What is the optimal strategy when driving to my university: Wait or take alternative route and (possibly) wait?

Let's start by representing the phase of a light by the full circle, of which the first $\theta$ arc corresponds to a green light, and let $\tau$ represent the phase shift between the two lights. Let $\phi$ represent the phase shift induced by the time to go from the first light to the second light.

The picture shows in red the time when the first light is green, and in blue the time when the second light is green. If you take a point $s$ not in $[0,\theta]$, you'll be at the second light at $s+\phi$. So if you wait in front of the first light you would wait $2\pi -s$, and if you go for the second light you would wait $w(s)$ (see below).

The wait function when going to the other light is in three cases, either you arrive before the second light is green, after, or you arrive during this time. $$ w(s) = \begin{cases} t - (s+\phi) & \mathrm{ if }\; s+\phi <\tau\\ 2\pi + \tau - (s+\phi) & \mathrm{ if }\;s+\phi >\tau + \theta\\ 0 & \mathrm{ otherwise} \end{cases} $$

Now the strategy "let's go to the second light" is better than the "wait at the first light strategy" if and only if $\mathbb{E}[2\pi -s]$ is greater than $\mathbb{E}[w(s)]$ over $s \in ]\theta,2\pi[$ (assuming you have no additional information besides the color of the light when you arrive in front of the first light).

Edit: I forgot some factors in my initial calculations, so the result was completely wrong. It seems that David K did things right anyway.

By the way, there's a very simple argument to see that going to the second light is always at least as good as waiting. The transformation $s \to s + \phi$ is measure-preserving, so what matters is how much of the circle is covered by its codomain; and since this area is at most $2\pi-\theta$ (which is attained when both lights are synchronized and with a phase shift of $2\pi$), this means that it cannot do worse than waiting at the first light.

(However this assumes that all lights stay green for the same amount of time, so this may not apply to the real-life example).

Let's choose a unit of time so that each light cycle is exactly one unit long.

As you noticed, if $v = \frac L\tau$ it does not matter which choice you make. In fact, since you will arrive at the second light exactly $\frac Lv$ units of time later than you arrived at the first light (if you choose to try the second light), the expected waiting time at the second light will be exactly the same as the expected waiting time of a light at the location of the first light, but phase-shifted $\frac Lv$ units earlier than the second light, which itself is phase-shifted $\tau$ units later than the first light. So choosing to go to the second light is equivalent to making an irrevocable choice to phase-shift the first light $\tau - \frac Lv$ units later. Moreover, since phase-shifting by any whole number of units of time has no effect, we can say that choosing the second light is equivalent to phase-shifting the first light by $\delta$, where $0 \leq \delta < 1$ and there is an integer $N$ such that $N + \delta = \tau - \frac Lv$.

Let's suppose you arrive at the first light at time $t$, where $t=0$ if the light is just turning red and $t=\Theta$ if the light is just turning green again. Assuming the first light is red when you arrive, $0 \leq t < \Theta$. In addition to that constraint, there are three possible cases to consider: \begin{align} t &< \Theta + \delta - 1 \tag1 \\ \Theta + \delta - 1 \leq t &< \delta \tag2 \\ \delta \leq t & \tag3 \\ \end{align} Which cases are even possible depends on $\Theta$ and $\delta$. Essentially, by shifting the light cycle by $\delta$, you turn the light green for some portion of the time when it would have been red. Suppose the part of the cycle that was red and is now green starts at time $\alpha$ and ends at time $\beta$. If $\delta > 1 - \Theta$, then $\alpha = \Theta + \delta - 1$; otherwise $\alpha = 0$, ruling out case $(1)$. If $\delta < \Theta$, then $\beta = \delta$; otherwise $\beta = \Theta$, ruling out case $(3)$. If $\delta = 0$ then Case $(2)$ is ruled out (and it makes no difference which path you take); but Case $(2)$ is possible whenever $\delta \neq 0$.

