- Email: [email protected]
Two-point boundary value problem of 2mth order differential equations
Applied Mathematics and Computation 138 (2003) 11–19 www.elsevier.com/locate/amc
Two-point boundary value problem of 2mth order differential equations Xiaojing Yang Department of Mathematics, Tsinghua University, Beijing 100084, People’s Republic of China
Abstract and uniqueness result for the boundary value problem xð2mÞ þ PmAn existence ð2k1Þ þ ð1Þm1 f ðt; xÞ ¼ eðtÞ, xðjÞ ð0Þ ¼ xðjÞ ðpÞ ¼ 0, j ¼ 0; 1; . . . ; m 1, is given. k¼1 Ak x Ó 2002 Published by Elsevier Science Inc. Keywords: Two-point boundary value problem; Min–max principle
1. Introduction Consider the two-point boundary value problem xð2mÞ þ
m X
Ak xð2k1Þ þ ð1Þ
m1
f ðt; xÞ ¼ eðtÞ;
x 2 Rn ;
ð1:1Þ
k¼1
xðjÞ ð0Þ ¼ xðjÞ ðpÞ ¼ 0;
j ¼ 0; 1; . . . ; m 1;
ð1:2Þ
where m 2 N , Ak is a constant symmetric n n matrix, k ¼ 1; 2; . . . ; m, f : ½0; p Rn ! Rn is a C 1 mapping and ðof =oxÞðt; xÞ is an n n symmetric matrix for all t 2 ½0; p and e : ½0; p ! Rn is continuous. When m ¼ 1, A1 ¼ 0, f ðt; xÞ ¼ grad GðxÞ, (1.1), (1.2) reduces to the following Dirichlet boundary value problem: x00 þ grad GðxÞ ¼ eðtÞ; xð0Þ ¼ xðpÞ ¼ 0:
E-mail address: [email protected] (X. Yang). 0096-3003/02/$ - see front matter Ó 2002 Published by Elsevier Science Inc. PII: S 0 0 9 6 - 3 0 0 3 ( 0 2 ) 0 0 0 9 6 - 6
ð1:3Þ ð1:4Þ
12
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
If D2 G exists, then the Hessian of G satisfies some nonresonance conditions 2 oG ðxÞ 6 B; ð1:5Þ A 6 D2 GðxÞ ¼ oxi oxj where A, B are two constant symmetric n n matrices such that if k1 6 k2 6 6 kn and l1 6 l2 6 6 ln are the eigenvalues of A and B, then there exist integers Nk , k ¼ 1; . . . ; n, satisfying the condition Nk2 < kk 6 lk < ðNk þ 1Þ2 :
ð1:6Þ
Then the existence and uniqueness of the solution satisfying (1.3), (1.4) have been obtained by many authors, see [1–6] and the references therein. By using an improved nonvariational version of a min–max theorem originally developed in [6], we will give some nonresonance conditions to guarantee the existence and uniqueness of a solution satisfying (1.1), (1.2).
2. A nonvariational version of a min–max principle The following nonvariational version of min–max theorem is due to Manasevich [6]. Theorem A. Let H be a real Hilbert space, X and Y be two closed subspaces of H such that H ¼ X Y , and T : H ! H be a C k mapping, k P 1. Assume that there exist two positive constants m1 and m2 such that hT 0 ðuÞx; xi 6 m1 kxk 0
hT ðuÞy; yi P m2 kyk
2
2
8u 2 H ; x 2 X ; 8u 2 H ; y 2 Y
and hT 0 ðuÞx; yi ¼ hx; T 0 ðuÞyi 8u 2 H ; x 2 X ; y 2 Y ; where h ; i is an inner product of H. Then T is a C k diffeomorphism. The following result will be used in our theorem. Theorem B. [7] Let E, F be Banach spaces and f : E ! F be a C 1 map. Suppose that (a) f 0 ðxÞ 2 Isom ðE; F Þ for every x 2 E. (b) kf 0 ðxÞ1 kR 6 wðkxkÞ for every x 2 E, where w : ½0; þ1Þ ! ð0; þ1Þ is continþ1 uous. If 1 ds=wðsÞ ¼ þ1, then f is a global C 1 diffeomorphism. Theorem 2.1. Let H, X, Y be defined in Theorem A. T : H ! H be a C 1 mapping. If there exist two continuous functions a, b : ½0; þ1Þ ! ð0; þ1Þ such that
