How do you derive Lagrange's equation of motion from a Routhian? 
*

*Given a Routhian $R(r,\dot{r},\phi,p_{\phi})$, how do you derive Lagrange's equation for $r$? Do you just solve the following for $r$?
$$\frac{d}{dt}\frac{∂R}{∂\dot{r}}-\frac{∂R}{∂r}=0.$$

*And as a related question, what is the motivation for using a Routhian?
 A: No. The coordinate $r$ stil follows the Euler-Lagrange equation, but $\phi$ and $p_\phi$ follow the Hamilton equations. But these are trivial, which is the whole point of the Routhian. The motivation is that the Routhian isn't really $R(r, \dot r, \phi, p_\phi)$ but just $R(r, \dot r)$ with a constant parameter $p_\phi$. $\phi$ isn't a coordinate because by definition, it was a cyclic coordinate in the Lagrangian and so it doesn't appear in the Routhian either. The momentum $p_\phi$, meanwhile, is conserved so it's really just some constant. With both of these conditions, we can simply take them both out, call $p_\phi$ a constant, and we wind up with a problem with 1 fewer effective dimensions. The same would be true using the complete Hamiltonian, but sometimes Lagrangians are easier to work with for the "hard" part of the problem (the non-cyclic coordinates), and the Routhian lets you stay "Lagrangian" for the hard part.
Here is a concrete example. Take the Lagrangian describing a harmonic potential in polar coordinates:
$$
\mathcal L = \frac{1}{2} m \left( \dot r^2 + r^2 \dot \phi^2 \right) - \frac{1}{2} k r^2
$$
Since $\phi$ does not appear in the Lagrangian, it is a cyclic coordinate. From the Euler-Lagrange equation
$$
\frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot \phi} = \frac{\partial \mathcal L}{\partial \phi} = 0
$$
it follows that
$$
p_{\varphi} =\frac{\partial \mathcal L}{\partial \dot \phi} = m r^2 \, \dot \phi
$$
is a conserved quantity. It would be nice to take $\dot\phi$ out of the set of coordinates and just replace it with the constant $p_\phi$. Then we'd essentially have a Lagrangian whose only coordinates are $r$ and $\dot r$, just with a constant parameter $p_\phi$. The way to do this correctly is to make the Routhian
$$
\mathcal R = \mathcal L - p_\phi \dot \phi = \frac{1}{2} m \dot r^2   - \frac{1}{2} k r^2- \frac{p_\phi^2}{2 mr^2}
$$
(Aside: Try solving $\dot \phi$ in terms of $p_\phi$ and substitute into the Lagrangian. You'll get something very different that will lead to very wrong results. If you want to remove the cyclic coordinate, you must do a Legendre transform w.r.t. that coordinate.).
Since the Legendre transform has been done w.r.t. $\theta$ and $\phi$, these coordinates follow Hamilton's equations (with an appropriate change of sign). But these is trivial:
$$
\dot \phi = - \frac{\partial \mathcal R}{\partial p_\phi} = \frac{p_\phi}{mr^2}, \quad \dot p_\phi = \frac{\partial \mathcal R}{\partial \phi} = 0
$$
What we've achieved here is essentially separation of variables. $\phi$ has its own equation of motion, which involves $r$, but the equation of motion of $r$ is independent of $\phi$. We can solve the equation of motion of $r$, which is an effectively one-dimensional problem, and then go back to find out how $\phi$ evolves.
The effective one-dimensional problem has an effective potential
$$
V(r) = \frac{1}{2} k r^2 + \frac{p_\phi^2}{2mr^2}
$$
This extra term, which blows up at small $r$, is called the centrifugal barrier, and accounts for the fact that angular momentum conservation makes it harder / impossible for the particle in question to reach the origin. The coordinates $r$ and $\dot r$ didn't have a Legendre transform applied, so they still follow the Euler-Lagrange equation
$$
\frac{d}{dt} \frac{\partial \mathcal R}{\partial \dot r} = \frac{\partial \mathcal R}{\partial r} \Rightarrow m \ddot r = -kr + \frac{p_\phi^2}{3mr^3}
$$
A follow up question might be "Yes, its convenient to not stay with a full Lagrangian, but why not just go full Hamiltonian?" The answer here probably depends on the context of the specific problem. With the Routhian, we have a 2nd order differential equation to solve, and then an integral to find the e.o.m. of $\phi$. With a full Hamiltonian, we'd have a system of two 1st order differential equations, followed by the same integral. One might be easier to solve, or better computationally, etc.
A: *

*Setting. Imagine that the configuration space consists of, say, both small & capital generalized positions $q^j$ and $Q^J$, with corresponding velocities $v^j$ and $V^J$, and momenta $p_j$ and $P_J$, respectively.


*Routhian. The Routhian
$$\begin{align}R(q,Q,v,P,t)
~=~&V^JP_J-L(q,Q,v,V,t)\cr
~=~& H(q,Q,p,P,t)-v^jp_j \end{align} \tag{R}$$
is a hybrid between [and a partial velocity-momentum Legendre transformation away from] the Lagrangian
$$L(q,Q,v,V,t)\tag{L}$$
and the Hamiltonian
$$H(q,Q,p,P,t),\tag{H}$$
such that the small velocity variables $v^j$ and the capital momentum variables $P_J$ are kept.


*Action principle. The Routhian equations are the Euler-Lagrange (EL) equations for the Routhian action
$$S_R[q,Q,v,P]~=~\int \! dt~L_R(q,Q,\dot{q},\dot{Q},P,t), \tag{SR}$$
with Routhian Lagrangian
$$ L_R (q,Q,\dot{q},\dot{Q},P,t)
~:=~\dot{Q}^JP_J-R(q,Q,\dot{q},P,t) ,\tag{LR}$$
which leads to Lagrange equations for the small variables and Hamilton's equations for the capital variables.


*The motivation is to harvest the usual benefits from both the Lagrangian and Hamiltonian sides. Examples:

*

*If the capital position variables $Q^J$ are cyclic variables [which means that $L$, $R$ and $H$ do not depend on $Q^J$], then the capital momentum variables $P_J$ are constants of motion. We can therefore demote the dynamical variables $(Q^J,P_K)$ to external parameters of the model. An action principle for the remaining dynamical variables (i.e. the small variables) is then given by the time integral of (minus) the Routhian
$$-\int \! dt~R(q,\dot{q},t). $$
For a simple application, see e.g. this Phys.SE post.


*If the Lagrangian $L$ depends affinely on $v^j$ and non-affinely on $V^J$, the Faddeev-Jackiw method to first-order formulations yields the Routhian action $S_R$.
