A note on iterative solutions of an iterative functional differential equation
UDC 517.9 We propose an iterative method for solving the iterative functional differential equation$$x\prime \prime (t) = \lambda_1x(t) + \lambda_2x^{[2]}(t) + . . . + \lambda_nx^{[n]}(t) + f(t).$$
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Institute of Mathematics, NAS of Ukraine
2020
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| author | Zhao, H. Y. Zhao, H. Y. |
| author_facet | Zhao, H. Y. Zhao, H. Y. |
| author_sort | Zhao, H. Y. |
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| description | UDC 517.9
We propose an iterative method for solving the iterative functional differential equation$$x\prime \prime (t) = \lambda_1x(t) + \lambda_2x^{[2]}(t) + . . . + \lambda_nx^{[n]}(t) + f(t).$$ |
| doi_str_mv | 10.37863/umzh.v72i11.6034 |
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DOI: 10.37863/umzh.v72i11.6034
UDC 517.9
H. Y. Zhao (School Math., Chongqing Normal Univ., China)
A NOTE ON ITERATIVE SOLUTIONS
OF AN ITERATIVE FUNCTIONAL DIFFERENTIAL EQUATION*
ЗАУВАЖЕННЯ ЩОДО IТЕРАЦIЙНИХ РОЗВ’ЯЗКIВ
IТЕРАТИВНИХ ФУНКЦIОНАЛЬНО-ДИФЕРЕНЦIАЛЬНИХ РIВНЯНЬ
We propose an iterative method for solving the iterative functional differential equation
x\prime \prime (t) = \lambda 1x(t) + \lambda 2x
[2](t) + . . .+ \lambda nx
[n](t) + f(t).
Запропоновано iтерацiйний метод знаходження розв’язкiв iтеративного функцiонально-диференцiального рiвняння
x\prime \prime (t) = \lambda 1x(t) + \lambda 2x
[2](t) + . . .+ \lambda nx
[n](t) + f(t).
1. Introduction. Second-order functional differential equation
x\prime \prime (t) = H
\Bigl(
t, x(t - \tau 0(t)), x
\bigl(
t - \tau 1(t)
\bigr)
, x[2](t), . . . , x
\bigl(
t - \tau n(t)
\bigr) \Bigr)
has been studied in [1] and [5]. If take \tau i(t) = t - x[i - 1](t), we obtain iterative functional differential
equations of the form
x\prime \prime (t) = H
\Bigl(
t, x[0](t), x[1](t), x[2](t), . . . , x[n](t)
\Bigr)
,
where x[0](t) = t, x[1](t) = x(t), x[2](t) = x(x(t)), . . . , x[n](t) = x(x[n - 1](t)). Petahov [9]
considers the iterative functional differential equation
x\prime \prime (t) = cx(x(t))
and obtains an existence theorem for solutions. Later, Si and Wang [11] study
x\prime \prime (x[r](t)) = c0t+ c1x(t) + c2x
[2](t) + . . .+ x[n](t),
and show the existence theorem of analytic solutions. Some various properties of solutions for several
second-order iterative functional differential equations, we refer the interested reader to [12 – 16].
In this paper, we intend to determine explicit approximate solutions, with given initial values, of
equations of the form
x\prime \prime (t) = \lambda 1x(t) + \lambda 2x
[2](t) + . . .+ \lambda nx
[n](t) + f(t). (1.1)
* This work was partially supported by the National Natural Science Foundation of China (Grant No. 11501069),
Science and Technology Research Program of Chongqing Municipal Education Commission (Grant No. KJQN201800502),
Foundation of Youth Talent of Chongqing Normal University (Grant No. 02030307-00039), the Natural Science Foundation
of Chongqing (Grant No. cstc2020jcyj-msxmX0857).
c\bigcirc H. Y. ZHAO, 2020
1564 ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
A NOTE ON ITERATIVE SOLUTIONS OF AN ITERATIVE FUNCTIONAL DIFFERENTIAL EQUATION 1565
To the best of our knowledge, there are little results about approximate solutions for iterative func-
tional differential equations. There exists several perturbative methods to determine explicit approx-
imate solutions [6, 8, 10], most of them require a small perturbative parameter. In this paper, our
iteration schemes inspired by [2 – 4, 7]. For convenience, we will make use C1(I, I) to denote the set
of all continuous differential functions from closed interval I to I with the norm \| x\| = \mathrm{s}\mathrm{u}\mathrm{p}t\in I | x(t)| .
For M > 0, define
C1
M (I) =
\Bigl\{
\varphi \in C1(I, I)
\bigm| \bigm| \bigm| | \varphi (t2) - \varphi (t1)| \leq M | t2 - t1| for all t, t1, t2 \in \BbbR
\Bigr\}
.
