This contribution analyzes the widely used and well-known “intelligent driver model” (briefly IDM), which is a second order car-following model governed by a system of ordinary differential equations. Although this model was intensively studied in recent years for properly capturing traffic phenomena and driver braking behavior, a rigorous study of the well-posedness of solutions has, to our knowledge, never been performed. First it is shown that, for a specific class of initial data, the  vehicles' velocities become negative or even diverge to \(-\infty\) in finite time, both undesirable properties for a car-following model. Various modifications of the IDM are then proposed in order to avoid such ill-posedness.  The theoretical remediation of the model, rather than post facto by ad-hoc modification of code implementations, allows a more sound numerical implementation and preservation of the model features.
Indeed, to avoid inconsistencies and ensure dynamics close to the one of the original model, one may need to inspect and clean large input data, which may result practically impossible for large-scale simulations.
Although well-posedness issues occur only for specific initial data, this may happen frequently when different traffic scenarios are analyzed, and especially in presence of lane-changing, on ramps and other network components as it is the case for most commonly used micro-simulators.
On the other side, it is shown that well-posedness can be guaranteed by straight-forward improvements, such as those obtained by slightly changing the acceleration to prevent the velocity from becoming negative.

IDM, Intelligent driver model, system of ODEs, discontinuous ODEs, traffic modelling, microscopic traffic modelling,
car following model,
well-posedness of ODEs, existence and uniqueness of solutions of ODE;
[2010]34A12, 34A38, 65L05, 65L08

§ INTRODUCTION

The field of car following modeling historically goes back to the early 1950' (and probably before) <cit.>. Most of the early work in this field focused on establishing the model equations, without paying much attention to the mathematical framework required to characterize solutions to the resulting ordinary differential equations (ODEs) describing the motion of the vehicles.
The models are mainly classified into acceleration models for longitudinal movement, lane-changing models for lateral movement and decisional models for discrete-choice situations.
Among all the car-following models introduced so far (<cit.> to just name a few) it is worth mentioning the Gazis-Herman-Rothery (GHR) model <cit.> which determines the relative velocity between two-lanes based vehicles, the Safe Distance Model <cit.>, the Optimal Velocity Model <cit.> in which the acceleration of the single vehicle is controlled according to the velocity of the leading vehicle, and the Intelligent Driver Model (IDM) which is subject of analysis in the present work.
For a comprehensive overview of the main car-following models we refer to <cit.>.

The IDM has been introduced in <cit.> and is a deterministic time-continuous model describing  the dynamics of the positions and velocities of every vehicle.  Similarly to any car-following model, the idea behind it is that drivers control their vehicles to react to the stimulus from preceding vehicles.
It aims to balance two different aspects, the necessity to keep safe separation with the vehicle in front and the desire to achieve “free flow” speed. This model presents some peculiarities which made it subject of intense research in the last two decades. Indeed, it is constructed to be collision-free, all the parameters can be interpreted and empirically measured, the stability of the model can be calibrated to empirical data, and there exists an equivalent macroscopic counterpart <cit.>.
In literature we find many extensions of the original IDM, each of which seeks to incorporate new realistic features.
The Enhanced IDM <cit.> presents an improved heuristic of the IDM useful for multi-lane simulations, which prevents the model from "over-reactions" even when the driver of the leading vehicle suddenly brakes with the maximum possible deceleration.

The Foresighted Driver Model (FDM) starts from the IDM and assumes that a driver acts in a way that balances predictive risk (due to possible collisions along his route) with utility (time required to travel, smoothness of the ride) <cit.>.

Other extensions of the IDM aim to improve the driver safety and to respect the vehicle capability <cit.>, to strengthen the power of each vehicle in proportion to the immediately preceding vehicle <cit.>, to incorporate the spatially varying velocity profile to account the variation in different types of maneuvers through intersection <cit.>.
Another natural extension is given by  Multi-anticipative IDM <cit.> which models the reaction of driver to several vehicles ahead just by summing up the corresponding vehicle-vehicle pair interactions with the same weight coefficients.

More recently, also stochastic versions of the IDM have been introduced: to describe a probabilistic motion prediction applicable for long term trajectory planning <cit.>, to study mechanisms behind traffic flow instabilities, indifference regions of finite human perception thresholds and external noise <cit.>, to incorporate context-dependent upper and lower bounds on acceleration <cit.>.

Throughout the decades, most of the engineering community worked on improving the ability of the models to capture specific behavioral phenomena, at the expense of the characterization of the solutions. Thus, to this day, only few articles use models, and corresponding solutions, that are well characterized in terms of existence, uniqueness, and regularity. An example of this practice is provided by the double integrator $\ddot{x}(t)=u\in U$, where $U$ is the input set, used abundantly as a canonical example in numerous control articles.
On the other hand, when models have inherent flaws leading to unbounded or undefined solutions, as in the case of unbounded acceleration, ad hoc methods have been traditionally applied post facto by engineering the numerical implementations. For instance, in commonly used microsimulation tools, such as SUMO <cit.>, Aimsun <cit.> and others, unbounded quantities are clipped, leading to “acceptable” numerical solutions. However, in the process, the fidelity to the original model is compromised, and the numerical simulations may not represent any instantiation of the model. Consequently, the properties of the considered continuous model might be lost as well. Finally, the process of clipping can introduce additional issues, not necessarily present in the originally model, and prevent the definition of any theoretical models corresponding to the obtained numerical simulations. The present article thus attempts to provide a full pipeline in which the model is first mathematically well defined (including existence, uniqueness and regularity characterization of the solutions), and then numerically implemented using appropriate numerical differentiation schemes. The final achievement is a thorough correspondence between theory and implementation.

§.§ The aim of this contribution

The introduced IDM has two mathematical and modelling drawbacks:

* The velocities of specific vehicles might become negative at specific times, which might not be desirable from a modelling point of view.

* The velocities of specific vehicles might diverge to \(-\infty\) in finite time, so that the solution of the system of ODE's ceases to exist.

We will discuss these drawbacks and determine under which conditions on the initial datum and parameters they can happen. Additionally, we will establish the well-posedness of the IDM for certain parameters. At last, we will present several improvements so that the solutions exist on every finite time horizon.

§.§ Structure of this article

The paper is organized in the following way. In <Ref> we review the classical Intelligent Driver Model (IDM) and describe briefly the physical meaning of the parameters involved. Well-posedness of the IDM for small time horizon is stated in <ref>. In <Ref> we analyze peculiar and possibly pathological behaviors of the model. Specifically, we provide explicit settings in which it produces negative velocities (<ref>), negative velocities and blow-up of the solution in finite time (<ref>), negative velocities and blow-up in finite time (<ref>). <Ref> collects the main results of this work. Existence and uniqueness of a solution for all times and "safe" initial configurations of the drivers are stated and proved <ref>. Moreover, these solutions show to have finite speed as long as they exist (<ref>) and to be “collision free” in the sense that the minimum distance between vehicles remains bounded, as proved in <ref>.

<Ref> is devoted to the exploration, analysis and comparison of adjustments to the classic IDM in order to avoid the problems mentioned in Section 3 for general initial data. To this end, we introduce four slightly modified versions of the IDM for which well-posedness is proved: the projected IDM, the acceleration projected IDM, the vra- IDM (velocity regularized acceleration) and the dra- IDM (distance regularized acceleration) defined respectively in  <ref>, <ref>, <ref> and <ref>.
A further and more drastic adjustment to the classic model is proposed in <ref> which involves a discontinuous acceleration, and therefore denoted as discontinuous improvement.

Finally, in <Ref> we draw conclusions from our work and mention possible research directions opened by this contribution.

§ THE INTELLIGENT DRIVER MODEL (IDM): DEFINITIONS AND BASIC RESULTS

In this section we introduce the intelligent driver model (IDM) as the following system of ordinary differential equations. To this end, we require to define the acceleration function as follows:

Let $T\in\R_{>0}$ be fixed. For a parameter set \((a,b,v_{\text{free}},\tau,s_{0},l,\delta)\in\R^{3}_{>0}\times(0,T)\times\R_{>0}^{2}\times \R_{>1}\) we define the following IDM car-following acceleration on the set

\begin{align*}
\mathcal{A}&\:\{(x,v,x_{\text{l}},v_{\text{l}})\in\R^{4}:\  x_{\text{l}}-x-l>0\},\\
\Acc&:\begin{cases}\mathcal{A}&\rightarrow\R\\
(x,v,x_{\text{l}},v_{\text{l}})&\mapsto a\bigg(1-\big(\tfrac{|v|}{v_{\text{free}}}\big)^{\delta}-\Big(\tfrac{2\sqrt{ab}(s_{0}+v\tau)+v(v-v_{\text{l}})}{2\sqrt{ab}(x_{\text{l}}-x-l)}\Big)^{2}\bigg).
\end{cases}
\end{align*}

It is worth mentioning that in most literature the parameter \(\delta\) is not precisely specified except that it is assumed to be positive. However, as we will show, velocities can become negative and this is why we assumed that the acceleration term in <ref> involves the absolute value of the velocity so that the term

\begin{equation}
\big(\tfrac{|v|}{v_{\text{free}}}\big)^{\delta}\label{eq:42}
\end{equation}

is well defined for all \(v\in\R\) and \(\delta\in\R_{>0}\). Obviously, this is only one choice and it might be more reasonable to replace this by

\begin{equation}
\sgn(v)\big(\tfrac{|v|}{v_{\text{free}}}\big)^{\delta}\label{eq:42_1}
\end{equation}

so that this term contributes positive to the acceleration for negative velocities and indeed counteracting a negative velocity. For \(\delta\in 2\N_{\geq1}+1\) <ref> can actually be replaced by the version without the absolute value and the same is true for \(\delta\in 2\N_{\geq1}\) with the drawback that this part of the acceleration will always remain negative as we also assume for now in <ref>.

An analysis similar as the one in this paper can then be carried out with mentioning that although in this case the solution's velocity can diverge to \(-\infty\). However, in the case that \(\delta\in\R_{>4}\) it can be shown that the solution always exists and the velocity cannot diverge. We do not go into details.

As we will require for the leader a specific acceleration, the “free-flow acceleration”, we define as follows

For \((a,v_{\text{free}},\delta)\in\R_{>0}^{2}\times\R_{>1}\) the free flow acceleration is defined by

\begin{align*}
\Accfront:&\begin{cases}\R^{2}&\rightarrow\R\\
(x,v)&\mapsto a\Big(1-\big(\tfrac{|v|}{v_{0}}\big)^{\delta}\Big).
\end{cases}
\end{align*}

Having defined the acceleration function, we are ready to present <ref>:

Given <ref>, we call the following system of ordinary differential equations in position \(\bx=(x_{\text{l}}, x):[0,T]\rightarrow\R^{2}\) and velocity \(\bv=(v_{\text{l}}, v):[0,T]\rightarrow\R^{2}\)

\begin{equation}
\begin{aligned}
\dot{x}_{\text{l}}(t) &=v_{\text{l}}(t), && t \in [0,T],\\
\dot{v}_{\text{l}}(t)&=\ul(t),&& t\in[0,T],\\
\dot{x}(t) &=v(t), && t \in [0,T],\\
\dot{v}(t)&=\Acc(x(t),v(t),x_{\text{l}}(t),v_{\text{l}}(t)), && t\in[0,T], \\
(\xl(0),x(0)) &=(\xl_{0},x_{0}),\\
(\vl(0),v(0)) &= (\vl_{0},v_{0}).
\end{aligned}
\end{equation}

with leading dynamics \(\ul:[0,T]\rightarrow\R\) the car-following IDM. \((x_0, \xl_{0}, v_0, \vl_{0})\in\R^{2}\times\R_{\geq 0}^{2}\) are initial positions and velocities.

This is schematically illustrated in <ref>

[->] (-0.5,-1) – (-0.5,2.5);
(-0.75,1) node \(t\);
[->] (-0.5,-1) – (7.25,-1);
[dashed,blue] (1.25,-0.5) – (4.5,1.75);
[dashed,red] plot [smooth] coordinates  (4,-0.5) (5.25,0.5) (6.25,1.5);

For \(t\in[0,T]\) the leader \(\xl(t)\) with its dynamics determined by the acceleration \(u_{\text{lead}}(t)\) and the follower \(x(t)\) with its dynamics governed by the classical IDM (acceleration \(\Acc\) as in <ref>). The overall dynamics is stated in <ref>. The follower approaches the leader. Will they collide?

