The new method for the searching periodic solutions of periodic differential systems

Abstract

In the paper we are giving the new method for searching periodic solutions of periodic differential systems. For this we construct a differential system with the same Reflecting Function as the Reflecting Function of the given system and with a known periodic solution. Then the initial data of the periodic solutions of this two systems coincide. In such a way the problem of existance periodic solutions goes to the Cauchy problem.

Full text

Journal of Applied Analysis and Computation Website:http://jaac-online.com/
Volume 6, Number 3, August 2016, 876–883 DOI:10.11948/2016056
THE NEW METHOD FOR THE SEARCHING
PERIODIC SOLUTIONS OF PERIODIC
DIFFERENTIAL SYSTEMS
Vladimir Ivanovich Mironenko1,†,
and Vladimir Vladimirovich Mironenko1
Abstract In the paper we are giving the new method for searching periodic
solutions of periodic differential systems. For this we construct a differential
system with the same Reflecting Function as the Reflecting Function of the
given system and with a known periodic solution. Then the initial data of the
periodic solutions of this two systems coincide. In such a way the problem of
existance periodic solutions goes to the Cauchy problem.
Keywords Differential equation, periodic solution, boundary value problem,
reflecting function, equivalence.
MSC(2010) 34A12.
1. Introduction
In this section of the paper we give the basic facts from the theory of Reflect-
ing Function [4–6], which are necessary to understand the main results of the
paper. For the quick acquaintance with the theory of Reflecting Function see
http://reflecting-function.narod.ru. The new important and interesting results in
the theory of Reflecting Function and its applications were obtained in the lust five
years [1,2,7–12].
For the system
dx
dt =X(t, x), t ∈R, x ∈D⊂Rn,(1.1)
with the continuously differentiable right-hand side X(t, x) and the general solution
x=φ(t, t0, x0) the Reflecting Function of the system (1.1) is defined by formula
F(t, x) := φ(−t;t, x).
If F(t, x) is Reflecting Function of (1.1) and x(t) is any solution of (1.1), which
exist on symmetric interval (−τ, τ ),then F(t, x(t)) ≡x(−t).
If X(t, x) is 2ω-periodic with respect to t, then F(−ω, x) = φ(ω, −ω, x) is the
in-period [−ω;ω] transformation for the system (1.1) (Poincare’ transformation).
So the solution x(t) := φ(t;−ω, x0) of the system (1.1) is 2ω-periodic if and only
if it is exist on [−ω;ω] and F(−ω, x0) = x0.A character of stability of the solution
is the same as the character of stability of the fixed point x0of the Poincare’ map
x→F(−ω, x).
†the corresponding author. Email address:vmironenko@tut.by(V.I. Miro-
nenko)
1Mathematical Department, Gomel State University by F. Scorina, Sovetskaya
street, 104, Gomel, 246019, Republic of Belarus

The new method for the searching periodic solutions of periodic differential systems 877
The differentiable function F(t, x) is the Reflecting Function of (1.1) if and only
if it is the solution of the Cauchy problem
∂F
∂t +∂F
∂x X(t, x) + X(−t, F )=0, F (0, x)≡x. (1.2)
Using the problem (1.2) we can sometimes to get Reflecting Function of system
(1.1) even then, when system (1.1) is no integrable in finite terms (no integrable
in quadratures). This case we have, for example, when X(−t, x) + X(t, x)≡0.In
this case the problem (1.2) has the solution F(t, x)≡x. So F=xis Reflecting
Function of (1.1) and all solutions of the (1.1), which are defined on symmetric
interval (−τ, τ ) are even, that is x(−t)≡x(t).It is interesting to say, that if every
solution of (1.1) is even, than F(t, x)≡xand X(−t, x) + X(t, x)≡0.
Suppose that a continuously differentiable function F(t, x) (or it’s restriction) is
defined in a domain of R1+n,which contained the hyper plain t= 0.Suppose also,
that F(−t, F (t, x)) ≡F(0, x)≡x, than this F(t, x) is Reflecting Function of any
system of the form
dx
dt =−1
2∂F
∂x −1∂F
∂t +∂F
∂x −1R(t, x)−R(−t, F ),(1.3)
where R(t, x) = (R1(t, x), R2(t, x), ..., Rn(t, x))Tis any differentiable vector-function.
Two differential systems of the form (1.1) we call equivalent if their Reflecting
Functions F1:G1→Rnand F2:G2→Rncoincide in the G1∩G2.So all systems
(1.3) form a class of equivalence.
Very often, when we cannot solve the problem (1.2), we can use the following
statement [6, p11]:
If ∆k(t, x), k ∈N, are solutions of the system
∂∆
∂t +∂∆
∂x X(t, x)−∂X
∂x ∆=0,(1.4)
and αk(t) are scalar continuous odd functions, then the system
dx
dt =X(t, x) + X
k
αk(t)∆k(t, x),
is equivalent to the system (1.1).
The Reflecting Function concept was used in works of Alsevich L.A., Biel-
skij V.V., Kastritsa O.A., Majorovskaya S.V., Musafirov E. V., Philiptsov V.F.,
Varenikova E.V., Zhang Shanlin, Zhou Zhengxin, Yan Yuexin, Yu Yuanhong and
others.
2. The main results
Suppose that system (1.1) is equivalent to the system
dy
dt =Y(t, y), t ∈R, y ∈D⊂Rn,(2.1)
and the system (2.1) has solution y(t) for which y(−ω) = y(ω).