It is easy to see C1
M (I) is closed convex and bounded subsets of C1(I, I).
2. Convergence of the sequence of approximate solutions. Now we will use iteration method
to solve Eq. (1.1), where f is a continuous function on a domain I = [\alpha - \delta , \alpha + \delta ].
Lemma 2.1. For any x, y \in C1
M (I), t1, t2 \in \BbbR , the following inequality holds:
\bigm\| \bigm\| x[k] - y[k]
\bigm\| \bigm\| \leq
k - 1\sum
j=0
M j\| x - y\| , k = 1, 2, . . . . (2.1)
Proof. It can be obtained by direct calculation by the definition of C1
M (I).
Noting the kth step equation for (1.1) is
x\prime \prime k+1(t) = \lambda 1xk+1(t) + \lambda 2x
[2]
k (t) + . . .+ \lambda nx
[n]
k (t) + f(t), t \in [\alpha - \delta , \alpha + \delta ], (2.2)
with xk+1(\alpha ) = \alpha , x\prime k+1(\alpha ) = \beta and \lambda 1 < 0, where x0(t) is an initial function, \alpha and \beta are given
real numbers. Integrating (2.2), we obtain
xk+1(t) = \alpha \mathrm{c}\mathrm{o}\mathrm{s}
\bigl( \sqrt{}
- \lambda 1(t - \alpha )
\bigr)
+
\beta \surd
- \lambda 1
\mathrm{s}\mathrm{i}\mathrm{n}
\bigl( \sqrt{}
- \lambda 1(t - \alpha )
\bigr)
-
- 1\surd
- \lambda 1
n\sum
i=2
\lambda i \mathrm{c}\mathrm{o}\mathrm{s}
\sqrt{}
- \lambda 1t
t\int
\alpha
x
[i]
k (s) \mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1sds -
- 1\surd
- \lambda 1
\mathrm{c}\mathrm{o}\mathrm{s}
\sqrt{}
- \lambda 1t
t\int
\alpha
f(s) \mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1sds+
+
1\surd
- \lambda 1
n\sum
i=2
\lambda i \mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1t
t\int
\alpha
x
[i]
k (s) \mathrm{c}\mathrm{o}\mathrm{s}
\sqrt{}
- \lambda 1sds+
+
1\surd
- \lambda 1
\mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1t
t\int
\alpha
f(s) \mathrm{c}\mathrm{o}\mathrm{s}
\sqrt{}
- \lambda 1sds =
=
\sqrt{}
\alpha 2 - \beta 2
\lambda 1
\mathrm{s}\mathrm{i}\mathrm{n}
\bigl( \sqrt{}
- \lambda 1(\nu + t - \alpha )
\bigr)
+
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
1566 H. Y. ZHAO
+
1\surd
- \lambda 1
n\sum
i=2
\lambda i
t\int
\alpha
x
[i]
k (s) \mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1(t - s)ds+
+
1\surd
- \lambda 1
t\int
\alpha
f(s) \mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1(t - s)ds, (2.3)
where
\mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1\nu =
\alpha \sqrt{}
\alpha 2 - \beta 2
\lambda 1
.
Next we will show sequence \{ xk(t)\} | \infty k=1 convergent to x(t) which is the solution of (1.1) if we
take any x0 satisfies x0(t) \in I for any t \in I.
Theorem 2.1. Let I = [\alpha - \delta , \alpha + \delta ], \lambda 1 < 0, and the following conditions hold:
(i)
\sqrt{}
\beta 2 - \alpha 2\lambda 1 + L\delta
n\sum
i=2
| \lambda i| + L\prime \delta \leq \mathrm{m}\mathrm{i}\mathrm{n}\{ M, 1\} , (2.4)
where L = \mathrm{m}\mathrm{a}\mathrm{x}
\bigl\{
| \alpha - \delta | , | \alpha + \delta |
\bigr\}
;
(ii)
\delta \surd
- \lambda 1
n\sum
i=2
i - 1\sum
j=0
M j | \lambda i| < 1. (2.5)
Then, for any \| f\| \leq L\prime , (1.1) has a solution in C1
M (I).
Proof. First, we need xk \in C1
M (I), k = 1, 2, . . . , for any t \in I. We will prove it by induction.
