上海理工大学学报  2019, Vol. 41 Issue (4): 313-320   PDF    
五次长短波共振方程初值问题解的存在性
王贝贝, 张卫国     
上海理工大学 理学院,上海 200093
摘要: 研究了五次长短波共振方程初值问题解的存在性。首先利用压缩映射原理证明了局部解的存在性,然后通过建立局部解具能量守恒性质,并利用先验估计方法,证明了具有高次非线性项的长短波共振方程的局部解可以延拓到整个定义域,最终证明了全局解的存在性。
关键词: 压缩映射原理     先验估计     能量守恒    
Existence of Solutions for the Initial Value Problems of a Five-Time Long-Short Wave Resonance Equation
WANG Beibei, ZHANG Weiguo     
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: The existence of the solutions to the initial value problem of a five-time long-short wave resonance equation was focused. First, the existence of the local solution was proved by using the compression mapping theorem. Then, the local solution with energy conservation properties was established and the prior estimation method was used to, It was also proved that the local solution of high-order nonlinear terms in the five-time long-short wave resonance equation can be extended to the whole domain. Finally, the existence of the global solution was testified.
Key words: compression mapping theorem     a prior estimate     energy conservation    
1 问题的提出

Djordjevic等[1] 在研究二维的毛细管重力波时,发现了描述长波和短波之间相互作用的演化方程

$\left\{ {\begin{array}{*{20}{l}} {i{u_t} + {u_{xx}} = \gamma uv} \\ {{v_t} + \beta {{\left( {|u{|^2}} \right)}_x} = 0} \end{array}} \right.$ (1)

该方程还出现在内波、Rossby波及等离子体波等许多物理问题中[2]

文献[3-4]分别用逆散射方法和先验估计方法证明了方程(1)初值问题整体解的存在性. 文献[5]研究了方程(1)的n孤子解。

在文献[6]中,Benney建立了长波与短波相互作用的一般理论,提出了长波与短波相互作用的模型方程

$\left\{ {\begin{array}{*{20}{l}} {i{u_t} + {u_{xx}} = uv + \gamma |u{|^2}u} \\ {{v_t} = {{\left( {|u{|^2}} \right)}_x}} \end{array}} \right.$ (2)

式中: $\gamma $ 为常数; $u$ 是复函数,表示长波; $v$ 是实函数,表示短波; $|u|$ $u$ 的模长。

文献[7]证明了方程(2)初值问题全局解的存在性。

1988年,Oikawa等[8]研究了双层流体中长波和短波在彼此分界面角度上的传播和共振作用,导出了(2+1)维长短波方程组,并将长短波方程拓展到高维空间,文献[914]分别研究了高维长短波方程及其推广形式解的存在唯一性。

还有学者研究了方程(2)在(1+1)维的推广形式,例如,文献[15]研究了广义LS型方程

$\left\{ {\begin{array}{*{20}{l}} {i{u_t} + {u_{xx}} - uv + \gamma q(|u{|^2})u = 0} \\ {{v_t} + {{\left( {|u{|^2}} \right)}_x} = \delta u + \gamma f(|u{|^2})} \end{array}} \right.$ (3)

的周期初值和初值问题。

2010年,Shang[16]研究了广义长短波方程

$\left\{ {\begin{array}{*{20}{l}} {i{u_t} + {u_{xx}} = uv + \gamma (|u{|^2})u + \delta |u{|^4}u} \\ {{v_t} = {{\left( {|u{|^2}} \right)}_x}} \end{array}} \right.$ (4)

的孤波解和若干特殊形式的周期解。

由于文献[15]中要求式(3)的 $q(s) \leqslant {c_1}{s^{2 - \alpha }} + {D_1}$ ,且 $\alpha > 0$ 。而方程(4)中包含了非线性项 $\delta |u{|^4}u$ ,故该问题未被文献[15]研究。作者也尚未见到其他文献研究过方程(4)的初值问题。本文研究方程(4)初值问题解的存在唯一性。

定义泛函空间

$\left\{\begin{split} & X_3^T = \{ f:[0,T] \times {\mathbb R} \to {\mathbf{C}}|f \in \left( {[0,T];{H^{5/2}}({\mathbb R})} \right),\\ &{\rm D}_x^kf \in {L^\infty }\left( {{\mathbb R};{L^2}[0,T]} \right)(k = 1,2,3),\\ &{\rm D}_x^{p - 2q}{\rm D}_t^qf\!\! \in\!\! {L^\infty }\left( {{\mathbb R};{L^2}[0,\!T]} \right)\left( {2 \!\!\leqslant\!\! p \!\!\leqslant \!\!3,q \!=\! [p/2]} \right)\} \end{split}\right.$ (5)

式中: ${{\rm D}_x}$ ${{\rm D}_t}$ 分别为关于 $x$ $t$ 的偏微分。

定义空间 $X_3^T$ 中元素的范数为

$\begin{split} &\quad\parallel f{\parallel _{X_3^T}} = \mathop {\sup }\limits_{0 \leqslant t \leqslant T} \parallel f(t){\parallel _{{H^{5/2}}}} +\\ &\qquad \mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\sum\limits_{k = 1}^3 {\int_0^T | } ({\rm D}_x^kf)(x,t){|^2}{\rm d}t} \right)^{1/2}} + \\ &\qquad \mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\sum\limits_{2 \leqslant p \leqslant 3,q = [p/2]} {\int_0^T | } ({\rm D}_x^{p - 2q}{\rm D}_t^q)f(x,t){|^2}{\rm d}t} \right)^{1/2}}\qquad \end{split} $

方程(4)初值问题的初值条件为

$\begin{gathered} u(0,x) = {u_0}(x),\quad x \in {\mathbb {R}} \\ v(0,x) = {v_0}(x),\quad x \in {\mathbb {R}} \\ \end{gathered} $

本文的主要结果可以归结为定理1。

定理1  对任意的 $({{u}_{0}},{{v}_{0}})\in {{H}^{5/2}}({\mathbb {R}})\times {{H}^{2}}({\mathbb{R}})$ ,只要 $\|{{u}_{0}}{{\|}_{{{L}^{2}}}}$ 足够小,方程(4)存在唯一的解 $(u,v)\in $ $C( [0,T];{{H}^{5/2}} )\times C\left( [0,T];{{H}^{2}} \right)$ ,且对任意的 $T>0$ ,有 $u\in X_{3}^{T}$

$s$ 是实数, 则Sobolev空间 ${H^s}({{\mathbb {R}}^n})$

${H^s}({{\mathbb {R}}^n}) = \{ f \in {L^2}({{\mathbb {R}}^n})\mid {\left( {1 + |\xi {|^2}} \right)^{s/2}}\hat f(\xi ) \in {L^2}({{\mathbb {R}}^2})\} $

$f \in {H^s}({{\mathbb{R}}^n})$

$\parallel f{\parallel _{{H^s}}} = {(2\pi )^{ - n/2}}{\left( {\int_{{{\mathbb{R}}^n}} {{{\left( {1 + |\xi {|^2}} \right)}^s}} |\hat f(\xi ){|^2}{\rm d}\xi } \right)^{1/2}}$

$s$ 是整数时,

$\begin{split} &\quad\|f\|_{{{H}^{s}}}^{2}={{(2\pi )}^{-n}}\int_{{{{\mathbb{R}}}^{n}}}{\sum\limits_{|\alpha |\leqslant s}{{{C}_{\alpha }}}}|{{\xi }^{\alpha }}\hat{f}(\xi ){{|}^{2}}{\rm d}\xi=\qquad\qquad\qquad\qquad \\ &\qquad\int_{{{{\mathbb{R}}}^{n}}}{\sum\limits_{|\alpha |\leqslant s}{{{C}_{\alpha }}}}|{{{\rm D}}^{\alpha }}f{{|}^{2}}{\rm d}x \end{split}$

式中,所有的系数 ${C_\alpha }$ 均为正数。

统一将 ${L^\infty }(0,T;X)$ 空间元素的范数记为 $\|\cdot {{\|}_{L_{T}^{\infty }(X)}}$ 。定义

$U(t) = {{\rm e}^{{{\rm{i}}t}({\partial ^2}/\partial {x^2})}} = {{\mathcal {F}}^{ - 1}}{{\rm e}^{ - {\rm i}t{\xi ^2}}}{\mathcal {F}}, \quad t \in {\mathbb {R}}$

式中: $\mathcal {F}$ ${{\mathcal {F}}^{ - 1}}$ 分别为傅里叶变换和逆傅里叶变换。

2 局部解的存在性

现利用压缩映射定理证明方程(4)在局部定义域上有解.

定理2  设 $({{u}_{0}},{{v}_{0}})\in {{H}^{5/2}}\times {{H}^{2}}({\mathbb R})$ ,存在 ${{T}_{0}}>0$ $0\leqslant T\leqslant {{T}_{0}}$ 和唯一的泛函对 $(u,v)\in C\left( [0,T];{{H}^{5/2}} \right)\times $ $C\left( [0,T];{{H}^{2}} \right)$ 满足方程(4),且 $u\in X_{3}^{T}$ .

