在本节中,我们主要讨论三个分数阶超扩散问题弱解的存在唯一性(1.1), (1.2)、和(1.3)在标准巴纳赫空间\(C([0,T];H_{0}^{1}(\Omega))\)首先,我们需要以下几个引理。
引理3.1
[11]
让
\(阿尔法>0).然后
$$I^{\alpha}{^{C}}D^{\alpha}u(t)=u(t$$
对一些人来说
\(R\中的c_{i}\),\(i=0,1,2,\ldot,n-1,n=[\alpha]+1).
引理3.2
让
\(y\在C[0,T]\中),\(T>0),\(1<α<2).然后是问题
$$\begin{aligned}D^{\alpha}u(t)=y(t),\quad t\in[0,t],\end{alinged}$$
(3.1)
有独特的解决方案
$$u(t)=u(0)+u'(0)t+\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}y(s)\,ds$$
引理3.3
请参见[15]
真的
\(θ>0)
和
\(在C_{0}^{\infty}(R)中为v\),我们有
$$\开始{aligned}\bigl({}_{-\infty}^{R} D类^{\θ}_{x} v(v)(x) ,{}{x}^{R} D类^{\theta}_{\infty}v(x)\bigr)_{L^{2}(R)}=&\cos(\pi\theta)\bigl\Vert{}_{-\infty}^{R} D类^{\θ}_{x} v(v)(x) \较大\垂直^{2}_{L^{2}(R)},\\bigl({}_{-\infty}^{R} D类^{\θ}_{x} v(v)(x) ,{}{x}^{R} D类^{\theta}_{\infty}v(x)\bigr)_{L^{2}(R)}=&\cos(\pi\theta)\bigl\Vert{}_{x}^{R} D类^{\theta}_{\infty}v(x)\bigr\Vert^{2}_{L ^{2}(R)}。\结束{对齐}$$
(3.2)
引理3.4
[36],厚度。4.1.2
让
秒
和
t吨
是实数,这样
\(宋体).然后
$$H^{s}\bigl(R^{n}\bigr)\hookrightarrow H^{t}\biggl(R^}\biger)$$
引理3.5
[37]
让
\((H^{s}(R^{n}))^{\素数}\)
表示的对偶空间
\(H^{s}(R^{n})\).然后
$$\bigl(H^{s}\bigle(R^{n}\biger)\bigr)$$
在代数和拓扑的意义上,对所有人来说
\(v\在H^{-s}(R^{n})中\),如果
\(在(H^{s}(R^{n}))^{prime}\中)
是连续线性函数运算符,那么我们有
$$\|v\|_{H^{-s}(R^{n})}=\|l\|_}(H^{s}(R^{n{))^{\prime}}\leq C$$
哪里
C类
是一个正常数.
3.1时空分数阶超扩散方程弱解的存在唯一性
按引理3.2问题(1.1)可以在分数阶积分算子下化简为等价积分方程\(I^{\alpha}\)如以下问题所示:
$$\left\{\textstyle\begin{array}{@{}l@{\quad}l}-\phi(x)-\psi(x)t+u(x,t)=\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}\epsilon(-(-\Delta)^{\ beta/2})u(x,t。\结束{array}\displaystyle\right$$
(3.3)
函数积分方程描述了自然科学、数学物理、力学和人口动力学各个领域的许多物理现象[38,39]. 在泛函分析、拓扑学和不动点理论工具的帮助下,积分方程理论正在迅速发展(参见,例如[40——42]),它反过来又是数学其他分支的有用工具,例如微分方程[43]. 现在,我们定义
$$\left\{\textstyle\begin{array}{@{}l@{\quad}l}\Phi(u)=\Phi(x)+\psi(x)t+\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}\epsilon(-(-\Delta)^{\ beta/2})u(x,t。\结束{array}\displaystyle\right$$
(3.4)
定义3.1
我们打电话给\(u在C([0,T];H_{0}^{1}(\Omega))中)时空分数阶超扩散方程的弱解(1.1)如果\(\int_{\Omega}(u-\Phi(u))v\,dx=0\)为所有人\(在[0,t]\中)以及每个\(v\在H_{0}^{1}(\Omega)中\)也就是说,
$$\int_{\Omega}uv\,dx=\int_}\Omega}\biggl[\phi(x)+\psi(x)t+\frac{1}{\Gamma(\alpha)}\int__{0}^{t}(t-s)^{\alpha-1}\epsilon\bigl(-(-\Delta)^{\ beta/2}\bigr)u(x,s)\,ds\biggr]v\,d x$$
我们知道\(B^{\alpha/2,\beta/2}(I\times\Omega)\子集C([0,T];H_{0}^{1}(\Omeca))\)根据的定义(2.15),并结合定理的结果2.5,我们有以下定理。
定理3.6
如果
\(1<α<2),\(1<\beta\leq2),和
\(u{t}(x,0)=0\),然后是操作员
\(\Phi(u):B^{\alpha/2,\beta/2}(I\ times\Omega)\rightarrow B^{\ alpha/2、\beta/2}(I \ times\O mega)\)
是完全连续的.
