 Solution

The intended dynamical assumption is steady inviscid Euler flow of constant density, with conservative body-force potential $\Phi$. Its equation is $(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla (p/\rho +\Phi )$. The vector identitythen givesWith no body force take $\Phi =0$. Taking the scalar product with $\mathbf{u}$ yields $\mathbf{u}\cdot \nabla H=0$, so the Bernoulli function is constant along streamlines, though not necessarily the same constant on different streamlines.

Choose the planar streamfunction convention $\mathbf{u}=({\psi }_y,-{\psi }_x,0)$. Then $\mathbf{\omega }=(0,0,\omega )$ with $\omega =-{\nabla }^2\psi$, andIf locally $H=H(\psi )$, comparison with $\nabla H=H^{\prime }(\psi )\nabla \psi$ giveson the nonstagnant part of that flow. It extends to isolated stagnation points by continuity. Where the streamfunction is constant on an entire open region, cancellation of its zero gradient alone imposes no derivative condition on an arbitrarily extended $H(\psi )$.

Steadiness and incompressibility alone, without the inviscid dynamical assumption, do not imply the stated identity. In a steady Newtonian viscous flow the correct form has $\mathbf{u}\times \mathbf{\omega }=\nabla H-\nu {\nabla }^2\mathbf{u}$. For example, a pressure-driven planar Poiseuille flow has a nonzero streamwise pressure gradient balanced by viscosity, while $\mathbf{u}\times \mathbf{\omega }$ has no streamwise component. This identifies the required qualification of the printed premise.