Recent content by gionole

@vanhees71 Yes, so with such variable endpoint problem, you consider paths that are with fixed endpoints and you also consider paths that are non-fixed points. It's like all of them are admissible, but ##J[y(x)]## must be 0 in first order with all of them. all I am saying now is that ##y(x) +...

@vanhees71 I actually understand what you're saying. I think we got a problem of misunderstanding my question. let me try to put it in another words. There're scenarios where you can apply 2 conditions. Imagine Brachistochrone problem where one of the end of the string/slide can be put...

@vanhees71 Hi. Thanks so much for your time. Can you read #16 reply ? I described my question much better.

Thanks so much @Orodruin . I think I have one last question remaining . Assume an action ##J[y(x)]## and assume that ##y(x)## is a true path. This means first order difference between the true path's action and perturbed path's action must be 0. Note that perturbed path can be represented by...

@Orodruin I tried my best, but somehow, I don't get the same thing. Can you give a hint ? that page contains lots of things(which I took a look at) but none of them help. Update: I figured it out. I went backwards, and integrated (1) instead of ##\frac{\partial F}{\partial y}dx##. Thank you.

Maybe that's where I'm stuck. what would be the integral out of it ? would it be ##\left[\frac{\partial F}{\partial y}x\right]_{x_0}^{x_1}## ? I don't think so.

how ? with the general formula and using ##\eta(x) = 1##, I ended up with: ##\delta J = \int_{x_0}^{x_1} (\frac{\partial F}{\partial y} - \frac{d}{dx}\frac{\partial F}{\partial y'}) dx + \frac{\partial F}{\partial y'}\Bigr|_{x_1} - \frac{\partial F}{\partial y'}\Bigr|_{x_0} ## (1) and with the...

@Orodruin If you mean, how I ended up with ##\int_{x_0}^{x_1} \frac{\partial F}{\partial y} dx##, then it's pretty simple. Assume ##y(x)## is a true path. Since we use transformation ##y(x) + \epsilon##, even in this case, since ##y(x)## is a true path, its action must still be the same as the...

@vanhees71 Take a look at this book (page 82 at the end). Note that at the beginning of page 83, author only makes ##\delta J## to be 0 on the hypothesis that actions ##J[y(x)]## and ##J[y(x) + \epsilon]## is invariant. So author replaces ##\delta x## and ##\delta y## by the transformation...

Yes, I understand this, btw, probably you should have said that ##\delta J = 0## instead of ##\delta J = 1##. I get that. I am just playing with it. If you hypothetically assume that ##\eta(x) = 1##, then ##\delta J = \int_{x_0}^{x_1} (\frac{\partial F}{\partial y} - \frac{d}{dx}\frac{\partial...

I had some several questions about variational calculus, but seems like I can't get an answer on math stackexchange. Takes huge time. Hopefully, this topic discussion can help me resolve some of the worries I have. Assume ##y(x)## is a true path and we do perturbation as ##y(x) + \epsilon...

I think you didn't get me. What I said was is this: we say ##I(0) = 0## because instead of ##\epsilon##, we insert 0 (even in the upper limit of the integral) and integral becomes from ##x_1## to ##x_1##. as for ##I'(0)##, this is basically saying: derivate ##I## with respect to ##\epsilon##...

I am not sure i understand it. It's mind boggling how ##I(0) = 0##, but ##I'(0) \neq 0##.

I think in my calculation, I was not making ##\epsilon## to be 0 in the upper limit of the integral which is what I meant by "what's the mistake in my calculation". I clearly see this was a mistake that I had. If you have ##I(\epsilon) = \int_{x_1}^{x_1 + \epsilon X_1} F(x, y + \epsilon \eta ...

Thanks for the answer @pasmith . 1. can you tell what was wrong in my analysis ? 2. can you tell why ##I(0) = 0## ? 3. I know that it's already an approximation, but assume that ##O(\epsilon^2)## is fully represented by the actual value. By approximation, I meant that when author makes it...