为证明定理2,需要用到下面的引理1~6。

引理1[17]  对任意的 ${{u}_{0}}\in {{L}^{2}}({\mathbb R})$ ,有

$\mathop {\sup }\limits_{ - \infty < x < \infty } \left( {\int_{ - \infty }^\infty | {\rm D}_x^{1/2}U(t){u_0}{|^2}{\rm d}t} \right) \leqslant C{\left( {\int_{ - \infty }^\infty | {u_0}(x){|^2}{\rm d}x} \right)^{1/2}}$

这里, ${\rm D}_x^{1/2} = {{\mathcal {F}}^{ - 1}}|\xi {|^{1/2}}{\mathcal {F}}.$

引理2[18]  对任意的 $f\in {{L}^{1}}\left({\mathbb R};{{L}^{2}}(0,T) \right)$ ,有

$\begin{split}&\quad\mathop {\sup }\limits_{0 \leqslant t \leqslant T} {\left( {\int_{ - \infty }^\infty | {\rm D}_x^{1/2}\int_0^t U (t - s)f(s,x){\rm d}s{ |^2}{\rm d}x} \right)^{1/2}} \leqslant\qquad\qquad\\ & \qquad C\int_{ - \infty }^\infty {{{\left( {\int_0^T | f(t,x){|^2}{\rm d}t} \right)}^{1/2}}} {\rm d}x\end{split}$

引理3[18]  对任意的 $f\in {{L}^{1}}\left( {\mathbb R};{{L}^{2}}(0,T) \right)$ ,有

$\begin{split} &\quad\mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T \left| {{\rm D}_x}\int_0^t U (t - s)f(s, \cdot ){\rm d}s\right|{^2}{\rm d}t} \right)^{1/2}} \leqslant\qquad\qquad \\ &\qquad C\int_{ - \infty }^\infty {{{\left( {\int_0^T | f(t,x){|^2}{\rm d}t} \right)}^{1/2}}} {\rm d}x\end{split}$

引理4[7]  对任意的 $f,g,h\in X_{3}^{T}$ ,有

$\begin{split}\int_0^T {{{\left\| {\int_0^t f (s){g_x}(s){\rm d}sh(t)} \right\|}_{{L^2}}}} {\rm d}t \leqslant C{T^{3/2}}\left\| f\right\|_{X_3^T}\left\|g\right\|_{X_3^T}\left\|h\right\|_{X_3^T}\end{split}$

引理5[7]  对任意的 $f,g,h\in X_{3}^{T}$ ,有

$\begin{split} &\quad \int_{-\infty }^{\infty }{{{\left[ \int_{0}^{T}{{{\left| {{\rm D}_{x}}\left( \int_{0}^{t}{f}(s){{g}_{x}}(s){\rm d}sh(t) \right) \right|}^{2}}}{\rm d}t \right]}^{1/2}}}{\rm d}x \leqslant \qquad\qquad\\ & \qquad T\underset{-\infty <x<\infty }{\mathop{\sup }}\,{{\left( \int_{0}^{T}{|}{{g}_{xx}}{{|}^{2}}{\rm d}t \right)}^{1/2}}\|f{{\|}_{L_{T}^{\infty }({{L}^{2}})}}\|h{{\|}_{L_{T}^{\infty }({{L}^{2}})}}+\\ &\qquad 2C{{T}^{3/2}}\|f{{\|}_{L_{T}^{\infty }({{H}^{1}})}}\|g{{\|}_{L_{T}^{\infty }({{H}^{1}})}}\|h{{\|}_{L_{T}^{\infty }({{H}^{1}})}} \leqslant\\ & \qquad C(T)\|f{{\|}_{X_{3}^{T}}}\|g{{\|}_{X_{3}^{T}}}\|h{{\|}_{X_{3}^{T}}} \end{split}$

引理 6  对任意的 $f,g,h\in X_{3}^{T}$ , 有

$\begin{split}&\quad \int_{-\infty }^{\infty }{{{\left[ \int_{0}^{T}{{{\left| {{\rm D}_{xx}}\left( \int_{0}^{t}{f}(s,x){{g}_{x}}(s,x){\rm d}sh(t,x) \right) \right|}^{2}}} {\rm d}t\right]}^{1/2}}}{\rm d}x\leqslant\qquad\\ &\qquad C(T)\|f{{\|}_{X_{3}^{T}}}\|g{{\|}_{X_{3}^{T}}}\|h{{\|}_{X_{3}^{T}}}\end{split}$

其中, $T \to 0$ 时, $C(T) \to 0$ .

证明  运用Cauchy不等式和Holder不等式,可得

$ \begin{split}&\quad\int_{ - \infty }^\infty {{{\left[ {\int_0^T {{{\left| {{{\rm D}_{xx}}\left( {\int_0^t f (s,x){g_x}(s,x){\rm d}sh(t,x)} \right)} \right|}^2}} {\rm d}t} \right]}^{1/2}}} {\rm d}x \leqslant \qquad\qquad\\ &\qquad \int_{ - \infty }^\infty {{{\left[ {\int_0^T {{{\left| {\int_0^t {(2{f_x}{g_{xx}})} {\rm d}sh} \right|}^2}} {\rm d}t} \right]}^{1/2}}} {\rm d}x + \\ &\qquad\int_{ - \infty }^\infty {{{\left[ {\int_0^T {{{\left| {\int_0^t {(f{g_{xxx}})} {\rm d}sh} \right|}^2}} {\rm d}t} \right]}^{1/2}}} {\rm d}x +\\ & \qquad \int_{ - \infty }^\infty {{{\left[ {\int_0^T {{{\left| {\int_0^t {({f_{xx}}{g_x})} {\rm d}sh} \right|}^2}} {\rm d}t} \right]}^{1/2}}} {\rm d}x + \\ &\qquad 2\int_{ - \infty }^\infty {{{\left[ {\int_0^T {{{\left| {\int_0^t {(f{g_{xx}})} {\rm d}s{h_x}} \right|}^2}} {\rm d}t} \right]}^{1/2}}} {\rm d}x+ \\ &\qquad 2\int_{ - \infty }^\infty {{{\left[ {\int_0^T {{{\left| {\int_0^t {({f_x}{g_x})} {\rm d}s{h_x}} \right|}^2}} {\rm d}t} \right]}^{1/2}}} {\rm d}x +\\ &\qquad\int_{ - \infty }^\infty {{{\left[ {\int_0^T {{{\left| {\int_0^t f {g_x}{\rm d}s{h_{xx}}} \right|}^2}} {\rm d}t} \right]}^{1/2}}} {\rm d}x\\[-20pt] \end{split}$ (6)

估计方程(6)不等式右边的第1项为

$\begin{split}& \quad\int_{ - \infty }^\infty {{{\left[ {\int_0^T {{{\left| {\left( {\int_0^t 2 {f_x}(s,x){g_{xx}}(s,x)} \right){\rm d}sh(t,x)} \right|}^2}} {\rm d}t} \right]}^{1/2}}} {\rm d}x \leqslant\qquad\qquad\\ &\qquad\int_{ - \infty }^\infty {{{\left[ {{{\left( {\int_0^T 2 {f_x}(s,x){g_{xx}}(s,x){\rm d}s} \right)}^2}\int_0^T | h{|^2}{\rm d}t} \right]}^{1/2}}} {\rm d}x\leqslant\\ &\qquad 2\mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T | {g_{xx}}{|^2}{\rm d}t} \right)^{1/2}}{\left( {\int_{ - \infty }^\infty {\int_0^T | } {f_x}{|^2}{\rm d}t{\rm d}x} \right)^{1/2}}{\text{·}}\\ &\qquad {\left( {\int_{ - \infty }^\infty {\int_0^T | } h{|^2}{\rm d}t{\rm d}x} \right)^{1/2}}\leqslant\\ &\qquad 2T\underset{-\infty <x<\infty }{\mathop{\sup }}\,{{\left( \int_{0}^{T}{|}{{g}_{xx}}{{|}^{2}}{\rm d}t \right)}^{1/2}}\|f{{\|}_{L_{T}^{\infty }({{H}^{1}})}}\|h{{\|}_{L_{T}^{\infty }({{L}^{2}})}}\leqslant\\ &\qquad {{C}_{1}}(T)\|f{{\|}_{X_{3}^{T}}}\|g{{\|}_{X_{3}^{T}}}\|h{{\|}_{X_{3}^{T}}}\end{split} $

同样地,可得式(6)余下的5项分别可以被 ${{C}_{i}}(T)\|f{{\|}_{X_{3}^{T}}}\|g{{\|}_{X_{3}^{T}}}\|h{{\|}_{X_{3}^{T}}}$ 控制,其中, $T \to 0$ 时, ${C_i}(T) \to $ $ 0$ $C(T) \geqslant 6{\rm {max}}\{ {C_i}(T)\} ,i = 1,2,3,4,5,6.$

