k,,, n, we suppose that Then n x k x n k n x k x k (t )A k x k (t )A k (t )B k+ (n )xn (t )A n x n 4 (t )A n (t )B n+ From (35) and (44) it results x n [ (n ) (t B C n ) x (t )x n 4 (t )A n (t ) From (4) and (45), it results x n [C B n + [ C n + B [ A dt B (4) x (t )A x (t )A (t )B 4 (4) x k (t )A k x k (t )A k (t )B k+ (43) (n )xn (t ) (t )A n (t ) x n k (t )A n k x n k (t )A n k (t )B n k+ x n (t )A n (t )A n (t )B n A n B B n [ A n [ A dt B B (44) dt (45) (46)

10 M CÎRNU EJDE-/ From this euqality for t t and (44), it results x n (t ) C n + xn (t ) hence C n and formula (46) becomes x n x n (t ), (t ) (t ) x n (t )A n (t )A n (t )B n+ (47) In conformity with induction axiom, formula (47) is satisfied for every natural number n For t t, from (47) it results x n (t ) xn (t ) (t ), hence (47) reduces to second formula (4) Reciprocally, if the functions x n, n,,,, are given by (4), and the initial values x n (t ), n,,, are in geometric progression, then we have x x(t) B and x n x n(t )A n A n (t )B n+ x n (t )A n (t )A n (t )B n+, hence for n,,, using (43), we have x n ax x n x n(t ) [ (n )A n A B n+ (n + )A n B n B A n (t )B n+ x (t )x n (t )aa n A n (t )B n+ x n(t )aa n [ (n )B + (n + )x A n (t )B n+ (t )A x (t )A (n )xn (t )aa n [ B + x (t (t )A n (t )B n+ )A (n )xn (t )aa n x n 4 (t )A n (t )B n+ n x k (t )A k x n k (t )A n k a x k (t )A k (t )B k+ x n k (t )A n k (t )B n k+ n a x k x n k ; therefore, the functions x n given by (4) satisfy () Lemma 4 This also results by Remark From (4) it results that in hypotheses of the Theorem 4, those of Theorem 3 are not satisfied The solutions of the second initial values problem are different from those of the first problem, because in the proof of Theorem 3 all arbitrary constants C n, n, that arise in solving the differential equations are zero, while in the proof of Theorem 4, the constant C is non-zero, given by formula (46) Corollary 43 If A(t ), and y (t ) y (t) y (t ) A(t ) functions y n, n,,,, with y n (t ) n!yn (t ) y n (t ), then the differentiable, n,,, are solutions

11 EJDE-/ DIFFERENTIAL RECURRENCE EQUATIONS of the recurrence equation with combinatorial auto-convolution () if and only if they are y y (t ) B, y n y n(t )A n A n (t )B n+, t I, n,, 5 Examples We give some examples that illustrate the above results, including those from [4 () x n n k x kx n k, n,, Here a, so A t (a) For the initial values x n (), n,,,, we have B + x ()A() x ()A t, hence we obtain the solutions x n xn() B n+ ( t) n+, n,,, Because, A(), second initial values problem can not be considered (b) If x n () n, we obtain x n n ( t) n+, n,,, The second initial values problem can not be considered when x n () n, n,,, because x () x () x () A (c) If x n (), n,,,, then x n (), B + x A x A 3 t, hence x n n (3 t), n,,, The second initial n+ values problem when x n, n,,, can also be considered and will be n given in the next example (d) Let x n (), n,,, and x n () x () x () A() Then B t, hence x x() B t and x n n,, () y n n ( n k k) yk y n k, n,,, n! (a) If y n () n!, then y n ( t), n,,, n+ (b) If y n () n! n, then y n n! n ( t) n+, n,,, xn()an A n ()B n+ t n n ( t), n+ (c) If y n () n! n!, then y n n n (3 t), n,,, n+ (d) If y () and y n () n!, n,,, then y n n!t n n ( t), n,, n+ (3) x n e t n t and y n ko x kx n k, n,,, Here a A e t If x n (), n,,,, we obtain the solutions x n ( e t ), n n+,,, The second problem can not be considered when x n (), n,,, because x () x () x () A() (4) x n sin t n k x kx n k, n,,, Here A cos t (a) Let x n (), n,,, Then B +x ()A() x ()A cos t, hence x n cos n+ t, n,,, (b) If x n (), n,,, and x () x () x () A() cos t, hence x x() B cos t and x n x n()a n A n ()B n+ cos n t, n,, ( cos t ) n+, then B (c) For x n ( π ), n,,,, we obtain xn (+cos t) n+, n,,, The second problem can not be considered, because A ( π )

