描述弱非线性水波的Stewartson系统形如:

$ \begin{equation*} \begin{cases} i\frac{\partial u}{\partial t} + \delta \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \chi |u|^2 u + b u\frac{\partial \varphi}{\partial x} \\ \frac{\partial^2 \varphi}{\partial x^2} + m \frac{\partial^2 \varphi}{\partial y^2} = \frac{\partial}{\partial x}(|u|^2) \end{cases} \end{equation*} $

式中$ \delta=\pm 1 $, 以及$ m $可以为负数^[1], 该方程具有非常重要的应用价值^{[2, 3]}, 该方程首次由Davey和Stewartson^[4]引入. 该方程最简单的形式为齐次Schrödinger方程:

$ \begin{equation} \begin{cases} i \partial_t u(t, x) + \Delta u(t, x) = 0, \\ u|_{t=0} = \varphi (x). \end{cases} \end{equation} $

(1.1)

众所周知, 上述Cauchy问题具有光滑性效应^[5]. 1985年, Ginibre J, Velo G^[6]通过紧凑性方法证明了非线性薛定谔方程全局弱解的存在, 并在更强的假设下证明了这些解的唯一性. 在1987年, Yajima K^[7]研究了拉普拉斯算子前系数为$ -\frac{1}{2} $, 且带有一般势函数的情况, 并论证了解的存在性、唯一性. Ohta M^{[8, 9]}在1994年和1995年分别讨论了广义Davey-Stewartson系统驻波的稳定性和不稳定性. 2001年, 王保祥和郭柏灵^[10]研究了椭圆-椭圆型的Davey-Stewartson方程初值问题的解和散射算子的存在性. 本文讨论$ \delta=-1 $的齐次Davey-Stewartson系统的方程, 即

$ \begin{equation} \begin{cases} i \partial_t u(t, x, y) + \Delta_x u(t, x, y) - \Delta_y u(t, x, y) = 0, \\ u|_{t=0} = \varphi (x, y), \quad x, y\in\mathbb{R}^N. \end{cases} \end{equation} $

(1.2)

这是一个非椭圆类的齐次Schrödinger方程的Cauchy问题, 本文的主要结论是:

定理1.1  对于方程(1.2), 假设$ m\in\mathbb{N}, \varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $以及$ (1+|(x, y)|^m) \varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $. 则Cauchy问题(1.2)存在唯一弱解满足:

$ \begin{equation} u \in \bigcap\limits_{0 \leq j \leq \left[\frac{m}{2}\right]} C^{j}(\mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H^{m-2j}_{loc}(\mathbb{R}^{N} \times \mathbb{R}^{N})), \end{equation} $

(1.3)

特别的, 若任给$ m\in\mathbb{N} $, 有$ (1+|(x, y)|^m)\varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 则$ u \in C^{\infty}(\mathbb{R} \ \backslash \ \lbrace 0 \rbrace \times \mathbb{R}^{N} \times \mathbb{R}^{N}) $.

因此我们证明了非椭圆类的齐次Schrödinger方程的Cauchy问题(1.2)也有完全类似于经典齐次Schrödinger方程的Cauchy问题的光滑性效应.

我们首先研究Cauchy问题(1.2)的弱解, 对于定义在$ \mathbb{R}^{2N}_{x, y} $的函数$ f(x, y) $定义其Fourier变换:

$ \cal{F}(f)(\xi, \eta)=\hat{f}(\xi , \eta)=\int_{\mathbb{R}^{2N}}e^{-ix \cdot \xi} e^{-i y \cdot \eta} f(x, y)dxdy $

和Fourier逆变换

$ \cal{F}^{-1}(f)(x, y)=\frac{1}{(2\pi)^{2N}}\int_{\mathbb{R}^{2N}} e^{ix\cdot \xi} e^{iy\cdot \eta} f(\xi, \eta) d\xi d\eta. $

我们有下面的命题.

命题2.1  假设$ \varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 则Cauchy问题(1.2)存在唯一弱解

$ \begin{equation} u(t, x, y)=\cal{T}(t)(\varphi) (x, y)\in C^0(\mathbb{R}, L^2(\mathbb{R}^{N} \times \mathbb{R}^{N})), \end{equation} $

(2.1)

其中$ \cal{T}(t) $满足

$ \cal{T}(t) (\varphi) = \cal{F}^{-1} (e^{-i (|\xi|^2 - |\eta|^2) t}\ \hat{\varphi}(\xi , \eta) )= K_t * \varphi, $

以及

$ K_t(x, y) = \bigg(\frac{1}{4\pi t}\bigg)^N e^{i\frac{|x|^2-|y|^2}{4t}}. $

命题的证明是通过Fourier变换将(1.2)转换成常微分方程的Cauchy问题求解, 以及在广义函数意义下

$ \begin{equation*} \cal{F}^{-1}(e^{-i(|\xi|^2 - |\eta|^2)t})=\left(\frac{1}{4\pi t}\right)^N\ e^{i\frac{|x|^2-|y|^2}{4t}} . \end{equation*} $

