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Here the zero-parameter solution is $y_0=1$, but stopping there erases a weakly excited growing mode. Expand $y=1+\epsilon y_1+\cdots$ at fixed $x$. The first correction satisfiesand thereforeThe $\epsilon x^2/2$ term eventually dominates the constant, so this is another singular perturbation on the unbounded domain.

Using the exact logarithmic-variable system from part (a), the initial values are now $z(0)=\epsilon$ and $w(0)=0$. Its linear growing/stable modes have matching coefficientsThus $w\sim Cx$ is order one only at $x=O({\epsilon }^{-2})$, rather than $O({\epsilon }^{-1})$. This is the nonlinear crossover delayed by a small unstable-mode coefficient. The first change in the size of $y$ occurs much earlier, at $x=O({\epsilon }^{-1/2})$, when its weak growing mode becomes comparable to the initial constant; that is not yet the derivative-nonlinearity crossover.

At $X=Cx=O(1)$ the same logistic crossover in a weakly nonlinear Euler equation gives $w=X/(1+X)$ and $H(X)=X-\log (1+X)$. The nonlinear-region solution is now $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-3}H({\epsilon }^{2}x)$ at the lowest order. A composite asymptotic expansion covering all these regions isIt gives $y(1)=1$ exactly, and its initial derivative is $O({\epsilon }^3)$ rather than zero, within leading accuracy. At fixed $x$ its expansion is precisely $1+\epsilon [(x^2-1)/2-\log x]+O({\epsilon }^2)$; for $Cx\ll 1$ it retains the weak $A{x}^2$ growth, and for $Cx\gg 1$ it gives $y\sim x/\epsilon$. The positive equilibrium and boundedness argument from part (a) applies after the initially zero derivative immediately becomes positive. Thus the same eventual slope saturation occurs, but much farther downstream. Both the seeding calculation and the changed scale are essential to a leading approximation on the entire half-line.

Write $T=\epsilon t$ and $q(T)=2e^{-T}$. The forcing acts on the fast initial-time scale, so first remove a localized particular solution. Expanding the coefficient there givesA particular choice is $p_1=(t+1/2)e^{-2t}$, henceIts initial value is $1+\epsilon /2$ and its initial derivative is $-2+O({\epsilon }^2)$. The localized forcing determines WKB phase constants; omitting it before imposing the initial conditions would miss the permanent free oscillation.

For the homogeneous part, the slow-time equation is ${\epsilon }^2{u}_{TT}+q^{2}u=0$. Write its WKB approximation as $A(T)[C\cos \Phi +B\sin \Phi ]$, normalizing $A(0)=1$. The leading amplitude transport equation gives $A=e^{T/2}$ and leading phase ${\epsilon }^{-1}{\int }_0^{T}q(s)ds=2(1-e^{-T})/\epsilon$.

To include every $O(\epsilon )$ contribution on bounded slow-time intervals, the first phase correction is needed. If the local phase wavenumber is $q+{\epsilon }^2{q}_2$, the real part of the amplitude-phase equation gives $q_2=A_{TT}/(2qA)=e^T/16$. The WKB phase correction for a slowly varying oscillator, anchored to zero at the initial time, is thereforeThere is no first-order bulk amplitude correction in this normalization. At $t=0$, the homogeneous initial value must be $-\epsilon /2+O({\epsilon }^2)$ and its initial derivative $1+O({\epsilon }^2)$. Since $A_t^{\prime }(0)=\epsilon /2$ and ${\Phi }_t^{\prime }(0)=2+O({\epsilon }^2)$, these give $C=-\epsilon /2+O({\epsilon }^2)$ and $B=1/2+O({\epsilon }^2)$. The requested forced approximation isfor $0\le \epsilon t\le T_0$ with fixed $T_0$. The approximate initial value is exactly one through this order and its derivative is $-1+O({\epsilon }^2)$. The particular-solution residual is uniformly localized and $O({\epsilon }^2)$; the corrected homogeneous residual is $O({\epsilon }^3)$ on the long interval, producing $O({\epsilon }^2)$ solution error after integration over times of order ${\epsilon }^{-1}$.