将方程(4)的第2式代入到第1式,并结合初值条件 $u(0,x) = {u_0}(x),x \in {\mathbb{R}}$ ,可得

$u(t) = U(t){u_0} - {\rm i}\int_0^t U (t - s){F}(u(s)){\rm d}s$ (7)

其中,

${F}\left( {u(t)} \right) = \left[ {\int_0^t {(|u(} x,s){|^2}{)_x}{\rm d}s} \right]u + {v_0}u + \gamma |u{|^2}u + \delta |u{|^4}u\!\!\!\!\!\!$ (8)

现用压缩映射定理

$B_{r}^{3}=\left\{ u\in X_{3}^{T}\mid \|u{{\|}_{X_{3}^{T}}}\leqslant r \right\},\quad r>0$

${u_0} \in {H^{5/2}}({\mathbb R}),{v_0} \in {H^2}({\mathbb R})$ , 定义算子 $\varPhi $ ,

$\left( {\varPhi (\omega )} \right)(t) = U(t){u_0} - {\rm i}\int_0^t U (t - s){ F}(\omega (s)){\rm d}s$ (9)

其中,

$\begin{split} &\quad{F}\left( {\omega (t)} \right) = \left[ {\int_0^t {(|\omega (} x,s){|^2}{)_x}{\rm d}s} \right]\omega + \qquad\qquad\qquad\qquad\qquad\\ &\qquad {v_0}\omega + \gamma |\omega {|^2}\omega + \delta |\omega {|^4}\omega\\[-10pt] \end{split}$ (10)

根据式(5)对空间 $X_3^T$ 的定义,可得

$\begin{split} &\quad\|\varPhi (\omega )(t){{\|}_{X_{3}^{T}}}=\underset{0\leqslant t\leqslant T}{\mathop{\sup }}\,\|\varPhi (\omega )(t){{\|}_{{{H}^{5/2}}}}+ \\ & \qquad\mathop {\sup }\limits_{ - \infty < x < \infty } \left( \int_0^T {{{\left| {{{\rm D}_x}\varPhi (\omega )(x,t)} \right|}^2}} {\rm d}t +\right.\\ &\left. \qquad\int_0^T {{{\left| {{\rm D}_x^2\varPhi (\omega )(x,t)} \right|}^2}} {\rm d}t + \int_0^T {{{\left| {{\rm D}_x^3\varPhi (\omega )(x,t)} \right|}^2}} {\rm d}t \right)^{1/2}+\qquad\qquad\\ & \qquad\mathop {\sup }\limits_{ - \infty < x < \infty } \left( \int_0^T | {{\rm D}_t}\varPhi (\omega )(x,t){|^2}{\rm d}t+ \right.\\ &\left. \qquad \int_0^T | {{\rm D}_x}{{\rm D}_t}\varPhi (\omega )(x,t){|^2}{\rm d}t \right)^{1/2}\\[-15pt]\end{split}$ (11)

现需要对式(11)中等式右边的每一项作出估计。

首先,对第1项的估计为

$\|\varPhi (\omega )(t){{\|}_{{{L}^{2}}}}\leqslant C\|{{u}_{0}}{{\|}_{{{L}^{2}}}}+\int_{0}^{t}{\|{ F}\left( \omega (s) \right){{\|}_{{{L}^{2}}}}}{\rm d}s$

由引理4可得

$\begin{split}& \quad\int_{0}^{T}{\|{F}\left( \omega (s) \right){{\|}_{{{L}^{2}}}}}{\rm d}t\leqslant C\left( {{T}^{3/2}}+T \right)\|\omega \|_{X_{3}^{T}}^{3}+\qquad\qquad\qquad\\ &\qquad CT\|{{v}_{0}}{{\|}_{{{L}^{\infty }}}}\|\omega {{\|}_{X_{3}^{T}}}+CT\|\omega \|_{X_{3}^{T}}^{5}\\[-15pt]\end{split}$ (12)

所以,有

$\begin{split}&\quad\|\varPhi (\omega )(t){{\|}_{{{L}^{2}}}}\leqslant C\|{{u}_{0}}{{\|}_{{{L}^{2}}}}+C\left( {{T}^{3/2}}+T \right)\|\omega \|_{X_{3}^{T}}^{3}+\qquad\qquad\\ &\qquad CT\|{{v}_{0}}{{\|}_{{{L}^{\infty }}}}\|\omega {{\|}_{X_{3}^{T}}}+CT\|\omega \|_{X_{3}^{T}}^{5}\\[-15pt]\end{split}$ (13)

由于当 $0 \leqslant s < t \leqslant T$ 时, 有

$\begin{split}&\quad |\omega (t){|^2} \leqslant |\omega (s){|^2} + \left| {\int_s^t {{{\rm D}_\tau }} |\omega (\tau ){|^2}{\rm d}\tau } \right|\leqslant\qquad\qquad\qquad\qquad\\ & \qquad|\omega (s){|^2} + 2{\left( {\int_0^T | {\omega _t}{|^2}{\rm d}t} \right)^{1/2}}{\left( {\int_0^T | \omega {|^2}{\rm d}t} \right)^{1/2}}\\[-15pt] \end{split}$ (14)

在式(14)中,取 $s = 0$ ,可得

$ \begin{split}&\quad\int_{ - \infty }^\infty {{{\left( {\int_0^T | \omega {|^6}{\rm d}t} \right)}^{1/2}}} {\rm d}x\leqslant\\ &\qquad C{{\left( \int_{-\infty }^{\infty }{|}\omega (s){{|}^{4}}{\rm d}x \right)}^{1/2}}{{\left( \int_{-\infty }^{\infty }{\int_{0}^{T}{|}}\omega {{|}^{2}}{\rm d}t{\rm d}x \right)}^{1/2}}+\\ &\qquad CT{{\left( \int_{-\infty }^{\infty }{\int_{0}^{T}{|}}{{\omega }_{t}}|{\rm d}t{\rm d}x \right)}^{1/2}}\|\omega \|_{L_{T}^{\infty }({{L}^{2}})}^{2}\leqslant\\ &\qquad C{{T}^{1/2}}\underset{-\infty <x<\infty }{\mathop{\sup }}\,|\omega (s)|{{\left( \int_{-\infty }^{\infty }{|}\omega (s){{|}^{2}}{\rm d}x \right)}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+ \qquad\qquad\qquad\\ &\qquad CT\|\omega \|_{X_{3}^{T}}^{3}\leqslant C{{T}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+CT\|\omega \|_{X_{3}^{T}}^{3}\\[-15pt]\end{split}$ (15)

$\begin{split}&\quad\int_{ - \infty }^\infty {{{\left( {\int_0^T | \omega {|^{10}}{\rm d}t} \right)}^{1/2}}} {\rm d}x\leqslant \\ &\qquad C{{T}^{1/2}}|\omega (s){{|}^{3}}\|\omega (s){{\|}_{{{L}^{2}}}}\|\omega {{\|}_{X_{3}^{T}}}+\\ &\qquad CT\underset{-\infty \!<\!x\!<\!\infty }{\mathop{\sup }}\,\int_{0}^{T}{|}{{\omega }_{t}}{{|}^{2}}{\rm d}t\int_{-\infty }^{\infty }{\underset{0\!\leqslant\! t\!\leqslant\! T}{\mathop{\sup }}\,}|\omega {{|}^{2}}{{\left( \int_{0}^{T}{|}\omega {{|}^{2}}{\rm d}t \right)}^{1/2}}{\rm d}x \!\leqslant \\ &\qquad C{{T}^{1/2}}|\omega (s){{|}^{3}}\|\omega (s){{\|}_{{{L}^{2}}}}\|\omega {{\|}_{X_{3}^{T}}}+\\ &\qquad CT\underset{-\infty <x<\infty }{\mathop{\sup }}\,\int_{0}^{T}{|}{{\omega }_{t}}{{|}^{2}}{\rm d}t\left( C{{T}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+CT\|\omega \|_{X_{3}^{T}}^{3} \right) \leqslant \\ &\qquad C{{T}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+C{{T}^{3/2}}\|\omega \|_{X_{3}^{T}}^{3}+C{{T}^{2}}\|\omega \|_{X_{3}^{T}}^{5}\\[-15pt] \end{split}$ (16)

