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Math Help - Problem 24

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    Problem 24

    Let f be continously differenciable on \mathbb{R}. Solve the integral equation:

    [f(x)]^2 = \int_0^x [f(t)]^2+[f'(t)]^2dt
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    Quote Originally Posted by ThePerfectHacker View Post
    Let f be continously differenciable on \mathbb{R}. Solve the integral equation:

    [f(x)]^2 = \int_0^x [f(t)]^2+[f'(t)]^2dt
    Nice.

    RonL
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    Bar0n janvdl's Avatar
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    So whats the answer CaptainBlack?
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    Quote Originally Posted by janvdl View Post
    So whats the answer CaptainBlack?
    Not telling; that would spoil the fun

    RonL
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    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let f be continously differenciable on \mathbb{R}. Solve the integral equation:

    [f(x)]^2 = \int_0^x [f(t)]^2+[f'(t)]^2dt
    I would much rather deal with a differential equation. Hopefully this works.

    \frac{d}{dx}[f(x)]^2=\frac{d}{dx}\int_0^x [f(t)]^2+[f'(t)]^2dt

    2[f(x)]f'(x) = [f(x)]^2+[f'(x)]^2

    [f(x)]^2-2 [f(x) \cdot f'(x)] + [f'(x)]^2 = 0

    [f(x) - f'(x)]^2=0

    f(x)-f'(x)=0
    f(x)=f'(x)
    f(x) = Ae^x

    Am I right?
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    Senior Member ecMathGeek's Avatar
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    Is there a way to apply color and font to LaTeX? I want to try to white-out my solution, but as far as I can tell that's impossible.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    I would much rather deal with a differential equation. Hopefully this works.

    \frac{d}{dx}[f(x)]^2=\frac{d}{dx}\int_0^x [f(t)]^2+[f'(t)]^2dt

    2[f(x)]f'(x) = [f(x)]^2+[f'(x)]^2

    [f(x)]^2-2 [f(x) \cdot f'(x)] + [f'(x)]^2 = 0

    [f(x) - f'(x)]^2=0

    f(x)-f'(x)=0
    f(x)=f'(x)
    f(x) = Ae^x

    Am I right?
    The LHS is
    A^2e^{2x}

    and the RHS is
    (A^2 - 1)e^{2x}

    so I would say "no."

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    No

    The right hand side is not what you say!

    RonL
    it would be \left( e^{2x} - 1\right)A^2 ?

    if so, it means there is something wrong with what ecMathGeek did, but i didn't really see anything wrong the first time i read it, i'll read it again
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    Quote Originally Posted by Jhevon View Post
    it would be \left( e^{2x} - 1\right)A^2 ?

    if so, it means there is something wrong with what ecMathGeek did, but i didn't really see anything wrong the first time i read it, i'll read it again
    ecMathGeeks proposed solution satisfies the integral equation (ish)

    RonL
    Last edited by CaptainBlack; May 30th 2007 at 08:57 PM.
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    Forum Admin topsquark's Avatar
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    Okay, I oopsied. But still:

    f(x) = Ae^x

    Thus
    f(t) = Ae^t

    \frac{df}{dt} = Ae^t

    Thus
    \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2

     = \int_0^x dt \, (A^2e^{2t} + A^2e^{2t} )

     = \int_0^x dt \, 2A^2e^{2t}

     = 2A^2 \int_0^x dt \, e^{2t}

     = 2A^2 \frac{1}{2}e^{2t}|_0^x

     = A^2(e^{2x} - 1)

    And
     [ f(x) ] ^2 = A^2e^{2x}

    So
     [ f(x) ] ^2 \neq \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2

    How can this then be the solution of the integral equation? I'm missing something here...

    -Dan
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    Quote Originally Posted by topsquark View Post
    Okay, I oopsied. But still:

    f(x) = Ae^x

    Thus
    f(t) = Ae^t

    \frac{df}{dt} = Ae^t

    Thus
    \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2

     = \int_0^x dt \, (A^2e^{2t} + A^2e^{2t} )

     = \int_0^x dt \, 2A^2e^{2t}

     = 2A^2 \int_0^x dt \, e^{2t}

     = 2A^2 \frac{1}{2}e^{2t}|_0^x

     = A^2(e^{2x} - 1)

    And
     [ f(x) ] ^2 = A^2e^{2x}

    So
     [ f(x) ] ^2 \neq \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2

    How can this then be the solution of the integral equation? I'm missing something here...

    -Dan
    So you have shown that if f(t) = Ae^t, then:

     A^2e^{2x}  = A^2(e^{2x} - 1)

    RonL
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    Quote Originally Posted by topsquark View Post
    Okay, I oopsied. But still:

    f(x) = Ae^x

    Thus
    f(t) = Ae^t

    \frac{df}{dt} = Ae^t

    Thus
    \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2

     = \int_0^x dt \, (A^2e^{2t} + A^2e^{2t} )

     = \int_0^x dt \, 2A^2e^{2t}

     = 2A^2 \int_0^x dt \, e^{2t}

     = 2A^2 \frac{1}{2}e^{2t}|_0^x

     = A^2(e^{2x} - 1)

    And
     [ f(x) ] ^2 = A^2e^{2x}

    So
     [ f(x) ] ^2 \neq \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2

    How can this then be the solution of the integral equation? I'm missing something here...

    -Dan
    If you don't like ecMathGeeks approach try a power series solution.

    RonL
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    Quote Originally Posted by ecMathGeek View Post
    f(x) = Ae^x
    Quote Originally Posted by CaptainBlack View Post
    ecMathGeeks proposed solution satisfies the integral equation (ish)
    Quote Originally Posted by CaptainBlack View Post
    So you have shown that if f(t) = Ae^t, then:

     A^2e^{2x}  = A^2(e^{2x} - 1)
    Then A = 0. So ecMathGeeks proposed solution does not satisfy the integral equation for any A, which the second post seems to imply, only when f(x) = Ae^x = 0. Is that what is going on here?
    Last edited by JakeD; May 30th 2007 at 11:46 PM.
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    Quote Originally Posted by JakeD View Post
    Then A = 0. So ecMathGeeks proposed solution does not satisfy the integral equation for any A, which the second post seems to imply, only when f(x) = Ae^x = 0. Is that what is going on here?
    The integral equation when converted to a diffrential equation is in fact
    an initial value problem as of necessity f(0)=0

    ecMathGeek gave a general solution to the DE that you get by differentiating
    the integral equation but omitted the initial condition that forces A=0.

    RonL
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    Quote Originally Posted by CaptainBlank View Post
    If you don't like ecMathGeeks approach try a power series solution.

    RonL
    Not every continously differenciable function is analytic.*


    *)By analytic I mean it has a power series expansion on \mathbb{R}.
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