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Math Help - Green's Function for first order Linear ODE

  1. #1
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    Green's Function for first order Linear ODE

    Hi there,
    I've been completing a problem of a question based on ODE and Green's Functions. The question asks me to come up with a general solution for the IVP ODE L[y] = y\prime + y\cdot sin(t)=\delta(t - \tau), y(0) = 0 by imposing a jump condition on y(t) at  t = \tau and solving the differential equation for t < \tau and  t > \tau. Then identify the Green's function G(t,\tau) for the IVP ODE L[y] = y\prime + y\cdot sin(t)=q(t), y(0) = 0
    I was wondering whether someone could read through my working and see whether I have the correct Green's function.

    Lets intergrate L[y] = y\prime + y\cdot sin(t)=\delta(t - \tau) from times \tau + \varepsilon to \tau - \varepsilon
    \delta(t - \tau) will equal 1 over this interval, as it is the step function. The area under it's curve will always equal 1.
    We take the limit as \varepsilon approaches 0. This makes the intergal of y\cdot sin(t) equal to zero. So we are left with [y] evaluated from \tau^+ to \tau^-. This equals 1.

    To find the general solution, I need to solve the IVP for t < \tau and  t > \tau. Start with t < \tau.
    The solution to the ODE is y(t) = A\cdot \exp^{cos(t)}. Since the IVP states y(0) = 0, 0 = A\cdot \exp^{cos(0)} This means that A=0. Therefore, y(t) = 0 for all t < \tau.

    For t > \tau, we need a new IVP. It is found from [y] = 1 from evaluating [y] from \tau^+ to \tau^-. Expanding this,  y(\tau^+) - y(\tau^-) = 1 . We found that  y(\tau^-) = 0 from above, therefore y(\tau^+) = 1. This is the new IVP for t > \tau.
    The ODE is y(t) = B\cdot \exp^{cos(t)} . From the IVP, 1 = B\cdot \exp^{cos(\tau)} . This means B =  \exp^{-cos(\tau)}.
    Therefore, y(t) = \exp^{-cos(\tau)} \cdot \exp^{cos(t)}.

    Finally, the solution to the ODE:
    y(t) = 0 for all t < \tau.
    y(t) = \exp^{cos(t) - cos(\tau)} for all t > \tau
    This means my Green's function is  \exp^{cos(t) - cos(\tau)}
    Am I right?

    Thank you for your time.
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  2. #2
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    Looks good!
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  3. #3
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    But you forgot the integration constant for cost!
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  4. #4
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    Yeah, I did... whoops. Thanks!
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