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
hT 0 ðuÞx; xi 6 aðkukÞkxk2 hT 0 ðuÞy; yi P bðkukÞkyk
2
13
8x 2 X ; u 2 H ;
ð2:1Þ
8y 2 Y ; u 2 H ;
ð2:2Þ
hT 0 ðuÞx; yi ¼ hx; T 0 ðuÞyi 8x 2 X ; y 2 Y ; u 2 H
ð2:3Þ
and Z
þ1
minfaðsÞ; bðsÞg ds ¼ þ1:
ð2:4Þ
0
Then T is a C 1 diffeomorphism of H. Proof. For any u 2 H fixed, if v1 ; v2 2 H , v1 6¼ v2 such that T 0 ðuÞv1 ¼ T 0 ðuÞv2 . Let v ¼ v1 v2 . Then v 2 H and v ¼ x þ y with x 2 X , y 2 Y . Then by (2.1)– (2.3), we have 0 ¼ hT 0 ðuÞðv1 v2 Þ; y xi ¼ hT 0 ðuÞv; y xi ¼ hT 0 ðuÞx; yi hT 0 ðuÞx; xi þ hT 0 ðuÞy; yi hT 0 ðuÞy; xi 2
2
¼ hT 0 ðuÞx; xi þ hT 0 ðuÞy; yi P aðkukÞkxk þ bðkukÞkyk > 0: A contradiction. Hence T 0 ðuÞ is one to one. Denote cðkukÞ ¼ minfaðkukÞ; bðkukÞg and by above analysis 2
2
hT 0 ðuÞv; y xi P aðkukÞkxk þ bðkukÞkyk ; we have 2
2
kT 0 ðuÞvkky xk P cðkukÞðkxk þ kyk Þ:
ð2:5Þ
Taking square on both sides of (2.5), we obtain 2
2 2
2
2
2
2
2
c2 ðkukÞðkxk þ kyk Þ 6 kT 0 ðuÞvk ky xk 6 2kT 0 ðuÞvk ðkxk þ kyk Þ hence cðkukÞkvk 6 2kT 0 ðuÞvk:
ð2:6Þ
Let v 2 T 0 ðuÞH . Then there exists a sequence fvn g such that vn 2 T 0 ðuÞH and vn ! v. Suppose zn 2 H such that T 0 ðuÞzn ¼ vn . Then for m; n 2 N , we have kvm vn k ¼ kT 0 ðuÞzm T 0 ðuÞzn k ¼ kT 0 ðuÞðzm zn Þk P
cðkukÞ kzm zn k: 2
This shows that fzn g is a Cauchy sequence. Let z 2 H and zn ! z. Then from the continuity of T 0 ðuÞ, we have T 0 ðuÞzn ! T 0 ðuÞz, this implies that v ¼ T 0 ðuÞz 2 T 0 ðuÞH . Hence T 0 ðuÞH is closed in H. We show next that T 0 ðuÞH ¼ H . To do this let us assume there exists ? a z 2 ðT 0 ðuÞH Þ , z 6¼ 0. Then hT 0 ðuÞv; zi ¼ 0 for all v 2 H . Since z can be
14
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
decomposed as z ¼ h þ k with h 2 X , k 2 Y . Take v ¼ k h. Then by (2.1)– (2.3), we have 0 ¼ hT 0 ðuÞv; zi ¼ hT 0 ðuÞðk hÞ; k þ hi ¼ hT 0 ðuÞk; ki þ hT 0 ðuÞk; hi hT 0 ðuÞh; ki hT 0 ðuÞh; hi 2
2
¼ hT 0 ðuÞk; ki hT 0 ðuÞh; hi P aðkukÞkhk þ bðkukÞkkk > 0: ?
Again a contradiction. Hence ½T 0 ðuÞH ¼ f0g and T 0 ðuÞ is onto H. From (2.6), T 0 ðuÞ is a linear operator from H onto H and satisfies kðT 0 ðuÞÞ1 k 6
2 : cðkukÞ
ð2:7Þ
Now (2.4), (2.7) and Theorem B imply that T is a diffeomorphism.
3. Main results Let us consider the two-point boundary value problem (1.1), (1.2). We have the following result. Theorem 3.1. Suppose that m 2 N , Ak is a constant symmetric n n matrix, k ¼ 1; . . . ; m, f : ½0; p Rn ! Rn is a C 1 mapping, ðof =oxÞðt; xÞ is an n n symmetric matrix for all t 2 ½0; p and e : ½0; p ! Rn is continuous. Assume moreover that there exist two commutative constant symmetric matrices A and B such that of ðt; xÞ 6 B bðkxkÞIn for all t 2 ½0; p ; x 2 Rn ; ð3:1Þ A þ aðkxkÞIn 6 ox where a, b are defined in Theorem 2.1 and satisfy (2.4), A and B have eigenvalues 2m Ni2m and ðNi þ 1Þ , i ¼ 1; 2; . . . ; n, respectively, fNi g are nonnegative integers and In is the unit n n matrix. Then problem (1.1), (1.2) has a unique solution. Proof. Let H ¼ fx j xðkÞ ð0Þ ¼ xðkÞ ðpÞ ¼ 0; x 2 C 2m ½0; p ; x 2 Rn and k ¼ 0; 1; . . . ; m 1g. Then H is a Hilbert space under the inner product Z p hx; yi ¼ ½ðxð2m1Þ ðtÞ; y ð2m1Þ ðtÞÞ þ ðxð2m2Þ ðtÞ; y ð2m2Þ ðtÞÞ þ 0
