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Use the same specified homogeneous Newtonian background and Jeans swindle, but describe stars by a mass-normalized galactic distribution function. The Collisionless Boltzmann equation is ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{x}f-\nabla \varphi \cdot {\nabla }_{v}f=0$. Let $f=f_0(\mathbf{v})+f_1{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ and similarly perturb the gravitational potential. To first order,Eliminating $f_1$ and the nonzero gravitational potential amplitude gives the collisionless Jeans dispersion relationFor growing modes $\mathrm{Im}\omega >0$ there is no real-velocity pole. Real and damped frequencies require a causal contour prescription or analytic continuation; one must not simply integrate through a pole without specifying it.

Take $b>0$ in the Cauchy velocity distribution. Its normalization is $\int f_0{d}^{3}v={\rho }_0$. Orient $\mathbf{k}$ along the $u=v_z$ axis and integrate over the perpendicular velocities:At the marginal mode $\omega =0$, $F^{\prime }(u)/u=-2{\rho }_{0}b/[\pi (u^2+b^2{)}^2]$, with its finite limiting value at zero. The provided Gamma function integral at power two givesConsequentlyTo check that this really separates growing modes, put $c=\omega /k$ in the upper half-plane. Residue integration, or integration by parts followed by the Cauchy resolvent, gives $\int F^{\prime }(u)/(u-c)du={\rho }_0/(c+ib{)}^2$. The upper-half-plane solution has $\omega =i(\sqrt{4\pi G{\rho }_0}-kb)$, which grows exactly for $0<k<k_J$. At the threshold it approaches the marginal mode continuously. Although $b$ plays the threshold role of a velocity scale, this distribution's second moment diverges: the radial integrand for $\int v^2{f}_0{d}^{3}v$ tends to a nonzero constant at large speed. It is therefore incorrect to identify $b^2$ with a finite velocity dispersion or to infer this result by substituting an rms speed into a fluid formula.

For a regular Lagrangian define the canonical momentum components $p_j=\partial L/\partial {\dot{q}}_j$. Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of $(q,p,t)$. The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is $dH={\sum }_j{\dot{q}}_{j}d{p}_j-{\sum }_j(\partial L/\partial q_j)d{q}_j-(\partial L/\partial t)dt$, since the $d{\dot{q}}_j$ terms cancel. The Euler-Lagrange equations give $\partial L/\partial q_j={\dot{p}}_j$, and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, $H$ is conserved. Its sign convention differs from the binding energy in question 2: here $E=H=v^2/2-\psi$ is the ordinary specific orbital energy.

The prescribed nondegenerate prolate spheroidal coordinates require $0<\alpha <\beta$; equality collapses the coordinate family. Their physical velocity components are $v_j=h_j{\dot{q}}_j$, and $h_{\varphi }=R$. Per unit stellar mass the Lagrangian is $L=\frac{1}{2}{\sum }_j{h}_j^2{\dot{q}}_j^2+\psi$, so $p_j=h_j^2{\dot{q}}_j=h_j{v}_j$. Its Hamiltonian isThis uses acceleration $+\nabla \psi$, as specified, rather than importing the opposite gravitational potential convention.

For the Stäckel potential, use the Hamilton-Jacobi equation with principal function $S=W_{\lambda }(\lambda )+W_{\mu }(\mu )+l\varphi -Et$. Axisymmetry makes $p_{\varphi }=l=R{v}_{\varphi }$ constant. Write $d=\lambda -\mu$ and $A(\tau )=(\tau +\alpha )(\tau +\beta )$. Multiplying $H=E$ by $d$ and usingseparates it as $F(\lambda )-E\lambda =F(\mu )-E\mu$, where $F(\tau )=2A(\tau )(W_{\tau }^{\prime }{)}^2-(\alpha -\beta )l^2/[2(\tau +\alpha )]-G(\tau )$. Each side depends on its own coordinate only and therefore equals a constant, chosen as $-I$. Here $G(\tau )$ is the generating function of the gravitational potential, not the gravitational constant. HenceThis proves the conserved axisymmetric Stäckel third integral. Eliminating $E$ between the two expressions and using $p_j=h_j{v}_j$ givesAt regular generic phase-space points the three separation constants $E,l,I$ are independent; special circular or degenerate orbits can reduce their local independence. Thus the model is an integrable Hamiltonian system.

