Strichartz 不等式在非线性偏微分方程的研究里扮演了重要角色. 并广泛出现于波方程, 薛定谔方程以及其它一些重要的偏微分方程之中. 关于这方面早期的一些工作, 可参阅^{[13, 17, 24, 28]}等.

在海森堡群上,文献^[3] 研究了和 sub-Laplacian 算子相关的波方程的 Strichartz 不等式,文献^[9] 则研究了海森堡群上的和 full-Laplacian 算子相关的波方程的相应估计. 文献^[26]得到了四元数海森堡群上和 sub-Laplacian算子相关的Strichartz估计.文献^[19] 研究了海森堡型群上的相关于full-Laplacian 算子的波方程的Strichartz 不等式.本文独立地探讨了四元数海森堡群上的相应估计, 在结果上与文献^[19]有些差异.

本文的目标是建立四元数海森堡群上如下波方程的 Cauchy 问题的解的Strichartz不等式 \begin{equation}\left\{ \begin{array}{l} \partial_{s}^{2}u+Lu=f\in L^{1}((0,T),L^{2}({\Bbb H}\times{\Bbb R}^{3})),\\ u|_{s=0}=u_{0}\in \dot{B}_{2,2}^{1}(L),\\ \partial_{s}u|_{s=0}=u_{1}\in L^{2}({\Bbb H}\times{\Bbb R}^{3}), \end{array}\right.\tag{1.1} \end{equation} 其中 ${\Bbb H}\times{\Bbb R}^{3}$ 表示四元数海森堡群,$L$ 表示群上的 full-Laplacian 算子而 $\dot{B}_{2,2}^{1}(L)$ 表示和 full-Laplacian 算子相关的 Besov 空间(它们的详细定义在第二节和第四节中给出).

本文的主要结果包括波方程的色散估计

定理 1.1  令 $\rho\in [\frac{13}{2},\frac{17}{2}]$. 假定 $u_{0}\in\dot{B}_{1,1}^{\rho},u_{1}\in\dot{B}_{1,1}^{\rho-1}$, $v$ 是齐次方程的解,那么存在着常数 $C>0$ 使得 \begin{equation} \| v(s,\cdot)\| _{L^{\infty}({\Bbb H}\times{\Bbb R}^{3})}\leq C|s|^{-\frac{1}{2}}(\| u_{0}\| _{\dot{B}_{1,1}^{\rho}}+ \| u_{1}\| _{\dot{B}_{1,1}^{\rho-1}}).\tag{1.2} \end{equation} 以及Strichartz估计

定理 1.2  令 $u$ 是波方程的Cauchy 问题的解,假定成立着如下条件 \[\left\{ \begin{array}{l} \rho' \in \Big[\frac{13}{2},\frac{17}{2}\Big],\\\ p>2\rho' ,\\\ \frac{1}{q}+\frac{1}{p}\Big(2-\frac{\rho' }{5}\Big)=\frac{2}{5}, \end{array}\right.\] 那么存在着常数 $C$,对任意的 $T>0$,有 \begin{equation}\label{tufy} \| u\| _{L^{p}([0,T],L^{q}({\Bbb H}\times{\Bbb R}^{3}))}\leq C\{\| f\| _{L^{1}([0,T];L^{2} ({\Bbb H}\times{\Bbb R}^{3}))}+E_{0}(u)^{\frac{1}{2}}\}, \tag{1.3} \end{equation} 这里 $E_{0}(u)=\| u_{0}\| _{\dot{H}^{1}({\Bbb H}\times{\Bbb R}^{3})}^{2} +\| u_{1}\| _{L^{2}}^{2}$ ($\dot{H}^1$ 表示 Sobelev 空间,其定义在第四节中给出).

本文的结构安排如下: 在第二节中给出一些基本的概念和结论; 第三节中讨论四元数海森堡群上 full-Laplacian 算子的性质以及相关的 Littlewood-Paley 理论; 第四节中研究与 full-Laplacian 算子相关的 Besov 空间和 Sobelev 空间的性质; 在第五节中, 我们首先讨论了波方程解的存在唯一性,然后给出了色散估计和Strichartz估计的证明.

本节给出文中所用到的一些定义、记号和一些重要的已知结论. 这些内容可以参阅文献^{[4, 7, 11, 16, 29, 30, 31]}等, 其中关于四元数海森堡群上的 sub-Laplacian算子、酉表示和 Fourier 变换的性质可参阅文献^[20].

记 ${\Bbb H}$ 为四元数代数. 作为一类特殊的 Clifford 代数, 它是非交换的、结合的、可除的. 四元数代数的基 $({e_{0},e_{1},e_{2},e_{3}})$ 满足 $$ e_{0}=1,e_{1}^{2}=e_{2}^{2}=e_{3}^{2}=-1, e_{1}e_{2}=-e_{2}e_{1}, e_{1}e_{2}=e_{3}. $$ 令 $a=\sum\limits_{k=0}^{3}a_{k}e_{k}$,$b=\sum\limits_{k=0}^{3}b_{k}e_{k}$ 表示两个四元数. 那么四元数的乘积公式由下式所给出 $$ a b=(a_{0}b_{0}-\Im a\cdot \Im b)+(a_{0}\Im b+b_{0}\Im a+\Im a\times \Im b), $$ 这里的 $\Im a=\sum\limits_{k=1}^{3}a_{k}e_{k}$,$ \Im b=\sum\limits_{k=1}^{3}b_{k}e_{k}$. $a$ 的共轭定义为 $\overline{a}=a_{0}-a_{1}e_{1}-a_{2}e_{2}-a_{3}e_{3}$. $a$ 的模定义为 $|a|^{2} = a\overline{a} = \sum\limits_{k=0}^{3} a_{k}^{2}$.

令 ${\Bbb H}^{n}\times {\Bbb R}^{3}=\{(x,u):x\in {\Bbb H}^{n} ,u\in {\Bbb R}^{3}\}$, 其中 ${\Bbb H}^{n}=\{(x_{1},x_{2},\cdots x_{n}),x_{i} \in {\Bbb H}\}$. 定义乘积 $$(x,u)(y,v) =\bigg (x+y,u+v-2\sum\limits_{j=1}^{n}\Im y_{j}\overline{x_{j}}\bigg),$$ 那么 ${\Bbb H}^{n}\times {\Bbb R}^{3}$ 构成了一个非交换李群, 它叫做四元数海森堡群. 在本文中,为了方便讨论,将只考虑 ${\Bbb H}\times{\Bbb R}^{3}$ 的情形.

${\Bbb H}\times{\Bbb R}^{3}$ 上的李代数由如下的左不变向量场生成 $$ X_{0}=\frac{\partial}{\partial x_{0}}-2x_{1}\frac{\partial}{\partial t_{1}} -2x_{2}\frac{\partial}{\partial t_{2}}-2x_{3}\frac{\partial}{\partial t_{3}}, $$ $$ X_{1}=\frac{\partial}{\partial x_{1}}+2x_{0}\frac{\partial}{\partial t_{1}} -2x_{3}\frac{\partial}{\partial t_{2}}+2x_{2}\frac{\partial}{\partial t_{3}}, $$ $$ X_{2}=\frac{\partial}{\partial x_{2}}+2x_{3}\frac{\partial}{\partial t_{1}} +2x_{0}\frac{\partial}{\partial t_{2}}-2x_{1}\frac{\partial}{\partial t_{3}}, $$ $$ X_{3}=\frac{\partial}{\partial x_{3}}-2x_{2}\frac{\partial}{\partial t_{1}} +2x_{1}\frac{\partial}{\partial t_{2}}+2x_{0}\frac{\partial}{\partial t_{3}}, $$ $$ T_{1}=\frac{\partial}{\partial t_{1}}, T_{2}=\frac{\partial}{\partial t_{2}}, T_{3}=\frac{\partial}{\partial t_{3}}.$$ 其李括号运算满足 $$ [X_{0},X_{1}]=[X_{3},X_{2}]=4T_{1}, $$ $$ [X_{0},X_{2}]=[X_{1},X_{3}]=4T_{2}, $$ $$ [X_{0},X_{3}]=[X_{2},X_{1}]=4T_{3}. $$ 四元数海森堡群上的伸缩由 $ \delta_{r}(x,t)=(rx,r^{2}t) r>0 $ 给出. sub-Laplacian算子有形式 \begin{equation} \triangle=-X_{0}^{2}-X_{1}^{2}-X_{2}^{2}-X_{3}^{2}, \tag{2.1} \end{equation} 它满足齐次性 $\triangle(f\circ \delta_{r})=r^{2}(\triangle f\circ \delta_{r}),r>0$. 而 full-Laplacian 算子 $$L=\triangle-T_{1}^{2}-T_{2}^{2}- T_{3}^{2}$$ 则不具备这种齐次性.

记 $L^{p}({\Bbb H}\times {\Bbb R}^{3})$ 为所有 ${\Bbb H}\times {\Bbb R}^{3}$ 上的 $L^{p},1\leq p\leq \infty$ 有界的复值函数.给定 $0 eq a\in \Im {\Bbb H}$. 取 $\tilde{a}=\frac{a}{|a|}$.那么 $\rho(\tilde{a}): q\mapsto q\tilde{a}$ 是一个 ${\Bbb H}\cong{\Bbb R}^{4} $ 的复结构. 下面考虑 $({\Bbb H},\rho(\tilde{a}))\cong {\Bbb C}^{2}$ 上的所有全纯函数 $F$ 构成的 Fock 空间 $\digamma_{a}$,其范数满足 \begin{eqnarray*} \| F\| _{\digamma_{a}}^{2}=\int_{{\Bbb H}}|F(q)|^{2}{\rm e}^{-2|a\| q|^{2}}{\rm d}q ＜\infty.\end{eqnarray*} 作用在 $\digamma_{a}$ 上的酉表示 $\pi_{a}(x,t) ((x,t)\in{\Bbb H}\times {\Bbb R}^{3})$ 为 \begin{eqnarray*} (\pi_{a}(x,t)F)(q)=F(q+x){\rm e}^{{\rm i}\langle a,t\rangle -|a|(|x|^{2} +2\langle q,x\rangle -2i\langle q\tilde{a},x\rangle )}.\end{eqnarray*}

选择一组满足如下条件的 ${\Bbb H}$ 的正交基 ($e_{0},e_{1},e_{2},e_{3}$)\begin{equation}\label{tuotuo} e_{0}=1,e_{1}=\tilde{a},e_{2}\tilde{a}=e_{3}.\tag{2.2} \end{equation} 并且记 \begin{equation} \label{ct} z_{1}=q_{0}+{\rm i}q_{1},z_{2}=q_{2}+{\rm i}q_{3},q=q_{0}+q_{1}\tilde{a}+q_{2}e_{2}+q_{3}e_{3} .\tag{2.3} \end{equation} 那么 \begin{equation} \{E_{k_{1},k_{2}}^{a}(q)=\pi^{-1}(k_{1}!k_{2}!)^{-\frac{1}{2}}(2|a|)^{\frac{k_{1}+k_{2}}{2}+1}z_{1}^{k_{1}}z_{2}^{k_{2}}: k_{1},k_{2}=0,1,2,\cdots\} \tag{2.4} \end{equation} 构成了 $\digamma_{a}$ 的一组正交基.由 Weyl 变换的性质知道 \begin{equation}\label{prop3.2} \sum\limits_{j=0}^{k}\langle \pi_{a}(x,t)E_{j,k-j}^{a}(q),E_{j,k-j}^{a}(q)\rangle _{\digamma_{a}} ={\rm e}^{{\rm i}\langle a,t \rangle}{\rm e}^{-|a\| x|^{2}}L_{k}^{1}(2|a\| x|^{2}), \tag{2.5} \end{equation} 上式中 $L_{k}^{\delta}$ 表示 Laguerre 多项式(关于 Laguerre 多项式的更多内容,参看文献^{[18, 21]}).

定义 2.1  对于 $f\in L^{1}({\Bbb H}\times {\Bbb R}^{3})\cap L^{2}({\Bbb H}\times {\Bbb R}^{3})$, $f$ 的群 Fourier 变换为 $$ {\cal F}(f)(a)=\int_{{\Bbb H}\times {\Bbb R}^{3}}f(x,t)\pi_{a}(x,t){\rm d}x{\rm d}t= \int_{{\Bbb H}}f^{a}(x)\pi_{a}(x,0){\rm d}x, $$ 这里的 $$ f^{a}(x)=\int_{\Im {\Bbb H}}f(x,t){\rm e}^{{\rm i}\langle a,t\rangle }{\rm d}t $$ 表示 $f$ 关于中心变量 $t$ 的 Fourier 逆变换.

由 Weyl 变换的经典理论可以得到下面的 Plancherel 公式 \begin{eqnarray}\label{l2} \|f\|_{L^{2}({\Bbb H}\times {\Bbb R}^{3})}^{2} &=&\frac{1}{2\pi^{5}}\sum\limits_{k=0}^{+\infty}\sum\limits_{j=0}^{k}\int_{\Im{\Bbb H}\backslash\{0\}} \| {\cal F}(f)(a)E_{j,k-j}^{a}\| _{\digamma_{a}}^{2}|a|^{2}{\rm d}a onumber\\ &=&\frac{1}{2\pi^{5}}\int_{\Im{\Bbb H}\backslash\{0\}}\| {\cal F}(f)(a)\| _{HS}^{2}|a|^{2}{\rm d}a, f\in L^{1}({\Bbb H}\times {\Bbb R}^{3})\cap L^{2}({\Bbb H}\times {\Bbb R}^{3}), \tag{2.6}\end{eqnarray} \begin{eqnarray}\label{sars} \langle f,g \rangle_{L^{2}({\Bbb H}\times {\Bbb R}^{3})} &=&\frac{1}{2\pi^{5}}\sum\limits_{k=0}^{+\infty}\sum\limits_{j=0}^{k}\int_{\Im{\Bbb H}\backslash\{0\}} \langle {\cal F}(f)(a)E_{j,k-j}^{a},{\cal F}(g)(a)E_{j,k-j}^{a}\rangle _{\digamma_{a}} |a|^{2}{\rm d}a, onumber\\ && f,g\in L^{1}({\Bbb H}\times {\Bbb R}^{3})\cap L^{2}({\Bbb H}\times {\Bbb R}^{3}).\tag{2.7}\end{eqnarray}

令 $S_{2}$ 表示 Fock 空间 $\digamma_{a}$ 上的 Hilbert-Schmidt 算子构成的 Hilbert 空间,配有内积 $(T,S)={\rm tr}(TS^{\ast})$. 令 ${\rm d}\mu(\lambda)=2^{-1}\pi^{-5}|a|^{2}{\rm d}a$, 且记 $L^{2}(\Im{\Bbb H}\backslash \{0\},S_{2},{\rm d}\mu)$ 为在 $S_{2}$ 中取值,在 $\Im{\Bbb H}\backslash \{0\}$ 上平方可积的函数, 积分测度为 ${\rm d}\mu$.

由于 $L^{1}({\Bbb H}\times {\Bbb R}^{3})\cap L^{2}({\Bbb H}\times {\Bbb R}^{3})$ 在 $L^{2}({\Bbb H}\times {\Bbb R}^{3})$ 中稠密,那么在范数相等的意义下可以将 Fourier 变换延拓至 $f\in L^{2}({\Bbb H}\times {\Bbb R}^{3})$ 上. 这样定义的 Fourier 变换满足 (2.6)式. 因此群 Fourier 变换就成为 $L^{2}({\Bbb H}\times {\Bbb R}^{3})$ 和 $L^{2}(\Im{\Bbb H}\backslash \{0\},S_{2},{\rm d}\mu)$ 之间的同构 (其证明与海森堡群上的情形相同).

