Available via license: CC BYNC 4.0
Content may be subject to copyright.
Available via license: CC BYNC 4.0
Content may be subject to copyright.
Electronic Journal of Diﬀerential Equations, Vol. 2009(2009), No. 51, pp. 1–8.
ISSN: 10726691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
ANALYTIC SOLUTIONS OF A FIRST ORDER FUNCTIONAL
DIFFERENTIAL EQUATION WITH A STATE DERIVATIVE
DEPENDENT DELAY
PINGPING ZHANG
Abstract. This article concerns the ﬁrstorder functional diﬀerential equa
tion
x
0
(z) = x(p(z) + bx
0
(z))
with the distinctive feature that the argument of the unknown function de
pends on the state derivative. An existence theorem is established for an alytic
solutions and systematic methods for deriving explicit solutions are also given.
1. Introduction
In the past few years there has been a growing interest in studying functional
diﬀerential equations with state dependent delay. We refer the readers to Eder [1],
Elbert [2], Feckan [3], Stanek [6], Wang [7]. Qiu [4] and Stanek [6] considered the
equation
x
0
(z) = x
p(z) + bx(z )
,
and establish suﬃcient conditions for the existence of analytic solutions. In this pa
per, we are concerned with analytic solutions of the ﬁrstorder functional diﬀerential
equation
x
0
(z) = x
p(z) + bx
0
(z)
, (1.1)
where b is a nonzero complex number, and p(z) is the given complex function of a
complex variable. A distinctive feature of (1.1) is that the argument of the unknown
function depends on the state derivative. To construct analytic solution of (1.1) in
a systematic manner, we ﬁrst let
y(z) = p(z) + bx
0
(z). (1.2)
Then for any number z
0
, we have
x(z) = x(z
0
) +
1
b
Z
z
z
0
(y(s) − p(s))ds (1.3)
and so
x(y(z)) = x(z
0
) +
1
b
Z
y (z)
z
0
(y(s) − p(s))ds.
2000 Mathematics Subject Classiﬁcation. 34K05, 34A25, 39B32.
Key words and phrases. Functional diﬀerential equation; analytic solution.
c
2009 Texas State University  San Marcos.
Submitted September 25, 2008. Published April 13, 2009.
1
2 P. ZHANG EJDE2009/51
Therefore, in view of (1.1) and x
0
(z) =
1
b
(y(z) − p(z)), we have
1
b
(y(z) − p(z)) = x(z
0
) +
1
b
Z
y (z)
z
0
(y(s) − p(s))ds. (1.4)
Furthermore, diﬀerentiating both sides of (1.4) with respect to z, we obtain
y
0
(z) − p
0
(z) = (y(y(z)) − p(y(z)))y
0
(z). (1.5)
To ﬁnd analytic solutions of (1.5), we ﬁrst seek an analytic solution g(z) of the
auxiliary equation
αg
0
(αz) − p
0
(g(z))g
0
(z) = α[g(α
2
z) − p(g(αz))]g
0
(αz), (1.6)
satisfying the initial value conditions
g(0) = 0, g
0
(0) = η 6= 0, (1.7)
where η is a complex number, and α satisﬁes the following conditions:
(H1) p(z) is analytic in a neighborhood of the origin, furthermore, p(0) = β 6=
−1 and p
0
(0) = α + αβ, where α, β are complex numbers;
(H2) 0 < α < 1;
(H3) α = 1, α is not a root of unity, and ln α
n
− 1
−1
≤ k ln n, n = 2, 3, . . . for
some positive constant k.
Then we show that (1.5) has an analytic solution of the form
y(z) = g(αg
−1
(z)) (1.8)
in a neighborhood of the origin.
2. Preparatory Lemmas
We begin with the following preparatory lemma, the proof of which can be found
in cites1.
Lemma 2.1. Assume that (H3) holds. Then there is a positive number δ such that
α
n
− 1
−1
< (2n)
δ
for n = 1, 2, . . . . Furthermore, the sequence {d
n
}
∞
n=1
deﬁned by
d
1
= 1 and
d
n
= α
n−1
− 1
−1
max
k
1
+···+k
m
=n
0≤k
1
≤···≤k
m
,m≥2
{d
k
1
. . . d
k
m
}, n = 2, 3, . . .
will satisfy
d
n
≤ N
n−1
n
−2δ
, n = 1, 2, . . . , (2.1)
where N = 2
5δ+1
.
