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A separated acoustic wave in a straight rigid channel $0<y<H$ has transverse Neumann boundary conditions, giving factors $\cos (n\pi y/H)$ for integers $n\ge 0$. The velocity potential satisfies the wave equation, and its axial factor obeysHere $c$ is the sound speed. Real $q_n$ permits propagation; positive imaginary $q_n$ produces an evanescent wave. A prescribed wall velocity proportional to one transverse factor excites that mode alone in the linear straight-channel model. The physical field is the real part of the complex expression.

An infinitely long circular cylinder of radius $a$ translating across its axis through otherwise stationary, inviscid incompressible fluid has added mass $m_a=\rho \pi a^2$ per unit length. The decaying, circulation-free velocity potential is $\varphi =-U{a}^2\cos \theta /r$. In moving centre coordinates, the fixed-laboratory derivative is ${\varphi }_t=-\dot{U}{a}^2\cos \theta /r-U^2{a}^2\cos (2\theta )/r^2$. The Unsteady Bernoulli equation gives $p-p_{\infty }=\rho a^2\dot{U}\cos \theta /r+\rho U^2{a}^2\cos (2\theta )/r^2-\rho U^2{a}^4/(2r^4)$. Integrating $-pa\cos \theta$ around $r=a$ yields $F_x=-\rho \pi a^2\dot{U}$; the transverse force vanishes. Taking a derivative at fixed centre coordinates would omit the necessary convective correction.

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.

A regular axisymmetric irrotational flow near a stagnation point on an impermeable tangent plane has equal tangential strain rates. The incompressibility condition forces the normal rate to be their negative double. Thus the leading velocity potential has the displayed quadratic form. Its streamline invariants are the azimuth and $z(x^2+y^2)$; for incoming flow from positive $z$, $W>0$.

Let $F(\zeta ,t)$ map the unit disk conformally to a planar Darcy flow region, with $F(0,t)=0$ at a point source of flux $2\pi Q$. With zero velocity potential on the boundary, the logarithmic source potential becomes $\varphi =Q\log |\zeta |$. Differentiating the inverse map at fixed physical position gives ${\varphi }_t=-Q\mathrm{Re}(F_t/(\zeta F_{\zeta }))$, while $|\nabla \varphi {|}^2=Q^2/|\zeta F_{\zeta }{|}^2$. The boundary law ${\varphi }_t+|\nabla \varphi {|}^2=0$ therefore gives the displayed equation, for nonzero $Q$.

The vector $\zeta F_{\zeta }d\theta$ is outward normal times arclength on $\zeta =e^{i\theta }$. Consequently the equation gives the area growth law $d|D|/dt=2\pi Q$. For a circular map $F=R(t)\zeta$, it reduces to $R\dot{R}=Q$, and hence $R(t{)}^2=R(0{)}^2+2Qt$.

A harmonic velocity potential decaying with depth, together with linearized kinematic and Bernoulli boundary conditions and capillary pressure, gives ${\omega }^2=gk+(T/\rho )k^3$. The corresponding phase speed combines long-wave gravity and short-wave capillarity contributions.

For bending stiffness $D$ and mass per area $m$, a plane wave of surface displacement $\eta$ satisfies $m{\eta }_{tt}+D{\nabla }_H^4\eta =p$. An inviscid deep-water velocity potential has vertical dependence $e^{kz}$; the kinematic boundary condition and linear Bernoulli equation give $p=({\rho }_w{\omega }^2/k-{\rho }_{w}g)\eta$. Eliminating $p$ gives the displayed dispersion relation. With water depth $d$, replace it by ${\omega }^2=(gk+D{k}^5/{\rho }_w)\tanh (kd)/[1+(m/{\rho }_w)k\tanh (kd)]$.

For a top-hat planar jet in stationary fluid of the same mass density, odd and even velocity potentials have the same temporal growth rate. Put $s=\tanh (|k|L)$. Their dispersion relations use $s$ and $1/s$, but $\sqrt{s}/(1+s)=\sqrt{1/s}/(1+1/s)$. Their real phase velocities are respectively $Vs/(1+s)$ and $V/(1+s)$. The common growth rate is strictly below $|kV|/2$ for finite $|k|L>0$, approaching the isolated vortex sheet value when the sheets decouple.

