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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.