It is easy to find x1 \in C1
M (I) if we take any x0 such that x0(t) \in I for any t \in I. Assume
xk \in C1
M (I), k \geq 2. By (2.3), it is obvious that xk+1(\alpha ) = \alpha , x\prime k+1(\alpha ) = \beta . From (2.4), we have\bigm| \bigm| xk+1(t) - \alpha
\bigm| \bigm| = \bigm| \bigm| xk+1(t) - xk+1(\alpha )
\bigm| \bigm| \leq
\leq
\Biggl( \sqrt{}
\beta 2 - \alpha 2\lambda 1 + L\delta
n\sum
i=2
| \lambda i| + L\prime \delta
\Biggr)
| t - \alpha | \leq \delta
and \bigm| \bigm| xk+1(t2) - xk+1(t1)
\bigm| \bigm| \leq
\leq
\Biggl( \sqrt{}
\beta 2 - \alpha 2\lambda 1 + L\delta
n\sum
i=2
| \lambda i| + L\prime \delta
\Biggr)
| t2 - t1| \leq M | t2 - t1| .
This proves that xk+1 belongs to C1
M (I).
By (2.1), it is obviously that
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
A NOTE ON ITERATIVE SOLUTIONS OF AN ITERATIVE FUNCTIONAL DIFFERENTIAL EQUATION 1567
\mathrm{s}\mathrm{u}\mathrm{p}
t\in [\alpha - \delta ,\alpha +\delta ]
\bigm| \bigm| xk+1(t) - xk(t)
\bigm| \bigm| \leq \delta \surd
- \lambda 1
n\sum
i=2
| \lambda i| \mathrm{s}\mathrm{u}\mathrm{p}
t\in [\alpha - \delta ,\alpha +\delta ]
\bigm| \bigm| \bigm| x[i]k (t) - x
[i]
k - 1(t)
\bigm| \bigm| \bigm| \leq
\leq \delta \surd
- \lambda 1
n\sum
i=2
i - 1\sum
j=0
M j | \lambda i| \| xk - xk - 1\| ,
i.e.,
\| xk+1 - xk\| \leq \Gamma \| xk - xk - 1\| ,
where
\Gamma =
\delta \surd
- \lambda 1
n\sum
i=2
i - 1\sum
j=0
M j | \lambda i| .
Therefore,
\| xk+1 - xk\| \leq \Gamma k - 1\| x2 - x1\| . (2.6)
Now, let us go back to Eq. (1.1) and its solution xk(t) as given in (2.3). Let
xm(t) = x1(t) +
m - 1\sum
k=1
(xk+1 - xk).
We shall show that
\sum \infty
k=1
\bigl(
xk+1(t) - xk(t)
\bigr)
converges on the interval [\alpha - \delta , \alpha + \delta ]. This would
imply that xm(t) has a limit on this interval as m \rightarrow \infty . Clearly to show the convergence of\sum \infty
k=1
\bigl(
xk+1(t) - xk(t)
\bigr)
. From (2.5), series
\infty \sum
k=1
\| xk+1 - xk\| \leq
\infty \sum
k=1
\Gamma k - 1\| x2 - x1\| =
1
1 - \Gamma
\| x2 - x1\|
converges.
This shows that \{ xm(t)\} is a Cauchy sequence under the supreme norm and, therefore, converges
uniformly to a continuous function x(t) on [\alpha - \delta , \alpha + \delta ]. Thus,
x(t) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
xk+1(t) =
\sqrt{}
\alpha 2 - \beta 2
\lambda 1
\mathrm{s}\mathrm{i}\mathrm{n}
\Bigl( \sqrt{}
- \lambda 1(\nu + t - \alpha )
\Bigr)
+
+
1\surd
- \lambda 1
n\sum
i=2
\lambda i
t\int
\alpha
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
x
[i]
k (s) \mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1(t - s)ds+
+
1\surd
- \lambda 1
t\int
\alpha
f(s) \mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1(t - s)ds =
=
\sqrt{}
\alpha 2 - \beta 2
\lambda 1
\mathrm{s}\mathrm{i}\mathrm{n}
\Bigl( \sqrt{}
- \lambda 1(\nu + t - \alpha )
\Bigr)
+
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
1568 H. Y. ZHAO
+
1\surd
- \lambda 1
n\sum
i=2
\lambda i
t\int
\alpha
x
[i]
k (s) \mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1(t - s)ds+
+
1\surd
- \lambda 1
t\int
\alpha
f(s) \mathrm{s}\mathrm{i}\mathrm{n}
\sqrt{}
- \lambda 1(t - s)ds. (2.7)
By direct substitution of (2.7) in (1.1), we show that x(t) satisfies this equation. In addition (2.7)
also shows that x(\alpha ) = \alpha and x\prime (\alpha ) = \beta . Then x(t) satisfies (1.1) along the required initial
conditions. In consequence, the sequence of functions given by S =
\bigl\{
x0(t), x1(t), . . . , xm(t) . . .
\bigr\}
can be considered as approximate solutions of Eq. (1.1).
Theorem 2.1 is proved.