þ ðxðtÞ; yðtÞÞ dt
for x; y 2 H :
Since A and B are symmetric and AB ¼ BA, by theory of matrices, there exists an orthogonal matrix P such that P > AP ¼ diagðN12m ; . . . ; Nn2m Þ;
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
P > BP ¼ diagððN1 þ 1Þ2m ; . . . ; ðNn þ 1Þ2m Þ:
15
ð3:2Þ
Let P ¼ ðg1 ; g2 ; . . . ; gn Þ, gk 2 Rn , k ¼ 1; 2; . . . ; n. Then (3.2) implies 2m
Agi ¼ Ni2m gi ; Bgi ¼ ðNi þ 1Þ gi ;
i ¼ 1; 2; . . . ; n:
ð3:3Þ
Define two subspaces U and V of H: ( ) Ni n X X fi ðtÞgi ; fi ðtÞ ¼ aik sin kt; t 2 ½0; p ; i ¼ 1; . . . ; n ; U ¼ uuðtÞ ¼ k¼1 i¼1 ð3:4Þ ( ) n 1 X X V ¼ vvðtÞ ¼ gi ðtÞgi ; gi ðtÞ ¼ aik sin kt; t 2 ½0; p ; i ¼ 1; . . . ; n : i¼1 k¼N þ1 i
ð3:5Þ
Then H ¼ U V . Define a mapping T : H ! H such that for all x; y 2 H , ! Z p" m X ð2mÞ ð2k1Þ hT ðxÞ; yi ¼ ð x ðtÞ; yðtÞÞ þ Ak x ðtÞ; yðtÞ 0
#
k¼1
m
þ ðð 1Þ f ðt; xÞ; yðtÞÞ dt: Integrating by parts, we obtain Z " p
ð 1Þm1 ðxðmÞ ðtÞ; y ðmÞ ðtÞÞ
hT ðxÞ; yi ¼
0
# m X k m ð 1Þ ðAk xðkÞ ðtÞ; y ðk1Þ ðtÞÞ þ ð 1Þ ðf ðt; xðtÞÞ; yðtÞÞ dt: þ k¼1 0
ð3:4 Þ From (3.40 ), it is easy to see T 2 C 1 and for all x; z; y 2 H , Z p" m X hT 0 ðxÞz; yi ¼ ð 1Þm1 ðzðmÞ ðtÞ; y ðmÞ ðtÞÞ þ ð 1Þk ðAk zðkÞ ðtÞ; y ðk1Þ ðtÞÞ 0
k¼1
þ ð 1Þ
m
of ðt; xÞzðtÞ; yðtÞ ox
#
dt:
0
ð3:5 Þ
Using Riesz representation theorem, there exists a unique vector d in H such that
16
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
hd; yi ¼
Z
p
ðeðtÞ; yðtÞÞ dt
for all y 2 H :
ð3:6Þ
0
It is easy to see that x is a solution of (1.1), (1.2) if and only if x satisfies the operator equation T ðxÞ ¼ d:
ð3:7Þ
Let u 2 U , v 2 V , x 2 H . Then by the definition of U and V, we have 0
0
hT ðxÞu; vi hu; T ðxÞi ¼
m X k¼1
ð1Þ
k
Z
p
ðAk uðkÞ ðtÞ; vðk1Þ ðtÞÞ dt ð1Þ
k
0
Z
p
ðuðk1Þ ðtÞ; Ak vðkÞ ðtÞÞ dt ¼ 0:
0 ðkÞ
ðk1Þ
Since Ak u is orthogonal to v and uðk1Þ is orthogonal to Ak vðkÞ in L ½0; p , k ¼ 1; . . . ; m, (2.3) is satisfied. Next let us note for k ¼ 1; 2; . . . ; m, x 2 H , we have Z p p 1 ðAk xðkÞ ðtÞ; xðk1Þ ðtÞÞ dt ¼ ðAk xðk1Þ ðtÞ; xðk1Þ ðtÞÞ0 ¼ 0: ð3:8Þ 2 0 2
From (3.5) and (3.8), for any u 2 U , v 2 V , x 2 H , we have Z p of m1 0 ðmÞ ðmÞ ðt; xÞuðtÞ; uðtÞ dt ðu ðtÞ; u ðtÞÞ hð1Þ T ðxÞu; ui ¼ ox 0 ð3:9Þ and hð1Þ
m1
0
T ðxÞv; vi ¼
Z 0
p
of ðmÞ ðmÞ ðt; xÞvðtÞ; vðtÞ dt: ðv ðtÞ; v ðtÞÞ ox ð3:10Þ
Moreover, we see Z p Z p n Z p n X X ðmÞ 2 2 ðuðmÞ ðtÞ; uðmÞ ðtÞÞ dt ¼ ðfi ðtÞÞ dt 6 Ni2m ðfi ðtÞÞ dt 0
i¼1
0
i¼1
0
ð3:11Þ and Z 0
p
ðvðmÞ ðtÞ; vðmÞ ðtÞÞ dt ¼
n Z X i¼1
p
ðmÞ
2
ðgi ðtÞÞ dt 0
Z p n X 2m 2 P ðNi þ 1Þ ðgi ðtÞÞ dt: i¼1
0
ð3:12Þ
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
17
From (3.1) and (3.11), we have Z p of ðt; xÞuðtÞ; uðtÞ dt ox 0 Z p Z p P ðAuðtÞ; uðtÞÞ dt þ ðaðkxkÞuðtÞ; uðtÞÞ dt 0 0 Z pX Z p n X n ¼ fi ðtÞfj ðtÞðAgi ; gj Þ dt þ aðkxkÞ ðuðtÞ; uðtÞÞ dt 0
¼
i¼1
n X
Ni2m
0
j¼1
Z
p 2
ðfi ðtÞÞ dt þ aðkxkÞ 0