由式(15)和式(16)可得

$\begin{split}&\quad\int_{ - \infty }^\infty {{{\left( {\int_0^T | {F}(\omega (t,x)){|^2}{\rm d}t} \right)}^{1/2}}} {\rm d}x\leqslant \\ &\qquad\int_{ - \infty }^\infty \left( \int_0^T | \left[ {\int_0^t {(|\omega (} s,x){|^2}{)_x}{\rm d}s} \right]\omega +{v_0}\omega + \right.\\ &\left.\qquad\gamma |\omega {|^2}\omega + \delta |\omega {|^4}\omega |^2 {\rm d}t \right)^{1/2} {\rm d}x \leqslant \\ &\qquad CT\underset{-\infty <x<\infty }{\mathop{\sup }}\,{{\left( \int_{0}^{T}{|}{{\omega }_{x}}{{|}^{2}}{\rm d}t \right)}^{1/2}}\|\omega \|_{X_{3}^{T}}^{2}+\\ &\qquad C{{T}^{1/2}}\|{{v}_{0}}{{\|}_{{{L}^{2}}}}\|\omega {{\|}_{X_{3}^{T}}}+C{{T}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+CT\|\omega \|_{X_{3}^{T}}^{3}+\qquad\qquad\qquad\\ &\qquad C{{T}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+C{{T}^{3/2}}\|\omega \|_{X_{3}^{T}}^{3}+C{{T}^{2}}\|\omega \|_{X_{3}^{T}}^{5} \leqslant\\ &\qquad C(T)\|\omega \|_{X_{3}^{T}}^{3}+C(T)\|\omega \|_{X_{3}^{T}}^{5}+C(T)\|\omega {{\|}_{X_{3}^{T}}}\\[-13pt] \end{split} $ (17)

结合引理1, 引理2 和式(17),可得

$\begin{split}&\|{\rm D}_{x}^{1/2}\varPhi (\omega )(t,x){{\|}_{{{L}^{2}}}}\leqslant C\|{{u}_{0}}{{\|}_{{{H}^{1/2}}}}+\\ &\qquad C(T)\|\omega \|_{X_{3}^{T}}^{3}+C(T)\|\omega \|_{X_{3}^{T}}^{5}+C(T)\|\omega {{\|}_{X_{3}^{T}}}\end{split}$ (18)

由引理3和式(17),可以估计式(11)的第2项为

$\begin{split}&\quad\underset{-\infty <x<\infty }{\mathop{\sup\limits }}\,{{\left( \int_{0}^{T}{|}{{\rm D}_{x}}\varPhi (\omega )(x,t){{|}^{2}}{\rm d}t \right)}^{1/2}}\leqslant \|{{u}_{0}}{{\|}_{{{H}^{1/2}}}}+\qquad\qquad\qquad\\ &\qquad C(T)\|\omega \|_{X_{3}^{T}}^{3}+C(T)\|\omega \|_{X_{3}^{T}}^{5}+C(T)\|\omega {{\|}_{X_{3}^{T}}}\\[-13pt] \end{split}$ (19)

其中, $T \to 0$ 时, $C(T) \to 0$ ,且 $C(T)$ 依赖于 $\|{{v}_{0}}{{\|}_{{{L}^{2}}}}$

估计式(11)余下的项之前, 先作出下面的估计。采用与式(15)和式(16)同样的估计方法,可得

$\begin{split}&\quad\int_{-\infty }^{\infty }{{{\left[ \int_{0}^{T}{{{\left| {{\rm D}_{x}}\left( |\omega (t,x){{|}^{2}}\omega (t,x) \right) \right|}^{2}}}{\rm d}t \right]}^{1/2}}}{\rm d}x\leqslant\qquad\qquad\qquad\\ &\qquad C{{T}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+CT\|\omega \|_{X_{3}^{T}}^{3}\\[-15pt]\end{split}$ (20)
$\begin{split}&\quad\int_{-\infty }^{\infty }{{{\left[ \int_{0}^{T}{{{\left| {{\rm D}_{x}}\left( |\omega (t,x){{|}^{4}}\omega (t,x) \right) \right|}^{2}}}{\rm d}t \right]}^{1/2}}}{\rm d}x\leqslant \qquad\qquad\qquad\\ &\qquad C{{T}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+C{{T}^{3/2}}\|\omega \|_{X_{3}^{T}}^{3}+C{{T}^{2}}\|\omega \|_{X_{3}^{T}}^{5}\\[-15pt]\end{split}$ (21)

结合引理5,式(20)和式(21),可得

$\begin{split}&\quad\int_{ - \infty }^\infty {{{\left( {\int_0^T | {{\rm D}_x}F(\omega (t,x)){|^2}{\rm d}t} \right)}^{1/2}}} {\rm d}x \leqslant \\ &\qquad C\left( T+{{T}^{3/2}} \right)\|\omega {{\|}_{X_{3}^{T}}}+C{{T}^{1/2}}\|{{v}_{0}}{{\|}_{{{H}^{1}}}}\|\omega {{\|}_{X_{3}^{T}}}+\\ &\qquad C{{T}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+C(T+{{T}^{3/2}})\|\omega \|_{X_{3}^{T}}^{3}+CT\|\omega \|_{X_{3}^{T}}^{5}\leqslant\qquad\qquad\qquad\\ & \qquad C(T)\left( \|\omega {{\|}_{X_{3}^{T}}}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega \|_{X_{3}^{T}}^{5} \right)\\[-13pt] \end{split}$ (22)

其中, $C(T)$ 依赖于 $\|{{v}_{0}}{{\|}_{{{H}^{1}}}}$ , 且 $T \to 0$ 时, $C(T) \to 0$ .

由引理1,引理2和式(22),可估计

$\begin{split}&\quad\|{\rm D}_{x}^{1/2}{{\rm D}_{x}}\varPhi (\omega )(t,x){{\|}_{{{L}^{2}}}}\leqslant\\ & \qquad C\|{{\rm D}_{x}}{\rm D}_{x}^{1/2}{{u}_{0}}{{\|}_{{{L}^{2}}}}+{{\left\| {\rm D}_{x}^{1/2}\int_{0}^{t}{U}(t-s){{\rm D}_{x}}{F}(\omega (s)){\rm d}s \right\|}_{{{L}^{2}}}}\leqslant\\ & \qquad C\|{{u}_{0}}{{\|}_{{{H}^{3/2}}}}+C\int_{-\infty }^{\infty }{{{\left( \int_{0}^{T}{{{\left| {{\rm D}_{x}}{F}(\omega (t,x)) \right|}^{2}}}{\rm d}t \right)}^{1/2}}}{\rm d}x\leqslant\\ & \qquad C\|{{u}_{0}}{{\|}_{{{H}^{3/2}}}}+C(T)\left( \|\omega {{\|}_{X_{3}^{T}}}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega \|_{X_{3}^{T}}^{5} \right)\\[-13pt] \end{split}$ (23)

同样地,结合引理1,引理3和式(22),可得

$ \begin{split} & \quad\mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T {{{\left| {{\rm D}_x^2\varPhi (\omega )(x,t)} \right|}^2}} {\rm d}t} \right)^{1/2}}\leqslant\\ & \qquad C\|{{u}_{0}}{{\|}_{{{H}^{3/2}}}}+C\int_{-\infty }^{\infty }{{{\left( \int_{0}^{T}{{{\left| {{\rm D}_{x}}F(\omega (t,x)) \right|}^{2}}}{\rm d}t \right)}^{1/2}}}{\rm d}x\leqslant\qquad\qquad\qquad\\ & \qquad C\|{{u}_{0}}{{\|}_{{{H}^{3/2}}}}+C(T)\left( \|\omega {{\|}_{X_{3}^{T}}}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega \|_{X_{3}^{T}}^{5} \right)\\[-15pt] \end{split} $ (24)

其中, $C(T)$ 依赖于 $\|{{v}_{0}}{{\|}_{{{H}^{1}}}}$ ,且 $T \to 0$ 时, $C(T) \to 0$

根据 $U(t)$ 的定义, 有

${\rm i}\;{{\rm D}_t}U(t)f + {{\rm D}_{xx}}U(t)f = 0$

可得

${{\rm D}_t}U(t)f = {\rm i}\;{{\rm D}_{xx}}U(t)f$

对式(9)的两边分别关于 $t$ 微分,可得

$\begin{split}&\quad{{\rm D}_t}\varPhi (\omega )(t) = {\rm {i\;D}}_x^2U(t){u_0} +\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\ &\qquad\int_0^t {{\rm D}_x^2} U(t \!-\! s){F}\left( {\omega (s)} \right){\rm d}s \!-\! {\rm {i\;}}F\left( {\omega (t)} \right)\\[-15pt]\end{split}$ (25)

由式(24)可得

$\begin{split} &\quad\mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T | {{\rm D}_t}\varPhi (\omega )(x,t){|^2}{\rm d}t} \right)^{1/2}}\leqslant\\[-2pt] & \qquad\mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T {{{\left| {{\rm D}_x^{1/2}U(t){\rm D}_x^{3/2}{u_0}(x)} \right|}^2}} {\rm d}t} \right)^{1/2}} +\\[-2pt] &\qquad C\int_{ - \infty }^\infty {{{\left( {\int_0^T | {{\rm D}_x}{F}(\omega (t,x)){|^2}{\rm d}t} \right)}^{1/2}}} {\rm d}x+\\[-2pt] &\qquad \mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T | {F}(\omega (t)){|^2}{\rm d}t} \right)^{1/2}}\leqslant\\[-2pt] &\qquad C\|{{u}_{0}}{{\|}_{{{H}^{3/2}}}}+C(T)\left( \|\omega {{\|}_{X_{3}^{T}}}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega \|_{X_{3}^{T}}^{5} \right)+\qquad\qquad\qquad\qquad\\[-2pt] &\qquad\underset{-\infty <x<\infty }{\mathop{\sup }}\,{{\left( \int_{0}^{T}{{{\left| {F}(\omega (t)) \right|}^{2}}}{\rm d}t \right)}^{1/2}}\\[-15pt] \end{split} $ (26)

