I'm trying to find the minimizer of the following functional: $$I(u)=\int_{0}^{1}u^2(2-u)^2dx,$$ where $u\in Y,$ and $Y=\{u\in C([0,1]):u(0)=1,u(1)=1\}.$
The Euler-Lagrange equation of this question is: $4u(u-2)(u-1)=0.$ After solving the Euler-Lagrange equation, I obtained three solutions: $u=0,u=1$ and $u=2$ on $[0,1]$. It could be noticed that the solution $u=1$ also satisfies the boundary conditions.
However, the solution showed that the infimum value of this functional is actually $0$, which could be obtained from constructing a minimizing sequence. I was kind of confused about how could we know that the solution $u=1$ is not a minimizer? Is it a maximizer instead?
Thank you so much in advance for any hints and helps!
Hint: If we Taylor expand $u=1+v$ around $u=1$, then we get
$$I[1+v]~=~ \int_0^1 \! \mathrm{d}x (1+v)^2(1-v)^2 ~=~ \int_0^1 \! \mathrm{d}x \left(1 \color{red}{-}2v^2 +v^4 \right).$$
The red minus shows that $u=1$ is a local maximum.