Potential flow

Potential flow is irrotational flow represented as the gradient of a scalar potential.

A uniform planar stream combined with an equal source and sink produces a closed dividing streamline enclosing them. Outside it the flow approaches the uniform stream; inside it streamlines connect the source and sink. For positive $U,m$ and separation one, the complex velocity is $U-m/[2\pi z(z-1)]$ and the stagnation points are $[1\pm \sqrt{1+2m/(\pi U)}]/2$. Both lie on the dividing streamline, which has zero streamfunction with consistent logarithm branches.

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.

For $0<a<1$, the Joukowski map $z=\zeta +a^2/\zeta$ maps the exterior unit circle to an ellipse of horizontal and vertical semiaxes $1+a^2$ and $1-a^2$. The Milne-Thomson circle theorem gives $W(\zeta )=U(e^{-i\alpha }\zeta +e^{i\alpha }/\zeta )$ for a uniform flow inclined by $\alpha$. The chain rule produces the displayed complex velocity in the physical plane. Since $z\sim \zeta$, the far-field flow retains its speed and direction. Its two stagnation points areobtained from $\zeta =\pm e^{i\alpha }$. The map derivative is nonzero on and outside the unit circle, so these zeros are preserved without introducing a boundary singularity.

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 $\psi$.

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 $\nabla \log r$.

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 ${\Psi }_0=U{r}^2{\sin }^2\theta /2$.

For a sphere translating through an unbounded inviscid incompressible fluid at rest far away, the laboratory fluid velocity written in coordinates centred on the moving sphere is the gradient of the displayed dipole potential. Its radial velocity at the surface equals $U\cos \theta$. The sphere-rest-frame velocity instead needs the additional uniform potential $-Ur\cos \theta$.

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 $f^{\prime }\ne 0$.

A planar potential dipole at $a$ with complex moment $m$ has complex potential of a two-dimensional potential flow $w(z)=m/(z-a)$. Its complex velocity is $w^{\prime }(z)=-m/(z-a{)}^2$, interpreted as $v_x-i{v}_y$. Unlike the three-dimensional potential dipole, this is a two-dimensional singularity of the logarithmic-source potential. For example, two opposite source coefficients $s$ at $a\pm \delta /2$ have potential $s[\log (z-a-\delta /2)-\log (z-a+\delta /2)]$, whose leading term is $-s\delta /(z-a)$ as the separation tends to zero with $s\delta$ fixed. This gives the dipole as a limiting source–sink pair, with the complex sign and orientation included in $m$.

A transported planar potential dipole follows the surrounding regular velocity, so its position satisfies the displayed equation after its own singular term is removed. The conjugation comes from the convention $w^{\prime }=v_x-i{v}_y$. Reflected image parameters are held fixed when taking the spatial derivative, then evaluated at the instantaneous dipole position.

For a unit circular wall, the Milne-Thomson circle theorem gives image potential $\bar{m}z/(1-\bar{a}z)$ and therefore $\dot{\bar{a}}=\bar{m}/(1-|a{|}^2{)}^2$. For a horizontal straight wall the image is $\bar{m}/(z-\bar{a})$, giving $\dot{\bar{a}}=-\bar{m}/(a-\bar{a}{)}^2$. For the line $y=-x$, reflected conjugation gives image $-i\bar{m}/(z+i\bar{a})$ and hence $\dot{\bar{a}}=i\bar{m}/(a+i\bar{a}{)}^2$. In each case the image lies outside the fluid region and makes the wall a streamline; differentiation of the regular image proves the motion formula. Several compatible straight-wall reflections can be summed before the same operation is applied.

For a complex potential $w(z)=m/(z-a)$, reflection in the straight wall through zero at angle $\theta$ is $Rz=e^{2i\theta }\bar{z}$. The image construction $W(z)=w(z)+\overline{w(Rz)}$ givesOn the wall $Rz=z$, this is real, so its stream function is constant there. This proves the impermeability condition. Thus the image position is $a^{\prime }=e^{2i\theta }\bar{a}$ and its moment is $m^{\prime }=e^{2i\theta }\bar{m}$. If the point dipole is advected by the regular image flow, the complex velocity convention $W^{\prime }=u-iv$ giveswhere $d=\mathrm{Im}(e^{-i\theta }a)$ is its signed distance from the wall. The last equality follows from $a-a^{\prime }=2i{e}^{i\theta }d$. It applies when $d\ne 0$.

