引言

Raman散射是基于光与物质作用后产生的对称分布在Rayleigh散射光两侧的非弹性光散射效应，可通过散射光与入射光相比频率的位移分析被散射分子的组成和结构。Raman光谱技术广泛应用于食品质量^[1]、生物医药^[2]、刑事侦查^[3]和环境保护^[4]等领域，因此对Raman散射的研究有着重要的实际意义。

Lie等^[5]提出受激Raman散射中的分子振动模型可以用阻尼和外部正弦场驱动的Morse振子的经典方程来描述

$ y^{\prime \prime}(t)+\alpha y^{\prime}(t)+\left(1-\mathrm{e}^{-y}\right) \mathrm{e}^{-y}=A \cos \omega t $

式中，t为时间变量，y代表振子的振幅, 也是时间t的函数，α为阻尼系数，A为调制强度，ω为驱动频率。他们利用四阶Runge-Kutta法得到数值解，当阻尼系数α在0.001~0.4之间(此时阻尼极限较弱)且驱动频率ω较大时，以ω为横坐标，振子的最大振幅y_max为纵坐标建立坐标系作图，发现图像中间部分会出现稳定的上下两个分支，即双稳态现象，解到底收敛于哪个分支取决于初始条件和驱动场的相位。如果驱动频率从小到大开始变化，随着驱动频率的增加，y_max的增长非常缓慢且一直收敛于下分支，直到某一个临界点，y_max会突然增大并跳到上分支，接下来又会迅速减小；如果驱动频率从大到小变化，y_max将从上分支开始迅速增大，直到某一个临界值，突然下降到下分支，体现了解的滞后现象。由于Stokes波强度的急剧增长与分子振幅的突然增加有关，一般的谐振子模型无法解释Stokes波强度的急剧增长现象，而Morse振子模型却可以很好地解释分子振幅的突然增加，因此该模型可以用来描述受激Raman散射中的分子振动，进而解释Stokes波强度的急剧增长现象。

本文旨在研究受激Raman散射分子振动数学模型在小阻尼且弱驱动下的初值问题

$ \left\{\begin{array}{l} y^{\prime \prime}+\varepsilon \alpha y^{\prime}+\left(1-\mathrm{e}^{-y}\right) \mathrm{e}^{-y}=\varepsilon^{2} \cos (t+\varepsilon \omega t) \\ y(0)=0, y^{\prime}(0)=0 \end{array}\right. $

(1)

其中ε是扰动系数，通常0 < ε≪1，α、ω为正常数。首先我们利用不动点定理证明初值问题(1)的解是存在唯一的，进而利用摄动方法求出渐近解的首项并给出余项估计，从而证明渐近解的一致有效性。

1 解的存在唯一性

为证明解的存在唯一性，作变量代换，令

$ y(t)=\varepsilon v(t) $

代入式(1)得

$ \left\{\begin{array}{l} v^{\prime \prime}+\varepsilon \alpha v^{\prime}+v=-\frac{1}{\varepsilon}\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v\right]+ \\ \ \ \ \ \ \ \ \ \varepsilon \cos (t+\varepsilon \omega t) \\ v(0)=0, v^{\prime}(0)=0 \end{array}\right. $

(2)

记

$ \|v\|_{C\left[0, \frac{T}{\varepsilon}\right]}=\max \limits_{\left[0, \frac{T}{\varepsilon}\right]}|v(t)| $

定理1  在0 < ε≤ε₀≪1，0 < εt≤T时，初值问题(2)存在唯一的解，且$\|v\|_{{C}\left[0, \frac{T}{\varepsilon}\right]} \leqslant M $。其中ε₀为足够小的正常数，并有

$ T=\frac{1}{32 M_{0}\left(\varepsilon_{0}\right)}, M=2 M_{0}\left(\varepsilon_{0}\right) $

$ \begin{aligned} &\ \ \ \ \ \ \ \ M_{0}\left(\varepsilon_{0}\right)=K_{0}\left[(\alpha+2 \omega)\left(2+\varepsilon_{0} \omega\right)+\alpha \cdot\right. \\ &\left.\left(1+\varepsilon_{0} \omega\right)^{2}\right] \end{aligned} $

