Journl of Funcon Spces nd Applcons Volume 2013, Arcle ID 968356, 5 pges hp://dx.do.org/10.1155/2013/968356 Reserch Arcle Oscllory Crer for Hgher Order Funconl Dfferenl Equons wh Dmpng Pegung Wng 1 nd H C 2 1 College of Elecronc nd Informon Engneerng, Hee Unversy, Bodng 071002, Chn 2 College of Mhemcs nd Compuer Scence, Hee Unversy, Bodng 071002, Chn Correspondence should e ddressed o Pegung Wng; pgwng@ml.hu.edu.cn Receved 19 Sepemer 2012; Acceped 25 Novemer 2012 Acdemc Edor: Felz Mnhós Copyrgh 2013 P. Wng nd H. C. Ths s n open ccess rcle dsrued under he Creve Commons Aruon Lcense, whch perms unresrced use, dsruon, nd reproducon n ny medum, provded he orgnl wor s properly ced. We nvesge clss of hgher order funconl dfferenl equons wh dmpng. By usng generlzed Rcc rnsformon nd negrl vergng echnque, some oscllon crer for he dfferenl equons re eslshed. 1. Inroducon In hs pper, we consder he followng hgher order funconl dfferenl equons wh dsrued devng rgumens of he form s follows: x (n) () +p() x (n 1) () + q (, ξ) f(x[g 1 (, ξ)],...,x[g m (, ξ)]) dμ (ξ) =0, 0, (1) where n 2 s n even numer, p() C 1 ([ 0, ), R + ), q(, ξ) C([ 0, ) [, ], R + ), g (, ξ) C([ 0, ) [, ], R), lm nf,ξ [,] g (, ξ) = for I m = {1,2,...,m},ndf(u 1,...,u m ) C(R m,r) hs he sme sgn s u 1,...,u m ; when hey hve he sme sgn, μ(ξ) C([, ], R) s nondecresng, nd he negrl of (1) s Seljes one. We resrc our enon o hose soluons x() of (1) whch exs on sme hlf lner [ μ, ) wh sup{x() : T} =0for ny T μ nd ssfy (1). As usul, soluon x() of (1) s clled oscllory f he se of s zeros s unounded from ove, oherwse, s clled nonoscllory. Equon (1) s clled oscllory f ll soluons re oscllory. In recen yers, here hs een n ncresng neres n sudyng he oscllon ehvor of soluons for he dfferenl equons wh dsrued devng rgumens, ndnumerofresulshveeenoned(refero[1 3] nd her references). However, o he es of our nowledge, very lle s nown for he cse of hgher order dfferenl equons wh dmpng. The purpose of hs pper s o eslsh some new oscllon crer for (1) ynroducng clss of funcons Φ(, s, r) defned n [2] nd generlzed Rcc echnque. Frsly, we defne he followng wo clss funcons. We sy h funcon Φ = Φ(,u,v) elongs o he funcon clss X, denoed y Φ X,fΦ C(E,R),where E = {(,u,v) : 0 v u < }, whch ssfes Φ(,, v) = Φ(, v, v) = 0, Φ(, u, v) =0, v<u<nd hs he prl dervve Φ/ u on E h s loclly negrle wh respec o u n E. Le D 0 = {(, u) : 0 u<}, D={(,u): 0 u }.We sy h funcon H=H(,u)elongs o he funcon clss Y, denoed y H Y,fH(, ) = 0 for 0, H(, u) =0n D 0, H hs connuous prl dervve n D 0 wh respec o nd s. In order o prove he mn heorems, we need he followng lemms. Lemm 1 (see [4]). Le x() C n ([ 0, ), R + ),fx (n) () s of consn sgn nd no denclly zero on ny ry [ μ,+ )for μ 0, hen here exss σ μ,nnegerl(0 l n),wh n+leven for x()x (n) () 0 or n+lodd for x()x (n) () 0; nd for σ μ, x()x () () > 0, 0 l,nd ( 1) +l x()x () () > 0, l n.
