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Green function

  1. Mar 30, 2007 #1
    1.
    The zero order bessel function J0(0+) = 1, J'0(0+) = 0. and J0(t) satisfies teh differential equation


    2. ty''(t) + y'(t) + ty(t) = 0, t>0
    Prove thata the Laplace transform of J0(t) is 1/(s + 1)^1/2

    3. The attempt at a solution

    Could anyone help on this. How to I go about with the question. Tried using a general solution that is J(t) = C1e^r1t + C2e^r2t..But im not sure if that right... I wasn't able to prove...Thx
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 30, 2007 #2

    Integral

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    Why not start by finding the Laplace transform of your DE?
     
  4. Mar 30, 2007 #3
    thx for replying...But how do I find the Laplace transform of the differential equation? Do i convert it to general solution J(t) = C1e^r1t + C2e^r2t. Is that right? Because i tried using that general equation and found two values of t and substituted in the general equation and solve simultaneously..but still wasn't able to prove...
     
  5. Mar 30, 2007 #4
    I've found websites and even read books. It all says the same thing about bessel equation having a general solution. http://eqworld.ipmnet.ru/en/solutions/ode/ode0213.pdf
    But what i dont get is that the in the general solution what is my v in my DE. I dont see any relationship. I am so confused..
     
  6. Mar 31, 2007 #5

    HallsofIvy

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    If you are working with Bessel functions and Green's functions (you titled this "Green function" but I don't see anything in the problem that has anything to do with Green's function) then you should have learned long ago that functions of the form [itex]C_1e^{r_1t}+ C_2e^{r_2t}[/itex] satisfy linear equations with constant coefficients. Bessel's equation, the equation you are given, obviously does NOT have integer coefficients. You are not expected to "find" a solution, you are expected to find the Laplace transform of the differential equation.

    Do you know how to find the Laplace transform of something like f(x) dy/dx in terms of the Laplace transform of y?
     
  7. Mar 31, 2007 #6
    I am not quite sure but is the Lapalce transform of f(x) dy/dx is sY(s) - Yo and the laplace transform of d2y/dx2 is s2Y(s) - sY0 - Y1.

    If I am wrong. Please kindly correct me and if you could kindly give me some hints to go about with the question. Thx
     
  8. Mar 31, 2007 #7
    ty''(t) + y'(t) + ty(t) = 0
    Divide the equation by t gives

    y''(t) + 1/ty'(t) + ty(t) = 0

    Using Laplace Transform it gives

    s2*Y - s*y0 - y1 + 1/t*(S*Y - y0) + t*Y = 0

    (s2 + 1/t*s + t)*Y - s - 1/t = 0

    so now Y = (s = 1/t)/(s2 + 1/t*s + t)

    but i still cant prove the equation.....:(
     
  9. Apr 1, 2007 #8

    HallsofIvy

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    The Laplace transform of tf(t) is NOT t times the Laplace transform of f(t)!
     
  10. Apr 1, 2007 #9
    oh there was an error

    ty''(t) + y'(t) + ty(t) = 0
    Divide the equation by t gives

    y''(t) + 1/t*y'(t) + y(t) = 0

    Using Laplace Transform it gives

    s2*Y - s*y0 - y1 + 1/t*(s*Y - y0) + Y = 0

    (s2 + 1/t*s + 1)*Y - s - 1/t = 0

    so now Y = (s + 1/t)/(s2 + 1/t*s + 1)

    I have tried completing the square but it doesnt help.....Is my Y correct?
     
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