The WKB approximation requires the relative frequency-change parameterIt therefore fails when $\epsilon e^{\epsilon t}=O(1)$, even though the coefficient has no finite-time zero. This is the Bessel transition for an exponentially decaying oscillator. Introduce the shifted slow time and amplitudeThe exact rescaled equation isFor fixed $\tau$ its right side is smaller than every algebraic power of $\epsilon$. The leading transition problem is consequentlyWith $Z=2e^{-\tau }$ this becomes $Z^2{U}_{0,ZZ}+Z{U}_{0,Z}+Z^2{U}_0=0$, the order-zero Bessel differential equation. This identifies the governing equation without solving it.

The amplitude-phase series also supplies overlap data while $\epsilon e^T\ll 1$, although its bounded-$T$ absolute error estimate is not uniform up to the transition. For matching, combine the earlier sine and cosine terms into $\frac{1}{2}\sin (\Phi -\epsilon )$ through first order. In the overlap $\epsilon \ll e^{\tau }\ll 1$, their phase isto the retained order. Thus amplitude and phase matching as $\tau \to -\infty$ requireThese are oscillatory amplitude-phase conditions, rather than pointwise ratio statements at zeros. The smaller $e^{\tau }/16$ phase term records the first overlap correction if required. The rapidly varying matching constant ${\alpha }_{\epsilon }$ must be retained as data; there is no single epsilon-independent limiting phase. The two matching conditions determine the leading transition solution, which is deliberately left unsolved.

Let $p$ be kinematic pressure, with the resting hydrostatic part subtracted, and let $\sigma$ be the buoyancy perturbation about background buoyancy $N^{2}z$. The nonrotating inviscid Boussinesq equations, with no buoyancy diffusion, areHere the third equation is mass conservation for incompressible flow, and the last expresses buoyancy conservation following a parcel. Linearization about rest removes the products of disturbances. For a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, the resulting Linearized Boussinesq equations becomeWith $K=(k^2+m^2{)}^{1/2}$ and $k\ne 0$, the first and third equations give $\hat{p}=-m\omega \hat{w}/k^2$. Substitution in vertical momentum and buoyancy gives the dispersion relationSince the velocity is the time derivative of the displacement field, $\hat{\mathbf{u}}=-i\omega \hat{\mathbf{\xi }}$. Thus the complete buoyancy polarization of a plane internal gravity wave isBoth velocity and displacement are perpendicular to the wavevector; pressure and velocity are in quadrature with buoyancy and displacement.

For upward/rightward phase velocity, choose $k,m>0$ and $\omega =Nk/K$. Differentiating this dispersion relation gives the group velocity, while the vector phase velocity points along the wavevector:Their dot product is zero and ${\mathbf{c}}_p+{\mathbf{c}}_g=(N/K,0)$. Draw ${\mathbf{c}}_g$ from the tip of ${\mathbf{c}}_p$: the right-triangle geometry of internal-wave velocities has horizontal hypotenuse $N/K$. This hypotenuse is distinct from the horizontal phase velocity $\omega /k$.

At an instant with real complex amplitude $\hat{\sigma }=A$, write $\varphi =kx+mz-\omega t$. The spatial fields areThe constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$. Buoyancy maxima lie a quarter wavelength from pressure maxima; the oscillating displacement and velocity run along the phase lines. The group velocity points down and right along them, despite upward/rightward motion of the wave phase.

