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数学物理学报, 2019, 39(6): 1334-1341 doi:

论文

一类混合型分数阶半线性积分-微分方程解的存在性

Existence of Mild Solutions for a Class of Fractional Semilinear Integro-Differential Equation of Mixed Type

Zhu Bo,1, Han Baoyan2, Liu Lishan3

通讯作者: 朱波, E-mail: zhubo207@163.com

收稿日期: 2018-08-29  

基金资助: 山东省高校科技计划项目.  J16LI14
国家自然科学基金.  11871302

Received: 2018-08-29  

Fund supported: the PSDPHESTP.  J16LI14
the NSFC.  11871302

摘要

该文利用非紧性测度、β-预解族、k-集压缩原理研究了一类混合型分数阶半线性积分-微分方程温和解的存在性.众所周知,利用k-集压缩原理证明解的存在性时需要单独给出附加条件来保证压缩系数小于1,而该文不需要单独附加保证压缩系数小于1的条件.在更一般的条件下证明了方程解的存在性.文章最后给出了一个例子说明该文主要结果的应用.

关键词: 混合型分数阶半线性积分-微分方程 ; k-集压缩 ; 非紧性测度 ; β-预解族

Abstract

In this paper, the authors studied the existence results of the mild solutions for a class of fractional semilinear integro-differential equation of mixed type by using the measure of noncompactness, k-set contraction and β-resolvent family. It is well known that the k-set contraction requires additional condition to ensure the contraction coefficient 0 < k < 1. We don't require additional condition to ensure the contraction coefficient 0 < k < 1. An example is introduced to illustrate the main results of this paper.

Keywords: Fractional semilinear integro-differential equation of mixed type ; k-Set contraction ; Measure of noncompactness ; β-Resolvent family

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本文引用格式

. 一类混合型分数阶半线性积分-微分方程解的存在性. 数学物理学报[J], 2019, 39(6): 1334-1341 doi:

Zhu Bo, Han Baoyan, Liu Lishan. Existence of Mild Solutions for a Class of Fractional Semilinear Integro-Differential Equation of Mixed Type. Acta Mathematica Scientia[J], 2019, 39(6): 1334-1341 doi:

1 引言

近几十年来,分数阶微分(发展)方程在工程、化学、物理和金融等领域得到成功应用.因而,大量科研人员正关注分数阶微分方程的研究并获得了丰硕成果,具体内容参见文献[1-9]及其参考文献.文献[10-11]中,作者研究了分数阶积分-微分系统的可控性.文献[12-13]中,作者研究了整数阶混合型常微分方程解的存在理论.而在文献[5, 7, 9, 14-17],作者研究了分数阶发展方程解的存在性.其中文献[7, 9, 15-17]研究的分数阶微分方程的阶数β(0,1],而文献[5, 14]研究的分数阶微分方程的阶数β(1,2].

然而,根据我们的了解,混合型分数阶积分-微分方程温和解的存在性还没有人研究.本文,我们研究如下的混合型分数阶积分-微分方程

cDβtu(t)=A(t)u(t)+f(t,u(t),Gu(t),Su(t)), tJ,
(1.1)

u(0)+g(u)=u0,
(1.2)

这里, cDβtβ阶Caputo分数阶导数(β(0,1]), J=[0,T0] (T0>0), A(t)是定义在Banach空间E上的闭线性算子,并且A(t)依赖于t. u0E,函数f,g在后面具体给出,算子GS定义如下

Gu(t)=t0k(t,s,u(s))ds,  Su(t)=T00h(t,s,u(s))ds,

这里k:D×EE,h:D0×EE是合适的函数. D={(t,s)R2:0stT0},D0={(t,s)R2:0t,sT0}.由于算子G是定义在区间[0, t]上的变上限积分,而算子S是定义在固定区间[0,T0]上的积分,因此,我们称方程(1.1)是混合型分数阶积分-微分方程.

