In what sense is $\int (u \cdot \nabla) u \cdot u dx$ an energy flux?

Question

Due to the nature of this question I have have cross-listed it on mathSE.

Let $u$ be either a solution to either the Euler equations or Navier-Stokes equations over a domain $\Omega$. In fluid dynamics it is common to define $\Pi$ as the total energy flux where $$\Pi = \int_\Omega (u \cdot \nabla) u \cdot u dx.$$

This definition is sometimes specified even further as: $$\Pi_q = -\int_{\Omega} (u \otimes u)_{\leq q} : \nabla u_{\leq q} dx$$ where $\Pi_q$ is the Littlewood-Paley energy flux through the wave number $\lambda_q$.

I have always thought of the nonlinear terms in fluid equations as an advection term, i.e. it transports the fluid. However from the above definitions for $\Pi$ or $\Pi_q$ it seems to also have an interpretation of energy flux from one Fourier mode to smaller ones. How can one see this interpretation from the definition of $\Pi$ or $\Pi_q$?

Answer 1

The kinetic energy density of a fluid with density $\rho$ is $T = \frac12 \rho \mathbf u\cdot\mathbf u$. However for an incompressible and homogeneous fluid, the density is constant in both space and time, so in the absense of a non-gravitational force the density is just an inconsequential factor and can be removed from the definition: $T = \frac12\mathbf u\cdot\mathbf u$. The kinetic energy flux density can be defined as $\mathbf J = T\mathbf u = (\frac12\mathbf u\cdot\mathbf u)\mathbf u$ and the total flux out of a closed surface $\partial\Omega$ is, $$\Pi = \oint_{\partial\Omega}\mathbf J\cdot\mathrm d\mathbf S$$ Using divergence theorem, $$\Pi = \int_\Omega \nabla\cdot\mathbf J\mathrm dV$$ Using product rule to expand the divergence, $$\nabla\cdot\mathbf J = \nabla\cdot(T\mathbf u) = (\mathbf u\cdot\nabla)T + T\nabla\cdot\mathbf u$$ For incompressible fluid, $\nabla\cdot\mathbf u = 0$ so the 2nd term goes away. Expanding $T$, $$\nabla\cdot\mathbf J = (\mathbf u\cdot\nabla)(\frac12\mathbf u\cdot\mathbf u) = [(\mathbf u\cdot\nabla)\mathbf u]\cdot\mathbf u$$ So the flux can be written as: $$\Pi = \int_\Omega[(\mathbf u\cdot\nabla)\mathbf u]\cdot\mathbf u\mathrm dV$$

Answer 2

The transfer from long wavelength (macroscopic, localizable mechanical wave objects) to short wavelengths of arbitrary phase correlations (noise) is accomplished by nonlinear terms by their representations via the Fourier convolution theorem:

$$\mathit F(x\mapsto v^2(x))(k)= \mathit F(x\mapsto \mathit F^{-1}(v)(k')^2(x))(k) = \mathit F\left(( \mathit F^{-1}(v)\star\mathit F^{-1}(v)(k) \right) $$

with the convolution integral

$$(f\star g)(k) = \int_{-\infty}^\infty f(u)g(k-u) du $$

By the fact, that products, more generally all powers of exponentials in wave numbers evaluate to exponentials of sums of wave numbers, the time evolution with a nonlinear term in the wave function generates waves of multiples of the wave numbers of the long waves, that typically constitute the simple start conditions at $t=0$ in an experiment in a box.

By energy conservation for short times a medium with low inner friction, the energy is distributed over modes of smaller wavelengths.

The term 'noise' for this long time evolution of 'choppy water' is not quite adequate, because the wave mode energy diffusion without friction is a classical point path in the Fourier mode representation and, without energy dissipation e.g. by friction is entropy conserving.