Let $P=\overline{{\xi }_{x}u+{\zeta }_{x}w}$, using horizontal averaging over a wavelength. This is the displacement pseudomomentum of an internal gravity wave, whose sign convention is opposite to $k\overline{E}/\omega$. For complex coefficients, $\overline{ab}=\frac{1}{2}\mathrm{Re}(\hat{a}{\hat{b}}^{\ast })$. The above wave polarization consequently givesSince $c_{gz}=-\omega m/K^2$, these obey $\overline{{\zeta }_{x}p}=c_{gz}P$. With weak kinematic viscosity, each leading monochromatic velocity component satisfies ${\nabla }^{2}u=-K^{2}u$, ${\nabla }^{2}w=-K^{2}w$. The dissipative term is therefore $-\nu K^{2}P$, so the supplied averaged balance reduces toFor constant $k,m,N$ and a steady wave envelope, $c_{gz}{P}_z=-\nu K^{2}P$. As $P$ is proportional to squared wave amplitude, viscous attenuation of an internal gravity wave givesHere $c_{gz}<0$, so amplitude decreases below a source at $z_s$: the source is above the receiving level. The same conclusion follows directly because downward group velocity carries energy from the source downward. Upward phase velocity does not locate the source.

For a slowly varying mean shear flow $U(z)$, the local intrinsic frequency is $\hat{\omega }=\omega -kU(z)$. A local WKB approximation uses ${\hat{\omega }}^2=N^2{k}^2/(k^2+m(z{)}^2)$, with varying vertical wavenumber and group velocity. The appropriate transported quantity is wave action, or horizontal wave pseudomomentum $kE/\hat{\omega }$; its vertical flux has viscous loss, and the changing polarization factors must be retained. In particular, $P/|\hat{\sigma }{|}^2$ is no longer constant.

Near a critical level of an internal gravity wave, $\hat{\omega }\to 0$, so $|m|\sim N|k|/|\hat{\omega }|$ and $|c_{gz}|\sim {\hat{\omega }}^2/(N|k|)$. The damping per vertical distance, of order $\nu K^2/|c_{gz}|$, grows rapidly. The waves are strongly attenuated over a narrow altitude range and transfer momentum to the mean flow there or before reaching the critical level. The inviscid WKB approximation eventually fails; viscosity and possibly internal-wave breaking regulate the small scales.

Let $h$ be layer thickness and $g$ gravitational acceleration. The unforced shallow water equations on the equatorial beta plane areLinearize about $u=v=0$, $h=h_0$. Put $c_0=\sqrt{g{h}_0}$ and $L=\sqrt{c_0/\beta }$, and scale $(x,y)=L(X,Y)$, $t=(L/c_0)T$, $(u,v)=c_0(U,V)$ and $h=h_0(1+\eta )$. Dropping capitals, the linearized shallow water equations becomeThe choice $\beta L^2/c_0=1$ makes both the wave speed and the coefficient of the dimensionless Coriolis parameter equal to one.

Take Fourier modes $\{u(y),v(y),\eta (y)\}{e}^{i(kx-\omega t)}$ that decay as $y\to \pm \infty$. Their equations areFor $v=0$ and nonzero $k$, the first two require ${\omega }^2=k^2$. The root $\omega =k$ gives $u=\eta$, and meridional balance becomes $u^{\prime }=-yu$. Hence the equatorial Kelvin wave isIt travels eastward at speed $c_0$ in dimensional units. The other sign gives $\eta =-u$ and $u^{\prime }=yu$, so $u\propto e^{y^2/2}$ is not trapped.

For $v\ne 0$, $\omega \ne 0$ and ${\omega }^2\ne k^2$, inversion of the first two equations gives the equatorial wave velocity polarizationSubstitute in the remaining momentum equation. Multiplication by $({\omega }^2-k^2)/i$ leavessoWriting $v=e^{-y^2/2}H(y)$ yields $H^{\prime \prime }-2y{H}^{\prime }+(\lambda -1)H=0$, the Hermite differential equation. The trapped solutions have the Hermite oscillator quantization $\lambda =2n+1$ and $H=H_n$.