A two-dimensional point sink of strength $Q>0$ removes volume flux $Q$ per unit depth at $z_0$. Its complex potential of a two-dimensional potential flow is the displayed expression, so $u-iv=-Q/[2\pi (z-z_0)]$. The radial velocity is $u_r=-Q/(2\pi r)$: integrating over a circle gives outward flux $2\pi r{u}_r=-Q$. This is the planar cross-section of a line sink. A passive particle satisfies $d(r^2)/dt=-Q/\pi$ and reaches the ideal point singularity in finite time. This is a flow singularity, distinct from a stable sink equilibrium of a regular dynamical system.

For a two-dimensional potential doublet of fixed complex strength $M$ outside a stationary circle of radius $a$, the Milne-Thomson circle theorem gives the image contribution $\bar{M}z/(a^2-\bar{b}z)$. If the singularity is stipulated to follow the regular fluid velocity at its position, differentiating this term and using the complex velocity convention gives the displayed motion law. Thus $\mathrm{Im}(b\bar{M})$ is constant: the trajectory is a straight line parallel to $M$. With $M=m{e}^{i\alpha }$ and initial location $b_0>a$ real, its polar coordinates satisfy $r\sin (\theta -\alpha )=-b_0\sin \alpha$. This statement assumes fixed orientation; the translational rule alone supplies no equation for orientation evolution.

For a steady two-dimensional incompressible flow and irrotational flow around an impermeable body, the displayed contour formula gives the force on the body per unit span. Here $W^{\prime }=u-iv$ is the complex velocity, $\rho$ is the constant mass density, and $C$ runs counterclockwise around the body, excluding other fluid singularities. To verify the sign, Bernoulli's principle gives $p=p_0-\rho |q{|}^2/2$. On a counterclockwise boundary, the complex pressure force is $d(X-iY)=-ipd\bar{z}$. Tangency gives $qdz=\bar{q}d\bar{z}$, and therefore $|q{|}^{2}d\bar{z}=q^{2}dz$. Integrating removes the constant $p_0$ term and gives the formula. The theorem is named after Heinrich Blasius.

The Milne-Thomson circle theorem places an image dipole inside a cylinder at the inverse point of an external dipole. In the residue of the squared complex velocity, only the cross term between the analytic external velocity and the image double pole contributes. The product of their conjugate dipole strengths is independent of orientation, so the cylinder force is radial toward the external dipole.

For a two-dimensional steady ideal flow outside a body, let $q=w^{\prime }$ be the complex velocity. Suppose $q$ is holomorphic in the exterior and $q^2$ has a Laurent series near infinity, with coefficient $a_{-1}$ of $z^{-1}$. Orient the body contour counterclockwise and assume the usual impermeability and Bernoulli hypotheses of the Blasius force theorem. Deforming the contour to a large circle gives $\oint q^{2}dz=2\pi i{a}_{-1}$. Substitution in Blasius' formula proves $F_x-i{F}_y=-\pi \rho a_{-1}$.

In particular, $q(z)=q_{\infty }+O(z^{-2})$ makes the force zero: its square has no reciprocal term. This criterion handles branch-cut obstacles by applying the contour in their analytic exterior, rather than separately assigning forces to endpoint singularities.

Place a held point vortex of circulation $\Gamma$ at real position $b>a$ outside a circular cylinder $|z|=a$, and prescribe circulation $-\Gamma$ about the cylinder. An opposite image at $c=a^2/b$ gives complex velocity $q(z)=-i\Gamma (1/(z-b)-1/(z-c))/(2\pi )$. On the positively oriented cylinder contour, the only pole of $q^2$ inside is at $c$, with residue $-{\Gamma }^2/(2{\pi }^2(b-c))$. The residue theorem and Blasius force theorem give $F_x-i{F}_y=\rho {\Gamma }^2/(2\pi (b-c))$. Thus $F_y=0$ and $F_x=\rho {\Gamma }^{2}b/(2\pi (b^2-a^2))>0$: the force on the cylinder points toward the vortex. The specified cylinder circulation is part of the hypothesis; a zero-circulation cylinder has an additional central image term.

