Existence and uniqueness results for periodic solution of nonlinear differential equations

Applied Mathematics and Computation 130 (2002) 213–223 www.elsevier.com/locate/amc

Existence and uniqueness results for periodic solution of nonlinear differential equations Xiaojing Yang Department of Mathematics, Tsinghua University, Beijing, People’s Republic of China

Abstract The existence and uniqueness of solutions for a class of 2nd-order differential equations with periodic conditions are discussed. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: Initial value problems; Periodic problems; Existence; Uniqueness

1. Introduction In this paper, we consider the following 2nth-order differential equation:
 xð2nÞ þ f t; x; x00 ; . . . ; x2ðn1Þ ¼ 0; xð0Þ ¼ xð2pÞ;

0

0

x ð0Þ ¼ x ð2pÞ; . . . ; x

ð1:1Þ ð2n1Þ

ð0Þ ¼ x

ð2n1Þ

ð2pÞ;

ð1:2Þ

where x 2 R; t 2 I ¼ ½0; 2p 

f t þ 2p; x; . . . ; x2ðn1Þ ¼ f t; x; . . . ; x2ðn1Þ 2 C 1 ðI Rn ; RÞ is periodic with respect to t with period 2p. For n ¼ 1, various results that give the existence and uniqueness of the solution have been obtained by many authors (see [1,2]). We shall give sufficient conditions that give the existence and uniqueness of the solution of (1.1) and (1.2). We shall also discuss the existence and uniqueness of the solution to the following problem:

E-mail address: [email protected] (X. Yang). 0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 0 1 ) 0 0 0 3 4 - 0

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X. Yang / Appl. Math. Comput. 130 (2002) 213–223

x00 þ f ðt; x; x0 Þ ¼ 0;

ð1:3Þ

0

xð0Þ ¼ xð2pÞ;

0

x ð0Þ ¼ x ð2pÞ:

ð1:4Þ

2. Initial value problem Consider the initial value problem u0 ðtÞ ¼ gðt; uðtÞÞ;

uð0Þ ¼ u0 :

ð2:1Þ

Lemma 2.1 [3]. Assume that g 2 CðI Rn ; Rn Þ and possesses continuous partial derivatives og=on on I Rn . Let the solution u0 ðtÞ ¼ uðt; t0 ; u0 Þ of (2.1) exist for t 2 I and let ogðt; uðt; t0 ; u0 ÞÞ : H ðt; t0 ; u0 Þ ¼ ou Then ouðt; t0 ; u0 Þ /ðt; t0 ; u0 Þ ¼ ou0 exists and is the solution of v0 ¼ H ðt; t0 ; u0 Þv ð2:2Þ such that /ðt0 ; t0 ; u0 Þ is the unit matrix. (See also the proof of Theorem 4.2 below). Suppose we rewrite (1.1) as vector value problem: X 00 ¼ F ðt; X Þ, where x1 ¼ x x2 ¼ x001 x3 ¼ x002 .. . xn ¼ x00n1 and


T T X ¼ ðx1 ; . . . ; xn Þ ¼ x; x00 ; . . . ; x2ðn1Þ ;
 f t; x; x00 ; . . . ; x2ðn1Þ ¼ f ðt; x1 ; . . . ; xn Þ; X 00 ¼ F ðt; X Þ ¼ AX þ bðt; X Þ

with

0

0 B0 A¼B @

1 0

0

0

0 1

... ... .. .

1 0 0C C; 1A

...

0

ð2:3Þ

bðt; X Þ ¼ ð0; . . . ; 0; f ðt; X T ÞÞ

T

X. Yang / Appl. Math. Comput. 130 (2002) 213–223

and let Y ¼ X 0 , Z ¼ differential equation:


X  Y

215

. Then (1.1) is equivalent to the following first-order

Z 0 ¼ Gðt; ZÞ;

ð2:4Þ

where Gðt; ZÞ ¼

Y : AX þ bðt; X Þ

0 Let Xð0Þ ¼ a, X ð0Þ ¼ b be the initial value vectors of the solution of (2.3) and

V ¼

a b

. Then the initial value problem
 xð2nÞ þ f t; x; x00 ; . . . ; x2ðn1Þ ¼ 0;