2 Journl of Funcon Spces nd Applcons Lemm 2 (see [5]). If he funcon x() s s n Lemm 1 nd x (n 1) ()x (n) () 0 for σ μ, hen here exss consn θ (0, 1) such h for suffcenly lrge, here exss consn M θ >0,ssfyng x (θ) M θ n 2 x (n 1) (). (2) Lemm 3 (see [3]). Suppose h x() s nonoscllory soluon of (1).If s lm exp { p (τ) dτ} ds =, T 0 0, (3) hen x()x (n 1) () > 0 for ny lrge. 2. Mn Resuls Theorem 4. Assume h (3) holds, nd (A 1 ) here exss funcon σ() C n ([ 0, ), R + ) such h σ() = mn 1 m {, nf ξ [,] g (, ξ)}, lm σ() =. σ () σ > 0, σ n 2 () ρ n 2 >0, (M θ /2)σρ n 2 =: > 0, whereσ, ρ,nd re consns. (A 2 )f(u 1,...,u m ) s nondecresng wh u, I m,nd here exs consns N>0nd λ>0such h f(u lm nf 1,...,u m ) λ, u u 0 u N, = 0. (4) 0 If here exss funcon Φ(, u, v) X,suchhfornyl() C 1 ([ 0, ), R + ), r() C 1 ([ 0, ),R)nd 0, lm sup {Φ 2 φ (s) (s, u, v) ψ (s) By Lemm 3, here exss 1 such h x (n 1) () > 0, 1.Thus,wehve x (n) () = p() x (n 1) () q (, ξ) f(x[g 1 (, ξ)],...,x[g m (, ξ)]) dμ(ξ) 0, 1. By Lemm 1, here exss 2 > 1 such h x () > 0, 2. Furher, y Lemm 2, here exs consn M θ >0nd 3 2,suchh Se hen x σ () [ 2 ] M θσ n 2 () x (n 1) [σ ()] M θ σ n 2 () x (n 1) (), 3. y () =φ() { x(n 1) () x [σ () /2] y () = φ () y () +φ() φ () (8) (9) p () +r() + }, (10) { q (, ξ) f(x[g 1 (, ξ)],...,x[g m (, ξ)]) x [σ () /2] dμ(ξ) p() x(n 1) () x [σ () /2] where [Φ s (s, u, v) + l 2 ] }ds>0, (5) φ () =l() exp { r (s) ds}, σ () x (n 1) () x [σ () /2] 2x 2 [σ () /2] +r () + p () }. (11) In vew of (A 1 ), (A 2 ) nd he defnon of y(), φ(),wehve ψ () =φ() {λ q (, ξ) dμ (ξ) +r 2 () Then (1) s oscllory. r () p () p2 () 4 }. Proof. Suppose o he conrry h (1) hs nonoscllory soluon x(). Whou loss of generly, we my suppose h x() s n evenully posve soluon. From he condons of g (, ξ) nd f(u 1,...,u m ), here exss 0,suchh x () >0, x[g (, ξ)] >0, f(x[g 1 (, ξ)],...,x[g m (, ξ)])>0,, I m. (6) (7) y () φ () y () +φ() φ () { p() x(n 1) () x [σ () /2] λ q (, ξ) dμ (ξ) = y2 () φ () [ x(n 1) 2 () x [σ () /2] ] +r () + p () } + l () y () l () ψ(), (12) where ψ() = φ(){λ q(, ξ)dμ(ξ)+r2 () r () p ()/ p 2 ()/4}.
Journl of Funcon Spces nd Applcons 3 Mulplyng (12) yφ 2 (, u, v) nd negrng from 3 o,wehve Φ 2 (s, u, v) ψ (s) ds 3 Φ 2 (s, u, v) [ y (s) + l (s) y (s)]ds 3 l (s) Φ 2 (s, u, v) y2 (s) 3 φ (s) ds. (13) Inegrng y prs nd usng negrl vergng echnque, we hve hus Φ 2 (s, u, v) ψ (s) ds 3 lm sup 3 φ (s) 3 { [Φ s (s, u, v) + l 2 ] ds { Φ (s, u, v) y (s) φ (s) φ (s) { 3 φ (s) [Φ s (s, u, v) + l ] } 2 } ds } [Φ s (s, u, v) + l 2 ] ds, {Φ 2 φ (s) (s, u, v) ψ (s) 3 (14) [Φ s (s, u, v) + l 2 ] }ds 0, (15) whch conrdcs (5). Ths complees he proof of Theorem 4. If we choose Φ(, u, v) = H 1 (, u)h 2 (u, v), whereh 1, H 2 Y.ByTheorem 4, we hve he followng resuls. Corollry 5. Assume h (3), (A 1 ),nd(a 2 ) hold. If here exs H 1,H 2 Ysuch h for ech 0, lm sup H 1 (s, u) H 2 (u, v) {ψ(s) [h 1 (s, u) +h 2 (u, v)] 2 }ds>0, 4 (16) where h 1 nd h 2 re defned y H 1 (, u)/ u = h 1 (, u)h 1 (, u), H 2 (u, v)/ u = h 2 (u, v)h 2 (u, v),nd ψ () =λ q (, ξ) dμ (ξ) p () p2 () 4. (17) Then (1) s oscllory. If we choose l() = 1, r() = 0, nd le Φ(, u, v) = ( u)(u v) α,α>1/2,ytheorem 4,wehvehefollowng corollry. Corollry 6. Assume h (3), (A 1 ),nd(a 2 ) hold. If here exss consn