Suppose we are given the followin where $f$,$u$, $g$ are given functions:

$-\Delta u = f$ in $\Omega$

$u=u_o$ on $\Gamma_1$

$\frac{du}{dn}=g$ on $\Gamma_2$

So in order for me to form the variational problem which is the first step, I did the following.

Assume $u_o$ is sufficiently smooth on $\hat{\Gamma}$ then, $\hat{u}=u-u_o$

This results in:

$-\Delta \hat{u}= f$ in $\hat{\Omega}$

$\hat{u}=0$ on $\hat{\Gamma_1}$


If I let $V$={$v \in H^1(\Omega): v|_{\Gamma_1} = 0$}

Then by applying my test function v and using Green's Formula, I can show that the new problem I've created is similar to the original problem and it results in:

$\int_{\Omega} \nabla u \nabla vdx$ = $\int_{\Omega}fvdx+\int_{\Gamma_2}gv ds $

where the bilinear case is $a<u,v> = \int_{\Omega} \nabla u \nabla vdx$

the linear case is $L(v) = \int_{\Omega}fvdx+\int_{\Gamma_2}gv ds $

So now that I have gotten the variational problem finished I need to implement some finite element method.

So this would mean I need to find a basis function which I would pick a triangulation in some smaller subspace $V_h=${$v_h$ is continous, linear.},

$\phi_j(x)= 1$ if $i=j$ and $0$ if $i\neq j$

So then,

$v_h=\sum_{j=1}^{M} \eta_j \phi_j(x)$ are the linear combinations

Now essentially I need to get to some sort of formulation of the finite element method where I would have

$A \zeta=b$ and then I could get some sort formulation of what the A matrix would look like. I know the A matrix will be symmetric and positive definite. I know the following,

$A_{ij}= <\phi_i' , \phi_j'>$

$b=<f, \phi_i>$

I'm stuck on the finite method part and I want to make sure I'm going about it the right way.


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