Deriving Calculus of Variations

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The discussion centers on a user's confusion regarding the derivation of the calculus of variations, specifically the application of the chain rule. The user derives a formula for the derivative with respect to alpha but finds it differs from the textbook's version, leading to uncertainty about the correct notation and interpretation. Clarification is provided that the function f is defined as an integral over a range, and that it depends on an entire interval of values for Y, not just a single value. The user is informed that the notation used in their derivation may be incorrect, as the partial derivative with respect to Y does not hold meaning in this context. Understanding the relationship between Y and f is crucial for correctly applying the calculus of variations.
cooev769
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Hey I'm having an issue deriving the calculus of variations because the chain rule i use ends up different to the one in the textbook. Firstly I assume we have some function of 3 variables Y=y+alpha eta with grad Y'=y'+alpha eta' and x. Secondly we have an integral of this function over x and want to minimise it, hence we want to differentiate with respect to alpha, and hence we need to use the chain rule. For me I end up with, and these d's are partial derivatives not normal:

df/dalpha = df/dY * eta + df/dY' eta'

But the textbook says the answer is with small y's instead of big and the rest of the class says that the book is correct and can't explain to me why, can anybody please enlighten me?

Thanks
 
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cooev769 said:
Secondly we have an integral of this function over x
and want to minimise it, hence we want to differentiate with respect to alpha, and hence we need to use the chain rule. For me I end up with, and these d's are partial derivatives not normal:
df/dalpha = df/dY * eta + df/dY' eta'

What is f ?

In the usual introduction to the calculus of variations, the problem can be stated as:
Minimize G(\alpha) = \int_a^b f(y,y',x) dx where y = y(x,\alpha) is a function of x and \alpha.

Are you are using the notation f(\alpha) = \int_a^b Y(y,y',x) dx?

if so, f is not a function of Y. A single value for Y does no determine a value for f. The function f is determined by an entire interval of vales for Y. So the symbol \frac{\partial f}{\partial Y} doesn't mean anything.
 
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