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In steady flow, the supplied Euler equations for an inviscid fluid givesTaking its scalar product with $\mathbf{u}$ gives $\mathbf{u}\cdot \nabla H=0$. Along a streamline with tangent $\mathbf{u}$, this is precisely zero derivative of $H$. Hence $H$ is constant along each streamline, with possibly different constants on different streamlines.

If the flow is also irrotational, $\mathbf{\omega }=0$ and thus $\nabla H=0$. Therefore $H$ has one constant value throughout each connected fluid region. The stronger conclusion uses vanishing of the entire gradient, not just its component along the velocity.

The momentum of a material volume $V(t)$ is ${\int }_{V(t)}\rho \mathbf{u}dV$. Newton's second law equates its time derivative to the pressure force and gravity:Apply the given Reynolds transport theorem componentwise to the left side, and the divergence theorem to the pressure term. ThenSince this holds for arbitrary material volumes and the fields are continuous, the integrand vanishes pointwise. Dividing by positive density gives the Euler equations for an inviscid fluidAt a fixed physical time $t_0$, a streamline is a curve tangent to $\mathbf{u}(\cdot ,t_0)$, which may be parametrized by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t_0)$. The chain rule gives $df(\mathbf{x}(s),t_0)/ds=\mathbf{u}\cdot \nabla f=0$, proving constancy along it. The time is held fixed; a time-dependent streamline statement does not by itself assert material conservation along particle trajectories.

For the proposed velocity field, $u_x+w_z=1-1=0$, so incompressibility holds. Its convective acceleration isand both components have zero Laplacian. The Navier-Stokes equation therefore requires $p_x=-x$ and $p_z=-z$. Hencewhere $p_{\ast }$ is an arbitrary constant reference pressure. The same pressure gradient balances the same convective acceleration in the Euler equations for an inviscid fluid. Since the viscous terms vanish identically, this is an exact solution for every positive Reynolds number, as well as for the inviscid limit.

At $z=0$, the normal velocity is zero, and the prescribed dimensionless tangential traction isThis stagnation point flow therefore has zero surface shear stress, although its strain-rate tensor and its viscous normal stresses are nonzero.

Write the density and pressure as ${\rho }_0(z)+{\rho }^{\prime }(x,z,t)$ and $p_0(z)+p^{\prime }(x,z,t)$, respectively. The background satisfies hydrostatic balance, $p_0^{\prime }=-g{\rho }_0$, and ${\rho }_0^{\prime }/{\rho }_0=-\alpha$. Linearizing the Euler equations for an inviscid fluid, incompressibility and advection of density givesDifferentiate the horizontal equation in $x$ and use incompressibility to obtain $p_{xx}^{\prime }={\rho }_0{w}_{zt}$. Differentiate the vertical equation in time and twice in $x$:Dividing by ${\rho }_0>0$ gives the equation for internal waves in an exponentially stratified incompressible fluid,The first-derivative term retains the variation of inertial density with height; no Boussinesq approximation has been made.

For a fluid in an expanding universe, the peculiar-velocity equation is the Euler equations for an inviscid fluid after decomposing physical velocity into Hubble flow and peculiar velocity. It contains the Hubble-drag term $H\mathbf{v}$ and force $-\nabla \Phi /a$ from the peculiar gravitational potential.

For steady flow of constant-density fluid in a uniformly rotating frame with conservative body forces, the displayed quantity is constant along each relative streamline. The centrifugal potential contributes the negative quadratic term, and the Coriolis force contributes no work because it is perpendicular to the relative velocity. The result follows by dotting the rotating Euler equations for an inviscid fluid with that velocity.

Consider steady isentropic flow toward a Newtonian gravitational potential $-GM/r$ through a tube of cross-sectional area $A(r)=C{r}^n$. Write $v>0$ for the inward speed and use a polytropic equation of state with specific-heat ratio $\gamma >1$. Mass conservation and the Euler equations for an inviscid fluid givewhere $c_s$ is the adiabatic sound speed. A regular sonic point therefore has $v_s=c_s$ and $r_s=GM/(n{c}_s^2)$. The Bernoulli equation, matched to a nearly stationary reservoir with sound speed $c_0$, givesFor $n>1/2$, a finite positive sonic point requires $1<\gamma <(2n+1)/(2n-1)$. Differentiating the flow equation at that point, with $x=r_s{v}_s^{\prime }/c_s$, givesIts discriminant is $4n[(2n+1)-(2n-1)\gamma ]$, so the same bound permits real regular slopes. The branch on which the Mach number rises inward selects the negative sign inFor a spherical tube $n=2$, this recovers the $5/3$ threshold of Bondi accretion; for a dipolar flux-tube area, $n=3$ gives $7/5$. The tube approximation must remain valid between the reservoir matching region and the accretor.

For inviscid planar incompressible flow, write velocity as $(-{\psi }_y,{\psi }_x)$ and vorticity as $\zeta ={\nabla }^2\psi$. Taking curl of the Euler equations for an inviscid fluid removes pressure and gives the displayed transport equation, with $J(\psi ,\zeta )={\psi }_x{\zeta }_y-{\psi }_y{\zeta }_x$. The Poisson equation then recovers the streamfunction at each instant. A velocity tangent to the plane cannot stretch its perpendicular vorticity vector.

Circular streamlines and constant vorticity imply $u_{\varphi }=\omega r/2+C/r$. Zero velocity at $2a$ and positive azimuthal velocity $V$ at $a$ giveThe radial Euler equations for an inviscid fluid give $p_r=\rho u_{\varphi }^2/r$, yielding outer-minus-inner pressure $(15-16\log 2)\rho V^2/18$. The pressure difference is independent of circulation orientation.

For time-dependent irrotational flow with $u=\nabla \varphi$, constant mass density, and no body force, the Euler equations for an inviscid fluid integrate toA time-dependent change of gauge in $\varphi$ may set $C(t)=0$.

This is the unsteady potential-flow version of Bernoulli's principle, with a time-dependent potential gauge.