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Math Help - di ffusion equation

  1. #1
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    di ffusion equation

    Good morning all, I am stuck on this question.

    Given u(x,t)=f(x-at) is a solution to the diffusion equation, find f and show that the speed is arbitrary.
    Thanks for any help at all guys and girls.
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  2. #2
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    Re: di

    Did you mean wave equation? If so, I'm not sure you can determine f uniquely. You're actually exhibiting one of the two d'Alembert solutions, where f is arbitrary, but a is not. The other d'Alembert solution looks like f(x+at).
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  3. #3
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    Re: di

    Quote Originally Posted by Ackbeet View Post
    Did you mean wave equation? If so, I'm not sure you can determine f uniquely. You're actually exhibiting one of the two d'Alembert solutions, where f is arbitrary, but a is not. The other d'Alembert solution looks like f(x+at).
    No, I mean the diffusion equation u_t-ku_{xx}=0. That's why this question confuses me.
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    Re: di

    Hmm. Well, I'm not sure I see my way through to the end of the problem, but plug in your f into the diffusion equation. What do you get?
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  5. #5
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    Re: di

    I get -a \cdot f'(x-at) -k \cdot f''(x-at)=0,
    so f'(x-at)= - \frac{a}{k}

    I think I am confusing myself.
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  6. #6
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    Re: di

    So now suppose s = x - at. You have an ODE for f, right? I agree with your first equation, but not the second. How did you get that?
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  7. #7
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    Re: di

    Sorry, that second was a mistake, makes no sense.

    -kf''(s)-af'(s)=0
    kf''(s)+af'(s)=0

    So the characteristic equation is kz^2+az=0
    which gives z=0 or z=-a/k

    Is this correct?
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    Re: di

    Quote Originally Posted by kittycat485 View Post
    Sorry, that second was a mistake, makes no sense.

    -kf''(s)-af'(s)=0
    kf''(s)+af'(s)=0

    So the characteristic equation is kz^2+az=0
    which gives z=0 or z=-a/k

    Is this correct?
    Yes, so far.

    So the solution is...
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  9. #9
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    Re: di

    Quote Originally Posted by Ackbeet View Post
    Yes, so far.

    So the solution is...
    got it. thanks.
    Last edited by kittycat485; October 3rd 2011 at 09:09 PM.
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  10. #10
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    Re: di

    You're very welcome. Have a good one!
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