I was reading a paper about the Navier-Stokes equation in $\mathbb{R}^2$ by Kato and at the beginning he affirmed that:

$$ ∇ × ({ \mathbb {P}}(u\cdot \nabla u) )=\nabla \cdot (\omega S\ast\omega)$$

where $u=u(t,x)$ is the velocity field ($t\in[0,\infty), x\in\mathbb{R}^2$), $ ∇$ is the gradient, $ ∇ × $ denotes the curl, $\omega$ is the vorticity (i.e. $\omega=∇ × u$) , ${ \mathbb {P}}$ is the Leray projection, $\cdot$ is the dot product, $\ast$ denotes the convolution, and


Is it easy to prove? Is there any book in which I can find something related to it?

  • $\begingroup$ So $u$, at a fixed time, is defined as a vector field over $\mathbb{R}^2$? $\endgroup$ – Hamed Jul 12 '18 at 14:25
  • $\begingroup$ Exactly @Hamed (sorry not to mentioned it before ) $\endgroup$ – Pires Dankan Jul 12 '18 at 22:26

So that I can use some 3D identities here, let $\mathbf u=(u_1,u_2,0)=(u;0)$ be a 3D velocity that does not depend on $x_3$, $$\mathbf u(x_1,x_2,x_3) = (u_1(x_1,x_2), u_2(x_1,x_2),0).$$ Define $\mathbf w:=\nabla\times \mathbf u = (0,0,\omega)$. In this answer, I reserve the notation $\nabla\times$ for the vector valued 3D curl, and the notation $\nabla^\perp\cdot:=(-\partial_2,\partial_1)\cdot$ for the scalar 2D curl. The choice of $\nabla = (\partial_1,\partial_2,\partial_3)$ or $(\partial_1,\partial_2)$ is clear from context.

Starting from the Leray projector $\mathbb P v = v + \nabla (-\Delta)^{-1} (\nabla \cdot v)$ we can write

$$\mathbb P(u\cdot \nabla u) = u\cdot \nabla u +\nabla (-\Delta)^{-1} (\nabla \cdot (u\cdot \nabla u))$$

And we want to take the curl of it. Well, the first good news is that curl annihilates gradients, so none of that messy second term for us here. Since we are trying to rewrite this in terms of the vorticity, the following identity from that Wikipedia link is useful- $$\mathbf u\cdot \nabla\mathbf u = \frac12 \nabla |\mathbf u|^2 - \mathbf u\times \mathbf w $$

Notice that $$-\mathbf u\times \mathbf w = \omega(- u_2,u_1,0) =: \omega (u^\perp;0).$$ Here a vector $U\in \mathbb R^2$ has its perpendicular $U^\perp := (-U_2,U_1)$. So, expressed in terms of $u$, we have,

$$u\cdot\nabla u = \frac12\nabla|u|^2 + \omega u^\perp \tag{*}\label{*} $$

We are fortunate again that $\nabla^\perp\cdot \nabla = 0$, so that on taking the 2D curl, $$\nabla^\perp \cdot (u\cdot \nabla u) = \nabla\times(\omega u^\perp) = \nabla^\perp \cdot( \omega u^\perp) = \nabla\cdot(\omega u)$$

So we have proved: $\nabla^\perp\cdot \mathbb P(u\cdot \nabla u) = \nabla\cdot(\omega u)$. To conclude, recall the Biot-Savart law in 2D, which allows us to recover the velocity from the vorticity with a convolution integral.

A quick remark to see how you get $S$: Recall that $$-\nabla^\perp (\nabla^\perp\cdot U) = - \Delta U + \nabla (\nabla\cdot U) \tag{**}\label{**}. $$ For divergence free fields $u$, this says $u= -(-\Delta)^{-1}(\nabla^\perp \omega)$. The kernel $S$ is obtained by (carefully) transferring the derivative (carefully) $\nabla^\perp$ onto the Green's function via integration by parts.

PS - this book of Majda & Bertozzi is a great book.

\eqref{*} Alternatively, one can avoid talking about 3D vectors by checking this identity directly, which follows from $$ u\cdot\nabla u = \begin{pmatrix}\color{red}{u_1\partial_1 u_1} + u_2\partial_2 u_1\\ u_1\partial_1u_2 + \color{red}{u_2\partial_2 u_2} \end{pmatrix}, \quad \frac12\nabla |u|^2 = \begin{pmatrix}\color{red}{u_1\partial_1 u_1} + u_2\partial_1 u_2\\ u_1\partial_2u_1 + \color{red}{u_2\partial_2 u_2} \end{pmatrix}$$

\eqref{**} Indeed,

$$ \nabla (\nabla\cdot U) = \begin{pmatrix} \partial_1\partial_1 U_1 + \partial_1\partial_2 U_2 \\ \partial_2\partial_1 U_1 + \partial_2\partial_2 U_2 \end{pmatrix}$$ and by replacing $(\partial_1,\partial_2)$ with $(-\partial_2,\partial_1)$, $$ \nabla^\perp (\nabla^\perp \cdot U) = \begin{pmatrix} \partial_2\partial_2 U_1 - \partial_1\partial_2 U_2 \\ -\partial_2\partial_1 U_1 + \partial_1\partial_1 U_2 \end{pmatrix}$$ Alternatively, one can use $\nabla\times(\nabla\times \mathbf U) = -\Delta \mathbf U + \nabla(\nabla\cdot \mathbf U)$ with $\mathbf U(x_1,x_2,x_3) = (U(x_1,x_2);0)$.

  • $\begingroup$ thanks a lot @calvin!! You have no idea how much you helped me right now $\endgroup$ – Pires Dankan Jul 13 '18 at 12:49
  • $\begingroup$ @PiresDankan I was unhappy with the abuse of notation so I tried to clean it up. Also I made the answer self contained with an direct proof of the 2D identity without using the 3D identities. You're welcome :) $\endgroup$ – Calvin Khor Jul 13 '18 at 12:57
  • $\begingroup$ youarebackkkkkkkkk $\endgroup$ – Chee Han Jul 13 '18 at 22:07

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