Positive Solutions of Operator Equations on Half-Line

Transcription

1 Int. Journl of Mth. Anlysis, Vol. 3, 29, no. 5, Positive Solutions of Opertor Equtions on Hlf-Line Bohe Wng 1 School of Mthemtics Shndong Administrtion Institute Jinn, 2514, P.R. Chin sdusuh@163.com Abstrct In this pper, under the weker conditions, we investigte the problem of the existence of positive solutions for opertor equtions on hlfline. We estblish some results on the existence of multiple positive solutions for opertor equtions on hlf-line by pplying the fixed-point theorem in specil function spce. Our results significntly extend nd improve mny known results. Mthemtics Subject Clssifiction: 34B15; 39A1 Keywords: Positive solutions; Fixed points; Opertor equtions; Hlf-line The purpose of this pper is to estblish the existence of positive solutions to the following opertor eqution on hlf-line: where A is n integrl opertor given by Ax(t) = x(t) =Ax(t), t R +. (1.1) G(t, s)m(s)f(s, x(s)ds, t R +. (1.2) nd R + =[, ),m( ),f( ) re given functions nd the kernel G(t,s) of (1.2) is given by { G(t, s) = 1 e ks (e kt e kt ), t s, (1.3) 2k e kt (e ks e ks ), s t, 1 Supported by the Stte Ministry of Eduction Doctorl Foundtion of Chin (267531).

2 212 Bohe Wng with k> constnt. In recent yers, mny uthors([1-7]) re interested in the existence of positive solutions for some boundry vlue problems on hlf-line. In fct, in [1,3-7], some concrete equtions of the specil cse where f is continuous t t= hve extensively been studied. Only in [2], the nonliner term is llowed to hve singulrities. To the uthor s knowledge, there is little reserch concerning (1.1), so it is worthwhile to investigte opertor eqution (1.1). The pper is orgnized s follows. In Section 2, we present some preliminries nd lemms tht will be used to prove our min results. Then, we give result of completely continuous opertor in Theorem 2.1. In Section 3, vrious conditions on the existence of multiple positive solutions to the opertor eqution (1.1) re discussed. u C 2 [, 1] is sid to be positive solution of opertor eqution (1.1) if nd only if u stisfies opertor eqution (1.1) nd u(t) >, for t R Preliminries nd Lemms In this Section, we present some definitions nd lemms tht will be used in the proof of our min results. For convenience the reders, we present here the definitions of cone nd completely continuous opertor. Definition 2.1. A nonempty subset K of Bnch spce E is clled cone if K is convex, closed, nd (i) αx Kfor ll x Kndα (ii)x, x Kimpliesx = θ. Definition 2.2. An opertor F : E E is sid to be completely continuous if F is continuous nd mps bounded sets into precompct sets. Obviously, we cn see tht the properties of the kernel given by (1.3) re s follows: G(t, s), (t, s) R + R +, (2.1) e μt G(t, s) e ks G(s, s), μ k(t, s) R + R +, (2.2) Let nd b be two numbers chosen t rndom from (, ). Without loss of generlity, we my ssume <b. We get G(t, s) m 1 G(s, s)e ks, (t, s) [, b] R +, (2.3) where m 1 := min{e kb,e k e k } < 1. (2.4) In the pper, We shll consider the following spce E: { } E := x C(R + ) : sup x(t) e λt <. (2.5) t R +

3 Positive solutions of opertor equtions 213 where λ>kis given. It is esy to testify tht E is Bnch spce equipped with following Bielecki s norm x := sup t R + x(t)e λt. (2.6) Menwhile,we define pone P s follows: P := { x E : x(t), } t (, ); min 1 x t [,b]. (2.7) For convenience, let us list the following ssumptions: (H 1 ) f : R + R + R + is continuous nd sup (t,x) R + R + f(t, x) <. (H 2 ) m :(, ) R + is continuous nd my be singulr t t =;m(t) on R +. (H 3 ) < e ks G(s, s)m(s)ds, e ks G(s, s)m(s)ds <. (H 4 ) f <L, l 1 <f. (H 5 ) f <L, l 1 <f.. (H 6 ) l 2 <f. (H 7 ) l 2 <f. (H 8 ) f <L. (H 9 ) f <L. (H 1 ) f : R + R + R + is continuous nd f(t, x) (t)+b(t)x, (t, x) R + R +. where, b re continuous functions. (H 11 ) The integrls M 1 = e ks (s)m(s)ds nd M 2 = e (λ k)s b(s)m(s)ds re convergent nd M 2 < 2K. In the bove ssumptions, we write f α f(t, x) = lim x α sup t R + x,α=, ; f f(t, x) β = lim x β inf,β =,, t R + x ( 1 L = e ks G(s, s)m(s)ds), ) 1 l 1 = (m 1 e ks G(s, s)m(s)ds,l 2 = l 1e λ(+b)/2. m 1 b Lemm 2.1. Let P be cone in Bnch spde X nd Ω X be bounded set nd A : Ω P P be completely continuous opertor. If Ax λx for ny x Ω P, λ 1, then the fixed point index i(a, Ω P, P) =1.

