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An axisymmetric Stokes streamfunction disturbance $\varphi (r)e^{ik(z-ct)}$ of an axial basic flow obeys the displayed ordinary differential equation. Regularity requires $\varphi =O(r^2)$ at the axis, and an impermeable wall requires $\varphi (a)=0$ for nonzero $k$. Multiplication by ${\varphi }^{\ast }/r$ and integration by parts shows that a growing real-wavenumber mode requires $(U^{\prime }/r{)}^{\prime }$ to change sign; if it vanishes identically, the positive energy integral rules out such a mode. The condition concerns this cylindrical vorticity gradient rather than $U^{\prime \prime }$ alone.

Substitute the given Stokes streamfunction representation into the divergence in spherical coordinates. The mixed derivatives cancel:Thus incompressibility is automatic for a smooth Stokes streamfunction. Since this is an axisymmetric flow with no azimuthal velocity, the only possibly nonzero component of its vorticity isDefine the differential operator in the Stokes streamfunction equation for irrotational flowThe stated equation is $\mathcal{L}\Psi =0$, and thereforeThe irrotational flow has a velocity potential locally, and globally on a regular simply connected domain. Regularity of the physical velocity field supplies the limiting interpretation on the polar axis, where the coordinate expressions contain $\sin \theta$ in the denominator.

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 $r^{-3}$.

The incident uniform flow $U{\mathbf{e}}_z$ has spherical components $u_r=U\cos \theta$, $u_{\theta }=-U\sin \theta$. Integrating the Stokes streamfunction formulas, with the additive constant chosen to vanish on the axis, givesIt is regular at the origin. Applying the sphere theorem for a Stokes streamfunction yieldsThe resulting velocity components areTo find the velocity potential, integrate ${\varphi }_r={\tilde{u}}_r$ and verify $r^{-1}{\varphi }_{\theta }={\tilde{u}}_{\theta }$. This givesThe first term is the incident uniform flow, and the second is a potential dipole located inside the excluded sphere. At $r=a$, the radial velocity vanishes, while the far-field velocity tends to $U{\mathbf{e}}_z$. The poles are stagnation points; the surface tangential speed reaches $3U/2$ at the equator. These formulas describe uniform potential flow around a fixed sphere.

Write the perturbation Stokes streamfunction as $\psi =\varphi (r)e^{ikz-i\omega t}$, with $\omega =kc$ and nonzero $k$. Its radial and axial velocity amplitudes are $u=-ik\varphi /r$ and $w={\varphi }^{\prime }/r$. These automatically obey incompressibility. Linearizing the radial and axial Euler equations, with pressure amplitude $P$, givesThusDifferentiating the first expression, the terms involving $U^{\prime }{\varphi }^{\prime }$ cancel. Equating the two expressions for $P^{\prime }$ proves the axisymmetric inviscid pipe stability equation:At the axis, finite smooth axisymmetric velocity requires $\varphi =O(r^2)$, after choosing the irrelevant constant in the streamfunction. At the wall, impermeability gives $\varphi (a)=0$. There is no tangential no-slip condition in this inviscid problem. A decoupled azimuthal disturbance satisfies $ik(U-c)u_{\theta }=0$ and does not introduce a growing discrete mode when $U\ne c$.

For real $k$, suppose $c=c_r+i{c}_i$ with $c_i\ne 0$, so division by $U-c$ is legitimate. Multiply the ordinary differential equation by ${\varphi }^{\ast }/[r(U-c)]$, integrate, and apply integration by parts. The endpoint terms vanish by the preceding regularity and wall conditions, leavingTaking its imaginary part yieldsA growing temporal mode therefore requires a sign change of $(U^{\prime }/r{)}^{\prime }$, rather than a sign change of $U^{\prime \prime }$ alone. This is the cylindrical counterpart of Rayleigh's inflection-point theorem. If this derivative vanishes identically, the positive first integral rules out a nontrivial growing real-wavenumber mode as well. Isolated zeros without a sign change do not make the second integral vanish for a nonzero regular mode.

