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Use the pressure scale $\rho {\hat{U}}^2$, and denote the scaled transverse velocity by $v$. The steady incompressible flow equations under the prescribed coordinate scaling areThe no-slip boundary condition is $u=v=0$ at $y=1$ and at the lower wall $y=ϵF(x)$, taking $F=0$ upstream. Far upstream, $u\to U_0=6y(1-y)$, $v\to 0$, and $p_x\to -12\lambda /R$. These are the dimensional Navier-Stokes equations divided by the streamwise and transverse inertial scales, respectively; the transverse equation's ${\lambda }^{-2}$ is essential.

A consistent nonlinear wall-layer regime can be chosen asNear a wall the oncoming plane Poiseuille flow is linear in distance, so a layer of thickness $ϵ$ has $u=O(ϵ)$ and $v=O({ϵ}^2)$. Its inertia is $O({ϵ}^2)$; its transverse viscous term is $(\lambda /R)O({ϵ}^{-1})=O({ϵ}^2)$, explaining the distinguished balance. The resulting pressure perturbation is $O({ϵ}^2)$. In the core the induced transverse pressure variation is $O(ϵ/{\lambda }^2)$, negligible compared with that pressure perturbation when ${\lambda }^2ϵ\gg 1$, equivalently $\lambda \gg R^{1/7}$. Also $\lambda \ll R$ makes both the indentation and the viscous boundary layer thin. Streamwise viscous derivatives are smaller still. Thus the core perturbation is inviscid at first order and pressure is common to the two wall layers at their leading order.

Write $u=U_0+ϵu_1+\cdots$ and $v=ϵv_1+\cdots$. The core pressure gradient only enters at order ${ϵ}^2$, so first-order continuity equation and streamwise momentum giveEliminating $u_{1x}$ shows $(v_1/U_0{)}_y=0$. Let that ratio be $-A^{\prime }(x)$; integration in $x$, with the unperturbed upstream condition, then givesThis displacement form automatically has zero first-order change in total core flux, because ${\int }_0^1{U}_0^{\prime }dy=0$.

For the lower boundary layer, usewhere $p_0^{\prime }=-12{ϵ}^3$ is the smaller upstream pressure gradient. The extra term in $v$ removes the velocity induced by the moving coordinate surface. The chain rule gives ${\partial }_x{|}_y={\partial }_x{|}_z-F^{\prime }{\partial }_z$ and ${\partial }_y={ϵ}^{-1}{\partial }_z$; the wall-slope terms cancel from both continuity equation and convection. At leading order the Prandtl boundary-layer equation isMatching to the core near the lower wall gives $U\sim 6[z+F+A]$, so $H_{\mathrm{lower}}=F+A$.

For the upper boundary layer, set $y=1-ϵz$, $u=ϵU$, $v=-{ϵ}^{2}V$. The same equations, wall data and common $P(x)$ result, but the core matching gives $U\sim 6[z-A]$, so $H_{\mathrm{upper}}=-A$. The two pressure-driven wall-layer problems have the same upstream shear, wall conditions and pressure forcing. Under the question's uniqueness assumption their solutions coincide, and their displacement limits must coincide. Thus the symmetric viscous interaction in an indented Poiseuille channel gives

For $F=b{x}^{1/3}$, one has $A=-(b/2)x^{1/3}$ and $H=(b/2)x^{1/3}$. Matching $U\sim 6[z+H(x)]$ in a similarity solution $U=x^{\beta }{G}^{\prime }(z{x}^{-\gamma })$ requires $\beta =\gamma =1/3$. Equating the powers in convection $x^{2\beta -1}$, viscosity $x^{\beta -2\gamma }$ and pressure gradient $x^{\sigma -1}$ gives $\sigma =2/3$. HenceUse the stream function $\psi =x^{2/3}G(\eta )$, with $\eta =z{x}^{-1/3}$. ThenDirect differentiation gives $U{U}_x+V{U}_z=x^{-1/3}[(G^{\prime }{)}^2/3-2G{G}^{\prime \prime }/3]$, whereas $-P^{\prime }+U_{zz}=x^{-1/3}[-2\bar{p}/3+G^{\prime \prime \prime }]$. The one-third-power similarity of an indented channel boundary layer is therefore governed byThe original PDF has $(G^{\prime }{)}^2$ in this equation; the converted TeX loses the derivative and gives $G^2$. Existence remains conditional on a value of $\bar{p}$ for which this boundary value problem has a solution, as stipulated in the question. The similarity describes the stated thin-indentation regime on $x>0$.

