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Thread: Problem 24

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    Problem 24

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

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

    $\displaystyle [f(x)]^2 = \int_0^x [f(t)]^2+[f'(t)]^2dt$
    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 $\displaystyle f $ be continously differenciable on $\displaystyle \mathbb{R}$. Solve the integral equation:

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

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

    $\displaystyle 2[f(x)]f'(x) = [f(x)]^2+[f'(x)]^2$

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

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

    $\displaystyle f(x)-f'(x)=0$
    $\displaystyle f(x)=f'(x)$
    $\displaystyle f(x) = Ae^x$

    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.

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

    $\displaystyle 2[f(x)]f'(x) = [f(x)]^2+[f'(x)]^2$

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

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

    $\displaystyle f(x)-f'(x)=0$
    $\displaystyle f(x)=f'(x)$
    $\displaystyle f(x) = Ae^x$

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

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

    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 $\displaystyle \left( e^{2x} - 1\right)A^2$ ?

    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 $\displaystyle \left( e^{2x} - 1\right)A^2$ ?

    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:

    $\displaystyle f(x) = Ae^x$

    Thus
    $\displaystyle f(t) = Ae^t$

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

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

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

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

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

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

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

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

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

    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:

    $\displaystyle f(x) = Ae^x$

    Thus
    $\displaystyle f(t) = Ae^t$

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

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

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

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

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

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

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

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

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

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

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

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

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

    $\displaystyle f(x) = Ae^x$

    Thus
    $\displaystyle f(t) = Ae^t$

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

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

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

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

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

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

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

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

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

    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
    $\displaystyle f(x) = Ae^x$
    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 $\displaystyle f(t) = Ae^t$, then:

    $\displaystyle A^2e^{2x}$$\displaystyle = A^2(e^{2x} - 1)$
    Then $\displaystyle A = 0.$ So ecMathGeeks proposed solution does not satisfy the integral equation for any $\displaystyle A$, which the second post seems to imply, only when $\displaystyle f(x) = Ae^x = 0.$ 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 $\displaystyle A = 0.$ So ecMathGeeks proposed solution does not satisfy the integral equation for any $\displaystyle A$, which the second post seems to imply, only when $\displaystyle f(x) = Ae^x = 0.$ 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 $\displaystyle f(0)=0$

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

    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 $\displaystyle \mathbb{R}$.
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