四元数海森堡群上的 Fourier 逆变换为 \begin{equation}\label{prop3.2 b} f(x,t)=\frac{1}{2\pi^{5}}\int_{\Im{\Bbb H}\backslash\{0\}} {\rm tr}(\pi_{a}^{*}(x,t){\cal F}(f)(a))|a|^{2}{\rm d}a,f\in S({\Bbb H}\times {\Bbb R}^{3}), \tag{2.8} \end{equation} 这里的 $\pi_{a}^{*}(x,t)=\pi_{a}(-x,-t)$,而 $S({\Bbb H}\times{\Bbb R}^{3})$ 表示 Schwartz 函数空间. 空间 $S({\Bbb H}\times{\Bbb R}^{3})$ 的 Schwartz 半范定义为 \begin{eqnarray}&& \| g(x,t)\| _{(N)}=\sup_{|I|\leq N,(x,t)\in{\Bbb H}\times{\Bbb R}^{3}}(1+|(x,t)|)^{11(N+1)},onumber\\ &&|(X_{0})^{i_{0}}(X_{1})^{i_{1}} (X_{2})^{i_{2}}(X_{3})^{i_{3}}T_{1}^{i_{4}}T_{2}^{i_{5}}T_{3}^{i_{6}}g(x,t)|,|I|=\sum\limits_{m=0}^{6}i_{m}.\tag{2.9}\end{eqnarray}

${\Bbb H}\times {\Bbb R}^{3}$ 上的卷积定义为 $$ (f\ast g)(w)=\int_{{\Bbb H}\times {\Bbb R}^{3}}f(wv^{-1})g(v){\rm d}v, $$ 卷积的 Fourier 变换满足 \begin{equation}\label{shu} {\cal F}(f\ast g)(a)={\cal F}(f)(a){\cal F}(g)(a).\tag{2.10} \end{equation}

记 $P$ 是四元数海森堡群上的左不变微分算子,那么对于 $f\in S' ({\Bbb H}\times {\Bbb R}^{3})$, $g\in S({\Bbb H}\times {\Bbb R}^{3})$,有 $$ P(f\ast g)=f\ast (P(g)).$$

在这一节里,我们首先考虑四元数海森堡群上 full-Laplacian 算子的一些性质.

定义 3.1 ^[25]定义 $$Sp(n)=\{g\in GL(n,{\Bbb H})|g^{*}g=I\}.$$ $Sp(1)$ 作用在 ${\Bbb H}\times{\Bbb R}^{3}$ 上有 $$ k\cdot (x,t):=(kx,t) \mbox{(对任意的$k\in Sp(1)$ 以及} (x,t)\in{\Bbb H}\times{\Bbb R}^{3} ).$$

令 $$L_{Sp(1)}^{1}({\Bbb H}\times{\Bbb R}^{3}):=\{f\in L^{1} ({\Bbb H}\times{\Bbb R}^{3})|f(kx,t)=f(x,t) \mbox{对任意的$k\in Sp(1) $成立}\}.$$ 由于 $k=a+be_{1}+ce_{2}+de_{3}\in Sp(1)$,根据 $Sp(1)$ 的定义,有 $\bar{k}k=1$, 也即 $a^{2}+b^{2}+c^{2}+d^{2}=1$. 那么容易得到 $$|kx|=(|kx|^{2})^{\frac{1}{2}}=(\overline{kx}kx)^{\frac{1}{2}}=(\bar{x}(\bar{k}k)x)^{\frac{1}{2}} =(\bar{x}x)^{\frac{1}{2}}=(|x|^{2})^{\frac{1}{2}}=|x|.$$ 因此,四元数海森堡群上的可积 $Sp(1)$ 不变的函数关于变量 $x$ 是径向的.

记 ${\Bbb D}({\Bbb H}\times{\Bbb R}^{3})$ 为四元数海森堡群上由 $\{X_{0},X_{1},X_{2},X_{3},T_{1},T_{2},T_{3}\}$ 生成的所有左不变微分算子的代数.我们把 $Sp(1)$ 不变的微分算子记做 \begin{eqnarray*} {\Bbb D}_{Sp(1)}({\Bbb H}\times{\Bbb R}^{3})&:=&\{D\in{\Bbb D}({\Bbb H}\times{\Bbb R}^{3})| D(f\circ k)=D(f)\circ k \\ &&\mbox{对一切} k\in Sp(1),f\in C^{\infty}({\Bbb H}\times{\Bbb R}^{3}) \mbox{成立}\}.\end{eqnarray*} 这样的算子构成了 ${\Bbb D}({\Bbb H}\times{\Bbb R}^{3})$ 的一个子代数.

引理 3.1  $\triangle$ 和 $L\in {\Bbb D}_{Sp(1)}({\Bbb H}\times{\Bbb R}^{3})$.

证  由于 $T_{1}^{2},T_{2}^{2},T_{3}^{2}\in {\Bbb D}_{Sp(1)}$,所以 $L\in {\Bbb D}_{Sp(1)}({\Bbb H}\times{\Bbb R}^{3})$ 是 $\triangle\in {\Bbb D}_{Sp(1)}({\Bbb H}\times{\Bbb R}^{3})$ 的一个直接结果. 因此只需要证明~ $\triangle\in {\Bbb D}_{Sp(1)}({\Bbb H}\times{\Bbb R}^{3})$. 首先 \begin{eqnarray*} \triangle&=&-X_{0}^{2}-X_{1}^{2}-X_{2}^{2}-X_{3}^{2}\\ &=& \bigg\{-\frac{\partial^{2}}{\partial x_{0}^{2}}-\frac{\partial^{2}}{\partial x_{1}^{2}} -\frac{\partial^{2}}{\partial x_{2}^{2}}-\frac{\partial^{2}}{\partial x_{3}^{2}}\bigg\} -\bigg\{4|x|^{2}(\frac{\partial^{2}}{\partial t_{1}^{2}}+\frac{\partial^{2}} {\partial t_{2}^{2}}+\frac{\partial^{2}}{\partial t_{3}^{2}})\bigg\} onumber\\ && +\bigg\{4x_{1}\frac{\partial}{\partial x_{0}}\frac{\partial}{\partial t_{1}}+ 4x_{2}\frac{\partial}{\partial x_{0}}\frac{\partial}{\partial t_{2}}+4x_{3} \frac{\partial}{\partial x_{0}}\frac{\partial}{\partial t_{3}}-4x_{0} \frac{\partial}{\partial x_{1}}\frac{\partial}{\partial t_{1}}\\ &&+4x_{3} \frac{\partial}{\partial x_{1}}\frac{\partial}{\partial t_{2}} -4x_{2} \frac{\partial}{\partial x_{1}}\frac{\partial}{\partial t_{3}} -4x_{3}\frac{\partial}{\partial x_{2}}\frac{\partial}{\partial t_{1}} -4x_{0}\frac{\partial}{\partial x_{2}}\frac{\partial}{\partial t_{2}} \\ &&+4x_{1}\frac{\partial}{\partial x_{2}}\frac{\partial}{\partial t_{3}} +4x_{2}\frac{\partial}{\partial x_{3}}\frac{\partial}{\partial t_{1}} -4x_{1}\frac{\partial}{\partial x_{3}}\frac{\partial}{\partial t_{2}} -4x_{0}\frac{\partial}{\partial x_{3}}\frac{\partial}{\partial t_{3}}\bigg\} \\ &=&\{I_{1}\}-\{I_{2}\}+\{I_{3}\}, onumber \end{eqnarray*} 这里的 $$ I_{1}=-\frac{\partial^{2}}{\partial x_{0}^{2}}-\frac{\partial^{2}}{\partial x_{1}^{2}}-\frac{\partial^{2}}{\partial x_{2}^{2}}-\frac{\partial^{2}}{\partial x_{3}^{2}}; $$ $$ I_{2}=4|x|^{2}\bigg(\frac{\partial^{2}}{\partial t_{1}^{2}} +\frac{\partial^{2}}{\partial t_{2}^{2}}+\frac{\partial^{2}}{\partial t_{3}^{2}}\bigg); $$ \begin{eqnarray*} I_{3}&=&4x_{1}\frac{\partial}{\partial x_{0}}\frac{\partial}{\partial t_{1}}+ 4x_{2}\frac{\partial}{\partial x_{0}}\frac{\partial}{\partial t_{2}}+4x_{3} \frac{\partial}{\partial x_{0}}\frac{\partial}{\partial t_{3}}-4x_{0} \frac{\partial}{\partial x_{1}}\frac{\partial}{\partial t_{1}}\\ &&+4x_{3} \frac{\partial}{\partial x_{1}}\frac{\partial}{\partial t_{2}} -4x_{2} \frac{\partial}{\partial x_{1}}\frac{\partial}{\partial t_{3}} -4x_{3}\frac{\partial}{\partial x_{2}}\frac{\partial}{\partial t_{1}} -4x_{0}\frac{\partial}{\partial x_{2}}\frac{\partial}{\partial t_{2}} \\ &&+4x_{1}\frac{\partial}{\partial x_{2}}\frac{\partial}{\partial t_{3}} +4x_{2}\frac{\partial}{\partial x_{3}}\frac{\partial}{\partial t_{1}} -4x_{1}\frac{\partial}{\partial x_{3}}\frac{\partial}{\partial t_{2}} -4x_{0}\frac{\partial}{\partial x_{3}}\frac{\partial}{\partial t_{3}}. \end{eqnarray*} 令 $k=a+be_{1}+ce_{2}+de_{3}\in S_{p}(1)$,那么容易得到 \begin{eqnarray} I_{1}f(kx)=(a^{2}+b^{2}+c^{2}+d^{2})(I_{1}f)(kx)=(I_{1}f)(kx)\tag{3.1}\end{eqnarray} 以及 \begin{eqnarray} I_{2}f(kx)&=&-4|x|^{2}\bigg(\frac{\partial^{2}}{\partial t_{1}^{2}}+ \frac{\partial^{2}}{\partial t_{2}^{2}}+\frac{\partial^{2}}{\partial t_{3}^{2}}\bigg)f(kx)=-4|kx|^{2}\bigg(\frac{\partial^{2}}{\partial t_{1}^{2}}+\frac{\partial^{2}} {\partial t_{2}^{2}}+\frac{\partial^{2}}{\partial t_{3}^{2}}f\bigg)(kx) onumber\\ &=&(I_{2}f)(kx).\tag{3.2}\end{eqnarray} 令 $kx=\{(kx)_{0},(kx)_{1},(kx)_{2},(kx)_{3}\}$,那么 $\{(kx)_{0},(kx)_{1},(kx)_{2},(kx)_{3}\}=\{ax_{0}-bx_{1}-cx_{2}-{\rm d}x_{3},ax_{1} +bx_{0}+cx_{3}-{\rm d}x_{2},ax_{2}+cx_{0}+{\rm d}x_{1}-bx_{3},ax_{3}+{\rm d}x_{0}+bx_{2}-cx_{1}\}$, 直接计算可以得到 \begin{eqnarray} I_{3}f(kx)&=&4(kx)_{1}\bigg(\frac{\partial}{\partial x_{0}}\frac{\partial}{\partial t_{1}} f\bigg)(kx)+ 4(kx)_{2}\bigg(\frac{\partial}{\partial x_{0}}\frac{\partial}{\partial t_{2}}f\bigg)(kx)+4(kx)_{3} \bigg(\frac{\partial}{\partial x_{0}}\frac{\partial}{\partial t_{3}}f\bigg)(kx) onumber\\&& -4(kx)_{0} \bigg(\frac{\partial}{\partial x_{1}}\frac{\partial}{\partial t_{1}}f\bigg)(kx)+4(kx)_{3} \bigg(\frac{\partial}{\partial x_{1}}\frac{\partial}{\partial t_{2}}f\bigg)(kx)-4(kx)_{2} \bigg(\frac{\partial}{\partial x_{1}}\frac{\partial}{\partial t_{3}}f\bigg)(kx) onumber\\&&-4(kx)_{3}\bigg(\frac{\partial}{\partial x_{2}}\frac{\partial}{\partial t_{1}}f\bigg)(kx)-4(kx)_{0}\bigg(\frac{\partial}{\partial x_{2}}\frac{\partial}{\partial t_{2}}f\bigg)(kx)+4(kx)_{1}\bigg(\frac{\partial}{\partial x_{2}}\frac{\partial}{\partial t_{3}}f\bigg)(kx) onumber\\&&+4(kx)_{2}(\frac{\partial}{\partial x_{3}}\frac{\partial}{\partial t_{1}}f)(kx)-4(kx)_{1}\bigg(\frac{\partial}{\partial x_{3}}\frac{\partial}{\partial t_{2}}f\bigg)(kx)-4(kx)_{0}\bigg(\frac{\partial}{\partial x_{3}}\frac{\partial}{\partial t_{3}}f\bigg)(kx) onumber\\&=&(I_{3}f)(kx).\tag{3.3}\end{eqnarray}

