Show that minimum exists (direct method)

Consider $$F(v):=\int\limits_0^1\lvert v'(x)\rvert^2\, dx,$$ with $$\left\{v\in H^{1,2}(0,1), v(0)=0=v(1)\right\}.$$

Show that a minimum exists and use the direct method.

When I am right informed, I have to show that:

1.) $H^{1,2}(0,1)$ is reflexive.

2.) $F$ is coercive.

3.) $F$ is weakly lower semi continiuos.

Is that right?

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Just take $v=0$. This is the minimum – Tomás Jan 23 '13 at 13:33
Hm, then this example was too easy. I wanted to have an example for the direct method (maybe on where one of the three points is not fullfilled). Do you have such an example for me? Because i want to train that. – math12 Jan 23 '13 at 13:42
Im gonna put some examples in the answer – Tomás Jan 23 '13 at 13:47
Great, thank you. – math12 Jan 23 '13 at 13:48
Well, you can generalize this to functional that computes distance along a curve in the plane $\sqrt{1 + (y')^2}dx$ or even further to a functional for a geodesic in a Riemannian manifold or a functional for motion of a point in a potential. The latter is just integral of the Lagrangian $L = T - V$ where $T = v'/2$ is kinetic energy and $V$ is the potential. Your functional corresponds to a free particle in 1D and the solution here is that it stands still. – Marek Jan 23 '13 at 13:57

The problem as it stated is to simple, because is easilty to verify that $v=0$ is the minimum. So I am gonna fulfill te request of math12 in the comments.

Let $\Omega\subset\mathbb{R}^N$ be a bounded open set.

1 - An functional that satisfies 1,2,3.

Let $f\in L^2(\Omega)$. Define $F:H_0^1(\Omega)\rightarrow\mathbb{R}$ by $$F(u)=\frac{1}{2}\int_\Omega|\nabla u|^2-\int fu$$

The functional $F$ is called the energy functional associated with the problem $$\tag{1} \left\{ \begin{array}{rl} -\Delta u =f&\mbox{in \Omega} \\ u=0 &\mbox{in \partial\Omega } \end{array} \right.$$

The minimum of $F$ is a wealy solution of (1).

2 - An Functional that not satisfies all yoru hypothesis.

This is a more general functional thats contains the first one

Let $\lambda_1$ be the first positive eigenvalue of the Dirichlet Laplacian, i.e. there exist $u\neq 0\in H_0^1(\Omega)$ such that (in the weak sense)$$\tag{2} \left\{ \begin{array}{rl} -\Delta u=\lambda_1u &\mbox{\Omega} \\ u=0 &\mbox{\partial\Omega} \end{array} \right.$$

and if $\lambda<\lambda_1$ satisfies (2) then $u=0$

Again let $f\in L^2(\Omega)$ and define $F:H_0^1(\Omega)\rightarrow\mathbb{R}$ by $$F(u)=\frac{1}{2}\int_\Omega |\nabla u|^2-\lambda\int_\Omega u^2-\int_\Omega fu$$

If $\lambda\geq\frac{\lambda_1}{2}$ then this problem is not coercive. On the other hand, if $\lambda<\frac{\lambda_1}{2}$ the problem satisfies 1,2,3.

Notes: If you want to understand these problems, you have to understand some things first.

I - In $H_0^1(\Omega)$ the number $(\int_\Omega |\nabla u|^2)^{\frac{1}{2}}$ is a norm in $H_0^1(\Omega)$ equivalent to the usual norm. Moreover, you have Poincare inequalitu $$\int_\Omega u^2\leq\frac{1}{\lambda_1}\int_\Omega |\nabla u|^2$$

II - I suggest you to take some books of functional analysis (like Brezis for example) to understand why $H_0^1$ is reflexive and why the norm is weakly sequentially lower semicontinuous.

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