For the parabolic Poiseuille flow, $U^{\prime }/r=-2U_0/a^2$ is constant. Under the specified exclusion of critical levels, division givesSetting $\varphi =rf$, or equivalently choosing $n=1$ in $r^{-n}\varphi$, turns this intoIt is the Modified Bessel differential equation of order one. Hence $\varphi =r[A{I}_1(kr)+B{K}_1(kr)]$. Since $r{K}_1(kr)$ tends to a nonzero constant at the axis, it gives a singular radial velocity and must be excluded. The admissible mode shape is thereforeFor $k=0$, direct solution gives $\varphi =C{r}^2$ after axis regularity, and the wall condition again forces the trivial solution. For nonzero $k$, the nonzero roots of $I_1$ are imaginary. Writing $j_{1,l}>0$ for the zeros of the ordinary Bessel function $J_1$, the evanescent potential modes of inviscid Poiseuille flow areTheir perturbation vorticity vanishes: ${\partial }_{z}u-{\partial }_{r}w=k^2\varphi /r-({\varphi }^{\prime }/r{)}^{\prime }=0$. These modes grow exponentially in one axial direction and decay in the other. They can describe end-forced or spatially localized potential fields, but neither sign gives a bounded normal mode on a whole infinite pipe. Moreover, the reduced radial equation contains no $c$: it supplies no temporal eigenvalue condition. An arbitrary imaginary part assigned to $\omega =kc$ here is consequently not evidence for a physical temporal instability.

The real-wavenumber calculation does exclude growing discrete axisymmetric modes of this inviscid profile. It does not establish every form of stability: singular neutral disturbances with $c$ in the range of $U$, transient dynamics, non-axisymmetric disturbances, and finite-amplitude transition require separate treatment. The exclusion $U\ne c$ specifically removes the inviscid continuous spectrum.

With viscosity, linearize the Navier-Stokes equations instead and eliminate pressure to obtain a fourth-order cylindrical counterpart of the Orr-Sommerfeld equation. Impose both $\varphi (a)=0$ and ${\varphi }^{\prime }(a)=0$, with appropriate regularity at the axis. The resulting eigenvalue problem involves the Reynolds number and determines complex frequencies, rather than leaving $c$ arbitrary. Viscosity supplies diffusion, no-slip boundary layers and critical-layer regularization, so the inviscid sign argument alone no longer decides its spectrum. That observation does not imply that viscous pipe flow must have a growing linear eigenmode; finite-amplitude pipe transition is a different question.

For an axisymmetric flow without azimuthal velocity, define the Stokes streamfunction byThese formulas enforce incompressibility. The only nonzero curl component is $R^{-1}[{\partial }_R(R{u}_{\theta })-{\partial }_{\theta }{u}_R]$. Substitution givesCurling the Stokes flow equations eliminates pressure and gives ${\nabla }^2(\nabla \times u)=0$. For an azimuthal field, its vector Laplacian is the scalar Laplacian minus $1/(R^2{\sin }^2\theta )$; applying that operator to $-D^2\Psi /(R\sin \theta )$ gives $-D^2(D^2\Psi )/(R\sin \theta )$. Hence the streamfunction equation is $\boxed{D^2{D}^2\Psi =0}$.

For the proposed sphere solution put $f(R)=R^2-3aR/2+a^3/(2R)$, so $\Psi =(U/2)f(R){\sin }^2\theta$. Its velocity isBoth components vanish at $R=a$ and tend to the components of $U{e}_z$ at infinity. Also $D^2\Psi =(U/2)(f^{\prime \prime }-2f/R^2){\sin }^2\theta =(3Ua/(2R)){\sin }^2\theta$, and applying $D^2$ once more gives zero. This verifies the differential equation and both physical boundary conditions.