878 V.I. Mironenko & V.V. Mironenko
Then the solution x(t) of (1.1) with the initial condition x(−ω) = y(−ω) has
the property x(ω) = y(ω) = x(−ω) if this x(t) is extendable on [−ω, ω].For this
reason if system (1.1) is 2ω-periodic in tthen this solution x(t) will be 2ω-periodic
too, even if y(t) is not periodic.
According to what has been said, it is important to construct systems (2.1)
equivalent to (1.1) and with the known solutions. We would like to construct
systems (2.1) equivalent to (1.1) with a constant solution, that is the systems (2.1),
for which Y(t, y0)≡0,where yo≡constant.
Lemma 2.1. Let the system (1.1) be equivalent to the system (2.1) with the constant
solution y(t)≡y0.Then X(0, y0) = 0.
Proof. Since system (1.1) is equivalent to the system (2.1) there exist vector-
function R(t, x) for which
X(t, x) = Y(t, x) + ∂F
∂x −1R(t, x)−R(−t, F (t, x)).
Therefore
X(0, y0) = Y(0, y0) + ∂F
∂x (0, y0)−1R(0, y0)−R(−0, F (0, y0)) = 0,
because Y(t, y0)≡0,F (0, x)≡xand ∂F
∂x (0, x)≡Eis the identity matrix.
Theorem 2.1. Suppose that for continuously differentiable vector-function X(t, x)
the following conditions are true:
1) X(0, x0) = 0;
2) X(t, xo)≡α(t)m0(t),where α(t)is an scalar odd continuous function, m0(t)
– differentiable vector-function;
3) there exist the solution ∆(t, x)of the problem ∂∆
∂t +∂∆
∂x X(t, x)−∂ X
∂x (t, x)∆ =
0,∆(t, x0)≡m0(t).
Then for any ωthe solution x(t), x(−ω) = x0,of the system (1.1) has the
property x(ω) = x(−ω),provided that it is extendible on [−ω, ω].
If, in addition, the system (1.1) is 2ω-periodic with respect to t, then the solution
x(t), x(−ω) = x0,is 2ω-periodic too.
Proof. As we already know, the system (1.1) is equivalent to the system
dx
dt =X(t, x)−α(t)∆(t, x) =: Y(t, x).
For this system
Y(t, x0) = X(t, x0)−α(t)∆(t, x0) = α(t)mo(t)−α(t)∆(t, x0)≡0.
It means that x0is constant solution of the equivalent system (2.1). So for the
common Reflecting Function of systems (1.1) and (2.1) we have F(t, x(t)) = x(−t)
and F(t, x0)≡x0,and therefore x(ω) = F(−t, x(−ω)) = F(−t, x0) = x0=x(−ω).
From now on we take x0= 0.If it is not so, we put in (1.1) x+x0instead of x
and will consider the system dx
dt =X(t, x +x0).In this case Theorem 2.1 will have
the form