需要特别解释的是关于时间变量$ t $在$ L^2(\mathbb{R}^{N} \times \mathbb{R}^{N}) $空间的连续性, 事实上, 任给$ t_0\in\mathbb{R} $,

$ \left|e^{-i (|\xi|^2 - |\eta|^2) t}-e^{-i (|\xi|^2 - |\eta|^2) t_0}\right|^2\ \left|\hat{\varphi}(\xi , \eta)\right|^2 \ \rightarrow \ 0, \quad \mbox{几乎处处的}\ (\xi, \eta)\in \mathbb{R}^{N} \times \mathbb{R}^{N}, $

以及

$ \left|e^{-i (|\xi|^2 - |\eta|^2) t}-e^{-i (|\xi|^2 - |\eta|^2) t_0}\right|^2\ \left|\hat{\varphi}(\xi , \eta)\right|^2\le 4\ \left|\hat{\varphi}(\xi , \eta)\right|^2\in L^1(\mathbb{R}^{N} \times \mathbb{R}^{N}), $

因此由Lebesgue控制收敛定理得到

$ \lim\limits_{t \rightarrow t_0}\|u(t, \cdot, \cdot)- u(t_0, \cdot, \cdot)\|_{L^2(\mathbb{R}^{2N})}=0, $

以及Cauchy初始条件:

$ \lim\limits_{t \rightarrow 0}\|u(t, \cdot, \cdot)- \varphi(\cdot, \cdot)\|_{L^2(\mathbb{R}^{2N})}=0, $

记$ \mathbb{S}(\mathbb{R}^n\times\mathbb{R}^n) $为Schwartz急减函数空间, 我们有下列解的光滑性结果.

命题2.2  假设$ \varphi \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 则Cauchy问题(1.2)解$ u\in C^\infty(\mathbb{R}_t; \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N})). $

证  由(2.1)知$ u(t, x, y)=\cal{T}(t)\varphi=\cal{F}^{-1} \left( \hat{\varphi}(\xi, \eta) e^{-i(|\xi|^2-|\eta|^2)t} \right) $.对于任意的重指标$ \alpha $以及正整数$ m $:

$ \begin{equation*} |\left( 1+|(\xi, \eta)|^2 \right)^{\frac{m}{2}} \partial_{\xi, \eta}^{\alpha} \left( \hat{\varphi} e^{-i(|\xi|^2-|\eta|^2)t} \right)| \leq C\left( 1+|(\xi, \eta)|^2 \right)^{\frac{m+|\alpha|}{2}}\sum\limits_{\beta \leq \alpha} |\partial_{\xi, \eta}^{\beta} \hat{\varphi}(\xi, \eta)|, \end{equation*} $

因为$ \varphi \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 故$ \hat{\varphi} \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $. 根据$ \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $的定义, 我们证明了

$ \hat{\varphi}(\xi, \eta)\ e^{-i(|\xi|^2-|\eta|^2)t} \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}), \quad \forall t\in\mathbb{R}, $

而关于$ t $的连续性可以立即由Lebesgue控制收敛定理得到, 故

$ u(t, x, y)=\cal{F}^{-1}\left( \hat{\varphi}(\xi, \eta)\ e^{-i(|\xi|^2-|\eta|^2)t}\right) \in C^0(\mathbb{R}, \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N})). $

此外, 任给$ k\in\mathbb{N} $,

$ \partial^k_t u(t, x, y)=\cal{F}^{-1}\left( \hat{\varphi}(\xi, \eta)\ (-i(|\xi|^2-|\eta|^2))^k\ e^{-i(|\xi|^2-|\eta|^2)t}\right), $

以及

$ \hat{\varphi}(\xi, \eta)\ (-i(|\xi|^2-|\eta|^2))^k\ e^{-i(|\xi|^2-|\eta|^2)t} \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}), \quad \forall t\in\mathbb{R}. $

因此任给$ k\in\mathbb{N} $,

$ \partial^k_t u(t, x, y)=\cal{F}^{-1}\left( \hat{\varphi}(\xi, \eta)\ (-i(|\xi|^2-|\eta|^2))^k\ e^{-i(|\xi|^2-|\eta|^2)t}\right) \in C^0(\mathbb{R}, \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N})). $

这就证明了$ u(t, x, y)\in C^{\infty} \left( \mathbb{R}, \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N})\right) $. 证毕.

最后特别指出乘子$ \cal{T}(t) $是$ L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $等距的.

命题2.3  假设$ \varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, $ u=\cal{T}(t)\varphi $, 是Cauchy问题(1.2)的解, 则任给$ t\in \mathbb{R} $, 有

$ \left\| u(t) \right\|_{L^2(\mathbb{R}^{2N})} = \left\| \varphi \right\|_{L^2(\mathbb{R}^{2N})}. $

现在研究解的高阶偏微.

