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For a straight singly quantized quantum vortex, use cylindrical distance $r$ from the line and writeThere is no dependence along the vortex line. Substitution into the normalized stationary equation gives, before separating real and imaginary parts,Consequently the wave amplitude and phase-gradient equations areThe second equation can also be written $(r{f}^{2}k{)}^{\prime }=-{\xi }_{I}r{f}^2(1-f^2)$, which remains convenient at the core.

To specify what is meant by radial velocity, use the condensate number current. The time-dependent equation has kinetic operator $-{\nabla }^2$ and therefore number current $\mathbf{j}=2|\Psi {|}^2\nabla S$. Its hydrodynamic superfluid velocity in these units is $\mathbf{u}=2\nabla S$, so $u_r=2k$ and $u_{\theta }=2/r$. In that convention the requested pair isIf instead velocity is defined as the complex argument gradient itself, the preceding $k$ pair is the corresponding convention; the distinction matters because the printed kinetic coefficient differs from that of the normalized conservative equation in Question 1.

Regularity of a unit-charge core gives $f(r)=ar+O(r^3)$ with $a>0$. Integrating the current equation from the origin, with no point source or singular radial flux, givesThe integral is $a^2{r}^4/4+O(r^6)$, so the radial flow near a driven vortex core isFor the phase-gradient convention, $k^{\prime }(0)=-{\xi }_I/4=\alpha /4$. Gain exceeds loss in the depleted core, hence this regular flow is outwards for $\alpha >0$. The real wave amplitude equation also gives $f=ar-{\xi }_{R}a{r}^3/8+O(r^5)$, independently consistent with the linear core behaviour. A nonzero integration constant in the radial-current identity would create a singular $u_r\sim r^{-3}$ and is excluded.

Finally, the literal boundary $\psi \to 1$ from the preceding part cannot apply to a multiplicity-one vortex: its complex argument changes by $2\pi$ around a large circle. Even $f(r)e^{i\theta }$ with $f\to 1$ has different limits along different rays. The usual intended vortex condition concerns the wave amplitude approaching its bulk value, with the winding complex argument retained; it is not a constant complex-field limit. The local equations and core slope derived here do not establish a global stationary vortex satisfying that literal boundary. In a driven system the far-field radial flow and oscillation frequency may also require selection, so no global zero-flow vortex is asserted.

The field $\varphi$ is a compositional order parameter in a binary fluid mixture. Its spatial integral is conserved in a closed sample, so advection and diffusion enter a continuity equation; the material derivative follows the incompressible fluid. The diffusive current $\mathbf{J}=-M\nabla \mu$, with positive order-parameter mobility $M$, moves material down the chemical potential gradient. The chemical potential of a composition field is the functional derivative of the symmetric quartic Landau free energyso $\mu =a\varphi +b{\varphi }^3-\kappa {\nabla }^2\varphi$. Positive $b,\kappa$ stabilize large amplitudes and penalize sharp interfaces; $a<0$ permits phase separation.

The momentum equation is the Navier-Stokes equation with dynamic viscosity $\eta$, constant mass density $\rho$, and the Korteweg force $-\varphi \nabla \mu$ arising from composition stresses. The pressure enforces incompressible flow, $\nabla \cdot \mathbf{v}=0$, and can absorb pure gradient contributions to that thermodynamic force. For example, $-\varphi \nabla \mu =\mu \nabla \varphi -\nabla (\varphi \mu )$ gives an equivalent force after redefining the pressure. These equations constitute deterministic Model H dynamics.

Their dissipative signs can be checked by the energy dissipation identity for Model H. Under periodic or no-flux, no-work boundary conditions, composition advection and capillary work cancel between $F$ and the kinetic energy, givingThus diffusion and viscosity dissipate energy, while the reversible advective coupling transfers it between composition and motion. Thermal noises are omitted at this deterministic hydrodynamic level.

