【应用数学】具因果算子的分数阶微分方程初值及边值问题解的存在性.pdf_第1页
【应用数学】具因果算子的分数阶微分方程初值及边值问题解的存在性.pdf_第2页
【应用数学】具因果算子的分数阶微分方程初值及边值问题解的存在性.pdf_第3页
【应用数学】具因果算子的分数阶微分方程初值及边值问题解的存在性.pdf_第4页
【应用数学】具因果算子的分数阶微分方程初值及边值问题解的存在性.pdf_第5页
已阅读5页,还剩43页未读 继续免费阅读

下载本文档

版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领

文档简介

?“10530201009031097 a o175?m a jf? 9k)?35 ? k)?35 ? ?“6o 4 ? ?o? ;a n9a ? ? on a ? j ? f 2013415 existence of solutions for initial value and boundary value problem of fractional diff erential equations involving causal operator candidatejiang jing fei supervisorprofessor c. f. li collegeschool of mathematics programapplied mathematics speciality theory and application of functional diff erential equations degreemaster ofscience universityxiangtan university dateapril 15th, 2013 ? ?m5( ?x-(?3?“?e?1 ?j.?aoi5?sn?,? ?j.? ?-?z?8n,3(i.? ?(?jd?. fc?f ? ? ?)?k?3! ?5, ?3ik?x?ef, #n?/?.? ? sn?k?1u,k?j?.?jf ?9k)?35.(?xe: 1?0?a?uyg,? i?a?. 1?3,?e,l?ecq)?,|%c, y?a3banachm?jf?k ). 1n31?:|i?jf? )?35 1o2?:k?,|ns“e? ajf?k4)?35, ?(j ?j,k:1:d?c?;2: ?. c: ?; jf; k; k; ); ) ?35. i abstract fractional calculus has been developed rapidly for recent years, which is involved in lots of application. the fractional system involves in many fi elds such as control, physics and mathematics. the extensive application has be- come the hot topics all over the world in past decades. because the integral part involved in fractional calculus has the function of memory and inherit, it has become the signifi cant tools in describing various complicated behaviors. the theory of fractional diff erential equation gains the subsequent developing. causal operator is adopted from engineering problem and has strong applica- tion background . with the develop of the causal operator theory of integral order diff erential equation, the causal operator has some results in initial-value problem and boundary problem. at present, the fractional diff erential equa- tion involved in causal operator has also many results and application, but very few in initial-value problem and boundary problem. this thesis concerns with existence of solutions for fractional diff erential equations involving causal operator, main content is organized as follows: in chapter 1, we give a general description on the fractional calculus, recall some preliminaries which are useful for the thesis. with constructing approximate solution and approximation technique, the existence of solution for initial value problem of fractional diff erential equation with causal operator in banach space is obtained in chapter 2. chapter 