Internat. J. Math. & Math. Sci.
VOL. 21 NO. 4 (1998) 767-774
767
ON THE EXISTENCE OF A PERIODIC SOLUTION OF A
NONUNEAR ORDINARY DIFFERENTIAL EQUATION
HSlN CHU
University of Marylnd
College Park, MD 20742
and
SUNHONG DING
Institute of Mathematical Sciences
Academia Sinlca
Chendu, China
(Received March 8, 1996 and in revised form December I0, 1996)
ABSTRACT.
Conslder a planar forced system of the followlng form
(x,y) + h(t)
-{x,y) + g(t},
where
h{t}
(y)
and
and
P{x,y}
g(t)
2x-perlodlc contlnuous functions, t
are
(-m,=)
are continuous and satlsfy local Llpschltz condltlons.
and
In this
paper, by uslng the Polnc&re’s operator we show that if we assume the condltlons,
{C2) and {C3) {see Section 2), then there is at least one 2=-perlodlc
{CI),
solutlon.
In concluslon, we provlde an expllclt example whlch is not in any class
of known results.
KEY WORDS AND PIIBASES:
Polnc&re mapplng, Polnc&re index, Nonlinear Jumpln dlffer-
entlal equation.
1980 AMS SIBJECT CLASSIFICATION CODES (1985 Revlslon}: 34C25;58F14.
1.
INTROIKICTION
We consider the following planar forced system of ordlnary dlfferentlal equa-
tlons:
----=
--d[t
Cx,y) + hCt)
(1.1)
-vCx,y) + gCt),
where h{t) and g(t) are 2-periodlc continuous functions, t e C-m,=) and
and
v{x,y)
are continuous and satisfy local Lipschltz conditions.
(x,y)
Our work is
inspired by previous work on the existence of periodic solutions of nonlinear
ordinary differential equations by Aguinaldo and Schmltt
[I], Cesari [4], Ding
768
H. CHU AND S. DING
[6-8], Dine [9], Lazer and Leach [12], Leach [13], Loud [14], Mawhln [15], [16],
[18], among others.
In this paper, we show
that if we assume the conditions, {CI), {C2) and {C3) {see Section 2), then there
is at least one 2=-perlodlc solution.
In conclusion, we provide an explicit
example which is not in any class of known results.
The authors wish to take this opportunity to thank the referee for polntlnE
out the wonderful paper Capietto et al. [2], which has results related to Theorem
2.1 as well as Lemma 2.1, and which both of us had overlooked.
Mawhln and Ward [17], Mawhin et al.
2.
NAINRITS
-
Consider the system (1.1}.
Let
x
p
cose
y
and
p sine.
Expressing
{1.1) in polar coordinates form, we have
k(p,e) +
(2.1)
de
where
g(e, t)
-(p, el
p
(p cose,p slne)cose
k(p,e)
I{ [p cose
(p,e)
(e,t)
h(e,t)
p(p cose,p sine)sine.
p sine)sine + P(p
cose
p slne}cose}
h(t)cose + g(t)sine, and
(e,t]
g(t)cose.
h(t)sine
We now shall establlsh the followlng theorem.
TI 2.1. Suppose the system of dlfferentlal equations (1.1) or (2.1)
satisfies the conditions:
(C1}
IA(p,e)I < k(p},
where
k(p)
is a continuous function and satlsfles a
local Lipschltz condition.
(C2} (a]
2x+,
some
(b)
< llm
5,
0"
de
OX
dB
(p,@) <
0 < 6 < mln {2/(2n+I),1}
0"
V+I-)
lira
2 + 6 < p+----
0 < 6 <
<Ii--
(pe) -p+=
(C3) lim
mln (p,B}p
p= 0sB<2x
> 0
de
(p,e)
for some
1t--p+(R)
2x-6
--n
for some
n e Z
+
and
or
0
de
(p,’e) <
for some
5.
in (0,1}.
Then the system has at least one 2x-periodic solutlon.
Take note that the assump-
tions of this theorem are comparable wlth the assumptions of Theorem 3 in
[2,
p.367].
We first establish two lemmas.