ð2:5Þ

ð2n1Þ

x0 ð0Þ ¼ x00 ; . . . ; xð2n1Þ ð0Þ ¼ x0

xð0Þ ¼ x0 ;

has an equivalent form Z 0 ¼ Gðt; ZÞ;

Zð0; V Þ ¼ V :

2n

ð2:6Þ

2n

Now define h : R ! R , hðV Þ ¼ Zð2p; V Þ; H ðV Þ ¼ V  hðV Þ:

ð2:7Þ

By Lemma 2.1, h and H are C 1 -differentiable. Therefore, solving the periodic solutions of (1.1) and (1.2) is equivalent to finding the fixed points of h or the zero points of H . 3. Eigenvalue problem Lemma 3.1. If the n n real matrix M is similar to the matrix N ¼ diag ðk1 ; . . . ; kn Þ and kk > 0, then the eigenvalues of the matrix

0 2pIn  ¼ M ð3:1Þ M 0 pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi are i 2px1 ; . . . ; i 2pkn .  . From Proof. It is easy to see that 0 is not an eigenvalue of M      2  kIn 0  kIn 2pIn   k In 2pkIn   ¼ k2n j k2 In þ 2pM j; ¼ 2  M kIn  M kIn   0 k In þ 2pM  we have   kIn  M

 2pIn  ¼j k2 In þ 2pM j; kIn 

 if and only if k2 ¼ 2pkk ; k ¼ 1; . . . ; n: and k is an eigenvalue of M

ð3:2Þ


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X. Yang / Appl. Math. Comput. 130 (2002) 213–223

Lemma 3.2 (Brown [4]). If A is an n n matrix, then k is an eigenvalue of A if and only if exp k is an eigenvalue of exp A. Lemma 3.3 (Brown [4]). If A is an n n matrix and there exists d > 0 such that n1 j k j> d for all eigenvalues of A, then kA1 k 6 dn kAk , where kAk ¼ max k1=2 ðA AÞ;

½A ¼ ð~ aji Þ if A ¼ ðaij Þ :

Lemma 3.4 (Lakshmikantham [5], Hadamard Inverse Function Theorem). Let X ; Y be Banach spaces and J 2 C 1 ðX ; Y Þ. Assume J is a local homeomorphic mapping and let 1

nðrÞ ¼ inf

x2X k½J 0 ðxÞ 1 k

:

ð3:3Þ

kxk 6 r

If Z

þ1

nðrÞ dr ¼ þ1;

ð3:4Þ

0

then J is a homeomorphism from X onto Y. Let F ; H be defined in (2.3) and (2.7), respectively. Lemma 3.5. If the eigenvalues of the matrix Z 2p oF ðt; xðt; vÞÞdt ox 0 are 0 < k1 ðvÞ 6 k2 ðvÞ 6    6 kn ðvÞ, then the eigenvalues of the matrix H 0 ðvÞ are pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi lk ¼ 1  cos 2pkk  i sin 2pkk ; k ¼ 1; 2; . . . ; n: ð3:5Þ Hence jlk j ¼ 2j sinð

pffiffiffiffiffiffiffiffiffiffi 2pkk =2Þj:

ð3:6Þ

Proof. Consider the variation equation of (2.4) with respect to Z, W0 ¼

oGðt; ZÞ W; oZ

where oGðt; ZÞ ¼ oz



0 In : oF ðt; X Þ 0 oX

ð3:7Þ

X. Yang / Appl. Math. Comput. 130 (2002) 213–223

Let Y ðtÞ ¼ exp

Z

t 0

217

oGðs; Zðs; V ÞÞ ds: oZ

Then Y ðtÞ is a fundamental solution matrix of (3.7) and Y ð0Þ ¼ I2n . Meanwhile, from Lemma 2.1 we know that 0 1 oX ðt; V Þ oX ðt; V Þ oZ B oa ob C C ¼B @ oY ðt; V Þ oY ðt; V Þ A oV oa ob is also a fundamental solution matrix of (3.7). Therefore, H 0 ðV Þ ¼ I2n  h0 ðV Þ oZð2p; V Þ oV Z 2p oGðs; Zðs; V ÞÞ ds : ¼ I2n  exp oZ 0

ð3:8Þ

¼ I2n 

By Lemmas 3.1 and 3.2, the eigenvalues of H 0 ðV Þ are pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi lk ¼ 1  cos 2pkk  i sin 2pkk ; k ¼ 1; . . . ; n: Hence jlk j ¼ 2j sin j