The dynamical scaling of binary-fluid coarsening assumes a statistically self-similar bicontinuous morphology with one growing length $L$, thin interfaces, curvature of order $1/L$, and typical fluid speed of order $\dot{L}$. Inertial acceleration then scales as $\ddot{L}$ and ${\dot{L}}^2/L$, viscous force density as $\eta \dot{L}/L^2$, and capillary force density as $\sigma /L^2$. The pressure is eliminated or estimated together with the interfacial force. The numerical coefficients summarize geometry and projected force directions; this is a scaling closure, not a spatially exact momentum equation.

The approximation further neglects diffusive growth relative to advective growth, assumes fixed volume fraction and material parameters, no external forcing, and no additional relevant macroscopic length. A finite initial domain size, interface width or finite container would introduce further ratios outside the asymptotic scaling regime. Velocity gradients need not stay on the domain scale in a genuinely multiscale turbulent flow, so this single-length argument is not a proof for every hydrodynamic morphology.

With $[\rho ]=\mathrm{M}{\mathrm{L}}^{-3}$, $[\eta ]=\mathrm{M}{\mathrm{L}}^{-1}{\mathrm{T}}^{-1}$ and $[\sigma ]=\mathrm{M}{\mathrm{T}}^{-2}$, dimensional analysis gives the unique scalesThus, within the stated asymptotic single-scale hypothesis, the hydrodynamic coarsening crossover scales give $L=L_{0}f(u)$, $u=t/t_0$. All four force-density terms have the same dimensional prefactor ${\rho }^2{\sigma }^3/{\eta }^4$. Dividing by it yieldsPrimes denote $d/du$. The PDF includes $\eta \gamma \dot{L}/L^2$; the TeX drops $\eta$. The PDF expression is dimensionally consistent. For a positive growth branch, a resisting viscous term and a driving capillary term have opposite signed coefficients; one may set $\gamma =-c_v<0$, $\delta =c_s>0$. The question calls these coefficients order one, not positive. Treating every term on the printed right-hand side as positive would lose the viscous-capillary balance.

The unit normal pointing out of the fluid at its lower boundary is $-{\mathbf{e}}_y$. Since impermeability gives $v(x,0)=0$, also $v_x(x,0)=0$, and the tangential viscous traction exerted on the fluid is $t_x=-\mu u_y(x,0)$. The cilia do positive work at rate $t_{x}u=\rho P$, soFor positive downstream velocity this requires $u_y<0$, as expected when the wall-driven flow slows away from the cilia. The wall velocity is not prescribed independently; the driving condition prescribes power.

In a laminar boundary layer, momentum diffuses over a thickness $\delta$ during the downstream advection time $x/U$, giving ${\delta }^2\sim \nu x/U$. The wall power condition gives $U^2/\delta \sim P/\nu$. With these balances the downstream Reynolds number grows and $\delta /x\propto x^{-3/5}$ tends to zero. This explains the thin layer for large $x$ within the laminar approximation. Pressure is uniform across it and has no imposed streamwise gradient because the outer fluid is at rest. The two-dimensional boundary-layer equations are

Choose the scales to satisfy the balances exactly:The stream function $\psi =U\delta f(\eta )$, $\eta =y/\delta$, enforces continuity equation and givesThus the advective term is $U^2[\frac{1}{5}{f}^{\prime 2}-\frac{3}{5}f{f}^{\prime \prime }]/x$, while $\nu u_{yy}=\nu U{f}^{\prime \prime \prime }/{\delta }^2=U^2{f}^{\prime \prime \prime }/x$. HenceImpermeability, imposed wall power and matching to resting outer fluid giveA regular decaying profile also has $f^{\prime \prime }\to 0$ and finite positive $f_{\infty }$; no arbitrary additional wall value for $f^{\prime }$ should be imposed.

The volume flux per unit span is $Q={\int }_0^{\infty }udy=U\delta f_{\infty }\propto x^{3/5}$, so continuity equation gives $v_{\infty }=-dQ/dx<0$: outer fluid is entrained toward the wall. In the physical positive decaying similarity profile, $f^{\prime }>0$ and $f^{\prime \prime }<0$. To justify monotonicity, $f^{\prime \prime }(0)<0$ by the power condition; if $f^{\prime \prime }$ first crossed zero with $f^{\prime }>0$, the equation would give $f^{\prime \prime \prime }=f^{\prime 2}/5>0$, after which $f^{\prime \prime }>0$ cannot cross back and $f^{\prime }$ could not decay to zero. Therefore $f(\eta )={\int }_0^{\eta }{f}^{\prime }(s)ds\ge \eta f^{\prime }(\eta )$, and the velocity formula gives $2\eta f^{\prime }-3f<0$ for $\eta >0$. Consequently the transverse velocity is negative throughout that physical layer, and zero at the wall. This pointwise conclusion uses the positive decaying branch; the flux argument already establishes entrainment of the exterior.

