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Finite Difference Approximation, Mathematica code

  1. Oct 27, 2010 #1
    1. The problem statement, all variables and given/known data

    I have to program a three component decay chain using finite difference approximation. I understand finite difference and have written my code, but I have an error I can not find which is giving me an erroneous answer. The curve is correct, but the magnitude of the y-axis values is too high. I've gone through it for hours but can not find the mistake.

    2. The attempt at a solution

    It should be pretty basic stuff, but there is something screwy in there. My problem starts with the second chain (N2a and N2b.)

    The governing differential equation for that section is:
    [tex]\frac{dN_{2}(t)}{dt}=-\lambda _{2}N_{2}(t)+\lambda _{1}N_{1}(t)[/tex]

    My forward approximation is:
    [tex]N_2b=(\lambda_{1} N_1a - \lambda_{2} N_2a) (tchange) + N_2a[/tex]

    For some reason result of this is far too large, but the curve looks to be correct. Here is an image of the real versus approx graph I get:

    Approx:
    [PLAIN]http://img19.imageshack.us/img19/9539/n2approxgraph.png [Broken]
    Real (analytical):
    [PLAIN]http://img222.imageshack.us/img222/5544/n2realgraph.png [Broken]

    Here is a portion of the code in question:

    Code (Text):
    lambdaA = .6931;
    lambdaB = .13863;
    tchange = .05;
    ttotal = 75;
    A0 = 100;
    B0 = 0;
    C0 = 0;

    timesteps = ttotal/tchange;
    list = Table[{i*tchange, 0}, {i, timesteps}];
    list[[1, 2]] = A0;
    list2 = Table[{i*tchange, 0}, {i, timesteps}];
    list2[[1, 2]] = B0;
    list3 = Table[{i*tchange, 0}, {i, timesteps}];
    list3[[1, 2]] = C0;
    For[i = 1, i < timesteps, i++,
     N1a = Evaluate[Extract[list, {i, 2}]];
     N1b = Evaluate[N1a (1 - tchange*lambdaA)];
     list[[i + 1, 2]] = N1b;
     N2a = Extract[list2, {i, 2}];
     N2b = Evaluate[(lambdaA N1a - lambdaB N2a) tchange + N2a];
     list2[[i + 1, 2]] = N2b;
     N3a = Extract[list3, {i, 2}];
     N3b = Evaluate[tchange*lambdaB*N2b + N3a];
     list3[[i + 1, 2]] = N3b;]
     
    Thanks a lot!
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 27, 2010 #2
    Oh my god! Its not wrong! I plotted the analytical solution incorrectly. How about that? The numerical solution being correct but the analytical being wrong. Ugh!

    Well, anyway.

    Anyone see any ways to streamline the code?
     
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