intuition behind the different discrete norms for Crank Nicolson

Question

I am solving a heat equation $u_t=Au$ with Crank-Nicolson finite-difference method and $A$ is a usual discretization matrix for $u_{xx}$ term. I want to tell something about the whole error vector over the entire grid. I can write the discrete iteration of the the method as $$u^{n+1}=Bu^n=\frac{I+\tau/2A}{I-\tau/2A}u^n$$ Since $B$ is normal I have $\rho(B)=||B||_2$ and I can see that having known eigenvalues of $A$ I can find eigenvalues of $B$ and thus I have $||B||_2\leq 1$ and $||e||_2=O(h^2)$ where $e$ is the vector error over the whole grid in the discrete $L^2$ norm.

Thus, my first question is whether I can imply $||e||_{\infty}=O(h^2)$ for Crank-Nicolson as well? For that I need to estimate $||B||_{\infty}=\max_{i,j}b_{i,j}$ and I don't know how to do that.

Second question, if the question one is a true statement, I have estimates for the error in two discrete norms: $L^2$ and $L^{\infty}$. What is the intuition of having $L^2$ discrete error of the error vector of order two, to me it looks that "on average" the error decreases by two, so if I pick the point on the grid I don't have to have quadratic convergence? Can I imply anything about how that will be converging at a particular point on the grid? The same question about $L^{\infty}$ norm. Which one should I use to measure the error, would one imply the other?

Answer 1

On your first question: I assume that by "usual discretization matrix" you mean either the 3-point finite difference discretization in 1d, or what you get using linear finite elements. In either case, it's not actually the Crank-Nicolson scheme that determines this. It's true that for the two spatial discretizations mentioned above, the spatial error is $O(h^2)$ both in the $L^2$ and $L^\infty$ norms and that if you choose $\tau=h$ then the overall error also has this convergence order. But this is only true in 1d.

On the other hand, if you were in a 2d, then you'd get $O(h^2)$ in the $L^2$ norm for the error, but $O(h^2 |\log h|)$ in the $L^\infty$ norm. What this shows is that it's not sufficient to simply look at properties of the Crank-Nicolson scheme -- it's in fact also necessary to look at properties of the underlying spatial discretization to prove estimates like the ones you are interested in. Your approach to just look at matrix norms of $B$ can therefore not be sufficient.

That also answers part of your second question. In any number of spatial dimensions, you always have that the $L^2$ error decays like $O(h^2)$, but that isn't true for the $L^\infty$ error. In other words, the first implies that on average, the spatial error decays quadratically, but that this does not have to be the case pointwise. On the other hand, if it were true that the $L^\infty$ error decayed as in $O(h^2)$, then it is trivial to show that that must also be true for the $L^2$ error.