3 studies the existence of maximal solution of fractional diff erential equations involving causal operator in use of the equality in cone. chapter 4 is devoted to study the existence of solution for two-point boundary value problems of fractional diff erential equa- tion with causal operator by extending the works on integer order two-point boundary value problem and monotone iterative method. key words: fractional diff erential equation; causal operator;initial- value problem;boundary-value problem; existence of solution. ii 8 . i abstract.ii 1x.1 1.1?9.1 1.2yg9? . 2 1.3?.3 1?banachmjf?k)? 35.7 2.1 . 7 2.2k)?35 . 7 1nbanachmjf?4)?35 18 3.1 . 18 3.24)?35.18 1ojf?k4)?35.23 4.1 . 23 4.2k4)?35 . 34 z.35 ?.40 am? . 41 1x 1.1?9 3?u?, leibnizlhospital31695c5the fractional calculus6-, uoldham spanier,k?sa, ? ?g,u?.m? 33x?n37. 20-v70c“?, b.b.mandelbort ?mm5#53, 54, 55, 56 m, ?n?ah5?u?,mandelbort3g ,.ne3x?9?nm3g qy, ?x?, ?nx naml?.y?x?41. ,? lg,.3?k$(fdw)riemann-liouville/? ?fk;x,?n7,x?u? ). x?ku?n3ak?y? ?,?5?na?d;-, x:1, 46, 47 ?a2, x?lagrange?hamiltion 48, 49, ?56n50, ?schr odinger51?; westerlandu1994c 3a;kamlars?44, j1?xe/ f = k0d tx(t), (1 ?1?9kj f?4, o?, jf3u 1 ,a,volterra,?40, ?j fkxd?5, ak?7?, ln? r?, la?r?n+?, x5?nd57. 1.2yg9? kkn?jl?+ ,ujf?kk,8ck z, o4339?zkv?v),3p2-e, z37. jf?z?, c. corduneanu j f?.)?35?1?3, ?3;4 akjf?1?x?. ?z. drici, f.a. mcrae, j.v. devijf?1?22, 15, 16, 17, j.v. devi?jf?8)?3556, 4) ?359y657,v.lakshmikantham, s.leela, z.drici f.a.mcrae 3;5jf?1?x?,/?1? ,n|?nk,v. lupulescu?a 3banachmjf?18, r.p.agarwal ? 3banachmjf?20, t.jankowski ? ajf?k21,f.geng ?ajf? 5k33,y.zhao ?ajf? k34, khaajf?)?35?1?40 j.jiang?a3banachmkp?jf?8 )?3532. ujf?,8cj f?,?,j?y,?j f5?usk4,kr?a?,r? jf?. ?sno,1?0?a? uyg,?i?a?. 1?3,? ?e,l?ecq)?,|%c,y?a3banach m?jf?k). 1n|i? ?jf?)?35. 1o2? 2 :k?,|ns“?ajf? k4)?35. 1.3? ?!ezy?eznn,yl 3z?. k?5. 1.1 1 f : t0,+) r ? ? i t0+f(t) = 1 () z t t0 f(s) (t s)1 ds, t t0, n 1 n, ()gamma , a?i t0+ ?f. 1.2 1 f : t0,+) r ? ?riemann-liouville ? l t0d tf(t) = 1 (n ) dn dtn z t t0 (t )n1f()d = dn dtn (i(n) t0+ f(t), n = + 1. a? l t0d t ?riemann-liouville f. 1.3 2 f : t0,+) r ? ?caputo ? c t0d tf(t) = l d ? f(t) n1 x k=0 (t t0)k k! f(k)(t0) ? , t t0, n 1 0. 5 2 1e k f(x) lpa,b (1 p ),k(dki t0+f)(x) = ikf(x) 3a,b a?. -j = t0,t r,x k|.| ?banachm, e = u : u c(t0,t,x), 3ekuke= sup t0tt ku(t)k,u e. lp(j,r)kv : j rkkp 0 : a ?l?k8cx. w, 0 diam(a) 0, 3lipschtizh(x) : x, ?|h(x) h(x)| ,x . 