Can we abuse traffic patterns to get home earlier?

Solution 1:

Let $\delta >0$ be the delay in your departure time from work. Also, we can model your remaining commute time $T$ as a function of the current time and your current position $(t,x_t): T(t,x_t)$. We will assume that $x_t$ is continuous (no teleportation allowed!) and monotonically increasing (no backtracking).

Your coworkers' conjecture (let's call it $H$) is that:

$$\exists \delta >0: \delta+T(\delta,0)<T(0,0)=2$$

This assumes that the two routes are independent (i.e., waiting to leave a little later lets you use an entirely different road). However, if we are using the same road, then $H$ implies a contradiction.

Let $x_t$ be the position of a car that leaves when $\delta=0$ and $y_t$ be the position of a car delayed by some $\delta>0$ such that $H$ is true (your two "shadow" cars). Then we get

$$ \delta+T(\delta,0)<T(0,0) \implies \exists t: x_t=y_t$$

This result is due to the fact that the paths $x_t,y_t$, are monotonically increasing and continuous:

Let $q=\delta+T(\delta,0)<2$

$$x_0=y_0=0, \;\;x_q=y_2=1 \implies \exists s\in [0,2]: x_s>y_s$$

Also, $$ \delta>0 \implies \exists s\in [0,2]: x_s<y_s$$

Taken together, we can see that:

$$\exists a,b\in [0,2]: y_a-x_a <0,\; y_b-x_b>0$$

The last step indicates that the function $z_t=y_t-x_t$ (which is continuous by virtue of $x_t,y_t$ being continuous) attains both positive and negative values over the domain $t\in[0,2]$.

By the Intermediate Value Theorem

$$\exists m\in[0,2]: z_m=0 \implies x_m=y_m$$

Now comes the contradiction:

$$x_m=y_m \implies T(m,x_m)=T(m,y_m) \implies x_{m+T(m,x_m)}=y_{m+T(m,y_m)}$$

In words, this says that if $x$ and $y$ ever meet, they will arrive home at the same time. Therefore, for all $\delta$ satisfying $H$, we have derived the following result:

$$\delta+T(\delta,0)=T(0,0) \;\forall \delta>0:H$$

This directly contradicts $H$, hence $\neg H$ is true:

$$\forall \delta>0 \;\;\delta+T(\delta,0)\geq T(0,0)$$

Paraphrasing Seinfeld, we could call this the No Secret Fast Lane Theorem ;-)

Existence of $\delta$ given $f(x,t)$

Let $f(x,t)$ be the speed function of position and time. Then we have:

$$x'=f(x,t)$$

This is a rather general differential equation. Not much more to be said until it gets made specific. Now, lets assume you can solve this differential equation to get $x(t;\delta)$. Then you need to verify the following:

$$ \exists (\delta,t)\in[0,2]: x(t;\delta)=x(t;0)$$

Again, without getting specific, we can't go much further. However, this is the general test for if such a $\delta$ exists given a general $f(x,t)$

Simple Traffic Model

The OP has provided a nice proof that smoothness of $x_t$ implies $\delta + T(\delta,0)>T(0,0)$. If we relax the smoothness requirement, we can achieve $\delta + T(\delta,0)=T(0,0)$ for some $\delta>0$. Below is a simple traffic model for a road starting at $x=0$ and ending at $x=1$.

Let $a>0,b \geq 1, c\leq a/b$

$$f(x,t)= \left\{ \begin{array}{lcc} a & x \leq ct \\ \\ a/b,& x>ct \\ \end{array} \right.$$

What is $T(t,x)$ here? It depends on the values of $a,b$ and $c$: Now, $b=1$ or $c<\frac{a}{b}$ will lead to scenarios where we never catch up, either because the traffic speed is constant, or because we will end up matching the ghost car' speed before we catch it.

Also, $a<c$ cannot happen by definition. Hence, we are left with the final case of $b>1,c=\frac{a}{b}$.

Case: $b>1,c=\frac{a}{b}$

If $x<ct$, then we we need to find how long it will take to hit the lower speed traffic (if ever), before we get home $(x=1)$:

$$x+a\Delta t_{x,t} = c(t+\Delta t_{x,t}) \implies \Delta t_{x,t} = \frac{ct-x}{a-c}$$

Our position at this point will be $x^*=x+a\Delta t_{x,t}$. If $x^*>1$, then $\Delta t_{x,t} = \frac{1-x}{a}$, so we can combine these requirements into:

$$\Delta t_{x,t}= \left\{ \begin{array}{lcc} \min\left\{\frac{1-x}{a}, \frac{ct-x}{a-c}\right\} & x \leq ct \\ \\ 0& x>ct \\ \end{array} \right.$$

Our remaining time will be:

$$T(t+\Delta t_{x,t},x^*)=\frac{1-x^*}{a/b}$$

Therefore, our total time is:

$$ T(t,x)=\Delta t_{x,t} + \frac{1-(x+a\Delta t_{x,t})}{a/b}$$

Lets calculate the arrival time for our ghost car:

$$T(0,0)=\frac{b}{a}$$

Now, the arrival time for our "delayed" car is:

$$\delta + T(\delta,0) = \delta+\min\left\{\frac{1}{a}, \frac{c\delta}{a-c}\right\}+\frac{1-a\min\left\{\frac{1}{a}, \frac{c\delta}{a-c}\right\}}{a/b}$$

Applying our assumptions, we get:

$$\delta + T(\delta,0) = \delta+\min\left\{\frac{1}{a}, \frac{\delta}{b-1}\right\}+\frac{1-a\min\left\{\frac{1}{a}, \frac{\delta}{b-1}\right\}}{a/b}$$ We need to demonstrate

$$\exists \delta>0: \delta + T(\delta,0) = \delta+\min\left\{\frac{1}{a}, \frac{\delta}{b-1}\right\}+\frac{1-a\min\left\{\frac{1}{a}, \frac{\delta}{b-1}\right\}}{a/b} = \frac{b}{a}$$

Scenario 1: $\frac{\delta}{b-1} > \frac{1}{a}$

$$T(\delta,0)=\delta+\frac{1}{a}+\frac{1-a\frac{1}{a}}{a/b} = \delta+\frac{1}{a}$$

Now

$$\frac{\delta}{b-1}>\frac{1}{a}\implies \delta > \frac{b-1}{a} \implies \delta + \frac{1}{a} > \frac{b}{a}$$

So, we can see that we will never meet up with other ghost car if $\delta > \frac{b-1}{a}$

Scenario 2: $\frac{\delta}{b-1} \leq \frac{1}{a}$

$$T(\delta,0)=\delta+\frac{\delta}{b-1}+\frac{1-a\frac{\delta}{b-1}}{a/b}=\frac{b}{a}$$

So, for $\delta \leq \frac{b-1}{a}$ we will catch up with the ghost car, provided that the ghost car is at the very end of the slow part of the traffic! (i.e, clear sailing behind the ghost car).