We give a short overview of some of the meanings of the parameters in the IDM.
The parameters $a$, $b$, $v_{\text{free}}$, $\tau$, $s_0$, $l$ and $\delta$, introduced in <ref>, are model parameters which have – according to  <cit.> – the following meaning:

acceleration $a$: the maximum vehicle acceleration;

comfortable braking deceleration $b$: a positive number;

desired velocity $v_{\text{free}}$: the velocity the vehicle would drive at in free traffic;

desired time headway $\tau$: the minimum possible time to the vehicle in front;

minimum spacing $s_0$: a minimum desired net distance;

the length of the vehicle $l$

the acceleration exponent $\delta$: Specifying how the acceleration decreases when approaching the desired velocity $v_{\text{free}}$.

<ref> shows some suggested values for the parameters already identified in <cit.>.

According to <cit.> typical and physical meaningful variables for the IDM

Parameters    Variable     Suggested value

Maximum acceleration    $a$       0.73 $m/s^2$

Desired deceleration    $b$         1.67 $m/s^2$

Desired velocity    $v_{\text{free}}$     120 $km/h$

Desired time headway    $\tau$       1.6 $s$

Minimum spacing     $s_0$     2 $m$

Length of the vehicle    $l$     5 $m$

Acceleration exponent     $\delta$     4

For the system to be physically reasonable we require some additional assumptions on the order of the initial position and other parameters for the acceleration functions. This is made precise in the following <ref>.
[Assumptions on input datum and more]
We assume that

Leading velocity:
\(
\ul\in \mathcal{U}_{\text{lead}}\:\Big\{u\in L^{\infty}((0,T)): \vl_{0}+\int_{0}^{t}u(s)\dd s\geq0\ \forall t\in[0,T]\Big\}.
\)

Input parameters for \(\Acc\): \((a,b,v_{\text{free}},\tau,s_{0},l,\delta)\in\R^{3}_{>0}\times(0,T)\times\R_{>0}^{2}\times\R_{>1}.\)

Physical relevant initial datum: \((x_{0},\xl_{0},v_{0},\vl_{0})\in\R^{2}\times\R_{\geq 0}^{2}:\ x_{0}<\xl_{0}-l\).

The previous assumption on the involved datum enables it to prove the well-posedness of solutions on a significantly small time horizon, i.e., that there exists a solution on the time horizon and that this solution is unique:

Given \(N\in\N_{\geq0}\) and <ref>, there exists a small enough time \(T^{*}\in\R_{>0}\) so that the IDM in <ref> admits a unique solution \((x,\xl)\in W^{2,\infty}([0,T^{*}])^{2}\).

The right hand side of <ref> is around the initial datum in <ref> locally Lipschitz-continuous. The existence and uniqueness on a small time horizon then follows by the Picard-Lindelöf Theorem (<cit.> or <cit.> ).

§ COUNTEREXAMPLES

Given <ref>, the next natural questions consist of whether the solution on the small time horizon can be extended to any finite time horizon and whether the model remains reasonable. As it turns out, neither points hold if we do not restrict our initial datum beyond <ref>. We present the shortcomings in the following.

§.§ Negative velocity

In this subsection, we show that the IDM can develop negative velocities for the following vehicle, although the leading vehicle might drive with positive speed. The reason for this is that if the following vehicle is too close to the leading vehicle, it needs to slow down. Assume now that it actually has already zero velocity, it will need to move backwards to make it to the “safety” distance \(s_{0}\) it aims for.
[Negative velocity]
Assume that
\(x_0=\xl_{0}-l-\eps\) for \(\eps\in\R_{>0}\) yet to be determined and \(v_0=0\). Then, we compute the change of velocity for the following vehicle and have for \(t\in[0,T]\) according to <ref>

\begin{align*}
\dot{v}(t)=a\bigg(1- \big(\tfrac{|v(t)|}{v_{\text{free}}}\big)^{\delta}-\Big(\tfrac{2\sqrt{ab}(s_{0}+v(t)\tau)+v(t)(v(t)-v_{\text{l}}(t))}{2\sqrt{ab}(x_{\text{l}}(t)-x(t)-l)}\Big)^{2}\bigg).
\end{align*}

Plugging in \(t=0\) leads to

\begin{align*}
\dot{v}(0)=a\bigg(1 -\Big(\tfrac{2\sqrt{ab}s_{0}}{2\sqrt{ab}\eps}\Big)^{2}\bigg)=a\Big(1-\big(\tfrac{s_{0}}{\eps}\big)^{2}\Big).
\end{align*}

Thus, whenever \(\epsilon<s_{0}\), the following vehicle has – at least for small time horizon –  a negative speed, although the leading vehicle drives with arbitrary speed \(\vl\in\R\).
This is also detailed in the following <ref> and in <ref> demonstrated that for larger spacing this does not occur:

legend pos=north west,
[color=red,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_3_1_11.csv;
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_3_1_11.csv;

legend pos=north west,
[color=red,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_3_1_21.csv;
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_3_1_21.csv;
[color=black, domain=0:3, line width=2pt,dotted]0;

legend pos=north west,
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_3_1_31.csv;

[color=black, domain=0:3, line width=2pt,dotted]0;

The IDM with parameters \(a=1,\ b=2,\ v_{\text{free}}=1,\ \tau=1.6,\ l=4,\ s_{0}=2,\ d=4\) and datum \(x_{0}=0,\ \xl_{0}=l+1.5<l+s_{0},\ v_0=0,\ \vl_{0}=0\). Left represents vehicles' positions, middle vehicles' velocities and right: followers acceleration. The leader follows the free flow acceleration profile as in <ref>.  As the initial distance between the two vehicles is smaller than \(s_{0}\), the following vehicle moves backwards to increase the space.

legend pos=north west,
[color=red,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_3_1_12.csv;
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_3_1_12.csv;

legend pos=north west ,
[color=red,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_3_1_22.csv;
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_3_1_22.csv;
[color=black, domain=0:3,, line width=2pt,dotted]0;

legend pos=north west,
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_3_1_32.csv;

[color=black, domain=0:3,, line width=2pt,dotted]0;

Continuation of <ref>: The IDM with parameters \(a=1,\ b=2,\ v_{\text{free}}=1,\ \tau=1.6,\ l=4,\ s_{0}=2,\ d=4\) and datum \(x_{0}=0,\ \xl_{0}=l+2=l+s_{0},\ v_0=0,\ \vl_{0}=0 \). The leader follows the free flow acceleration as in <ref>. The initial distance between the two vehicles is large enough so that the following vehicle does not attain negative velocity.

A more reasonable approach for avoiding this type of behavior is that the following car just waits until the leading car has moved farther away. This can be achieved by adjusting the model accordingly as done in <Ref>.

§.§ Velocity exploding in finite time

In this subsection, we show that the solution can cease to exist in finite time. We first present an example, with fixed parameters, to explain the reasons behind this phenomenon. Then, we generalize the example and illustrate how this phenomenon may occur for parameters in a whole region of the space.
[Negative velocity and a blow-up of the solution in finite time]
Assume the constants and initial data are as in <ref>, with
\(0<\epsilon<1\) and positive initial velocity of the leading vehicle $\vl_{0} \geq 0$.
The leading vehicle's position is given by

\begin{align*}
\xl(t)= \xl_{0}+\vl_{0}t+\tfrac{u}{2}t^{2},
\end{align*}

and the leading vehicle's velocity by

\begin{align*}
\vl(t) = \vl_{0} + u t.
\end{align*}

Plugging this into the equations for the follower, we obtain the following system of ODEs

\begin{align*}
\dot{x}(t)&=v(t)\\
\dot{v}(t)&=a\bigg(1- \big(\tfrac{|v(t)|}{v_{\text{free}}}\big)^{\delta}-\Big(\tfrac{2\sqrt{ab}(s_{0}+v(t)\tau)+v(t)(v(t)-v_{\text{l}}(t))}{2\sqrt{ab}\cdot( x_{\text{l}}(t)-\xl_{0}+\epsilon-\int_{0}^{t}v(s)\dd s)}\Big)^{2}\bigg)\\
x(0)&= \xl_{0}-l-\epsilon\\
\end{align*}

Now we fix \(v_{\text{free}}=1=a\); \(s_{0}=16\); \(b=\tfrac{1}{4a}\); \(\tau=8\); \(\vl_{0}=0\); \(u=0\); \(\delta=4\); \(\xl_{0}=0\), which gives

\begin{align*}
\dot{x}(t)&=v(t)\\
\dot{v}(t)&=1- \big(v(t)\big)^{4}-\Big(\tfrac{(4+v(t))^{2}}{\epsilon-\int_{0}^{t}v(s)\dd s}\Big)^{2}.
\end{align*}

We now show that there exists $\bar{t}$ such that $v(\bar{t})< -1$.
Assume, by contradiction, that $v(t)\geq -1$ on $t\in[0,1]$,
then since $\epsilon<1$ for $t\in[0,1]$ we get

\begin{equation}
\dot{v}(t) \leq
1-\Big(\tfrac{(4+v(t))^{2}}{\epsilon-\int_{0}^{t}v(s)\dd s}\Big)^{2}\leq 1-\big(\tfrac{4}{\epsilon+1}\big)^2\leq
1 - \big(2\big)^2 = - 3,
\end{equation}

therefore $v(1)\leq - 3<-1$ reaching a contradiction.

Blow-up of the solution in finite time
Let \(t^{*}\in(0,T]\) denote the first time such that \(v(t^{*})<-1\).
Then, recalling the semi-group property of ODEs, we can consider the initial value problem in \(\tilde{v}:[t^{*},T]\rightarrow\R\)

\begin{align*}
\tilde{v}(t^{*})&=v(t^{*})<-1\\
\dot{\tilde{v}}(t)    &=1- \big(\tilde{v}(t)\big)^{4}-\Big(\tfrac{(4+\tilde{v}(t))^{2}}{\epsilon-\int_{0}^{t}\tilde{v}(s)\dd s}\Big)^{2} && t\in[t^{*},T].
\end{align*}

Estimating \(\dot{\tilde{v}},\) we have

\begin{align*}
\dot{\tilde{v}}(t)\leq 1-\big(\tilde{v}(t)\big)^{4}\quad \forall t\in[t^{*},T].
\end{align*}

As the initial value \(\tilde{v}(t^{*})<-1\) we obtain that \(\tilde{v}\) is monotonically decreasing
and we can thus estimate

\begin{align*}
\dot{\tilde{v}}(t)\leq 1-\big(\tilde{v}(t)\big)^{2}\quad\forall t\in[t^{*},T]
\end{align*}

which can be solved explicitly to obtain

\begin{align*}
\tilde{v}(t)\leq \tfrac{v(t^{*})\exp(2t)+v(t^{*})+\exp(2t)-1}{\exp(2t)+1+v(t^{*})(\exp(2t)-1)}.
\end{align*}

However, the right hand side goes to \(-\infty\) when \(t\rightarrow \tfrac{1}{2}\ln\Big(\tfrac{-1+v(t^{*})}{1+v(t^{*})}\Big)\) and thus, also the solution for \(v\) ceases to exist for \(t\geq t^{
This is illustrated in <ref>.

legend pos= north west,
[color=red,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_3_3_1.csv;
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_3_3_1.csv;

legend pos=north west ,
[color=red,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_3_3_2.csv;
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_3_3_2.csv;
[color=black, domain=0:3, line width=2pt,dotted]0;

legend pos=north west,
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_3_3_3.csv;
[color=black, domain=0:3, line width=2pt,dotted]0;

Same parameters as in <ref>, but \(\xl_{0}=l+1<l+s_{0}\). Left: Positions of the vehicles;
Middle: Velocities of the vehicles, Right: Acceleration of the follower. Velocity of the follower diverging to \(-\infty\) around \(t\approx 1.06\). The solution ceases to exist for larger \(t\).

In this example, we chose very specific parameters for convenience. However, the same behavior and divergence of the speed to $-\infty$ in finite time can be shown for a rather general range of parameters. This is presented in <ref>. To prove this, we start by showing the following Lemma.