定义3.1  对$ k, j \in \{ 1, ..., N \} $, 对于$ u(t, \cdot, \cdot) \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 定义$ \mathbb{R}^{2N+1} $上的偏微分算子:

$ \begin{equation} P_{x_k}u(t, x, y)=(x_k+2it\partial_{x_k})u(t, x, y), \quad P_{y_j}u(t, x, y)=(-y_j+2it\partial_{y_j})u(t, x, y), \end{equation} $

(3.1)

对多重指标$ \alpha , \beta $, 定义:

$ \begin{equation} P_{\alpha , \beta} = \prod\limits_{k=1}^N P_{x_{k}}^{\alpha_k} P_{y_{k}}^{\beta_k} \end{equation} $

(3.2)

引理3.2  定义3.1中的偏微分算子和算子$ i\partial_t + \Delta_x - \Delta_y $是可交换的:

$ [P_{\alpha, \beta}, i\partial_t + \Delta_x - \Delta_y] u=P_{\alpha , \beta}((i\partial_t + \Delta_x - \Delta_y)u) - (i\partial_t + \Delta_x - \Delta_y) P_{\alpha , \beta}u= 0. $

证   为证明引理中的结论, 仅需证明下式:

$ \begin{equation} P_{\alpha , \beta}((i\partial_t + \Delta_x - \Delta_y)u) = (i\partial_t + \Delta_x - \Delta_y) P_{\alpha , \beta}u, \end{equation} $

(3.3)

下面证明

$ \begin{equation} (i\partial_t + \Delta_x - \Delta_y)(P_{x_k} P_{y_j} u) = P_{x_k} P_{y_j} ((i\partial_t + \Delta_x - \Delta_y)u). \end{equation} $

(3.4)

等式(3.4)左边等于

$ \begin{equation*} \begin{aligned} &\quad (i\partial_t + \Delta_x - \Delta_y) (P_{x_k} P_{y_j} u)\\ &= (i\partial_t + \Delta_x - \Delta_y) (-x_k y_j u + 2it \partial_{y_j}u - 2it\partial_{y_j} u - 2iy_j t \partial_{x_k}u - 4t^2\partial_{x_k}\partial_{y_j}u)\\ &= -ix_k y_j \partial_t u - 2x_k t \partial_t \partial_{y_j} u + 2y_j t \partial_t \partial_{x_k} u-4it^2 \partial_{t} \partial_{x_k} \partial_{y_j} u - x_k y_j \Delta_x u \\ &\quad + 2ix_k t \Delta_x \partial_{y_j} u - 2iy_j t \Delta_x (\partial_{x_k} u) -4t^2 \Delta_x (\partial_{x_k} \partial_{y_j} u) + x_k y_j \Delta_y u - 2ix_k t \Delta_y (\partial_{y_j} u)\\ &\quad + 2iy_j t \Delta_y \partial_{x_k} u + 4t^2\Delta_y (\partial_{x_k} \partial_{y_j}u), \end{aligned} \end{equation*} $

等式(3.4)右边等于

$ \begin{equation*} \begin{aligned} &\quad P_{x_i}P_{y_j} ((i\partial_t + \Delta_x - \Delta_y)u)\\ &= (x_k + 2it\partial_{x_k}) (-y_j+2it\partial_{y_j})(i\partial_t u + \Delta_x u - \Delta_y u)\\ &= (x_k + 2it\partial_{x_k}) (-iy_j \partial_t u -y_j \Delta_x u + y_j \Delta_y u - 2t\partial_{y_j}\partial_t u + 2it\partial_{y_j}\Delta_x u - 2it\partial_{y_j} \Delta_y u)\\ &=-ix_k y_j \partial_t u -x_k y_j \Delta_x u + x_k y_j \Delta_y u - 2x_k t \partial_{y_j} \partial_t u + 2ix_k t \partial_{y_j}\Delta_x u - 2ix_k t \partial_{y_j} (\Delta_y u)\\ & \quad +2y_j t \partial_{x_k} \partial_t u -2iy_j t \partial_{x_k} (\Delta_x u) + 2iy_j t \partial_{x_k} (\Delta_y u) - 4it^2 \partial_{x_k} \partial_{y_j} \partial_t u - 4t^2\partial_{x_k} \partial_{y_j} \Delta_x u\\ & \quad + 4t^2 \partial_{x_k} \partial_{y_j} \Delta_y u, \end{aligned} \end{equation*} $

故(3.4)的左边减去右边等于0.

因此(3.4)成立. 高阶微分的情形由数学归纳法证明.