由 (2.5) 式,定义 \begin{eqnarray} \zeta_{\pi_{a},k}(x,t)&=&\frac{1}{k+1}\sum\limits_{j=0}^{k} \langle \pi_{a}(x,t)E_{j,k-j}(q),E_{j,k-j}(q)\rangle _{\digamma_{a}} onumber\\ &=& \frac{1}{k+1}{\rm e}^{{\rm i}\langle a,t \rangle}{\rm e}^{-|a\| x|^{2}}L_{k}^{1}(2|a\| x|^{2}).\tag{3.4}\end{eqnarray} 根据文献^[20],$\zeta_{\pi_{a},k}(x,t)$ 是 $\triangle$ 的特征值为 $8(k+1)|a|$ 的特征函数. 因此可以得到 \begin{equation}\label{333} L\zeta_{\pi_{a},k}(x,t)=\bigg(\triangle-\frac{\partial^{2}}{\partial t_{1}^{2}} -\frac{\partial^{2}}{\partial t_{2}^{2}}-\frac{\partial^{2}}{\partial t_{3}^{2}}\bigg)\zeta_{\pi_{a},k}(x,t)=(8(k+1)|a|+|a|^{2})\zeta_{\pi_{a},k}(x,t).\tag{3.5} \end{equation} 对任意的 $f\in L_{Sp(1)}^{1}({\Bbb H}\times{\Bbb R}^{3})$,成立着如下事实(见文献^[26])\begin{eqnarray}\label{21} {\cal F}(f)(a)E_{j,k-j}^{a}(q)&=&\int_{{\Bbb H}\times{\Bbb R}^{3}}f(x,t)\frac{1}{k+1}{\rm e}^{{\rm i}\langle a,t \rangle }{\rm e}^{-|a\| x|^{2}}L_{k}^{1}(2|a\| x|^{2}){\rm d}x{\rm d}tE_{j,k-j}^{a}(q) onumber\\ &=&\int_{{\Bbb H}\times{\Bbb R}^{3}}f(x,t)\zeta_{\pi_{a},k}(x,t){\rm d}x{\rm d}tE_{j,k-j}^{a}(q).\tag{3.6}\end{eqnarray} 由于 $L\in {\Bbb D}_{Sp(1)}({\Bbb H}\times{\Bbb R}^{3})$,如果 $f$ 是 $Sp(1)$ 不变的,那么 $Lf$ 也是 $Sp(1)$ 不变的. 对于 $f\in S({\Bbb H}\times{\Bbb R}^{3})$ 且 $f$ 是 $Sp(1)$ 不变的,有 \begin{eqnarray}\label{kolo} {\cal F}(Lf)(a)E_{j,k-j}^{a}(q)&=&\int_{{\Bbb H}\times{\Bbb R}^{3}}Lf(x,t)\zeta_{\pi_{a},k}(x,t){\rm d}x{\rm d}tE_{j,k-j}^{a}(q) \ \mbox{(由 (3.6)式)} onumber\\& =&\int_{{\Bbb H}\times{\Bbb R}^{3}}f(x,t)L\zeta_{\pi_{a},k}(x,t){\rm d}x{\rm d}tE_{j,k-j}^{a}(q)(\mbox{分部积分}) onumber\\ & =&(8(k+1)|a|+|a|^{2}){\cal F}(f)(a)E_{j,k-j}^{a}(q)\mbox{(由 (3.5)式) .} \tag{3.7}\end{eqnarray} 在海森堡群上,(3.7) 式对一切 $f\in S({\Bbb H}^{n})$ 成立(见文献 ^[2]). 我们断言对于 $f\in S({\Bbb H}\times{\Bbb R}^{3})$,(3.7) 式在四元数海森堡群上也成立.下面证明该断言. 首先成立着 \[\begin{align} & \text{F}(\vartriangle f)(a)E_{j,k-j}^{a}(q)=\int_{\mathbb{H}\times {{\mathbb{R}}^{3}}}{\vartriangle }f(x,t){{\pi }_{a}}(x,t)E_{j,k-j}^{a}(q)\text{d}x\text{d}tonumber \\ & =\int_{\mathbb{H}\times {{\mathbb{R}}^{3}}}{f}(x,t)\vartriangle [{{\pi }_{a}}(x,t))E_{j,k-j}^{a}(q)]\text{d}x\text{d}t. \\ \tag{3.8}\end{align}\] 考虑第二节中 $e_{1},e_{2},e_{3}$ 的定义($e_{1}=\tilde{a},e_{2}\tilde{a}=e_{3}$, 见(2.2)式),记 $X_{j}^{a}$ 是和 $e_{j}$,$j=1,2,3$ 相关的左不变向量场,那么 $$ X_{1}^{a}f([x,t])=\frac{\rm d}{{\rm d}s}|_{s=0}f([x,t][se_{1},0]), $$ $$ X_{2}^{a}f([x,t])=\frac{\rm d}{{\rm d}s}|_{s=0}f([x,t][se_{2},0]), $$ $$ X_{3}^{a}f([x,t])=\frac{\rm d}{{\rm d}s}|_{s=0}f([x,t][se_{3},0]). $$ 直接计算,得到 \begin{eqnarray} X_{0}[\pi_{a}(x,t)E_{j,k-j}^{a}(q)]&=&(2j|a|)^{\frac{1}{2}}\pi_{a}(x,t)E_{j-1,k-j}^{a}(q) onumber\\ && -(2(j+1)|a|)^{\frac{1}{2}}\pi_{a}(x,t)E_{j+1,k-j}^{a}(q), onumber\\ X_{1}^{a}[\pi_{a}(x,t)E_{j,k-j}^{a}(q)]&=&i(2j|a|)^{\frac{1}{2}}\pi_{a}(x,t)E_{j-1,k-j}^{a}(q) onumber\\ &&+i(2(j+1)|a|)^{\frac{1}{2}}\pi_{a}(x,t)E_{j+1,k-j}^{a}(q), onumber\\ X_{2}^{a}[\pi_{a}(x,t)E_{j,k-j}^{a}(q)]&=&(2(k-j)|a|)^{\frac{1}{2}}\pi_{a}(x,t)E_{j,k-j-1}^{a}(q) onumber\\&&-(2(k-j+1)|a|)^{\frac{1}{2}}\pi_{a}(x,t)E_{j,k-j+1}^{a}(q), onumber\\ X_{3}^{a}[\pi_{a}(x,t)E_{j,k-j}^{a}(q)]&=&i(2(k-j)|a|)^{\frac{1}{2}}\pi_{a}(x,t)E_{j,k-j-1}^{a}(q) onumber\\&& +i(2(k-j+1)|a|)^{\frac{1}{2}}\pi_{a}(x,t)E_{j,k-j+1}^{a}(q).\tag{3.9}\end{eqnarray} 因此,由 $\triangle$ 的不变性,有 \begin{eqnarray} \triangle[\pi_{a}(x,t)E_{j,k-j}^{a}(q)]&=&(-X_{0}^{2}-(X_{1}^{a})^{2}-(X_{2}^{a})^{2} -(X_{3}^{a})^{2})[\pi_{a}(x,t)E_{j,k-j}^{a}(q)] onumber\\& =&8(k+1)|a|\pi_{a}(x,t)E_{j,k-j}^{a}(q).\tag{3.10}\end{eqnarray} 那么 (3.8) 式就可以写成 \begin{eqnarray}\label{bobo} {F}(\triangle f)(a)E_{j,k-j}^{a}(q)=8(k+1)|a| {F}( f)(a)E_{j,k-j}^{a}(q),\forall f\in S({\Bbb H}\times{\Bbb R}^{3}).\tag{3.11}\end{eqnarray} 由上式很容易知道 \begin{eqnarray}\label{soso} {F}(L f)(a)E_{j,k-j}^{a}(q)=(8(k+1)|a|+|a|^{2}){F}( f)(a)E_{j,k-j}^{a}(q),\forall f\in S({\Bbb H}\times{\Bbb R}^{3}).\tag{3.12}\end{eqnarray}

取 $f\in L^{2}({\Bbb H}\times{\Bbb R}^{3})$. 对于 $g\in S({\Bbb H}\times{\Bbb R}^{3})$, 则有下式成立 \begin{eqnarray} && \langle Lf,g\rangle_{L^{2}({\Bbb H}\times{\Bbb R}^{3})} onumber\\ &=&\langle f,Lg\rangle_{L^{2}({\Bbb H}\times{\Bbb R}^{3})} onumber\\ &=&\frac{1}{2\pi^{5}}\sum\limits_{k=0}^{+\infty}\sum\limits_{j=0}^{k}\int_{\Im{\Bbb H}\backslash\{0\}} \langle {\cal F}(f)(a)E_{j,k-j}^{a},(8(k+1)|a|+|a|^{2}){\cal F}(g)(a)E_{j,k-j}^{a} \rangle _{\digamma_{a}} |a|^{2}{\rm d}a onumber\\ & =&\frac{1}{2\pi^{5}}\sum\limits_{k=0}^{+\infty}\sum\limits_{j=0}^{k}\int_{\Im{\Bbb H}\backslash\{0\}} \langle (8(k+1)|a|+|a|^{2}){\cal F}(f)(a)E_{j,k-j}^{a},{\cal F}(g)(a)E_{j,k-j}^{a} \rangle _{\digamma_{a}} |a|^{2}{\rm d}a. \tag{3.13}\end{eqnarray} 根据 $f\in L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 的 Fourier 变换的定义和稠密性,有 \begin{eqnarray}\label{fou}&& {\cal F}(Lf)(a)E_{j,k-j}^{a}(q)=(8(k+1)|a|+|a|^{2}){\cal F}(f)(a)E_{j,k-j}^{a}(q), \forall f \in L^{2}({\Bbb H}\times{\Bbb R}^{3}).\tag{3.14}\end{eqnarray}

由于 $L$ 是 $L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 上的正自伴稠定算子, 由谱定理,对任意定义在 $[0,+\infty)$ 上的有界 Borel 函数可以定义算子 $m(L)$, 它是 $L^{2}$ 有界的. 由于 $0$ 在谱分解中可以忽略 (见文献^{[1, 5]}), 考虑定义在 ${\Bbb R}_{+}$ 上的 $m$,则成立 \begin{eqnarray}\label{mfunc} {\cal F}(m(L)f)(a)E_{j,k-j}^{a}(q)=m(8(k+1)|a|+|a|^{2}){\cal F}(f)(a)E_{j,k-j}^{a}(q).\tag{3.15}\end{eqnarray} 此外,若记 $D({\Bbb H}\times{\Bbb R}^{3})$ 为有紧支集的 $C^{\infty}$ 函数, 记 $D' ({\Bbb H}\times{\Bbb R}^{3})$ 为 $D({\Bbb H}\times{\Bbb R}^{3})$ 的对偶, 那么对任意的 $f\in D({\Bbb H}\times{\Bbb R}^{3})$,算子 $m(L)$ 存在核 $M\in D' ({\Bbb H}\times{\Bbb R}^{3})$ 且满足 $m(L)f=f\ast M$.

以下结论在 Littlewood-Paley 分解的讨论中非常有用.

定理 3.1  令 $m$ 是某个Schwartz 函数在 ${\Bbb R}_{+}$ 上的限制. 那么算子 $m(L)$ 的核 $M$ 属于 $S({\Bbb H}\times{\Bbb R}^{3})$ 而且是 $S_{p}(1)$ 不变的.

证  核属于 $S({\Bbb H}\times{\Bbb R}^{3})$ 的断言由文献[10,命题6]得到, 径向性质则是由于 full-Laplacian 的 $S_{p}(1)$ 不变性得到(引理 3.1).

令 $R(\cdot)={\cal R}(|\cdot|)$ 是一个非负径向函数,且属于 $C^{\infty}({\Bbb R}^{3}\backslash\{0\})$,其支集满足 supp${\cal R}\subset[\frac{1}{4},4]$ 并且有 $$ \sum\limits_{j\in{\Bbb Z}}R(2^{-2j}a)=1,a\in{\Bbb R}^{3}\backslash\{0\}.$$ 对任意的 $j\in{\Bbb Z}$,记 $\psi_{j}$ 为算子 ${\cal R}(2^{-2j}L)$ 的核.由定理3.1,$\psi_{j}\in S({\Bbb H}\times{\Bbb R}^{3})$ 并且是 $S_{p}(1)$ 不变的.由 (2.10),(3.15)式 \begin{equation}\label{ss} {\cal F}(\psi_{j})(a)E_{k_{1},k-k_{1}}^{a}(q)={\cal R}(2^{-2j}(8(k+1)|a|+|a|^{2}))E_{k_{1},k-k_{1}}^{a}(q).\tag{3.16} \end{equation} 根据 Fourier 逆变换公式 (2.8),(2.5) 和 (3.16),有 \begin{eqnarray*} \psi_{j}&=&\frac{1}{2\pi^{5}}\int_{\Im{\Bbb H}\backslash\{0\}} {\rm tr}(\pi_{a}^{*}(x,t){\cal F}(\psi_{j})(a))|a|^{2}{\rm d}a onumber\\& =&\frac{1}{2\pi^{5}}\int_{\Im{\Bbb H}\backslash\{0\}} \sum\limits_{k=0}^{+\infty}\sum\limits_{k_{1}=0}^{k}{\cal F}(\psi_{j})(a)\langle \pi_{a}(-x,-t)E_{k_{1},k-k_{1}}^{a}(q),E_{k_{1},k-k_{1}}^{a}(q)\rangle _{\digamma_{a}}|a|^{2}{\rm d}a onumber\\& =&\frac{1}{2\pi^{5}}\sum\limits_{k=0}^{+\infty}\int_{\Im{\Bbb H}\backslash\{0\}} {\rm e}^{-{\rm i}\langle a,t\rangle}{\cal R}(2^{-2j}(8(k+1)|a|+|a|^{2}))L_{k}^{1}(2|a\| x|^{2}){\rm e}^{-|a\| x|^{2}}|a|^{2}{\rm d}a onumber.\end{eqnarray*}

令 $ \tilde{\psi}_{j}=\psi_{j-1}+\psi_{j}+\psi_{j+1}.$ 由 $\psi$ 的定义知道~对 $\forall f\in S' ({\Bbb H}\times{\Bbb R}^{3})$,$ f\ast\psi_{j}=f\ast\psi_{j}\ast\tilde{\psi}_{j} $ 成立.

令 $\triangle_{j}f=f\ast\psi_{j}$. 对于任意的 $f\in L^{2}({\Bbb H}\times{\Bbb R}^{3})$, 下面的Littlewood-Paley分解成立 \begin{equation} f=\sum\limits_{j\in{\Bbb Z}}\triangle_{j}f \mbox{在} L^{2}({\Bbb H}\times{\Bbb R}^{3})\mbox{意义下}.\tag{3.17} \end{equation} 由文献[10,命题 6],存在 $C>0$ 使得 \begin{equation}\label{fmv} \| \psi_{j}\| _{L^{1}({\Bbb H}\times{\Bbb R}^{3})}\leq C,j\in{\Bbb Z}.\tag{3.18} \end{equation} 通过文献^[10] 中对多项式增长李群情形的讨论,可由 (3.18) 式得到 \begin{eqnarray}\label{impo} \| L^{\frac{\sigma}{2}}\triangle_{j}u\| _{L^{p}({\Bbb H}\times{\Bbb R}^{3})} &\leq& C2^{j\sigma}\| \triangle_{j}u\| _{L^{p}({\Bbb H}\times{\Bbb R}^{3})},\sigma\in{\Bbb R}, j\in{\Bbb Z},1\leq p\leq \infty, onumber\\&& u\in S' ({\Bbb H}\times{\Bbb R}^{3}).\tag{3.19}\end{eqnarray} 证毕.

下面给出 Littlewood-Paley 定理,它的证明和 ${\Bbb R}^{n}$ 是类似的.通过文献^[27] 中的方法和 $L$ 的乘子定理(见文献 ^[23]),有

定理 3.2  令 $1 ＜p＜\infty$ 且 $u\in S' ({\Bbb H} \times{\Bbb R}^{3})$. 那么 $u\in L^{p}({\Bbb H} \times{\Bbb R}^{3})$ 当且仅当~ $(\sum\limits_{j\in{\Bbb Z}}|\Delta_{j}u|^{2})^{\frac{1}{2}}\in L^{p}({\Bbb H}\times{\Bbb R}^{3})$. 且存在 $C_{p}$,使得 \begin{equation} C_{p}^{-1}\| u\| _{L^{p}}\leq \bigg\|\bigg (\sum\limits_{j\in{\Bbb Z}}|\Delta_{j}u|^{2} \bigg)^{\frac{1}{2}}\bigg\| _{L^{p}}\leq C_{p}\| u\| _{L^{p}}.\tag{3.20} \end{equation}

记算子 ${\cal R}(2^{-2j}\triangle)$ 的核为 $\varphi_{j}$,那么有 $\varphi_{j}\in S({\Bbb H} \times{\Bbb R}^{3})$ 且 $${\cal F}(\varphi_{j})(a)E_{k_{1},k-k_{1}}^{a}(q)={\cal R}(2^{-2j}8(k+1)|a|)E_{k_{1},k-k_{1}}^{a}(q)$$ (见(3.11)式和文献 ^[26]), 下面的结果给出了 $\varphi_{j}$ 和 $\psi_{j}$ 的关系.

引理 3.2  对任意的 $j\in{\Bbb Z}$ 集合~ $U_{j}=\{j_{0}\in{\Bbb Z}: \psi_{j}\ast\varphi_{j_{0}} eq 0\}$ 是有限集.

证  取定 $j\in{\Bbb Z}$ 和 $j_{0}\in U_{j}$.$$ {\cal R}(2^{-2j}(8(k+1)|a|+|a|^{2})){\cal R}(2^{-2j_{0}}8(k+1)|a|) eq 0.$$ 令 $\xi=8(k+1)|a|$ 和 $\eta=|a|^{2}$. $(\xi,\eta)$ 满足下面的不等式组 \[\left\{ \begin{array}{l} \frac{1}{4}\leq 2^{-2j}(\xi+\eta)\leq 4,\\[3mm] \frac{1}{4}\leq 2^{-2j_{0}}\xi\leq 4,\\[3mm] 0\leq \eta \leq \frac{\xi^{2}}{64}.\end{array}\right.\]

由于 $2^{2j_{0}-2}\leq \xi \leq \xi+\eta \leq 2^{2j+2}$,那么 $j_{0}\leq j+2$.

因为 $\frac{1}{4}\leq 2^{-2j}(\xi+\eta)$,那么有 $\frac{1}{8}\leq 2^{-2j}\xi$ 或者 $\frac{1}{8}\leq 2^{-2j}\eta$.

如果 $\frac{1}{8}\leq 2^{-2j}\xi$,那么 $2^{2j-3}\leq \xi \leq 2^{2j_{0}+2}$,这说明 $j_{0}\geq j-\frac{5}{2}$.

如果 $\frac{1}{8}\leq 2^{-2j}\eta$,那么 $2^{2j-3}\leq\eta\leq\frac{\xi^{2}}{64} \leq\frac{2^{4j_{0}+4}}{64}=2^{4j_{0}-2}$,这说明 $j_{0}\geq\frac{j}{2}-\frac{1}{4}$;

此外 $2^{2j-3}\leq\eta\leq\frac{\xi^{2}}{64} \leq\frac{2^{4j+4}}{64}$,这说明 $j\geq-\frac{1}{2}$.

这样,我们证明了当 $j<0$ 时,$j-\frac{5}{2}\leq j_{0}\leq j+2$. 当 $j\geq 0$ 时,$\frac{j}{2}-\frac{1}{4}\leq j_{0} \leq j+2$.从而引理得到了证明.

这一节讨论齐次 Besov 空间和齐次 Sobelev 空间的性质.

