Dynamic Equilibria in Time-Varying Networks^††thanks: Funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’s Excellence Strategy – The Berlin Mathematics Research Center MATH+ (EXC-2046/1, project ID: 390685689).

Let $\phi\in\mathbb{R}_{\geq 0}$ be a particle such that for all arcs $e=uv\in E$ the derivatives of $x_{e}$, $\ell_{u}$, $T_{e}\circ\ell_{u}$ and $\tau_{e}$ exist and $x_{e}^{\prime}(\phi)=f_{e}^{+}(\ell_{u}(\phi))\cdot\ell^{\prime}_{u}(\phi)=f_{e}^{-}(\ell_{v}(\phi))\cdot\ell^{\prime}_{v}(\phi)$ as well as $1+\tau^{\prime}_{e}(\ell_{u}(\phi))=\gamma_{e}(\ell_{u}(\phi))$. This is given for almost all $\phi$.

By (4) we have $\int_{0}^{\ell_{s}(\phi)}r(\xi)\mathop{}\!\mathrm{d}\xi=\phi$ and taking the derivative by applying the chain rule, yields $r(\ell_{s}(\phi))\cdot\ell^{\prime}_{s}(\phi)=1$, which shows (TF1).

Taking the derivative of (4) at time $\ell_{u}(\phi)$ by using the differentiation rule for a minimum (Lemma 7 in the appendix) yields $\ell^{\prime}_{v}(\phi)=\min_{e=uv\in E^{\prime}}T^{\prime}_{e}(\ell_{u}(\phi))\cdot\ell^{\prime}_{u}(\phi)$. By using Lemma 2 (vii) we obtain

which shows (TF2).

Finally, in the case of $f_{e}^{-}(\ell_{v}(\phi))\cdot\ell^{\prime}_{v}(\phi)=x^{\prime}_{e}(\phi)>0$ we have by (3) that

This shows (TF3) and finishes the proof. ∎

By Lemma 4 we have that $\ell_{v}(\phi)-\ell_{u}(\phi)>\tau_{e}(\phi)$ for all $e\in E^{*}_{\phi}$ and $\ell_{v}(\phi)-\ell_{u}(\phi)<\tau_{e}(\phi)$ for all $e\in E\setminus E^{\prime}_{\phi}$. Since $\tau_{e}$ is continuous there is an $\alpha_{1}>0$ such that (5) and (6) are satisfied for all $\xi\in[0,\alpha_{1})$. Since $\nu_{e}$, $r$ and $\lambda_{e}$ are right-constant so is $\gamma_{e}$, and hence, there is an $\alpha_{2}>0$ such that (7), (8) and (9) are fulfilled for all $\xi\in[0,\alpha_{2})$. Clearly, $\alpha\coloneqq\min\Set{\alpha_{1},\alpha_{2}}>0$ is feasible. ∎

Lemma 3 yields $F_{e}^{+}(\ell_{u}(\varphi))=F_{e}^{-}(\ell_{v}(\varphi))$ for all $\varphi\in[0,\phi)$, so for $\xi\in[0,\alpha)$ it holds that

It follows again by Lemma 3 together with Lemma 6 that the $\alpha$-extension is a restricted Nash flow over time on $[0,\phi+\alpha)$. ∎