如果β=1并且算子A(t)=0,则方程(1.1)就变成文献[12-13]中作者研究的整数阶混合型常微分方程.也就是说,文献[12-13]中作者研究的方程是本文的特例.众所周知,利用k -集压缩证明解的存在性时需要附加条件保证压缩系数0<k<1,文献[7, 16-17]就用了k -集压缩原理证明了解的存在性.本文中,我们成功克服了这一限制条件,不需要附件条件来保证压缩系数0<k<1.另外,上述提到的文献中的线性算子A大部分是不依赖于t的,而本文中的线性算子A(t)依赖于t.因此,我们在更一般的条件下得到了问题(1.1)-(1.2)温和解的存在性.我们的结果推广和改进了前人的一些经典结论.

2 预备知识和引理

α表示非紧性测度, (E,是一个Banach空间. C[J, E]=\{u:J\rightarrow E连续\}是一个Banach空间,具有范数\|u\|_C=\max \{\|u(t)\|:t\in J\}.对任意的B\subset C[J, E], t\in J, B(t)=\{u(t):u\in B\}.对任意实数R>0, B_R=\{u\in C[J, E]: \|u\|_C\leq R\}, T_R=\{x\in E:\|x\|\leq R\}, B(E)表示Banach空间E上的有界线性算子的集合.

定义2.1[18-19]  设A(t):D(A)\subset E是一个闭线性算子, \beta>0, \rho[A(t)]A(t)的预解集,如果存在\omega\geq 0和一个强连续函数U_\beta:\mathbb{R} _+^2\rightarrow B(E)使得\{\lambda^\beta: Re \lambda >\omega\}\subset \rho (A)

(\lambda^\beta I-A(s))^{-1}u=\int_0^\infty {\rm e}^{-\lambda (t-s)}U_\beta(t, s)u{\rm d}t, \ \ \ {\rm Re}(\lambda)>\omega, u\in E

成立,我们称A(t)生成一个\beta -预解族.这里, U_\beta(t, s)称为由A(t)生成的\beta -预解族.

引理2.1[13]  F是Banach空间E的一个凸闭子集,算子{\cal A}:F\rightarrow F连续且{\cal A}(F)有界.对任意的有界子集D\subset F,记

\widetilde{{\cal A}}^1(D)={\cal A}(D), \ \ \ \ \widetilde{{\cal A}}^n(D)={\cal A}(\overline{Co}(\widetilde{{\cal A}}^{n-1}(D))), \ \ \ n=2, 3, \cdot\cdot\cdot.
(2.1)

若存在一个实数0\leq k < 1和一个正整数n_0,使对所有的有界子集D\subset F,都有

\alpha(\widetilde{{\cal A}}^{n_0}(D))\leq k\alpha(D)
(2.2)

成立,则{\cal A}F中存在一个不动点.

引理2.2[13, 22]  D\subset C[J, E]有界且等度连续,则\overline{Co}D \subset C[J, E]有界且等度连续, m(t)=\alpha(D(t))J上也连续且有

\alpha\left(\int_JD(s){\rm d}s\right)\leq \int_J\alpha(D(s)){\rm d}s.
(2.3)

引理2.3[22]  如果函数g : J\times T_R\times T_R\to E有界且一致连续, D \subset C[J, E]有界且等度连续,那么\{g(t, u(t), u(t)): u\in D\}C[J, E]上有界且等度连续.

引理2.4[23]  设常数0 < \varrho < 1, \gamma>0.

S=\varrho^n+C_n^1\varrho^{n-1}\gamma+\frac{C_n^2\varrho^{n-2}\gamma^2}{2!} +\cdot\cdot\cdot+\frac{\gamma^n}{n!}, \ \ n\in {\Bbb N},

则对固定的常数0 < \xi < 1和任意常数s>1,有

S\leq O\left(\frac{\xi^n}{\sqrt{n}}\right)+o\left(\frac{1}{n^s}\right)=o\left(\frac{1}{n^s}\right), \ \ n\rightarrow +\infty.

