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The Bernoulli principle relates pressure, kinetic energy per unit volume and Newtonian gravitational potential along a streamline of steady flow with inviscid flow. For constant mass density, the Bernoulli equation gives $p/\rho +u^2/2+\Phi$ constant along that streamline. Constancy across different streamlines requires further hypotheses, such as irrotational flow.

Precisely, suppose a fluid of constant positive mass density obeys the Euler equations for an inviscid fluid, its velocity field and pressure are steady, and the body force per unit mass is $-\nabla \Phi$. ThenEvery term has units of energy per unit mass. For a uniform gravitational field, Newtonian gravitational potential is $\Phi =gz$, up to an additive constant. The pressure term represents work by surrounding fluid, so acceleration can exchange pressure work with kinetic energy. Dotting the steady Euler equations with $\mathbf{u}$ gives the displayed directional derivative: this is why constancy follows a streamline. ^[ref]

Continue the horizontal water constriction example from incompressible flow. Assume negligible viscous losses, steady flow and approximately uniform flow at both measurement sections, so the streamline speeds there equal the mean speeds $0.20$ and $0.80\mathrm{m}{\mathrm{s}}^{-1}$. Equal height cancels the Newtonian gravitational potential terms. The ideal pressure drop isThis is the Venturi effect in its ideal limit: the gain in kinetic energy density is supplied by pressure work. Cambridge's treatment uses this relation to connect a measured pressure difference to volumetric flow rate. ^[ref]

The $300\mathrm{Pa}$ result is conditional, rather than a prediction of the actual viscous pipe. Mass conservation fixes $Q$ and the ratio of mean speeds even with viscosity, but nonuniform velocity profiles require an integrated energy balance, and wall friction dissipates mechanical energy. A pump would supply additional work. In particular, constant mean speed in a straight pipe does not imply constant pressure: Poiseuille flow needs a continuing pressure gradient to overcome viscosity. ^[ref]

A complex-valued function is complex differentiable at $z_0$ ifexists. The same limit must hold for every approach, including curved paths. Equivalently,The error divided by $|h|$ tends to zero. Thus the derivative is a single complex multiplier, not an arbitrary real linear map of the plane. If the real coordinate map is differentiable, complex differentiability is equivalent to its satisfying the Cauchy-Riemann equations at that point.

The simplest failure is complex conjugation, $f(z)=\overline{z}$. At every $z_0$ its increment quotient is $\overline{h}/h$. For real $h=t\ne 0$ it is $1$, while for $h=it$ it is $-1$. More generally, $h=t{e}^{i\theta }$ gives $e^{-2i\theta }$. Although complex conjugation is a smooth real reflection and has every real directional derivative, these incompatible quotients rule out a complex derivative everywhere.

Agreement on two axes is necessary, but does not by itself establish a limit over the whole complex plane. The Cauchy-Riemann equations need the real differentiability condition above, or a sufficient regularity condition such as continuous partial derivatives near the point. There is a second distinction: $f(z)=|z{|}^2$ has complex derivative zero at zero because its quotient there is $\overline{h}\to 0$, but its Cauchy-Riemann equations fail at every nonzero point. It is complex differentiable at zero without being a holomorphic function on any neighbourhood of zero.

A real convex function locally bounded on an open convex set has a continuous supporting linear functional at every interior point. Apply the dominated Hahn-Banach theorem to its sublinear directional derivative $p_x$ to obtain $l\le p_x$. Increasing secant slopes imply $p_x(z)\le f(x+z)-f(x)$ whenever the endpoint is in the domain. A local Lipschitz continuity bound gives both $l(y)\le L\Vert y\Vert$ and $-l(y)=l(-y)\le L\Vert y\Vert$, proving continuity.

For a smooth function $f:U\to \mathbb{R}$ on an open set $U\subseteq {\mathbb{R}}^n$, a critical point is an input $x_{\ast }$ where the gradient vanishes: $\nabla f(x_{\ast })=0$. Equivalently, every first partial derivative is zero. Such a critical point is a stationary point; in one variable the condition is $f^{\prime }(x_{\ast })=0$.