Now, we shall give the result for \lambda 1 > 0. Integrating (2.2), we obtain
xk+1(t) =
1
2
\Biggl( \biggl(
\alpha +
\beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1(t - \alpha ) +
\biggl(
\alpha - \beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1(\alpha - t)+
+
1\surd
\lambda 1
n\sum
i=2
\lambda ie
\surd
\lambda 1t
t\int
\alpha
x
[i]
k (s)e -
\surd
\lambda 1sds+
1\surd
\lambda 1
e
\surd
\lambda 1t
t\int
\alpha
f(s)e -
\surd
\lambda 1sds -
- 1\surd
\lambda 1
n\sum
i=2
\lambda ie
-
\surd
\lambda 1t
t\int
\alpha
x
[i]
k (s)e
\surd
\lambda 1sds - 1\surd
\lambda 1
e -
\surd
\lambda 1t
t\int
\alpha
f(s)e
\surd
\lambda 1sds
\Biggr)
=
=
1
2
\Biggl( \biggl(
\alpha +
\beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1(t - \alpha ) +
\biggl(
\alpha - \beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1(\alpha - t)+
+
1\surd
\lambda 1
n\sum
i=2
\lambda i
t\int
\alpha
x
[i]
k (s)e
\surd
\lambda 1(t - s)ds+
1\surd
\lambda 1
t\int
\alpha
f(s)e
\surd
\lambda 1(t - s)ds -
- 1\surd
\lambda 1
n\sum
i=2
\lambda i
t\int
\alpha
x
[i]
k (s)e
\surd
\lambda 1(s - t)ds - 1\surd
\lambda 1
t\int
\alpha
f(s)e
\surd
\lambda 1(s - t)ds
\Biggr)
. (2.8)
Theorem 2.2. Let I = [\alpha - \delta , \alpha + \delta ], \lambda 1 > 0, and the following conditions hold:
(i)\sqrt{}
\lambda 1
\biggl(
\alpha +
\beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1\delta +
1
2
\surd
\lambda 1
\Bigl(
e
\surd
\lambda 1\delta - e -
\surd
\lambda 1\delta
\Bigr) \Biggl(
L\prime + L
n\sum
i=2
| \lambda i|
\Biggr)
\leq \mathrm{m}\mathrm{i}\mathrm{n}\{ M, 1\} , (2.9)
where L = \mathrm{m}\mathrm{a}\mathrm{x}
\bigl\{
| \alpha - \delta | , | \alpha + \delta |
\bigr\}
;
(ii)
1
2\lambda 1
\Bigl(
e
\surd
\lambda 1(\alpha +\delta ) - e -
\surd
\lambda 1(\alpha +\delta )
\Bigr) n\sum
i=2
i - 1\sum
j=0
M j | \lambda i| < 1. (2.10)
If we take x0 \in C1(I, I), then for any \| f\| \leq L\prime , Eq. (1.1) has a solution in C1
M (I).
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
A NOTE ON ITERATIVE SOLUTIONS OF AN ITERATIVE FUNCTIONAL DIFFERENTIAL EQUATION 1569
Proof. Similar as Theorem 2.1, if we take any x0 \in C1(I, I), easy to see that x1 belongs to
C1
M (I). Assume xk(\alpha ) = \alpha and x\prime k(\alpha ) = \beta , by (2.8), it is obviously xk+1(\alpha ) = \alpha and x\prime k+1(\alpha ) =
= \beta . Furthermore, by (2.9),
| xk+1(t) - \alpha | =
\bigm| \bigm| xk+1(t) - xk+1(\alpha )
\bigm| \bigm| \leq
\leq
\Biggl( \sqrt{}
\lambda 1
\biggl(
\alpha +
\beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1\delta +
1
2
\surd
\lambda 1
\Bigl(
e
\surd
\lambda 1\delta - e -
\surd
\lambda 1\delta
\Bigr) \Biggl(
L\prime + L
n\sum
i=2
| \lambda i|
\Biggr) \Biggr)
| t - \alpha | \leq \delta
and
| xk+1(t2) - xk+1(t1)| \leq
\leq
\Biggl( \sqrt{}
\lambda 1
\biggl(
\alpha +
\beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1\delta +
1
2
\surd
\lambda 1
\Bigl(
e
\surd
\lambda 1\delta - e -
\surd
\lambda 1\delta
\Bigr) \Biggl(
L\prime + L
n\sum
i=2
| \lambda i|
\Biggr) \Biggr)
| t2 - t1| \leq
\leq M | t2 - t1| .
This proves that xk+1 belongs to C1
M (I).
