Review of the notes growth main (070).

Motivation

There are at least two models in continuum mechanics where mass balance is violated, that is,

\dot\varrho+\varrho\operatorname{div}\mathbf v\neq 0\label{eq:1}

$\dot\varrho=0$ $\operatorname{div}\mathbf v\neq 0$ $\eqref{eq:1}$ is replaced by

\dot\varrho+\varrho\operatorname{div}\mathbf v=h,

$h$ is the mass supply. In both cases, the characterization of inertia is problematic.

$\operatorname{div}\mathbf T+\mathbf b=\mathbf 0$ $\mathbf{b}=\mathbf{b}_{0}+\boldsymbol\iota$ , where

\iota=-\varrho\dot{\mathbf v}\label{eq:3}

$\Omega_t$ $\Omega_t$ is, according to Reynolds' theorem,

\frac{d}{dt}\int_{\Omega_t}\frac 12 \varrho \mathbf v\cdot\mathbf v=\int_{\Omega_t}\left(\frac 12 {\dot\varrho}\mathbf v+\varrho\dot{\mathbf v}\right)\cdot\mathbf v+\frac 12 \varrho\mathbf v\cdot\mathbf v\operatorname{div}\mathbf v=\int_{\Omega_t}\left(\varrho \dot{\mathbf v}+\frac 12 h\mathbf v\right)\cdot\mathbf v.

Important remark

incompressible fluid $\varrho$ $\varrho\operatorname{div}\mathbf v$ .

Growing materials

Prompted by these considerations, the note considers a growing body, where the growth rate

\omega=\frac h \varrho

$\int_{\mathcal{P}} \check{s}\underline\omega$ $\int_{\mathcal{P}} s \underline{\omega}$ $\underline\omega$ is the virtual accretion rate. This leads to the balance equation

\check s=s

which complements the standard force balance. The dissipation inequality takes the form:

\int_{\mathcal{P}} \mathbf{b} \cdot \mathbf{v} d V+\int_{\partial \mathcal{P}} \mathbf{t} \cdot \mathbf{v} d A+\int_{\mathcal{P}} s \omega d V+\int_{\mathcal{P}} \check{p} \frac{h}{\rho} d V-\frac{d}{d t} \int_{\mathcal{P}} \rho \psi d V=\mathcal{D}(\mathcal{P}) \geq 0 \quad \forall \mathcal{P}

$\check p$ is understood as chemical potential. In its local form, the inequality reads:

(\operatorname{dev} \mathbf{T}-\operatorname{dev} \hat{\mathbf{T}}(\mathbf{F})) \cdot \dot{\mathbf{F}} \mathbf{F}^{-1}+(p-\hat{p}(\rho)) \frac{\dot{\rho}}{\rho}+(\check{p}+\hat p_a(\beta)+\check{s}-p-\rho \psi) \omega \geq 0,

where the constitutive equations for the deviatoric part of the stress are given by the requirement that:

\rho \dot{\psi}=\rho \frac{d}{d t} \hat{\psi}_{\mathrm{a}}(\beta, \mathbf{F})=\operatorname{dev} \hat{\mathbf{T}}(\mathbf{F}) \cdot \dot{\mathbf{F}} \mathbf{F}^{-1}+\hat{p}(\rho) \frac{\dot{\rho}}{\rho}-\hat{p}_{\mathbf{a}}(\beta) \omega\label{eq:19}

$\eqref{eq:3}$ $s$ $s=s_0+s^{\rm in}$ , where

s^{\rm in}=-\varrho a\left(\dot\omega+\frac 12\omega^2\right)\label{eq:8}

$a>0$ is a coefficient. The motivation for this assumption is that if we take the kinetic energy

\kappa=\varrho k,\qquad k=\frac{1}{2}\left(\mathbf{v} \cdot \mathbf{v}+a \omega^{2}\right),\label{eq:9}

$\dot\beta=0$ ,

\frac{d}{dt} \int_{{\Omega_t}} \frac{1}{2} \varrho \mathbf{v} \cdot \mathbf{v}=\int_{\Omega_t}\varrho\dot{\mathbf v}\cdot\mathbf v,

$\dot\varrho=0$ ,

\frac{d}{dt} \int_{{\Omega_t}} \frac{1}{2} \varrho a \omega^2=\int_{\Omega_t}\varrho a\left(\dot{\omega}+\frac{1}{2} \omega^{2}\right) \omega

$\eqref{eq:9}$ $\dot\varrho$ $\dot\omega$ equal to null. Clearly, without the splitting assumption, the time derivative of the kinetic energy would contain an extra coupling term:

\frac{1}{2} \rho \omega \mathbf{v} \cdot \mathbf{v}=\frac{1}{2}(\dot{\rho}+\rho \operatorname{div} \mathbf{v}) \mathbf{v} \cdot \mathbf{v}.

$\eqref{eq:8}$ $s=s_0-\varrho a\left(\dot\omega+\frac 12\omega^2+\frac 12\omega \mathbf v\cdot\mathbf v\right)$ $\eqref{eq:3}$ , thus giving $\boldsymbol\iota=

For a viscous fluid the evolution equations are:

\left\{ \begin{aligned} &\varrho\dot{\mathbf v}+\nabla p-\mu_{\rm vis}\left(\Delta\mathbf v+\frac 13 \nabla\operatorname{div}\mathbf v\right)=\mathbf b_0,&&\textrm{balance of forces}\\ &\operatorname{div}\mathbf v=-\frac{\dot\varrho}\varrho+\omega, &&\textrm{kinematic splitting}\\ &\varrho\omega=h,&&\textrm{balance of mass}\\ &p-\check p-\hat p_a(\beta)+\varrho\psi+\mu_{\rm a}\omega=-\varrho a \left(\dot\omega+\frac 12 \omega^2\right),&& \textrm{microforce balance law} \end{aligned} \right.

$p$ is related to the free energy by the equation

p=\hat p(\varrho)=\varrho^2\widehat\psi'(\varrho).

$\check p$ $h=\varrho\omega$ $\check p$ $\omega$ $h$ .

$J_0$ . The evolution equations are:

\frac{1}{\operatorname{vol}(\mathcal{R})} \mathbf{M}^{i n}-p^{e x t} \mathbf{I}=\mathbf{T}

where

\frac{1}{\operatorname{vol}(\mathcal{R})} \mathbf{M}^{i n}=-\rho \frac{2}{15} r_0^2 J_{\mathrm{o}}^{2 / 3}\left(\frac{\ddot{J}_{\mathrm{o}}}{J_{\mathrm{o}}}-\frac{2}{3}\left(\frac{\dot{J}_{\mathrm{o}}}{J_{\mathrm{o}}}\right)^{2}\right) \mathbf{I}

the constitutive equation for the Cauchy stress follows from the choice of the free energy:

\rho \frac{d}{d t} \hat{\psi}_{\mathrm{sph}}(\beta, \rho)=\hat{p}(\rho) \frac{\dot{\rho}}{\rho}-\hat{p}_{\mathrm{a}}(\beta) \omega

Comments

$\operatorname{dev}\widehat{\mathbf T}(\mathbf F)$ is missing, hence it is difficult to reconstruct the precise form of the evolution equations.
$\hat p_a(\beta)$ in the final equations is hidden. Of course, one can work out the expressions, but it would be way easier to read these notes if these expressions were written explicitly.
It would be useful to have written the ODE system for the homogeneous deformations, so that one co

Questions

what is the explicit form of the balance of energy? I guess that with the splitting we use the balance of energy does not hold. Am I correct?
$a$ ?