Assume that the initial velocity of the following vehicle $v_{0}=0$, the initial velocity of the leading vehicle $\vl_{0}>0$, and the initial positions of the two vehicles are separated by $l+\varepsilon$ for some $\varepsilon\in (0, s_0)$, i.e., $\xl_{0}-x_{0}-l =\varepsilon<s_0$. Choose parameters $s_0$, $\tau$ and $v_{\text{free}}$ such that $-\frac{s_0}{1.01 \tau} < -v_{\text{free}}$, let $\delta>1$

and let $v_{\max}>0$ be an upper bound for the vehicles' velocities. Then there exists $t^{**} \in [0, T]$, such that $v(t^{**}) < -v_{\text{free}}$.

We consider two different cases:

If there exists $t^{*} \in [0, T]$, such that $v(t^{*}) \leq -\frac{s_0}{1.01\tau} < -v_{\text{free}}$, then we chose $t^{**} = t^{*}$.

Otherwise, we assume that for every $t \in [0, T]$, $v(t) > -\frac{s_0}{1.01 \tau}$, i.e.,

\begin{equation}
\label{exm2_0}
s_0 + v(t) \tau >\tfrac{s_0}{101}>0
\end{equation}

and $v(t) \geq -v_{\text{free}}$. By the definition of $v_{\max}$ and $v_{\text{free}}$, we have for every $t \in [0, T]$, $v_{\text{l}}(t)-v(t) \leq v_{\max} + v_{\text{free}}$. Furthermore, since $v_{\text{l}}(t)>0$ and $v(t)<0$, we have $v(t)(v(t)-v_{\text{l}}(t))>0$. Then one can find an upper bound for the distance between the two vehicles.  That is, for every  $t \in [0,T]$,
\begin{equation}
\label{upper_bounded_distance}
x_{\text{l}}(t)-x(t) -l \leq \varepsilon+(v_{\text{max}}+v_{\text{free}}) t.
\end{equation}

Therefore, for every $t \in [0, T]$,
\begin{align}
\label{exm2_1}
\dot{v}(t) &= a\bigg(1-\big(\tfrac{|v(t)|}{v_{\text{free}}}\big)^{\delta}-\Big(\tfrac{2\sqrt{ab}(s_{0}+v(t)\tau)+v(t)(v(t)-v_{\text{l}}(t))}{2\sqrt{ab}(x_{\text{l}}(t)-x(t)-l)}\Big)^{2}\bigg)\\ \label{exm2_2}
&\leq  a\bigg(1-\Big(\tfrac{2\sqrt{ab}(s_{0}+v(t)\tau)+v(t)(v(t)-v_{\text{l}}(t))}{2\sqrt{ab}(x_{\text{l}}(t)-x(t)-l)}\Big)^{2}\bigg)\\ \label{exm2_3}
&\leq a\bigg(1-\Big(\tfrac{s_{0}+v(t)\tau}{x_{\text{l}}(t)-x(t)-l}\Big)^{2}\bigg)\\
\label{exm2_4}
&\leq a\bigg(1-\Big(\tfrac{s_{0}+v(t)\tau}{\varepsilon+(v_{\max}+v_{\text{free}})t}\Big)^{2}\bigg)\\ \label{exm2_5}
&\leq a\bigg(1-\Big(\tfrac{\frac{s_0}{101}}{\varepsilon+(v_{\max}+v_{\text{free}})t}\Big)^{2}\bigg).
\end{align}

Note that inequality (<ref>) is true since $\frac{|v(t)|}{v_{\text{free}}}$ is non-negative, inequality  (<ref>) is due to the fact that for every $t \in [0, T]$, $v(t)(v(t)-v_{\text{l}}(t)) >0$,
inequality (<ref>) is because of the upper bounded for the distance between the two vehicles given by inequality (<ref>), and inequality (<ref>) is due to inequality (<ref>).

Hence, for every $t \in [0,T]$,

\begin{align*}
v(t) &\leq a\int_0^{t} \bigg(1-\Big(\tfrac{\frac{s_0}{101}}{\varepsilon+(v_{\max}+v_{\text{free}})s}\Big)^{2}\bigg) \dd s \\
&= a \bigg( t + \tfrac{s_0^2}{101^2 (v_{\max}+v_{\text{free}})}\left( \tfrac{1}{\varepsilon+(v_{\max}+v_{\text{free}})t} -\tfrac{1}{\varepsilon}\right)   \bigg)
= a t\bigg(1-\tfrac{s_0^2}{101^2}\tfrac{1}{\varepsilon(\varepsilon+(v_{\max}+v_{\text{free}})t)}    \bigg).
\end{align*}

Setting $t = \varepsilon$, we have,

\begin{align*}
v( \varepsilon) &\leq  a  \varepsilon\bigg(1-\tfrac{s_0^2}{101^2}\tfrac{1}{\varepsilon(\varepsilon+(v_{\max}+v_{\text{free}}) \varepsilon)}    \bigg)= a\tfrac{101^2\varepsilon(\varepsilon+(v_{\text{max}}+v_{\text{free}})\varepsilon)-s_0^2}{101^2(\varepsilon+(v_{\text{max}}+v_{\text{free}})\varepsilon)}\rightarrow -\infty \text { as } \varepsilon \to 0.
\end{align*}

Note that the upper bound for the velocity at time $t=\varepsilon$ goes to negative infinity as $\varepsilon$ goes to zero.
Therefore, for $\varepsilon>0$ small enough, we have,

\begin{align*}
v(\varepsilon) \leq -v_{\text{free}}.
\end{align*}

Hence, there exists $t^{**}=\varepsilon \in [0, T]$, such that $v(t^{**}) < -v_{\text{free}}$.

The previous Lemma guarantees that for properly chosen initial datum and velocity the follower's velocity can become more negative than \(-v_{0}\), the negative free-flow velocity. This enables us to prove that the solution ceases to exist in finite time. This is related to the famous example of a blowup of ODEs in finite time:
[Negative velocity and a blow-down of the velocity in finite time] Again we assume the parameters and initial data as in <ref>. Then by <ref>, there exists $t^{**} \in [0, T]$, such that $v(t^{**}) < -v_0$.
Recall that $\delta>1$. For any time $t$ such that $v(t) < -v_0$,

\begin{align*}
\dot{v}(t) \leq a\Big(1- \big(\tfrac{|v(t)|}{v_{0}}\big)^{\delta}\Big) <0
%\leq a\Big(1- \big(\tfrac{v_2(t)}{v_{0}}\big)^{2}\Big) <0.
\end{align*}

Thus $v$ is strictly decreasing on the time interval $[t^{**}, T]$.
In addition, we are going to show that $v$ cease to exist in finite time and that there exists $t_{1}>0$ such that $\lim\limits_{t\rightarrow t_{1}} v(t) =-\infty$. Assume by contradiction that this is not the case. Then as $v(T)<-v_{0}$ the solution remains strictly decreasing as long as it exists and as does not reach $-\infty$ in finite time the solution can be extended on $[0,+\infty)$ and is strictly decreasing on $[T,+\infty)$. Therefore we have the following

\begin{equation}
\int_{T}^{t}\tfrac{\dot v(s)}{1-\left(\frac{|v(s)|}{v_{\text{free}}}\right)^{\delta}}ds\geq a (t-T),\ \forall t\in [T,+\infty).\end{equation}

As $v$ is strictly decreasing we can perform a change of variable in the integral by setting $y = -v(s)$ to get

\begin{equation}\label{eq-interm}
\int_{-v(T)}^{-v(t)}\tfrac{1}{\left(\frac{y}{v_{\text{free}}}\right)^{\delta}-1}dy\geq a (t-T),\ \forall t\in [T,\infty).
\end{equation}

Note that $-v(T)>v_{0}$. We denote $\eta \coloneqq -v(T)-v_{\text{free}}>0$, <ref> implies

\begin{equation}
\int_{v_{\text{free}}+\eta}^{+\infty}\tfrac{1}{\left(\frac{y}{v_{\text{free}}}\right)^{\delta}-1}dy\geq a(t-T),\ \forall t\in [T,\infty).
\end{equation}

Letting $t\rightarrow+\infty$ this implies that

\begin{equation}
\int_{v_{\text{free}}+\eta}^{+\infty}\tfrac{1}{\left(\frac{y}{v_{\text{free}}}\right)^{\delta}-1}dy = \infty,
\end{equation}

but because $\delta >1$ we have  \(\int_{v_{\text{free}}+\eta}^{+\infty}\tfrac{1}{\left(\frac{y}{v_{\text{free}}}\right)^{\delta}-1}dy\in \R\), which gives a contradiction. Therefore, $v$ ceases to exist and converges to $-\infty$ in finite time.

As we have seen from the previous <ref> the velocity can blow up in finite time. However, what is not clear is whether the position of the car can consequently also explode. Thanks to the relation between position and velocity, i.e., \(x'(t)=v(t),\ t\in[0,T]\) this is a matter of whether \(v\in L^{1}((0,t^{*}))\) if \(t^{*}\) is the time where the velocity goes to \(-\infty\). And indeed, it can be shown that this holds true and the position remains bounded:

Let <ref>, \(\delta\in\R_{>2}\) and assume that – as investigated in <ref> – there exists a time horizon \(t^{*}\in\R_{>0}\) so that

\[
\lim_{t\nearrow t^{*}} v(t)=\lim_{t\nearrow t^{*}} \dot{x}(t)=-\infty.
\]

Then, the position at the time of the blow up remains finite, i.e.

\[
\exists c\in\R: \lim_{t\nearrow t^{*}}x(t)=c
\]

or equivalently stated

\[
v\in L^{1}((0,t^{*})).
\]

The proof consists of showing that the \(L^{1}\) mass of the velocity remains bounded. To this end, we estimate the acceleration from
Choose \(t_{1}\in [0,t^{*}]\) so that

\begin{align*}
\left((v(s)^{2}-v(s)v_{\text{l}}(s))+2\sqrt{ab} (s_{0}+v(s)\tau)\right)^{2}> 0 \wedge  v(s)\leq -\max( 2 v_{\text{free}},1)\qquad \forall s\in[t_{1},t^{*}).
\end{align*}

Such a \(t_{1}\) always exists as \(v\) diverges to \(-\infty\) so that for \(s\) close enough to \(t^{*}\) the quadratic term in the previous estimate will always outnumber the affine linear term and \(v_{\text{l}}\) is essentially bounded.
Then, recalling the definition of the IDM in <ref> we have for \(s\in [t_{1},t^{*})\)

\begin{equation}
\begin{split}
\dot v(t) \leq -\frac{a}{2}\left(\tfrac{|v(t)|}{v_{\text{free}}}\right)^{\delta}.
\end{split}
\end{equation}

Assuming \(\delta\in\R_{>2}\) divide by $|v(t)|^{\delta-1}$ (this is possible because $v$ is strictly decreasing and $v(t_{1})<-1$), and integrating between $t_{1}$ and $t\in[t_{1},t^{*})$, one has

\begin{equation}
\begin{split}
\int_{t_{1}}^{t} \tfrac{\dot v}{|v(\tau)|^{\delta-1}}d\tau &\leq -\tfrac{a}{2v_{\text{free}}^{\delta}}\int_{t_{1}}^{t}|v(\tau)|d\tau,\\
\tfrac{(-v)^{2-\delta}(t)}{\delta-2}-\tfrac{(-v)^{2-\delta}(t_{1})}{\delta-2} &\leq -\tfrac{a}{2v_{\text{free}}^{\delta}}\int_{t_{1}}^{t}|v(\tau)|d\tau.
\end{split}
\end{equation}

Dividing by $-a/2v_{\text{free}}^{\delta}<0$ and letting $t\rightarrow t^{*}$, this gives

\begin{equation}
\begin{split}
\left(\tfrac{a}{2v_{\text{free}}^{\delta}}\right)^{-1}\tfrac{(-v)^{2-\delta}(t_{1})}{\delta-2} &\geq \|v\|_{L^{1}(t_{1},t^{*})}.
\end{split}
\end{equation}

Hence, $\|v\|_{L^{1}(t_{1},t^{*})}<+\infty$.

