Consider the Functional $$J(y)=\int_0^2(1-y’^2)^2dx$$ defined on $$\{y\in C[0,2]\mid \text{y is piecewise $C^1$ and $y(0)=y(2)=0$}\}$$ Let $y_e$ be a minimizer of the above functional. Then $y_e$ has
$1.$ A unique corner point.
$2.$ Two corner point .
$3.$ More than two corner points.
$4.$ No corner point.
As I applied Euler equation with Weierstrass-Erdmann conditions, I got corner point $c$ with $y’(c-0)=1$ and $y’(c+0)=-1$ or $y’(c-0)=-1$ and $y’(c+0)=1$ . So according to this there can be many corner points by shifting and making slopes 1 and -1 in left or right accordingly, like
$ y(x)= \begin{cases} x & \text{if } 0 \leq x \leq \frac{1}{4} \\ -(x-\frac{1}{2})& \text{if } \frac{1}{4}\leq x \leq \frac{5}{4} \\ x-2& \text{if } \frac{5}{4} \leq x \leq 2\\ \end{cases} $
But answer is given option $1$ i.e. unique corner. According to me any zig-zag curve consisting of segments of slopes +1 and -1 and joining the two end-points would work. Please suggest. Thank you.