$ K_{0}=\frac{1}{\alpha^{2}+4 \omega^{2}} $

故T、M、M₀(ε₀)、K₀也为正常数。

接下来将证明定理1，将式(2)写成积分方程的形式

$ v=F(v)=F_{1}(v)+F_{2}(t) $

(3)

其中，

$ F_{1}(v)=\int_{0}^{t}-\frac{1}{\varepsilon} \phi(t, \tau)\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v\right] \mathrm{d} \tau $

(4)

$ F_{2}(t)=\int_{0}^{t} \phi(t, \tau) \varepsilon \cos (\tau+\varepsilon \omega \tau) \mathrm{d} \tau $

(5)

记

$ \phi(t, \tau)=\frac{1}{K_{\varepsilon}} \mathrm{e}^{-\frac{\varepsilon \alpha(t-\tau)}{2}} \sin \left[K_{\varepsilon}(t-\tau)\right] $

$ K_{\varepsilon}=\sqrt{1-\frac{\varepsilon^{2} \alpha^{2}}{4}} $

取$\varepsilon_{1}=\frac{\sqrt{3}}{\alpha} $，当0 < ε≤ε₁时，有

$ \begin{aligned} &\frac{1}{2} \leqslant K_{\varepsilon} \leqslant 1 \\ &|\phi(t, \tau)|=\left|\frac{1}{K_{\varepsilon}} \mathrm{e}^{-\frac{\alpha \varepsilon(t-\tau)}{2}} \sin \left[K_{\varepsilon}(t-\tau)\right]\right| \leqslant 2 \end{aligned} $

(6)

由于

$ \lim \limits_{\varepsilon v \rightarrow 0}\left|\frac{1}{(\varepsilon v)^{2}}\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v\right]\right|=\frac{3}{2} $

故∃ε₂ > 0，当0 < ε≤ε₂≪1时

$ \left|\frac{1}{(\varepsilon v)^{2}}\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v\right]\right| \leqslant 2 $

即

$ \begin{aligned} &\ \ \ \ \ \ \ \ \left|\frac{1}{\varepsilon v}\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v\right]\right| \leqslant \left| \frac{1}{(\varepsilon v)^{2}}[(1-\right. \\ &\left.\left.\left.\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v\right]\right|| \varepsilon v|\leqslant 2| \varepsilon v | \end{aligned} $

(7)

由于

$ \lim \limits_{\varepsilon \xi \rightarrow 0}\left|\left(2 \mathrm{e}^{-2 \varepsilon \xi}-\mathrm{e}^{-\varepsilon \xi}-1\right) /(\varepsilon \xi)\right|=3 $

故∃ε₃ > 0，当0 < ε≤ε₃≪1时

$ \left|\left(2 \mathrm{e}^{-2 \varepsilon \xi}-\mathrm{e}^{-\varepsilon \xi}-1\right) /(\varepsilon \xi)\right| \leqslant 6 $

即

$ \begin{aligned} &\ \ \ \ \ \ \ \ \left|2 \mathrm{e}^{-2 \varepsilon \xi}-\mathrm{e}^{-\varepsilon \xi}-1\right| \leqslant \mid\left(2 \mathrm{e}^{-2 \varepsilon \xi}-\mathrm{e}^{-\varepsilon \xi}-1\right) /(\varepsilon \xi) \mid\cdot \\ &|\varepsilon \xi| \leqslant 6 \varepsilon|\xi| \end{aligned} $

(8)

由于

$ \begin{aligned} &\ \ \ \ \ \ \ \ \lim \limits_{\varepsilon v \rightarrow 0}\left|-\frac{1}{(\varepsilon v)^{3}}\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v+\frac{3}{2} \varepsilon^{2} v^{2}\right]\right|= \\ &\frac{7}{6} \end{aligned} $

故∃ε₄ > 0，当0 < ε≤ε₄≪1时

$ \begin{aligned} &\ \ \ \ \ \ \ \ \left|-\frac{1}{\varepsilon}\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v+\frac{3}{2} \varepsilon^{2} v^{2}\right]\right| \leqslant 2 \varepsilon^{2} \cdot \\ &\left|v^{3}\right| \end{aligned} $