This is particularly interesting as it also illustrates that the model behaves still reasonable (even in the case of a diverge of the velocity to \(-\infty\)) and underlines the fact that a change in the acceleration to prevent the velocity to diverge might be enough to “improve” the model (compare <Ref>).

§ WELL-POSEDNESS FOR SPECIFIC INITIAL DATUM AND ALL TIMES

In this section we state results guaranteeing existence and uniqueness of solutions on every finite time horizon under (suitable) conditions on the initial datum and the parameters.

Assume that for the dynamics in <ref> with parameters as in <ref>, the leading vehicle follows the free flow velocity, that is,

\begin{equation}
\dot{v}_{\text{l}}(t) = a\left(1-\left(\tfrac{v_{\text{l}}(t)}{v_{\text{free}}}\right)^{\delta}\right),\qquad t\in[0,T]
\label{free_flow_dynamics}
\end{equation}

with $\delta>0$ being even and in addition the initial positions and velocities of the vehicles satisfy

\begin{align}
\label{initial_condition_wellposedness}
v_0>\vl_{0}>0 \ \wedge \xl_{0} - x_0 -l =s_0.
\end{align}

Then, there exists a certain time $t_1 \in [0, T]$ such that $v(t_1) = v_l(t_1)$.

Assume that such a \(t_{1}\in[0,T]\) would not exist. Then, by the continuity of the velocity and the initial velocity, for every $t\in [0,T]$, we have $v(t)>v_{\text{l}}(t)$. Furthermore, for every $t \in [0, T]$, $x_{\text{l}}(t)-x(t)-l < s_0$. Then for $t=\frac{v_{0}-\vl_{0}+1}{a}>0$ we obtain, using the acceleration profile in <ref>

\begin{align*}
&\quad +\int_{0}^{t} a\left(\tfrac{v_{\text{l}}(s)^{\delta}-v(s)^{\delta}}{v_{\text{free}}^{\delta}}-\left(\tfrac{2\sqrt{ab}(s_0+v(s)\tau)+v(s)(v(s)-v_{\text{l}}(s))}{2\sqrt{ab}(x_{\text{l}}(s)-x(s)-l)}\right)^2\right) \dd s\\
&\leq v_{0}-\vl_{0}-\int_{0}^{t} a\left(\left(\tfrac{2\sqrt{ab}(s_0+v(s)\tau)+v(s)(v(s)-v_{\text{l}}(s))}{2\sqrt{ab}(x_{\text{l}}(s)-x(s)-l)}\right)^2\right) \dd s\\
&\leq v_{0}-\vl_{0}-\int_{0}^{t} a\left(\tfrac{ s_0}{x_{\text{l}}(s)-x(s)-l}\right)^2 \dd s\\
&\leq v_{0}-\vl_{0}-at\\
&= -1
\end{align*}

which is a contradiction.
Therefore, there exists $t_1 \in [0, T]$, such that $v(t)>v_1(t)$ for every $t\in [0, t_1)$ and $v(t_1)-v_{\text{l}}(t_1)=0$.

Let the assumptions of <ref> hold, and in addition that the model parameters and initial datum satisfy

\begin{equation}
s_0 -\tfrac{(v_{0}-\vl_{0})^2}{2a}>0.\label{eq:safe_distancing}
\end{equation}
Then, the vehicles do not collide before they reach to the same velocity.

Recall from the proof of <ref> that $t_1$ is the first time when the two vehicles reach to the same velocity. That is, for every $t \in [0, t_1)$, $v(t) > v_{\text{l}}(t)$ and $v(t_1)=v_{\text{l}}(t_1)$. Furthermore, for every $t \in [0, t_1]$,

\begin{equation}
\label{distance_lm2}
\begin{aligned}
x_{\text{l}}(t)-x(t)&=\xl_{0}-x_{0}+\int_0^{t} v_{\text{l}}(s)-v(s) \dd s\\
&= \xl_{0}-x_{0} +(\vl_{0}-v_{0}) t\\
&\quad + \int_0^{t} \left(\int_0^s a\left(\tfrac{v(\tilde{t})^{\delta}-v_{\text{l}}(\tilde{t})^{\delta}}{v_{\text{free}}^{\delta}}+\left(\tfrac{2\sqrt{ab}(s_0+v(\tilde{t})\tau)+v(\tilde{t})(v(\tilde{t})-v_{\text{l}}(\tilde{t}))}{2\sqrt{ab}(x_{\text{l}}(\tilde{t})-x(\tilde{t})-l)}\right)^2\right) \dd \tilde{t}
\right)\dd s \\
&\geq \xl_{0}-x_{0} +(\vl_{0}-v_{0}) t
+ \int_0^{t} \left(\int_0^s\!\! a\left(\tfrac{2\sqrt{ab}(s_0+v(\tilde{t})\tau)+v(\tilde{t})(v(\tilde{t})-v_{\text{l}}(\tilde{t}))}{2\sqrt{ab}(x_{\text{l}}(\tilde{t})-x(\tilde{t})-l)}\right)^2\!\!\dd \tilde{t}
\right)\dd s \\
&\geq \xl_{0}-x_{0} +(\vl_{0}-v_{0}) t+ a\int_0^{t} \left(\int_0^s \left(\tfrac{s_0}{(x_{\text{l}}(\tilde{t})-x(\tilde{t})-l)}\right)^2\dd \tilde{t}
\right)\dd s \\
& \geq \xl_{0}-x_{0} +(\vl_{0}-v_{0}) t+ a\int_0^{t} \left(\int_0^s 1 \dd \tilde{t}
\right)\dd s \\
&= \xl_{0}-x_{0} +(\vl_{0}-v_{0}) t+ \tfrac{t^2}{2} a\\
&=l+s_0 + (\vl_{0}-v_{0}) t + \tfrac{t^2}{2}a\\
&\geq l+s_0 -\tfrac{(v_{0}-\vl_{0})^2}{2a}>l.
\end{aligned}
\end{equation}

Thereby, we applied to condition on the initial velocities in relation to \(s_{0}\) and the acceleration \(a\) in <ref>.

Assume that for the dynamics in <ref> with parameters as in <ref>, the leading vehicle follows the free flow velocity as defined in (<ref>), and the model parameters and initial datum satisfy
(<ref>) and (<ref>), then there exists $\varepsilon_0 \in (0, s_0)$ such that

\[
x_{\text{l}}(t)-x(t)-l \geq \varepsilon_0\quad \forall t\in[0,T].
\]

We again consider $t_1$ as defined in the proof of <ref>. That is, for every $t\in [0, t_1)$,
$v(t)>v_{\text{l}}(t)$ and $v(t_1)-v_{\text{l}}(t_1)=0$.
Note that

\begin{align*}
& \dot{v}(t_1)-\dot{v}_{\text{l}}(t_1)=-a\left(\tfrac{s_0+v(t_1)\tau}{x_{\text{l}}(t_1)-x(t)-l}\right)^2<0.
\end{align*}

Therefore, there exist $t_2 >0$, such that $v(t)<v_{\text{l}}(t)$ for every $t\in (t_1, t_2)$.

Hence, it is enough to show that $x_{\text{l}}(t_1)-x(t_1)-l \geq \varepsilon_0$ for $0<\varepsilon_0=s_0 +(\vl_{0}-v_{0}) t_1+ \frac{t_1^2}{2}a <s_0$.
By the previous <ref> and in particular by (<ref>), it holds that

\begin{align*}
x_{\text{l}}(t_1)-x(t_1) &\geq \xl_{0}-x_{0} +(\vl_{0}-v_{0}) t_1+ \tfrac{t_1^2}{2} a,\\
&= s_{0}+l +(\vl_{0}-v_{0}) t_1+ \tfrac{t_1^2}{2} a.
\end{align*}

Thus, the only thing left to show is that $0<s_0 +(\vl_{0}-v_{0}) t_1+ \frac{t_1^2}{2}a<s_{0}$. To this end, recall that  $v_{\text{l}}(t_1) - v(t_1)=0$ implies

\begin{align*}
\vl_{0}-v_{0} =\int_{0}^{t_1} a\left(\tfrac{v_\text{l}(t)^{\delta}-v(t)^{\delta}}{v_{\text{free}}^{\delta}}-\left(\tfrac{2\sqrt{ab}(s_0+v(t)\tau)+v(t)(v(t)-v_{}(t))}{\text{l}2\sqrt{ab}(x_{\text{l}}(t)-x(t)-l)}\right)^2\right) \dd t\leq -at_1.
\end{align*}

Therefore, we obtain

\begin{align*}
(\vl_{0}-v_{0}) t_1+ \tfrac{t_1^2}{2} a= t_1\left((\vl_{0}-v_{0})+\tfrac{t_1}{2}a \right) \leq t_1\left(-at_1 +\tfrac{t_1}{2}a\right)<0
\end{align*}

and thus, $\varepsilon_0=s_0 +(\vl_{0}-v_{0}) t_1+ \frac{t_1^2}{2}a <s_0$.

Furthermore, we claim that $\varepsilon_0=s_0 +(\vl_{0}-v_{0}) t_1+ \frac{t_1^2}{2}a>0$. Indeed, for every $t \in [0, t_1]$, then

\begin{align*}
\dot{v}_{\text{l}}(t) - \dot{v}(t)= a\left(\tfrac{v(t)^{\delta}-v_{\text{l}}(t)^{\delta}}{v_{\text{free}}^{\delta}}+\left(\tfrac{2\sqrt{ab}(s_0+v(t)\tau)+v(t)(v(t)-v_{\text{l}}(t))}{2\sqrt{ab}(x_{\text{l}}(t)-x(t)-l)}\right)^2\right) >a.
\end{align*}

Additionally, by <ref>, the vehicles do not collide before time $t_1$.
Thus $ (\vl_{0}-v_{0})t_1 +\frac{t_1^2}{2}a <-s_0$. Therefore, $\varepsilon_0=s_0 +(\vl_{0}-v_{0}) t_1+ \frac{t_1^2}{2}a>0$.

The next lemma shows that the drivers velocity remains bounded from above as long as a solution of the IDM exists.

As long as the solution to the IDM in <ref> exists, the velocity of the following vehicle satisfies

\begin{equation}
v(t)\leq \max\{v_{0},v_{\text{free}}\}\quad \forall t\in[0,T].\label{eq:boundedness_velocity}
\end{equation}

We prove this result in two steps:

Velocity above \(v_{\text{free}}\):
Assume that the velocity of the follower satisfies at any given time \(t\in[0,T]\)

\begin{align*}
v(t)\geq v_{\text{free}}.
\end{align*}

Then, plugging this into the acceleration function of the IDM gives

\begin{align*}
\dot{v}(t)=a\bigg(1-\Big(\tfrac{|v(t)|}{v_{\text{free}}}\Big)^{\delta} - \Big(\dots\Big)^{2}\bigg)<0
\end{align*}

so that the velocity is monotonically decreasing and <ref> is satisfied.

Velocity below \(v_{\text{free}}\): Then, obviously <ref> is satisfied and the proof is complete.

The following <ref> states that whenever a solution to the IDM for a general acceleration profile  \(u_{\text{lead}}\) for the follower exists and the initial datum is chosen according to <ref> that a minimal distance between two cars is ensured. <Ref> is thus analogue to <ref>, however the minimal distance \(\eps_{0}\) is replaced by another smaller lower bound \(\eps_{0}^{*}\).

Recall <ref> and the corresponding parameters in <ref>. Let $v_{\text{max}}\in\R_{>0}$ as instantiated in <ref> be an upper bound for the velocity. Then, there exists $\varepsilon_0^{*}\in\R_{>0}$ such that if the initial position satisfies

\[\xl_{0} - x_{0}-l > \varepsilon_0^{*},\]

then the distance of the vehicles remains bounded from below over all times, i.e.

\[x_{\text{l}}(t) - x(t)-l \geq \varepsilon_0^{*} \qquad \forall t\in[0,T]\]

as long as the solution \(\x\) exists.
In particular, one may choose
\[\varepsilon_0^{*} = \tfrac{as_0^2\delta^{*}}{8\left(a\delta^{*}+2v_{\text{max}}^2\right)}\ \text{  with } \ \delta^{*} =\tfrac{-3v_{\text{max}}^2+\sqrt{9v_{\text{max}}^4 +4a^2s_0^2}}{2a}.\]
Note that $\varepsilon_0^{*}$ is independent on the vehicles' initial velocities.