i¼1
Z
p
ðuðtÞ; uðtÞÞ dt: 0
Since 2
kuk ¼ hu; ui ¼
2X m1
Z
j¼0
p
ðuðjÞ ðtÞ; uðjÞ ðtÞÞ dt 6
0
2X m1 j¼0
M 2j
!Z
p
ðuðtÞ; uðtÞÞ dt; 0
where M ¼ max1 6 i 6 n fNi g, we have hð1Þ
m1
T 0 ðxÞu; ui 6
aðkxkÞ 2 kuk ; M1
ð3:13Þ
bðkxkÞ 2 kvk ; M2
ð3:14Þ
P2m1 where m1 ¼ j¼0 M 2j . Similarly, from (3.1) and (3.12), we obtain hð1Þ
m1
T 0 ðxÞv; vi P
where M2 ¼
2m1 X
2j
ðM þ 1Þ :
j¼0
Define a1 ðsÞ ¼ aðsÞ=M1 , b1 ðsÞ ¼ bðsÞ=M2 . Then cðsÞ ¼ minfa1 ðsÞ; b1 ðsÞg P
1 minfaðsÞ; bðsÞg: M2
ð3:15Þ
Then from (2.4), we have Z þ1 cðsÞ ds ¼ þ1: 1
By Theorem 2.1, T is a C 1 diffeomorphism of H, hence there exists a unique x 2 H satisfying (1.1), (1.2). Corollary 3.2. If there exist 2n positive constants e1 ; e2 ; . . . ; e2n such that condition (3.1) is replaced by
18
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
A þ diagðe1 ; . . . ; en Þ 6
of ðt; xÞ 6 B diagðenþ1 ; . . . ; e2n Þ; ox
then the conclusion of Theorem 3.1 still holds. Remark. If we replace the Dirichlet boundary conditions by Neumann boundary conditions for high order differential equations or by periodic boundary conditions when f ðt; xÞ and eðtÞ are periodic in t, then the conclusions of Theorem 3.1 still hold and only minor changes are required in the proofs. Example 1. Consider the boundary value problem uð4Þ ðsin2 t þ 1 þ eÞu ¼ sin t; uð0Þ ¼ uðpÞ ¼ 0; u0 ð0Þ ¼ u0 ðpÞ ¼ 0; where 0 < e < 3. Then fu ðt; uÞ ¼ sin2 t þ 1 þ e satisfies 12 < 1 þ e 6 fu ðt; uÞ 6 2 þ e < 4 ¼ 22 : From Corollary 3.2, the above problem has a unique solution. In the next example we show that condition (2.4) is sometimes essential for periodic boundary conditions. Example 2 u00 þ 4n2 u þ arctg u ¼ 4 cos 2nt; uð0Þ ¼ uðpÞ; u0 ð0Þ ¼ u0 ðpÞ:
ð3:16Þ
Clearly 2
ð2nÞ < fu ðt; uÞ ¼ 4n2 þ
1 2 < ð2n þ 1Þ 1 þ u2
but condition (2.4) is not satisfied. We claim that no solution of (3.16) exists. In fact, if u ¼ u ðtÞ is a solution of (3.16), then u ¼ u ðtÞ is also a solution of the following problem: u00 þ 4n2 u ¼ 4 cos 2nt arctg u ðtÞ; uð0Þ ¼ uðpÞ;
u0 ð0Þ ¼ u0 ðpÞ:
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
19
Hence 2p ¼
Z
p
4 cos2 2nt dt ¼
Z
0
¼
Z
p
cos 2nt ðu00 þ 4n2 u þ arctg u ðtÞÞ dt
0 p
cos 2nt arctg u ðtÞ dt 6
0
Z
p 0
p p2 dt ¼ ; 2 2
that is, 4 6 p, a contradiction.
References [1] Z. Shen, On the periodic solution to the Newtonian equations of motion, Nonlinear Anal. 13 (1989) 145–150. [2] R. Kaman, J. Locker, On a class of nonlinear boundary value problems, J. Differential Equations 26 (1977) 1–8. [3] A.C. Lazer, D.A. Sanchez, On periodically perturbed concervative systems, Michigan Math. J. 16 (1969) 193–200. [4] J. Mawhin, Contractive mappings and periodically perturbed concervative systems, Arch. Math. 12 (1976) 67–73. [5] W. Li, Solving the periodic boundary value problem with the initial value problem method, J. Math. Anal. Appl. 226 (1998) 259–270. [6] R.F. Manasevich, A nonvariational version of a max–min principle, Nonlinear Anal. 7 (1983) 565–570. [7] R. Plastock, Homeomorphisms between Banach spaces, Trans. Amer. Math. Soc. 200 (1974) 169–183.