(9)

下面证明F(v)=F₁(v)+F₂(t)是定义在连续函数空间上的压缩映像，从而用不动点定理证明定理1。

取ε₀=min {ε₁, ε₂, ε₃, ε₄}，由于

$ \begin{aligned} &\ \ \ \ \ \ \ \ |F(v)| \leqslant \int_{0}^{t}|\phi(t, \tau)| \left| \frac{1}{\varepsilon v}\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\right.\right. \\ &\left.\varepsilon v]\right|v| \mathrm{d} \tau \end{aligned} $

(10)

由式(6)、(7)可知，当|v|≤M，0 < εt≤T时

$ \left|F_{1}(v)\right| \leqslant 2 \times 2|\varepsilon v| \times|v| \times t \leqslant 4 M^{2} T $

(11)

通过计算F₂(t)可知，当0 < ε≤ε₀≪1时

$ \begin{aligned} &\ \ \ \ \ \ \ \ F_{2}(t)=\frac{1}{\alpha^{2}(1+\varepsilon \omega)^{2}+\omega^{2}(2+\varepsilon \omega)^{2}}\{\alpha(1+ \\ &\varepsilon \omega) \sin (t+\varepsilon \omega t)-\omega(2+\varepsilon \omega) \cos (t+\varepsilon \omega t)-\mathrm{e}^{-\frac{\alpha \varepsilon t}{2}} \frac{\alpha}{2 K_{\varepsilon}}\cdot \\ &{\left[1+(1+\varepsilon \omega)^{2}\right] \sin \left(K_{\varepsilon} t\right)+\mathrm{e}^{-\frac{\alpha \varepsilon t}{2}} \omega(2+\varepsilon \omega) \cdot} \\ &\left.\cos \left(K_{\varepsilon} t\right)\right\} \leqslant K_{0}\left[(\alpha+2 \omega)(2+\varepsilon \omega)+\alpha(1+\varepsilon \omega)^{2}\right] \leqslant \\ &K_{0}\left[(\alpha+2 \omega)\left(2+\varepsilon_{0} \omega\right)+\alpha\left(1+\varepsilon_{0} \omega\right)^{2}\right] \end{aligned} $

即

$ F_{2}(t) \leqslant M_{0}\left(\varepsilon_{0}\right) $

(12)

将式(11)、(12)代入式(3)得

$ |F(v)| \leqslant\left|F_{1}(v)\right|+\left|F_{2}(t)\right| \leqslant M_{0}\left(\varepsilon_{0}\right)+4 M^{2} T $

令

$ M_{0}\left(\varepsilon_{0}\right)+4 M^{2} T \leqslant M $

(13)

当式(13)成立时，满足$\|v \quad\|_{{C}\left[0, \frac{T}{\varepsilon}\right]} \leqslant M $, $\|F(v)\|_{C\left[0, \frac{T}{\varepsilon}\right]} \leqslant M $，则F：v→F(v)是在连续空间$ C\left[0, \frac{T}{\varepsilon}\right]$上的映射。

接下来证明F：v→F(v)为压缩映射，由式(3)可得

$ \begin{aligned} &\ \ \ \ \ \ \ \ \left|F\left(v_{1}\right)-F\left(v_{2}\right)\right|=\left| \int_{0}^{t} \phi(t, \tau)\left[g\left(v_{2}\right)-\right.\right. \\ &\left.\left. g\left(v_{1}\right)\right] \mathrm{d} \tau\right|\leqslant \int_{0}^{t} 2\left| g\left(v_{2}\right)-g\left(v_{1}\right) \right|\mathrm{d} \tau \end{aligned} $

(14)

其中，

$ g(v)=\frac{1}{\varepsilon}\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v\right] $

由于v₁∈v, v₂∈v，故|v₁|≤M, |v₂|≤M，则有

$ g^{\prime}(v)=2 \mathrm{e}^{-2 \varepsilon v}-\mathrm{e}^{-\varepsilon v}-1 $

根据不等式(8)和拉格朗日中值定理∃ξ∈(v₁, v₂)，|ξ|≤M，有

$ \begin{aligned} &\ \ \ \ \ \ \ \ \lg \left(v_{2}\right)-g\left(v_{1}\right)|=| g^{\prime}(\xi)\left(v_{2}-v_{1}\right)|\leqslant| 2 \mathrm{e}^{-2 \varepsilon \xi}- \\ &\mathrm{e}^{-\varepsilon \xi}-1|| v_{2}-v_{1}|\leqslant 6 \varepsilon M| v_{2}-v_{1} \mid \end{aligned} $

