Ytukyg

This is one of the exercise questions in the book Numerical Analysis by Richard L.Burden

Show that the difference method
$$y_0 = alpha \
y_{i+1} = y_i + a_1 f(t_i,y_i)+a_2 f(t_i+alpha_2, y_i+delta_2 f(t_i,y_i)$$
for each $i=0,1,2,cdots,N-1$, cannot have local truncation error $mathcal{O}(h^3)$ for any choice of constants $a_1,a_2,alpha_2,$ and $delta_2$

Answer:

Using the notation $y_{i+1}=y(t_{i+1}),y_i = y(t_i), $ and $f_i=f(t_i,y(t_i))$ , and also the definition of the local truncation error we have
$$htau_{i+1}=y_{i+1}-y_i-a_1f_i-a_2f(t_i+alpha_2,y_i+delta_2 f_i))$$
If we expand $y_{i+1}$ and $f(t_i+alpha_2,y_i+delta_2 f_i)$ in Taylor series about $t_i$ and $f(t_i,y_i)$:
$$begin{equation}begin{aligned} htau_{i+1} &= left( y_i+hf_i + frac{h^2}{2}f'_i+frac{h^3}{6}f''_i+cdots right) - y_i - a_1f_i \&- a_2 left( f_i+alpha_2 f_t(t_i,y_i)+delta_2 f_i f_y (t_i,y_i) + alpha_2 delta_2 f_i f_{ty}(t_i,y_i) +frac{alpha^2}{2}f_{tt}(t_i,y_i) \+ frac{delta^2 f_i^2}{2}f_{yy}(t_i,y_i)+cdotsright) \&= (h-a_1-a_2)f_i+ left( frac{h^2}{2}-a_2 alpha_2 right)f_t (t_i,y_i) + left( frac{h^2}{2}-a_2 delta_2 right)f_i f_y (t_i,y_i) \&+left( frac{h^3}{6}-a_2frac{alpha^2}{2} right)f_{tt}(t_i,y_i)+left( frac{h^3}{3}-a_2 alpha_2 delta_2 right)f_i f_{ty} (t_i, y_i) \&+ left( frac{h^3}{6} -a_2 frac{delta_2^2}{2}right)f_i^2 f_{yy} (t_i,y_i)+color{red}{frac{h^3}{6}left( f_t (t_i,y_i) f_y (t_i,y_i) + f_i f_y^2 (t_i,y_i) right)}+cdots
end{aligned}end{equation}$$

I can only expand the equation until the black part, I couldn't get where the red part comes from, and it's the essence to show that the $h^3$ term cannot be cancelled regardless of the choice of the constants $a_1,a_2,alpha_2,delta_2$. The $frac{h^3}{6}$ in the $color{red}{frac{h^3}{6}left( f_t (t_i,y_i) f_y (t_i,y_i) + f_i f_y^2 (t_i,y_i) right)}$ hinting that it comes from the Taylor expansion of $frac{h^3}{6}f_i''$, but when I expand it I only get three terms:
$$frac{h^3}{6}f''_i = frac{h^3}{6} left( f_{tt}+f_{yy}f_i^2+2f_{ty}f_i right)$$

$f''$ denotes the second derivative of $f(t, y(t))$ as function in $t$. With $y'(t)=f(t,y(t))$ it computes as
$$
frac{d^2}{dt^2}fbigl(t,y(t)bigr)
=frac{d}{dt}Bigl(f_tbigl(t,y(t)bigr)+f_ybigl(t,y(t)bigr)fbigl(t,y(t)bigr)Bigr)
=f_{tt}+2f_{ty}f+f_{yy}f^2+f_y(f_t+f_yf).
$$
So indeed there are two additional parts more than you expected. They result conceptually from the fact that the offset $y(t+s)-y(t)$ for the Taylor expansion at time $t$ is not constant but itself a function that contributes higher order terms. Directly they result from the application of the product rule to the second term in the first derivative.