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2pkk =2j:


From Lemma 3.4, let X ¼ Y ¼ Rm . We obtain the following. Lemma 3.6. Assume that F : Rm ! Rm is continuously differentiable on Rm and 1 k½F 0 ðX Þ k 6 M < þ1 for all X 2 Rm . Then F is a homeomorphism of Rm onto Rm . 4. Main results Theorem 4.1. If for any V 2 R2n , the matrix Z 2p oF ðt; X ðt; V ÞÞ dt Dv ¼ oX 0 is similar to a diagonal matrix diag ðk1 ; k2 ; . . . ; kn Þv and for each k 2 f1; . . . ; ng, there exists an Nk 2 N [ f0g; Pk ; Qk > 0 such that 2pNk2 < Pk 6  kk ðV Þ 6 Qk < 2pðNk þ 1Þ2 ; then Eqs. (1.1) and (1.2) has a unique solution. Moreover, fkk ðV of the following equation:

ð4:1Þ Þgn1

are the roots

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X. Yang / Appl. Math. Comput. 130 (2002) 213–223

kn þ fn kn1 þ ð2pÞfn1 kn2 þ    þ ð2pÞn2 f2 k þ ð2pÞn1 f1 ¼ 0;

ð4:2Þ

where fk ¼

Z

2p

0

of ðt; X ðt; V ÞÞ dt; oxk

k ¼ 1; . . . ; n:

Proof. From (2.3) and (3.7), we have

Z 2p oG 0 2pIn 0 R dt ¼ ¼ 2p oF dt 0 D oZ v 0 oX 0

ð4:3Þ

2pIn 0



and 0 Z 0

2p

B oF ðt; X ðt; V ÞÞdt B ¼B B oX @



0 0 ... R 0

2p of 0 ox1

dt

2p 0 ... 0 ...

0 2p ... 0 ...

... ... ... ... ...



0 0 ... R 2p2pof 0

oxn

1

dt

C C C C A

¼ Dv ; which implies n2 n1 j kIn  Dv j¼ kn þ fn kn1 þ ð2pÞfn1 kn2 þ    þ ð2pÞ f2 k þ ð2pÞ f1

with fk ¼ fk ðV Þ ¼

Z 0

2p

of ðt; X ðt; V ÞÞ dt; oxk

k ¼ 1; 2; . . . ; n:

R 2p From Lemmas 3.1, 3.5 and (4.1), the eigenvalues flk g of 0 ðoG=oZÞdt satisfy  pffiffiffiffiffiffiffiffiffiffi   jlk j ¼ 2 sinð 2pkk =2Þ n pffiffiffiffiffiffiffiffiffiffi   pffiffiffiffiffiffiffiffiffiffiffi o  P 2 min  sinð 2ppk =2Þ;  sinð 2pQk =2Þ ¼ d > 0; ð4:4Þ 16k6n

where Nk p <

pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 2pPk =2 6 2pQk =2 < ðNk þ 1Þp:

Therefore H 0 ðV Þ is invertible, and from Lemmas 3.3 and 3.6,

2 1 þ exp 2 max ðNk þ 1Þ p 16k6n 1 k½H 0 ðV Þ k 6 : dn H ðV Þ is a homeomorphism of R2n onto R2n , that is, there exists a unique V0 2 R2n such that H ðV0 Þ ¼ 0, or hðV0 Þ ¼ V0 . From the definition of h and H in

X. Yang / Appl. Math. Comput. 130 (2002) 213–223

219

(2.7), we obtain the unique initial vector V0 of the solution Zðt; V0 Þ which satisfies (1.1) and (1.2).
Example 1. If n ¼ 1, then (1.1) and (1.2) are the scalar second-order ODE, x00 þ f ðt; xÞ ¼ 0;

f ðt; þ2p; xÞ ¼ f ðt; xÞ 0

xð0Þ ¼ xð2pÞ;

8t 2 I; x 2 R:

0

x ð0Þ ¼ x ð2pÞ:

ð4:5Þ ð4:6Þ

In this case Z 2p of ðt; xðt; V ÞÞdt ¼ 0; kþ ox 0 hence 2pN12 < P 6

Z

2p

0

of 2 ðt; xðt; V ÞÞdt 6 Q < 2pðN1 þ 1Þ ox

for N1 2 N ;

