一类具连续偏差变元的非线性中立双曲型偏泛函微分方程的振动性_英文_

一类具连续偏差变元的非线性中立双曲型偏泛函微分方程的振动性_英文_ Oscillation of Cetain Nonlinea Neutal Hypebolic rrrr Partial Functional Differential Equations with Continuous Distribution Arguments LIN Wen-xian ( Department M oaft hematics and Information Technology，Hanshan Normal University，Chaozhou 521041，China) Abstract: In this article，the oscillation of a lcass of nonlinear neutral hyperbolic partial differential equations with continuous deviating argumentiss studied, By employing the generalized Riccati transformation and thete ch- nique of differential inequalities，some new suffcient conditions for oscillation of all solutions of such queations are obtained underR obin andD irichlet boundaryv alue conditions, The results fully indicate that theo scillation is causedby delay, The results generalize some thela stest results, Key word word文档格式规范word作业纸小票打印word模板word简历模板免费word简历 s: hyperbolic; partial functional differential equation; oscillation; generalized Riccati transformation CLC Number: O175, 27 Document Code: A Article ID: 1007 ) 855X( 2012) 01) 0090 ) 05 一类具连续偏差变元的非线性中立 双曲型偏泛函微分方程的振动性 林文贤 ( 521041)，韩山师范学院 数学与信息技术系广东 潮州 : ，摘要研究了一类具有连续偏差变元的非线性中立双曲型偏泛函微分方程的振动性借助广义 Riccati ，Robin Dirichlet 变换和微分不等式技巧获得了这类方程分别在 和 边值条件下所有解振 ，,动的若干新的充分性条件所得结果推广了最近文献的相关结果 : ; ; ; Riccati 关键词双曲型偏泛函微分方程振动性广义 变换 0 Introduction In the last few years，the osc illatory theory of edlay partial functional differential equations has attracted a ,1 ) 8, great deal of attention by many authors, For their applications in many fields，such as engineering，physics， chemistry，medicine and biology and soon , In this paper，wec onsider a class of nonlinear neutral hyperbolic partial functional differential equations with continuous deviating arguments of theorm f bn b( t)+ q( x，t，) f( u,x，g( t，) ,) d( )ξξσξ, , u( x，t)p( t) u(x ，( t) ) + , μ, ? ?i i t ta i = 1 m = a( t) h( u)u ( x，t) + a( t) h( u(x ，( t) ) ) u( x，( t) ，) ΔτΔτ j j j j ?j = 1 ) 01 ) 17,:2011 收稿日期 : ( x，t)× R G( E)? Ω ? + subject to theR obin boundaryv alue conditions u + r( x，t) u( x，t) = 0， ( x，t) × ，( B )? Ω +1R ν or Dirichlet boundaryv alue conditions× R ( B)? Ω 2 u( x，t) = 0， ( x，t) N ， + where is a boundedo mdain with a piecewise smooth boundary is the Laplacian operator and is R，Ω Ω? ν ,a unit exterior normal vector on R = ,0，)， r C( × R ，R ，Ω ? ? Ω + + ) + It is easy to see thatequ ation( E) includes the following neutral hyperbolic differential equation n s + b( x，t) f( u(x ，t ) ( t) ) )b( t)σ, , u( x，t)p( t) u(x ，t ) ( t) ) ?k k + μ, , ?i i t t k = 1 i = 1 m ( E* )= a( t) h( u)u ( x，t) + a( t) h( u(x ，t ) ( t) ) ) u( x，t ) ( t) ，) ΔτΔτj j j j ? j = 1 For equatonE(* ) ，the special case b( t) = 1 of (E * ) has beenstu died in ,5,，so that ther esults of this paperg eneralize and include the results in ,5,, Throughout，wwei ll assume that theoll ofwing conditions hold: n ? 