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I am learning calculus of variations and met the following example:

Find the extremal of the functional v[y]= ∫ (12ty+(y’)^2)dt, where y=y(t)

let F= 12ty+(y’)^2, using the Euler rule, we need to calculate the second order derivative of F with respect first to y' and then to t. It is clear that the derivative of F with respect to y' is 2y', why when derive 2y' with respect to t, the answer is 0, rather than 2y'' (since y' should also be a function of t). Some may say that here we treat y as a variable as t, but since y is a function of t, when we change t, won't if affect y'' and thus F ?

thisisjayliu
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I think you're confusing the partial derivative $\frac{\partial}{\partial t}$ with the total derivative $\frac{d}{dt}$. You would have $\frac{\partial}{\partial t} 2y' = 0$ and $\frac{d}{dt} 2y' = 2y''$.

The partial derivative only takes into account variables that directly appear in the equation, but not variables on which these quantities depend.

You might find this link helpful: https://en.wikipedia.org/wiki/Partial_derivative

johnny
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  • Thank you! I think you are right...I will take a closer look at the definition of partial derivative. Thanks a lot for your help ! – thisisjayliu Feb 10 '20 at 02:53