The new method for the searching periodic solutions of periodic differential systems 879
Theorem 2.2. Suppose that for continuously differentiable vector-function X(t, x)
the following conditions are true:
1) X(t, 0) = α(t)m0(t),where α(t)is an scalar odd continuous function, m0(t)
is continuous vector-function;
2) there exist solution ∆(t, x)of the problem
∂∆
∂t +∂∆
∂x X(t, x)−∂∆
∂x X(t, x) = 0,∆(t, 0) = m0(t).(2.2)
Then for any ωthe solution x(t), x(−ω)=0of the system (1.1) has the property
x(ω) = x(−ω)=0,provided that this solution is extendible on [−ω, ω].
If, in addition, the system (1.1) is 2ω-periodic with respect to t, then x(t)is
2ω-periodic too.
To apply this theorem in a concrete case we will seek ∆(t, x) in the appropriate
form. If, for example, X(t, x) is polynomial in xof degree kthen ∆(t, x) we will
seek in the polynomial form of the same degree. Below we give some examples.
Consider first the Riccati equation
dx
dt =a0(t) + a1(t)x+a2(t)x2,(2.3)
where ai(t), i = 0,1,2 are continuous.
The function ∆(t, x) for this equation we will seek in the form
∆ = m0(t) + m1(t)x+m2(t)x2.
Putting function ∆ in the (1.4) for equation (2.3) and equalizing coefficients at xi,
i= 0,1,2,we get system
m0
0+m1a0−a1m0= 0,
m0
1+ 2m2a0−2a2m0= 0,
m0
2−a2m1+a1m2= 0.
(2.4)
From here we obtain 2m0
0m2+ 2m0m0
2−m1m0
1= 0 and therefore
4m0m2−m2
1=c0=const. (2.5)
From the first and second equations in (2.4) we get m1a0=a1m0−m0
0,2a3
0m2=
(2a2
0a2−a0a0
1+a0
0a1)m0−(a0a1+a0
0)m0
0+a0m00
0.
Then multiplying (2.5) by a3
0and using previous relations, we obtain
(4a2
0a2−2a0a0
1+2a0
0a1−a0a2
1)m2
0−2a0
0m0m0
0−a0m02
0−2a0m0m00
0=c0a3
0.(2.6)
If we can guess any solution m0(t) of (2.6) such that α(t) = a0(t)
m0(t)is an odd con-
tinuous function, then we will be sure that for the solution x(t), x(−ω) = 0 of the
equation (2.3), x(ω) = x(−ω)=0,provided, that this solution is extendible on the
[−ω, ω].If, in addition, the equation (2.3) is 2ω-periodic, then this solution x(t),
x(−ω)=0,will be periodic too.
In particular case, when m0= 1,we get the

880 V.I. Mironenko & V.V. Mironenko
Theorem 2.3. Suppose that a0(t)is an odd continuously differentiable and a1(t),
a2(t)are continuous on Rfunctions, for which the identity (4a2
0a2−2a0a0
1+2a0
0a1−
a0a2
1)/a3
0≡const is hold. Then for every solution x(t), x(−ω) = 0,of equation
(2.3), which exist on [−ω, ω ],we have x(ω) = x(−ω)=0.If, in addition, the
equation (2.3) is 2ω-periodic in t, then the solution x(t), x(−ω)=0,is 2ω-periodic,
provided it is extendible on the [−ω, ω].
The question is: how often this procedure brings us to our aim? To answer this
question we have to observe, that if we take any odd continuous function α(t),any
continuous function m1(t) and any constant c0,than the solution x(t), x(−ω)=0,
of the equation
dx
dt =α(t)m0(t) + m0
0(t) + m0(t)m1(t)α(t)
m0(t)x+m0
1(t)+2m0(t)m2(t)α(t)
2m0(t)x2,
where m0(t) = exp Rt
0(a1(τ)−m1(τ)α(τ)) dτ, m2(t) = c0+m2
1(t)
4m0(t),has the property
x(ω) = x(−ω)=0,provided that this solution is extendible on [−ω, ω ].
The similar results we can get for an Abel equation
dx
dt =a0(t) + a1(t)x+a2(t)x2+a3(t)x3, a0(t) = α(t)m0(t).
In this case for ∆ = m0(t) + m1(t)x+m2(t)x2+m3(t)x3we have system
m0
0+m1a0−a1m0= 0, m0
2−m1a2+m2a1+ 3m3a0−3a3m0= 0,
m0
1+ 2m2a0−2a2m0= 0, m0
3−2a3m1+ 2m3a1= 0,
m3a2=m2a3.
It follows from this system, that for every Abel equation we have m0d
dt (4m0m2−
m2
1)−6m0m3dm1
dt = 0,and m0m2m0
3−m1m3m0
1+ 2m2m3m0
0= 0.
Reader can easily get some result, when m0(t)≡1.
If we have the equation
dx
dt =b0(t) + b1(t)x+... +bn(t)xn
a0(t) + a1(t)x+... +an(t)xn,
where b0(t) = α(t)m0(t), α(t) is odd, then it is advisable to seek the solution ∆(t, x)
of the equation (2.2) in the form
∆(t, x) = m0(t) + m1(t)x+... +mn(t)xn
a0(t) + a1(t)x+... +an(t)xn.
The application of the method to differential systems we consider first on the
example of the system


x0
y0

=

0 1
(sin2t−cos t)x0



x
y

+

sin t
−sin2t

.(2.7)
This system has the form x0=p(t)x+q(t) sin t. We want to seek ∆ in the form
∆ = m(t)q(t) = m(t)