There is a broader convention in elementary one-variable calculus: a "critical number" is an interior input where the derivative is zero or does not exist, provided the function itself is defined there. Thus a corner can be a critical number without being a stationary point. Here the smooth meaning of critical point is used unless the broader convention is explicitly stated. ^[ref]

For the polynomial $f(x)=x^4-2x^2$ on the real line,The stationary points at $x=\pm 1$ are strict local minima and attain the global minimum; the stationary point at $x=0$ is a strict local maximum. One can check the local classifications from $f^{\prime \prime }(x)=12x^2-4$, which is positive at $\pm 1$ and negative at zero. The identity $f(x)=(x^2-1{)}^2-1\ge -1$ proves the global minimum directly. Since $f(x)\to +\infty$ as $|x|\to \infty$, there is no global maximum on $\mathbb{R}$.

A zero derivative alone does not guarantee an extremum. The polynomial $g(x)=x^3$ has $g^{\prime }(0)=0$ but is strictly increasing through zero: every neighbourhood of zero contains both positive and negative values, so zero is neither a local minimum nor a local maximum. Conversely, the absolute value function $h(x)=|x|$ has a strict local minimum and a unique global minimum at zero, but its left and right derivatives are $-1$ and $+1$. It is not differentiable there, so zero is a critical number under the broader convention and is not a stationary point.

The usefulness of critical points is that they reduce the search for smooth interior extrema to equations. Every local minimum or local maximum of a differentiable function at an interior point has zero gradient; this condition is necessary, and $x^3$ shows that it is not sufficient. The striking feature is that all the first directional derivatives must vanish together, even though the surrounding values can form a valley, a peak, or a saddle point of a scalar function. ^[ref]

Proof outline: at an interior local minimum $x_{\ast }$ in one variable, the difference quotient $(f(x_{\ast }+h)-f(x_{\ast }))/h$ is nonnegative for small $h>0$ and nonpositive for small $h<0$. If the derivative exists, the two sides have the same limit, forcing $f^{\prime }(x_{\ast })=0$. Reverse the inequalities for a local maximum. In several variables, apply this argument along each coordinate line to obtain zero partial derivatives and hence zero gradient. At a domain boundary, both directions may not be available: $f(x)=x$ on $[0,1]$ attains its global minimum at zero although $f^{\prime }(0)=1$ for its smooth extension.

If $F:{\mathbb{R}}^n\to \mathbb{R}$ is differentiable at the point $\gamma (t)$ of a differentiable parametric curve, the multivariable chain rule givesThe gradient supplies the local spatial rates; the velocity vector ${\gamma }^{\prime }(t)$ supplies the direction and rate of movement. Their dot product is the change in the sampled scalar field per unit parameter. If the parameter is time, this is a rate per unit time; a directional derivative per unit distance instead uses a unit vector.

Take $F(x,y)=x^2+2y^2$ and the unit circle path $\gamma (t)=(\cos t,\sin t)$. Along the path,At $t=\pi /4$, the gradient is $(\sqrt{2},2\sqrt{2})$, the velocity vector is $(-1/\sqrt{2},1/\sqrt{2})$, and their dot product is $-1+2=1$. The sampled value is $3/2$ and increases at rate one, even though one coordinate is decreasing. This is why a moving observation must combine every coordinate's contribution.

For a level curve, $F(\gamma (t))$ is constant, so the derivative along a curve is zero and its velocity vector is orthogonal to the gradient. Along other paths, the same formula combines a field's geometry with the observer's motion. ^[ref].