Two-point boundary value problem of 2mth order differential equations Xiaojing Yang Department of Mathematics, Tsinghua University, Beijing 100084, People’s Republic of China
Abstract and uniqueness result for the boundary value problem xð2mÞ þ PmAn existence ð2k1Þ þ ð1Þm1 f ðt; xÞ ¼ eðtÞ, xðjÞ ð0Þ ¼ xðjÞ ðpÞ ¼ 0, j ¼ 0; 1; . . . ; m 1, is given. k¼1 Ak x Ó 2002 Published by Elsevier Science Inc. Keywords: Two-point boundary value problem; Min–max principle
1. Introduction Consider the two-point boundary value problem xð2mÞ þ
m X
Ak xð2k1Þ þ ð1Þ
m1
f ðt; xÞ ¼ eðtÞ;
x 2 Rn ;
ð1:1Þ
k¼1
xðjÞ ð0Þ ¼ xðjÞ ðpÞ ¼ 0;
j ¼ 0; 1; . . . ; m 1;
ð1:2Þ
where m 2 N , Ak is a constant symmetric n n matrix, k ¼ 1; 2; . . . ; m, f : ½0; p Rn ! Rn is a C 1 mapping and ðof =oxÞðt; xÞ is an n n symmetric matrix for all t 2 ½0; p and e : ½0; p ! Rn is continuous. When m ¼ 1, A1 ¼ 0, f ðt; xÞ ¼ grad GðxÞ, (1.1), (1.2) reduces to the following Dirichlet boundary value problem: x00 þ grad GðxÞ ¼ eðtÞ; xð0Þ ¼ xðpÞ ¼ 0:
E-mail address: [email protected] (X. Yang). 0096-3003/02/$ - see front matter Ó 2002 Published by Elsevier Science Inc. PII: S 0 0 9 6 - 3 0 0 3 ( 0 2 ) 0 0 0 9 6 - 6
ð1:3Þ ð1:4Þ
12
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
If D2 G exists, then the Hessian of G satisfies some nonresonance conditions 2 oG ðxÞ 6 B; ð1:5Þ A 6 D2 GðxÞ ¼ oxi oxj where A, B are two constant symmetric n n matrices such that if k1 6 k2 6 6 kn and l1 6 l2 6 6 ln are the eigenvalues of A and B, then there exist integers Nk , k ¼ 1; . . . ; n, satisfying the condition Nk2 < kk 6 lk < ðNk þ 1Þ2 :
ð1:6Þ
Then the existence and uniqueness of the solution satisfying (1.3), (1.4) have been obtained by many authors, see [1–6] and the references therein. By using an improved nonvariational version of a min–max theorem originally developed in [6], we will give some nonresonance conditions to guarantee the existence and uniqueness of a solution satisfying (1.1), (1.2).
2. A nonvariational version of a min–max principle The following nonvariational version of min–max theorem is due to Manasevich [6]. Theorem A. Let H be a real Hilbert space, X and Y be two closed subspaces of H such that H ¼ X Y , and T : H ! H be a C k mapping, k P 1. Assume that there exist two positive constants m1 and m2 such that hT 0 ðuÞx; xi 6 m1 kxk 0
hT ðuÞy; yi P m2 kyk
2
2
8u 2 H ; x 2 X ; 8u 2 H ; y 2 Y
and hT 0 ðuÞx; yi ¼ hx; T 0 ðuÞyi 8u 2 H ; x 2 X ; y 2 Y ; where h ; i is an inner product of H. Then T is a C k diffeomorphism. The following result will be used in our theorem. Theorem B. [7] Let E, F be Banach spaces and f : E ! F be a C 1 map. Suppose that (a) f 0 ðxÞ 2 Isom ðE; F Þ for every x 2 E. (b) kf 0 ðxÞ1 kR 6 wðkxkÞ for every x 2 E, where w : ½0; þ1Þ ! ð0; þ1Þ is continþ1 uous. If 1 ds=wðsÞ ¼ þ1, then f is a global C 1 diffeomorphism. Theorem 2.1. Let H, X, Y be defined in Theorem A. T : H ! H be a C 1 mapping. If there exist two continuous functions a, b : ½0; þ1Þ ! ð0; þ1Þ such that
