Show that the explicit standard Runge-Kutta method with the process function
$\varphi(t,y,h)=\frac16 (f_1+2f_2+2f_3+f_4)$
$f_1=f(t,y)$
$f_2=f(t+\frac{h}2,y+\frac{h}2f_1)$
$f_3=f(t+\frac{h}2,y+\frac{h}2f_2)$
$f_4=f(t+h,y+hf_3)$
has at least consistency order 4, if $f$ is sufficiently enough differentiable.
So I have to show, that for
$\tau_h(h,\stackrel{\sim}{y}(t)):=\frac1h(\stackrel{\sim}{y}(t+h)-(\stackrel{\sim}{y}(t)+h\varphi(t,\stackrel{\sim}{y}_{|I_h}, I_h)))$
it is $\|\tau_h\|_\infty =\mathcal{O}(h^4)$
I guess, I need Taylor's formula.
We have from the process function, that $f_1, f_2, f_3, f_4$ are expressed in terms of $f$, by:
$f_1=f(t,y)$
$f_2=f(t+\frac{h}{2}, y+\frac{h}{2}f(t,y))$
$f_3=f(t+\frac{h}{2}, y+\frac{h}{2}f(t+\frac{h}{2}, y+\frac{h}{2}f(t,y)))$
$f_4=f(t+h, y+hf(t+\frac{h}{2}, y+\frac{h}{2}f(t+\frac{h}{2}, y+\frac{h}{2}f(t,y))))$
Now I could write down $\varphi(t,y,h)=\frac16(f_1+f_2+f_3+f_4)$, but I do not think, it helps at this point.
First I should be clear, what the route is.
I have to show, that $\tau_h(t,y(t))=\frac{1}{h}(y(t+h)-(y(t)+h\varphi(t,y(t),h)))=\mathcal{O}(h^4)$
For that, I would use the (higher dimensional) Taylor formula up to degree 4. But this seems to get out of controll pretty quick.
Or I might just need the Taylor series to degree 1. Since we have compositions of the function $f$ four times, this might do it.
Am I correct up to this point? Is there a simple way, or can you guide/help me through this?
Thanks in advance.
Simplify to the autonomous case. For $k\ge 2$ denote $f^{(k)}[v_1,...,v_k]$ the $k$th derivative of $f$ in $(t,y)$ evaluated in directions $v_1,...,v_k$. Then the Taylor expansion of the exact solution $y(x+h)$ has the coefficients resp. involves the derivatives \begin{align} \dot y&=f\\ \ddot y&=f'\dot y=f'f\\ y^{(3)}&=f''[f,f]+f'^2f\\ y^{(4)} &= f'''[f,f,f]+3f''[f'f,f]+f'f''[f,f]\\ y^{(5)} &= f^{(4)}[f,f,f,f]+3f'''[f'f,f,f]+4f''[f''[f,f],f]+3f''[f'f,f'f] +f'f'''[f,f,f]+2f'f''[f'f,f] \end{align} while the stages of RK4 have the expansion \begin{align} k_1&=f\\ k_2&=f(y+\tfrac12hf)&&=f+\tfrac12f'fh+\tfrac18f''[f,f]h^2+\tfrac1{48}f'''[f,f,f]h^3 +... \\ k_3&=f(y+\tfrac12hk_2) &&\color{lightblue}{=f + \tfrac12f'\left(f+\tfrac12f'fh+\tfrac18f''[f,f]h^2\right)h + \tfrac18f''[f+\tfrac12f'fh,f+\tfrac12f'fh]h^2 + \tfrac1{48}f'''[f,f,f]h^3 + ...}\\ &&&=f+\tfrac12f'fh+\left(\tfrac14f'^2f+\tfrac18f''[f,f]\right)h^2 + \left(\tfrac1{16}f'f''[f,f]+\tfrac18f''[f'f,f]+\tfrac1{48}f'''[f,f,f]\right)h^3 + ... \\ k_4 &= f(y+hk_3) &&\color{lightblue}{= f + f'\left(f+\tfrac12f'fh+\left(\tfrac14f'^2f+\tfrac18f''[f,f]\right)h^2\right)h + \frac12f''\left[f+\tfrac12f'fh,f+\tfrac12f'fh\right]h^2 + \frac16f'''[f,f,f]h^3 + ...} \\ &&&=f+f'fh+\left(\tfrac12f'^2f+\tfrac12f''[f,f]\right)h^2 + \left(\tfrac1{8}f'f''[f,f]+\tfrac12f''[f'f,f]+\tfrac1{6}f'''[f,f,f]\right)h^3 + ... \end{align} Combining the terms of matching degree you should find that they are the same up to degree $4$.