Volume 5 No. 9 207, 47-64 ISSN: 3-8080 printed verion; ISSN: 34-3395 on-line verion url: http://www.ijpm.eu ijpm.eu Ocilltion of neutrl frctionl order prtil differentil eqution with dmping term V. Sdhivm, J. Kvith 2 nd N. Ngjothi 3,2,3 PG nd Reerch Deprtment of Mthemtic Thiruvlluvr Government Art College Affli. to Periyr Univerity, Ripurm - 637 40, Nmkkl Dt. Tmil N, Indi. ovdh@gmil.com, kvikhit@gmil.com 2 ngjothi006@gmil.com 3 Abtrct In thi rticle, we invetigte ocilltion criteri for the frctionl order neutrl prtil differentil eqution with dmping term of the form ] ptd,t ux, t b i tux, τ i t t rtd,t ux, t b i tux, τ i t qx, t, ζfux, gt, ζ]dσζ = t ux, t l c j t, ζ ux, h j t, ζdσζ ex, t, x, t G = R, j= ubject to Robin boundry condition ux, t γ ψx, tux, t = 0, x, t R. 47
Uing the generlized Riccti Technique nd integrl verging method, new ocilltion criteri re etblihed. AMS Subject Clifiction: 35R, 34K, 26A33 Key Word nd Phre:Frctionl prtil differentil eqution, neutrl, ocilltion, dmping term. Introction Frctionl differentil eqution hve ttrcted coniderble ttention becue of their frequent occurrence in vriou field uch Vicoelticity, Signl nd imge proceing, Electro mgntic, Robotic nd o on. It i lo ued vluble tool for decription of hereditry propertie of vriou mteril nd procee. In the lt two decde, the qulittive behvior of olution of frctionl differentil eqution nd their ppliction hve been invetigted extenively, for exmple ee the monogrph nd pper, 2, 4, 6-0, 7, 20, 2]. The tudy of differentil eqution with deviting rgument i one of the importnt nd ignificnt brnch of nonliner nlyi nd vriety of reult cn be found in the pper 3, 5, 3-6]. However, to the bet of uthor knowledge only very few reult hve ppered regrding the ocilltory behvior of frctionl prtil differentil eqution up to now. See, 2, 8, 9] nd the reference cited there in. Motivted by thi, we conider the neutrl frctionl prtil differentil eqution with continuou ditributed deviting rgument nd dmped term of the form ] ptd,t ux, t b i tux, τ i t t rtd,t ux, t b i tux, τ i t qx, t, ζfux, gt, ζ]dσζ = t ux, t l c j t, ζ ux, h j t, ζdσζ ex, t, x, t G = R, j= 48
with the Robin boundry condition ux, t γ ψx, tux, t = 0, x, t R 2 where i bounded domin of R N with piecewie mooth boundry ; 0, i contnt; G = R, R = 0,, D,tu i the Riemnn- Liouville frctionl derivtive of order of u with repect to t; i the Lplcin opertor; γ i the unit exterior norml vector to nd ψx,t i nonnegtive continuou function on R. Throughout thi pper, we ume tht the following condition hold: A pt C R ; 0,, d =, P t where Pt=exp p r d p ; rt CR ; R; A 2 b i t C R ; R, i =, 2,..., m, t CR ; R ; c j t, ζ CR, b], R ; A 3 qx, t, ζ C R, b], R, Qt, ζ = min x qx, t, ζ, fu CR; R i convex in R, ufu > 0 nd fu λ > 0 for u u 0; A 4 h j t, ζ CR, b], R, h j t, ζ t for ζ, b], j=,2,...,l; gt, ζ CR, b], R, gt, ζ t for ζ, b]; gt, ζ nd h j t, ζ re nondecreing with repect to t nd ζ repectively nd lim inf gt, ζ = lim inf h jt, ζ =, j =, 2,..., l; t,ζ,b] t,ζ,b] A 5 σζ, b], R i nondecreing nd the integrl of eqution i tieltje one; A 6 e CG, R i continuou function uch tht ex, tdx 0. Motivted by the opertor defined in 9], we introce the following new liner integrl opertor. Definition. For ny function kt, C,, t; R, 0, we define the liner integrl opertor L ρ, t t n L ρ kt, = ρ kt, d Θu where n > i n integer, nd Θ :, R i continuou t function tifying condition lim =, ρ t Θu C,, R 49