定义 4.1  我们称 $f$ 有阶数为 $N$ 的消失性,如果对 \begin{eqnarray*} \forall p\in P_{N-1},\int_{{\Bbb H}\times{\Bbb R}^{3}}f(x,t)p(x,t){\rm d}x=0, \end{eqnarray*} 且其积分绝对收敛. 这里的 $P_{N-1}$ 是阶数为 $N-1$ 的齐次多项式空间^[8].

记 $Z({\Bbb H}\times{\Bbb R}^{3})$ 是具有任意阶消失性的全体Schwartz 函数构成的空间.

引理 4.1  如果 $M\in S({\Bbb H}\times{\Bbb R}^{3})$,$|\alpha|$ 为多项式 $p(x,t)$ 的齐次阶数且 $|\alpha|<2K$,那么 \begin{equation} \int_{{\Bbb H}\times{\Bbb R}^{3}}p(x,t)L^{K}M{\rm d}x{\rm d}t=0.\tag{4.1} \end{equation}

证  这个引理是分部积分的直接结果(见文献[12,命题4.2]).

注 4.1  令 $g_{K}(|a|)=R(2^{-2j}|a|)|a|^{-K}$,这里 $K$ 是任意的正整数. 取 $g_{K}(|\cdot|)=G_{K}(\cdot)$. 那么有 $G_{K}(\cdot)\in S({\Bbb R}_{+})$ 和 ${\cal R}(2^{-2j}L)f=(L)^{K} G_{K}(L)f$. 由于 $G_{K}(L)$ 的核是一个 Schwartz 函数(记做 $M$) 且 $\psi_{j}=(L)^{K}M$,结合引理4.1, 有 $\psi_{j}\in Z({\Bbb H}\times{\Bbb R}^{3})$. 同样的讨论可以用于 ${\cal R}(2^{-2j}\triangle)$ 的核 $\varphi_{j}$ 从而可知 $\varphi_{j}\in Z({\Bbb H}\times{\Bbb R}^{3})$ (也可参看文献[26,注4.3]).

下面两个引理给出了空间 $Z({\Bbb H}\times{\Bbb R}^{3})$ 的一些基本性质.

引理 4.2  (参见文献[8,引理3.3]) $Z({\Bbb H}\times{\Bbb R}^{3})$ 是 $S({\Bbb H}\times{\Bbb R}^{3})$ 的一个闭子空间, 且有 $S({\Bbb H}\times{\Bbb R}^{3})\ast Z({\Bbb H}\times{\Bbb R}^{3})\subset Z({\Bbb H}\times{\Bbb R}^{3})$.$Z({\Bbb H}\times{\Bbb R}^{3})$ 的拓扑对偶空间, $Z' ({\Bbb H}\times{\Bbb R}^{3})$,等价于 $S' ({\Bbb H}\times{\Bbb R}^{3})/P$, 这里 $P$ 表示多项式.

引理 4.3  (参见文献[8,引理3.4]) 对每一个 $\varphi\in S({\Bbb H}\times{\Bbb R}^{3})$,映射 $u\mapsto u\ast \varphi$ 是一个~ $S' ({\Bbb H}\times{\Bbb R}^{3})/P\rightarrow S' ({\Bbb H}\times{\Bbb R}^{3})/P$ 的有定义的连续算子. 如果 $\varphi\in Z({\Bbb H}\times{\Bbb R}^{3})$, 相应的卷积算子则是一个 $S' ({\Bbb H}\times{\Bbb R}^{3})/P\rightarrow S' ({\Bbb H}\times{\Bbb R}^{3})$ 的连续算子.

四元数海森堡群上和 full-Laplacian 算子相关的齐次 Besov 空间的定义为

定义 4.2  假定 $\rho\in{\Bbb R},1\leq p,r\leq+\infty$, 记 $\dot{B}_{p,r}^{\rho}(L)$ 为齐次 Besov 空间, 它的定义为 \begin{equation} \dot{B}_{p,r}^{\rho}(L)=\{u\in S' ({\Bbb H}\times{\Bbb R}^{3})/P:\{2^{j\rho}\| u\ast\psi_{j}\| _{L^{p}}\}_{j\in{\Bbb Z}}\in\ell^{r}\}, \tag{4.2} \end{equation} 范数为 \begin{equation} \| u\| _{\dot{B}_{p,r}^{\rho}(L)}= \bigg(\sum\limits_{j\in{\Bbb Z}}2^{jr\rho}\| \Delta_{j}u\| _{L^{p} ({\Bbb H}\times{\Bbb R}^{3})}^{r}\bigg)^{\frac{1}{r}} ＜+\infty.\tag{4.3} \end{equation}

注 4.2  这里定义的 Besov 空间和文献^[26] 中定义的和 sub-Laplacian算子相关的 Besov 空间不同. 为了方便起见,在下面的讨论中令 $\dot{B}_{p,r}^{\rho}=\dot{B}_{p,r}^{\rho}(L)$.

定义 4.3 ^[8] 令 $N\in{\Bbb N}$. 如果存在 $C>0$ 使得对所有 $x\in{\Bbb H}\times{\Bbb R}^{3}$ 有 $$ |f(x)|\leq C (1+|x|)^{-N}.$$ 那么称函数 $f : {\Bbb H}\times{\Bbb R}^{3}\rightarrow{\Bbb C}$ 有阶数为 $N$ 的多项式衰减.

令 $$\tilde{g}(x)=g(x^{-1}), ~~ I=\{I_{i},J_{j}\in{\Bbb N}; i=1,2,3,4; j=1,2,3\},~~ {\rm d}(I)=\sum\limits_{i}I_{i}+\sum\limits_{j}2J_{j}.$$

引理 4.4  (参见文献[8,引理3.2]) 令 $N,k\in{\Bbb N}$. $X^{I}=\prod\limits_{i,j}X_{i}^{I_{i}}T_{j}^{J_{j}}$,这里的 $I_{i},J_{j}\in{\Bbb N}$.

(a)~ 如果 $f\in C^{k}$,对所有满足 ${\rm d}(Ⅰ)\leq k$ 的 $X^{I}(f)$ 有 $N$ 阶的多项式衰减. $g$ 有 $k$ 阶的消失性并且有 $N+k+10+1$ 阶的多项式衰减. 那么存在着一个常数, 只依赖于 $X^{I}(f)$ 和 $g$ 的衰减性,使得 \begin{equation} \forall x\in {\Bbb H}\times{\Bbb R}^{3},\forall 0 ＜t<1 ,|g\ast (D_{t}f)(x)|\leq C t^{k+10}(1+|tx|)^{-N}.\tag{4.4} \end{equation} 特别的,如果 $p>\frac{10}{N}$, \begin{equation}\label{3.21} \forall 0＜t<1 ,\| g\ast (D_{t}f)\| _{p}\leq C't^{k+10(1-\frac{1}{p})}.\tag{4.5} \end{equation}

(b)~ 现在假定 $g\in C^{k}$,如果对 ${\rm d}(Ⅰ)\leq k$,$X^{I}\tilde{g}$ 有阶数为 $N$ 的多项式衰减. 令 $f$ 有 $k$ 阶的消失性且有 $N+k+10+1$ 阶的多项式衰减. 那么存在一个常数,只依赖于 $f$ 和 $X^{I}\tilde{g}$ 的衰减性,使得 \begin{equation} \forall x\in {\Bbb H}\times{\Bbb R}^{3},\forall 1＜t＜+\infty ,|g\ast (D_{t}f)(x)|\leq C t^{-k}(1+|x|)^{-N}.\tag{4.6} \end{equation} 特别的,如果 $p>\frac{10}{N}$, \begin{equation}\label{3.22} \forall 1＜t＜+\infty ,\| g\ast (D_{t}f)\| _{p}\leq C't^{-k}.\tag{4.7} \end{equation}

引理4.4将在下面的引理的证明中发挥重要作用.

引理 4.5  对于 $\forall 1\leq p,r\leq\infty$ 及 $\forall \rho\in{\Bbb R}$, 存在连续嵌入映射 $Z({\Bbb H}\times{\Bbb R}^{3})\hookrightarrow\dot{B}_{p,r}^{\rho}\hookrightarrow S' ({\Bbb H}\times{\Bbb R}^{3})/P$. 对于 $p,r＜\infty,Z({\Bbb H}\times{\Bbb R}^{3})\subset\dot{B}_{p,r}^{\rho}$ 是稠密的.

证  令 $g\in Z({\Bbb H}\times{\Bbb R}^{3})$. 根据和 sub-Laplacian 算子相关的 Littlewood-Paley 分解的性质 (参见文献^[26]), 有 $\Delta_{j}g=\sum\limits_{i\in{\Bbb Z}}\Delta_{j}g\ast\varphi_{i}$. 这样由引理3.2,有 \begin{equation}\label{4.2} \Delta_{j}g=\sum\limits_{i\in U_{j}}\Delta_{j}g\ast\varphi_{i}.\tag{4.8} \end{equation} 由事实 $\psi_{j}\ast\varphi_{i}=\varphi_{i}\ast\psi_{j}$ (该事实可由群 Fourier 变换和 (3.18)式得到) 可得到下面的结果 \begin{equation}\label{4.3} \| \triangle_{j}g\ast\varphi_{i}\| _{L^{p}}=\| g\ast\psi_{j}\ast\varphi_{i}\| _{L^{p}} \leq \| g\ast\varphi_{i}\| _{L^{p}}\| \psi_{j}\| _{L^{1}}\leq C\| g\ast\varphi_{i}\| _{L^{p}}\forall i\in U_{j}.\tag{4.9} \end{equation} 由于 $\varphi_{j}(\cdot)=2^{10j}\varphi(2^{j}\cdot)$ (参见文献^[26]) 且 $\varphi_{j}\in Z({\Bbb H}\times{\Bbb R}^{3})$ (注4.1),那么 (4.5) 和 (4.7) 式可以使得下面的估计对所有 $g\in Z({\Bbb H}\times{\Bbb R}^{3})$ 和 $k\in{\Bbb N}$ 成立 \begin{equation}\label{4.1} \sum\limits_{i\in U_{j}}\| g\ast\varphi_{i}\| _{L^{p}}\leq\sum\limits_{i\in U_{j}}C_{k}2^{-|i|k} . \tag{4.10} \end{equation} 考察引理3.2 的证明和结论知,对于 $j<0$ 有 $|j-i|\leq 4$,对于 $j\geq 0$ 有$|j-i|\leq |\frac{j}{2}|+\frac{9}{4}$. 那么 (4.10) 式就变成 \begin{equation}\label{4.4} \sum\limits_{i\in U_{j}}C_{k}2^{-|i|k}\leq\sum\limits_{i\in U_{j}} C_{k}2^{-|j|k}2^{|j-i|k}\leq C_{k}\bigg(4+\frac{|j|}{2}+\frac{9}{4}\bigg) 2^{-\frac{|j|}{2}k}. \tag{4.11} \end{equation} 然后由 (4.8)-(4.11)式,可得到 \begin{eqnarray} \| \Delta_{j}g\| _{L^{p}}\leq C \sum\limits_{i\in U_{j}}\| g\ast\varphi_{i}\| _{L^{p}}\leq C_{k} \bigg(4+\frac{|j|}{2}+\frac{9}{4}\bigg) 2^{-\frac{|j|}{2}k}.\tag{4.12}\end{eqnarray} 这表明 $Z({\Bbb H}\times{\Bbb R}^{3})\subset \dot{B}_{p,q}^{r}$.

对另一个嵌入,有 \begin{eqnarray*} |\langle f,g\rangle |&=& \bigg|\sum\limits_{j\in{\Bbb Z}}\langle f,g\ast\psi_{j}\rangle \bigg| =\bigg|\sum\limits_{j\in{\Bbb Z}}\langle \sum\limits_{j' \in{\Bbb Z}}f\ast\psi_{j' },g\ast\psi_{j}\rangle \bigg| onumber\\ & \leq& \sum\limits_{j\in{\Bbb Z}}\sum\limits_{|j' -j|\leq 1}| \langle f\ast\psi_{j' },g\ast\psi_{j}\rangle | \\ &\leq&\sum\limits_{j\in{\Bbb Z}}\sum\limits_{|j' -j|\leq 1}\| f\ast\psi_{j' }\| _{L^{p' }}\| g\ast\psi_{j}\| _{L^{p}} onumber\\ &\leq &C\sum\limits_{j\in{\Bbb Z}}\sum\limits_{|j' -j|\leq 1}2^{-j' s}\| f\ast\psi_{j' }\| _{L^{p' }}2^{js}\| g\ast\psi_{j}\| _{L^{p}} \\ &\leq& C\| f\| _{\dot{B}_{p' ,q' }^{-s}}\| g\| _{\dot{B}_{p,q}^{s}} onumber \end{eqnarray*} 对一切 $f\in Z({\Bbb H}\times{\Bbb R}^{3})\subset \dot{B}_{p' ,q' }^{-s}$ 和 $g\in \dot{B}_{p,q}^{s}$ 成立. 这表明另一个嵌入也是正确的.

关于稠密性的断言,令 $u\in \dot{B}_{p,q}^{s}$,$\varepsilon>0$. 由于 $q ＜\infty$,那么存在着 $N\in{\Bbb N}$ 使得 \begin{eqnarray*} \sum\limits_{|j|>N-1}2^{jsq}\| \Delta_{j}u\| _{L^{p}}^{q}＜\varepsilon.\end{eqnarray*} 定义 $K_{N}=\sum\limits_{|j|\leq N}\psi_{j}$ 并且令 $\omega=u\ast K_{N}$,则 $\omega\in L^{p}$ ($p＜\infty$). 由稠密性知存在 $g\in S({\Bbb H}\times{\Bbb R}^{3})$ 使得 $\| \omega-g\| _{L^{p}}<2^{-|s|N}\varepsilon^{\frac{1}{q}}$. 令 $f=g\ast K_{N}$,那么 $f\in Z({\Bbb H}\times{\Bbb R}^{3})$. 对于 $j\in{\Bbb Z}$, \begin{eqnarray*} \| \Delta_{j}(u-f)\| _{L^{p}}&=&\| (u-f)\ast\psi_{j}\| _{L^{p}}= \bigg\| u\ast\psi_{j}-g\ast\sum\limits_{|j' |\leq N}\psi_{j' }\ast\psi_{j}\bigg\| _{L^{p}} onumber\\ & \leq& \bigg\| u\ast\psi_{j}-u\ast\sum\limits_{|j' |\leq N}\psi_{j' }\ast\psi_{j} \bigg\| _{L^{p}} \\ &&+ \bigg\| u\ast\sum\limits_{|j' |\leq N}\psi_{j' }\ast\psi_{j}- g\ast\sum\limits_{|j' |\leq N}\psi_{j' }\ast\psi_{j}\bigg\| _{L^{p}}. onumber \end{eqnarray*} 对于 $|j|\leq N-1$,可知 $\sum\limits_{|j' |\leq N}\psi_{j' }\ast\psi_{j}= \psi_{j}\ast\tilde{\psi}_{j}=\psi_{j}$. 因此 $$ \| \Delta_{j}(u-f)\| _{L^{p}}\leq \| \omega-g\| _{L^{p}}\| \psi_{j}\| _{L^{1}}\leq C2^{-|s|N}\varepsilon^{\frac{1}{q}}.$$ 对于 $|j|>N+1$,可知 $\sum\limits_{|j' |\leq N}\psi_{j' }\ast\psi_{j}=0$. 因此 $$ \| \Delta_{j}(u-f)\| _{L^{p}}=\| u\ast\psi_{j}\| _{L^{p}}.$$ 对于 $j=\pm N,j=\pm (N+1)$,我们断言 $$ \| \Delta_{j}(u-f)\| _{L^{p}}\leq C2^{-|s|N}\varepsilon^{\frac{1}{q}}.$$ 例如,对于 $j=N$,有 \begin{eqnarray*} &&\| \Delta_{N}(u-f)\| _{L^{p}} \\ &\leq& \| u\ast\psi_{N}-u\ast(\psi_{N}+\psi_{N-1})\ast\psi_{N}\| _{L^{p}} +\| \omega\ast\psi_{N}-g\ast(\psi_{N} +\psi_{N-1})\ast\psi_{N}\| _{L^{p}} \\ &\leq& \| u\ast\psi_{N}\| _{L^{p}}(1+\| \psi_{N-1}+\psi_{N}\| _{L^{1}})+ \| (\omega-g)\ast\psi_{N}\| _{L^{p}} onumber\\&& +\| g\ast\psi_{N}\| _{L^{p}}(1+\| \psi_{N-1}+\psi_{N}\| _{L^{1}}) onumber\\ & \leq& \| u\ast\psi_{N}\| _{L^{p}}(1+\| \psi_{N-1}+\psi_{N}\| _{L^{1}}+ \| \psi_{N-1}+\psi_{N}\| _{L^{1}}(1+\| \psi_{N-1}+\psi_{N}\| _{L^{1}})) onumber\\ && +\| (\omega-g)\ast\psi_{N}\| _{L^{p}}(2+\| \psi_{N-1}+\psi_{N}\| _{L^{1}}) onumber\\ & \leq& C\| u\ast\psi_{N}\| _{L^{p}}+C2^{-|s|N}\varepsilon^{\frac{1}{q}} \leq C2^{-|s|N}\varepsilon^{\frac{1}{q}}. onumber \end{eqnarray*} 同样的讨论可以用于 $j=-N,j=\pm (N+1)$.

