Velocity Dependent Potentials and Dissipation function

When we derive Euler Lagrange equation, we considered only simple potentials that depends only on the positional coordinates. But there are other kinds of potentials, for e.g. Potential which depends on the velocity of the particle. Let us deal with those potential to derive a general Euler Lagrange Equation.  

Let us denote velocity dependent potentials $V(q_j, \dot{q_j}) $ where generalized force is defined by, $$ Q_j = -\frac{\partial{V}}{\partial{q_j}} + \frac{d}{dt}\left(\frac{\partial{V}}{\partial{q_j}}\right) \,\,\,...eq.(1) $$  

But we know that, if  L = T - V the generalized force is given by, $$ Q_j =  \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q_j}}} - \frac{\partial{L}}{\partial{q_j}} \,\,\,...eq.(2)$$ 

where L contains only the Potential that arise only from the gradient of generalized coordinates i.e. only conservative potential, and Q_j is the generalized force that doesn't include any part from the conservative forces. It includes only forces that are not derivable from a conservative potentials, Eg. Frictional forces. 

But if we could denote the frictional forces itself in a perfect differential form, there is a hope we could make it simpler looking with Euler Lagrange equations. 

If frictional forces could be written in the form, $$ F^{frictional}_{j} = -k_{j}v_{j} $$

where "j" the generalized coordinate.

These kind of frictional forces can be derived from a general function U, also known as Rayleigh's dissipation function. 

It is given by, $$ U = \frac{1}{2} \sum_i \sum_j k_j v_{ij}^2 $$

where "i" is the summation over each particle.

From this definition, the forces are derived from the potential as, $$ F^{frictional}_j = -\frac{\partial{U}}{\partial{v_j}} $$ 

Then the generalized force arising from this frictional force is given by, 

$$ Q_j =  \sum_i \vec{F^{frictional}_i}\cdot \frac{\partial{\vec{r_i}}}{\partial{q_j}} = - \sum_i \frac{\partial{U}}{\partial{v_i}} \frac{\partial{\vec{r_i}}}{\partial{q_j}} $$

From eq.(4) in D'Alemberts principle and derivation of Euler Lagrange equation, we know that, 

$$ \frac{\partial{\vec{v_i}}}{\partial{\dot{q_j}}} = \frac{\partial{\vec{r_i}}}{\partial{q_j}} $$ 

Applying it in our equation, we get, 

$$ Q_j = -\sum_i \frac{\partial{U}} {\partial{v_i}} \frac{\partial{\vec{v_i}}}{\partial{\dot{q_j}}} \\~\\ = - \frac{\partial{U}}{\partial{\dot{q_j}}} $$   

Substituting in eq.(2) we get our desired result, 

$$ \frac{d}{dt} \left(\frac{\partial{L}}{\partial{\dot{q_j}}}\right) - \frac{\partial{L}}{\partial{q_j}} + \frac{\partial{U}}{\partial{\dot{q_j}}} = 0 \,\,\,...eq.(3) $$

If we are initially given with L and U i.e. Lagrangian and Rayleigh dissipation function, then everything about the problem can be easily solved using the above Euler - Lagrange equation (with dissipation).