证明
放置\(F(u)=\epsilon(-(-\Delta)^{\beta/2})u(x,t)\)。那么
$$\Phi(u)=\Phi(x)+\psi(x)t+\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}F(u)\,ds,\quad\对于所有u,v\在B^{\alpha/2,\beta/2}(I\ times\Omega)中$$
对于每个\(v\在H_{0}^{1}(\Omega)中\)根据定义中的Riesz分数导数,使用分部积分2.4和引理3.3,自\(1<\beta\leq2),通过引理3.4结合实阶Sobolev嵌入定理,我们得到\(H_{0}^{1}(\Omega)\hookrightarrow H_{0}^{\beta/2}(\ Omega因此,\(\|u\|_{L^{2}}(\Omega)\leq\|u\ |_{H_{0}^{beta/2}(\ Omega。我们进一步表示\(\|u\|_{H_{0}^{1}(\Omega)}\)和\(\|u\|_{H_{0}^{-1}(\Omega)}\)通过\(\|u\|_{H_{0}^{1}}\)和\(\|u\|_{H_{0}^{-1}}\)分别是。因此,通过Cauchy-Schwarz和Hölder不等式,标准的Sobolev嵌入定理2.5、和引理3.5,自\(1<α<2),对于任何\(v\在H_{0}^{1}(\Omega)中\)令人满意的\(\|v\|_{H_{0}^{1}(\Omega)}\leq1\),我们获得
$$\begin{aligned}\bigl\vert\bigl\langle F(u),v\bigr\rangle\bigr\ vert=&\biggl\vert\int_{\Omega}\epsilon\bigl(-(-\Delta)^{\beta/2}\bigr)u(x,t)v(x,t)\,ds\biggr\vert\=&\bigl\vert_bigl x,t)\biger)\bigr\vert=\bigl\vert\bigl(\epsilon^{R} D类^{\beta}_{|x|}u(x,t),v(x,t)\biger)\bigr\vert\\=&\bigl\vert-\epsilon C_{\beta}\bigl[\bigl({}^{右}_{a} D类^{\beta/2}_{x} u个, {}^{右}_{x} D类^{\beta/2}_{b} v(v)\bigr)_{L^{2}(U)}+\bigl({}^{右}_{x} D类^{\beta/2}_{b} u个, {}^{右}_{a} D类^{\beta/2}_{x} v(v)\bigr)_{L^{2}(U)}\bigr]\bigr\vert\\\leq&\epsilon C_{beta}\bigl[\bigl\vert{}^{右}_{a} D类^{\beta/2}_{x} u个\bigr\Vert_{L^{2}(U)}\bigl\Vert{}^{右}_{x} D类^{\beta/2}_{b} v(v)\bigr\Vert_{L^{2}(U)}+\bigl\Vert{}^{右}_{x} D类^{\beta/2}_{b} u个\bigr\Vert_{L^{2}(U)}\bigl\Vert{}^{右}_{a} D类^{\beta/2}_{x} v(v)\bigr\Vert_{L^{2}(U)}\bigr]\\leq&2\epsilon C_{beta}\Vert U\Vert_{L^}(I,H^{beta/2}(\Omega))}\Vertv\Vert{H^{alpha/2}_{0},\beta/2}(I\times\Omega)}\Vert v\Vert_{H^{1}_{0}(I\times\Omega)}\\leq&2\epsilon C_{\beta}\Vert-u\Vert_{B^{\alpha/2,\beta/2}(I \times\ Omega^{1}_{0}(\Omega)}\\leq&2\epsilon C_{\beta}\bigl\Vert\phi(x)\bigr\Vert_{L^{2}(U)}\bigle\Vert t^{-\alpha}\bigr\ Vert_{L ^{q}(U)}=M_{1}。\结束{对齐}$$
在这里\(C_{\β}=\压裂{1}{2\cos(\压裂{\pi\β}{2})}\)、和\(q=\压裂{2}{\α},M_{1}\)是一个正常数。因此,
$$\开始{aligned}\bigl\Vert\Phi(u)\bigr\Vert_{H^{-1}}=&\sup_{\|v\|H^{1}_{0}\leq1}\bigl\vert\bigl\langle\Phi(u),v\bigr\rangle\bigr\ vert\=&\sup_{\|v\|H^{1}_{0}\leq1}\biggl\vert\bigl\langle\phi(x),v\bigr\rangle+\bigl\angle\psi(x)^{t}(t)_{0}(t-s)^{\alpha-1}\bigl\langle F(u),v\bigr\rangle\,ds\biggr\vert\\leq&\bigl\vert\phi(x)\bigr\vert_^{1}_{0}(\Omega)}+\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\欧米茄)}\|v\|_{H^{1}_{0}(\Omega)}T\\&{}+\bigl\vert\bigl\langle F(u),v\bigr\rangle\bigr\ vert\biggl\vert\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}\,ds\biggr\vert\\\leq&\bigl\vert\phi(x)\bigr\vert_{L^{infty}(\Omega)}\vert v\vert_{H^{1}_{0}(\Omega)}+\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\欧米茄)}\|v\|_{H^{1}_{0}(\Omega)}T\\&{}+2\epsilon C_{\beta}\bigl\Vert\phi(x)\bigr\Vert_{L^{2}(\ Omega^{t}(t)_{0}(t-s)^{\alpha-1}\,ds\biggr\vert\\\leq&\bigl\vert\phi(x)\bigr\vert_{L^{infty}(\Omega)}+\bigl\ vert\psi(x)\ bigr\vert_{L_{infty}(\ Omega(x)\bigr\vert_{L^{infty}(\Omega)}+\bigl\vert\psi.\end{对齐}$$
因此,\(\Phi(u)\)是有界运算符。
另一方面,给定\(epsilon>0\),套
$$\delta=\min\biggl\{\frac{\epsilon}{4\|\psi(x)\|_{L^{\infty}(|\Omega)}},\frac}{1}{2}\biggl(\ frac{\ epsilon\Gamma(\alpha)}{2M_{1}}\bigcr)^{\frac{1}{\alpha}\bigr\}$$
