Advanced numerical solution of differential equations

Question

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.

Answer 1

Your formula for the step is missing some closing parantheses at the correct places. The correct form of the Heun resp. explicit trapezoidal method is

k1 = f(t,y)
k2 = f(t+h, y+h*k1)
ynext = y + h*(k1+k2)/2 

With your given function this translates as \begin{align} k1 &= y(1-y)\\ k2 &= (y+hy(1-y))\,(1-y-hy(1-y))=y(1+h-hy)(1-y)(1-hy)\\ y_+&=y+\tfrac12 hy(1-y)\bigl(1+(1+h-hy)(1-hy)\bigr)\\ &=y+\tfrac12 hy(1-y)\bigl((1-hy+\tfrac12h)^2+1-\tfrac14h^2\bigr) \end{align}

Thus for $h>2$ the last quadratic factor contributes the fixed points \begin{align} y&=\tfrac1h+\tfrac12\pm\sqrt{\tfrac14-\tfrac1{h^2}} \end{align}