To proceed, we state and prove two preparatory lemmas which will be used in
the proof of our main result.
Lemma 2.2. Suppose (H1)–(H2) hold. Then for any nonzero complex number η,
equation (1.6) has an analytic solution g(z) in a neighborhood of the origin such
that g(0) = 0 and g
0
(0) = η.
Proof. Because p(z) satisﬁes (H1),we assume
p(z) =
∞
X
n=0
p
n
z
n
, p
0
= β, p
1
= α + αβ. (2.2)
EJDE2009/51 ANALYTIC SOLUTIONS 3
Then there exists a positive constant ρ such that
p
n
 ≤ ρ
n−1
, n = 2, 3, . . . .
Introducing new functions
G(z) = ρg(ρ
−1
z), P (z) = ρp(ρ
−1
z),
we obtain from g(0) = 0 and g
0
(0) = η that G(0) = 0 and G
0
(0) = η respectively,
and by (1.6) we have
α[G(α
2
z) − P(G(αz))]G
0
(αz) = αG
0
(αz) − P
0
(G(z))G
0
(z),
which is again an equation of the form (1.6). Here P is of the form
P (z) =
∞
X
n=0
P
n
z
n
,
but P
n
 = p
n
ρ
1−n
 ≤ 1 for n ≥ 2. Then, without loss of generality,we assume
p
n
 ≤ 1, n = 2, 3, . . . . (2.3)
Next, we assume that (1.6) has a formal power series solution
g(z) =
∞
X
n=1
c
n
z
n
(2.4)
and substitute (2.2) and (2.4) into (1.6), we see that the se quence {c
n
}
∞
n=1
is suc
cessively determined by the condition
αc
1
+ αp
0
c
1
− p
1
c
1
+
∞
X
n=1
[(n + 1)c
n+1
α
n+1
+ p
0
(n + 1)c
n+1
α
n+1
− p
1
(n + 1)c
n+1
]z
n
=
∞
X
n=1
n
X
k=1
c
k
c
n−k+1
(n − k + 1)α
n+k+1
z
n
+
∞
X
n=1
h
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
((m + 1)p
m+1
− p
m
α
n+1
)c
l
1
. . . c
l
m
(n − k + 1)c
n−k+1
i
z
n
,
where n = 1, 2, . . . , in a unique manner.
By comparing coeﬃcients in both sides, it is easy to see that the coeﬃcient
sequence {c
n
}
∞
n=1
satisﬁes
(α + αp
0
− p
1
)c
1
= 0, (2.5)
and
(n + 1)(α
n+1
+ p
0
α
n+1
− p
1
)c
n+1
=
n
X
k=1
c
k
c
n−k+1
(n − k + 1)α
n−k+1
+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
[(m + 1)p
m+1
− p
m
α
n+1
]c
l
1
. . . c
l
m
(n − k + 1)c
n−k+1
,
(2.6)
for n=1,2,. . . .
4 P. ZHANG EJDE2009/51
By (H1), we can ensure α + αp
0
− p
1
= α + αβ − α − αβ = 0. Hence we know it
is suitable for any complex number c
1
= η 6= 0. We may now see by (2.6) that the
resulting relation deﬁnes c
n
(n = 2, 3, . . . ) in a unique manner. Next, we want to
prove that the power series (2.4) is convergent in a suﬃcient small neighborhood
of the origin. With lemma conditions, we can prove α
n+1
+ p
0
α
n+1
− p
1
= α(α
n
−
1)(1 + β) 6= 0 for n = 1, 2, . . . . If not, assuming α(α
n
− 1)(1 + β) = 0, we can get
α
n
= 1, so α = 1 which contradicts condition (H2). To see this, note that
lim
n→∞
1
α(α
n
− 1)(1 + β)
= −
1
α(1 + β)
, 0 < α < 1,
thus there exist some positive number M such that 
1
α(α
n
−1)(1+β)
 ≤ M for n ≥ 1.
