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irrotational flow has $\nabla \times \mathbf{v}=0$, so locally $\mathbf{v}=\nabla \Phi$; the simply connected fluid domain here permits a globally single-valued velocity potential. Incompressibility gives $\nabla \cdot \mathbf{v}=0$, hence ${\nabla }^2\Phi =0$. In a multiply connected domain, zero local vorticity alone would not rule out circulation and a multivalued potential.

Take $k>0$ so the disturbance decays at depth. Differentiating the given candidate potential yieldsThe $x$ and $z$ second derivatives of $\Phi$ cancel in the Laplacian. At $z=0$, $v_z=h_0\omega \cos (\omega t)\sin (kx)={\partial }_{t}Z$, so the linearized kinematic free-surface condition holds. This uses $k{h}_0\ll 1$; advection of the displaced surface is higher order.

Since ${\cos }^2(kx)+{\sin }^2(kx)=1$, the local squared speed is $h_0^2{\omega }^2{\cos }^2(\omega t)e^{2kz}$. The depth integral is thereforeIt is independent of horizontal position; no further spatial average is needed for this kinetic energy. Replacing the upper boundary $Z$ by zero changes it only beyond quadratic order in amplitude.

Relative to the flat surface, the gravitational energy at each horizontal position changes by $\rho g{\int }_0^{Z}zdz=\rho g{Z}^2/2$. Averaging over one wavelength givesThe energy of a deep-water standing gravity wave, averaged over wavelength, is $T+⟨U{⟩}_x$ and must be constant for a free inviscid standing wave, which requires equal coefficients of the complementary squared sine and cosine:As an independent check, the linearized dynamic boundary condition ${\partial }_t\Phi +gZ=0$ at $z=0$ gives exactly $g-{\omega }^2/k=0$. Infinite depth, negligible surface tension and no dissipative forcing are part of this gravity-wave model.

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.

Vorticity and potential flow. Vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. A small fluid element rotates locally with angular velocity $\mathbf{\omega }/2$ after its deformational part is separated. The vector identitygives $\nabla B=\mathbf{u}\times \mathbf{\omega }$ for steady constant-density inviscid flow. Thus an irrotational region has one common Bernoulli constant on each connected region. In a simply connected irrotational domain, introduce a velocity potential $\mathbf{u}=\nabla \varphi$. Incompressibility then gives Laplace's equation, ${\nabla }^2\varphi =0$. For unsteady irrotational flow,An impermeable stationary wall fixes ${\partial }_n\varphi =0$, but ideal potential flow does not enforce no slip. In a multiply connected domain, irrotational flow may still have nonzero circulation around an excluded vortex core, so a globally single-valued potential requires care.

For example, uniform two-dimensional flow past a circular cylinder has $\varphi =U(r+a^2/r)\cos \theta$ outside $r=a$. Its normal velocity vanishes at the cylinder, and its surface tangential velocity is $-2U\sin \theta$. Bernoulli's equation predicts a symmetric pressure distribution and zero drag. This is D'Alembert's paradox: the assumptions omit the viscous boundary layers, separation and wake responsible for real drag.

For an incompressible fluid of constant kinematic viscosity $\nu$, the curl of the motion equation givesThe first term stretches and tilts vortex lines; the second diffuses vorticity. In two-dimensional incompressible inviscid flow the stretching term is absent and vorticity is carried with the fluid. In a three-dimensional inviscid flow it can be amplified by stretching. Density gradients not parallel to pressure gradients provide an additional source in a general compressible fluid.

Viscous flow. A Newtonian fluid's shear stress is proportional to rate of strain. For constant viscosity and incompressibility the Navier-Stokes equation iswith dynamic viscosity $\eta$ and kinematic viscosity $\nu =\eta /\rho$. At an ordinary solid wall, the no-slip boundary condition sets fluid velocity equal to the wall velocity. Viscosity transfers momentum between neighbouring layers and converts mechanical energy into heat; unlike the pressure force, its shear contribution depends on velocity gradients.