引理3.3  当$ t \ne 0 $时, 对于$ u(t, \cdot, \cdot) \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 以及多重指标$ \alpha, \beta \in \mathbb{N}^{N} $, 有

$ \begin{equation} P_{\alpha, \beta} u(t, x, y) = (2it)^{|\alpha| + |\beta|} e^{i \frac{|x|^2 - |y|^2}{4t}} \partial_{x, y}^{\alpha, \beta} (e^{-i \frac{|x|^2 - |y|^2}{4t}}u). \end{equation} $

(3.5)

证   首先计算:

$ \begin{equation} \begin{aligned} &(2it)e^{i \frac{|x|^2 - |y|^2}{4t}} \partial_{y_j} (e^{-i \frac{|x|^2 - |y|^2}{4t}}u) \\ =&(2it)e^{i \frac{|x|^2 - |y|^2}{4t}}(\frac{i}{2t}\cdot e^{-i \frac{|x|^2 - |y|^2}{4t}}\cdot y_j u + e^{-i \frac{|x|^2 - |y|^2}{4t}} \partial_{y_j}u)\\ =&-y_j u + 2it\partial_{y_j}u=(-y_j + 2it\partial_{y_j})u, \end{aligned} \end{equation} $

(3.6)

因此由(3.1) (3.6):

$ \begin{equation} \begin{aligned} P_{x_{k}} P_{y_{j}} u(t, x, y) &= (x_k + 2it\partial_{x_k})(-y_j + 2it\partial_{y_j}) u(t, x, y) \\ &= (x_k + 2it \partial_{x_k}) ((2it)e^{i \frac{|x|^2 - |y|^2}{4t}} \partial_{y_j} (e^{-i \frac{|x|^2 - |y|^2}{4t}}u))\\ &=(2it)e^{i \frac{|x|^2 - |y|^2}{4t}}\partial_{x_k} (e^{-i \frac{|x|^2 - |y|^2}{4t}} (2it) e^{i \frac{|x|^2 - |y|^2}{4t}} \partial_{y_j} (e^{-i \frac{|x|^2 - |y|^2}{4t}}u)) \\ &=(2it)(2it)e^{i \frac{|x|^2 - |y|^2}{4t}}\partial x_k \partial y_j (e^{-i \frac{|x|^2 - |y|^2}{4t}}u). \end{aligned} \end{equation} $

(3.7)

高阶微分公式由数学归纳法证明.

引理3.4  如果$ u(0) \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 则对多重指标$ \alpha, \beta \in \mathbb{N}^{N} $, 有

$ \begin{equation} \left(P_{\alpha, \beta}u(t, x, y)\right)|_{t=0}=x^{\alpha}(-y)^{\beta}u(0, x, y). \end{equation} $

(3.8)

证   因为$ u(0) \in \cal{S} \subset C^{\infty} $, 我们仅仅需要证明, 对于$ \alpha_k, \beta_j\in\mathbb{N}, \; k, \; j=1, 2, \cdots, N $, 有

$ \begin{equation} \left(P_{x_k}^{\alpha_k}P_{y_j}^{\beta_j}u(t, x, y)\right)|_{t=0}=x_k^{\alpha_k}(-y_j)^{\beta_j}u(0, x, y). \end{equation} $

(3.9)

当$ \alpha_k=\beta_j=1 $时, 根据(3.1)定义计算可得

$ \begin{equation*} P_{x_k}P_{y_j}u=-x_k y_j u -y_j \cdot 2it \partial_{x_k}u + x_k \ 2it\partial_{y_j}u-4t^2\partial_{y_j}\partial_{x_k}u, \end{equation*} $

因此

$ \left(P_{x_k}P_{y_j}u(t, x, y)\right)|_{t=0}=x_k (-y_j)u(0, x, y). $

假设$ \alpha_k = m, \beta_j = l $时, 有

$ \begin{equation*} \left(P_{x_k}^{m}P_{y_j}^{l}u(t, x, y)\right)|_{t=0}=\left((x_k + 2it\partial_{x_k})^{m}(-y_j+2it\partial_{y_j})^{l} u(t, x, y) \right)|_{t=0}=x_k^m (-y_j)^l u(0, x, y). \end{equation*} $

则当$ \alpha_k = m+1, \beta_j = l + 1 $时:

$ \begin{equation*} \begin{aligned} &\quad \left(P_{x_k}^{m+1}P_{y_j}^{l+1}u(t, x, y)\right)|_{t=0}\\ &=\left(P_{x_k}P_{x_k}^{m}P_{y_j}P_{y_j}^{l}u(t, x, y)\right)|_{t=0}\\ &=\left(P_{x_k}P_{y_j}P_{x_k}^{m}P_{y_j}^{l}u(t, x, y)\right)|_{t=0}\\ &=\left[(x_k+2it\partial_{x_k})(-y_j+2it\partial_{y_j})(x_k+2it\partial_{x_k})^{m} (-y_j+2it\partial_{y_j})^{l}u(t, x, y)\right] |_{t=0}\\ &=-x_k y_j \left[(x_k+2it\partial_{x_k})^{m} (-y_j+2it\partial_{y_j})^{l}u(t, x, y)\right] |_{t=0} \\ &\quad + \left[2it \cdot x_k \cdot \partial_{y_j} (x_k+2it\partial_{x_k})^{m} (-y_j+2it\partial_{y_j})^{l}u(t, x, y)\right] |_{t=0} \\ &\quad -\left[2it\cdot y_j \cdot \partial_{x_k}(x_k+2it\partial_{x_k})^{m} (-y_j+2it\partial_{y_j})^{l}u(t, x, y)\right] |_{t=0}\\ &\quad + \left[(2it)^{2} \partial_{x_k}\partial_{y_j}(x_k+2it\partial_{x_k})^{m} (-y_j+2it\partial_{y_j})^{l}u(t, x, y)\right] |_{t=0}\\ &=x_k (-y_j) x_k^{m} (-y_j)^{l} u(0, x, y)\\ &=x_k^{m+1}(-y_j)^{l+1}u(0, x, y), \end{aligned} \end{equation*} $

因此(3.9)成立.

