The independent-column representation gives $T(x)=c_1(x_1+3x_3)+c_2(x_2+2x_4)$. Thus a vector lies in the kernel of a linear map represented by this matrix precisely when $x_1=-3x_3$ and $x_2=-2x_4$. Writing the two free coordinates separately givesThe two displayed vectors are linearly independent, so they form a basis of the kernel. Its dimension two also agrees with the rank-nullity theorem, $4=2+2$.

Let $c_1=(1,1,1{)}^T$ and $c_2=(2,-2,6{)}^T$. The remaining columns are $3c_1$ and $2c_2$, so these two vectors span the column space. They are linearly independent: the determinant of their first two coordinates is $-4\ne 0$. Therefore $\{(1,1,1{)}^T,(2,-2,6{)}^T\}$ is a basis of the range, and the matrix rank is two.

Equivalently the range is the plane $y_3=2y_1-y_2$: both displayed basis vectors satisfy this equation, and their two-dimensional span fills that plane.

For fixed $F>0$, the left side $L(v)=F^2{v}^2$ is a strictly increasing function on $v>1$, with limit $F^2$ at $1$ and unbounded growth. The right side $R(v)=1+3/[2(v-1)]+1/[2(v-1{)}^2]$ is a strictly decreasing function, sinceIt tends to infinity as $v\downarrow 1$ and to one as $v\to \infty$. Their continuity and the intermediate value theorem give a crossing, and strict opposite monotonicity makes it unique. There is exactly one admissible bore speed $V>U$ for every $F>0$.

The illustrated case is $F=1$; changing $F>0$ moves the increasing curve without changing the existence or uniqueness argument. The vertical asymptote at $v=1$ expresses the growing depth ratio $H/h_0=v/(v-1)$.

Let $H$ be the depth behind the bore. The wall's kinematic boundary condition makes the velocity there $U$; ahead of the bore the depth and velocity are $h_0$ and zero. Across a rightward hydraulic bore moving at $V$, the Rankine-Hugoniot conditions areThe first gives $R=H/h_0=V/(V-U)=v/(v-1)>1$. Substituting into the second gives $F^{2}v=(R^2-1)/2$, where $v=V/U$ and $F=U/\sqrt{g{h}_0}$. Hence the piston-driven shallow-water bore obeysThe positive depth increase is consistent with the energy-loss condition in ii, now viewed in the bore's own shock frame.

Put $R=h_2/h_1>1$. The squared Froude numbers follow directly from the discharge relation:Thus the jump connects supercritical flow to subcritical flow. Long surface gravity waves have speed $\sqrt{gh}$ relative to the fluid, so the characteristic speeds in the stationary frame are $u\pm \sqrt{gh}$. Upstream both point with the flow; downstream one points against it. The jump changes whether long-wave information can propagate upstream through the local flow.

Orient the flow so $q>0$, and let subscripts 1 and 2 denote upstream and downstream. The mechanical energy flux is $\rho qB$, where $B=u^2/2+gh$. Inserting the relation for $q^2$ in i gives the hydraulic jump head loss,An admissible jump dissipates mechanical energy, so $\rho q(B_1-B_2)>0$. The denominator is positive, henceThe zero depth-change case is continuous flow. A depth-decreasing discontinuity would require an energy input at the jump; it is not the dissipative hydraulic jump selected by this model.

Take a unit-width control volume spanning the stationary hydraulic jump. Mass conservation makes the discharge $q=hu$ constant. The horizontal momentum flux is $\rho h{u}^2$, and the integrated hydrostatic pressure force on a vertical cross-section is ${\int }_0^h\rho g(h-z)dz=\rho g{h}^2/2$. Momentum conservation therefore givesThese are the Rankine-Hugoniot conditions for the shallow water equations in the stationary shock frame. With depths $h_1\ne h_2$ on the two sides, eliminating the velocity gives

Write the cosine in i as two complex exponentials. On a ray $x=vt$, the oscillatory phases are $kv\pm \omega (k)$. The stationary phase method requires $v=\mp {\omega }^{\prime }(k)$. By ii, the largest absolute horizontal group velocity for the occupied $n=1$ mode isInside this region there are stationary wavenumbers supporting appreciable wave packet contributions. Outside it no stationary wavenumber exists; oscillatory cancellation suppresses the large-time response. This explains "mainly confined": the initial Lorentzian profile already has nonzero algebraic tails, so the displayed region is not the exact support of the solution. Near its boundary the stationary points meet $k=0$ and the usual nondegenerate stationary-phase formula needs a transition approximation.

