I have recently started calculus of variations and face the following task:
Consider the Lagrangian $$L(y',y):=|y'|$$ on the space of curves $$U = \{y\in C^2([0;T];\mathbb R): y(0)=a, y(T=1)=b \}$$ For which values of $a$ and $b$ does the functional $$A=\int_0^TL(y',y)dx$$ admit a unique minimizer? For which $a$ and $b$ does it admit infinitely many minimizers?
I tried to use Euler-Lagrange equation, but it gave nothing ($0=0$). Also $L(y',y)=|y'|$ is not strictly convex, so the unique theorem can't be applied.
Then I tried to calculate the integral directly
$$\int_0^T|y'|=\dfrac{y|y|}{2}\bigg|_0^T=\dfrac{y(T)|y(T)|}{2}-\dfrac{y(0)|y(0)|}{2}=\dfrac{b|b|-a|a|}{2}$$
so it doesn't really depend on curve, but only on endpoints, so I should have infinitely many minimizers for any $a,b$, but it doesn't really corresponds with question.
Thank for any help.
You are correct that, in general, there are an infinite number of minimizes for $U = \{y\in C^2([0;T];\mathbb R): y(0)=a, y(T=1)=b \}, $ $\int_0^T |y'|$ . These correspond to the smooth, monotonic functions over $[0,T]$ with $y(0)=a, y(T) = b$. The value taken by the integral at the minimizer is $|a-b|$.
However, you obtain a unique minimizer when $a=b$, the constant function $y(x)=a$. The value taken by the integral of the minimizer is 0. If you take a non-constant $y$, when $a=b$, there will certainly be a point at which $y'$ changes sign. This will "mess" with the integral $\int_0^T |y'|$, and have its result be greater than 0. So the solution is to set $a=b$ and have the constant function as the only minimizer.