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??

  • $\begingroup$ Isn't 1st order ODE easy enough to solve analytically? Do you have a specific equation in mind? $\endgroup$
    – Pu Zhang
    Apr 24, 2015 at 11:44
  • $\begingroup$ agree but the reason for our inability to handle it are probably same as for 3 order, for example. $\endgroup$
    – Sohail
    Apr 24, 2015 at 16:49

1 Answer 1


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.


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