定义2.2  如果u\in C[J, E]满足下面的方程

u(t)=U_\beta(t, 0)(u_0-g(u))+\int_0^tU_\beta(t, s)f(s, u(s), {\cal G}u(s), {\cal S}u(s)){\rm d}s, \ \ t\in J,
(2.4)

u(t)是问题(1.1)-(1.2)的温和解.

3 主要结论

定义算子{\cal Q}:C[J, E]\rightarrow C[J, E]如下

\begin{equation}{\cal Q}u(t)=U_\beta(t, 0)(u_0-g(u))+\int_0^tU_\beta(t, s)f(s, u(s), {\cal G}u(s), {\cal S}u(s)){\rm d}s.\end{equation}
(3.1)

定理3.1  假设下列条件成立.

(H_1)f:J\times E\times E\times E\rightarrow E,存在非负Lebesgue可积函数L_i\in L(J, \mathbb{R} _+)(i=1, 2, 3),使对所有的u, v \in E t\in J,都有

\|f(t, u_1, u_2, u_3)-f(t, v_1, v_2, v_3)\|\leq L_1(t)\|u_1-v_1\|+L_2(t)\|u_2-v_2\|+L_3(t)\|u_3-v_3\|.

(H_2)k:J\times J\times E\rightarrow E,存在非负Lebesgue可积函数L_4\in L(J, \mathbb{R} _+),使对所有的u, v \in E t\in J,都有

\|k(t, s, u)-k(t, s, v)\|\leq L_4(t)\|u-v\|.

(H_3)h:J\times J\times E\rightarrow E,存在非负Lebesgue可积函数L_5\in L(J, \mathbb{R} _+),使对所有的u, v \in E t\in J,都有

\|h(t, s, u)-h(t, s, v)\|\leq L_5(t)\|u-v\|.

(H_4)g:C(J, E)\rightarrow E,存在非负常数N,使对所有的u, v \in C(J, E) t\in J,都有

\|g(u)-g(v)\|\leq N\|u-v\|_C.

(H_5)M^*\left(N+\int_{0}^{T_0} L_1(s){\rm d}s+\int_{0}^{T_0}\left(L_2(s)\int_{0}^{T_0}L_4(\tau){\rm d}\tau \right){\rm d}s+\int_{0}^{T_0}L_3(s)\int_{0}^{T_0}L_5(\tau){\rm d}\tau {\rm d}s\right) < 1, 这里M^*=\max\limits_{0\leq s < t\leq {T_0}}\|U_\beta(t, s)\| < +\infty.

那么问题(1.1)-(1.2)在C[J, E]上存在唯一温和解.

  由(3.1)式,对所有的t\in J, u, v \in C[J, E],我们有

\begin{eqnarray*} &&\|({\cal Q}u)(t)-({\cal Q}v)(t)\|\\ &\leq &M^*\|g(u)-g(v)\|+M^*\int_0^t\|f(s, u(s), {\cal G}u(s), {\cal S}u(s))-f(s, v(s), {\cal G}v(s), {\cal S}v(s))\|{\rm d}s\\ &\leq& M^*N\|u-v\|_C+M^*\int_0^tL_1(s){\rm d}s\|u-v\|_C\\ &&+M^*\left(\int_0^t\left(L_2(s)\int_0^sL_4(\tau){\rm d}\tau \right){\rm d}s+\int_0^t\left(L_3(s)\int_0^{T_0}L_5(\tau)\right){\rm d}\tau {\rm d}s\right)\|u-v\|_C\\ &\leq& \left(M^*N+M^*\int_0^tL_1(s){\rm d}s+ M^*\int_0^tL_2(s)\int_0^sL_4(\tau){\rm d}\tau {\rm d}s\right.\\ &&\left.+M^*\int_0^tL_3(s)\int_0^{T_0}L_5(\tau){\rm d}\tau {\rm d}s\right)\|u-v\|_C. \end{eqnarray*}

根据(H_5),算子{\cal Q}是压缩映像.因此,算子{\cal Q}存在唯一不动点u^*\in C[J, E],从而问题(1.1)-(1.2)存在唯一温和解.

定理3.2  假设下列条件成立.

