王培光,李志芳
(1.河北大學 電子信息工程學院,河北 保定 071002;2.河北大學 數學與計算機學院,河北 保定 071002)
研究報告
含causal算子分數階非線性微分方程的擬線性方法
王培光1,李志芳2
(1.河北大學 電子信息工程學院,河北 保定 071002;2.河北大學 數學與計算機學院,河北 保定 071002)
采用擬線性化方法討論了含causal算子的分數階非線性微分方程初值問題,通過構造2個單調迭代序列,證明了它們一致且平方收斂于給出問題的解.
擬線性方法;causal算子;分數階微分方程;平方收斂
在非線性微分方程解的定性問題的研究中, 擬線性化方法得到了廣泛的使用[1].由于含causal算子微分方程系統模型可描述現實世界的許多問題, 因而引起了人們的廣泛關注.文獻[2]利用上下解結合單調迭代方法, 給出了一類一致收斂于含causal算子微分方程解的迭代序列.近年來分數階微分方程引起了人們的廣泛關注[3-7].然而關于用擬線性化方法研究含causal算子分數階微分方程初值解的結果并未見到.本文將利用擬線性化方法對含causal算子的分數階非線性微分系統兩項和的初值問題(簡稱IVP)
cDqu(t)=(Qu)(t)+(Pu)(t),u(0)=u0
(1)
進行研究,得到解的一致且平方收斂的結果.這里Q,P∶E→E是連續causal算子,cDq是Caputo分數階導數,0lt;qlt;1,E=C(J×R,R)和t∈J=[0,T].
利用下面定義和引理證明主要定理.
定義1 如果對于E=C[J×R,R]中每對元素(x,y),使x(s)=y(s),有(Qx)(t)=(Qy)(t),其中0≤s≤t,tlt;T,T是任意正實數,則稱Q∶E→E是causal算子.
式(1)等價的Volterra分數階積分方程[9]為

其中Γ是Gamma函數.
定義2 考慮初值問題式(1),若α,β∈Cq[J,R],滿足
cDqα(t)≤(Qβ)(t)+(Pβ)(t),α(0)≤u0,
cDqβ(t)≥(Qα)(t)+(Pα)(t),β(0)≥u0,
則稱α,β分別是式(1)的耦合下解和耦和上解.
引理1[2]若v,w∈Cq[J,R]分別是式(1)的耦合下解和耦合上解,且v(t)≤w(t),t∈J和Q,P∈[Ω,R],其中Ω=[(t,x)∶v(t)≤x≤w(t),t∈J],則存在式(1)的唯一解x(t)滿足v(t)≤x(t)≤w(t),t∈J.
引理2[10]設v,w∈Cq[J,R],Q∈C[J×R2,R],若cDqv(t)≤Q(t,w,w),cDqw(t)≥Q(t,v,v),
Q(t,x1,y1)-Q(t,x2,y2)≥-L[(x1-x2)+(y1-y2)],L≥0,
其中Q是causal算子,x1≥x2,y1≥y2,且v(0)≤w(0),則有v(t)≤w(t),t∈J.
證明設w0=w+εEq(3Ltq),v0=v-εEq(3Ltq),其中Eq(3Ltq)表示e3Lt的0lt;qlt;1階導數,εgt;0是任意小的實數,則有w0gt;w,v0lt;v和w0(0)gt;w(0)gt;v0(0).由已知條件可得
cDqv0(t)=cDqv(t)-3LεEq(3Ltq)≤Q(t,w,w)-3LεEq(3Ltq)≤
Q(t,w,w0)+LεEq(3Ltq)-3LεEq(3Ltq)≤
Q(t,w0,w0)+2LεEq(3Ltq)-3LεEq(3Ltq)lt;
Q(t,w0,w0).
同理可得
cDqw0(t)=cDqw(t)+3LεEq(3Ltq)≥Q(t,v,v)+3LεEq(3Ltq)≥
Q(t,v,v0)-LεEq(3Ltq)+3LεEq(3Ltq)≥
Q(t,v0,v0)-2LεEq(3Ltq)+3LεEq(3Ltq)gt;
Q(t,v0,v0).
下面證明v0(t)lt;w0(t),t∈J.假設不然,則存在t0∈(0,T]使v0(t0)=w0(t0),v0(t)lt;w0(t),0≤tlt;t0成立,由此可得cDqv0(t0)≥cDqw0(t0),則進一步有
Q(t,w0(t0),w0(t0))gt;cDqv0(t0)≥cDqw0(t0)gt;Q(t,v0(t0),v0(t0)),
此不等式與
Q(t,w0(t0),w0(t0))=Q(t,v0(t0),v0(t0))
矛盾,其中v0(t0)=w0(t0).因此v0(t)lt;w0(t),t∈J成立.
在v(t)-εEq(3Ltq)=v0(t)lt;w0(t)=w(t)+εEq(3Ltq)中令ε→0,則有v(t)≤w(t),t∈J.證畢.
引理3[7-8]對于Caputo線性分數階微分方程
cDqu=λu+(Qu)(t),u(0)=u0,
其中Q∈Cq[J,R],且對q為H?lder連續,其唯一解

