Derivation of Lagrangian in the calculus of variations

[Joggl]

Homework Statement

$\frac {d} {d\epsilon}\left. L(q+\epsilon \psi) \right|_{\epsilon = 0}$

where the index $\epsilon = 0$ means the derivative of the function evaluated at this point.

Relevant Equations

$q=q(t)$
$\psi = \psi (q(t), \dot{q}(t), t)$

Hello. In a chapter of a book I just read it is given that

$\frac {d} {d\epsilon}\left. L(q+\epsilon \psi) \right|_{\epsilon = 0} = \frac {\partial L} {\partial q} \psi$

While trying to get to this conclusion myself I've stumbled over some problem.
First I apply the chain rule:

$\left.(\frac {\partial L(q+\epsilon \psi)} {\partial (q+\epsilon \psi)} \frac {d (q+\epsilon \psi)} {d \epsilon}) \right|_{\epsilon = 0}$

The second part of the product can be evaluated since $q$ and $\psi$ do not depend on $\epsilon$:

$\frac {d (q+\epsilon \psi)} {d \epsilon} = \psi$

This leads to:
$\frac {d} {d\epsilon}\left. L(q+\epsilon \psi) \right|_{\epsilon = 0} = \left.(\frac {\partial L(q+\epsilon \psi)} {\partial (q+\epsilon \psi)} \frac {d (q+\epsilon \psi)} {d \epsilon}) \right|_{\epsilon = 0} = \left.(\frac {\partial L(q+\epsilon \psi)} {\partial (q+\epsilon \psi)} \psi) \right|_{\epsilon = 0}$

And now, I don't know how to continue since I think it is not allowed to evaluate the arguments of the Operator $\frac {\partial L(q+\epsilon \psi)} {\partial (q+\epsilon \psi)}$ at $\epsilon = 0$
In my opinion this evaluation has to be done after the derivation which I can't calculate since it's not clear how $L$ depends on $(q+\epsilon \psi)$

 

[jambaugh]

As you write it, the Lagrangian is a function only of the one variable. You shouldn't be writing a partial derivative. But to answer your question the two functions $L'(q+\epsilon\psi)$ and $L'(q)$ agree at $\epsilon=0$ where $L'(y)=\frac{dL(y)}{dy}$.

 

[jack476]

Can you think of a way to write $L(q+\epsilon \psi)$ in terms of derivatives of $L$ with respect to $q$, given that $\epsilon \psi$ is very small?

 

[jambaugh]

Yes... look up Taylor-Maclaurian series expansion of a function in your Calculus 1 textbook.