For the intended regular, phase-mixed steady model, Jeans theorem permits a galactic distribution function $f=F(E,l,I)$. At any fixed point, $E$ and $I$ depend on $v_{\lambda }$ and $v_{\mu }$ only through their squares, and $l$ depends only on $v_{\varphi }$. Thus $f$ is even separately in $v_{\lambda }$ and $v_{\mu }$, whether or not it is even in $v_{\varphi }$. Odd integration proves velocity-ellipsoid alignment in a Stäckel potential:The argument allows azimuthal streaming: it does not require zero mean $v_{\varphi }$. The means in the two meridional directions vanish, so the off-diagonal velocity covariance also vanishes. Second moments must exist for this conclusion.

There is a hypothesis behind the printed universal wording. On a generic nonresonant regular invariant torus, a continuous steady distribution is constant along a dense orbit and hence depends only on the torus integrals; this justifies the phase-mixed form. Bare stationarity alone is not sufficient in a degenerate resonant gravitational potential with additional integrals. For example $H=\frac{1}{2}(v^2+{\omega }^2{r}^2)$ is a Stäckel Hamiltonian: choose $G(\tau )=-\frac{1}{2}{\omega }^2[{\tau }^2+(\alpha +\beta )\tau ]$, up to a constant vacuum shift. The additional conserved quantity $K_{xy}=v_x{v}_y+{\omega }^{2}xy$ gives a positive normalizable steady tracer $f\propto \exp (-aH+ϵK_{xy})$ for $a>|ϵ|>0$. Its local $⟨v_x{v}_y⟩=ϵ/(a^2-{ϵ}^2)\ne 0$. At azimuth zero and a regular point where the $\lambda$ direction has a radial component, this gives nonzero $⟨v_{\lambda }{v}_{\varphi }⟩$. This tracer need not itself source the harmonic gravitational potential. It exhibits why the stronger, ordinary phase-mixed three-integral assumption is necessary instead of claiming that every imaginable steady distribution has the asserted alignment.

The geometry follows by multiplying the coordinate-defining equation and evaluating at $\tau =-\alpha ,-\beta$:Constant $\lambda$ surfaces are prolate spheroids, with equatorial semiaxis $\sqrt{\lambda +\alpha }$ and polar semiaxis $\sqrt{\lambda +\beta }$. Constant interior $\mu$ surfaces are two-sheeted hyperboloids. Their common foci lie at $z=\pm \sqrt{\beta -\alpha }$. The orbit volume in an axisymmetric Stäckel potential is determined byIts allowed coordinate intervals have turning boundaries or coordinate endpoints. A typical bound orbit with $l\ne 0$ oscillates between inner and outer spheroids, oscillates vertically between hyperboloidal limits, and circulates about the symmetry axis with $\dot{\varphi }=l/R^2$. The equatorial fold $\mu =-\beta$ joins the positive and negative $z$ branches; it is not necessarily a physical turning point in $z$.

For three incommensurate frequencies, the regular trajectory densely samples its projected allowed volume, with nonuniform occupation determined by its speeds. It does not follow one coordinate surface, and it need not close. Resonant orbits may be closed or lie in a smaller projected set; $l=0$ confines azimuthal motion to a meridional plane (with the usual axis coordinate qualification), and circular or equatorial limits are also lower-dimensional. Smooth Stäckel separation removes the generic chaotic behavior of nonintegrable potentials; it does not assert that every orbit fills a three-dimensional volume.

