Euler-Lagrange equations
\[S = \int_{t_i}^{t_f} L(q,\dot{q}) \text{d}t\] \[\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{q}}= \frac{\partial L}{\partial q}\]Proof
\[\delta S = S(q,\dot{q}) - S(q+\delta q , \dot{q}+\delta(\dot{q}))=0\]where $\delta q = \epsilon(t)$ is a function of time.
The Lagrangian transforms as:
\[L(q+\delta q , \dot{q}+\delta(\dot{q})) = L(q,\dot{q}) + \delta L\] \[\delta L = \frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot{q}}\delta(\dot{q})\]We assume the following:
\[\delta(\dot{q})= \dot{\delta q}\] \[\delta q(t_i)= \delta q(t_f)=0\]Now, we can compute:
\[\begin{align} \delta S &= \int_{t_1}^{t_2} \delta L\\ &= \int_{t_1}^{t_2} \frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot{q}}(\dot{\delta q})\text{d}t \end{align}\]We can integrate by part, (i.e. integrate the product rule). We recall here that:
\[\frac{\text{d}}{\text{d}t}(uv) = u\frac{\text{d}v}{\text{d}t} + \frac{\text{d}v}{\text{d}t}u\]Thus:
\[[uv]_{t_a}^{t_b} = \int_{t_a}^{t_b} u\frac{\text{d}v}{\text{d}t} \text{d}t + \int_{t_a}^{t_b}\frac{\text{d}v}{\text{d}t}u \text{d}t\]and
\[\int_{t_a}^{t_b} u\frac{\text{d}v}{\text{d}t} \text{d}t = [uv]_{t_a}^{t_b} - \int_{t_a}^{t_b}\frac{\text{d}v}{\text{d}t}u \text{d}t\]Applied here with $u = \frac{\partial L}{\partial \dot{q}} $ $v= \delta q$, we obtain
\[\begin{align} \delta S &= \int_{t_1}^{t_2} \frac{\partial L}{\partial q}\delta q \text{d}t + \int_{t_1}^{t_2} \frac{\partial L}{\partial \dot{q}}(\dot{\delta q})\text{d}t\\ &= \int_{t_1}^{t_2} \frac{\partial L}{\partial q}\delta q \text{d}t + [\frac{\partial L}{\partial \dot{q}} \delta q]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{q}}\delta q\text{d}t\\ &= [\frac{\partial L}{\partial \dot{q}} \delta q]_{t_1}^{t_2} + \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}- \frac{\partial L}{\partial \dot{q}}\right)\delta q\text{d}t\\ &= \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}- \frac{\partial L}{\partial \dot{q}}\right)\delta q\text{d}t =0 \end{align}\]Where we used the boundary conditions ($\delta q (t_1) = \delta q (t_2)=0$).
We use the theorem:
If
\[\int f(t)g(t) \text{d}t =0 f(t) =0\]for all values of $g(t)$, then $f(t)=0$$.
Hence, we find the Euler-Lagrange equations:
\[\frac{\partial L}{\partial q}- \frac{\partial L}{\partial \dot{q}}=0\]