(15)

将式(15)和0 < εt≤T代入式(14)得

$ \begin{aligned} &\ \ \ \ \ \ \ \ \left|F\left(v_{1}\right)-F\left(v_{2}\right)\right| \leqslant 12 \varepsilon M\left|v_{1}-v_{2}\right| t \leqslant 12 M T \mid v_{1}- \\ &v_{2} \mid \end{aligned} $

令

$ 12 M T<1 $

(16)

取$ T=\frac{1}{32 M_{0}\left(\varepsilon_{0}\right)}, M=2 M_{0}\left(\varepsilon_{0}\right)$即可同时满足式(13)和式(16)，此时F：v→F(v)为连续空间${C}\left[0, \frac{T}{\varepsilon}\right] $上的压缩映射，根据Banach不动点定理，v=F(v)存在唯一的解v(t)，即初值问题(2)存在唯一的解，故定理1得证。

2 多重尺度法求近似解

根据文献[6]中对多重尺度法的介绍，可取两时间尺度分别为t₁=t，t₂=εt，代入式(1)得

$ \left\{\begin{array}{l} \left(\partial_{t 1}^{2}+2 \varepsilon \partial_{t_{1}} \partial_{t_{2}}+\varepsilon^{2} \partial_{t 2}^{2}\right) y+\varepsilon \alpha\left(\partial_{t_{1}}+\varepsilon \partial_{t_{2}}\right) y+ \\ \ \ \ \ \ \ \ \ \left(1-\mathrm{e}^{-y}\right) \mathrm{e}^{-y}=\varepsilon^{2} \cos \left(t_{1}+\omega t_{2}\right) \\ y(0, 0)=0, \left.\left(\partial t_{1}+\varepsilon \partial t_{2}\right) y\right|_{(0, 0)}=0 \end{array}\right. $

(17)

令

$ y=\varepsilon y_{0}+\varepsilon^{2} y_{1}+\varepsilon^{3} y_{2}+\cdots $

(18)

故

$ \begin{array}{l} \ \ \ \ \ \ \ \ \left(1-\mathrm{e}^{-y}\right) \mathrm{e}^{-y}=\left(y-\frac{1}{2} y^{2}+\frac{1}{6} y^{3}-\cdots\right)(1-y+ \\ \left.\frac{1}{2} y^{2}-\frac{1}{6} y^{3}+\cdots\right)=\varepsilon y_{0}+\varepsilon^{2}\left(y_{1}-\frac{3}{2} y_{0}^{2}\right)+\cdots \end{array} $

(19)

将式(18)、(19)代入到式(17)，并令ε的一次幂系数相等得

$ \left\{\begin{array}{l} \frac{\partial^{2} y_{0}}{\partial t_{1}^{2}}+y_{0}=0 \\ y_{0}(0, 0)=0, \left.\frac{\partial y_{0}}{\partial t_{1}}\right|_{(0, 0)}=0 \end{array}\right. $

解得

$ \left\{\begin{array}{l} y_{0}=A\left(t_{2}\right) \cos t_{1}+B\left(t_{2}\right) \sin t_{1} \\ A(0)=0, B(0)=0 \end{array}\right. $

(20)

令ε的二次幂系数相等得

$ \left\{\begin{array}{l} \frac{\partial^{2} y_{1}}{\partial t_{1}^{2}}+y_{1}=\frac{3}{2} y_{0}^{2}-2 \frac{\partial^{2} y_{0}}{\partial t_{1} \partial t_{2}}-\alpha \frac{\partial y_{0}}{\partial t_{1}}+\cos \left(t_{1}+\omega t_{2}\right) \\ y_{1}(0, 0)=0, \left.\frac{\partial y_{1}}{\partial t_{1}}\right|_{(0, 0)}=-\left.\frac{\partial y_{0}}{\partial t_{2}}\right|_{(0, 0)} \end{array}\right. $

