With $U=U_0$ and $\beta =0$, the Orr-Sommerfeld equation factors into constant-coefficient operators:For distinct nonzero characteristic roots, the general solution isEither square-root choice for $\gamma$ gives the same solution space. The no-slip boundary condition and impermeability impose $\hat{v}=D\hat{v}=0$ at each wall.

The displayed exponential basis needs the usual repeated-root qualification. If ${\gamma }^2={\alpha }^2\ne 0$, replace it by $(a_1+a_{2}y)e^{\alpha y}+(a_3+a_{4}y)e^{-\alpha y}$. If just one root pair is zero, that pair contributes $1,y$; if both pairs are zero, the basis is $1,y,y^2,y^3$. These are the complete limiting cases, not four independent copies of a repeated exponential.

An Orr-Sommerfeld mode has $\hat{v}\ne 0$ satisfying the Orr-Sommerfeld equation, with accompanying vorticity satisfying the forced Squire equation. A Squire mode has $\hat{v}=0$ and $\hat{\eta }\ne 0$, so its Squire equation is homogeneous.

For a Squire mode multiply that homogeneous equation by $\overline{\eta }$ and integrate over the channel. Under the no-slip condition $\eta =0$ at the walls, integration by parts givesSince $U,\alpha$ are real, taking the real part provesThe numerator is strictly positive for a nontrivial finite-channel mode, and $Re>0$. Thus Squire modes are exponentially damped in the convention $e^{-i\omega t}$. This does not rule out transient growth from non-normal modes in the coupled system.

Choose a length $\ell$, velocity $U_s$, time $\ell /U_s$, and pressure $\rho U_s^2$, giving Reynolds number $Re=U_s\ell /\nu$. Write perturbation velocity as $(u,v,w)$ and let $\mathcal{L}={\partial }_t+U{\partial }_x-R{e}^{-1}{\nabla }^2$. Linearizing the incompressible Navier-Stokes equations givesTaking the divergence gives ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying ${\nabla }^2$ to the wall-normal momentum equation and usingthen cancels the pressure derivatives and yields $\mathcal{L}{\nabla }^{2}v-U^{\prime \prime }{v}_x=0$. Define wall-normal vorticity $\eta =u_z-w_x$. Applying ${\partial }_z$ to the streamwise equation minus ${\partial }_x$ to the spanwise equation gives $\mathcal{L}\eta =-U^{\prime }{v}_z$.

For the assumed normal modes, set $D=d/dy$ and ${\kappa }^2={\alpha }^2+{\beta }^2$. The two equations become the Orr-Sommerfeld equation and Squire equation:At rigid no-slip walls the conditions are $\hat{v}=D\hat{v}=\hat{\eta }=0$ for nonzero horizontal wavenumber.

For fixed $h\in (0,1)$, $X=\alpha h+O({\alpha }^3)$ and $Y=\alpha (1-h)+O({\alpha }^3)$. The factored dispersion relation therefore giveswhere the latter follows from $X,Y\to 1$ and $c^2\sim (1-1/(2\alpha h){)}^2$. A real negative $c^2$ yields a growing branch $c=i\sqrt{-c^2}$ for positive $k$. Thus $h<1/2$ gives instability at sufficiently small nonzero wavenumber.

For the converse, monotonicity toward the limiting value one would suffice, as permitted in the question. In fact a direct sign argument avoids this extra assumption. With $q=\alpha h>0$, $X=\tanh q<q$, so $J-XY=qX+(q-X)Y>0$. If $h\ge 1/2$, then $q\ge d=\alpha (1-h)$ and $Y=\tanh d<d\le q$. Hence $H-Y=q(1+XY)-Y>0$, so $c^2>0$ for every $\alpha >0$. We obtain the exact instability threshold for a bounded piecewise-linear shear layer:At ${\zeta }_L=1/2$ the long-wave limiting value is zero, but every finite positive wavenumber has $c^2>0$.