For two resting incompressible inviscid layers between impermeable horizontal walls, the upper layer has density ${\rho }_1$ and depth $L_1$, and the lower layer has ${\rho }_2<{\rho }_1$ and depth $L_2$. Solving the Laplace equation for the velocity potential and imposing the linearized kinematic and capillary pressure conditions gives the displayed dispersion relation. With positive surface tension $\gamma$, exponential Rayleigh-Taylor instability occurs for $0<k<k_c$, where $k_c=\sqrt{g({\rho }_1-{\rho }_2)/\gamma }$. The depths change the growth rate, but not this cutoff.

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.

A semi-infinite horizontal plate at unit height and a floor at zero give a velocity potential strip of height $a$ and an upper unit semicircle in the complex velocity $q$. Their matching half-plane maps give $(1+e^{\pi w/a}{)}^{-1}=((q-1)/(q+1){)}^2$. Along the fluid free boundary put $q=e^{-i\theta }$, $-\pi <\theta <0$. Then $dz/d\theta =-(a/\pi )e^{i\theta }\cot (\theta /2)$ and $y=1-(a/\pi )(\theta +\pi +\sin \theta )$. Its downstream limit is $1-a$, which must equal the downstream depth $a$. Thus the contraction is exactly one half.

An incompressible point source of volume flux $Q$ at height $a$ above an impermeable plane has velocity potential $\varphi =-Q(1/r_{+}+1/r_{-})/(4\pi )$, with an equal image source below the plane. On the wall, $u_s=Qs/[2\pi (s^2+a^2{)}^{3/2}]$. The Bernoulli equation gives $p-p_{\infty }=-\rho Q^2{s}^2/[8{\pi }^2(s^2+a^2{)}^3]$. Integrating the pressure deficit over the plane gives the displayed attraction force relative to ambient pressure. If source strength instead denotes the coefficient $Q/(4\pi )$ of radial inverse-square velocity, the force is $\pi \rho m^2/a^2$.

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]

Two deep layers of incompressible flow with upper mass density ${\rho }_1$, lower mass density ${\rho }_2$, and tangential velocities $U_1,U_2$ have this dispersion relation when surface tension is absent. Decaying velocity potentials and the linearized kinematic boundary condition determine their amplitudes; the dynamic boundary condition for an inviscid interface gives the displayed relation. For $k>0$ the squared departure of the phase velocity $c$ from the density-weighted mean velocity is $g({\rho }_2-{\rho }_1)/[k({\rho }_1+{\rho }_2)]-{\rho }_1{\rho }_2(U_1-U_2{)}^2/({\rho }_1+{\rho }_2{)}^2$. A negative value gives Kelvin-Helmholtz instability despite stable density stratification.

For a small-amplitude surface gravity wave described by velocity potential $\varphi$ and surface elevation $\eta$, linearizing the kinematic boundary condition for a free-surface graph and the Bernoulli equation at the mean surface $z=0$ givesHere $p$ is the applied excess pressure above constant atmospheric pressure. With $p=0$, differentiate the second equation in time and use the first to obtain ${\varphi }_{tt}+g{\varphi }_z=0$. These surface conditions apply at finite depth as well as in the deep-water gravity wave limit; depth enters through the bottom condition. The quadratic kinetic energy term is omitted at this order, and surface tension is neglected.

For outgoing velocity potential $\varphi =F(r-ct)/r$, the radial velocity is $F^{\prime }/r-F/r^2$ and the pressure perturbation is ${\rho }_{0}c{F}^{\prime }/r$. The acoustic energy density separates into kinetic density ${\rho }_0(F^{\prime }/r-F/r^2{)}^2/2$ and compressional density ${\rho }_0(F^{\prime }/r{)}^2/2$. They are generally unequal near the source but agree at leading far-field order. The additional $r^{-2}$ velocity term is the spherical near field.