5 1.6 35 8k x i,ek k, 0,k+k k,k = k k tk = , banach mx ?, k k ?4 . i?xex: x y y x k, x y, y x k0, x, y x. 1.7 4fq : c(t0,t),x) l loc(t0,t),x) jf xj zt (t0,t) k?u,v c(t0,t),x)?t0 s t, u(s) = v(s) k (qu)(s) = (qv)(s)a.e. t0 s t 0,3() l loc(t0,t),r+), ?uu c(t0,t),x), sup t0t 0, 0,a.e. t t0, t). 7 y ?f(,) = (t )1( )1(qu)(), k5yf(,) 3t0,t t0,tbochner ?. f(,) 3t0,t t0,t?k,da1-a2,? ? ? ? ? ? ? ? ? z t t0 z t t0 (t )1( )1(qu)()dd ? ? ? ? ? ? ? ? z t t0 z t t0 |(t )1( )1|()dd = z t t0 (t )1d z t t0 |( )1|()d = z t t0 (t )1d z t0 ( )1()d + z t t0 (t )1d z t ( )1()d k(t)kl(j)(t t0)+ ?()( + 1) ( + + 1) + 1 + ? , dn1.1?:f(,)bochner ?,2da1-a2, ? ? ? ? z t t0 (t s)1(qu)(s)ds ? ? ? ? k(t)kl(j) (t t0) . dn1.1, r t t0(t s) 1(qu)(s)ds bochner ?.? (i t0+i t0+(qu)(t) = 1 () z t t0 (t )1(i t0+(qu)()d = 1 ()() z t t0 (t )1 ?z t0 ( )1(qu)()d ? d = 1 ()() z t t0 z t0 (t )1( )1(qu)()dd = 1 ()() z t t0 d z t (t )1( )1(qu)()d. - = + s(t ),k (i t0+i t0+(qu)(t) = 1 ()() z t t0 (qu)()d z t (t )1( )1d = 1 ()() z t t0 (qu)()d z 1 0 (t )(1 s)1s(t )1(t )ds = 1 ( + ) z t t0 (t )+1(qu)()d = i+ t0+ (qu)(t). : (i t0+i t0+(qu)(t) = (i + t0+ (qu)(t).y. 8 n 2.1 eb?a1-a2 .kk2.1.1 ?due? u(t) = n1 x j=0 bj j! (t t0)j+ 1 () z t t0 (t s)1(qu)(s)ds,(2.2.1) t t0,t1,t1 (t0,t). y ?r 0u br= u cn1(t0,t1,x);kuke r, d n2.1?yl, (t s)(qu)(s),s t0,t),t (t0,t1)bochner ?. ky5. eu(t) = n1 p j=0 bj j!(t t0) j + 1 () r t t0(t s) 1(qu)(s)ds, d1.3, 51 n2.1? cdu(t) =cd ?n1 x j=0 bj j! (x t0)j+ 1 () z t t0 (t s)1(qu)(s)ds ? = cd(i t0+(qu)(t) = ld ? i t0+(qu)(t) n1 x j=0 i t0+(qu)(t) (j)|t=t 0 j! (t t0)j ? = ldi t0+(qu)(t) n1 x j=0 i t0+(qu)(t) (j)|t=t 0 (j + 1) (t t0)j = dn dtn in t0+ (i t0+(qu)(t) n1 x j=0 i t0+(qu)(t) (j)|t=t 0 (j + 1) (t t0)j = dn dtn in t0+(qu)(t) n1 x j=0 i t0+(qu)(t) (j)|t=t 0 (j + 1) (t t0)j. d1.152? dn dtn in t0+(qu)(t) = (qu)(t) dj dtj i t0+(qu)(t) = i j t0+(qu)(t). qu?s t0,t),(t s)j1(qu)(s) lebesgue , i t0+(qu)(t) (j)|t=t 0 = ij t0+(qu)(t)|t=t0 = 0. k?: cdu(t) = (qu)(t). eyu(k)(t0) = bk, 2.2.1k(k = 0,1,2, ,n 1)g? ?: u(k)(t) = n1 x j=k bj (j k)!(t t0) (jk) + 1 ( k) z t t0 (t s)k1(qu)(s)ds. 9 -s = t0+ x(t t0),k u(k)(t) = n1 x j=k bj (j k)!(tt0) (jk)+(t t0) k ( k) z 1 0 (1x)k1(qu)(t0+x(tt0)dx. da1-a2,?u?x 0,1,(1x)k1(qu)(t0+x(tt0) lebesgue ?, -t t0,?u(k)(t0) = bk. ?(ci,di)n i=1 j?xy?mm n p i=1(c i di) 0. k n x i=1 kun1(di) un1(ci)k = n x i=1 ? ? ? ? 1 ( n + 1) z di t0 (di s)n(qu)(s)ds 1 ( n + 1) z ci t0 (ci s)n(qu)(s)ds ? ? ? ? 