I was reading the research paper Homogenization of peristaltic flows in piezoelectric porous media and came across the hydrodynamic equation:

$$\mu \nabla^2 v^f -\underline{ \rho_f (\dot{v}^f + w \cdot \nabla) v^f)} = \nabla p -f $$

where $\mu$ is the dynamic viscosity, $v^f$ is the fluid velocity, $w$ is the seepage velocity, $p$ is the pressure, and $f$ is the body force.

What confuses me is the time derivative term of the velocity (the underlined term in the equation above), especially when I compare it with the general form of the Navier Stokes equation:

$$\underline{\rho (\frac{\partial v^f}{\partial t} + (v^f \cdot \nabla) v^f)} = - \nabla p + \mu \nabla^2 v^f + \frac{1}{3} \mu \nabla(\nabla \cdot v^f) + \rho g $$

where $u$ here is the fluid velocity, $\rho$ is its density, $p$ is pressure of fluid, $\mu$ is the dynamic viscosity of fluid, $\rho g$ is the body force (gravity).

Comparing the time derivative terms from both equations (the underlined terms), I don't understand why the seepage velocity is used in the material time derivative in the first equation. The seepage velocity is the velocity of fluid passing between the pores: I don't understand why it is used in the material time derivative that its multiplied by the fluid velocity. Why are the two velocities are used together?

Also, the term of partial time derivative w.r.t. time in the first equation is $\rho_f \dot{v}^f v^f$ where as in the second equation its $\rho \frac{\partial u}{\partial t}$, and so I don't understand why in the first equation we have it multiplied with the fluid velocity.

Note: In both equations, the fluid is a Newtonian fluid.

  • 1
    $\begingroup$ They're investigating heterogenous media, that have macroscopic different behavior from Newtonian fluids (governed by Navier--Stokes equations) $\endgroup$
    – basics
    Commented Oct 1, 2023 at 10:14
  • 1
    $\begingroup$ The fluid that is flowing in the porous media is considered to be Newtonian fluid. (I will add this information above) $\endgroup$
    – user134613
    Commented Oct 1, 2023 at 10:30
  • $\begingroup$ I'll take a look at the paper later $\endgroup$
    – basics
    Commented Oct 1, 2023 at 11:14
  • $\begingroup$ @user134613 may you comment on why the existing answer is not satisfactory? $\endgroup$
    – Quillo
    Commented Oct 3, 2023 at 15:58

1 Answer 1


Here my guess: the authors are solving fluid equations on a grid moving with the solid matrix, implicitly using an ALE (arbitrary Lagrangian Eulerian) approach.

To add some details, in continuum mechanics there are three main approaches:

  • Lagrangian: every material point is labelled with a coordinate $\mathbf{x}_0$
  • Eulerian: the problem is described using a stationary set of coordinates $\mathbf{x}$
  • arbitrary Lagrangian Eulerian: the problem is described following points labelled with coordinates $\mathbf{x}_b$ in an arbitrary motion.