LENI 2.1. If the system {1.1) or (2.1) satisfies (CI), then for each positive number M
there exists a positive number H, with M
so that for
M
1,
each solution
(p(t},e(t}) of (2.1}, with an initial condition p(t 0} > M for
t e [0,2}, we have p(t} > M
for 0
t
2.
O
This lemma is very close to the "elastic property" in [2,p.351].
NONLINEAR ORDINARY DIFFERENTIAL EQUATION
From [2.1
PROOF.
769
and (C1), we have
dp
< k(p) + c
k(p) < -[
-c
where
c
lCe,t}
max
+
where
0’
Os8s2x
0
> O.
O<_t<_2x
Let
p*{t)
p-{t)
and
be solutlons of the equations
dtd’-P
k(p} + c
dtd- P
-k(p)
(+)
and
p+(tO)
with the Inltial condltlons
p+(t)
Observe that if
of (-}.
p(t)
c
M
for (+)
is a solution of (+) then
and
p+(2=}
p-(t O}
p+(2+to-t}
for (-}.
will be a solution
p+(2X+to-t}
p+(2x). Let M p+{2=}.
with p-(t O)
< O, from (-}, we have p (2) s p (t) for all 0 s t s 2.
p-{t)
Thus
cause -k{p)
(-}
c
dp
k(p,8) +
be a solution of -[
(8,t)
BeLet
p(t O) > M.
with an initial condition
Then, according to the comparison theorem {e.g., see [I0,p.26]) of ordinary differentlal equations, we have p (t} < p(t), as 0 s t s 2.
Together, we have
M!
p {2} s p (t} < p{t}
M
t, 0 s t s 2=.
for all
Observe that
M
p (t
o
z p(2}
> O, from the above inequality. The lemma is now proved.
LEMMA 2.2. If the system (2.1} satisfies (CI}, (C2), (a) or {b), and {C3}
then there exists a constant M > 0 such that for every solution
(2.1} with an initial condition p(t O} > M, for t e [0,2],
o
2n= < 8(0}
8{2=} < 2(n+1}=,
when
n e Z
{p(t),8(t)} of
the inequality:
+
(2.2a}
holds if (C2)(a} is assumed, and the inequality
e[2] < 2
0 < 8[0)
(2.2b)
holds if (C2)(b) is assumed.
PROOF.
that for all
By (C3} and (C2), (a) or (b), there is a
p > M2, we have {p,8) > O,
(p,e)p
for any
=
c, llm mln (p,e)p
p oe<2
n+l
n e Z
+
,
C
constant
M2
such
{2.3}
> c > O, and
2+C/2)
for some
>
posl{Ive
<
de
Jo
(p,e)
<
2x-(/2}
n
(2.4a)
if (C2)(a) is assumed, or
. ]0
2
2 +
if [C2](b] is assumed.
Let
<
/
de
,(p,e)
<
(2.4b)
770
H. CHU AND S. DING
c
l(e,
max
2
oe<2
o<t2
and
(2.5)
By Lemma 2.1, there exists H > 0 such that for every solution (p,(t,e(t]] of
p(t 0) > H we have p(t) > H for 0 s t s 2=. Divide the second equa3
tion of (2.1) by -(p,8) and tntesrate between 8(0) and 8(2=):
(2.1) with
de
g(e, t
(p, e p dt
dt +
(p,e]
(2.6]
(o)
Assume
(2.2a
(or
(2.2b))
2(n+1) z e(2=),
e(0)
false.
+
is
nEZ
where
Then,
n
or
e(0)
say
e(2] z 2(n+1 ],
By usin8 (2.3),
0.
or
(2.4a) or
(2.4b), (2.5), (2.6), we have:
.e(0)
2=+
e(0)
< (n+l)
m(0)
o(,O)
Cp,(9)
(0)-2=
<
(0)-2 (n+l)=
(2)
However,
u(p,e)
u(p’e)
(2)
We must have
Thls ls a contradiction.
0(2=) < 2=).
O(O)
0(0)- 0(2=) > 0).
2= +
..’
(p,e)p
dt <2=+2
e(o)
Slmllarly
we
e(0)-9(2=) <2(n+1)=, where
have
(0)
0(2=] > 2n=,
where
n eZ
n
Z
+
+
(or
(or
The lemma is now proved.
