paper_on_navier

The compressible Navier-Stokes system

The compressible Navier-Stokes system is:

\begin{aligned} &\varrho{\boldsymbol v}_t+\varrho\nabla\boldsymbol v\,\boldsymbol v+\mu\Big(\Delta\boldsymbol v+\frac 2 3\nabla\operatorname{div}\boldsymbol v\Big)-\nabla p=0\\ &\varrho_t+\operatorname{div}(\varrho\boldsymbol v)=0\\ &p=P(\varrho) \end{aligned}

The first equation follows from the balance of momentum, the second is the balance of mass, and the third is the equation of state.

$\Omega$ $\boldsymbol v=\boldsymbol 0$ $\partial\Omega$ (which entail null working from the exterior).

Balance of energy for the compressible Navier-Stokes system

By definition, the kinetic energy is:

T=\frac 12 \int_\Omega \varrho|\boldsymbol v|^2.

Then:

\begin{aligned} \frac{{\rm d}}{{\rm d}t}T&=\int_\Omega \frac 12 \varrho_t|\boldsymbol v|^2+\boldsymbol v\cdot(\varrho\boldsymbol v_t)\\ &=-\int_\Omega \color{blue}\frac 12|\boldsymbol v|^2\operatorname{div}(\varrho\boldsymbol v)+\varrho\boldsymbol v\cdot\nabla\boldsymbol v \,\boldsymbol v\color{black}+\nabla p\cdot\boldsymbol v-\mu\Delta\boldsymbol v\cdot\boldsymbol v-\frac \mu 3 (\nabla\operatorname{div}\boldsymbol v)\cdot\boldsymbol v. \end{aligned}

Integration by parts shows that [^1:]

\int_\Omega \color{blue}\frac 12|\boldsymbol v|^2\operatorname{div}(\varrho\boldsymbol v)+\varrho\boldsymbol v\cdot\nabla\boldsymbol v \,\boldsymbol v\color{black}=0.\label{eq:6}

Furthermore, the null-slip boundary condition entails

\int_\Omega\nabla p\cdot\boldsymbol v=-\int_\Omega p\operatorname{div}\boldsymbol v.

As a result, we obtain

\frac{{\rm d}}{{\rm d}t}T-\int_\Omega p\operatorname{div}\boldsymbol v+\underbrace{\int_\Omega \mu\big(|\nabla\boldsymbol v|^2+\frac \mu 3 (\operatorname{div}\boldsymbol v)^2\big)}_{\displaystyle D}=0,\label{eq:5}

where the last term on the right-hand side is the dissipation.

$\psi(\varrho)$ a function such that

\psi'(\varrho)=-\frac {P(\varrho)}{\varrho^2},

$-\int_\Omega p \operatorname{div}\boldsymbol v=-\int_\Omega P(\varrho)\frac{\dot\varrho}\varrho=\int_\Omega \varrho\dot\psi$ $\boldsymbol v=\boldsymbol 0$ at the boundary, by a standard transport theorem, we obtain

-\int_\Omega p\operatorname{div} \boldsymbol v=\frac{\mathrm d}{{\mathrm d}t}\int_\Omega\varrho\psi=:\dot F,\label{eq:2}

$F$ the total Helmholtz free energy. We therefore obtain the balance of energy

\frac{{\rm d}}{{\rm d}t}T+\frac{{\rm d}}{{\rm d}t}F+D=0.\label{eq:1}

The quasi-compressible Navier-Stokes system

$\varrho_*$ $p_*=P(\varrho_*)$ $K=\varrho_*P'(\varrho_*)$ $\varrho=\varrho_*$ . As a result the flux has null divergence and we obtain the incompressible Navier-Stokes system:

\begin{aligned} &\varrho_*{\boldsymbol v}_t+\varrho_*\nabla\boldsymbol v\,\boldsymbol v+\mu\Delta\boldsymbol v-\nabla p=\boldsymbol 0,\\ &\operatorname{div}\boldsymbol v=0. \end{aligned}

$\eqref{eq:2}$ $\eqref{eq:1}$ vanishes. Clearly the balance of energy holds in the simpler form

\frac{{\rm d}}{{\rm d}t}T_*+D=0,

where the kinetic energy can be written as:

T=\frac 1 2\varrho_*\int_\Omega |\boldsymbol v|^2.\label{eq:4}

A variant of the incompressible system, referred to as quasi-incompressible is

\begin{aligned} &\varrho_*{\boldsymbol v}_t+\varrho_*\nabla\boldsymbol v\,\boldsymbol v+\mu\Delta\boldsymbol v-\nabla p=\boldsymbol 0,\\ &\frac{{p}_t}K+\operatorname{div}\boldsymbol v=0. \end{aligned}\tag{Q}

$\varrho=\varrho_*$ $\varrho\simeq\varrho_*$ $p_t=P’(\varrho_*)\varrho_t=K\varrho_t/\varrho_*$ $\varrho_t=\varrho_* p_t/K$ $\rm (Q)$ . ¹

Originally, it was introduced as a regularization of the incompressible system for numerical applications. However, the model has some value per se. In fact, the incompressible system is not capable to describe the propagation of pressure waves, whereas the quasi-compressible variant is able to do so.