然后,每\(v\在H_{0}^{1}(\Omega)中\)和\([0,t]\中的t_{1},t_{2}\)这样的话\(0<t_{2} -吨_{1} <\增量\),我们有
$$\bigl\Vert\Phi u(t_{2})-\Phi u$$
也就是说,\(\Phi(u)\)是等连续的。事实上,
$$\开始{对齐}&\bigl\Vert\Phi u(t_{2})-\Phi u sup_{\|v\|{H_{0}^{1}}\leq1}\biggl|\bigl\langle\psi(x),v\bigr\rangle(t_{2} -吨_{1} )+\frac{1}{\Gamma(\alpha)}\int_{0}^{t{2}}(t_{2} -秒)^{\alpha-1}\bigl\langle F(u),v\bigr\rangle\,ds\\&\qquad{}-\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{1}}(t_{1} -秒)^{\alpha-1}\bigl\langle F(u),v\bigr\rangle\,ds\biggr|\\&\quad\leq\bigl\ Vert\psi(x)\bigr\ Vert_{L^{infty}(\Omega)}\|v\|_{H_{0}^{1}}|t_{2} -吨_{1} |+\bigl\vert\bigl\langle F(u),v\bigr\rangle\bigr\ vert\biggl\vert\frac{1}{\Gamma(\alpha)}\int_{t_{1}}^{t_{2}}(t_{2} -秒)^{\alpha-1}\,ds\biggr\vert\\&\qquad{}+\bigl\vert\bigl\langle F(u),v\bigr\rangle\bigr\ vert\biggl\vert\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{1}}\bigl((t_{2} -秒)^{\alpha-1}-(t_{1} -秒)^{\alpha-1}\biger)\,ds\biggr\vert\\&\quad\leq\bigl\vert\psi(x)\bigr\vert_{L^{\infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{1}}{\Gamma(\alpha)}\int_{t_{1}}^{t_{2}}\bigl\vert(t_{2} -秒)^{\alpha-1}\bigr\vert\,ds\\&\qquad{}+\frac{M_{1}}{\Gamma(\alpha)}\int_{0}^{t_{1}{\bigl\vert(t_{2} -秒)^{\α-1}-(t_{1} -秒)^{\alpha-1}\bigr\vert\,ds\\&\quad\leq\bigl\vert\psi(x)\bigr\ vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{1}}{\alpha\Gamma(\alpha)}t_{2}^{\alpha}-\frac{M_{1}}{\alpha\Gamma(\alpha)}t_{1}^{\alpha}\&&quad\leq\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{1}}{\alpha\Gamma(\alpha)}\bigl(t_{2}^{\alpha}-t_{1}^{\alpha}\bigr)。\结束{对齐}$$
在下文中,我们将证据分为两种情况。
案例1:\(δt_{1}<t_{2}<t\).自\(1<α<2),我们得到
$$\begin{aligned}和\bigl\Vert\Phi u(t_{2})-\Phi u q\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{1}}{\alpha\Gamma(\alpha)}\bigl(t_{2}^{\alfa}-t_{1{^{\alpha}\bigr)\\&\quad=\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{1}}{\alpha\Gamma(\alpha)}\alpha-t^{\alfa-1}(t_{2} -吨_{1} )\\&\quad\leq\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{1}}{\Gamma(\alpha)}\delta^{\alpha}\\&&\dquad<\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}\delta+\frac{M_{1}}{\Gamma(\alpha)}\delta^{\alpha}\\&&\dquad<\frac{1}{2}\times\frac{\epsilon}{2}+\biggl(\frac{1}{2}\biggr)^{\alpha}\frac ilon}{2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon。\结束{对齐}$$
在这里\(t{1}<t<t{2}\),我们应用了中值定理\(t{2}^{\alpha}-t{1}^{\ alpha}=\alphat^{\alpha-1}(t_{2} -吨_{1})\).
案例2:\(0\leq t_{1}<\delta,t_{2}<2\delta\).我们有
$$\begin{aligned}\bigl\Vert\Phi u(t_{2})-\Phi u \psi(x)\bigr\Vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{1}}{\alpha\Gamma(\alpha)}\bigl(t_{2}^{\alalpha}-t_{1{^{\alpha}\bigr)\\leq&\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}\delta+\frac{M_1}{\阿尔pha(\alpha)垂直\psi(x)\bigr\Vert_{L^{infty}(\Omega)}\delta+\frac{M_{1}}{\Gamma(\alpha+1)}(2\delta)^{\alpha}\\<&\frac}\epsilon}{2}+\frac{\epsilon}{2}=\epsilen。\结束{对齐}$$
因此,Φ是等度连续且一致有界的。Arzelá-Ascoli紧性定理暗示Φ是紧的\(B^{\alpha/2,β/2}(I\times\Omega)\),所以操作员\(\Phi(u):B^{\alpha/2,\beta/2}(I\ times\Omega)\rightarrow B^{\ alpha/2、\beta/2}(I \ times\O mega)\)是完全连续的。这就完成了定理的证明3.6. □
自\(B^{\alpha/2,\beta/2}(I\times\Omega)\子集C([0,T];H_{0}^{1}(\Omeca))\),结合定理的结果3.6利用Schauder不动点定理(引理2.4),我们得到了时空分数阶超扩散问题(1.1)有一个独特的弱解决方案\(u在C([0,T];H_{0}^{1}(\Omega))中).