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
hT 0 ðuÞx; xi 6 aðkukÞkxk2 hT 0 ðuÞy; yi P bðkukÞkyk
2
13
8x 2 X ; u 2 H ;
ð2:1Þ
8y 2 Y ; u 2 H ;
ð2:2Þ
hT 0 ðuÞx; yi ¼ hx; T 0 ðuÞyi 8x 2 X ; y 2 Y ; u 2 H
ð2:3Þ
and Z
þ1
minfaðsÞ; bðsÞg ds ¼ þ1:
ð2:4Þ
0
Then T is a C 1 diffeomorphism of H. Proof. For any u 2 H fixed, if v1 ; v2 2 H , v1 6¼ v2 such that T 0 ðuÞv1 ¼ T 0 ðuÞv2 . Let v ¼ v1 v2 . Then v 2 H and v ¼ x þ y with x 2 X , y 2 Y . Then by (2.1)– (2.3), we have 0 ¼ hT 0 ðuÞðv1 v2 Þ; y xi ¼ hT 0 ðuÞv; y xi ¼ hT 0 ðuÞx; yi hT 0 ðuÞx; xi þ hT 0 ðuÞy; yi hT 0 ðuÞy; xi 2
2
¼ hT 0 ðuÞx; xi þ hT 0 ðuÞy; yi P aðkukÞkxk þ bðkukÞkyk > 0: A contradiction. Hence T 0 ðuÞ is one to one. Denote cðkukÞ ¼ minfaðkukÞ; bðkukÞg and by above analysis 2
2
hT 0 ðuÞv; y xi P aðkukÞkxk þ bðkukÞkyk ; we have 2
2
kT 0 ðuÞvkky xk P cðkukÞðkxk þ kyk Þ:
ð2:5Þ
Taking square on both sides of (2.5), we obtain 2
2 2
2
2
2
2
2
c2 ðkukÞðkxk þ kyk Þ 6 kT 0 ðuÞvk ky xk 6 2kT 0 ðuÞvk ðkxk þ kyk Þ hence cðkukÞkvk 6 2kT 0 ðuÞvk:
ð2:6Þ
Let v 2 T 0 ðuÞH . Then there exists a sequence fvn g such that vn 2 T 0 ðuÞH and vn ! v. Suppose zn 2 H such that T 0 ðuÞzn ¼ vn . Then for m; n 2 N , we have kvm vn k ¼ kT 0 ðuÞzm T 0 ðuÞzn k ¼ kT 0 ðuÞðzm zn Þk P
cðkukÞ kzm zn k: 2
This shows that fzn g is a Cauchy sequence. Let z 2 H and zn ! z. Then from the continuity of T 0 ðuÞ, we have T 0 ðuÞzn ! T 0 ðuÞz, this implies that v ¼ T 0 ðuÞz 2 T 0 ðuÞH . Hence T 0 ðuÞH is closed in H. We show next that T 0 ðuÞH ¼ H . To do this let us assume there exists ? a z 2 ðT 0 ðuÞH Þ , z 6¼ 0. Then hT 0 ðuÞv; zi ¼ 0 for all v 2 H . Since z can be
14
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
decomposed as z ¼ h þ k with h 2 X , k 2 Y . Take v ¼ k h. Then by (2.1)– (2.3), we have 0 ¼ hT 0 ðuÞv; zi ¼ hT 0 ðuÞðk hÞ; k þ hi ¼ hT 0 ðuÞk; ki þ hT 0 ðuÞk; hi hT 0 ðuÞh; ki hT 0 ðuÞh; hi 2
2
¼ hT 0 ðuÞk; ki hT 0 ðuÞh; hi P aðkukÞkhk þ bðkukÞkkk > 0: ?
Again a contradiction. Hence ½T 0 ðuÞH ¼ f0g and T 0 ðuÞ is onto H. From (2.6), T 0 ðuÞ is a linear operator from H onto H and satisfies kðT 0 ðuÞÞ1 k 6
2 : cðkukÞ
ð2:7Þ
Now (2.4), (2.7) and Theorem B imply that T is a diffeomorphism.
3. Main results Let us consider the two-point boundary value problem (1.1), (1.2). We have the following result. Theorem 3.1. Suppose that m 2 N , Ak is a constant symmetric n n matrix, k ¼ 1; . . . ; m, f : ½0; p Rn ! Rn is a C 1 mapping, ðof =oxÞðt; xÞ is an n n symmetric matrix for all t 2 ½0; p and e : ½0; p ! Rn is continuous. Assume moreover that there exist two commutative constant symmetric matrices A and B such that of ðt; xÞ 6 B bðkxkÞIn for all t 2 ½0; p ; x 2 Rn ; ð3:1Þ A þ aðkxkÞIn 6 ox where a, b are defined in Theorem 2.1 and satisfy (2.4), A and B have eigenvalues 2m Ni2m and ðNi þ 1Þ , i ¼ 1; 2; . . . ; n, respectively, fNi g are nonnegative integers and In is the unit n n matrix. Then problem (1.1), (1.2) has a unique solution. Proof. Let H ¼ fx j xðkÞ ð0Þ ¼ xðkÞ ðpÞ ¼ 0; x 2 C 2m ½0; p ; x 2 Rn and k ¼ 0; 1; . . . ; m 1g. Then H is a Hilbert space under the inner product Z p hx; yi ¼ ½ðxð2m1Þ ðtÞ; y ð2m1Þ ðtÞÞ þ ðxð2m2Þ ðtÞ; y ð2m2Þ ðtÞÞ þ 0