These approaches are related with derivative of composite functions. As an example, the position in space of the material point labelled with $\mathbf{x}_0$ and the arbitrary point labelled with $\mathbf{x}_b$ can be described by functions \begin{equation} \mathbf{x}(\mathbf{x}_0, t) \qquad , \qquad \mathbf{x} (\mathbf{x}_b,t) \ , \end{equation} and their velocity by \begin{equation} \mathbf{v}_0 = \dfrac{\partial \mathbf{x}}{\partial t}\bigg|_{\mathbf{x}_0}=:\mathbf{u} \qquad , \qquad \mathbf{v}_b = \dfrac{\partial \mathbf{x}}{\partial t}\bigg|_{\mathbf{x}_b} \ . \end{equation} Given a function $f(\mathbf{x},t)$ of the Eulerian coordinates, it can be represented with Lagrangian or arbitrary coordinates as \begin{equation} f(\mathbf{x},t) = f(\mathbf{x}(\mathbf{x}_0,t),t) =: f_0(\mathbf{x}_0,t) \\ f(\mathbf{x},t) = f(\mathbf{x}(\mathbf{x}_b,t),t) =: f_b(\mathbf{x}_b,t) \ , \end{equation} and the time derivatives of these functions are related by \begin{equation} \dfrac{\partial f}{\partial t}\bigg|_{\mathbf{x}_0} = \dfrac{\partial f}{\partial t}\bigg|_{\mathbf{x}} + \mathbf{u} \cdot \nabla f \\ \dfrac{\partial f}{\partial t}\bigg|_{\mathbf{x}_b} = \dfrac{\partial f}{\partial t}\bigg|_{\mathbf{x}} + \mathbf{v}_b \cdot \nabla f \ . \end{equation} If we apply these relations to the time derivative of the velocity field, it's possible to write \begin{equation} \mathbf{a} = \dfrac{\partial \mathbf{u}}{\partial t}\bigg|_{\mathbf{x}_0} = \dfrac{\partial f}{\partial t}\bigg|_{\mathbf{x}} + ( \mathbf{u} \cdot \nabla ) \mathbf{u} =\dfrac{\partial f}{\partial t}\bigg|_{\mathbf{x}_b} + ( ( \mathbf{u} - \mathbf{v}_b ) \cdot \nabla ) \mathbf{u} \ . \end{equation} If we mean the operator $\dot(f)$ as the time derivative w.r.t. the points of the solid mesh labelled with $\mathbf{x}_b$, $\frac{partial}{\partial t}|_{\mathbf{x}_b}$, we get the expression of the paper for the fluid equation \begin{equation} \dot{\mathbf{u}} + ( ( \mathbf{u} - \mathbf{v}_b ) \cdot \nabla ) \mathbf{u} \ . \end{equation}

  • $\begingroup$ The original question seems more like a theoretical one, not a matter of numerical methods, namely why are the hydrodynamic equations written that way (i.e. how are they derived?). $\endgroup$
    – Quillo
    Commented Oct 1, 2023 at 13:17
  • $\begingroup$ It's not numerical. It's just the definition of $\dot{()}$ operator in the equation. To me, it looks like they mean the time derivative of the quantities at the points that can move. Nothing special about the physics there, if not for useless extra-definitions that can be avoided if one knows how to approach a problem in continuum mechanics (and I suspect that many authors and reviewers of scientific papers know very little math, for the papers they're writing or reviewing) $\endgroup$
    – basics
    Commented Oct 1, 2023 at 13:24
  • $\begingroup$ To be more explicit, the operator $\dot{()}$ means time derivative following the points of the solid mesh, that is equivalent to the material derivative for the solid domain, and to an ALE approach for the fluid domain. $\endgroup$
    – basics
    Commented Oct 1, 2023 at 13:30
  • $\begingroup$ I didn't downvote but I personally find that this answer does not touch the real problem. The Eulerian or Lagrangian descriptions are just different (equivalent) kinematic formulations, here we have to justify why the dynamics has that form. The point of the question is: why does the dynamics is different from Navier Stoker. I think that this can only be understood by reviewing the derivation of the dynamical laws (not an easy task, unless one finds a reference for the theoretical foundations of Newtonian fluids in porous media). $\endgroup$
    – Quillo
    Commented Oct 1, 2023 at 19:27
  • $\begingroup$ +1 After checking the paper it seems that the idea is correct. The answer would be clearer by specifying that there is a solid component described by a displacement field $\mathbf{u}$ and a fluid with velocity $\mathbf{v}_f$ (i.e. the velocity of the solid part is $\dot{\mathbf{u}}$. The solid is treated in a Lagrangian way, as suggested in the answer. What is unclear from the question is that this is a two-field model and the nature of these two fields (one field is a displacement and the other a velocity). $\endgroup$
    – Quillo
    Commented Oct 2, 2023 at 11:54

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