4 214 Bohe Wng Lemm 2.2. Let P be cone in Bnch spde X nd Ω X be bounded set nd A : Ω P P be completely continuous opertor. If there exists u >θsuch tht x Ax tu x, x Ω P, t. then we hve i(a, Ω P, P) =. Lemm 2.3 Let X be Bnch spce, nd let P be cone in X. Assume Ω 1, Ω 2 re open subsets of X with Ω 1, Ω 1 Ω 2. Let A : P (Ω 2 \ Ω 1 P ) be completely continuous opertor, stisfying either (i) Ax x, x P Ω 1 ; Ax x, x P Ω 2. or (ii) Ax x, x P Ω 1 ; Ax x, x P Ω 2. Then A hve fixed point in P (Ω 2 \ Ω 1 ). Lemm 2.4 Let E be the spce given by { } E := x C(R + ) : sup x(t) p(t) <. t R + equipped with the norm x := sup t R + { x(t) p(t)}. where p; let Ω E. If the function x Ω re lmost equicontinuous on R + (i.e., they re equicontinuous in ech intervl [,T],T (, )nd uniformly bounded in the sense of the norm x q := sup t R + { x(t) q(t)}. where the function q is positive nd continuous on R + nd then is Ω reltively compct in E. p(t) lim t q(t) =. In the following section, We will give result of completely continuous opertor.

5 Positive solutions of opertor equtions 215 Theorem 2.1 Assume tht (H 1 ) (H 3 ) hold. Then for ny bounded set Ω E, we know tht A : Ω P P is completely continuous. Proof. Let us choose ny bounded set Ω E. Firstly, we prove tht A : Ω P P. From (H 2 ), we know tht there exists t (, ) such tht m(t ) > orh(t ) >. Since m(t) orh(t) re continuous t t = t,(1.2),(2.1) nd (H 1 ) imply tht Ax(t), x Ω P. (3.1) By (H 1 ), (H 3 ) nd (2.2),we get for ny x Ω P nd t R +, Ax(t) e λt = Inequlities (3.2)imply tht sup { Ax(t) e λt } t R + e λt G(t, s)m(s)f(s, x(s))ds e ks G(s, s)m(s)ds sup f(s, x). (s,x) R + R + e ks G(s, s)m(s)ds e ks G(s, s)m(s)f(s, x(s))ds sup f(s, x) <. (s,x) R + R + Thus Ax E, x Ω P. (3.3) Moreover, for ny x Ω P nd σ R +, we know by (2.2) nd (2.3)tht min t [,b] Ax(t) = min t [,b] G(t, s)m(s)f(s, x(s))ds m 1 e λσ G(σ, s)m(s)f(s, x(s))ds = m 1 e λσ Ax(σ) m 1 e ks G(s, s)m(s)f(s, x(s))ds Therefore, min Ax(t) m 1 Ax, x Ω P. (3.4) t [,b] (3.1),(3.3), nd (3.4) tell us tht A(Ω P ) P s desired. Now let us prove thta : Ω P P is completely continuous when m :[, ) [, ),h :[, ) [, )is continuous. For ny λ 1 such tht k<λ 1 <λnd x Ω P, t R +, we see similrly from the proof of (3.2)tht functions {Ax : x Ω P } re uniformly bounded with respect to the norm x λ1 := sup t R + { x(t) e λ 1t }. Moreover, for ny T (, ), the fct tht G(t, s) C(R + R + ), m(t) C(R + ), f(t, x) C(R + R + ),