For plane Couette flow between plates $y=0,h$, with the upper plate moving at $U$, the steady solution is $u_x=Uy/h$, with shear stress $\eta U/h$. For pressure-driven flow through a pipe of radius $a$, a constant pressure gradient gives the parabolic Poiseuille profileThe Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares convective inertia with viscous force. Small Reynolds number often permits the Stokes-flow approximation; large Reynolds number can produce thin boundary layers and, under suitable conditions, turbulence. It does not justify neglecting the wall conditions. Bernoulli's inviscid constant is generally not conserved along a viscous streamline, and pipe flow instead shows a pressure loss corresponding to dissipation.

For a cylinder of radius $R_c$ in unbounded two-dimensional incompressible potential flow, a velocity potential is $\phi =u(r+R_c^2/r)\cos \theta$. It givesOn the surface $v_r=0$. The points nearest the banks are the sides perpendicular to the stream, $\theta =\pi /2,3\pi /2$, hence the speed is $2u$ at each. This is the inviscid outer-flow speed; a real no-slip boundary layer has zero velocity at the material surface. The small-radius/wide-river assumption neglects bank effects.

For incompressible planar flow define the streamfunction by $u={\psi }_y$, $v=-{\psi }_x$. A velocity potential satisfies $u={\varphi }_x$, $v={\varphi }_y$ when the flow is irrotational; the differential condition is $v_x-u_y=0$. The complex potential of a two-dimensional potential flow is $w=\varphi +i\psi$, with derivative $w^{\prime }=u-iv$.

For $w=m\log (z-z_0)/(2\pi )$, its derivative is $m/[2\pi (z-z_0)]$. This gives outward radial velocity $m/(2\pi r)$, whose flux across a circle is $2\pi r\cdot m/(2\pi r)=m$. It is therefore the stated line source.

Superposition givesFor positive $U,m$, the stagnation points areThey lie on the real axis, one upstream of the source and the other downstream of the sink. Streamlines are level sets of $\psi =Uy+(m/2\pi )[\arg z-\arg (z-1)]$. The dividing streamline passes through both stagnation points and forms a Rankine oval enclosing the source and sink; outside, streamlines tend to uniform horizontal flow.

For potential flow, each velocity potential satisfies Laplace's equation in its fluid. The exact kinematic boundary condition on $y=\eta$ is ${\eta }_t+{\varphi }_{j,x}{\eta }_x={\varphi }_{j,y}$. Keeping only first-order terms givesThe linearized Bernoulli equation is $p_j=p_0-{\rho }_j({\varphi }_{j,t}+g\eta )$ at the interface, so pressure continuity givesFor a decaying disturbance away from the interface, put $\psi =kx-\omega t$ and takeThey satisfy Laplace's equation and the kinematic boundary conditions for $\eta =A\cos \psi$. Substituting in the dynamic condition gives the dispersion relationIf the heavier fluid is on top, ${\omega }^2<0$. Instead of oscillation, a Fourier disturbance has an exponentially growing component with growth rate $\sqrt{gk({\rho }_2-{\rho }_1)/({\rho }_1+{\rho }_2)}$. This is the Rayleigh-Taylor instability in the absence of surface tension.

Irrotationality permits a velocity potential with $\mathbf{u}=({\varphi }_x,{\varphi }_z)$. Incompressibility gives Laplace's equation in the moving domain $z<\eta (x,t)$. Neglecting surface tension and taking the free-surface pressure as constant atmospheric pressure, the full surface conditions areThe kinematic condition makes the free surface material; the dynamic condition is Bernoulli's equation on that surface. A time-dependent additive gauge in $\varphi$ has absorbed the spatially uniform Bernoulli constant. Deep below the surface, impose $\nabla \varphi \to 0$ as $z\to -\infty$, together with suitable horizontal decay. Initially $\eta (x,0)={\eta }_0(x)$ and the fluid is at rest, so $\nabla \varphi (x,z,0)=0$; its initial constant may be chosen as zero.

