Find the stability function of $y_{n+1}-y_{n-1} = 2h\lambda y_n$

Question

I'm asked to find the stability function $R(z)$ in the expression $y_{n+1} = R(z)y_n$ given the method $$y_{n+1}-y_{n-1} = 2h\lambda y_n.$$ How do I do this? I can't do it when three stages is present in the equation.

Answer 1

Here you need two time steps in the state vector of the method, so that $$ \pmatrix{y_{n+1}\\y_n}=R(z{=}λh)\pmatrix{y_n\\y_{n-1}} =\pmatrix{2λh&1\\1&0}\pmatrix{y_n\\y_{n-1}}. $$ For the stability you need to explore the eigenvalues of the matrix $R(z)$.

The two-step Nyström or central Euler method is only weakly stable, in that the stability region (where $R(z)$ is non-expanding) is the segment $[-i,i]$ on the imaginary axis.

• Stability region for two step Nyström method
• Deriving the central Euler method and intuition

Answer 2

I'll give a general answer that you can specialize.

The region of absolute stability is found by applying a linear multi-step method to the problem $u'=\lambda u.$ Since the general form of such a method is $$\sum\limits_{j=0}^r\alpha_jU^{n+j}=k\sum\limits_{j=0}^r\beta_j f(U^{n+j},t_{n+j}),$$ applying it to the ODE gives $$\sum\limits_{j=0}^r\alpha_jU^{n+j}=k\sum\limits_{j=0}^r\beta_j \lambda U^{n+j}.$$ If we take $z=k\lambda,$ then we can re-write this as $$\sum\limits_{j=0}^r(\alpha_j-z\beta_j)U^{n+j}=0.$$ This is a linear difference equation with a solution of the form $$U^n=\sum\limits_{j=0}^r c_j\zeta^n_j,$$ where $\zeta_j$ are roots of the characteristic polynomial $$\pi(\zeta;z)=\sum\limits_{j=0}^r(\alpha_j-z\beta_j)\zeta^j.$$ If you're familiar with the characteristic polynomials $$p(\zeta)=\sum\limits_{j=0}^r \alpha_j\zeta^j\ \ \ \ \text{ and } \ \ \ \ \sigma(\zeta)=\sum\limits_{j=0}^r \beta_j\zeta^j,$$ then we can write $$\pi(\zeta;z)=p(\zeta)-z\sigma(\zeta).$$ The region of absolute stability will be the $z$ for which the absolute polynomial satisfies the root condition.

Answer 3

The difference equation

$$ y_{n+1}-2h\lambda y_n-y_{n-1} = 0 $$

has as solution

$$ y_n = C_1 \left(h\lambda-\sqrt{1+h^2\lambda^2}\right)^n+C_2\left(h\lambda+\sqrt{1+h^2\lambda^2}\right)^n $$

and is is stable as long as

$$ \left\{\left|h\lambda-\sqrt{1+h^2\lambda^2}\right| < 1\right\}\cap \left\{\left|h\lambda+\sqrt{1+h^2\lambda^2}\right| < 1\right\} $$

hence it is unstable.