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J. Math. Anal. Appl. 317 (2006) 320–330
www.elsevier.com/locate/jmaa

On the functional-differential equation of advanced type
2 with 0
0
Tsuyoshi Yoneda
Department of Mathematics, Osaka Kyoiku University, Kashiwara, Osaka 582-8582, Japan
Received 25 August 2004
Available online 22 December 2005
Submitted by T. Krisztin

Abstract
In this paper we construct solutions for the equation
2
0

0

where is a constant with
0. The solutions are infinitely differentiable and bounded on . Using our
method, we can get numerical data easily with a computer. Applying one of the solutions we show that the
derivative of order
0 of a function
or
coincides lim 0
, where
is the Sobolev space and
is
a
family
of
-functions.
0
2005 Elsevier Inc. All rights reserved.
Keywords: Functional-differential equation; Advanced type

1. Introduction
Frederickson [1,2] (1971) investigated functional-differential equations of advanced type
(1)
where
1, and provided several properties of solutions. Kato and McLeod [4] (1971) and Kato
[3] (1972) studied asymptotic behavior of solutions of (1).
Frederickson [1] provided a global existence theorem for equations
2
E-mail address: bkaoj300@rinku.zaq.ne.jp.
0022-247X/$ – see front matter
doi:10.1016/j.jmaa.2005.11.034

2005 Elsevier Inc. All rights reserved.

T. Yoneda / J. Math. Anal. Appl. 317 (2006) 320–330

321

where is an odd, continuous function with
0 for
0. In [1] he applied the Schauder
fixed point theorem to the proof. He showed that the absolute value of the solution
is periodic for
0. Frederickson [2] also provided a constructive method of solutions for equations
of advanced type

where

and

1. He gave solutions in the form of a Dirichlet series
0

where is allowed to vary as a parameter. In the case of
0 and
, the solution is analytic
in the upper half plane
0, continuous on
0, and the line
0 is a natural boundary.
In this paper, using another method, we construct solutions for the equation
2
0

(2)

0

where is a constant with
0. The solution is not unique. If is a solution, then a constant
times is also a solution. Our solutions are infinitely differentiable and bounded on . Using
our method we can get numerical data easily with a computer. We also give the Fourier transform
of one of our solutions and show the uniqueness of the solution of (2) with a certain condition.
Applying one of the solutions we show that the derivative of order
0 of a function
or
coincides lim 0
, where
is the Sobolev space and
0 is a
family of
-functions.
2. Main results
Our main results are the following:
Theorem 1. If
0, then there exists a solution
of (2) with
0 such
that vanishes on
0 and
has period 4 on 0
. If
0, then there exists a
solution
of (2) with
0 such that vanishes on 0
and
has
period 4
on
0.
If there exists a solution
0

4
0

of

2

(3)

such that vanishes on
0 and
has period 1 on 0
, then
4 is a solution
of (2). If
0, then
4 vanishes on
0 and
4 has period 4 on 0
.
If
0, then
4 vanishes on 0
and
4 has period 4
on
0.
To construct a solution of (3), we use the function in the next lemma.
Lemma 2. There exist a nonnegative function
(i)
(ii)
(iii)

,
satisfies (3) for 0
1
for 0

1 2,
1,

with supp

0 1 such that

322

T. Yoneda / J. Math. Anal. Appl. 317 (2006) 320–330

Fig. 1.

(iv)
(v)
(vi)
(vii)

.

1 2
2 for 0
1 2,
is increase for 0
1 2,
0
1
0 for
0 1 2
, and
1.

1
0

Remark 3. The function
then
for 0

in Lemma 2 is unique. If satisfies (ii), (iii) and
1 (see Theorem 11 and Remark 12).

