ANALYTIC SOLUTIONS OF A FUNCTIONAL DIFFERENTIAL EQUATION WITH STATE DEPENDENT ARGUMENT. Jian-Guo Si and Sui Sun Cheng

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1 TWIWANESE JOURNAL OF MATHEMATICS Vol., No. 4, pp , December 997 ANALYTIC SOLUTIONS OF A FUNCTIONAL DIFFERENTIAL EQUATION WITH STATE DEPENDENT ARGUMENT Jian-Guo Si and Sui Sun Cheng Abstract. This paper is concerned with a functional differential equation x (z) = x(az + bx(z)), where a and b 0. By constructing a convergent power series solution y(z) of a companion equation of the form βy (βz) = y (z)[y(β 2 z) ay(βz) + a], analytic solutions of the form (y(βy (z)) az)/b for the original differential equation are obtained. Functional differential equations of the form x (t) = x(t σ(t)) have been studied to some extent by many authors. However, when the function σ(t) is state dependent, say, σ(t) = ( a)t bx(t), relatively little is known. Indeed, to the best of our knowledge, there are only a few reports (see [, 3-0]) on functional differential equations with state dependent arguments. In this note, we will be concerned with a class of functional differential equation of the form () x (z) = x(az + bx(z)). When a = 0 and b =, equation () reduces to the iterative functional differential equation x (z) = x(x(z)) which has been investigated by Eder [] and analytic solutions are shown to exist by means of the Banach fixed point theorem. When b = 0 and a, equation () reduces to the functional differential equation x (z) = x(az), 0 Received March, 997; revised September 25, 997. Communicated by S.-B. Hsu. 99 Mathematics Subject Classification: 30D05, 34K25. Key words and phrases: Functional differential equation, analytic solution. Presented at the International Mathematics Conference 96, National Changhua University of Education, December 3 6, 996. The Conference was sponsored by the National Science Council and the Math. Soc. of the R.O.C. 47

2 472 Jian-Guo Si and Sui Sun Cheng which has an entire solution of the form (see Elbert [3]) a (n(n )/2) x(z) = ηz n. n! n=0 Indeed, if we seek a power series solution of the form x(z) = b n z n, n=0 then substituting it into the above equation leads to Taking b 0 = η, we see that and that (n + )b n+ = a n b n, n = 0,, 2,.... b n+ lim n b n b n = a(n(n )/2) η n! a n = lim s n + = 0, as required. When a and b 0, by means of the reasoning just used and several more involved ideas, we will be able to construct analytic solutions for our equations in a neighborhood of the complex number ()/( a), where β satisfies either one of the following conditions: (H) 0 < β < ; or (H2) β =, β is not a root of unity, and log for some positive constant µ. β n µ log n, n = 2, 3,... The technique for obtaining such solutions is as follows. We first seek a formal power series solution for the following initial value problem (2) y (βz) = β y (z){y(β 2 z) ay(βz) + a}, (3) y(0) = a.

3 Analytic Solutions 473 Then we show that such a power series solution is majorized by a convergent power series. Then we show that (4) x(z) = b y(βy (z)) a b z is an analytic solution of () in a neighborhood of ()/( a). Finally, we make use of a partial difference equation to show how to explicitly construct such a solution. We begin with the following preparatory lemma, the proof of which can be found in [2, Chapter 6]. Lemma. Assume that (H2) holds. Then there is a positive number δ such that β n < (2n) δ for n =, 2,... Furthermore, the sequence {d n } defined by d = and d n = β n max {d n..., d nt }, n = 2, 3,..., n=n +...+n t, 0<n... n t,t 2 will satisfy d n (2 5δ+ ) n n 2δ, n =, 2,.... Lemma 2. Suppose (H) holds. Then for any nontrivial complex number η, equation (2) has an analytic solution of the form (5) y(z) = a + ηz + n=2 b n z n in a neighborhood of the origin, and there exists a positive constant M such that for z in this neighborhood, y(z) a + 2M. Proof. We seek a solution of (2) in a power series of the form (5). By defining b 0 = ()/( a) and b = η and then substituting (5) into (2), we see that the sequence {b n } n=2 is successively determined by the condition (β n+ β) (n + )b n+ (6) n = (k + ) (β 2(n k) aβ n k )b k+ b n k, n =, 2,

