Positive Periodic Solutions For A Kind Of Prescribed Mean Curvature Du¢ ng-Type Equation
Fan Chao Kong
y, Shi Ping Lu
zReceived 15 May 2015
Abstract
In this paper, we study the existence of periodic solutions to the following pre- scribed mean curvature Du¢ ng-type equation with a singularity and a deviating argument:
u0(t) p1 + (u0)2
!0
+cu0(t) +g(t; u(t )) =p(t);
whereghas a strong singularity atx= 0and satis…es a small force condition at x=1, which are di¤erent from the known literatures.
1 Introduction
In recent years, the problems of periodic solution have been studied widely for some types of di¤erential equations with a singularity, see [3, 6–8, 13–16] and references therein. For example, Wang [15] studied periodic solutions for the Liénard equation with a singularity and a deviating argument of the form
x00(t) +f(x(t))x0(t) +g(t; x(t )) = 0;
where0 < T is a constant,f :R!R,g:R (0;+1)!Ris anL2-Carathéodory function, g(t; x)is a T-periodic function in the …rst argument and can be singular at x= 0,i:e:; g(t; x)can be unbounded asx!0+.
Nowadays, the prescribed mean curvature equation u0(t)
p1 + (u0)2
!0
=f(u(t));
and its modi…ed forms, which arises from some problems associated to di¤erential geometry and combustible gas dynamics, were studied extensively, see [1, 2, 11, 12] and the references therein. Moreover, we note that the existence of periodic solutions for the prescribed curvature mean equations has attracted much attention from researchers.
Mathematics Sub ject Classi…cations: 34B16, 34K13.
yCollege of Mathematics and Computer Science, Anhui Normal University, Wuhu 241000, P. R.
China
zCollege of Mathematics and Statistics, NUIST, Nanjing 210044, P. R. China
304
However, it is not easy to study the periodic solutions for the prescribed curvature mean equations. The main di¢ culty lies in the nonlinear term(pu0(t)
1+(u0)2)0, the existence of which obstructs the usual method of …nding a priori bounds for the Liénard or the Rayleigh equations from working. Until, in [4] , Feng considered a kind of prescribed mean curvature Liénard equation
u0(t) p1 + (u0)2
!0
+f(u(t))u0(t) +g(t; u(t (t))) =e(t); (1) where ; e 2 C(R;R) are T-periodic, and g 2 C(R R;R) is T-periodic in the …rst argument, T >0is a constant. Through the transformation, Feng asserts that Eq.(1) is equivalent to the following system
( u0(t) ='(v(t)) = pv(t)
1 v2(t);
v0(t) = f(t; '(v(t))) g(t; u(t (t))) +e(t):
Then by applying Mawhin’s continuation theorem under some su¢ cient conditions, the author show that Eq.(1) has at least one periodic solution.
On the basis of Feng’s work, various types of prescribed curvature mean equations have been studied, see [9, 10, 17] and the references therein.
However, to the best of our knowledge, the study of positive periodic solutions for the prescribed mean curvature equation with a singularity is relatively infrequent. This is due to the fact that the mechanism on which how the solution is in‡uenced by the singularity and the nonlinear term(pu0(t)
1+(u0)2)0associated to prescribed mean curvature equation is far away from clear.
Inspired by the above facts, in this paper, we consider the following prescribed mean curvature Du¢ ng-type equation with a singularity and a deviating argument
u0(t) p1 + (u0)2
!0
+cu0(t) +g(t; u(t )) =p(t); (2) where c is a constant, 0 < T; g : [0; T] (0;+1)!R is a continuous function.
g can be singular atu= 0, p(t)is continuous and T-periodic with RT
0 p(t)dt = 0:By applying Mawhin’s continuation theorem, we prove that Eq.(2) has at least one positive T-periodic solution.
The structure of the rest of this paper is as follows. In Section 2, we state some necessary de…nitions and lemmas. In Section 3, we prove the main result. Finally, we give an example of an application in Section 4.
2 Preliminary
In order to use Mawhin’s continuation theorem, we …rst recall it.
