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Math Help - [SOLVED] unique point

  1. #1
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    [SOLVED] unique point

    How do we find the fixed point of the map F:X-->X
    for F(f)(x)=1+ integration of f(x-y)dy from 0 to x
    X=C[0,1/2]

    Thank you
    Last edited by jerad; January 15th 2010 at 01:17 PM.
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  2. #2
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    Quote Originally Posted by jerad View Post
    How do we find the fixed point of the map F:X-->X
    for F(f)(x)=1+ integration of f(x-y)dy from 0 to x
    X=C[0,1/2]

    Thank you
    You want f such that f(x)=1+\int_0^x f(x-y)dy. Put z=x-y in the integral, then differentiate the equation you get, this gives you a differential equation satisfied by f.
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  3. #3
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    thank you

    I will try now
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  4. #4
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    Hi LAURENT

    IS THAT GIVING f '(x)= -f(x)?

    and the unique fixed point is e^(-x)

    am I right?
    Last edited by jerad; January 14th 2010 at 11:57 AM.
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  5. #5
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    Quote Originally Posted by jerad View Post
    Hi LAURENT

    IS THAT GIVING f '(x)= -f(x)?

    and the unique fixed point is e^(-x)

    am I right?
    No, it is f'(x)=f(x). The integral becomes \int_x^0 f(z)(-dz)=\int_0^x f(z) dz, there's no minus sign.
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  6. #6
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    why is the limits from x to 0?
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  7. #7
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    Quote Originally Posted by jerad View Post
    why is the limits from x to 0?
    It comes from the change of variable: if y=0, then z=x, and vice-versa.
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  8. #8
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    oh yes. I substitute wrong. Thank you very much.

    And how do we show F is a contraction on X?
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  9. #9
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    Quote Originally Posted by jerad View Post
    oh yes. I substitute wrong. Thank you very much.

    And how do we show F is a contraction on X?
    | Ff(x)-Fg(x) | \leq  \int_{0}^{x} |f(x-y)-g(x-y) |dy \leq \frac{1}{2} \sup_{z\in [0, \frac{1}{2} ]} \{ |f(z)-g(z) |\}
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