Advanced numerical solution of differential equation

[rickyflair]

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.

 

[pasmith]

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.

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

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

For stability: a fixed point y of y_{n+1} = g(y_n) is stable if |g&#039;(y)| &lt; 1, unstable if |g&#039;(y)| &gt; 1 and indeterminate at linear order if |g&#039;(y)| = 1.

 

[rickyflair]

Thank you sir. that was really helpful.