The velocity potential gives horizontal and vertical velocity components $({\varphi }_x,{\varphi }_z)$. The kinematic boundary condition at each fixed impermeable wall is zero normal fluid velocity, henceLet $\eta (x,t)$ be the surface elevation. To first order in its amplitude, the kinematic boundary condition for a free-surface graph gives ${\eta }_t={\varphi }_z$ at $z=0$. The dynamic boundary condition for an inviscid interface fixes the surface pressure to atmospheric pressure. The unsteady Bernoulli equation, after choosing the time-dependent gauge of $\varphi$, then gives ${\varphi }_t+g\eta =0$ at $z=0$: the quadratic velocity term is second order in amplitude. Differentiate this last equation in time and eliminate ${\eta }_t$ to obtain the linearized free-surface boundary conditions in combined form,

Uniform irrotational flow of speed $U$ past a circular cylinder of radius $a$, with zero circulation, has velocity potentialOn the cylinder, the tangential speed is $-2U\sin \theta$, so the Bernoulli equation gives

Adding circulation $\kappa$ gives the possibly multivalued velocity potentialThe velocity remains single-valued, and on the cylinder its tangential component is $-2U\sin \theta +\kappa /(2\pi a)$.

In an unbounded incompressible flow with spherical symmetry, the kinematic boundary condition on a sphere of radius $R(t)$ gives $u_r=R^2\dot{R}/r^2$ and velocity potential $\varphi =G(t)-R^2\dot{R}/r$. The Unsteady Bernoulli equation at infinity has $F(t)=\dot{G}+p_{\infty }/\rho$. Subtracting it from the finite-radius equation gives the displayed pressure, independently of $G$. Time differentiation is at fixed $r$. Evaluating at $r=R(t)$ yields the Rayleigh equation for an inviscid spherical bubble when the boundary pressure is prescribed.

A velocity potential obeying the wave equation with constant speed of sound $c_0$ has spherical normal modes $\varphi =e^{-i\omega t}[A\cos kr+B\sin kr]/r$, $k=\omega /c_0$. Zero radial velocity at radii $a,b$ gives Neumann conditions on the potential, not on its numerator. The inner condition is built into $F(r)=C[\cos k(r-a)+(ka{)}^{-1}\sin k(r-a)]/r$. The outer condition gives $k(b-a)\cos k(b-a)-(1+k^{2}ab)\sin k(b-a)=0$, equivalent to the displayed equation for nonzero oscillatory modes. The formal zero root is a constant potential gauge, rather than a nonzero-frequency acoustic wave. Applying a one-dimensional sine condition to the numerator would miss the spherical geometric factors.

A source dipole is the limit of an equal source and sink brought together with their source-strength times separation vector fixed at $\mathbf{M}$. Differentiating the three-dimensional point source field gives $\mathbf{u}=-(\mathbf{M}\cdot \nabla )[\mathbf{r}/(4\pi r^3)]$. It has velocity potential $-\mathbf{M}\cdot \mathbf{r}/(4\pi r^3)$, zero total volume flux, and $r^{-3}$ far-field decay. Unlike a stresslet, it is an irrotational potential singularity rather than a force-dipole flow.

For an axisymmetric flow without azimuthal velocity, the spherical Stokes streamfunction gives $u_r={\Psi }_{\theta }/(r^2\sin \theta )$ and $u_{\theta }=-{\Psi }_r/(r\sin \theta )$. These formulas enforce incompressibility. The only possibly nonzero component of vorticity isThus the displayed equation is precisely the condition for irrotational flow, away from coordinate poles, with limits interpreted through regularity of the physical velocity field. It differs from the scalar spherical Laplace equation: $\Psi$ is a Stokes streamfunction, not a velocity potential.

Consider fluid at rest with mean depth $d$ and a bottom displacement $a\sin (k(x-ct))$. The linearized free-surface boundary conditions at $y=0$ and moving-bottom kinematic boundary condition at $y=-d$ give a periodic particular response $\eta =b\sin (k(x-ct))$ and velocity potentialThe bottom derivative fixes the signed response coefficient to the title formula. Here $k>0$, $c\ne 0$, and the forcing must avoid $k^2{c}^2=gk\tanh (kd)$. This equality is resonance with the natural finite-depth surface gravity wave; nonzero undamped resonant forcing has no bounded periodic solution. The response reverses phase below resonance. General initial data can add free surface gravity waves, which do not decay in this inviscid model. Near resonance the resulting wave steepness and particle excursions, rather than the bottom displacement alone, determine whether the linearization remains valid.

