Basic question about variational calculus

[Kashmir]

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 $\frac{d J}{d \alpha}$ ?

 

[wrobel]

as

Shouldn't it have been ?

for a function of a single variable that is the same

 

[Kashmir]

for a function of a single variable that is the same

After the Integral, the function is of only one variable $\alpha$

 

[PhDeezNutz]

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.

 

[PhDeezNutz]

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.

 

[Stephen Tashi]

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.

 

[wrobel]

those notation indicate derivatives that are real valued functions of a real variables

yes, and that is exactly the case under consideration.

 

[Kashmir]

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 $J=J(\alpha)$ ?

 

[PhDeezNutz]

Then isn't it somewhat lazy for the authors to write $J=J(\alpha)$ ?

I think so as well but I’m no mathematician.