解得

$ \begin{aligned} &\ \ \ \ \ \ \ \ y_{1}=-\frac{1}{2} t_{1} \cos t_{1}\left[2 A^{\prime}\left(t_{2}\right)+\alpha A\left(t_{2}\right)-\right. \\ &\left.\sin \left(\omega t_{2}\right)\right]-\frac{1}{2} t_{1} \sin t_{1}\left[2 B^{\prime}\left(t_{2}\right)+\alpha B\left(t_{2}\right)-\right. \\ &\left.\cos \left(\omega t_{2}\right)\right]+F_{1}\left(t_{2}\right) \cos t_{1}+F_{2}\left(t_{2}\right) \sin t_{1}+ \\ &\frac{1}{2} \cos t_{1} \cos \left(\omega t_{2}\right)+\frac{1}{4}\left[-A^{2}\left(t_{2}\right)+B^{2}\left(t_{2}\right)\right] \cdot \\ &\cos \left(2 t_{1}\right)-\frac{1}{2} A\left(t_{2}\right) B\left(t_{2}\right) \sin \left(2 t_{1}\right)-B^{\prime}\left(t_{2}\right) \cos t_{1}- \\ &\frac{1}{2} \alpha B\left(t_{2}\right) \cos t_{1}+\frac{3}{4}\left[A^{2}\left(t_{2}\right)+B^{2}\left(t_{2}\right)\right] \end{aligned} $

为消除长期项，需满足

$ \left\{\begin{array}{l} 2 A^{\prime}\left(t_{2}\right)+\alpha A\left(t_{2}\right)-\sin \left(\omega t_{2}\right)=0 \\ 2 B^{\prime}\left(t_{2}\right)+\alpha B\left(t_{2}\right)-\cos \left(\omega t_{2}\right)=0 \end{array}\right. $

解得

$ \left\{\begin{array}{l} A\left(t_{2}\right)=K_{0}\left[2 \mathrm{e}^{-\frac{\alpha t_{2}}{2}} \omega-2 \omega \cos \left(\omega t_{2}\right)+\right. \\ \ \ \ \ \ \ \ \ \left.\alpha \sin \left(\omega t_{2}\right)\right] \\ B\left(t_{2}\right)=K_{0}\left[-\mathrm{e}^{-\frac{\alpha t_{2}}{2}} \alpha+\alpha \cos \left(\omega t_{2}\right)+\right. \\ \ \ \ \ \ \ \ \ \left.2 \omega \sin \left(\omega t_{2}\right)\right] \end{array}\right. $

(21)

将式(21)以及t₁=t，t₂=εt代入到式(20)得

$ \begin{array}{l} \ \ \ \ \ \ \ \ y_{0}(t)=\sqrt{K_{0}}\left[\mathrm{e}^{-\frac{\alpha \varepsilon t}{2}} \cos (t+\varphi)-\cos (t+\varepsilon \omega t+\right.\\ \varphi) ] \end{array} $

(22)

记$ \cos \varphi=\frac{2 \omega}{\sqrt{\alpha^{2}+4 \omega^{2}}}, \sin \varphi=\frac{\alpha}{\sqrt{\alpha^{2}+4 \omega^{2}}}$，因此初值问题(1)渐近解的首项为

$ \begin{array}{l} \ \ \ \ \ \ \ \ \varepsilon y_{0}(t)=\varepsilon \sqrt{K_{0}}\left[\mathrm{e}^{-\frac{\alpha \varepsilon t}{2}} \cos (t+\varphi)-\cos (t+\varepsilon \omega t+\right.\\ \varphi) ] \end{array} $

3 余项估计

定理2  在0 < ε≤ε₀≪1，0 < εt≤T时，有|y(t)-εy₀(t)|=ε|v(t)-y₀(t)|≤ε²M₃(ε₀, T)，其中M₃(ε₀, T)是与ε₀和T有关的正常数。