The directional derivative tests a function along a chosen straight line, including lines that follow neither coordinate axis. For a fixed vector $v$ and a point $a$ in an open subset of ${\mathbb{R}}^n$, defineif the limit exists. With a unit vector $v$, the parameter $t$ measures signed distance in the input Euclidean space, so the directional derivative is change per unit distance. For a general $v$, it is change per unit parameter; the norm of $v$ also sets the rate of travel.^[ref]

For a differentiable real-valued function, the multivariable chain rule givesThus one gradient predicts the local response in every direction. This formula requires differentiability, rather than merely separate partial derivatives.^[ref]

Return to $f(x,y)=x^2+xy+2y^2$ at $a=(1,1)$. Its gradient is $(3,5)$. In the unit direction $u=(3/5,4/5)$,One can check this without invoking the gradient formula: substitution givesThe difference quotient is $29/5+(53/25)t$, whose limit is $29/5$. This calculation separates the exact change along a finite path from its first-order rate at the starting point.

Changing the travel speed changes the parameter rate. Taking $v=(3,4)=5u$ gives $D_{v}f(a)=29$, because the input travels five distance units per parameter unit. For a differentiable function, $D_{cv}f(a)=c{D}_{v}f(a)$ for any real $c$: negative $c$ reverses the orientation as well as scaling the rate. Normalizing $v$ before comparing different slopes avoids confusing direction with speed.

The Cauchy-Schwarz inequality bounds the rate in any unit direction by $\Vert \nabla f(a)\Vert$. Here the largest rate is $\sqrt{34}\approx 5.831$, attained in the direction $(3,5)/\sqrt{34}$; the chosen $(3/5,4/5)$ gives the slightly smaller rate $5.8$. The perpendicular unit direction $(-5,3)/\sqrt{34}$ gives zero. That means no first-order height change, rather than no change at all: along this direction the example has $f(a+tu)=4+(28/34)t^2$.

This is useful for following a height or temperature scalar field, choosing a local search direction in mathematical optimization, and distinguishing a coordinate sensitivity from a response along an actual path. For a curved path, derivative along a curve uses its instantaneous tangent vector. Even existence of all directional derivatives does not guarantee one common linear approximation; directional derivatives do not imply differentiability gives a concrete counterexample.

Define $f(x,y)=x^3/(x^2+y^2)$ away from the origin and zero at the origin. Every directional derivative exists. Along a unit vector $n$, the derivative at the origin is $n_1^3$, which is not linear in $n$. Thus these directional derivatives cannot be the restrictions of one Fréchet derivative, and the function is not Fréchet differentiable at that point.

In Cartesian coordinates on Euclidean space, the gradient of a scalar-valued differentiable function is the vector of its first partial derivatives. It converts a scalar field, such as temperature or a height, into a vector field describing its local change.

For a unit vector $u$, the directional derivative is $D_{u}f=u\cdot \nabla f$. The Cauchy-Schwarz inequality gives $|D_{u}f|\le |\nabla f|$. At a point where $\nabla f\ne 0$, the greatest increase per unit distance occurs along $u=\nabla f/|\nabla f|$, while the opposite direction gives the greatest decrease. These are first-order statements at the current point; they do not predict how the function behaves far along a straight path.

For $f(x,y)=x^2+2y^2$ at $(1,1)$, $\nabla f=(2,4)$. Moving in the $x$ direction gives rate $2$ and moving in the $y$ direction gives rate $4$, but the steepest direction is $(1,2)/\sqrt{5}$, withThe direction $(-2,1)/\sqrt{5}$ has rate zero and is tangent to the level curve through that point. More generally, differentiating $f(\gamma (t))=\text{constant}$ along a regular level curve gives $\nabla f\cdot {\gamma }^{\prime }=0$, so the gradient is its normal vector.

This is useful when navigating a measured scalar field or applying gradient descent: one collection of partial derivatives predicts every infinitesimal directional change. The derivative assumes a differentiable function, not merely the existence of all directional derivatives. ^[ref].

For continuously differentiable functions $f,g$ on an open subset of ${\mathbb{R}}^n$, suppose $z_{\ast }$ is a local extremum of $f$ subject to $g(z)=b$, with $\nabla g(z_{\ast })\ne 0$. Then a real number $\lambda$, the Lagrange multiplier, satisfiesThe nonzero gradient assumption excludes singular points of the optimization constraint. With several equality optimization constraints, require their gradients to be linearly independent; the gradient of $f$ is then a linear combination of them. These are necessary conditions, not automatic certificates of a global maximum or global minimum.