For steady incompressible flow and irrotational flow of constant mass density, with no body force, this formula gives the hydrodynamic torque about the origin per unit span on an impermeable body. Orient its boundary counterclockwise, and let positive torque be counterclockwise. The proof uses $M=\mathrm{Re}{\oint }_{C}pzd\bar{z}$, the Bernoulli equation $p=P_{\ast }-\rho |w^{\prime }{|}^2/2$, and the tangency identity $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. The real part of the constant-pressure term vanishes because $\mathrm{Re}(zd\bar{z})=d|z{|}^2/2$. The full complex integral of $zd\bar{z}$ need not vanish.

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 $b=(m+\sqrt{m^2+4U^2})/(2U)$.

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.

A two-dimensional potential doublet is a localized potential flow singularity with complex potential of a two-dimensional potential flow $W=M/(z-b)$, where $M=m{e}^{i\alpha }$ specifies strength and orientation. It is the limiting field of an equal point source and sink as their separation tends to zero with strength times separation fixed, with the sign absorbed in $M$. Its complex velocity is $-M/(z-b{)}^2$, so the velocity decays as inverse distance squared. The total volumetric flow rate around the singularity is zero. Its two-dimensional decay differs from that of the three-dimensional source dipole.

For a complex potential of a two-dimensional potential flow $W=\varphi +i\psi$ in the coordinate $z=x+iy$, the Cauchy-Riemann equations give $W^{\prime }={\varphi }_x-i{\varphi }_y=u-iv$. A material point with physical velocity $(u,v)$ therefore satisfies $\dot{z}=\overline{W^{\prime }(z)}$. Complex conjugation matters when deriving a pathline from a complex potential of a two-dimensional potential flow.

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.

A unit-speed free layer turning downward around a reentrant wall corner has complex velocity $q=u-iv$ in the first-quadrant exterior of the unit disc. Upstream $q=1$, downstream $q=i$, and the corner has infinite speed because its fluid angle is $3\pi /2$. Squaring $q$, applying $(s-1)/(s+1)$, and squaring again maps this domain to the upper half-plane with upstream, corner and downstream at $0,1,\infty$. The potential strip map $e^{\pi w}$ has the same normalization, proving the formula. The sign of the turn and the fluid corner angle determine whether the relevant hodograph is an exterior or interior quarter disc.

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.

For a planar reservoir draining below a vertical lip of height one into a unit-speed stream, the potential strip $0<\mathrm{Im}w<a$ and first-quadrant unit velocity disk satisfyOn the fluid free boundary put $q=e^{-i\theta }$, $-\pi /2\le \theta <0$. Then $dz/d\theta =-2a{e}^{i\theta }\cot \theta /\pi$, so the downstream height is $1-2a/\pi$. Equating it to $a$ gives $a=\pi /(\pi +2)$.

For an inviscid irrotational unit-speed layer turning through a right angle, the complex potential of a two-dimensional potential flow occupies the strip $0<\mathrm{Im}w<1$ and its complex velocity $q$ occupies a fourth-quadrant quarter disk. The maps $r=q^2$, $s=(r-1)/(r+1)$ and $s^2$ take this disk to the upper half-plane, sending upstream velocity one, corner stagnation zero, and downstream velocity $-i$ to $0,1,\infty$. The map $e^{\pi w}$ has the same normalization, proving the displayed relation.

The conformal map of a vertical half-strip by sine sends $-\pi /2<\mathrm{Re}z<\pi /2$, $\mathrm{Im}z>0$ to the upper half-plane. The complex potential of a two-dimensional potential flow $W(z)=\log (a+\sin z)$ therefore pulls back the radial logarithmic flow originating at $\zeta =-a$. Its complex velocity is $W^{\prime }=\cos z/(a+\sin z)$, with a boundary singularity at $z=-\pi /2+i\mathrm{arcosh}a$. For $z=x+iy$, define $D=(a+\sin x\cosh y{)}^2+{\cos }^{2}x{\sinh }^{2}y$. ThenThe bottom boundary carries rightward flow, the right boundary upward flow, and the left boundary flow away from the singular point. The velocity tends to $(0,1)$ at large height. The finite bottom corners are stagnation points, not singularities of $W$.