By (2.3) and (2.6), we have
c
n+1
 ≤ M(
n
X
k=1
c
k
c
n−k+1
 +
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)c
l
1
 . . . c
l
m
c
n−k+1
), (2.7)
where n = 1, 2, . . . . If we now deﬁne a positive sequence {q
n
}
∞
n=1
by q
1
= η and
q
n+1
= M[
n
X
k=1
q
k
q
n−k+1
+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)q
l
1
. . . q
l
m
q
n−k+1
], n = 1, 2, . . . ,
then it is easily seen that
c
n+1
 ≤ q
n+1
, n = 0, 1, 2, . . . .
In other words, the series
P
∞
n=1
q
n
z
n
is a majorant s eries of
P
∞
n=1
c
n
z
n
. So next
we want to show that the power series
P
∞
n=1
q
n
z
n
is convergent in a suﬃcient small
neighborhood of the origin. Now if we deﬁne Q(z) =
P
∞
n=1
q
n
z
n
, then
Q(z) =
∞
X
n=1
q
n
z
n
=
∞
X
n=0
q
n+1
z
n+1
= ηz +
∞
X
n=1
q
n+1
z
n+1
= ηz + M
∞
X
n=1
(
n
X
k=1
q
k
q
n−k+1
)z
n+1
+ M[
∞
X
n=1
(
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)q
l
1
. . . q
l
m
q
n−k+1
)z
n+1
]
= ηz + M(
∞
X
n=1
q
n
z
n
)
2
+ M[
∞
X
n=1
(n + 2)(Q(z))
n
](
∞
X
n=1
q
n
z
n
)
= ηz + M[Q(z)]
2
+ M[
∞
X
n=1
(n + 1)(Q(z))
n
]Q(z) + M
∞
X
n=1
(Q(z))
n+1
= ηz + M[Q(z)]
2
+ M(
∞
X
n=1
(Q(z))
n+1
)
0
Q(z) + M
Q
2
(z)
1 − Q(z)
= ηz + M
4Q
2
(z) − 4Q
3
(z) + Q
4
(z)
[1 − Q(z)]
2
.
EJDE2009/51 ANALYTIC SOLUTIONS 5
So the function Q = Q(z) is the implicit function which deﬁned by the function
Q = ηz + M
4Q
2
− 4Q
3
+ Q
4
(1 − Q)
2
,
or
F (z, Q) ≡ Q − ηz − M
4Q
2
− 4Q
3
+ Q
4
(1 − Q)
2
= 0.
Because the function F (z, Q) is continuous in a neighborhood of the origin, fur
thermore, F (0, 0) = 0 and F
0
Q
(0, 0) = 1 6= 0. By the implicit function theorem,
we see that the Q = Q(z) is analytic on a disk with the origin as the center and
with a positive radius. The proof is completed.
Lemma 2.3. Suppose (H1), (H3) hold. Then equation (1.6) has an analytic solu
tion g (z) in a neighborhood of the origin such that g(0) = 0 and g
0
(0) = η 6= 0.
Proof. Similar to the proof of Lemma 2.2, we seek a formal power series solution
(2.4) of equation (1.6) with c
1
= η and
(n + 1)α(α
n
− 1)(1 + β)c
n+1
=
n
X
k=1
c
k
c
n−k+1
(n + 1 − k)α
n+k+1
+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
[(m + 1)p
m+1
− p
m
α
n+1
]c
l
1
. . . c
l
m
(n − k + 1)c
n−k+1
(2.8)
for n ≥ 1. So next we want to prove that the power series
P
∞
n=1
c
n
z
n
is convergent
in a suﬃcient small neighborho od of the origin.
The formulation (2.8) can be written as
c
n+1
 ≤
1
1 + β
α
n
− 1
−1
[
n
X
k=1
c
k
c
n+1−k

+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)c
l
1
 . . . c
l
m
c
n−k+1
], n = 1, 2, . . . .
(2.9)
If we now deﬁne a positive sequence {v
n
}
∞
n=1
by v
1
= η and
v
n+1
= Mα
n
− 1
−1
h
n
X
k=1
v
k
v
n−k+1
+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)v
l
1
· · · v
l
m
v
n−k+1
i
,
where M = 1/(1 + β) > 0, n ≥ 1, then it is not diﬃcult to show by induction that
c
n+1
 ≤ v
n+1
, n = 0, 1, 2, . . . . (2.10)
In other words, V (z) =
P
∞
n=0
v
n
z
n
is a majorant series of g(z) =
P
∞
n=0
c
n
z
n
. we
now only need to show that V (z) has a positive radius of convergence. To see this,
we deﬁne the positive recursive sequence {q
n
} by q
1
= η and
q
n+1
= M
h
n
X
k=1
q
k
q
n−k+1
+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)q
l
1
. . . q
l
m
q
n−k+1
i
, n = 1, 2, . . . .