现在对 $j$ 求和就得到了最后的结论.

引理 4.6  对于 $\rho\in{\Bbb R},1\leq p,r\leq+\infty$,$\dot{B}_{p,r}^{\rho}$ 是一个 Banach 空间.

证  这个引理的证明和文献[8,定理3.16] 基本一致,只有很小的差别, 只需要用本文中的 $\psi$ 替换掉该证明中的 $\psi^{*}$ 就可以了.

如果 $a+b=1,a\geq 0,b\geq 0$,那么 $\frac{1}{2}\leq a^{2}+b^{2}=2a^{2}-2a+1\leq 1$.于是 \begin{equation} \frac{1}{2}\leq \sum\limits_{j\in{\Bbb Z}}|{\cal R}(2^{-2j}a)|^{2}\leq 1.\tag{4.13} \end{equation} 通过 Plancherel 公式,很容易证明 $L^{2}({\Bbb H}\times{\Bbb R}^{3})=\dot{B}_{2,2}^{0}$.

下面的定理给出了 Besov 空间的一些基本性质.

定理 4.1  以下结论在四元数海森堡群上成立 \begin{equation} L^{p}({\Bbb H}\times{\Bbb R}^{3})\subset \dot{B}_{p,\infty}^{0}, \tag{4.14} \end{equation} \begin{equation} \dot{B}_{p,1}^{0}\subset L^{p}({\Bbb H}\times{\Bbb R}^{3}), \tag{4.15} \end{equation} \begin{equation} \mbox{当}r\leq r'\mbox{时,} \dot{B}_{p,r}^{\rho}\subset \dot{B}_{p,r' }^{\rho}, \tag{4.16} \end{equation} \begin{equation} \label{kwer} \mbox{当}p_{1}\leq p_{2} \mbox{时,} \dot{B}_{p_{1},r}^{\rho}\subset \dot{B}_{p_{2},r}^{\rho-10(\frac{1}{p_{1}}-\frac{1}{p_{2}})}.\tag{4.17} \end{equation}

证  证明与 ${\Bbb R}^{n}$ 上类似.

下面的两个结论将会在Strichartz不等式的证明中用到.

定理 4.2  \begin{equation}\label{kwee} \dot{B}_{p,2}^{0}\subset L^{p}({\Bbb H}\times{\Bbb R}^{3}) ;p\geq 2, \tag{4.18} \end{equation} \begin{equation} \dot{B}_{p,2}^{\rho}\subset L^{a}({\Bbb H}\times{\Bbb R}^{3}),0\leq\rho ＜\frac{10}{p},\frac{1}{a}=\frac{1}{p}-\frac{\rho}{10}.\tag{4.19} \end{equation}

证  当 $p\geq 2$,由定理3.2,有 \begin{eqnarray*} \| f\| _{L^{p}}^{2}&\leq& C \bigg\|\bigg (\sum\limits_{j\in{\Bbb Z}}|\Delta_{j}f|^{2}\bigg)^{\frac{1}{2}}\bigg\| _{L^{p}}^{2} =\bigg\| \sum\limits_{j\in{\Bbb Z}}|\Delta_{j}f|^{2}\bigg\| _{L^{\frac{p}{2}}} onumber\\ & \leq&\sum\limits_{j\in{\Bbb Z}}\| |\Delta_{j}f|^{2}\| _{L^{\frac{p}{2}}}=\sum\limits_{j\in{\Bbb Z}}\| \Delta_{j}f\| _{L^{p}}^{2} =\| u\| _{\dot{B}_{p,2}^{0}}^{2}. onumber \end{eqnarray*} 定理中第二个结论是 (4.17) 和 (4.18)式的自然推论.

现在讨论四元数海森堡群上的 Sobolev 空间.

定义 4.4  令 $m\in{\Bbb N}$,记 $\dot{H}^{m}({\Bbb H}\times{\Bbb R}^{3})$ 为四元数海森堡群上的 Sobelev 空间,它是 Schwartz 函数在如下范数下的完备化 \begin{equation} \| u\| _{\dot{H}^{m}({\Bbb H}\times{\Bbb R}^{3})}^{2}=\sum\limits_{l= m}\| Y_{1}\cdots Y_{l}u\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})}^{2}＜\infty.\tag{4.20} \end{equation} $Y_{i}\in \{X_{0},X_{1},X_{2},X_{3},T_{1},T_{2},T_{3}\}$.

在四元数海森堡群上成立着如下两个结论 \begin{equation}\label{sobi1} u\in \dot{H}^{m}({\Bbb H}\times{\Bbb R}^{3})\Leftrightarrow(L)^{\frac{m}{2}}u\in L^{2},m\mbox{是偶数}, \tag{4.21} \end{equation} \begin{equation}\label{sobo2} u\in \dot{H}^{m}({\Bbb H}\times{\Bbb R}^{3})\Leftrightarrow(L)^{\frac{m-1}{2}}u\in \dot{H}^{1},m\mbox{是奇数}.\tag{4.22} \end{equation}

在下面的情形中,Besov 空间和 Sobelev 空间是等价的.

定理 4.3  对于 $\forall m\in{\Bbb N}$,$\dot{B}_{2,2}^{m}=\dot{H}^{m}({\Bbb H}\times{\Bbb R}^{3})$.

证  假定 $m$ 是偶数. 由 (4.21)式,有 $$ \| u\| _{\dot{H}^{m}({\Bbb H}\times{\Bbb R}^{3})}^{2}=\frac{1}{2\pi^{5}}\int_{\Im{\Bbb H}\backslash\{0\}}\sum\limits_{k=0}^{+\infty} \sum\limits_{j=0}^{k}(8(k+1)|a|+|a|^{2})^{m}\| \hat{u}(a)E_{j,k-j}^{a}\| ^{2}|a|^{2}{\rm d}a.$$ 根据 Besov 空间的定义,有 \begin{eqnarray}\label{turbo} \| u\| _{\dot{B}_{2,2}^{m}}^{2}&=& \sum\limits_{j' \in{\Bbb Z}}2^{2j' m}\| u\ast\psi_{j' }\| _{L^{2} ({\Bbb H}\times{\Bbb R}^{3})}^{2} onumber\\ &=&\frac{1}{2\pi^{5}} \int_{\Im{\Bbb H}\backslash\{0\}}\sum\limits_{k=0}^{+\infty}\sum\limits_{j=0}^{k}\sum\limits_{j' \in{\Bbb Z}}2^{2j' m}|{\cal R}(2^{-2j' }(8(k+1)|a|+|a|^{2}))|^{2} \| \hat{u}(a)E_{j,k-j}^{a}\| ^{2}|a|^{2}{\rm d}a. onumber\\ \tag{4.23}\end{eqnarray} 由于函数 ${\cal R}$ 是有紧支集的,所以 \begin{equation}\label{ciyt} \sum\limits_{j' \in{\Bbb Z}}2^{2j' m}|{\cal R}(2^{-2j' }(8(k+1)|a|+|a|^{2}))|^{2}\sim (8(k+1)|a|+|a|^{2})^{m}.\tag{4.24} \end{equation} 那么当 $m$ 是偶数时,$u\in \dot{H}^{m}({\Bbb H}\times{\Bbb R}^{3})\Leftrightarrow u\in \dot{B}_{2,2}^{m}$.

假定 $m$ 是奇数,则 $$ \| u\| _{\dot{H}^{1}({\Bbb H}\times{\Bbb R}^{3})}^{2}=(Lu,u), $$ 那么 $\| (L)^{\frac{m-1}{2}}u\| _{\dot{H}^{1}({\Bbb H}\times{\Bbb R}^{3})}^{2} =((L)^{\frac{m+1}{2}}u,(L)^{\frac{m-1}{2}}u)$.由 (2.7)式,可知 \begin{equation} \| u\| _{\dot{H}^{m}({\Bbb H}\times{\Bbb R}^{3})}^{2}=\frac{1}{2\pi^{5}}\int_{\Im{\Bbb H}\backslash\{0\}}\sum\limits_{k=0}^{+\infty} \sum\limits_{j=0}^{k}(8(k+1)|a|+|a|^{2})^{m}\| \hat{u}(a)E_{j,k-j}^{a}\| ^{2}|a|^{2}{\rm d}a.\tag{4.25} \end{equation} 由于 (4.23) 和 (4.24)式对 $\forall m\in {\Bbb N}$ 成立,我们完成了对定理的证明.

本节研究如下波方程的色散估计和Strichartz估计.\begin{eqnarray}\label{111}\left\{ \begin{array}{l} \partial_{s}^{2}u+Lu=f\in L^{1}((0,T),L^{2}({\Bbb H}\times{\Bbb R}^{3})),\\ u|_{s=0}=u_{0}\in \dot{B}_{2,2}^{1}(L),\\ \partial_{s}u|_{s=0}=u_{1}\in L^{2}({\Bbb H}\times{\Bbb R}^{3}).\end{array}\right.\tag{5.1}\end{eqnarray} 首先讨论方程 (5.1)的解的存在性和唯一性.

定义算子族 $U_{s}$,$A_{s}$ 和 $\frac{{\rm d}A_{s}}{{\rm d}s}$ 如下 $$U_{s}={\rm e}^{{\rm i}Ls}, A_{s}=\frac{\sin(L^{\frac{1}{2}}s)}{L^{\frac{1}{2}}}, \frac{{\rm d}A_{s}}{{\rm d}s}=\cos(L^{\frac{1}{2}}s).$$

那么对任意的 $f\in L^{1}({\Bbb H}\times{\Bbb R}^{3})\bigcap L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 有 \begin{equation}\label{5.1} ({\cal F}(U_{s}f))(a)E_{j,k-j}^{a}={\rm e}^{{\rm i}b_{k}(a)s}{\cal F}(f)(a)E_{j,k-j}^{a}, \tag{5.2} \end{equation} \begin{equation} \label{5.2}({\cal F}(A_{s}f))(a)E_{j,k-j}^{a}=a_{k}(a,s){\cal F}(f)(a)E_{j,k-j}^{a}, \tag{5.3} \end{equation} \begin{equation} \label{5.3}({\cal F}(\frac{{\rm d}A_{s}}{{\rm d}s}f))(a)E_{j,k-j}^{a}=\cos(b_{k}(a)s){\cal F}(f)(a)E_{j,k-j}^{a}, \tag{5.4} \end{equation} 这里的 $b_{k}(a)=(8(k+1)|a|+|a|^{2})^{\frac{1}{2}}$,$a_{k}(a,s)=\frac{\sin(b_{k}(a)s)}{b_{k}(a)}$.

Cauchy 问题 (5.1) 的解可以写成 $u=v+w$,其中 $v$ 是如下的齐次波方程的解 \begin{eqnarray}\label{5.4}\left\{ \begin{array}{l} \partial_{s}^{2}v+Lv=0,\\ v|_{s=0}=u_{0},\\ \partial_{s}v|_{s=0}=u_{1}.\end{array}\right.\tag{5.5}\end{eqnarray} 而 $w$ 是非齐次的波方程的解 \begin{eqnarray}\label{5.5}\left\{ \begin{array}{l} \partial_{s}^{2}w+Lw=f,\\ w|_{s=0}=0,\\ \partial_{s}w|_{s=0}=0.\end{array}\right.\tag{5.6}\end{eqnarray}

取 $f\in S([0,T]\times{\Bbb H}\times{\Bbb R}^{3})$, $u_{0},u_{1}\in S({\Bbb H}\times{\Bbb R}^{3})$.首先考虑方程(5.5). 由于 ${\cal F}(L f)(a)$ $E_{j,k-j}^{a}= (8(k+1)|a|+|a|^{2}){\cal F}(f)(a)E_{j,k-j}^{a}$ 对任意的 $f\in S({\Bbb H}\times{\Bbb R}^{3})$ 成立 (见 (3.12)式),那么 \begin{equation} {\cal F}(L v(s,\cdot))(a)E_{j,k-j}^{a}=(8(k+1)|a|+|a|^{2}){\cal F}(v(s,\cdot))(a)E_{j,k-j}^{a}.\tag{5.7} \end{equation} 记 $g(s)=\langle{\cal F}(v(s,\cdot))(a)E_{j,k-j},F\rangle_{\digamma_{a}}$ ($a,j,k$ 给定,$F\in\digamma_{a}$),那么 $g$ 必定满足 $$ g"(s)+(8(k+1)|a|+|a|^{2})g(s)=0; $$ $$ g(0)=\langle{\cal F}(u_{0})(a)E_{j,k-j},F\rangle_{\digamma_{a}},\ g' (0)=\langle{\cal F}(u_{1})(a)E_{j,k-j},F\rangle_{\digamma_{a}}. onumber $$ 根据常微分方程的经典理论,该方程存在唯一解 \begin{eqnarray} g(s)&=&\cos((8(k+1)|a|+|a|^{2})^{\frac{1}{2}}s)\langle{\cal F}(u_{0})(a)E_{j,k-j},F\rangle_{\digamma_{a}} onumber\\&&+\frac{\sin((8(k+1)|a|+|a|^{2})^{\frac{1}{2}}s)}{(8(k+1)|a|+|a|^{2})^{\frac{1}{2}}} \langle{\cal F}(u_{1})(a)E_{j,k-j},F\rangle_{\digamma_{a}}.\tag{5.8}\end{eqnarray} 那么根据 (5.2)-(5.4)式和群 Fourier 逆变换 (2.8),齐次波方程的解为 \begin{equation} u(s,\cdot)=\frac{{\rm d}A_{s}}{{\rm d}s}u_{0}+A_{s}u_{1}.\tag{5.9} \end{equation} 对非齐次波方程 (5.6) 使用同样的方法,可以得到 \begin{equation} w(s,\cdot)=\int_{0}^{s}A_{s-s' }f(s' ,\cdot){\rm d}s' .\tag{5.10} \end{equation} 那么波方程的解为 \[\left\{ \begin{array}{ll} v(s,\cdot)=\frac{{\rm d}A_{s}}{{\rm d}s}u_{0}+A_{s}u_{1}; onumber\\[3mm] \partial_{s}v(s,\cdot)=-A_{s}Lu_{0}+\frac{{\rm d}A_{s}}{{\rm d}s}u_{1}; onumber\\[3mm] \partial^{2}_{s}v(s,\cdot)=-\frac{{\rm d}A_{s}}{{\rm d}s}Lu_{0}-A_{s}Lu_{1}; onumber\\ [3mm] w(s,\cdot)=\int_{0}^{s}A_{s-s' }f(s' ,\cdot){\rm d}s' ; onumber\\[3mm] \partial_{s}w(s,\cdot)=\int_{0}^{s}\frac{{\rm d}A_{s-s' }}{{\rm d}s}f(s' ,\cdot){\rm d}s' onumber\\ [3mm] \partial^{2}_{s}w(s,\cdot)=-\int_{0}^{s}A_{s-s' }Lf(s' ,\cdot){\rm d}s' +f(s,\cdot).\end{array}\right.\] 可以看出,对于 $f(s,\cdot)\in L^{2}({\Bbb H}\times{\Bbb R}^{3}),u_{0}\in \dot{B}_{2,2}^{1}(L),$ $u_{1}\in L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 上式都有定义并且很明显的是波方程的解.

