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?


Let $I$ denote the integral.

  1. 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, $$f(x) - 2x= \int_0^x \int_0^s f''(t) dt ds,$$


$$ f(x) -2x = \int_0^x \left( f''(t) \int_{t}^x s ds\right) dt=\int_0^x f''(t) (x-t)dt.$$

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

$I \ge 12$.

  1. 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, $$\int_0^1 f''(t)^2 dt = \int_0^1 36 (t-1)^2 dt = 36 \int_0^1 t^2 dt = 12.$$

  1. 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 $$ \delta I[f]~=~ \int_0^1\! dx~ f^{\prime\prime}(x) ~\delta f^{\prime\prime}(x)$$ $$~\stackrel{\text{int. by parts}}{=}~ f^{\prime\prime}(1)~\delta f^{\prime}(1) -f^{\prime\prime}(0)~\delta \underbrace{f^{\prime}(0)}_{=0} -f^{\prime\prime\prime}(1)~\delta \underbrace{f(1)}_{=0} +f^{\prime\prime\prime}(0)~\delta \underbrace{f(0)}_{=0} $$ $$ +\int_0^1\! dx~ f^{\prime\prime\prime\prime}(x) ~\delta f(x).\tag{2}$$

  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 $$ f^{\prime\prime}(1)~=~0\quad\text{and}\quad\forall x\in [0,1]:f^{\prime\prime\prime\prime}(x)~=~0.\tag{4}$$

  3. 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 $$ f(x) ~=~x(x-1)(x-2). \tag{6}$$

  4. 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.


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