I'm trying to derive the incompressible Navier-Stokes equation in Fourier space, given in e.g. Kraichnan & Montgomery (1980), Rep. Prog. Phys. 43 547 (PDF) (but probably countless other places), as $$ (\partial_t + \nu k^2)u_i(\mathbf{k}) = -\mathrm{i}\left(\delta_{ij}- \frac{k_i k_j}{k^2}\right) k_m \sum_{\mathbf{k}=\mathbf{k}_1+\mathbf{k}_2} u_j(\mathbf{k}_1) u_m(\mathbf{k}_2) \tag{0}\label{0} $$ where roman indices are tensor components, repeated indices are summed, $\mathbf{k}_1$ and $\mathbf{k}_2$ are dummy wavevectors in the sum over resonant wave triads, and I have omitted hats on the Fourier variables (i.e. $u_i(\mathbf{k}) = \hat{u}_i(\mathbf{k})= \int u_i(\mathbf{x})e^{-\mathrm{i}\mathbf{k}\cdot\mathbf{x}}d^dx$, modulo some normalising constants).

The left-hand side of $\eqref{0}$ is easy enough but I'm having trouble making the right-hand side nice and symmetric:

Navier-Stokes (with density normalised so that $\rho=1$) is $$ \partial_t\mathbf{u} + (\mathbf{u}\cdot\nabla)\mathbf{u}=-\nabla p + \nu\nabla^2\mathbf{u} \tag{1}\label{1} $$ and incompressibility ($\nabla\cdot\mathbf{u}=0$) gives for the pressure $$ \nabla^2p = -\nabla\cdot[(\mathbf{u}\cdot\nabla)\mathbf{u}] .\tag{2}\label{2} $$

I put $\eqref{2}$ in index notation and write $p, \mathbf{u}$ in Fourier series, e.g. $u_i(\mathbf{x})=\sum_\mathbf{k'}u_i(\mathbf{k'})e^{\mathrm{i}\mathbf{k'}\cdot\mathbf{x}}$. I then multiply by $e^{-\mathrm{i}\mathbf{k}\cdot\mathbf{x}}$, integrate over space and use $\int e^{\mathrm{i}(\mathbf{k'}-\mathbf{k})\cdot\mathbf{x}}d^dx=\delta_{\mathbf{k'}\mathbf{k}}$(modulo constants) to get $$ p(\mathbf{k})=-\frac{1}{k^2}k_j \sum_{\mathbf{k}=\mathbf{k}_1+\mathbf{k}_2} k_{1m} u_j(\mathbf{k}_1)u_m(\mathbf{k}_2). \tag{3}\label{3} $$ The $k_j$ on the RHS of $\eqref{3}$ comes from the outer del $\nabla\cdot[\ldots]$ on the RHS of $\eqref{2}$, and from the $j$-th component of the resonance condition $\mathbf{k}=\mathbf{k}_1+\mathbf{k}_2$.

Now I write $\eqref{1}$ in index notation and put the variables in Fourier series, using $\eqref{3}$ to eliminate the pressure. Using all the same tricks I get $$ (\partial_t + \nu k^2)u_i(\mathbf{k}) = -\mathrm{i}\left(\delta_{ij}- \frac{k_i k_j}{k^2}\right) \sum_{\mathbf{k}=\mathbf{k}_1+\mathbf{k}_2} k_{1m} u_j(\mathbf{k}_1) u_m(\mathbf{k}_2) \tag{4}\label{4}. $$ (Again, roman indices are components, numeric indices represent dummy wavevectors in the sum over wave triads).

The RHS of my eq. $\eqref{4}$ is similar to, but not the same as the desired RHS $\eqref{0}$. Specifically I just need to show that $$ P_{ij} \sum_{\mathbf{k}=\mathbf{k}_1+\mathbf{k}_2} k_{1m} u_j(\mathbf{k}_1) u_m(\mathbf{k}_2) = P_{ij}k_m\sum_{\mathbf{k} = \mathbf{k}_1+\mathbf{k}_2} u_j(\mathbf{k}_1) u_m(\mathbf{k}_2) $$ where $P_{ij}=\left(\delta_{ij}- \frac{k_i k_j}{k^2}\right)$. So I need to turn the $k_{1m}$ inside the sum into a $k_m=k_{1m}+k_{2m}$ that can be brought outside the sum. Now I believe the trick is to do some index manipulation with the dummy indices $j,m$ and the dummy wavevectors $\mathbf{k}_1,\mathbf{k}_2$. But I don't see it.

Any advice?


1 Answer 1


Worked it out. For the benefit of the next person who struggles with it, the answer has nothing to do with index trickery (and $P_{ij}$ is a red herring in the final equation in my question).

The trick is to use incompressibility again: $0=\nabla\cdot\mathbf{u}(\mathbf{x})=\partial_m u_m(\mathbf{x})=\sum_{\mathbf{k'}} \mathrm{i}k'_mu_m(\mathbf{k'})e^{\mathrm{i}\mathbf{k'}\cdot\mathbf{x}}$.

In particular for mode $\mathbf{k}_2$ this reads $k_{2m}u_m(\mathbf{k}_2)=0$ and so $-\mathrm{i}P_{ij}k_{2m}u_m(\mathbf{k}_2)u_j(\mathbf{k}_1)=0$ for any $\mathbf{k}_1,\mathbf{k}_2$. Choosing all $\mathbf{k}_1,\mathbf{k}_2$ that are in resonance triads with $\mathbf{k}$ and adding all such terms (each of which are identically zero) to eq. (4) the RHS becomes $$ -\mathrm{i}P_{ij}\sum_{\mathbf{k} = \mathbf{k}_1+\mathbf{k}_2} (k_{1m}+k_{2m})u_j(\mathbf{k}_1) u_m(\mathbf{k}_2) $$ which is exactly what we want.

  • $\begingroup$ Brilliant. I think a shorcut is to realize you may substitute $k_{1m} \to k_{1m} + k_{2m}$ , because the sum over the new term is zero, due to incompressibility. You may actually write $k_{1m} + 99999 k_{2m}$ , it would still be right, but you want just the sum since then you can just write $k_m$, and pull that $k_m$ from the summation. $\endgroup$ Commented Jan 10, 2022 at 12:56

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