α>1/2such h for ech 0, lm sup 1 2α+1 (s u) 2 (u v) 2α ψ(s) ds > α (2α 1)(2α + 1), where ψ() s defned s n Corollry 5.Then(1) s oscllory. Theorem 7. Assume h (3) holds, nd (18) (A 3 ) here exs funcons γ () C n ([ 0, ), R + ),suchh γ () mn 1 m {, nf ξ [,] g (, ξ)}, lm γ () =, γ () γ >0,whereγ re consns, I m ; (A 4 ) here exs consns λ [0, 1] nd λ>0,suchh γ n 2 m =1 λ γ () >0, f(u 1,...,u m ) λ 1 u 1 + +λ m u m λ. (19) () ρ n 2 0, (M θ /2) m =1 λ γ ρ n 2 ()=:>0,where ρ nd re consns, I m. If here exss funcon Φ X,suchhfornyl() C 1 ([ 0, ), R + ), r() C 1 ([ 0, ), R), nd 0,nd(5) holds, where φ() s defned s n Theorem 4: ψ () =φ() {λ q (, ξ) dμ (ξ) Then (1)soscllory. +r 2 () r () p () p2 () 4 }. (20) Proof. Suppose o he conrry h (1) hs nonoscllory soluon x(). Whou loss of generly, we my suppose h x() s n evenully posve soluon. Smlr o he proof of Theorem 4, here exss 0,suchhx[g (, ξ)] > 0, x[γ ()] > 0, x () > 0, f(x[g 1 (,ξ)],...,x[g m (, ξ)]) > 0, x (n 1) () > 0,ndx (n) () 0,for, I m.se x (n 1) () y () =φ() { m =1 λ x[γ () /2] p () +r() + }, (21)
4 Journl of Funcon Spces nd Applcons hen y () = φ () y () +φ() φ () { p() x (n 1) () m =1 λ x[γ () /2] q (, ξ) f(x[g 1 (, ξ)],...,x[g m (, ξ)]) m =1 λ x[γ () /2] dμ(ξ) m =1 x (n 1) () 2{ m =1 λ x[γ () /2]} 2 λ x [γ ()]γ () +r () + p () }. (22) If here exss funcon Φ X,such h for ny l() C 1 ([ 0, ), R + ), r() C 1 ([ 0, ), R), nd(5) holds, where φ() sdefnedsntheorem 4: ψ () =φ() { q (, ξ) dμ (ξ) Then (1) s oscllory. +r 2 () r () p () p2 () 4 }. (25) Proof. Suppose o he conrry h (1) hs nonoscllory soluon x(). Whou loss of generly, we my suppose h x() s n evenully posve soluon. Smlr o he proof of Theorem 4, here exss 0,when, nd we hve x[g (, ξ)] > 0, x[γ ()] > 0, x () > 0, f(x[g 1 (,ξ)],...,x[g m (, ξ)]) > 0, x (n 1) () > 0,ndx (n) () 0, I m.se In vew of (A 3 ), (A 4 ) nd he defnon of y(), φ(),wehve y () y2 () φ () + l () y () l () ψ(). (23) The followng proof s smlr o Theorem 4,ndweomhe dels. Ths complees he proof of Theorem 7. hen x (n 1) () y () =φ() { f(x[γ 1 () /2]+ +x[γ m () /2]) y () +r() + p () }, (26) Smlr o Corollres 5 nd 6, wehvehefollowng corollres. Corollry 8. Assume h (3), (A 3 ),nd(a 4 ) hold. If here exs H 1,H 2 Ysuch h for ech 0,nd(16) holds, where h 1,h 2 re defned s n Corollry 5: ψ () = λ q (, ξ) dμ (ξ) p () p2 () 4. (24) Then (1) s oscllory. Corollry 9. Assume h (3), (A 3 ),nd(a 4 ) hold. If here exss consn α > 1/2 such h for ech 0,nd (18) holds. where ψ() s defned s n Corollry 8, hen(1) s oscllory. For he cse of he funcon f(u 1,...,u m ) wh monooncy, we hve he followng heorem. Theorem 10. Assume h (3), (A 3 ) hold, nd (A 5 ) here exs f/ u nd f/ u λ 0,suchh m =1 λ γ () > 0, whereλ s consns. γ n 2 ρ n 2 0, (M θ /2) m =1 λ γ ρ n 2 () =: > 0,nwhchρ nd re consns, I m. = φ () y () +φ() φ () { p() x (n 1) () f(x[γ 1 () /2]+ +x[γ m () /2]) q (, ξ) f(x[g 1 (, ξ)],...,x[g m (, ξ)]) f(x[γ 1 () /2]+ +x[γ m () /2]) dμ(ξ) m =1 x (n 1) () 2{f (x [γ 1 () /2]+ +x[γ m () /2])} 2 λ x [γ ()]γ () +r () + p () }. (27) In vew of (A 3 ), (A 5 ) nd he defnon of y(), φ(),wehve y () y2 () φ () + l () y () l () ψ(). (28) The followng proof s smlr o Theorem 4, weomhe dels. Ths complees he proof of Theorem 10. Smlr o Corollres 5 nd 6, wehvehefollowng corollres.
