1D Burgers euqation with 4th Runge Kutta

dokeun · August 4, 2011, 03:57

Hello

Now i'm tring to apply 4 Stage Runge Kutta scheme to 1D Burgers euqation as a model equation like this.

$u^{n+1}_{i} = u^{n}_{i} - \Delta t \left( \frac{\partial E}{\partial x}\right)^{n}$

hear, $E_{i} = \frac{1}{2} {u_{i}}^2$

At this point, the slop is suggested as central differencing.

$\frac{\partial E}{\partial x} = \frac{E_{i+1} - E_{i-1}}{2 \Delta x}$

The Question is..

How to handle at the point i=2,3,IM-2,IM-1?

Because 4 Stage Runge Kutta requires i-2 ~ i+2 points for next step value at i, I can't get ${u^{n+1}_{i}}$ at i=2,3,IM-2,IM-1.

I got the answer using 3 stage RK to i=3,IM-2 and 2 stage RK to i=2,IM-1. 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.

cfdnewbie · August 5, 2011, 03:18

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....

dokeun · August 8, 2011, 02:45

Dear cfdnewbie

Thank you for your reply.

Would please review my spreading out of the equations.

$u^{(1)}_{i} = u^{n}_{i}$

$u^{(2)}_{i} = u^{n}_{i} - \frac{1}{2}\Delta t \left( \frac{\partial E}{\partial x}\right)^{(1)}_{i}$

$u^{(3)}_{i} = u^{n}_{i} - \frac{1}{2}\Delta t \left( \frac{\partial E}{\partial x} \right)^{(2)}_{i}$

$u^{(4)}_{i} = u^{n}_{i} - \frac{1}{2}\Delta t \left( \frac{\partial E}{\partial x}\right)^{(3)}_{i}$

$u^{n+1}_{i} = u^{n}_{i} - \Delta t \left[ \frac{1}{6} \left( \frac{\partial E}{\partial x} \right)^{(1)} + \frac{1}{3} \left( \frac{\partial E}{\partial x} \right)^{(2)} + \frac{1}{3} \left( \frac{\partial E}{\partial x} \right)^{(3)} + \frac{1}{6} \left( \frac{\partial E}{\partial x} \right)^{(4)} \right]$

Becasue $\frac{\partial E}{\partial x}$ is defined as $\frac{E_{i+1} - E_{i}}{2 \Delta x} = \frac {u^{2}_{i+1} - u^{2}_{i-1}}{4 \Delta x}$ , $u^{(2)}_{i}$ refers $u^{n}_{i+1},u^{n}_{i-1}$ .

And, in turn, $u^{(3)}_{i}$ refers $u^{(2)}_{i+1},u^{(2)}_{i-1}$ .

Here, $u^{(2)}_{i+1}$ comes from $u^{n}_{i+2},u^{n}_{i}$ and $u^{(2)}_{i-1}$ is derived from $u^{n}_{i},u^{n}_{i-2}$ .

As a result, to get $u^{(3)}_{i}$ , I need the values of $u^{(n)}$ at i+2~i-2.

As the same manner, $u^{(4)}_{i}$ requires the value of $u^{(n)}$ at i-3~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.

dokeun · August 8, 2011, 06:34

dear cfdnewbie

I understood my problem.

my original program was...

DO i=2,IM-1
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,NM-1
DO i=2,IM-1
u1(i) = function of u(n,i)
END DO

DO i=2,IM-1
u2(i) = function of u1(i)
END DO

DO i=2,IM-1
u3(i) = function of u2(i)
END DO

DO i=2,IM-1
u4(i) = function of u3(i)
END DO

DO i=2,IM-1
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)

August 4, 2011, 03:57	1D Burgers euqation with 4th Runge Kutta	#1
dokeun Member Dokeun, Hwang Join Date: Apr 2010 Location: Korea, Republic of Posts: 98 Rep Power: 16	Hello Now i'm tring to apply 4 Stage Runge Kutta scheme to 1D Burgers euqation as a model equation like this. $u^{n+1}_{i} = u^{n}_{i} - \Delta t \left( \frac{\partial E}{\partial x}\right)^{n}$ hear, $E_{i} = \frac{1}{2} {u_{i}}^2$ At this point, the slop is suggested as central differencing. $\frac{\partial E}{\partial x} = \frac{E_{i+1} - E_{i-1}}{2 \Delta x}$ The Question is.. How to handle at the point i=2,3,IM-2,IM-1? Because 4 Stage Runge Kutta requires i-2 ~ i+2 points for next step value at i, I can't get ${u^{n+1}_{i}}$ at i=2,3,IM-2,IM-1. I got the answer using 3 stage RK to i=3,IM-2 and 2 stage RK to i=2,IM-1. 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.

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August 5, 2011, 03:18		#2
cfdnewbie Senior Member cfdnewbie Join Date: Mar 2010 Posts: 557 Rep Power: 20	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
dokeun Member Dokeun, Hwang Join Date: Apr 2010 Location: Korea, Republic of Posts: 98 Rep Power: 16	Dear cfdnewbie Thank you for your reply. Would please review my spreading out of the equations. $u^{(1)}_{i} = u^{n}_{i}$ $u^{(2)}_{i} = u^{n}_{i} - \frac{1}{2}\Delta t \left( \frac{\partial E}{\partial x}\right)^{(1)}_{i}$ $u^{(3)}_{i} = u^{n}_{i} - \frac{1}{2}\Delta t \left( \frac{\partial E}{\partial x} \right)^{(2)}_{i}$ $u^{(4)}_{i} = u^{n}_{i} - \frac{1}{2}\Delta t \left( \frac{\partial E}{\partial x}\right)^{(3)}_{i}$ $u^{n+1}_{i} = u^{n}_{i} - \Delta t \left[ \frac{1}{6} \left( \frac{\partial E}{\partial x} \right)^{(1)} + \frac{1}{3} \left( \frac{\partial E}{\partial x} \right)^{(2)} + \frac{1}{3} \left( \frac{\partial E}{\partial x} \right)^{(3)} + \frac{1}{6} \left( \frac{\partial E}{\partial x} \right)^{(4)} \right]$ Becasue $\frac{\partial E}{\partial x}$ is defined as $\frac{E_{i+1} - E_{i}}{2 \Delta x} = \frac {u^{2}_{i+1} - u^{2}_{i-1}}{4 \Delta x}$ , $u^{(2)}_{i}$ refers $u^{n}_{i+1},u^{n}_{i-1}$ . And, in turn, $u^{(3)}_{i}$ refers $u^{(2)}_{i+1},u^{(2)}_{i-1}$ . Here, $u^{(2)}_{i+1}$ comes from $u^{n}_{i+2},u^{n}_{i}$ and $u^{(2)}_{i-1}$ is derived from $u^{n}_{i},u^{n}_{i-2}$ . As a result, to get $u^{(3)}_{i}$ , I need the values of $u^{(n)}$ at i+2~i-2. As the same manner, $u^{(4)}_{i}$ requires the value of $u^{(n)}$ at i-3~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
dokeun Member Dokeun, Hwang Join Date: Apr 2010 Location: Korea, Republic of Posts: 98 Rep Power: 16	dear cfdnewbie I understood my problem. my original program was... DO i=2,IM-1 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,NM-1 DO i=2,IM-1 u1(i) = function of u(n,i) END DO DO i=2,IM-1 u2(i) = function of u1(i) END DO DO i=2,IM-1 u3(i) = function of u2(i) END DO DO i=2,IM-1 u4(i) = function of u3(i) END DO DO i=2,IM-1 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)