Euler-Lagrange equations YouTube

Small deviations revisited

Figure 2.4 - Small deviation to a trajectory

Figure 2.4 - Small deviation to a trajectory

Recall, we were considering functionals of the form $$A = \int_{t_1}^{t_2} L(q_i, \dot{q_i}) \mathrm{d}t$$ under the variations $$\begin{align*}\delta q_i &= \epsilon f_i(t)\\ \delta \dot{q_i}&= \epsilon \dot{f_i}(t)\end{align*}$$

We want to find equations involving the $q_i, \dot{q_i}$ that minimise $A$. To do this, we assume that we have already found the solution trajectory, $q(t)$, then show that small deviations to this trajectory don't result in small changes in $A$ - equivalent to setting $\delta A = 0$.

Recall the functions. $f_i(t)$ are arbtrary - meaning that we want the result to hold for any deviation. The only constraint is that the deviations are zero at the endpoints, which just means we set $f_i(t_1) = f_i(t_2) = 0$. Figure 2.4 shows the situation (although we have dropped the indices).

Earlier on in the lecture, we saw that setting $\delta A = 0$ led to the integral $$\delta A = \int_{t_1}^{t_2} \sum_{i = 1}^{N} \left\{ \frac{\partial L}{\partial q_i} f_i(t) + \frac{\partial L}{\partial \dot{q_i}} \dot{f_i}(t) \right\} \mathrm{d}t = 0$$

Finding the stationary trajectory

Using integration by parts on each of the $\: i \:$ factors in the second term in the integrand, with $$\begin{align*}x & \to t\\ f(x) & \to \frac{\partial L}{\partial \dot{q_i}}\\ g(x) & \to f_i(t)\end{align*}$$ the integral becomes $$\begin{align*}0 &= \delta A\\&= \int_{t_1}^{t_2} \sum_{i = 1}^{N} \left\{ \frac{\partial L}{\partial q_i} f_i(t)+ \frac{\partial L}{\partial \dot{q_i}} \dot{f_i}(t) \right\} \mathrm{d}t\\&= \int_{t_1}^{t_2} \sum_{i = 1}^{N} \left\{ \frac{\partial L}{\partial q_i} f_i(t) - \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L}{\partial \dot{q_i}} \right) f_i(t) \right\} \mathrm{d}t+ \left [ \sum_{i = 1}^{N} \frac{\partial L}{\partial \dot{q_i}} f_i(t) \right ]_{t_1}^{t_2}\\&= \int_{t_1}^{t_2} \sum_{i = 1}^{N} \left\{ \left( \frac{\partial L}{\partial q_i} - \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L}{\partial \dot{q_i}} \right) \right) f_i(t) \right\} \mathrm{d}t+ \sum_{i = 1}^{N} \left. \frac{\partial L}{\partial \dot{q_i}} \right |_{t_2} f_i(t_2)- \sum_{i = 1}^{N} \left. \frac{\partial L}{\partial \dot{q_i}} \right |_{t_1} f_i(t_1)\\&= \int_{t_1}^{t_2} \sum_{i = 1}^{N} \left\{ \left( \frac{\partial L}{\partial q_i} - \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L}{\partial \dot{q_i}} \right) \right) f_i(t) \right\} \mathrm{d}t\end{align*}$$ since we insisted that $f_i(t_1) = f_i(t_2) = 0$ for each $i$.

Because the $f_i(t)$'s are arbitrary, we can now use the (generalised) zero integral theorem, replacing $$\begin{align*}x & \to t\\ f_i(x) & \to \frac{\partial L}{\partial q_i} - \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L}{\partial \dot{q_i}} \right)\\ g_i(x) & \to f_i(t)\end{align*}$$ to show that $$ \frac{\partial L}{\partial q_i} - \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L}{\partial \dot{q_i}} \right) = 0 \quad \forall i = 1, \dots, N$$

These are called the Euler-Lagrange equations. We can solve these equations for the $q_i, \dot{q_i}$ to find the trajectory that minimises our integrated quantity, $A$.