6 216 Bohe Wng nd stndrd rguments tell us tht {Ax : x Ω P } re equicontinuous in intervl [,T]. So {Ax : x Ω P } re lmost equicontinuous on R +. If we set p(t) =e λt, q(t) =e λ1t, then Lemm 2.1 implies tht A(Ω P )is precompct set in E. Hence A is completely continuous. In the end, we clim tht opertor A is lso completely continuous if m, h : [, ) [, ) re singulr t t=. For ech n 1, denotes the opertor A n by A n (x)(t) = 1/n It follows from the bove proof tht G(t, s)m(s)f(s, x(s))ds, x Ω P, t R +. (3.5) A n : Ω P P is completely continuous, for ech n 1. (3.6) By (H 1 ) nd (2.2), we hve nd so A(x)(t) A n (x)(t) e λt = 1/n 1/n 1/n sup { A(x)(t) A n (x)(t) e λt } t R + e λt G(t, s)m(s)f(s, x(s))ds e ks G(s, s)m(s)f(s, x(s)) e ks G(s, s)m(s)ds 1/n e ks G(s, s)m(s)ds sup f(s, x). (s,x) R + R + sup f(s, x). (s,x) R + R + (3.7) Assumption (H 3 ) nd the bsolute continuity of integrl imply tht lim n 1/n G(s, s)m(s)ds =. Conditions (3.6)-(3.8)justify tht is A is lso completely continuous. To sum up, the conclusion of Lemm 2.1 follows. Remrk 2.1 Since m(t) is llowed to hve singulrity t t = nd t is in [, ), the proof of Theorem 2.1 hs lrger difference with those of the finite intervls. 3. Multiplicity results In this section, we re concerned on the existence of t lest two positive solutions of opertor eqution (1.1). We obtin the following existence results.

7 Positive solutions of opertor equtions 217 Theorem 3.1 Assume tht (H 1 ) (H 4 )hold, Then opertor eqution (1.1) hs t lest one positive solution. Proof. The conditions f <L,imply tht there exist s 1 > nd ɛ 1 > such tht sup f(t, x) (L ɛ 1 ) x, x s 1, t R + Hence f(t, x) (L ɛ 1 ) x, x s 1,t R +, (3.1) Set Ω 1 = {x E : x <s 1 }, Then we get from (3.1) nd (H 3 ), for ny x Ω 1 P Ax = sup t R + { (L ɛ 1 )r 1 sup = r 1 L G(t, s)m(s)f(s, x(s))ds e λt } { t R + e λt G(t, s)m(s)ds} (L ɛ 1 )r 1 e ks G(s, s)m(s)ds e ks G(s, s)m(s)ds ɛ 1 r 1 e ks G(s, s)m(s)ds < s 1 Thus Ax <s 1, x Ω 1 P. (3.2) Under such circumstnces, we my conclude tht Ax λx, x Ω 1 P, λ 1, (3.3) Otherwise, there would existx 1 Ω 1 P nd λ 1 1 such tht Ax 1 = λ 1 x 1. Thus Ax 1 = λ 1 x 1 x 1 = s 1. (3.4) Obviously, (3.4) is in contrdiction with (3.2). This implies tht (3.3)holds. Furthermore, ssumptions (H 1 ) (H 3 )tell us tht Theorem 2.1 holds. Thus A : Ω 1 P P is completely continuous. Therefore, Lemm 2.1 nd (3.3) men i(a, Ω 1 P, P) =1. (3.5) The conditions l 1 <f imply tht there exist η 1 >m 1 s 1 > nd ɛ 2 > such tht inf f(t, x) (l t R ɛ 2 )x, x η 1, Hence f(t, x) (l 1 + ɛ 2 )x, x η, t [, b]. (3.6) Write s 2 = η 1 m 1 >s 1, Ω 2 = {x E : x <s 2 }. (3.7)

8 218 Bohe Wng Let u =1 P, then x Ax λu, x Ω 2 P, λ. (3.8) Suppose tht (3.8)were flse, then there would exist x 2 Ω 2 P nd λ 2 such tht x 2 Ax 2 = λ 2. Condition (3.6) nd the fct tht x 2 = s 2 = η m 1 > η 1 imply tht f(t, x 2 (t)) (l 1 + ɛ 2 )x 2 (t),t [, b]. (3.9) Set C 2 = min{x 2 (t) :t [, b]}. (3.1) By virtue of (2.3),(3.9)nd (3.1), we hve for ny t [, b], x 2 (t) = G(t, s)m(s)f(s, x 2 (s))ds + λ 2 G(t, s)m(s)f(s, x 2 (s)))ds m 1 e ks G(s, s)m(s)(l 1 + ɛ 2 )x 2 (s)ds G(t, s)m(s)f(s, x 2 (s))ds m 1 e ks G(s, s)m(s)f(s, x 2 (s))ds m 1 (l 1 + ɛ 2 ) min x 2(s) e ks G(s, s)m(s)ds t [,b] = Cm 1 l 1 e ks G(s, s)m(s)ds + Cɛ 2 m 1 e ks G(s, s)m(s)ds = C + Cɛ 2 m 1 e ks G(s, s)m(s)ds (3.11) nd (H 3 ) imply tht (3.11) x 2 (t) >C, t [, b]}. (3.12) Obviously, (3.12) is in contrdiction with (3.1). This implies tht (3.8)holds. Therefore, Lemm 2.1 implies i(a, Ω 2 P, P) =. (3.13) Noting (3.5),(3.13)nd the fct tht Ω 1 Ω 2, we hve i(a, (Ω 2 \ Ω 1 ) P, P) =i(a, Ω 2 P, P) (A, Ω 1 P, P), = 1= 1, (3.14) (3.14) nd the solution property of the fixed point index imply tht the opertor A hs fixed points x which belongs to (Ω 2 \ Ω 1 ) P such tht <s 1

