3
$\begingroup$

I was trying to solve a question of an entrance exam. I have taken help for a similar problem from MSE but again I am stuck in the problem. Please help me.

Let $ D= \{f \in [0,1] : f \text{continuous and} \displaystyle \int_0^1f(x)dx = 1\}$. Find the value of $$\text{min}_{f\in D} \displaystyle \int_0^1 (1+x^2)f^2(x)dx$$

I have tried the same eay as described in this problem but I failed to apply Euler-Lagrange equation.I can not find any other way to proceed. Please help me. Thnx in advance.

$\endgroup$
  • $\begingroup$ Just a question does, $f^2(x)$ means second derivative $f''(x)$? $\endgroup$ – Mann May 7 '15 at 14:06
  • $\begingroup$ No it is the square of $f$ $\endgroup$ – usermath May 7 '15 at 14:13
  • $\begingroup$ Uh, thanks I guess this problem beyond high school anyway :), well tried anyway. Found hints from other answer however , seems interesting. $\endgroup$ – Mann May 7 '15 at 14:15
3
$\begingroup$

Here's a way:

Write $F(x) = \int_0^x f(t)dt$.

$D$ now corresponds to those $F$ for which $F \in \mathcal C^1$, $F(0)=0$ and $F(1)=1$. And you're looking for $\min_F \int_0^1(1+x^2)(F')^2(x)dx$.

Your Lagrangian can be $\mathcal L = (1+x^2)(y')^2$. Can you take it from there?

$\endgroup$
  • $\begingroup$ Thank you very much sir, I will try now. I think now I can do it. $\endgroup$ – usermath May 7 '15 at 14:04
  • $\begingroup$ FYI, I get $f(x) = \frac4{\pi}\arctan(x)$ if that helps $\endgroup$ – Alexandre Halm May 7 '15 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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