For a sphere translating upward through otherwise stationary fluid, subtract the uniform background in the sphere frame and reverse its far-flow direction. Its leading disturbance is the upward StokesletPlace an image at $z=+d$ with the opposite, downward velocity $-U$, while the real sphere is at $z=-d$. The sum has $u_z$ odd under $z\mapsto -z$ and $u_r$ even. Thus at $z=0$, $u_z=0$ and ${\partial }_z{u}_r=0$; moreover ${\partial }_r{u}_z=0$. The tangential stress ${\sigma }_{rz}=\mu ({\partial }_z{u}_r+{\partial }_r{u}_z)$ is therefore zero, as required by the stress-free boundary condition.

At a point on the plane with radial coordinate $r$, the real Stokeslet gives $u_r=3aUrd/[4(r^2+d^2{)}^{3/2}]$. The image gives the same radial contribution, soThe surface flow is outward. Although the isolated-sphere dipole term is only $O(U(a/d{)}^3)$, the next correction is $O(U(a/d{)}^2)$: the image's velocity at the real sphere is downward and of size $3aU/(4d)$, so satisfying the real sphere's prescribed velocity requires a relative $O(a/d)$ correction to its Stokeslet strength. Since the leading surface flow is $O(Ua/d)$, the first reflection already changes it at the displayed second order. Ignoring this boundary-interaction correction would give the wrong next power.

For axisymmetric incompressible flow without swirl, the Stokes streamfunction satisfiesSubstitution into the divergence in spherical coordinates makes incompressibility an identity. Its only vorticity component isTaking the curl of the Stokes equation eliminates pressure and gives ${\nabla }^2\mathbf{\omega }=0$. For an axisymmetric azimuthal vector, its scalar component obeys $(\Delta -1/(R^2{\sin }^2\theta )){\omega }_{\varphi }=0$. Direct differentiation verifiesApply this to $h={\mathcal{D}}^2\Psi$ to obtain

Fix the scaling constants as $U=A/z$, $\delta =Bz$, with $A=F/(\rho \nu )$ and $B=\nu \sqrt{\rho /F}$, so $A{B}^2=\nu$. The Stokes streamfunction is then $\psi =\nu zf(\eta )$. Write $g=f^{\prime }/\eta$. Differentiation givesSubstitution in the boundary-layer equation, cancelling $A^2/z^3$, gives $-(f/\eta )g^{\prime }-g^2=g^{\prime \prime }+g^{\prime }/\eta$. Multiplication by ${\eta }^3$ therefore gives the required similarity equationAxis regularity requires zero radial velocity and a finite smooth axial velocity: choose the streamfunction gauge $f(0)=0$ and require $f(\eta )=c{\eta }^2+O({\eta }^4)$. Equivalently $f^{\prime }(0)=0$, $f^{\prime }/\eta$ has a finite limit, and $(f^{\prime }/\eta {)}^{\prime }\to 0$ at the axis. The constant $c$ is not specified in advance. In the quiescent ambient fluid require $f^{\prime }/\eta \to 0$ and $f^{\prime }-f/\eta \to 0$ as $\eta \to \infty$, so both velocity components vanish. The localized physical jet has finite axial volume flux, corresponding to finite $f(\infty )$.

Finally the prescribed force fixes the amplitude through the finite momentum-flux normalizationThe axis conditions express regularity rather than independently prescribed arbitrary derivatives; the far-field conditions express matching to the resting fluid, and the integral condition supplies the jet strength.

Suppose the incident axisymmetric flow is an incompressible flow and an irrotational flow, with Stokes streamfunction $\Psi$ regular inside $r<a$. The displayed transformation adds an impermeable sphere at $r=a$ without new finite exterior singularities. Writing $s=a^2/r$ and $G=(r/a)\Psi (s,\theta )$, direct differentiation showsHence the Stokes streamfunction equation for irrotational flow is preserved. At $r=a$, $\tilde{\Psi }=0$, making the radial velocity zero. Exterior points sample only the regular interior under inversion. If $\Psi$ vanishes at the origin and is $O(r)$ there with smooth regular derivatives, the induced velocity decays at infinity. This is an inversion construction for an axisymmetric streamfunction, distinct from a scalar Kelvin transform.

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.

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