Given $y(t)$ the initial-valu problem:
$y'=f(t,y),$ $y(0)=y_0$
I need to find the order of the local truncation error for the following numerical integration method:
$w_{n+1}=w_n+hf(t_n+\alpha h,w_n+\alpha hf(t_n,w_n)), \alpha ∈ [0,1]$
I know I have to assume $w_n=w(t_n)$, and then Taylor expand the exact value $y(t_{n+1})$ and everything that isn't $w(t_n)$ in the right hand side and then calculate the difference. I get stuck at Taylor expanding the right hand side. Can somebody help me?
Best Answer
Following the great book by Vuik, Vermolen, van Gijzen and Vuik(Numerical Methods for Ordinary Differential Equations).
We are interested in the local truncation error \begin{equation} \tau_{n+1} := \frac{y_{n+1}-w_{n+1}}{\Delta t} \end{equation}, where $y_n$ is the exact solution and $w_n$ the approximation at $t_n$. (I take $h=\Delta t$). Then for "simple" methods you can take the Taylor series of $y_{n+1}$ at $t_n$ and plug in the method for $w_{n+1}$. In this case the $f(\cdot,f(\cdot,\cdot))$ part complicates the story. Again, Taylor expansion will help! Take the expansion about $(t_n,y_n)$ and see that we can write: \begin{align} f(\bar{t},\bar{w}) &= f(t_n+\alpha\Delta t, w_n+\alpha \Delta t f(t_n, w_n))\\ & =f(t_n,y_n) + (t_n + \alpha \Delta t -t_n) \frac{\partial f}{\partial t} + (y_n+\alpha \Delta t f(t_n,y_n)-y_n)\frac{\partial f}{\partial y} + \mathcal{O}(\Delta t^2)\\ &= f_n + \alpha \Delta t \left(\frac{\partial f}{\partial t} + f_n \frac{\partial f}{\partial y} \right) + \mathcal{O}(\Delta t^2) \implies \\ &w_{n+1} = y_n + \Delta t f_n + \alpha \Delta t^2 \left(\frac{\partial f}{\partial t} + f_n \frac{\partial f}{\partial y} \right) + \mathcal{O}(\Delta t^3) \end{align} Then by the chain-rule we can continue and write \begin{align} &y''_n =\frac{\partial f}{\partial t} + \frac{\partial f}{\partial y}y_n' = \frac{\partial f}{\partial t} + f_n\frac{\partial f}{\partial y} \implies\\ &w_{n+1} = y_n + \Delta t y'_n + \alpha \Delta t^2 y''_n + \mathcal{O}(\Delta t^3). \end{align} However, the Taylor series of $y_{n+1}$ at $t_n$ can be written as \begin{align} y_{n+1} = y_n + \Delta t y'_n + \frac{1}{2} \Delta t^2y''_n + \mathcal{O}(\Delta t^3). \end{align} From there you see that the selection of $\alpha$ determines your local truncation error being either $\mathcal{O}(\Delta t)$ or $\mathcal{O}(\Delta t^2)$.