其中
分別是含1個參數和2個參數的Mittag-Leffler方程.
定理1 假設下列條件成立:
(A1)α0,β0∈Cq[J,R],α0(t)≤β0(t),t∈J,滿足
cDqα0(t)≤(Qβ0)(t)+(Pβ0)(t),
cDqβ0(t)≥(Qα0)(t)+(Pα0)(t);
(A2)Q,P∈C[Ω,R],Frechet導數Qu,Pu,Quu和Puu存在,連續且滿足(Quuu)(t)≥0,(Puuu)(t)≤0,(t,u)∈Ω.
(A3)(Quu)(t)≤0和(Puu)(t)≤0,(t,u)∈Ω,
則存在2個單調序列{αn}和{βn}一致且平方收斂于式(1)的唯一解.
證明由(Quuu)(t)≥0,(Puuu)(t)≤0,對u≥v有下列不等式成立
(Qu)(t)≤(Qv)(t)+(Quu)(u-v)(t),
(2)
(Pu)(t)≤(Pv)(t)+(Puv)(u-v)(t).
(3)
對于任意的α0(t)≤u2≤u1≤β0(t),t∈J;Q,P滿足
L(u1-u2)≥(Qu1)(t)-(Qu2)(t)≥-L(u1-u2),Lgt;0,
(4)
L(u1-u2)≥(Pu1)(t)-(Pu2)(t)≥-L(u1-u2).
(5)
考慮下面初值問題
(6)
(7)
其中α0(0)≤u0≤β0(0).
由不等式(2),(3)和(A1)可知
cDqα0(t)≤(Qβ0)(t)+(Pβ0)(t)≡F(t,α0,β0;β0);
cDqβ0(t)≥(Qα0)(t)+(Pα0)(t)≥(Qβ0)(t)+(Quβ0)(α0-β0)(t)+
(Pβ0)(t)+(Puα0)(α0-β0)(t)≡F(t,α,β0;α0);
cDqα0(t)≤(Qβ0)(t)+(Pβ0)(t)≤(Qα0)(t)+(Quβ0)(β0-α0)(t)+
(Pα0)(t)+(Quα0)(β0-α0)(t)≡G(t,α0,β0;β0)
cDqβ0(t)≥(Qα0)(t)+(Pα0)(t)≡G(t,α0,β0;α0).
由于(Quu)(t)≤0和(Puu)(t)≤0,可得F(t,α0,β0;v)和G(t,α0,β0;u)分別對于v和u是非增的.由引理1可知式(6)和(7)存在唯一解(α1,β1),滿足α0≤α,β1≤β0,t∈J.
即
cDqα1(t)=F(t,α0,β0;β1);
cDqβ1(t)=G(t,α0,β0;a1).
由不等式(2)和(3)可得
cDq(α1)(t)=(Qβ0)(t)+(Quβ0)(β1-β0)(t)+(Pβ0)(t)+(Puα0)(β1-β0)≤
(Qβ1)(t)+(Quβ0)(β0-β1)(t)+(Quβ0)(β1-β0)(t)+
(Pβ1)(t)+(Puβ1)(β0-β1)(t)+(Puα0)(β1-β0)(t)=
(Qβ1)(t)+(Pβ1)(t)+[(Puβ1)(t)-(Puα0)(t)](β0-β1)(t)≤
(Qβ1)(t)+(Pβ1)(t);
cDqβ1(t)=(Qα0)(t)+(Quβ0)(α1-α0)(t)+(Pα0)(t)+(Puα0)(α1-α0)(t)≥
(Qα1)(t)+(Quα1)(α0-α1)(t)+(Quβ0)(α1-α0)(t)+
(Pα1)(t)+(Puα0)(α0-α1)(t)+(Puα0)(α1-α0)(t)=
(Qα1)(t)+(Pα1)(t)+[(Quβ0)(t)-(Quα1)(t)](α1-α0)(t)≥
(Qα1)(t)+(Pα1)(t).
因為(Quu)(t)關于u是非減的,(Puu)(t)關于u是非增的,因此應用引理2,可得α1(t)≤β1(t),t∈J即
α0(t)≤α1(t)≤β1(t)≤β0(t),t∈J.
(8)
考慮下面一組初值問題
cDqu(t)=F(t,α1,β1;v),u(0)=u0,
(9)
cDqv(t)=G(t,α1,β1;u),v(0)=u0,
(10)
可推出下列不等式
cDqα1(t)≤(Qβ1)(t)+(Pβ1)(t)≡F(t,α1,β1;β1);
cDqβ1(t)≥(Qα1)(t)+(Pα1)(t)≥(Qβ1)(t)+(Quβ1)(α1-β1)(t)+
(Pβ1)(t)+(Puα1)(α1-β1)(t)≡F(t,α1,β1;α1);
cDqα1(t)≤(Qβ1)(t)+(Pβ1)(t)≤(Qβ1)(t)+(Quβ1)(β1-α1)(t)+
(Pα1)(t)+(Puα1)(β1-α1)(t)≡
G(t,α1,β1;β1);
cDqβ1(t)≥(Qα1)(t)+(Pα1)(t)≡G(t,α1,β1;α1).
由引理1可知式(9)和式(10)存在唯一解α2,β2使α1≤α2,β2≤β1,t∈J成立.同樣由于
cDqα2(t)≤(Qβ2)(t)+(Pβ2)(t);
cDqβ2(t)≥(Qα2)(t)+(Pα2)(t).
則應用引理2可得α2(t)≤β2(t),t∈J.綜上可知
α0≤α1≤α2≤β2≤β1≤β0,
如此繼續下去,可得
α0≤α1≤α2≤…≤αn≤βn≤…≤β2≤β1≤β0,
(11)
其中單調序列{αn(t)},{βn(t)}是下列線性方程初值問題
cDqαn+1(t)=F(t,αn,βn;βn+1),αn+1(0)=u0,
(12)
cDqβn+1(t)=G(t,αn,βn;αn+1),βn+1(0)=u0
(13)
的解.綜上所述很容易得知{αn(t)},{βn(t)}序列一致收斂于式(1)的唯一解.
下面證明收斂速度是2次的.為了證明收斂速度是2次, 設pn(t)=u(t)-αn(t),qn(t)=βn(t)-u(t), 其中u(t)是式(1)的唯一解.利用αn,βn的定義, 中值定理以及(A2), 有
cDqpn(t)=cDqu(t)-cDqαn(t)=
(Qu)(t)+(Pu)(t)-[(Qβn-1)(t)+(Quβn-1)(βn-βn-1)(t)+
(Pβn-1)(t)+(Puαn-1)(βn-βn-1)(t)]=
-(Quξ)qn-1(t)-(Puσ)qn-1(t)-(Quβn-1)[qn-qn-1](t)-
(Puαn-1)[qn-qn-1](t)≤
[(Quβn-1)(t)-(Quu)(t)]qn-1(t)+
[(Puαn-1)(t)-(Puβn-1)(t)]qn-1(t)+Mqn(t)=

