# Confusion with Euler-Lagrange Derivation

This is mostly a re-hash of this thread, but it did not receive an adequate answer. In the derivation that I am reading, there is one step that is not justified. Perhaps obvious, but it is not clear to me. Here is the build-up:

Suppose we have the following (one-dimensional) integral that we want to minimize by finding a suitable function $Y(x)$: $\int_{x_1}^{x_2} f(Y(x),Y'(x),x) dx$. Assume we already know the function that minimizes this to be $y(x)$. Define $Y(x,\alpha) = y(x)+\alpha \eta(x)$ and require $\eta(x)$ to be a differentiable function with $\eta(x_1) = \eta(x_2) = 0$ so that this is a suitable path between the points of interests, and assume $\alpha$ is real. From this, define

$$S(\alpha) = \int_{x_1}^{x_2} f(Y(x,\alpha),Y'(x,\alpha),x) dx$$

It is clear that the minimum path is simply $Y(x,0)$, which implies that $\frac{\partial S}{\partial \alpha}\Big|_{\alpha = 0} = 0$. Now, here is where my confusion is:

With no dependence on $\alpha$ for the bounds, we know that

$$\frac{\partial S}{\partial \alpha}\Big|_{\alpha = 0} = \Big[\int_{x_1}^{x_2} \frac{\partial f}{\partial \alpha} dx \Big]_{\alpha = 0} = \Big[\int_{x_1}^{x_2} \Big(\eta(x)\frac{\partial f}{\partial Y}+\eta'(x)\frac{\partial f}{\partial Y'}\Big) dx\Big]_{\alpha = 0} = 0$$

and my author, without justification, immediately makes the jump to

$$\Big[\int_{x_1}^{x_2} \Big(\eta(x)\frac{\partial f}{\partial Y}+\eta'(x)\frac{\partial f}{\partial Y'}\Big) dx\Big]_{\alpha = 0} = \int_{x_1}^{x_2} \Big(\eta(x)\frac{\partial f}{\partial y}+\eta'(x)\frac{\partial f}{\partial y'}\Big) dx$$

Perhaps in clearer form, the following step is taken:

$$\frac{\partial f(y+\alpha \eta,y'+\alpha \eta',x)}{\partial \alpha} = \eta \frac{\partial f}{\partial y} + \eta'\frac{\partial f}{\partial y'}$$

Can someone provide justification for this step? My intuition tells me that

$$\frac{\partial f}{\partial Y(x,\alpha)}\Big|_{\alpha = 0} = \frac{\partial f}{\partial Y(x,0)} = \frac{\partial f}{\partial y(x)}$$

but some essential part of that just feels wrong. It reminds me of the silly mistake that a careless calculus student can make, saying that

$$\frac{\partial f(x,y)}{\partial x}\Big|_{(x,y) = (x_o,y_o)} = \frac{\partial f(x_o,y_o)}{\partial x}$$

Which is clearly incorrect by noting that $f(x_o,y_o)$ is simply some constant. In the above case I think the issue is much subtler.

• Your problem is the part, where the $\partial Y$ and $\partial Y'$ are replaced by $\partial y$ and $\partial y'$? The line above can be taken as known? – Thomas Sep 3 '14 at 6:24
• That is the chain rule. – James S. Cook Sep 3 '14 at 6:32
• Indeed, take the line above as known, as you can work out by bringing in the partial derivative and applying the chain rule. – dsm Sep 3 '14 at 6:33
• the intuitive step is wrong because $f$ is a function of both $Y$ and $Y'$ which are technically independent variables here. Moreover, both depend on $\alpha$ so ignoring the $Y'$ does matter. – James S. Cook Sep 3 '14 at 6:47

$$\frac{\partial f(y+\alpha \eta,y'+\alpha \eta',x)}{\partial \alpha} = \eta \frac{\partial f}{\partial y} + \eta'\frac{\partial f}{\partial y'}$$ Why? Well, if $f$ is a function with arguments $f(Y,Y',x)$ and $Y,Y',x$ are thought of as functions of $\alpha$ then the chain-rule is: $$\frac{\partial}{\partial \alpha}f(Y,Y',x) = \frac{\partial f}{\partial Y}\frac{\partial Y}{\partial \alpha}+ \frac{\partial f}{\partial Y'}\frac{\partial Y'}{\partial \alpha}+ \frac{\partial f}{\partial x}\frac{\partial x}{\partial \alpha}.$$ But, for you variation about the optimal path $y$, $$\frac{\partial Y}{\partial \alpha} = \eta \qquad \& \qquad \frac{\partial Y'}{\partial \alpha} = \eta'$$ whereas $\frac{\partial x}{\partial \alpha}=0$. Finally, setting $\alpha =0$ changes the evaluation of those partials from arbitrary $Y$ to $y$ and writing the partial derivative w.r.t. $y$ merely expresses that evaluation at the optimal path.
• Thanks for the response. Yes, I understand how the expression with $Y$ is generated with the chain rule. It's the switch from $\partial Y$ to $\partial y$ in the denominator that I wanted justification for (your very last comment). Sure, it may seem obvious that along the optimal path that $\partial Y$ = $\partial y$ since we are holding $\alpha$ constant at zero. For me, though, this lacks rigor, and kind of masks the somewhat clear calculus that I imagine goes into it's justification. – dsm Sep 3 '14 at 14:28
• This is a standard notation, we write $df/du$ to denote $f'(u)$. To be explicit, we could write $(df/dx)(u)$, but $df/du$ compactly expresses the derivative evaluated at $u$. It is the same with $Y$ verses $y$. – James S. Cook Sep 3 '14 at 19:54