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?


1 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.

  • $\begingroup$ I think I am very confused, so I split that into a few questions. Yes, I meant central discretization matrix for the second derivative. I saw the proof of estimates for C.N. method in $L^2$(stability is either using eigenvalue argument or energy methods for the case of variable coefficient equation). However, what is the approach for the $L^{\infty}$ norm? If you point me to a book or a paper that would be great, I wonder where is this $log(h)$ shows up, is that from the stability estimate? Is the result $||u^{n+1}||_{\infty}\leq C ||u^{n}||_{\infty}$ valid for c.N.? $\endgroup$
    – Kamil
    Jan 7, 2013 at 4:17
  • $\begingroup$ Why looking at the matrix $B$ is not sufficient? Once I put the restriction on $B$ that would put the restriction on $A$ through the eigen values. In fact, in general, if $B$ is not normal, I would have to estimate $||B||$ in either $L^2$ or $L^{\infty}$ in either 1 or 2 dimensions. $\endgroup$
    – Kamil
    Jan 7, 2013 at 4:29
  • $\begingroup$ From above, it looks that if I have convergence pointwise, that would imply everything. So, why do people write errors for the vector $e$, and not just in one point? Because as I can see, it is the same discrete equation at each point on the grid. Is that because this is only the case with constant coefficient? $\endgroup$
    – Kamil
    Jan 7, 2013 at 4:36
  • $\begingroup$ Last one: Why convergence in $L^{\infty}$ implies convergence in $L^{2}$. There is no norm equivalence between discrete $L^p$ norms as the mesh size goes to zero. So, in that case if I observe $L^2$ that implies $L^1$? So, in that case should always start measuring $L^1$, then $L^2$ and then $L^{\infty}$ to see how far I can go with my stability estimates? $\endgroup$
    – Kamil
    Jan 7, 2013 at 4:42
  • $\begingroup$ Hm, I think I managed to confuse myself as well: the claim about the $O(h^2 |\log h|)$ convergence order in $L^\infty$ is for the Laplace equation, not the heat equation you are considering here. I do not recall what the correct orders for the heat equation would be, but would look into the papers by Vidar Thomee from a long time ago. Maybe also those of Lars Wahlberg. $\endgroup$ Jan 7, 2013 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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