How can I prove that $\inf\limits_{u\in X} I(u)=0 $

Question

Please leave just hints not the whole solution in the below.

Problem Consider the following space $X = \{u \in C^1([−1, 1]): u(−1) = 0; u(1) = 1 \},$ and the functional $ I : X \to [0,1) $ defined as follows: $$ I(u) :=\int_{-1}^{1}(u'(x))^2(1 − u'(x))^2dx ; $$ prove that $\inf\limits_{u\in X} I(u)=0 $ and that such that $ \inf $ is not attained inside $X.$

Answer 1

The main idea is that the contribution to $I$ is small provided that $u'$ is close to $0$ or $1$. The average slope of $u$ is forced to be $1/2$ by the boundary conditions. So a function that has derivative $1$ on half the interval and $0$ on the other half of the interval satisfies the boundary conditions and makes $I$ zero, which is certainly as small as it can be. Such a function is not $C^1$; the point of the exercise is to approximate such a function by $C^1$ functions.

Answer 2

Here is a hack answer:

Define $u_n(x) = \begin{cases} 0, & x < -{ 1\over n} \\ p_n(x), & x \in [-{1 \over n} , {1 \over n}] \\ x , & x > { 1\over n} \end{cases} $, where $p_n(x) = (x+{1 \over n})^2 (x-{1 \over n})(x-{ n^3-4\over 4 n})$.

We have $p_n(-{1 \over n} ) = p_n'(-{1 \over n} ) = p_n({1 \over n} ) = 0$ and $p_n'({1 \over n} ) = 1$, hence $u_n$ is $C^1$. (Also, note that $\lim_n u_n(x) = \max(0,x)$).

Then $I(u) = \int_{-{1 \over n}}^{1 \over n} p_n'(x)^2 (1 - p_n'(x)^2 )^2 dx = {16 -5 n^3 \over 15 n^5}$.