Let $\mathbf{x}(\mathbf{X},t)$ be the particle flow map from the fixed reference volume $V_0$ to the material volume, and let $J=\det (\partial \mathbf{x}/\partial \mathbf{X})$. Change variables and differentiate on the fixed domain:Use the two permitted identities, $DJ/Dt=J\nabla \cdot \mathbf{u}$ and $Df/Dt=f_t+\mathbf{u}\cdot \nabla f$. Changing back givesThis is the Reynolds transport theorem: both the value carried by each particle and the changing volume element contribute.

For $U<0$ the same positive algebraic root remains real and satisfies $0<V<c_0$, because $V^2-c_0^2=(\gamma +1)UV/2<0$. The jump relations then give a density decrease, ${\rho }_1/{\rho }_0=V/(V-U)<1$, and a pressure decrease. Where both states remain thermodynamically positive, this is an expansion shock: its upstream shock-frame flow is subsonic and its entropy change has the forbidden sign. At sufficiently large negative $U$ the formula even predicts nonpositive pressure, an additional reason it cannot represent a physical gas state.

The observed outgoing wave is instead a continuous isentropic rarefaction wave. Its plus-family rays spread from the initial piston corner rather than coalescing. The fan head enters the undisturbed gas at $x=c_{0}t$; for moderate recession the tail is $x=(U+c_1)t$, with the uniform gas next to the piston moving at $U$ and $c_1=c_0+(\gamma -1)U/2>0$. The appropriate Riemann invariant is $u-2c/(\gamma -1)=-2c_0/(\gamma -1)$, so inside the fan, with $\xi =x/t$,These expressions show the spreading characteristics in the right panel of the preceding figure. If $U\le -2c_0/(\gamma -1)$, the gas cannot follow the piston with positive sound speed; a vacuum gap forms, bounded by the limiting rarefaction wave edge. Thus the fan description also identifies the strong-recession limit.

The second law requires the specific entropy to increase along the fluid's passage through a nontrivial physical shock: $s_1>s_0$. For a perfect gas this selects a compressive shock, whose upstream shock-frame Mach number is greater than one and whose downstream shock-frame flow is subsonic.

Here the upstream Mach number is $M_0=V/c_0$. The quadratic gives $V^2-c_0^2=(\gamma +1)UV/2>0$, so $\boxed{M_0>1}$. Density and pressure both rise. With $c_1^2=\gamma p_1/{\rho }_1$, one also has $V-U<c_1$. For example, put $M=V/c_0$ and use the normal-shock identityThe characteristic geometry is therefore $c_0<V<U+c_1$. The plus-family characteristics from the gas ahead are overtaken by the shock, while those behind catch up with it; they converge into the discontinuity. The minus-family speeds are $-c_0$ ahead and $U-c_1$ behind. The piston is the line $x=Ut$ and the shock is the steeper-in-$x/t$ line $x=Vt$.

Let state zero be the undisturbed gas and state one the uniform gas between piston and shock. Their laboratory velocities are zero and $U$. In the shock frame their speed magnitudes are $V$ and $V-U$. Mass and momentum jumps giveEnergy conservation gives $\gamma [p_1/{\rho }_1-p_0/{\rho }_0]/(\gamma -1)=VU-U^2/2$. Substitute the preceding relations and $\gamma p_0/{\rho }_0=c_0^2$, then divide by the nonzero $U$. The result isThe right-moving root isThe other root is negative and does not describe the outgoing shock. For $U>0$, the quadratic is negative at $V=U$, hence its positive root obeys $V>U$: the shock stays ahead of the piston and the density ratio is positive.

Integrate each steady one-dimensional conservation law across a thin region enclosing the shock. Mass conservation gives constant mass flux $j=\rho u$. Momentum conservation includes both advected momentum and pressure force, so $p+\rho u^2$ is constant. The total energy flux is $u(\rho e+p)=j(e+p/\rho )$, where $e=c_{v}T+u^2/2$. For a genuine shock $j\ne 0$, division by the mass flux and the equation of state $c_{v}T=p/[(\gamma -1)\rho ]$ give the conserved stagnation enthalpyThus the Rankine-Hugoniot conditions for a perfect gas areThe velocities here are measured in the stationary shock frame; a moving-shock problem must first be transformed to that frame.