对式(26)右边最后一项的估计为

$\begin{split} &\!\!\!\!\!\! \mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T | {F}(\omega (t)){|^2}{\rm d}t} \right)^{1/2}}\leqslant\\[-3pt] &\!\!\!\!\!\!\quad \mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T {{{\left| {\int_0^t | \omega (s,x)||{\omega _x}(s,x)|{\rm d}s\omega (t,x)} \right|}^2}} {\rm d}t} \right)^{1/2}} +\\[-3pt] &\!\!\!\!\!\!\quad \mathop {\sup }\limits_{ - \infty \!< \!x \!< \!\infty } {\left( {\int_0^T | \omega (t,x){|^6}{\rm d}t} \right)^{1/2}}\!\!\!\!+\!\!\!\!\mathop {\sup }\limits_{ - \infty \! <\! x \!<\! \infty } {\left( {\int_0^T | \omega (t,x){|^{10}}{\rm d}t} \right)^{1/2}} \!\!+\\[-3pt] &\!\!\!\!\!\!\quad \mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T | {v_0}(x)\omega (t,x){|^2}{\rm d}t} \right)^{1/2}} \leqslant \\[-3pt] &\!\!\!\!\!\!\quad C{{\left[ \underset{-\infty <x<\infty }{\mathop{\sup }}\,{{\left( \int_{0}^{T}{|}\omega {{|}^{2}}{\rm d}t \right)}^{1/2}} \right]}^{2}}\underset{-\infty <x<\infty }{\mathop{\sup }}\,{{\left( \int_{0}^{T}{|}{{\omega }_{x}}{{|}^{2}}{\rm d}t \right)}^{1/2}}+\\[-3pt] &\!\!\!\!\!\!\quad C{{T}^{1/2}}\|\omega \|_{L_{T}^{\infty }({{H}^{1}})}^{3}+C{{T}^{1/2}}\|\omega \|_{L_{T}^{\infty }({{H}^{1}})}^{5}+C{{T}^{1/2}}{\text{·}}\\[-3pt] &\!\!\!\!\!\!\quad \|{{v}_{0}}{{\|}_{{{L}^{\infty }}}}\|\omega {{\|}_{L_{T}^{\infty }({{H}^{1}})}} \!\leqslant\! C(T)\left( \|\omega \|_{X_{3}^{T}}^{5}\!+\!\|\omega \|_{X_{3}^{T}}^{3}\!+\!\|\omega {{\|}_{X_{3}^{T}}} \right)\\[-15pt] \end{split} $ (27)

所以,由式(26)和式(27)可得

$ \begin{split}& \quad\mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T | {{\rm D}_t}\varPhi (\omega )(x,t){|^2}{\rm d}t} \right)^{1/2}}\leqslant\\ &\qquad C\|{{u}_{0}}{{\|}_{{{H}^{3/2}}}}+C(T)\left( \|\omega \|_{X_{3}^{T}}^{5}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega {{\|}_{X_{3}^{T}}} \right)\qquad\qquad \\[-15pt]\end{split} $ (28)

最后,用引理6可以估计出

$ \begin{split} &\quad\int_{ - \infty }^\infty {{{\left( {\int_0^T | {\rm D}_x^2F(\omega (t,x)){|^2}{\rm d}t} \right)}^{1/2}}} {\rm d}x\leqslant\\[-3pt] & \qquad C(T)\|\omega \|_{X_{3}^{T}}^{3}+C{{T}^{1/2}}\|{{v}_{0}}{{\|}_{{{H}^{2}}}}\|\omega {{\|}_{X_{3}^{T}}}+\qquad\qquad\qquad\qquad\\[-3pt] &\qquad C\int_{-\infty }^{\infty }{{{\left( \int_{0}^{T}{{{\left| {\rm D}_{x}^{2}\left( |\omega {{|}^{2}}\omega \right) \right|}^{2}}}{\rm d}t \right)}^{1/2}}}{\rm d}x+\\[-3pt] & \qquad C\int_{ - \infty }^\infty {{{\left( {\int_0^T {{{\left| {{\rm D}_x^2\left( {|\omega {|^4}\omega } \right)} \right|}^2}} {\rm d}t} \right)}^{1/2}}} {\rm d}x\\[-15pt] \end{split}$ (29)

对式(29)右边第3项的估计为

$ \begin{split} & \quad\int_{ - \infty }^\infty {{{\left( {\int_0^T {{{\left| {{\rm D}_x^2\left( {|\omega {|^2}\omega } \right)} \right|}^2}} {\rm d}t} \right)}^{1/2}}} {\rm d}x\leqslant\\ & \qquad \int_{ - \infty }^\infty {{{\left( {\int_0^T {{{\left| {6|\omega ||{\omega _x}{|^2} + 3|\omega {|^2}|{\omega _{xx}}|} \right|}^2}} {\rm d}t} \right)}^{1/2}}} {\rm d}x\leqslant\qquad\qquad\qquad\\ & \qquad C\int_{ - \infty }^\infty {\mathop {\sup }\limits_{0 \leqslant t \leqslant T} } |{\omega _x}{|^2}{\left( {\int_0^T | \omega {|^2}{\rm d}t} \right)^{1/2}}{\rm d}x + \\ &\qquad C\int_{ - \infty }^\infty {\mathop {\sup }\limits_{0 \leqslant t \leqslant T} } |\omega {|^2}{\left( {\int_0^T | {\omega _{xx}}{|^2}{\rm d}t} \right)^{1/2}}{\rm d}x \leqslant\\ &\qquad C{{T}^{1/2}}\|\omega \|_{X_{3}^{T}}^{3}\\[-13pt] \end{split} $ (30)

由式(15)和式(21)可得对式(29)右边最后一项的估计为

$ \begin{split} &\quad\int_{ - \infty }^\infty {{{\left( {\int_0^T {{{\left| {{\rm D}_x^2\left( {|\omega {|^4}\omega } \right)} \right|}^2}} {\rm d}t} \right)}^{1/2}}} {\rm d}x \leqslant\\ & \qquad C\int_{ - \infty }^\infty \!\! {{{\left( {\int_0^T | \omega {|^6}|{\omega _x}{|^4}{\rm d}t} \right)}^{1/2}}} \!{\rm d}x +\\ &\qquad C\int_{ - \infty }^\infty {{{\left( {\int_0^T | \omega {|^8}|{\omega _{xx}}{|^2}{\rm d}t} \right)}^{1/2}}} {\rm d}x\! \leqslant\\ & \qquad C{{T}^{1/2}}\int_{-\infty }^{\infty }{\underset{0\leqslant t\leqslant T}{\mathop{\sup }}\,}|\omega {{|}^{2}}{{\left( \int_{0}^{T}{|}\omega {{|}^{2}}{\rm d}t \right)}^{1/2}}{\rm d}x\|\omega \|_{X_{3}^{T}}^{2}+\\ &\qquad \int_{-\infty }^{\infty }{\underset{0\leqslant t\leqslant T}{\mathop{\sup }}\,}|\omega {{|}^{4}}{{\left( \int_{0}^{T}{|}{{\omega }_{xx}}{{|}^{2}}{\rm d}t \right)}^{1/2}}{\rm d}x \leqslant C{{T}^{1/2}}\|\omega {{\|}_{X_{3}^{T}}}+\qquad\qquad\\ &\qquad C(T+{{T}^{3/2}})\|\omega \|_{X_{3}^{T}}^{3}+ C({{T}^{3/2}}+{{T}^{2}})\|\omega \|_{X_{3}^{T}}^{5}\\[-13pt]\end{split} $

因此,由式(30)和式(31)可得

$\begin{split}&\quad \int_{-\infty }^{\infty }{{{\left( \int_{0}^{T}{|}{\rm D}_{x}^{2}F(\omega (t,x)){{|}^{2}}{\rm d}t \right)}^{1/2}}}{\rm d}x\leqslant \qquad\qquad\qquad\qquad\qquad\\[-2pt] &\qquad C(T)\left( \|\omega {{\|}_{X_{3}^{T}}}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega \|_{X_{3}^{T}}^{5} \right)\\[-15pt]\end{split}$ (32)