Assume that we have $\xl_{0}-x_0 -l > \varepsilon_0$, for some  $\varepsilon_0>0$. Let $t^{*} \in [0,T]$ be the first time such that
\[x_{\text{l}}(t^{*})-x(t^{*})-l=\varepsilon_0.\]
This means that for any $s \in [0, t^{*}]$: $x_{\text{l}}(s)-x(s)-l>\varepsilon_0$. It thus suffices to show that $v_{\text{l}}(t^{*})-v(t^{*})\geq 0$.

We prove by contradiction and assume that $v_{\text{l}}(t^{*})-v(t^{*})< 0$. Then, by the continuity of the velocity, there exists $t_0<t^{*}$, such that for every $s \in [t_0, t^{*})$, $v_{\text{l}}(s)-v(s)<0$. In addition, the position difference of the two vehicles, $x_{\text{l}}-x$, decreases over the time interval $[t_0, t^{*})$.

Assume that $x_{\text{l}}(t_0)-x(t_0) -l =\varepsilon_0+\delta_0$ for some $\delta_0 >0$.
Note that $t^{*}-t_0 \geq \tfrac{\delta_0}{2v_{\text{max}}}$. By the continuity of the position, $\delta_0 \to 0$ as $t_0 \to t^{*-}$. Thus, one can choose $t_0>0$ such that $\delta_0 < \min\left\{\varepsilon_0, \frac{-3v_{\text{max}}^2+\sqrt{9v_{\text{max}}^4 +4a^2s_0^2}}{2a}\right\}$. Furthermore, we have,

\begin{align}
& v_{\text{l}}(t^{*})-v(t^{*})\notag\\
&= v_{\text{l}}(t_0) - v(t_0)+\int_{t_0}^{t^{*}}\!\!\!\! u(s) \dd s-\int_{t_0}^{t^{*}}\!\!\!\! a\left(1-\left(\tfrac{|v(s)|}{v_{\text{free}}}\right)^{\delta}-\bigg(\tfrac{2\sqrt{ab}(s_{0}+v(s)\tau)+v(s)(v(s)-v_{\text{l}}(s))}{2\sqrt{ab}(x_{\text{l}}(s)-x(s)-l)}\bigg)^2\right) \dd s\notag\\
\label{4.3_1}
&\geq  v_{\text{l}}(t_0) - v(t_0)+\int_{t_0}^{t^{*}} u(s) \dd s-\int_{t_0}^{t^{*}} a \dd s +  a\int_{t_0}^{t^{*}} \bigg(\tfrac{2\sqrt{ab}(s_{0}+v(s)\tau)+v(s)(v(s)-v_{\text{l}}(s))}{2\sqrt{ab}(x_{\text{l}}(s)-x(s)-l)}\bigg)^2 \dd s\\
\label{4.3_2}
& \geq  - v(t_0)-\int_{t_0}^{t^{*}} a \dd s +  a\int_{t_0}^{t^{*}} \tfrac{s_0^2}{(x_{\text{l}}(s)-x(s)-l)^2} \dd s\\
&\geq  - v(t_0) -a(t^{*}-t_0)  +  a s_0^2 \int_{t_0}^{t^{*}} \tfrac{1}{(x_{\text{l}}(s)-x(s)-l)^2} \dd s\notag\\
\label{4.3_4}
&\geq  - v(t_0) -a(t^{*}-t_0)  +  a s_0^2 \int_{t_0}^{t^{*}} \tfrac{1}{(\varepsilon_0+\delta_0)^2} \dd s\\
&\geq  - v(t_0) -a(t^{*}-t_0)  +
\tfrac{a s_0^2}{(\varepsilon_0+\delta_0)^2}(t^{*}-t_0)\notag\\
&\geq - v(t_0) +\left(
\tfrac{a s_0^2}{(\varepsilon_0+\delta_0)^2}-a\right)(t^{*}-t_0)\notag\\
\label{4.3_7}
&\geq - v(t_0) +\left(
\tfrac{a s_0^2}{(\varepsilon_0+\delta_0)^2}-a\right)\tfrac{\delta_0}{2v_{\text{max}}}.
\end{align}

The above inequality (<ref>) is due to the positivity of $\left(\frac{|v(s)|}{v_0}\right)^{\delta}$, inequality (<ref>) is because of that the velocity of the leading vehicle, $v_{\text{l}}(t)>0$ for every $t \in [0,T]$, inequality (<ref>) is true since for every $s \in [t_0, t^{*})$, $x_{\text{l}}(s)-x(s)-l < \varepsilon_0+\delta_0$, and the last inequality (<ref>) is derived by the fact that $t^{*}-t_0>\tfrac{\delta_0}{2v_{\text{max}}}$.

Note that since $\delta_0 \in \left(0,\frac{-3v_{\text{max}}^2+\sqrt{9v_{\text{max}}^4 +4a^2s_0^2}}{2a}\right)$,

\begin{align*}
\lim \limits_{\varepsilon_0 \to 0}  \left(
\tfrac{a s_0^2}{(\varepsilon_0+\delta_0)^2}-a\right)\delta_0 = a\left(\tfrac{s_0^2}{\delta_0^2} -1 \right)\delta_0>3v_{\text{max}}^2.
\end{align*}

For convenience, we set $\delta^{*} =\frac{-3v_{\text{max}}^2+\sqrt{9v_{\text{max}}^4 +4a^2s_0^2}}{2a}$. Then $\delta_0 \in \left(0, \min\left\{\varepsilon_0, \delta^{*}\right\}\right)$, and

\begin{align*}
\left(
\tfrac{a s_0^2}{(\varepsilon_0+\delta_0)^2}-a\right)\delta_0 >
\tfrac{a s_0^2 \delta_0}{4 \varepsilon_0^2}-a\delta^{*}.
\end{align*}

Let $\varepsilon_0^{*}= \sqrt{\frac{as_0^2\delta^{*}}{4\left(a\delta^{*}+2v_{\text{max}}^2\right)}}$.
In particular, one can choose $\varepsilon_0 =\varepsilon_0^{*}$, then $ \left(\frac{a s_0^2}{(\varepsilon_0^{*}+\delta_0)^2}-a\right) \delta_0>2v_{\text{max}}^2$.
Therefore, using this in (<ref>), $v_1(t^{*})-v(t^{*})~>0$ which is a contradiction.

Let the assumption of <ref> hold
and in addition for the parameters that

\begin{equation}s_{0}\leq \tfrac{as_0^2\delta^{*}}{8\left(a\delta^{*}+2v_{\text{max}}^2\right)}\ \text{ with }\ \delta^{*} =\tfrac{-3v_{\text{max}}^2+\sqrt{9v_{\text{max}}^4 +4a^2s_0^2}}{2a}.\label{eq:s_0}
\end{equation}

Then, we obtain

\[\dot{x}(t)=v(t) \geq 0\quad \forall t\in[0,T],\]

and in particular, the well-posedness of the solution on every finite time horizon.

Let $t^{0}\in[0,T]$ be a time at which $v(t^{0})=0$. It is enough to show that $\dot{v}(t^{0}) \geq 0$. Recalling the change of velocity according to <ref> and inserting \(v(t^{0})=0\) we have

\begin{align*}
\dot{v}(t^{0}) = a\bigg(1-\Big(\tfrac{s_{0}}{(x_{\text{l}}(t^{0})-x(t^{0})-l)}\Big)^{2}\bigg).
\end{align*}

However, following <ref>, we know that \(\xl(t)-x(t)-l\geq \epsilon^{*}_{0}\) and as the parameters are chosen according to <ref> we have

\begin{align*}
x_{\text{l}}(t^{0})-x(t^{0})-l \geq s_0,
\end{align*}

and thus

\[
\dot{v}(t^{0})=a\bigg(1-\Big(\tfrac{s_{0}}{(x_{\text{l}}(t^{0})-x(t^{0})-l)}\Big)^{2}\bigg)\geq 0.
\]

This means that the velocity cannot drop below zero. As \(t^{0}\in[0,T]\) was arbitrarily chosen to satisfy \(v(t^{0})=0\), we obtain the claim.

§ SEVERAL IMPROVEMENTS FOR THE IDM

In this section, we present several improvements of the IDM to fix the problems illustrated in
<Ref> for general initial datum.
Before doing this, however, we present some other numerics on how the classical IDM behaves for specific data. This will serve as a comparison to the proposed improvements later:
[Some additional numerical results for the classical IDM]
All examples – except those who are physically unreasonable (compare <Ref>) – will be tested on three different scenarios:

* As the set of parameters where one can observe a negative velocity of the follower in the original IDM – see <ref> and <ref>.

* As the set of parameters where one can observe that the velocity of the follower diverges to \(-\infty\) in finite time in the original IDM – see <ref> and <ref>.

* A heavy stop and go wave traffic situation with the leader's acceleration satisfying \(u_{\text{lead}}\equiv a\cdot\mathds{1}_{\{t\in[0,T]:\ \sin(t/4)\geq 0.8\}} - a\cdot\mathds{1}_{\{t\in[0,T]:\ \sin(t/4)\leq -0.8\}}\),  \(a=0.73,\ b=1.67,\ v_{\text{free}}=\frac{120}{36},\ \tau=1.6,\ l=4,\ s_{0}=2,\ d=4\) illustrated in <ref>.

legend pos=north west       ]
[color=blue,line width=1.5pt,color=red,domain=0:3.71] 0;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (3.71,x);
[color=blue,line width=1.5pt,color=red,domain=3.71:8.86] 0.73;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (8.86,0.73-x);
[color=blue,line width=1.5pt,color=red,domain=8.86:16.28] 0;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (16.28,-x);
[color=blue,line width=1.5pt,color=red,domain=16.28:21.43] -0.73;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (21.43,-0.73+x);
[color=blue,line width=1.5pt,color=red,domain=21.43:28.85] 0;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (28.85,x);

[color=blue,line width=1.5pt,color=red,domain=28.85:34.00] 0.73;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (34.00,0.73-x);
[color=blue,line width=1.5pt,color=red,domain=34.00:41.42] 0;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (41.42,-x);

[color=blue,line width=1.5pt,color=red,domain=41.42:46.57] -0.73;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (46.57,-0.73+x);
[color=blue,line width=1.5pt,color=red,domain=46.57:53.99] 0;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (53.99,x);

[color=blue,line width=1.5pt,color=red,domain=53.99:59.14] 0.73;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (59.14,0.73-x);
[color=blue,line width=1.5pt,color=red,domain=59.14:66.56] 0;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (66.56,-x);

[color=blue,line width=1.5pt,color=red,domain=66.56:71.71] -0.73;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (71.71,-0.73+x);
[color=blue,line width=1.5pt,color=red,domain=71.71:79.13] 0;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (79.13,x);

[color=blue,line width=1.5pt,color=red,domain=79.13:84.28] 0.73;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (84.28,0.73-x);
[color=blue,line width=1.5pt,color=red,domain=84.28:91.70] 0;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (91.70,-x);

[color=blue,line width=1.5pt,color=red,domain=91.70:96.85] -0.73;
[color=blue,line width=1.5pt,color=red,domain=0:0.73] (96.85,-0.73+x);
[color=blue,line width=1.5pt,color=red,domain=96.85:100] 0;

The acceleration profile \(u_{\text{lead}}\) of the leader. This is intended to specify a leader that repeats the pattern “accelerate, constant velocity, decelerate, constant velocity”.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_positions.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_positions.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_velocity.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_acc.csv;

The original IDM, with parameters \(x_{0}=0,\ \xl_{0}=l+\textcolor{red}{1}<l+s_{0},\ v_{0}=0=\vl_{0}=0\). Both vehicles start with \(0\) velocity, and the leader follows the acceleration profile in <ref>.
Left is the position of both leader and follower, middle the velocity of the follower and right is the acceleration of the follower. We leave the velocity and acceleration profile of the leader out as it is fully determined by the given \(u_{\text{lead}}\).

