# Advanced numerical solution of differential equation

Show that the explicit Runge-Kutta scheme
\frac {y_{n+1} -y_{n}}{h}= \frac{1}{2} [f(t,y_{n} + f(t+h, y_{n}+hk_{1})]

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
\frac {y_{n+1} -y_{n}}{h}= \frac{1}{2} [f(t,y_{n} + f(t+h, y_{n}+hk_{1})]

y_{n+1} = y_{n} + \frac{h}{2} [f(t,y_{n} + f(t+h, y_{n}+hf(t,y_{n})]

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
Homework Helper
Show that the explicit Runge-Kutta scheme
\frac {y_{n+1} -y_{n}}{h}= \frac{1}{2} [f(t,y_{n} + f(t+h, y_{n}+hk_{1})]

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
\frac {y_{n+1} -y_{n}}{h}= \frac{1}{2} [f(t,y_{n} + f(t+h, y_{n}+hk_{1})]

y_{n+1} = y_{n} + \frac{h}{2} [f(t,y_{n} + f(t+h, y_{n}+hf(t,y_{n})]

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
$$0 = f(y + f(y + hf(y))).$$ 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 $|g'(y)| < 1$, unstable if $|g'(y)| > 1$ and indeterminate at linear order if $|g'(y)| = 1$.

Thank you sir. that was really helpful.