1 21 ( H1) a( t) ，a( t)，C( R，R) ，p( t) ( R; R) ，p( t) 1，b( t) ( R; R)C C dt = ，? ? ? ? ? j + + i + + ?i + + ? b( t) 0 i = 1 f( u) f C( R，R)is convex， f( )u) = )f ( u) ， const , 0 (= u ) ,i I= { 1，2，…，n} ，j I;0 ? ? λ ? ? ? n m u ( H2) ( t) ，( t) C( R ，R ) ，( t) t，( t) t ，andl im( t) = lim( t) = I，j I;，i μτ? μ? τ? μτ? ? ? i j + + i j i j n m tt?? ?? ( H3)g C( R × ,a，b,，R) ，g( t，) t，，a，b,，g is nondecreasing with t and espectively，,，r? ξ? ξ? ξ + 珔 g'( t，a) ,0 ，and limmin g( t，) = + ，q C( G× ,a，b,，R ) ，Q( t，) = minq( x，t，);ξ? ξξ ? + t+ a，b,,? ? ξ? x?Ω ,0 ，uh( u)( H4) h( u)， h( u)C( R，R) ，and uh'( u) 0，uh'( u)0，uh( u) , 0 ，foru I,0，j ? ? ??? j j j m 1 Main Results 1 ) ，such th atTheorem 1 If there exists a function( t) C( ,t ，)， R φ? ? 0+ ?n2 b ,'( t) ,b( g(t ，a) ) φ ( 1)dt= ， ? ( t) Q( t，) d( ))φξσξ, , 1 ) p,g( t，) ,ξ, ,? ? i?t4( t) g'(t ，a) φ a0 i = 1 then eachso lution of the rpoblem ( E) ，( B) ocsillates in the domain G, 1 Proof Suppose that thero bplem( E) ，( B) hasn onoscillatory solution u( x，t) in G, Without loss of gener- 1 ality，we may assume thatx ，ut() ,0 ，for x( ，t),0 ) ,F rom (H 2) ，theree xists some× ,t，+ ) (t ? Ω ? 0 0 ,0 ，for x( ，t)tt，such that ux(， ( t) ) , 0 ，u,x，g( t，) , ,0 ，u( x，( t) )×, t，+ ) ，μξσ? Ω ? ξ ? ? 1 0 i k 1 with respect txo over the odmain ，we have,a，b,，i I，j I,Integrating both of E() ? ? Ω n m n b d d + q( x，t，) f( u,x，g( t，) ,) d( ) dxb( t),u( x，( t) )d x,ξξσξu( x，t) dx + p( t)μ ?i i , , ? ? ?? Ω dt Ω dtΩ a i = 1 m = a( t) h( u) u( x，t) dx + a( t) h( u(x ，( t) ) ) u( x，( t) ) dx，t t,( 2)ΔτΔτ? ?j j j j 1 ?? Ω Ω j = 1 By Greens formula and the boundarycon dition ( B) ，it follows’ 1 2 h( u) udx) h( u) r(x ，t) u( x，t) dS ) h'( u)dS 0( 3)= Δ? grad u? ? ? Ω Ω Ω where dSis the sufaceel ement on ,Ω h( u(x ，( t) ) ) u( x，( t) ) dx 0，j ，t t,I( 4)τΔτ? ? ? j j j m 1 ?Ω And by Jensens nequaty ，yeds iliil’ bb， ，， ， q( x，t，) fd( ) dx =q( x，t，) f d( ) dx ξσξξσξu,x，g( t，) , u,x，g( t，) , ξξ?? ? ? Ω aa Ω b )1 Q( t，) ( 5)d( )? ξ,σξ fdx u,x，g( t，) ,dx dx ξ,, , ， ， ， ，, ? ? ??Ω aΩ Ω )1 Let V( t) = u( x，t) dx( dx) ，it is easy too btain V( t)0，t t,T heref ore，from ( 2 )) (5 ) ? ? 1?? Ω Ω we have s b ( 6)' + Q( t，) f( U,g( t，) ,) d( ) 0， t t,, , ξξσξ? ? b( t), V( t)+ p( t) V( ( t) ), ' μ1 ?i i ? a i = 1 n Let Z( t) = V( t)by( H1) ，we have，+ p( t) V( ( t) )μ?i i i = 1 Z( t) ,0 ，Z( t)( 7)V( t)， t t,? ? 