§.§ Projection on nonnegative velocities and restricting the maximal deceleration

A straight forward improvement consists of projecting the velocity to nonnegative values. This is detailed in the following <ref>:

Given <ref>, we replace the acceleration in <ref> and velocity for the IDM  model in <ref> by

\begin{align}
\dot{x}(t) &= \max\{v(t),0\},&& t\in[0,T]\\\label{eq:4242}
\dot{v}(t) &= \Acc\big(x(t),\max\{v(t),0\},\xl(t),\vl(t)\big), && t\in[0,T]
\end{align}

and call the model the velocity projected IDM.

Given <ref> the velocity projected IDM in <ref> admits on a sufficiently small time horizon \(T^{*}\in\R_{>0}\) a unique solution \((x,v)\in W^{1,\infty}((0,T^{*}))^{2}\).

The proof is almost identical to the proof of <ref> when recalling that the right hand side is still locally Lipschitz-continuous. We do not go into details.

However, as we will see the model's solution will not exist for general initial datum:
[The following vehicle waits for too long to start driving] Assume that the constants and initial data as in <ref> with $\varepsilon < s_0$. Then, the leading vehicle's trajectory can be computed as

\begin{align*}
\xl(t)= \xl_{0}+\vl_{0}t+\tfrac{u}{2}t^{2}, t\in [0, T],
\end{align*}

and the leading vehicle's velocity can be computed as

\begin{align*}
v_{\text{l}}(t) = \vl_{0} + u t, t\in [0, T].
\end{align*}

Plugging this into the change of the vehicle's velocity we obtain the following system of ODEs

\begin{align*}
\dot{x}(t)&=\max\{v(t),0\}\\
\dot{v}(t)&=a\Big(1- \big(\tfrac{\max\{v(t),0\}}{v_{0}}\big)^{\delta}-\Big(\tfrac{2\sqrt{ab}(s_{0}+\max\{v(t),0\}\tau)+\max\{v(t),0\}(\max\{v(t),0\}-v_{\text{l}}(t))}{2\sqrt{ab}( x_{\text{l}}(t)-\xl_{0}+\epsilon-\int_{0}^{t}\max\{v(s),0\}\dd s)}\Big)^{2}\Big)\\
x(0)&= \xl_{0}-l-\epsilon\\
\end{align*}

Note that

\begin{align*}
\dot{v}(0) = a\left(1-\left(\tfrac{s_0}{\varepsilon}\right)^2\right)<0,
\end{align*}

therefore, there exists some small time interval $[0, t_1]$ such that for every $t \in [0, t_1]$, $v(t) <0$. During the time interval $[0, t_1]$, the distance between the two vehicles is

\[l+\varepsilon+x_l(t)-\xl_{0} = l+\varepsilon+\vl_{0}t+\frac{u}{2}t^2, t \in [0, t_1].\]

Thus, for every $t\in [0, t_1]$,

\begin{align*}
\dot{v}(t)=a\bigg(1 -\Big(\tfrac{s_{0}}{\varepsilon+\vl_{0} t +\tfrac{u}{2}t^2}\Big)^{2}\bigg).
\end{align*}

Note that $\dot{v} \colon [0, t_1] \mapsto \mathbb{R}$ is strictly increasing.
Without loss of generality, we assume that the initial velocity of the leading vehicle is $\vl_{0}=0$ and the acceleration of the leading vehicle is $u=2$. Then for every $t \in [0, t_1]$,

\begin{align*}
v(t)=&\int_0^t  a\Big(1 -\Big(\tfrac{s_{0}}{\varepsilon +s^2}\Big)^{2}\Big)  \dd s = at - 4as_0^2 \int_{0}^{t} \left(\tfrac{1}{s^2+\varepsilon} \right)^2\dd s
= at - 4as_0^2\left(\tfrac{\frac{\sqrt{\varepsilon}t}{\varepsilon+t^2}+\arctan\left(\frac{t}{\sqrt{\varepsilon}}\right)}{2\varepsilon^{\frac{3}{2}}}\right).
\end{align*}

In particular, as illustrated in <ref>, we have, $t_1$ increases as $\varepsilon$ decreases. That is, the smaller the initial distance between the two vehicles, the longer it takes the following vehicle to recover its positive velocity.

Velocity dependent on epsilon and time. The dark blue indicates values less or equal \(-5\) and the white area positive function values, so that the red curve separating the white and colored region can be seen as the values where the velocity is actually zero. In particular, as the initial distance between two vehicles $\varepsilon$ increases, the time when the following vehicle recovers its positive velocity $t_1$ decreases.

Another example illustrates the projected velocity model numerically with regard to other scenarios:
[Velocity projected IDM]
As can be observed the actual velocity in all the three different scenarios is bounded from below by zero and the solution exists on the entire time horizon considered. However, the projection operator leads to the problem that the follower waits too long until they speed up. This can be observed in particular in <ref> where the distance of the two vehicles after both have started speeding up (\(t\approx 5\)) is approximately around \(8.5\) which is quite far from the comfortable vehicle distance \(s_{0}\) and thus leading to a too large distance. Same can be observed in <ref> for smaller time.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m1_positions_negative_v.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m1_positions_negative_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m1_velocity_negative_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m1_acc_negative_v.csv;

The IDM with projection as in <ref> and same parameters as in <ref> with \(x_{0}=0,\ \xl_{0}=l+1.5<l+s_{0}\). Left positions of the vehicles, middle the velocity of the follower, and right the acceleration of the follower. The leader follows the free-flow acceleration as in <ref>. The follower stays still and waits until there is a safe space to speed up.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m1_positions_inf_v.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m1_positions_inf_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m1_velocity_inf_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m1_acc_inf_v.csv;

The IDM with projection, with the same parameters as in <ref> and \(x_{0}=0,\ \xl_{0}=l+0.5<l+s_{0},\ v_{0}=0=\vl_{0}=0\). Left vehicles' positions, middle the follower's velocity and right the follower's acceleration. The leader follows the free flow acceleration as in <ref>.  The follower stays still and waits until there is a safe space to speed up.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m1_positions.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m1_positions.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m1_velocity.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m1_acc.csv;

The IDM with projection as in <ref> and initial data \(x_{0}=0,\ \xl_{0}=l+\textcolor{red}{1}<l+s_{0},\ v_{0}=\vl_{0}=0\). Left position of the vehicles, middle the velocity of the follower and right the acceleration of the follower.
The leader follows the acceleration profile in <ref>. The velocities remain positive, however the waiting time for the follower until speeding up is large.

§.§ Projection to nonnegative velocities with bounded deceleration

Another improvement for this is projecting the acceleration to prevent it from becoming too negative.
Then, the corresponding model reads as

Given <ref>, we replace the acceleration in <ref> and velocity for the IDM  model in <ref> by

\begin{align*}
\dot{x}(t) &= \max\{v(t),0\},&& t\in[0,T]\\
\dot{v}(t) &= \max\{\Acc(x(t),\max\{v(t),0\},\xl(t),\vl(t)),-a_{\min}\}, && t\in[0,T]
\end{align*}

with a parameter \(a_{\min}\in\R_{>0}\) be given and call the model the acceleration projected IDM.

Given <ref> the acceleration projected IDM in <ref> admits for every \(T\in\R_{>0}\) a unique solution \((x,v)\in W^{1,\infty}((0,T))\).

The proof of existence and uniqueness for small time is almost identical to the proof of <ref> when recalling that the right hand side is still locally Lipschitz-continuous. We do not go into details.

So it remains to show that we can find uniform estimates for \((x(t),v(t)),\ t\in[0,T]\).
\[v(t)\geq v_{\text{free}}-a_{\min}t\quad \forall t\in[0,T].\]
Thanks to the structure of \(\Acc\) (see <ref>) we also obtain as a bound from above

\[v(t)\leq v_{\text{free}}+at\quad \forall t\in[0,T].\]

As \(v\) is uniformly bounded on every finite time horizon, so is \(x\) and we are done.

However, although the previous change of the acceleration profile in <ref> looks promising as according to <ref> a solution exists on every finite time horizon, the physical representation, the model itself is unreasonable as the car behind can overtake the leading car – or differently put, the car behind can bump into the leading car without the model noticing it. This is detailed in the following <ref>.
[Physical unreasonability]
As the deceleration of the following vehicle is bounded from below by \(-a_{\min},\) we can always chose an initial velocity of the follower which leads to the fact that \(\xl(t)-x(t)-l\rightarrow 0\) in finite time.

In formulae, assume for simplicity that the leading vehicle has the following trajectory

\[
\xl(t)={\xl}_{0}+{\vl}_{0}t + \tfrac{1}{2}u_{\text{lead}}t^{2} ,\qquad t\in[0,T].
\]

with \(u_{\text{lead}}\in\R_{\geq0}\) and \(({\xl}_{0},{\vl}_{0})\in \R\times\R_{>0}\). Then, we take the difference of the vehicles position with car length \(l\in\R_{>0}\) and have for \(t\in[0,T]\)

\begin{align*}
&\leq {\xl}_{0}+{\vl}_{0}t-l + \tfrac{1}{2}u_{\text{lead}}t^{2}-x_{0}-\int_{0}^{t}\max\{v(s),0\}\dd s\\
&\leq {\xl}_{0}+{\vl}_{0}t-l + \tfrac{1}{2}u_{\text{lead}}t^{2}-x_{0}-\int_{0}^{t}v(s)\dd s\\
&\leq {\xl}_{0}+{\vl}_{0}t-l + \tfrac{1}{2}u_{\text{lead}}t^{2}-x_{0}\\
&\qquad -\int_{0}^{t}v_{\text{free}}+\int_{0}^{s} \max\{\Acc(x(\tau),\max\{0,v(\tau)\},\xl(\tau),\vl(\tau)),-a_{\min}\}\dd\tau\dd s\\
&\leq{\xl}_{0}+{\vl}_{0}t-l + \tfrac{1}{2}u_{\text{lead}}t^{2}-x_{0}-tv_{\text{free}}+\int_{0}^{t}\int_{0}^{s} a_{\min}\dd\tau\dd s\\
&\leq {\xl}_{0}-x_{0}-l +({\vl}_{0}-v_{\text{free}})t +\tfrac{1}{2}(u_{\text{lead}}+a_{\min})t^{2}.
\end{align*}

Obviously, for \(v_{0}\) sufficiently large, we obtain for small time that \(\xl(t)-x(t)<l\).
An extreme case for this is illustrated in <ref> where not only the vehicles get closer than \(l\) but the follower (in blue) entirely overtakes the leader (in red).

legend pos=north west       ]
[color=red,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_6_2_1.csv;
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_6_2_1.csv;

legend pos=north west       ]
[color=red,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_6_2_2.csv;
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_6_2_2.csv;

legend pos=north west       ]
[color=red,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/Example_6_2_3.csv;
[color=blue,line width=2pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/Example_6_2_3.csv;

Illustration of the improvement in <ref> and its physical unreasonability as shown in <ref>. The parameters are \(a=a_{\min}=1,\ b=2,\ v_{\text{free}}=1,\ \tau=1.6,\ l=4,\ s_{0}=2,\ d=4\) and datum \(x_{0}=0,\ \xl_{0}=l+1.5,\ v_0=5,\ \vl_{0}=0 \).
Left Vehicles' positions, middle vehicles' velocities, right vehicles' acceleration.
The follower overtakes the leader in finite time, although the follower breaks constantly.

§.§ Velocity regularized acceleration

Another improvement of the IDM is to add a regularization term which will make the third term in the acceleration function <ref> of the IDM in <ref> become zero if the corresponding velocity approaches zero.

Given <ref>, we replace acceleration in <ref> in <ref> by

\begin{align*}
\Acc_{\text{vra}}:\begin{cases}\mathcal{A}&\rightarrow\R\\
(x,v,x_{\text{l}},v_{\text{l}})&\mapsto a\bigg(1-\big(\tfrac{|v|}{v_{\text{free}}}\big)^{\delta}-h(v)\Big(\tfrac{2\sqrt{ab}(s_{0}+v\tau)+v(v-v_{\text{l}})}{2\sqrt{ab}(x_{\text{l}}-x-l)}\Big)^{2}\bigg)
\end{cases}\
\end{align*}

with \(\mathcal{A}\) as in <ref> and regularization \(h\in W^{1,\infty}(\R;\R_{\geq0})\) be a monotonically increasing function satisfying \(h(0)=0\) and – for a given \(\epsilon\in\R_{>0}\) – \(h(v)=1\ \forall v\in\R_{\geq\epsilon}\). We call this the velocity regularized acceleration IDM.

