I want to minimize the functional $$I=\int_{-1}^1u^2(x)|2x-u'(x)|^2dx$$
Here i applied and found the euler langrange equation and found the differential equation
$$u'^2+2uu'-4u=4x^2$$ given is $u\in C^1 $ and $u(-1)=0 , u(1)=1$
but the minimizer is given as $$u(x)=0:x\in[-1,0], x^2:x\in[0,1]$$
Can anyone help me how to go about with this problem .
I need some idea on this functional as well to see that if we choose approperiate $h\in (-1,1)$ i get a circle arc as a minimizer with radium $\frac{1}{h}$
Functional is $$I=\int_0^1\sqrt{1+y'^2} +h y dx$$
Thank you for your kind guidance .
For the first functional. Since the integrand is non-negative, we have that $I\ge0$. If $u'=2\,x$ then $I=0$. Thus $$ u=x^2+C $$ is a minimizer for any $C\in\mathbb{R}$. Are these the only ones? No. A function $u$ will be a minimizer if $u^2\,|2\,x-u'|\equiv0$. This, and the regularity condition gives two new minimizers: $$ u(x)=\begin{cases} 0 & \text{if }-1\le x<0 \\ x^2 &\text{if }0\le x\le1,\end{cases}\text{ and } u(x)=\begin{cases} x^2 & \text{if }-1\le x<0,\\ 0 & \text{if }0\le x\le1.\end{cases} $$ Only the first one satisfies the boundary conditions $u(-1)=0$, $u(1)=1$.