min $\int_0^1 (f''(x))^2 \, dx$

Question

Let A be the set of twice continuously differentiable functions on the interval $[0, 1]$,

and let $B = \{f\in A : f (0) = f (1) = 0, f ′ (0) = 2\}$. What is

min $\int_0^1 (f''(x))^2$.

There is one given hint consider $(1-x)f''(x)$.

Any idea how to start solving this problem?

This was already asked the other day.. Try looking for it in the search bar. — Matthew Cassell, May 03 '15 at 14:15
@Mattos Another option is for you to close the question as a duplicate; that is preferable. — Emily, May 03 '15 at 14:48
@Arkamis I considered it, but I couldn't find the previous link and hence thought the OP might not be able to find it, in which case the OP wouldn't have got any help. So I decided against it. — Matthew Cassell, May 03 '15 at 15:01
@Arkamis, why do you want to close the question? what about all the time i put into answering? — abel, May 03 '15 at 16:22
@abel For the same reason duplicates should always be closed. You can always copy the answer to the basis question. — Emily, May 03 '15 at 17:15

score 6 · Accepted Answer · edited Oct 10 '15 at 15:34

6

we have $$\begin{align}\int_0^1(1-t)f''(t) \, dt &= {(1-t)f'(t)}\Big|_0^1+\int_0^1f'(t)dt\\ &=2 +f(1) - f(0) = 2 \end{align}$$ now by Cauchy-Schwarz, $$\begin{align}\Big|\int_0^1(1-t)f''(t) \, dt\Big|^2 &\le \int_0^1(1-t)^2 \, dt \int_0^1(f''(t))^2 \, dt \\ 4 &\le \frac 13 \int_0^1(f''(t))^2 \, dt\end{align}$$

therefore $$\int_0^1\left(f''(t)\right)^2 \, dt \ge 12. $$

$\bf p.s.$ the minimum is achieved for $$f'' = 6(t-1), f' = 3(t-1)^2 - 1, f=(t-1)^3-(t-1) $$

edited Oct 10 '15 at 15:34

user26857

53,190

answered May 03 '15 at 14:26

abel

29,612

So minimum is 12 right? – Sankha May 05 '15 at 19:41
1

@radha, yes. $12$ is the minimum. – abel May 05 '15 at 20:02
What is the idea behind the solution? – Henry Sep 22 '24 at 07:57

min $\int_0^1 (f''(x))^2 \, dx$

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