Why can the gradient be nonzero at an optimum? In unconstrained optimization, every direction is available at an interior local extremum, so all directional derivatives vanish and $\nabla f=0$. On a regular equality feasible set, only directions in its tangent space are available to first order. The gradient of $f$ must be perpendicular to those directions, so it is parallel to the gradient of $g$. In two dimensions this explains the tangency of their level sets. See MIT, Lagrange multipliers and contour geometry.

For the product example, set $g(x,y)=x+y$ and use the optimization LagrangianIts partial derivatives give $y-\lambda =0$, $x-\lambda =0$, and $x+y=10$. Hence $x=y=\lambda =5$. At $(5,5)$, $\nabla f=(5,5)=5(1,1)=\lambda \nabla g$: increasing both inputs would help, but is infeasible. Along the allowed direction $(1,-1)$, the directional derivative is zero. The endpoints must still be checked because of the nonnegativity optimization constraints; completing the square establishes the unique global maximum.

The sign convention matters: writing $L=xy+\mu (x+y-10)$ instead gives $\mu =-5$. Neither equality Lagrange multiplier has a universal nonnegativity requirement. Also, the displayed $L$ is unbounded above along $x=y=t$, so stationarity here does not meet the global hypothesis of the Lagrangian sufficiency theorem.

The Lagrange multiplier also measures the value of extra resources. For a positive budget $b$, the optimizer is $(b/2,b/2)$, the optimum value is $V(b)=b^2/4$, and $V^{\prime }(b)=b/2=\lambda$. At $b=10$, an extra $0.1$ units predicts a gain of $0.5$; the exact gain is $V(10.1)-V(10)=0.5025$. This is the envelope theorem in a computed sensitivity analysis: the multiplier translates a binding budget constraint into a marginal benefit.

The normal derivative of a differentiable function $u$ along a hypersurface with unit normal $N$ is the directional derivativeHigher normal derivatives at a point $x$ may be defined by differentiating the one-variable function $s\mapsto u(x+sN(x))$ repeatedly at $s=0$.

Use suffix notation, summing repeated indices. The contraction of two Levi-Civita symbols givesAdding the directional derivative $(F\cdot \nabla )G_i=F_j{\partial }_j{G}_i$ cancels the second term. Interchanging $F,G$ gives another pair whose sum is $G_j{\partial }_i{F}_j$. The total right-hand side is thereforethe $i$th component of the gradient of the dot product. Since this holds for each component, the dot-product gradient identity is proved:It holds for any continuously differentiable vector fields; no irrotationality assumption is used.

A Nash bargaining problem consists of a compact convex set $S$ of feasible two-player payoff vectors and a disagreement point $d\in S$. If bargaining fails, the players receive $d$; agreement selects a feasible vector. The convex set allows jointly agreed correlated payoff lotteries over outcomes. Assume essentiality: some feasible vector strictly improves both coordinates of $d$. The admissible answers satisfy bargaining individual rationality, $u_i\ge d_i$.

The four characterizing axioms are Pareto efficiency (no feasible vector weakly improves both chosen payoffs and strictly improves one), bargaining symmetry (interchanging identical player roles leaves the answer unchanged), positive affine invariance in bargaining (independently replacing utilities by $a_i{u}_i+b_i$, $a_i>0$, transforms the answer by the same maps), and bargaining independence of irrelevant alternatives (restricting to an admissible smaller feasible set that still contains the chosen vector leaves the choice unchanged). The last axiom keeps the disagreement point fixed.

The arbitration procedure maximizes the Nash product over individually rational vectors:A maximizer exists by compactness and has positive gains by essentiality. It is unique, since the logarithm of the product is a strictly concave function of the two gains on their positive domain. The product is strictly increasing in either positive gain, giving Pareto efficiency. A positive affine utility change multiplies the product by $a_1{a}_2$, giving positive affine invariance in bargaining. Uniqueness gives bargaining symmetry and bargaining independence of irrelevant alternatives.