我们首先研究Cauchy问题(1.2)中的初始值$ \varphi\in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $的情形.

命题4.1  假设$ \varphi \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}), u $是Cauchy问题(1.2)的解, 则任给$ \alpha, \beta\in \mathbb{N}^N $, 有

$ \begin{equation*} (2|t|)^{|\alpha|+|\beta|} ||\partial_{x, y}^{\alpha, \beta} (e^{-i \frac{|x|^2 - |y|^2}{4t}}u)||_{L^2} = ||x^{\alpha} y^{\beta} \varphi||_{L^2}, \quad t \ne 0. \end{equation*} $

证   根据命题2.2, 因为$ \varphi \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 所以Cauchy问题(1.2)的解$ u\in C^\infty(\mathbb{R}_t; \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $. 另一方面, 对于$ \alpha, \beta\in\mathbb{N}^N $, 考虑Cauchy问题

$ \begin{equation} \begin{cases} i \partial_t v(t, x, y) + \Delta_x v(t, x, y) - \Delta_y v(t, x, y) = 0, \\ v|_{t=0} = x^{\alpha} (-y)^{\beta} \varphi (x, y), \quad x, y\in\mathbb{R}^N. \end{cases} \end{equation} $

(4.1)

因为仍然有$ x^{\alpha} (-y)^{\beta} \varphi \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 所以Cauchy问题(4.1)存在唯一解

$ v_{\alpha, \beta}=\cal{T}(t) (x^{\alpha} (-y)^{\beta} \varphi)\in C^\infty(\mathbb{R}_t; \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N})). $

利用命题2.3, 则任给$ t\in \mathbb{R} $, 有

$ \left\| v_{\alpha, \beta}(t) \right\|_{L^2(\mathbb{R}^{2N})} = \|x^{\alpha} (-y)^{\beta} \varphi \|_{L^2(\mathbb{R}^{2N})}= \|x^{\alpha} y^{\beta} \varphi \|_{L^2(\mathbb{R}^{2N})}. $

同时

$ \begin{equation*} (i\partial_t +\Delta_x - \Delta_y)(P_{\alpha, \beta}u)=P_{\alpha, \beta}((i\partial_t + \Delta_x - \Delta_y)u(t, x, y)) = 0. \end{equation*} $

利用引理3.4, 有

$ \begin{equation} \begin{cases} (i\partial_t +\Delta_x - \Delta_y)(P_{\alpha, \beta}u)= 0, \\ (P_{\alpha, \beta}u)|_{t=0} = x^{\alpha} (-y)^{\beta} \varphi (x, y), \quad x, y\in\mathbb{R}^N. \end{cases} \end{equation} $

(4.2)

因为Cauchy问题(1.2)的解是唯一的, 所以

$ \begin{equation} v_{\alpha, \beta}(t, x, y)=P_{\alpha, \beta}u(t, x, y). \end{equation} $

(4.3)

因此

$ \begin{equation} \|P_{\alpha, \beta}u(t, \cdot, \cdot)\|_{L^2(\mathbb{R}^{2N})}=\| x^{\alpha} y^{\beta} \varphi\|_{L^2(\mathbb{R}^{2N})}. \end{equation} $

(4.4)

根据(3.5)有

$ \begin{equation} \|P_{\alpha , \beta}u(t, \cdot, \cdot)\|_{L^2(\mathbb{R}^{2N})}=\left\| (2it)^{|\alpha|+|\beta|} e^{i\frac{|x|^2-|y|^2}{4t}} \partial_{x, y}^{\alpha, \beta} (e^{-i\frac{|x|^2-|y|^2}{4t}}u)\right\|_{{L^2(\mathbb{R}^{2N})}} \end{equation} $

(4.5)

故由(4.5)知,

$ \begin{equation*} \left\|(2it)^{|\alpha|+|\beta|} e^{i\frac{|x|^2-|y|^2}{4t}} \partial_{x, y}^{\alpha, \beta} (e^{-i\frac{|x|^2-|y|^2}{4t}}u) \right\|_{L^2(\mathbb{R}^{2N})}= \| x^{\alpha} y^{\beta} \varphi \|_{L^2(\mathbb{R}^{2N})}. \end{equation*} $

因此得到

$ \begin{equation*} (2|t|)^{|\alpha|+|\beta|} \left\| \partial_{x, y}^{\alpha, \beta} (e^{-i\frac{|x|^2-|y|^2}{4t}}u) \right\|_{L^2(\mathbb{R}^{2N})} = \| x^{\alpha} y^{\beta} \varphi \|_{L^2(\mathbb{R}^{2N})}, \quad \forall t\not=0. \end{equation*} $

现在研究$ L^2 $初始值的Cauchy问题(1.2).

