Linearization
=============

In balance:

 :math:`W(u,\delta u)=0` [:ref:`7 <7>`]

If, within one tolerance, the previous equation is not satisfied, we look for a :math:`\Delta u` correction of :math:`u` such that:

:math:`W(u,\delta u)+\text{DW}(u,\delta u)\text{.}\Delta u=0`

:math:`\text{DW}(u,\delta u)\text{.}\Delta u` being the directional derivative of :math:`\text{W}(u,\delta u)` in the :math:`\Delta u` [bib] and [bib] direction. According to [eq], we obviously have:

:math:`\text{DN}\text{.}\Delta u={E}_{a}A\mathrm{Dg}\text{.}\Delta u={E}_{a}AB\text{.}\Delta u`

According to [eq]:

:math:`DB\text{.}\Delta u={(\left[\frac{d}{{\text{ds}}_{o}}1\right]\Delta u)}^{T}\left[\frac{d}{{\text{ds}}_{o}}1\right]`.

So:

 :math:`\begin{array}{c}{\mathrm{DW}}_{\text{int}}(u,\delta u)\text{.}\Delta u={\int }_{{s}_{1}}^{{s}_{2}}\left\{{(B\delta u)}^{T}{E}_{a}AB\Delta u\right\}\text{ds}\\ +{\int }_{{s}_{1}}^{{s}_{2}}\left\{{(\left[\frac{d}{{\text{ds}}_{o}}1\right]\delta u)}^{T}N\left[\frac{d}{{\text{ds}}_{o}}1\right]\Delta u\right\}\text{ds}\end{array}` [:ref:`8 <8>`]

According to [eq]:

 :math:`{\mathrm{DW}}_{\text{iner}}(\ddot{u},\delta u)\text{.}\Delta u=-{\int }_{{s}_{1}}^{{s}_{2}}(\delta {u}^{T}\rho A\Delta \ddot{u})\text{ds}` [:ref:`9 <9>`]