Assume the holonomic constraints have linearly independent gradients, so their common level set is locally a smooth constraint manifold parametrized by $q_1,\dots ,q_n$. The multiplier form of the Euler-Lagrange equations isAt fixed time ${\sum }_A{\partial }_A{f}_{\alpha }{\partial }_i{x}_A=0$. The chain rule for the reduced Lagrangian givesConversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. Linear independence of the constraint gradients implies that it is a linear combination of those gradients, giving the required Lagrange multipliers. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.

For the bead, $\dot{y}=f^{\prime }(x)\dot{x}$, soThe autonomous conservation of energy gives $E=\frac{1}{2}(1+f^{\prime 2}){\dot{x}}^2$. If $E>0$, the motion has a fixed sign $s=\mathrm{sgn}\dot{x}$ andThe integral is the signed arc length divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a function of its unchanging position is not meaningful.

Work per unit axial length, with $y$ vertically upward and $x$ across the thin gap. In the frame of the downward-translating cylinder, the wall moves upward at speed $U$ and the cylinder surface is stationary. Let $\eta$ be viscosity and use the leading lubrication theoryThe transverse pressure gradient is negligible; solving $\eta u_{xx}=p_y$ with $u(0)=U$, $u(h)=0$ givesThe flux $q$ is constant by incompressibility. Equal ambient pressure at both ends implies $\int p_{y}dy=0$. With $I_j={\int }_{-\infty }^{\infty }{h}^{-j}dy$ and $I_2/I_3=4h_0/3$, this yieldsThe tangential traction on the cylinder isSince $I_2=I_1/(2h_0)$, its integrated force is $-2\eta U{I}_1+4\eta U{h}_0{I}_2=0$. Pressure acts radially on a circular cylinder and contributes no torque; the tangential traction has moment arm $a$ to leading order. HenceFor clarity, the total vertical hydrodynamic force is not zero. Its leading pressure contribution isupward, while the leading shear contribution just calculated vanishes. Thus this model predicts no induced rolling direction: a freely rotating cylinder can remain nonrotating, rather than rolling like a wheel in contact with the wall. This is the torque-free cylinder in a lubrication gap cancellation; finite ends or different boundary conditions would require a different model.

Set $F(z,w)=w^2-z^n+1$. Its complex gradient is $(-n{z}^{n-1},2w)$, never zero on $F=0$: simultaneous zero would give $z=w=0$, which is not on the curve. The complex implicit function theorem therefore gives a one-dimensional complex manifold, hence a Riemann surface. Where $w\ne 0$ use $z$ as coordinate, and where $z\ne 0$ use $w$; the transition maps and both projections are holomorphic.

For $\pi$, a ramification point of a holomorphic map has $w=0$ and $z=\zeta$ with ${\zeta }^n=1$. In the coordinate $w$, $z-\zeta =w^2/(n{\zeta }^{n-1})+O(w^4)$, giving ramification index $2$. Thus there are $n$ such points, and their branch points are the $n$th roots of unity.

For $\tau$, a ramification point of a holomorphic map has $z=0,w=\pm i$. In the coordinate $z$, $w-w_0=z^n/(2w_0)+O(z^{2n})$, giving ramification index $n$ at both points and branch points $\pm i$.

Write $n=2m$. At infinity set $u=1/z$, $v=w{u}^m$; then $v^2=1-u^n$. At $u=0$ there are two smooth points, $v=\pm 1$, and $u$ is a local coordinate. The extended $\pi$ has a simple pole at each, so neither is ramified. The degree-two Riemann-Hurwitz formula for the map to the Riemann sphere gives $2g-2=2(-2)+n$, henceThe list for $\tau$ above concerns the original affine Riemann surface; on the compactification its poles at the two added points have order $m$ and are additionally ramified if $m>1$.

