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How are they differentiating this ODE?

  1. May 28, 2012 #1
    They give a differential equation: [itex] x' = f_a(x) = ax(1-x) [/itex]. In determining if the equilibrium points are sources or sinks, they say: We may also determine this information analytically. We have [itex] f'_a(x) = a - 2ax [/itex]

    How can they differentiate with respect to x? x is a function, it doesn't represent a point on the real line. I tried assuming that they really mean [itex] x'(t) = f_a(x(t)) = ax(t)(1 - x(t)) [/itex], but that would mean that [itex] x''(t) = f'_a(x(t)) = ax'(t) - 2ax(t)x'(t) [/itex], which according to the book is wrong.

    What's going on here?
     
  2. jcsd
  3. May 28, 2012 #2

    Simon Bridge

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    Differentiating with respect to a function is not a problem.

    1st pass - mistake in notation?
    Well x'=1 if we take the primed notation to indicate differentiation in x. Perhaps that's supposed to be a dot? Then I can put v = x' = dx/dt right?

    That would mean that v = ax(1 - x) and and you can certainly differentiate speed with respect to space to give: dv/dx = a - 2ax

    Of course this means that the notation is inconsistent.
    I think it's pretty clear that they are differentiating f with respect to x. For the first it's not so clear from the example what is intended ... I mean where they came from is something like x=(0.5)ax²(1-x)-ax+c ... which is only true for at most three values of x. So there is something missing from the description here.

    2nd pass: your analysis holds but recalling that x'=dx(t)/dx=1 then x''=0 and your equation simplifies to:

    0 = f'(x(t)) = a - 2ax(t)

    ... isn't that what they have?

    3rd pass ... If the prime implies d/dt (JIC) always then ...

    f = ax - ax² = x'
    f' = ax' -2ax.x' = af -2axf = ax(1-x-2x²) (check - not what they have)

    I think we need context but it really looks like they have differentiated f wrt x not t.
     
    Last edited: May 28, 2012
  4. May 28, 2012 #3
    I know they're differentiating with respect to x, that's exactly what I have a problem with. As far as I know, the usual derivative of a map requires the domain to be a subset of [itex] \mathbb{R}^n [/itex], but the x that they are differentiating with respect to is a function, it's not a real number, or an n-tuple of real numbers. Shouldn't [itex] x in \mathbb{R}^n [/itex] if we're differentiating with respect to x?
     
  5. May 29, 2012 #4

    Simon Bridge

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    So you are telling me that x is not in [itex]\mathbb{R}^n[/itex]?
    What is it's domain then? You could, in principle, plot a graph of x vs t right?

    Are you saying that if y=f(x(t)) you can't do dy/dx?
     
  6. May 29, 2012 #5
    Wow, now I feel silly. I confused the hell outta myself, I should've known better. Thanks for the replies.
     
  7. May 29, 2012 #6

    Simon Bridge

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    No worries - you've been thinking too hard go have a drink.
     
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