# I Partial derivative used in Calc of Variation

1. Sep 11, 2016

### ibkev

I'm working through the discussion of calculus of variations in Taylor's Classical Mechanics today. There's a step where partial differentiation is involved that I don't understand.

Given:
$$S(\alpha)=\int_{x_1}^{x_2} f(y+\alpha\eta, y'+\alpha\eta', x)\,dx$$

The goal is to determine $y(x)$ when the derivative of $S(\alpha)=0$ and ensure that $\alpha=0$ at this point.

So this means:
$$\frac {dS(\alpha)} {d\alpha}=\int_{x_1}^{x_2} \frac {\partial f} {\partial \alpha}\,dx=0$$

and now here is the step that I don't understand. It's apparently an application of the chain rule but I'm just not seeing it...
$$\frac {dS(\alpha)} {d\alpha}=\int_{x_1}^{x_2} \left( \eta\frac {\partial f} {\partial y}+\eta'\frac {\partial f} {\partial y'} \right)\,dx=0$$

Suggestions are appreciated! :)

Last edited: Sep 11, 2016
2. Sep 11, 2016

### PeroK

First, Taylor is being a bit lazy (although in a conventional way) and using $y$ and $y'$ to stand for two different things. In $\frac {\partial f} {\partial y}$, he's using $y$ as the first argument of the the function $f$ and, likewise, $y'$ as its second argument.

But, the arguments are not simply $y$ and $y'$ so technically some additional notation is required.

There's a previous post here for the same question:

3. Sep 11, 2016

### Twigg

The integrand f is a function of two variables $y$ and $y'$. The notation used is a little confusing because you end up re-using the same variables $y$ and $y'$ to describe the unperturbed trajectory and its unperturbed velocity.

To make it a little easier to see what's going on, you can change the notation a bit. Let $u_{1}$ and $u_{2}$ be coordinates for the position and velocity of a trajectory (any trajectory), and reserve the functions $y(x)$ and $y'(x)$ for a trajectory which makes the action integral S stationary. That makes f be a function of $u_{1}$ and $u_{2}$ in general, so it's written $f(u_{1},u_{2})$ in the new notation. That's how the chain rule comes into play. You want to find the trajectory $(u_{1},u_{2}) = (y,y')$. You get the expression for $S(\alpha)$ by substituting $(u_{1},u_{2}) = (y + \alpha \eta, y' + \alpha \eta')$. The final expression really says: $$\frac{dS(\alpha)}{d\alpha} = \int_{x_1}^{x_2} (\eta \frac{\partial f}{\partial u_{1}} + \eta' \frac{\partial f}{\partial u_{2}})dx$$

4. Sep 11, 2016

### ibkev

Ah ok - I'm starting to make sense of this now. Thanks!

The thing I find strange about how Taylor does this is that he actually sets up the problem at first similar to the $(u_1, u_2)$ in Twigg's post but calls them $(Y, Y')$ and then deliberately drops them.

He writes:
\begin{align} S(\alpha)&=\int_{x_1}^{x_2} f(Y, Y', x),dx \\ &=\int_{x_1}^{x_2} f(y+\alpha\eta, y'+\alpha\eta', x),dx \end{align}
But then he next writes:
$$\frac {\partial} {\partial\alpha} f(y+\alpha\eta, y'+\alpha\eta', x) =\eta\frac {\partial f} {\partial y} + \eta'\frac {\partial f} {\partial y'}$$
Do I understand right that he actually means for that previous equation to say:
$$\frac {\partial} {\partial\alpha} f(Y, Y' x)=\eta\frac {\partial f} {\partial Y} + \eta'\frac {\partial f} {\partial Y'}$$

>Taylor is being a bit lazy (although in a conventional way)
Honestly it seems like it should be an errata, no?

Last edited: Sep 11, 2016
5. Sep 11, 2016

### Twigg

Yeah, this is what he means, but it's also implied that $Y = y + \alpha \eta$ inside the argument of the function f. The variables $Y$ and $Y'$ represent coordinates in a geometric space whose axes are the position and velocity of the trajectory, so the substitution $(Y,Y') = (y + \alpha \eta, y' + \alpha \eta')$ is just a way of parametrizing that space the same way we could parametrize ordinary 3D space with $\vec{r} = \vec{r_{0}} + \alpha \vec{u}$ where $\vec{u}$ is an undetermined unit vector. In that picture, the expression you mentioned and I quoted above is just the directional derivative of f along the "direction" $(\eta, \eta')$ in the $Y$-$Y'$ plane.

I think Taylor intentionally dropped the Y,Y' notation because he probably didn't want to detract from the important points of this optimization procedure by making the theoretical distinction between y and Y, at the expense of confusing more cautious readers. While his notation is mathematically misleading, it still works and is a convenient shorthand.

6. Sep 12, 2016

### ibkev

That's a way of looking at it that I hadn't considered - thanks.

Incidentally, I checked how the Morin Classical Mechanics text does it and he carries the Y, Y' through to the very end (although he calls them $(y_\alpha, y_\alpha')$ and only then does he replace them with y, y' (albeit magically and without any explanation.)

After thinking about this, I came up with another way to look at this from the fact that there are TWO goals of our solution not one:
1. solve $\frac {dS(\alpha)} {d\alpha}=0$ to find the local minimum
2. require that $\alpha=0$ at that point as well
So the final step of the derivation recognizes that although we defined $(Y,Y') = (y + \alpha \eta, y' + \alpha \eta')$, setting $\alpha=0$ at the end as our problem requires, results in $(Y,Y')=(y,y')$

Does that seem like a reasonable way to look at it as well?

Last edited: Sep 12, 2016
7. Sep 12, 2016

### Twigg

You are correct, and if you hadn't mentioned it I would've missed a second oversight in Taylor's math. Hamilton's principle does not say $$\frac{dS_{\alpha}}{d\alpha} = 0$$. That would would mean that all functions of the form $y + \alpha \eta$ where $y$ is a trajectory that makes the action integral stationary, are all stationary trajectories. What Hamilton's principle really says is this: $$\frac{dS_{\alpha}}{d\alpha}|_{\alpha = 0} = 0$$. In other words, it says that $y$ is the stationary trajectory, not just any trajectory of the form $y + \alpha \eta$.