The process starts with the empty flow over time, i.e., a restricted Nash flow over time for $[0,0)$. We apply Theorem 5.1 with a maximal feasible $\alpha$. If one of the $\alpha$ is unbounded we are done. Otherwise, we obtain a sequence $(f_{i})_{i\in\mathbb{N}}$, where $f_{i}$ is a restricted Nash flow over time for $[0,\phi_{i})$, with a strictly increasing sequence $(\phi_{i})_{i\in\mathbb{N}}$. In the case that this sequence has a finite limit, say $\phi_{\infty}<\infty$, we define a restricted Nash flow over time $f^{\infty}$ for $[0,\phi_{\infty})$ by using the point-wise limit of the $x$- and $\ell$-labels, which exists due to monotonicity and boundedness of these functions. Note that there are only finitely many different thin flows, and therefore, the derivatives $x^{\prime}$ and $\ell^{\prime}$ are bounded. Then the process can be restarted from this limit point. This so called transfinite induction argument works as follows: Let $\mathcal{P}_{G}$ be the set of all particles $\phi\in\mathbb{R}_{\geq 0}$ for which there exists a restricted Nash flow over time on $[0,\phi)$ constructed as described above. The set $\mathcal{P}_{G}$ cannot have a maximal element because the corresponding Nash flow over time could be extended by using Theorem 5.1. But $\mathcal{P}_{G}$ cannot have an upper bound either since the limit of any convergent sequence would be contained in this set. Therefore, there exists an unbounded increasing sequence $(\phi_{i})_{i=1}^{\infty}\in\mathcal{P}_{G}$. As a restricted Nash flow over time on $[0,\phi_{i+1}]$ contains a restricted Nash flow over time on $[0,\phi_{i}]$ we can assume that there exists a sequence of nested restricted Nash flow over time. Hence, we can construct a Nash flow over time $f$ on $[0,\infty)$ by taking the point-wise limit of the $x$- and $\ell$-labels, completing the proof. ∎

a

1. (i)

   Consider two points in time $\theta_{1}<\theta_{2}$, then $\tau\coloneqq\theta_{1}-\theta_{2}+\tau_{e}(\theta_{1})$ is strictly smaller than $\tau_{e}(\theta_{2})$ since

   $$\int_{\theta_{2}}^{\theta_{2}+\tau}\lambda_{e}(\xi)\mathop{}\!\mathrm{d}\xi=\int_{\theta_{2}}^{\theta_{1}+\tau_{e}(\theta_{1})}\lambda_{e}(\xi)\mathop{}\!\mathrm{d}\xi<\int_{\theta_{1}}^{\theta_{1}+\tau_{e}(\theta_{1})}\lambda_{e}(\xi)\mathop{}\!\mathrm{d}\xi=1,$$

   where the strict inequality holds, since $\lambda_{e}$ is always strictly positive. The last equality follows by the definition of $\tau_{e}(\theta_{1})$. Hence, with the definition of $\tau_{e}(\theta_{2})$ we have $\theta_{1}+\tau_{e}(\theta_{1})<\theta_{2}+\tau_{e}(\theta_{2})$.

2. (ii)

   Since $\theta\mapsto\theta+\tau_{e}(\theta)$ is monotone, Lebesgue’s theorem for the differentiability of monotone functions implies that it is almost everywhere differentiable. The same is then true for $\tau_{e}$. The continuity follows directly from the definition since $\lambda_{e}$ is always strictly positive.

3. (iii)

   By the definition of $\tau_{e}(\theta)$ we have

   $$\int_{0}^{\theta+\tau_{e}(\theta)}\lambda_{e}(\xi)\mathop{}\!\mathrm{d}\xi-\int_{0}^{\theta}\lambda_{e}(\xi)\mathop{}\!\mathrm{d}\xi=1.$$

   Taking the derivatives of both sides and using Lebesgue’s differentiation theorem together with the chain rule, we obtain

   $$\lambda_{e}(\theta+\tau_{e}(\theta))\cdot(1+\tau^{\prime}_{e}(\theta))-\lambda_{e}(\theta)=0.$$

   Since $\lambda_{e}$ is always strictly positive, we get

   $$1+\tau^{\prime}_{e}(\theta)=\frac{\lambda_{e}(\theta)}{\lambda_{e}(\theta+\tau_{e}(\theta))}.$$

∎

a

1. (i)

   This follows directly from the definition of the waiting time $q_{e}$.

