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For a smooth constant-density exterior potential flow satisfying the finite-energy condition for planar exterior potential flow, use $\mathbf{u}=\nabla \varphi$ and the planar streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Since ${\nabla }^2\varphi =0$, $\nabla \cdot (\varphi \nabla \varphi )=|\mathbf{u}{|}^2$. Apply the divergence theorem to a truncated exterior region. The outer term vanishes; on a counterclockwise body boundary the body's outward normal satisfies ${\partial }_n\varphi ds=d\psi$, while the fluid's outward normal points the opposite way. This gives the displayed kinetic energy per unit span. The formula yields the added mass of a circular cylinder directly from its surface values of $\varphi$ and $\psi$.

In planar potential flow, encode a physical force $(F_x,F_y)$ by $F_x-i{F}_y$, using the same negative-imaginary-component convention as the complex velocity $u-iv$. The Blasius force theorem then expresses the force per unit span as a single contour integral. The encoding is a convention: the physical vertical component is the negative imaginary part of this complex number.

For an incompressible irrotational two-dimensional potential flow, an analytic complex potential of a two-dimensional potential flow combines the velocity potential $\Phi$ and streamfunction $\Psi$. The Cauchy-Riemann equations give $u={\Phi }_x={\Psi }_y$ and $v={\Phi }_y=-{\Psi }_x$, hence $W^{\prime }(z)=u-iv$. Curves of constant $\Psi$ are streamlines. An analytic coordinate change $z=f(\zeta )$ transforms the complex velocity by $dW/dz=(dW/d\zeta )/f^{\prime }(\zeta )$ wherever $f^{\prime }\ne 0$.

For a clean bubble of size $a$ and speed $U$ at large Reynolds number, the outer potential flow violates the stress-free condition by a strain of order $U/a$. A boundary layer of thickness $\delta \sim a{\mathrm{Re}}^{-1/2}$ needs a velocity correction only of order $U\delta /a$, so its dissipation is smaller than the outer-flow dissipation by order $\delta /a$. The leading drag power is therefore obtained from

An equipotential curve is a planar level curve of a scalar potential. Where the gradient is nonzero, its tangent is perpendicular to that gradient. In potential flow, it is therefore perpendicular to the velocity field; this distinguishes equipotential curves from streamlines.

For a planar potential flow outside a bounded body, suppose its single-valued complex velocity $q=u-iv$ is holomorphic near infinity and tends to zero. Its exterior Laurent series begins $q=b_1/z+O(z^{-2})$. The outward volume flux $Q$ and counterclockwise circulation $\Gamma$ satisfy $\Gamma +iQ=2\pi i{b}_1$. Thus both vanish precisely when the leading term is absent. A nonzero $b_1$ gives $|q{|}^2\sim |b_1{|}^2/r^2$ and logarithmically divergent exterior kinetic energy. When $b_1=0$, $q=O(r^{-2})$ and the far-field energy integral converges. Zero circulation alone does not exclude the radial source field $\nabla \log r$.

In a planar potential flow, superpose a positive-$x$ uniform flow of speed $U$ and a line source of strength $Q=2\pi m$ at the real point $b>1$. Introduce an impermeable unit circular cylinder with zero circulation. The Milne-Thomson circle theorem gives the complex potential of a two-dimensional potential flowThe displayed horizontal force per unit length acts on the cylinder, and $F_y=0$ by reflection symmetry. To prove it, let $q=w^{\prime }$ be the complex velocity. At infinity, $q=U+m/z+O(z^{-2})$, so the sum of all finite residues of $q^2$ is $2Um$. At the external pole $b$, write $q=m/(z-b)+H(z)$; thenThe residue theorem and the counterclockwise Blasius force theorem give $F_x-i{F}_y=2\pi \rho m[H(b)-U]$, proving the formula. With $m>0$ and $U>0$, its sign changes at $b=(m+\sqrt{m^2+4U^2})/(2U)$.

