Tải bản đầy đủ - 0 (trang)
5 Landesman–Lazer Conditions: The Asymmetric Case

# 5 Landesman–Lazer Conditions: The Asymmetric Case

Tải bản đầy đủ - 0trang

146

6 Playing Around Resonance

Theorem 6.5.1 Assume that

g.t; x/ D xC

where

> 0,

x C h.t; x/ ;

> 0 are such that

p

Cp D

T

;

N

and h is a bounded function, i.e., there is a C > 0 such that

jh.t; x/j Ä C ;

for every .t; x/ 2 Œ0; T

R:

If, moreover, for any non-zero v satisfying

.Plim /

v 00 C v C

v D 0;

v.0/ D v.T/ ; v 0 .0/ D v 0 .T/ ;

one has

Z

Z

lim sup h.t; x/v.t/ dt C

fv<0g x! 1

lim inf h.t; x/v.t/ dt > 0 ;

fv>0g x!C1

then problem (P) has a solution.

Proof We fix a small " > 0 and consider, for

.P /

2 Œ 12 ; 1, the problem

x00 C .2 2 /.. C "/xC . C "/x / C .2

x.0/ D x.T/ ; x0 .0/ D x0 .T/ :

1/g.t; x/ D 0 ;

As usual, we assume by contradiction that there are two sequences . n /n in Œ 12 ; 1,

and .xn /n , with xn solution of (P n ), such that kxn k1 ! 1. Then, vn D xn =kxn k1

verifies

8 00

v .t/ C Œ.2 2 n /. C "/ C .2 n 1/  vnC

ˆ

ˆ

< n

h.t; xn /

D 0;

Œ.2 2 n /. C "/ C .2 n 1/  vn C .2 n 1/

ˆ

kxn k1

ˆ

:

0

0

vn .0/ D vn .T/ ; vn .0/ D vn .T/ :

There are some subsequences, which we denote by . n /n , and .vn /n , a real number

N 2 Œ 12 ; 1, and a function v 2 C1 .Œ0; T/ such that n ! N , and vn ! v in

C1 .Œ0; T/. So, kvk1 D 1, and v satisfies

v 00 C Q v C Q v D 0 ;

v.0/ D v.T/ ; v 0 .0/ D v 0 .T/ ;

6.5 Landesman–Lazer Conditions: The Asymmetric Case

147

with Ä Q Ä C ", Ä Q Ä C ". Therefore, it has to be N D 1, Q D , Q D ,

and v is a solution of .Plim /.

Let us write .xn ; x0n / in modified polar coordinates, as follows:

if xn 0,

1

xn D p

n

cos Ân ; x0n D

n

sin Ân ;

1

xn D p

n

cos Ân ; x0n D

n

sin Ân :

if xn Ä 0,

For n large enough, we have that

xn .t/2 C x0n .t/2 > 0 ;

for every t 2 Œ0; T ;

and we can see that

8

p x00n xn .x0n /2

ˆ

ˆ

; if xn > 0 ;

<

x2n C .x0n /2

Ân0 D

00

0

2

p xn xn .xn /

ˆ

ˆ

:

; if xn < 0 :

x2n C .x0n /2

Integrating on fxn > 0g and fxn < 0g, respectively, we obtain

N D

p

p

N D

Z

Z

p

Z

p

Z

Œ.1

fxn >0g

fxn >0g

Â

1C

Œ.1

fxn <0g

fxn <0g

n /.

n h.t; xn /xn

x2n C .x0n /2

n /.

Â

1C

C "/ C

Ã

C "/ C

n h.t; xn /xn

x2n C .x0n /2

n

x2n

Ã

 x2n C n h.t; xn /xn C .x0n /2

C .x0n /2

;

n

x2n

 x2n C n h.t; xn /xn C .x0n /2

C .x0n /2

:

Hence,

Z

Z

fxn >0g

n h.t; xn /xn

x2n C .x0n /2

N

Äp

measfxn > 0g ;

N

Äp

fxn <0g

n h.t; xn /xn

x2n C .x0n /2

measfxn < 0g :

148

6 Playing Around Resonance

Since v has only simple zeros, and vn ! v in C1 .Œ0; T/, also vn has only simple

zeros, for n sufficiently large, so that the set of points where xn vanishes has zero

measure. Therefore,

Z

T

0

N

N

n h.t; xn /xn

Ä p Cp

2 C .x /2 C .x0 /2

.xC

/

n

n

n

T D 0:

Hence also

Z

h.t; xn /vn

Ä 0:

.vnC /2 C .vn /2 C .vn0 /2

T

0

and, by the Fatou Lemma,

Z

h.t; xn /vn

Ä 0:

.vnC /2 C .vn /2 C .vn0 /2

T

lim inf

0

n

Since v C .t/2 C v .t/2 C v 0 .t/2 is constant in t, and

lim. .vnC /2 C .vn /2 C .vn0 /2 / D .v C /2 C .v /2 C .v 0 /2 ;

n

uniformly in Œ0; T, it has to be

Z

T

0

lim inf h.t; xn /vn Ä 0 ;

n

and, therefore,

Z

Z

lim sup h.t; x/v.t/ dt C

fv<0g x! 1

lim inf h.t; x/v.t/ dt Ä 0 ;

fv>0g x!C1

For 2 0; 12 , one proceeds as in the proof of Theorem 5.5.1, connecting the

point . C "; C "/ to the diagonal by the use of a curve which does not touch the

set †. The proof is thus completed.

Symmetrically, we also have the following.

Theorem 6.5.2 Assume that

g.t; x/ D xC

x C h.t; x/ ;

6.6 Lazer–Leach Conditions for the Asymmetric Oscillator

> 0,

where

149

> 0 are such that

Cp D

p

T

;

N

and h is a bounded function, i.e., there is a C > 0 such that

jh.t; x/j Ä C ;

for every .t; x/ 2 Œ0; T

R:

If, moreover, for any non-zero v satisfying

v 00 C v C

v D 0;

v.0/ D v.T/ ; v 0 .0/ D v 0 .T/ ;

.Plim /

one has

Z

Z

lim inf h.t; x/v.t/ dt C

fv<0g x! 1

lim sup h.t; x/v.t/ dt < 0 ;

fv>0g x!C1

then problem (P) has a solution.

6.6 Lazer–Leach Conditions for the Asymmetric Oscillator

We consider again problem (Q), and assume that

g.x/ D xC

with

> 0,

x C h.x/ ;

> 0 such that

Cp

D p

D

T

;

N

and that h has finite limits

h. 1/ D lim h.x/ ;

x! 1

h.C1/ D lim h.x/ :

x!C1

As in Sect. 5.4, let

.t/ D

8

1

p

ˆ

ˆ

t/ ;

< p sin.

p

1

ˆ

ˆ

: p sin

p

Ä

if t 2 0 ; p

Ä

ÁÁ

t ; if t 2 p ;

;

;

150

6 Playing Around Resonance

extended by -periodicity to the whole R. Let us define the -periodic continuous

function

Ã Z T

Â

h.C1/ h. 1/

ˆ.Â/ D 2N

e.t/ .t C Â/ dt :

(6.3)

0

The Lazer–Leach condition is generalized in the following corollary.

Corollary 6.6.1 (Dancer, 1976) If

./ Ô 0 ;

for every 2 0; ;

then problem (Q) has a solution.

Proof Assume, for instance, that ˆ.Â/ > 0 for every Â 2 Œ0; . Writing v.t/ D

.t C Â/, it can be seen that

Z

fv>0g

vD

2N

Z

;

fv<0g

. v/ D

2N

:

(6.4)

Hence,

Z

lim sup .h.x/

e.t//v.t/ dt D

fv<0g x! 1

D

2N

Z

h. 1/

fv<0g

e.t/ .t C Â/ dt ;

and

Z

lim inf .h.x/

e.t//v.t/ dt D

fv>0g x!C1

D

2N

Z

h.C1/

fv>0g

whence, setting h.t; x/ D h.x/

e.t/ .t C Â/ dt ;

e.t/,

Z

Z

lim sup h.t; x/v.t/ dt C

fv<0g x! 1

Â

D 2N

h.C1/

lim inf h.t; x/v.t/ dt D

fv>0g x!C1

h. 1/

Ã

Z

T

0

e.t/ .t C Â/ dt > 0 ;

so that Theorem 6.5.1 applies.

In the case when ˆ.Â/ < 0 for every Â 2 Œ0; , one sees analogously that

Theorem 6.5.2 can be applied.

### Tài liệu bạn tìm kiếm đã sẵn sàng tải về

5 Landesman–Lazer Conditions: The Asymmetric Case

Tải bản đầy đủ ngay(0 tr)

×