Q:

Examining $\inf \int_0^1(x')^{2}(t)\mathrm dt$

I need to examine 
$$\inf \int_0^1(x')^{2}(t)\mathrm dt$$
with such conditions:
\begin{cases} x(0)=0, x(1)=1 & \\
\int_0^1x(t)dt=0 &
\end{cases}
I started with writing Euler-Lagrange equation for $F[t,x(t),x'(t)]=\int_0^1(x')^{2}(t)\mathrm dt$:
$$\frac{d}{dx}2x'(t)=0.$$
Obviously, the solution of this differential equation is $x(t)=Ct+D$. Because of the conditions from $(1)$ I assumed that $C=1, D=0$. But then, function $x(t)=t$ does not fulfill the second condition. What did I do wrong? Where should I take this second condition into account? Thanks for any help!

A:

You cannot use the Euler Lagrange equation directly as you have another constraint
$$L(x) = \int_0^1 x(t) dt = 0.$$
Instead, you need to use the Lagrangian multiplier: You want to minimize 
$$G(x) = \int_0^1 (x')^2(t) dx$$
subject to $L(x) = 0$. So by Lagrangian multiplier, there is $\lambda\in \mathbb R$ so that 
$$\nabla G = \lambda \nabla L,$$
that is, for all test functions 
$$2 \int_0^1 x'(t) \phi'(t) dt = \lambda \int_0^1 \phi(t) dt.$$
Using integration by part, we have 
$$\int_0^1 (2x''(t) + \lambda)\phi(t) dt = 0$$
for all test functions $\phi$. So 
$$x''(t) = C$$
and we can write $x(t) = at^2 + bt +c$. Using now the boundary conditions and $L(x) = 0$, we find 
$$x(t) = 3t^2 - 2t.$$
Thus the infimum of $G(x)$ is (!)
$$G(3t^2 - 2t) = \int_0^1 (6t - 2)^2 dt = 4.$$
(I am bypassing a hard question: How do I know that this is a minimum? In general the Eular-Lagrange method give us only the critical points)