2. (ii)

   By equation (3) we have that $f_{e}^{-}(\xi)\leq\nu_{e}(\xi)$ almost everywhere. Hence, we have by definition that $q_{e}(\theta)$ is the minimal value such that

   $$\int_{\theta+\tau_{e}(\theta)}^{\theta+\tau_{e}(\theta)+q_{e}(\theta)}\nu_{e}(\xi)\mathop{}\!\mathrm{d}\xi=z_{e}(\theta+\tau_{e}(\theta)).$$

   Thus, we obtain for $\xi\in[0,q_{e}(\theta))$ that

   	$\displaystyle F_{e}^{-}(\theta+\tau_{e}(\theta)+\xi)-F_{e}^{-}(\theta+\tau_{e}(\theta))$	$\displaystyle=\int_{\theta+\tau_{e}(\theta)}^{\theta+\tau_{e}(\theta)+\xi}f_{e}^{-}(\xi)\mathop{}\!\mathrm{d}\xi$
   		$\displaystyle\leq\int_{\theta+\tau_{e}(\theta)}^{\theta+\tau_{e}(\theta)+\xi}\nu_{e}(\xi)\mathop{}\!\mathrm{d}\xi$
   		$\displaystyle<z_{e}(\theta+\tau_{e}(\theta))$
   		$\displaystyle=F_{e}^{+}(\theta)-F_{e}^{-}(\theta+\tau_{e}(\theta)).$

   Or in short: $F_{e}^{+}(\theta)-F^{-}_{e}(\theta+\tau_{e}(\theta)+\xi)>0$ for $\xi\in[0,q_{e}(\theta))$. Since $F_{e}^{+}$ is non-decreasing we obtain for all $\xi\in[0,q_{e}(\theta))$ that

   $$z_{e}(\theta+\tau_{e}(\theta)+\xi)\!=\!F_{e}^{+}(\theta+\xi)-F^{-}_{e}(\theta+\tau_{e}(\theta)+\xi)\!\geq\!F_{e}^{+}(\theta)-F^{-}_{e}(\theta+\tau_{e}(\theta)+\xi)\!>\!0.$$

3. (iii)

   By (3) and (ii) we obtain for almost all ${\xi\in[\theta+\tau_{e}(\theta),\theta+\tau_{e}(\theta)+q_{e}(\theta))}$ that $f_{e}^{-}(\xi)=\nu_{e}(\xi)$. By the definition of $q_{e}$ we have

   	$\displaystyle F_{e}^{-}(\theta+\tau_{e}(\theta)+q_{e}(\theta))-F_{e}^{-}(\theta+\tau_{e}(\theta))$	$\displaystyle=\int_{\theta+\tau_{e}(\theta)}^{\theta+\tau_{e}(\theta)+q_{e}(\theta)}f_{e}^{-}(\xi)\mathop{}\!\mathrm{d}\xi$
   		$\displaystyle=\int_{\theta+\tau_{e}(\theta)}^{\theta+\tau_{e}(\theta)+q_{e}(\theta)}\nu_{e}(\xi)\mathop{}\!\mathrm{d}\xi$
   		$\displaystyle=z_{e}(\theta+\tau_{e}(\theta))$
   		$\displaystyle=F_{e}^{+}(\theta)-F_{e}^{-}(\theta+\tau_{e}(\theta)).$

   Hence, $F_{e}^{-}(T_{e}(\theta))=F_{e}^{+}(\theta)$.

4. (iv)

   Since $F_{e}^{+}(\theta_{1})=F_{e}^{+}(\theta_{2})$ we obtain with the monotonicity of $F_{e}^{-}$ together with Lemma 1 (i) that

   	$\displaystyle z_{e}(\xi+\tau_{e}(\xi))$	$\displaystyle=F_{e}^{+}(\xi)-F_{e}^{-}(\xi+\tau_{e}(\xi))$
   		$\displaystyle\geq F_{e}^{+}(\theta_{2})-F_{e}^{-}(\theta_{2}+\tau_{e}(\theta_{2}))=z_{e}(\theta_{2}+\tau_{e}(\theta_{2}))>0,$

   hence, (3) provides $f_{e}^{-}(\xi)=\nu_{e}(\xi)$ for almost all ${\xi\in[\theta_{1}+\tau_{e}(\theta_{1}),\theta_{2}+\tau_{e}(\theta_{2})]}$.