Journl of Funcon Spces nd Applcons 5 Corollry 11. Assume h (3), (A 3 ),nd(a 5 ) hold. If here exs H 1,H 2 Ysuch h for ech 0,nd(16) holds, where h 1,h 2 re defned s n Corollry 5: ψ () = q (, ξ) dμ (ξ) p () p2 () 4. (29) Then (1) s oscllory. Corollry 12. Assume h (3), (A 3 ),nd(a 5 ) hold. If here exss consn α > 1/2 such h for ech 0,nd (18) holds. where ψ() s defned s n Corollry 11. Then(1) s oscllory. 3. Exmples Exmple 13. Consder he followng equon x (4) () + 1 π/2 x(3) () 0 ξ 2 2x (+sn ξ) dξ = 0, 2 exp ( x 2 (+cos ξ)) References [1] P. Wng nd M. Wng, Oscllon of clss of hgher order neurl dfferenl equons, Archvum Mhemcum, vol. 40,no.2,pp.201 208,2004. [2] Y. G. Sun, Oscllon of second order funconl dfferenl equons wh dmpng, Appled Mhemcs nd Compuon,vol.178,no.2,pp.519 526,2006. [3] P. G. Wng nd X. Lu, Inervl oscllon for some nds of even-order nonlner dfferenl equons, Chnese Journl of Engneerng Mhemcs,vol.22,no.6,pp.1031 1038,2005. [4] I. T. Kgurdze, On he oscllory chrcer of soluons of he equon d m u/d m + () u n sgnu = 0[J], Memchesl Sorn,vol.65,pp.172 187,1964(Russn). [5] C. Phlos, A new creron for he oscllory nd sympoc ehvor of dely dfferenl equons, L Acdéme Polonse des Scences. Bullen. Sére des Scences Mhémques,vol.29, no.7-8,pp.367 370,1981. 1, (30) where u 1 = x( + cos ξ), u 2 = x( + sn ξ), ovously f(u 1,u 2 )/u 2 1=λ.Choosngσ=1, ρ=1,hen=m θ /2, nd lm sup 1 2α+1 =( M θ 2 + 1 ) lm sup 4 0 (s u) 2 (u v) 2α ( M θ 2s 2 + 1 4s 2 )ds 1 2α+1 0 (s u) 2 (u v) 2α 1 s 2 ds (31) =( M θ 2 + 1 4 ) 1 α (2α + 1)(2α 1). Thus, here exss consn α>1/2,suchh(m θ /2+1/4) > α 2,hs, ( M θ 2 + 1 4 ) 1 α (2α + 1)(2α 1) > α (2α + 1)(2α 1). (32) By Corollry 6,hen(30)soscllory. Exmple 14. Consder he followng equon x (4) () + 1 1 ξ x(3) () + 0 2 [x( ξ) +x(+ξ) +x 3 ( ξ) +x 5 (+ξ)]dξ=0, (33) where f(u 1,u 2 ) = u 1 +u 2 +u 3 1 +u5 2,ovously f/ u 1 = 1+3u 2 1 1, f/ u 2 =1+5u 4 2 1.Choosngλ =1, γ =1, ρ =1,nd=1,2,hen=M θ.bycorollry 12, hen(33) s oscllory. Acnowledgmens The uhors would le o hn he revewers for her vlule suggesons nd commens. The reserch ws suppored y he Nurl Scence Foundon of Chn (11271106).
Advnces n Operons Reserch Advnces n Decson Scences Journl of Appled Mhemcs Alger Journl of Proly nd Sscs The Scenfc World Journl Inernonl Journl of Dfferenl Equons Sum your mnuscrps Inernonl Journl of Advnces n Comnorcs Mhemcl Physcs Journl of Complex Anlyss Inernonl Journl of Mhemcs nd Mhemcl Scences Mhemcl Prolems n Engneerng Journl of Mhemcs Dscree Mhemcs Journl of Dscree Dynmcs n Nure nd Socey Journl of Funcon Spces Asrc nd Appled Anlyss Inernonl Journl of Journl of Sochsc Anlyss Opmzon