Use $\hat{\eta }(k,t)=\int \eta (x,t)e^{-ikx}dx$ and the inverse factor $1/(2\pi )$. Fourier transformation gives ${\hat{\eta }}_{tt}+{\lambda }^2{k}^4\hat{\eta }=0$. The supplied Lorentzian transform and zero initial velocity therefore implyAlong $x=Vt$, split the cosine into two exponentials. Their phases are ${\psi }_{\pm }=Vk\pm \lambda k^2$, with stationary points $k_{\pm }=\mp V/(2\lambda )$, phase values $\mp V^2/(4\lambda )$ and second derivatives $\pm 2\lambda$. Both amplitudes equal $a{e}^{-ϵ|V|/(2\lambda )}$. The two stationary-phase contributions are conjugate, givingThus the finite-width pulse retains a ray-dependent attenuation. In the further regime $ϵ|V|/(2\lambda )\ll 1$, it may be replaced by one, recoveringThe large-time ray limit means $V=x/t$ is fixed. Merely making both $x$ and $t$ large does not justify dropping the attenuation if $|x|/t$ becomes large compared with $\lambda /ϵ$. The leading oscillation decays as $t^{-1/2}$ because each saddle samples a Fourier band of width $t^{-1/2}$.

Split $I$ into the two tails and the interval $|k-k_{\ast }|<\delta$. On each tail ${\psi }^{\prime }$ has a fixed nonzero sign, so changing variable to $s=\psi (k)$ produces a nonstationary Fourier integral. Under the usual decay and regularity needed for integration by parts, these pieces are $O(t^{-1})$.

Near the nondegenerate stationary point, make the signed local changeIt has $dk/dz=1$ at zero and makes the phase exactly $\psi (k_{\ast })+{\psi }^{\prime \prime }(k_{\ast })z^2/2$. The transformed amplitude tends to $f(k_{\ast })$. Scale $z$ by $t^{-1/2}$ and use the Fresnel integral to obtainThe $O(t^{-1})$ tails are smaller than this leading stationary contribution. If ${\psi }^{\prime \prime }(k_{\ast })<0$, complex conjugation of the Gaussian phase changes the phase correction to $-\pi /4$ and replaces the second derivative by its absolute value:When the leading coefficient is nonzero this may be written as an asymptotic equivalence. For full rigor, continuity of an arbitrary amplitude alone is not sufficient for the stated nonstationary $O(t^{-1})$ rate, nor does it ensure convergence of an infinite integral. Piecewise differentiability with integrable derivative, bounded variation, or an equivalent localization/decay hypothesis supplies the usual missing assumption. The exponentially decaying beam amplitude below has the required piecewise regularity.

Retain the rigid-wall cosine in $x$ and the lower-wall cosine in $y$:The wave equation gives ${\omega }^2=c_0^2[(n\pi /a{)}^2+(\beta /b{)}^2]$. The upper-wall condition gives $-k\beta \sin \beta /b-{\omega }^2\cos \beta =0$, henceFor nonzero real $\beta$, eliminating $\omega$ gives $-\tan \beta =M(\beta +{\kappa }_n^2/\beta )$ with $M=c_0^2/(kb)>0$, ${\kappa }_n=n\pi b/a$.

On every interval $((j+1/2)\pi ,(j+1)\pi )$, the left side decreases from positive infinity to zero, while the right side is finite and positive. Continuity gives at least one intersection per interval for each $n$. The functions are analytic between their poles, so roots are isolated there; their union over countably many intervals is a countably infinite set. Reflection gives the redundant negative-$\beta$ solutions. For $n=0$, the limiting formal root $\beta =0$ is the zero-frequency gauge mode, not one of these positive-frequency branches.

The graph uses representative dimensionless values $M=1$ and $b/a=1$; the interval argument applies for all positive values. For fixed $n$ and fixed branch $j$, take $M\to \infty$. The positive right side becomes large, forcing the intersection towards the tangent pole from the right:More precisely, if ${\beta }_j^0=(j+1/2)\pi$, then ${\beta }_j-{\beta }_j^0\sim {\beta }_j^0/[M(({\beta }_j^0{)}^2+{\kappa }_n^2)]$. Therefore the requested real-branch frequencies satisfyThe limiting upper condition is effectively pressure release for these acoustic branches. No claim is made here about imaginary or other complex $\beta$ branches, which were excluded from the requested analysis.

