The solution of the nonlinear equation $\delta W=0$ is obtained by linearizing this relation as $$\begin{equation} \delta W+D\delta W\left[\Delta\mathbf{u}\right]+D\delta W\left[\Delta\mathbf{w}\right]+D\delta W\left[\Delta J^{f}\right]\approx0\,,\label{eq:fsi-linearized-virtual-work} \end{equation}$$ where the operator $D\delta W\left[\cdot\right]$ represents the directional derivative of $\delta W$ at $\left(\mathbf{u}^{s},\mathbf{w},J^{f}\right)$ along an increment $\Delta\mathbf{u}$ of $\mathbf{u}^{s}$ , $\Delta\mathbf{w}$ of $\mathbf{w}$ , or $\Delta J^{f}$ of $J^{f}$ .

To linearize the virtual work, we need to express the integrals appearing in $\delta W_{int}$ and $\delta W_{ext}$ over the material frame of the finite element solid domain. For notational convenience, we let $J\equiv J^{s}$ and $\mathbf{F}\equiv\mathbf{F}^{s}$ . Thus, $$\begin{equation} \begin{aligned}\int_{b}\boldsymbol{\sigma}^{s}:\grad\delta\mathbf{v}^{s}\,dv & =\int_{B}\mathbf{F}\cdot\mathbf{S}^{s}:\Grad\delta\mathbf{v}^{s}\,dV\end{aligned} \,,\label{eq:fsi-work-material-1} \end{equation}$$ where $\mathbf{S}^{s}=J\cdot\mathbf{F}^{-1}\cdot\boldsymbol{\sigma}^{s}\cdot\mathbf{F}^{-T}$ is the second Piola-Kirchhoff stress for the solid material. Similarly, $$\begin{equation} \int_{b}\boldsymbol{\tau}:\grad\delta\mathbf{w}\,dv=\int_{B}\boldsymbol{\tau}\cdot J\mathbf{F}^{-T}:\Grad\delta\mathbf{w}\,dV\,.\label{eq:fsi-work-material-2} \end{equation}$$ Next, $$\begin{equation} \begin{aligned} & \int_{b}\delta\mathbf{w}\cdot\left(\grad p+\rho^{f}\mathbf{a}^{f}\right)\,dv\\ & =\int_{B}\delta\mathbf{w}\cdot\left(J\mathbf{F}^{-T}\cdot\Grad p+J\rho^{f}\left(\dot{\mathbf{w}}+\dot{\mathbf{v}}^{s}\right)+\rho^{f}\left(\Grad\mathbf{w}+\Grad\mathbf{v}^{s}\right)\cdot\mathbf{W}\right)\,dV \end{aligned} \label{eq:fsi-work-material-3-4} \end{equation}$$ $$\begin{equation} \int_{b}\delta J^{f}\left[\frac{1}{J^{f}}\left(\dot{J}^{f}+\grad J^{f}\cdot\mathbf{w}\right)-\frac{\dot{J}}{J}\right]\,dv=\int_{B}\delta J^{f}\left[\frac{1}{J^{f}}\left(J\dot{J}^{f}+\Grad J^{f}\cdot\mathbf{W}\right)-\dot{J}\right]\,dV\label{eq:fsi-work-material-5} \end{equation}$$ $$\begin{equation} \int_{b}\mathbf{w}\cdot\grad\delta J^{f}\,dv=\int_{B}\Grad\delta J^{f}\cdot\mathbf{W}\,dV\label{eq:fsi-work-material-6} \end{equation}$$ where $\mathbf{W}=J\mathbf{F}^{-1}\cdot\mathbf{w}$ .