Cauchy 问题的解的唯一性是由波半群 $\cos(s\sqrt{L})$ 和 $\frac{\sin(s\sqrt{L})}{\sqrt{L}}$ 的有限传播速度所保证的 (参见文献^[22]). 它的意思是,如果记波半群的核为 $G_{s}$, 那么 \begin{equation}\label{unique} {\rm supp}(G_{s})\subset\{(x,y)\in({\Bbb H}\times{\Bbb R}^{3})\times({\Bbb H}\times{\Bbb R}^{3}): |x^{-1}y|\leq s \}.\tag{5.11} \end{equation} 如果 Cauchy 问题还存在另一个解 $u' $,那么 $u-u' $ 是如下方程的解 \[\left\{ \begin{array}{l} \partial_{s}^{2}\mu+L\mu=0, onumber\\ \mu|_{s=0}=0, onumber\\ \partial_{s}\mu|_{s=0}=0.\end{array}\right.\] 根据解的有限的传播速度 (5.11),由于在 $B((x_{0},t_{0}),s_{0})\times\{s=0\}, \forall s_{0}>0$ 上,有 $u-u' \equiv 0$ 成立,所以在 $\{((x,t),s):0\leq s\leq s_{0},|(x_{0},t_{0})^{-1}(x,t)|\leq s_{0}-s\}$ 上,有 $u-u' \equiv 0$ 成立. 这隐含了 Cauchy 问题的解的唯一性.

下面证明波方程的色散估计. 首先给出色散估计.

定理 5.1  } 令 $\rho\in [\frac{13}{2},\frac{17}{2}]$. 假定 $u_{0}\in\dot{B}_{1,1}^{\rho},u_{1}\in\dot{B}_{1,1}^{\rho-1}$, $v$ 是齐次方程的解,那么存在着常数 $C>0$ 使得 \begin{eqnarray} \| v(s,\cdot)\| _{L^{\infty}({\Bbb H}\times{\Bbb R}^{3})}\leq C|s|^{-\frac{1}{2}}(\| u_{0}\| _{\dot{B}_{1,1}^{\rho}}+ \| u_{1}\| _{\dot{B}_{1,1}^{\rho-1}}).\tag{5.12}\end{eqnarray}

为了证明它,给出如下一些引理.

引理 5.1  (参见文献[27,p332-334]) 假定 $g,h\in C^{\infty}([a,b])$,其中 $g$ 是实值的并且有 $h(b)=0$.$|g^{k}(x)|\geq \delta$ 对任意的 $x\in [a,b]$ 成立,其中 $k\in{\Bbb N}_{+}$ 且 $\delta>0$. 如果 $k=1$,还要求 $g' $ 在 $[a,b]$ 上是单调的. 那么存在着一个常数 $C_{k}>0$,只依赖于 $k$ 但是不依赖于 $a,b,g,h,\delta$,使得 $$ \bigg|\int_{a}^{b}{\rm e}^{-{\rm i}g(x)}h(x) {\rm d}x\bigg|\leq C_{k}\delta^{-\frac{1}{k}}\int_{a}^{b}|h' (x)|{\rm d}x.$$

引理 5.2  (参见文献[14,附录B.7,B.8]) 令 $J_{v}(r)$ 表示 Re$v>-\frac{1}{2}$ 的 Bessel 函数并且有 $0 ＜r＜\infty$. 对于$0＜r\leq 1$ 有 $$ |J_{v}(r)|\leq C_{0}{\rm e}^{C_{0}|{\rm Im} v|^{2}}r^{{\rm Re} v}.$$ 对于 $r\geq 1$ 有 $$ |J_{v}(r)|\leq C_{0}({\rm Re} v){\rm e}^{\pi|{\rm Im} v|+\pi^{2}|{\rm Im} v|^{2}}r^{-\frac{1}{2}}.$$

引理 5.3  (参见文献[3,引理4.3]) 对于 $\forall \alpha\in{\Bbb N}$,存在着一个常数 $C_{\alpha}$ 使得下面的结论对 $\forall y\geq 0,m\in{\Bbb N}$ 成立 \begin{equation} |L_{m}^{\alpha}(y){\rm e}^{-\frac{y}{2}}|\leq C_{\alpha}(m+1)^{\alpha}, \tag{5.13} \end{equation} \begin{equation} \bigg|y\frac{\rm d}{{\rm d}y}(L_{m}^{\alpha}(y){\rm e}^{-\frac{y}{2}})\bigg|\leq C_{\alpha}(m+1)^{\alpha}.\tag{5.14} \end{equation}

通过把求和与积分做比较,很容易得到以下估计

引理 5.4  取 $\beta\in{\Bbb R}$. 存在着常数 $C_{\beta}>0$ 使得对于 $0 ＜a＜b$ 有 \begin{equation} \sum\limits_{k\in{\Bbb Z}_{+},k+1\geq a}(k+1)^{\beta}\leq C_{\beta}a^{\beta+1},\beta＜-1; \tag{5.15} \end{equation} \begin{equation} \sum\limits_{k\in{\Bbb Z}_{+},k+1\leq b}(k+1)^{\beta}\leq C_{\beta}b^{\beta+1},\beta>-1; \tag{5.16} \end{equation} \begin{equation} \sum\limits_{k\in{\Bbb Z}_{+},a\leq k+1\leq b}(k+1)^{-1}\leq\log(C\frac{b}{a}).\tag{5.17} \end{equation}

下面引入一个非常有用的引理.

引理 5.5  存在着一个常数 $C>0$,使得对任意的 $\rho\in [\frac{13}{2},\frac{17}{2}]$, $j\in{\Bbb Z}$ 和 $s\in{\Bbb R},$ $s eq0$ 有 \begin{equation}\label{cccccc} \| {\rm e}^{-{\rm i}s\sqrt{L}}\psi_{j}\| _{L^{\infty}_{{\Bbb H}\times{\Bbb R}^{3}}}\leq C|s|^{-\frac{1}{2}}2^{j\rho}.\tag{5.18} \end{equation}

证  首先 \begin{eqnarray*} {\rm e}^{-{\rm i}s\sqrt{L}}\psi_{j}(x,t)&=&\frac{1}{2\pi^{5}}\sum\limits_{k=0}^{+\infty}\int_{\Im{\Bbb H}\backslash\{0\}} {\rm e}^{-{\rm i}\langle a,t\rangle }{\rm e}^{-{\rm i}s(8(k+1)|a|+|a|^{2})^{\frac{1}{2}}} onumber\\ &&\times {\cal R}(2^{-2j}(8(k+1)|a|+|a|^{2})){\rm e}^{-|a\| x|^{2}}L_{k}^{1}(2|a\| x|^{2})|a|^{2}{\rm d}a onumber\\ & =&\frac{1}{2\pi^{5}}\sum\limits_{k=0}^{+\infty}\int_{0}^{\infty} \int_{S^{2}}{\rm e}^{-2\pi{\rm i}\langle \tilde{a},\frac{|a|}{2\pi}t\rangle } {\rm d}\tilde{a}{\rm e}^{-{\rm i}s(8(k+1)|a|+|a|^{2})^{\frac{1}{2}}} onumber\\ &&\times {\cal R}(2^{-2j}(8(k+1)|a|+|a|^{2})) {\rm e}^{-|a\| x|^{2}}L_{k}^{1}(2|a\| x|^{2})|a|^{4}{\rm d}|a| onumber\\ & =&\frac{1}{2\pi^{5}}\sum\limits_{k=0}^{+\infty}\int_{0}^{\infty}\frac{2\pi}{|\frac{at}{2\pi}|^{\frac{1}{2}}}J_{\frac{1}{2}} (|at|){\rm e}^{-{\rm i}s(8(k+1)|a|+|a|^{2})^{\frac{1}{2}}}\\ &&\times{\cal R}(2^{-2j}(8(k+1)|a|+|a|^{2})) {\rm e}^{-|a\| x|^{2}}L_{k}^{1}(2|a\| x|^{2})|a|^{4}{\rm d}|a|, onumber \end{eqnarray*} $J_{k}(r)$ 表示 Bessel 函数. 令 $|a|=r$,有 \begin{eqnarray}\label{os} {\rm e}^{-{\rm i}s\sqrt{L}}\psi_{j}(x,t)&=&\sqrt{2}(\pi^{-\frac{7}{2}})\sum\limits_{k=0}^{+\infty}\int_{0}^{\infty} (r|t|)^{-\frac{1}{2}}J_{\frac{1}{2}}(r|t|){\rm e}^{-{\rm i}s(8(k+1)r+r^{2})^{\frac{1}{2}}} onumber\\&& \times {\cal R}(2^{-2j}(8(k+1)r+r^{2})){\rm e}^{-r|x|^{2}}L_{k}^{1}(2r|x|^{2})r^{4}{\rm d}r onumber\\ & =&\sqrt{2}(\pi^{-\frac{7}{2}})\sum\limits_{k=0}^{+\infty}\int_{a_{k}}^{b_{k}}{\rm e}^{-{\rm i}g_{k}(r)}h_{k}(r){\rm d}r.\tag{5.19}\end{eqnarray} 由于 supp$ {\cal R}(\cdot)\subset [\frac{1}{4},4]$ 且 $$ g_{k}(r)=s(8(k+1)r+r^{2})^{\frac{1}{2}}, $$ $$ h_{k}(r)=(r|t|)^{-\frac{1}{2}}J_{\frac{1}{2}}(r|t|){\cal R}(2^{-2j}(8(k+1)r+r^{2})){\rm e}^{-r|x|^{2}}L_{k}^{1}(2r|x|^{2})r^{4}, onumber $$ 可以得到 $\{a_{k}=-4(k+1)+\frac{\sqrt{64(k+1)^{2}+2^{2j}}}{2}; b_{k}=-4(k+1)+\frac{\sqrt{64(k+1)^{2}+2^{2j+4}}}{2}\}$.

直接计算有 $$ g_{k}' (r)=s(8(k+1)r+r^{2})^{-\frac{1}{2}}(4(k+1)+r), $$ $$ g_{k}"(r)=s(8(k+1)r+r^{2})^{-\frac{1}{2}}-s(8(k+1)r+r^{2})^{-\frac{3}{2}}(4(k+1)+r)^{2}. onumber $$ 那么在 $[a_{k},b_{k}]$ 上有 \begin{eqnarray}\label{5.13} |g_{k}"(r)|=|s|(8(k+1)r+r^{2})^{-\frac{3}{2}}16(k+1)^{2}\geq |s|2^{-3j+1}(k+1)^{2}.\tag{5.20}\end{eqnarray}

此外 \begin{eqnarray*} h_{k}' (r)&=&-(r|t|)^{-\frac{1}{2}}J_{\frac{3}{2}}(r|t|)r|t|{\cal R} (2^{-2j}(8(k+1)r+r^{2})){\rm e}^{-r|x|^{2}}L_{k}^{1}(2r|x|^{2})r^{3} onumber\\ && +(r|t|)^{-\frac{1}{2}}J_{\frac{1}{2}}(r|t|){\cal R}' (2^{-2j} (8(k+1)r+r^{2}))(2^{-2j}8(k+1)r\\ &&+2^{1-2j}r^{2}){\rm e}^{-r|x|^{2}} L_{k}^{1}(2r|x|^{2})r^{3}\\ &&+(r|t|)^{-\frac{1}{2}}J_{\frac{1}{2}}(r|t|){\cal R}(2^{-2j}(8(k+1)r+r^{2}))[({\rm e}^{-r|x|^{2}}L_{k}^{1}(2r|x|^{2}))' 2r|x|^{2}]r^{3} onumber\\ && +(r|t|)^{-\frac{1}{2}}J_{\frac{1}{2}}(r|t|){\cal R}(2^{-2j} (8(k+1)r+r^{2})){\rm e}^{-r|x|^{2}}L_{k}^{1}(2r|x|^{2})4r^{3}. onumber \end{eqnarray*} 通过 Bessel 函数的性质 (引理 5.2),对于 $k\geq-\frac{1}{2}$, $(r|t|)^{\pm\frac{1}{2}}J_{k}(r|t|)$ 在 $0＜r|t|＜\infty$ 是有界的.考虑到事实 ${\cal R}\in C_{c}^{\infty}$,$|{\cal R}(\cdot)|\leq 1$, supp${\cal R}(\cdot)\subset [\cdot,\frac{1}{4}\leq|\cdot|\leq4]$ 和引理5.3, 可以得到 \begin{eqnarray}\label{os1} |h_{k}' (r)|\leq C(k+1)r^{3}.\tag{5.21}\end{eqnarray} 由于 $$b_{k}=-4(k+1)+\frac{\sqrt{64(k+1)^{2}+2^{2j+4}}}{2}= \frac{2^{2j+2}}{4(k+1)+\frac{\sqrt{64(k+1)^{2}+2^{2j+4}}}{2}}, $$ 则 \begin{eqnarray}\label{os2} |b_{k}|&\leq &\frac{2^{2j+2}}{2\sqrt{2(k+1)\sqrt{64(k+1)^{2}+2^{2j+4}}}}\leq \frac{2^{2j+2}}{2\sqrt{16(k+1)\sqrt{(k+1)2^{j}}}} onumber\\ &=&2^{\frac{7}{4}j-1}(k+1)^{-\frac{3}{4}}.\tag{5.22}\end{eqnarray} 然后由 (5.19)-(5.22)式和引理5.1,可知 \begin{equation}\label{13} \| {\rm e}^{-{\rm i}s\sqrt{L}}\psi_{j}\| _{L^{\infty}_{{\Bbb H}\times{\Bbb R}^{3}}}\leq C\sum\limits_{k=0}^{\infty}(k+1)^{-3}2^{\frac{17}{2}j}\leq C 2^{\frac{17}{2}j}.\tag{5.23} \end{equation} 此外,如果 $k+1\geq 2^{j}$,那么 $b_{k}\leq C\frac{2^{2j}}{k+1}$; 如果 $k+1\leq 2^{j}$,那么 $b_{k}\leq C 2^{j}$. 这样由引理5.4 就得到 \begin{eqnarray}\label{17} \| {\rm e}^{-{\rm i}s\sqrt{L}}\psi_{j}\| _{L^{\infty}_{{\Bbb H}\times{\Bbb R}^{3}}} &\leq & C |s|^{-\frac{1}{2}} \bigg(\sum\limits_{k+1\geq 2^{j}}2^{(8+\frac{3}{2})j}(k+1)^{-4} +\sum\limits_{k+1\leq 2^{j}}2^{4j+\frac{3}{2}j}\bigg) onumber\\ & \leq& C|s|^{-\frac{1}{2}}2^{\frac{13}{2}j}.\tag{5.24}\end{eqnarray} 由 (5.23)和 (5.24)式,我们得到对任意的 $\rho\in [\frac{13}{2},\frac{17}{2}], j\in{\Bbb Z},s\in{\Bbb R},s eq 0$, \begin{equation} \| {\rm e}^{-{\rm i}s\sqrt{L}}\psi_{j}\| _{L^{\infty}_{{\Bbb H}\times{\Bbb R}^{3}}}\leq C |s|^{-\frac{1}{2}}2^{j\rho}.\tag{5.25} \end{equation}

