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In this application of the Euler-Lagrange equation, it is said that there is no $y$ in the function $\sqrt{1 + (y')^2}$. I see that the algorithm in progress treats $y'$ as unusually autonomous, as in the expression $\frac{\partial F}{\partial y'}$ but it still seems that it should count as an appearance of $y$ and also contribute to $\frac{\partial F}{\partial y}$. After all, an appearance of $\frac{1}{y}$, a binary operator with $y$ as an operand, would contribute to $\frac{\partial F}{\partial y}$. Why shouldn't a unary operator operating on $y$ contribute to $\frac{\partial F}{\partial y}$ (making $\frac{\partial F}{\partial y}$ nonzero in this case)?

To flesh this out, how should $\frac{\partial F}{\partial y}$ of a function such as $\sqrt{1 + (y^2)'}$ be interpreted?

user10478
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  • Possible duplicates: https://math.stackexchange.com/q/580858/11127 , https://physics.stackexchange.com/q/885/2451 and links therein. – Qmechanic Apr 11 '19 at 18:01

2 Answers2

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See $\mathcal{L}:\mathbb{R}\times\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ as a function with three arguments, say $t,u,p$ and $\mathcal{L}:(t,u,p)\mapsto\mathcal{L}(t,u,p)$. So when we write $\partial \mathcal{L}/\partial y$ we actually mean $\mathcal{L}$'s partial derivative with respect to its second argument, $u$. Note that this has nothing to do with what the other arguments are, we are simply saying the partial derivative of $\mathcal{L}$ with respect to its second argument.

I agree that the notation you have mentioned is a little messy and I have seen people using $\partial \mathcal{L}/\partial u$, $\partial \mathcal{L}/\partial p$ (e.g. Jurgen Jost's book on Calculus of Variations) to denote partial derivatives with respect to the second and third arguments of the Lagrangian.

Semih Kara
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  • But even renamed, the relation between $u$ and $p$ would still hold, would it not? I assume this bears some connection to the Lagrangian of constrained optimization (https://youtu.be/m-G3K2GPmEQ?t=470), except the dependencies are completely different. Interestingly, $\lambda *$ turns out to be a derivative, but not with respect to another argument of the Lagrangian. – user10478 Apr 12 '19 at 16:09
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As an answer to your comment (which I couldn't fit it as a comment to my former answer):

What I was trying say by introducing $u$ and $p$ was that they should be regarded as independent. So even though we write partial derivative with respect to $y$ and $y'$ what we mean is actually the derivative with respect to the second and third arguments, where these arguments are regarded as independent. Partial derivative of the Lagrangian with respect to $y$ is the usual partial derivative with respect to some argument so it just has to do with the form of the Lagrangian (i.e. how $y$ appears in it) and nothing else.

To illustrate my point consider the following. Let $\mathcal{L}(x,y(x),y'(x)) = x^2 + y^2(x) +y'^2(x)$. To evaluate $\partial\mathcal{L}/\partial y$ you can do the following. Just replace $y$ with $u$, and $y'$ with $p$ and regard $u$ and $p$ as independent. Then $\mathcal{L}(x,u,p) = x^2 + u^2 + p^2$ and $\partial\mathcal{L}/\partial u = 2u$. So what you have in terms of the original variables is $\partial\mathcal{L}/\partial y = 2y$.

This is just what we mean by the partial derivative of $\mathcal{L}$ with respect to $y$ or $y'$, it is simply the partial derivatives of $\mathcal{L}$ with respect to its second and third arguments, but it is chosen to be written this way.

Semih Kara
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