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Boundary Value problem and ODE

  1. May 10, 2012 #1
    Hi, I'm not sure if this is on the right thread but here goes. It's a perturbation type problem.

    Consider the boundry value problem

    $$\epsilon y'' + y' + y = 0$$
    Show that if $$\epsilon = 0$$ the first order constant coefficient equation has
    the solution
    $$y_{outer} (x) = e^{1-x} $$
    I have done this fine.
    Find a suitable rescaling $$X = x/\delta$$ so that the highest derivative is important and balances another term and find the solutions $y_{inner}(X)$ of
    this equation, containing one free parameter, satisfying the boundary
    condition at $$x = X = 0$$

    So I am at the rescaling part and solved the differential equation (after neglecting the δy part of the full equation)
    $$\frac{d^2y}{dX^2} + \frac{dy}{dX} = 0 $$
    yielding $$ y = Ae^{-X} + B$$

    imposing the boundry condition $$x = X = 0$$

    gives $$A = B$$

    so is $$y_{inner} = Ae^{-X}$$ ??
    I think I covered that the highest derivative is important (Although again I was unsure about the wording)

    When I continue on I think I either have this part wrong or the matching is wrong as I am not getting the right answer. I am supposed to be doing intermediate scaling. If anyone has any comments about this that would also be much appreciated.
    Thank you in advance!

    P.S. This is posted on a different thread. A mistake on my part. Not sure how to delete it so please could someone tell me!
     
  2. jcsd
  3. May 10, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Why do you call this a "boundary value problem" when there is no boundary condition given?

    If a condition is that y(0)= e, yes, that would be correct.

    Yes, that's good.

    ??? at X= 0, y(0)= A+ B but if you are given no specific value of y there, it gives you nothing. If you are taking y(0)= 0, then you would have A= -B, not A= B.

    Why? If you do have A= B, then it would be y= A(e^{-X}+ 1). If A= -B, it would be y= A(e^{-X}- 1)
    Now, as X goes to infinity (x is not infinitesmal) this would give y= -A and you want to match that to y= e^{1- x} which is e at x= 0.

     
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