1 is decreasing in t ，and we can provethat By ( 6) ，weh ave ,b( t)Z '( t) ,' 0，fort t,T hen b( tZ) '( t) ?? 1 Z'( t)In fact，if Z'( t)0，fort t,,0 ，for t t，then thereex ists someT ,t such that b( Zt)' ( t) ,? ? ? 1 1 1 0，and b( Zt)' ( t)b( T)Z '( T) for t ,T , foows thatIt ll? t 1 Z( t)Z( T)ds，t T , + b( T)Z '( T) ? ? ? b( s)T ,0 , Therefore，limZ( t) = ) which contradicts the factZ ( t)，? t?? Furthermore，from( 6) Han1d) ，( we have b, , 0 ,( 8)' + Q( t，) V,g( t，) ,d( )? λξξσξb( t)Z '( t) ? a Therefore nb ,b( t)Z '( t) ,' + Q( t，) d( )0 ,( 9)λξσξ ? , , Z,g( t，) , ) p,g( t，) ,V( ,g( t，) ,ξξμξ? ?ii a i = 1 Noticing that ( 7，) Z'( t) 0 and (H 2) ，from ( 9) ，wobet ain ? n b ( 10),b( t)Z '( t) ,' + Q( t，) Z,g( t，) ,d( ) 0 ,λξ, , ξσξ? 1 ) p,g( t，) ,ξ? ?i a i = 1 From (H 3)we have gt(， ) g( t，a) ，then Z,g( t，) ,g( t，a) ,，t t，a，b,,Z,,ξ? ξ? ? ξ ? 1 Now，( 10) y ields nb ( 11),b( t)Z '( t) ,' + Z,g( t，a) ,Q( t，) d( ) 0 ,λξ, , σξ? 1 ) p,g( t，) ,ξ? i ? a i = 1 ( t) b( Zt)' ( t)φ Let w(t )= ，t t ,It is easy to see W( t,) 0 ，noticing ,b( t)Z '( t) ,' 0 and (11 ) ， ? ? 1 Z( g( t，a) ) we have ' ( t) b( t) Z' ( t) φ( t) ,b( tZ) ' ( t), ' ( t) b( t) Z' ( t) Z' ( g( t，a) ) g' ( t，a) φφW'( t) = +)2 Z( g( t，a) )Z( g( t，a) ) ( g(t ，a) )Z 2 ( t) ,b( tZ) ' ( t), 'φ ' ( t) b( t) Z' ( t) ( t) ,b( t) Z' ( t) , g' ( t，a) φφ +)? 2 Z( g( t，a) )Z( g( t，a) ) q( g(t ，a) ) Z( g(t ，a) ) nb2 ,'( t) ,b( g(t ，a) ) φ ,+) ( t ) Q( t，) d( )? φξ, , σξ1 ) p,g( t，) ,ξ? ?i 4( t) g'(t ，a) φa = 1i 2 b( t)Z ' ( t) ( t) g' ( t，a) ' ( t) b( g(t ，a) ) φφ 槡槡) ) , , b( g(t ，a) ) Z( g( t，a) )( t) g'(t ，a)2 φ槡 槡 nb2 ,'( t) ,b( g(t ，a) ) φ ) ( t ) Q( t，) d( ),+ φξ, , σξ?1 ) p,g( t，) ,ξ? ?i 4( t) g'(t ，a) aφ i = 1 Integrating boths ides of the abovein equality fromt to t ( t)，we have? 1 1 tn 2 b ,'( s) ,b( g(s ，a) ) φ ds ,W( t)W( t)) ) ( s) Q( s，) d( ) ? φξσξ+ , , 1 ) p,g( s，) ,1 ξ, , ? ? i?t4( s)g '( s，a) φ a1 i = 1 This completes the proof ofth eTaking t + ( 1) ，we getlim W( t)= ) which contradicts the fact,，by ，? ? ? t?? theorem1 , Corollary 1 If the differential ineqality( 6) has noev entually positive solution，then eachs olution of the problem ( E) ，( B) ocsillates in the domain G, 1 heoem 2 n theorem1 ，if thec ondton( 1) s repaced by the foowng condtonTrIiiillliii n b?Q( t，) d( ) dt = ，ξσξ( 12), , ? 