First suppose $c_i\ne 0$ and put $F=\hat{w}/(U-c)$. The wall condition gives $F(\pm L)=0$. Substitution into the Rayleigh equation for inviscid shear flow givesMultiply by $\overline{F}$ and use integration by parts. The resulting weighted identity for Howard's semicircle theorem isIts imaginary part gives $\int UQ=c_{r}I$, and its real part then gives $\int U^{2}Q=(c_r^2+c_i^2)I$. Set $a=U_{\min }$ and $b=U_{\max }$. Since $(U-a)(b-U)\ge 0$,Completing the square proves Howard's semicircle theorem:In particular every growing normal mode lies in the upper semicircle for $k>0$. If $c$ is real and outside $[a,b]$, $F$ is again regular, but the same identity has a strictly positive integrand unless the mode vanishes. Such a neutral mode is impossible. Consequently neutral regular modes also satisfy the bound; no division by $c_i$ is needed for this last conclusion. The nonreal proof avoids critical-level singularities altogether.

For the horizontal film, part (a) gives the lubrication theory equationThe length scale is fixed at $L$, so balancing $h/t$ against $K{h}^4/L^2$ gives $h\sim (L^2/(Kt){)}^{1/3}$. The draining viscous film on a finite horizontal plate therefore has the long-time similarity solutionAn additive shift of the time origin can represent the initial transient without changing the long-time law. The positive profile is even, with $H^{\prime }(0)=0$ and $H(\pm 1)=0$. Substitution yieldsTo integrate, put $q=H^3{H}^{\prime }$ and use $q^{\prime }=qdq/dH/H^3$. ThenFor $0<\eta <1$ the decreasing branch has $q<0$, so inversion givesThe negative half of the profile follows by symmetry.

At the edge, $\eta (0)=1$. Set $u=(s/H_0{)}^5$ in the integral to obtainHere $B$ is the beta function. Thus $H_0^3=10/(3C^2)$, and the centre thickness is

For $H\ll H_0$, the implicit integral remaining between $0$ and $H$ givesConsequently the edge region of a draining viscous film hasIts slope satisfies $|h_x|\sim (h_0/L)(1-x/L{)}^{-3/4}$ up to a constant factor, so it becomes order one atThe local height is then also $O(\ell )$, violating the shallow geometry required by lubrication theory. A full local flow is needed to turn the fluid over the edge.

The imposed zero thickness is nevertheless a consistent leading outer boundary condition: $\ell /h_0\sim (h_0/L{)}^{1/3}\ll 1$, so the unresolved edge height is small compared with the film's bulk height. It should not be interpreted as an exact pointwise prediction inside the edge region. Moreover, $H^3{H}^{\prime }\to -\sqrt{2/15}{H}_0^{5/2}$ at the right edge, so the outward volume flux remains finite even as the outer thickness tends to zero. The singular slope is what allows this outer solution to describe drainage.

For an axisymmetric viscous gravity current on a cone, the cylindrical radius at slant distance $x$ is $x\cos \alpha$. A surface annulus therefore has area $2\pi x\cos \alpha dx$. The local downslope volume flux is the same as in part (a), but conservation of mass must account for this growing circumference. ThusThis is a leading thin-film description away from the apex, with $0<\alpha <\pi /2$ and thickness small relative to the local curvature and streamwise length scales.

The original PDF uses the long-time criterion $x_N\gg (V\csc \alpha {)}^{1/3}$; the local TeX incorrectly reads $V\cos \alpha$. To see why the PDF criterion is the relevant one, scale $h\sim V/(x_N^2\cos \alpha )$. In the bulk, the hydrostatic-gradient term relative to direct downslope gravity isHence the leading outer evolution, with $\beta =K\sin \alpha$, isFor a fixed-volume similarity solution, let $x_N\propto t^b$ and $h\propto t^a$. Conservation of mass gives $a+2b=0$, while the evolution equation gives $a-1=3a-b$. Thus $b=1/5$, $a=-2/5$. Writing $h=t^{-2/5}H(\xi )$, $\xi =x/t^{1/5}$, yieldsThe positive branch $H=c\sqrt{\xi }$ satisfies this equation when $c^2=1/(5\beta )$. It givesThe ideal outer current has zero thickness beyond its front. Its mass integral isThereforeThe reduced equation is a conservation law after multiplication by $x$. Its front obeys the Rankine-Hugoniot condition, ${\dot{x}}_N=\beta h_N^2=x_N/(5t)$, consistent with the mass-derived front position. The outer profile has a finite jump at the front; the omitted hydrostatic-gradient term resolves a narrow front region. There is also a small apex region where the outer gradient and conical curvature violate the local approximations. Neither region sets the leading conserved-volume scaling above.