Using (2.1), we see that
\mathrm{s}\mathrm{u}\mathrm{p}
t\in [\alpha - \delta ,\alpha +\delta ]
\bigm| \bigm| xk+1(t) - xk(t)
\bigm| \bigm| \leq
\leq 1
2\lambda 1
\Bigl(
e
\surd
\lambda 1t - e -
\surd
\lambda 1t
\Bigr) n\sum
i=2
| \lambda i| \mathrm{s}\mathrm{u}\mathrm{p}
t\in [\alpha - \delta ,\alpha +\delta ]
\bigm| \bigm| \bigm| x[i]k (t) - x
[i]
k - 1(t)
\bigm| \bigm| \bigm| \leq
\leq 1
2\lambda 1
\Bigl(
e
\surd
\lambda 1(\alpha +\delta ) - e -
\surd
\lambda 1(\alpha +\delta )
\Bigr) n\sum
i=2
i - 1\sum
j=0
M j | \lambda i| \| xk - xk - 1\| ,
i.e.,
\| xk+1 - xk\| \leq \Gamma 1\| xk - xk - 1\| ,
where
\Gamma 1 =
1
2\lambda 1
\Bigl(
e
\surd
\lambda 1(\alpha +\delta ) - e -
\surd
\lambda 1(\alpha +\delta )
\Bigr) n\sum
i=2
i - 1\sum
j=0
M j | \lambda i| .
Therefore,
\| xk+1 - xk\| \leq \Gamma k - 1
1 \| x2 - x1\| . (2.11)
Now, let us go back to Eq. (1.1) and its solution xk(t) as given in (2.8). Let
xm(t) = x1(t) +
m - 1\sum
k=1
(xk+1 - xk).
We shall show that
\sum \infty
k=1
\bigl(
xk+1(t) - xk(t)
\bigr)
converges on the interval [\alpha - \delta , \alpha + \delta ]. This would
imply that xm(t) has a limit on this interval as m \rightarrow \infty . Clearly to show the convergence of
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
1570 H. Y. ZHAO\sum \infty
k=1
\bigl(
xk+1(t) - xk(t)
\bigr)
. From (2.10), series
\infty \sum
k=1
\| xk+1 - xk\| \leq
\infty \sum
k=1
\Gamma k - 1
1 \| x2 - x1\| =
1
1 - \Gamma 1
\| x2 - x1\|
converges.
This shows that \{ xm(t)\} is a Cauchy sequence under the supreme norm and, therefore, converges
uniformly to a continuous function x(t) on [\alpha - \delta , \alpha + \delta ]. Thus,
x(t) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
xk+1(t) =
=
1
2
\Biggl( \biggl(
\alpha +
\beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1(t - \alpha ) +
\biggl(
\alpha - \beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1(\alpha - t)+
+
1\surd
\lambda 1
n\sum
i=2
\lambda i
t\int
\alpha
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
x
[i]
k (s)e
\surd
\lambda 1(t - s)ds+
1\surd
\lambda 1
t\int
\alpha
f(s)e
\surd
\lambda 1(t - s)ds -
- 1\surd
\lambda 1
n\sum
i=2
\lambda i
t\int
\alpha
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
x
[i]
k (s)e
\surd
\lambda 1(s - t)ds - 1\surd
\lambda 1
t\int
\alpha
f(s)e
\surd
\lambda 1(s - t)ds
\Biggr)
=
=
1
2
\Biggl( \biggl(
\alpha +
\beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1(t - \alpha ) +
\biggl(
\alpha - \beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1(\alpha - t)+
+
1\surd
\lambda 1
n\sum
i=2
\lambda i
t\int
\alpha
x[i](s)e
\surd
\lambda 1(t - s)ds+
1\surd
\lambda 1
t\int
\alpha
f(s)e
\surd
\lambda 1(t - s)ds -
- 1\surd
\lambda 1
n\sum
i=2
\lambda i
t\int
\alpha
x[i](s)e
\surd
\lambda 1(s - t)ds - 1\surd
\lambda 1
t\int
\alpha
f(s)e
\surd
\lambda 1(s - t)ds
\Biggr)
. (2.12)
By direct substitution of (2.12) in (1.1), we show that x(t) satisfies this equation. In addition
(2.12) also shows that x(\alpha ) = \alpha and x\prime (\alpha ) = \beta . Then x(t) satisfies (1.1) along the required initial
conditions. In consequence, the sequence of functions given by S =
\bigl\{
x0(t), x1(t), . . . , xm(t), . . .
\bigr\}
can be considered as approximate solutions of Eq. (1.1).
Theorem 2.2 is proved.
3. Examples. In this section, some examples will be showed.
Example 3.1. Now, we will show that the conditions in Theorem 2.1 do not self-contradict.