6 P. ZHANG EJDE2009/51
Then
Q(z) =
∞
X
n=1
q
n
z
n
=
∞
X
n=0
q
n+1
z
n+1
= ηz +
∞
X
n=1
q
n+1
z
n+1
= ηz +
∞
X
n=1
M(
n
X
k=1
q
k
q
n+1−k
+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)q
l
1
. . . q
l
m
q
n−k+1
)z
n+1
= ηz +
4Q
2
(z) − 4Q
3
(z) + Q
4
(z)
[1 − Q(z)]
2
.
(2.11)
Hence by induction, we easily see by Lemma 2.1 that
v
n+1
≤ q
n+1
d
n+1
, n = 0, 1, 2, . . . , (2.12)
where the sequence {d
n
}
∞
n=1
is deﬁned by Lemma 2.1.
In fact, if k = 0, it holds. Assume by induction that v
k
≤ q
k
d
k
for k =
1, 2, . . . , n − 1. Then
v
n+1
= Mα
n
− 1
−1
h
n
X
k=1
v
k
v
n+1−k
+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)v
l
1
. . . v
l
m
v
n−k+1
i
≤ Mα
n
− 1
−1
h
n
X
k=1
q
k
d
k
q
n+1−k
d
n+1−k
+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)q
l
1
. . . q
l
m
d
l
1
. . . d
l
m
q
n−k+1
d
n−k+1
i
≤ M
h
n
X
k=1
q
k
q
n−k+1
+
n
X
k=1
X
l
1
+···+l
m
=k
m=1,2,...,k
(n + 2)q
l
1
. . . q
l
m
q
n−k+1
i
× α
n
− 1
−1
max
l
1
+···+l
m
=n+1
m=1,2,...,n+1
{d
l
1
. . . d
l
m
}
= q
n+1
d
n+1
, n = 0, 1, 2, . . . .
By equation (2.11), the implicit function of Q = Q(z) is deﬁned by
F (z, Q) = Q − ηz −
4Q
2
− 4Q
3
+ Q
4
(1 − Q
2
)
.
In view of F (0, 0) = 0 and F
0
Q
(0, 0) = 1 6= 0, by virtue of the implicit function
theorem there exists a positive constant δ such that the function Q(z) =
P
∞
n=1
q
n
z
n
converges for z < δ. So there exists k > 0 such that q
n
≤ R
n
for n = 1, 2, . . . .
By Lemma 2.1 and (2.12), we ﬁnally see that
v
n
≤ R
n
N
n−1
n
−2δ
, n = 1, 2, . . . ,
which implies that the series
P
∞
n=1
v
n
z
n
converges for z < (RN)
−1
, therefore, the
series (2.4) also converges for z < (RN)
−1
. This completes the proof.
EJDE2009/51 ANALYTIC SOLUTIONS 7
3. Existence of Analytic Solutions to (1.1)
In this section, we state and prove our main result in this article.
Theorem 3.1. Suppose the conditions of Lemma 2.2 or lemma 2 .3 are satisﬁed.
Then (1.5) has an analytic solution g(z) of the form (1.8) in a neighborhood of the
origin, where g(z) is an analytic solution of (1.6).
Proof. In view of Lemma 2.2 or lemma 2.3, we may ﬁnd a sequence {c
n
}
∞
n=1
such
that the function g(z) of the form (2.4) is analytic solution of (1.6) in a neigh
borhood of the origin. Since g
0
(0) = η 6= 0, the function g
−1
(z) is analytic in a
neighborhood of the point g(0) = 0. If we now deﬁne y(z) by (1.8), then
y
0
(z) − p
0
(z) =
αg
0
(αg
−1
(z))
g
0
(g
−1
(z))
− p
0
(z) =
αg
0
(αg
−1
(z)) − p
0
(z)g
0
(g
−1
(z))
g
0
(g
−1
(z))
,
and
[y(y(z)) − p(y(z))]y
0
(z) = [g(αg
−1
(gαg
−1
(z))) − p(g(αg
−1
(z)))]
αg
0
(αg
−1
(z))
g
0
(g
−1
(z))
=
αg
0
(αg
−1
(z)) − p
0
(z)g
0
(g
−1
(z))
g
0
(g
−1
(z))
as required. The proof is completed.