Temam’s force

The quasi-compressible Navier-Stokes system is often considered in the following modified form:

\begin{aligned} &\varrho_*{\boldsymbol v}_t+\varrho_*\nabla\boldsymbol v\,\boldsymbol v+\mu\Delta\boldsymbol v-\nabla p+\color{red}\boldsymbol f_{\mathrm T}\color{black}=\boldsymbol 0,\\ &\frac{{p}_t}K+\operatorname{div}\boldsymbol v=0, \end{aligned}\tag{Q'}

where

\color{red}\boldsymbol f_{\mathrm T}=-\frac 1 2 \varrho_*(\operatorname{div}\boldsymbol v)\boldsymbol v

is the so-called Temam’s force. Most sources say that the rationale behind this modification is the need of preserving energy balance.

Energy balance for the quasi-compressible system with Temam’s force

$\eqref{eq:4}$ as the expression for the kinetic energy.

$\eqref{eq:5}$ we obtain:

\begin{aligned} \frac{{\rm d}}{{\rm d}t}T_*&=\int_\Omega\varrho_*\boldsymbol v\cdot\boldsymbol v_t\\ &=-\int_\Omega \Big(\color{blue}\varrho_*\boldsymbol v\cdot\nabla\boldsymbol v \,\boldsymbol v\color{black}+\nabla p\cdot\boldsymbol v-\mu\Delta\boldsymbol v\cdot\boldsymbol v-\frac \mu 3 (\nabla\operatorname{div}\boldsymbol v)\cdot\boldsymbol v\Big)+\int_\Omega\color{red}\frac 1 2 \varrho_*(\operatorname{div}\boldsymbol v)|\boldsymbol v|^2. \end{aligned}

Since

\int_\Omega\frac 12\varrho_*(\operatorname{div} \boldsymbol v)|\boldsymbol v|^2=-\int_\Omega\varrho_*\boldsymbol v\cdot\nabla\boldsymbol v^{\mathsf T}\boldsymbol v=-\int_\Omega\varrho_*\boldsymbol v\cdot\nabla\boldsymbol v\,\boldsymbol v,

we see that the extra force plays a crucial role to obtain the energy balance

\frac{{\rm d}}{{\rm d}t}T_*+\frac{{\rm d}}{{\rm d}t}F_*+D=0,\label{eq:3}

$F_*=\frac 12 p^2/K$ . Note that as an alternative for the extra force one could choose,

\tilde{\boldsymbol f}_{\rm T}=\frac 1 2\varrho_*\nabla |\boldsymbol v|^2=\varrho_*\nabla\boldsymbol v^{\mathsf T}\boldsymbol v.

$\int_\Omega \tilde{\boldsymbol f}_{\rm T}\cdot\boldsymbol v$ $\boldsymbol f_{\rm T}$ $\tilde{\boldsymbol f}_{\rm T}$ My paper on arXiv $\boldsymbol f_{\rm T}$ guarantees balance of kinetic energy in a “pointwise sense”.

The correct choice of the kinetic energy

As a final remark, we should observe that the extra force is not invariant under Galilean transformations. Thus, is to be considered unphysical. As a matter of fact, it appears that the best way to restore energy balance is to take the following expression of the kinetic energy:

\widetilde T=\int_\Omega\tilde\varrho|\boldsymbol v|^2,\qquad \text{where }\tilde\varrho=\varrho_*\left(\frac p K-1\right).

Then the calculation becomes

\begin{aligned} \frac{{\rm d}}{{\rm d}t}T&=\int_\Omega \frac 12 \varrho_t|\boldsymbol v|^2+\boldsymbol v\cdot(\varrho\boldsymbol v_t)=\int_\Omega \frac 12 \frac {p_t}K\varrho_*|\boldsymbol v|^2+\boldsymbol v\cdot(\varrho_*\boldsymbol v_t)\\ &=-\int_\Omega \color{blue}\frac 12\varrho_*|\boldsymbol v|^2\operatorname{div}(\boldsymbol v)+\varrho_*\boldsymbol v\cdot\nabla\boldsymbol v \,\boldsymbol v\color{black}+\nabla p\cdot\boldsymbol v-\mu\Delta\boldsymbol v\cdot\boldsymbol v-\frac \mu 3 (\nabla\operatorname{div}\boldsymbol v)\cdot\boldsymbol v. \end{aligned}

so that again energy balance is restored. In essence, it appears that Temam’s force is not needed, and the quasi-compressible system has nice mathematical properties even without Temam’s force.

\int_\Omega\frac 12 |\boldsymbol v|^2\operatorname{div}(\varrho\boldsymbol v)=-\int_\Omega\nabla\left(\frac 12|\boldsymbol v|^2\right)\cdot\varrho\boldsymbol v=-\int_\Omega\varrho\nabla\boldsymbol v^\top\,\boldsymbol v\cdot\boldsymbol v=-\int\varrho\boldsymbol v\cdot\nabla\boldsymbol v\,\boldsymbol v.

1 Note that if in the procedure that leads to the second of

$\rm (Q)$ we start from the mass balance equation in the form

$\dot\varrho+\varrho\operatorname{div}\boldsymbol v$ we obtain the material time derivative

$\dot p$ instead of

$p_t$ . ↩