Absolute stability of finite difference scheme

[ra_forever8]

The Finite difference scheme:
\begin{equation} y_{n+3}-y_{n+1}= \frac {h}{3}(f_{n}-2f_{n+1}+7f_{n+2})
\end{equation}
Deduce that the scheme is convergent and find its interval of absolute stability(if any)
=> the first characteristic polynomial is then
\begin{equation} ρ(r)= r^3 -r \end{equation}
\begin{equation} r^3 -r =0 \end{equation}
\begin{equation} r(r^2-1) =0 \end{equation}
\begin{equation} r=0,-1,1 \end{equation}
These have magnitude less than or equal to 1 and there are no repeated zero with magnitude equal to 1, $|r|<1$. So the method is zero-stable and is convergent.
For its interval of absolute stability, i was thinking to substitute $$y'=λy$$ into finite scheme but because of $y_{n+3}$involve, its not good idea to do so. Can anyone help how to calculate the interval of absolute stability?

 

[jvicens]

Hello! The finite difference scheme that you have provided is known as the "three-point backward difference scheme". This scheme is used to approximate the solution of a differential equation at a given point based on the values of the function at three previous points.

To show that this scheme is convergent, we need to show that as the step size h approaches 0, the error between the exact solution and the numerical solution also approaches 0. This can be done by using the Taylor series expansion of the function f at the point xn+2, which is given by:

\begin{equation} f(x_{n+2}) = f(x_n) + 2hf'(x_n) + \frac{4h^2}{2}f''(x_n) + \frac{8h^3}{6}f'''(x_n) + O(h^4) \end{equation}

Similarly, we can expand the function at the point xn+1 and substitute it in the finite difference scheme. After simplifying, we get:

\begin{equation} y_{n+3} = y_n + 3hy_n' + \frac{9h^2}{2}y_n'' + \frac{27h^3}{6}y_n''' + O(h^4) \end{equation}
\begin{equation} y_{n+1} = y_n + hy_n' + \frac{h^2}{2}y_n'' + \frac{h^3}{6}y_n''' + O(h^4) \end{equation}
\begin{equation} f_n = f(x_n) \end{equation}
\begin{equation} f_{n+1} = f(x_n) + hf'(x_n) + \frac{h^2}{2}f''(x_n) + \frac{h^3}{6}f'''(x_n) + O(h^4) \end{equation}
\begin{equation} f_{n+2} = f(x_n) + 2hf'(x_n) + \frac{4h^2}{2}f''(x_n) + \frac{8h^3}{6}f'''(x_n) + O(h^4) \end{equation}

Substituting these values in the finite difference