For unit channel width, incompressible flow towards a corner withdrawal point has $\psi =0$ on the lower wall and $\psi =-m$ on the upper and left walls, with $\psi \to -my$ far downstream. For irrotational flow, separation of variables gives the displayed harmonic correction: the Fourier sine coefficients of $-m(1-y)$ are $-2m/(n\pi )$. Near the corner, $\psi \sim -2m\theta /\pi$ and radial velocity is $-2m/(\pi r)$, whose integrated inward flux across a quarter-circle is $m$.

The region bounded by the dividing streamlines of a two-dimensional point source superposed on uniform flow. Its stagnation point is $-Q/(2\pi U)$ upstream and its far downstream width is $Q/U$. The boundary encloses fluid originating at the source rather than a rigid obstacle; the upstream stagnation point is reached only asymptotically by the upstream source trajectory.

A two-dimensional incompressible line source with area flux $Q$ has radial velocity $Q/(2\pi r)$, obtained from flux conservation around a circle. It is irrotational away from the source, with potential $(Q/(2\pi ))\ln r$. For a circular expanding interface, $Q=2\pi a{a}_t$ and $u_r=a{a}_t/r$, whose radial strain is $-a{a}_t/r^2$. That strain supplies a normal viscous stress even though the velocity is a potential flow.

A potential dipole in three dimensions is the potential flow $\mathbf{u}=\nabla \varphi$ with $\varphi =\mathbf{d}\cdot \mathbf{x}/r^3$. Away from the origin its potential is a harmonic function, so it is also a constant-pressure Stokes flow. Its velocity decays as $r^{-3}$ and its vorticity is zero. It is distinct from a stresslet, whose leading velocity decays as $r^{-2}$.

A two-dimensional point source of strength $q$ has velocity potential $\varphi =q\log r/(2\pi )$ and outward velocity $q{\mathbf{e}}_r/(2\pi r)$. Every closed contour enclosing it has volume flux $q$.

A line sink is a planar line source with negative volume flux: fluid moves radially inward towards its location. For withdrawal rate $S>0$, its complex potential of a two-dimensional potential flow is $-S\log (z-b)/(2\pi )$.

In a planar incompressible flow, a line source injects volume flux $Q$ per unit length perpendicular to the plane. Its radial velocity is $u_r=Q/(2\pi r)$, since integrating the outward velocity around a circle gives $2\pi r{u}_r=Q$. Its complex potential of a two-dimensional potential flow is $w=Q\log (z-b)/(2\pi )$ and its complex velocity is $w^{\prime }=Q/[2\pi (z-b)]$. Here $Q>0$ means outward flow; a line sink has negative $Q$. Away from $b$, this is an irrotational flow as well as an incompressible flow.

In the upper half-plane, a line source of outward volume flux $Q$ at height $a$ above an impermeable wall has an equal image source below it. On the wall $u=Qx/[\pi (x^2+a^2)]$ and the normal velocity vanishes. The steady Bernoulli equation gives $p-p_{\infty }=-\rho Q^2{x}^2/[2{\pi }^2(x^2+a^2{)}^2]$. Sincethe ambient-relative force per unit length pulls the wall toward the source with the displayed magnitude. The torque about the source's projection onto the wall is zero in the symmetric Cauchy principal value sense. Individual one-sided moments diverge logarithmically. This planar result differs from the three-dimensional hydrodynamic attraction of a plane wall to a point source.

In the impermeable strip $x>0$, $0<y<a$, a line source of strength $Q$ at an interior point $c$ has the displayed complex potential of a two-dimensional potential flow, with locally consistent complex logarithm and $f(z)=\cosh (\pi z/a)$. The hyperbolic-cosine map of a half-strip sends the walls to the real axis. An equal image line source at $\overline{f(c)}$ makes that axis a streamline. Since $f^{\prime }(c)\ne 0$, the local singular part remains $Q\log (z-c)/(2\pi )$, proving preservation of strength. With no extra imposed flow, the complex velocity tends to $Q/a$ as $x\to \infty$, so the outgoing volume flux is $Q$. The source position $c$ has positive imaginary part; assuming it real would place it on a wall.

