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Computers and Mathematics with Applications 43 (2002) 81-90
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Analytic Solutions
of a Second-Order Iterative
Functional Differential E q u a t i o n
J I A N - G U O S I AND K I N - P I N G W A N G
Department of Mathematics, Binzhou Normal College
Binzhou, Shandong 256604, P.R. China
(Received August 2000; acceptedJanuary 2001)
A b s t r a c t - - T h i s paper is concerned with a second-order iterative functional differential equation x'(z) = (xm(z)) 2. Its analytic solutions are discussed by locally reducing the equation to
another functional differentialequation with proportional delays ~2y'(l.tz)y~(z) = i.Ly~(l.~z)y'(z)+
[~It(Z)]3[~([J,
mZ)] 2 and by constructing a convergent power series solutions for the latter equation.
© 2001 Elsevier Science Ltd. All rights reserved.
K e y w o r d s - - F u n c t i o n a l differential equation, Analytic solution.
1. I N T R O D U C T I O N
In the last few years, there has been a growing interest in studying the existence of solutions of
functional differential equations with state dependent delay. We refer the reader to the papers
by Cooke [1], Eder [2], Feckan [3], Stanek [4], Grimm [5], Driver [6], Oberg [7], Jackiewicz [8],
Dunkel [9], Wang [10], and the authors of [11-17]. However, there are only few papers dealing
with a second-order iterative functional differential equation. In [18], Peta.hov considered the
existence of solutions of the equation
x"(t) = a ~ ( x ( t ) ) .
In [12], the authors studied the existence of analytic solutions of the equation
x"(z) =x(az+bx'(z)).
The purpose of this paper is to discuss the existence of analytic solutions of a second-order
iterative functional differential equation of the form
where
xm(z)
x"(z)=(xm(z))
2,
denotes the rr~th iterate of the function
x(z).
(1)
Project supported by the Natural Science Foundation of Shandong Province.
0898-1221/01/$ - see front matter (~) 2001 Elsevier Science Ltd. All rights reserved.
PII: S0898-1221(01)00273-5
Typeset by ~4A~-TEX
82
J.-G. Sl AND X.-P. WANG
The general idea for our discussion is similar to that used in [12]. To find analytic solutions
of (1), we locally reduce equation (1) to another functional differential equation with proportional
delays (the research of such equations with proportional delay is in itself very interesting, see [19221)
~ y " ( ~ z ) ~ ' ( z ) = ~y'(~z)y"(z) + [y'(z)] 3 [y ( ~ z ) ] ~
(2)
satisfying the condition
y(0) = s,
y'(0) = ~ # 0,
(3)
called the auxiliary equation of (1), where # satisfies one of the following conditions:
(H1) 0 < I~1 < 1;
(H2) I~l = 1,# is not a root of unity, and
log - -
1
I~n - 11
n=2,3,...,
_< K l o g n ,
for some positive constant K. Then we show that
X ( Z ) = y (/zy--I(z))
is an analytic solution of (1) in a neighborhood of s. Here y - l ( z ) denotes the inverse
function of y(z).
2. S O M E
PREPARATORY
LEMMAS
In this section, we will state and prove some preparatory lemmas which will be used in the
proof of our main result.
LEMMA 1. Assume that (HI) holds. Then for the initial conditions (3), equation (2) has an
analytic solution of the form
oo
y(z) = s + rlz + E
bnzn
(4)
n=2
in a neighborhood of the origin.
PROOF. Rewrite (2) in the form
, y " ( , z ) y ' ( z ) - y ' ( , z ) y " ( z ) = ! y'(z) [y ( , ~ z ) ] 2
[y'(z)] ~
or
( y ' ( # z ) ~'
1 ,
\ y,(z) ] = ~ y (z) [y ( , ~ z ) ] 2
Therefore, in view of yr(0) = 7/• 0, we have
y ' ( . z ) = y'(z)
1 + -~
y'(t) (y (.mt)) 2 at .