3.2时空分数阶非线性超扩散方程弱解的存在唯一性
在这一部分中,我们采用了与第节中相同的Banach空间及其范数和性质以及相同的思想3.1.我们还需要以下引理。
引理3.7
假设存在一个常数
\(L>0\)
这样的话
$$\bigl\vert f(u)-f(v)\bigr\vert\leq L|u-v|,\quad\forall t\in I,u,v\in\Omega$$
然后
\(f(u)\)
以为界Ω,因此,存在一个正常数
N个
这样的话
$$\bigl\vert f(u)\bigr\vert\leq N,\quad\forall t\in I,u\in \Omega$$
与第节类似3.1,按定义2.1和2.3我们有这个
$$I^{\alpha}D^{\alpha}u=\frac{1}{\Gamma(\alpha)\Gamma(1-\{\alfa\})}\int_{0}^{t}\int_0}^{\tau}(t-\tau)^{\α-1}(\tau-s)^{-\{\α\}}u^{([\alpha]+1)}(s)\,ds \,D\tau$$
拿\(τ=(t-s)\lambda+s),我们有
$$开始{对齐}[b]I^{\alpha}D^{\alpha}u&=\frac{1}{\Gamma[\alpha]+1)},ds\\&=\frac{1}{\伽马(\alpha)\Gamma(1-\{\alpha\})}\int_{0}^{t} B类\bigl(\alpha,1-\{\alpha\}\bigr)(t-s)^{[\alpha]}u^{([\alfa]+1)}(s)\,ds\\&=\frac{1}{[\alpha]!}\int_{0}^{t}(t-s$$
通过分块反复积分,我们得到
$$开始{对齐}I^{\alpha}D^{\alpha}u=&-\sum_{k=1}^{[\alpha]}\frac{u^{([\alfa]-k)}(0)}{([\alpha]-k)!}t^{[\ alpha]-k}+\int_{0}^{t} u个'(s)\,ds,\\=&-\sum_{k=0}^{[\alpha]}\frac{u^{([\alfa]-k)}(0)}{([\alpha]-k)!}t^{[\ alpha]-k}+u(t)。\结束{对齐}$$
什么时候?\((1,2)中的α),\(I^{\alpha}D^{\alpha}u=-u(0)-u^{'}(0)t+u(t)\).与引理组合3.2,我们可以减少问题(1.2)分数阶积分算子下的等价积分方程\(I^{\alpha}\)如以下问题所示:
$$\left\{\textstyle\begin{array}{@{}l@{\quad}l}-\phi(x)-\psi(x)t+u(x,t)\\quad=\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}(-\frac{\partial}{\partitlex}f(u)+\epsilon(-(-\Delta)^{\ beta/2}))u。\结束{array}\displaystyle\right$$
(3.5)
现在,我们定义
$$\left\{\textstyle\begin{array}{@{}l@{\quad}l}\Phi(u)=\Phi(x)+\psi(x)t\\\hphantom{\Phi(u)=}{}+\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}(-\frac{\partial}{\partitlex}f(u)+\epsilon(-(-\Delta)^{\ beta/2}))u。\结束{array}\displaystyle\right$$
(3.6)
定义3.2
我们打电话给\(u在C([0,T];H_{0}^{1}(\Omega))中)时空分数阶非线性超扩散方程的弱解(1.2)如果\(\int_{\Omega}(u-\Phi(u))v\,dx=0\),\(对于[0,t]\中的所有t\)对于每个\(v\在H_{0}^{1}(\Omega)中\)也就是说,
$$\int_{\Omega}uv\,dx=\int_}\Omega}\biggl[\phi(x)+\psi(x)t+\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}\bigl(\epsilon\bigle(-(-\Delta)^{\ beta/2}\biger)\bigr)u(x,s)-\frac{\部分f(u)}{\部分x})\,ds\biggr]v\,dx$$
根据的定义(2.15)我们有\(B^{\alpha/2,\beta/2}(I\times\Omega)\子集C([0,T];H_{0}^{1}(\Omeca))\),并结合定理的结果2.5再次,我们得到以下定理。
定理3.8
让
\(1<α<2),\(1<\beta\leq2),和
\(u{t}(x,0)=0\).然后是操作员
\(\Phi(u):B^{\alpha/2,\beta/2}(I\ times\Omega)\rightarrow B^{\ alpha/2、\beta/2}(I \ times\O mega)\)
是完全连续的.
证明
放置\(F(u)=\epsilon(-(-\Delta)^{\beta/2})u(x,s)+\frac{\partial}{\paratilx}F(u)\).我们可以重写
$$\Phi(u)=\Phi(x)+\psi(x)t+\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}F(u)\,ds,\quad\对于B^{\alpha/2,\beta/2}(I\times\Omega)中的所有u$$
对于任何\(v \in H)^{1}_{0}(\Omega)\)令人满意的\(\|v\|_{H^{1}_{0}(\Omega)}\leq1\),我们有
$$\begin{aligned}\bigl|\bigl\langle F(u),v\bigr\rangle\bigr|=&\biggl|\biggl \langle\epsilon\bigl(-(-\Delta)^{\beta/2}\bigr)u(x,t)+\frac{\partial}{\partic x}F(u)^{R} D类^{\beta}_{|x|}u(x,t)+\frac{\partial}{\particalx}f(u),v\biggr\rangle\biggr|\\leq&\bigl|\bigl\langle-\epsilon C_{\beta}\bigl(^{右}_{a} D类^{\beta}_{x}+^{右}_{x} D类^{\beta}_{b}\bigr),v\bigr\rangle\bigr|+\biggl|\biggl \langle\frac{\partial f(u)}{\parial x},v\biggr \rangle\ biggr|\\leq&\bigl|-\epsilon C_{\beta}\bigl[\bigl(^{右}_{a} D类^{\beta/2}_{x} u个, ^{右}_{x} D类^{\beta/2}_{b} v(v)\biger)_{L^{2}(U)}+\bigl(^{右}_{x} D类^{\beta/2}_{b} u个, ^{右}_{a} D类^{\beta/2}_{x} v(v)\bigr)_{L^{2}(U)}\bigr]\bigr|+\bigl|f(U)\cdot\Delta v\bigr|\\leq&&epsilon C_{\beta}\bigl[\bigl\|{}^{右}_{a} D类^{\beta/2}_{x} u个\bigr\|_{L^{2}(U)}\bigl\|{}^{右}_{x} D类^{\beta/2}_{b} v(v)\bigr\|_{L^{2}(U)}\\&{}+\bigl\|{}^{右}_{x} D类^{\beta/2}_{b} u个\bigr\|_{L^{2}(U)}\bigl\|{}^{右}_{a} D类^{\beta/2}_{x} v(v)\bigr\|_{L^{2}(U)}\bigr]+\bigl|f(U)\bigr|\cdot\|\Delta v \|{L^}(\Omega)}\\leq&\epsilon C_{beta}\|U\|{L^{2](I,H^{beta/2}(\ Omega{β}(I,H^{β/2}(\Omega))}^{1}_{0}(\Omega)}\\leq&2\epsilon C_{\beta}\|u\|_{\beta^{\alpha/2,\beta/2}(I\times\Omeca))}^{1}_{0}(I\times\Omega)}+N\|v\|_{H^{1}_{0}(\Omega)}\\leq&2\epsilon C_{\beta}\|u\|_{\beta^{\alpha/2,\beta/2}(I\times\Omeca))}^{1}_{0}(\Omega)}+N\|v\|_{H^{1}_{0}(\Omega)}\\leq&2\epsilon C_{\beta}\bigl\|\phi(x)\bigr\|_{L^{2}$$
