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1 change: 1 addition & 0 deletions docs/source/Formula/index.rst
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- :doc:`RT`
- :doc:`RT_liquid`
- :doc:`uiz`

**静力学震源参数**

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73 changes: 73 additions & 0 deletions docs/source/Formula/uiz.rst
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求解位移对 :math:`z` 的偏导
==============================

:Author: Zhu Dengda
:Email: zhudengda@mail.iggcas.ac.cn

-----------------------------------------------------------

求解应力、应变、旋转张量时需要求解偏导 :math:`\dfrac{\partial \mathbf{u}}{\partial z}` ,其本质只需要修改最终表达式中的“接收函数矩阵”即可。
推导过程其实也是推导z平面牵引力表达式的一部分。为了讲清推导过程,这里把一些公式再写一遍。

在 :ref:`初稿 <yao_init_manuscripts>` 以及 :doc:`RT` 中已经提到,三分量位移可展开在柱面谐矢量坐标系下,

.. math::
:label:

\def \Bm{\mathbf{B}_m}
\def \Cm{\mathbf{C}_m}
\def \Pm{\mathbf{P}_m}

\mathbf{u} &= \sum_{m=0}^{\infty} \int_0^{\infty} \left( q_m \Bm + v_m \Cm + w_m \Pm \right) k dk \\

根据势函数 :math:`(\phi, \psi, \chi)` 与位移的关系,以及柱面谐矢量的表达式,可以得到位移 :math:`\mathbf{u}` 各分量的表达式,

.. math::
:label:

q_m &= k \phi_m + \dfrac{d \psi_m}{d z} \\
w_m &= \dfrac{d \phi_m}{d z} + k \psi_m \\
v_m &= k \chi_m \\

上式两边对 :math:`z` 求偏导,并根据垂直波函数的性质,同样引入 :math:`\epsilon` 这个符号变量(:math:`z` 正则正,:math:`z` 负则负)

.. math::
:label:

\def \parz#1{\dfrac{\partial #1}{d z}}

\parz {q_m} &= -\epsilon ak \phi_m + b^2 \psi_m \\
\parz {w_m} &= a^2 \phi_m - \epsilon bk \psi_m \\
\parz {v_m} &= -\epsilon bk \chi_m \\

从而得到位移对 :math:`z` 的偏导与垂直波函数之间的关系,

.. math::
:label:

\def \parz#1{\dfrac{\partial #1}{d z}}

\begin{bmatrix}
\parz {q_m} \\
\parz {w_m}
\end{bmatrix} &=
\begin{bmatrix}
ak & b^2 & -ak & b^2 \\
a^2 & bk & a^2 & -bk \\
\end{bmatrix}
\begin{bmatrix}
\phi_m^- \\
\psi_m^- \\
\phi_m^+ \\
\psi_m^+ \\
\end{bmatrix} \\
\parz {v_m} &=
\begin{bmatrix}
bk & -bk \\
\end{bmatrix}
\begin{bmatrix}
\chi_m^- \\
\chi_m^+ \\
\end{bmatrix}

这样在计算位移对应的垂直波函数时,只需改用上式的接收函数矩阵,即可得到偏导 :math:`\dfrac{\partial \mathbf{u}}{\partial z}` 。