12 M CÎRNU EJDE-/ (5) x n n t k x kx n k, x n (), n,,, Then x n ( ln t), n,,, The second problem can not be considered, because n+ A() (6) x n n t k x kx n k, n,,, Here a t, so A t (a) If x n (), then B t, hence x n t n+, n,,, (b) Let x n (), n,,, and x () x () x t () A() Then B t, hence the solutions are x x() B t t, x n xn()an A n ()B n+ t ( t), n+ n,, (c) If x n () ( ) n+ 4, then B 3t 4 t, hence x n ( )n+ 4t n+ (3t 4), n n+,,, (d) Let x n () ( ) n+ 4, n,,, x () x () x () A() Then B (t ) t, hence the solutions are x t t, x t n ( t), n,, n+ (7) y n n ( n t k k) yk y n k, n,,, (a) If y n () n!, then y n n!t n+, n,,, (b) If y (), and y n () n!, n,,, then y n!t ( t) n+, n,, (c) If y n () ( ) n+ n!4, then y n ( )n+ n!4t n+ (3t 4) n+, n,,, (d) If y (), and y n () ( ) n+ n!4, n,,, then y t t, and y n t t, and y n n!t ( t), n,, n+ (8) tz n + z n n k z kz n k, n,,, We make the change of unknown functions x n tz n, n,,, (a) If z n (), then z n t n, n,, This example was given in [4, Theorem 4 (a) (b) Let z (), z n (), n,, Then z t, z t n ( t), n+ n,, (c) If z n () ( ) n+, then z n ( )n+ 4t n (3t 4), n,,, n+ (d) Let z (), and z n () ( ) n+, n,, Then z t and t z n ( t), n,, This example was given in [4, theorem 4 (b) n+ (9) tu n + u n n ( n k k) uk u n k, n,,, (a) If u n () n!, n,,, Then u n n!t n, n,,, This example was given in [4, corollary of theorem 4 (a) (b) If u (), u n () n!, n,,, then u t, u n!t n ( t), n+ n,, (c) If u n () ( ) n+ n!, n,,,, then u n ( )n+ n!4t n (3t 4), n n+,,, (d) If u () and u n () ( ) n+ n!, n,,, then u t, and u n n!t ( t), n,, This example was given in [4, corollary of theorem n+ 4 (b), with some mistakes corrected here Acknowledgments The proof of the Theorem 3 by the generating function method, included in the remark in section 3, was suggested by the anonymous referee, to whom I want to give many thanks

13 EJDE-/ DIFFERENTIAL RECURRENCE EQUATIONS 3 References [ M Cîrnu; Solving difference and differential equations by discrete deconvolution, UPB Scientific Bulletin, Series A, vol 69, no, pp 3-6, 7 [ M Cîrnu; Determination of particular solutions of nonhomogeneous linear differential equations by discrete deconvolution, UPB Scientific Bulletin, Series A, vol 69, no, pp 3-6, 7 [3 M Cîrnu; Approximate calculus by deconvolution of the polynomial roots, UPB Scientific Bulletin, Series A, no 4, pp 9-, 7 [4 M Cîrnu; First order differential recurrence equations with discrete auto-convolution, International Journal of Mathematics and Computation, vol 4, no S9, pp 4-8, 9 [5 M A Cuenod; Introduction a l Analyse Impulsionelle Principes et Applications, Dunod, Paris, 97 [6 I P Dimitrov; The stability of the solutions of differential-recurrence equations, (Russian), Differencialnye Uravnenia, vol 8 (97), Part I, pp 68-75, Part II, pp [7 N M Flaisher; A certain differential recurrence equation, (Russian) Arch Math, vol 4 968, pp [8 W Koepf, D Schmersau; Solution properties of the de Branges differential recurrence equation, Complex Variable Theory and Applications, vol 5 (5), no 7-, pp [9 H S Wilf; Generating functionology, Academic Press, 99 Mircea Cîrnu Department of Mathematics III, Faculty of Applied Sciences, University Politehnica of Bucharest, Romania address: cirnumircea@yahoocom