(H_6)f:J\times T_R\times T_R\times T_R\rightarrow E有界且一致连续,

\begin{equation}\limsup\limits_{R\rightarrow\infty}\frac{M(R)}{R}<\frac{1}{a_0T_0M^*} , \end{equation}
(3.2)

这里M(R)=\sup\{\|f(t, u_1, u_2, , u_3)\|:(t, u_1, u_2, , u_3)\in J\times T_R^3\}, k_0=\max\{|k(t, s)|:(t, s)\in D\}, h_0=\max \{|h(t, s)|:(t, s)\in D_0\}, a_0=\max \{1, T_0h_0, T_0k_0\}.

(H_7)存在非负Lebesgue可积函数L_i\in L(J, \mathbb{R} _+)(i=1, 2),对任意有界集B\subset E t\in J,都有

\alpha(f(t, B_1, B_2, B_3))\leq \sum\limits_{i=1}^2L_i(t)\alpha(B_i).

(H_8)g:C[J, E]\rightarrow E是连续的紧函数, \|g(u)\|\leq d\|u\|+e,这里常数d>0, e>0.

那么问题(1.1)-(1.2)在C[J, E]存在一个温和解.

  根据(H_6)知由(3.1)式定义的算子{\cal Q}:C[J, E]\rightarrow C[J, E]有界且连续.由(3.2)式,取\limsup\limits_{R\rightarrow\infty}\frac{M(R)}{R} < r < \frac{1}{a_0T_0M^*}R_0>0,对任意R\geq a_0R_0,都有

\begin{equation}M(R)<rR.\end{equation}
(3.3)

记集合R^*=\max\{R_0, M^*(\|u_0\|+d\|u\|+e)(1-a_0T_0rM^*)^{-1}\}.对所有u\in B_{R^*}=\{u\in C[J, E] : \|u\|_C\leq R^*\},我们可得

\|u\|_C\leq R^*\leq a_0R^*, \ \ \ \|{\mathcal G}u\|_C\leq T_0k_0\|u\|_C\leq T_0k_0R^*\leq a_0R^*, \\\ \ \ \ \ \ \ \ \ \ \|{\mathcal S}u\|_C\leq T_0h_0\|u\|_C\leq T_0h_0R^*\leq a_0R^*.

根据(3.3)式,我们有

\begin{eqnarray*}\|Fu\|_C&\leq& M^*\|u_0\|+M^*(d\|u\|+e)+M^*T_0M(a_0R^*)\\&\leq &M^*\|u_0\|+M^*(d\|u\|+e)+M^*T_0ra_0R^*\leq R^*.\end{eqnarray*}

因此, {\cal Q}:B_{R^*}\rightarrow B_{R^*}是连续有界算子.

下面证明{\cal Q}(B_{R^*})是等度连续的.对所有的u\in B_{R^*}, t_1, t_2 \in J \ (t_1 < t_2),根据算子{\cal Q}的定义和(3.3)式,可得

\begin{eqnarray} \|({\cal Q}u)(t_2)-({\cal Q}u)(t_1)\|&\leq&\|(U_\beta(t_2, 0)-U_\beta(t_1, 0))(u_0-g(u))\|\\ &&+\int_{0}^{t_1}\|(U_\beta(t_2, s)-U_\beta(t_1, s))f(s, u(s), {\cal G}u(s), {\cal S}u(s))\|{\rm d}s\\ &&+\int_{t_1}^{t_2}\|U_\beta(t_2, s)f(s, u(s), {\cal G}u(s), {\cal S}u(s))\|{\rm d}s\\ &\leq& \|(U_\beta(t_2, 0)-U_\beta(t_1, 0))(u_0-g(u))\|\\& &+ M(a_0R^*)t_1\sup\limits_{s\in J}\|U_\beta(t_2, s)-U_\beta(t_1, s)\|+ M^*M(a_0R^*)(t_2-t_1).\\ \end{eqnarray}
(3.4)

由于算子U_\beta(t, s)是强连续的,因而对任意实数\varepsilon>0,存在实数\delta>0,对所有u\in B_{R^*}, t_1, t_2\in J, t_2-t_1 < \delta,都有\|({\cal Q}u)(t_2)-({\cal Q}u)(t_1)\| < \varepsilon成立.因此, {\cal Q}(B_{R^*})等度连续.