LetX andY be two Banach spaces, a linear operatorL:D(L) X !Y is said to be a Fredholm operator of index zero provided that
(a) ImLis a closed subset of Y, (b) dim kerL=codim ImL <1:
Let X and Y be two Banach spaces, X be an open and bounded set, and L : D(L) X ! Y be a Fredholm operator of index zero. A continuous operator N : X !Y is said to beL-compact in provided that
(c) Kp(I Q)N( ) is a relative compact set ofX, (d) QN( ) is a bounded set ofY,
where we de…ne X1= kerL, Y2= ImL, and X =X1M
X2 andY =Y1M Y2:
LetP :X ! X1, Q :Y !Y1 be continuous linear projectors (meaningP2 =P andQ2=Q), andKp=Ljker1P\D(L).
LEMMA 1 ([5]). Let X and Y be two real Banach spaces, and be an open and bounded set ofX, andL:D(L) X!Y be a Fredholm operator of index zero. The operator N : X !Y is said to be L-compact in . In addition, if the following conditions hold:
(1) Lx6= N x;8(x; )2@ (0;1);
(2) QN x6= 0;8x2kerL\@ ;
(3) degfJ QN; \kerL;0g 6= 0whereJ : ImQ!kerLis a homeomorphism.
ThenLx=N xhas at least one solution inD(L)\ . In order to use Lemma 1, let us consider the problem
8<
:
u0(t) = (v(t)) = pv(t)
1 v2(t);
v0(t) = c (v(t)) g(t; u(t )) +p(t):
(3)
Obviously, if(u(t); v(t))> is a solution of (3), thenu(t)is a solution of (2). Let X =Y =fx:x(t) = (u(t); v(t))>2C1(R;R2); x(t) =x(t+T)g; where the normal
kxk= maxfjjujj0;jjvjj0g,jjujj0= max
t2[0;T]juj, and jjvjj0= max
t2[0;T]jvj: It is obvious thatX andY are Banach spaces.
Now we de…ne the operator
L:D(L) X!Y; Lx=x0= (u0(t); v0)>; where
D(L) = x:x= (u(t); v(t))> 2C1(R;R2)andx(t) =x(t+T) : Let
X0= x= (u(t); v(t))>2C1(R;R ( 1;1)) :x(t) =x(t+T) : De…ne a nonlinear operatorN : (X\X0) X!Y as follows:
N x= v(t)
p1 v2(t); cv(t)
p1 v2(t) g(t; u(t )) +p(t)
!>
;
where X0 X and is an open and bounded set. Then problem (3) can be written asLx=N xin . We know
kerL= x:x2X; x0= (u0(t); v0(t))>= (0;0)> :
Then we have u0(t) = 0,v0(t) = 0 fort2R. Obviouslyu2R,v2R, thuskerL=R2, and it is also easy to prove that
ImL= (
y2Y : Z T
0
y(s)ds= 0 )
:
Therefore,Lis a Fredholm operator of index zero. Let P :X !kerL; P x= 1
T Z T
0
x(s)ds;
Q:Y !ImQ; Qy= 1 T
Z T 0
y(s)ds:
LetKp=Ljker1L\D(L). Then it is easy to see that (Kpy)(t) =
Z T 0
Gk(t; s)y(s)ds;
where
Gk(t) = ( s T
T for0 t s;
s
T fors t T:
For all with (X\X0) X, we see thatKp(I Q)N( ) is a relative compact set of X andQN( )is a bounded set ofY. So the operatorN isL-compact in .
For the sake of convenience, we list the following assumptions
[H1] There exist positive constantsA1 andA2 withA1< A2 such that
(1) For each positive continuousT-periodic functionx(t)satisfying Z T
0
g(t; x(t))dt= 0;
there exists a positive point 2[0; T]such that A1 x( ) A2:
(2) g(x)<0for allx2(0; A1)andg(x)>0for allx > A2where
g(x) = 1 T
Z T 0
g(t; x)dt; x >0:
[H2] g(t; x) = g1(t; x) +g0(x) where g1 : [0; T] (0;+1) ! R is a continuous function and
(1) There exist positive constantsaandbsuch that
g(t; x) ax+b for all(t; x)2[0; T] (0;+1):
(2) R1
0 g0(x)dx= 1:
Throughout this paper, de…ne B:=
Z T
0 jp(t)j2dt)12 + sup
t2[0;T]jp(t)j
!
