 Solution

Let $R_s$ be the finite nonempty set of routes for source-sink pair $s$, with fixed nonnegative demand $f_s$. A Wardrop equilibrium is a feasible route-flow vector such that every route carrying positive flow has minimum total delay among the routes for its pair. Ifthis means that, for each $s$, some ${\lambda }_s$ satisfies $c_r(y)\ge {\lambda }_s$ for $r\in R_s$, with equality whenever $x_r>0$.

The link-route incidence matrix has $A_{jr}=1$ when route $r$ uses link $j$ and zero otherwise, or the traversal multiplicity if repeated traversals are permitted. The source-sink route incidence matrix has $H_{sr}=1$ when route $r$ serves pair $s$ and zero otherwise. Thus $Ax=y$ gives link throughputs, and $Hx=f$ gives the fixed demands.

The feasible route-flow set is a product of finite simplexes, hence nonempty and both a compact set and a convex set. The Beckmann potentialis continuous and convex, so it attains a minimum. Its route derivative is ${\partial }_r\Phi =c_r(Ax)$. At a minimum, moving a small amount of flow from any used route to any other route for the same pair cannot decrease the potential. Hence every used route has minimum route delay, exactly the Wardrop equilibrium condition. Conversely, at a Wardrop equilibrium $x$, for any feasible $\tilde{x}$,because the used flow has cost ${\lambda }_s$ and competing routes cost at least ${\lambda }_s$. The first-order inequality for a convex function makes $x$ a global minimum. This proves both existence and the specified optimization characterization.

Since each $D_j$ is strictly increasing, its primitive is a strictly convex function. Therefore the objective is strictly convex as a function of link loads $y$, and the equilibrium throughputs $y$ are unique. The mapping $x\mapsto Ax$ need not be injective on $Hx=f$, so route flows need not be unique.

For a concrete route-flow nonuniqueness at a Wardrop equilibrium example, use one unit-demand pair with two successive stages, each containing two parallel links. There are four routes, indexed by their two link choices. Give every link delay $D(u)=u$. For every $0\le t\le 1/2$,produces load $1/2$ on all four links. Every route has delay one, so each vector is a Wardrop equilibrium with the same unique link loads.

To implement minimum average delay, minimize the total delayThe average differs only by the fixed total demand ${\sum }_s{f}_s$, provided it is positive. Choose any global minimizer $x^{\ast }$, which exists by compactness, and write $y^{\ast }=A{x}^{\ast }$. Its first-order route-exchange conditions are the Wardrop equilibrium conditions for marginal link costsA genuinely traffic-dependent choice guaranteed under exactly the printed hypotheses is the calibrated toll functionHere $v_{+}=\max (v,0)$. The calibrated base charges are nonnegative because differentiable increasing delays have nonnegative derivatives, and the added penalties are nonnegative and nondecreasing in $u$. The perceived link costs $D_j(u)+T_j(u)$ remain strictly increasing, and at $y^{\ast }$ they equal $m_j^{\ast }$. Hence $x^{\ast }$ is a tolled Wardrop equilibrium. The same strictly convex Beckmann potential argument makes all its equilibrium link loads equal to $y^{\ast }$; every tolled equilibrium therefore has the globally minimum average delay. This is optimal-flow calibrated congestion tolling.

The familiar marginal external cost toll is $T_j(u)=u{D}_j^{\prime }(u)$, whose perceived-cost primitive is exactly $u{D}_j(u)$. It implements the optimum directly when the total-delay objective is convex, for example when each $u{D}_j(u)$ is convex. That extra hypothesis does not follow from the printed strict increase of $D_j$: for two identical parallel links with $D(u)=1-e^{-u}$ and total demand six, equal loads of three form a tolled Wardrop equilibrium, but the second derivative of $uD(u)$ at three is $-e^{-3}<0$, so an unequal nearby split lowers total delay. The calibrated traffic-dependent toll above establishes the requested existence without silently assuming this additional convexity.