$\newcommand{\Dev}{\operatorname{Dev}}$ $\newcommand{\dev}{\operatorname{dev}}$ $\newcommand{\grad}{\operatorname{grad}}$ $\newcommand{\Grad}{\operatorname{Grad}}$ $\newcommand{\divg}{\operatorname{div}}$ $\newcommand{\Ei}{\operatorname{Ei}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\Divg}{\operatorname{Div}}$ $\newcommand{\cay}{\operatorname{cay}}$ $\newcommand{\rot}{\operatorname{rot}}$

The BFGS method updates the stiffness matrix (or rather its inverse) to provide an approximation to the exact matrix. A displacement increment is defined as $$\begin{equation} \mathbf{d}_{k}=\mathbf{x}_{k}-\mathbf{x}_{k-1}\,,\label{eq357} \end{equation}$$ and an increment in the residual is defined as $$\begin{equation} \mathbf{G}_{k}=\mathbf{R}_{k-1}-\mathbf{R}_{k}\,.\label{eq358} \end{equation}$$ The updated matrix $\mathbf{K}_{k}$ should satisfy the quasi-Newton equation: $$\begin{equation} \mathbf{K}_{k}\mathbf{d}_{k}=\mathbf{G}_{k}\,.\label{eq359} \end{equation}$$ In order to calculate this update, as displacement increment is first calculated: $$\begin{equation} \mathbf{u}=\mathbf{K}_{k-1}^{-1}\mathbf{R}_{k-1}\,.\label{eq360} \end{equation}$$ This displacement vector defines a “direction” for the actual displacement increment. A line search (see next section) can now be applied to determine the optimal displacement increment: $$\begin{equation} \mathbf{x}_{k}=\mathbf{x}_{k-1}+s\mathbf{u}\,,\label{eq361} \end{equation}$$ where $s$ is determined from the line search. With the updated position calculated, $\mathbf{R}_{k}$ can be evaluated. Also, using equations (3.8.2-1) and (3.8.2-2), $\mathbf{d}_{k}$ and $\mathbf{G}_{k}$ can be evaluted. The stiffness update can now be expressed as $$\begin{equation} \mathbf{K}_{k}^{-1}=\mathbf{A}_{k}^{T}\mathbf{K}_{k-1}^{-1}\mathbf{A}_{k}\,,\label{eq362} \end{equation}$$ where the matrix $\mathbf{A}$ is an $n\times n$ matrix of the simple form: $$\begin{equation} \mathbf{A}_{k}=\mathbf{1}+\mathbf{v}_{k}\mathbf{w}_{k}^{T}\,.\label{eq363} \end{equation}$$ The vectors $\mathbf{v}$ and $\mathbf{w}$ are given by $$\begin{equation} \mathbf{v}_{k}=-\left(\frac{\mathbf{d}_{k}^{T}\mathbf{G}_{k}}{\mathbf{d}_{k}^{T}\mathbf{K}_{k-1}\mathbf{d}_{k}}\right)^{1/2}\mathbf{K}_{k-1}\mathbf{d}_{k}-\mathbf{G}_{k}\,,\label{eq364} \end{equation}$$ $$\begin{equation} \mathbf{w}_{k}=\frac{\mathbf{d}_{k}}{\mathbf{d}_{k}^{T}\mathbf{G}_{k}}\,.\label{eq365} \end{equation}$$ The vector $\mathbf{K}_{k-1}\mathbf{d}_{k}$ is equal to $s\mathbf{R}_{k-1}$ and has already been calculated.