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Let $\ell =(H_m-H_0)/\sin \alpha$ be the along-slope distance to the snowline, so $H(x)=H_m-x\sin \alpha$ and $a(x)=A(1-x/\ell )$. Define $D=g\sin \alpha /(3\nu )$.

Treat ice as an incompressible Newtonian fluid, neglect inertia, and use lubrication theory with thickness measured normal to the slope. The bed has a no-slip boundary condition, the free surface has zero tangential stress, and normal pressure is hydrostatic. The small thickness slope allows its pressure-gradient contribution to be neglected against gravity along the mountain. With normal coordinate $y$, the tangential equation and boundary conditions areIntegration givesThe horizontal radius of a ring is $x\cos \alpha$, so mass conservation gives the gravity-driven ice flow on a conical slope equationNegative accumulation is ice ablation and applies only where ice exists; the ice-free region has $h=0$.

In a steady state, regularity and zero total flux at the apex require $xD{h}^3\to 0$ as $x\to 0$. Integrating givesRequiring a continuous zero-thickness steady terminus yields the steady conical ice cap with a linear accumulation gradient:The terminus lies below the snowline, allowing the ablation region to balance snowfall. The maximum thickness occurs at $x=3\ell /4$.

The volume follows from integrating the ring areas. With $s=x/\ell$,Substituting $D$ and $\ell =\Delta H/\sin \alpha$ givesThe ideal outer profile has steep slopes very close to the apex and terminus. Those small regions require local corrections to the assumed slope balance, while the bulk profile and leading volume follow from the stated approximation.

Use the Cartesian streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Taking the curl of the incompressible Stokes equation eliminates pressure and gives the biharmonic stream function for planar Stokes flow equationThe material points have no horizontal velocity in the swimming frame and have vertical velocity $-ϵ\cos \theta$. The no-slip boundary condition is thereforeAt infinity the laboratory fluid is at rest, so in this translating frame it moves with $U{\mathbf{e}}_x$:Velocity and its perturbations are periodic in $x$, with period $2\pi$, and the pressure has no imposed mean gradient. The Taylor swimming sheet is force-free; the unbounded problem's bounded far-field velocity enforces the absence of mean shear. An additive constant in the streamfunction has no physical effect.

The wall is stationary in the laboratory and moves with velocity $+U{\mathbf{e}}_x$ in the sheet frame. Its no-slip boundary condition is thereforeThe first equation prescribes wall tangential velocity; the second imposes zero normal velocity. The streamfunction may have a constant mean value on the wall, set by the fluid transport, and that constant need not equal its value at the sheet.

Away from the rim's $O(h)$ edge region, the geometry and uniform injection have no radial scale other than the factor required by continuity. It is therefore consistent to take $w=w(z)$. Regularity at $r=0$ then lets the continuity equation integrate toThe no-slip boundary condition and prescribed normal velocities areSince $p$ is independent of $z$, the radial equation implies that $w^{\prime \prime \prime }$ is constant. The four boundary conditions giveand henceUsing $w^{\prime \prime \prime }=12V/h^3$ in $p_r=\mu u_{zz}$ givesTaking the pressure at the rim to be atmospheric, $p(R)=p_0$, produces the porous-plate lubrication cushion pressureThe upward pressure force balances the disc's weight:Since $\epsilon =h/R$,

Let $\mathbf{u}$ be the Stokes flow and let $\mathbf{v}$ be any other admissible incompressible flow with the same no-slip boundary condition on the sphere and the same decay at infinity. PutThen $\nabla \cdot \mathbf{w}=0$ and $\mathbf{w}=0$ on the sphere. The viscous dissipation satisfiesFor the Stokes stressincompressibility of $\mathbf{w}$ givesThe divergence theorem, the homogeneous boundary data, decay at infinity, and $\nabla \cdot \mathbf{\sigma }=0$ giveThereforeThis proves the Minimum-dissipation theorem for Stokes flow.

Let the sphere have radius $a$; set $a=1$ for the unit sphere implicit in the question. Sinceone findsOn $r=a$, the no-slip boundary condition requires $\mathbf{u}=\mathbf{U}$ for every surface normal. The coefficient of
$(\mathbf{U}\cdot \mathbf{n})\mathbf{n}$ must vanish, while the coefficient of $\mathbf{U}$ must equal one:Solving,For a unit sphere,Substitution reproduces the standard translating sphere in Stokes flow.