For a tank with side lengths $a,b$, depth $h$, impermeable walls and negligible surface tension, the small-amplitude surface gravity wave normal modes have horizontal factors $\cos (m\pi x/a)\cos (n\pi y/b)$ and $k_{mn}^2=(m\pi /a{)}^2+(n\pi /b{)}^2$. The velocity potential has vertical factor $\cosh (k_{mn}(z+h))$, satisfying the bottom Neumann boundary condition. Combining the linearized free-surface boundary conditions gives ${\omega }_{mn}^2=g{k}_{mn}\tanh (k_{mn}h)$. The uniform $m=n=0$ surface elevation is excluded when the total water volume is fixed. Taking one horizontal index zero gives the two-dimensional tank modes.

For $k_{mn}h\ll 1$, the shallow-water approximation gives ${\omega }_{mn}\sim \sqrt{gh}{k}_{mn}$; for $k_{mn}h\gg 1$, the deep-water gravity wave relation ${\omega }_{mn}^2\sim g{k}_{mn}$ follows. An oscillating-wall forcing of a rectangular tank couples to these same normal modes and becomes resonant at their natural frequencies.

For a tank $0<x<L$, $-h<z<0$, let the left wall move by $\mathrm{Re}(c{e}^{-i\omega t})$ with small amplitude, while the right wall and bottom remain fixed. A periodic velocity potential $\mathrm{Re}(f{e}^{-i\omega t})$ has $f_x(0,z)=-i\omega c$. Add the harmonic functionto obtain $G=f+Q$ with homogeneous wall and bottom Neumann boundary conditions. Thus $G=a_0/2+{\sum }_{n\ge 1}{a}_n\cos (k_{n}x)\cosh (k_n(z+h))$, where $k_n=n\pi /L$. The linearized free-surface boundary conditions give $G_z-\lambda G=H$ at $z=0$, with $\lambda ={\omega }^2/g$ andIts Fourier cosine coefficients $H_n=(2/L){\int }_0^{L}H(x)\cos (k_{n}x)dx$ give the title expression for $n\ge 1$ and $a_0=-H_0/\lambda$. In particular $H_n=2i\omega c\lambda /(L{k}_n^2)$ for $n\ge 1$. The denominator vanishes at the corresponding normal modes of surface gravity waves in a rectangular tank, giving resonance. Exact undamped resonance produces secular growth instead of a bounded periodic response. Free transients do not generally decay in this inviscid model.

Let $\zeta =\zeta (z,t)$ be the inverse conformal map. With $Q$ real and nonzero, the function $Q\log |\zeta |$ is harmonic away from the source, vanishes on the moving boundary, and has the required logarithmic singularity. Indeed $F(\zeta ,t)=a_1(t)\zeta +O({\zeta }^2)$ with $a_1\ne 0$, so $\log |\zeta |=\log |z|-\log |a_1|+o(1)$ near the origin. Subtracting this candidate from the physical velocity potential leaves a bounded harmonic function near the source. Its singularity is removable, and the difference has zero boundary values. The maximum principle for harmonic functions makes it identically zero. ThereforeOnly the real part is needed, so the multivalued imaginary part of the logarithm causes no ambiguity in $\varphi$.

Differentiate $F(\zeta (z,t),t)=z$ while holding the physical point $z$ fixed:Also the local analytic complex potential of a two-dimensional potential flow has derivativeThe distinction between differentiating at fixed $z$ and fixed $\zeta$ is essential because the boundary map depends on time. Substituting into the boundary evolution law and multiplying by $|\zeta F_{\zeta }{|}^2/Q$ givesTaking the conjugate inside a real part leaves it unchanged, so the desired conformal map equation for planar Darcy growth isAs a normalization check, parameterize the boundary by $\zeta =e^{i\theta }$. Its outward normal vector times arclength is $\zeta F_{\zeta }d\theta$, so its area growth rate is ${\int }_0^{2\pi }\mathrm{Re}(F_t\overline{\zeta F_{\zeta }})d\theta =2\pi Q$, exactly the specified source flux. For an expanding circular map $F=R(t)\zeta$, the equation reduces to $R\dot{R}=Q$.

An irrotational flow satisfies $\nabla \times \mathbf{u}=0$. On a simply connected region, or locally on any small regular region, this gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$. The incompressibility condition is $\nabla \cdot \mathbf{u}=0$, soThus the potential is a harmonic function. The locality qualification matters: a curl-free circulation around an excluded vortex core need not admit a globally single-valued potential. No such circulation is needed for the sphere flow below.