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I am having a hard time getting my head around Functionals and Calculus of Variations,

My question is: Given a functional and using the Euler-Lagrange equation to find an extremal, how do we show that the extremal provides a min/max (if it does)

The question I am working on is

$J(y) = \int_{0}^{1} ((y')^2 -y)dx$ with $y(0)=0, y(1)=1$

I found the extremal to be: $y(x) = \frac{-1}{4}x^2 +\frac{5}{4}x$ which I am told is a minimum to the functional problem.

However I am unsure on what is sufficient to show this, in the notes I have it is shown that:

$J(y+f) = J(y) + \int_{0}^{1}(f')^2dx \geq J(y)$ where f is continuously differentiable on the interval 0,1 with $y(0)=y(1)=0$

Thanks in advance!

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1 Answer 1

Find the necessary condition for a sufficiently smooth curve $y (x)$ to be an extremal of the functional $$ J(y)=\int_a^b L((x,y(x),y'(x),y''(x))dx $$ With boundary conditions $y(a)=y_0$, $y(b)=y_1$, $y'(a)=y'_0$, $y'(b)=y'_1$

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how is this an answer? I don't get it.. – Franco Mar 14 at 22:28

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