<+1:
3 Main Results
THEOREM 1. Suppose the conditions [H1]–[H2] hold, jcj> aT and aA2T+bT+Bp
T
jcj aT (c+ 2aT) +T(2aA2+ 2b+B)<1:
Then Eq.(2) has at least one positiveT-periodic solution.
PROOF. Let
1=fz2 :Lz= N zand 2(0;1)g: Ifz2 1, we have
8<
:
u0(t) = (v(t)) = p v(t) 1 v2(t);
v0(t) = c (v(t)) g(t; u(t )) + p(t):
(4)
Integrating the second equation of (4) from 0 to T, we have Z T
0
g(t; u(t ))dt= 0: (5)
It follows from [H1](1) that there exist positive constantsA1, A2 and 2[0; T]such that
A1 u( ) A2: (6)
Then, we can have
kuk0 = max
t2[0;T]ju(t)j max
t2[0;T] u( ) + Z t
u0(s)ds (7)
A2+ Z T
0 ju0(s)jds A2+p
Tku0k2:
Multiplying the second equation of (4) by u0(t)and integrating on the interval [0; T], we have
0 = Z T
0
v0(t)u0(t)dt= Z T
0
c(u0)2dt
Z T 0
g(t; u(t ))u0(t)dt
+ Z T
0
p(t)u0(t)dt:
Combining with [H2], we get jcj
Z T 0 ju02dt
Z T 0
(aju(t )j+b)ju0(t)jdt+ Z T
0 jp(t)jju0(t)jdt ap
Tkuk0ku0k2+bp
Tku0k2+Bku0k2; which, combining with (7), gives
jcj ku0k22 akuk0
pTku0k2+bp
Tku0k2+Bku0k2
a h
A2+p Tku0k2
ip
Tku0k2+bp
Tku0k2+Bku0k2
=aTku0k22+ (aA2p T+bp
T+B)ku0k2: Then byjcj> aT, we obtain
ku0k2
aA2
pT+bp T+B
jcj aT : (8)
Substituting (8) into (7), we obtain
kuk0 A2+aA2T+bT+Bp T
jcj aT :=M1: (9)
From the second equation of (4), we can get Z T
0 jv0(t)jdt Z T
0 jcjju0(t)jdt+ Z T
0 jg(t; u(t ))jdt+ Z T
0 jp(t)jdt: (10) Write
I+=ft2[0; T] :g(t; u(t )) 0g and I =ft2[0; T] :g(t; u(t )) 0g: Then, combining with (5) and [H2](1), we have
Z T
0 jg(t; u(t ))jdt= Z
I+
g(t; u(t ))dt Z
I
g(t; u(t ))dt
= 2 Z
I+
g(t; u(t ))dt
2a Z T
0
u(t )dt+ 2 Z T
0
bdt 2aT kuk0+2bT:
(11)
Substituting (11) into (10) and in view of (8) and (9), we obtain Z T
0 jv0(t)jdt jcjp
T ku0 k2+ (2aT kuk0+2bT) + BT aA2T +bT+Bp
T
jcj aT (c+ 2aT) +T(2aA2+ 2b+B):
(12)
Integrating the …rst equation of (4) on the interval[0; T], we can get Z T
0
p v(t)
1 v2(t)dt= 0:
Then we can see that there exists 2[0; T]such thatv( ) = 0. It implies that jv(t)j=
Z t
v0(s)ds+v( ) Z T
0 jv0(s)jds;
which, combining with (12), gives jv(t)j
Z T
0 jv0(s)jds aA2T +bT +Bp
T
jcj aT (c+ 2aT) +T(2aA2+ 2b+B) := :
(13)
Since
aA2T+bT+Bp T
jcj aT (c+ 2aT) +T(2aA2+ 2b+B)<1;
we obtain
kvk0= max
t2[0;T]jv(t)j <1: (14)
By (4), we can also have
ku0k0 max
t2[0;T]
jv(t)j
p1 v2(t) 1 2: (15)
