- #1

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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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- Thread starter cooev769
- Start date

- #1

- 114

- 0

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

- #2

Science Advisor

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- 1,591

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 [itex] f [/itex] ?

In the usual introduction to the calculus of variations, the problem can be stated as:

Minimize [itex] G(\alpha) = \int_a^b f(y,y',x) dx [/itex] where [itex] y = y(x,\alpha) [/itex] is a function of [itex] x [/itex] and [itex] \alpha [/itex].

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

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

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