Write $\mathbf{g}=-g{\mathbf{e}}_z$. The vertical component of the balance gives hydrostatic pressure,Its horizontal gradient is ${\nabla }_{h}p=\rho g{\nabla }_{h}h$. The horizontal balance is consequentlyThus $\mathbf{u}\cdot {\nabla }_{h}h=0$: the velocity is tangent to every level set of $h$, so those contours are streamlines of the geostrophic flow.

Across a vertical section joining the contours $h_0$ and $h_0+\Delta h$, an element of horizontal distance normal to the contours is $dn=dh/|{\nabla }_{h}h|$, while $|\mathbf{u}|=(g/f)|{\nabla }_{h}h|$. The volumetric flow rate is thereforeFor dimensional analysis, $[g]=L{T}^{-2}$, $[f]=T^{-1}$ and $[h_0]=[\Delta h]=L$, so the result has dimension $L^3{T}^{-1}$, exactly that of a volume flux.

Assume uniform ellipticity $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ with $\lambda >0$, bounded coefficients, and $c\le 0$. The weak maximum principle for elliptic operators isWhen $c=0$, zero can be omitted. The sign restriction is necessary, as shown by the failure of the weak maximum principle with a positive zeroth-order coefficient.

Let $M=\max (0,{\sup }_{\partial \Omega }u)$ and $w=u-M$, so $Lw=Lu-cM\ge 0$ and $w\le 0$ on the boundary. For sufficiently large $k$, $q=e^{k{x}_1}$ satisfiesIf $w$ is positive somewhere, $w+\epsilon q$ retains a positive interior maximum above all its boundary values for small $\epsilon >0$. At that point its gradient is zero and its Hessian matrix is negative semidefinite; ellipticity and $c\le 0$ give $L(w+\epsilon q)\le 0$, contradicting its strict positivity. This exponential perturbation proof of the weak maximum principle with drift also works with $M={\sup }_{\partial \Omega }u$ of either sign when $c=0$.

The cone condition implies $u(\lambda x)=\lambda u(x)$, so $Du(\lambda x)=Du(x)$. The gradient is homogeneous of degree zero, and $v=|Du{|}^2$ attains its global maximum on the compact unit sphere. For $n\ge 2$ the punctured space is connected. Part (b) gives $\mathcal{L}v\ge 0$ for a locally uniformly elliptic operator with no zeroth-order term. The strong maximum principle for elliptic operators makes $v$ constant. HencePositive definiteness gives $D^{2}u=0$. Thus $u=a\cdot x+b$, and degree-one homogeneity forces $b=0$: $\boxed{u(x)=a\cdot x}$. This proves that a minimal graphical cone is a plane. Strictly, the graph restricted to $x\ne 0$ is the plane with its origin omitted; its closure is the full plane through the origin.

For real test functions $f,g\in C_c^{\infty }(D)$, the Dirichlet inner product in the normalization compatible with the logarithmic Green kernel isIts norm measures gradient energy rather than the values of the functions. A change of variables shows that this energy is invariant under planar conformal maps: the squared derivative in the transformation of the gradient cancels the area Jacobian determinant.

First consider unit requirements, $A_{jr}\in \{0,1\}$. The Erlang loss formula for a resource with capacity $C$ and offered traffic $a$ isThe Erlang fixed point approximation assumes that resources block independently and that the traffic retained after screening at other resources can be treated as a Poisson process. If $B_j$ is the approximate blocking probability and $q_j=1-B_j$, the reduced-load approximation givesThe link whose load is being calculated is excluded from the screening product. Otherwise one would confuse its offered load with its carried load.

For existence, these equations define a continuous map from $[0,1{]}^J$ into itself. The Brouwer fixed-point theorem gives a fixed point. Assume $C_j\ge 1$; a zero-capacity resource forces rejection of every call needing it and can be removed together with those call types.

For uniqueness, let $K$ have probabilities proportional to $a^k/k!$, $0\le k\le C$, and write $m_C(a)=\mathbb{E}K$. Direct differentiation givesThese identities hold for $a>0$, and $m_C(0)=0$. Also $E(a,C)$ increases from zero to one and $m_C(a)$ increases from zero to $C$.