由引理2.2可知, \overline{Co}{\cal Q}(B_{R^*})\subset B_{R^*}有界且等度连续,算子{\cal Q}:\overline{Co}{\cal Q}(B_{R^*})\rightarrow \overline{Co}{\cal Q}(B_{R^*})有界且连续.对任意的B\subset \overline{Co}{\cal Q}(B_{R^*}),根据(2.2)式、(3.1)式、引理2.2和2.3知\widetilde{{\cal Q}}^n(B)(n=1, 2, \cdot\cdot\cdot)有界且等度连续,这里

\begin{equation}\alpha(\widetilde{{\cal Q}}^n(B))=\max\limits_{t\in J}\alpha((\widetilde{{\cal Q}}^n(B))(t)), \ \ \ n=1, 2, \cdot\cdot\cdot.\end{equation}
(3.5)

下面,我们证明对任意的B\subset \overline{Co}{\cal Q}(B_{R^*}),存在实数0\leq k < 1和一个自然数n_0,使得

\begin{equation}\alpha(\widetilde{{\cal Q}}^{n_0}(B))\leq k\alpha(B).\end{equation}
(3.6)

由条件(H_7)和(H_8),我们得到

\begin{eqnarray} \alpha((\widetilde{{\cal Q}}^1(B))(t))&= &\alpha(({\cal Q}(B)(t))\\ &\leq &M^*\int_{0}^{t}[L_1(s)\alpha(B(s))+L_2(s)\alpha(({\cal G}B)(s))]{\rm d}s\\ &\leq& M^*\int_{0}^{t}[L_1(s)+T_0k_0L_2(s)]\alpha(B(s)){\rm d}s\\ &= &\int_{0}^{t}L(s)\alpha(B(s)){\rm d}s\\ &\leq& \int_{0}^{t}L(s){\rm d}s\alpha(B), \ \ t\in J, \end{eqnarray}
(3.7)

这里, L(t)=M^*[L_1(t)+T_0k_0L_2(t)]J上是Lebesgue可积的.对任意常数0 < \varepsilon < 1,存在一个连续函数\varphi:J\rightarrow \mathbb{R} ^1使得

\begin{equation}\int_{0}^{T_0}|L(s)-\varphi(s)|{\rm d}s<\varepsilon.\end{equation}
(3.8)

由(3.7)式和(3.8)式得

\alpha((\widetilde{{\cal Q}}^1(B))(t))\leq \left[\int_{0}^{t}|L(s)-\varphi(s)|{\rm d}s+\int_{0}^{t}|\varphi(s)|{\rm d}s\right]\alpha(B)\leq (\varepsilon+Ht)\alpha(B),

这里H=\max\{|\varphi(t)|:t\in J\}.对每一个t\in J, 我们假设下面的方程成立

\alpha((\widetilde{{\cal Q}}^k(B))(t))\leq \left[\varepsilon^k+C_k^1\varepsilon^{k-1}(Ht)+\frac{C_k^2\varepsilon^{k-2}(Ht)^2}{2!} +\cdot\cdot\cdot+\frac{(Ht)^k}{k!}\right]\alpha(B),

