弹性波_Rayleigh波的推导

\begin{equation}
\mathrm{d}u_i=u_{i,j}\mathrm{d}x_j
\end{equation}

\begin{equation}
\mathrm{d}u_i=\dfrac{1}{2}\left[\left(u_{i,j}+u_{j,i}\right)+\left(u_{i,j}-u_{j,i}\right)\right]\mathrm{d}x_j
\end{equation}

\begin{equation}
e_{ij}=\dfrac{1}{2}\left(u_{i,j}+u_{j,i}\right)
\end{equation}

\begin{equation}
\omega_{ij}=\dfrac{1}{2}\left(u_{i,j}-u_{j,i}\right)
\end{equation}

\begin{equation}
t_i=\tau_{ij}n_j
\end{equation}
Conservation of mass
\begin{equation}
m=\int_{V}\rho \mathrm{d}V
\end{equation}

\begin{equation}
\dfrac{\mathrm{D}\rho}{\mathrm{D}t}=\dfrac{\partial\rho}{\partial t}+\dfrac{\partial\rho}{\partial x_j}\dfrac{\partial x_j}{\partial t}
\end{equation}

\begin{equation}
\dfrac{\partial\rho}{\partial t}+\left(\rho v_{i,i}\right)=0
\end{equation}
Conservation of momentum

\begin{equation}
\dfrac{\mathrm{D}}{\mathrm{D}t}\int_{V}v_i\rho\mathrm{d}V=
\int_{S}t_i\mathrm{d}S+\int_{V}f_i\rho\mathrm{d}V
\end{equation}

\begin{equation}
\int_{V}\left(\rho \ddot{u}_i-\tau_{ji,j}-\rho f_i\right)\mathrm{d}V=0
\end{equation}

\begin{equation}
\tau_{ji,j}+\rho f_i=\rho \ddot{u}_i
\end{equation}
Conservation of moment of momentum

\begin{equation}
\dfrac{\mathrm{D}}{\mathrm{D}t}\int_{V}e_{ijk}\ddot{u}_kx_j\rho\mathrm{d}V=\int_{S}e_{ijk}t_kx_j\mathrm{d}S+\int_{V}e_{ijk}f_kx_j\rho\mathrm{d}V
\end{equation}

\begin{equation}
\int_{S}e_{ijk}t_kx_j\mathrm{d}S=\int_{S}e_{ijk}x_j\left(\tau_{lk}n_l\right)\mathrm{d}S
\end{equation}

\begin{equation}
\nabla^2\vec{\psi}=\sum_{i=1}^{3}\dfrac{1}{h_i}\dfrac{\partial}{\partial u_i}\vec{e}_i
\end{equation}

\begin{equation}
\nabla \cdot \vec{A}=\dfrac{1}{h_1h_2h_3}\left[\dfrac{\partial}{\partial u_1}\left(A_1h_2h_3\right)+\dfrac{\partial}{\partial u_2}\left(h_1A_2h_3\right)+\dfrac{\partial}{\partial u_3}\left(h_1h_2hA_3\right)\right]
\end{equation}

\begin{equation}
\nabla \cdot \vec{A}=\dfrac{1}{r^2\sin\theta}\left[\dfrac{\partial}{\partial r}\left(r^2\sin \theta A_r\right)+\dfrac{\partial}{\partial \theta}\left(r\sin \theta A_{\theta}\right)+\dfrac{\partial}{\partial \varphi}\left(rA_{\varphi}\right)\right]
\end{equation}
\\
约定：应力张量为$T$，每个分量的下标分别表示作用面与作用方向。
\begin{equation}
T=\left(\begin{array}{ccc}
\sigma_{xx} & \sigma_{yx} & \sigma_{zx} \\
\sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\
\sigma_{xz} & \sigma_{yz} & \sigma_{zz} \\
\end{array}\right)
=\left(\begin{array}{ccc}
T_x \\ T_y \\ T_z
\end{array}\right)
\end{equation}
\\约定：$\sigma$表示正应力，$\tau$表示切应力。应力张量$T$为：
\begin{equation}
T=\left(\begin{array}{ccc}
\sigma_x  & \tau_{xy} & \tau_{xz} \\
\tau_{xy} & \sigma_y  & \tau_{yz} \\
\tau_{xz} & \tau_{yz} & \sigma_z
\end{array}\right)+
\left(\begin{array}{ccc}
0           & \varpi_{xy} & \varpi_{xz} \\
\varpi_{xy} & 0           & \varpi_{yz} \\
\varpi_{xz} & \varpi_{yz} & 0
\end{array}\right)
\end{equation}
\\
其中：
\begin{equation}
\begin{array}{ll}
\tau_{ij}=\dfrac{1}{2}\left(\sigma_{ij}+\sigma_{ji}\right)\\
\varpi_{ij}=\dfrac{1}{2}\left(\sigma_{ij}-\sigma_{ji}\right)
\end{array}
\end{equation}
\\
约定：
当试块在拉伸试验中，$x$方向单位伸长可以是：
\begin{equation}
\epsilon_x=\dfrac{\sigma_x}{E}
\end{equation}
同时试块也会在侧向发生形变：
\begin{equation}
\epsilon_y=-\nu\dfrac{\sigma_x}{E}
\end{equation}
\begin{equation}
\epsilon_z=-\nu\dfrac{\sigma_x}{E}
\end{equation}
可以写成：
\begin{equation}
\begin{array}{ccc}
\epsilon_x=\dfrac{1}{E}\left[\sigma_x-\nu\left(\sigma_y+\sigma_z\right)\right]\\
\epsilon_y=\dfrac{1}{E}\left[\sigma_y-\nu\left(\sigma_z+\sigma_x\right)\right]\\
\epsilon_z=\dfrac{1}{E}\left[\sigma_z-\nu\left(\sigma_x+\sigma_y\right)\right]
\end{array}
\end{equation}