From the second equation of (4) and by [H2], we can have
v0(t+ ) = cu0(t+ ) [g1(t+ ; u(t)) +g0(u(t))] + p(t+ ): (16) Multiplying both sides of Eq.(16) byu0(t), we can see that
v0(t+ )u0(t) = cu0(t+ )u0(t) [g1(t+ ; u(t)) +g0(u(t))]u0(t)
+ p(t+ )u0(t): (17)
Let 2[0; T]be as in (6). For anyt2[ ; T], integrating Eq.(17) on the interval[ ; T], we obtain
Z u(t) u( )
g0(u)du= Z t
g0(u(t))u0(t)dt
= Z t
v0(t+ )u0(t)dt Z t
cu0(t+ )u0(t)dt Z t
g1(t+ ; u(t))u0(t)dt+ Z t
p(t+ )u0(t)dt:
Then from the inequality above and combining with (12), we get Z u(t)
u( )
g0(u)du =
Z t
g0(u(t))u0(t)dt Z T
0 jv0(t+ )jju0(t)jdt+ Z T
0 jcjju0(t+ )jju0(t)jdt +
Z T
0 jg1(t+ ; u(t))jju0(t)jdt+ Z T
0 jp(t+ )jju0(t)jdt ku0k0
"
aA2T+bT+Bp T
jcj aT (c+ 2aT) +T(2aA2+ 2b+B)
#
+ Z T
0 jcjju0(t+ )jju0(t)jdt+ Z T
0 jg1(t+ ; u(t))jju0(t)jdt +
Z T
0 jp(t+ )jju0(t)jdt: (18)
SetGM1 = max
juj M1jg1(t; u)j, we have Z T
0
cju0(t+ )jju0(t)jdt jcjTku0k20 (19)
and Z T
0 jg1(t+ ; u(t))jju0(t)jdt GM1Tku0k0: (20) Substituting (19) and (20) into (18), we can obtain
Z u(t) u( )
g0(u)du
ku0k0
"
aA2T +bT +Bp T
jcj aT (c+ 2aT) +T(2aA2+ 2b+B)
#
+jcjTku0k20+ GM1Tku0k0+ BTku0k0
1 2
"
aA2T+bT+Bp T
jcj aT (c+ 2aT) +T(2aA2+ 2b+B)
#
+jcjT
1 2
2
+GM1T
1 2 +BT 1 2; which, combining with (15), gives
Z u(t) u( )
g0(u)du
1 2
"
aA2T +bT+Bp T
jcj aT (c+ 2aT) +T(2aA2+ 2b+B)
#
+jcjT
1 2
2
+GM1T
1 2 + BT
1 2
<+1:
According to [H2](2), fort 2[ ; T], we can see that there exists a constant M2 > 0 such that
u(t) M2: (21)
For the case t2[0; ], we can handle it similarly.
De…ne
0< D1= minfA1; M2gandD2= maxfA2; M1g: Then by (6), (9) and (21) we obtain
D1 u(t) D2: (22)
Set
= x= (u; v)> 2X: D1
2 < u(t)< D2+ 1;kvk0< 1< + 1
2 :
Then the condition (1) of Lemma 1 is satis…ed.
Suppose that there existsx2@ \kerLsuch thatQN x=T1RT
0 N x(s)ds= (0;0)>,
i:e; 8
>>
<
>>
:
1 T
RT 0
pv(t)
1 v2(t)dt= 0;
1 T
RT
0 cpv(t)
1 v2(t) g(t; u(t )) +p(t) dt= 0:
(23)
SincekerL=R2, and u,v2Rare constant, by the …rst equation of (23), we have v= 0< 1:
Then from the second equation of (23), we get 1
T Z T
0
g(t; u(t ))dt= 0:
It follows from [H1](1) that D1
2 < D1< A1 u(t) A2< D2< D2+ 1;
which is contrary to the assumption x 2 @ . So for all x 2 kerL\@ , we have QN x6= 0. Then, we can see that the condition (2) of Lemma 1 is satis…ed.