The graph of the function
Theorem 4. A solution

1 4

(

),

is in Fig. 1.
of (3) is expressed by

1

1

(4)

1

where

is in Lemma 2 and
0
1
1 2
0
2 1
0 2
1
2 1

1

2

if
if

1
0

2
2

Remark 5. Remark 3 shows that, if
is a solution of (3) with
1
for 0
1
and 1 4
, then
for 0
1, where is in (4). If is a solution of (3), then
the value of
on 2 2 1 is determined by the value of
on 2 1 2 ,
0 1 2
.
Therefore,
for
0.
The solution
2

1

satisfies
1

2

1

0 2

1

1 2

(5)

and
has period 1 on 0
as Frederickson mentioned in [1]. Actually,
1
(
0 1 2 ) implies 1 2
1 2
(
0 1 2 ). Then 4 1 2
1 2
1 2
4 1 2 (
0 1 2 ), i.e. 1
1
(
0 1 ). This is the
equality (5) for
1. In the same way we have (5) for
2 3
. Moreover,
.
The graph of the solution
of (3) is in Fig. 2.
The next theorem is another expression of the solution . Let

T. Yoneda / J. Math. Anal. Appl. 317 (2006) 320–330

Fig. 2. The solution

of (3).

and sinc
sin
. Let be the space of all rapidly decreasing functions and
space of all tempered distributions.
Theorem 6. The Fourier transform of the solution
1

1

323

be the

in (4) is expressed by

in

1

where
2

lim

sinc

2

1

sinc

2

1

uniformly on

1

For a nonnegative integer
in (4), we have the following.
Theorem 7. Let

, let

be the Sobolev space. Applying

be the solution in (4) and
2

If

and 1

or

1 2

1

1

02

0 1

, then

lim

0

uniformly on each compact subset in
Remark 8.
molifier

is in

or in

, respectively.

with compact support. To prove the theorem, we use Friedrichs’
, where
,
0, and is the function in Lemma 2.

We construct a solution of (3) and prove Lemma 2 and Theorem 4 in the next section. We
prove Theorems 6 and 7 in Sections 4 and 5, respectively.

324

T. Yoneda / J. Math. Anal. Appl. 317 (2006) 320–330

3. Construction of a solution of (3)
In this section we prove Lemma 2 and Theorem 4 to construct a solution of (3). The initial
value problem (3) is equivalent to the integral equation
2

2

(6)
0

If

is a solution of (3) and satisfies

1

0

1 , then we have

2

2

1
0
21

21

2

2
0

: supp
2
0
1
2

2
2

1

2

and an operator

0 1

1
2

2

0

Then we define a function space
1

1

1

1

as follows:

1
0 1 2
1 2 1

1

0

0 1

We construct a function
such that
in the proof of Lemma 2. Then
satisfies
(6) for
0 1 2 . The function
satisfies (6) for
0 clearly. In the proof of
Theorem 4 the function
is extended uniquely to the right (increasing value of ) by using
the equality
4 2 .
3.1. Proof of Lemma 2 (Step 1)
We state two lemmas. Let
Lemma 9. The operator
2

01

be the characteristic function of the interval

.

is expressed by
2

(7)

and satisfies
1

1

(8)
0

0

0 if

0

(9)

1

2

(10)
0

T. Yoneda / J. Math. Anal. Appl. 317 (2006) 320–330

Proof. From supp

0 1 and

2

01

2

2

2

01

it follows that

12

1

325

1

2

01

2

2

2

0

12

0

Then we have (7). The properties (8)–(10) follows from (7) and the definition of .

in

Let 0
and
01
1
. It follows from Lemma 9 that

,

0 1 2

. Then

0

is a sequence of functions

1

1

0

2

(11)

0

We note that
1 2

2

0 1 2

0 1 2

(12)

Actually,
2

1 2

2

2

2

1

2

0

0

2
0

Lemma 10. There exists a function
Proof. Let

1,

2

such that

1 2

1 2
1
3 2

2

. Then

0

as
and

1

0 1 2
0 1
1 2 1

0

uniformly on .