哪里\(q=\压裂{2}{\α}\)、和\(M_{2}\)是一个正常数。
因此,
$$\begin{aligned}\bigl\Vert\Phi(u)\bigr\Vert=&&sup_{\Vert v\Vert_{H^{1}_{0}\leq1}}\bigl|\bigl\langle\Phi(u),v\bigr\rangle\bigr|\\=&\sup_{\Vert v\Vert_{H^{1}_{0}\leq1}}\biggl|\bigl\langle\phi(u),v\bigr\rangle+\bigl\ langle\psi(x),v\ bigr\Ranglet+\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}\bigl\langle F(u),v\bigr\rangle\,ds\biggr|\\leq&\bigl\ Vert\phi(x)\bigr\ Vert_{L^{infty}(\Omega)}\Vert v\Vert_{H^{1}_{0}(\Omega)}+\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\欧米茄)}\Vert v\Vert_{H^{1}_{0}(\Omega)}T\\&{}+\bigl|\bigl\langle F(u),v\bigr\rangle\bigr|\biggl|\frac{1}{\Gamma(\alpha)}\int^{t}(t)_{0}(t-s)^{\alpha-1}\,ds\biggr|\\leq&\bigl\Vert\phi^{t}(t)_{0}(t-s)^{\alpha-1}\,ds\biggr|\\leq&\bigl\Vert\phi \bigr\Vert_{L^{infty}(\Omega)}+\bigl\Vert\psi(x)\bigr\ Vert_{L ^{inffy}。\结束{对齐}$$
因此,\(\Phi(u)\)有界。
另一方面,给定\(epsilon>0\),套
$$\delta=\min\biggl\{\frac{\epsilon}{4\|\psi(x)\|_{L^{\infty}(|\Omega)}},\frac}{1}{2}\biggl(\ frac{\ epsilon\Gamma(\alpha)}{2M{2}\ biggr)^{\frac{1}}{\alpha}\bigr\}$$
然后,每\(v\在H_{0}^{1}(\Omega)中\)以及所有\([0,t]\中的t_{1},t_{2}\)这样的话\(0<t_{2} -吨_{1} <\增量\),我们获得
$$\bigl\|\Phi u(t_{2})-\Phi u(t_{1}$$
也就是说,\(\Phi(u)\)是等连续的。事实上,
$$\begin{aligned}\bigl\|\Phi u(t_{2})-\Phi u(t_{1})\bigr\|_{H^{-1}}=&\sup_{\|v\|{H_0}^{1}}\leq1}\bigl|\bigl\ langle\Phi u[t_{2{)-\ Phi u H_{0}^{1}}\leq1}\biggl|\bigl\langle\psi(x),v\bigr\rangle(t_{2} -吨_{1} )+\frac{1}{\Gamma(\alpha)}\int_{0}^{t{2}}(t_{2} -秒)^{\alpha-1}\bigl\langle F(u),v\bigr\rangle\,ds\\&{}-\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{1}}(t_{1} -秒)^{\alpha-1}\bigl\langle F(u),v\bigr\rangle\,ds\biggr|\\leq&\bigl\|\psi(x)\bigr\|_{L^{infty}(\Omega)}\|v\|_{H_{0}^{1}}|t_{2} -吨_{1} |+\bigl|\bigl\langle F(u),v\bigr\rangle\bigr|\biggl|\frac{1}{\Gamma(\alpha)}\int_{t_{1}}^{t_{2}}(t_{2} -秒)^{\alpha-1}\,ds\biggr|\\&{}+\bigl|\bigl\langle F(u),v\bigr\rangle\bigr|\biggl|\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{1}}\bigl((t_{2} -秒)^{\alpha-1}-(t_{1} -秒)^{\alpha-1}\biger)\,ds\biggr|\\leq&\bigl\|\psi(x)\bigr\|_{L^{\infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{2}}{\Gamma(\alpha)}\int_{t_{1}}^{t_{2}}\bigl|(t_{2} -秒)^{\alpha-1}\bigr|\,ds\\&{}+\frac{M_{2}}{\Gamma(\alpha)}\int_{0}^{t_{1}}\bigl|(t_{2} -秒)^{\α-1}-(t_{1} -秒)^{\alpha-1}\bigr|\,ds\\leq&\bigl\|\psi(x)\bigr\|_{L^{\infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{2}}{\alpha\Gamma(\alpha_{2} -吨_{1} |+\frac{M_{2}}{\alpha\Gamma(\alpha)}\bigl(t_{2}^{alpha}-t_{1}^{alpha}\bigr)。\结束{对齐}$$
在下文中,我们将证据分为两种情况。
案例1:\(δt_{1}<t_{2}<t\).自\(1<α<2),我们得到
$$\begin{aligned}\bigl\Vert\Phi u(t_{2})-\Phi u \psi(x)\bigr\Vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{2}}{\alpha\Gamma(\alpha)}\bigl(t_{2}^{\alfa}-t_{1}^{\ alpha}\bigr)\\=&\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{2}}{\alpha\Gamma(\alpha)}\alpha-t^{\alfa-1}(t_{2} -吨_{1} )\\leq&\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{2}}{\Gamma(\alpha)}\delta^{\alpha}\\<&\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}\delta+\frac{M_2}}{\ Gamma 2}\biggr)^{\alpha}\frac{\epsilon}{2}\\<&\ frac{\ epsilon{2}+\ frac}\epsilen}{2{=\ epsilen。\结束{对齐}$$
在这里\(t{1}<t<t{2}\),我们应用了中值定理\(t{2}^{\alpha}-t{1}^{\ alpha}=\alphat^{\alpha-1}(t_{2} -吨_{1})\).