Consider the equation
x\prime \prime (t) = - 25x(t) + x(x(t)) + \mathrm{s}\mathrm{i}\mathrm{n} t, (3.1)
where \lambda 1 = - 25, \lambda 2 = 1, f(t) = \mathrm{s}\mathrm{i}\mathrm{n} t. Here, \alpha = 0, \beta =
1
3
. Take L = \delta =
1
10
, M = 1,
L\prime = 1, x0 =
1
12
, a simple calculation yields
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
A NOTE ON ITERATIVE SOLUTIONS OF AN ITERATIVE FUNCTIONAL DIFFERENTIAL EQUATION 1571
\sqrt{}
\beta 2 - \alpha 2\lambda 1 + L\delta | \lambda 2| + L\prime \delta =
133
300
\leq 1 = \mathrm{m}\mathrm{i}\mathrm{n}\{ M, 1\} ,
and
\delta \surd
- \lambda 1
| \lambda 2|
1\sum
j=0
M j =
1
25
< 1.
Then (2.4) and (2.5) are satisfied. By Theorem 2.1, equation (3.1) has sequence of approximate
solutions \{ xk\} , k \geq 0, such that
\bigm| \bigm| xk(t2) - xk(t1) = | \leq | t2 - t1| \forall t1, t2 \in
\biggl[
- 1
10
,
1
10
\biggr]
. Here,
xk+1(t) =
1
15
\mathrm{s}\mathrm{i}\mathrm{n} 5t+
1
5
t\int
0
x
[2]
k (s) \mathrm{s}\mathrm{i}\mathrm{n} 5(t - s)ds+
1
5
t\int
0
\mathrm{s}\mathrm{i}\mathrm{n} s \mathrm{s}\mathrm{i}\mathrm{n} 5(t - s)ds =
=
7
120
\mathrm{s}\mathrm{i}\mathrm{n} 5t+
1
24
\mathrm{s}\mathrm{i}\mathrm{n} t+
1
5
t\int
0
x
[2]
k (s) \mathrm{s}\mathrm{i}\mathrm{n} 5(t - s)ds.
Moreover, we can find xk(0) = 0 and x\prime k(0) =
1
3
, then Eq. (3.1) has a solution in
C1
M
\biggl( \biggl[
- 1
10
,
1
10
\biggr] \biggr)
.
Example 3.2. Now, we will show that the conditions in Theorem 2.2 do not self-contradict.
Consider the equation
x\prime \prime (t) = 25x(t) + x(x(t)) + \mathrm{s}\mathrm{i}\mathrm{n} t, (3.2)
where \lambda 1 = 25, \lambda 2 = 1, f(t) = \mathrm{s}\mathrm{i}\mathrm{n} t. Here, \alpha = 0, \beta =
1
3
. Take L = \delta =
1
10
, M = 1, L\prime = 1,
x0 =
1
12
, a simple calculation yields
\sqrt{}
\lambda 1
\biggl(
\alpha +
\beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1\delta +
1
2
\surd
\lambda 1
\Bigl(
e
\surd
\lambda 1\delta - e -
\surd
\lambda 1\delta
\Bigr) \bigl(
L\prime + L| \lambda 2|
\bigr)
< 0.665 \leq 1 = \mathrm{m}\mathrm{i}\mathrm{n}\{ M, 1\}
and
1
2\lambda 1
\Bigl(
e
\surd
\lambda 1(\alpha +\delta ) - e -
\surd
\lambda 1(\alpha +\delta )
\Bigr)
| \lambda 2| (1 +M) < 0.209 < 1.
Then (2.9) and (2.10) are satisfied. By Theorem 2.2, equation (3.2) has sequence of approximate
solutions \{ xk\} , k \geq 0 such that
\bigm| \bigm| xk(t2) - xk(t1)
\bigm| \bigm| \leq | t2 - t1| \forall t1, t2 \in
\biggl[
- 1
10
,
1
10
\biggr]
. Here,
xk+1(t) =
1
30
\Bigl(
e5t - e - 5t
\Bigr)
+
1
5
t\int
0
\Bigl(
x
[2]
k (s) + \mathrm{s}\mathrm{i}\mathrm{n} s
\Bigr) \Bigl(
e5(t - s) - e5(s - t)
\Bigr)
ds.
Moreover, we can find xk(0) = 0 and x\prime k(0) =
1
3
, then Eq. (3.2) has a solution in
C1
M
\biggl( \biggl[
- 1
10
,
1
10
\biggr] \biggr)
.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
1572 H. Y. ZHAO
Example 3.3. Consider the equation
x\prime \prime (t) = \lambda x(t) + x(x(t)) + \mathrm{s}\mathrm{i}\mathrm{n} t, (3.3)
where \lambda 1 = \lambda , \lambda 2 = 1, f(t) = \mathrm{s}\mathrm{i}\mathrm{n} t. Here, \alpha = 0, \beta =
1
3
, take L =
1
10
, \delta = \delta , M = M, L\prime = 1.