Use the Fourier transform convention $\hat{W}(k,z,t)={\int }_{\mathbb{R}}W(x,z,t)e^{-ikx}dx$, with inverse factor $1/(2\pi )$. The initial profile occupies only the vertical mode $\sin (\pi z/h)$. Its transform isThe transformed evolution is a harmonic oscillator for each $k$, withTherefore inversion givesUsing the nonnegative even frequency instead gives the same cosine. At $t=0$, ${\int }_{\mathbb{R}}{e}^{-a|k|}{e}^{ikx}dk=2a/(a^2+x^2)$ with $a=ϵh$, recovering the specified initial profile and its zero time derivative.

Choose the rightgoing signed dispersion relation $\omega (k)=Nk/\sqrt{k^2+b_n^2}$. The horizontal phase velocity and group velocity areThe leftgoing branch has their negatives. For a convention using positive frequency at both signs of $k$, multiply each displayed speed by $\mathrm{sgn}k$. For every nonzero $k$, $0<c_g<c_p$ on the rightgoing branch; the horizontal group-speed bound for a confined internal wave is $c_g\le N/b_n$, approached as $k\to 0$.

Substitute $W=f(z)e^{i(kx-\omega t)}$ into the transformed equation. It yieldsThe nonzero eigenfunctions of this Dirichlet boundary value problem are $f=A\sin (n\pi z/h)$, $n\ge 1$. Hence, with $N=\sqrt{\alpha g}$ and $b_n^2=(n\pi /h{)}^2+{\alpha }^2/4$,The physical vertical velocity includes the factor $e^{\alpha z/2}\sin (n\pi z/h)$. The zero horizontal wavenumber is the zero-frequency limiting case, not a propagating oscillation.

For $w=e^{\alpha z/2}W$, direct differentiation givesThus the transformed partial differential equation and its impermeable-wall boundary conditions areThe positive exponential factor does not change where the vertical velocity vanishes.

Write the density and pressure as ${\rho }_0(z)+{\rho }^{\prime }(x,z,t)$ and $p_0(z)+p^{\prime }(x,z,t)$, respectively. The background satisfies hydrostatic balance, $p_0^{\prime }=-g{\rho }_0$, and ${\rho }_0^{\prime }/{\rho }_0=-\alpha$. Linearizing the Euler equations for an inviscid fluid, incompressibility and advection of density givesDifferentiate the horizontal equation in $x$ and use incompressibility to obtain $p_{xx}^{\prime }={\rho }_0{w}_{zt}$. Differentiate the vertical equation in time and twice in $x$:Dividing by ${\rho }_0>0$ gives the equation for internal waves in an exponentially stratified incompressible fluid,The first-derivative term retains the variation of inertial density with height; no Boussinesq approximation has been made.

For the $n$th normal mode, the denominator just obtained isIts forcing coefficient is nonzero for $c\ne 0$ and $\omega >0$. Therefore the periodic response grows like $({\omega }_n^2-{\omega }^2{)}^{-1}$ as $\omega$ approaches ${\omega }_n$: this is resonance. At exact resonance there is no bounded periodic particular solution in the linear inviscid model; the resonant amplitude instead contains a secular term proportional to $t\sin ({\omega }_{n}t)$ or $t\cos ({\omega }_{n}t)$. Physical damping and nonlinear effects eventually matter. Away from resonance, undamped free oscillations can persist alongside the periodic particular solution.

Put $\lambda ={\omega }^2/g$ andThis polynomial is a harmonic function: $Q_{xx}+Q_{zz}=0$. Also $Q_x(0,z)=i\omega c$, $Q_x(L,z)=0$ and $Q_z(x,-h)=0$. Thus $G$ satisfies Laplace's equation with homogeneous Neumann boundary conditions on both side walls and the bottom. Its Fourier cosine series consequently has the formThe surface condition becomes $G_z-\lambda G=H(x)$, whereFor a nonzero forcing frequency away from the natural frequencies, the coefficients of the oscillating-wall forcing of a rectangular tank areHere $k_0=0$ in the definition of $H_0$. In fact $H_n=2i\omega c\lambda /(L{k}_n^2)$ for $n\ge 1$, since ${\int }_0^L(L-x{)}^2\cos (k_{n}x)dx=2L/k_n^2$. The zero-mode coefficient ensures that the mean surface elevation is $hc/L$ in complex amplitude, exactly the first-order volume conservation adjustment to the moving wall.

