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Dealing with boundary conditions in system of ODEs

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Homework Statement


I'm trying to plot the steady state concentration of yA vs. x, yB vs x and yu vs x using centered finite difference method.
7bf7ff66226dc782690fe0f4ed0d511a.png

fc9d4e25a6988d7185732bc227e8441b.png


Homework Equations


1953a53dd7b94b8f9a1a3dff7d3d322b.png



The Attempt at a Solution


τ represents the dimensionless time variable, so steady state would mean that the left hand side of each of the differential equations is 0.

I began this problem by replacing the derivatives that appear in the differential equations with the finite difference approximations for them. For example, yA:
gif.gif

I also used the boundary conditions:
286a851dbe10ba7c4bf2a865c10f7be4.png

but I don't know at which point, I am taking yA on the right side from. In the below equation should yA on the right side be yA,1 or something else (since the below equation estimates the derivative at the node n=1?
gif.gif


More importantly, how do I deal with this type of boundary condition in Matlab? I am used to dealing with boundary conditions such as yA(0,t)=0 for which I can simply initialize Y(1)=0 in matlab but in this case I'm given a derivative boundary condition.
 

Answers and Replies

  • #2
SteamKing
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These aren't ODEs; they're PDEs.
 
  • #3
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What you do is add a fictitious point at -1 such that:

[tex]\frac{y_{1}-y_{-1}}{2h}=-(1-y_0)[/tex]

Combining this with ##\frac{(y_{-1}+2y_0+y_1)}{h^2}## gives:

[tex]\frac{(y_{-1}-2y_0+y_1)}{h^2}=\frac{2(y_1-y_0)}{h^2}-\frac{2(1-y_0)}{h}[/tex]

You solve the time-dependent equation at x = 0, and use this in that equation. This is a standard way of handling that type of flux boundary condition.

Chet
 

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