For completeness, quantization follows without assuming that every acceptable solution is a polynomial. On decaying smooth functions, define $a=d/dy+y$, $a^{†}=-d/dy+y$ and $\mathcal{H}=-d^2/d{y}^2+y^2=a^{†}a+1$. Integration by parts gives $\lambda \ge 1$ and $\Vert av{\Vert }^2=(\lambda -1)\Vert v{\Vert }^2$. If $av\ne 0$, it is an eigenfunction of $\mathcal{H}$ with eigenvalue $\lambda -2$. Repeated lowering must terminate: otherwise it eventually produces an eigenvalue below one. At termination $av=0$, so the terminal eigenfunction is $e^{-y^2/2}$ and has eigenvalue one. Repeated raising gives $H_n(y)e^{-y^2/2}$ and $\lambda =2n+1$; uniqueness of a decaying solution at one end of this second-order equation fixes each mode up to scale.

Thus, away from the exceptional quotients, all fields are determined byInsert this $v$ and its derivative in the polarization formulae to obtain $u$ and $\eta$. For $n\ge 1$, the cubic ${\omega }^3-(k^2+2n+1)\omega -k=0$ supplies the low-frequency equatorial Rossby wave and two equatorial inertia--gravity wave branches. Their meridional structures are Gaussian-weighted Hermite polynomials.

For $n=0$, multiplication by $\omega$ factorizes the equatorial shallow-water dispersion relation asThe physically admissible quadratic gives the mixed Rossby-gravity wave, equivalently the Yanai wave:These fields satisfy all three original equations directly and decay at both infinities, including where the quotient formulae are singular.

To test the other factor, set $\omega =-k\ne 0$. The original equations give $u+\eta =-iyv/k$ and $v^{\prime }=-yv$. Put $\delta =u-\eta$; meridional momentum then givesDecay of $u,\eta$ at both infinities requires the integral of the right side to vanish. Hence $C=0$ unless $k^2=1/2$. Ordinarily the extra factor supplies no trapped mode. At the exceptional root of the mixed Rossby-gravity mode, $k^2=1/2$, it coincides with the quadratic root and gives the same physical fields, not an additional branch. This distinction avoids discarding a regular physical mode at the crossing.

These arguments concern nonzero wave angular frequency. At $k=0$, the oscillatory nonzero-$v$ modes have $\omega =\pm \sqrt{2n+1}$, while zero-frequency $v=0$ geostrophic states must be considered in the original equations rather than in expressions divided by $\omega$.

Potential-vorticity inversion is the diagnostic recovery of the flow from its advected vortical scalar, using the balance relation and specified boundary conditions. Its usefulness is a separation of tasks: transport the scalar by the current velocity, invert to find the new velocity, and repeat. The inverse is spatially nonlocal even though the transport law is local. The four models differ chiefly in the operator being inverted and the information needed at boundaries.

In two-dimensional vortex dynamics, planar incompressibility permits $\mathbf{u}=(-{\psi }_y,{\psi }_x)$, and vertical vorticity is $\zeta ={\nabla }^2\psi$. The curl of inviscid momentum givesThere is no vortex stretching in strictly planar motion, so vorticity is materially conserved. The two-dimensional potential-vorticity inversion is the Poisson equation ${\nabla }^2\psi =\zeta$. On the plane, with suitable integrability and far-field behavior, its Green's function is $(2\pi {)}^{-1}\log |\mathbf{x}-{\mathbf{x}}^{\prime }|$, givingDifferentiation gives the planar Biot-Savart kernel. Here ${\psi }_h$ is a harmonic function fixed by boundary, circulation and far-field data. A rigid impermeable wall is a streamline, so $\psi$ is constant on it; constants or circulations around distinct holes require specification. The vorticity alone cannot distinguish two flows differing by a permitted harmonic contribution, such as a uniform current on an unbounded plane. A constant added to $\psi$ has no physical effect.

