How to find the minimum value of following integral? Let $A$ be the set of twice differentiable functions on the interval $[0,1]$, and
$$B=\{f \varepsilon A: f(0)=f(1)=0, f'(0)=2\}.$$ 
What is 
$${\rm Min}_{f\varepsilon B}\int_{0}^{1} (f''(x))^2 dx~?$$
Now I apply Euler-Lagrange equation to evaluate this but for that I need $4$ boundary conditions?
How to evaluate this?
 A: Let $I$ denote the integral. 


*

*A lower bound. 


$$f(x) = \int_0^x f'(s) ds = \int_0^x \left(2 + \int_0^s f''(t)  dt\right)ds.$$
Thus, $$f(x) - 2x= \int_0^x \int_0^s f''(t) dt ds,$$ 
or 
$$ f(x) -2x = \int_0^x \left( f''(t) \int_{t}^x s ds\right) dt=\int_0^x f''(t) (x-t)dt.$$ 
Apply Cauchy-Schwarz to the right-hand side obtain  
$$(*)\quad  |f(x)-2x| \le \left (\int_0^x f''(t)^2 dt \right)^{1/2} \left(\int_0^x (x-t)^2 dt\right)^{1/2}= \sqrt{I} \times \frac{x^{3/2}}{\sqrt{3}}.$$
In particular, letting $x=1$, we have that 
$I \ge 12$.


*A function attaining the lower bound. 


We do this by choosing a function for which Cauchy-Schwarz $(*)$  is an equality for $x=1$. That is, we need  $f''$ equal to a constant multiple of $1-t$. From this it follows that $f$ must be of the form 
$$f(t)= c_1(1-t)^3 + c_2 t+ c_3.$$
The constraints determine the constants, and once this is settled, we automatically know that for our chosen $f$, the inequality in $(*)$ is an equality for $x=1$,  and therefore the bound obtained in 1. is attained by $f$. 
Let's find the constans.  Setting $f(0)=0$, we have that $c_1+c_3=0$. Setting $f(1)=0$, we have $c_2+c_3=0$. Therefore $f(t) = c_1 (1-t)^3+c_1t - c_1$. Now $f'(0)= -3 c_1+c_1= -2c_1$. Therefore $c_1=-1$. We have found that the function
$$f(t)=(t-1)^3 -(t-1)=(t-1)((t-1)^2 - 1) = (t-1)t(t-2).$$
satisfies the constraints.
The next step is redundant, but let's include it for the sake of completeness: $f''(t) = 6(t-1)$. Thus, 
$$\int_0^1 f''(t)^2 dt = \int_0^1 36 (t-1)^2 dt = 36 \int_0^1 t^2 dt = 12.$$   
A: *

*Assume first that $f\in C^4([0,1])$. If one varies infinitesimally the functional 
$$I[f]~:=~ \frac{1}{2}\int_0^1\! dx~ f^{\prime\prime}(x)^2\tag{1}$$ 
without discarding boundary contributions, one finds
$$ \delta I[f]~=~ \int_0^1\! dx~ f^{\prime\prime}(x) ~\delta f^{\prime\prime}(x)$$
$$~\stackrel{\text{int. by parts}}{=}~  f^{\prime\prime}(1)~\delta f^{\prime}(1) -f^{\prime\prime}(0)~\delta \underbrace{f^{\prime}(0)}_{=0}
-f^{\prime\prime\prime}(1)~\delta \underbrace{f(1)}_{=0}  
+f^{\prime\prime\prime}(0)~\delta \underbrace{f(0)}_{=0}   
$$ $$ +\int_0^1\! dx~ f^{\prime\prime\prime\prime}(x) ~\delta f(x).\tag{2}$$

*Besides the given boundary conditions 
$$f(0)~=~0~=f(1)\quad\text{and}\quad f^{\prime}(0)~=~2,\tag{3}$$ 
one concludes from formula (2) that a stationary configuration must obey 
$$ f^{\prime\prime}(1)~=~0\quad\text{and}\quad\forall x\in [0,1]:f^{\prime\prime\prime\prime}(x)~=~0.\tag{4}$$

*Eqs. (3) & (4) imply that the stationary solution is a third-order polynomial of the form
$$ f(x) ~=~ x(x-1)(Ax+B) .\tag{5}$$
The condition $f^{\prime}(0)=2$ implies $B=-2$.
The condition $f^{\prime\prime}(1)=0$ then leads to
$$ f(x) ~=~x(x-1)(x-2). \tag{6}$$

*To confirm that the found stationary solution (6) is a minimum configuration, write an arbitrary function $f\in C^2([0,1])$ as 
$$f(x) ~=~x(x-1)(x-2)+g(x),\tag{7} $$
where $g\in C^2([0,1])$. Then 
$$f^{\prime\prime}(x) ~=~6(x-1)+g^{\prime\prime}(x),\tag{8} $$
and
$$ \frac{1}{6}I[f] 
- \underbrace{\frac{1}{12}\int_0^1\! dx~ g^{\prime\prime}(x)^2}_{\geq 0} 
- \underbrace{\frac{1}{12}\int_0^1\! dx~36(x-1)^2}_{=1} 
~\stackrel{(1)+(8)}{=}~ \int_0^1\! dx~(x-1)g^{\prime\prime}(x)$$ 
$$~\stackrel{\text{int. by parts}}{=}~ 
- \underbrace{g^{\prime\prime}(0)}_{=0} 
- \underbrace{g^{\prime}(1)}_{=0} 
+ \underbrace{g^{\prime}(0)}_{=0} ~=~0.\tag{9} $$
End of proof.