A free streamline is a streamline that forms an unknown boundary of a fluid region, typically against a region of prescribed pressure. In steady inviscid potential flow with gravity and surface tension neglected, the Bernoulli equation makes its speed constant when its pressure is constant. It is simultaneously a level set of the stream function. These two boundary conditions take particularly simple forms in the complex potential of a two-dimensional potential flow and the hodograph plane.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

The hodograph plane records velocity rather than position. For a two-dimensional potential flow, the complex coordinate $q=u-iv$ is the complex velocity. A straight impermeable wall fixes the velocity direction and hence maps to a radial line in this plane. A constant-speed free streamline maps to a circle centered at zero. Thus a free-boundary problem in physical space can become a fixed geometric domain in velocity space. Recovering the physical map then uses $dz/dw=1/q(w)$, wherever $q$ is nonzero.

For $a>0$, this conformal map takes $\mathrm{Re}z>0$, $0<\mathrm{Im}z<a$ bijectively onto the upper half-plane. Its imaginary part is $\sinh (\pi x/a)\sin (\pi y/a)>0$, and its derivative is nonzero in the interior. The end wall maps to $[-1,1]$ and the two long walls to the complementary real rays. To prove the map is onto and injective, first put $s=e^{\pi z/a}$, which maps onto $|s|>1$, $\mathrm{Im}s>0$. The equation $s+s^{-1}=2\zeta$ has exactly one root outside the unit circle when $\mathrm{Im}\zeta >0$: the roots have product one and neither has unit modulus. The identity $\mathrm{Im}\zeta =(1-|s{|}^{-2})\mathrm{Im}s/2$ puts that root in the correct half-plane. This permits planar potential flow problems in a closed-end channel to be solved by the method of images.

Use irrotational flow to identify where a velocity potential can describe the velocity field locally. An irrotational flow has $\mathbf{\omega }=\nabla \times \mathbf{u}=0$ throughout the fluid region. For a continuously differentiable velocity field, the Poincaré lemma supplies a velocity potential on sufficiently small neighborhoods. A global single-valued velocity potential requires zero circulation around every closed curve; a simply connected domain is a sufficient domain assumption.

The potential line vortex $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$ is an irrotational flow on the punctured plane, or on an annulus $a<r<b$ with $a>0$, even though its pathlines are circular. Every sufficiently small loop bounding a disk in that fluid has zero circulation, by Stokes theorem. A loop winding once counterclockwise around the excluded origin has circulation $2\pi K$. Thus local irrotational flow and nonzero circulation around a hole are compatible: the enclosing disk passes outside the domain of the velocity field.

An irrotational flow is a kinematic condition on a velocity field, independent of whether a particular model neglects viscosity or assumes incompressible flow. Adding incompressible flow makes the local velocity potential satisfy the Laplace equation. This reduces the number of unknowns and makes potential flow a powerful starting point for flow around bodies and for gravity waves. ^[ref]

The Kutta condition selects the circulation of an idealized potential flow past a body with a sharp trailing edge by requiring finite limiting velocity at that edge. For the Joukowski map $z=\zeta +a^2/\zeta$, the trailing edge $z=2a$ corresponds to a simple zero of $dz/d\zeta$ at $\zeta =a$. If $W$ is the complex potential in the circle plane, then $dw/dz=W^{\prime }(\zeta )/(1-a^2/{\zeta }^2)$ is finite there only if $W^{\prime }(a)=0$. For a uniform flow of speed $U$ at angle $\alpha$ and a point vortex of circulation $\Gamma$, this gives $\Gamma =-4\pi Ua\sin \alpha$, and cancellation of the simple zeros gives the tangential trailing-edge velocity $U\cos \alpha$. The opposite endpoint can retain a leading-edge singularity.