9 Positive solutions of opertor equtions 219 x 1 s 2,it is cler tht x is positive solution of opertor eqution (1.1). The proof is completed. Theorem 3.2 Assume tht (H 1 ) (H 3 ) nd (H 5 )hold, Then opertor eqution (1.1) hs t lest one positive solution. Proof. The proof is similr to tht of Theorem 3.1 nd we omit it. Theorem 3.3 Assume tht (H 1 ) (H 3 ) nd (H 6 ), (H 1 ), (H 11 )hold, Then opertor eqution (1.1) hs t lest two positive solutions. Proof. Firstly, set r 1 = M 1 (2k M 2 ) 1,Ω r1 = {x P : x <r 1 }. Then we get from (H 1 ) nd (H 11 ) tht for ny t R + nd x Ω r1 Ax(t) e λt = 1 2k e λt G(t, s)m(s)f(s, x(s))ds e ks G(s, s)((s)+b(s)x(s))m(s)ds e ks G(s, s)(s)m(s)ds + e ks (s)m(s)ds + 1 2k x e ks G(s, s)m(s)f(s, x(s))ds e ks G(s, s)b(s)x(s)m(s)ds e (λ k)s b(s)m(s)ds = 1 2k (M 1 + r 1 M 2 )=r 1 = x which indictes Ax x, x Ω r1. (3.24) secondly, the condition l 2 <f in (H 6 ) tells us tht there exists r 2 >r 1 such tht f(t, x) l 2 x, t R +, x r 2. (3.25) Set Ω r2 = {x P : x <r 2 }. Then we cn see from (2.3)(2.7)nd (3.25)tht,for ny x Ω r2, Ax( + b 2 ) e λ(+b)/2 = e λ(+b)/2 G( + b,s)m(s)f(s, x(s))ds 2 e λ(+b)/2 G( + b 2,s)m(s)l 2x(s)ds e λ(+b)/2 G( + b 2,s)m(s)l 2m 1 x ds l 2 m 1 e λ(+b)/2 m 1 e ks G(s, s)m(s)ds x = x (3.26)

10 22 Bohe Wng Hence Ax x, x Ω r2. (3.27) Finlly,(3.24),(3.27) nd Lemm 2.3 imply tht the opertor A hs fixed points x which belongs to (Ω r2 \ Ω r1 ) P such tht <r 1 x 3 r 2. It is cler tht x is positive solutions of opertor eqution (1.1). The proof is completed. Theorem 3.4 Assume tht (H 1 ) (H 3 ) nd (H 7 ), (H 1 ), (H 11 )hold, Then opertor eqution (1.1) hs t lest two positive solutions. Proof. The proof is similr to tht of Theorem 3.3 nd we omit it. Remrk 3.1 In this pper, some results for positive solutions of opertor eqution (1.1) re obtined. Obviously, the conditions used in this pper is more extensive thn the superliner nd subliner conditions. Therefore, the pper generlizes nd includes some known results. References [1] M.Zim. On positive solutions of boundry vlue problems on the hlfline, J. Mth. Anl. Appl. 259(21): [2] Zho-ci Ho, Jin Ling, Ti-jun. Positive solutions of opertor eqution on hlf-line, J. Mth. Anl. Appl. 314(26): [3] J.W.Bebernes,L.K.Jckson, Infinite intervl boundry vlue problems for y = f(x, y). Duke Mth.J. 34(1967): [4] A.Grns.R.B.Guenther.J.W.Lee.D.O Regn. Boundry vlue problems on infinite intervls nd semiconductor devices, J. Mth. Anl. Appl. 116(1986): [5] P.A.Mrkowich, Anlysis of boundry vlue problems on infinite intervls. SIAM J. Mth. Anl. 14(1983): [6] B.Przerdzki. On two-point boundry vlue problems for equtions on the hlf-line, Ann. Polon. Mth. 5(1989): [7] J.V.Bxley, Existence nd uniqueness for nonliner boundry vlue problems on infinite intervls. J. Mth. Anl. Appl. 147(199): [8] Z.C. Ho, A.M. Mo. A necessry nd sufficient condition for the existence of positive solutions to clss of singulr second-order boundry vlue problems, J. Sys. Sci. Mth. Scis.. 21(21):93-1. Received: August, 28