(5)
We now seek a solution of (2) in the form a power series (4). By defining b0 = s and bl = ~
and substituting (4) into (5), we see that the sequence {bn}~=~ is successively determined by the
condition
~_~ ~
(/zn+l - 1) (n + 2)bn+2 = ~ ~
i=0 j=O
n-~-j (i + 1)(j + 1)It re(n-i-j)
E
bi+ l bj+ l bkbn-i-j-k,
n
i+1
k=O
n = 0,1,2,...,
(6)
Analytic Solutions
83
in a unique manner. We need to show that the power series (4) converges in a neighborhood of
the origin. First of all, since
I (i + 1)(j + 1)Ure(n-i-j)
I<
(/~-+T - - ~ - ~ ~
1
1) - I#"+1 - 11 < M,
oo
for some positive number M, thus, if we define a sequence {Dn},~=0
by Do = lsl, D1 = JrlJ, and
~
n-i n--i--j
Di+ I Dj+ I Dk D n - i - j - k ,
i=0 j=0
n=0,1,...,
k----0
then in view of (6),
Ib~l ~ Dn,
Now, if we define
n = O, 1,....
oo
G(z) = E Dnzn,
r~=O
(7)
then
n=0
k=0
n=0
j=0
k=0
and
+ E
Dj+lDkDn-j-k
n=0
j=0
= Isl2 ~
DkDn-k
n=0
zn
k=0
+21sl ~
Dj+lDkDn-j-k
n=0
j=0
qi=0 j=0
z n+l
k=0
E
n=0
z n+l
k=0
Di+lDj+lDkDn-i-j-k
k=0
)
z n+2
1 oo
= Isl2G2(z) + 2lsl (G3(z) - Isla2(z)) + - ~ ~_. Dn+2Zn+2
rl.~O
1
= I s l 2 a 2 ( z ) + 21sl (G3(z) - IslG2(z)) + ~ ( G ( z ) - Isl - Irllz)
= 21slGa(z) - Isl2G2(z) + M G ( z ) - - ~1
(l~lz + Isl),
84
J.-G. Sl ANDX.-P. WANG
t h a t is,
a 4 ( z ) - 2]s[a3(z) + Is[~a2(z) Let
1
G(z) + ~ ([~l z + Isl) = O.
M
(8)
1
R(z,03) = 034 _ 21s[033 + 1812032_ 1 0 3 + ~ (l~lz + 181),
for (z,w) from a neighborhood of (0, Is D. Since R(0, [sl) = 0,R~(0, Is[) = - ( l / M ) ¢ 0, there
exists a unique function 03(z), analytic on a neighborhood of zero, such t h a t 03(0) = [s[,03'(0) = 171
and satisfying the equality R(z,03(z)) = 0. According to (7) and (8), we have G(z) = 03(z). It
follows t h a t the power series (7) converges on a neighborhood of the origin, which implies t h a t
the power series (4) also converges in a neighborhood of the origin. T h e proof is complete.
1
Now we introduce the following lemma, the proof of which can be found in [23, C h a p t e r 6]
or [24, pp. 166-174].
LEMMA 2. Assume that (H2) holds. Then there is a positive number 5 such that 1#~ - 11-1 <
(2n) ~ for n
1, 2 , . . . . Furthermore, the sequence { ,~}n=l defined by dl
1 and
1
dn
l# ~-1 - 1]
n=nl+...+n,max {d~, . . . d ~ } ,
n = 2,3,...
O<nl <...<nt,t>2
will satisfy
d~z <~ (255+1) n - i n -25,
n= 1,2,....
LEMMA 3. Suppose (H2) holds. Then for 0 < [~71 -< 1, equation (2) has an analytic solution of
the form
oo
y(z) = s + ~z + ~
b~z ~
(9)
n=2
in a neighborhood of the origin.
PROOF. As in the proof of L e m m a 1, we seek a power series solution of form (9). T h e n defining b0 = s and bl = 7/, (6) holds again. So t h a t
n
n - i n-i--j
[bn+2[_<l#n+'- I1-1 E E E
i=O j=0
Ibe+~Ilbj+l [[bkl[bn-~-j-k[,
n = 0,1,....