In nonrotating layerwise-two-dimensional vortex dynamics, strong stable density stratification constrains motion to almost horizontal layers. In the ideal independent-layer limit, vertical velocity vanishes at leading order, $(u,v)=(-{\psi }_y,{\psi }_x)$ andhold separately at each fixed $z$. Height is a parameter: the horizontal inverse Laplacian contains no vertical derivative. With background buoyancy $N^{2}z$, the leading Ertel potential vorticity is $Q=\mathbf{\omega }\cdot \nabla b\simeq N^2\zeta$, so the materially advected normalized scalar is $Q/N^2\simeq \zeta$. The same horizontal Poisson equation is inverted layer by layer, using each layer's boundary data. This is an asymptotic constrained model, not arbitrary three-dimensional Euler motion with vertical velocity set to zero: its validity requires small Froude number and controlled vertical scales. At finite buoyancy frequency, the weak vertical motion and tilted isopycnals restore coupling. The inversion may be independent by layer while the reconstructed fields vary with height. The horizontal Euler equations for an inviscid fluid determine the pressure gradient, while hydrostatic balance relates its vertical derivative to buoyancy.

In shallow-water quasi-geostrophic inversion, begin with the exact shallow-water potential vorticity $(f+\zeta )/h$. Linearizing its anomaly about depth $H$ gives, after multiplying by $H$, $\zeta -f_0\eta /H$. For small Rossby number, geostrophic balance gives $\psi =g\eta /f_0$, while $\zeta ={\nabla }_h^2\psi$. Keeping the beta plane background gradient givesThus $q-\beta y$ is inverted with the modified Helmholtz equation $({\nabla }_h^2-L_D^{-2})\psi =q-\beta y$, under the specified boundary and far-field conditions. The velocity follows from $\psi$ and surface elevation from $\eta =f_0\psi /g$. A horizontal Fourier mode of the anomaly satisfiesThe Rossby deformation radius therefore limits the reach of the balanced response. Layer stretching, represented by $-\psi /L_D^2$, is absent from ordinary planar vorticity inversion. Unlike the pure Poisson equation, the finite-deformation operator also responds to the constant part of $\psi$, which specifies mean elevation rather than an arbitrary streamfunction offset. Specified bottom topography adds a known source to $q$; it is subtracted along with the background before inversion.

In stratified quasi-geostrophic inversion, continuously varying vertical displacement replaces the shallow layer's single stretching term. For positive $N^2(z)$ and constant nonzero $f_0$, the three-dimensional quasi-geostrophic potential vorticity isGeostrophic balance and hydrostatic balance give velocity $(-{\psi }_y,{\psi }_x)$ and buoyancy perturbation $b=f_0{\psi }_z$. Invert the anisotropic elliptic boundary value problem using interior $q$, prescribed top and bottom buoyancy as Neumann boundary conditions, and the lateral, far-field and global-flow data. Boundary buoyancy evolves by its own material conservation when the boundary is impermeable. It cannot be discarded: the Eady edge wave in the preceding solution has zero interior potential vorticity anomaly but nonzero boundary buoyancy and a nonzero induced flow.

For constant $N,f_0$, the stretched coordinates for quasi-geostrophic inversion $Z=Nz/|f_0|$ turn the inversion operator into the three-dimensional Laplacian. Its infinite-space Green's function is $-1/(4\pi |\mathbf{X}-{\mathbf{X}}^{\prime }|)$ in $(x,y,Z)$ coordinates. Hence a horizontal disturbance of scale $L$ communicates vertically over scale $|f_0|L/N$, explicitly coupling layers. In a bounded domain with homogeneous difference data, multiplying the difference of two inversions by its streamfunction and applying integration by parts givesThis proves uniqueness up to a constant when the relevant boundary terms vanish; a mean or pressure reference fixes the remaining constant. Compatible boundary fluxes are needed for a purely Neumann boundary-value problem.

Across all four models, inversion provides the induced flow, while scalar transport supplies time evolution. The independent horizontal inverse in the first two models becomes a finite-deformation inverse in shallow water and a vertically coupled elliptic inverse in continuous stratification. The principle applies within the stated balance or constrained model. In the unrestricted Boussinesq equations or shallow water equations, gravity waves and divergent motion can share the same potential vorticity, so the vortical scalar alone does not reconstruct every possible flow.