命题5.1  令$ \varphi \in L^2(\mathbb{R}^{N} \times \mathbb{R}^{N}), \alpha, \beta\in\mathbb{N}^N $, 满足$ x^{\alpha} y^{\beta} \varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $.若$ u(t) = T(t)\varphi \in C(R, L^2(\mathbb{R}^{N} \times \mathbb{R}^{N})) $, 则

$ \begin{equation} \partial^{\alpha, \beta}_{x, y} \left(e^{-i \frac{|x|^2 - |y|^2}{4t}} u\right) \in C(\mathbb{R} \ \backslash \ \lbrace0\rbrace , L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N})). \end{equation} $

(5.1)

证   利用命题4.1, 如果$ \varphi \in \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}), $则(5.1)成立, 以及

$ \begin{equation*} \left\| \partial_{x, y}^{\alpha, \beta} (e^{-i\frac{|x|^2-|y|^2}{4t}}u) \right\|_{L^2(\mathbb{R}^{2N})} =(2|t|)^{-|\alpha|-|\beta|} \| x^{\alpha} y^{\beta} \varphi \|_{L^2(\mathbb{R}^{2N})}, \quad \forall t\not=0. \end{equation*} $

现在假设$ \varphi, x^{\alpha} y^{\beta} \varphi \in L^2(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 则存在$ \{\varphi_j\} \subset \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 满足

$ \begin{equation} \lim\limits_{j \rightarrow +\infty}\|\varphi_j-\varphi\|_{ L^2(\mathbb{R}^{2N})}=0, \ \ \lim\limits_{j \rightarrow +\infty}\|x^{\alpha} y^{\beta}(\varphi_j-\varphi)\|_{ L^2(\mathbb{R}^{2N})}=0. \end{equation} $

(5.2)

这个结果可以由$ \cal{S}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $在$ H^m(\mathbb{R}^{N} \times \mathbb{R}^{N}) $中稠密得到. 因此$ \{x^{\alpha} y^{\beta}\varphi_j\} $是$ L^2(\mathbb{R}^{N} \times \mathbb{R}^{N}) $的一个Cauchy序列. 记$ u_j(t) = T(t)\varphi_j $, 则由Cauchy问题(1.2)的解的唯一性

$ \lim\limits_{j \rightarrow +\infty}\|u_j-u\|_{ L^2(\mathbb{R}^{2N})}=0. $

另一方面, 任给$ j, k\in\mathbb{N} $,

$ \begin{equation*} \left\| \partial_{x, y}^{\alpha, \beta} (e^{-i\frac{|x|^2-|y|^2}{4t}}(u_j-u_k)) \right\|_{L^2(\mathbb{R}^{2N})} =(2|t|)^{-|\alpha|-|\beta|} \| x^{\alpha} y^{\beta} (\varphi_j-\varphi_k) \|_{L^2(\mathbb{R}^{2N})}, \quad \forall t\not=0. \end{equation*} $

由此导出$ \{\partial_{x, y}^{\alpha, \beta} (e^{-i\frac{|x|^2-|y|^2}{4t}}u_j)\} $是$ L^2(\mathbb{R}^{N} \times \mathbb{R}^{N}) $的一个Cauchy序列, 由唯一性有

$ \begin{equation} \lim\limits_{j \rightarrow +\infty}\left\| \partial_{x, y}^{\alpha, \beta} (e^{-i\frac{|x|^2-|y|^2}{4t}}(u_j-u)) \right\|_{L^2(\mathbb{R}^{2N})} =0, \quad \forall t\not=0. \end{equation} $

(5.3)

结合(5.2), (5.3), 我们就证明了, 如果$ \varphi, x^{\alpha} y^{\beta} \varphi \in L^2(\mathbb{R}^{N} \times \mathbb{R}^{N}), $则

$ \begin{equation} \left\| \partial_{x, y}^{\alpha, \beta} (e^{-i\frac{|x|^2-|y|^2}{4t}}u) \right\|_{L^2(\mathbb{R}^{2N})} =(2|t|)^{-|\alpha|-|\beta|} \| x^{\alpha} y^{\beta} \varphi \|_{L^2(\mathbb{R}^{2N})}, \quad \forall t\not=0. \end{equation} $

(5.4)

这就证明了(5.1). 关于$ t\not=0 $的连续性由上式直接给出.