For completeness, these axioms also force the procedure. Normalize the product-maximizing vector to $(1,1)$ and $d$ to zero using positive affine invariance in bargaining. For any normalized feasible vector $z$, the directional derivative of the product along the feasible segment from $(1,1)$ to $z$ is $z_1+z_2-2$. The initial part of that segment has positive gains, so optimality implies $z_1+z_2\le 2$. Choose $M$ sufficiently large that the transformed feasible set is contained in the symmetric triangleOn this triangle, bargaining symmetry requires equal coordinates and Pareto efficiency then forces $(1,1)$. By bargaining independence of irrelevant alternatives, restricting from $T_M$ to the original normalized set still selects $(1,1)$. Undoing the normalization proves the Nash bargaining solution characterization. This is the supporting triangle for Nash bargaining argument.

For the numerical game, the maximin disagreement point is the vector of the players' separate security level payoffs. If player I uses its first action with probability $p$, its worst expected payoff isThe two lines meet at $p=1/4$; before that point the increasing second line is smaller and afterwards the decreasing first line is smaller. Hence $d_1=7/2$. If player II uses its first action with probability $q$, its worst payoff ismaximized at $q=1$, giving $d_2=4$. Thus $d=(7/2,4)$. These are separate worst-case guarantees; they need not equal the actual payoffs when both security strategies are used together.

The cooperative feasible set is the convex hull of the four outcome vectors. The outcomes $(2,4)$ and $(2,3)$ are dominated by $C=(4,5)$. Its Pareto frontier is the segment joining $C$ to $B=(8,2)$, with equation $y=8-3x/4$. Any maximizer of the positive Nash product is on this frontier. Bargaining individual rationality restricts its relevant portion to $4\le x\le 16/3$. There we maximizeThe stationary point lies strictly inside that interval and is the unique maximum. Therefore the Nash bargaining solution isThe gains are $11/12$ and $11/16$, with Nash product $121/192$. The arbitration can implement the vector by the agreed correlated payoff lottery assigning probability $5/48$ to outcome $B$ and $43/48$ to outcome $C$. Indeed $4+4(5/48)=53/12$ and $5-3(5/48)=75/16$. This is a correlated payoff lottery; independent mixed strategies are not required to implement the cooperative agreement.

For a scalar, the Lie derivative of a function is its directional derivative:The Lie derivative of a vector field is their Lie bracket of vector fields:For a torsion-free connection,The last term vanishes because the connection coefficients are symmetric in $b,c$. This proves ${\mathcal{L}}_{\xi }X={\nabla }_{\xi }X-{\nabla }_X\xi$.

For the metric tensor statements use the Levi-Civita connection, which also has metric compatibility. The tensor derivation property of the Lie derivative of a tensor field givesExpanding the first term using $\nabla g=0$ and using the vector identity cancels the derivatives of $X$ and $Y$. What remains isSince $X,Y$ are arbitrary, the Killing equation isMetric compatibility is needed here, even though symmetry alone sufficed for the vector identity. In general the Lie derivative of a metric with nonmetricity contains a term ${\xi }^c{\nabla }_c{g}_{ab}$. For example, with the metric tensor $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the Lie derivative of a tensor field is $2e^{2x}$, whereas $2{\nabla }_x{\xi }_x=4e^{2x}$. The usual metric connection interpretation of the question supplies the required compatibility.

For the Riemann curvature tensor calculation, keep the original PDF's convention. With $R_{abcd}=g_{ae}{R}^e{}_{bcd}$ and $R_{abc}{}^d=g^{de}{R}_{abce}$, it meansThus the commutator on a covector has the opposite sign. Write $K_{abc}={\xi }_{a;bc}={\nabla }_c{\nabla }_b{\xi }_a$. Taking a covariant derivative of the Killing equation makes $K$ antisymmetric in its first two indices, while the Ricci identity givesCombining this with the first-two-index antisymmetry gives the three relationsAdd the first and third and subtract the second. Pair symmetry and the first Bianchi identity imply $R_{adbc}+R_{bdca}-R_{cdab}=-2R_{abcd}$. ConsequentlyThis is the second covariant derivative of a Killing vector in the paper's convention; its sign is tied to the declared curvature-sign convention in Killing derivative identities, not to raising a particular index.