At leading order the two wall conditions may be evaluated on the undeformed line $y=b$: shifting the evaluation point multiplies small perturbations and creates only second-order errors. The dynamic relation and the linear Bernoulli relation give $\eta =p^{\prime }/({\rho }_{0}k)=-{\varphi }_t/k$. Differentiate this in time and use ${\varphi }_y={\eta }_t$:The sign follows from the convention that positive pressure displaces the upper wall upward.

Impermeability imposes homogeneous Neumann data: ${\varphi }_x=0$ at $x=0,a$ and ${\varphi }_y=0$ at $y=0,b$. Separation of variables therefore givesOne may take $n,m\ge 0$ without duplication. For a nontrivial oscillation they are not both zero. A time-independent spatially constant potential is a gauge mode at zero frequency, rather than an additional sound oscillation.

The inviscid equations are ${\rho }_t+\nabla \cdot (\rho \mathbf{u})=0$ and $\rho [{\mathbf{u}}_t+(\mathbf{u}\cdot \nabla )\mathbf{u}]=-\nabla p$. Write $\rho ={\rho }_0+{\rho }^{\prime }$, $p=P({\rho }_0)+p^{\prime }$ and retain only linear terms. With $c_0^2=P^{\prime }({\rho }_0)>0$, barotropy gives $p^{\prime }=c_0^2{\rho }^{\prime }$. Irrotational perturbations allow ${\mathbf{u}}^{\prime }=\nabla \varphi$. The momentum equation integrates to $p^{\prime }=-{\rho }_0{\varphi }_t$, after using the time-dependent gauge of $\varphi$ to remove the spatially constant integration term. ThusSubstitution into ${\rho }_t^{\prime }+{\rho }_0\Delta \varphi =0$ gives the acoustic wave equation $\boxed{{\varphi }_{tt}=c_0^2\Delta \varphi }$.

For spatially uniform $h(t)$, the gap equation gives $\Delta p=12\mu \dot{h}/h^3$. Axisymmetry makes this $r^{-1}(r{p}_r{)}_r=12\mu \dot{h}/h^3$. Its general integral contains a logarithmic term; bounded pressure at zero excludes that term. Enforcing zero gauge pressure at the outer edge gives the axisymmetric squeeze-film pressureThe mean radial velocity is ${\overline{u}}_r=-h^2{p}_r/(12\mu )=-r\dot{h}/(2h)$. At the material edge this must equal $\dot{R}$, so $2h\dot{R}+R\dot{h}=0$. HenceIt is conservation of the incompressible liquid volume, with no liquid entering or leaving through the plates or material rim.

When the plates approach, $\dot{h}<0$ and $p>0$ inside: the liquid is squeezed outward. Separation requires $p<0$ relative to the surrounding air and inward motion of the edge. Negative gauge pressure is not intrinsically impossible, but substantial suction can cause cavitation or air invasion, and an advancing air/liquid interface can develop fingering. Thus the same mathematical lubrication solution exists for either sign, whereas its opening-plate physical validity needs these additional failure mechanisms to be absent; the closing case is the robust regime of the stated idealized model.

Scale $x,y$ by $L$, $z$ by $\delta L$, tangential velocities by $U$, normal velocity by $\delta U$, and time by $L/U$. Continuity balances all three derivatives. Tangential viscous forces scale as $\mu U/({\delta }^2{L}^2)$, hence their pressure balance requiresThe inertial/normal-viscous ratio is ${\delta }^2\rho LU/\mu$, so it is negligible under the given smallness condition. Tangential viscous derivatives are smaller by ${\delta }^2$; the normal pressure balance gives $p_z=0$ at leading order. In dimensional variables the remaining equations are continuity, $\mu u_{zz}=p_x$, $\mu v_{zz}=p_y$, $p_z=0$.