with ρ > 0. If kt, kt, L ρ C,, t, R we get t = ρ n L ρ Θ n kt, Θu t ρ ρ Θu ] kt,. We etblih the ufficient condition for the ocilltion of olution of nd 2 by uing generlized Riccti technique nd integrl verging method. 2 Preliminrie The following nottion will be ued for our convenience. Ut = ux, tdx, F t = λ Qt, ζ m b igt, ζ dσζ. Definition 2. By olution of we men function u C t, ; R C t, ; R tht tifie, where { { } { }} t = min 0, min inf τ it, min inf h jt, ζ, i m t 0 ζ,b]; j l j 0 { { }} t = min 0, min inf gt, ζ. ζ,b] t 0 Definition 3. The olution ux,t of the problem nd 2 i id to be ocilltory in the domin G if for ny poitive number µ there exit point x 0, µ, uch tht ux 0, = 0 hold. Definition 4. The Riemnn-Liouville frctionl prtil derivtive of order 0 < < with repect to t of function ux, t i given by D,tux, t := t Γ t 0 t ξ ux, ξdξ 3 provided the right hnd ide i pointwie defined on R, where Γ i the gmm function. 50
Definition 5. The Riemnn-Liouville frctionl integrl of order > 0 of function y : R R on the hlf-xi R i given by I yt := Γ t 0 t ξ yξdξ for t > 0 4 provided the right hnd ide i pointwie defined on R. Definition 6. The Riemnn-Liouville frctionl derivtive of order > 0 of function y : R R on the hlf-xi R i given by Dyt := d I dt y t for t > 0 5 provided the right hnd ide i pointwie defined on R, where i the ceiling function of. Lemm 7. Let y be the olution of nd Et := t 0 t ξ yξdξ for 0, nd t > 0. 6 Then E t = Γ D yt. 3 Min Reult In thi ection, we etblih ome ufficient condition for ocilltion behvior of, 2. Theorem 8. If the functionl differentil inequlity ptd wt rtd wt F twθt 0, t t 7 h no eventully poitive olution, then every olution of, 2 i ocilltory in G. Proof. Aume to the contrry tht there i nonocilltory olution ux,t to the problem,2. Without lo of generlity we my ume tht ux, t > 0, x, t, ; 0. By the umption tht there exit t > uch tht gx, t 5
, h j t, ζ j=,2,...,l for t, ζ t,, b] nd τ i t for t t then ux, gt, ζ > 0 for t, x, ζ t,, b], ux, τ i t > 0 for x, t t, nd ux, h j t, ζ > 0 for t, x, ζ t,, b]. Integrting with repect to x over the domin, we hve d ] ptd,t ux, tdx b i t ux, τ i tdx dt rtd,t ux, tdx b i t ux, τ i tdx qx, t, ζfux, gt, ζ]dσζdx = t ux, tdx l c j t, ζ ux, h j t, ζdσζdx j= ex, tdx 8 Uing Green formul nd 2, it i obiviou tht ux, t ux, tdx = ds 0, t t 9 γ ux, h j t, ζdx = ux, h j t, ζ ds 0, t t, 0 γ j=,2,...,l, where ds i urfce element on. Moreover, uing Jenen inequlity nd from A 3, it follow tht qx, t, ζf ux, gt, ζ] dσζdx Qt, ζ f ux, gt, ζ] dx dσζ = λqt, ζ Ugt, ζ]dσζ 52
Combining 8 nd uing A 6, we hve ] ptd Ut b i tuτ i t d dt rtd Ut b i tuτ i t λqt, ζugt, ζ]dσζ 0. Set wt = Ut m b ituτ i t. Then, we get the inequlity ptd wt rtd wt λ Qt, ζ Ugt, ζ]dσζ 0. 2 It i ey to clculte tht wt > 0 for t t nd hence Et = t 0 t ξ wξdξ > 0. Next we prove tht D wt > 0 for t t 2. Suppoe we ume tht there exit T t 2 uch tht D wt 0. ptd wt rtd wt 0, t t 2. ptd wt p t rt Dwt 0, t t 2. 3 From A, we hve P t = P t nd P t > 0, P t 0 for t t 2. Multiply P t pt p trt pt on both ide of 3, we hve P td wt P td wt = P td wt 0, t t2. 4 From 4, we hve P td wt P T D wt 0, t T. 5 By Lemm 7, from 5 we hve E t Γ = D wt C P t, where C = P T D wt, t T. Integrting 6 from T to t, 6 Et ET Γ C t T d P 7 53
