The only velocity component is axial, so incompressibility gives ${\partial }_{z}w=0$. Thus $w=w(x,y)$, and the steady convective acceleration is $w{\partial }_z(w\mathbf{k})=0$. The transverse momentum equations give $p_x=p_y=0$, so $p=p(z)$. The axial momentum equation is thereforeIts left side depends only on $x,y$, while its right side depends only on $z$. Equality throughout the pipe forces both to be one constant, say $G$. Consequently the axial pressure gradient is constant, and $p(z)=Gz+p_{\ast }$. This is absence of dependence on a coordinate, rather than probabilistic independence.

Insert the traction formula into the material-volume balance and apply the divergence theorem to each row of the stress tensor:The volume is arbitrary, so continuity of the fields gives the local Cauchy momentum equationFor an incompressible Newtonian fluid, the constitutive law isHere the pressure enforces incompressibility, and the viscous stress is twice the viscosity times the symmetric rate-of-strain tensor. With constant $\mu$,Using $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ proves the stated Navier-Stokes equations.

Define ${\sigma }_{ij}={\mathbf{e}}_i\cdot \mathbf{t}({\mathbf{e}}_j)$: the second index labels the normal direction of a coordinate face, and the first labels the force component. These nine entries form the Cauchy stress tensor.

Take a small tetrahedron with oblique face of area $S$ and outward normal $\mathbf{n}$. First suppose $n_j>0$; its other three faces have normals $-{\mathbf{e}}_j$ and areas $S_j=n_{j}S$, by orthogonal projection of the oblique face. The total surface force isIf its linear size is $\ell$, its inertia is $O({\ell }^3)$ while its surface forces are $O({\ell }^2)$. Divide the momentum balance by $S$ and shrink the tetrahedron to the point. Smoothness of the stress and the odd-normal relation then give $\mathbf{t}(\mathbf{n})={\sum }_j{n}_j\mathbf{t}({\mathbf{e}}_j)$. Choosing coordinate faces with the appropriate reversed normals gives the same identity in the other octants. In components,The face-area relations used are exactly $S_j=|n_j|S$, with each coordinate face oriented opposite the corresponding component of the oblique normal.

For a material volume moving with the fluid, the Reynolds transport theorem gives, componentwise,For a scalar $q$ this is also ${\int }_{V(t)}(Dq/Dt+q\nabla \cdot \mathbf{u})dV$. Apply the component formula to momentum density $q=\rho u_i$. Constant density and incompressibility make the integrand $\rho D{u}_i/Dt$.

The traction $\mathbf{t}(\mathbf{n})$ is the force per unit area exerted by the material outside a surface on the material inside, with outward unit normal $\mathbf{n}$. Reversing the normal exchanges the two sides of the same internal surface. Thus $\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})$ expresses equal and opposite internal contact forces, the local form of Newton's third law.

With no body force, Newton's second law equates the rate of change of the material volume's total momentum to its surface force. Combining this balance with the transport theorem givesThis is a vector identity, since the component argument applies to each direction.

The head of the rarefaction reaches the wall at $(-a,t_0)$ with $t_0=a/c_0$. The right-running characteristic through that event is obtained by setting $\tau =t_0$ in the preceding formula:It bounds the region reached by the first wall-induced change in $Q_{+}$. To its right the characteristic information remains the original dam-break information, soBehind it, the impermeable wall must keep $u(-a,t)=0$. Continuing the original fan there would instead give $2(c_0-a/t)/3>0$ for $t>t_0$. Reflected characteristic data therefore change $Q_{+}$ and invalidate the simple fan formula. The shaded region in the characteristic diagram represents this changed domain of dependence; it includes $0<x<x_b(t)$ once that interval becomes nonempty. Its first positive extent occurs when $t/t_0>(3/2{)}^{3/2}$.

