How can i find a solution of this optimization problem?

Question

I need to give a solution for this variational problem $$\min_{x\in C^2[0,1]}\int_0^1\left(1+|x''(t)|^2\right)dt$$ $$x(0)=0\hspace{0.2cm},\hspace{0.2cm} x(1)=1\hspace{0.2cm},\hspace{0.2cm}x'(0)=1\hspace{0.2cm},\hspace{0.2cm} x'(1)=1$$ I know I have to use Euler-Lagrange equation but I'm having a problem with the second order derivative of $x$. Can someone enlighten me plis?

Answer 1

The integral yields a minimum value of $1$ precisely when $x''(t)=0$ uniformly on $[0,1]$. Otherwise, there would exist non-zero continuous $y'':t\mapsto y''(t)$ such that $$\int_0^1\big(1+0^2\big)dt\geq\int_0^1\big(1+|y''(t)|^2\big)dt$$ implying $\int_0^1|y''(t)|^2dt\leq0$. However, since $y''$ is non-zero and continuous we may fix $\epsilon>0$ and a non-empty open sub-interval $(a,b)$ of $[0,1]$ on which $y''(t)^2>\epsilon$ yielding the contradiction $$\int_0^1|y''(t)|^2dt=\int_a^b|y''(t)|^2dt+\int_{[0,1]-(a,b)}|y''(t)|^2dt>(b-a)\epsilon+0>0$$ because every Riemann sum approximation of $\int_{[0,1]-(a,b)}|y''(t)|^2dt$ is a non-negative real number. Therefore, because $x'(1)=1$ then $x'$ is the constant function defined by $x'(t)=1$. Since $x(0)=0$ then $x(t)=t$ which also satisfies $x(1)=1$.