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Advanced numerical solution of differential equation

  1. Apr 10, 2014 #1
    Show that the explicit Runge-Kutta scheme
    \begin{equation} \frac {y_{n+1} -y_{n}}{h}= \frac{1}{2} [f(t,y_{n} + f(t+h, y_{n}+hk_{1})]
    \end{equation}
    where $k_{1} = f(t,y_{n})$

    applied to the equation $y'= y(1-y)$ has two spurious fixed points if $h>2$.

    Briefy describe how you would investigate their stability.

    => my attempt so far
    from $y'= y(1-y)$

    $y'= 0$

    $y=0$ or
    $y=1$ which are the true fixed points.
    after that i rearranged the runge kutta scheme
    \begin{equation} \frac {y_{n+1} -y_{n}}{h}= \frac{1}{2} [f(t,y_{n} + f(t+h, y_{n}+hk_{1})]
    \end{equation}
    \begin{equation} y_{n+1} = y_{n} + \frac{h}{2} [f(t,y_{n} + f(t+h, y_{n}+hf(t,y_{n})]
    \end{equation}
    i try to put the fixed points into above scheme and try to get two two Spurious fixed point for $y_{n}$ but i got struck. for the stability to describe i need to get two Spurious fixed point first. but in general please help to describe stability too because this part i really get confuse. Anyone please help me, it will be really helpful for my other problems too if i got this answer correctly.
     
  2. jcsd
  3. Apr 10, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    Your [itex]f(t,y)[/itex] doesn't actually depend on [itex]t[/itex], so you may as well drop it and write [itex]f(y) = y(1 - y)[/itex].

    A fixed point of the iteration will satisfy [itex]y_{n+1} = y_n = y[/itex], which gives you
    [tex]
    0 = f(y + f(y + hf(y))).
    [/tex] Now for [itex]y = 0[/itex] and [itex]y = 1[/itex] you have [itex]f(y) = 0[/itex], so they satisfy the above and are fixed points of the iteration. But [itex]f(y + f(y + hf(y)))[/itex] is here an eighth-order polynomial in [itex]y[/itex], so it might have other real roots aside from [itex]y = 0[/itex] and [itex]y = 1[/itex]. But determining that really requires the aid of a CAS.

    For stability: a fixed point [itex]y[/itex] of [itex]y_{n+1} = g(y_n)[/itex] is stable if [itex]|g'(y)| < 1[/itex], unstable if [itex]|g'(y)| > 1[/itex] and indeterminate at linear order if [itex]|g'(y)| = 1[/itex].
     
  4. Apr 10, 2014 #3
    Thank you sir. that was really helpful.
     
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