其中, $C(T)$ 依赖于 $\|{{v}_{0}}{{\|}_{{{H}^{2}}}}$ , 且 $T \to 0$ 时, $C(T) \to 0$

结合引理1,引理2和式(32),可得

$\begin{split} &\quad\|{\rm D}_{x}^{1/2}{\rm D}_{x}^{2}\varPhi (\omega )(t,x){{\|}_{{{L}^{2}}}}\leqslant \\[-2pt] &\qquad C\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}+C\int_{-\infty }^{\infty }{{{\left( \int_{0}^{T}{{{\left|{\rm D}_{x}^{2}{F}(\omega (t,x)) \right|}^{2}}}{\rm d}t \right)}^{1/2}}}{\rm d}x\leqslant \qquad\quad\\[-2pt] &\qquad C\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}+C(T)\left( \|\omega {{\|}_{X_{3}^{T}}}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega \|_{X_{3}^{T}}^{5} \right)\\[-15pt]\end{split} $ (33)

由引理1, 引理3和式(32), 可得对式(11)的第4项的估计为

$\begin{split} &\quad \mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T | {\rm D}_x^3\varPhi (\omega )(x,t){|^2}{\rm d}t} \right)^2}\leqslant \\[-2pt] &\qquad C\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}+C\int_{-\infty }^{\infty }{{{\left( \int_{0}^{T}{|}{\rm D}_{x}^{2}{F}(\omega (t,x)){{|}^{2}}{\rm d}t \right)}^{1/2}}}{\rm d}x\leqslant\qquad\quad\\[-2pt] & \qquad C\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}+C(T)\left( \|\omega \|_{X_{3}^{T}}^{5}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega {{\|}_{X_{3}^{T}}} \right)\\[-15pt] \end{split} $ (34)

同样地,

$\begin{split} & \quad \mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T | {{\rm D}_x}{{\rm D}_t}\varPhi (\omega )(x,t){|^2}{\rm d}t} \right)^{1/2}}\leqslant\\[-2pt] &\qquad \mathop {\sup }\limits_{ - \infty < x < \infty } \left( \int_0^T \left| {{\rm D}_x}[ {\rm {iD}}_x^2U(t){u_0} + \int_0^t {{\rm D}_x^2} U(t - s){\text{·}}\right.\right.\qquad\qquad\qquad\\[-2pt] &\qquad \left.\left. F(\omega (s)){\rm d}s -{\rm {i}}F(\omega (t)) ] \right|^2 {\rm d}t \right)^{1/2} \leqslant\\[-2pt] &\qquad C\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}+C\int_{-\infty }^{\infty }{{{\left( \int_{0}^{T}{|}{\rm D}_{x}^{2}{F}(\omega ){{|}^{2}}{\rm d}t \right)}^{1/2}}}{\rm d}x+\\[-2pt] &\qquad\underset{-\infty <x<\infty }{\mathop{\sup }}\,{{\left( \int_{0}^{T}{|}{{\rm D}_{x}}F(\omega ){{|}^{2}}{\rm d}t \right)}^{1/2}}\\[-15pt] \end{split} $ (35)

由方程(27), 可得对式(35)的第3项的估计为

$\begin{split} &\quad\underset{-\infty <x<\infty }{\mathop{\sup\limits }}\,{{\left( \int_{0}^{T}{|}{{\rm D}_{x}}{F}(\omega ){{|}^{2}}{\rm d}t \right)}^{1/2}}\leqslant \\ &\qquad C(T)\|\omega {{\|}_{X_{3}^{T}}}+C(T)\|\omega \|_{X_{3}^{T}}^{3}+C(T)\|\omega \|_{X_{3}^{T}}^{5}\qquad\qquad\qquad\qquad\end{split}$

因此,

$\begin{split}&\quad\mathop {\sup }\limits_{ - \infty < x < \infty } {\left( {\int_0^T | {{\rm D}_x}{{\rm D}_t}\varPhi (\omega )(x,t){|^2}{\rm d}t} \right)^{1/2}}\leqslant\\ &\qquad C\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}+C(T)\left( \|\omega \|_{X_{3}^{T}}^{5}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega {{\|}_{X_{3}^{T}}} \right)\qquad\qquad\qquad\\[-15pt]\end{split} $ (36)

结合以上的估计, 可得

$\begin{split} &\quad\|\varPhi (\omega ){{\|}_{X_{3}^{T}}}\leqslant {{C}_{7}}\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}+\\ &\qquad{{C}_{8}}(T)\left( \|\omega \|_{X_{3}^{T}}^{5}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega {{\|}_{X_{3}^{T}}} \right)\qquad\qquad\qquad\qquad\\[-15pt]\end{split}$ (37)

其中, ${C_8}(T)$ 是常数, 且 $T \to 0$ 时, ${C_8}(T) \to 0$

选定 $r$ ,且有

${{C}_{7}}\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}\leqslant \frac{r}{2}$

取足够小的 $T$ , 使得

${{C}_{8}}(T)\left( \|\omega \|_{X_{3}^{T}}^{5}+\|\omega \|_{X_{3}^{T}}^{3}+\|\omega {{\|}_{X_{3}^{T}}} \right)<\frac{r}{2}$

可知

$\varPhi :B_r^3 \to B_r^3$

$\varPhi $ 是在 $B_r^3$ 上的一个压缩。采用与估计不等式(37)同样的方法,可得

$\begin{split} &\quad\|\varPhi ({{\omega }_{1}})-\varPhi ({{\omega }_{2}}){{\|}_{X_{3}^{T}}}\leqslant{{C}_{9}}(T)\left( \|{{\omega }_{1}}\|_{X_{3}^{T}}^{4}+\|{{\omega }_{2}}\|_{X_{3}^{T}}^{4}+\right.\qquad\qquad\qquad\quad\\ &\qquad\left.\|{{\omega }_{1}}\|_{X_{3}^{T}}^{2}+\|{{\omega }_{2}}\|_{X_{3}^{T}}^{2}+1 \right)\|{{\omega }_{1}}-{{\omega }_{2}}{{\|}_{X_{3}^{T}}}\end{split}$

$T$ 足够小,满足

${C_9}(T)(2{r^4} + 2{r^2} + 1) < 1$

因此,存在唯一的泛函 $u \in X_3^T$ ,使得 $\varPhi (u) = u$

3 全局解的存在性

首先证明方程(4)的解的守恒性质和2个先验估计结果,然后运用反证法,证明方程(4)初值问题的解在空间 $X_3^T$ 中的全局存在性。

引理 7  设 ${{u}_{0}}\in {{H}^{5/2}}(\mathbb{R})$ u是方程(4)的解,则有

$\qquad\qquad\quad\int_{ - \infty }^\infty | u(t,x){|^2}{\rm d}x = \int_{ - \infty }^\infty | {u_0}(x){|^2}{\rm d}x$ (38)

证明  对方程(4)的第1个式子的两边关于 $u$ 作内积,可得

$\begin{split}&\quad{\rm i}\langle {u_t},u\rangle + \langle {u_{xx}},u\rangle = \langle vu,u\rangle + \langle \gamma |u{|^2}u,u\rangle +\qquad\qquad\qquad\qquad\qquad\\ &\qquad\langle \delta |u{|^4}u,u\rangle\\[-10pt]\end{split}$ (39)

可知 $\langle u,u\rangle = \int_{\mathbb{R}} {\bar u} u{\rm d}x$ 是实的。取方程(39)的虚部,可得

$\langle {{u}_{t}},u\rangle =\frac{1}{2}\frac{{\rm d}}{{\rm d}t}\|u\|_{{{L}^{2}}}^{2}=0$ (40)

对方程(40)的两边关于 $t$ 积分,可得

$ \int_{ - \infty }^\infty | u(t,x){|^2}{\rm d}x = \int_{ - \infty }^\infty | {u_0}(x){|^2}{\rm d}x $

引理 8  设 $\left( u,v \right)$ 是方程(4)的解,有

$ \begin{split}&\quad\int_{ - \infty }^\infty {\left( {v(t,x)|u(t,x){|^2} + |{u_x}(t,x){|^2} + \dfrac{\gamma }{2}|u(t,x){|^4} + \dfrac{\delta }{3}|u(t,x){|^6}} \right)} {\rm d}x=\\ &\qquad \int_{ - \infty }^\infty {\left( {{v_0}(t,x)|{u_0}(t,x){|^2} +|{u_{{0_x}}}(t,x){|^2} + \dfrac{\gamma }{2}|{u_0}(t,x){|^4} + \dfrac{\delta }{3}|{u_0}(t,x){|^6}} \right)} {\rm d}x\end{split}$ (41)
$\int_{ - \infty }^\infty {\left( {{v^2}(t,x) + 2{\rm {Im}}\left( {u(t,x)\overline {{u_x}(t,x)} } \right)} \right)} {\rm d}x= \int_{ - \infty }^\infty {\left( {v_0^2(t,x) + 2{\rm{Im}}\left( {{u_0}(t,x)\overline {{u_{{0_x}}}(t,x)} } \right)} \right)} {\rm d}x $ (42)

