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In this paper, I can not reproduce the results obtained equation 62.

I have tried to reproduce it using Wolfram alpha but the results are different.

However, using equation (40) instead of the one present in the text just before equation (52), I was able to reproduce the results and confirm equation (62). There is only a change of sign between both expressions. Therefore I am confused.

My question is : what is the correct resolvant condition associated with the inner scheme for the inflow and outflow cases.

Thank you

EDIT1 : Modified the link to the paper EDIT2 : A few equations below trying to make the post self-explanatory.

We consider a 1D transport equation : $$\partial_t U - \partial_x U =0\mbox{, }x \ge 0\mbox{, }t\ge 0$$

No boundary condition is needed at $x=0$. This is the outflow case. The inflow case is similar but a boundary condition at $x=0$ must be provided : $$\partial_t U + \partial_x U =0\mbox{, }x \ge 0\mbox{, }t\ge 0$$

The equations are approximated using finite-difference schemes. Inside the domain : $$f'_{j-1}+4f'_j+f'_{j+1} = \frac{1}{\Delta x}\left( 3f_{j+1}-3f_{j-1} \right)$$

At the boundary : $$2f'_0 + 4f'_1 = \frac{1}{\Delta x}\left( -5f_0 + 4f_1 +f_2 \right)$$

Combining the boundary scheme with the one at $j=1$ leads to : $$4f'_1 + 2 f'_2 = \frac{1}{\Delta x}\left( -f_0 - 4f_1 + 5f_2 \right)$$

For the inflow and outflow equations, we are looking for solutions $U_j(t) = U_0 e^{St} k^j$ and we assume $\Delta x =1$. Substitution of $U_j(t)$ in the PDE of the outflow/inflow leads to : $$\left( k + 4 + \frac{1}{k} \right) S = \pm 3 \left( k - \frac{1}{k} \right)$$

For the outflow, the boundary condition is the scheme at $j=0$ : $\left( 2+4k \right)S=-5+4k+k^2$

For the inflow, the article assumes $f_0$ and the boundary condition is the combined scheme at $j=1$ : $\left( 4+2k \right)S=-4+5k$

This is close to the article. I think the boundary condition for the inflow should be $\left( 4+2k \right)S=4-5k$ but I would like to have this checked seriously.

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  • $\begingroup$ Well, it's behind a paywall for those people who do not have institutional access. Why don't you write down those equations here, and give some background where appropriate? $\endgroup$ – Todd Trimble Jan 13 '15 at 20:50
  • $\begingroup$ @ToddTrimble : I have modified the link so that everybody can access the paper. I will try to add the equations in my post but my understanding of the paper is limited. $\endgroup$ – user1824346 Jan 14 '15 at 10:13

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