= Solution
For a material fluid volume, <mass conservation> yields $\rho_t+\nabla\cdot(\rho\mathbf u)=0$. Constant <mass density> reduces this to $\nabla\cdot\mathbf u=0$. For <inviscid flow>, the <Cauchy stress tensor> is $-pI$; its net surface <force> is the volume integral of $-\nabla p$ by the <divergence theorem>. <Newton's second law> for each material volume therefore gives $\rho D\mathbf u/Dt=-\nabla p$. Since $D/Dt=\partial_t+\mathbf u\cdot\nabla$, these are exactly the two stated equations.
In two dimensions they read
$$
u_t+uu_x+vu_y=-p_x/\rho,\qquad v_t+uv_x+vv_y=-p_y/\rho,\qquad u_x+v_y=0.
$$
Differentiate the second momentum equation in $x$ and subtract the $y$ derivative of the first. The <pressure> terms cancel. For $\omega=v_x-u_y$, collecting terms gives
$$
\omega_t+u\omega_x+v\omega_y+(u_x+v_y)\omega=0.
$$
<Incompressibility> removes the last term, giving
$$
\boxed{\frac{D\omega}{Dt}=0.}
$$
Thus \b[each fluid parcel retains its scalar <vorticity>] as it moves. In particular an initially <irrotational flow> remains irrotational in this smooth two-dimensional inviscid evolution. There is no <vortex stretching> in this two-dimensional setting.
Back to article page
