Interpreting common PDE in lagrangian mechanics? I'm learning Lagrangian mechanics, and more than a few times you find expressions of the form 
$$-\frac{\partial{Z}}{\partial{q_j}} + \frac{\mathrm d}{\mathrm dt}\left(\frac{\partial{Z}}{\partial{\dot{q_j}}}\right)=0.$$
Obviously it comes up where $Z=L$, and $Z=Q$ where it gives the components of generalised force, but I have trouble intuiting its meaning.  I can obviously understand it term by term, but I don't fully understand how to interpret the whole. 
The first part can be seen almost as being like $-\nabla Z$ except the derivative is with respect to state-space, not space proper, but the second half is conceptually mysterious to me. Can anyone give a good explanation of what this means in general?
 A: Intuitively, the second term is just the time derivative of the momentum.
For simple point mass in some potential, we have $\mathscr{L}=\frac{1}{2}m  v^2-V(x)$. Then, the Euler-Lagrange equation becomes
$$0=-\frac{\partial V}{\partial x} -\frac{\text{d}}{\text{d} t}\left(\frac{\partial }{\partial v}\frac{1}{2}mv^2\right)\,.$$
The first term you have already identified with the force as gradient of the potential. The second term becomes 
$$\frac{\text{d}}{\text{d} t}\left(mv\right)=\dot{p}\,.$$
In other words, we have $\dot{\vec p}=\vec F$, which is just Newton's second axiom.
For more complicated situations, e.g. motion with constraints, your second term is still interpreted as a "generalised momentum" conjugate to the respective coordinate, and the first term is the "generalised force".
A: We can construct the Euler-Lagrange operator as follows:
$$
\bigg[\frac{d}{dt}\bigg(\frac{\partial}{\partial \dot q^i}\bigg) - \frac{\partial }{\partial q^i }\bigg] L = 0
$$
As you mention in your question, the second portion can be thought of as the force. We can recover this conceptually if we let $L = T(\dot q) - U(q)$ and hence:
$$
-F_i =  \frac{\partial U }{\partial q^i } 
$$
We then have: 
$$
\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot q^i}\bigg) - F_i = 0
$$
If we define the bracketed term as a momentum, $p = \partial L / \partial \dot q$, we recover Newton's second law: 
$$
\dot p^i = F_i
$$
Therefore, we can interpret the first portion as the rate of change of momentum and the second portion as the force, if we must insist on paralleling the Newtonian formulation.
In addition, the Euler-Lagrange equation is a second order ODE in $q^i$ rather than a PDE. 
Edit
In light of your comment, the first portion of Euler-Lagrange operator on some function $Z(q,\dot q)$ defines the rate of change of the conjugate momentum to $q^i$, where the definition of the momentum is:
$$
p_i = \frac{\partial Z}{\partial \dot q^i}
$$
It happens, that this then satisfies the Legendre transform relationship, linking $p_i$ and $\dot q^i$ as conjugate variables to a phase space description. As mentioned above and by @Toffomat, the rate of change of the conjugate momentum is by definition the generalised force $F(q,\dot q)$. Whether this coincides with the classical notion of force depends on whether $Z=L$. 
In another framework, one cannot abstract further than this for a given $Z$. It is unclear, without further specification of $Z, q $ and $p$, how to interpret what the generalised momentum/force actually is. 