Given <ref> the velocity regularized IDM in <ref> admits for every finite time horizon \(T\in\R_{\geq0}\) a unique solution  satisfying \(x\in W^{2,\infty}((0,T))\) and additionally

\[
\dot{x}\equiv v\geqq 0\ \text{ on } [0,T].
\]

To show the well-posedness it suffices to make sure that

* \(\exists C\in\R_{>0}: \xl_{0} - x_0 -l \geq C \text{ implies that } x_{\text{l}}(t)-x(t)-l\geq C\ \forall t\in[0,T]\). However, this is a direct consequence of <ref> as the regularization $h$ is bounded from above by $1$ and
for any $t \in [0, T]$,

\[\Acc_{\text{vra}}(x(t), v(t), v_{\text{l}}(t), v_{\text{l}}(t)) \leq \Acc(x(t), v(t), v_{\text{l}}(t), v_{\text{l}}(t)).\]

* \(\dot{x}(t)\geq 0\ \forall t\in[0,T]\). However, we know by the Picard-Lindelöf theorem that there exists a solution on a significantly small time horizon \([0,T^{*}]\) with \(T^{*}\in\R_{>0}\). Assume that the velocity of a given vehicle \(\dot{x}\) becomes zero at a given time \(T^{**}\in\R_{>0}\). Then, by continuity of the velocity and due to the nonnegativity of the initial velocity there exists a first time \(t\in[0,T^{**})\) so that \(v(t)=\dot{x}(t)=0\). Plugging this into the corresponding acceleration we obtain at that time

\begin{align*}
\ddot{x}(t)=\dot{v}(t)&=\Acc_{\text{vra}}(x(t),\dot{x}(t),x_{\text{l}}(t),\dot{x}_{\text{l}}(t))=\Acc_{\text{vra}}(x(t),0,x_{\text{l}}(t),\dot{x}_{\text{l}}(t))=a>0,
\end{align*}

thanks to the assumption on \(h\) in <ref>, namely \(h(0)=0\). This means that whenever the velocity becomes zero, the derivative is strictly positive so that the velocity can never become negative. This gives the claim.

A proper choice for \(h\) consists for \(\epsilon\in\R_{>0}\) of

\begin{equation}
\R&\rightarrow [0,1]\\
v&\mapsto \tfrac{v}{\eps}\mathds{1}_{\R_{\geq0}}(v)\cdot \mathds{1}_{\R_{\leq\eps}}(v).
\end{cases}\label{eq:h}
\end{equation}

This is illustrated in the following <ref>. As can be seen this “saturation” function is only continuous, resulting in an acceleration function which is not differentiable. Obviously, this could be changed by smoothing \(h_{\eps}\). We do not go into details.

legend pos=north west       ]
[color=blue,line width=1.5pt,domain=-1:0,dashed] 0;
[color=blue,line width=1.5pt,domain=0:1,dashed,forget plot] x;
[color=blue,line width=1.5pt,domain=1:1.5,dashed,forget plot] 1;
[color=red,line width=1.5pt,domain=-1:0,dotted] 0;
[color=red,line width=1.5pt,domain=0:0.5,dotted,forget plot] 2*x;
[color=red,line width=1.5pt,domain=0.5:1.5,dotted,forget plot] 1;
[color=green,line width=1.5pt,domain=-1:0,dash dot] 0;
[color=green,line width=1.5pt,domain=0:0.1,dash dot, forget plot] 10*x;
[color=green,line width=1.5pt,domain=0.1:1.5,dash dot,forget plot] 1;
\(\epsilon=1\);
\(\epsilon=0.5\);
\(\epsilon=0.1\);

Different choices of the “saturation” function \(h_{\epsilon}\) for \(\eps\in\{1,0.5,0.1\}\) as suggested in <ref>.

We illustrate the model in the following
[Velocity regularized acceleration]
As can be seen in fig:velocity_regularized_acceleration_0fig:velocity_regularized_acceleration_2, with this fix the spacing between the two vehicles is not getting large as had been observed in fig:projected_IDM_large_spacing_0fig:projected_IDM_large_spacing_2 but the follower speeds up immediately when there is enough safe distance to do so. The clipping due to the function \(h_{\epsilon}\) can be observed in particular in the acceleration profile which is nonsmooth. In all fig:velocity_regularized_acceleration_0fig:velocity_regularized_acceleration_2 we choose $\epsilon = 0.1$.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m2_positions_negative_v.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m2_positions_negative_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m2_velocity_negative_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m2_acc_negative_v.csv;

The IDM with velocity regularized acceleration as in <ref> and parameters as in <ref>. Initial data are \(x_{0}=0,\ \xl_{0}=l+1.5<l+s_{0},\ v_{0}=\vl_{0}=0.\)
Left the vehicles' position, middle the vehicles' velicities, and right the vehicles acceleration.
The leader follows the free flow acceleration.
The follower stays still and waits until there is a safe space to speed up.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m2_positions_inf_v.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m2_positions_inf_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m2_velocity_inf_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m2_acc_inf_v.csv;

The IDM with velocity regularized acceleration, with initial gap equal 0.5 and same parameters as in <ref>. Both vehicles start with 0 velocity, and the leader follows the free flow IDM dynamics. Left: \(x_{0}=0,\ \xl_{0}=l+0.5<l+s_{0},\ v_{0}=0=\vl_{0}=0\). The follower stays still and waits until there is a safe space to speed up.
Right: the follower remains still in the initial phase.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m2_positions.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m2_positions.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m2_velocity.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m2_acc.csv;

The IDM with velocity regularized acceleration, with initial gap equal 1. Both vehicles start with 0 velocity, and the leader follows the acceleration profile in figure <ref>. Left: \(x_{0}=0,\ \xl_{0}=l+\textcolor{red}{1}<l+s_{0},\ v_{0}=0=\vl_{0}=0\).
Right: again, the velocity always stays nonnegative.

§.§ IDM with distance regularized acceleration

Similar to the improvement in <Ref> we can also regularize the third term in the acceleration function in <ref> in space which then reads as

Given <ref>, we replace acceleration in <ref> of the IDM in <ref> by

\begin{align}
\label{defi:IDM_distance_regularized}
\Acc_{\text{dra}}:\begin{cases}\mathcal{A}&\rightarrow\R\\
(x,v,x_{\text{l}},v_{\text{l}})&\mapsto a\bigg(1-\big(\tfrac{|v|}{v_{0}}\big)^{\delta}-\tilde{h}(\xl-x-l)\Big(\tfrac{2\sqrt{ab}(s_{0}+v\tau)+v(v-v_{\text{l}})}{2\sqrt{ab}\cdot(x_{\text{l}}-x-l)}\Big)^{2}\bigg)
\end{cases}\
\end{align}

with \(\mathcal{A}\) as in <ref> and regularization \(\tilde{h}\in W^{1,\infty}(\R;\R_{\geq0})\) be a monotonically increasing function satisfying – for a given \(\epsilon\in(0,s_0)\) –  \(\tilde{h}(x) = \frac{\epsilon}{s_0} \ \forall x\in\R_{\leq\epsilon}\) and \(\tilde{h}(x) = 1 \ \forall x\in\R_{> s_0}\).

We call this the distance regularized acceleration IDM.

A possible choice for \(\tilde{h}\) and \(s_{0}\in\R_{>0}\) as in <ref> in <ref> is for \(\eps\in(0,s_{0})\)

\begin{equation}
\tilde{h}_{\epsilon}:\begin{cases} \R &\rightarrow [\tfrac{\eps}{s_{0}},1]\\
x&\mapsto  \frac{\varepsilon}{s_0} + \left(x-\varepsilon\right)\frac{1-\frac{\varepsilon}{s_0}}{s_0-\varepsilon}\mathds{1}_{(\varepsilon,s_0)}(x) + \left(1-\frac{\varepsilon}{s_0}\right)\mathds{1}_{[s_{0},\infty)}(x).
\end{cases}
\label{eq:tilde_h}
\end{equation}

This function is illustrated in <ref>:

legend pos=north west]
[color=blue,line width=1.5pt,domain=-1:1,dashed] 1/2;
[color=blue,line width=1.5pt,domain=1:2,dashed,forget plot] 1/2+(x-1)*(1/2);
[color=blue,line width=1.5pt,domain=2:4,dashed,forget plot] 1;
[color=red,line width=1.5pt,domain=-1:0.5,dotted] 1/4;
[color=red,line width=1.5pt,domain=0.5:2,dotted,forget plot] 1/4+(x-0.5)*(1/2);
[color=red,line width=1.5pt,domain=2:4,dotted,forget plot] 1;
[color=green,line width=1.5pt,domain=-1:0.1,dash dot] 0.05;
[color=green,line width=1.5pt,domain=0.1:2,dash dot,forget plot] 0.05+(x-0.1)*(0.95/1.9);
[color=green,line width=1.5pt,domain=2:4,dash dot,forget plot] 1;
\(\epsilon=1\);
\(\epsilon=0.5\);
\(\epsilon=0.1\);

Different choices of the “saturation” for \(s_{0}=2\) function \(h_{\epsilon}\) for \(\eps\in\{1,0.5,0.1\}\) as suggested in <ref>.

Given <ref> the distance regularized IDM in <ref> admits for every finite time horizon \(T\in\R_{>0}\) a unique solution satisfying \(x\in W^{2,\infty}((0,T))\) and additionally

\[
\dot{x}\geqq0.
\]

The proof is similar to the proof to
<Ref>. We decide to omit it.

Some numerical results for the distance regularized acceleration are illustrated in the following
[Distance regularized acceleration]
Very similar to the velocity regularized accelerations are the results for the distance regularized acceleration in fig:distance_regularized_0fig:distance_regularized_2 at least with regard to the position of the vehicles. In fig:distance_regularized_0fig:distance_regularized_2 we choose $\epsilon = 0.5$.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m3_positions_negative_v.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m3_positions_negative_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m3_velocity_negative_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m3_acc_negative_v.csv;

The IDM with distance regularized acceleration, with initial gap equal 1.5 and same parameters as in <ref>. Both vehicles start with 0 velocity, and the leader follows the acceleration profile in figure <ref>. Left: \(x_{0}=0,\ \xl_{0}=l+1.5<l+s_{0},\ v_{0}=0=\vl_{0}=0\).
Right: again, the velocity always stays nonnegative.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m3_positions_inf_v.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m3_positions_inf_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m3_velocity_inf_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m3_acc_inf_v.csv;

The IDM with distance regularized acceleration, with initial gap equal 0.5 and same parameters as in <ref>. Both vehicles start with 0 velocity, and the leader follows the acceleration profile in figure <ref>. Left: \(x_{0}=0,\ \xl_{0}=l+0.5<l+s_{0},\ v_{0}=0=\vl_{0}=0\).
Right: again, the velocity always stays nonnegative.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m3_positions.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m3_positions.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m3_velocity.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m3_acc.csv;

The IDM with distance regularized acceleration, with initial gap equal 1. Both vehicles start with 0 velocity, and the leader follows the acceleration profile in figure <ref>. Left: \(x_{0}=0,\ \xl_{0}=l+\textcolor{red}{1}<l+s_{0},\ v_{0}=0=\vl_{0}=0\).
Right: again, the velocity always stays nonnegative.

§.§ A discontinuous improvement to prevent negative velocity

Our last potential improvement for the IDM which had also been suggested a lot (see for instance <cit.>) is a improvement which will become active only if the velocity becomes zero and the corresponding acceleration at the time where the velocity is zero is negative. In this way, the following improvement is the most natural one. We state it in the following <ref>.

Let <ref> and \(\Acc\) as in <ref> be given.
Then, replacing in <ref> the acceleration of the follower in the following discontinuous way as

\begin{equation}
\begin{split}
\dot{v}(t)&=\begin{cases}
\Acc(x(t),v(t),\xl(t),\vl(t))   & \text{if } v(t)> 0 \\
\Acc(x(t),v(t),\xl(t),\vl(t))     &\text{if }v(t)=0\ \wedge\ \xl(t) -x(t)-l\geq s_{0}\\
0 & \text{if }v(t)= 0 \ \wedge\ x_{\text{l}}(t)     -x(t)-l< s_{0}
\end{cases},
\end{split}
\label{eq:IDM_discontinuous}
\end{equation}

we call this the discontinuous IDM.

