In some artictle I've found Navier-Stokes Momentum Equation written in a following form $$\frac{\partial u}{\partial t}+\text{div}(u\otimes u)+\nabla p - \nu\Delta u =f$$ Since usually it appears with $u\cdot \nabla u $ term instead of $\text{div}(u\otimes u)$. It all came up from using following form of the Newton Law $$\underbrace{\frac{d}{dt}\int_B(\rho u)(t,\cdot)dx}_{\text{change of the linear momentum}}=\underbrace{-\int_{\partial B} (\rho u\otimes u )(t,\cdot)n(\cdot)dS}_{\text{flux of the momentum through boundary}}+\underbrace{F_B}_{\text{applied force}}\\ F_B=\underbrace{\int_B(\rho f)(t,\cdot)dx}_{\text{volume forces}}+\underbrace{\int_{\partial B}\text{T}(t,\cdot)n(\cdot)dS}_{\text{surface forces (the tension)}}$$ Then we end up with integral form $$\int_B\Big(\frac{\partial}{\partial t}(\rho u)(t,\cdot)+\text{div}(\rho u\otimes u)(t,\cdot)-(\rho f)(t,\cdot)-\text{divT}(t,\cdot)\Big) dx=0.$$ Defining stress tensor $T$ and assuming constant density we end up with the first equation, but since I'm not really fluent with tensors I could use some help understanding the difference between this formulation and the original one, and how to make calculations to get one form another.

Thank you!


1 Answer 1


The most generic form of the Navier-Stokes equation is $$ \frac{\partial (\rho \mathbf{u})}{\partial t} + \nabla \cdot(\rho \mathbf{u} \otimes \mathbf{u}) = \nabla p - \mathbf{f} + \nabla \cdot \mathbf{S}, \tag{1} $$ in which $\mathbf{S}$ is the shear stress tensor ($\mathbf{S}=\mu\nabla \mathbf{u}$ in your case). The continuity equation is $$ \frac{\partial \rho}{\partial t}+\nabla \cdot (\rho \mathbf{u})=0. \tag{2} $$ Using the product rule for derivatives in equation $(1)$ leads to $$ \rho \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u}\frac{\partial \rho}{\partial t} + \rho \mathbf{u} \cdot \nabla \mathbf{u} + \mathbf{u} \nabla \cdot(\rho \mathbf{u}) = \nabla p - \mathbf{f} + \nabla \cdot \mathbf{S}, \tag{3} $$ and using the continuity equation we see that the second and fourth terms in LHS cancel each other, leading to $$ \rho\left( \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} \right) = \nabla p - \mathbf{f} + \nabla \cdot \mathbf{S}. \tag{4} $$ Since it's usually assumed that the continuity equation holds, equation $(4)$ is completly equivalent to equation $(1)$. In practice, however, there are some "differences":

  1. Equation $(1)$ is called the conservative form of Navier-Stokes equation, while equation $(4)$ is called the non-conservative form. These names are a bit misleading: both equations are the momentum conservation equations. However, when solving numerically the governing equations of fluid dynamics, it's sometimes more useful to use equation $(1)$. It's basically due to the fact that accross a shock wave the velocity $\mathbf{u}$ is discontinuous (and, therefore, equation $(4)$ has the gradient of a discontinuous function), while $\rho \mathbf{u} \otimes \mathbf{u}$ is continuous even accross the shock. See that equation $(1)$ is called conservative form because it has derivatives of a conserved quantity (the momentum, i.e., $\rho \mathbf{u}$) while equation $(4)$ has derivatives of a non-conserved quantity (the velocity).

  2. Equation $(4)$ explicitly shows the transport of momentum in the term $\mathbf{u} \cdot \nabla \mathbf{u}$. Notice that the conservation equation of the property $\phi$ (which can be enthalpy, vorticity, chemical species, etc.) will have a term $\mathbf{u} \cdot \nabla \phi$. Therefore, equation $(4)$ "looks like" every other conservation equation. It's usually defined the material derivative of property $\phi$ as $$ \frac{D \phi}{Dt} = \frac{\partial\phi}{\partial t} + \mathbf{u} \cdot \nabla\phi, $$ which can be interpreted as the rate of change of the property $\phi$ along time in a particle of fluid while this particle is transported by the flow. Then, the conservation equation of any property $\phi$ can be written generically as $$ \frac{D \phi}{Dt} = \text{source terms}, $$ in which the source terms for the case $\phi=\mathbf{u}$ (i.e., the Navier-Stokes equation) are $(\nabla p - \mathbf{f} + \nabla \cdot \mathbf{S})/\rho$.

Summarizing: both equations are the same. When you need to solve them numerically, equation $(1)$ can be more suitable. If you want to interpret the physical meaning of the terms of Navier-Stokes equation, equation $(4)$ is more suitable.

  • 1
    $\begingroup$ Thank you very much, really well explained! $\endgroup$
    – user396656
    Dec 18, 2018 at 16:06

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