Take $z$ increasing upwards and write $s=\varphi /{\varphi }_{\max }$. The particles have downward settling velocity; their upward volume flux is $\varphi V$. Conservation of particle volume in any interval givesThus the kinematic sedimentation equation and its characteristic speed areFor $0<s<1$ this is a scalar hyperbolic partial differential equation: its one characteristic speed is real. The method of characteristics gives $ds/dt=0$ and $dz/dt=a(s)$. A concentration disturbance need not descend at the particle settling velocity; $a(s)$ is the derivative of the transported flux, not $V(s)$. In particular $a(s)$ can be positive even though every particle settles downwards. At $\alpha =0$, $a=-V_0$ and the equation is ordinary translation. At the packing endpoint $s=1$ the differentiability of the chosen constitutive law depends on $\alpha$; none of the initial values used below is at that endpoint.

For $\alpha =1$, $f=V_0(s^2-s)$, so the characteristic speed is $a=V_0(2s-1)$ and the conservation law flux is strictly convex. Let $\tau =V_{0}t$ and let $\xi$ be the initial vertical position of a characteristic curve.

For the increasing ramp, the characteristic curves within its sloping part satisfy $z=\xi (1+0.6\tau )$. They spread apart and never cross. The outer plateau characteristic speeds are $-0.6V_0$ and $0.6V_0$, so the complete entropy solution isThe transition widens linearly and its gradient decreases. At large time it approaches a rarefaction wave, with $s\simeq (1+z/\tau )/2$ inside its expanding transition.

For the decreasing ramp, $z=\xi (1-0.6\tau )$ and the characteristic curves converge. Before they meet,All the ramp characteristic curves meet at $z=0$ atThereafter a stationary sedimentation shock separates $s_L=0.8$ below from $s_R=0.2$ above. Indeed the Rankine-Hugoniot condition givesand the incoming characteristic speeds satisfy $a(0.8)=0.6V_0>0>a(0.2)=-0.6V_0$, which makes it an admissible entropy shock. The discontinuity transports no net excess particle flux because both plateau fluxes equal $-0.16V_0$. These are local solutions before a finite container's boundaries affect the transition.

The dependence on $\alpha$ is controlled by the hindered-settling flux inflection:For any smooth initial profile, the characteristic flow map has derivativeThis formula gives the first characteristic crossing: minimize the positive values of $-1/[f^{\prime \prime }(s_0)s_0^{\prime }]$. At $\alpha =0$ neither ramp changes shape; both move down at $V_0$. For $0<\alpha \le 3/2$, the flux is convex throughout $0.2\le s\le 0.8$: the increasing ramp spreads and the decreasing ramp steepens, although its entire ramp generally no longer focuses simultaneously. For $\alpha \ge 9$, the flux is concave throughout that interval and these roles reverse. At the endpoint values the curvature vanishes at one plateau; it does not reverse sign inside the interval.

For $3/2<\alpha <9$, the inflection $s_i=2/(\alpha +1)$ lies between the plateau concentrations. Below $s_i$ the increasing ramp spreads, while above it that ramp compresses; the decreasing ramp has the opposite local behavior. After characteristic crossing, the entropy solution can contain both a shock wave and a rarefaction wave. Quantitatively, an increasing step is selected by the lower convex envelope of $f$ on the interval between its states: curved portions supply rarefaction waves, and straight portions supply shock waves. A decreasing step uses the upper concave envelope. A join between a fan and a shock wave at a state $s_{\ast }$ satisfies the tangent-chord relation $f^{\prime }(s_{\ast })=[f(s_b)-f(s_{\ast })]/(s_b-s_{\ast })$, where $s_b$ is the other shock state. Thus curvature, not simply the magnitude of $\alpha$, decides the wave pattern; the envelope may reduce to one shock if the tangent lies outside the state interval.

It is helpful to solve using the nonsingular total flux $J$ and volume flux $Q$ before recovering $M$. The relation $J=M/(1-Q^2/(R^{2}M))$ is quadratic:The plus sign selects the small-area pure-source branch. Its discriminant vanishes when $J=4Q^2/R^2$, givingThereforeThe specific upward plume and ambient momentum-flux contributions are equal, not opposite. The ambient's transported momentum has the opposite sign from its transported volume, so their product remains positive.

