Runge-Kutte: stability of fixed points

[ra_forever8]

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.

 

[chisigma]

Re: Advanced numerical solution of differential equations

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.

In Your case $\displaystyle f(t,x) = x\ (1-x)$ is function of the x alone so that the fixed points are the zeroes of the function...

$\displaystyle g(x)= f(x) + f \{ x + h\ f(x)\} = x\ (1-x) + x\ (1-x) -h\ x^{2} (1-x) + h\ x\ (1-x)^2 - h^{2}\ x^{2}\ (1-x)^{2}\ (1) $

Clearly x=0 and x=1 are both zeroes of (1) but other two zeroes are solution of the equation... $\displaystyle h^{2}\ x^{2} - (2\ h + h^{2})\ x + 2 + h =0\ (2)$

The (2) admits real solutions only if the discriminant is positive and that happens if $h^{2} > 4 \implies h > 2$... Kind regards $\chi$ $\sigma$

 

[ra_forever8]

Re: Advanced numerical solution of differential equations

how did you apply $\displaystyle f(t,x) = x\ (1-x)$ into explicit runge kutta scheme to get the following

$\displaystyle g(x)= f(x) + f \{ x + h\ f(x)\} = x\ (1-x) + x\ (1-x) -h\ x^{2} (1-x) + h\ x\ (1-x)^2 - h^{2}\ x^{2}\ (1-x)^{2}\ (1) $

Clearly x=0 and x=1 are both zeroes of (1) but other two zeroes are solution of the equation...

how did you get this equation when you substitute x=0 and x=1.
$\displaystyle h^{2}\ x^{2} - (2\ h + h^{2})\ x + 2 + h =0\ (2)$

The (2) admits real solutions only if the discriminant is positive and that happens if $h^{2} > 4 \implies h > 2$...

how did you know $h^{2} > 4 \implies h > 2$...

can you please go little slowly, so i can understand clearly sir

so stability depends upon $y_n$. if $h>2$ means $y_n$ is stable? what can i say more about stability?
Kind regards

 

[ra_forever8]

Re: Advanced numerical solution of differential equations

after long thinking and practice ,finally got it what you have done sir. Thank you