The Milne-Thomson circle theorem constructs a circular impermeable boundary in a two-dimensional potential flow. Suppose an incident complex potential of a two-dimensional potential flow $f$ is analytic on and inside $|z|=a$, with its physical singularities outside that disk. Then the displayed $W$ has constant stream function on $|z|=a$. The proof is the identity $a^2/\bar{z}=z$ on the circle: there $W=f(z)+\overline{f(z)}$ is real. Hence the circle is a streamline and satisfies the no-penetration boundary condition. The reflected term has its image singularities inside the obstacle. Additional imposed circulation must be specified separately. The theorem is named after Louis Melville Milne-Thomson.

Consider uniform fields $B_{-}\hat{\mathbf{y}}$ below and $B_{+}\hat{\mathbf{x}}$ above an interface, with local constant densities ${\rho }_{-},{\rho }_{+}$. For ideal magnetohydrodynamics and local incompressible potential flow, the interface displacement $h{e}^{st+ikx+imy}$ has displacement potentials ${\chi }_{-}=h{e}^{Kz}/K$ and ${\chi }_{+}=-h{e}^{-Kz}/K$, $K=\sqrt{k^2+m^2}$. Magnetic flux freezing gives ${\mathbf{b}}_i=({\mathbf{B}}_i\cdot \nabla )\nabla {\chi }_i$. The linear magnetohydrodynamic total pressure is $-[{\rho }_i{s}^2+({\mathbf{k}}_h\cdot {\mathbf{B}}_i{)}^2/{\mu }_0]{\chi }_i$. Matching Lagrangian total pressure gives the displayed dispersion relation. For ${\rho }_{+}>{\rho }_{-}$ the most unstable orientation has its wavevector along the weaker field.

// Target: continuum-mechanics.bigb

Assume an inviscid barotropic fluid acted on by a conservative body force $-\nabla \psi$. Define its barotropic enthalpy function byThe Euler equations for an inviscid fluid are then ${\partial }_t\mathbf{v}+(\mathbf{v}\cdot \nabla )\mathbf{v}=-\nabla (H+\psi )$. Using the vorticity $\mathbf{\omega }=\nabla \times \mathbf{v}$ and the vector identitygives ${\partial }_t\mathbf{v}+\nabla B=\mathbf{v}\times \mathbf{\omega }$, where $B=v^2/2+H+\psi$.

For steady flow, dotting this equation with $\mathbf{v}$ yields $\mathbf{v}\cdot \nabla B=0$. Thus Bernoulli's equation isThe constant need not agree between different streamlines. If the flow is also irrotational, the vector equation gives $\nabla B=0$, so one common constant applies throughout a connected flow region.

For an isentropic process with $p=K{\rho }^{\gamma }$, $H=\gamma p/[(\gamma -1)\rho ]$ up to an additive constant, equal to the thermodynamic specific enthalpy. For an externally maintained isothermal gas, $H=a^2\ln (\rho /{\rho }_{\mathrm{ref}})$; this pressure potential is not the constant thermodynamic enthalpy of a fixed-temperature ideal gas.

If steady flow is not assumed, a useful additional case is unsteady potential flow $\mathbf{v}=\nabla \varphi$. The same vector equation integrates spatially toGeneral unsteady barotropic flow does not possess the steady streamline constant without further conditions.

The Milne-Thomson circle theorem constructs a circular impermeable boundary in planar potential flow. Suppose $f(\zeta )$ is an incident complex potential of a two-dimensional potential flow with no singularities on or inside $|\zeta |=R$. Thenis a complex potential of a two-dimensional potential flow outside the circle, with the same exterior singularities and their strengths as the incident flow; the added image singularities lie inside the excluded disk. On $|\zeta |=R$, $R^2/\bar{\zeta }=\zeta$, so $W=f+\bar{f}$ is real. Its imaginary part, the streamfunction, is constant on the circle, proving the no-penetration boundary condition. The theorem gives the usual zero-circulation construction; a separately prescribed circulation can be added by its logarithmic vortex potential.