Although one might expect that the introduction of the discontinuity in <ref> might prohibit a solution to exist for all times or also might destroy uniqueness, it actually does not as the following <ref> states:

Given <ref> and assuming that the velocity of the leader is only zero at finitely many intervals, the discontinuous improvement of the IDM as in <ref> admits a unique solution on every finite time horizon \(T\in\R\) and satisfies

\[
x\in W^{2,\infty}((0,T)):\ \dot{x}\geqq0
\]

We consider several different cases:

* Assume that at \(t=0\) we have \(v(0)=c > 0\). Obviously, for small time the solution of the system is unique as the right hand side is locally Lipschitz-continuous then. Either, \(v(t)>0\ \forall t\in[0,T]\). Then, there is nothing more to do as we never run into the discontinuity or \(\exists t^{*}\in(0,T]: v(t^{*})=0\). We distinguish two cases:

\(\xl(t^{*})-x(t^{*})-l\geq s_{0}\): However, in this case the right hand side has not changed so that the solution still exists and is unique. As at that time, the velocity is zero and as the leading car never moves backwards, we can never go into the third case where it would hold \(\xl(t)-x(t)-l<s_{0}\) but only back into the first case with strictly positive velocity. In this case, the solution exists and is unique and there is no more to do.

\(\xl(t^{*})-x(t^{*})-l< s_{0}\): Then, the velocity of the follower is zero and the solution then can only go into the second case when the leader's position \(\xl\) increases, we are automatically left with the case that either we stay in the third case or that we move back to the second case. The second case, however, was already treated previously.

* Assume that we have \(v(0)=0\). Then, we are either in item one or two of the previous case and are done.

As all of these changes only depend on the leader's trajectory \(\xl\) which is given independent on the status of the follower, we can conclude the existence and uniqueness of a solution.

[The discontinuous improvement]
In fig:discontinuous_0fig:discontinuous_2 we illustrate the dynamics of leader and follower for the discontinuous improvement proposed in <ref>. Here, the velocity can become zero and stay zero for an amount of time dependent on the leaders velocity/position, but remains nonnegative. If the spacing between the two cars is large enough, the follower immediately speeds up and does not let the gap increase too much.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m4_positions_negative_v.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m4_positions_negative_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m4_velocity_negative_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m4_acc_negative_v.csv;

The discontinuous improvement of the IDM, with initial gap equal 1.5 and same parameters as in <ref>. Both vehicles start with 0 velocity, and the leader follows the acceleration profile in figure <ref>. Left: \(x_{0}=0,\ \xl_{0}=l+1.5<l+s_{0},\ v_{0}=0=\vl_{0}=0\).
Right: again, the velocity always stays nonnegative.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m4_positions_inf_v.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m4_positions_inf_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m4_velocity_inf_v.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m4_acc_inf_v.csv;

The discontinuous improvement of the IDM, with initial gap equal 0.5 and same parameters as in <ref>. Both vehicles start with 0 velocity, and the leader follows the acceleration profile in figure <ref>. Left: \(x_{0}=0,\ \xl_{0}=l+0.5<l+s_{0},\ v_{0}=\vl_{0}=0\).
Right: again, the velocity always stays nonnegative.

legend pos=north west       ]
[color=red,line width=1.5pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m4_positions.csv;
[color=blue,line width=1.5pt, mark=none]  table [y index=2, x index=0, col sep=comma] Tikz_data/IDM_m4_positions.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m4_velocity.csv;

legend pos=north west       ]
[color=blue,line width=2pt, mark=none]  table [y index=1, x index=0, col sep=comma] Tikz_data/IDM_m4_acc.csv;

The discontinuous improvement of the IDM, with initial gap equal 1. Both vehicles start with 0 velocity, and the leader follows the acceleration profile in figure <ref>. Left: \(x_{0}=0,\ \xl_{0}=l+\textcolor{red}{1}<l+s_{0},\ v_{0}=\vl_{0}=0\).
Right: again, the velocity always stays nonnegative.

§.§ Comparisons among modifications

We herein summarize the strength and weakness of each modification.

* The acceleration projected IDM in <ref> is the most straightforward modification to fix the negative velocities. However as shown in <ref>, it suffers from the fact that the follower waits too long to start driving. It may also lead to physical unreasonability issue with certain initial datum as we point out in <ref>.

* The velocity regularized IDM in <ref> and distance regularized IDM in <ref> are both capable of preventing negative velocity, and the follower speeds up immediately when it observes safe distance. However in both modifications one needs to specify the regularization functions $h$ or $\tilde{h}$, which might require extra parameter tuning.

* The discontinuous improvement of the IDM as in <ref> prohibits negative velocities, without the need to introduce extra saturation functions.

In order to quantitatively compare the modifications, we compute the average distance (as well as the variance) between the leader and follower over time in <ref> on three different scenarios we tested.

The average distance between the leader and follower (and the variance reported in the parenthesis)

Average distance (variance)     <ref>    <ref>    <ref>

acceleration projected IDM    7.99 (1.19)       8.09 (3.55)       14.94 (54.99)

velocity regularized IDM    7.76 (1.00)         7.24 (1.70)     12.81 (25.51)

distance regularized IDM    7.73 (1.02)     7.29 (1.77)    12.79 (25.87)

discontinuous improvement of the IDM    7.75 (1.01)    7.31 (1.75)     12.39 (25.18)

§.§ Generalization to many cars

In the proposed framework, we have only studied the case where we have one leader and one follower and the leader (in most cases) satisfies an acceleration profile where their velocity is nonnegative. However, as this is somewhat arbitrary, all proposed results and all “improvements” remain valid as long as the velocity of the follower remains non-negative when we generalize this to more than two cars. We make this precise in the following <ref> but first introduce the number of cars as well as some physical reasonable assumption on the input datum:
[Input datum for multiple cars]
Let \(N\in\N_{\geq1}\) be given. We assume that the parameters of the \(\Acc\) satisfy what we have assumed in <ref> as well as the leaders acceleration \(u_{\text{lead}}\). Additionally, we assume for the initial datum (position and velocity)

\[
\bx_{0}\in\R^{N}: \bx_{0,i}-\bx_{0,i+1}> l\ \forall i\in\{1,\ldots,N-1\}\ \wedge\ \bv_{0}\in\R_{\geq0}^{N}.
\]

Given <ref> and <ref>. Then, the dynamics for the IDM with many vehicles \(N\in\N_{\geq1}\) read in \((\bx,\bv):[0,T]\rightarrow\R^{N}\times\R^{N}\) as

\begin{equation}
\begin{aligned}
\dot{\bx}_{1}(t)&=\bv_{1}(t),&&&& t\in[0,T]\\
\dot{\bv}_{1}(t)&=\ul(t),&&&& t\in[0,T]\\
\dot{\bx}_{i}(t)&=\bv_{i}(t),&& i\in\{2,\ldots,N\} && t\in[0,T]\\
\dot{\bv}_{i}(t)&=\Acc(\bx_{i}(t),\bv_{i}(t),\bx_{i-1}(t),\bv_{i-1}(t)), && i\in\{2, \ldots, N\}&& t\in[0,T] \\
\bx(0) &=\bx_{0},\\
\bv(0) &= \bv_{0}.
\end{aligned}
\end{equation}

The system for many cars is illustrated in <ref>.

[->] (-6.5,-1) – (-6.5,2.5);
(-6.75,1) node \(t\);
[->] (-6.5,-1) – (7.25,-1);
[dashed,blue] (1.25,-0.5) – (4.5,1.75);
[dashed,blue!75!white] (-.75,-0.5) – (2.5,1.75);
[dashed,blue!60!white] (-2.75,-0.5) – (.5,1.75);
[dashed,blue!45!white] (-4.75,-0.5) – (-1.5,1.75);
[dashed,red] plot [smooth] coordinates  (4,-0.5) (5.25,0.5) (6.25,1.5);
(3.5,-1) node[below] \(\bx_{1}(t)\);
(0.5,-1) node[below] \(\bx_{2}(t)\);
(-1.5,-1) node[below] \(\bx_{3}(t)\);
(-3.5,-1) node[below] \(\bx_{4}(t)\);
(-5.5,-1) node[below] \(\bx_{5}(t)\);

The leader \(\bx_{1}(t)\) with its dynamics determined by the acceleration \(u_{\text{lead}}(t)\) and the following cars \(\bx_{2}(t)\), \(\bx_{3}(t)\),  \(\bx_{4}(t)\), \(\bx_{5}(t)\) with its dynamics governed by the classical IDM <ref>. The overall dynamics is stated in <ref>.

Having the definition, we obtain the following general result on the well-posedness when changing the right hand side of the ODEs as suggested in the corresponding improvements:

Let <ref> hold and
Consider as “improvement” of the IDM either the

* the velocity regularized acceleration in <ref>

* the distance regularized acceleration in <ref>

* the discontinuous improvement in <ref>.

Then, the system of \(2N\) coupled initial value problems (\(N\in\N_{\geq1}\)) admits a unique solution on every given time horizon \(T\in\R_{>0}\) and the solutions satisfy

\begin{align*}
\bx\in W^{2,\infty}((0,T);\R^{N}):\ \dot{\bx}\equivd\bv\geqq\boldsymbol{0} \text{ on } [0,T].
\end{align*}

The proof consists of recalling that the proofs of the corresponding improvements all worked for any leaders acceleration proposed in <ref> as long as the corresponding velocity would remain nonnegative.
As the dynamics are only “one-directionally” coupled, i.e.  the dynamics of the follower depend on the leader but not vice versa, we can use an
inductive argument and first look on the dynamics for \(\bx_{1},\bx_{2}\). According to <ref>, <ref>, <ref> we obtain the well-posedness. However, additionally also that the velocity \(\bv_{2}\) is nonnegative. As a next step we can thus consider the dynamics of \(\bx_{2},\bx_{3}\). As \(\bx_{2}\) satisfies the non-negativity of velocities we can again apply the stated well-posedness results for two cars. This can be iterated until we have reached \(\bx_{N}\) and we conclude with the existence and uniqueness of solutions and the nonnegativity of the velocities.

§ CONCLUSIONS AND FUTURE WORK

In this contribution, we have demonstrated the ill-posedness of the rather often used intelligent driver model (IDM) for specific initial datum and have presented some improvements to avoid these problems.
The proposed work builds a solid foundation for future work on
1) Multi-lane traffic with lane-changing. The lane-changing requires to know under which condition a car can change lane and how the well-posedness of solutions is affected by the lane change. With the proposed improvements, the well-posedness can be guaranteed easily. For optimizing lane-changing to smooth traffic and avoid stop and go waves (thus saving energy) we will next consider a suitable hybrid optimal control problem based on the proposed dynamics.
2) Signalized junctions/intersection modelled with a form of the IDM also requires the proposed well-posedness in particular as a red traffic light necessitates the vehicles in front of it to stop (velocity becomes zero).
3) Stability of solutions with regard to the model parameters and comparison of stability for the different suggested improvements.
Implementations of the models presented in this article. While the numerical work presented above results from  <cit.> implementations and the use of  and similar routines, it would be interesting to also theoretically study the discretization of these equations with standard finite difference schemes to see what guarantees can be provided for the numerical solutions (for example order of the numerical schemes, error bounds on the numerical solution etc.).
5) Finally, it would also be of great interest for the improvements of the IDM model presented here to validate them against field data, or to see if the resulting microsimulation implementations (SUMO <cit.>, Aimsun <cit.>, etc.) match experimental data for specific choices of numerical parameters. In particular, it would be interesting to measure discrepancies with the prior IDM improvements in their implementation, and compare their respective performances.

§ ACKNOWLEDGEMENTS

This material is based upon work supported by the National Science Foundation under Grant Numbers CNS-1837244 (A. Bayen), CNS-1837481 (B. Piccoli).
The research is based upon work supported by the U.S. Department of Energy’s Office of Energy Efficiency and Renewable Energy (EERE) under the Vehicle Technologies Office award number CID DE-EE0008872.
The views expressed herein do not necessarily represent the views of the U.S. Department of Energy or the United States Government.