How do I read the simple, but contradictory, Lagrangian ($\mathcal{L} = x + v$)? I understand the lagrangian formulation of classical mechanics, to a degree. I can derive the Euler-Lagrange equations from the "least" action principle, and equivalently can determine the equations of motion from a given lagrangian. I can handle lagrangian exercises in textbooks with ease.
I don't quite grok it, though. If you erased it from my mind, I wouldn't be able to re-invent it. So I'm going back through it.
The way I understand the Euler-Lagrange equation is thus: In classical physics, by observation, there is a stationary quantity. This stationary quantity is called the "action", and it is the sum of energies over time (alternatively, a product of energy and time). Again by observation, energies can be calculated given the positions and velocities of all elements of the system. Call such a function $\mathcal{L}(x, \dot{x})$, treat it as a black box. Then we have $Action = \int dt \mathcal{L}(x, \dot{x})$. Making this stationary, we derive
$$\frac{\partial\mathcal{L}}{\partial x} = \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x}}$$
Informally, this says to me that the paths taken in actual physics are ones where small perturbations in a particle's position are canceled out precisely by the changes in velocity before and after that position (where the velocity changes are caused by the position perturbation). This has an almost economic elegance to it.
But it still doesn't completely make sense to me what a Lagrangian is. Its unit is energy, sure, but I also don't quite grok energy beyond the abstract. So I figured I'd play with a few simple Lagrangians, hoping to break the formulation and learn something from how the pieces fell. Take, for example, this trivial function:
$$\mathcal{L}(x, v) = x + v$$
It describes an unphysical world, surely. Energy is far from conserved. But I figured constructing a weird but simple lagrangian would give me insight as to the nature of the formulation. Let's derive the equations of motion:
$$\frac{\partial\mathcal{L}}{\partial x} = 1$$
$$\frac{\partial\mathcal{L}}{\partial v} = 1$$
$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial v} = 0$$
So the path taken is the path satisfying
$$1 = 0$$
...Huh. The laws of motion are unsatisfiable. I'm not sure how to take that.
What, precisely, went wrong here? I'm looking for a geometric or intuitive explanation -- the algebraic troubles I understand.
What sort of world was I trying to construct? Whence the contradiction?
More generally, when you are handed a Lagrangian, what does it really signify? I can integrate it to get the action (an abstract number assigned to a path) or I can plug it into the Euler-Lagrange equations to figure out motion, but what does it mean in its native form? How do I read a Lagrangian without twisting its arm?
 A: When you get a nonsensical equation of motion like this, it means that the action has no local extrema over the space of all possible paths. (Like how the curve $y = x$ has no local minimum or maximum.) There's no path $x(t)$ such that varying the path produces only a second-order (or higher) variation in the action.
In fact, you can see this explicitly for your Lagrangian $\mathcal{L} = \alpha x + \beta \dot{x}$ (constants inserted to make the units work): start with an arbitrary path $x(t)$, construct the path
$$y(t) = x(t) + \epsilon\sin\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr)$$
(for example), and computing the action, you get
$$\begin{align}S[y]
&= \int_{t_1}^{t_2} \left[\alpha x(t) + \alpha\epsilon\sin\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr) + \beta \dot{x}(t) + \frac{\beta\pi\epsilon}{t_2 - t_1}\cos\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr)\right]\mathrm{d}t \\
&= \int_{t_1}^{t_2} \left[\alpha x(t) + \beta \dot{x}(t)\right]\mathrm{d}t + \int_{t_1}^{t_2} \alpha\epsilon\sin\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr)\mathrm{d}t + \int_{t_1}^{t_2} \frac{\beta\pi\epsilon}{t_2 - t_1}\cos\biggl(\pi\frac{t - t_1}{t_2 - t_1}\biggr)\mathrm{d}t \\
&= S[x] + \frac{2\alpha\epsilon(t_2 - t_1)}{\pi} + 0
\end{align}$$
which is strictly greater than $S[x]$ if $\alpha\epsilon > 0$ and strictly less if $\alpha\epsilon < 0$. In other words, given any path $x(t)$ with action $S[x]$, you can make another, arbitrarily close path with a larger (or smaller) action. That means, by definition, there is no extremal path.
That, in turn, means that this Lagrangian is meaningless, to put it bluntly. Remember, the whole validity of the stationary action principle depends on that local extremum of the action existing in the first place. Without that, the derivation of the Euler-Lagrange equations doesn't even really work. It's kind of like how the validity of division in algebra depends on the denominator not being zero - if you break that underlying assumption, your derivation loses logical consistency. And similarly, when you try to use action extremization on an action which has no extremum, you just get illogical nonsense.
A: First I want to remind you what is going on behind the scenes. You know where the particle is at some initial time $t_1$, and you know where the particle is at some final time $t_2$, and the question you are asking is, which path will get me from the initial position at the initial time to the final position at the final time in a way that minimizes the action. Mathematically you want the function $x(t)$ that satisfies the conditions $x(t_1)=x_1$ and $x(t_2)=t_2$ and that minimizes $S$.
You wrote two terms, let's consider them separately. The easy one to deal with is the $v$ term. That term is a total derivative:
$$S=\int_{t_1}^{t_2} dt \frac{dx}{dt} = x(t_2)-x(t_1) = x_2 - x_1$$
But we don't have any freedom to change $x_2$ or $x_1$, they are part of the known information. We don't vary the path at the end points. So this term is actually irrelevant, and it is why the Euler Lagrange equations give zero for this term. This is a crucial property of lagrangians: I can add a total derivative to a lagrangian without changing the equations of motion.
So if we throw out that irrelevant term really the action you wrote down was
$$S=\int_{t_1}^{t_2} dt x(t)$$
You want the function that minimizes this integral.
But there is no function that will do that. For example, imagine a parabola in the x,t plane connecting $x_1$ and $x_2$. By changing the height of the parabola I can make the integral arbitrarily positive, negative, or zero. The key point though is that there is nothing that makes any of the parabolas special. 
Explicitly for parabolas of the form $x(t)=3/2 a t^2$, and choosing the origin of time so that $t_1=-T$ and $t_2=T$, we have
$$S = \int_{-T}^{T} 3/2 a t^2 = a T^3$$
Then $dS/da=T^3$. In other words as I vary a I always get the same answer for the action. There is nothing here picking out a special parabola to minimize the action.
It's really exactly the same as trying to find the minimum of the function $f(u)=u$. In calculus you're told to take the derivative and set it equal to zero, that gives you the critical points. But if you do that here you get $f'(u)=1=0$, so what's up? The point is that there are no values of $u$ where $1=0$ so there is no minimum. It's the same problem here.
What this illustrates is that you really need the kinetic term $v^2$ in the lagrangian. The kinetic term penalizes paths that have unnecessary wild changes (the kinetic term punishes parabolas with large a in the example above). In the example above where $x_1=x_2=0$ adding the kinetic term would pick out one parabola that the particle actually follows. where you threw the particle in the air and watched it come back down (your potential $x$ is just the potential of the earth's gravitational field). The kinetic term is really the crucial thing. One thing you can try is to only consider parabolic paths in your action where you add a $v^2$ term, compute the action explcitly for those paths, and then just find the value of $a$ that minimizes the action. You don't have to use the Euler Lagrange equations at all here. (of course this only works since you know that the final path is a parabola, but it's proving a conceptual point)
By the way, to respond to some comments above the fact that the dimensions don't line up isn't actually that big a deal in this case, you can just put a parameter $\tau$ in front of the velocity with units of time and then work in units with $\tau=1$. The problem you are finding isn't affected by keeping $\tau$.