即

(14)
其中ult;ξ,σlt;βn-1,αn-1lt;σ1lt;βn-1,ult;ξ1lt;βn-1,|(Quu)(t)|≤M1,|(Puu)(t)|≤M2,|(Quuu)(t)|≤N1,|(Puuu)(t)|≤N2和M=M1+M2.
同理,
cDqqn(t)=cDqβn(t)-cDqu(t)=
(Qαn-1)(t)+(Quβn-1)(αn-αn-1)(t)+(Pαn-1)(t)+
(Puαn-1)(αn-αn-1)(t)-(Qu)(t)-(Pu)(t)=
(Quξ)(αn-1-u)(t)+(Puσ)(αn-1-u)(t)+
(Quβn-1)(αn-αn-1)(t)+(Puαn-1)(αn-αn-1)(t)=
-(Quξ)pn-1(t)-(Puσ)pn-1(t)+(Quβn-1)pn-1(t)-
(Quβn-1)pn(t)+(Puαn-1)pn-1(t)-(Puαn-1)pn(t)=
[-(Quξ)(t)+(Quβn-1)(t)]pn-1(t)+[-(Puσ)(t)+
(Puαn-1)(t)]pn-1(t)-(Quβn-1)pn(t)-(Puαn-1)pn(t)≤
[(Quβn-1)(t)-(Quαn-1)(t)]pn-1(t)+[(Puαn-1)(t)-(Puu)(t)]pn-1(t)-
[(Quβn-1)(t)+(Puαn-1)(t)]pn(t)=
[(Quβn-1)(t)+(Puαn-1)(t)]pn(t),
其中αn-1lt;ξ,σlt;u,αn-1lt;ξ1lt;βn-1,αn-1lt;σ1lt;u.但

因此

(15)




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(責任編輯:王蘭英)
Quasilinearizationforsolutionofnonlinearcausalfractionaldifferentialequations
WANGPei-guang1,LIZhi-fang2
(1.College of Electronic and Information Engineering, Hebei University, Baoding 071002, China; 2.College of Mathematics and Computer Science, Hebei University, Baoding 071002, China)
By using the quasilinearization method for causal fractional differential equations, the authors construct two monotone sequences, then prove that they both converge uniformly and quadratically to the solution of the given problem.
quasilinearization method; causal operator; fractional differential equations; quadratic convergence
O175.1
A
1000-1565(2012)01-0001-06
2011-09-21
國家自然科學基金資助項目(10971045);河北省自然科學基金資助項目(A2009000151)
王培光(1963-), 男, 黑龍江哈爾濱人, 河北大學教授, 博士生導師, 主要從事微分方程與控制理論方面的研究.
E-mail:pgwang@hbu.edu.cn
MSC2010: 34A34