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A system of differential equations not having a analytical solution

  1. May 18, 2014 #1
    I am going to quote an article below and there is a part I would like clarification.I do not understand why chaotic differential systems do not have an analytical solutions. In the simplest form,an equation such as this one [sin(x) + x - 0.5 = 0] does not have an analytical solution i.e it can be only solved numerically.
    1.What would be its equivalent for a differential system not having a analytical solution?
    2.When the differential equation/system gives a different response for different initial conditions,how does it imply "not having a analytical solution"? I'm a novice.


     
    Last edited: May 18, 2014
  2. jcsd
  3. May 18, 2014 #2

    lurflurf

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    What equations do not have analytical solution depends on what functions are available.
    usual examples would be
    x''(t)+sin(x(t))=0
    h x''(t)+t x'(t)-x(t)=0
    h x''(t)+x'(t)-t x(t)=0

    different response for different initial conditions is a different thing, but they often occur together
    If you have the analytical solution it is (hopefully) easier to deal with the different responses
    If we have an approximate solution we hope it is not to far from the actual solution
    If we have very different solution we cannot distinguish we have a problem

    We might want to use the differential equation to make a prediction. If tiny changes in the conditions make a huge difference in the response we cannot predict very well.
     
  4. May 18, 2014 #3

    pasmith

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    Consider the one-dimensional system [tex]
    \frac{dx}{dt} = f(x)
    [/tex] with continuous [itex]f[/itex]. This has implicit solution [tex]
    \int_{x(0)}^{x(t)} \frac{1}{f(u)}\,du = t.
    [/tex] There is then a question as to whether the integral can actually be done analytically (which it almost always can't) and whether the resulting equation can be solved analytically for [itex]x(t)[/itex] (which it almost always can't). The only case we can always solve is the case where [itex]x(0)[/itex] is such that [itex]f(x(0)) = 0[/itex], and the solution is then [itex]x(t) = x(0)[/itex].

    "Sensitive dependence on initial conditions" is a phenomenon which marks a system out as being chaotic.

    However even with non-chaotic systems the behaviour of a system may be drastically different depending on the initial conditions: consider [tex]
    \frac{dx}{dt} = x(1-x),
    [/tex] which falls into the vanishingly small category of systems for which a solution can be found analytically:
    [tex]
    x(t) = \begin{cases}
    \frac{x_0 e^t}{1 + x_0(e^t - 1)} & 0 \leq t < \log \left(1 + \frac{1}{|x_0|}\right), x_0 < 0 \\
    0 & 0 \leq t < \infty, x_0 = 0 \\
    \frac{x_0 e^t}{1 + x_0(e^t - 1)} & 0 \leq t < \infty, 0 < x_0 < 1 \\
    1 & 0 \leq t < \infty, x_0 = 1 \\
    \frac{x_0 e^t}{1 + x_0(e^t - 1)} & 0 \leq t < \infty, x_0 > 1
    \end{cases}
    [/tex]
    The salient points are firstly that if [itex]x_0 = x(0) < 0[/itex] then the trajectory diverges to [itex]-\infty[/itex] in finite time, secondly that if [itex]x_0 = 0[/itex] then [itex]x_0 = 0[/itex] for all time, and thirdly that if [itex]x_0 > 0[/itex] then [itex]x_0 \to 1[/itex] as [itex]t \to \infty[/itex]. Thus if all we know is that [itex]-\epsilon < x_0 < \epsilon[/itex] for some [itex]\epsilon > 0[/itex], then we cannot predict the future behaviour of the system.
     
    Last edited: May 18, 2014
  5. May 19, 2014 #4
    The above solution of x(t) does not work for x(0)=0. Is there an inherent assumption that I am missing? I wrote the solution I got for the D.E x'=x(1-x) below. After substituting for x(0)=0,I get a fraction that tends to 1.
    [tex]
    x(t)=\frac{1}{1+C_1*e^{-t}} \\
    x(0)=\frac{1}{1+C_1} \\
    C_1=x(0)-1 \\
    \Rightarrow x(t)=\frac{1}{1+(x(0)-1)e^{-t}} \\
    [/tex]
     
  6. May 19, 2014 #5

    pasmith

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    The solution in the case [itex]x(0) = 0[/itex] is [itex]x(t) = 0[/itex].

    Separation of variables should give you [tex]
    \log \left| \frac{x}{1-x}\right| = t + C[/tex] so that [tex]
    \left| \frac{x}{1-x}\right| = Ae^{t}[/tex] from which it follows that [tex]
    A = \left| \frac{x_0}{1 - x_0}\right|.[/tex] If at this point you set [itex]x_0 = 0[/itex] then you obtain [itex]A = 0[/itex], and hence [tex]
    \frac{x}{1-x} = 0.[/tex]
     
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