wrobel said:
Because this is not a variational problem. Variational problem is a problem about extremums of a functional which is defined on some manifold with constraints. When you solve a variational problem you can not choose variations as you want. The constraints determine the space of variations.
So the phrase:
is meaningless in the mathematical sense. The constraints ##a_i\dot x^i=0## imply the following space of variations ##\{\delta x^j\mid a_i\frac{d}{dt}(\delta x^i)+\frac{\partial a_i}{\partial x^k}\dot x^k\delta x^i=0\}##. This corresponds to the vaconomic mechanics. But the Largange-D'Alembert equations do not correspond to conditional extremum of any functional in nonholonomic case.
And please never tell to specialists in dynamical systems that Landau&Lifshitz vol. 1 is a good book they will die laughing
I don't bother, how you call what I did. I find this very straight forward, and I never understood, why non-holnomic constraints are more complicated than holonomic ones. Again the logic is the following. One defines the action functional as usual
$$A[q]=\int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t).$$
Then one defines the (in general nonholonomic) constraints as constraints on the allowed variations
$$a_{ik}(q,t) \delta q^k=0.$$
These have to be included by the introduction of Lagrange multipliers in the usual way, leading to
$$\delta A[q] = \int_{t_1}^{t_2} \mathrm{d} t [\delta L-\lambda^{i} a_{ik} \delta q^k] \stackrel{!}{=}0.$$
Thanks to the Lagrange multipliers now we can vary the ##\delta q^k## independently and get the usual equations of motion as from d'Alembert's principle,
$$\frac{\partial L}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\dot{q}^k}=\lambda^{i} a_{ik}.$$
That's, how it's derived in Landau&Lifshitz vol. I, Section 38 too, as I've just checked.