4 474 Jian-Guo Si and Sui Sun Cheng in a unique manner. Furthermore, since 0 k n, (7) β 2(n k) aβ n k β n+ β + a β n M, n 2 for some positive number M, thus if we define a sequence {B n } by B = η and B n+ = M n B k+ B n k, n =, 2,..., then b n B n for n =, 2,.... Now if we define then Hence G(z) = B n z n, G 2 (z) = (B B n + B 2 B n B n B )z n n=2 = (B B n + B 2 B n + + B n B )z n+ = B n+ z n+ = M M G(z) M η z. n G(z) = { } ± 4M η z. 2M But since G(0) = 0, only the negative sign of the square root is possible, so that G(z) = { } 4M η z. 2M It follows that the power series G(z) converges for z /(4M η ), which implies that (5) is also convergent for z /(4M η ). Next, note that for z /(4M η ), or Thus G( z ) = 2M 4M η z = + 4M η z 2 η z G( z ) 2 η z 2 η 4M η = 2M. 2 η z,

5 Analytic Solutions 475 y(z) a + b n z n a + B n z n = a + G( z ) a + 2M as required. The proof is complete. Next, we consider the case when (H2) holds. Lemma 3. Suppose (H2) holds. Then equation (2) has an analytic solution of the form (8) y(z) = a + z + n=2 b n z n in a neighborhood of the origin, and there exists a positive constant δ such that y(z) a + 2 5δ+ n 2δ. Proof. As in the proof of Lemma, we seek a power series solution of the form (8). Then defining b 0 = ()/( a) and b =, (6) and (7) again hold so that (9) b n+ + a n β n b k+ b n k = + a β n Let us now consider the function G(z) = n +n 2 =n+; n,n 2 n 2( + a ) b n b n2, n =, 2,.... { } 4( + a )z which, in view of the binomial series expansion, can also be written as G(z) = z + C n z n n=2 for z < /4( + a ). Since G(z) satisfies the equation ( + a )G 2 (z) + z = G(z), 475

6 476 Jian-Guo Si and Sui Sun Cheng thus, by the method of undetermined coefficients, it is not difficult to see that the coefficient sequence {C n } n=2 will satisfy C = and n C n+ = ( + a ) C k+ C n k = ( + a ) n +n 2 =n+; n,n 2 n C n C n2, n =, 2,.... Hence by induction, we easily see from Lemma that b n C n d n, n =, 2,..., where the sequence {d n } is defined in Lemma. Since G(z) converges on the open disc z < /4( + a ), there exists a positive constant T such that C n T n for n =, 2,.... In view of this and Lemma, we finally see that b n T n Q n n 2δ, n =, 2,..., where Q = 2 5δ+, which shows that the series (5) converges for z < (T Q). Finally, when z (T Q), we have y(z) a + b n z n a + C n d n z n a + T n Q n n 2δ z n a + T n Q n n 2δ (T Q) n = a + Q n 2δ, as required. The proof is complete. We now state and prove our main result in this note. Theorem. Suppose the complex number β satisfies either (H) or (H2). Then equation () has an analytic solution x(z) of the form (4) in a neighborhood of ()/( a), where y(z) is an analytic solution of equation (2). Furthermore, when (H) holds, there is a positive constant M such that x(z) ( b a + ) + a 2M b z

7 Analytic Solutions 477 in a neighborhood of ()/( a); and when (H2) holds, there is a positive number δ such that x(z) ( b a + ) Q n 2δ + a b z, Q = 25δ+, in a neighborhood of ()/( a). Proof. In view of Lemmas 2 and 3, we may find a sequence {b n } n=2 such that the function y(z) of the form by (8) is an analytic solution of (2) in a neighborhood of the origin. Since y (0) =, the function y (z) is analytic in a neighborhood of the point y(0) = ()/( a). If we now define x(z) by means of (4), then x (z) = b βy (βy (z)) (y ) (z) a b = β b y (βy (z)) y (y (z)) a b = b {y(β2 y (z)) ay(βy (z)) + a} a b = b {y(β2 y (z)) ay(βy (z))}, and ( [ x(az + bx(z)) = x az + b b y(βy (z)) a ]) b z = x(y(βy (z))) = b y(βy (y(βy (z)))) a b y(βy (z)) = b {y(β2 y (z)) ay(βy (z))} as required. Next, if (H) holds, then in view of Lemma 2, x(z) = b y(βy (z)) az b ( y(βy (z)) + a z ) ( b a + ) + a 2M b z ; and if (H2) holds, then in view of Lemma 3, x(z) = b y(βy (z)) az b ( y(βy (z)) + a z ) ( b a + ) Q n 2δ + a b z. 477