Let the positive undisturbed surface tension be $\bar{\sigma }={\sigma }_0-A{\Gamma }_0>0$. An insoluble surfactant perturbation $g$ produces ${\sigma }_1=-Ag$. The tangential Marangoni stress changes the surface condition to $\mu u_z(h)={\sigma }_x$. Solving the lubrication theory momentum equation now givesAt leading order, conservation of insoluble surfactant on a moving interface reduces to ${\Gamma }_t+(\Gamma u_s{)}_x=0$. Corrections from the changing circumference are smaller by $h_0/a$. Likewise the normal-pressure contribution $-{\sigma }_1/a$ gives a coupling smaller by $h_0/a$ than the retained tangential Marangoni stress, so it is omitted at the same order.

For thickness and concentration amplitudes $H,g$ proportional to $e^{\beta t+ikx}$, put $D=a^{-2}-k^2$ and use $p_1=-\bar{\sigma }DH$. The two conservation equations becomeSet $s=k^2{a}^2$, $\lambda =h_0^3\bar{\sigma }/(3\mu a^4)$ and $\alpha =h_{0}A{\Gamma }_0/(\mu a^2)$. The uncoupled capillary growth and elastic relaxation rates are $L=\lambda s(1-s)$ and $S=\alpha s$. The off-diagonal product is $3LS/4$, so the determinant is $(\beta -L)(\beta +S)+3LS/4$. Thus the surfactant stabilization of an annular liquid lining relation isThis proves the dispersion relation before its lettered deductions. The physical assumption $\bar{\sigma }>0$ is needed for the claimed capillary instability band.

Measure $z$ from the rigid tube wall towards the air core, so $0<z<h$. To leading order in $h_0/a$ and small interface slope, the curvature entering the Young–Laplace equation isThe air pressure is spatially uniform at the order retained. Thus the linear liquid pressure perturbation associated with thickness amplitude $H$ is $p_1=-{\sigma }_0(a^{-2}-k^2)H$. The circumferential curvature favors growth at long wavelength; the curvature opposes it at short wavelength.

The lubrication theory momentum equation is $\mu u_{zz}=p_x$. With the wall no-slip boundary condition $u(0)=0$ and zero surface shear stress, $u_z(h)=0$, its solution and flux areLeading thin-film mass conservation is $h_t+q_x=0$. For a mode $e^{\beta t+ikx}$, this gives the capillary instability of an annular liquid liningTherefore the unstable band is $\boxed{0<k^2<a^{-2}}$. Maximization with respect to $k^2$ gives $k_{\ast }^2=1/(2a^2)$ andThese curvature coefficients use $a$ instead of $a-h_0$ consistently at leading order in the thin-lining approximation.

Let the surface be at $R$ with vanishing external pressure. Multiply hydrostatic equilibrium by $4\pi r^3$ and integrate:The other side is $-{\int }_0^{R}4\pi Gm(r)\rho (r)rdr=-{\int }_0^{M}Gm/r(m)dm$. Building the star one shell at a time identifies this as its negative gravitational potential energy $E_{\mathrm{grav}}$. ThereforeIf the star has nonzero surface pressure $P_s$, the integration instead gives $3⟨P⟩V-3P_{s}V=-E_{\mathrm{grav}}$; the requested form assumes $P_s=0$.