命题5.2  假设$ m\in\mathbb{N}, \varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $以及$ (1+|(x, y)|^m) \varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $. 则Cauchy问题(1.2)存在唯一弱解满足:

$ \begin{equation} e^{-i\frac{|x|^2-|y|^2}{4t}}u \in C^{0}(\mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H^{m}(\mathbb{R}^{N} \times \mathbb{R}^{N})). \end{equation} $

(5.5)

证   首先, 对于$ m\in\mathbb{N} $,

$ \begin{equation*} (1 + |(x, y)|^m)^2 \ge (|x|^2 +|y|^2)^{m} \ge C \sum\limits_{|(\alpha, \beta)| = m} (x^\alpha y^\beta)^2 . \end{equation*} $

故由命题5.1可知, 当$ |(\alpha, \beta)| = m $时, $ \partial^{\alpha, \beta}_{x, y} e^{-i \frac{|x|^2-|y|^2}{4t}} u(t) \in C(\mathbb{R} \ \backslash \ \lbrace0\rbrace , L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N})) $, 进而利用(5.4), 任给$ |t|\ge \delta_0>0 $, 有

$ \begin{align*} \left\| e^{-i \frac{|x|^2-|y|^2}{4t}} u(t) \right\|^2_{H^m(\mathbb{R}^{2N})} &= \sum\limits_{|\alpha, \beta| \leq m} \left\| \partial^{\alpha, \beta}_{x, y} e^{-i \frac{|x|^2-|y|^2}{4t}} u(t) \right\|_{L^2(\mathbb{R}^{2N})}^2\\ &=\sum\limits_{|\alpha, \beta| \leq m} (2|t|)^{-2|\alpha|-2|\beta|} \| x^{\alpha} y^{\beta} \varphi \|^2_{L^2(\mathbb{R}^{2N})}\\ &\le C_0 \delta_0^{-2m}\|(1+|(x, y)|^m \varphi \|^2_{L^2(\mathbb{R}^{2N})}. \end{align*} $

因此$ e^{-i \frac{|x|^2-|y|^2}{4t}} u(t) \in C^0(\mathbb{R}\ \backslash \ \lbrace 0 \rbrace, H^{m}(\mathbb{R}^{N} \times \mathbb{R}^{N})) $. 这里关于$ t\not=0 $的连续性和命题5.1的证明一样.

引理5.3  假设$ 0<a<b<+\infty, 0<R<\infty $, 任给$ \alpha, \beta\in\mathbb{N}^N $, 存在$ C(\alpha, \beta, a, b, R)>0 $使得

$ \begin{equation} \left|\partial_{x}^{\alpha}\partial_{y}^{\beta} e^{i \frac{|x^2-|y|^2}{4t}}\right|\le C(\alpha, \beta, a, b, R), \quad \forall\ a\le |t|\le b, \ \ |(x, y)|\le R. \end{equation} $

(5.6)

事实上

$ \partial_{x}^{\alpha}\partial_{y}^{\beta}\left( e^{i \frac{|x|^2-|y|^2}{4t}}\right)=P_{\alpha, \beta}\left(\frac{1}{t}, x, y\right) e^{i \frac{|x|^2-|y|^2}{4t}}, $

这里$ P_{\alpha, \beta}(\frac{1}{t}, x, y) $是变量$ \frac{1}{t}, x, y $的$ |(\alpha, \beta)| $阶多项式. 因此

$ \left|\partial_{x}^{\alpha}\partial_{y}^{\beta}\left( e^{i \frac{|x|^2-|y|^2}{4t}}\right)\right|= \left|P_{\alpha, \beta}\left(\frac{1}{t}, x, y\right)\right|\le C(\alpha, \beta, a, b, R), \quad \forall\ a\le |t|\le b, \ \ |(x, y)|\le R. $

命题5.4  假设$ m\in\mathbb{N}, \varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $以及$ (1+|(x, y)|^m) \varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $. 则Cauchy问题(1.2)的唯一弱解满足:

$ \begin{equation} u \in C^{0}(\mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H^{m}_{loc}(\mathbb{R}^{N} \times \mathbb{R}^{N})). \end{equation} $

(5.7)

证   我们需要证明, 任给$ 0<R<+\infty, 0<a<b<\infty $, $ \|u(t)\|_{H^m(B_R)}<+\infty, $以及关于$ t $在[a, b]上连续, 这里