Inside the fan, $u+c=(4c_0+x/t)/3$. Hence a right-running characteristic satisfiesMultiplication by the integrating factor $t^{-1/3}$ givesIntegration yields $x_{+}=2c_{0}t+C{t}^{1/3}$. Requiring $x_{+}(\tau )=-c_0\tau$ determines $C=-3c_0{\tau }^{2/3}$, soThe invariant $Q_{+}=2c_0$ stays constant along it until it enters a region affected by new boundary data.

Before a signal traveling left at speed $c_0$ can reach the wall, the portion $-a<x<-c_{0}t$ remains at rest with $c=c_0$, $h=h_0$. In the rarefaction fan, right-running characteristics bring the unchanged invariant $Q_{+}=2c_0$. The left-running family originates at the removed dam and has speed $u-c=x/t$. Solving these two relations givesAt the dry front $c=0$, so its speed and location are $2c_0$ and $2c_{0}t$, consistent with the stated constant-speed condition. Therefore, for $0<t<a/c_0$,The velocity is zero in the untouched region and $2(c_0+x/t)/3$ in the fan. Ahead of the front there is no fluid, so $h=c=0$ and a fluid velocity need not be assigned. In particular the wall remains at $u=0,c=c_0$ until time $a/c_0$.

For positive surface tension the threshold is even in $k$, tends to infinity as $k\to 0$ and as $|k|\to \infty$, and has its two equal minima atMinimizing $\Delta g/q+\gamma q$ for $q>0$ givesThus the initial state has some exponentially growing mode exactly when $|U|>U_{\min }$. If $D={\rho }_1{\rho }_2{U}^2/\Sigma$, the unstable band isGravity suppresses long wavelengths and surface tension suppresses short ones; intermediate wavelengths become unstable when the shear is large enough.

Let $\Sigma ={\rho }_1+{\rho }_2$, $\Delta ={\rho }_2-{\rho }_1>0$. Completing the square in the dispersion relation givesThe coefficients are real. There is an exponentially growing temporal mode exactly when the right side is negative, because one root then has positive imaginary part in $e^{-i\omega t}$. Rearrangement givesEquality is the marginal threshold rather than exponential growth.

Put $q=|k|>0$ and ${\omega }_1=\omega -kU/2$, ${\omega }_2=\omega +kU/2$. Harmonicity and decay selectThe kinematic conditions give $B_1=i{\omega }_{1}A/q$ and $B_2=-i{\omega }_{2}A/q$. Inserting them into the dynamic condition and cancelling the nonzero interface amplitude givesExpansion proves

Each perturbation potential is harmonic in its own half-space:Write $D_1={\partial }_t+(U/2){\partial }_x$ and $D_2={\partial }_t-(U/2){\partial }_x$. Linearizing the material-interface condition at $z=0$ givesBernoulli's equation at the displaced interface gives $p_i^{\prime }=-{\rho }_i(D_i{\tilde{\varphi }}_i+g\eta )$. Using the supplied pressure jump therefore yieldsThe potentials decay toward their respective infinities. The hydrostatic displacement terms are retained in this dynamic condition.

Use $\varphi =F(r)\Theta (\theta )Z(z)e^{-i\omega t}$. Periodicity gives $\Theta =\cos m\theta$ or $\sin m\theta$, $m\ge 0$, with only the cosine branch for $m=0$. The end-wall conditions give $Z=\cos (\ell \pi z/h)$, $\ell \ge 0$. With radial separation constant ${\kappa }^2$, the remaining equation isFor $\kappa >0$, boundedness on the axis selects $F=J_m(\kappa r)$. The cylindrical wall then requires $J_m^{\prime }(\kappa a)=0$. Thus the modes are real and imaginary parts ofHere take the positive derivative zeros and include the separate constant radial mode $m=0$, $k=0$. For zero radial separation constant, regularity gives $r^m$ when $m>0$, which fails the wall condition, and a constant for $m=0$. This avoids counting a formal derivative zero at the origin that would make $J_m(0r)$ identically zero. The completely constant mode has zero frequency and no velocity; the remaining modes are acoustic oscillations. The self-adjoint Neumann Laplacian has these separated modes as a complete system, so no additional nonzero natural frequencies occur.

