Why do we impose |ρ(ξ)|^2 ≤ 1 in the Crank-Nicholson scheme?

Question

I'm being introduced to numerical methods for PDE's. I've learned that in the Crank Nicholson scheme, we can map the discrete Fourier transform of the current space lattice to the FT of the same lattice for the next time interval:

$$ \hat U^n = \rho (\xi) \, \hat U^{n+1} $$

Additionally, Parseval's identity relates the fourier transforms of the two adjacent transforms (i.e. adjacent in time):

$$||\hat U ^{n+1} ||_{{L^2}[-\pi, \pi]} = ||\rho (\xi) \, \hat U ^{n} ||_{{L^2}[-\pi, \pi]}$$

In all the examples I have seen so far, we apply the constraint

$$|\rho (\xi)|^2 \leq 1 $$

when we are proving stability. Why do we constrain $|\rho (\xi)|^2$ to 1 when proving stability? Qualitatively, I understand this to mean that the norm of the discrete FT's does not increase as time increases, but I'm having trouble understanding how this constraint relates to stability for the scheme solution, which exists in the spatial (not $\xi$) domain.

Answer 1

In this context Parseval's identity says that the Fourier transform preserves the 2 norm. So requiring that $$\|\hat U ^{n+1} \|_{{L^2}[-\pi, \pi]} = \|\rho (\xi) \hat U ^{n} \|_{{L^2}[-\pi, \pi]} \implies \frac{\|U ^{n+1} \|_{{L^2}}}{\| \, U ^{n} \|_{{L^2}}} = \frac{\|\hat U ^{n+1} \|_{{L^2}[-\pi, \pi]}}{\| \, \hat U ^{n} \|_{{L^2}[-\pi, \pi]}} \leq \|\rho (\xi)\|.$$ In other words, no matter what Fourier mode $\xi$ you consider, by bounding $\|\rho(\xi)\|$ above by 1, the norm of your solution vector (in either domain, spacial or Fourier) does not grow as time evolves. If this was not the case the norm of your solution vector would be growing as time evolves, and if you have a large number of time steps your solution will blow up towards infinity.