8 478 Jian-Guo Si and Sui Sun Cheng The proof is complete. We now show how to explicitly construct an analytic solution of () by means of (4). Since x(z) = b y(βy (z)) a b z, thus ( ) x = a b y(0) a b a = b a a b a =. b Furthermore, ( ) ( x = x a ( )) a a + bx a ( = x a a + b ) = x b ( ) a =. b By calculating the derivatives of both sides of (), we obtain successively x (z) = x (az + bx(z)) (a + bx (z)), x (z) = x (az + bx(z))(a + bx (z)) 2 + x (az + bx(z)) (bx (z)), so that ( ) ( x = x a ( a a + bx a ( ) x a = βx ( a ) = ( ) = x a β(), b )) ( ( ) β 2 + x a = b [β() (β2 + )]. ( )) a + bx a ( ) bx a It seems from the above calculations that the higher derivatives x (m) (z) at z = ξ ()/( a) can be determined uniquely in similar manners. To see this, let us denote the derivative (x (i) (az + bx(z))) (j) at z = ξ by λ ij, where i, j 0. Note that the two derivatives x (k) (z) and x (k) (az + bx(z)) are equal at the point z = ξ since aξ + bx(ξ) = ξ. In other words, x (k) (ξ) = λ k0.

9 Analytic Solutions 479 Furthermore, in view of (), we see that x (k+) (z) = (x(az + bx(z))) (k) which implies λ k+,0 = λ 0,k. Finally, since (x (i) (az + bx(z))) (j+) = (x (i+) (az + bx(z)) (a + bx (z))) (j) we see also that λ i,j+ = j = j ( j k ) (a + bx (z)) (k) ( x (i+) (az + bx(z))) (j k), ( j k ) λ i+,j k (a + bx (z)) (k) z=ξ = βλ i+,j + b j k= ( j k ) λ i+,j k λ 0,k, i = 0,,... ; j = 0,,..., where we have used the fact that λ k+,0 = λ 0,k in obtaining the last equality. Clearly, if we have obtained the derivatives x (0) (ξ) = λ 00,..., x (m) (ξ) = λ m0 = λ 0,m, then by means of the above partial difference equation, we can successively calculate λ m,, λ m 2,, λ m 2,2,..., λ, λ 2,..., λ,m, λ 0m in a unique manner. x (m+) (ξ). This shows that In particular, λ 0m = λ m+,0 is the desired derivative x(z) = b + ( b () z a ( + β() (β2 + ) 3!b ) β() + 2!b z a ( z ) 2 a ) 3 ( λ i,0 + i! i=4 z a ) i. References. E. Eder, The functional differential equation x (t) = x(x(t)), J. Differential Equations 54 (984), M. Kuczma, Functional Equations in a Single Variable, Polish Scientific Publishers, Warszawa,

10 480 Jian-Guo Si and Sui Sun Cheng 3. A. Elbert, Asymptotic behaviour of the analytic solution of the differential equation y (t) + y(qt) = 0 as q, J. Comput. Appl. Math. 4 (992), Wang Ke, On the equation x (t) = f(x(x(t))), Funkcial. Ekvac. 33 (990), H. Z. Wu, On the existence and asymptotic behavior of strong solutions of equation x (t) = f(x(x(t))), Ann. Differential Equations, 9 (993), B. Shi and Z. X. Li, Existence of solutions and bifurcation of a class of firstorder functional differential equations, Acta Math. Appl. Sinica 8 (995), B. Shi, Z. C. Wang and Z. X. Li, Asymptotic behavior of the solutions of a class of new type of functional differential equations, J. Hunan University 23 (2) (996), H. Z. Wu, A class of functional differential equations with deviating arguments depending on the state, Acta Math. Sinica 38 (6) (995), B. H. Stephen, On the existence of periodic solutions of z (t) = az(t r + µk(t, z(t))) + F (t), J. Differential Equations 6 (969), K. L. Cooke, Functional differential systems: some models and perturbation problems, Inter. Symp. Diff. Eqs. Dynamical Systems, Puerto Rico, 965. Jian-Guo Si Department of Mathematics, Binzhou Normal College Binzhou, Shandong , P. R. China Sui Sun Cheng Department of Mathematics, Tsing Hua University Hsinchu, Taiwan