Modified equation for KdV using 2.Order Discretization scheme
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The modified equation of linear PDEs can be found in a systematic manner (https://www.sciencedirect.com/science/article/pii/0021999174900114). However, it does not seem to be that easy for nonlinear PDEs. The references I found about modified equations for nonlinear PDEs are not very specific and just state that the process involves lots of algebra.
I'm trying to find the modified equation of the KdV equation
$u_x + uu_x + u_{xxx} = 0$
using the standard discretitzation scheme:
$(u_j^{k+1}-u_j^{k-1})/2tau + (u_{j+1}^k+u_j^k+u_{j-1}^k)(u_{j+1}^k-u_{j-1}^k)/6h + (u_{j-2}^k-2u_{j+1}^k+2u_{j-1}^k-u_{j-2}^k)/2h^3=0 $
When I start eliminating the 3rd time-derivative with the same approach as in the linear case, the number of terms starts blowing up. Trying to eliminate those creates even more inconvenient terms and I haven't found a systematic way to solve this problem.
Any comments are very much appreciated!
pde numerical-methods mathematical-physics fluid-dynamics finite-differences
asked Jan 12 at 20:37
The_wild_ponyThe_wild_pony
63
$endgroup$
add a comment |
$begingroup$
The modified equation of linear PDEs can be found in a systematic manner (https://www.sciencedirect.com/science/article/pii/0021999174900114). However, it does not seem to be that easy for nonlinear PDEs. The references I found about modified equations for nonlinear PDEs are not very specific and just state that the process involves lots of algebra.
I'm trying to find the modified equation of the KdV equation
$u_x + uu_x + u_{xxx} = 0$
using the standard discretitzation scheme:
$(u_j^{k+1}-u_j^{k-1})/2tau + (u_{j+1}^k+u_j^k+u_{j-1}^k)(u_{j+1}^k-u_{j-1}^k)/6h + (u_{j-2}^k-2u_{j+1}^k+2u_{j-1}^k-u_{j-2}^k)/2h^3=0 $
When I start eliminating the 3rd time-derivative with the same approach as in the linear case, the number of terms starts blowing up. Trying to eliminate those creates even more inconvenient terms and I haven't found a systematic way to solve this problem.
Any comments are very much appreciated!
pde numerical-methods mathematical-physics fluid-dynamics finite-differences
asked Jan 12 at 20:37
The_wild_ponyThe_wild_pony
63
$endgroup$
$begingroup$
Why did you decide to get rid of the 3rd derivative? I mean, you starting point is 3rd order pde.
$endgroup$
– VorKir
Feb 3 at 5:37
add a comment |
$begingroup$
The modified equation of linear PDEs can be found in a systematic manner (https://www.sciencedirect.com/science/article/pii/0021999174900114). However, it does not seem to be that easy for nonlinear PDEs. The references I found about modified equations for nonlinear PDEs are not very specific and just state that the process involves lots of algebra.
I'm trying to find the modified equation of the KdV equation
$u_x + uu_x + u_{xxx} = 0$
using the standard discretitzation scheme:
$(u_j^{k+1}-u_j^{k-1})/2tau + (u_{j+1}^k+u_j^k+u_{j-1}^k)(u_{j+1}^k-u_{j-1}^k)/6h + (u_{j-2}^k-2u_{j+1}^k+2u_{j-1}^k-u_{j-2}^k)/2h^3=0 $
When I start eliminating the 3rd time-derivative with the same approach as in the linear case, the number of terms starts blowing up. Trying to eliminate those creates even more inconvenient terms and I haven't found a systematic way to solve this problem.
Any comments are very much appreciated!
pde numerical-methods mathematical-physics fluid-dynamics finite-differences
asked Jan 12 at 20:37
The_wild_ponyThe_wild_pony
63
$endgroup$
The modified equation of linear PDEs can be found in a systematic manner (https://www.sciencedirect.com/science/article/pii/0021999174900114). However, it does not seem to be that easy for nonlinear PDEs. The references I found about modified equations for nonlinear PDEs are not very specific and just state that the process involves lots of algebra.
I'm trying to find the modified equation of the KdV equation
$u_x + uu_x + u_{xxx} = 0$
using the standard discretitzation scheme:
$(u_j^{k+1}-u_j^{k-1})/2tau + (u_{j+1}^k+u_j^k+u_{j-1}^k)(u_{j+1}^k-u_{j-1}^k)/6h + (u_{j-2}^k-2u_{j+1}^k+2u_{j-1}^k-u_{j-2}^k)/2h^3=0 $
When I start eliminating the 3rd time-derivative with the same approach as in the linear case, the number of terms starts blowing up. Trying to eliminate those creates even more inconvenient terms and I haven't found a systematic way to solve this problem.
Any comments are very much appreciated!
pde numerical-methods mathematical-physics fluid-dynamics finite-differences
pde numerical-methods mathematical-physics fluid-dynamics finite-differences
asked Jan 12 at 20:37
The_wild_ponyThe_wild_pony
63
asked Jan 12 at 20:37
The_wild_ponyThe_wild_pony
63
asked Jan 12 at 20:37
The_wild_ponyThe_wild_pony
63
asked Jan 12 at 20:37
The_wild_ponyThe_wild_pony
63
asked Jan 12 at 20:37
The_wild_ponyThe_wild_pony
63
63
$begingroup$
Why did you decide to get rid of the 3rd derivative? I mean, you starting point is 3rd order pde.
$endgroup$
– VorKir
Feb 3 at 5:37
add a comment |
$begingroup$
Why did you decide to get rid of the 3rd derivative? I mean, you starting point is 3rd order pde.
$endgroup$
– VorKir
Feb 3 at 5:37
$begingroup$
Why did you decide to get rid of the 3rd derivative? I mean, you starting point is 3rd order pde.
$endgroup$
– VorKir
Feb 3 at 5:37
$begingroup$
Why did you decide to get rid of the 3rd derivative? I mean, you starting point is 3rd order pde.
$endgroup$
– VorKir
Feb 3 at 5:37
add a comment |
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$begingroup$
Why did you decide to get rid of the 3rd derivative? I mean, you starting point is 3rd order pde.
$endgroup$
– VorKir
Feb 3 at 5:37