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

 

[blue_raver22]

Firstly, it is important to note that the explicit Runge-Kutta scheme is a numerical method used to approximate the solution of a differential equation. In this case, the differential equation is $y'=y(1-y)$.

By rearranging the given scheme, we get:
\begin{equation}
y_{n+1} = y_{n} + \frac{h}{2} \left[f(t,y_{n}) + f(t+h, y_{n}+hf(t,y_{n}))\right]
\end{equation}

Now, let's substitute the fixed points into this scheme. For $y=0$, we have $y'=0$, which gives us $f(t,y_{n})=0$. Substituting this into the scheme, we get:
\begin{equation}
y_{n+1} = y_{n} + \frac{h}{2} \left[0 + f(t+h, 0)\right]
\end{equation}
Since $f(t,y)=y(1-y)$, we get $f(t+h,0)=0$, and hence, $y_{n+1}=y_{n}$ for $y=0$. This means that $y=0$ is a fixed point of the scheme.

Similarly, for $y=1$, we have $y'=0$, which gives us $f(t,y_{n})=0$. Substituting this into the scheme, we get:
\begin{equation}
y_{n+1} = y_{n} + \frac{h}{2} \left[0 + f(t+h, h)\right]
\end{equation}
Since $f(t,y)=y(1-y)$, we get $f(t+h,h)=h(1-h)$, and hence, $y_{n+1}=y_{n}+h(1-h)$ for $y=1$. This means that $y=1$ is also a fixed point of the scheme.

Now, to investigate the stability of these fixed points, we can use linear stability analysis. This involves studying the behavior of small perturbations around the fixed points. In this case, we can consider the perturbation $u_{n}=y_{n}-y$, where $y$ is the fixed point. Substituting this into the scheme, we get:
\begin{equation