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I consider the Navier-Stokes equation for uniformly incompressible, force-free, Newtonian fluids with constant viscosity. The equations describing the situation are:

$\partial_tv-v\times \text{curl}(v)-\nu \Delta v+\nabla f=0$,

$\text{div}(v)=0$,

where $\nu>0$ is the constant viscosity and $f=\frac{v^2}{2}+\frac{p}{\rho}$ is the corresponding Bernoulli function, $\rho>0$ the constant density and $p$ the pressure. What I am interested in is the change of the kinetic energy over time:

$\frac{d}{dt}E(t)=\frac{d}{dt}\frac{\rho}{2}\int_{\Omega}v^2d^3x$,

where $\Omega\subset \mathbb{R}^3$ is some bounded domain in $\mathbb{R}^3$ with smooth boundary in which the fluid is contained. I've often seen that people consider the 'no slip' condition, that is $v|_{\partial \Omega}=0$. In that case one can compute:

$\frac{d}{dt}E(t)=-\nu \rho \int_{\Omega}\omega^2 d^3x\leq 0$,

where $\omega=\text{curl}(v)$ is the vorticity of $v$. Hence the kinetic energy dissipates over time, due to the non-vanishing viscosity.

However to me it is far more intuitive to consider the boundary condition $v\cdot \vec{n}=0$ on $\partial \Omega$, where $\vec{n}$ is the unit outward normal vector, i.e. $v$ is tangent to the boundary. Intuitively this would imply that no kinetic energy can be gained from the 'outside' since no particle from the ouside enters our domain of interest. Due to the non-vanishing viscosity I would again expect that the kinetic energy will be decreasing over time, that is I still expect $\frac{d}{dt}E(t)\leq 0$ to hold true. However if I compute the change of energy rate I obtain:

$\frac{d}{dt}E(t)=-\nu \rho \int_{\Omega}\omega^2 d^3x-\nu \rho\int_{\partial \Omega}\left(\omega\times v \right)\cdot \vec{n}dS$,

where $\times$ denotes the standard cross product. In case of the no-slip condition the second term vanishes. But in our case the term $\omega \times v$ in general can be parallel to $\vec{n}$ so that the additional term doesn't always vanish. Neither does the surface term seem to have a definit sign. So my questions are:

1) Can we still show that in this situation $\frac{d}{dt}E(t)\leq 0$ holds true?

2) If not, where does my physical intuition fail me?

3) What is the physical interpretation of the surface term in case of tangent boundary conditions?

Thanks a lot in advance

PS: On a side note, I am a mathematician and not a physicists. However I always try to grasp the intuition behind the equations I'm working with and here it appears to me that my intuition is leading me astray.

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Please first of all find my conclusions here:

(1) It is not always true that $\dot{E}\le 0$,

(2) because the impenetrability condition $\mathbf{v}\cdot\mathbf{n}=0$ does not complete the problem, i.e., this does not guarantee the uniqueness of the solution (note that the no-slip condition $\mathbf{v}=\mathbf{0}$ is a vector-valued condition, while the impenetrability condition is merely a scalar condition).

(3) As is mentioned, $\mathbf{v}\cdot\mathbf{n}=0$ is called the impenetrability condition, meaning that the fluid inside $\Omega$ may never flow outside $\Omega$. For this reason, the normal component of $\mathbf{v}$ must vanish, and this is how the condition comes into being.

Let me explain with more details. Since you are a mathematician, let us deal with Navier-Stokes equations in a more conventional way. Consider \begin{align} \rho\left(\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla\right)\mathbf{v}&=-\nabla p+\nabla\cdot\mathbb{T},\\ \nabla\cdot\mathbf{v}&=0, \end{align} where the viscous stress tensor $$ \mathbb{T}=\mu\left(\nabla\otimes\mathbf{v}+\left(\nabla\otimes\mathbf{v}\right)^{\top}\right). $$ You may check that these equations are identical to your proposal.

Thanks to the divergence theorem as well as the impenetrability condition, you will obtain $$ \dot{E}=-\frac{1}{2\mu}\int_{\Omega}\left\|\mathbb{T}\right\|_F^2\,{\rm d}V+\int_{\partial\Omega}\mathbf{n}\cdot\mathbb{T}\cdot\mathbf{v}\,{\rm d}S, $$ where $E$ is the kinetic energy as you defined, $\left\|\cdot\right\|_F$ denotes the Frobenius norm of a matrix, and $\mathbf{n}$ stands for the unit outward normal of $\Omega$.

As can be seen, the first term gives you the energy dissipation rate inside $\Omega$. This dissipation is caused by the viscosity of the fluid. It is of course negative. The second term, however, depicts the rate of changing energy on the boundary. It is not always negative. In fact, the kinetic energy of the fluid could increase if $\partial\Omega$ does positive work on it (consider, say, the lid-driven cavity setup).

In the case you expect that the system is everywhere dissipative, an additional condition must be imposed, among which the most popular one is the Navier-slip condition, i.e, $$ \mathbf{n}\cdot\mathbb{T}\cdot\mathbf{t}=-\beta\,\mathbf{v}\cdot\mathbf{t} $$ on $\partial\Omega$ with some $\beta>0$ for all tangent vectors $\mathbf{t}$ on $\partial\Omega$. Intuitively, this condition states that kinetic friction appears on the boundary. Due to this friction, the system is definitely dissipative. Another choice is the free-shear condition, i.e., $$ \mathbf{n}\cdot\mathbb{T}\cdot\mathbf{t}=0 $$ on $\partial\Omega$ for all tangent vectors $\mathbf{t}$ on $\partial\Omega$. This condition means that $\partial\Omega$ merely provides normal support for the fluid, without any static or kinetic friction in tangential directions. In this case, there is no dissipation on $\partial\Omega$.

Finally, recall the no-slip condition $\mathbf{v}=\mathbf{0}$. This condition also makes the second term vanish. Physically, this means that $\partial\Omega$ provides static friction, as large as needed, so that fluid particles at the boundary always stick to it and are not able to move.

Hope my explanation could be helpful for you.

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  • $\begingroup$ I'm really sorry for the late reply! Your answer has really helped to get a better understanding! $\endgroup$ – Dennis Apr 13 at 12:03

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