I Basic question about variational calculus

Kashmir
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We've a functional

##J(\alpha)=\int_{x_{1}}^{x_{2}} f\left\{y(\alpha, x), y^{\prime}(\alpha, x) ; x\right\} d x##

It's derivative with respect to the parameter ##\alpha## is given in textbook Thornton Marion as ##\frac{\partial J}{\partial \alpha}##

Shouldn't it have been ##\frac{d J}{d \alpha}## ?
 
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Kashmir said:
as

Shouldn't it have been ?
for a function of a single variable that is the same
 
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wrobel said:
for a function of a single variable that is the same
After the Integral, the function is of only one variable ##\alpha##
 
Kashmir said:
After the Integral, the function is of only one variable ##\alpha##
If I remember correctly the derivation of the Euler-Lagrange equation requires you to differentiate inside the integral and that’s why it’s still a partial derivative imo.
 
Kashmir said:
After the Integral, the function is of only one variable ##\alpha##

I realize my last answer was sort of superficial. I think I have a better one now.

the functional integrand is a function f(y,y’,x) when you integrate with respect to x it’s still a function of y and y’. Remember y and y’ are not fixed (you’re trying to optimize J with the correct y) so J is still a function of y and y’ after integration. I think this is why you have a partial derivative with respect to ##\alpha## instead of a full derivative.
 
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Kashmir said:
Shouldn't it have been ##\frac{d J}{d \alpha}## ?

Neither ##\frac{dJ}{d \alpha}## nor ##\frac{\partial J}{\partial\alpha}## are correct notation if you intend those notations to denote what they do in elementary calculus. In elementary calculus, those notation indicate derivatives that are real valued functions of a real variables. By contrast, a point in the domain of a functional is defined both by any parameter ##\alpha## involved and also by the function where the functional is evaluated. As @PhDeezNutz points out, the derivative of a functional has to be a mapping of the form: (function, parameter) -> number.

For example, if you consider the task of finding "where" the derivative of a functional is zero, you typically have a different job that finding where the derivative of a real valued function is zero. The task for where the derivative of a functional is zero can involve finding conditions on a function. Those conditions can involve the global behavior of the function, not simply the value of the function evaluated at one point in its domain.I notice the article https://en.wikipedia.org/wiki/Functional_derivative prefers notation like ##\frac{ \delta J}{\delta \alpha}## for the derivative of a functional with respect to a parameter.
 
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Stephen Tashi said:
those notation indicate derivatives that are real valued functions of a real variables
yes, and that is exactly the case under consideration.
 
PhDeezNutz said:
I realize my last answer was sort of superficial. I think I have a better one now.

the functional integrand is a function f(y,y’,x) when you integrate with respect to x it’s still a function of y and y’. Remember y and y’ are not fixed (you’re trying to optimize J with the correct y) so J is still a function of y and y’ after integration. I think this is why you have a partial derivative with respect to ##\alpha## instead of a full derivative.
Then isn't it somewhat lazy for the authors to write ##J=J(\alpha) ## ?
 
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Kashmir said:
Then isn't it somewhat lazy for the authors to write ##J=J(\alpha) ## ?
I think so as well but I’m no mathematician.
 

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