I find myself having trouble with an exercise from Gautschi's Numerical Analysis on finding the truncation error of Euler's method. I have been given a system:
$$y' = Ay, \quad y \in \mathbb{R}^{d}, A \in \mathbb{R} $$
for this, I believe I should use this formula for the "upper bound" on the truncation error:
$$ \frac{1}{2}h^2y''(\xi_j), \quad t_j \le \xi_j \le t_j+h $$
However, I am having trouble wrapping my head around what exactly $y''(\xi_j)$ is. Is it $\frac{d^2y}{dt^2}$, and how do I find this?
Yes, it is the second derivative. $$ y'=f(x,y)\implies y''=\partial_xf(x,y)+\partial_yf(x,y)y' =\partial_xf(x,y)+\partial_yf(x,y)f(x,y) $$ In the concrete case of a linear system $$ y'=Ay\implies y''=Ay'=A^2y. $$
The formula in question is semi-wrong, it is only valid for a scalar equation. It is based on the mean value theorem, which in the general multivariate case does not give the same mid-point in all components of the system.
For instance, using the Taylor formula with integral remainder term giving for degree 1 gives $$ y(x+h)=y(x)+y'(x)h+h^2\int_0^1(1-s)y''(x+sh)\,ds. $$ For each component of $y$ one can apply the mean value theorem for integrals. However, there is no way to force the midpoints to be the same. What is possible is to get upper bounds on the error from the upper bounds of the second derivative.