瑞利波边界条件：
\begin{align}
\sigma_{zz}=&\left(\lambda+2\mu \right)\dfrac{\partial w}{\partial z} +\lambda\dfrac{\partial u}{\partial x}  \\
\sigma_{zx}=&\mu\left(\dfrac{\partial w}{\partial x} +\dfrac{\partial u}{\partial z}\right)
\end{align}
速度势与位移关系：
\begin{align}
u =& \dfrac{\partial \varphi}{\partial x}+\dfrac{\partial \psi}{\partial z} \\
w =& \dfrac{\partial \varphi}{\partial z}-\dfrac{\partial \psi}{\partial x}
\end{align}
用速度势表示边界条件：
\begin{align}
\sigma_{zz}=&\left(\lambda+2\mu \right)\dfrac{\partial^2\varphi}{\partial z^2} + \lambda\dfrac{\partial^2 \varphi}{\partial x^2} -2\mu\dfrac{\partial^2 \psi}{\partial z\partial x} \\
\sigma_{zx}=&\mu\left(\dfrac{\partial^2 \psi}{\partial z^2} -\dfrac{\partial^2 \psi}{\partial x^2} + 2\dfrac{\partial^2 \varphi}{\partial x \partial z}\right)
\end{align}
假设两个速度势的为：
\begin{align}
\varphi = A& \exp\left[-\sqrt{\left(\dfrac{\omega}{c_r}\right)^2-\left(\dfrac{\omega}{c_p}\right)^2}z+\mathrm{i}\left(\omega t -\dfrac{\omega}{c_r}x\right)\right] \\
\psi = A& \exp\left[-\sqrt{\left(\dfrac{\omega}{c_r}\right)^2-\left(\dfrac{\omega}{c_s}\right)^2}z+\mathrm{i}\left(\omega t -\dfrac{\omega}{c_r}x\right)\right]
\end{align}
带入假设解到边界条件：
\begin{align}\label{ray1}
A \left[ \left( \lambda+2\mu \right) \left( \left(\dfrac{\omega}{c_r}\right)^2-\left(\dfrac{\omega}{c_p}\right)^2\right) -\lambda\left(\dfrac{\omega}{c_r}\right)^2\right]
+& B\left[-2\mathrm{i}\mu\dfrac{\omega}{c_r} \sqrt{\left(\dfrac{\omega}{c_r}\right)^2-\left(\dfrac{\omega}{c_s}\right)^2}\right] = 0 \\
A\left[2\mathrm{i}\dfrac{\omega}{c_r}\sqrt{\left(\dfrac{\omega}{c_r}\right)^2-\left(\dfrac{\omega}{c_p}\right)^2}\right]
+& B\left[2\left(\dfrac{\omega}{c_r}\right)^2-\left(\dfrac{\omega}{c_s}\right)^2\right]=0
\end{align}
化简：
\begin{align}
A =& \dfrac{\mathrm{i}}{2} B \dfrac{2-x^2}{\sqrt{1-\sigma x^2}} \\
A =&  2\mathrm{i} B \dfrac{\sqrt{1-x^2}}{2-x^2}
\end{align}
其中：
\begin{equation}
\sigma = \left(\dfrac{c_s}{c_p}\right)^2 = \dfrac{\mu}{\lambda+2\mu} =\dfrac{2\nu-1}{2\nu-2}
\end{equation}
\begin{equation}
x = \dfrac{c_r}{c_s}
\end{equation}
由速度势表示位移：
\begin{align}
u_1 =& -\mathrm{i}A \dfrac{2\pi}{\Lambda_r}\left[ \exp\left(-2\pi\sqrt{1-\sigma x^2}\dfrac{z}{\Lambda_r} \right)
-\dfrac{2-x^2}{2} \exp\left(-2\pi\sqrt{1-x^2}\dfrac{z}{\Lambda_r} \right) \right] \\
