I'm trying to prove that this function:
$$R[x]= \int_{0}^{1} (x'^2(t)+4x(t))\, dt$$
defined as: $R:X\longrightarrow \mathbb{R}$ where $X = \{x\in C^1[0,1],\mathbb{R}|x(0)=0,x(1)=1\}$ has a minimum at $x_0 = t^2$.
But I don't know how to do it properly. I've tried to use the Euler-Lagrange equation, but it doesn't seem to work.
Another way I've found is to use a theorem that states for a strictly convex funtion (which I've already proved that it is), if for $x_0$, the first order variation $\delta R(x-x_0)=0$ equals to cero, then $x_0$, is the only minimum of $R[x]$. But I'm confused because the definition for the first order variation states that: $$\delta R(x_0,h)=\lim_{x \to{0}}(\frac{R(x_0+\lambda h)-R(x_0)}{\lambda})$$
But for $\delta R(x-x_0)=0$, where do I substitute that value$(x-x_0)$ in the first order variation? How can I solve it? I'm confused with the algebra, please, help. Does anybody knows a better way to solve this problem?
You can certainly use E-L equations $$ \frac{d}{dt}\frac{\partial L}{\partial x'}-\frac{\partial L}{\partial x}=0 $$ which in this case give you the ODE $$ 2x''(t)-4=0. $$ The general solution is then $$ x(t)=t^2+Ct+D $$ The given boundary conditions force $D=0$ and $1+C+D=1$ that is $C=D=0$. Summing up, the only critical point is $$ x(t)=t^2. $$