þ ðxðtÞ; yðtÞÞ dt
for x; y 2 H :
Since A and B are symmetric and AB ¼ BA, by theory of matrices, there exists an orthogonal matrix P such that P > AP ¼ diagðN12m ; . . . ; Nn2m Þ;
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
P > BP ¼ diagððN1 þ 1Þ2m ; . . . ; ðNn þ 1Þ2m Þ:
15
ð3:2Þ
Let P ¼ ðg1 ; g2 ; . . . ; gn Þ, gk 2 Rn , k ¼ 1; 2; . . . ; n. Then (3.2) implies 2m
Agi ¼ Ni2m gi ; Bgi ¼ ðNi þ 1Þ gi ;
i ¼ 1; 2; . . . ; n:
ð3:3Þ
Define two subspaces U and V of H: ( ) Ni n X X fi ðtÞgi ; fi ðtÞ ¼ aik sin kt; t 2 ½0; p ; i ¼ 1; . . . ; n ; U ¼ uuðtÞ ¼ k¼1 i¼1 ð3:4Þ ( ) n 1 X X V ¼ vvðtÞ ¼ gi ðtÞgi ; gi ðtÞ ¼ aik sin kt; t 2 ½0; p ; i ¼ 1; . . . ; n : i¼1 k¼N þ1 i
ð3:5Þ
Then H ¼ U V . Define a mapping T : H ! H such that for all x; y 2 H , ! Z p" m X ð2mÞ ð2k1Þ hT ðxÞ; yi ¼ ð x ðtÞ; yðtÞÞ þ Ak x ðtÞ; yðtÞ 0
#
k¼1
m
þ ðð 1Þ f ðt; xÞ; yðtÞÞ dt: Integrating by parts, we obtain Z " p
ð 1Þm1 ðxðmÞ ðtÞ; y ðmÞ ðtÞÞ
hT ðxÞ; yi ¼
0
# m X k m ð 1Þ ðAk xðkÞ ðtÞ; y ðk1Þ ðtÞÞ þ ð 1Þ ðf ðt; xðtÞÞ; yðtÞÞ dt: þ k¼1 0
ð3:4 Þ From (3.40 ), it is easy to see T 2 C 1 and for all x; z; y 2 H , Z p" m X hT 0 ðxÞz; yi ¼ ð 1Þm1 ðzðmÞ ðtÞ; y ðmÞ ðtÞÞ þ ð 1Þk ðAk zðkÞ ðtÞ; y ðk1Þ ðtÞÞ 0
k¼1
þ ð 1Þ
m
of ðt; xÞzðtÞ; yðtÞ ox
#
dt:
0
ð3:5 Þ
Using Riesz representation theorem, there exists a unique vector d in H such that
16
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
hd; yi ¼
Z
p
ðeðtÞ; yðtÞÞ dt
for all y 2 H :
ð3:6Þ
0
It is easy to see that x is a solution of (1.1), (1.2) if and only if x satisfies the operator equation T ðxÞ ¼ d:
ð3:7Þ
Let u 2 U , v 2 V , x 2 H . Then by the definition of U and V, we have 0
0
hT ðxÞu; vi hu; T ðxÞi ¼
m X k¼1
ð1Þ
k
Z
p
ðAk uðkÞ ðtÞ; vðk1Þ ðtÞÞ dt ð1Þ
k
0
Z
p
ðuðk1Þ ðtÞ; Ak vðkÞ ðtÞÞ dt ¼ 0:
0 ðkÞ
ðk1Þ
Since Ak u is orthogonal to v and uðk1Þ is orthogonal to Ak vðkÞ in L ½0; p , k ¼ 1; . . . ; m, (2.3) is satisfied. Next let us note for k ¼ 1; 2; . . . ; m, x 2 H , we have Z p p 1 ðAk xðkÞ ðtÞ; xðk1Þ ðtÞÞ dt ¼ ðAk xðk1Þ ðtÞ; xðk1Þ ðtÞÞ0 ¼ 0: ð3:8Þ 2 0 2
From (3.5) and (3.8), for any u 2 U , v 2 V , x 2 H , we have Z p of m1 0 ðmÞ ðmÞ ðt; xÞuðtÞ; uðtÞ dt ðu ðtÞ; u ðtÞÞ hð1Þ T ðxÞu; ui ¼ ox 0 ð3:9Þ and hð1Þ
m1
0
T ðxÞv; vi ¼
Z 0
p
of ðmÞ ðmÞ ðt; xÞvðtÞ; vðtÞ dt: ðv ðtÞ; v ðtÞÞ ox ð3:10Þ
Moreover, we see Z p Z p n Z p n X X ðmÞ 2 2 ðuðmÞ ðtÞ; uðmÞ ðtÞÞ dt ¼ ðfi ðtÞÞ dt 6 Ni2m ðfi ðtÞÞ dt 0
i¼1
0
i¼1
0
ð3:11Þ and Z 0
p
ðvðmÞ ðtÞ; vðmÞ ðtÞÞ dt ¼
n Z X i¼1
p
ðmÞ
2
ðgi ðtÞÞ dt 0
Z p n X 2m 2 P ðNi þ 1Þ ðgi ðtÞÞ dt: i¼1
0
ð3:12Þ
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
17
From (3.1) and (3.11), we have Z p of ðt; xÞuðtÞ; uðtÞ dt ox 0 Z p Z p P ðAuðtÞ; uðtÞÞ dt þ ðaðkxkÞuðtÞ; uðtÞÞ dt 0 0 Z pX Z p n X n ¼ fi ðtÞfj ðtÞðAgi ; gj Þ dt þ aðkxkÞ ðuðtÞ; uðtÞÞ dt 0
¼
i¼1
n X
Ni2m
0
j¼1
Z
p 2
ðfi ðtÞÞ dt þ aðkxkÞ 0
i¼1
Z
p
ðuðtÞ; uðtÞÞ dt: 0
Since 2