Put $s=a^2/r$ and define the image term $G(r,\theta )=(r/a)\Psi (s,\theta )$. The key step is to verify that inversion preserves the Stokes streamfunction equation for irrotational flow. Direct differentiation giveswhere the quantities on the right are evaluated at $(s,\theta )$. HenceBy linearity, $\tilde{\Psi }=\Psi -G$ obeys the same equation and describes an incompressible flow and an irrotational flow wherever the original exterior flow is regular.

On the sphere, $s=a$ and the two terms cancel identically:Thus the sphere is a stream surface and satisfies the no-penetration boundary condition. Tangential slip is allowed in this potential flow model. For every finite $r>a$, the image argument satisfies $0<s<a$, where $\Psi$ has no singularities. The prefactor $r/a$ is regular there. Therefore the construction creates no new finite exterior singularities; any original exterior singularities remain those of the prescribed incident flow. This is the sphere theorem for a Stokes streamfunction.

For the far-field claim, the stated bound givesOne must compare velocities, rather than require the correction to the Stokes streamfunction to tend to zero. Smoothness at the original origin, together with $\Psi (0,\theta )=0$, gives ${\Psi }_s=O(1)$ and ${\Psi }_{\theta }=O(s)$ there. Thus $G_r=O(r^{-1})$ and $G_{\theta }=O(1)$, and the correction hasThese estimates hold away from the coordinate poles, with the regular-axis limits supplied by the original smooth physical flow. ConsequentlyThe far-field velocity field is unchanged. For a fully smooth incident velocity field at the origin, the usual regular expansion is stronger and makes the sphere disturbance decay as $r^{-3}$.

For planar incompressible flow and irrotational flow, combine a velocity potential and a streamfunction into the complex potential of a two-dimensional potential flow:The chosen conventions giveThese are the Cauchy-Riemann equations, so $w$ is a holomorphic function wherever the flow is regular and a consistent local potential is chosen. Its complex derivative is the complex velocityA spatially uniform flow has $w=(U-iV)z+$ a constant. The following contour integral identities also apply to nonuniform potential flow, including the specified line vortex. A complex potential of a two-dimensional potential flow may be multivalued around a source or vortex even though its derivative is single-valued.

For inviscid flow that is also incompressible flow and irrotational flow with velocity potential $\mathbf{u}=\nabla \varphi$, the Euler equations integrate to the Unsteady Bernoulli equationThe same $F(t)$ applies throughout a connected irrotational fluid region; an additive time-dependent potential can absorb it. With gravity absent, fluid at rest at infinity and $\varphi \to 0$, it is $p_{\infty }(t)/\rho$.

Take the cylinder centre at laboratory position $(X(t),0)$ with $\dot{X}=U(t)$, and define $r,\theta$ relative to this moving centre. The circulation-free potential flow satisfies Laplace's equation, decays at infinity, and has surface condition ${\varphi }_r(a,\theta )=U\cos \theta$. The $\cos \theta$ harmonic solution is consequentlyThe velocities are $u_r=U{a}^2\cos \theta /r^2$ and $u_{\theta }=U{a}^2\sin \theta /r^2$, hence $|\mathbf{u}{|}^2=U^2{a}^4/r^4$.

The time derivative must be taken at fixed laboratory coordinates, not fixed $r,\theta$. At a fixed fluid point $\dot{r}=-U\cos \theta$, $\dot{\theta }=U\sin \theta /r$, soSubstituting into the Unsteady Bernoulli equation gives the entire pressure fieldAt the cylinder surface this is $p_{\infty }+\rho a\dot{U}\cos \theta +\rho U^2(\cos 2\theta -1/2)$. Pressure acts against the outward radial normal of the cylinder, so the force per unit length isAll steady terms cancel by trigonometric orthogonality. ThereforeThis identifies the added mass of a circular cylinder as the mass of displaced fluid per unit length. The solution assumes the ambient fluid is initially at rest and has no separately prescribed circulation; these select the usual pure translating-cylinder flow.