Put $p_j=-\log q_j$. Define $a_C(p)$ by $E(a_C(p),C)=1-e^{-p}$, and define $h_C(p)=m_C(a_C(p))$. This is a continuous, strictly increasing function on $[0,\infty )$, starting at zero and tending to $C$. Multiplying each Erlang fixed point approximation load equation by $q_j$ transforms it intoThese are exactly the zero-gradient conditions ofEach integral is a strictly convex function; the exponential terms are convex functions. Thus $\Phi$ is a strictly convex function. It is also a coercive function, since $h_{C_j}(p)\to C_j>0$, so it has a unique minimizer. At $p_j=0$, its partial derivative is negative if some positive-traffic route uses $j$, so that coordinate of the minimizer is positive. If no route uses $j$, the unique minimizing coordinate is zero. The minimizer therefore satisfies the equations in every coordinate. Conversely any fixed point has $B_j<1$, finite $p$, and these zero-gradient equations, so must equal that unique minimizer. Hence the Erlang fixed point approximation exists and is unique for fixed routing.

For integer requirements, the common generalized Erlang fixed point approximation usesMultiplication by $q_j$ gives the same equations for $\Phi$, so the existence and uniqueness argument also covers this generalized approximation. It remains an approximation, rather than the exact blocking law for a call requesting several units.

Let $R_s$ be the finite, nonempty set of available routes for source-sink pair $s$, and write $x_r$ for a route flow. The link-route incidence matrix has $A_{jr}=1$ when route $r$ uses link $j$, and zero otherwise. The source-sink route incidence matrix has $H_{sr}=1$ when $r\in R_s$. Thus $y=Ax$ gives link throughputs and $f=Hx$ gives aggregate source-sink flows. A route's delay isA Wardrop equilibrium uses only routes of minimum delay for their source-sink pair:An infinitesimal user cannot improve its delay by changing route.

For fixed $f_s\ge 0$, the feasible route flows form a nonempty compact convex set: for each $s$, $x_r\ge 0$ and ${\sum }_{r\in R_s}{x}_r=f_s$. Define the Beckmann potentialIt is continuous, so the extreme value theorem gives a minimizer. Since every $D_j$ is strictly increasing, $V$ is a strictly convex function of $y$. Its route-flow gradient is $\partial V(Ax)/\partial x_r=d_r(x)$.

If an occupied route $r\in R_s$ has greater delay than another route $q\in R_s$, moving a small amount of flow from $r$ to $q$ gives a negative directional derivative, contradicting minimality. Conversely, at a Wardrop equilibrium, for any feasible $\tilde{x}$,The first-order condition for convex optimization therefore proves optimality. ConsequentlyIf two minimizers had different $y$, their midpoint would give a strictly smaller Beckmann potential. Thus the equilibrium link throughputs are unique. The route flows need not be unique, since $V(Ax)$ need not be a strictly convex function of $x$.

For a concrete example, take two successive stages of parallel links: $a,b$ in the first stage and $c,d$ in the second, with one source-sink pair and four routes $ac,ad,bc,bd$. Put $D_j(y)=y$ on every link and total flow $f=2$. All allocationshave the same link throughputs $(1,1,1,1)$ and every route has delay two. These distinct allocations are all Wardrop equilibria.

For elastic demand, use the standard physical convention that $D_j\ge 0$ and $B_s:[0,\infty )\to [0,\infty )$ is finite, continuous, and strictly decreasing. Set $M_s=B_s(0)$ and ${\ell }_s={\lim }_{\lambda \to \infty }{B}_s(\lambda )$, and define the inverse demand function $P_s=B_s^{-1}$ on $({\ell }_s,M_s]$. It is continuous and strictly decreasing, with $P_s(M_s)=0$ and $P_s(u)\to \infty$ as $u\downarrow {\ell }_s$. Choose any reference value $f_s^0\in ({\ell }_s,M_s)$ and setEach summand of $-G$ is a strictly convex function. The reference values only add a constant to the objective. In the usual case ${\ell }_s=0$, one may use a lower limit zero whenever the improper integral is finite; that integrability is not guaranteed merely by continuity of $B_s$.

