What are the pros of Fourier-Galerkin spectral methods while solving PDEs?

Here's the one that came in my mind first:

  1. Easy implementation: using this method, differentiation operator computation is really simple since $$\partial^p\hat{u}_k=(\imath k)^p\hat{u}_k$$
  2. Exponential convergence: let $u \in C^m$ the exact solution, $u_N$ the numerical solution and $\epsilon=||u-u_N||_p$. $$\epsilon\leq \alpha N^{-m}||u^{(m)}(x)||$$ Therefore the convergence is exponential if $m=\infty$.
  • 1
    $\begingroup$ You already have the two big pros. You should also think about the cons :-) $\endgroup$ – Wolfgang Bangerth Jul 22 at 16:34


  1. With trigonometric basis functions your problem size is $N \text{log}(N)$ instead of $N^2$.
  2. Stabilization techniques are easy to implement and cheap:
  • Filtering in the modal space.
  • Zero padding in the modal space.
  1. No aliasing due to the Galerkin ansatz.
  2. Energy/Entropy stable disctretizations, e.g. via a skew symmetric implementation, are quite easy.


  1. You are restricted to periodic boundary conditions.
  • However, you may use a different basis, e.g. the Chebyshev expansion, with a fast DCT (Discrete cosine transform). This allows also computations with $N \text{log}(N)$ and also Dirichlet or Neumann BC's.
  • However, then you are restricted to a Chebyshev grid.
  1. You are restricted to smooth problems.
  2. You are restricted to structured meshes.

But to be honest, it is hard to talk about pros and cons if you do not consider your application/problem. This should be the first step.

| cite | improve this answer | |
  • $\begingroup$ Thanks! I work on a turbulent flow (with variable density) problem. $\endgroup$ – Dave Jul 23 at 14:12
  • $\begingroup$ I am curious about the "no aliasing due to the Galerkin ansatz", I never heard about that. Any references on the subject? $\endgroup$ – Loïc Reynier Jul 24 at 10:15
  • $\begingroup$ Dear @LoïcReynier you are correct in the sense that without additional effort you still have aliasing. The point implicitly assumes that all integrals are solved (quasi) exact. Here, you may use an overintegration strategy, see point 2 in Pros. $\endgroup$ – ConvexHull Jul 24 at 10:55
  • 1
    $\begingroup$ @ConvexHull For polynomial-based Galerkin methods, complete elimination of aliasing is generally not possible for something like the compressible Navier-Stokes equations due to the fluxes being rational functions of the conservative variables. Would this not also be the case for Fourier spectral methods? $\endgroup$ – Tristan Montoya Jul 30 at 15:05
  • 1
    $\begingroup$ @TristanMontoya Yes that's true, e.g., due the use of the EOS or the Riemann solver. Theoretically you would have to oversample with $N=\infty$. That's the reason i mentioned "quasi" exact. $\endgroup$ – ConvexHull Jul 30 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.