# Derivation of Lagrange's eqs

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1. May 28, 2016

### lampCable

1. The problem statement, all variables and given/known data
So I'm deriving Lagrange's equations using Hamilton's principle which states that the motion of a dynamical system follows the path, consistent with any constraints, that minimise the time integral over the lagrangian $L = T-U$, where $T$ is the kinetic energy and $U$ is the potential energy.

We define the lagrangian as $L = L(q_j,\dot{q}_j,t)$. Now I want to let $q_j = q_j^{(0)}+\delta q_j$, where $\delta q_j$ is the variation of $q_j$. We also define
$$S=\int_{t_1}^{t_2}L(q_j,\dot{q}_j,t)dt$$

So according to Hamilton's principle we would now like to minimise $S$. At extremum we have $\delta S = 0$, i.e. the variation of S is zero.

Now, my problem:

My first experience with calculus of variations was to find Euler's equation. We considered then the functional
$$J = \int_{x1}^{x2}f(y,y',x)dx$$
and our goal was to find the function $y$ that minimise S. To do this we let $y(x,\alpha) = y(x)+\alpha\eta(x)$, and set $\frac{\partial J}{\partial\alpha}|_{\alpha=0}=0$, where $\eta$ is some arbitrary function. This would give us an equation in $y(x)$ since we evaluate at $\alpha=0$.

So, returning to the derivation of Lagrange's equations. I set to find $q_j$ that minimise $S$ in similar fashion as we did for $J$. But this time I do not have any $\alpha$ that I can put to zero. Should I instead take $\delta S|_{\delta q_j = 0}=0$? For unless I have understood it wrong, it is actually $q_j^{(0)}$ that we want to find, right? I mean it seems strange to find $q_j$ to be some curve $q_j^{(0)}$ added by some arbitrary variations.

At the same time, I am not sure whether evaluating $\delta q_j$ at makes sense. What increase my doubts is that neither in my text book (Marion and Thornton) nor at any lecture has it been emphasized that we evaluate with $\delta q_j = 0$.

Anyhow, I am thankful for answers.

2. May 29, 2016

### ShayanJ

Remember the definition of a derivative ? $\displaystyle f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$. And to find an extremum of the function, you just set $f'(x)=0$ .
Its the same here, the only difference being that here we're talking about a functional, a function in the space of functions defined on a given interval with given boundary conditions. There is just one difference though. What we do here to derive the EL equations is, after doing the calculations and arriving at the following equation:
$\delta S=\int_{t_1}^{t_2} \left[ \frac{\partial L}{\partial q_j}-\frac{d}{dt} \frac{\partial L}{\partial \dot{q_j}} \right]\delta q_j dt=0$
We reason that this equation should be true for any non-zero $\delta q_j$ so we should have $\frac{\partial L}{\partial q_j}-\frac{d}{dt} \frac{\partial L}{\partial \dot{q_j}} =0$. The point is, although $\delta q_j$ is supposed to be very small, its not exactly zero. But there should be somewhere that we actually use the fact that $\delta q_j$ is very small, and its actually in the derivation of the above equation were we expand the Lagrangian in a Taylor series and treat $\delta q_j$ as the small displacement from $q_j^{(0)}$.
Anyway, I think you still can use the $y(x,\alpha)=y(x)+\alpha \eta(x)$ approach here. But even here, its actually $\alpha\to 0$ not $\alpha=0$.