1 ) p,g( t，) ,ξ?i ? ? ta 0 i = 1 then thec onclusion of theorem1s till holds, ,9, LemmaLet be thes mallest eigenvalue of the ofllowing Dirichlet eigenvalue problemα 0 ，x ( x) + ( x)= 0 ΔΩ? ( 13), x ，= 0( x) ? Ωand ( x) be toherre scponding eigenfunction of ,T hen , 0，( x) , 0 for x ,αα? Ω 0 0 Theorem 3 If the all conditions of therem1 hold，and h( u) = h( u)1 ，j I，then eachs olution? ? j m of the rpoblem ( E) ，( B) ocsillates in the domain G, 2 Proof Suppose that thero bplem( E) ，( B) hasn onoscillatory solution u( x，t) in G, Without loss of gener- 2 ality，we may assume thatx ，ut() ,0 ，for x( ，t),0 ) ,F rom (H 2) ，theree xists some× ,t，+ ) (t ? Ω ? 0 0 tt，such that ux(， ( t) ) , 0 ，u,x，g( t，) , ,0 ，u( x，( t) ),0 ，for x( ，t)×, t，+ ) ，μξσ? ? Ω ? ξ ? 1 0 i k 1 ,a，b,，i I，j I, ? ? n m Multilying boths ides of (E ) by ( x) andinte grating with respect txo over the odmain ，we have Ω n d d = a( t) ( x) u( x，t) dxq( t),Δu( x，t) ( x)d x + p( t) u( x，( t) ) ( dxx), μ , ?i i , ? ?? Ω dt dtΩ Ω i = 1 m b + a( t) ( x) u( x，( t) d) x ) ( x) dx, ( 14)Δτq( x，t，) f( u,x，g( t，) ,) d( ) ξξσξ?j j , , ?? ? Ω Ω j = 1 a By Green’s formula and bondaryco ndition ( B) ，it follws 2 ( 15)( x) u( x，t) dx = u( x，t) ( x) dx = ) u( x，t) ( x) dx， ΔΔα 0 ???Ω Ω Ω ( x) u( x，( t) )d x = u( x，( t) ) ( x) dx = ) u( x，( t) ) x)( dx，j I,( 16)ΔττΔατ? j j 0 j m ???Ω Ω Ω And by Jensens inequality and (H 2) ，yields ’ bb， ， ， ， ( x) d( ) dxq( x，t，) fσξq( x，t，) f ( x) d( ) dx= ξξσξu,x，g( t，) , ξu,x，g( t，) , ξ? ? ?? ΩΩ aa ， ， bu,x，g( t，) ,( x) dx ξ? Ω , , f ,( 17)Q( t，)( x)d x d( )? ξσξ? ? Ω a, , , , ( x)d x ， ， ? Ω )1 ，then U( t) , 0 ，,T heref ore，cobining ( 14 ) )Denote U( t) = u( x，t) ( x) dx( ( x) dx)t tΦΦ? 1?? Ω Ω ( 17) ，we get m n ' + a( t) U(t) +αa( t) U(t ) ( t) )+ , , ατb( t), U( t) + p( t) U( ( t) ), ' μ0 0 ?j j ?i i j = 1 i = 1 b Q( t，) f( U,g( t，) ,) d( ) 0 ，t t,( 18)ξξσξ? ? 1 ? a From (H 1) ，we have n b ( 19)' + Q( t，) U,g( t，) ,d( ) 0 ，t t,, , λξξσξ? ? b( t), U( t)+ p( t) U( ( t) ), ' μ1 ?i i ? a i = 1s Let Y( t) = U( t)，it is easy to see that t)Y( ,0 ，,b( t) Y'( t), ' and Y'(t ),0 ，for0 + p( t) U( ( t) ) μ ??i i i = 1 t t,Form ( 19) ，woebt ain? 1 b ,b( t) Y'( ,t)' + Q( t，) U,g( t，) ,d( ) 0，t t,λξξσξ? ? 1 ? a The remainder of the proofis similar to that of theorm 1，and om wit ei t, This completes the proof ofth etheorem2 , By the differential ineqality ( 18) ，we have m n ' + a( t) U(t ) ( t) )ατ0 ，t t,? ? , , b( t), U( t) + p( t) U( ( t) ), ' μ0 j j ?1 ?i i j = 1 i = 1 Similar to the proof of theorem 1，we haveollo wtheing fresult: 1Theorem 4 If ' ( t),0 ，h( u) = h( u)1 ，and theree xist a function ( t) C ( ,t，)， R ) ，τη? ? ? j j 0 + sucht hat 2 ? n,'( t) ,b( ( t) )ητ jdt= I，，j ? ? 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