则对所有的t\in J,我们有

\begin{eqnarray*} \alpha((\widetilde{{\cal Q}}^{k+1}(B))(t)) &=&\alpha\bigg(U_\beta(t, 0)(u_0-g(\widetilde{{\cal Q}}^k(B)))\\ &&+\int_{0}^{t}U_\beta(t, s)f(s, (\overline{Co}\widetilde{{\cal Q}}^k(B))(s), (G\overline{Co}\widetilde{{\cal Q}}^k(B))(s), (S\overline{Co}\\ \widetilde{{\cal Q}}^k(B))(s)){\rm d}s\bigg)\\ &\leq& M^*\int_{0}^{t}[L_1(s)+T_0k_0L_2(s)]\alpha ((\widetilde{{\cal Q}}^k(B))(s)){\rm d}s\\ &=& \int_{0}^{t}L(s)\alpha((\widetilde{{\cal Q}}^k(B))(s)){\rm d}s\\& \leq& \bigg(\int_{0}^{t}[|L(s)-\varphi(s)|+|\varphi(s)|]{\rm d}s\alpha((\widetilde{{\cal Q}}^k(B))\bigg)\\ &=&\bigg[\varepsilon^{k+1}+C_{k+1}^1\varepsilon^k(Ht)+\frac{C_{k+1}^2\varepsilon^{k-1}(Ht)^2}{2!} +\cdot\cdot\cdot+\frac{(Ht)^{k+1}}{(k+1)!}\bigg]\alpha(B).\end{eqnarray*}

因而,对任意正整数nt\in J,可得

\alpha((\widetilde{{\cal Q}}^n(B))(t))\leq \left[\varepsilon^n+C_n^1\varepsilon^{n-1}(Ht)+\frac{C_n^2\varepsilon^{n-2}(Ht)^2}{2!} +\cdot\cdot\cdot+\frac{(Ht)^n}{n!}\right]\alpha(B).

根据(3.5)式,对任意正整数n,我们有

\alpha(\widetilde{{\cal Q}}^n(B))\leq \left[\varepsilon^n+C_n^1\varepsilon^{n-1}(HT_0)+\frac{C_n^2\varepsilon^{n-2}(HT_0)^2}{2!} +\cdot\cdot\cdot+\frac{(HT_0)^n}{n!}\right]\alpha(B).
(3.9)

由引理2.4可得

\alpha(\widetilde{{\cal Q}}^n(B))\leq \left[O\left(\frac{\lambda^n}{\sqrt{n}}\right)+o\left(\frac{1}{n^s}\right)\right]\alpha(B)=o\left(\frac{1}{n^s}\right)\alpha(B), \ \ \ n\rightarrow\infty,

这里0 < \lambda < 1, s>1.因此,存在常数0\leq k < 1和自然数n_0使得(3.6)式成立.由引理2.1可知算子{\cal Q} \overline{Co}{\cal Q}(B_{R^*})里至少存在一个不动点,从而问题(1.1)-(1.2)在 \overline{Co}{\cal Q}(B_{R^*})\subset C[J, E]里至少存在一个温和解u^*(t).

4 应用

考虑如下的混合型分数阶偏微分方程

\left\{\begin{array}{ll} {}^cD_t^\beta z(x, t)=t^2\frac{\partial^2}{\partial x^2}z(x, t) +\frac{t}{1+t^2}z(x, t)+\frac{1}{1+t^2}\int_0^ta(s)z(x, s){\rm d}s\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{1}{1+t^2}\left(\int_0^{T_0}b(s)z(x, s){\rm d}s\right)^{\frac{1}{2}}, 0<t\leq T_0, 0< x<\pi, \\ z(0, t)=z(\pi, t)=0, \\ z(x, 0)+\frac{1}{2}z=z_0(x), \ \ 0< x<\pi, \end{array} \right.
(4.1)

这里0 < \beta\leq 1, z_0(x)\in E=L^2[0, \pi],定义算子A(t)(z)=t^2\frac{\partial^2z}{\partial x^2}, A:D(A)\subset E\rightarrow E,其中

D(A)=\{z\in E:z''\in E, \ \ z(0)=z(\pi)=0\}.

显然算子A(t)E上生成一个\beta -预解族.

z(t)(\cdot)=z(t, x), \ \ g(z)=\frac{1}{2}z,

f(t, x, y, \kappa)=\frac{t}{1+t^2}x+\frac{1}{1+t^2}y+\frac{1}{1+t^2}\kappa^{\frac{1}{2}},

则方程(4.1)变为(1.1)-(1.2)的形式.如果满足定理3.1和3.2的假设条件,则问题(4.1)存在一个温和解.

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