The inner cylinder is $x^2+(y-a{)}^2=a^2$. With $x=r\cos \theta$ and $y=r\sin \theta$, this becomes $r^2-2ar\sin \theta =0$, so away from the touching pointAt this boundary the outward radial vector from the cylinder center is $(\sin 2\theta ,-\cos 2\theta )$, so the counterclockwise unit tangent is $\mathbf{t}=(\cos 2\theta ,\sin 2\theta )$. Since ${\mathbf{e}}_r\cdot \mathbf{t}=\cos \theta$ and ${\mathbf{e}}_{\theta }\cdot \mathbf{t}=\sin \theta$, the no-slip boundary condition givesFor a streamfunction in polar coordinates, no penetration through a stationary material curve makes $\psi$ constant along that curve. Both cylinders touch at the origin, so continuity forces their constants to agree. Since adding a constant does not alter velocity, choose their common value as $\boxed{\psi =0}$.

Let $x$ point down the plane and $z$ normal to it, with velocity $(u,w)$. The two-dimensional incompressible Navier-Stokes equation isThe lubrication theory limit has film thickness much smaller than its streamwise length. Transverse viscous derivatives dominate, inertia is negligible at late times, and the leading equations areAt the wall, $u=w=0$ by the no-slip boundary condition. At the free surface $z=h(x,t)$, take $p=p_{\mathrm{atm}}$ and $u_z=0$, neglecting surface tension. The normal balance gives the hydrostatic pressureIf the current has thickness scale $H$ and length scale $L$, thenThus the streamwise pressure gradient is asymptotically smaller than the downslope body force for a fixed nonzero inclination. The streamwise equation becomesUsing $u(0)=0$ and $u_z(h)=0$ givesThe volume flux isIntegrating incompressibility across the film and using the kinematic boundary condition gives $h_t+q_x=0$, hence

Conservation of volume flux through a circular arc of radius $r$ givesso the outer flow is the sink flow in a sectorThis is a potential flow and already satisfies the no-penetration boundary condition on the radial walls. It has nonzero tangential speed $Q/r$ there, however, and therefore violates the no-slip boundary condition. Thin vorticity-bearing boundary layers are needed to reduce that speed to zero at each wall.

The axial velocity $u(r)$ decreases toward the wall, so $\dot{\gamma }=-du/dr\ge 0$. For a power-law fluid, the stress magnitude isHenceIntegrating inward from the no-slip boundary condition $u(R)=0$ givesFor $n=1$ this reduces to the parabolic Hagen--Poiseuille profile.

Both candidate fields are divergence-free. For solid-body rotation,each component is linear in $\mathbf{x}$, so ${\nabla }^2\mathbf{u}=0$. Also,is harmonic for $r>0$, because ${\nabla }^2(1/r)=0$ there and constant-coefficient derivatives commute with the Laplacian. Thus both solve the Stokes equations with $p=0$.

The rotationally symmetric solution is consequentlyThe no-slip boundary condition givesThereforeandThis is the Rotational Stokes flow between concentric spheres.

For the steady, fully developed ansatz $\mathbf{u}=w(r){\mathbf{e}}_z$, both ${\partial }_t\mathbf{u}$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanish. The axial Navier-Stokes equation reduces toTwo integrations giveRegularity at the axis forces $C_1=0$, and the no-slip boundary condition $w(R)=0$ gives the Hagen-Poiseuille flowFor flow in the positive $z$ direction, $\partial p/\partial z<0$.

Use the Papkovich–Neuber representation in the formwhere $\mathbf{\Phi }$ and $\chi$ are harmonic functions. A torque is a axial vector, so rotational covariance and decay select the harmonic vector fieldHere $\mathbf{x}\cdot \mathbf{\Phi }=0$ and $\nabla \cdot \mathbf{\Phi }=0$, soThis rotlet equals the rotating sphere in Stokes flowwhen $\mathbf{\Omega }=\mathbf{G}/(8\pi \mu a^3)$. It satisfies the no-slip boundary condition on $r=a$, decays at infinity, and its Newtonian fluid stress tensor transmits the applied couple $\mathbf{G}$.