令v(t)=y₀(t)+R(t)，代入式(2)得

$ \left\{\begin{array}{l} R^{\prime \prime}+\varepsilon \alpha R^{\prime}+R=g_{1}(t)+\frac{3}{2} \varepsilon R^{2}+3 \varepsilon y_{0} R+g_{2}(t) \\ R(0)=0, R^{\prime}(0)=0 \end{array}\right. $

(23)

其中

$ g_{1}(t)=-\frac{1}{\varepsilon}\left[\left(1-\mathrm{e}^{-\varepsilon v}\right) \mathrm{e}^{-\varepsilon v}-\varepsilon v+\frac{3}{2} \varepsilon^{2} v^{2}\right] $

$ g_{2}(t)=\frac{3}{2} \varepsilon y_{0}^{2}-y_{0}^{\prime \prime}-\varepsilon \alpha y_{0}^{\prime}-y_{0}+\varepsilon \cos (t+\varepsilon \omega t) $

将初值问题(23)写成积分方程的形式

$ \begin{array}{l} \ \ \ \ \ \ \ \ R=\int_{0}^{t} \phi(t, \tau)\left[g_{1}(\tau)+\frac{3}{2} \varepsilon R^{2}+3 \varepsilon y_{0} R+\right. \\ \left.g_{2}(\tau)\right] \mathrm{d} \tau \end{array} $

由式(6)知|ϕ(t, τ)|≤2，则

$ |R| \leqslant s_{1}(t)+s_{2}(t)+s_{3}(t) $

(24)

其中，$s_{1}(t)=2 \int_{0}^{t}\left|g_{1}(\tau)\right| \mathrm{d} \tau $, $s_{2}(t)=\int_{0}^{t}\left| 3 \varepsilon R^{2}+ 6 \varepsilon y_{0} R\right|\mathrm{~d} \tau $, $ s_{3}(t)=\left| \int_{0}^{t} \phi(t, \tau) g_{2}(\tau) \mathrm{d} \tau\right|$。

根据不等式(9)

$ s_{1}(t) \leqslant 2 \int_{0}^{t} 2 \varepsilon^{2}\left|v^{3}\right| \mathrm{d} \tau \leqslant 4 \varepsilon^{2} M^{3} t \leqslant 4 \varepsilon M^{3} T $

(25)

由式(22)可知，$ \left|y_{0}(t)\right| \leqslant 2 \sqrt{K_{0}}$，则

$ s_{2}(t) \leqslant \varepsilon A \int_{0}^{t}\left(\left|R^{2}\right|+|R|\right) \mathrm{d} \tau $

(26)

记$ A=\max \left\{12 \sqrt{K_{0}}, 3\right\}$，故A为正常数。

利用Maple软件计算s₃(t)可知

$ \begin{array}{l} \qquad s_{3}(t)=\left|\int_{0}^{t} \phi(t, \tau) g_{2}(\tau) \mathrm{d} \tau\right|=\mid f_{1}(t)+f_{2}(t)+ \\ f_{3}(t)+f_{4}(t)+f_{5}(t)+f_{6}(t) \mid \end{array} $

其中

$ \begin{array}{l} \qquad f_{1}(t)=\int_{0}^{t} \varepsilon^{2} K_{0} \phi(t, \tau)\left(\frac{1}{2} \alpha^{2} \omega \mathrm{e}^{-\frac{\varepsilon \alpha \tau}{2}} \cos \tau-\right. \\ \left.\frac{1}{4} \alpha^{3} \mathrm{e}^{-\frac{\varepsilon \alpha \tau}{2}} \sin \tau\right) \mathrm{d} \tau \end{array} $

$ \begin{aligned} &\qquad f_{2}(t)=\int_{0}^{t} \varepsilon^{2} K_{0} \phi(t, \tau)\left[-\left(\alpha^{2}+2 \omega^{2}\right) \omega \cos (\tau+\right. \\ &\left.\varepsilon \omega \tau)-\alpha \omega^{2} \sin (\tau+\varepsilon \omega \tau)\right] \mathrm{d} \tau \end{aligned} $

$ \begin{aligned} &\qquad f_{3}(t)=\int_{0}^{t} \varepsilon K_{0}^{2} \phi(t, \tau)\left[-3 \alpha \omega \mathrm{e}^{-\varepsilon \alpha \tau} \sin (2 \tau)+\right. \\ &\left.\frac{3}{4}\left(4 \omega^{2}-\alpha^{2}\right) \mathrm{e}^{-\varepsilon \alpha \tau} \cos (2 \tau)\right] \mathrm{d} \tau \end{aligned} $

