Take the 2-minute tour ×
Computational Science Stack Exchange is a question and answer site for scientists using computers to solve scientific problems. It's 100% free, no registration required.

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.

share|improve this question
    
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 $h\rightarrow 0$. –  Hui Zhang Jan 22 '13 at 22:05
    
Norm equivalence holds only in finite-dimensional spaces. –  David Ketcheson Jan 24 '13 at 11:24
    
@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. –  Kamil Jan 24 '13 at 12:23
    
You claim convergence in different norms "at the same rate", which isn't generally true. –  David Ketcheson Jan 27 '13 at 8:38
    
@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? –  Kamil Jan 27 '13 at 14:53

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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? –  Kamil Jan 22 '13 at 11:44
    
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. –  chris Jan 22 '13 at 14:31
    
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? –  Kamil Jan 23 '13 at 3:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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