案例2:\(0\leq t_{1}<\delta,t_{2}<2\delta\).我们有
$$\begin{aligned}\bigl\Vert\Phi u(t_{2})-\Phi u \psi(x)\bigr\Vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{2}}{\alpha\Gamma(\alpha)}\bigl(t_{2}^{\alalpha}-t_{1}^{\ alpha}\bigr)\\leq&\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}\delta+\frac{M_2}{垂直\psi(x)\bigr\Vert_{L^{infty}(\Omega)}\delta+\frac{M_2}}{\Gamma(\alpha+1)}(2\delta)^{\alpha}\\<&\frac}\epsilon}{2}+\frac{\epsilon}{2}=\epsilen。\结束{对齐}$$
因此,Φ是等度连续且一致有界的。根据Arzelá-Ascoli定理,Φ在空间上是紧的\(B^{\alpha/2,β/2}(I\times\Omega)\),所以操作员\(\Phi(u):B^{\alpha/2,\beta/2}(I\ times\Omega)\rightarrow B^{\ alpha/2、\beta/2}(I \ times\O mega)\)是完全连续的。这就完成了定理的证明3.8. □
自\(B^{\alpha/2,\beta/2}(I\times\Omega)\子集C([0,T];H_{0}^{1}(\Omeca))\),结合定理的结果3.8,通过Schauder不动点定理(引理2.4)我们得到了时空分数阶非线性超扩散方程(1.2)有一个独特的弱解决方案\(u在C([0,T];H_{0}^{1}(\Omega))中).
3.3时空分数阶漂移超扩散方程弱解的存在唯一性
在这一部分中,我们考虑了时空分数阶漂移超扩散方程弱解的存在唯一性。类似地,我们需要一些函数空间及其性质来分析多维时空分数漂移超扩散方程。通过引理2.2和2.3我们将以下空间重新定义如下[15,16]:
$$\begin{aligned}B^{\alpha/2,\gamma/2}(I\times\Omega):=H^{\alpha/2}\bigl(I,L^{2}(\Omeca)\bigr)\cap L^{2\bigl$$
(3.7)
符合规范
$$\开始{aligned}\|v\|_{L^{\alpha/2,\gamma/2}}(I\times\Omega):=&\bigl(\|v\ |^{2}_{H^{\alpha/2}(I,L^{2}(\Omega))}+\|v\|^{2}_{L^{2}(I,H_{0}^{\gamma/2}(\Omega))}\biger)^{\frac{1}{2}},对于B^{\alpha/2中的所有v,\quad\,\gamma/2}(I\ times\Omega,\end{aligned}$$
(3.8)
哪里
$$H^{\alpha/2}\bigl(I,L^{2}(\Omega)\biger$$
被赋予了规范
$$\|v\|_{H^{\alpha/2}(I,L^{2}(\Omega))}:=\bigl\|\bigl\垂直v(\cdot,t)\bigr\Vert_{L^{2](\O mega)}\bigr\ |_{H^{\ alpha/2{(I)}$$
(3.9)
显然,\(B^{\alpha/2,\gamma/2}(I\次\Omega)\)是标准Banach空间的闭有界凸子空间(子集)\(C([0,T];H_{0}^{1}(\Omega))\)类似地,我们回忆起中定理3.1的相同论点[15]得到以下有用的不等式。
定理3.9
假设
\(1<α<2),\(0<\gamma\leq1\),和
\(θ{t}(x,0)=0).然后,系统(1.3)在空间中有唯一的弱解
\(B^{\alpha/2,\gamma/2}(I\次\Omega)\).此外,
$$\begin{aligned}\|\theta\|_{B^{\alpha/2,\gamma/2}(I\times\Omega)}\leq\bigl\Vert\theta(x,0)\bigr\Vert_{L^{2}(\Omeca)}\bigl\ Vert-t^{-\alpha}\bigr\ Vert_{L^{q}(\ Omega)}\bigl\Vertt^{-\alpha}\bigr\Vert_{L^{q}(\Omega)},\quad q=\frac{2}{\alpha}。\结束{对齐}$$
(3.10)
然后通过引理3.2问题(1.3)可以在分数阶积分算子下化简为等价积分方程\(I^{\alpha}\)如以下问题所示:
$$\left\{\textstyle\begin{array}{@{}l@{quad}l}-\phi(x)-\psi(x)t+\theta(x,t)\\quad=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}(-\operatorname{div}(u\cdot\theta Delta)^{\Gamma}\θ(x,s)),ds&\mbox{in}\Omega_{t},\\theta(x,t)=0&\mbax{on}\partial\Omega{t}。\结束{array}\displaystyle\right$$
(3.11)
现在,我们定义
$$\left\{\textstyle\begin{array}{@{}l@{\quad}l}\Phi(θ)=\Phi(x)+\psi(x)t\\\hphantom{\Phiθ)=}{}+\epsilon\Delta\θ(x,s)-(-\Delta)^{\Gamma}\theta(x,s)),ds&\mbox{in}\Omega_{t},\\theta(x,t)=0&&mbox{on}\部分\ Omega_{T}。\结束{array}\displaystyle\right$$
(3.12)
定义3.3
我们打电话给\(C([0,T];H_{0}^{1}(\Omega))中的θ)时间分数阶漂移超扩散方程的弱解(1.3)如果\(\int_{\Omega}(\theta-\Phi(\theta))v\,dx=0\)为所有人\(在[0,t]\中)以及每个\(v\在H_{0}^{1}(\Omega)中\)也就是说,
$$开始{aligned}\int_{\Omega}\theta v\,dx={}&\int_}\Omega}\biggl[\phi(x)+\psi(x)t+\frac{1}{\Gamma(\alpha)}\int_0}^{t}(t-s)^{\alpha-1}\bigl(-\operatorname{div}\bigle(u\cdot\theta(x,s)\biger)\\&{}+\epsilon\Delta\theta(x,s)-(-\Delta)^{\Gamma}\theta(x,s)\bigr)\,ds\biggr]v\,dx。\结束{对齐}$$
根据的定义(3.7)我们得到\(B^{\alpha/2,\gamma/2}(I\次\Omega)\子集C([0,T];H_{0}^{1}(\Omeca))\),以及定理的结果3.9,我们有以下内容:
定理3.10
如果
\(1<α<2),\(0<\gamma\leq1\),和
\(θ{t}(x,0)=0),然后是操作员
\(\Phi(\theta):B^{\alpha/2,\gamma/2}(I\times\Omega)\rightarrow B^{\ alpha/2
是完全连续的.