Letting t in 7, we hve lim Et =, which contrdict t with the fct tht Et > 0. Thu Dwt > 0 nd τ i t t,2,...,m for t t, we hve Ut = wt b i tuτ i t b i t wt, t t. Therefore from 2, we hve ptdwt rtdwt λ Qt, ζ b i gt, ζ wgt, ζdσζ 0, t t. From A 4 nd D wt > 0, we hve wgt, ζ] wgt, ] > 0, ζ, b] ptd wt rtd wt λwgt, ] Qt, ζ b i gt, ζ dσζ 0. From the umption, we hve θt gt,, h j t, t, j =, 2,..., l thu wθt wgt, ], wθt wt nd wθt wh j t, ]j =, 2,..., l for t t. Therefore, ptdwt rtdwt F twθt 0, t t. Theorem 9. lim up t t Aume tht Lρ Θu n F pd w 4w θθ n Θ t r p ρ ρ Θu then every olution ux,t of, 2 i ocilltory in G. 2 ] = 8 Proof. To prove the olution of, 2 re ocilltory in, by bove theorem it i enough to prove tht the functionl differentil inequlity 7 h no eventully poitive olution. Suppoe tht wt > 0 i olution of the inequlity 7. Define 54
zt = ptd wt wθt, t t. Then, z t rtzt pt F t w θtθ t ptd wt z2 t 9 Denote u = w θθ z r pd w. pd w 2 p w θθ Apply the opertor L ρ to 9, with t replced by, we get, t ρ n ] n z L ρ Θu Θ t ρ z ρ Θu w L ρ θθ pdw z2 r ] z F p ] n L ρ Θ t ρ z ρ Θu L ρ w θθ 2 pdw z w 2 θθ pdw z r pdw 2 p w θθ 2 ] r pdw L ρ 4 p w θθ 2 ] r 4 p t pdw w θθ F t n ρ z, Θu n L ρ Θ t ρ ρ Θu r pd 2 ] w L ρ 2 p w θθ t n ρ z, Θu n L ρ Θ t ρ ρ Θu L ρ F r 4 p ] z L ρ w θθ pdw z 2 pd w w θθ F 4 ] z L ρ u 2 ] 2 ] r pdw p w θθ t n ρ z, Θu 55
n L ρ Θ t ρ pdw ρ w θθ u ] r pdw 2 p w θθ Θu L ρ u 2 L ρ F 2 ] r pdw t n ρ z, 4 p w θθ Θu L ρ u n 2 Θ t ρ pd 2 ] w ρ w θθ Θu ] 2 ] L ρ F pd w r n 4w θθ p Θ t ρ ρ Θu t n ρ z. 20 Θu The firt term i nonnegtive, o 2 ] L ρ F pd w r n 4w θθ p Θ t ρ ρ Θu t n ρ z, t 2 Θu Set = nd divide 2 by t, nd tke lim up in 2 t n 0 Θu t, we get lim up t t Lρ Θu n F pd w r n 4w θθ p Θ t Θu 2 ] ρ z < which contrdict 8. ρ ρ Corollry 0. lim up t lim up t t t Θu n Θu If 8 in Theorem 8 i replced by n L ρ F = nd pd L ρ w r n 4w θθ p Θ t ρ ρ Θu 2 ] <. 56
Then every olution ux,t of, 2 i ocilltory in G. Theorem. tht lim up t t n Θu pd L ρ w n w θθ Θ Suppoe tht there exit f C, uch t nd there exit A C, uch tht nd lim up t ρ p t n Θ Θu t n L ρ Θu ρ 2 ] < 22 ρ F pd w r 4w θθ p ρ ρ Θu A w θθ r ρ Dw 2 2 ] A, 23 D w w θθ ] 2 d =, 24 A = mxa, 0, then every olution ux,t of the problem,2 i ocilltory in G. Proof. Aume to the contrry tht there i nonocilltory olution ux, t to the problem, 2. A in the proof of Theorem 9, 20 hold for ll t. Hence for t, we hve lim inf t t Lρ Θu n lim up t t u 2 Lρ Θu n n Θ t F pd w 4w θθ Θu ρ ρ r p ] 2 ] ρ ρz, for ll t ρ 0. Hence by 23 2 ] pd w w θθ n Θ t Θu 57
ρz A lim inf t t Lρ Θu n u n 2 Θ for ll. Thi how tht, t Θu ρ ρ 2 ] pd w, w θθ ρz A, nd 25 lim inf t n t Θu L ρ u n 2 Θ t ρ ρ Θu Define χ t = t Lρ Θu n u 2 ], n χ 2 t = t Lρ Θu n L ρ u 2 Θ χ t χ 2 t t n Θ t Θu n t From 26 nd 27 we get, Now, to prove tht ρ Θu ρ ρ ρ Θu pdw 2 ] <. w θθ 26 pd w w θθ u ]. Then ] pdw 2. 27 w θθ lim inf t χ t χ 2 t] <. 28 ρu 2 d <. 29 Aume to contrry tht ρu 2 d =. Thu lim χ t =. 30 t Now uppoe {t n } n=, tifying lim n t n = nd lim χ t n χ 2 t n ] = lim inf χ t χ 2 t]. 3 n t 58
