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I'm reading Simmons's Differential Equations and find that I have to obtain Euler-Lagrange equations of the following integrand (integrand of the action functional):

$L = \frac 1 2 m \Big\lbrace \Big(\frac{dx}{dt}\Big)^2+\Big(\frac{dy}{dt}^2\Big)+\Big(\frac{dz}{dt}\Big)^2 \Big\rbrace - V(x,y,z)$

the text claims that the Euler-Lagrange equations are:

$m\Big(\frac{d^2x}{dt^2}\Big)+\frac{\partial V}{dx} = 0$

$m\Big(\frac{d^2y}{dt^2}\Big)+\frac{\partial V}{dy} = 0$

$m\Big(\frac{d^2z}{dt^2}\Big)+\frac{\partial V}{dz} = 0$

I'm trying to see how this holds from Wikipedia's list of Euler-Lagrange equations. I think my case corresponds to 'Several functions of single variable with single derivative'.

But there I find this notation $\frac{\partial L}{\partial f_i} - \frac{d}{dx}\frac{\partial L}{\partial f_i'}$ where $f_i' = \frac{df_i}{dx}$. The correspondence with my example is $x \to t$ and $f_1 \to x,f_2 \to y,f_3 \to z$.

What does exactly, $\frac{\partial L}{\partial f_i}$ and $\frac{\partial L}{\partial f_i'}$ so that I can obtain the above equations?

user1868607
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    You need to make a distinction between $\partial$ ($\TeX$-command \partial) and $d$ in your derivatives. Especially in the Euler Lagrange equations, where at least in the general form, we use both simultaneously. – Arthur Mar 31 '18 at 19:29
  • @Arthur i corrected it, so what is the difference between the two? – user1868607 Mar 31 '18 at 19:32
  • When using $\partial$, you treat all the variables as independent, and differentiate with respect to the one you're supposed to differentiate with respect to. With $d$ (which I think it's only used with $\frac{d}{dt}$ if you read it carefully), it means to consider every single variable as the function of $t$ that it is, and then differentiate the result with respect to $t$ (using chain rule where necessary). As an example, take $f(t,x,x')=t+x^2+x'$. Then $\frac{\partial f}{\partial x}=2x$, while $\frac{df}{dt}=1+2x'+x''$. – Arthur Mar 31 '18 at 19:44
  • (Note: in some of the wikipedia examples, it is $x$, not $t$, which is the "free" variable, and every other variable is a function of $x$. In those cases, $\frac{d}{dx}$ is used instead.) – Arthur Mar 31 '18 at 19:47

1 Answers1

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Thanks to @Arthur's comment I came to these computations:

$\frac{\partial L}{\partial x'} = \frac{d}{dt}\Big(m \frac{dx}{dt} \Big) = m \frac{d^2x}{dt^2}$

$\frac{\partial L}{\partial x} = - \frac{\partial V}{\partial x}$

Thus,

$\frac{\partial L}{\partial x} - \frac{\partial L}{\partial x'} = - \frac{\partial V}{\partial x} - m \frac{d^2x}{dt^2} = 0$

or equivalently, the equation:

$m\Big(\frac{d^2x}{dt^2}\Big)+\frac{\partial V}{dx} = 0$

For the difference between $\partial$ and $d$ I would refer to this.

user1868607
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