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Differential operator

  1. Oct 17, 2005 #1
    why is the answer to a differential operator the same as the answer to the original O.D.E. equation?
     
  2. jcsd
  3. Oct 17, 2005 #2

    HallsofIvy

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    That question doesn't even makes sense. A "differential operator" is neither an equation not a "problem" and doesn't have an answer!

    Could you please rephrase your question.
     
  4. Oct 18, 2005 #3

    dextercioby

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    I think he means the "fundamental solution" of the differential operator. :rolleyes:

    Daniel.
     
  5. Oct 18, 2005 #4
    for example:
    (D^2-d-2)y=0
    if you solve D, which is D=2,-1
    it's the same as solving the equation~
     
  6. Oct 19, 2005 #5

    dextercioby

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    In the light of your latest discoveries, your question still doesn't make too much sense.

    Daniel.
     
  7. Oct 19, 2005 #6

    HallsofIvy

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    Actually, I would argue that that doesn't make sense either- an operator is not an equation. An equation may have a solution, but not the operator!

    Okay, I can understand that, although your terminology is still odd!
    I presume you are referring to the differential equation:
    (D2- D- 2)y= 0 where "D" is the differential operator d/dx (or d/dt). D2- D- 2 would also be a linear differential operator.

    However, solving the equation D2- D- 2= 0 to get D= 2 or D= -1 is what mathematicians call "abuse of terminology". If you intend D to be "d/dx", it clearly doesn't make sense to turn around and say that D= 2!
    It is, though, convenient shorthand and we do it all the time. It is convenient shorthand for the "characteristic equation". If we were to look for a solution of the form [itex]y= e^{\lambda x}[/itex], putting that into the equation would give [itex]\lambda^2 e^{\lambda x}- \lambda e^{\lambda x}- 2e^{\lambda x}= 0[/itex] or [itex]\left(\lambda^2- \lambda- 2\right)e^{\lambda x}= 0[/itex]. Since [itex]y= e^{\lambda x}[/itex] is never 0, we must have [itex]\lambda^2- \lambda- 2= 0[/itex], the characteristic equation.
    That "looks like" the original equation, especially in operator notation (which is the main reason for using it) because of that very nice property of exponentials:
    [tex]\frac{d^n(e^{\lambda x}}{dx^n}= \lambda^n e^{\lambda x}[/tex]

    Caution!! This is only true for a very limited (though important) class of differential equations: linear equations with constant coefficients.
     
    Last edited: Oct 19, 2005
  8. Oct 19, 2005 #7
    sorry, i didn't make myself clear~
    for the example above, usually the normal way is to suppose that
    y=ce^(namda)x
    and find namda
    but the answer to namda is the same as finding the answer to the differential operator~
     
  9. Oct 21, 2005 #8

    HallsofIvy

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    Yes, that was what I just said. Oh, by the way, the Greek letter is "lambda", not "nambda".
     
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