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First Order Differential Equation with Reflected Argument

  1. Sep 7, 2012 #1
    I am trying to solve:

    (x + 1 + f(-x) )(1 - f ' (x) ) = x+1
    f(0) = x_0
    x in (-1,1)

    I approximated it numerically but any analytic method I try fails. Any ideas?
     
  2. jcsd
  3. Sep 7, 2012 #2

    haruspex

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    Hint: What do you get if you differentiate (x + 1 + f(-x) )? It should make you think of the chain rule.
     
  4. Sep 7, 2012 #3
    Ok, so I can make a substitution:

    y(x) = x + 1 + f(-x). Then

    y ' (x) = 1 - f ' (-x).

    I don't see where to go from there, since y ' (x) does not appear in the original equation. However

    y ' (-x) = 1 - f ' (x) does appear in the original equation. If i make that substitution I get

    y(x) * y ' (-x) = x+1

    But I still do now know how to solve that. Is my reasoning correct so far?
     
  5. Sep 8, 2012 #4

    haruspex

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    Hmm.. yes I was too quick - I thought the minus sign would not be a problem.
    However, you can at least try making an assumption about y(-x) and see if it produces a solution consistent with the assumption. I.e., try y(-x) = - y(x), then try y(-x) = y(x).
     
  6. Sep 8, 2012 #5
    Judging from the numeric approximation, neither of those seem to be the case. I will try it and see what happens though.

    I have tried many things. One thing I was considering was to use the cross correlation of f(x) with a conveniently chosen function, make a substitution, and use the Laplace transform to force the negative sign inside, then hope the inverse Laplace transform works out. That sounds really horribly complicated though.

    I have tried a series solution but it did not seem to result in anything worthwhile.

    Perhaps a solution does not exist.
     
  7. Sep 15, 2012 #6

    haruspex

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    I finaly realised that f(x) might as well be separate functions for x> 0 and x < 0. so write g( x) = f(-x). obtain (1-x+f(x))(-g'(x)) = 1-x by swapping sign of x. Write orig equation as g(x) = expression in x and f'(x), then differentiate that to obtain expression for g'(x) and substitute in above. Now have 2nd ord (nasty) equation in only f and x. Not sure it helps much though.
     
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