8V 2 R2 is a sufficient condition for the existence and uniqueness of the solution to Eqs. (4.5) and (4.6). Example 2. If n ¼ 2, we have xð4Þ þ f ðt; x; x00 Þ ¼ 0; ðkÞ

ðkÞ

x ð0Þ ¼ x ð2pÞ;

f ðt þ 2p; x; x00 Þ ¼ f ðt; x; x00 Þ; k ¼ 0; 1; 2; 3:

ð4:7Þ ð4:8Þ

In this case, we have k2 þ f2 k þ 2pf1 ¼ 0; where f1 ¼ f1 ðV Þ ¼ f2 ¼ f2 ðV Þ ¼

Z

2p

Z0 2p 0

of ðt; xðt; V Þ; x00 ðr; V ÞÞ dt; ox of ðt; xðt; V Þ; x00 ðt; V ÞÞ dt: ox00

V 2 R2

Then a sufficient condition for the existence and uniqueness of the solution to Eqs. (4.7) and (4.8) is that f1 > 0; f22 > 8pf1 and there exist two nonnegative integers N1 ; N2 and constants P1 ; P2 ; Q1 ; Q2 > 0 such that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4pN12 < 2P1 6 f2  f22  8pf1 6 2Q1 < 4pðN1 þ 1Þ ; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4pN22 < 2P2 6 f2 þ f22  8pf1 6 2Q2 < 4pðN2 þ 1Þ : In the remaining part of this section, we consider the following second-order differential equation:

220

X. Yang / Appl. Math. Comput. 130 (2002) 213–223

x00 þ f ðt; x; x0 Þ ¼ 0;

ð4:9Þ

with periodic condition x0 ð0Þ ¼ x0 ð2pÞ:

xð0Þ ¼ xð2pÞ;

ð4:10Þ

Theorem 4.2. If f ðt; x; x0 Þ satisfies the following assumptions: (a) f ðt; x; yÞ 2 C 1 ð½0; 2p R R; RÞ; f ðt þ 2p; x; yÞ ¼ f ðt; x; yÞ; 8t 2 ½0; 2p ; (b) There exist a nonnegative integer N and positive constants P, Q, R such that for any V ¼ ða; bÞ 2 R2 , ðxðt; aÞ; x0 ðt; bÞÞ is the solution of (4.9) satisfying ðx; ð0; aÞ; x0 ð0; bÞÞ ¼ ða; bÞ, we have Z

2

2N p < P 6

2p

of 2 ðt; xðt; aÞ; x0 ðt; bÞÞdt 6 Q < 2ðN þ 1Þ p ox 0  Z 2p    of 0  ðt; xðt; aÞ; x ðt; bÞÞdt 6 R:  oy 0

Then Eqs. (4.9) and (4.10) has a unique solution. Proof. Let z0 ¼ ðx0 ; y0 Þ 2 R2 . Consider the initial value problem x00 þ f ðt; x; x0 Þ ¼ 0; xð0Þ ¼ x0 ; x0 ð0Þ ¼ y0 :

ð4:11Þ

Let zðt; z0 Þ ¼ ðxðt; z0 Þ; yðt; z0 ÞÞ be the solution of (4.11) satisfying zð0; z0 Þ ¼ z0 ¼ ðx0 ; y0 Þ: Define H : R2 ! R2 ;

H ðz0 Þ ¼ z0  zð2p; z0 Þ;

then from f 2 C 1 , we know H 2 C 1 ðR2 ; R2 Þ. It is evident that we need only to prove that H ðz0 Þ has a unique zero point. 1 Similar to the proof of Theorem 4.1, we show that ½H 0 ðz0 Þ exists and there 1 0 exists a positive constant C0 , such that k½H ðz0 Þ k 6 C0 . T Let y ¼ x0 ; z ¼ ðx; yÞ; sðt; zÞ ¼ ðy; f ðt; x; yÞÞ . Then (4.11) has the following form: z0 ¼ Sðt; zÞ; zð0Þ ¼ z0 :