For constant $P=w\rho$ and velocity potential $v_i=i{k}_i\theta$, stress-energy conservation gives the displayed density equation in synchronous gauge in cosmology. Here $h$ is the trace of the spatial metric perturbation, and primes denote conformal time. In particular cold baryons obey ${\delta }_b^{\prime }=k^2{\theta }_b-h^{\prime }/2$. A positive sign in front of $k^2{\theta }_b$ on the left contradicts this potential convention.

A point injection of volume at rate $Q$ produces the displayed radial potential flow in three dimensions. Its velocity potential is $-Q/(4\pi r)$, and every enclosing sphere has outward volume flux $Q$. The divergence vanishes away from the source and is $Q$ times the three-dimensional Dirac delta distribution at it. The field is also a Stokes flow with constant pressure away from its singularity.

For an inviscid flow of constant mass density $\rho$ and depth $h$, impose excess pressure $p_1\cos (k(x-ct))$ with $k>0$ on a free surface. Neglect surface tension. A travelling solution of the linearized free-surface boundary conditions isThe velocity potential satisfies the Laplace equation and the bed no-penetration boundary condition. Its surface time derivative is $-A{c}^{2}k\coth (kh)\cos (k(x-ct))$, so the dynamic condition gives the displayed signed response. The nonnegative amplitude is $|A|$. It is finite except atthe squared phase speed of a free surface gravity wave at that wavenumber. Equality produces resonance. This describes a travelling forced component; free surface gravity waves must be added to meet general initial conditions. At exact resonance, a solution of the initial value problem generally grows with time rather than having a bounded periodic amplitude.

A uniform flow has the same velocity at every position. A constant velocity field has zero divergence and zero vorticity; a velocity potential is $\varphi (\mathbf{x})=\mathbf{U}\cdot \mathbf{x}$. It describes the remote ambient flow in idealized obstacle and line vortex problems.

For a fixed impermeable sphere of radius $a$ and incident uniform flow $U{\mathbf{e}}_z$, the velocity potential is the displayed expression and the Stokes streamfunction is $\Psi =(U/2)(r^2-a^3/r){\sin }^2\theta$. Differentiation gives $u_r=U(1-a^3/r^3)\cos \theta$ and $u_{\theta }=-U(1+a^3/(2r^3))\sin \theta$. Thus the no-penetration boundary condition holds, the poles are stagnation points, and the equatorial surface speed is $3U/2$. The disturbance is a potential dipole decaying as $r^{-3}$ in velocity. The sphere theorem for a Stokes streamfunction constructs this solution from ${\Psi }_0=U{r}^2{\sin }^2\theta /2$.

Use a velocity potential to reconstruct an irrotational flow from one scalar field. A velocity potential $\varphi (\mathbf{x},t)$ satisfiesIt has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is unchanged as a description of the velocity field when any function of time alone is added. For twice continuously differentiable $\varphi$, the identity curl of a gradient proves zero vorticity. In incompressible flow, taking the divergence also gives ${\nabla }^2\varphi =0$, the Laplace equation. This turns a three-component velocity field into a single harmonic function with appropriate boundary conditions. ^[ref]

For uniform flow $\mathbf{u}=(U,0,0)$, take $\varphi =Ux$. With $U=2\mathrm{m}{\mathrm{s}}^{-1}$, the change from $x=0$ to $x=3\mathrm{m}$ is $\Delta \varphi =6{\mathrm{m}}^2{\mathrm{s}}^{-1}$. By the fundamental theorem for line integrals, the line integral of velocity between these points equals that change along any path in the domain; a closed curve gives zero circulation.

For the potential line vortex, choose a local angle branch in polar coordinates and set $\varphi =K\theta$. Then $u_r={\partial }_r\varphi =0$ and $u_{\theta }=r^{-1}{\partial }_{\theta }\varphi =K/r$. One counterclockwise turn changes the continued velocity potential by $2\pi K$, although the velocity field returns to its original value. With $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$, this change is $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$, exactly the circulation. A full annulus therefore admits local angle branches and a multivalued velocity potential, but no global single-valued velocity potential when $K\ne 0$. Cutting the annulus along a radial line allows a single-valued angle branch on the remaining domain. The obstruction is path independence, not a failure of the local gradient formula.