Crocco's Equation Problem

Question

For $2$-D steady, incompressible inviscid flow, subject to a conservative body force $\underline{X} = -\nabla\chi$, show that Crocco's equation reduces to $$-\Omega\underline{\nabla}\psi = \underline{\nabla}H\quad\text{where}\quad \underline{\omega} = \underline{\nabla}\times\underline{u} = \Omega\underline{\hat k},\quad H = \frac{1}{\rho}p + \frac{1}{2}||\underline{u}||^2 + \chi$$ where $\psi$ is the $2$-D stream function, $\underline{u} = \underline{\nabla}\psi\times\underline{\hat k}$. Hence show that $$\frac{\mathrm{d}H}{\mathrm{d}\psi} = -\Omega(\psi)$$

I am truly completely confused about all of this. All I have is that $$\underline{\omega} = \underline{\nabla}\times\underline{u} = \underline{\omega} = \underline{\nabla}\times(\underline{\nabla}\psi\times\underline{\hat k}) = \underline{\nabla}\psi(\underline{\nabla}\cdot\underline{\hat k}) - \underline{\hat k}(\underline{\nabla}\cdot\underline{\nabla}\psi) = \Omega\underline{\hat k}$$ But I don't know what else to dod from here or even if this is right. I struggle with vector calculus so much...

For incompressible flow with a conservative body-force $\underline{X}=-\underline{\nabla}\chi$, the equation of motion (1) reduces to the incompressible form of Crocco's equation. $$\frac{\partial\underline{u}}{\partial t}-\underline{u}\times\underline{\omega}=-\underline{\nabla}\left[\frac{1}{\rho}p+\frac{1}{2}||\underline{u}||^2 + \chi\right]-\nu\underline{\nabla}\times\underline{\omega}$$ where $\underline{\omega} = \underline{\nabla}\times\underline{u}$ is the vorticity vector.

Answer 1

The Crocco's equation is $$ \frac{du}{dt} - u\times\omega = -\nabla H - \nu\nabla\times\omega. $$ Suppose $u = (u_1,u_2,0)$ is the 2D steady, incompressible, inviscid flow field. With $\nu = 0$, $\frac{du}{dt} = 0$ and $\omega = (0,0,\Omega)$, Crocco's equation reduces to $$ -u\times\omega = -\nabla H $$ or $$ u\times\omega = \nabla H. $$ We only need to show that $u\times\omega = -\Omega\nabla\psi$, where $\psi$ is the 2D stream function. First, $$ u\times\omega = \begin{pmatrix} u_1 \\ u_2 \\ 0 \end{pmatrix}\times\begin{pmatrix} 0 \\ 0 \\ \Omega \end{pmatrix} = \begin{pmatrix} \Omega u_2 \\ -\Omega u_1 \\ 0 \end{pmatrix}. $$ From the definition of stream function, we find that $$ u = \nabla\psi\times\hat{k} = \begin{pmatrix} \partial_x\psi \\ \partial_y\psi \\ 0 \end{pmatrix}\times\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \partial_y\psi \\ -\partial_x\psi \\ 0 \end{pmatrix} = \begin{pmatrix} u_1 \\ u_2 \\ 0 \end{pmatrix}. $$ The desired result follows from combining these two computations: $$ \nabla H = u\times\omega = \Omega\begin{pmatrix} -\partial_x\psi \\ -\partial_y\psi \\ 0 \end{pmatrix} = -\Omega\nabla\psi. $$