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Given $\int_0^1(y')^3dx$ functional and $y(0) = 0 ,y(1)=1$ conditions. Using Euler–Lagrange equation I have got $y(x)=x$. So $y$ is a stationary point of the functional. How to check if it is the minimum for $y \in C^2[0,1]$ ?

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Compute a second variation, see if its positive definite at $y(x) = x$. –  muzzlator Mar 17 '13 at 14:35

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up vote 2 down vote accepted

Consider the function $$ y(t) = \begin{cases} -\lambda t & \text{ for $t <1/\lambda$} \\\\ \frac{2}{\lambda-1} (\lambda t-1) -1 & \text{for $t\ge 1/\lambda$}. \end{cases} $$ You have $$ \int_0^1 (y')^3\, dx = -\lambda^2 + (1-1/\lambda)\left(\frac{2\lambda}{\lambda-1}\right)^3 \to -\infty \qquad \lambda\to +\infty $$ hence your functional does not have an absolute minimum.

It is possible to smooth out the function $y(t)$ to get a $C^\infty$ function with the same property.

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Thanks, what about $y \in C^2[0,1]$. I think $y=x$ is minimum there –  Ashot Mar 17 '13 at 15:46
    
No, you can find a counterexample with a $C^2$ function (just smooth out the angle point) –  Emanuele Paolini Mar 17 '13 at 15:54
    
Is there a option to check whether a stationary point is extermum ? –  Ashot Mar 19 '13 at 13:50
    
If your functional is strictly convex you know that the stationary point is a minimum. If the second variation is positive you know that the point is a local minimum. –  Emanuele Paolini Mar 19 '13 at 18:04
    
Isn't the second derivative positive for the above example?. Please see this question math.stackexchange.com/questions/334546/… –  Ashot Mar 19 '13 at 18:07

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