
minimising a functional
Hi, if u belongs to C2(0,1)=space of twice differentiable continuous functions on (0,1). Also u(1)=u(0)=0 are boundary conditions with u(x)=u double prime =f (ie 2nd derivative of u wrt x). Prove that u minimizes the functional F where domain is sobolev space with m=1 on (0,1) (subscript 0) and range is real line.
F(v)=1/2*int(v prime)^2  int(f(x)v(x))dx (Integrating over 0 and 1)
From above we know u is solution of strong problem which implies it is the solution of the variational problem.
Now i've managed to represent F(v) as 1/2a(v,v)a(u,v) now to show this is minimized i was thinking i could bound it above by something with a constant in and then choose a large constant. Looking at coercivity and continuity these could be applied but still not sure which way of going about it. Any suggestions would be much appreciated. Thanks.(Nod)

You mean $\displaystyle F(v)=a(u',v')/2a(f,v)$, where $\displaystyle a(v,w)=\int vw$ is the dot product on $\displaystyle W^{1,2}_0=H^1_0$.
You need to prove the following:
 A minimizer for F is a solution of the DE.
 A solution of the DE is a minimizer for F.
For 1, notice that the DE is the EulerLagrange equation of F.
For 2, calculate $\displaystyle F(u+tv)F(u)$ and use the DE to show this is equal to $\displaystyle \int t^2v'^2/2$. So F attains a minimum on u.