How to they get to this equation? (Integral representation for differential equation)

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  1. TheEd's Avatar
    1 11-04-2012  18:42
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    How to they get to this equation? (Integral representation for differential equation)
    I'm trying to understand how to use Laplace method to find solutions to ODEs. How do they use the correspondence rules (6.24) below to get from f''-2zf' + 2\nu f=0 to t^2 v + 2(tv)' + 2\nu v=0 ?

    Last edited by TheEd; 11-04-2012 at 18:43.
  2. sexbo's Avatar
    2 11-04-2012  18:56
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    Re: How to they get to this equation? (Integral representation for differential equat
    **** I should be revising this.
    Post rating:
    1
  3. steve10's Avatar
    3 11-04-2012  20:17
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    Re: How to they get to this equation? (Integral representation for differential equat
    Apply the "trick" in (6.23) to the product z f'(z). Then use that result in your 3rd-last eqn.

    You should be able to get the middle term, 2(tv)'.

    Does that help?
    Last edited by steve10; 11-04-2012 at 20:20.
  4. TheEd's Avatar
    4 11-04-2012  20:38
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    Re: How to they get to this equation? (Integral representation for differential equat
    (Original post by steve10)
    Apply the "trick" in (6.23) to the product z f'(z). Then use that result in your 3rd-last eqn.

    You should be able to get the middle term, 2(tv)'.

    Does that help?
    How have they obtained (6.23)? What have they integrated by parts?
  5. steve10's Avatar
    5 11-04-2012  20:58
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    Re: How to they get to this equation? (Integral representation for differential equat
    For (6.23), the IBP rule is,

    \int u\frac{dv}{dt}dt=uv-\int v\frac{dv}{dt}dt

    Call the zf(z) that you are working with in your post as uv in the IBP formula, with the uv term being (I imagine) the boundary term(s) that vanishes.

    Can you see now how (6.23) is got ?

    I'm off to watch the Apprentice now. Back in an hour.
  6. TheEd's Avatar
    6 11-04-2012  22:02
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    Re: How to they get to this equation? (Integral representation for differential equat
    If you put u=z and v=f(z)=\int e^{zt} v(t) dt in the IBP formula don't you get:

    \displaystyle \int z e^{zt} v(t) dt = zf(z) - \int (\int e^{zt} v(t) dt) \frac{dz}{dt} dt
  7. steve10's Avatar
    7 11-04-2012  22:52
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    Re: How to they get to this equation? (Integral representation for differential equat
    zf(z)=\int ze^{zt} v(t)dt
    zf(z)=\int v(t)\frac{d(e^{zt})}{dt}dt

    Writing the IBP formula as,

    \int u\frac{dw}{dt}dt=uw-\int w\frac{du}{dt}dt

    Now let u=v(t) and w=e^{zt}, then we get

    \int v(t)\frac{d(e^{zt})}{dt}dt=[v(t)\cdot e^{zt}]_{t_1}^{t_2}-\int e^{zt} \frac{dv}{dt}dt

    Here the boundary term, [v(t)\cdot e^{zt}]_{t_1}^{t_2} is assumed to be zero, giving,

    zf(z)=\int v(t)\frac{d(e^{zt})}{dt}dt=-\int e^{zt} \frac{dv}{dt}dt
  8. steve10's Avatar
    8 11-04-2012  23:06
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    Re: How to they get to this equation? (Integral representation for differential equat
    Sorry, I think I must have gotten mixed up with u's and v's and IBP. I gave you some bad info on post #5. Please ignore that post.

    My bad
  9. TheEd's Avatar
    9 11-04-2012  23:27
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    Re: How to they get to this equation? (Integral representation for differential equat
    (Original post by steve10)
    zf(z)=\int ze^{zt} v(t)dt
    zf(z)=\int v(t)\frac{d(e^{zt})}{dt}dt

    Writing the IBP formula as,

    \int u\frac{dw}{dt}dt=uw-\int w\frac{du}{dt}dt

    Now let u=v(t) and w=e^{zt}, then we get

    \int v(t)\frac{d(e^{zt})}{dt}dt=[v(t)\cdot e^{zt}]_{t_1}^{t_2}-\int e^{zt} \frac{dv}{dt}dt

    Here the boundary term, [v(t)\cdot e^{zt}]_{t_1}^{t_2} is assumed to be zero, giving,

    zf(z)=\int v(t)\frac{d(e^{zt})}{dt}dt=-\int e^{zt} \frac{dv}{dt}dt
    Thanks. I get where the equation comes from now but I don't really get the 'correspondence' things \leftrightarrow.
  10. steve10's Avatar
    10 11-04-2012  23:37
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    Re: How to they get to this equation? (Integral representation for differential equat
    Well, to be honest, I don't fully understand (6.24), but I can get the 2nd last eqn in your post. Give me few minutes.
  11. steve10's Avatar
    11 11-04-2012  23:51
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    Re: How to they get to this equation? (Integral representation for differential equat
    Using the method in my post #7, we can get,

    zf'(z)=-\int e^{zt}(tv)'dt

    Now, f''=\int e^{zt}t^2vdt and f=\int e^{zt}vdt

    So, f''-2zf'+2\nu f=0 becomes

    \int e^{zt}t^2vdt+2\int e^{zt}(tv)'dt+2\nu\int e^{zt}vdt=0

    \int e^{zt}\left(t^2v+2(tv)'+2\nu v\right)dt=0

    i.e. t^2v+2(tv)'+2\nu v=0
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