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Odd Differencing Method for Solving DE

  1. Jul 12, 2011 #1
    Hey all,

    I have a friend working in Earth Sciences who appears to be doing something with DEs. If I understand his data correctly, it appears as though he's solving a differential equation of the form
    [tex] \frac{dy}{dx} = f(x) [/tex]
    Let [0,L] be the interval over which this is to be solved, [itex] y(0) = y_0 [/itex] and take a partition [itex] 0 =x_0 < x_1 < \cdots < x_{n-1} < x_n = L [/itex]. Then it appears that his differencing method is giving the approximation of the [itex] (i+1)^{st} [/itex] value as
    [tex] y(x_{i+1}) = \frac{ f(x_{i+1}) (x_{i+1} - x_i) - \left(\sum_{j=0}^i y(x_j)\right) (x_i - x_{i-1}) }{x_{i+1}} [/tex]
    I don't recognize the formula. The summation term would be theoretically reminiscent of an integral yes? Does anyone recognize this?

    Edit: Sorry, that summation might actually only be [itex] y(x_i) [/itex]. I'm not quite sure yet since the data is a little fuzzy. If that's the case, this is almost an Euler method right? But it still doesn't quite seem there.
     
  2. jcsd
  3. Jul 12, 2011 #2
    Are you sure it's a y(x_j) in the sum and not f(x_j)? Given the DE, y(x_j) makes no sense at all.

    EDIT: No, actually, it doesn't make sense either way.
     
  4. Jul 12, 2011 #3
    Just checked the data. That summation is wrong, but it does represent the solution at the previous partition point. This would be [itex] y(x_i) [/itex].

    It almost looks like an Euler method with some sort of bizarre scaling. We can re-arrange it to get

    [tex] y(x_{i+1}) x_{i+1} = y(x_i)(x_i -x_{i-1}) + f(x_{i+1}) (x_{i+1} - x_i) [/tex]

    My friend now tells me that if we drop the [itex] x_{i-1} [/itex] term from the y(x_i) he gets better results. This would leave

    [tex] y(x_{i+1}) x_{i+1} = y(x_i)x_i + f(x_{i+1}) (x_{i+1} - x_i) [/tex]

    Which looks exactly like an Euler method with some sort of weird scaling on the solution.
     
  5. Jul 12, 2011 #4
    Looks more like Adams-Bashforth than Euler to me.
     
  6. Jul 13, 2011 #5
    Isn't one-step AB just Euler?
     
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