To show that the pure branch actually reaches this fold in finite height and that it is a genuine pole, introduceThen $s=q^2/m$ andThe pure-source condition is $j\sim (5/8{)}^{2/5}{q}^{4/5}$. Before the fold, $s<1/2$ and $m>2q^2$. Thus for every fixed positive $q_1$, the derivative $dj/dq$ is bounded above by a constant times $q^{-2}$ on $q\ge q_1$. If the branch continued for arbitrarily large $q$, $j$ would remain bounded, contradicting $j\ge 4q^2$. Hence it reaches $j=4q^2$ at a finite $q_o>0$. Its height is finite because $d\zeta /dq=(1-s)/(2\sqrt{m})$ is finite away from the source and behaves like a constant times $q^{-2/5}$ at the source, an integrable singularity.

For the numerator, write $m=j(1-s)$. ThenAt the fold $j=4q_o^2$, so $q_o^3\ge 5/256$. On the other hand the dimensionless numerator in the $m$ equation at $s=1/2$ isIt is strictly negative because $5/256>1/(64\sqrt{2})$. The denominator $1-2s$ approaches zero from above. Consequently $dM/dz\to -\infty$ at finite $z_o$, although the fluxes and velocities themselves remain finite. This proves the confined-plume momentum fold rather than merely noticing a possible zero denominator.

The dimensionless pure-source branch uniquely determines the remaining amplitudes. They can be specified without an elementary antiderivative by integrating the displayed scalar equation for $j(q)$ until $j=4q^2$, and settingA direct numerical integration gives $q_o\simeq 0.281610$, ${\zeta }_o\simeq 0.400049$, so the dimensional results for this specified entrainment convention areA convention using $\alpha w$ instead of $\alpha (w-v)$ would produce different numerical amplitudes; the relative-speed convention was stated in part (a).

Physically $z_o$ is a turnover or breakdown height for a slender steady plume surrounded by uniform counterflow. The opposing streams then occupy equal areas and have equal speeds; the single-valued small-area branch cannot continue smoothly. Above it one expects lateral spreading, stronger recirculation, mixing and unsteady buoyant structures rather than a continuation of the same slender top-hat column. A detailed flow or upper-layer evolution requires restoring pressure and stress dynamics and, eventually, the modified ambient stratification. The model does not predict an infinite physical velocity, nor does it require that all buoyant fluid forever remain below $z_o$.

For a polygonal domain with finite real prevertices $a_j$ and interior angles $\pi {\alpha }_j$, the Schwarz-Christoffel formula has the formwith branches chosen in the complex upper half-plane. A prevertex at infinity is interpreted by the corresponding limiting version of the formula. The exponent records the angle because locally $F-F(a_j)$ behaves as $(z-a_j{)}^{{\alpha }_j}$.

For a slit from $0$ to $\tau$ in the upper half-plane, put $\vartheta =\arg \tau \in (0,\pi )$. The two approaches to its base are separate boundary vertices with angles $\pi -\vartheta$ and $\vartheta$, and the tip has angle $2\pi$. With ordered prevertices $a<c<b$ mapping respectively to the base, tip and base, the Schwarz-Christoffel mapping therefore hasThe tip's factor is a simple zero of the boundary derivative, not an interior critical point.

For the specified power product, the principal branches givethroughout the complex upper half-plane, so $g$ maps into that half-plane. Logarithmic differentiation givesThere are no zeros of this derivative in the upper half-plane, so the map is locally conformal there. Its boundary values are negative real for $x<-1$, positive real for $x>1$, andThe positive scalar factor increases from zero to its maximum at $c$ and then decreases to zero, since its logarithmic derivative is $-k/(1-x)+(1-k)/(1+x)$. Therefore the tip isThe real boundary runs from the negative real axis to zero, out to $\tau$, back to zero, and then along the positive real axis.