1
−sin t

=

m(t)
−m(t) sin t

,

The new method for the searching periodic solutions of periodic differential systems 881
where m(t) is even scalar non-zero function. Then α(t) = sin t
m(t)will be odd function.
The equation (2.2) in this case has the form


m0(t)
−m0(t) sin t−m(t) cos t

−

0 1
sin2t−cos t0



m(t)
−m(t) sin t

= 0.
From this we get m0(t)+m(t) sin t= 0; −m0(t) sin t−m(t) cos t−(sin2t−cos t)m(t) =
0,and finally m(t) = ecos t.It means that system (2.7) is equivalent to the system


x0
y0

=

0 1
sin2t−cos t0



x
y

+

1
−sin t

sin t−

ecos t
−ecos tsin t

e−cos tsin t.
That is to the system
x0=y, y0= (sin2t−cos t)x, (2.8)
with the solution x(t)≡0, y(t)≡0.So the solution x(t), y(t), x(π) = y(π) = 0,of
the system (2.8) is 2π-periodic.
The general solution of the system (2.8) is
x=c1ecos t+c2ecos tZexp(−2ecos t)dt, y =x0.
It implies that system (2.8) has one-dimensional set of 2π-periodic solutions.
One simple generalization of the example gives
Theorem 2.4. Suppose that for system
dx
dt =P(t)x+α(t)q(t), x ∈Rn, t ∈R,
with the continuous n×nmatrix P(t),the scalar odd function α(t)and the contin-
uous vector-function q(t),there exist an scalar even function m(t),for which
m0(t)q(t) = m(t)[P(t)q(t)−q0(t)].
Then for every ωsolution x(t), x(−ω)=0,has the property x(ω) = x(−ω) = 0.
Such solution is 2ω-periodic, provided that P(t), α(t)and q(t)are 2ω-periodic.
Proof. For ∆ = m(t)q(t) we have
∂∆
∂t +∂∆
∂x X−∂X
∂x ∆ = m0(t)q(t) + m(t)q0(t)−P(t)m(t)q(t)=0.
Therefore the given for us system is equivalent to the system dx
dt = (P(t)x+
α(t)q(t)) −α(t)
m(t)m(t)q(t),i.e. the system dx
dt =P(t)x.
Appropriate reference [6, p173] complete the proof.
Theorem 2.5. Suppose that for the system
dxi
dt =α(t)∆i(t) + pi1(t)x1+... +pin(t)xn+fi(t, ai1(t)x1+... +ain (t)xn), i = 1; n,
with the continuously differentiable right-hand side the following conditions are true:

882 V.I. Mironenko & V.V. Mironenko
1) α(t)is continuous odd function;
2) ai1(t)∆1(t) + ai2(t)∆2(t) + ... +ain(t)∆n(t)≡0for every i= 1; n;
3) d∆(t)
dt ≡P(t)∆(t),where ∆(t) = (∆1(t),∆2(t), ..., ∆n(t))Tand matrix P(t) =
(pij (t)), i = 1; n,j= 1; n.
Then for every ω > 0the solution x(t) = (x1(t), x2(t), ..., xn(t))T, x(−ω) = 0,
of the system has the property x(ω) = x(−ω) = 0,provided that the solution is
extendible on [−ω, ω].If, in addition, this system is 2ω-periodic in t, then this
solution is 2ω-periodic as well.
Proof. In this case we put ∆(t, x) = ∆(t) = (∆1(t),∆2(t), ..., ∆n(t))T.Then
∂∆
∂t +∂∆
∂x X−∂X
∂x ∆ = d∆
dt −∂X
∂x ∆ = d∆
dt −P(t)∆ −∂f
∂x ∆,
where matrix P(t) = (pij (t)), i = 1; n,j= 1; n, f = (f1(t, z1), f2(t, z2), ..., fn(t, zn))T,
zi=ai1(t)x1+ai2(t)x2+... +ain(t)xn.Here ∂ f
∂x ∆ is the vector-function with the
components
∂fi
∂x1∆1+∂ fi
∂x2∆2+... +∂ fi
∂xn∆n=∂ fi
∂zi[ai1(t)∆1(t) + ai2(t)∆2(t) + ...+ain(t)∆n(t)] ≡0.
So ∂∆
∂t +∂∆
∂x X−∂ X
∂x ∆≡0, in accordance with the conditions of the theorem.
Therefore, in accordance with the theorem 2.2 we get the conclusion of the Theorem
2.5.
Example 2.1. The solution (x1(t), x2(t)), x1(ω) = x2(ω) = 0,of the system dxi
dt =
α(t)∆i(t) + pi(t)x1+qi(t)x2+ai(t) sin[bi(t)(∆2(t)x1−∆1(t)x2)], i= 1,2,where
∆0
1=p1∆1+q1∆2,∆0
2=p2∆1+q2∆2,is 2ω-periodic if the right-hand side of the
system is continuous 2ω-periodic in t. It follows from the Theorem 2.1 and from the
fact that all solutions of the system is extendible on R[3, p61].
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