   Thus, the definition of $q_{e}$ implies that $q_{e}(\theta_{1})$ equals

   		$\displaystyle\min\Set{q\geq 0}{\begin{array}[]{l}\displaystyle\int_{\theta_{1}+\tau_{e}(\theta_{1})}^{\theta_{2}+\tau_{e}(\theta_{2})}f_{e}^{-}(\xi)\mathop{}\!\mathrm{d}\xi+\int_{\theta_{2}+\tau_{e}(\theta_{2})}^{\theta_{1}+\tau_{e}(\theta_{1})+q}\nu_{e}(\xi)\mathop{}\!\mathrm{d}\xi\\[15.00002pt] \hfill\qquad\qquad=F_{e}^{+}(\theta_{1})-F_{e}^{-}(\theta_{1}+\tau_{e}(\theta_{1}))\end{array}}$
   	$\displaystyle=$	$\displaystyle\min\Set{p\geq 0}{\int_{\theta_{2}+\tau_{e}(\theta_{2})}^{\theta_{2}+\tau_{e}(\theta_{2})+p}\!\!\!\!\nu_{e}(\xi)\mathop{}\!\mathrm{d}\xi=F_{e}^{+}(\theta_{2})\!-\!F_{e}^{-}(\theta_{2}\!+\!\tau_{e}(\theta_{2}))}$
   		$\displaystyle\hskip 184.9429pt+\theta_{2}+\tau_{e}(\theta_{2})\!-\!\theta_{1}\!-\!\tau_{e}(\theta_{1})$
   	$\displaystyle=$	$\displaystyle\;q_{e}(\theta_{2})+\theta_{2}+\tau_{e}(\theta_{2})-\theta_{1}-\tau_{e}(\theta_{1}).$

   Here, we substitute $q$ by ${p+\theta_{2}+\tau_{e}(\theta_{2})-\theta_{1}-\tau_{e}(\theta_{1})}$ in order to obtain the first equation. Note that the condition $p\geq 0$ is always satisfied since the right hand side $F_{e}^{+}(\theta_{2})-F_{e}^{-}(\theta_{2}+\tau_{e}(\theta_{2}))$ equals $z_{e}(\theta_{2}+\tau_{e}(\theta_{2}))$ and is therefore strictly positive by assumption. Hence, we obtain

   $$T_{e}(\theta_{1})=\theta_{1}+\tau_{e}(\theta_{1})+q_{e}(\theta_{1})=\theta_{2}+\tau_{e}(\theta_{2})+q_{e}(\theta_{2})=T_{e}(\theta_{2}).$$

5. (v)

   Considering two points in time $\theta_{1}<\theta_{2}$, we show that $T_{e}(\theta_{1})\leq T_{e}(\theta_{2})$. Since $F_{e}^{+}$ is non-decreasing, (iii) implies that

   $$F_{e}^{-}(T_{e}(\theta_{2}))=F_{e}^{+}(\theta_{2})\geq F_{e}^{+}(\theta_{1})=F_{e}^{-}(T_{e}(\theta_{1})).$$

   (10)

   If this holds with strict inequality, we obtain by monotonicity of $F_{e}^{-}$ that $T_{e}(\theta_{1})<T_{e}(\theta_{2})$. If equation (10) holds with equality we have two cases. If ${z_{e}(\theta_{2}+\tau_{e}(\theta_{2}))>0}$, (iv) states that $T_{e}(\theta_{1})=T_{e}(\theta_{2})$. If $z_{e}(\theta_{2}+\tau_{e}(\theta_{2}))=0$, (ii) applied to $\theta_{1}$ implies that $\xi\coloneqq\theta_{2}+\tau_{e}(\theta_{2})-\theta_{1}-\tau_{e}(\theta_{1})\not\in[0,q_{e}(\theta_{1}))$. Since $\xi\geq 0$ by Lemma 1 (i) we have $\xi\geq q_{e}(\theta_{1})$, and thus,

   $$T_{e}(\theta_{2})\stackrel{{\scriptstyle\text{\ref{it:q_equiv_z:timedep}}}}{{=}}\theta_{2}+\tau_{e}(\theta_{2})\geq\theta_{1}+\tau_{e}(\theta_{1})+q_{e}(\theta_{1})=T_{e}(\theta_{1}).$$

