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August 4, 2011, 03:57 
1D Burgers euqation with 4th Runge Kutta

#1 
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Dokeun, Hwang
Join Date: Apr 2010
Posts: 71
Rep Power: 9 
Hello
Now i'm tring to apply 4 Stage Runge Kutta scheme to 1D Burgers euqation as a model equation like this. hear, At this point, the slop is suggested as central differencing. The Question is.. How to handle at the point i=2,3,IM2,IM1? Because 4 Stage Runge Kutta requires i2 ~ i+2 points for next step value at i, I can't get at i=2,3,IM2,IM1. I got the answer using 3 stage RK to i=3,IM2 and 2 stage RK to i=2,IM1. inspite of complicate euqations used. But I guess there manybe more simple way for near the boundary points. Please let me know simple solution for this. Thank you in advance. 

August 5, 2011, 03:18 

#2 
Senior Member
cfdnewbie
Join Date: Mar 2010
Posts: 557
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maybe I'm misunderstanding you, but the classical RK schemes use stages/nodes that are INSIDE the interval you are trying to integrate....so you are integrating in time, and you need the evaluation at (say) 1/3 delta t, 1/2 delta t, 2/3 delta t and so on...
how you discretize your spatial domain is up to you, since you are using a method of lines approach.... 

August 8, 2011, 02:45 

#3 
Member
Dokeun, Hwang
Join Date: Apr 2010
Posts: 71
Rep Power: 9 
Dear cfdnewbie
Thank you for your reply. Would please review my spreading out of the equations. Becasue is defined as , refers . And, in turn, refers . Here, comes from and is derived from . As a result, to get , I need the values of at i+2~i2. As the same manner, requires the value of at i3~i+3. I feel that something is wrong with my equations but I'm in trouble to figure out it cleary. I attached my F90 code for fixing..(it works any way) I appreciate that again. 

August 8, 2011, 06:34 

#4 
Member
Dokeun, Hwang
Join Date: Apr 2010
Posts: 71
Rep Power: 9 
dear cfdnewbie
I understood my problem. my original program was... DO i=2,IM1 u1(i) = function of u(n,i) u2(i) = function of u(n,i) u3(i) = function of u(n,i) u4(i) = function of u(n,i) END DO but the right one should be DO n=1,NM1 DO i=2,IM1 u1(i) = function of u(n,i) END DO DO i=2,IM1 u2(i) = function of u1(i) END DO DO i=2,IM1 u3(i) = function of u2(i) END DO DO i=2,IM1 u4(i) = function of u3(i) END DO DO i=2,IM1 u(n+1,i) = u(n,i) + ~~~ END DO END DO at first i tred to write all u1, u2, u3, u4 as functions of u(n) along just one iteration, but now there are each iteration for u1~u4, and in another iteration I calculate u(n+1) 

Tags 
burgers equation, runge kutta 
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