The wall position is $x_w(t)=\mathrm{Re}(c{e}^{-i\omega t})$, so its velocity is $\mathrm{Re}(-i\omega c{e}^{-i\omega t})$. The kinematic boundary condition equates the normal fluid velocity to the normal wall velocity. Linearizing the evaluation point to the mean wall position givesThe other wall and bottom retain $f_x(L,z)=0$ and $f_z(x,-h)=0$, while the free surface gives $f_z(x,0)=({\omega }^2/g)f(x,0)$.

Apply separation of variables to Laplace's equation. The side-wall Neumann boundary conditions give horizontal factors $\cos (k_{n}x)$ with $k_n=n\pi /L$. The bottom condition fixes the vertical factor to $\cosh (k_n(z+h))$. Substitution into the surface condition in i gives the normal modes of surface gravity waves in a rectangular tank:The corresponding surface elevation follows from ${\eta }_n=-{\partial }_t{\varphi }_n(x,0,t)/g$. The spatially uniform $n=0$ mode has zero frequency; its constant elevation changes total fluid volume, so it is absent when that volume is fixed. An additive constant in $\varphi$ changes no fluid velocity.

The velocity potential gives horizontal and vertical velocity components $({\varphi }_x,{\varphi }_z)$. The kinematic boundary condition at each fixed impermeable wall is zero normal fluid velocity, henceLet $\eta (x,t)$ be the surface elevation. To first order in its amplitude, the kinematic boundary condition for a free-surface graph gives ${\eta }_t={\varphi }_z$ at $z=0$. The dynamic boundary condition for an inviscid interface fixes the surface pressure to atmospheric pressure. The unsteady Bernoulli equation, after choosing the time-dependent gauge of $\varphi$, then gives ${\varphi }_t+g\eta =0$ at $z=0$: the quadratic velocity term is second order in amplitude. Differentiate this last equation in time and eliminate ${\eta }_t$ to obtain the linearized free-surface boundary conditions in combined form,

At the interface, the tangential velocity is $u(x,0)=x{f}^{\prime }(0)$ and the normal velocity remains zero. The preceding error function expression givesHence, within the leading boundary layer model,Allowing the formally discarded finite-Reynolds number terms gives an additional order-${\delta }^2$ error, smaller than the order-$\gamma$ correction under the assumed ordering. The coefficient is $\sqrt{\pi /8}\approx 0.626657$.

Thus positive surface shear increases the outward surface speed linearly: for $x>0$ it increases $u$, and for $x<0$ it makes $u$ more negative. The stagnation point remains at $x=0$. A negative coefficient would result from using the printed $+\infty$ condition literally, and would describe the different half-line problem identified in part (ii).

Set $g=f_1^{\prime }$. The boundary value problem becomesFor $y_1=1+{\zeta }^2$, direct substitution gives $y_1^{\prime \prime }+\zeta y_1^{\prime }-2y_1=2+2{\zeta }^2-2(1+{\zeta }^2)=0$. This growing polynomial solution is useful for reduction of order.

Write $g=y_{1}v$. Substitution into the linear ordinary differential equation cancels the terms proportional to $v$ and givesAn integrating factor for this first-order equation for $v^{\prime }$ yieldsThe far-field boundary condition selects the convergent integral from $-\infty$:Any additional multiple of $1+{\zeta }^2$ would grow at $-\infty$. The chosen expression tends to zero: for $r=-\zeta \ge 1$,Here $(1+t^2{)}^{-2}\le (1+r^2{)}^{-2}$ and ${\int }_r^{\infty }{e}^{-t^2/2}dt\le r^{-1}{\int }_r^{\infty }t{e}^{-t^2/2}dt$. Thus the far-field boundary condition is verified without a cancellation of large terms.

At $\zeta =0$, differentiating gives $g^{\prime }(0)=C$, so $C=1$. Using the antiderivative supplied in the paper gives the explicit answerHere $\mathrm{erf}$ is the error function. Finally,recovers the unique correction satisfying $f_1(0)=0$. The decay bound also shows that $f_1$ approaches a finite constant at $-\infty$, as allowed by the normal velocity matching. This is the weak surface-shear perturbation of stagnation flow.

If the printed $+\infty$ condition is instead imposed literally on a positive half-line, reduction of order givesIt satisfies $g_{+}^{\prime }(0)=1$ but grows as $\zeta \to -\infty$, so it cannot describe the fluid domain specified in this paper. This makes the consequence of the printed sign error explicit.