In the last section, we have shown that under the conditions of lemma 2.2 or
lemma 2.3, equation (1.5) has an analytic solution y(z) = g(α g
−1
(z)) in a neigh
borhood of the point, where g is an analytic solution of (1.6). We now show how
to explicitly construct an analytic solution of (1.1) by (1.5). In view of
x
0
(z) =
1
b
[y(z) − p(z)],
we have x
0
(0) =
1
b
[y(0) − p(0)] = −β/b. Furthermore,
x
0
(0) = x(p(0) + bx
0
(0)) = x(β − β) = x(0) = −
β
b
.
By calculating the derivatives of both sides of (1.1), we obtain successively
x
00
(z) = x
0
(p(z) + bx
0
(z))(p
0
(z) + bx
00
(z)),
x
000
(z) = x
00
(p(z) + bx
0
(z))(p
0
(z) + bx
00
(z))
2
+ x
0
(p(z) + bx
0
(z))(p
00
(z) + bx
000
(z)),
so that
x
00
(0) = x
0
(p(0) + bx
0
(0))(p
0
(0) + bx
00
(0)),
x
000
(0) = x
00
(0)(p
0
(0) + bx
00
(0))
2
+ x
0
(0)(p
00
(0) + bx
000
(0));
that is,
x
00
(0) = −
αβ
b
, x
000
(0) = −
β(α
3
+ p
2
)
b(1 + β)
.
In general, we can show by induction that
(x(p(z) + bx
0
(z)))
(m)
=
m
X
i=1
p
im
(p
0
(z) + bx
00
(z), p
00
(z) + bx
000
(z), . . . , p
(m)
(z)
+ bx
(m+1)
(z))x
(i)
(p(z) + bx
0
(z)),
8 P. ZHANG EJDE2009/51
where m = 1, 2, . . . , and p
im
is a polynomial with nonnegative coeﬃcients. Hence
x
(m+1)
(0) =
m
X
i=1
p
im
(p
0
(0) + bx
00
(0), p
00
(0) + bx
000
(0), . . . , p
(m)
(0)
+ bx
(m+1)
(0))x
(i)
(0) =: Γ
m
for m = 1, 2, . . . . It is then easy to write out the explicit form of our solution
x(z) = −
β
b
−
β
b
z −
αβ
2!b
z
2
−
β(α
3
+ p
2
)
3!b(1 + β)
z
3
+
∞
X
m=3
Γ
m
(m + 1)!
z
m+1
.
References
[1] E. Eder; The functional diﬀerential equation x
0
(z) = x(x(t)), J .Diﬀerential Equations.
54(1984): 390400.
[2] A. Elbert; Asymptotic behavior of the analytic solution diﬀerential equation y
0
(t) + y(qt) = 0
as q → 1
−1
, J. Comput. Appl. Math. 41(1992): 522.
[3] E. Feckan; On certain type of functional diﬀerential equations, Math. Slovaca, 43(1993):
3943.
[4] Fang Qiu, HanZe Liu; Analytic Solutions of a First Order Iterative Diﬀerential Equation
x
0
(z) = x(p(z) + bx(z)), Demonstratio Mathematica, 2006, 39(1): 8190.
[5] C. L.Siegel; Iterative of analytic function, Ann of Math. 1942, 43 (4): 607612.
[6] S. Stanek; On global properties of solutions of functional diﬀerential equation x
0
(t) =
x(x(t)) + x(t), Dynamic Sys. Appl., 4(1995): 263278.
[7] K. Wang; On the equation x
0
(t) = f(x(x(t))), Funkcialaj Ekvacioj, 33(1990): 405425.
Pingping Zhang
Department of Mathematics, Binzhou University, Shandong 256603, China
Email address: zhangpingpingmath@163.com