# Finite element method for odd order DE

What are theoretical hurdles in applying Galerikin method on, say, first order time dependent ODE?

Is there no way we can form an inner product??

• Isn't 1st order ODE easy enough to solve analytically? Do you have a specific equation in mind? – Pu Zhang Apr 24 '15 at 11:44
• agree but the reason for our inability to handle it are probably same as for 3 order, for example. – Sohail Apr 24 '15 at 16:49

## 1 Answer

I do not think that you can form an inner product there. Nevertheless, there are other methods that do not use the same space for test and trial functions, e.g., Petrov-Galerkin methods. The problem here is that the matrix is not symmetric anymore.

You can also use Least Squares FEM. For this you have $$\mathcal{L}u=f\quad \text{in }\Omega\quad \text{and}\quad \mathcal{R}u=g\quad \text{on } \partial\Omega$$ where, $\mathcal{L}u=g$ and $\mathcal{R}u=g$ is the boundary condition. Then, you form the functional $$J(u;f,g) =\Vert \mathcal{L}u-f\Vert^2 + \Vert \mathcal{R}u-g\Vert^2$$ and solve the minimization problem $$\min_{u\in S} J(u;f,g)$$ and you end up with a system of the form $$(\mathcal{L}v,\mathcal{L}u) + (\mathcal{R}v,\mathcal{R}u) = (\mathcal{L}v,f) + (\mathcal{R}v,\mathcal{L}f) \forall\ v \in S \enspace .$$ A caveat here is that the norms and inner products are not in the same space. I have seen this method for Quantum Electrodynamics, where the equation is first order, example here.