For completeness, these boundary facts give global injectivity and surjectivity. At infinity,Apply the argument principle on a large upper half-disc, indenting the real prevertices by small semicircles. For a fixed $w$ in the upper half-plane off $[0,\tau ]$, the limiting image contour is the real boundary with the slit traversed once in each direction, followed by a large upper semicircle. Its winding number about $w$ is one. The two slit traversals cancel, so $g(z)-w$ has exactly one zero in the upper half-plane. There are no poles.

Nor can an interior point map onto the open slit. Each point of that slit has two real preimages, one on either side of $c$, with nonzero boundary derivatives. Local continuation across the real line maps the upper half-neighborhoods at these preimages onto the two different sides of the slit. An additional interior preimage would persist for nearby off-slit points and contradict the already proved one-preimage count. The tip cannot be an interior image either: the open mapping theorem would then place nearby open-slit points in the interior image. Finally, $g$ never vanishes inside the half-plane. Thus $g$ is a conformal bijection onto $S(\tau )$.

This is the power-product map to a slit half-plane. Its derivative is exactly the Schwarz-Christoffel formula with $a=-1$, $b=1$, $c=1-2k$, $\vartheta =\pi k$ and $C=1$. The additive constant is fixed by the boundary value $g(1)=0$. Thus the power-product expression is an explicit antiderivative of the corresponding Schwarz-Christoffel mapping, with this normalization.

The equilibria have $y=0$ and $x(\lambda +x^2)=0$. ThusAt $O$, the linearization has trace $\kappa$ and determinant $\lambda$. Hence $O$ is a saddle equilibrium when $\lambda <0$, a sink equilibrium when $\lambda >0,\kappa <0$, and a source equilibrium when $\lambda >0,\kappa >0$. The sink equilibrium/source equilibrium may be a node equilibrium or a focus equilibrium; the node-focus discriminant is not a bifurcation curve. At $P_{\pm }$, the trace and determinant are $\kappa +\lambda$ and $-2\lambda$. These points are sink equilibria for $\kappa <-\lambda$ and source equilibria for $\kappa >-\lambda$.

The local bifurcation curves are thereforeFor $\kappa \ne 0$ on the pitchfork curve, a leading centre-manifold equation is $\dot{x}=(\lambda x+x^3)/\kappa +\cdots$. For $\kappa <0$, with unfolding parameter $-\lambda$, the pitchfork is supercritical and creates two sink equilibria from a sink equilibrium that becomes a saddle equilibrium. For $\kappa >0$ it is subcritical in the centre direction; the nonzero branches are unstable, and the transverse eigenvalue is also positive. The origin of parameter space is a reflection-symmetric double-zero degeneration.

To determine Hopf criticality at the origin, use the energy $E=y^2/2+\lambda x^2/2+x^4/4$, whose derivative is $(\kappa -x^2)y^2$. For a small nearly harmonic oscillation $x=R\cos (\sqrt{\lambda }t)$, averaging gives $\dot{R}=\kappa R/2-R^3/8+\cdots$. Thus this Hopf bifurcation is supercritical and creates an attracting periodic orbit for $\kappa >0$. By the supplied opposite-criticality fact, the simultaneous Hopf bifurcations at the two nonzero equilibria are subcritical. A direct calculation confirms this: writing $x=\sqrt{-\lambda }+X$, $Y=-y/\sqrt{-2\lambda }$ puts the cubic radial coefficient at $1/4>0$, including the contribution from the quadratic terms.

The figure includes these local curves and the global curves derived below. All lines emerging from the double-zero point are local asymptotic sketches.

The relevant saddle equilibrium is the origin, with eigenvaluesFor $\lambda <0$, they have opposite signs. Its saddle index isnear the homoclinic curve, because that curve has $\kappa >0$. Along its leading asymptote, $\delta =1-\frac{4}{5}\sqrt{-\lambda }+o(\sqrt{-\lambda })$.