So I am working through Peter Olver's book Application of Lie Groups to Differential Equations and there is this example on page 445 with the Euler Equations for inviscid ideal fluid flow in two dimensions that I am having trouble with. I am mainly having trouble with calculating the vorticity equation and then using that to compute the total derivative. To start the Euler Equation with 3 independent spacial variables $\bf{x}$ $ = (x,y,z)$ and time-variable $t$ and 3 dependent variables $\bf u$ $ = (u,v,w)$ is
they also use curl to determine the total derivative of our functional (which is the energy functional in this case), i.e. we want to determine the $\psi$ this way, $\psi$ is also the vector stream function
Now I tried working through the example in two dimensions and I am just not getting what they are getting for the vorticity equation or $\psi$
first I am confused about $7.21$, because the formula should be $\frac{d \omega}{dt} = \omega \cdot \nabla \bf{u} – \bf{u} \cdot \nabla \omega$, now it seems that the term $\omega \cdot \nabla \bf{u}$ is 0, but I though that term would be $(v_x-u_y)\cdot (u_x +v_y) = u_xv_x -u_yu_x – u_yv_y +v_xv_y $
giving me
$$\frac{d \omega}{dt} = (v_x-u_y)\cdot (u_x +v_y) – (u,v) \cdot (w_x,w_y) = u_xv_x -u_yu_x – u_yv_y +v_xv_y – uw_x – vw_y $$. Unless I am understandsing this wrong somehow…
And then second with respect to finding $\psi$, I am confused because we are looking for $\psi$ such that $$\nabla \times \psi = \bf{u}$$
this confuses me because they are trying to find a function $\psi$ whose curl is a 2-dim vector.. and I am confused because I can't seem to figure out how they calculated that from my knowledge of vectors and curl. I thought you can only take the curl of a vector of 2-dim and higher, and I've never done the curl and gotten a 2 dim vector, because this tells us that $\nabla \times \psi = (-\psi_y, \psi_x) = (u,v) = \bf u$, which I am confused about.
Best Answer
It's a bit of a cheat of notation. You can think of the "two-dimensional" vector $\pmb{u}$ as being given by $\pmb{u}=(u,v,0)$ (i.e. a three-dimensional vector which is zero in one component). You should then find that \begin{align*} \pmb{\omega}\,=\,\nabla\times\pmb{u} \,=\, (\partial_{x},\partial_{y},\partial_{z})\times(u,v,0) \,=\, (0,0,\omega) \qquad\text{where}\qquad \omega\,=\, v_{x}-u_{y} \end{align*} by recalling $u\equiv u(x,y)$, $v\equiv v(x,y)$ and the notation $v_{x}=\partial_{x}v$ and $u_{y}=\partial_{y}u$ is used. As for the "stream function", notice it is originally a vector (confusing right?) so we can define $\pmb{\psi}=-\psi\,\pmb{e}_{z}$ in terms of the stream function $\psi$ (the minus sign is to match Olver's conventions) and so $\nabla\times\pmb{\psi}$ can now be calculated to give what Olver has. You can equivalently think of a two-dimensional operator $\nabla^{\perp}\equiv(-\partial_{y},\partial_{x})$ with $\pmb{u}=\nabla^{\perp}\psi$ being a 2D analogue of the 3D curl.
As for the $\pmb{\omega}\cdot \nabla\pmb{u}$, I think your problems might again be notational: I would always write it $(\pmb{\omega}\cdot\nabla)\pmb{u}$ so that you see acting on the vector $\pmb{u}$ is the differential operator $\pmb{\omega}\cdot\nabla=(0,0,\omega)\cdot(\partial_{x},\partial_{y},\partial_{z})=\omega\partial_{z}$ for the case in question (once you work out $\pmb{\omega}$). It should be clear why this gives $(\pmb{\omega}\cdot\nabla)\pmb{u}=0$. A similar calculation can be done for the other term.
Hope some of this helps. Feel free to ask follow-up questions!