I have seen a number of papers that propose a finite-difference method and then show the numerical results for it. Without providing a rigorous analysis(can be some summary or note or whatever, just no proofs involved) one can say method is of "second order". So to establish convergence the author would pick a point on the grid and measure the ratio of the differences between solutions, that is $$R:=\frac{u_h(h/2)-u_h(h)}{u_h(h/4)-u_h(h/2)}$$ then, he would say that we observe $2$ or $4$, so it confirms the expectations. It is in essence based on a pointwide convergence, thus if I measure that ratio at every single point on the grid(the initial grid with size $h$, the coarsest one) and observe a certain rate, then it would converge "on average" in the discrete $L^2$-norm, so as in the maximum $L^{\infty}$ error norm with the same rate as well. So what are the disadvantages of checking the convergence this way? I don't have an analytical solution to compare to but I want to verify somehow I have $L^2$ convergence that I have theoretically established. It might be "overkill" to do it this way as I would in fact be checking pointwise convergence, but it would still imply the needed type if convergence if I happen to observe the desired rate. Do I have any other options to establish that or the way I described is valid attempt? What if I don't observe it in some cases, how I can in fact check $L^2$ or $L^{\infty}$ convergence in practice? I have tested the method by using method of manufactured solution but there is an equation that I would like to check the convergence as well and I am not able to construct a solution(it is a weak solution so some of the derivatives don't exist and initial data is only continuous), therefore I need some "brute force" approach.

  • $\begingroup$ Usually, one solves on a very fine mesh to approximate the exact solution. If it is a good approximation, we could see how the error (to the finest mesh solution) decreases as $h\rightarrow 0$. $\endgroup$
    – Hui Zhang
    Jan 22, 2013 at 22:05
  • $\begingroup$ Norm equivalence holds only in finite-dimensional spaces. $\endgroup$ Jan 24, 2013 at 11:24
  • $\begingroup$ @David, I agree that there is no norm equivalence here but the embeddings still hold. Thus, I claim only that if I have convergence in $L^{\infty}$ I have it in $L^2$ as well, not the other way around. However, if I have convergence at every point, then it also implies convergence in the maximum norm. That's all. $\endgroup$
    – Kamil
    Jan 24, 2013 at 12:23
  • $\begingroup$ You claim convergence in different norms "at the same rate", which isn't generally true. $\endgroup$ Jan 27, 2013 at 8:38
  • $\begingroup$ @David: Asume I proved that for a finite difference scheme I have $||error||_{\infty}=O(h^2+k^2)$ where $error$ is the vector and is the difference between the numerical solution and the analytical solution, does that imply that $||error||_2=O(h^2+k^2)$, if not can you please hint on why and if there any other type of convergence I can deduce from it? $\endgroup$
    – Kamil
    Jan 27, 2013 at 14:53

1 Answer 1


It doesn't have to be $u_h$ at a grid point. You can apply this using any functional of $u_h$ over the domain, including the 2-norm. Assume a second order method, so $\| u_h - u \| \leq Ch^2$. Then $\| u_{2h} - u_h \| \leq \| u_{2h} - u\| + \| u - u_h \| = 5Ch^2$. Similarly $\| u_{4h} - u_{2h} \| \leq 20Ch^2$ and you get the ratio 4 if the code is correct.

  • $\begingroup$ did not get an answer completely. I can follow you with inequalities but what I can see is that they have different constants $5C$ and $20C$. Are you saying I should keep track of the difference between two subsequent solutions in order to establish convergence? Since they are of a different size, shall I project on a coarser or finer grid? $\endgroup$
    – Kamil
    Jan 22, 2013 at 11:44
  • $\begingroup$ yes, you'll need 3 grids. The constants come from the difference size, for example on the grid with spacing $2h$ you get $C(2h)^2=4Ch^2$. Use the coarser of the two grids to evaluate the norm in nominator and denominator of R. $\endgroup$
    – chris
    Jan 22, 2013 at 14:31
  • $\begingroup$ thanks for that suggestion, I will try that. I wonder though are there advantages of projecting a coarser solution to a finer grid or the opposite, taking the finer grid solution and picking only the points that fall onto a coarser grid? $\endgroup$
    – Kamil
    Jan 23, 2013 at 3:37

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