1. ## 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.

2. You mean $\displaystyle F(v)=a(u',v')/2-a(f,v)$, 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:

1. A minimizer for F is a solution of the DE.
2. A solution of the DE is a minimizer for F.

For 1, notice that the DE is the Euler-Lagrange 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.