No-slip boundary condition at each plate imposes $u=v=0$, while their normal velocities are $w(0)=0$ and $w(h)=\dot{h}$. Integrating the two tangential equations twice givesIntegrate continuity through the gap. Since $h$ is independent of in-plane position and the tangential velocity vanishes at the upper wall, the flux divergence is $\nabla \cdot (h\overline{\mathbf{u}})$ and the normal contribution is $\dot{h}$. Therefore

The Von Mises boundary-layer transformation uses $\Psi$ as transverse coordinate. With $u=U(x,\Psi )$ and ${\Psi }_x=-V$, ${\Psi }_Y=u=U$, the chain rule givesThe streamwise advection terms reduce to $u{u}_x+V{u}_Y=U{U}_x$, so, for $Y>0$ where $U>0$, division by $U$ givesContinuity was already enforced by the stream function. Substitute $U=g(\eta )$ with $\eta =\Psi /\sqrt{x}$. Then $U_x=-\eta g^{\prime }/(2x)$ and ${\partial }_{\Psi }(U{U}_{\Psi })=(g{g}^{\prime }{)}^{\prime }/x$, yieldingThe wall coordinate is zero because $\Psi$ was set to zero there, and the far condition matches unit external speed. This is the transformed Blasius similarity problem. One must not additionally prescribe a finite $g^{\prime }(0)$: the coordinate change degenerates at the no-slip wall, and the regular physical shear corresponds to $g\propto \sqrt{\eta }$ near zero. This similarity reduction is obtained for the attached monotone boundary layer assumed in the question.

Existence of a positive normalized profile can also be seen directly. Put $q=g{g}^{\prime }$ and choose $q(0)=a>0$. In $g$ coordinates the regular initial-value system is $dq/dg=-\eta /2$, $d\eta /dg=g/q$, with $\eta (0)=0$. It starts as $g\sim \sqrt{2a\eta }$. For finite positive $\eta$, $q^{\prime }/q=-\eta /(2g)$ keeps $q$ positive and $q\le a$, while $(g^2{)}^{\prime }=2q$ gives $g\le \sqrt{2a\eta }$. Consequently $q^{\prime }/q\le -\sqrt{\eta }/(2\sqrt{2a})$, so $q$ is integrable at infinity. It follows that $g$ increases to a finite positive limit $M$. The rescaled profile $\tilde{g}(\eta )=M^{-1}g(\sqrt{M}\eta )$ satisfies the same equation and has limit one. This supplies the requested similarity solution with both boundary conditions, rather than only its formal differential equation.

Let $\delta =R{e}^{-1/2}$ and use the inner variables $\boxed{Y=y/\delta =R{e}^{1/2}y,V=v/\delta =R{e}^{1/2}v}$. Continuity balances $u_x$ with $V_Y$. In the tangential momentum balance, advection is order one while the normal viscous term is $R{e}^{-1}{\delta }^{-2}{u}_{YY}=u_{YY}$. Streamwise viscosity is smaller by $R{e}^{-1}$. The normal momentum balance forces the leading pressure to be independent of $Y$. ThusOn the stationary impermeable plate, no-slip boundary condition and no-penetration boundary condition give $u=V=0$ at $Y=0$, $x>0$. Matching to the uniform outer stream gives $u\to 1$ as $Y\to \infty$, and the leading-edge/upstream data are $u\to 1$ as $x\downarrow 0$ at fixed $Y>0$. There is no requirement that the scaled transverse velocity $V$ vanish at infinity; its physical value $v=\delta V$ matches a smaller outer correction. The outer streamwise Euler equation with constant speed one gives $p_x=0$, and $p_Y=0$ carries this pressure gradient throughout the layer.

For $\mathbf{u}=u(y,t){\mathbf{e}}_x$, incompressibility is automatic and the nonlinear advection term is zero. With no pressure gradient the streamwise momentum equation is $\boxed{\rho u_t=\mu u_{yy}}$. Seek its periodic response as $u=\mathrm{Re}[Uf(y)e^{i\Omega t}]$. The amplitude satisfies $f^{\prime \prime }=i\rho \Omega f/\mu$ with $f(\pm d)=1$. Since $[k(1+i){]}^2=i\rho \Omega /\mu$ for $k=\sqrt{\rho \Omega /(2\mu )}$, even symmetry givesOdd components vanish because the two boundary amplitudes are equal. This is the unique periodic solution. Without a specified initial profile, decaying zero-wall diffusion transients may be added, so the formula should not be read as a claim that every transient is absent.