证明
放置
$$F(θ)=-\operatorname{div}\bigl(u\cdot\theta(x,t)\bigr)+\epsilon\Delta\theta$$
我们可以重写
$$\Phi(θ)=\Phi(x)+\psi(x)t+\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}F(\theta)\,ds$$
对于每个\(v\在H_{0}^{1}(\Omega)中\)这样的话\(\|v\|_{H_{0}^{1}(\Omega)}\leq 1\),按部分积分,根据定义中的Riesz分数导数2.4和引理3.3,自\(0<伽马<1),通过引理3.4结合实阶Sobolev嵌入定理,我们得到了\(H_{0}^{1}(\Omega)\hookrightarrow H_{0}^{\gamma/2}(\ Omega,因此\(\|\theta\|_{L^{2}}(\Omega)\leq\|\ttheta\|{H_{0}^{gamma/2}(\ Omega。在下面,我们表示\(\|\theta\|_{H_{0}^{1}(\Omega)}\)和\(\|\theta\|_{H_{0}^{-1}(\Omega)}\)通过\(\|\theta\|_{H_{0}^{1}}\)和\(\|\theta\|_{H_{0}^{-1}}\)分别是。因此,通过Cauchy-Schwarz和Hölder不等式以及标准的Sobolev嵌入定理,定理3.9、和引理3.5,自\(1<\alpha<2\),我们获得
$$\begin{aligned}\bigl\vert\bigl\langle F(\theta),v\bigr\rangle\bigr\fort=&\biggl\vert\int_{\Omega}(u\cdot\theta+\epsilon\nabla\theta |u\|_{L^{2}(\Omega)}\biggl(\int_{\Omega}|\theta|^{2{,dx\biggr)^{\frac{1}{2}}+|\epsilon|\biggl(\int_{\Omega}|\nabla\theta|^{2}\,dx\biggr)\theta(x,t),v(x,t)\bigr\vert\\leq&\bigl(\Omega)}\biger)\|\nabla v\|_{L^{2}(欧米茄)}+|\epsilon|\|\nabla\theta\|_{L^{2}(\Omega)}\biger)^{右}_{a} D类^{\gamma}_{x}\θ^{右}_{x} D类^{\伽马}_{b} v(v)\bigr)_{L^{2}(U)}+\bigl({}^{右}_{x} D类^{\gamma}_{b}\theta^{右}_{a} D类^{\伽马}_{x} v(v)bigr)_{L^{2}(U)}\bigr]\bigr\vert\\leq&C_{1}\bigl^{右}_{a} D类^{\gamma}_{x}\theta\bigr\|_{L^{2}(U)}\bigl\|{}^{右}_{x} D类^{\伽马}_{b} v(v)\bigr\|_{L^{2}(U)}+\bigl\Vert{}^{右}_{x} D类^{\gamma}_{b}\theta\bigr\Vert_{L^{2}(U)}\bigl\|{}^{右}_{a} D类^{\伽马}_{x} v(v)\bigr\|_{L^{2}(U)}\bigr]\\leq&C_{1}\bigl(\|U\|{L^}(\Omega)}+\max|\epsilon|\bigr)\|\theta\|_}L^{2](I,H^{\gamma/2}(\ Omega{L^{2}(I,H^{\gamma/2}(\Omega))}}+\max|\epsilon|\biger)\|\theta\|_{B^{\alpha/2,\gamma/2}(I\times\Omega)}^{1}_{0}(I\times\Omega)}\\leq&C_{3}\bigl(\|u\|_{L^{2}(\Omeca)}+\max|\epsilon|\biger)\|\theta\|__{B^{\frac{\alpha}{2},\frac}\gamma}{2{}(I \times\ Omega{\gamma}\|\theta\|{B^{\alpha/2,\gamma/2}(I\times\Omega)}\|v\|{H^{1}_{0}(\Omega)}\\leq&C_{3}\bigl(\|u\|_{L^{2}(\ Omegaφ(x)\bigr\Vert_{L^{2}(\Omega)}\bigl\|t^{-\alpha}\bigr\ |_{L^}(\ Omega(\Omega)}\leq M_{3}。\结束{对齐}$$
在这里,\(C_{1},C_{2},C_{3}\)分别表示最佳Sobolev常数,\(C=\max\{C_{3}(\|u\|_{L^{2}(\欧米茄)}+\max|\epsilon|)+2C_{gamma}\}\)、和\(M_{3}\)是一个正常数。
因此,通过Cauchy-Schwarz不等式,我们得到
$$\begin{aligned}\bigl\Vert\Phi(\theta)\bigr\Vert_{H^{-1}}=&&\sup_{\|v\|_{H_{0}^{1}}}\leq1}\bigl\Vert\bigl\langle\Phi(\theta),v\bigr\rangle\bigr\Vert\\=&&\sup_{\|v\|_{H_{0}^{1}}\leq1}\biggl\Vert\bigl\langle\Phi(x),v\bigr\rangle+\bigl\langle\psi(x),v\bigr\rangle t+\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\bigl\langle F(θ),v\bigr\rangle,ds\biggr\vert\\\leq&\bigl\vert\bigl\langle\phi(x)\theta),v\bigr\rangle\,ds\biggr\vert\\\leq&\bigl\vert\phi(x)\bigr\Vert_{L^{\infty}(\Omega)}\Vert v\Vert_{H_{0}^{1}}+\bigl\Vert\psi(x)\bigr\Vert_{L^}\infty}(\ Omega c{1}{\Gamma(\alpha)}\int_{0}^{T}(T-s)^{\alpha-1}\,ds\biggr\Vert\\\leq&\bigl\Vert\phi(x)\bigr\Vert_{L^{infty}(\Omega)}+\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\Omega)}T+\frac{M_{3}}{\Gamma(\alpha)}\biggl|\int_{0}^{T}(T-s)^{\alpha-1}\,ds\biggr|\\leq&\bigl\ Vert\phi \bigr\Vert_{L^{infty}(\Omega)}T+\frac{M_{3}}{\alpha\Gamma(\alpha)}T^{\alfa}\\leq&\bigl\Vert\phi(x)\bigr\ Vert_{L ^{inffy}(欧米茄)}+\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\Omega)}T+\frac{M_{3}}{alpha\Gamma(\alpha)}T^{alpha}。\结束{对齐}$$