w_1 =& -A\dfrac{2\pi}{\Lambda_r}\left[ \sqrt{1-\sigma x^2}\exp\left(-2\pi\sqrt{1-\sigma x^2}\dfrac{z}{\Lambda_r} \right)
-\dfrac{2-x^2}{2\sqrt{1-x^2}} \exp\left(-2\pi\sqrt{1-x^2}\dfrac{z}{\Lambda_r} \right) \right] \\
u_2 =& -A\dfrac{2\pi}{\Lambda_r}\left[ \exp\left(-2\pi\sqrt{1-\sigma x^2}\dfrac{z}{\Lambda_r} \right)
-\dfrac{2\sqrt{1-\sigma x^2}\sqrt{1-x^2}}{2-x^2} \exp\left(-2\pi\sqrt{1-x^2}\dfrac{z}{\Lambda_r} \right) \right] \\
w_2 =& -A\dfrac{2\pi}{\Lambda_r}\sqrt{1-\sigma x^2}\left[ \exp\left(-2\pi\sqrt{1-\sigma x^2}\dfrac{z}{\Lambda_r} \right)
-\dfrac{2}{2-x^2} \exp\left(-2\pi\sqrt{1-x^2}\dfrac{z}{\Lambda_r} \right) \right]
\end{align}
设$\alpha=\dfrac{2\pi z}{\Lambda_r}$。可以先求出表面$z=0$处的位移值：
\begin{align}
u_1\left(0\right) =& -\mathrm{i}Ak\dfrac{x^2}{2}\\
w_1\left(0\right) =& -Ak\dfrac{2\sqrt{1-\sigma x^2}\sqrt{1-x^2}-\left(2-x^2\right)}{2\sqrt{1-x^2}}\\
u_2\left(0\right) =& -Ak\dfrac{2-x^2-2\sqrt{1-\sigma x^2}\sqrt{1-x^2}}{2-x^2}\\
w_2\left(0\right) =& -Ak\sqrt{1-\sigma x^2}\dfrac{x^2}{x^2-2}
\end{align}
归一化：
\begin{align}
\dfrac{u_1\left(\alpha\right)}{u_1\left(0\right)} = &
\dfrac{
    2\exp\left(-\alpha\sqrt{1-\sigma x^2} \right)-
    \left(2-x^2\right)\exp\left(-\alpha\sqrt{1-x^2} \right)}
 {x^2}\\
 \dfrac{w_1\left(\alpha\right)}{w_1\left(0\right)} = &
\dfrac{
    2\sqrt{1-x^2}\sqrt{1-\sigma x^2}\exp\left(-\alpha\sqrt{1-\sigma x^2} \right)-
    \left(2-x^2\right)\exp\left(-\alpha\sqrt{1-x^2} \right)}
 {2\sqrt{1-\sigma x^2}\sqrt{1-x^2}-\left(2-x^2\right)}\\
 \dfrac{u_2\left(\alpha\right)}{u_2\left(0\right)} = &
\dfrac{
    2\sqrt{1-x^2}\sqrt{1-\sigma x^2}\exp\left(-\alpha\sqrt{1-x^2}\right)
    -\left(2-x^2\right)\exp\left(-\alpha\sqrt{1-\sigma x^2} \right) }
 {2\sqrt{1-\sigma x^2}\sqrt{1-x^2}-\left(2-x^2\right)}\\
 \dfrac{w_2\left(\alpha\right)}{w_2\left(0\right)} = &
\dfrac{
    2\exp\left(-\alpha\sqrt{1-x^2} \right)
    -\left(2-x^2\right)\exp\left(-\alpha\sqrt{1-\sigma x^2} \right)}
 {x^2}\\
\end{align}\par
    这里的解是半无限大固体中无衰减的瑞利波，波沿$x$方向传播，没有$y$方向的位移分量。该结果反映了波的幅度在$z$方向的衰减，省略了时间简谐项。
\end{document}

简化[\ref{ray1}]得到：
\begin{align}
\left[ 2 - \left( \dfrac{c_r}{c_s} \right)^2 \right]^2 - 4 \sqrt{1-\left( \dfrac{c_r}{c_p} \right)^2} \sqrt{1-\left( \dfrac{c_r}{c_s} \right)^2} =0
\end{align}
 设：
\begin{align}
x = \dfrac{c_p}{c_s}
\end{align}
有：
\begin{align}
\left( 2-y^2 \right)^2 -4 \sqrt{1-y^2}\sqrt{1-\left(xy\right)^2} = 0
\end{align}