[sachinsiitkgp]

My coursework involves studying the implementation of eno and weno schemes. I was wondering if anyone could tell if there are any pre-requisitive topics which i should look into before going for the ENO and WENO schemes or is it okay if I directly dive into that?
Regards,

[mprinkey]

ENO/WENO schemes are pure interpolation methods in a numerical analysis sense. The algorithms are very simple--selecting from (in ENO) or blending (in WENO) several possible quadratic interpolations based on a nonlinear smoothness operator with the goal to simply avoid interpolation across discontinuities. If you understand basic interpolation techniques, it should not be too difficult to get started on ENO/WENO.

Other flux interpolation/limiting schemes (like the large variety of TVD schemes) in my opinion, require more insight, especially into the underlying transport equations.

[Prixy11]

I implemented the ENO scheme for hyperbolic eqaution but it gives results more accurate (upon comparing with true solution) than the upwind method only for cfl less than 0.5. And I could obtain the theoritical accuracy of 2nd degree at cfl less than 0.1. I dont understand this behavior. Can anyone help me in explaining it. Thank you.

[FMDenaro]

Originally Posted by Prixy11

I implemented the ENO scheme for hyperbolic eqaution but it gives results more accurate (upon comparing with true solution) than the upwind method only for cfl less than 0.5. And I could obtain the theoritical accuracy of 2nd degree at cfl less than 0.1. I dont understand this behavior. Can anyone help me in explaining it. Thank you.

for a theoretical accuracy analysis you can use a smooth solution to check the accuracy on several refined grids. Be careful, the solution must be smooth otherwise there is no sense in checking accuracy

[Prixy11]

Thanks for the vital information. Actually, I am using a smooth function (sine curve) but observe accurate results for CFL below 0.10. And I read in some papers also they are assuming CFL below 0.40. I am tryng to figure out why the behavior is like that. Is it because the scheme is more depdndent on temporal accuracy thatsy the result improves with low CFL having less dt. Can you tell me?

[FMDenaro]

Are you solving df/dt+u df/dx=0 (u=constant) on a periodic domain?
when you perform the accuracy analysis you are taking constant cfl=0.1 and use several refined meshes?

[Prixy11]

Yes it is that equation. No, I am not taking constant cfl for accuracy analysis. I am using a constant dt=0.005 and dx = 0.628,0.0628,0.00628 (0 to 2pi domain) and use periodic boundary conditions. The error for each space step = analytical sol - result at that space point. Then, to find overall error, I am using rooot mean square error = sqrt(sum of error at all points.^2/n+1 ) and then I am plotting - log( root mean square ) vs log dx . Till dx =0.0628 i get a slope of 2v:1H but after that it become 1V:2H. I dont know why. I have checked my implementation it seems to be correct. I get very less error for low cfl values of 0.1 which are good comapred to upwind errors. Can you help me with it?

[Prixy11]

But i dont refine my dx. dx is constant for each time step. Is it a problem?

[FMDenaro]

ok, then the problem is quite clear... the accuracy analysis is based on the fact that you reduce the local truncation error (LTE). The LTE is a function of dt and dx, for this reason the analysis is performed by taking dt/dx= constant so that dt and dx go to zero at the same rate.
What you are observing is due to the fact that taking constant dt=0.005 you have some terms of the LTE that do not tend to zero as the dx is diminished but remain constant. Therefore, when the error diminishes, it will reach the point at which the slope is altered due to the constant dt.
If my guess is correct, you can simply fix a smaller dt, for example 0.00005 and you see the correct slope, at least until the error becomes of the O(dt).

[FMDenaro]

of course, try also the test a cfl= constant

[Prixy11]

Thanks a lot! I got a slope of 2v:1h for dt=0.00005 and t-0.1. It shows that the method is second order accurate in space. But I have some more final queries.
I checked for a const cfl of 0.4 with dx=0.628, ... 0.00628, but a got a slope of less than 1 for final time of t=5.

If the slope of error = order this implies the code implementation is correct , rite?

For a scheme like RUnge Kutta 3rd order, it is 3rd order in time and 2nd order in space, To check the implemntation for RK, to check if it is 3rd order in time, I shall fix dx = 0.00005 or so and then find error.

Sorry, I am troubling you a lot. But if I will understand this today I will be very clear . Thanks

[FMDenaro]

Originally Posted by Prixy11

Thanks a lot! I got a slope of 2v:1h for dt=0.00005 and t-0.1. It shows that the method is second order accurate in space. But I have some more final queries.
I checked for a const cfl of 0.4 with dx=0.628, ... 0.00628, but a got a slope of less than 1 for final time of t=5.

If the slope of error = order this implies the code implementation is correct , rite?

For a scheme like RUnge Kutta 3rd order, it is 3rd order in time and 2nd order in space, To check the implemntation for RK, to check if it is 3rd order in time, I shall fix dx = 0.00005 or so and then find error.

Sorry, I am troubling you a lot. But if I will understand this today I will be very clear . Thanks

ok, that sounds strange ...
1) Use the L_inf norm for the error so that you can also note the index of the max error
2) Use cfl=0.1 and repeat the test
3) try to do the analysis by computing the errors after only 1 time step

[hityangsir]

Dear Professor:

I am trying to understand WENO reconstruction scheme in this paper bu Dr Shu: http://epubs.siam.org/doi/abs/10.1137/070679065

Could you please give me some instruction on the WENO Reconstruction. part, how to derive (2.11)

u(1)_(i+ 1/2) = 1/3 * u_(i−2) − 7/6*u_(i−1) + 11/6*u_i.

I can derive the WENO interpolation part WENO Interpolation by applying the technique similar to Lagrange interpolation to get the equation of (2.1).

u(1)_(i+ 1/2) = 3/8 * u_(i−2) − 5/4*u_(i−1) + 15/8*u_i.

Could you give me some instruction,
Thank you very much