1 ( n + 1) n x i=1 z di ci (di s)n(s)ds + 1 ( n + 1) n x i=1 z ci t0 (ci s)n (di s)n)(s)ds 1 ( n + 1) n x i=1 z di ci (di s)ndsk(t)kl(j1) + 1 ( n + 1) n x i=1 z ci t0 (ci s)n (di s)n)dsk(t)kl(j) = k(t)kl(j1) ( n + 2) n x i=1 (di ci)n+1 + k(t)kl(j1) ( n + 2) n x i=1 (ci t0)n+1+ (di ci)n+1 (di t0)n+1 2k(t)kl(j1) ( n + 2) n x i=1 (di ci)n+1 0. u(n1)(t)3t0,t1y?. , eu c(t0,t1,x) k2.1.1?), kuv2.2.1.y. ekk2.1.1)35?a(j. n 2.2 ea1-a3,k3 0xe/g%cs? um(t) = n1 x j=0 bj j! (t t0)j+ 1 () z t t0 (t s)1(qum1)(s)ds 10 t j = t0,t0+ ,kum(t)uk2.1.1?). y -b = u cn1(t0,t),x) : ku u0ke b,u0= n1 p j=0 bj j!(t t0) j, = min ?( (+1)b k(t)kl ) 1 ,(t t0)?. u?t j, da2? kum(t) u0k 1 () z t t0 (t s)1k(qum1)(s)kds k(t)kl(j) 1 () z t t0 (t s)1ds k(t)kl(j) (t t0) ( + 1) b. ut j,um,m 1?. ?t1,t2 j t1 0, 3v?1, ? 2 1 (+1)k(t)kl (j) . d, u zt t0,t0+ ), dn1.3, ? ? 1 () z t t1 (t s)1(qum)(s)ds,m 1 ? 2 () z t t1 (t s)1(s)ds . (a(t) ? 1 () z t1 t0 (t s)1(qum)(s)ds,m 1 ? + , t t0,t0+ ). dn1.3a3,? (a(t) 1 () z t1 t0 (t s)1(qa)(s)ds + 1 () z t t0 (t s)1g(s, sup t0us (a(u)ds + , ?, (a(t) 1 () z t t0 (t s)1g(s, sup t0us (a(u)ds,t t0,t0+ ). q(a(0) = 0, g(t,x)?5, ?(a(t) = 0,t t0,t0+ ). d ?,dn1.4, (a) =sup t0tt0+ (a(t),(a) = 0. ax? ;8. 2dn1.1,3?f?,”e,pun(t)n1, 3t0,t0+?yu buvk2.1.1.xd, u(t) k2.1.1?). y. n 2.3 ea1-a4. kk2.1.13mt0,t0+ 13), 1= min ?,( (+1)l k(t)kl ) 1 ?. y dn2.2,k2.1.1)3,|y,b?),”b? k2.1.13)u(t), v(t),t t0,t0+ 1,k |u(t) v(t)| = | 1 () z t t0 (t s)1(qu)(s) (qv)(s)ds| 1l ( + 1)|u(t) v(t)| |u(t) v(t)|, b?g.b?,=)?, y. n 2.4 ea1 a3, kk2.1.1)?3mt0,t). 12 y -u : t0,) xk2.1.1?),t0 t,|y ,b? u7. ut0 t1 t2 0, 33, + ?)u(t),ub?g ,b?,y. pb = u cn1(t0,t),x) : ku u0ke b,u0= n1 p j=0 bj j!(t t0) j, k2.1.1)3?%c,ken. n 2.5 b? (a) jfq : b l loc(t0,t),x)y?,3() l loc(t0,t),r+) ?|(qu)(t)| (t),u b; (b)s?mv0 m 1 lim m m= 0. k,m z+,k cdu(t) = (qu)(t),u(k)(t0) = bk,bk r (k = 0,1, ,n 1;n = + 1) 3%c)um(t),t0,t0+, = min ? (+1)b k(t)kl ?1 ,(t t0)?,= v: (i) um(t) acnt0,t), (ii) kcdum(t) (qum)(t)k m, t0 t t0+ . 13 y ddugundjisn, q 3yq : e l loc(t0,t),x) vkqu(t)k (t), u e. dn1.5, u?0 m 1, 3qm: e l loc(t0,t),x) vk qmu(t) qu(t)k m.w,?: kqmu(t) qu(t)k m,kqmu(t)k k(t)kl+ 1,u(t) b. -um(t)lcd(t)u(t) = (qmu)(t),u(k)(t0) = bk?),3mt0,t0+ , = min ? (+1)b k(t)kl ?1 ,(t t0)?. ?: kcd(t)um(t) (qum)(t)k = k(qmum)(t) (qum)(t)k m, t0 t t0+ . um(t)y?cq),y. n 2.6 b?n2.5?(a): (c): g c(t0,t) 0,2b,r), g(t,0) 0 zk.8(t) 0 cd(t) = g(t,), t0,t) (t0) = 0. (2.2.2) ?). (d): (qa)(t) g(t,(a),a(s) b. kk2.1.13t0,t0+ )3, = min ? (+1)b k(t)kl ?1 ,(t t0)?. y dn2.5, 3cq)?um(t),t t0,t0+ v cdum(t) = (qum)(t) + ym(t), u(k) m(t0) = bk, bk r (k = 0, 1, , n 1; n = + 1), |ym(t)| m, t t0,t0+ ,(m = 1,2,3,.). u?t1, t2 j = t0, t0+ t1 t2, |um(t1) um(t2)| ? ? ? ? n1 x j=0 bj j! (t1 t0)j n1 x j=0 bj j! (t2 t0)j ? ? ? ? + 1 () ? ? ? ? z t1 t0 (t1 s)1 (t2 s)1(qum)(s)ds 14 z t2 t1 (t2 s)1(qum)(s)ds ? ? ? ? (2.2.3) + ? ? ? ? 