The Reynolds number isWhen it is large, viscosity is negligible in most of the flow, but the inviscid outer solution cannot generally satisfy the no-slip boundary condition at a rigid wall. A thin boundary layer resolves this mismatch: tangential gradients remain of outer scale $\mathcal{L}$, normal gradients have the much shorter scale $\delta$, and the small viscous coefficient is offset by $u_{yy}=O(\mathcal{U}/{\delta }^2)$. Matching a viscous inner solution to an inviscid outer solution retains the leading viscous effects without solving the full equations uniformly everywhere.

The Papkovich–Neuber representation writes a homogeneous incompressible Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and harmonic scalar $\chi$:with ${\nabla }^2\mathbf{\Phi }=0$ and ${\nabla }^2\chi =0$. The representation satisfies incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and substitution then verifies the Stokes equation.

For a sphere translating with constant vector velocity $\mathbf{U}$, rotational covariance and decay at infinity suggest a harmonic vector monopole and scalar dipole:Substitution gives the translating sphere in Stokes flowAt $r=a$ the radial tensor terms cancel and $\mathbf{u}=\mathbf{U}$, while $\mathbf{u}\to 0$ as $r\to \infty$, so the no-slip boundary condition and far-field condition hold. The resulting traction integrates to the Stokes drag law $6\pi \mu a\mathbf{U}$ in magnitude.

A rigid body obeys the no-slip boundary condition. Its outer potential flow generally has an $O(U)$ tangential slip velocity at the surface, so the boundary layer must make an $O(U)$ velocity correction across thickness $\delta$. Its strain is therefore $O(U/\delta )$, and in two dimensions its dissipation per unit axial length iswhich exceeds the $O(\mu U^2)$ outer dissipation by $a/\delta \sim {\mathrm{Re}}^{1/2}$.

The dominant drag consequently depends on the detailed boundary-layer velocity field and possibly on boundary-layer separation. It cannot be obtained by integrating the outer irrotational strain alone.

The leading streamwise momentum equation in lubrication theory iswith $p=p(x)$. The no-slip boundary condition isTwice integrating in $y$ gives the local Couette-Poiseuille flow in a thin gapIts volume flux per unit span is

Let $z$ measure height above the sponge. Under the long-wave approximation, hydrostatic pressure gives $p_x=\rho g{H}_x$, and the horizontal balance for viscous fluid flow isThe no-slip boundary condition $u(0)=0$ and stress-free boundary condition $u_z(H)=0$ giveIntegrating this lubrication theory profile gives the horizontal volume flux per unit span

Local mass conservation includes the downward loss found in part i:Consequently the required nonlinear partial differential equation isIf the imposed inlet flux is $Q_0{t}^{\alpha }$ and the moving front is $x=x_N(t)$, sufficient boundary conditions areFor a current released onto a dry substrate one also takes the initial condition $H(x,0)=0$ away from the source. The front position is part of this moving-boundary problem.

Let $d(z,t)$ be the local water-film thickness. To first order in the interface amplitudes,The lubrication theory flux down the vertical surface, with a no-slip boundary condition at the ice and a stress-free boundary condition at the water-air interface, isFor the unperturbed film $p_z=0$, so

The linearized Young–Laplace equation gives the capillary-pressure perturbationBecause the prescribed volume flux is uniform, its first-order perturbation must vanish. Linearization of the flux law givesThus, with $\Gamma =\gamma /(3\rho g)$,and henceThe complex amplitude ratio records the phase shift caused by surface tension in the long-wave approximation.

Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In lubrication theory, radial velocity and radial pressure variation are negligible. Axisymmetric incompressible flow is obtained frombecause ${\partial }_{\theta }(u_{\theta }\sin \theta )=0$. The tangential Stokes equation then separates:and hence, after choosing an irrelevant pressure constant,

In the sphere frame, $u_z=-u_{\theta }\sin \theta =-\varphi$. The no-slip boundary condition gives $\varphi (0)=0$ on the sphere and $\varphi (\delta )=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^{2}U$, soUnder the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral givesChanging the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.

The pressure scale is $p\sim \mu UR/{\delta }^2$, whereas the viscous shear scale is $\tau \sim \mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta \gg 1$, an instance of lubrication pressure dominates shear stress. Put $c=\cos \Delta \theta$. The axial pressure force isAs $\Delta \theta \to 0$, the bracket tends to $-2$, and thereforeThe resulting confined-sphere drag coefficient isThus it exceeds the free Stokes drag law coefficient by $2R^2/{\delta }^2$. The Stokes–Einstein relation then gives