Let $\mathbf{e}=\mathbf{X}/R$, $s=\mathbf{e}\cdot \mathbf{Be}$, and $A_{\parallel }=\mathbf{A}\cdot \mathbf{e}$. Applied forces and couples below are forces and torques on the solid; by force balance their values are also the strengths exerted by that solid on the fluid. This fixes the signs in Faxén's first law and Faxén's rotational law. To hold the passive sphere fixed, set its translation and angular velocity to zero in those laws. The incident two-mode tensorial squirmer flow givesOnly the stresslet part has vorticity:Consequently, for fixed $\mathbf{A},\mathbf{B}$ as $a/R\to 0$,The leading force is the radial stresslet force. Its next correction, $O(\mu a^4|\mathbf{A}|/R^3)$, comes from the incident potential dipole. The next $\mathbf{B}$ correction, $O(\mu a^5\Vert \mathbf{B}\Vert /R^4)$, contains both the swimmer's $r^{-4}$ field and the Laplacian term in Faxén's first law. If the relevant coefficient vanishes, these next terms must be retained instead of calling the vanished term a nonzero leading approximation. Further exchanges in the method of reflections for Stokes flow are smaller still.

The passive sphere reflects primarily a Stokeslet with strength ${\mathbf{F}}_h$. At the swimmer, its velocity is $(\mathbf{I}+\mathbf{ee}){\mathbf{F}}_h/(8\pi \mu R)$. The unchanged surface slip velocity adds the free swimming velocity to the incident-flow contribution in Faxén's first law. ThusThe leading change is $O(a^3\Vert \mathbf{B}\Vert /R^3)$; its next correction is the displayed $O(a^4|\mathbf{A}|/R^4)$ term. Finite-radius Faxén's first law corrections, the passive sphere's reflected stresslet, and its rotlet contribute at order $a^5\Vert \mathbf{B}\Vert /R^5$ to translation.

For a Stokeslet at the origin,The curl of the first term is $\mathbf{F}\times \mathbf{x}/r^3$; that of the second is also $\mathbf{F}\times \mathbf{x}/r^3$, before multiplying by $1/(8\pi \mu )$. Hence ${\mathbf{\omega }}_F=\mathbf{F}\times \mathbf{x}/(4\pi \mu r^3)$. At the swimmer use displacement $-\mathbf{X}$. The leading $\mathbf{B}$ force is parallel to $\mathbf{X}$, so its Stokeslet has zero vorticity there. This eliminates the putative $O(a^3\Vert \mathbf{B}\Vert /R^4)$ rotation. The potential dipole itself has no vorticity, but the holding force it induces is not radial. Applying Faxén's rotational law to its reflected Stokeslet yieldsThe passive sphere's rotlet and reflected stresslet can affect that last order, so they cannot restore the larger missing rotation. When $\mathbf{A}\times \mathbf{X}=0$, the boxed coefficient is zero and the higher-order terms determine any rotation.

Take $z$ positive downwards. The leading axial velocity of a slender viscous thread is independent of radius. Incompressible flow then gives $u_r=-r{w}^{\prime }/2$, while steady mass conservation gives $(a^{2}w{)}^{\prime }=0$. The Newtonian fluid stress tensor has ${\sigma }_{rr}=-p-\mu w^{\prime }$ and ${\sigma }_{zz}=-p+2\mu w^{\prime }$. The Young–Laplace equation, neglecting axial curvature at slender order, gives ${\sigma }_{rr}=-p_e-\gamma /a$ and hence $p=p_e+\gamma /a-\mu w^{\prime }$.

The axial cut transmits both the integrated Newtonian fluid stress tensor and the circumferential surface tension. After subtracting the ambient pressure, its tensile force isThe coefficient three is the Trouton ratio for axial extension. The last term is essential: the capillary tensile force of a slender thread is $\pi \gamma a$, not $2\pi \gamma a$, because the capillary pressure subtracts half the interfacial line contribution. Axial force balance is $T^{\prime }+\pi \rho g{a}^2=0$. Together the governing equations areDefine the dimensionless variablesThen $a=a_0{W}^{-1/2}$ and the axial equation becomesFor the increasing power-law branch, all three terms have the same dependence when $\alpha =2$. Substitution leaves $1-6k^2-\Gamma k=0$. ThusThis remains valid at $\Gamma =0$, when $k=1/\sqrt{6}$. The other root is negative and describes a decreasing local branch up to its singular endpoint, not a thread accelerating downwards indefinitely. Positive surface tension decreases $k$, so it slows the fall at a given height on this branch. Thinning reduces the capillary tensile force $\pi \gamma a$. Its negative axial derivative supplies an upward capillary force on a thread segment, balancing part of the weight. The viscous tensile force therefore supplies a smaller remaining balance; on the quadratic family this reduces the stretching rate.