In the following, we prove that the condition (3) of Lemma 1 is also satis…ed. De…ne z=Kx=K u
v = u A1+A2 2
v :
Then we have that
x=z+
A1+A2
2
0 :
De…ne J : ImQ!kerLis a linear isomorphism with J(u; v) = v
u ; and de…ne
H( ; x) = Kx+ (1 )J QN x; 8(x; )2 [0;1];
Then,
H( ; x) = u (A12+A2)
v +1
T RT
0 [pcv
1 v2 +g(t; u)]dt RT
0 p v
1 v2dt
!
: (24)
Now we claim thatH( ; x)is a homotopic mapping. Assume, by way of contradiction, that there exist
02[0;1]andx0= u0 v0 2@
such thatH( 0; x0) = 0:Substituting 0 andx0into (24), we have
H( 0; x0) = 0
@ 0u0 0(A12+A2)+ (1 0)pcv0
1 v20 + (1 0)g(u0)
0v0+ (1 0)pv0
1 v02
1
A: (25)
SinceH( 0; x0) = 0, we can see that
0v0+ (1 0) v0
p1 v02 = 0;
which combining with 02[0;1], we obtainv0= 0:Thusu0=A1orA2:Ifu0=A1, it follows from [H1](2) thatg(u0)<0. Then substitutingv0= 0into (25), we can have
0u0 0(A1+A2)
2 + (1 0)g(u0)< 0(u0 A1+A2
2 )<0: (26) If u0=A2, it follows from [H1](2) thatg(u0)>0, then substitutingv0= 0into (25), we can have
0u0 0(A1+A2)
2 + (1 0)g(u0)> 0(u0
A1+A2
2 )>0: (27) Combining with (26) and (27), we can see that H( 0; x0)6= 0, which contradicts the assumption. ThereforeH( ; x)is a homotopic mapping and
x>H( ; x)6= 0; 8(x; )2(@ \kerL) [0;1]:
Then
deg(J QN; \kerL;0) = deg(H(0; x); \kerL;0)
= deg(H(1; x); \kerL;0)
= deg(Kx; \kerL;0)
= X
x2K 1(0)
sgn(detK0(x))
= 16= 0:
Thus, the condition (3) of Lemma 1 is also satis…ed. Therefore, by applying Lemma 1, we can conclude that Eq.(2) has at least one positiveT-periodic solution.
4 Example
In this section, we provide an example to illustrate results from the previous sections.
Example 4.1. As an application, we consider the following example:
u0(t) p1 + (u0)2
!0
+ 7u0(t) + 1
32(1 + sin 8t)u(t ) 1
u(t ) = 1
64sin 8t: (28)
Conclusion. The Problem (28) has at least one positive 4-periodic solution. Cor- responding to Theorem 1 and (2), we have
g(t; u(t )) = 1
32(1 + sin 8t)u(t ) 1
u(t ); p(t) = 1 64sin 8t:
Then we can choose T=
4; a= 1
16; b= 1
32; c= 7; A1= 1; A2= 4;
and
B:=
Z T
0 jp(t)j2dt
!12
+ sup
t2[0;T]jp(t)j< 1
32 <+1: Then we can see that [H1] and [H2] hold. Moreover,jcj> aT and
aA2T+bT +Bp T
jcj aT (c+ 2aT) +T(2aA2+ 2b+B) 0:7202<1:
Hence, by applying Theorem 1, we can see that Eq.(28) has at least one positive 4- periodic solution.
REMARK 1. Since all the results in [1]-[17] and the references therein are not applicable to Eq.(28) for solving positive periodic solutions with periodic =4, Theorem 1 in this paper is essentially new.
Acknowledgement. The authors express their thanks to the referee for his (or her) valuable suggestions. The work was supported by the National Natural Science Foundation of China (Grant No. 11271197).
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