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How to handle the Dirac delta function as a boundary condition

  1. Jul 6, 2009 #1
    Using perturbation theory, I'm trying to solve the following problem

    [tex]\frac{\partial P}{\partial \tau} = \frac{1}{2}\varepsilon^2 \alpha^2 \frac{\partial^2 P}{\partial f^2} + \rho \varepsilon^2 \nu \alpha^2 \frac{\partial^2 P}{\partial f \partial \alpha} + \frac{1}{2}\varepsilon^2 \nu^2 \alpha^2 \frac{\partial^2 P}{\partial \alpha^2}, \quad \mbox{for } \tau>0,
    with initial condition [tex]P = \alpha^2~\delta(f-K), \quad \mbox{for } \tau=0. [/tex]

    Expanding [tex] P_\varepsilon=P_0 + \varepsilon^2 P_1 + \ldots[/tex] the [tex]\mathcal{O}(1)[/tex] equation is given by
    [tex]\frac{\partial P_0}{\partial \tau} = 0, \quad \mbox{for } \tau>0,[/tex]
    with boundary condition [tex]P_0 = \alpha^2~\delta(f-K) \mbox{ for } \tau=0[/tex].

    Obviously, this gives [tex]P_0 = \alpha^2~\delta(f-K).[/tex]

    Now I would like to solve the [tex]\mathcal{O}(\varepsilon^2)[/tex] problem
    [tex]\frac{\partial P_1}{\partial \tau} = \frac{1}{2} \alpha^2 \frac{\partial^2 P_0}{\partial f^2} + \rho \nu \alpha^2 \frac{\partial^2 P_0}{\partial f \partial \alpha} + \frac{1}{2} \nu^2 \alpha^2 \frac{\partial^2 P_0}{\partial \alpha^2}, \quad \mbox{for } \tau>0[/tex]
    with initial condition [tex]P_1 = 0 \mbox{ for } \tau=0[/tex].

    Does anyone of you know how to handle the Dirac Delta function in the initial condition and O(1) solution here?
    Last edited: Jul 6, 2009
  2. jcsd
  3. Jul 6, 2009 #2
    There shouldn't be any epsilon in the equations.
  4. Jul 6, 2009 #3
    It is kind of 2D diffusion (or heat conduction) equation with an initial condition, not boundary.

    Physically the exact solution should describe the "relaxation" of initial non uniformity of P.

    I am afraid it cannot be solved by the perturbation theory in powers of epsilon - you neglect the derivative terms that are responsible for the space relaxation.

    Consider a simpler 2D equation - with constant coefficients and analyse the exact solution, if it is expandable (analytical in epsilon at epsilon=0).
  5. Jul 6, 2009 #4
    Sorry, that's my mistake.. think it's a copy-paste error. I corrected it in the previous message.
  6. Jul 6, 2009 #5
    Have you tried a numerical approach?
  7. Jul 6, 2009 #6
    That's always possible, but the assignment here is to do it analytically... Tomorrow I'll ask my supervisor if he thinks there's another way to solve this analytically.
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