1 2

(13)

For example,
1

2

1

0

4

4

1

01 2

16

1

16

4

4

3

1 21

1 4

01 4

01 2

4
1 8

5 8

2

2

1 4

1 23 4

1 21

01

16
16

3 8
7 8

2

2

1 4
1 4

1 41 2
3 41

We show
0
0
by induction. At first,
0 1 4 , and

0 1 4
3 4 1
1 4 3 4
1

1 2

satisfies (14) clearly. Assume that
satisfies (14). From
1 2
0,
0 1 2 , it follows that
2

1

2

0
0

and

0 1 8

(14)
0,

326

T. Yoneda / J. Math. Anal. Appl. 317 (2006) 320–330

Fig. 3. 1

2

2
0

1 4

1,

2

1 2 2

We note that supp

sup
Then we have

1 instead of .
(15)

0 1 . From (13) for
sup
sup
sup

01 4

. For

1

2

1

2
0

since

0(

0 1 4 ) and
1

1 8

0(
2
0

Therefore, we obtain (15). Thus

1

1 8

0

1 4
1

0 1 4 we have

1 4

2

0

it follows that

1

1 41 2
1

1

1

1 21

1

01 4

1 8 1 4

1 2

1

01 2

0
0

we have (14) for

1 2

sup

1 2 2

2
0

1 2

1 4

1 4

2

1

2

2
0

Using (13) for
Next we show

.

1 4

2
1

and 2

1 4 1 2 ). Hence
1
2

converge uniformly on

.

By Lemma 10, we have
. Moreover, from Lemma 9, (11) and (12) it follows that
satisfies (ii)–(v) and (vii) in Lemma 2.
3.2. Proof of Lemma 2 (Step 2)
for

We show that satisfies (i) and (vi) in Lemma 2, i.e.,
0 1 2
by induction.

and

0

1

0

T. Yoneda / J. Math. Anal. Appl. 317 (2006) 320–330

The equality

327

, i.e.,
2
2

2
0
1
2

0 1 2
(16)

1 2 1

1

0

0 1

0
implies that
and 0
1
0.
Assume that
and
0
1
0. Then the equality (16) implies that
1
0 1 2 1 . From
0
0 and the continuity of
it follows that
1

0
4 2

1

0

2

2

2

0

0

i.e.,
lim

1

0

0

By the mean value theorem, we have that
0
As
that

1

0 we have that
1 1
0 and
1

1
1

1 2
1
0.

1

1 is continuous at 0. By
0
0 and
is continuous at 1. By

4 2
4 2

1

we have that
1 0

1
1

0

2

2

2

1

2
1

0 and

2

0
2

1

1
1 2
1 2

0

we have

0
0
1

is continuous at 1 2. Therefore

and

3.3. Proof of Theorem 4
If

is a solution of (3), then the value of
on 2 1 2 .
We define a function
as follows:
0

on 2 2

1

1 2

By induction we show that is a solution of (3) and that
0
1
0 for
0 1 2
it follows that
First we show that
1
From
for

1

is determined by the value of

0
0 1
1 2
2 2

2 4
2 4

4

1

0 2
0 1

2
,

is expressed by (4). Then from
.

1

(17)
and

2 4 for

0 1 , it follows that,

1 2,
2 4

2 4

1

2 4

2

1

328

T. Yoneda / J. Math. Anal. Appl. 317 (2006) 320–330

Then we have (17). By (17) we have
4

2

4 2

Assume that, for general
2

4

1

1

4 2

1

0 1

1

0 2
0

1

(18)

2,
1

2

(19)

1

Then
1
1
If

2

1

1 2

, then

1
4

2

2
1
4

1

2

1

2

1

This shows that (19) holds for
4

2

4

2

1

1
1

2

2

1

1

0 1 . By (20) and (18) we have

1

1
1

(20)

2

2

2
2

and

1
2

1
1

1

1

1

2
1

instead of

1
1
2

2

1
1

1. Therefore we have (4) and

0

Clearly
Then

0

is a solution of (3).

4. Fourier transform of the solution
Let

01

. Then

in (4)

2
2

2

1

for
2

2

with supp

sinc

0 1 . Hence

2

By
1

2

2

we have
2

2

2

1

1

sinc

2

1

2

1

this is
1 2 1 2

1

sinc

2

1

2

(21)

1

Proof of Theorem 6. Using (21), we have
01

2

sinc
1

2

1

sinc

2

1


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