Within the prescribed uniform-density approximation, $n=3M/(4\pi m_p{R}^3)$. Treating the given approximate pressure coefficient as one, the degeneracy pressure and specified kinetic-energy relation giveFor a uniform sphere, $m(r)=M(r/R{)}^3$, so shell assembly givesThusAt fixed mass, this energy has its minimum at $R_{\ast }=2\alpha /\beta$, with $E^{\prime \prime }(R_{\ast })=2\alpha /R_{\ast }^4>0$. Expansion beyond this radius raises the energy, while strong compression costs positive energy growing as $R^{-2}$, faster than the attractive $R^{-1}$ term. This explains radial stability of the model white dwarf. The numerical coefficient in $\alpha$ inherits the approximate pressure law; a relativistic degeneracy law would change this stability argument.

Define source strength $m$ here as the volume flux through a small sphere around the source. For incompressible potential flow, $\mathbf{u}=\nabla \varphi$ and the governing conditions areEquivalently the Laplace equation holds away from the source, with local behavior $\varphi \sim -m/(4\pi |\mathbf{x}-{\mathbf{x}}_s|)$. The boundary condition expresses impermeability. The method of images uses an equal source below the plane:The image singularity lies outside the fluid, the two vertical derivatives cancel on the wall, and the source has the specified outward flux.

Let $s=\sqrt{x^2+y^2}$. On the wall $u_z=0$ and the tangential radial velocity is $u_s=ms/[2\pi (s^2+a^2{)}^{3/2}]$. With ambient pressure $p_{\infty }$, the steady Bernoulli equation in this irrotational flow givesThe fluid pressure is reduced everywhere on the wall except at its axis and in the infinite-distance limit. The finite hydrodynamic force is the excess over the uniform ambient-pressure force. Integrating its magnitude over the plane yieldsPutting $u=s^2$ gives the last integral as $\frac{1}{2}{\int }_0^{\infty }u(u+a^2{)}^{-3}du=1/(4a^2)$. HenceRelative to ambient pressure on the other side, the boundary is attracted toward the source. This is the hydrodynamic attraction of a plane wall to a point source.

There is an alternative source-strength convention: if $m$ denotes the coefficient in $u_r=m/r^2$ rather than total volume flux, replace the flux above by $4\pi m$. Then $\varphi =-m(1/r_{+}+1/r_{-})$, $p-p_{\infty }=-2\rho m^2{s}^2/(s^2+a^2{)}^3$, and $F=\pi \rho m^2/a^2$. The physical result is identical once the normalization is fixed.

For an inviscid flow of constant density $\rho$, driven by a conservative body force $-\nabla \Phi$, the Euler equation isThe vector identity $(u\cdot \nabla )u=\nabla (|u{|}^2/2)-u\times (\nabla \times u)$ and irrotationality give the Unsteady Bernoulli equation. Specifically, wherever a single-valued potential $u=\nabla \varphi$ is available,On a connected potential-flow region this meansThe integration function is spatially constant but need not be constant in time; it can be removed by changing the potential's time-dependent gauge. A simply connected irrotational region provides the requisite global potential. For a barotropic fluid the pressure term is replaced by $\int dp/\rho (p)$; here the liquid density is constant.

For the tube, let $U(t)$ be its approximately uniform axial velocity. The large reservoir's surface velocity is negligible and gravity is neglected as stipulated. Entrance-region inertia contributes an effective length of order $a$, negligible against $L$. The potential difference between the outlet and the reservoir is consequently $LU$ to leading order. Subtract Bernoulli at the reservoir surface, where pressure is $P+p_a$, from its value at the outlet, where pressure is $p_a$ and speed is $U$. This givesThe inertial term cannot be dropped during startup. With $U_{\infty }=\sqrt{2P/\rho }$, separate variables:Hence the inviscid startup in a pressure-driven tube givesInitially $U\sim Pt/(\rho L)$, the acceleration of the stationary liquid column. At long times $U\to \sqrt{2P/\rho }$, the inviscid pressure-to-kinetic-energy limit. The result applies while the tube remains full and reservoir dimensions are large enough that surface motion and the neglected entrance inertia remain small.