Set $L_{ab}={\xi }_{a;b}$. Along any smooth curve with tangent vector $V$, the chain rule and the just-proved identity give the Killing transport equationsChoose a parallel-propagated orthonormal frame along the curve. These become a homogeneous linear system of ordinary differential equations for the components of $(\xi ,L)$. Uniqueness of its initial value problem implies that zero initial data remain zero. Any point in the same connected component as $P$ can be joined to $P$ by a piecewise smooth curve. Therefore zero value and zero first derivative at $P$ force the Killing field and its derivative to vanish throughout that component, with no completeness assumption. On a manifold with one connected component this is the requested global result. Connectedness cannot be dropped: on two disjoint flat spacetimes, take the zero field on one and a nonzero translation on the other.

On a pseudo-Riemannian manifold of dimension $n$ with one connected component, the map $\xi \mapsto ({\xi }_a(P),L_{ab}(P))$ is therefore injective. There are $n$ independent components of the value and $n(n-1)/2$ of its antisymmetric derivative. The dimension bound for Killing vector fields isIt is attained in flat space by ${\xi }_a=A_a+B_{ab}{x}^b$ with constant antisymmetric $B$, giving translations and infinitesimal rotations or Lorentz transformations. The four-dimensional maximum is 10. For a disconnected manifold the bound applies componentwise; there is no uniform global bound depending only on $n$ if the number of components is unrestricted.

Use finite sets of links $J$, source-sink pairs $S$, and admissible routes $R$, with at least one route for each pair. Route $r$ carries aggregate flow $x_r$. The link-route incidence matrix has $A_{jr}=1$ if route $r$ uses link $j$ and zero otherwise; repeated traversals, if allowed, are counted with their multiplicity. The source-sink route incidence matrix has $H_{sr}=1$ if $r$ serves pair $s$. Thus $y=Ax$ gives link throughputs and $f=Hx$ gives source-sink demands. Write $R_s$ for the routes serving $s$ andA Wardrop equilibrium routes positive flow only on minimum-delay routes:This is a nonatomic equilibrium: an individual infinitesimal user's route change does not alter the aggregate link delays.

For fixed nonnegative demand $f$, the feasible route-flow set $\{x\ge 0:Hx=f\}$ is a nonempty product of finite-dimensional simplexes and hence a compact set. The Beckmann potentialis continuous, so it attains a minimum over this feasible set. Each primitive is a strictly convex function, since its derivative $D_j$ is continuous and strictly increasing. Its route-flow derivative is exactly the corresponding route delay:If a minimizer had positive flow on $r$ and a lower-delay alternative $q\in R_s$, transferring a small amount from $r$ to $q$ would give directional derivative $L_q-L_r<0$, contradicting minimality. Hence every minimizer is a Wardrop equilibrium.

Conversely, let $x$ be a Wardrop equilibrium and $z$ any feasible competitor. Convexity givesFor each $s$, subtract ${\lambda }_s$ inside this sum. The demand constraint makes ${\sum }_{r\in R_s}{\lambda }_s(z_r-x_r)=0$, and the remaining terms are nonnegative: used routes have $L_r-{\lambda }_s=0$, while unused routes have $x_r=0$, $z_r\ge 0$ and $L_r-{\lambda }_s\ge 0$. Thus $V(Az)\ge V(Ax)$. We have proved both existence and the optimization characterization.