Minimize $\Psi$ over the compact feasible set $x\ge 0$, ${\ell }_s\le (Hx{)}_s\le M_s$, interpreting the objective at $f_s={\ell }_s$ by its one-sided limit, which can be $+\infty$. This extension is sequentially lower semicontinuous, and an interior-demand feasible point has finite value, so a minimizer exists. It cannot have $f_s={\ell }_s$: adding a small amount on any route for $s$ has bounded marginal link cost, whereas its marginal demand benefit $P_s(f_s)$ tends to infinity. Thus $f_s>{\ell }_s$.

For ${\ell }_s<f_s<M_s$, the route-flow partial derivative is $d_r-P_s(f_s)$. The first-order conditions for convex optimization giveIf $f_s=M_s$, reducing any occupied route cannot improve the objective only if that route has zero delay; nonnegative link delays then give ${\lambda }_s=0=P_s(M_s)$. Therefore the conditions in every case areConversely these conditions give the nonnegative directional derivative for every feasible competitor and hence minimize the convex function $\Psi$. Allowing $f$ explicitly yields the stated form with $Hx=f$ and $Ax=y$, with inverse-demand endpoint conventions understood. In particular, $G$ is defined on its demand range, rather than evaluating $B_s^{-1}$ outside that range.

Strict convexity in the pair $(y,f)$ proves that the equilibrium aggregate source-sink flows and link throughputs are unique. Individual route flows may still be nonunique. Nonnegative, finite demand, nonnegative delays, finite route sets, and a usable route for each source-sink pair are the implicit modeling assumptions needed for the existence assertions.

For one observation, let ${\ell }_i={\sum }_{c=0}^1{t}_{ic}\log p_{ic}$ and define ${\delta }_{ic}=t_{ic}-p_{ic}$. Differentiating the softmax function and its log-likelihood gives $\partial {\ell }_i/\partial o_{ic}={\delta }_{ic}$. The sigmoid-softmax network gradients areThese equations are backpropagation: use the chain rule through the output affine map, then ${\sigma }^{\prime }(u)=\sigma (u)(1-\sigma (u))$ through the hidden units, then the input affine map.

For a learning rate $\eta >0$, select one observation and update every parameter by $\theta \leftarrow \theta +\eta \nabla {\ell }_i(\theta )$, computing all entries of the gradient from the old parameters before applying the update. A uniformly selected observation gives an unbiased gradient of the average log-likelihood; multiplying by 3000 would instead estimate the gradient of the sum. Equivalently perform stochastic gradient descent on the negative log-likelihood. Batch size one makes one update per observation, and five epochs make five passes through the data, typically with a fresh random ordering on each pass.

Use the Euler equations for an inviscid fluid with no magnetic or gravitational force. Let $w$ denote specific enthalpy. For a homentropic flow, $dw=dp/\rho$, so the pressure acceleration is $-\nabla w$. The velocity identitytherefore givesTaking the curl, using that the curl of a gradient is zero and that differentiation commutes for smooth fields, proves barotropic vorticity transport:More generally the same proof works for any barotropic fluid, replacing $w$ by a pressure potential ${\int }^{\rho }{p}^{\prime }(s)ds/s$. Here “isentropic” must supply this barotropic closure, for example through one common specific entropy throughout the fluid. Merely imposing entropy advection equation $Ds/Dt=0$ allows spatial specific entropy gradients; in that case the vorticity equation contains the additional baroclinic vorticity generation term ${\rho }^{-2}\nabla \rho \times \nabla p$.

The Iterated Tikhonov regularization update is the proximal operator stepThe objective has coercivity and is weakly sequentially lower semicontinuous and strongly convex, so the direct method in the calculus of variations and uniqueness of a minimizer of a strictly convex function give existence and uniqueness. Its gradient equation iswhich is exactly $(I+\tau K^{\ast }K)u=u_{\delta }^{(k)}+\tau K^{\ast }{f}^{\delta }$. The second term penalizes the change from the previous iterate, rather than the norm of $u$ relative to zero.