Consider the following numerical method for approximating the solution to y′=f(y,t)y’ = f(y,t):

yn+1=yn−1+2hf(yn,tn) y^{n+1} = y^{n-1} + 2hf(y^n,t^n)

Where superscripts refer to indices not exponentiation. I want to find the stability of this method so I consider the model problem y′=kyy’ = ky and set λ=kh\lambda = kh. This gives me the homogeneous linear recurrence relation

yn=2λyn−1+yn−2y^n = 2\lambda y^{n-1} + y^{n-2}

Which has general solution

yn=α1(λ+√λ2+1)+α2(λ−√λ2+1)y^n = \alpha_1 (\lambda + \sqrt{\lambda^2 + 1}) + \alpha_2 (\lambda – \sqrt{\lambda^2 + 1})

Assuming the initial conditions give both α1,α2≠0\alpha_1, \alpha_2 \ne 0 then the method is stable iff both

|λ+√λ2+1|<1|\lambda + \sqrt{\lambda^2 + 1}| < 1, and |λ−√λ2+1|<1|\lambda - \sqrt{\lambda^2 + 1}| < 1. Multiplying these together gives |λ+√λ2+1||λ−√λ2+1|=|−1|=1<1 \left|\lambda + \sqrt{\lambda^2 + 1}\right| \: \left|\lambda - \sqrt{\lambda^2 + 1}\right|= \left|-1\right| = 1 <1 Does this mean this method is unstable for all λ\lambda (where α1,α2≠0\alpha_1, \alpha_2 \ne 0) or have I made some error? ================= Already the Viete equations tell you that the product of roots of z2−2خ»z−1=0z^2-2 خ»z-1=0 will always be −1-1. So yes, your conclusion is correct. – LutzL 16 hours ago ================= =================