Strict convexity in the link vector makes $y$ unique: two different minimizing link vectors would have a feasible midpoint of strictly smaller potential. Thus link delays and the minimum route delays at the Wardrop equilibrium are also unique. The route vector need not be unique, because different route decompositions may have the same $Ax$ and $Hx$. For example, two successive stages with two parallel links at each stage have four routes. With unit demand and link delay $D(u)=u$, the vectorsall give throughput $1/2$ on each of the four links and delay one on every route. This exhibits route-flow nonuniqueness at a Wardrop equilibrium. The equilibrium link throughputs are unique; the individual route flows need not be.

For elastic demand, make the usual physical assumptions explicit: delays are nonnegative on $[0,\infty )$, every route is nonempty, and $B_s:[0,\infty )\to [0,\infty )$ is finite, continuous and strictly decreasing. PutThe limit exists by monotonicity; it need not be zero. The inverse demand function $P_s$ is continuous, strictly decreasing, satisfies $P_s(M_s)=0$, and tends to infinity as $v\downarrow {\ell }_s$.

Choose any reference $b_s\in ({\ell }_s,M_s)$ and use elastic-demand utility with an interior reference:Each utility is strictly concave, with derivative $P_s$. A reference inside the range avoids a potentially divergent integral from zero. For instance, $B(\lambda )=(1+\lambda {)}^{-1/2}$ has inverse $v^{-2}-1$, whose integral from zero is infinite, although the displayed reference-point primitive is finite at every interior demand.

Give $G_s$ its limiting value at ${\ell }_s$, which may be $-\infty$, and extend it by $-\infty$ outside $[{\ell }_s,M_s]$. This encodes the physical demand range in the objective. After substituting $y=Ax$, $f=Hx$, minimizeIts effective feasible set is contained in $\{x\ge 0:{\ell }_s\le (Hx{)}_s\le M_s\}$, a nonempty compact set. The objective is lower semicontinuous, including a possible value $+\infty$ at a lower endpoint, and is finite when all demands equal their interior reference values. It therefore has a finite-valued minimizer.

The boundary exclusion for elastic Wardrop demand shows that every minimizing demand is interior. If $f_s={\ell }_s$ and the utility there is finite, adding a small flow $h$ on any route serving $s$ costs at most a bounded constant times $h$, while the utility gain is at least $h{P}_s({\ell }_s+h)$, whose ratio to $h$ tends to infinity. This decreases the objective. If that endpoint utility is $-\infty$, a finite minimizer cannot occur there in the first place. At $f_s=M_s>0$, choose a used route. Its delay is strictly positive, since its nonempty links have positive throughput and strictly increasing nonnegative delays. Removing a small amount of its flow saves a positive first-order cost, whereas the utility loss is $o(h)$ because $P_s(M_s)=0$. This also decreases the objective. Hence ${\ell }_s<f_s<M_s$ for every pair.

The interior-demand route derivatives are nowAt a minimizer they are zero on positive route flows and nonnegative on zero route flows, by increasing or decreasing a single route flow. Since $f_s>0$, some route is used, soThese are precisely the elastic-demand Wardrop equilibrium conditions. Conversely, those conditions give the same nonnegative directional derivative for every competitor; convexity then proves global minimality. The equivalent constrained form is the requested minimization of $V(y)-G(f)$ with $Hx=f$, $Ax=y$, $x\ge 0$. Strict convexity determines both $y$ and $f$ uniquely, while their route decomposition can still be nonunique.

Only active routes, those with $n_r>0$, enter the proportional fairness inequality and the weighted logarithmic utility. Their rates must be positive; inactive routes have no per-flow rate to determine and may be assigned zero. This convention removes otherwise meaningless terms $0\log 0$ and $0/0$.

For positive active rates, concavity of the logarithm gives, for any positive feasible competitor $y$,Thus the proportional fairness inequality proves that $x$ maximizes the log-utility objective. A competitor with an active zero rate has utility $-\infty$ and cannot improve it. Conversely, if $x$ maximizes utility, the right directional derivative along $x+\epsilon (y-x)$ at $\epsilon =0$ is nonpositive, giving exactly that inequality. Positive resource capacities and finitely many active routes allow a feasible allocation with every active rate positive, so an optimizer cannot have an active zero rate. Strict concavity gives uniqueness of the active rates. Hence proportional fairness is equivalent to the stated log-utility maximization.

