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ODE with Dirac Delta and conditions at infinity

  1. Mar 9, 2015 #1
    I'm trying to solve the following equation (even if I'm not sure if it's well posed)

    [tex]\partial_{x} \, y(x) + a(x)\, y(x) = \delta(x) [/tex]

    with ##\quad \lim_{x \rightarrow \pm \infty}y(x) = 0##

    It would be a classical first order ODE If it were not for the boundary conditions and the Dirac Delta. How should I take account of them?
     
  2. jcsd
  3. Mar 9, 2015 #2

    RUber

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    One idea might be to integrate over x.
     
  4. Mar 9, 2015 #3
    I'm not sure, what you meant....I would get some kind of integral equation ..
     
  5. Mar 9, 2015 #4

    RUber

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    Right - often solutions will involve an integral of some sort.
    Can you solve the homogeneous problem ## y_x + ay = 0?##
    Then, for the dirac delta, you enforce continuity in the function at x=0, and a jump condition in the first derivative at x=0.
     
  6. Mar 9, 2015 #5

    Dick

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    It is a classical ODE except for what must happen at x=0. y(x) must have a jump discontinuity at x=0 assuming a(x) is regular function. Just solve y'(x)+a(x)*y(x)=0 in the domains x>0 and x<0 and patch in the discontinuity.
     
  7. Mar 9, 2015 #6

    RUber

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    Right...thanks Dick. Sorry for the error in my last post. First order ODE has jump in the 0th derivative, i.e. the function itself. I have been spending too much time in 2nd order problems.
     
  8. Mar 9, 2015 #7
    Thanks, following your hint this is I would go:

    if ##y_{\pm}## solve ##y'_{\pm}(x)+a(x)*y_{\pm}(x)=0## in ## \pm x> 0## then ##y_{\pm} = C_{\pm} \, e^{-A_{\pm}(x)} ## and a solution is given by

    [tex] y(x) = y_{-}(x) + H(x) (u_{+}(x) - u_{-}(x)) [/tex]

    where

    [tex] A_{+}(x) = \int_{0^{+}}^{x} a(z) dz [/tex]

    [tex] A_{-}(x) = \int_{x}^{0^{-}} a(z) dz [/tex]

    and ##H(x)## is the Heaviside step function.

    Now I need a jump of one in 0, so I impose ##C_{+} - C_{-} = 1##

    How do I fix the remaining constant? Moreover I guess that at infinity ##A_{\pm}(x)## has to blow up
     
  9. Mar 10, 2015 #8

    Dick

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    I don't think you want to reverse the limits in your definition of ##A_{-}(x)##. But otherwise, I generally agree. I don't think you have enough information to fix the constants any better than that.
     
    Last edited: Mar 10, 2015
  10. Mar 10, 2015 #9

    Ray Vickson

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    The Dirac delta right-hand-side is exactly what you would use when calculating a Green's Function. Since ##\delta(x) = 0## when ##x \neq 0##, you need ##y'(x) + a(x) y(x) = 0## for ##x < 0## and for ##x > 0##. If you integrate the de from ##-\epsilon## to ##+\epsilon##, you get
    [tex] \int_{-\epsilon}^{+\epsilon} y'(x) \, dx + \int_{-\epsilon}^{+\epsilon} a(x) y(x) \, dx = 1.[/tex]
    If ##y(x)## is bounded around ##x = 0##, the second term above vanishes in the limit ##\epsilon \to 0##, while the first term is ##y(+\epsilon) - y(-\epsilon)##, which approaches ##y(+0) - y(-0)## as ##\epsilon \to 0##. In other words, ##y(x)## has a jump discontinuity of magnitude 1 as ##x## passes from left to right through 0.
     
  11. Mar 10, 2015 #10
    Thanks, that's what I thought!
     
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