In case $(1)$, you get a green light at time $\Theta + \delta - 1$, which is $1 - \delta$ units of time earlier than you would have gotten without the phase shift.

In case $(2)$, the phase shift turns the light green, and you can cross $\Theta - t$ units of time earlier than you would have without the phase shift.

In case $(3)$, you arrived at the first light after the portion of the light cycle that is turned green by the phase shift; you end up crossing $\delta$ units of time later than you would have without the phase shift.

Assuming that the distribution of your arrival time within the light cycle is uniform, the expected change in waiting time (positive for increased waiting, negative for decreased waiting), given that you arrive at the first light when it is red, is therefore \begin{align} E[\Delta T] &= \frac 1\Theta \int_0^\alpha (\delta - 1) \,dt + \frac 1\Theta \int_\alpha^\beta (t - \Theta) \,dt + \frac 1\Theta \int_\beta^\Theta \delta \,dt \\ &= \frac \alpha\Theta(\delta - 1) + \frac 1\Theta \left(\frac12 \beta^2 - \frac12 \alpha^2\right) - \frac{\beta - \alpha}{\Theta} \Theta + \frac{\Theta - \beta}{\Theta} \delta \\ &= \frac{1}{2\Theta}(\beta^2 - \alpha^2) + \frac \alpha\Theta(\delta - 1) + \alpha - \beta + \frac{\Theta - \beta}{\Theta} \delta \\ &= \frac{1}{2\Theta}(\beta^2 - \alpha^2) + \frac \alpha\Theta(\Theta + \delta - 1) + \frac{\Theta - \beta}{\Theta} (\Theta + \delta) - \Theta \end{align}

The reason for the seemingly peculiar rearrangement of terms on the last line of the equation above is that we know that either $\alpha = 0$ (in which case the term involving $\alpha$ disappears) or $\alpha = \Theta + \delta - 1$ (in which case the term becomes $\frac1\Theta (\Theta + \delta - 1)^2$); we also know that either $\beta = \Theta$ (in which case the term involving $\beta$ disappears) or $\beta = \delta$ (in which case that term becomes $\Theta - \frac1\Theta \delta^2$). We then have the following four possible answers, depending on whether $\delta \leq 1 - \Theta$ (that is, $\alpha = 0$) and on whether $\delta \geq \Theta$ (that is, $\beta = \Theta$):

Case $\Theta \leq \delta \leq 1 - \Theta$: $$E[\Delta T] = -\frac{\Theta}{2}.$$

Case $\delta \leq 1 - \Theta, \delta < \Theta$: $$E[\Delta T] = -\frac{\delta^2}{2\Theta}.$$

Case $\delta > 1 - \Theta, \delta \geq \Theta$: $$E[\Delta T] = \frac{(\Theta + \delta - 1)^2 - \Theta^2}{2\Theta}.$$

Case $1 - \Theta < \delta < \Theta$: $$E[\Delta T] = \frac{(\Theta + \delta - 1)^2 - \delta^2}{2\Theta}.$$

You should wait at the first light if $E[\Delta T]$ is positive, but go to the second light if $E[\Delta T]$ is negative. If $E[\Delta T] = 0$ either choice is equally good.

In the first two formulas for $E[\Delta T]$, where $\delta \leq 1 - \Theta$, clearly it is always advantageous to try the second light if the first is red, unless $\delta = 0$ (in which case it makes no difference). In the other formulas, where $\delta > 1 - \Theta$, we have $\Theta + \delta - 1 > 0$, so $(\Theta + \delta - 1)^2 - \Theta^2$ is negative if and only if $\Theta > \Theta + \delta - 1$, that is, if and only if $\delta < 1$, which is true. For similar reasons, $(\Theta + \delta - 1)^2 - \Theta^2$ is negative if and only if $\Theta < 1$, which presumably is also true.

So it seems that except for the case $\delta = 0$ (where it obviously makes no difference which path you take), it is always advantageous to go to the second light if the first is red.