1 () z t1 t0 (t1 s)1ym(s)ds 1 () z t2 t0 (t2 s)1ym(s)ds ? ? ? ? n1 x j=1 bj (j 1)!(t2 t0)j1(t2 t1) + 2k(t)kl(j) ( + 1) (t2 t1) + ? ? ? ? 1 () z t1 t0 (t1 s)1 (t2 s)1ym(s)ds 1 () z t2 t1 (t2 s)1ym(s)ds ? ? ? ? n1 x j=1 bj (j 1)!(t t0) j1(t2 t1) + 2k(t)kl(j) ( + 1) (t2 t1) + 2 m( + 1)(t2 t1), um(t); n 1 ?. eyum(t); n 1;8, =(um(t); m 1) = 0. -bk= um(t)|m = k,k + 1,., w, (bk(t) = (b1(t)(k = 1,2,3,.). kym(t) = (b1(t) = (bk(t) = 0. dn1.2,ut0 s t t0+ ,? |m(t) m(s)| = |(um(t);m k) (um(s);m k)| |(um(t) um(s);m k)| = ? ? ? ? ?n1 x j=0 bj j! (t t0)j n1 x j=0 bj j! (s t0)j+ 1 () z t t0 (t )1(qum)() + ym()d 1 () z s t0 (s )1(qum)() + ym()d;m k ? ? ? ?, - xm(t) = 1 () z t t0 (t)1(qum)()+ym()d 1 () z s t0 (s)1(qum)()+ym()d, 15 ut0 s t t0+ ,? |xm(t)| = ? ? ? ? 1 () z t t0 (t )1(qum)() + ym()d 1 () z s t0 (s )1(qum)() + ym()d ? ? ? ? 2k(t)kl(j) ( + 1) (t s)+ 2 m( + 1)(t s) , -ak= xm(t);m k,k sup xmak |xm| 2k(t)kl(j) ( + 1) (t s)+ 2 m( + 1)(t s) (ak) 4k(t)kl(j) ( + 1) (t s)+ 4 m( + 1)(t s) . d |m(t) m(s)| 4k(t)kl(j) ( + 1) (t s)+ 4 m( + 1)(t s) , m(t)3t0,t0+ y. gycdm(t) g(t,m(t),t t0,t0+ . um(t) k2.1.1?cq), m(t) = (bk(t) = ?n1 x j=0 bj j! (t t0)j+ ? 1 () z t t0 (t s)1(qum)(s)ds + 1 () z t t0 (t s)1ym(s)ds;m k ? ? 1 () z t t0 (t s)1(qum)(s)ds;m k ? + ? 1 () z t t0 (t s)1ym(s)ds;m k ? 1 () z t t0 (t s)1(qbk)(s)ds + ? 1 () z t t0 (t s)1(ym)(s)ds;m k ? 1 () z t t0 (t s)1g(s,(bk(s)ds + 1 () z t t0 (t s)1 2 m ds, xd,? cdm(t) g(t,m(t) + 2 m . = m(t) = (um(t); m k) rm(t, t0, 0), t t0,t0+ , 16 rm(t, t0, 0)xek?4). cdu(t) = g(t,u) +2 m u(t0) = 0. d(c)z5?lemma 5.6.3,? lim m rm(t, t0, 0) = r(t, t0, 0),t t0, t0+ . r(t, t0, 0) (2.2.2)?). xd, m(t) = (um(t); m k) = 0,t t0, t0+,um(t); m 1 ;?, ?f?umk(t) uyu(t), qqy?, u(t) = n1 x j=0 bj j! (t t0)j+ 1 () z t t0 (t s)1(qu)(s)ds.k . y. 5: xj(d)u?k.8a(s) b, (qa)(t) g(t, sup t0tt (a). ?(?. 17 1nbanachmjf? 4)?35 3.1 3? cdu(t) = (qu)(t) t0 t t, 0 0,3() l loc(t0,t,r+) ?u c(t0,t,x), sup t0tt kuk r kk(qu)(t)k (t). a3 u?k.8a c(t0,t,x), (qa)(t) g(t, sup t0st (a(s),t t0,ta.e. g : t0,t r+kamke, a(t) = u(t);u a. 3.2)?35 xen: n 3.1 ea1 a2 , ku c(t0,t,x) x(3.1.1) ?)? =? u(t) = u0+ 1 () z t t0 (t s)1(qu)(s)ds, t t0

温馨提示

  • 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
  • 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
  • 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
  • 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
  • 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
  • 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
  • 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。

评论

0/150

提交评论