The condition $W(0)=1$ alone does not select a unique solution of this second-order ordinary differential equation. The boxed power law is the requested verified branch, with tensile force tending to zero at large height; an additional nozzle tension or downstream condition is required to select it physically. The slender, negligible-inertia approximation must remain valid over the height range considered.

Put the fixed outer centre at the origin, and let the moving inner centre be $\mathbf{d}=(x,z)$. This reverses the sign of the vertical offset used in the previous parts: now $h=\Delta -\mathbf{d}\cdot \mathbf{n}$. At concentricity rotation produces no force and translation has isotropic resistance $R_0=12\pi \mu a^3/{\Delta }^3$. Balancing the external load givesThere is no inertial acceleration in this Stokes flow model; velocity is determined instantaneously by force balance.

For the subsequent motion, the full-film journal bearing assumption matters: pressure is single-valued around the whole annulus, with no specified cavitation boundary. Let $r=|\mathbf{d}|$, $e=r/\Delta$, $\varphi =\arg (x+iz)$, and ${\mathbf{e}}_r,{\mathbf{e}}_{\varphi }$ be its radial and tangential unit vectors. The squeeze resistance of an eccentric journal bearing has radial and tangential coefficientsFor completeness, rotate coordinates until $\mathbf{d}$ points along $x$, so $h=\Delta (1-e\cos \theta )$. Radial translation gives squeezing flux $q=a{v}_r\sin \theta$ and resistance proportional to $\int {\sin }^2\theta /(1-e\cos \theta {)}^3=\pi /(1-e^2{)}^{3/2}$. Tangential translation gives $q=a{v}_t[-\cos \theta +3e/(2+e^2)]$; the constant enforces periodic pressure. Its resistance usesThese expressions supply both translation directions; retaining only the vertical squeezing formula would miss the moving axle's horizontal dynamics.

The rotational fluid force from the earlier calculation points along ${\mathbf{e}}_{\varphi }$, with magnitude $R_t\Omega r/2$. Superposing rotation and translation by Linearity of Stokes flow, and resolving the load, givesFor positive $\Omega$ the axle initially moves down, then to the right because the rotation-generated force is perpendicular to the downward eccentricity. The equilibrium is displaced horizontally to the right, at the same height as the outer axis. Its eccentricity $0<e_{\ast }<1$ is the unique solutionThe rotational force then points upwards and balances the load. For negative $\Omega$ the displacement is to the left, using $|\Omega |$ in the magnitude equation. Uniqueness follows because the logarithmic derivative of $e/[\sqrt{1-e^2}(1+e^2/2)]$ has positive numerator $1-e^2/2+e^4$ after a common positive denominator. If $\Omega =0$ and $F>0$, no separated static equilibrium exists in this ideal model. If $F=0$, concentricity remains a solution.

One should not assert that this full-film model necessarily settles to the equilibrium. It predicts full-film journal-bearing whirl. For $F,\Omega >0$, eliminate time from the two boxed evolution equations and put $K=6\pi \mu \Omega a^3/{\Delta }^2$. The orbit starting at concentricity obeysso, with its zero initial horizontal displacement,The orbit begins on the lower branch, turns at $z=0$ before contact, and returns along the upper branch, forming a closed orbit through concentricity. The ratio of the right-hand side to $e$ increases from zero to infinity, giving a unique turning eccentricity below one. Linearizing at equilibrium likewise gives $\delta \dot{r}=-(F/R_{r,\ast })\delta \varphi$ and $\delta \dot{\varphi }=c_{\ast }\delta r$ with $c_{\ast }>0$: the eigenvalues are imaginary, not damped. Thus the equilibrium position exists, but the initially concentric axle circulates around it in the stated ideal full-film approximation. Cavitation or additional mechanical constraints would change the model; none were specified.

The normalized centre trajectory below illustrates the full-film prediction for $K/F=1$. The dashed circle is the contact locus of the axle centre, not either cylinder wall.

