Velocity potential ($\mathbf{u}=\nabla \varphi$)

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.

Adding a spatially constant function of time to a velocity potential leaves its velocity $\mathbf{u}=\nabla \varphi$ unchanged. When integrating the Euler equations in space, this freedom can absorb the time-only integration term in the potential form of momentum balance. A spatially constant zero-frequency potential represents no acoustic velocity and must not be counted as a propagating normal mode merely because a frequency determinant vanishes at zero. Pressure conventions and time dependence must still be fixed consistently when the potential is used to recover physical fields.

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 $-2U\sin \theta +\kappa /(2\pi a)$.

For complex potential of a two-dimensional potential flow $w=U(z+a^2/z)-i\Gamma \log z/(2\pi )$, zeros of the complex velocity solve a quadratic. If $0<\gamma <1$, the displayed two roots are distinct and both have modulus $a$; positive circulation places them above the centreline. Their polar angles satisfy $\sin \theta =\gamma$. At the threshold $\gamma =1$ they merge at the top of the cylinder. This follows also from the surface tangential velocity $-2U\sin \theta +\Gamma /(2\pi a)$. The shifted stream function $U(r-a^2/r)\sin \theta -\Gamma \log (r/a)/(2\pi )$ is constant on the cylinder.

For steady two-dimensional inviscid flow of density $\rho$, far-field velocity $\mathbf{U}$, and circulation $\kappa$, the force per unit span is perpendicular to the flow. With counterclockwise circulation taken positive and $\mathbf{U}=U{\mathbf{e}}_x$, it is $-\rho \kappa U{\mathbf{e}}_y$.

For uniform flow $\mathbf{U}$ plus a line vortex of counterclockwise circulation $\Gamma$, the outward advective momentum flux around a large circle is $\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The Bernoulli equation supplies an equal pressure contribution. Their sum is the force on the fluid; the force on an obstacle per unit span is its negative, $\rho \mathbf{U}\times \mathbf{\Gamma }$. Retaining only the advective term loses half the lift and obscures its sign.

For liquid occupying $a(t)<r<b(t)$, radial symmetry and incompressible flow implyThe kinematic boundary condition at the outer interface giveswhich expresses conservation of the liquid's area.

If the pressure at $r=b(t)$ is $p_{\infty }$, the Unsteady Bernoulli equation gives

For $a=a_0(1+ϵ)$, equilibrium outer radius $b_0$, and a bubble gas obeying $p_0\pi a^2=\text{constant}$, the linearization isThe bubble therefore undergoes simple harmonic motion with