因此,\(\Phi(\theta)\)有界。
另一方面,给定\(epsilon>0\),套
$$\delta=\min\biggl\{\frac{\epsilon}{4\|\psi(x)\|_{L^{\infty}(\Omega)}},\frac}1}{2}\biggl(\frac[\epsilen\Gamma(\alpha)}{2M_{3}}\bigr)^{\frac{1}{\alpha}\bighr}$$
然后,每\(v\在H_{0}^{1}(\Omega)中\)以及所有\([0,t]\中的t_{1},t_{2}\)这样的话\(0<t_{2} -吨_{1} <\增量\),我们有
$$\bigl\Vert\Phi\theta(t_{2})-\Phi\ttheta$$
也就是说,\(\Phi(\theta)\)是等连续的。事实上,
$$\begin{aligned}\bigl\Vert\Phi\theta(t_{2})-\Phi\ttheta=&\sup_{\Vert v\Vert_{H_{0}^{1}}\leq1}\biggl|\bigl\langle\psi(x),v\bigr\rangle(t_{2} -吨_{1} )+\frac{1}{\Gamma(\alpha)}\int_{0}^{t{2}}(t_{2} -秒)^{\alpha-1}\bigl\langle F(θ),v\bigr\rangle\,ds\&{}-\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{1}}(t_{1} -秒)^{\alpha-1}\bigl\langle F(\theta),v\bigr\rangle\,ds\biggr|\\leq&\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}\Vert v\Vert_{H_{0}^{1}}|t_{2} -吨_{1} |\\&{}+\bigl\vert\bigl\langle F(\theta),v\bigr\rangle\bigr\ vert\biggl\vert\frac{1}{\Gamma(\alpha)}\int_{t_{1}}^{t_{2}}(t_{2} -秒)^{\alpha-1}\,ds\biggr\vert\\&{}+\bigl\vert\bigl\langle F(\theta),v\bigr\rangle\bigr\ vert\biggl\vert\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{1}}\bigl((t_{2} -秒)^{\alpha-1}-(t_{1} -秒)^{\alpha-1}\biger)\,ds\biggr\vert\\\leq&\bigl\vert\psi(x)\bigr\vert_{L^{\infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{3}}{\Gamma(\alpha)}\int_{t{1}}^{t{2}}\bigl\vert(t_{2} -秒)^{\alpha-1}\bigr\vert\,ds\\&{}+\frac{M_{3}}{\Gamma(\alpha)}\int_{0}^{t_{1}}\bigl|(t_{2} -秒)^{\α-1}-(t_{1} -秒)^{\alpha-1}\bigr|\,ds\\leq&\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{3}}{\alpha\Gamma(\alpha_{2} -吨_{1} |+\frac{M_{3}}{\alpha\Gamma(\alpha)}\bigl(t_{2}^{alpha}-t_{1}^{alpha}\bigr)。\结束{对齐}$$
在下文中,我们将证据分为两种情况。
案例1:\(δt_{1}<t_{2}<t\).自\(1<α<2),我们得到
$$\begin{aligned}\bigl\Vert\Phi\theta(t_{2})-\Phi\tea(t_{1}\bigl\|\psi(x)\bigr\|_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{3}}{\alpha\Gamma(\alpha)}\bigl(t_{2}^{\alfa}-t_{1}^{\ alpha}\bigr)\\=&\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{3}}{\alpha\Gamma(\alpha)}\alpha-t^{\alfa-1}(t_{2} -吨_{1} )\\leq&\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{3}}{\Gamma(\alpha)}\delta^{\alpha}\\<&\bigl\Vert\psi(x)\bigr\Vert_{L^{\infty}(\Omega)}\delta+\frac{M_3}}{\ Gamma 2}\biggr)^{\alpha}\frac{\epsilon}{2}<\frac{\epsilon}{2}+\frac{\ epsilon{2}=\epsilen。\结束{对齐}$$
在这里\(t{1}<t<t{2}\),我们应用了中值定理\(t{2}^{\alpha}-t{1}^{\ alpha}=\alphat^{\alpha-1}(t_{2} -吨_{1})\).
案例2:\(0\leq t_{1}<\delta,t_{2}<2\delta\).我们有
$$\begin{aligned}\bigl\Vert\Phi\theta(t_{2})-\Phi\tea(t_{1}\bigl\Vert\psi(x)\bigr\Vert_{L^{infty}(\Omega)}|t_{2} -吨_{1} |+\frac{M_{3}}{\alpha\Gamma(\alpha垂直\psi(x)\bigr\Vert_{L^{infty}(\Omega)}\delta+\frac{M_{3}}{\Gamma(\alpha+1)}(2\delta)^{\alpha}\\<&\frac}\epsilon}{2}+\frac{\epsilon}{2}=\epsilen。\结束{对齐}$$
因此,Φ是等度连续且一致有界的。根据Arzelá-Ascoli紧性定理,Φ在空间上是紧的\(B^{\alpha/2,\gamma/2}(I\次\Omega)\),因此操作员\(\Phi(u):B^{\alpha/2,\gamma/2}(I\次\Omega)\rightarrow B^{\ alpha/2、\gamma/2}(I \次\O mega)\)是完全连续的。这就完成了定理的证明3.10. □
自\(B^{\alpha/2,\gamma/2}(I\次\Omega)\子集C([0,T];H_{0}^{1}(\Omeca))\),根据定理3.10,使用Schauder不动点定理(引理2.4),我们得到了多维分数阶漂移超扩散方程(1.3)有一个独特的弱解决方案\(C([0,T];H_{0}^{1}(\Omega))中的θ).