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The functional $$\int_{0}^{1}(1+x)(y')^{2}dx,y(0)=0,y(1)=1$$ Possesses

$1.$ Strong maxima.

$2.$ Strong minima.

$3.$ Weak maxima but not a strong maxima.

$4.$ Weak minima but not a strong minima.

I tried it as $F=(1+x)(y')^{2}$ so $F_{y'y'}=2(1+x)>0$ for any $y,y',$ so strong minima by Legendr's conditions. Am i right? Please suggest me. Thanks a lot.

neelkanth
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  • To understand how to answer this question, you need to understand how such a functional is determined to be an extrema, which is the Euler-Lagrange equation. It is determined that, for $\mathcal{L}(y + \eta \lambda, \dot{y} + \eta \dot{\lambda}, t)$, $\frac{d\mathcal{L}}{d\eta} = 0$, that $y$ is an extrema of the Laplacian. The question now is: if $\frac{d\mathcal{L}}{d\eta} = 0$ then what is $\frac{d^2\mathcal{L}}{d\eta^2}$. – Jared Jun 02 '16 at 03:13

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