We shall study (3.3) with \lambda < 0 or \lambda > 0.
If \lambda < 0, then \sqrt{}
\beta 2 - \alpha 2\lambda 1 + L\delta | \lambda 2| + L\prime \delta =
1
3
+
11
10
\delta \leq \mathrm{m}\mathrm{i}\mathrm{n}\{ M, 1\}
and
\delta \surd
- \lambda 1
| \lambda 2| (1 +M) =
\delta (1 +M)\surd
- \lambda
< 1
or
0 < \delta \leq 10
11
M - 10
33
, \delta <
\surd
- \lambda
1 +M
, 0 < M < 1,
0 < \delta \leq 20
33
, \delta <
\surd
- \lambda
1 +M
, M \geq 1.
We see that the range of \delta depend on the value of M and \lambda , i.e.,
0 < \delta \leq 10
11
M - 10
33
, if \lambda < - 100
1089
(1 +M)2(3M - 1)2, 0 < M < 1,
0 < \delta <
\surd
- \lambda
1 +M
, if - 100
1089
(1 +M)2(3M - 1)2 \leq \lambda < 0, 0 < M < 1,
(3.4)
0 < \delta \leq 20
33
, if \lambda < - 400
1089
(1 +M)2, M \geq 1,
0 < \delta <
\surd
- \lambda
1 +M
, if - 400
1089
(1 +M)2 \leq \lambda < 0, M \geq 1.
Then (2.4) and (2.5) are satisfied. By Theorem 2.1, equation (3.3) has sequence of approximate
solutions \{ xk\} , k \geq 0, such that
\bigm| \bigm| xk(t2) - xk(t1)
\bigm| \bigm| \leq M | t2 - t1| \forall t1, t2 \in [ - \delta , \delta ]. Here take
x0 \in C1
\bigl(
[ - \delta , \delta ], [ - \delta , \delta ]
\bigr)
and
xk+1(t) =
1
3
\surd
- \lambda
\mathrm{s}\mathrm{i}\mathrm{n}(
\surd
- \lambda t) +
1\surd
- \lambda
t\int
0
x
[2]
k (s) \mathrm{s}\mathrm{i}\mathrm{n}
\surd
- \lambda (t - s)ds+
+
1\surd
- \lambda
t\int
0
f(s) \mathrm{s}\mathrm{i}\mathrm{n}
\surd
- \lambda (t - s)ds.
Moreover, we see that xk(0) = 0 and x\prime k(0) =
1
3
. Then Eq. (3.3) has a solution in C1
M
\bigl(
[ - \delta , \delta ]
\bigr)
.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
A NOTE ON ITERATIVE SOLUTIONS OF AN ITERATIVE FUNCTIONAL DIFFERENTIAL EQUATION 1573
If \lambda > 0, then\sqrt{}
\lambda 1
\biggl(
\alpha +
\beta \surd
\lambda 1
\biggr)
e
\surd
\lambda 1\delta +
1
2
\surd
\lambda 1
\Bigl(
e
\surd
\lambda 1\delta - e -
\surd
\lambda 1\delta
\Bigr) \bigl(
L\prime + L| \lambda 2|
\bigr)
=
=
\biggl(
1
3
+
11
20
\surd
\lambda
\biggr)
e
\surd
\lambda \delta - 11
20
\surd
\lambda
e -
\surd
\lambda \delta \leq \mathrm{m}\mathrm{i}\mathrm{n}\{ M, 1\}
and
1
2\lambda 1
\Bigl(
e
\surd
\lambda 1(\alpha +\delta ) - e -
\surd
\lambda 1(\alpha +\delta )
\Bigr)
| \lambda 2| (1 +M) =
1
2\lambda
\Bigl(
e
\surd
\lambda \delta - e -
\surd
\lambda \delta
\Bigr)
(1 +M) < 1
or
\delta <
1\surd
\lambda
\mathrm{l}\mathrm{n}
\sqrt{}
\lambda 2 + (1 +M)2 + \lambda
1 +M
= H1(\lambda ,M)
and
0 < \delta \leq 1\surd
\lambda
\mathrm{l}\mathrm{n}
\Biggl(
30
20 + 33
\surd
\lambda
\Biggl( \sqrt{}
\surd
\lambda M2 +
11
15
+
121
100
\surd
\lambda
+ 1
\Biggr) \Biggr)
= H2(\lambda ,M), 0 < M < 1,
0 < \delta \leq 1\surd
\lambda
\mathrm{l}\mathrm{n}
\Biggl( \sqrt{}
33
33 + 20
\surd
\lambda
+
90\lambda
(20
\surd
\lambda + 33)2
+
30
\surd
\lambda
20
\surd
\lambda + 33
\Biggr)
= H3(\lambda ), M \geq 1.
