We know that one can treat nonholonimic (but differential) constraints in the same manner as holonimic constraints. With a given Lagrange Function $L$, the equations of motion for a holonomic constraint $g(\vec{x}) = 0$ follow as: $ \frac{d}{dt} \frac{\partial L }{\partial \dot{\vec{x}}} - \frac{\partial L}{\partial \vec{x}} = - \lambda \frac{\partial g}{\partial \vec{x}} $
You can derive these either by looking at the constraint-forces, or by a somewhat enhanced least-action principle: The physical Path $\vec{x}(t)$ is the one with stationary action, with respect to all variations $\delta \vec{x}_g(t)$ that satisfy the boundary condition $g$.
For non-holonomic constraints, given as $\vec{a}(\vec{x})\dot{\vec{x}} = 0$, the equations of motion are given as: $ \frac{d}{dt} \frac{\partial L }{\partial \dot{\vec{x}}} - \frac{\partial L}{\partial \vec{x}} = - \lambda \vec{a} $
My question now is wether here also is an underlying principle of least action. My theory is that the physical path $\vec{x}(t)$, that we would obtain solving the above equation, is the one with stationary action with respect to variations $\delta \vec{x}(t)$ that satisfy the (now nonholonomic) boundary-conditions. Is that true?
To show it, I formulated my "Assumptions" mathematically: The equations of motion are given by $\vec{x}_l(t)$, with $S[\vec{x}_l + \delta \vec{x}_a]-S[\vec{x}_l] = 0$, with $\delta \vec{x}_a$ being a variation that satisfies $\vec{a} \cdot (\dot{ \vec{x}}_l + \delta \dot{\vec{x}}_a) = 0 $
I however fail to derive the above given equations of motion from this assumptions. Am I wrong about my assumptions?