Let $F(x,z,t)=z-\eta (x,t)$. The interface is a material surface for both fluids: its material derivative vanishes on $F=0$. With $u_j={\partial }_x{\varphi }_j$ and $w_j={\partial }_z{\varphi }_j$, this givesThus the kinematic boundary condition says that particles stay on the moving interface. The tangential velocities can differ; each side has its own material derivative. Replacing both derivatives by one background speed would be incorrect.

The dynamic boundary condition for an inviscid interface is pressure continuity, because the fluids have no viscous normal stress and there is no surface tension. The unsteady Bernoulli equation in each layer of irrotational flow readsEquating these pressures at $z=\eta$ gives the required dynamic boundary condition for an inviscid interface. Choose the potential gauges so that ${\rho }_1{C}_1={\rho }_2{C}_2$, or set both constants to zero relative to a common reference pressure. For example, ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ removes the distinct uniform-stream Bernoulli constants without changing either velocity. Material-interface kinematics and pressure continuity supply the two matching conditions, with the Bernoulli gauge understood.

For an even velocity potential, take ${\Phi }_I=C\cosh (kz)$ and the decaying exterior amplitudes equal at the upper and lower interfaces. At $z=L$, kinematics gives $A=ic{B}_{+}$ and $C\sinh (kL)=i(V-c)B_{+}$. The dynamic condition becomes $-cA=(V-c)C\cosh (kL)$. ThereforeWriting $s=\tanh (kL)$, its two roots areThe interior vertical velocity is odd, so $B_{-}=-B_{+}$; this is a varicose mode of a planar jet. As with the odd potential, $k>0$, finite $L>0$ and $V\ne 0$ give a growing root. The parity of the potential and the parity of the interface displacements are opposite because the vertical velocity is its derivative.

For an odd velocity potential, chooseAt $z=L$ the two kinematic conditions give $A=ic{B}_{+}$ and $C\cosh (kL)=i(V-c)B_{+}$. The dynamic condition is $-cA=(V-c)C\sinh (kL)$. Eliminating $A,C,B_{+}$ gives, with $s=\tanh (kL)$,The lower interface gives the same equation by odd parity of the velocity potential. The interior vertical velocity is even, so $B_{-}=B_{+}$: the displacement is a sinuous mode of a planar jet even though the potential is odd.

For $L>0$, $V\ne 0$ and $k>0$, $0<s<1$ and one root has $c_i>0$. Thus every nonzero wavenumber is unstable in this ideal top-hat planar jet. The phrase “all choices” is understood with these assumptions: $k=0$ has no exponential wave growth, and $V=0$ has no shear.

Write ${\varphi }^{\prime }=\Phi (z)e^{ik(x-ct)}$ and let $B_{\pm }$ be the displacement amplitudes at $z=\pm L$. Use subscripts $I$ for the jet interior and $O$ for the adjoining exterior. Each interface is a material vortex sheet, so its linearized kinematic boundary conditions areThe equal mass densities make the hydrostatic displacement terms cancel. The dynamic boundary condition for an inviscid interface therefore givesThe division form assumes $c\ne 0,V$; the preceding undivided equations remain the proper conditions in those exceptional cases. Potentials themselves need not be continuous, and their vertical derivatives need not be equal: the same material displacement is advected by different tangential base velocities. At infinity the exterior potentials decay. These displacement and pressure conditions, rather than continuity of $\Phi$ and ${\Phi }^{\prime }$, determine the equal-density top-hat planar-jet dispersion relation.

Let $D=\overline{U}-c$ and $F=\hat{w}/D^{1/2}$. For $c_i>0$, $D$ never vanishes, so an analytic branch of a square root exists along the real flow domain. Take real smooth velocity and buoyancy frequency coefficients, nonzero $k$ (chosen positive for the growing-wave convention), a nontrivial regular normal mode, and boundary decay strong enough to remove the integration-by-parts term.