kuk ¼ hu; ui ¼
2X m1
Z
j¼0
p
ðuðjÞ ðtÞ; uðjÞ ðtÞÞ dt 6
0
2X m1 j¼0
M 2j
!Z
p
ðuðtÞ; uðtÞÞ dt; 0
where M ¼ max1 6 i 6 n fNi g, we have hð1Þ
m1
T 0 ðxÞu; ui 6
aðkxkÞ 2 kuk ; M1
ð3:13Þ
bðkxkÞ 2 kvk ; M2
ð3:14Þ
P2m1 where m1 ¼ j¼0 M 2j . Similarly, from (3.1) and (3.12), we obtain hð1Þ
m1
T 0 ðxÞv; vi P
where M2 ¼
2m1 X
2j
ðM þ 1Þ :
j¼0
Define a1 ðsÞ ¼ aðsÞ=M1 , b1 ðsÞ ¼ bðsÞ=M2 . Then cðsÞ ¼ minfa1 ðsÞ; b1 ðsÞg P
1 minfaðsÞ; bðsÞg: M2
ð3:15Þ
Then from (2.4), we have Z þ1 cðsÞ ds ¼ þ1: 1
By Theorem 2.1, T is a C 1 diffeomorphism of H, hence there exists a unique x 2 H satisfying (1.1), (1.2). Corollary 3.2. If there exist 2n positive constants e1 ; e2 ; . . . ; e2n such that condition (3.1) is replaced by
18
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
A þ diagðe1 ; . . . ; en Þ 6
of ðt; xÞ 6 B diagðenþ1 ; . . . ; e2n Þ; ox
then the conclusion of Theorem 3.1 still holds. Remark. If we replace the Dirichlet boundary conditions by Neumann boundary conditions for high order differential equations or by periodic boundary conditions when f ðt; xÞ and eðtÞ are periodic in t, then the conclusions of Theorem 3.1 still hold and only minor changes are required in the proofs. Example 1. Consider the boundary value problem uð4Þ ðsin2 t þ 1 þ eÞu ¼ sin t; uð0Þ ¼ uðpÞ ¼ 0; u0 ð0Þ ¼ u0 ðpÞ ¼ 0; where 0 < e < 3. Then fu ðt; uÞ ¼ sin2 t þ 1 þ e satisfies 12 < 1 þ e 6 fu ðt; uÞ 6 2 þ e < 4 ¼ 22 : From Corollary 3.2, the above problem has a unique solution. In the next example we show that condition (2.4) is sometimes essential for periodic boundary conditions. Example 2 u00 þ 4n2 u þ arctg u ¼ 4 cos 2nt; uð0Þ ¼ uðpÞ; u0 ð0Þ ¼ u0 ðpÞ:
ð3:16Þ
Clearly 2
ð2nÞ < fu ðt; uÞ ¼ 4n2 þ
1 2 < ð2n þ 1Þ 1 þ u2
but condition (2.4) is not satisfied. We claim that no solution of (3.16) exists. In fact, if u ¼ u ðtÞ is a solution of (3.16), then u ¼ u ðtÞ is also a solution of the following problem: u00 þ 4n2 u ¼ 4 cos 2nt arctg u ðtÞ; uð0Þ ¼ uðpÞ;
u0 ð0Þ ¼ u0 ðpÞ:
X. Yang / Appl. Math. Comput. 138 (2003) 11–19
19
Hence 2p ¼
Z
p
4 cos2 2nt dt ¼
Z
0
¼
Z
p
cos 2nt ðu00 þ 4n2 u þ arctg u ðtÞÞ dt
0 p
cos 2nt arctg u ðtÞ dt 6
0
Z
p 0
p p2 dt ¼ ; 2 2
that is, 4 6 p, a contradiction.
References [1] Z. Shen, On the periodic solution to the Newtonian equations of motion, Nonlinear Anal. 13 (1989) 145–150. [2] R. Kaman, J. Locker, On a class of nonlinear boundary value problems, J. Differential Equations 26 (1977) 1–8. [3] A.C. Lazer, D.A. Sanchez, On periodically perturbed concervative systems, Michigan Math. J. 16 (1969) 193–200. [4] J. Mawhin, Contractive mappings and periodically perturbed concervative systems, Arch. Math. 12 (1976) 67–73. [5] W. Li, Solving the periodic boundary value problem with the initial value problem method, J. Math. Anal. Appl. 226 (1998) 259–270. [6] R.F. Manasevich, A nonvariational version of a max–min principle, Nonlinear Anal. 7 (1983) 565–570. [7] R. Plastock, Homeomorphisms between Banach spaces, Trans. Amer. Math. Soc. 200 (1974) 169–183.