令 $2\leq r\leq \infty$ 且有 $\frac{1}{r}+\frac{1}{\bar{r}}=1$. 记 \begin{equation}\label{wudiansi} \alpha(r)=\frac{1}{2}-\frac{1}{r}.\tag{5.26} \end{equation} 由于 $U_{s}$ 和 ${\cal R}(2^{-2j}L)$ 可交换的,并且由 $U_{s}$ 的左不变性, 对任意的 $f\in S({\Bbb H}\times{\Bbb R}^{3})$, \begin{equation} U_{s}(f\ast\psi_{j})=(U_{s}f)\ast\psi_{j}=f\ast(U_{s}\psi_{j} ),\tag{5.27} \end{equation} \begin{equation}\label{te} (U_{s}f)\ast\psi_{j}=f\ast\psi_{j}\ast(U_{s}\tilde{\psi}_{j})\tag{5.28} \end{equation} 成立,这里$ \tilde{\psi}_{j}=\psi_{j-1}+\psi_{j}+\psi_{j+1}$.根据Plancherel 定理很容易得到 \begin{equation}\label{ffffff} \| (U_{s}f)\ast\psi_{j}\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})}=\| U_{s}(f\ast\psi_{j})\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})} =\| f\ast\psi_{j}\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})}.\tag{5.29} \end{equation} 通过 (5.18),(5.28)式可知 \begin{equation}\label{dis} \| (U_{s}f)\ast\psi_{j}\| _{L^{\infty}({\Bbb H}\times{\Bbb R}^{3})}\leq C|s|^{-\frac{1}{2}}2^{j\rho}\| f\ast\psi_{j}\| _{L^{1}({\Bbb H}\times{\Bbb R}^{3})}, \tag{5.30} \end{equation} 并且由 (5.29)式,再使用插值定理,那么对任意的 $ 2\leq r\leq \infty$ 有 \begin{equation}\label{chs} \| (U_{s}f)\ast\psi_{j}\| _{L^{r}({\Bbb H}\times{\Bbb R}^{3})}\leq C|s|^{-\alpha(r)}2^{2j\rho\alpha(r)} \| f\ast\psi_{j}\| _{L^{\bar{r}}({\Bbb H}\times{\Bbb R}^{3})} (\frac{1}{r}+\frac{1}{\bar{r}}=1).\tag{5.31} \end{equation} 因此 \begin{eqnarray}\label{sihu} \| U_{s}f\| _{\dot{B}_{r,2}^{-\rho\alpha(r)}({\Bbb H}\times{\Bbb R}^{3})}&=& \bigg(\sum\limits_{j\in{\Bbb Z}}2^{-2j\rho\alpha(r)} \| (U_{s}f)\ast\psi_{j}\| _{L^{r}({\Bbb H}\times{\Bbb R}^{3})}^{2}\bigg)^{\frac{1}{2}} onumber\\ & \leq& C\bigg(\sum\limits_{j\in{\Bbb Z}}2^{2j\rho\alpha(r)} |s|^{-2\alpha(r)}\| f\ast\psi_{j}\| _{L^{\bar{r}} ({\Bbb H}\times{\Bbb R}^{3})}^{2}\bigg)^{\frac{1}{2}} onumber\\& \leq &C|s|^{-\alpha(r)}\| f\| _{\dot{B}_{\bar{r},2}^{\rho\alpha(r)}}.\tag{5.32}\end{eqnarray} 由于 $$ v(s,\cdot)=\frac{{\rm d}A_{s}}{{\rm d}s}u_{0}+A_{s}u_{1}, $$ 把它写成 $$ v=\frac{{\rm d}A_{s}}{{\rm d}s}u_{0}+A_{s}u_{1}=\frac{U_{s}+U_{-s}}{2}u_{0}+L^{-\frac{1}{2}} \frac{U_{s}-U_{-s}}{2i}u_{1}.$$ 又因为 $Z({\Bbb H}\times{\Bbb R}^{3})$ 在 $\dot{B}_{p,r}^{\rho} $ 稠密, 通过 (5.30) 式可以得到 \begin{eqnarray} \| U_{s}u_{0}\| _{L^{\infty}}&=& \bigg\| \sum\limits_{j\in{\Bbb Z}}U_{s}u_{0}\ast\psi_{j}\bigg\| _{L^{\infty}} onumber\\ &\leq& \sum\limits_{j\in{\Bbb Z}}\| U_{s}u_{0}\ast\psi_{j}\| _{L^{\infty}} onumber\\ & \leq&\sum\limits_{j\in{\Bbb Z}}C|s|^{-\frac{1}{2}}2^{ j\rho}\| u_{0}\ast\psi_{j}\| _{L^{1}} onumber\\ &=& C|s|^{-\frac{1}{2}}\| u_{0}\| _{B_{1,1}^{\rho}} \tag{5.33}\end{eqnarray} 以及 \begin{eqnarray} \| L^{-\frac{1}{2}}U_{s}u_{1}\| _{L^{\infty}}&\leq&\sum\limits_{j\in{\Bbb Z}} \| L^{-\frac{1}{2}} U_{s}u_{1}\ast\psi_{j}\| _{L^{\infty}} onumber\\ &\leq & \sum\limits_{j\in{\Bbb Z}}C|s|^{-\frac{1}{2}}2^{ j\rho}\| L^{-\frac{1}{2}}u_{1}\ast\psi_{j}\| _{L^{1}} onumber\\ &\leq& C|s|^{-\frac{1}{2}}\sum\limits_{j\in{\Bbb Z}}2^{j(\rho-1)}\| u_{1}\ast\psi_{j}\| _{L^{1}} onumber\\ &=&C|s|^{-\frac{1}{2}} \| u_{1}\| _{B_{1,1}^{\rho-1}}.\tag{5.34}\end{eqnarray} 这样就证明了色散估计. 上式最后一个不等式是由 (3.19) 式得到的.

现在证明Strichartz估计. 首先需要一些引理 (参见文献^[13]).

引理 5.6  (参见文献[13,引理2.2]) 令 $H$ 是一个 Hilbert 空间,$X$ 是一个 Banach 空间,$X^{*}$ 是 $X$ 的对偶, 并且 $D$ 是一个在 $X$ 中稠密的向量空间. 令 $A \in {{\text{£}}_a}(D,H),{A^*} \in {{\text{£}}_a}(H,D_a^*)$ 是它的伴随算子,满足 \begin{equation} \langle A^{*}v,f\rangle_{D}=\langle v,Af\rangle_{H},\forall f\in D,\forall v\in H, \tag{5.35} \end{equation} 这里的 ${{\text{£}}_a}(Y,Z)$ 是由向量空间 $Y$ 到向量空间 $Z$ 的线性映射, $D_{a}^{*} $ 是 $D$ 的代数对偶,$\langle \varphi,f \rangle _{D}$ 是 $D_{a}^{*}$ 对 $D$ 中元素的作用,$\langle ,\rangle _{H}$ 是 $H$ 中的内积. 那么下面三个条件是等价的.

(1)~ 存在着 $a$,$0\leq a ＜ \infty$ 使得对所有的 $f\in D$ \begin{equation} \| Af\| \leq a\| f;X\| .\tag{5.36} \end{equation}

(2)~ $\Re(A^{*})\subset X^{*}$,并且存在着 $a,0\leq a＜\infty$,使得对所有的 $v\in H$ \begin{equation} \| A^{*}v; X^{*}\| \leq a\| v\| .\tag{5.37} \end{equation}

(3)~ $\Re(A^{*}A)\subset X^{*}$ 并且存在着 $a,0\leq a＜\infty$,使得对所有的 $f\in D$ \begin{equation} \| A^{*}Af; X^{*}\| \leq a^{2}\| f;X\| , \tag{5.38} \end{equation} 这里 $\| \cdot\| $ 表示 $H$ 的范数. 常数 $a$ 在三个结论中一致. 如果其中有一个成立, 算子 $A$ 和 $A^{*}A$ 可以分别连续延拓到 $X$ 到 $H$ 上的和 $X$ 到 $X^{*}$ 上的有界线性算子.

引理 5.7  (参见文献[13,推论2.1]) 令 $H,D$ 和 $(X_{i},A_{i},a_{i}),i=1,2$,都满足引理5.6 中的任何一个条件. 那么对 $i,j=1,2$,$\Re(A_{i}^{*}A_{j})\subset X_{i}^{*}$,和所有 $f\in D$, \begin{equation} \| A_{i}^{*}A_{j}f;X_{i}^{*}\| \leq a_{i}a_{j}\| f;X_{j}\| .\tag{5.39} \end{equation}

我们将在下面的情形中应用以上引理. $L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 是一个 Hilbert 空间. 定义在本节开始的 $U_{s}$ 是一个 $L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 中的强连续单参数酉群. 令 $I$ 是 ${\Bbb R}$ 中的一个区间,可以取 ${\Bbb R}$ 自身. 定义 $A$ 是从 $L^{1}(I,L^{2}({\Bbb H}\times{\Bbb R}^{3}))$ 到 $L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 的有界线性算子,其形式如下 $$ Af=\int_{I}U_{-s}f(s,\cdot){\rm d}s.$$ 那么它的伴随 $A^{*}$ 为 $$ A^{*}v(s)=U_{s}v, $$ 并将 $L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 映到 $L^{\infty}(I,L^{2}({\Bbb H}\times{\Bbb R}^{3}))$,这里的对偶是通过 $L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 和 $L^{2}(I,L^{2}({\Bbb H} \times{\Bbb R}^{3}))$ 上的内积来定义的. 因此 $A^{*}A$ 是一个从 $L^{1}(I,L^{2}({\Bbb H}\times{\Bbb R}^{3}))$ 到 $L^{\infty}(I,L^{2}({\Bbb H}\times{\Bbb R}^{3}))$ 的有界线性算子,其形式为 $$ A^{*}Af=\int_{I}U_{s-s' }f(s' ,\cdot){\rm d}s' .$$ 很明显我们还需要 retarded 算子 $(A^{*}A)_{R}$ $$ (A^{*}A)_{R}f(s)=\int_{I}\chi_{+}(s-s' )U_{s-s' }f(s' ,\cdot){\rm d}s' .$$

设 $X$ 是一个和时间空间相关的分布空间,如果其元素与区间 $J$ 上的特征函数相乘后是 $X$ 上的有界算子并且关于 $J$ 一致有界,那么称 $X$ 在时间限制下是稳定的.

引理 5.8  令 $I$ 是 ${\Bbb R}$ 上的区间,而 $X\subset S' (I\times({\Bbb H}\times{\Bbb R}^{3}))$ 是一个 Banach 空间, 令 $X$ 在时间限制下是稳定的,$A$ 的定义如上且 $X$ 和 $A$ 满足引理5.6 中的任何一个条件.那么算子 $(A^{*}A)_{R}$ 是一个由$L_{s}^{1}(I,L^{2}({\Bbb H}\times{\Bbb R}^{3}))$ 到 $X^{*}$ 以及由 $X$ 到 $L_{s}^{\infty}(I,L^{2}({\Bbb H}\times{\Bbb R}^{3}))$ 的有界算子.

证  证明与文献^[13] 中的引理2.3 相同.

下面的定理在Strichartz估计中非常有用.

定理 5.2  令 $\rho_{1},\mu\in{\Bbb R}$,$2\leq \frac{p_{1}}{2},r_{1}\leq\infty$, $\rho\in [\frac{13}{2},\frac{17}{2}]$,且有 \begin{equation}\label{5.32} \frac{2}{p_{1}}=\alpha(r_{1}); \tag{5.40} \end{equation} \begin{equation} \label{5.33} \rho_{1}-1=-\rho\alpha(r_{1}); \tag{5.41} \end{equation} 那么有下面的结论成立

(1)~ 如果 $u_{0}\in\dot{H}^{1}({\Bbb H}\times{\Bbb R}^{3})$,$u_{1}\in L^{2}({\Bbb H}\times{\Bbb R}^{3})$, 那么 \begin{equation}\label{dyuj} \| v\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}})}+\| \partial_{s}v\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1})} \leq C\{\| u_{0}\| _{\dot{H}^{1}}+\| u_{1}\| _{L^{2}}\}.\tag{5.42} \end{equation}

(2)~ 对任意区间 $[0,T]$,$0＜T\leq\infty$ 有 \begin{equation}\label{ww} \| w\| _{L^{p_{1}}([0,T],\dot{B}_{r_{1},2}^{\rho_{1}})}+\| \partial_{s}w\| _{L^{p_{1}}([0,T],\dot{B}_{r_{1},2}^{\rho_{1}-1})} \leq C\| f\| _{L^{1}([0,T],L^{2})}.\tag{5.43} \end{equation}

证  由 (5.32)式,首先 \begin{eqnarray} \| U_{s}f\| _{\dot{B}_{r,2}^{-\rho\alpha(r)}} \leq C|s|^{-\alpha(r)}\| f\| _{\dot{B}_{\bar{r},2}^{\rho\alpha(r)}}.\tag{5.44}\end{eqnarray} 此外成立着 $$ v=\frac{{\rm d}A_{s}}{{\rm d}s}u_{0}+A_{s}u_{1}=\frac{U_{s}+U_{-s}}{2}u_{0}+L^{-\frac{1}{2}} \frac{U_{s}-U_{-s}}{2{\rm i}}u_{1} $$ 和 $$ \partial_{s}v=-A_{s}Lu_{0}+\frac{{\rm d}A_{s}}{{\rm d}s}u_{1} =-\frac{U_{s}-U_{-s}}{2{\rm i}}L^{\frac{1}{2}}u_{0}+\frac{U_{s}+U_{-s}}{2}u_{1}, $$ 这里的 $L^{\frac{1}{2}}u_{0}$ 和 $u_{1}$ 同属于 $L^{2}({\Bbb H}\times{\Bbb R}^{3})$.

由 (3.19)式, $$ \| v\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}})}+\| \partial_{s}v\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )} \leq C(\| L^{\frac{1}{2}}v\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )} +\| \partial_{s}v\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )}).$$ 那么 \begin{eqnarray}&& \| v\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}})}+\| \partial_{s}v\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )} onumber\\& \leq &C(\| U_{s}L^{\frac{1}{2}}u_{0}\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )} +\| U_{-s}L^{\frac{1}{2}}u_{0}\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )} onumber\\&& +\| U_{s}u_{1}\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )}+\| U_{-s}u_{1}\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )}).\tag{5.45}\end{eqnarray}

我们断言 \begin{equation}\label{dee} \| U_{s}g\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )}\leq C\| g\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})}.\tag{5.46} \end{equation}

为了证明 (5.46)式,只需对 $\varphi\in L^{2}({\Bbb H}\times{\Bbb R}^{3})$,$\zeta\in L^{\bar{p}_{1}}({\Bbb R},\dot{B}_{\bar{r}_{1},2}^{\rho\alpha(r_{1})} )$ 证明如下结论成立.\begin{equation} |\langle\zeta,U_{s}\varphi\rangle_{L^{2}({\Bbb R}\times({\Bbb H}\times{\Bbb R}^{3}))}| \leq C\| \varphi\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})}\| \zeta\| _{L^{\bar{p}_{1}}({\Bbb R},\dot{B}_{\bar{r}_{1},2}^{\rho\alpha(r_{1})} )}.\tag{5.47} \end{equation} 根据稠密性,进一步的只要对 $\varphi\in L^{2}({\Bbb H}\times{\Bbb R}^{3})$ 和 $\zeta\in S({\Bbb R}\times{\Bbb H}\times{\Bbb R}^{3})$ (并且对 $\forall s\in{\Bbb R}$ 在 ${\Bbb H}\times{\Bbb R}^{3}$ 有任意消失性) 证明下式即可.\begin{equation} |\langle\zeta,U_{s}\varphi\rangle_{L^{2}({\Bbb R}\times({\Bbb H}\times{\Bbb R}^{3}))}| \leq C\| \varphi\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})}\| \zeta\| _{L^{\bar{p}_{1}}({\Bbb R},\dot{B}_{\bar{r}_{1},2}^{\rho\alpha(r_{1})} )}.\tag{5.48} \end{equation}