As $kd\to \infty$, the wall motions penetrate only a distance $k^{-1}=\sqrt{2\mu /(\rho \Omega )}$. Near either wall, with inward distance $s=d-|y|$, the leading velocity isThe two oscillatory viscous layers are separated by an almost motionless core; the phase lag increases inward. This is in-phase oscillating parallel-plate flow.

For a material volume $V(t)$ moving with velocity $\mathbf{u}$, the Reynolds transport theorem isIt applies componentwise to vector quantities. Cauchy's stress theorem represents traction on a surface with outward unit normal $\mathbf{n}$ as $t_i={\sigma }_{ij}{n}_j$. Conservation of mass gives $d{\int }_{V(t)}\rho dV/dt=0$ and hence ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Constant density reduces this to $\boxed{\nabla \cdot \mathbf{u}=0}$.

With no body force, conservation of momentum and the divergence theorem giveApplying transport and then mass conservation cancels the terms proportional to $u_i$, leaving $\rho ({\partial }_t{u}_i+u_j{\partial }_j{u}_i)={\partial }_j{\sigma }_{ij}$. For the stated Newtonian stress, constant viscosity gives ${\partial }_j{\sigma }_{ij}=-{\partial }_{i}p+\mu \Delta u_i+\mu {\partial }_i(\nabla \cdot \mathbf{u})$. The last term vanishes. Since the material volume was arbitrary,A body force, if present, would have to be added to the momentum balance; it is absent from the specified model.

Integrate once and use $h=e^{-\alpha x}$ to obtain $p^{\prime }=6e^{2\alpha x}+C{e}^{3\alpha x}$. The two zero endpoint pressures determineConsequentlyThe constant $M={\int }_0^1{h}^{-2}dx/{\int }_0^1{h}^{-3}dx$ is a weighted average of $h$ with positive weight $h^{-3}$, so $e^{-\alpha }<M<1$. Since $e^{-\alpha x}$ decreases strictly, $p^{\prime }$ changes sign exactly once, from positive to negative. Thus the unique maximum is atFor small $\alpha$, $\log M=-\alpha /2-5{\alpha }^2/24+O({\alpha }^4)$, so $\boxed{x_m=\frac{1}{2}+\frac{5}{24}\alpha +O({\alpha }^3)\sim \frac{1}{2}}$. The exponential-gap slider bearing provides a particularly simple example where a converging gap produces positive load-bearing pressure.

Let $X=x_{\dim }/L$, $Y=y/h_0$, scale horizontal and vertical velocities by $U$ and $U{h}_0/L$, and define $P=(p_{\dim }-p_0)h_0^2/(\mu UL)$. The small aspect ratio makes $P$ independent of $Y$ at leading order. The second small parameter, $U{h}_0^2/(\nu L)$, makes inertia negligible compared with transverse viscous stress. Thus the lubrication equations are $u_{YY}=P_X$ and $u_X+V_Y=0$.

No slip gives $u(X,0)=1$, $u(X,h)=0$, soNo penetration at the two solid walls and continuity make $q$ constant. Therefore, writing the dimensionless variables again as $x,p$,The pressure scale is set by balancing the longitudinal pressure gradient with shear across the thin gap, not by the inertial dynamic-pressure scale.

The normalization denominator vanishes when $\sin 2\alpha -2\alpha \cos 2\alpha =0$, namely $\boxed{\tan 2{\alpha }_c=2{\alpha }_c}$. The relevant first root above $\pi /2$ is ${\alpha }_c\simeq 2.24670$ radians. The zero root is not the critical angle of interest.

For $\pi /2<\alpha <{\alpha }_c$ the denominator is positive. The profile is even, with outward flow near the bisector and reverse flow adjacent to each wall:It vanishes at the walls and at $\theta =\pm (\pi -\alpha )$; its minima are at $\theta =\pm \pi /2$. The following sketch shows the leading approximation to $G$ for small $ϵ$.

As $\alpha$ approaches ${\alpha }_c$ from below, the large positive central and negative wall contributions nearly cancel while the prescribed net flux remains one. The leading Stokes solution diverges, so the supposedly small nonlinear inertial term can no longer be treated as a perturbation. This is the critical angle of a flux-normalized Stokes wedge flow. It demonstrates breakdown of this regular expansion; it does not establish that the exact viscous flow must have infinite speed or that a global steady branch continues through the critical angle.