(10)
k=O
Let us now consider the equation
W(z, 03)
=
03 4
--
2lS]03 3 -[- IS12032 -- 03 + Isl + Z = O,
(11)
for (z, 03) from a neighborhood of (0, Isl). Since W(0, Is[) = 0 and W~' (0, Isl) = - ( l / M ) ¢ 0, there
exists a unique function 03(z), analytic on a neighborhood of zero, such t h a t 03(0) = Is[, 03'(0) = 1
and satisfying the equality W(z,03(z)) = O. Now if
oo
03(z) = ]s[ + z + E
B~z~'
(12)
n~2
where the coefficient sequence {B}~=o satisfies B0 = [sl, B1 = 1 and
~
.°+2=
n--i n--i--j
EE
i=O j=0
Bi+ l Bj+ l Bk B n - i - j - k ,
k=O
n=O, 1,...,
(13)
Analytic Solutions
85
then
w2(z) = ~
n=0
BkB.-k
z n,
k=0
023(Z) : (]SI"[" ~n~oBn"['lZn-bl) (n=~O(k=~oSkBn-k) zn )
n=O k=O
)
n=O j=O k=O
and
w4(z)
=
Z
Isl-[-
Bn+lzn+l
+~
n=O
B~Bn-k
Isl
n----0
Bj+lBkBn-j-k
j--0
BkBn-k
z"
k--0
Bj+lBkBn-j-k
+ 21Sl Z
n=O
z n+l
j----0
+Z
~
n=0
z '~+1
k=O
= Isl2 ~
n=0
z n
k:0
i=0 j=0
Bi+lBj+lBkB,~-i-j-k
z '~+2
k=0
oo
= Isl~2(z)
+ 21sl (~3(z)
-
Isl~2(z)) + ~ Sn+2zn+2
n-----0
= Is12~2(z) + 21sl (~3(z) -Isl~2(z)) + ~(z) - I s l -
z
= 21sl~o3(z) - Isl2~o~(z) + ~o(~) - Isl - z,
t h a t is, w(z) satisfies equation (11). It follows t h a t the power series (12) converges in a neighb o r h o o d of zero, a n d there is a positive constant T such t h a t
Br, <_T n,
n = 1,2,....
Now by induction, we prove t h a t
Ibn[ <_Bndn,
where t h e sequence
CO
{dn}n=l
n = 1,2,...,
is defined in L e m m a 2. In fact,
Ibll = I~/I <
1 = Cldl,
[b21= I# - l[-Xlblllbll[bo[Ibol
< ]#- l l - l B 1 d l . B i d 1 • Bo" Bo
= B21/~ - 1[ -1
m a x {dn,dn2}
0~nl~n2
= B2d2.
(14)
86
J.-G. Si AND X.-P. WANG
A s s u m e t h a t the above inequality holds for n -- 1, 2 , . . . , l. T h e n
l-1 l-l--i l - - l - i - j
[bt+ll -< Itzt - I I - 1 E E
E
i=o j=0
Ibi+ l [Ib~+ l l lbk l lbt- l - i - 3 - k l
k=0
~
l-1 l - 2 - i
-[t'l- 11-1 \ 2 ~=o Ib~+lllbj+~llbollbol+2 ~~=o j=o
-
+
Ibi+lllbj+lllbollbl-l-~-jl
[b~+xllbj+lllbkllbt-l-~-3-kl
i=o
j=o
k=l
<_[#l-- l[-1 ( 2 ~
k
Bi+ldi+lBj+ldj+lBoBo
i=0
l-1 l - 2 - i
+2 E
E
i=o j=o
Bi+ldi+lBj+ldj+lBoBl-l-i_jdl_l_i_j
l-1 l--l-i l - 2 - - i - j
+EE
i=0
Bi +l di+l B j +l dj +l Bkda Bt- l-i-j_adt_ l_i_j_k )
j=0
k=l
l-1
< [~z _
11-1
max
nl+n2+...+nt=l+l;
O<nl <_...<nt ,t> 2
{d,~...dm}
l-1 l - 2 - i
+2EE
i=0
2 E Bi+IBj+IBoBo
/=0
l-1 l - l - i l - 2 - i - j
Bi+lBj+lBoBl-l-i_j + E
j=0
i=0
E
E
j=0
k=l
Bi+lBj+lBkBl-l-i-j_k)
= Bl+ldl+l
as desired. In view of (14) and L e m m a 2, we finally see t h a t
[bnl < T n (25~+1) ~-1 n -2~,
n = 1,2,...,
t h a t is,
lim sup (Ib~l) 1/n _< lim sup T
~-'*OO
(255+1) (n--1)/nn_26/n
=
T (255+1) ,
n--* OO
This implies t h a t the series (9) converges for [z[ < (T(25~+1)) -1. T h e proof is complete.