We see that the range of \delta depend on the value of M and \lambda , i.e.,
0 < \delta < H1(\lambda ,M), if H1(\lambda ,M) < H2(\lambda ,M), 0 < M < 1,
0 < \delta \leq H2(\lambda ,M), if H2(\lambda ,M) < H1(\lambda ,M), 0 < M < 1,
(3.5)
0 < \delta \leq H3(\lambda ), if H3(\lambda ) < H1(\lambda ,M), M \geq 1,
0 < \delta < H1(\lambda ,M), if H1(\lambda ,M) < H3(\lambda ), M \geq 1.
Then (2.9) and (2.10) are satisfied. By Theorem 2.2, equation (3.3) has sequence of approximate
solutions \{ xk\} , k \geq 0, such that | xk(t2) - xk(t1)| \leq M | t2 - t1| \forall t1, t2 \in [ - \delta , \delta ]. Here take
x0 \in C1
\bigl(
[ - \delta , \delta ], [ - \delta , \delta ]
\bigr)
and
xk+1(t) =
1
6
\surd
\lambda
\Bigl(
e
\surd
\lambda t - e -
\surd
\lambda t
\Bigr)
+
1
2
\surd
\lambda
t\int
0
\Bigl(
\lambda 2x
[2]
k + f(s)
\Bigr) \Bigl(
e
\surd
\lambda (t - s) - e
\surd
\lambda (s - t)
\Bigr)
ds.
Moreover, we see that xk(0) = 0 and x\prime k(0) =
1
3
. Then Eq. (3.3) has a solution in C1
M
\bigl(
[ - \delta , \delta ]
\bigr)
.
Remark 3.1. It is easy to see that M = 1, \lambda = - 25 and 0 < \delta =
1
10
\leq 20
33
in Example 3.1,
satisfies the third line of (3.4). In Example 3.2,
M = 1, \lambda = 25,
H1(\lambda ,M) =
1
5
\mathrm{l}\mathrm{n}
\Biggl(
25 +
\surd
629
2
\Biggr)
> 0.644 > 0.10026 >
1
5
\mathrm{l}\mathrm{n}
\biggl(
1
133
(
\surd
4839 + 150)
\biggr)
= H3(\lambda )
and \delta = 0.1 < H3, satisfies the third line of (3.5).
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
1574 H. Y. ZHAO
References
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Received 03.01.18
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 11
|
| id | umjimathkievua-article-6034 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T03:25:35Z |
| publishDate | 2020 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/f7/87015efbd1f7117890ff7719480c3ef7.pdf |
| spelling | umjimathkievua-article-60342025-03-31T08:49:35Z A note on iterative solutions of an iterative functional differential equation A note on iterative solutions of an iterative functional differential equation Zhao, H. Y. Zhao, H. Y. UDC 517.9 We propose an iterative method for solving the iterative functional differential equation$$x\prime \prime (t) = \lambda_1x(t) + \lambda_2x^{[2]}(t) + . . . + \lambda_nx^{[n]}(t) + f(t).$$ UDC 517.9 Зауваження щодо ітераційних розв'язків ітеративних функціонально-диференціальних рівнянь Запропоновано ітераційний метод знаходження розв'язків ітеративного функціонально-диференціального рівняння $$x''(t)=\lambda_1x(t)+\lambda_2x^{[2]}(t)+\ldots+\lambda_nx^{[n]}(t)+f(t).$$ Institute of Mathematics, NAS of Ukraine 2020-11-20 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6034 10.37863/umzh.v72i11.6034 Ukrains’kyi Matematychnyi Zhurnal; Vol. 72 No. 11 (2020); 1564-1574 Український математичний журнал; Том 72 № 11 (2020); 1564-1574 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6034/8774 |
| spellingShingle | Zhao, H. Y. Zhao, H. Y. A note on iterative solutions of an iterative functional differential equation |
| title | A note on iterative solutions of an iterative functional differential equation |
| title_alt | A note on iterative solutions of an iterative functional differential equation |
| title_full | A note on iterative solutions of an iterative functional differential equation |
| title_fullStr | A note on iterative solutions of an iterative functional differential equation |
| title_full_unstemmed | A note on iterative solutions of an iterative functional differential equation |
| title_short | A note on iterative solutions of an iterative functional differential equation |
| title_sort | note on iterative solutions of an iterative functional differential equation |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6034 |
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