Substitution into the Taylor–Goldstein equation gives the half-power case of the power-transformed Taylor–Goldstein energy identity:For completeness, differentiating $\hat{w}=D^{1/2}F$ yields ${\hat{w}}^{\prime \prime }=D^{1/2}{F}^{\prime \prime }+{\overline{U}}^{\prime }{D}^{-1/2}{F}^{\prime }+[\frac{1}{2}{\overline{U}}^{\prime \prime }{D}^{-1/2}-\frac{1}{4}({\overline{U}}^{\prime }{)}^2{D}^{-3/2}]F$, which explains the coefficient $1/4$.

Multiply by $F^{\ast }$ and integrate over $z$. The boundary term $[F^{\ast }D{F}^{\prime }]$ is zero for the given homogeneous endpoint conditions, or for sufficiently decaying finite-energy modes at infinity. ThereforeSince $\mathrm{Im}D=-c_i$, $\mathrm{Im}(1/D)=c_i/|D{|}^2$, and $S,{\overline{U}}^{\prime \prime }$ are real, taking the imaginary part and dividing by $-c_i$ givesThe first two terms have strictly positive integral for a nonzero mode. Thus $S$ cannot be nonnegative everywhere:This proves the Miles–Howard theorem in contrapositive form. Where ${\overline{U}}^{\prime }\ne 0$, the corresponding gradient Richardson number must fall below $1/4$ somewhere. Failure of this sufficient-stability criterion does not prove instability. The derivation uses $c_i>0$; it cannot be applied unchanged to a neutral singular critical layer.

Use $\nu =R{e}^{-1}>0$, $\kappa =P{e}^{-1}>0$, $\mathbf{U}=\overline{\mathbf{u}}+\mathbf{u}$, and $\Theta =\overline{\theta }+\theta$. Work along a sufficiently regular direct trajectory satisfying the given constraints. The zero spatial mean of the scalar field is conserved by incompressible flow, impermeable walls and zero scalar flux, so minimizing $J=(\Theta (T),\Theta (T))$ is equivalent to minimizing scalar variance, up to the fixed domain volume.

The printed functional fixes the initial state to a candidate ${\mathbf{u}}_0$; it contains no term that enforces its kinetic energy. For the optimization over that candidate, add the real Lagrange multiplier constraintEquivalently, one can restrict all control variations to the sphere in a normed vector space of fixed kinetic energy. This term changes the initial-control optimality condition, not the interior adjoint equations.

Let $\mathbf{v}=\delta \mathbf{u}$, $\sigma =\delta \theta$ and $\pi =\delta p$. Linearization of the momentum and scalar transport residuals givesBoth appearances of the perturbation velocity in the nonlinear momentum term have been differentiated. In particular, the coefficient is the gradient of the total velocity, not just the base shear.

For the negative-constraint convention of the functional, integration by parts gives the interior coefficients of $\mathbf{v},\sigma ,\pi$ asThus the adjoint equations for Boussinesq scalar mixing areThe transpose is essential: the $j$th component of $(\nabla \mathbf{U}{)}^T{\mathbf{u}}^{†}$ is ${\sum }_i{u}_i^{†}{\partial }_j{U}_i$. The coupling $-{\theta }^{†}\nabla \Theta$ transposes advection of the scalar by a velocity perturbation; $-R{i}_B{u}_y^{†}$ transposes buoyancy feedback. Dropping the latter would give a passive scalar adjoint, not the active scalar problem.

These equations are integrated backward, not forward. If $\tau =T-t$, they readwith direct coefficients evaluated at $t=T-\tau$. Both terms from the diffusion equation now have the usual forward sign in $\tau$. A direct-adjoint looping method stores or reconstructs the forward trajectory, solves these equations backward, and uses the initial adjoint as the control gradient. The endpoint and fixed-energy conditions below give necessary conditions for a local optimizer, not a global optimality theorem.

The temporal integration by parts terms areThe terminal states are free and the objective has no explicit terminal-velocity dependence. ThereforeThe factor two follows from the displayed objective without a $1/2$ prefactor. The initial scalar field perturbation is fixed, so $\delta \theta (0)=0$ and there is no independently prescribed adjoint-scalar condition at $t=0$. Its value there is obtained by backward integration. Independent variation of the initial velocity state gives ${\mathbf{u}}^{†}(0)={\mathbf{u}}_0^{†}$.