3. E X I S T E N C E
OF ANALYTIC
SOLUTIONS
OF EQUATION
II
(1)
In this section, we s t a t e and prove our main result in this note.
THEOREM 1. Suppose the conditions of Lemma 1 or Lemma 3 are satisfied. Then equation (1) has
an analytic solution x(z) in a neighborhood of s. This solution has the form x(z) = y(#y-1 (z)),
where y(z) is an anMytic solution of the initial problem (2),(3).
PROOF. B y L e m m a 1 and L e m m a 3, we can find an analytic solution y(z) of the auxiliary
equation (2) in the from of (4) such t h a t y(0) -- s and y'(0) -- y ¢ 0. Clearly, the inverse y-l(z)
exists and is analytic in a neighborhood of y(0) = s. Let
x(z) = y ( , y - i ( z ) ) ,
(15)
Analytic Solutions
87
which is also analytic in a neighborhood of s. From (2), it is easy to see
X'(Z) -~ ~y' (~y-l(z)) (y-l(z))! ---- ~y' (]2Y-I(z))
y' (y-~(z)) '
~2y,, (]£y-l(z)) __ ~y, (~y-l(z)) y,, (y--l(z)) " (l/y/ (y-l(z)))
• "(z) =
( ¢ (Y-~(Z))):
[~y" ( ~ y - ~ ( Z ) ) ¢ (y-~(Z)) -- ¢ ( ~ y - ~ ( z ) ) ¢ ' (y-~(z))]
b' (y-~(z))f
= [y ( ] A m y - l ( z ) ) ] 2 = (xrn(z)) 2
as required. The proof is complete.
|
In the above theorem, we have shown that under the conditions of Lemma 1 or Lemma 3,
equation (1) has an analytic solution x(z) = y(py-l(z)) in a neighborhood of the number s,
where y is an analytic solution of (2). Since the function y(z) in (11) can be determined by (6),
it is possible to calculate, at least in theory, the explicit form of x(z). However, knowing t h a t an
analytic solution of (1) exists, we can take an alternative route as follows. Assume that x(z) is
of the form
-~"(~)
~ . ( z - ~)~ + . . ,
• (z) = ~(~) + ~'(s)(z - s) +
we need to determine the derivatives
x(n)(s),
n = O, 1 , . . . . First of all, in view of (15), we have
• (s) = y ( , y - ~ ( ~ ) ) = y ( , . 0) = y(0) =
and
x'(s) = Èy' ( , y - l ( s ) )
y'(y-~(s))
=
, y ' ( , o)
y'(0)
= "'
respectively. Furthermore,
• "(s) = ( ~ ( s ) ) ~
= ~.
x"(~) = 2x~(~)~ ' ( ~ m - l ( ~ ) ) .
~- 2s# m.
x'(~(s))~'(s)
For convenience, we take the following notations:
• i~(z) = ~ ( ~ ) ( x J ( z ) )
By induction, we may prove that
(Xm(Z)) (k) "~ Pm,k (Xlo(Z),.-.,Xl,m-I(Z);.-.;Xko(Z),...
,xk,m_l(z)),
where Pm,k is a uniquely defined multivariate polynomial with nonnegative coefficients. Next by
calculating the higher derivatives of both sides of (1), we can obtain
x(3+n)(z) = 2 (zm(z)(zm(z))') (n)
n
= 2 ~
c~ (~m(zll(k)(~(zl)
(~-k÷l)
k=0
n
=
2~
[CknPm,k(Xio(Z) .... ,Xl,m-I(Z);... ;Xko(Z),... ,Xk,rn_l(Z))
k=O
X Pm,n-k+l (XlO(Z),... ,Xl,m-l(Z);... ;Xn-k+l,O(Z),... ,Xr~-k+l,m-l(Z))]
J.-G. $I AND X.-P. WANG
88
for n = 1, 2 , . . . . Hence,
~-~ [cknpm,k ( X l 0 ( S ) , . . . , X l , m - l ( S ) ; . . . ; X k 0 ( S ) , . . . , X k , m - l ( S ) )
k=O
x Pro,n-k+1 (Xl0(S),..., Xl,rn-1 ( s ) ; . . . ;Xn-k+l,O(S),... , Xn_k+l,m_ 1(8)) 3
x (a+~)(s) = 2
Cknpmk
=2
...
x(k)(s),..
k=O
X Pm,n-k+l (Xt(S),''',Xt(8);''';X(n-k+l)(8),
=
kp,
k=O
X Pm,n-k+l
...,
.