ð4:12Þ

Consider the variation equation W0 ¼

oSðt; zÞ W; oz

ð4:13Þ

X. Yang / Appl. Math. Comput. 130 (2002) 213–223

where

0 0 oSðt; zÞ @ of ¼  ðt; zÞ oz ox

and let Y ðtÞ ¼ exp

Z

t

0

1 1 A of  ðt; zÞ oy

221

ð4:14Þ

oSðs; zÞ ds: oz

Then Y ðtÞ is the Standard Fundamented Matrix (SFM) solution of (4.13). From the following equation:   



d oz o dz o oS oz ðSðt; zðt; z0 ÞÞÞ ¼ ðt; zðt; z0 ÞÞ ðt; z0 Þ ; ¼ ¼ dt oz0 oz0 dt oz0 oz0 oz0 we know that 0

ox B ox0 ðt; z0 Þ oz ¼B oz0 @ oy ðt; z0 Þ ox0

1 ox ðt; z0 Þ C oy0 C; A oy ðt; z0 Þ oy0

oz ð0; z0 Þ ¼ I2 oz0

is also the SFM of (4.13). Therefore H 0 ðz0 Þ ¼ I2 

oz ð2p; z0 Þ ¼ I2  exp oz0

Z

2p 0

oS ðS; zðs; z0 ÞÞds: oz0

ð4:15Þ

Let k be an eigenvalue of Z 2p os ðs; zðs; z0 ÞÞds; oz 0 0 and l be the corresponding eigenvalue of H 0 ðz0 Þ. Then l ¼ 1  ek ¼ 1  ea ðcos b isinbÞ; where a ¼ Rek; b ¼ Imk. Hence 2

2

jlj ¼ 1 þ e2a  2ea cos b ¼ ð1  ea Þ þ 2ea ð1  cos bÞ:

ð4:16Þ

From (b) and (4.14) pffiffiffiffi 1 k1;2 ¼ ðK þ DÞ; 2 where D ¼ K 2  8pJ and Z 2p of ðt; x; yÞdt; k¼ oy 0 is any solution of (4.11).

ð4:17Þ

J¼

Z 0

2p

of ðt; x; yÞdt; ox

x ¼ xðt; z0 Þ;

y ¼ ðt; z0 Þ

222

X. Yang / Appl. Math. Comput. 130 (2002) 213–223

pffiffiffiffiffiffiffiffi pffiffiffiffi (i) If K ¼ 0, then k1;2 ¼ D ¼ i 2pJ and by (b), jk1;2 j P 2N p. (ii) If K 6¼ 0, then there are two cases: (1) D P 0. If K > 0, then pffiffiffiffi 1 1 D  K2 8pJ 4N 2 p2 P Rek1;2 ¼ ReðK þ DÞ 6 pffiffiffiffi 6 < 0: 6 2 2 DþK 4K R If K < 0, similarly, 4N 2 p2 P > 0: R pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) D < 0. Let  ¼ ðP  2N 2 pÞ > 0. If jKj P e, then Rek1;2 P

1 1 jRek1;2 j ¼ jKj P e > 0: 2 2 If jKj < e, then from (4.17) and (b) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2N p ¼ 2pP  e2 < 2pP  e2 =4 6 2pJ  e2 =4 < Imk1 1 pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi D 6 2pJ 6 2pQ < 2ðN þ 1Þp; ¼ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffiffi  2ðN þ 1Þp <  2pQ 6 Imk2 ¼ D 6  2pP  e2 =4 < 2N p: 2 From the above analysis, there exists a constant C1 > 0 such that 2

jlj P C1 > 0; which implies that ðH 0 ðz0 ÞÞ1 exists and there is a constant C0 > 0, such that 1

kðH 0 ðz0 ÞÞ k 6 C0 : From Lemma 3.6, the mapping H : R2 ! R2 is homeomorphic, and hence H ðz0 Þ has a unique zero point, that is, (4.9) and (4.10) has a unique solution.
Remark. By choosing any z0 2 R2 , we define a mapping G : R2 ! R2 , 1

Gðz0 Þ ¼ z0  ðH 0 ðz0 ÞÞ H ðz0 Þ: If z0 is a fixed point of G, then z0 is the initial vector of the unique solution of (4.9) and (4.10). By using Newtons’s method of approximation, we can apply the following iteration process to obtain z0 , beginning with z0 : znþ1 ¼ zn  ðH 0 ðzn ÞÞ1 H ðzn Þ;

n ¼ 0; 1; . . . :

X. Yang / Appl. Math. Comput. 130 (2002) 213–223

223

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