证明  令 $\vec S = (u,v)$ ,将方程(4)写成哈密尔顿系统

$\frac{{{\rm d}\vec S}}{{{\rm d}t}} = JE'(\vec S)$

其中, $J$ 是斜对称算子,且有

$J = \left( {\begin{array}{*{20}{c}} { - {\rm i}}&{} \\ {}&{2\dfrac{\partial }{{\partial x}}} \end{array}} \right)$
$E'(\vec S) = \left( {\begin{array}{*{20}{c}} { - {u_{xx}} + uv + \gamma |u{|^2}u + \delta |u{|^4}u} \\ {\dfrac{1}{2}|u{|^2}} \end{array}} \right)$

可知

$\begin{split} &\quad E(\vec S) = \frac{1}{2}\int_{ - \infty }^\infty \left( v(t,x)|u(t,x){|^2} + |{u_x}(t,x){|^2} + \right.\qquad\qquad\qquad\qquad\qquad\qquad\\ &\qquad\left.\frac{\gamma }{2}|u(t,x){|^4} + \frac{\delta }{3}|u(t,x){|^6} \right) {\rm d}x\end{split} $

$u = {u_1} + {\rm i}{u_2}$ ,所以,有

$ \begin{split}&\quad\frac{{{\rm d}E(\vec S)}}{{{\rm d}t}} = {\rm {Re}}\int_{\mathbb{R}} {\frac{1}{2}} |u{|^2}{\left( {|u{|^2}} \right)_x}{\rm d}x=\\ & \qquad\int_{\mathbb{R}} {\left( {u_1^2{u_2}{u_{2x}} + {u_1}{u_{1x}}u_2^2} \right)} {\rm d}x {\rm{ = 0}}\qquad\qquad\qquad\qquad\qquad\\[-15pt]\end{split}$ (43)

其中, $E'$ 是Frechet导数。由式(43)可知 $E(\vec S)$ 是守恒量, 故可得式(41)和式(42)。

引理 9  设u是方程(4)初值问题的解, 则有

$\|u{{\|}_{L_{T}^{\infty }({{L}^{2}})}}\leqslant \|{{u}_{0}}{{\|}_{{{L}^{2}}}}$ (44)

$\|u{{\|}_{L_{T}^{\infty }({{H}^{1}})}}\leqslant {{M}_{1}}$ (45)

其中, ${M_1}$ 是大于零的常数,且仅依赖于 $\|{{u}_{0}}{{\|}_{{{H}^{1}}}}$ $\|{{v}_{0}}{{\|}_{{{L}^{2}}}}$

证明  由式(41)和式(42)可得

$\begin{split} &\quad\int_{ - \infty }^\infty | {u_x}{|^2}{\rm d}x \leqslant {\left( {\int_{ - \infty }^\infty | v{|^2}{\rm d}x} \right)^{1/2}}{\left( {\int_{ - \infty }^\infty | u{|^4}{\rm d}x} \right)^{1/2}} +\\ &\qquad \frac{{|\gamma |}}{2}\int_{ - \infty }^\infty | u{|^4}{\rm d}x + \frac{{|\delta |}}{3}\int_{ - \infty }^\infty | u{|^6}{\rm d}x + C\leqslant\\ & \qquad C{\left( {\int_{ - \infty }^\infty | v{|^2}{\rm d}x} \right)^{1/2}}{\left( {\int_{ - \infty }^\infty | {u_x}{|^2}{\rm d}x} \right)^{1/4}}{\left( {\int_{ - \infty }^\infty | {u_0}{|^2}{\rm d}x} \right)^{3/4}} +\qquad\qquad\\ &\qquad C{\left( {\int_{ - \infty }^\infty | {u_x}{|^2}{\rm d}x} \right)^{1/2}}{\left( {\int_{ - \infty }^\infty | {u_0}{|^2}{\rm d}x} \right)^{3/2}} +\\ & \qquad C\frac{{|\delta |}}{3}\left( {\int_{ - \infty }^\infty | {u_x}{|^2}{\rm d}x} \right){\left( {\int_{ - \infty }^\infty | {u_0}{|^2}{\rm d}x} \right)^2} +C \\ & \quad \int_{ - \infty }^\infty | v{|^2}{\rm d}x \leqslant2{\left( {\int_{ - \infty }^\infty | {u_0}{|^2}{\rm d}x} \right)^{1/2}}{\left( {\int_{ - \infty }^\infty | {u_x}{|^2}{\rm d}x} \right)^{1/2}} + C\qquad \end{split}$

因此,

$\begin{split} &\quad \int_{-\infty }^{\infty }{|}{{u}_{x}}{{|}^{2}}{\rm d}x\leqslant C{{\left( \int_{-\infty }^{\infty }{|}{{u}_{x}}{{|}^{2}}{\rm d}x \right)}^{1/2}}+\\ & \qquad C{{\left( \int_{-\infty }^{\infty }{|}{{u}_{x}}{{|}^{2}}{\rm d}x \right)}^{1/4}}\!+\!C\frac{|\delta |}{3}\|{{u}_{0}}\|_{{{L}^{2}}}^{4}\int_{-\infty }^{\infty }{|}{{u}_{x}}{{|}^{2}}{\rm d}x\!+\!C\qquad\qquad\\[-15pt]\end{split}$ (46)

由式(46)可知,只要 $C\dfrac{|\delta |}{3}\|{{u}_{0}}\|_{{{L}^{2}}}^{4}<1$ ,可以作出下面的估计:

$\begin{split} &\quad\left(1-C\frac{|\delta |}{3}\|{{u}_{0}}\|_{{{L}^{2}}}^{4}\right)\int_{-\infty }^{\infty }{|}{{u}_{x}}{{|}^{2}}{\rm d}x\leqslant\\ & \qquad C{{\left( \int_{-\infty }^{\infty }{|}{{u}_{x}}{{|}^{2}}{\rm d}x \right)}^{1/2}}+C{{\left( \int_{-\infty }^{\infty }{|}{{u}_{x}}{{|}^{2}}{\rm d}x \right)}^{1/4}}+C\leqslant \qquad\qquad\qquad\qquad\qquad\\ &\qquad \frac{1-C\dfrac{|\delta |}{3}\|{{u}_{0}}\|_{{{L}^{2}}}^{4}}{2}\int_{-\infty }^{\infty }{|}{{u}_{x}}{{|}^{2}}{\rm d}x+C\end{split}$

其中, $\|{{u}_{0}}{{\|}_{{{L}^{2}}}}$ $\delta $ 必须足够小, $\delta $ 不是小常数时, $\|{{u}_{0}}{{\|}_{{{L}^{2}}}}$ 必须足够小才能满足式(45)。

引理 10  设u是方程(4)初值问题的解, 则有

$\|{{u}_{t}}{{\|}_{L_{T}^{\infty }({{L}^{2}})}}\leqslant {{M}_{2}}$ (47)

$\|u{{\|}_{L_{T}^{\infty }({{H}^{2}})}}\leqslant {{M}_{2}}$ (48)

其中, ${M_2}$ 是大于零的常数,且仅依赖于 $\|{{u}_{0}}{{\|}_{{{H}^{2}}}}$ $\|{{v}_{0}}{{\|}_{{{L}^{2}}}}$

证明  令 $\psi = {u_t},u \in L_T^\infty ({H^3}({\mathbb{R}}))$ ,由式(4)可得

$\begin{split} &\quad{\rm i}{\psi _t} + {\psi _{xx}} = {(|u(x,s){|^2})_x}u + \left[ {\int_0^t {(|u(} x,s){|^2}{)_x}{\rm d}s} \right]\psi +\qquad\qquad\\ & \qquad{v_0}(x)\psi + 2\gamma |u{|^2}\psi +\gamma {u^2}\bar \psi + 2\delta |u{|^2}{u^2}\bar \psi + 3\delta |u{|^4}\psi\quad\\[-10pt] \end{split}$ (49)

分别对式(49)的两边关于 $\psi $ 作内积,可得

$ \begin{split} &\quad{\rm i}\langle {\psi _t},\psi \rangle + \langle {\psi _{xx}},\psi \rangle = \langle {(|u(x,s){|^2})_x}u,\psi \rangle +\\ &\qquad\left\langle {\left[ {\int_0^t {(|u(} x,s){|^2}{)_x}{\rm d}s} \right]\psi ,\psi } \right\rangle \!+\! \langle {v_0}(x)\psi ,\psi \rangle \!+\! \langle 2\gamma |u{|^2}\psi ,\psi \rangle +\quad\\ &\qquad \langle \gamma {u^2}\bar \psi ,\psi \rangle + \langle 2\delta |u{|^2}{u^2}\bar \psi ,\psi \rangle + \langle 3\delta |u{|^4}\psi ,\psi \rangle\\[-10pt] \end{split}$ (50)