Keep in mind that we are solving for the motions $\boldsymbol{\chi}_{n+1}^{\iota}$ at $t_{n+1}$ . Therefore, $$\begin{aligned}D\mathbf{u}^{s}\left[\Delta\mathbf{u}\right] & =\alpha_{f}\Delta\mathbf{u}\\ D\mathbf{F}\left[\Delta\mathbf{u}\right] & =\alpha_{f}\Grad\Delta\mathbf{u}\\ DJ\left[\Delta\mathbf{u}\right] & =\alpha_{f}J\left(\divg\Delta\mathbf{u}\right)\\ D\dot{J}\left[\Delta\mathbf{u}\right] & =D\left(J\mathbf{F}^{-T}:\Grad\mathbf{v}^{s}\right)\left[\Delta\mathbf{u}\right]\\ & =\alpha_{f}J\left[\left(\divg\mathbf{v}^{s}+\frac{\gamma}{\beta\Delta t}\right)\left(\divg\Delta\mathbf{u}\right)-\left(\grad\Delta\mathbf{u}\right)^{T}:\mathbf{L}^{s}\right]\\ D\mathbf{v}^{s}\left[\Delta\mathbf{u}\right] & =\frac{\alpha_{f}\gamma}{\beta\Delta t}\Delta\mathbf{u}\\ D\mathbf{w}\left[\Delta\mathbf{w}\right] & =\alpha_{f}\Delta\mathbf{w}\\ DJ^{f}\left[\Delta J^{f}\right] & =\alpha_{f}\Delta J^{f}\\ D\rho^{f}\left[\Delta J^{f}\right] & =-\alpha_{f}\frac{\rho^{f}}{J^{f}}\Delta J^{f} \end{aligned}$$ In general, $\boldsymbol{\sigma}^{s}$ (and thus, $\mathbf{S}^{s}$ ) is only a function of the solid strain, such as the right Cauchy-Green tensor $\mathbf{C}=\mathbf{F}^{T}\cdot\mathbf{F}$ or the Green-Lagrange strain $\mathbf{E}=\left(\mathbf{C}-\mathbf{I}\right)/2$ . $$D\mathbf{E}\left[\Delta\mathbf{u}\right]=\frac{\alpha_{f}}{2}\left(\Grad^{T}\Delta\mathbf{u}\cdot\mathbf{F}+\mathbf{F}^{T}\cdot\Grad\Delta\mathbf{u}\right)$$ Therefore, following the standard approach in solid mechanics, the linearization of $\mathbf{S}^{s}$ is $$\begin{equation} \begin{aligned}D\mathbf{S}^{s} & =D\mathbf{S}^{s}\left[\Delta\mathbf{u}\right]=\frac{\partial\mathbf{S}^{s}}{\partial\mathbf{E}}:D\mathbf{E}\left[\Delta\mathbf{u}\right]\\ & =\alpha_{f}\boldsymbol{\mathbb{C}}^{s}:\frac{1}{2}\left(\Grad^{T}\Delta\mathbf{u}\cdot\mathbf{F}+\mathbf{F}^{T}\cdot\Grad\Delta\mathbf{u}\right)\\ & =\alpha_{f}\boldsymbol{\mathbb{C}}^{s}:\left(\mathbf{F}^{T}\,\underline{\otimes}\,\mathbf{F}^{T}\right):\Delta\boldsymbol{\varepsilon} \end{aligned} \label{eq:fsi-solid-stress-linearization} \end{equation}$$

In general, $\boldsymbol{\tau}$ is a function of the fluid volume ratio $J^{f}$ and the rate of deformation of the fluid, $\mathbf{D}^{f}=\frac{1}{2}\left(\mathbf{L}^{f}+\left(\mathbf{L}^{f}\right)^{T}\right)$ . However, since $\mathbf{v}^{f}$ is not a degree of freedom, we need to let $\mathbf{D}^{f}=\mathbf{D}^{w}+\mathbf{D}^{s}$ , where $\mathbf{D}^{w}=\frac{1}{2}\left(\mathbf{L}^{w}+\left(\mathbf{L}^{w}\right)^{T}\right)$ and $\mathbf{D}^{s}=\frac{1}{2}\left(\mathbf{L}^{s}+\left(\mathbf{L}^{s}\right)^{T}\right)$ . Thus, $$\begin{equation} D\boldsymbol{\tau}=\boldsymbol{\mathcal{C}}^{\tau}:\left(D\mathbf{D}^{f}\left[\Delta\mathbf{w}\right]+D\mathbf{D}^{f}\left[\Delta\mathbf{u}\right]\right)+\frac{\partial\boldsymbol{\tau}}{\partial J^{f}}DJ^{f}\left[\Delta J^{f}\right]\label{eq:fsi-fluid-stress-linearization} \end{equation}$$ where $$\begin{equation} \boldsymbol{\mathcal{C}}^{\tau}=\frac{\partial\boldsymbol{\tau}}{\partial\mathbf{D}^{f}}\label{eq:fsi-viscous-tangent-tensor} \end{equation}$$