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Thread: Fibonacci differential equation

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Fibonacci differential equation

    Hello!
    This is just something I've asked myself, and not something I need help with for school.

    Does the differential equation $\displaystyle f'(x)=f(x+1)$ have any nontrivial solutions?
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  2. #2
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    I believe it does. If you seek solutions of the form $\displaystyle f(x) = c e^{kx}, $ then we obtain that $\displaystyle k$ satisfies $\displaystyle k = e^k$.
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  3. #3
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    i think that
    $\displaystyle
    k=ln \; k
    $
    there is no solution.


    $\displaystyle
    f(x)=sin(wx+a)
    $

    $\displaystyle
    f'(x)=w \; cos(wx+a)
    $

    Solving

    $\displaystyle
    w \; cos(wx+a)=sin(w(x+1)+a)
    $

    gives if I am not mistaken

    $\displaystyle
    w=\pm 1
    $

    $\displaystyle
    tg^2 \; a =-1
    $
    there is no solution.
    Last edited by zzzoak; Oct 24th 2010 at 04:39 PM.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Danny View Post
    I believe it does. If you seek solutions of the form $\displaystyle f(x) = c e^{kx}, $ then we obtain that $\displaystyle k$ satisfies $\displaystyle k = e^k$.
    Yeah but $\displaystyle x<e^x$ always! So there's no solution of that form. What about other forms?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Hello!
    This is just something I've asked myself, and not something I need help with for school.

    Does the differential equation $\displaystyle f'(x)=f(x+1)$ have any nontrivial solutions?
    If $\displaystyle \mathcal {L} \{f(t)}\}= \varphi(s)$, for the basic properties of the Laplace Tranform is...

    $\displaystyle \displaystyle \mathcal {L} \{f^{'}(t)}\}= s\ \varphi(s) - f(0)$

    $\displaystyle \displaystyle \mathcal {L} \{f(1+t)}\}= e^{s}\ \varphi(s) $ (1)

    Now from the (1) it follows that a solution of the equation $\displaystyle f^{'} (t) = f(1+t)$ is [if any inverse LT exists]...

    $\displaystyle \displaystyle f(t)= \mathcal{L}^{-1} \{\frac{c}{s-e^{s}}\}$ (2)

    ... being $\displaystyle c=f(0)$ a constant...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  6. #6
    MHF Contributor chisigma's Avatar
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    If it is not requested that $\displaystyle f^{'}(t) = f(1+t)$ is valid for every t but for 'almost every t' we have several 'candidate functions'. One of the most simple is the periodic function with period T=1 defined as...

    $\displaystyle f(t)= e^{t}$ , $\displaystyle 0<t<1$

    Very suggestive is also the 'PSK modulated' periodic function with period T=4 defined as...

    $\displaystyle f(t)=\left\{\begin{array}{ll}\sin 2 \pi t ,\,\,0 < t <1\\{}\\ \cos 2 \pi t ,\,\, 1 < t <2\\{}\\ -\sin 2 \pi t,\,\, 2<t<3 \\{}\\ -\cos 2 \pi t,\,\, 3<t<4\end{array}\right.$

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  7. #7
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    Quote Originally Posted by Bruno J. View Post
    Yeah but $\displaystyle x<e^x$ always! So there's no solution of that form. What about other forms?
    What about complex $\displaystyle x$? Big Picard's gives us infinitely many solutions.
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