A standard example of calculus of variation that I don't understand

Let us consider the problem $$\min \left\{ \int_0^1 \psi\left(\dot{u}\right) \ dx: u \in \mathcal{C}^1([0,1]), u(0) =0, u(1)=2 \right\}$$ where $$\psi \colon \mathbb{R} \longrightarrow \mathbb{R}$$ is the function $$\psi (x) = (x^2-1)^2.$$ Hence by my professor is it easy to conjecture that the minimum of the functional is $$u_0 (x) = 2x$$. Anyway I do not find it "easy to conjecture". To convince myself of this conjecture I tried to work as it follows:

1. Find with the first variation of the integral a condition that assures the second derivative to be zero

2. Notice that there is only one linear function that satisfies the requirements of the ambient space of the problem. (By linear I improperly mean $$f(x) = mx + q$$).

3. The function $$u_0(x)$$ is the only minimum of the functional

Let us start with point 1. Let us consider $$v \in V = \left \{ v \in \mathcal{C}^1([0,1]) : v(0)=v(1) = 0\right\}$$ evaluating the first variation I find the condition: $$\int_0^1 \left( 4 \dot{u}(x)^3\dot{v}(x) - 2 \dot{u}(x) \dot{v}(x)\right) \ dx$$ which I find pretty strange since given the function $$\psi$$ I would have expected a condition that will allow me to deduce $$\dot{u} = 1$$ where the $$\inf$$ of the function $$\psi$$ is taken, even tough is not in the ambient space of the problem. Maybe I am doing something wrong, suggestions accepted.

Let us focus on point 2 so that let us consider the convexified function of $$\psi$$ that coincides with $$\psi$$ for $$|x| \geq 1$$ and is zero otherwise. This is a convex function and, for my professor, $$u_0(x) = 2x$$ is the minimum of the function driven by this convexified function. Why? Anyway if it is true, if I consider the functional $$G(u)$$ the one driven by the convexified function we should have $$F(u) \geq G(u) \geq G(u_0) = F(u_0)$$ hence this shall prove that $$u_0$$ is the minimum of the functional of the problem.

Finally the uniqueness of the minimum is given by the convexity of the functional $$G$$ hence it is a unique minimum also for $$F$$.

• I think the first variation is $$\int_0^1 \left( 4 \dot{u}(x)^3\dot{v}(x) - \color{red}{4} \dot{u}(x) \dot{v}(x)\right)\,dx=4\int_0^1(\dot u^2-1)\dot u\dot v\,dx.$$ – A.Γ. Jan 12 at 19:52
• Let $u\in\mathcal{C}^1[0;1]$ such that $u(0)=0$ and $u(1)=2$. Let $v\in V$. Then $\psi(u+v)=((\dot{u}+\dot{v})^2-1)^2=(\dot{u}^2-1+2\dot{u}\dot{v}+\dot{v}^2)^2=\psi(u)+4(\dot{u}²-1)\dot{u}\dot{v}+O(\dot{v}^2)$. This $O(\dot{v})$ is actually a $O(v^2)$ by Roll's theorem. So, if a minimum is reached at $u$, for all $v\in V$: $$0=\int_{0}^14(\dot{u}^2-1)\dot{u}\dot{v}=-4\int_{0}^1(3\dot{u}^2-1)\ddot{u}v$$ after integrating by part. It follows that $\ddot{u}(3\dot{u}^2-1)=0$ and it's easy to show that this implies $\ddot{u}=0$ and therefore that a minimum, if it exists, is affine. – Ayoub Jan 15 at 3:11