Rigid impermeable walls require zero normal fluid velocity, so the potential satisfies a homogeneous Neumann condition. In the cylinder,It is also single-valued and periodic in $\theta$, and regular on the axis. No tangential no-slip boundary condition is imposed on this inviscid acoustic potential.

Linearization about the uniform resting state givesFor irrotational motion write ${\mathbf{u}}^{\prime }=\nabla \varphi$. The momentum equation integrates to $p^{\prime }=-{\rho }_0{\varphi }_t$ after choosing the time-dependent potential gauge. Substituting into the pressure equation yieldsThe constant is the adiabatic sound speed, also given by $(\partial p/\partial \rho {)}_S^{1/2}$ at the base state.

Conservation of the parcel's entropy makes $p/{\rho }^{\gamma }$ constant along its trajectory. Its initial value is $p_0/{\rho }_0^{\gamma }$, soThis is a relation following the specified parcel; parcels with different initial entropy can have different constants.

Continuity gives $D\log \rho /Dt=-\nabla \cdot \mathbf{u}$. Using $p=\rho RT$, the temperature equation gives $D\log T/Dt=-(R/c_v)\nabla \cdot \mathbf{u}$. ThereforeConsequentlyEach smooth parcel evolves isentropically under these adiabatic inviscid equations; the argument does not cover entropy-producing shocks.

The material derivative $\boxed{D/Dt={\partial }_t+\mathbf{u}\cdot \nabla }$ measures change while following a moving gas parcel. The four equations respectively express mass conservation, momentum conservation, internal-energy balance, and the ideal gas law. The first accounts for density change under expansion or compression. The second is Newton's law with pressure as the only force. In the third the specific internal energy is $c_{v}T$ and pressure work $-p\nabla \cdot \mathbf{u}$ heats compression and cools expansion. The fourth relates thermodynamic pressure, density and temperature; $R$ is the gas constant per unit mass.

Put $\rho =|z|$. The triangle inequality gives $1-\rho \le |z-e^{i\theta }|\le 1+\rho$, so the Poisson kernel for the unit disk obeysIf $\rho \le r$, replace the lower bound by $(1-r)/(1+r)$ and the upper bound by $(1+r)/(1-r)$. Integrating against the positive measure $h(e^{i\theta })d\theta /(2\pi )$ givesHere $u(0)$ is exactly the uniform boundary average from part (a). Positivity permits multiplication of the pointwise kernel bounds; continuity on the compact circle ensures all integrals are finite. This is the Harnack inequality on the unit disk.

By Brownian scaling, $r^{-1}{Z}_{r^{2}t}$ is planar Brownian motion started at $w_0/r$. Its unit-disk exit angle is exactly the original radius-$r$ exit angle. Substitute $z_0=w_0/r$ into the supplied Poisson kernel for the unit disk:This density is with respect to angle $d\theta$. Its density with respect to arc length would be smaller by a factor $r$.

By conformal invariance of planar Brownian motion, $w(Z_{T_D})$ has the uniform boundary law from part (a), because the image process starts at zero. A disk automorphism extends to an orientation-preserving boundary diffeomorphism. Choose a continuous angular lift $\varphi :\mathbb{R}\to \mathbb{R}$ satisfyingThe principal-angle version may have a branch jump, which is harmless for the density. The change of variables for probability densities yields the exit-angle density ${\varphi }^{\prime }(\theta )/(2\pi )$. Differentiating the boundary relation givesTaking absolute values proves ${\varphi }^{\prime }(\theta )=|w^{\prime }(e^{i\theta })|$. ThereforeThe angular lift explains rigorously the derivative appearing in the printed hint despite the principal-angle branch cut.