Edit: replaced all c's with y's as the c just denotes replacing a series of coupled linear equations Ay with uncoupled equations $\Lambda c$ no biggie.
Im working through the lecture notes for a course on computational modelling of flow and transport. The first Section is on Numerical Methods for IVPs. The text wants to derive order of accuracy by calculating truncation error as
$$ \tau_{\Delta t} = \frac{y^{n+1} – y^{n-1}}{2 \Delta t} – \lambda\big(\frac{1}{6}y^{n+1} + \frac{2}{3}y^n + \frac{1}{6}y^{n-1}\big)\ with\ \lambda\ being\ a\ complex\ number $$
so I tried to calculate the truncation error by expanding taylor series of $ y^{n+1} $ and $ y^{n-1} $
$$ \tau_{\Delta t} = \Big(y + \Delta t y' +\frac{\Delta t^2 y''}{2} +\frac{\Delta t^3 y^{(3)}}{6} +\frac{\Delta t^4 y^{(4)}}{24} +\frac{\Delta t^5 y^{(5)}}{120} – y + \Delta t y' -\frac{\Delta t^2 y''}{2} +\frac{\Delta t^3 y^{(3)}}{6} -\frac{\Delta t^4 y^{(4)}}{24} +\frac{\Delta t^5 y^{(5)}}{120} \Big) \div 2 \Delta t – \lambda \Big( \frac{y}{6} + \frac{\Delta t y'}{6} +\frac{\Delta t^2 y''}{12} +\frac{\Delta t^3 y^{(3)}}{36} +\frac{\Delta t^4 y^{(4)}}{144} +\frac{\Delta t^5 y^{(5)}}{720} + \frac{2 y}{3} + \frac{y}{6} – \frac{\Delta t y'}{6} +\frac{\Delta t^2 y''}{12} -\frac{\Delta t^3 y^{(3)}}{36} +\frac{\Delta t^4 y^{(4)}}{144} -\frac{\Delta t^5 y^{(5)}}{720} \Big)$$
so cancelling obvious terms:
$$ \tau_{\Delta t} = \Big(2\Delta t y' +\frac{\Delta t^3 y^{(3)}}{3} +\frac{\Delta t^5 y^{(5)}}{60} \Big) \div 2\Delta t – \lambda \Big( \frac{y}{3} + \frac{\Delta t^2 y''}{6} + \frac{\Delta t^4 y^{(4)}}{72} + \frac{2y}{3} \Big) $$ $$<=> \tau_{\Delta t} = y' +\frac{\Delta t^2 y^{(3)}}{6} +\frac{\Delta t^4 y^{(5)}}{120} – \lambda y – \frac{\lambda \Delta t^2 y''}{6} – \frac{\lambda \Delta t^4 y^{(4)}}{72} $$ grouping powers of $ \Delta t $
$$ \tau_{\Delta t} = (y' – \lambda y) + \Delta t^2(\frac{y^{(3)}}{6} – \frac{\lambda y''}{6}) + \Delta t^4(\frac{y^{(5)}}{120} – \frac{\lambda y^{(4)}}{72}) + H.O.T.$$ this is where im stuck. I googled it but all simpson's rule stuff is about integration?? and they just state the truncation error not prove it?? Any Help?
Edit: I think it makes sense if I can assume $ y'' = \lambda y'$ Does anyone know if this is a valid assumption??
Best Answer
The method you are exploring is $$ \frac{y^{n+1} - y^{n-1}}{2 \Delta t} = \big(\frac{1}{6}f(y^{n+1}) + \frac{2}{3}f(y^n) + \frac{1}{6}f(y^{n-1})\big) $$ for differential equation $y'=f(y)$. For linear multistep-methods, it is sufficient to explore the approximation error of the most simple non-trivial ODE $y'=f(y)=λy$ with solution $y(t)=C·e^{λt}$. This reduces the error term to $$ τ_{Δt}=y_n·\left(\frac{e^{λΔt}-e^{-λΔt}}{2Δt}-λ\frac{e^{λΔt}+4+e^{-λΔt}}6\right) $$ and by inserting the Taylor series of the exponential resp. hyperbolic functions you will see that indeed the lowest remaining power is $(λΔt)^4$.