首先 \begin{eqnarray}\label{sh} |\langle\zeta,U_{s}\varphi\rangle_{L^{2}({\Bbb R}\times({\Bbb H}\times{\Bbb R}^{3}))}| &=& \bigg|\langle\int_{{\Bbb R}}U_{-\tau}\zeta(\tau,\cdot){\rm d}\tau,\varphi\rangle_{L^{2}({\Bbb H}\times{\Bbb R}^{3})}\bigg| onumber\\ &\leq& \| \varphi\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})} \bigg\| \int_{{\Bbb R}}U_{-\tau}\zeta(\tau,\cdot){\rm d}\tau\bigg\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})}.\tag{5.49}\end{eqnarray} 然后 \begin{eqnarray}\label{sk} \bigg\| \int_{{\Bbb R}}U_{-\tau}\zeta(\tau,\cdot){\rm d}\tau\bigg\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})}^{2} &=&\int_{{\Bbb R}}\langle\zeta(s,\cdot),\int_{{\Bbb R}}U_{s-\tau}\zeta(\tau,\cdot)\rangle_{L^{2}({\Bbb H}\times{\Bbb R}^{3})}{\rm d}s onumber\\ & \leq& \| \zeta\| _{L^{\bar{p}_{1}}({\Bbb R},\dot{B}_{\bar{r}_{1},2}^{\rho\alpha(r_{1})} )} \bigg\| \int_{{\Bbb R}}U_{s-\tau}\zeta(\tau){\rm d}\tau\bigg\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{-\rho\alpha(r_{1})} )}.\tag{5.50}\end{eqnarray} 进一步有 \begin{eqnarray}&& \bigg\| \int_{{\Bbb R}}U_{s-\tau}\zeta(\tau,\cdot){\rm d}\tau\bigg\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{-\rho\alpha(r_{1})} )} =\bigg(\int_{{\Bbb R}}\bigg\| \int_{{\Bbb R}}U_{s-\tau}\zeta(\tau,\cdot){\rm d}\tau \bigg\| _{\dot{B}_{r_{1},2}^{-\rho\alpha(r_{1})}}^{^{p_{1}}}{\rm d}s\bigg)^{\frac{1}{p_{1}}}. onumber \end{eqnarray} 由于 \begin{eqnarray} \bigg\| \int_{{\Bbb R}}U_{s-\tau}\zeta(\tau,\cdot){\rm d}\tau\bigg\| _{\dot{B}_{r_{1},2}^{-\rho\alpha(r_{1})}} =\bigg[\sum\limits_{j\in{\Bbb Z}}2^{2j(-\rho\alpha(r_{1}))} \bigg\| \int_{{\Bbb R}}U_{s-\tau}\zeta(\tau,\cdot)\ast \psi_{j}{\rm d}\tau \bigg\| _{L^{r_{1}}}^{2}\bigg]^{\frac{1}{2}}, \tag{5.51}\end{eqnarray} 使用 Minkowski 不等式得到 \begin{eqnarray*} \bigg\| \int_{{\Bbb R}}U_{s-\tau}\zeta(\tau,\cdot)\ast \psi_{j}{\rm d}\tau\bigg\| _{L^{r_{1}}} &\leq &\bigg\| \int_{{\Bbb R}}|U_{s-\tau}\zeta(\tau,\cdot)\ast \psi_{j}|{\rm d}\tau\bigg\| _{L^{r_{1}}} onumber\\ & \leq& \int_{{\Bbb R}}\| U_{s-\tau}\zeta(\tau,\cdot)\ast \psi_{j}\| _{L^{r_{1}}}{\rm d}\tau \\ &\leq &\int_{{\Bbb R}} C|s-\tau|^{-\alpha(r)}2^{2j\rho\alpha(r_{1})}\| \zeta\ast\psi_{j}\| _{L^{\bar{r}_{1}}}{\rm d}\tau onumber.\end{eqnarray*} 因此下式成立 \begin{eqnarray}&& \bigg\| \int_{{\Bbb R}}U_{s-\tau}\zeta(\tau,\cdot)\ast\psi_{j}{\rm d}\tau\bigg\| _{L^{p_{1}}({\Bbb R},L^{r_{1}})} onumber\\ &\leq & C2^{2j\rho\alpha(r_{1})} \bigg(\int_{{\Bbb R}}\bigg(\int_{{\Bbb R}}|s-\tau|^{-\alpha(r)}\| \zeta(\tau,\cdot)\ast\psi_{j}\| _{L^{\bar{r}_{1}}}{\rm d}\tau\bigg)^{p_{1}}{\rm d}s\bigg)^{\frac{1}{p_{1}}}.\tag{5.52}\end{eqnarray} 由 Hardy-Littlewood-Sobelev 不等式(见文献 ^[27]),则 \begin{equation} \bigg(\int_{{\Bbb R}}\bigg(\int_{{\Bbb R}}|s-\tau|^{-\alpha(r_{1})}\| \zeta(\tau,\cdot)\ast\psi_{j}\| _{L^{\bar{r}_{1}}}{\rm d}\tau\bigg)^{p_{1}}{\rm d}s\bigg)^{\frac{1}{p_{1}}}\leq \| \zeta\ast\psi_{j}\| _{L^{\bar{p}_{1}}({\Bbb R},L^{\bar{r}_{1}})}.\tag{5.53} \end{equation} 那么 \begin{eqnarray} \bigg\| \int_{{\Bbb R}}U_{s-\tau}\zeta(\tau,\cdot)\ast\psi_{j}{\rm d}\tau\bigg\| _{L^{p_{1}}({\Bbb R},L^{r_{1}})}\leq C 2^{2j\rho\alpha(r_{1})}\| \zeta\ast\psi_{j}\| _{L^{\bar{p}_{1}}({\Bbb R},L^{\bar{r}_{1}})}.\tag{5.54}\end{eqnarray} 取 $\ell^{2}$ 范数并且使用 Minkowski 不等式得到 \begin{equation}\label{sj} \bigg\| \int_{{\Bbb R}}U_{s-\tau}\zeta(\tau,\cdot){\rm d}\tau\bigg\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{-\rho\alpha(r_{1})})}\leq \| \zeta\| _{L^{\bar{p_{1}}}({\Bbb R},\dot{B}_{\bar{r}_{1},2}^{\rho\alpha(r_{1})})}.\tag{5.55} \end{equation} 由 (5.49),(5.50)和 (5.55) 式知 \begin{equation} |\langle\zeta,U_{s}\varphi\rangle_{L^{2}({\Bbb R}\times{\Bbb H}\times{\Bbb R}^{3})}| \leq C\| \varphi\| _{L^{2}({\Bbb H}\times{\Bbb R}^{3})}\| \zeta\| _{L^{\bar{p_{1}}}({\Bbb R},\dot{B}_{\bar{r}_{1},2}^{\rho\alpha(r_{1})})}, \tag{5.56} \end{equation} 这样 (5.46)式成立.最后得到 \begin{equation} \| U_{s}L^{\frac{1}{2}}u_{0}\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )} \leq C \| L^{\frac{1}{2}}u_{0}\| _{L^{2}} \leq C\| u_{0}\| _{H^{1}}; \tag{5.57} \end{equation} \begin{equation} \| U_{s}u_{1}\| _{L^{p_{1}}({\Bbb R},\dot{B}_{r_{1},2}^{\rho_{1}-1} )}\leq C\| u_{1}\| _{L^{2}}.\tag{5.58} \end{equation} 同样可以对 $U_{-s}$ 进行讨论. 这样就得到了 (5.42)式.

接下来证明 (5.43)式.$$ w=\int_{0}^{s}A_{s-s' }f(s' ,\cdot){\rm d}s' =\int_{0}^{s}L^{-\frac{1}{2}} \frac{U_{s-s' }-U_{s' -s}}{2{\rm i}}f(s' ,\cdot){\rm d}s' , $$ $$ \partial_{s}w=\int_{0}^{s}\frac{U_{s-s' }+U_{s' -s}}{2}f(s' ,\cdot){\rm d}s' .$$ 首先成立着 \begin{eqnarray*} &&\| w\| _{L^{p^{1}}([0,T],\dot{B}_{r_{1},2}^{\rho_{1}})} \\ &\leq& C\| L ^{\frac{1}{2}}w\| _{L^{p_{1}}([0,T],\dot{B}_{r_{1},2}^{\rho_{1}-1} )} onumber\\ & \leq& C\bigg(\bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s' \bigg \| _{L^{p_{1}}([0,T],\dot{B}_{r_{1},2}^{\rho_{1}-1} )} +\bigg\| \int_{0}^{s}U_{s' -s}f(s' ,\cdot){\rm d}s' \bigg\| _{L^{p_{1}} ([0,T],\dot{B}_{r_{1},2}^{\rho_{1}-1} )}\bigg) onumber \end{eqnarray*} 以及 \begin{eqnarray*}&& \| \partial_{s}w\| _{L^{p_{1}}([0,T],\dot{B}_{r_{1},2}^{\rho_{1}-1} )} onumber\\ &\leq &C\bigg(\bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s' \bigg \| _{L^{p_{1}}([0,T],\dot{B}_{r_{1},2}^{\rho_{1}-1} )} +\bigg\| \int_{0}^{s}U_{s' -s}f(s' ,\cdot){\rm d}s' \bigg\| _{L^{p_{1}} ([0,T],\dot{B}_{r_{1},2}^{\rho_{1}-1} )}\bigg) onumber.\end{eqnarray*} 现在只需要证明 \begin{equation} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s' \bigg\| _{L^{p_{1}} ([0,T],\dot{B}_{r_{1},2}^{\rho_{1}-1} )} \leq C\| f\| _{L^{1}([0,T],L^{2})}, \tag{5.59} \end{equation} 这里的指标满足假定 (5.40) 和 (5.41)式.可以知道 \begin{eqnarray} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s' \bigg\| _{L^{p_{1}}([0,T],\dot{B}_{r_{1},2}^{\rho_{1}-1} )} =\bigg(\int_{0}^{T}\bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s' \bigg \| _{\dot{B}_{r_{1},2}^{\rho_{1}-1}}^{^{p_{1}}}\bigg)^{\frac{1}{p_{1}}}.onumber \end{eqnarray} 由于 \begin{eqnarray} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s'\bigg \| _{\dot{B}_{r_{1},2}^{\rho_{1}-1}} =\bigg[\sum\limits_{j\in{\Bbb Z}}2^{2j(\rho_{1}-1)} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot)\ast \psi_{j}{\rm d}s'\bigg \| _{L^{r_{1}}}^{2}\bigg]^{\frac{1}{2}}, \tag{5.60}\end{eqnarray} 使用 Minkowski 不等式, \begin{eqnarray*} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot)\ast \psi_{j}{\rm d}s'\bigg \| _{L^{r_{1}}} &\leq&\bigg\| \int_{0}^{s}|U_{s-s' }f(s' ,\cdot)\ast \psi_{j}|{\rm d}s'\bigg \| _{L^{r_{1}}} onumber\\ &\leq& \int_{0}^{s}\| U_{s-s' }f(s' ,\cdot)\ast \psi_{j}\| _{L^{r_{1}}}{\rm d}s' \\ &\leq& \int_{0}^{s} C|s-s' |^{-\alpha(r_{1})}2^{2j\rho\alpha(r_{1})} \| f\ast\psi_{j}\| _{L^{\bar{r}_{1}}}{\rm d}s' onumber.\end{eqnarray*} 那么 \begin{eqnarray}&& \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot)\ast\psi_{j}{\rm d}s' \bigg\| _{L^{p_{1}}([0,T],L^{r_{1}})} onumber\\ &\leq& C2^{2j\rho\alpha(r_{1})}\bigg(\int_{0}^{T}\bigg(\int_{0}^{s} |s-s' |^{-\alpha(r_{1})}\| f(s' ,\cdot)\ast\psi_{j}\| _{L^{\bar{r}_{1}}}{\rm d}s'\bigg )^{p_{1}}{\rm d}s\bigg)^{\frac{1}{p_{1}}}.\tag{5.61}\end{eqnarray} 使用 Hardy-Littlewood-Sobelev 不等式 (见文献 ^[27]),则 \begin{equation} \bigg(\int_{0}^{T}\bigg(\int_{0}^{s}|s-s' |^{-\alpha(r_{1})}\| f(s' ,\cdot)\ast\psi_{j}\| _{L^{\bar{r}_{1}}}{\rm d}s'\bigg )^{p_{1}}{\rm d}s\bigg)^{\frac{1}{p_{1}}}\leq \| f\ast\psi_{j}\| _{L^{\bar{p}_{1}}([0,T],L^{\bar{r}_{1}})}.\tag{5.62} \end{equation} 那么有 \begin{eqnarray} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot)\ast\psi_{j}{\rm d}s' \bigg\| _{L^{p_{1}}([0,T],L^{r_{1}})}\leq C 2^{2j\rho\alpha(r_{1})}\| f\ast\psi_{j}\| _{L^{\bar{p_{1}}}([0,T],L^{\bar{r}_{1}})}.\tag{5.63}\end{eqnarray} 取 $\ell^{2}$ 范数并应用 Minkowski 不等式 \begin{equation}\label{cv} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s' \bigg\| _{L^{p_{1}}([0,T],\dot{B}_{r_{1},2}^{-\rho\alpha(r_{1})})}\leq C \| f\| _{L^{\bar{p_{1}}}([0,T],\dot{B}_{\bar{r}_{1},2}^{\rho\alpha(r_{1})})}.\tag{5.64} \end{equation} 令 $p_{1}=\infty,r_{1}=2$,由 (5.64) 式知 \begin{eqnarray*} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s' \bigg\| _{L^{\infty}([0,T];L^{2})}\leq C\| f\| _{L^{1}([0,T],L^{2})}.\end{eqnarray*} 由Ginibre 和Velo 的结果 (参见文献[13,引理 2.2] 或者本节引理 5.8), 则 \begin{equation}\label{kchess} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s' \bigg\| _{L^{p_{1}}([0,T],\dot{B}_{r_{1},2}^{-\rho\alpha(r_{1})})}\leq C\| f\| _{L^{1}([0,T],L^{2})}, \tag{5.65} \end{equation} \begin{equation} \bigg\| \int_{0}^{s}U_{s-s' }f(s' ,\cdot){\rm d}s' \bigg\| _{L^{\infty}([0,T];L^{2})}\leq C\| f\| _{L^{\bar{p_{1}}}([0,T],\dot{B}_{\bar{r}_{1},2}^{\rho\alpha(r_{1})})}, \tag{5.66} \end{equation} 这里 $\frac{1}{p_{1}}=\frac{\alpha(r_{1})}{2}$,$\rho_{1}-1=-\rho\alpha(r_{1})$.定理得证.

最后证明Strichartz估计.

定理 5.3  令 $u$ 是波方程的 Cauchy 问题的解,假定成立着如下条件 \[\left\{ \begin{array}{l} \rho' \in \Big[\frac{13}{2},\frac{17}{2}\Big],\\\ p>2\rho' ,\\\ \frac{1}{q}+\frac{1}{p}\Big(2-\frac{\rho' }{5}\Big)=\frac{2}{5}.\end{array}\right.\] 那么存在着常数 $C$,对任意的 $T>0$,有 \[\|u{{\|}_{{{L}^{p}}([0,T],{{L}^{q}}(\mathbb{H}\times {{\mathbb{R}}^{3}}))}}\le C\{\|f{{\|}_{{{L}^{1}}([0,T];{{L}^{2}}(\mathbb{H}\times {{\mathbb{R}}^{3}}))}}+{{E}_{0}}{{(u)}^{\frac{1}{2}}}\},\tag{5.67}\] 这里 $E_{0}(u)=\| u_{0}\| _{\dot{H}^{1}({\Bbb H}\times{\Bbb R}^{3})}^{2}+\| u_{1}\| _{L^{2}}^{2}$.

证  令 $r,\rho$ 满足定理5.2 中的指标关系. 也即, $\frac{2}{p}=\alpha(r)$ 和 $\rho=1-\rho' \alpha(r)$.

由 $\frac{1}{q}+\frac{1}{p}(2-\frac{\rho' }{5})=\frac{2}{5}$, 我们知道 $\frac{1}{q}=\frac{1}{r}-\frac{\rho}{10}$.

由于 $p>2\rho' >5-\frac{\rho' }{2}$,那么 $1-\rho' \frac{2}{p}>0$ 和 $p-2\rho' <5p-20$ 成立.这表明 $0 ＜\rho＜\frac{10}{r}$.

当 $\frac{1}{q}=\frac{1}{r}-\frac{\rho}{10}$,$0＜\rho＜\frac{10}{r}$,有 $\dot{B}_{r,2}^{\rho}\subset L^{q} $. 那么由定理5.2, $$ \| v\| _{L^{p}([0,T],L^{q}({\Bbb H}\times{\Bbb R}^{3}))}\leq C\| v\| _{L^{p}({\Bbb R},\dot{B}_{r,2}^{\rho})}\leq C E_{0}(u)^{\frac{1}{2}}.$$ 结合 (5.43) 式,那么 (5.67)式得证.