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Dirac delta function

  1. May 20, 2008 #1
    [SOLVED] Dirac delta function

    1. The problem statement, all variables and given/known data
    Prove that [tex]\delta(cx)=\frac{1}{|c|}\delta(x)[/tex]


    2. Relevant equations



    3. The attempt at a solution

    For any function f(x), [tex]
    \int_{-\infty}^{\infty}f(x)\delta(cx) dx = \frac{1}{c}\int_{-\infty}^{\infty}f(t/c)\delta(t) dt
    [/tex]

    where I have used t=cx.
    [tex]
    =\frac{1}{c}f(0)
    [/tex]
    This is fine and matches RHS for c>0. But how do we get the mod sign for c<0. Why isn't the above procedure valid for c<0 as well?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 20, 2008 #2
    You have implicitly assumed that c>0 when you changed variables. Either that, or you forgot what else needs to be changed...
     
  4. May 20, 2008 #3

    nrqed

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    Science Advisor
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    Gold Member

    What happens to the limits of integration when you make the change of variable and c is negative?
     
  5. May 20, 2008 #4
    thanks nrged!!
    wonder how it didn't strike me after pulling my hair off for one hour
     
  6. Jul 11, 2008 #5
    Re: [SOLVED] Dirac delta function

    Consider a spring-mass system with a time-dependent force f(t) applied to the mass.
    The situation is modelled by the second-order differential equation
    mx"(t) + cx0'(t) + kx(t) = f(t)
    where t is time and x(t) is the displacement of the mass from equilibrium. Now suppose
    that for t greater than or equal to 0 the mass is at rest in its equilibrium position, so x'(0) = x(0) = 0. At
    t = 0 the mass is struck by an “instantaneous” hammer blow. This situation actually occurs
    frequently in practice—a system sustains an forceful, almost-instantaneous input. Our goal
    is to model the situation mathematically and determine how the system will respond.
    In the above situation we might describe f(t) as a large constant force applied on a very
    small time interval. Such a model leads to the forcing function
     
  7. Jul 11, 2008 #6
    Re: [SOLVED] Dirac delta function

    The Dirac delta function is a special kind of object - it is not a
    function in the usual sense, although we sometimes think of it that
    way. Basically, it can be thought of as the "function" d(x) such that:

    d(x) = 0 for x <> 0
    = Infinity for x = 0

    and the "size" of the infinity is such that the integral of d(x) with
    respect to x from a to b is zero if a and b are of the same sign, and
    the integral is exactly unity if a < 0 < b. More precisely, it can be
    thought of as the limit of a series of functions that get more and
    more peaked near zero, and whose area is always unity.

    It is always to be implicitly used inside an integral, even though we
    carry out operations such as:

    f(x) = d(x) + d(x-2) (when integrated, this has area 2)
     
    Last edited: Jul 11, 2008
  8. Jul 11, 2008 #7
    Re: [SOLVED] Dirac delta function

    Consider a spring-mass system with a time-dependent force f(t) applied to the mass.
    The situation is modelled by the second-order differential equation
    mx"(t) + cx0'(t) + kx(t) = f(t)
    where t is time and x(t) is the displacement of the mass from equilibrium. Now suppose
    that for t greater than or equal to 0 the mass is at rest in its equilibrium position, so x'(0) = x(0) = 0. At
    t = 0 the mass is struck by an “instantaneous” hammer blow. This situation actually occurs
    frequently in practice—a system sustains an forceful, almost-instantaneous input. Our goal
    is to model the situation mathematically and determine how the system will respond.
    In the above situation we might describe f(t) as a large constant force applied on a very
    small time interval. Such a model leads to the forcing function
    f (t) = [tex]\left\{\stackrel{^{\frac{A}{\epsilon},}}{0,else}[/tex] 0 [tex]\leq[/tex] t [tex]\leq[/tex] [tex]\epsilon[/tex]

    Here A is some constant and [tex]\epsilon[/tex] is a “small” positive real number. When [tex]\epsilon[/tex] is close to zero the applied force is very large during the time interval 0 [tex]\leq[/tex] t [tex]\leq[/tex] [tex]\epsilon[/tex]
    In this case it’s easy to see that for any choice of [tex]\epsilon[/tex] we have
    [tex]\int[/tex][tex]\stackrel{\infty^}{f (t) dt = A}[/tex]
    [tex]-\infty[/tex]
     
    Last edited: Jul 11, 2008
  9. Jul 13, 2008 #8
    Re: [SOLVED] Dirac delta function

    k..thanx for the info Horbacz !!
    :)
     
  10. Aug 19, 2008 #9
    Re: [SOLVED] Dirac delta function

    your welcome raze 2 dust. and by the way, im only 13...
     
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