6. (vi)

   The continuity of $q_{e}$ follows since $\nu_{e}$ is always strictly positive and $z_{e}$ is continuous, as it is the difference of two continuous functions. Finally, $T_{e}$ is continuous since it is the sum of three continuous functions.

   By (v) the function $T_{e}$ is non-decreasing for all $e\in E$, and hence, Lebesgue’s theorem for the differentiability of monotone functions states that $T_{e}$ is almost everywhere differentiable. Since ${\theta\mapsto\theta+\tau_{e}(\theta)}$ is monotone this also holds for $\tau_{e}$ since it is the difference of two almost everywhere differentiable functions. As a sum of almost everywhere differentiable functions, $q_{e}(\theta)=T_{e}(\theta)-\tau_{e}(\theta)-\theta$ has this property as well.

7. (vii)

   The definition of $q_{e}(\theta)$ states that

   $$\int_{0}^{T_{e}(\theta)}\nu_{e}(\xi)\mathop{}\!\mathrm{d}\xi-\int_{0}^{\theta+\tau_{e}(\theta)}\nu_{e}(\xi)\mathop{}\!\mathrm{d}\xi=z_{e}(\theta+\tau_{e}(\theta))=F_{e}^{+}(\theta)-F_{e}^{-}(\theta+\tau_{e}(\theta)).$$

   Taking the derivative on both sides we obtain by using the chain rule that

   $$\nu_{e}(T_{e}(\theta))\cdot T^{\prime}_{e}(\theta)-\nu_{e}(\theta+\tau_{e}(\theta))\cdot(1+\tau^{\prime}_{e}(\theta))=f_{e}^{+}(\theta)-f_{e}^{-}(\theta+\tau_{e}(\theta))\cdot(1+\tau^{\prime}_{e}(\theta)).$$

   If $q_{e}(\theta)>0$ we have by equation (3) that $f_{e}^{-}(\theta+\tau_{e}(\theta))=\nu_{e}(\theta+\tau_{e}(\theta))$, and therefore, dividing by $\nu_{e}(T_{e}(\theta))$ (which is strictly positive by assumption) yields

   $$T^{\prime}_{e}(\theta)=\frac{f_{e}^{+}(\theta)}{\nu_{e}(T_{e}(\theta))}.$$

   For $q_{e}(\theta)=0$ we have $f_{e}^{-}(\theta+\tau_{e}(\theta))=\min\Set{\frac{f_{e}^{+}(\theta)}{\gamma_{e}(\theta)},\nu_{e}(\theta+\tau_{e}(\theta))}$ and $T_{e}(\theta)=\theta+\tau_{e}(\theta)$. Hence, dividing by $\nu_{e}(\theta+\tau_{e}(\theta))=\nu_{e}(T_{e}(\theta))$ and using Lemma 1.(iii) provides

   	$\displaystyle T^{\prime}_{e}(\theta)$	$\displaystyle=\gamma_{e}(\theta)+\frac{f_{e}^{+}(\theta)}{\nu_{e}(T_{e}(\theta))}-\min\Set{\frac{f_{e}^{+}(\theta)}{\gamma_{e}(\theta)},\nu_{e}(T_{e}(\theta))}\cdot\frac{\gamma_{e}(\theta)}{\nu_{e}(T_{e}(\theta))}$
   		$\displaystyle=\max\Set{\gamma_{e}(\theta),\frac{f_{e}^{+}(\theta)}{\nu_{e}(T_{e}(\theta))}},$

   which finishes the proof.