Quantization of a waveguide: he only has one of the two E.O.M in his Lagrangian. How can the quantization be correct? I am following this reference, appendinx C and D, around page 61
The goal is to quantize electric of current of propagating wave in a waveguide.
Classical E.O.M:
We model a waveguide by a succession of LC oscillators:

We will study electric wave propagating on this line, thus we do a differential reasoning: $c_0$ is the capacitance per unit length, $l_0$ the inductance per unit length.
Using Kirchoff laws on an element of length $dx$, we then find the telegraph equations:
$$\partial_x V(x,t)=-l_0 \partial_t I(x,t) $$
$$\partial_x I(x,t)=-c_0 \partial_t V(x,t) $$
Writing down Lagrangian and Hamiltonian:
We define the flux variable $\phi$ as:
$$ \phi(x,t)=\int_{-\infty}^t V(x,t') dt'$$
Thus, we have by definition: $V(x,t)=\partial_t \phi(x,t)$
We also have: $I(x,t)=-\frac{1}{l_0} \partial_x \phi(x,t) $, it comes from the voltage drop around an inductance ($U=L \dot{I}$) and the fact $\phi$ is the time integral of $V$. (I can add precision if necessary).
Now, the author says that the Lagrangian density for our system is (I guess it is inspired from the Lagrangian of an LC oscillator):
$$\mathcal{L}(\dot{\phi},\partial_x \phi)=\frac{c}{2} \dot{\phi}^2-\frac{1}{2l}(\partial_x \phi)^2 $$
This Lagrangian leads to the following E.O.M:
$$\frac{1}{c_0 l_0}\partial_x^2 \phi - \partial_t^2 \phi=0 \Rightarrow \partial_x I(x,t)=-c_0 \partial_t V(x,t)$$
We notice that this Lagrangian only implies one of the two E.O.M
Then, he finds the momentum associated to $\phi$, he writes down the Hamiltonian, he imposes commutation relation between position and momentum to quantize the theory.
My question
Here the Lagrangian density only contains one of the two E.O.M of the system. Why is the quantization then correct? For me you must have the full dynamic encoded in the Lagrangian then Hamiltonian to be able to quantize. How can the quantization then be correct? I am confused. 
 A: The definitions
$$V(x,t)=\partial_t \phi(x,t), \quad I(x,t)=-\frac{1}{l_0} \partial_x \phi(x,t) $$
automatically imply, by the equality of mixed partial derivatives ($\partial_x \partial_t \phi = \partial_t \partial_x \phi$), the result
$$\partial_x V = - \frac{1}{l_0} \partial_t I$$
which is your "missing" equation. This holds independent of the equations of motion.
This is not uncommon when formalizing things in Lagrangian mechanics. For example, the electromagnetic field strength can be defined as $F = dA$. In that case, the result $dF = 0$ follows by definition, independent of the equations of motion, and contains two of Maxwell's equations.

Edit: it seems the real question is, how can $I(x,t)=-\frac{1}{l_0} \partial_x \phi(x,t)$ be a definition when it is derived from Faraday's law? The point is that derivations in one context can be laws in another context, or postulates or definitions in yet another. Within the context of classical electromagnetism, starting from Maxwell's equations, this result is derived. But in the context of modeling a waveguide with an extremely simple Lagrangian, it must be a definition, because your Lagrangian doesn't even know what letter $I$ is or means. That's fine, because the two contexts are logically independent.