由于 $\langle \psi ,\psi \rangle = \int {\bar \psi } \psi dx$ 是实的,所以,取方程(50)的虚部为

$\begin{split} &\quad 2\langle {{\psi }_{t}},\psi \rangle =\frac{{\rm d}}{{\rm d}t}\|\psi \|_{{{L}^{2}}}^{2}= 2{\rm {Im}}\langle {(|u(x,s){|^2})_x}u,\psi \rangle +\\ & \qquad 2{\rm {Im}}\langle \gamma {u^2}\bar \psi ,\psi \rangle + 2{\rm {Im}}\langle 2\delta |u{|^2}u\bar \psi ,\psi \rangle \leqslant\\ &\qquad C\|u\|_{L_{T}^{\infty }({{H}^{1}})}^{3}\|\psi {{\|}_{{{L}^{2}}}}\!+\!C\|u\|_{L_{T}^{\infty }({{L}^{2}})}^{2}\|\psi \|_{{{L}^{2}}}^{2}\!+\!C\|u\|_{L_{T}^{\infty }({{L}^{2}})}^{4}\|\psi \|_{{{L}^{2}}}^{2}\quad \end{split} $

由Gronwall不等式可得

$\|\psi {{\|}_{L_{T}^{\infty }({{L}^{2}})}}\leqslant {{M}_{2}}$

另外,由式(4)可得式(48)。

现采用反证法。假设 $[0,{T_1}]$ 是初值问题(4)的唯一解 $u$ 的最大时间间隔,即对任意选定的 $T$ ,需满足 $T < {T_1}$ ,才能使得初值问题(4)在 $X_3^T$ 中有唯一的解。对于 $p > 0$ ,记

$\begin{split}&\quad\left\| \left| u \right| \right\|_{p}^{T}=\underset{-\infty <x<\infty }{\mathop{\sup\limits}}\,{{\left( \int_{0}^{T}{\sum\limits_{k=1}^{p}{|}}{\rm D}_{x}^{k}u(t,x){{|}^{2}}{\rm d}t \right)}^{1/2}}+\qquad\qquad\qquad\qquad\\ &\qquad\underset{0\leqslant t\leqslant T}{\mathop{\sup\limits}}\,\|{\rm D}_{x}^{1/2}{\rm D}_{x}^{p-1}u(t){{\|}_{{{L}^{2}}}}\end{split}$

参照对方程(11)等式右边各项的估计,可以对 $\left\| {\left| u \right|} \right\|_p^T(p \leqslant 3)$ 作出估计,这里只估计 $\left\| {\left| u \right|} \right\|_3^T$ 。用与式(19),(23),(24),(33),(34)相同的估计方法,可得

$\begin{split} &\quad\left\| \left| u \right| \right\|_{3}^{T}=\underset{-\infty <x<\infty }{\mathop{\sup\limits}}\,\left( \int_{0}^{T}{|}{{\rm D}_{x}}u(x,t){{|}^{2}}+|{\rm D}_{x}^{2}u(t,x){{|}^{2}}+\right. \\ &\qquad\left.|{\rm D}_{x}^{3}u(t,x){{|}^{2}}{\rm d}t \right)^{1/2}+\underset{0\leqslant t\leqslant T}{\mathop{\sup\limits}}\,\|{\rm D}_{x}^{1/2}{\rm D}_{x}^{2}u(t){{\|}_{{{L}^{2}}}} \leqslant \\ &\qquad\mathop {\sup\limits }\limits_{ - \infty \!<\! x \!<\! \infty } \!\!{\left( {\int_0^T \!\!\!| {{\rm D}_x}u(t,x){|^2}{\rm d}t} \right)^{1/2}}\!\!\!\!\!+\!\!\!\!\! \mathop {\sup\limits }\limits_{ - \infty \!<\! x \!<\! \infty } {\left(\! {\int_0^T \!\!\!| {\rm D}_x^2u(t,x){|^2}{\rm d}t}\! \right)^{1/2}}\!\!\!\!+\\ &\qquad\underset{-\infty <x<\infty }{\mathop{\sup\limits }}\,{{\left( \int_{0}^{T}{|}{\rm D}_{x}^{3}u(t,x){{|}^{2}}{\rm d}t \right)}^{1/2}}\!\!+\!\!\underset{0\leqslant t\leqslant T}{\mathop{\sup\limits }}\,\|{\rm D}_{x}^{1/2}{{\rm D}_{x}}u(t){{\|}_{{{L}^{2}}}}\leqslant\\ &\qquad C\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}\!\!+\!\!CT\|u{{\|}_{L_{T}^{\infty }({{H}^{1}})}}\||u|\|_{3}^{T}\!\!+\!\!C(T)\|{{v}_{0}}{{\|}_{{{H}^{2}}}}\|u{{\|}_{L_{T}^{\infty }({{H}^{2}})}}\!\!+\\ &\qquad C(T)\|u\|_{L_{T}^{\infty }({{H}^{2}})}^{3}+C(T)\|u\|_{L_{T}^{\infty }({{H}^{2}})}^{5} \leqslant\\ &\qquad {{C}_{10}}\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}+{{C}_{11}}TM_{1}^{2}\||u|\|_{3}^{T}+\\ &\qquad{{C}_{12}}(T)\|{{v}_{0}}{{\|}_{{{H}^{2}}}}{{M}_{2}}+{{C}_{13}}(T)M_{2}^{3}+{{C}_{14}}(T)M_{2}^{5} \\[-10pt] \end{split}$ (51)

由式(51)可得

$\begin{split}&\quad(1-{{C}_{11}}TM_{1}^{2})\||u|\|_{3}^{T}\leqslant {{C}_{10}}\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}+\\ &\qquad{{C}_{12}}(T)\|{{v}_{0}}{{\|}_{{{H}^{2}}}}{{M}_{2}}+{{C}_{13}}(T)M_{2}^{3}+{{C}_{14}}(T)M_{2}^{5}\qquad\qquad\qquad\qquad\qquad\qquad\end{split}$

$\hat T > 0$ ,使得

$\hat T = {\rm {min}}\;\left\{ {\frac{1}{{2{C_{11}}M_1^2}},{T_3}} \right\}$

可得

$\left\| {\left| u \right|} \right\|_3^{\hat T} \leqslant {K_1}$ (52)

其中, ${K_1}$ ${C_i}(T)$ ( $T \to 0$ 时, ${C_i}(T) \to 0$ $i = 11,12,13,$ $14$ )是大于零的常数,且仅依赖于 $\|{{u}_{0}}{{\|}_{{{H}^{5/2}}}}$ $\|{{v}_{0}}{{\|}_{{{H}^{2}}}}$

假设 $\hat T < {T_1}$ ,考虑初值问题

$\begin{split}&\quad {\rm i}u_t^{(1)} + u_{xx}^{(1)} = \left[ {\int_0^T {{{\left( {{{\left| {{u^{(1)}}(x,s)} \right|}^2}} \right)}_x}} {\rm d}s} \right]{u^{(1)}} + {v_0}{u^{(1)}} +\qquad\qquad\qquad\qquad\qquad\qquad\\ &\qquad\gamma {\left| {{u^{(1)}}} \right|^2}{u^{(1)}} + \delta {\left| {{u^{(1)}}} \right|^4}{u^{(1)}}\end{split}$

其中,初值条件 ${u^{(1)}}(\hat T,x) = u(\hat T,x),x \in {\rm R}$

同样地,可估计

$\left\| {\left| u \right|} \right\|_3^{2\hat T} \leqslant K_1^{(1)}$ (53)

其中, $K_1^{(1)}$ 是仅依赖于 ${K_1}$ $\|{{v}_{0}}{{\|}_{{{H}^{2}}}}$ 的正常数。

$\quad\begin{split}u(t,x)& =\left\{ {\begin{array}{*{20}{l}} {u(t,x), 0 \leqslant t \leqslant \hat T} \\ {{u^{(1)}}(t,x), \hat T \leqslant t \leqslant 2\hat T} \end{array}} \right.\\ &\left\| {\left| u \right|} \right\|_3^{2\hat T} \leqslant {\rm {max}}\;\{ {K_1},K_1^{(1)}\}\end{split}$

$n \in {N^ + },n\hat T \geqslant {T_1}$ ,采用与估计式(52)和式(53)同样的方法,将区间 $\hat T$ 逐步延拓到 $n\hat T$ ,存在正常数 $K$ ,使得

$\left\| {\left| u \right|} \right\|_3^{n\hat T} \leqslant K$

这样就引出了矛盾,故全局解的存在性得到证明。

本文运用压缩映射原理及先验估计,并通过证明所研究的方程的解满足能量守恒定律,克服了方程(4)中具高次非线性项带来的困难,得到了五次的长短波共振方程全局解存在的结果。另外,若将方程式(4)推广成下列形式:

$\left\{ {\begin{array}{*{20}{l}} {{\rm i}{u_t} + {u_{xx}} = uv + \gamma |u{|^2}u + \delta |u{|^{2m}}u} \\ {{v_t} = {{(|u{|^2})}_x}} \end{array}} \right.$ (54)

通过对引理9的证明过程可知,在式(54)中,当 $m > 2$ 时,不能估计出 $\|u{{\|}_{L_{T}^{\infty }({{H}^{1}})}}$ 被常数控制,故用本文方法难以证明方程(54)初值问题全局解的存在性。

致谢:在本文的研究过程中,得到了汪文军老师很大的帮助,在此表示衷心的感谢!

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