$ \begin{aligned} &\qquad f_{4}(t)=\int_{0}^{t} \varepsilon K_{0}^{2} \phi(t, \tau)\left[6 \alpha \omega \mathrm{e}^{-\frac{\varepsilon \alpha \pi}{2}} \sin (2 \tau+\varepsilon \omega \tau)+\right. \\ &\left.\frac{3}{2}\left(\alpha^{2}-4 \omega^{2}\right) \mathrm{e}^{-\frac{\varepsilon \alpha \tau}{2}} \cos (2 \tau+\varepsilon \omega \tau)\right] \mathrm{d} \tau \end{aligned} $

$ \begin{aligned} &\qquad f_{5}(t)=\int_{0}^{t} \varepsilon K_{0}^{2} \phi(t, \tau)[-3 \alpha \omega \sin (2 \tau+2 \varepsilon \omega \tau)+ \\ &\left.\frac{3}{4}\left(4 \omega^{2}-\alpha^{2}\right) \cos (2 \tau+2 \varepsilon \omega \tau)\right] \mathrm{d} \tau \end{aligned} $

$ \begin{aligned} &\qquad f_{6}(t)=\int_{0}^{t} \varepsilon K_{0} \phi(t, \tau)\left[-\frac{3}{2} \mathrm{e}^{-\frac{\varepsilon \alpha \tau}{2}} \cos (\varepsilon \omega \tau)+\right. \\ &\left.\frac{3}{4}\left(1+\mathrm{e}^{-\varepsilon \alpha \tau}\right)\right] \mathrm{d} \tau \end{aligned} $

通过Maple软件计算可知，存在常数ε和T，当0 < ε≤ε₀≪1，0 < εt≤T时，有

$ s_{3}(t) \leqslant \varepsilon M_{1}\left(\varepsilon_{0}, T\right) $

(27)

其中M₁(ε₀, T)是与ε₀和T有关的正常数。

将式(25)~(27)代入到式(24)得

$ \begin{aligned} &\qquad |R| \leqslant 4 \varepsilon M^{3} T+\varepsilon A \int_{0}^{t}\left(\left|R^{2}\right|+|R|\right) \mathrm{d} \tau+ \\ &\varepsilon M_{1}\left(\varepsilon_{0}, T\right) \leqslant \varepsilon\left[M_{2}\left(\varepsilon_{0}, T\right)+A \int_{0}^{t}\left(\left|R^{2}\right|+|R|\right) \mathrm{d} \tau\right] \end{aligned} $

(28)

其中M₂(ε₀, T)也是与ε₀和T有关的正常数, 且

$ \begin{aligned} &\qquad M_{2}\left(\varepsilon_{0}, T\right)=M_{1}\left(\varepsilon_{0}, T\right)+4 M^{3} T=M_{1}\left(\varepsilon_{0}, T\right)+ \\ &32 M_{0}^{3}\left(\varepsilon_{0}\right) T \end{aligned} $

由文献[7-8]可知，当0 < ε≤ε₀≪1，0 < εt≤T时，存在与ε₀和T有关的正常数M₃(ε₀, T)，使得

$ |R| \leqslant \varepsilon M_{3}\left(\varepsilon_{0}, T\right) $

根据第2节求得的渐近解首项为εy₀，故

$ \left|y-\varepsilon y_{0}\right| \leqslant \varepsilon|R| \leqslant \varepsilon^{2} M_{3}\left(\varepsilon_{0}, T\right) $

由此可知第2节求得的渐近解首项是一致有效的。

4 结束语

本文利用多重尺度法求解了受激Raman散射分子振动数学模型中初值问题的渐近解首项，并证明了初值问题的解的存在性和渐近解的一致有效性，但不足之处是只求解了渐近解的首项，且在选取时间尺度时仅选取了两个时间尺度，故求得的近似解不够精确。若要得到更高的精确度，可计算渐近解的后几项或者选取多重时间尺度，当然计算量也会随之增加。