Forward substitution in this case? Is it as simple as I think it is?

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SUMMARY

The discussion centers on the application of forward substitution in a system of equations for 1-dimensional advection, specifically using finite difference calculations. The user seeks clarification on whether it is correct to substitute previously calculated values of uik+1 into subsequent steps of the equations. The equations provided are u1k+1 = (u1k - a(0, tk+1)/d, u2k+1 = (u2k - a(0, u1k+1)/d, and uik+1 = (uik - a(0, ui-1k+1)/d, where a and d are constants. The user is encouraged to utilize standard techniques from linear algebra to proceed with their calculations.

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  • Understanding of linear algebra concepts, particularly forward substitution.
  • Familiarity with finite difference methods for numerical analysis.
  • Basic knowledge of 1-dimensional advection equations.
  • Proficiency in coding for implementing mathematical models.
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  • Study the application of forward substitution in solving linear systems.
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This discussion is beneficial for students and professionals in computational mathematics, numerical analysts, and software developers working on simulations involving advection and linear algebra techniques.

bzz77
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Hi everyone:

I am very rusty on linear algebra, so apologies if this is a silly question. The question is, in the system below, is it correct to take the calculated value of uik+1 from each PREVIOUS step and simply plug it in at the NEXT step where (a * ui-1k+1 is required.

I need to incorporate a finite difference calculation for 1-dimensional advection in some code I'm writing. I have derived the system of equations that I need, but my linear algebra is so rusty that I'm stuck on what to do next/how to actually use them. I'm hoping that the solution I proposed will be sufficient. If it isn't, what method from linear algebra should I use?

I would very much appreciate any help. Thank you all.

Here is the system (a and d are constants that I can specify):

u1k+1 = (u1k - a(0, tk+1)/d

u2k+1 = (u2k - a(0, u1k+1)/d

uik+1 = (uik - a(0, ui-1k+1)/d
 
Last edited:
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Yes, that is a standard technique.
 

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