Index the four adjacent-pair routes in cyclic order and put $z_i=n_i{x}_i$, the aggregate service on route $i$. The resource constraints areThey involve every odd-even route pair, so they are equivalent toFor given maxima, increasing each active odd service to $p$ and each active even service to $q$ remains feasible and improves utility. Apart from the constant $-{\sum }_i{n}_i\log n_i$, the objective therefore reduces toWhen both groups are present the constraint is tight, and differentiation gives $P/p=Q/(1-p)$. If one group is empty, the other receives unit aggregate service on each of its active routes. Thus the opposite-route reduction for a proportionally fair four-cycle yields, with $T=P+Q>0$,Inactive rates can be set to zero, including every rate in the empty state. In particular $n_{\{1,2\}}{x}_{\{1,2\}}=(n_{\{1,2\}}+n_{\{3,4\}})/{\sum }_r{n}_r$ whenever that route is active. This proves the required formula even when other routes are empty.

For the flow-level network model, an active document's residual size retains an exponential distribution with parameter ${\mu }_i$. Receiving service at rate $x_i(n)$ gives completion intensity ${\mu }_i{x}_i(n)$, so $n_i$ independent active documents give total completion intensity ${\mu }_i{n}_i{x}_i(n)$. Consequently the population continuous-time Markov chain hasAll death intensities are zero at $T=0$. Their total is bounded by ${\sum }_i{\mu }_i$, and the total arrival rate is constant, proving nonexplosion.

Put ${\rho }_i={\nu }_i/{\mu }_i$. The required stationary weight uses a binomial coefficient, as printed in the PDF; the converted TeX incorrectly rendered it as an ordinary fraction. DefineFor an odd route, the neighboring binomial coefficient ratio iswhile its death intensity at the neighboring state is ${\mu }_i(P+1)/(T+1)$. Their product is ${\nu }_i$, so $w(n)q(n,n+e_i)=w(n+e_i)q(n+e_i,n)$. For an even route, replace $P$ by $Q$ in the same calculation. This proves detailed balance and the stationary law of a proportionally fair four-cycle:A finite $B$ is essential for this expression to be a probability distribution. We determine its exact convergence condition rather than asserting stationarity for arbitrary arrival rates. LetGrouping the states by odd and even population givesWith $A=\max (a,c)$ and $E=\max (b,d)$, the bounds $A^p\le h_p(a,c)\le (p+1)A^p$ and $E^q\le h_q(b,d)\le (q+1)E^q$ show that convergence holds exactly when $A+E<1$. For necessity, the lower bound sums to ${\sum }_{k\ge 0}(A+E{)}^k$, which diverges at or above one. For sufficiency, choose slightly larger $A^{\prime },E^{\prime }$ with $A^{\prime }+E^{\prime }<1$; the polynomial factors are absorbed into constants times $(A^{\prime }{)}^p(E^{\prime }{)}^q$, giving a convergent geometric generating series. Thus the stability region for the reversible four-cycle flow network isEquivalently, the offered load on each resource, ${\rho }_1+{\rho }_4$, ${\rho }_1+{\rho }_2$, ${\rho }_2+{\rho }_3$, or ${\rho }_3+{\rho }_4$, is strictly below its unit capacity. For positive arrival rates this finite normalizing constant gives the unique stationary distribution of the irreducible process. At or beyond the boundary, the displayed invariant weights are not summable and do not define a stationary probability law.

If an explicit normalizing constant is desired, $h_p(a,c)=(a^{p+1}-c^{p+1})/(a-c)$ for $a\ne c$, and ${\sum }_{p,q}\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)u^p{v}^q=(1-u-v{)}^{-1}$. Hence for $a\ne c$, $b\ne d$ in the stable region,Equal-parameter cases are obtained by continuity or directly from the positive series. In particular the empty-traffic limit is $B=1$.