By 28 nd 3, there exit contnt B > 0 uch tht χ t n χ 2 t n B n=,2,... Thi together with 30 men lim n χ 2 t n =. So for n lrge enough, χ 2t n <, i.e., χ 2t n >. Moreover, χ t n 2 χ t n 2 χ 2 lim 2t n n χ t n =. 32 On the other hnd, by chwrz inequlity, for n lrge enough, ] 2 χ 2 2t n = tn Lρ n t0 Θu 2n Θ pd w u tn Θu ρ ρ w θθ 2 ] tn Lρ t0 Θu n u 2 ] tn Lρ n t0 Θu n Θ pd w tn Θu ρ ρ w θθ ] n tn = χ t n L ρ n. Θu By 30 nd 32 we get, lim n tn Θu Θ n L ρ n Θ tn tn ρ Θu ρ ρ Θu ρ 2 pd w w θθ 2 ] pd w =, w θθ which contrdict with 22. So we obtin 29. Then by 25 nd in view of ut = A ρ w θθ r pd w 2 So by 29 we get w θtθ t zt ptd wt 2 ρ p D w rt pt ptd wt w θtθ t we hve, 2 w θθ p] u 2 for ll t 0. 2 A w θθ r D w ρ D w 2 w θθ ] d ρu 2 d <, which contrdict with 24. Thi complete the proof of Theorem. Theorem 2. Suppoe tht there exit f C, tifying 22 nd there exit A C, uch tht F lim inf t t Lρ Θu n pd w 4w θθ r n p Θ t ρ ρ Θu 2 ] A,.Then every olution ux, t of problem, 2 i ocilltory in G provided 24 hold. 59
Proof. Similr to the proof of Theorem, we obtin lim up t χ t χ 2 t] <. Suppoe 29 i not true, we get 30. Now tke {t n }, tifying lim k {t k } = nd lim k χ t k χ 2 t k ] = lim up t χ t χ 2 t]. Similr method led to contrdiction, o we get 29. The reminder prt re me to Theorem, o we complete the proof. Theorem 3., uch tht lim up t t Θu n L ρ F pd w r 2 2r n 4w θθ p 2 p Θ lim up t t Lρ Θu n Suppoe tht there exit f C,, A, A 2 pd w 4w θθ n Θ t t ρ Θu ρ ρ Θu ρ ] A, 2 ] A 2 nd ρ p A w θθ A 2 ρ Dw 4ρ r 2 w θθ Dw D w w θθ 2 <, d =. 33 Then every olution ux,t of the problem,2 i ocilltory in G. t Proof. Followed the proof of Theorem 9, we get 2. So F pd w r 2 2r n 4w θθ p 2 p Θ t 2 ] Lρ Θu n ρz 4 t Lρ Θu n Uing the condition we get, pd w 4w θθ n Θ t ρ Θu ρ ρ Θu ρ ] A ρz 4 A 2, 34 Similr to the proof of Theorem, we obtin 29. Condition 29 nd 34 led to contrdiction with 33. So we complete the proof. 60
4 Exmple In thi ection,we give n exmple to illutrte our min theorem in ection 4. Exmple. we conider the frctionl prtil differentil eqution t t D 2,t ux, t 4 ux, t 2π ] t 2 D 2,t ux, t 4 ux, t 2π π ux, t 3π 5 π ζdζ = ux, t 0 2 8 ux, t 3π 2 0 2 ζdζ 5 4 in t e x e x co t 2e x co t, forx, t G, 2 t where G = 0, π 0,, 35 u x 0, t u0, t = u x π, t uπ, t = 0, t 0. 36 Here =, m =, l =, n = 2, P t =, rt =, b 2 t t 2 t =, τ 4 t = t 2π, qx, t, ζ =, fu = u, gt, ζ, h t, ζ = t 3π ζ,, b] = 0, π], t =, C 2 t = 5 8, σζ = ζ, ex, t = 2 5 4 in t e x e x co t 2e x co t, Qt, ζ = min 2 t x = min x 0,π] =, fu = > 0 nd F t = π 3 dζ = 3π. Let Θu =, nd u 0 4 4 ρt =, θt = t. All the condition A t t A 6 hold. Ut = ux, tdx = e π co nd wt = 5 4 e π co. Conider, t = = = Θu n Lρ t 2 Lρ F pd w 4w θθ t 3π 4 3π t 2 4 t 2 d 3π t 4 t 2 0 t 2 t4 4 t 2 4 2 4 r n p Θ t ] 4 co in 4 2 in t 2 2 ] 2t4 tt 3 0 3 ρ ρ Θu which how tht ll the condition of Theorem 9 re tified. Thu every olution of problem 35,36 ocillte in 0, π 0,. Therefore ux, t = e x co t i one uch olution. 2 ] 6
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