Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $y \in C^2[0,1]$ ?

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

share|cite|improve this answer
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… – Ashot Mar 19 '13 at 18:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.