$ B_R=\{(x, y)\in\mathbb{R}^{N} \times \mathbb{R}^{N}; |(x, y)|<R\}. $

利用莱布尼茨公式, 当$ a\le |t|\le b $时有

$ \begin{equation*} \begin{aligned} &\left\| u(t) \right\|^2_{H^m (B_R)} = \sum\limits_{|(\alpha, \beta)| \leq m} \left\| \partial^{\alpha, \beta}_{x, y} \left(e^{i \frac{|x|^2-|y|^2}{4t}} \cdot e^{-i \frac{|x|^2-|y|^2}{4t}} u(t, x, y)\right)\right\|_{L^2}^2\\ =& \sum\limits_{|(\alpha, \beta)| \leq m} \int_{B_R} \Big| \sum\limits_{(\tilde{\alpha}, \tilde{\beta}) \leq (\alpha, \beta)}\dbinom{\alpha, \beta}{\tilde{\alpha}, \tilde{\beta}} \big(\partial_{x, y}^{\tilde{\alpha}, \tilde{\beta}} e^{i \frac{|x|^2-|y|^2}{4t}}\big) \times \partial_{x, y}^{\alpha-\tilde{\alpha}, \beta - \tilde{\beta}} \big(e^{-i \frac{|x|^2-|y|^2}{4t}} u(t, x, y)\big)\Big|^2 dxdy\\ \le& C(a, b, R, m) \sum\limits_{|(\alpha, \beta)| \leq m} \int_{B_R} \Big| \sum\limits_{(\tilde{\alpha}, \tilde{\beta}) \leq (\alpha, \beta)} \left| \partial_{x, y}^{\alpha-\tilde{\alpha}, \beta - \tilde{\beta}} \big(e^{-i \frac{|x|^2-|y|^2}{4t}} u(t, x, y)\big)\right|\Big|^2 dxdy \\ \le& \tilde{C}(a, b, R, m) \left\| e^{-i\frac{|\ \cdot\ |^2 - |\ \cdot\ |^2}{4t}} u(t, \cdot, \cdot) \right\|^2_{H^{m}(\mathbb{R}^{2N})}. \end{aligned} \end{equation*} $

因此$ u \in C(\mathbb{R} \backslash \{ 0\}, H_{loc}^m (\mathbb{R}^{N} \times \mathbb{R}^{N})) $.

完成定理1.1的证明

现在证明(1.3), 即:

$ \begin{equation*} u \in \bigcap\limits_{0 \leq j \leq \left[\frac{m}{2}\right]} C^{j}(\mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H^{m-2j}_{loc}(\mathbb{R}^{N} \times \mathbb{R}^{N})). \end{equation*} $

命题5.4就是$ j=0 $的情形. 如果$ m\ge 2 $, 由$ u \in C^0( \mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H_{loc}^{m} (\mathbb{R}^{N} \times \mathbb{R}^{N})) $, 有

$ \Delta_x u, \Delta_y u \in C^0( \mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H_{loc}^{m-2} (\mathbb{R}^{N} \times \mathbb{R}^{N})), $

因此, 利用方程(1.2)

$ \partial_t u(t, x, y) =i \Delta_x u(t, x, y) - i \Delta_y u(t, x, y)\in C^0( \mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H_{loc}^{m-2} (\mathbb{R}^{N} \times \mathbb{R}^{N})), $

所以

$ u\in C^1( \mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H_{loc}^{m-2} (\mathbb{R}^{N} \times \mathbb{R}^{N})). $

同理, 如果$ m\ge 4 $, 有

$ \Delta_x u, \Delta_y u \in C^1( \mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H_{loc}^{m-4} (\mathbb{R}^{N} \times \mathbb{R}^{N})), $

因此利用方程(1.2)

$ \partial_t u(t, x, y) =i \Delta_x u(t, x, y) - i \Delta_y u(t, x, y)\in C^1( \mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H_{loc}^{m-4} (\mathbb{R}^{N} \times \mathbb{R}^{N})), $

所以

$ u\in C^2( \mathbb{R}\ \backslash \ \lbrace 0 \rbrace , H_{loc}^{m-4} (\mathbb{R}^{N} \times \mathbb{R}^{N})). $

重复上述过程, 对于$ k\le \left[\frac m 2\right] $, 最后得到:

$ \begin{equation*} u \in \bigcap\limits_{0 \leq j \leq k} C^{j}(\mathbb{R} \backslash \{ 0 \} , H_{loc}^{m-2j}(\mathbb{R}^{N} \times \mathbb{R}^{N})). \end{equation*} $

现在假设任给$ m\in\mathbb{N} $, 有$ (1+|x, y|^{m})\varphi \in L^{2}(\mathbb{R}^{N} \times \mathbb{R}^{N}) $, 则

$ \begin{equation*} u \in \bigcap\limits_{0 \leq j \leq \left[\frac m 2\right]} C^{j}(\mathbb{R} \backslash \{ 0 \} , H_{loc}^{m-2j}(\mathbb{R}^{N} \times \mathbb{R}^{N})), \quad \forall m\in\mathbb{N}. \end{equation*} $

利用Soblev嵌入定理, 任给$ p\in\mathbb{N}, $

$ H_{loc}^{p+N+1}(\mathbb{R}^{N} \times \mathbb{R}^{N})\subset C^p(\mathbb{R}^{N} \times \mathbb{R}^{N}). $

这就证明了

$ \begin{equation*} u \in \bigcap\limits_{0 \leq j \leq \left[\frac {(m-N-1)} 2\right]} C^{j}(\mathbb{R} \backslash \{ 0 \} , C^{m-2j-N-1}(\mathbb{R}^{N} \times \mathbb{R}^{N})), \quad \forall m >N. \end{equation*} $

因此得

$ \begin{equation*} u(t) \in C^{\infty} (\mathbb{R} \backslash \{ 0\} \times \mathbb{R}^{N} \times \mathbb{R}^{N}), \end{equation*} $

这就完成了定理1.1的证明.