As a direct consequence of Lemma 2.2, we have the followlnE Corollary.
COROLLARY 2.1.
If, in addition to the conditions of Lemma 2.2, we assume
h(t)
g(t)
O, then all 2=-perlodlc solutions of (2.1} are uniformly bounded,
that is, there is a positive number M such that p(t) s M for every 2-perlodlc
solution
(p(t),e(t))
of (2.1).
Let us denote
PROOF OF THEOREM 2.1.
of (1.1)with the initial condition
point
(x0,Y0),
Since
u(x,y), v(x,y), h(t)
define the Poincre
inE P(x0,Y0)
v(x0’Y0) Y(2=;x0’Y0)
of
(1.1]
and
(x(t;x0,Y0), y(t;x0,Y0]) the solution
(x(0;x0,Y0), y(0;x0,Y0)) (x0,Y0). For every
mapping P(x0,Y0)
(x(Z=;x0,Y0) y(2=;x0,Y0)).
g(t) of (1.1) are continuous, the Poincre mappLet u(x0,Y0)
x(Z=;x0,Y0) x and
is continuous.
Y0"
0
The pair (u,v) is a continuous vector field In the
plane
Let M be the positive constant Euaranteed by Lemma 2.2.
Choose a positive
that
>
M.
to
M’
M’
the Lemma 2.2, for all
According
(Xo,YO) such
+
(S’) 2, we have 2n= < 0(0)
8(2=) < 2(n+I)= for some nonnegatlve
that
v(x 0, y ))
=(x
Thus (u(x y
integer n
O’ 0
o’Yo for any real constant and 2for
0
+ 2
(H’) 2
all (x0,Y0) such that
As
(x0,Y0) moves alone the circle x0 +
x0 Y0
2
,2 in counter
(H)
clockwise direction, the vector u(x0,Y0), v(x0,Y0) rotates
Y0
constant ysuch
x
=
2’
throuEh
an
anEle
of
2=.
continuous vector field
The sum of the Poincre index of a
(u(x0,Y0), v(x0,Y0))
sinsular point of the
in the compact domain
{(x0’Y0)lxu
+
NONLINEAR ORDINARY DIFFERENTIAL EQUATION
y
2 s (M’
is equal to
There must exist at least one singular point of the
I.
We denote it by
vector field in this domain.
s (M’)
+
Hence
that
2 2
771
2.
(,).
P(,y)
(,y).
(x,y),
(x(2=;,), y(2=;x,))
It is clear
This is a 2=-perlodlc
solution.
The following two theorems (2.2 and 2.3) are almost the same as two well-known
However,
theorems.
_
these
two
particular
theorems
require
slightly different
Furthermore, these two,theorems can
assumptions than the two well-known theorems.
be proved using Theorem 2.1.
TIl 2.2.
where
+
and
is a continuous
Consider the following nonlinear ordinary differential equation:
d2x
dr2
(+)2x+
()2x-
are positive constants, x
<
< n2
+
+
H(t)
max {-x,O} and
max {x,O}, x
If
2=-perlodic function.
2
n+l
+
(2.7)
H(t)
for some
neZ
+
(2.8a)
or
2 <
<
+
(2 8b)
+
then the equation (2.7) has a
PROOF.
2-perlodic solution.
Let us define
u(x)
(+ )2
if
x z 0
[(_12
if
x < O.
Then
u(x}-x
(+)2x+
)2x
()2x- f (+
[ (_)2x
if
xzO
if
x< O.
Clearly, u(x).x is a continuous functlon of x.
dx
y. Then the equation (2.7) is equivalent to:
Let
(2.9)
=
which is of the same
-u(xl’x + H(t}
It satisfies a local Lipschltz condition.
type as (I.I).