". ,x(n-k+l)(s))]
..
([2,...,#;...;x(n-k+l)(s),...,x(n-k+l)(8))]
for n ----1, 2 , . . . . It is t h e n easy to write out the explicit form of our solution x(z):
oo
s~ (z - s) 2 + ~2s~'~ (z - s) 3 + ~
x ( z ) = s + # ( z - s) + -~.
( n ~'~
+ 3)-----~(z - s) '~+3 '
n=l
THEOREM 2. E q u a t i o n (1) has an a n a l y t i c s o l u t i o n o f the form
X(Z) = S(1-fl)Z fl
(16)
in t h e region [ z - s[ < s s a t i s f y i n g
x(s) = s,
x'(s) = Z,
(17)
where s = I31
fl(X/-fi~ - 1) a n d fl is a r o o t o f algebraic e q u a t i o n
2fl m - fl + 2 = O.
(i8)
PROOF. Let us seek for a solution of (1) in the form
x(z) = ~z ~
(19)
Substituting (19) into (1) and comparing the coefficients, we obtain
A f l ( f l - 1) -- A2( I+fi+'''+z .... 1),
fl-2
= 2 ~ m,
t h a t is,
A = [ f l ( f l - 1)1(1/3)(1-fl) = sl-fl,
2Z TM - fl + 2 = 0
(20)
(21)
Therefore, in view of (20) and (21), we see t h a t x ( z ) -- s l - ~ z ~ is a solution of (1) and show easily
t h a t (17). In fact, we have
x ( s ) = As ~ = s 1 - ~ . s fi = s,
x ' ( s ) = flAs ~-1 = f l s l - ~ s ~-1 = ft.
Analytic Solutions
89
Moreover, since
x(z) = si-~z~ = si-Z " s~ (l + Z - S)
=s
(
1+
[
=s
s
1+~
+...+/3(/~-l)...(/3-n+l)n[
(~_~__~)n + ' " ]
s) 2
f~(f~~ - 1) (z -
=~+~(z-s)+
+ . . - + ~ ( ~ - i)... ( ~ - n + i) ( z - ~)~ + . . .
n!sn-1
we see that (16) is analytic in the region [z - s I < s. The proof is complete.
We remark that the equation 2ft 2 - / ~ + 2 = 0 has two roots
i+iv~
~+="
4
'
if m = 2 in (21). Thus, equation (1) has two analytic solutions
x+(z) = [l~+(fl+_ l)](1/3)(1-~+) z~+ __ ( -9
_q=_S!~)(3:]=iq"~)/12z(l:kix/T5)/4,
such that
x~ ([z~(e~ - 1)1 '/3) = [z~(e~ - 1)11/3
and
/3+ is not a root of unity.
If m = 3 in (21), the equation 2/~3 - / ~ + 2 = 0 has three roots
.. 'V108-
'V,0 +
/ 3 , = - ~ 1 ( g 2 I V l 0 8 _ 63 V / ~ + g l l v l 0 8 + 6
where
-i
~I
-
-
+ i~
2
iv~
-1 '
C2 --~
3V/~)
2
Thus, equation (1) has three analytic solutions
x~(z) = s~'-" z~, ,(s' = [zj(z~ - 1)] lj3, ~ = 1,2,a),
such that
(22)
xj(sj) --- sj(j = 1, 2, 3) and Ixi(sx)l = lt311 > 1, Ix'j(sj)l -- I/~jl < l(j -- 2, 3). Obviously,
the solution Xl(Z) = s 1-~1
1 z~l is not included in Theorem 1.
90
J.-G. SI AND X.-P. WANG
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