# ENO and WENO schemes

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 January 6, 2015, 09:30 ENO and WENO schemes #1 New Member   West Bengal Join Date: Dec 2014 Location: Kharagpur Posts: 4 Rep Power: 11 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, __________________ Sachin Shivakumar B.Tech in Mechanical Engineering IIT Kharagpur

 January 6, 2015, 10:06 #2 Senior Member   Michael Prinkey Join Date: Mar 2009 Location: Pittsburgh PA Posts: 363 Rep Power: 24 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.

 April 18, 2015, 13:21 Connection between CFL number and accuracy of ENO schemes #3 New Member   Priyam Join Date: Apr 2015 Posts: 6 Rep Power: 10 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.

April 18, 2015, 13:39
#4
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Filippo Maria Denaro
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Quote:
 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

 April 18, 2015, 14:59 #5 New Member   Priyam Join Date: Apr 2015 Posts: 6 Rep Power: 10 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?

 April 18, 2015, 15:11 #6 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,738 Rep Power: 71 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?

 April 18, 2015, 15:22 #7 New Member   Priyam Join Date: Apr 2015 Posts: 6 Rep Power: 10 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?

 April 18, 2015, 15:23 #8 New Member   Priyam Join Date: Apr 2015 Posts: 6 Rep Power: 10 But i dont refine my dx. dx is constant for each time step. Is it a problem?

 April 18, 2015, 15:33 #9 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,738 Rep Power: 71 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).

 April 18, 2015, 15:36 #10 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,738 Rep Power: 71 of course, try also the test a cfl= constant

 April 18, 2015, 17:09 #11 New Member   Priyam Join Date: Apr 2015 Posts: 6 Rep Power: 10 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

April 18, 2015, 18:22
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Filippo Maria Denaro
Join Date: Jul 2010
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Quote:
 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

 February 18, 2016, 14:03 WENO scheme coefficient derivation #13 New Member     Y. Yang Join Date: Mar 2010 Location: Miami, United States Posts: 28 Rep Power: 15 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 __________________ PhD candidate

 Tags eno, numerical schemes, weno

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