Lipschltz constant is equal to Max
{(e+ )2
(
12
The
Expressing (2.9) in the polar
1}
coordinates, we have
dp
X(p,e) + H(t)slne
(z. zo)
-_(-H{t)cos e),
-(p,e}
when
sine
cose
u(x}’x sine
-
p sine cose(l-u(x})
pslnecose(1
=+]2)
if
pslnecose(1
_}2)
if
e <
e < 2
772
H. CH-U AND S. DING
nd
(p,e]
sln2e + (x).x
cose
_
2
2
sln2e + +cos
e,
sln2e + 2cos2e_
sln2e + g(x)cos2e
Now,
de
+
(p e)
+
-x
if
e < x
e <
If
-.
It Is easy to verlfy that the equatlon (2.7) satisfles the condltlons (C1), (C2)
and (C3).
THEOREN 2.3.
Consider
d2x
dx
+ G(x)
2
(2.11)
H(t)
where G(x) is Lipschitz continuous with respect to
continuous with respect to
x, and H(t) is 2=-periodic and
Suppose (2.11) satisfies the followin8 condition:
t.
G(x____)
(n+6) 2 < lim
x
Ix-]-;-
ml
2,
(2. IZ]
and some
-=
-/,-,
-
(n+l-6)
Ixl(R)
for the some non-negatlve Integer n
0 < 6 <
<
1-- G(x----!
x
_<
ml
2x
Then the equation (2.11) has a 2-perlodlc solution.
F.
Let
dx
y, then the equation (2.11) is equivalent to the system
-[
y
(2.131
dy
-G(x) + H(t).
-[
the system
Observe
that
(x,y)
y, u(x,y)
(2.13)
is
G(x), h(t]
a
0
polar coordinates form (2.1), we have:
sin2e + p!G(p
cose)
cose
(e,t)
special
case
g(t)
A(p,e]
H(t) sine
of
p
and
the
system
(1.1), where
Expressin8 (2.11) in the
H(t).
and
cose
G(p cose) sine, (p,e)
(e,t)
-H(t)
cose.
Now, one can easily verify that the equation (2.13) satisfies the conditions
(C1), (C), and (C3).
3.
Consider the system
dx
(3.1)
=-f(x,y} + H(t)
where
f(x,y)
(x,y}
O
if
k2x3
otherwise,
x2+y2’
(0,0)
NONLINEAR ORDINARY DIFFERENTIAL EQUATION
k >
and H(t) is a continuous 2=-periodlc function.
773
PROOF.
f(x,y)
Observe that the function
--
and (C3).
Express (3.1), in polar coordinate form.
dp
A(p,e)
and
6f
+ H(t)slne
(3.2)
(slnZe
+
kZcos4e)
+ H(t)cose
k2cos2e)
e cos e 8(1
p sin
6f
and
Then
e(1-k2cos2e)
p sine cos
de
with
is continuous,
We now show that (3.1) satisfies the conditions (CI), (C2),
x,y.
exist, for all
-
Then (3.1) has a 2=-perlodlc
solution.
sinZe + kZcos4e.
w(p,e}
and
Using the Residue Theorem of complex analysis we find
jo
Let
tan
e
w(p,e)
j0 slne+k
2
cos
2
e
d tan
2
tan2e +
e
-/2
k
+
2
tan2e
x then
e
d tan
2
2
.tan2
-/2
+
(l+x2)dx
=2
k2
+
x4+xZ+k2
=2
(x2+
(l+xZ)dx
x+k) (x2-2V-1
x+k)
tan2e
There are two complex roots:
Xl
2
x2
2
and
in the upper-half complex plane.
Thus the above
4(2k-1)(k+1)
4i[2i
Since
follows that
integer
n
k
and
(4k{Zk-1))
k+----such that
]
inteEral
is equal to
2(k+1)
kV2k+1
are relatively prime and
k +
k2kV-f
+ i/k+
k +
> v2k+l
k z
,
it
cannot be an integer, and thus there exists a non-negatlve
n <
k2kV
< n + 1.
k+l
The condition (CZ},
satisfied.
It is evident that (Cl} and (C3) are true.
IX(p,e}l
as
plsine
cose(1-kZcos2e)l
In fact"
< p(1 + k2)
and
li___m
min
wCpe)p=
p-(R) O_<e<2
lim[l-4-l =
Hence, by Theorem 2.1, there is a 2-periodlc solution.
k(p}
(a} or (b}, is
774
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u(O}
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non-
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