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.

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.

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 $\pi \rho m^2/a^2$.

A line sink of strength $\alpha Q$ at the vertex of a sector $0<\theta <\alpha$ produces the irrotational outer flowIt satisfies no penetration at the radial walls but violates no slip there, so viscous boundary layers form along both walls.

Near a radial wall, let $x$ measure distance from the sink along the wall and $y$ point into the fluid. Withthe Prandtl boundary-layer equation reduces toOne physical solution has

For a sphere of radius $a$ translating at speed $U$ along the polar axis through fluid at rest at infinity, the laboratory-frame potential isIt gives $u_r=U{a}^3\cos \theta /r^3$ and $u_{\theta }=U{a}^3\sin \theta /(2r^3)$.

For radial flow outside a sphere, incompressible flow makesindependent of radius. Its velocity potential is $\varphi =-Q/(4\pi r)$.

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.

In radial potential flow outside a spherical bubble, incompressibility and the moving-boundary condition give $u_r=R^2\dot{R}/r^2$ and $\varphi =-R^2\dot{R}/r$. The Unsteady Bernoulli equation, with its gauge fixed at infinity, yields the displayed radius equation when surface tension and viscosity are neglected. Pressure continuity at the interface identifies the liquid pressure with the gas pressure $p_g$. The vacuum special case is Rayleigh collapse of a spherical cavity.

For spherical expansion in an unbounded incompressible flow, continuity gives $u_r=R^2\dot{R}/r^2$. Integrating the kinetic energy outside the bubble gives $K=2\pi \rho R^3{\dot{R}}^2$. The Rayleigh equation for an inviscid spherical bubble then yields $\dot{K}=(p_b-p_{\infty })\dot{V}$, where $V=4\pi R^3/3$. With $p_b=p_{\infty }{V}_0/V$ and constant $p_{\infty }$, integration gives $K-K_0=p_{\infty }[V_0\log (V/V_0)-V+V_0]$. This neglects viscosity, surface tension and external body-force differences, which contribute additional terms in other bubble models.

For gas pressure depending only on radius and initial speed zero, multiplying the Rayleigh equation for an inviscid spherical bubble by $2R^2\dot{R}$ integrates to the displayed identity. A polytropic pressure $p_g=K/R^{3\kappa }$ gives elementary power integrals when $\kappa \ne 1$, and a logarithm when $\kappa =1$. Turning radii are the positive zeros of this first integral. At $\kappa =2$ and $p_g(R_0)=p_{\infty }/2$, they are $R_0$ and $R_0/2^{1/3}$; the inward and outward accelerations at them have opposite signs, giving bounded undamped oscillation in the ideal model.

The Rayleigh-Plesset equation balances the radial inertia of an incompressible liquid around a spherical bubble against internal and ambient pressure, surface tension and viscous stress. Removing the last two effects and setting the bubble pressure to zero gives the following vacuum-collapse case.

A vacuum cavity of radius $a(t)$ in an infinite inviscid incompressible fluid of density $\rho$ and far-field pressure $p_0$ obeysIf $a(0)=a_0$ and $\dot{a}(0)=0$, its collapsing branch satisfies

For a spherical vacuum cavity initially at rest with radius $a$, in an inviscid liquid of constant mass density $\rho$ and positive constant far-field pressure $p_{\infty }$, the Rayleigh equation for an inviscid spherical bubble has the first integral $R^3{\dot{R}}^2=2p_{\infty }(a^3-R^3)/(3\rho )$. To obtain it, multiply the radius equation by $2R^2\dot{R}$ and integrate the resulting exact derivative. Choosing the negative square root and integrating $dt=dR/\dot{R}$ givesThis improper integral converges at both endpoints, proving finite-time Rayleigh collapse of a spherical cavity in the ideal model. It has numerical value approximately $0.914681$. The prediction neglects viscosity, surface tension and liquid compressibility.

Inviscid flow between closing hinged plates is a quadratic potential flow that expels fluid radially.