Variation of the candidate ${\mathbf{u}}_0$ in the energy-augmented functional gives the fixed-energy initial-condition optimality condition. For the usual divergence-free no-slip control space $H$, write $P_H$ for its orthogonal projection, the Leray-Helmholtz projection in the usual incompressible velocity space. This is the standard formal control condition when pressure is determined by incompressible flow and normal momentum. If the independently imposed direct pressure flux is retained as an additional constraint, the admissible control variations must also satisfy its compatibility conditions along the trajectory; they need not fill the usual space $H$. The standard projection formula alone does not prove stationarity for that more restricted problem. ThenFor the natural solenoidal adjoint space, $P_H{\mathbf{u}}^{†}(0)={\mathbf{u}}^{†}(0)$. Equivalently, its component tangent to the sphere in a normed vector space of fixed kinetic energy vanishes:Here $E_0>0$ is needed for the sphere in a normed vector space to be regular and for division by $2E_0$. The multiplier is real and can have either sign; normalizing the initial adjoint with a prescribed positive sign is not a general necessary condition for minimization. If $E_0=0$, the only feasible initial velocity is zero and this tangent formula for the sphere in a normed vector space is inapplicable. First-order stationarity alone also allows maxima or saddles; a local minimum requires the appropriate nonnegative constrained second variation.

The spatial boundary terms vanish with periodic adjoints in $x,z$, homogeneous no-slip boundary conditions ${\mathbf{u}}^{†}=0$ at $y=\pm 1$, and ${\partial }_y{\theta }^{†}=0$ there. To see this, velocity variations vanish at the wall while their normal derivatives need not, so the viscous boundary term forces the adjoint velocity to vanish. Scalar variations have zero normal derivative but free values, forcing the adjoint scalar's normal derivative to vanish. Pressure variation gives adjoint incompressible flow; no independent terminal or initial datum is assigned to $p^{†}$. Its additive time-dependent constant may be fixed by a zero spatial mean.

In particular, a homogeneous Neumann condition for the direct pressure does not by variational transposition require ${\partial }_y{p}^{†}=0$. For a smooth adjoint solution, the wall-normal adjoint momentum equation instead suppliesat the flat walls, because the direct scalar normal derivative is zero there. A zero adjoint-pressure derivative would be an additional compatibility choice, valid only when this right-hand side vanishes.

There are also literal direct-data compatibility qualifications in the printed setup. The initial total scalar field is $\overline{\theta }=-\mathrm{erf}(30y)$, whose wall derivative is $-60e^{-900}/\sqrt{\pi }\ne 0$. It is extraordinarily small but mathematically not zero. Thus exact initial scalar data and exact zero wall flux are incompatible for a classical solution smooth at $t=0$. One may use a parabolic mild solution of an abstract Cauchy problem with the boundary condition enforced for $t>0$, or replace the initial profile by an exactly compatible smooth one; these are conventions, not an equality $e^{-900}=0$. The reference scalar field is also not a stationary profile of the diffusion equation at finite $Pe$: the full $\Delta \Theta$ term must be retained.

The pressure compatibility at a no-slip wall is similarly important. Direct normal momentum gives ${\partial }_{y}p=\nu \Delta u_y-R{i}_B\theta$. The separately imposed zero pressure derivative requires this right side to vanish. It is not automatic for arbitrary fixed-energy initial velocities. For example, the divergence-free no-slip field generated by the periodic stream function $S=A(1-y^2{)}^2\sin (\alpha x)$ with $\alpha =\pi /L_x>0$ has $u_x={\partial }_{y}S$, $u_y=-{\partial }_{x}S$ and $\Delta u_y{|}_{y=\pm 1}=-8A\alpha \cos (\alpha x)\ne 0$, while $\theta (0)=0$. With $A\ne 0$ adjusted to any positive energy, no classical solution smooth up to the initial wall can satisfy that extra pressure condition. The displayed adjoints are the formal necessary equations along admissible smooth trajectories; a well-posed physical formulation normally determines pressure from incompressible flow and normal momentum, rather than imposing independent homogeneous pressure flux for every control.