= Solution
For a <material volume>, <conservation of mass> and the <Reynolds transport theorem> give $\rho_t+\nabla\cdot(\rho u)=0$. Constant material <density>, or equivalently <incompressibility> with the stated constant initial <density>, reduces this to $\nabla\cdot u=0$.
In an fluid undergoing <inviscid flow> the stress is the isotropic <pressure> stress $-pI$. <Momentum conservation> for a <material volume> then gives
$$
\frac d{dt}\int_{V(t)}\rho u\,dV=-\int_{\partial V(t)}pn\,dS+\int_{V(t)}\rho g\,dV.
$$
Transport and the <divergence theorem> localize this to
$$
\boxed{\rho Du/Dt=-\nabla p+\rho g,\qquad D/Dt=\partial_t+u\cdot\nabla.}
$$
For a steady flow, $u_t=0$. For an <irrotational flow> the <vorticity> is zero, so the supplied identity gives $(u\cdot\nabla)u=\nabla(|u|^2/2)$. Uniform downward gravity is $g=-\nabla(gz)$, and constant <density> permits pulling $1/\rho$ inside the <pressure> gradient. Therefore
$$
\boxed{\nabla\left(p/\rho+|u|^2/2+gz\right)=0.}
$$
The <Bernoulli invariant> is constant throughout each <connected> fluid region. Without irrotationality the corresponding steady statement generally gives a constant only along each <streamline>.
Back to article page
