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Delta function

  1. Oct 31, 2009 #1
    I'm working on discretizing pde with boundary and initial conditions. The assumption of the method is that functions must be at least twice differentiable. I have an intial condition given by the following function
    [tex]u(x,0)=f(x)=\left\{\begin{array}{cc}x,&\mbox{ } 0 \leq x <1\\2-x, & \mbox{ } 1 \leq x <2 \end{array}\right.[/tex]

    I can calculate the first derivative as
    [tex]f'(x)=\left\{\begin{array}{cc}1,&\mbox{ } 0 \leq x <1\\-1, & \mbox{ } 1 \leq x <2 \end{array}\right.[/tex]
    Nevermind that the derivative at x=1 is not correct. We use discrete variable.

    What about the second derivative. Is it equal to [tex]\delta (x-1) [/tex] or zero? How do we discretize a delta function?
  2. jcsd
  3. Oct 31, 2009 #2
    "Must be at least twice differentiable" ... this means the second derivative must be an actual function, not a generalized function like the delta function. But to answer your question, yes the second derivative is a delta function. It's whatever you have to integrate to recover the previous derivative.
  4. Oct 31, 2009 #3
    Why do you differentiate the initial conditions? Simply use it as it is, which is continuous. No need for a delta functional.

    Anyway, when a delta functional appears in the pde, you will have to solve it in some weak sense.
  5. Nov 1, 2009 #4
    It's clear that f"(x) is not zero. Is it a Dirac delta function or an impulse function ? A Dirac delta function would be difficult to write a program code because it involves infinity.

    As you pointed out, we need to recover the previous derivative. How is this possible? As I see it (for unit impulse)
    [tex]\int_0^x \delta(t-1) dt =\left\{\begin{array}{cc}0,&\mbox{ } 0 \leq x <1\\1, & \mbox{ } 1 \leq x <2 \end{array}\right.[/tex]

    I do not know how to do it for Dirac delta function.
  6. Nov 1, 2009 #5
    I'm trying to solve a simple wave equation utt=uxx. One method that I read from a paper is to use discretize iteration to approximate the solution (the paper claim the method works for Burger equation and Sine-Gordon equation)

    [tex] \ddot{U}_{s+1,i} = U''_{s,i} [/tex]

    I need the initial u"(x,0) to proceed.
  7. Nov 17, 2009 #6
    [tex]f''(x) = -2 \delta(x-1) [/tex]. Discrete delta is a box with area equal to 1, with a width that depends on the size of your step.
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