I have the following question.
I have shown that for coefficients $\alpha_k, \beta_k$ of a linear multi-step process with characteristic polynomial $\rho$ and
$$C_j=\sum_{k=0}^m \alpha_k\frac{k^j}{j!}-\sum_{k=0}^m \beta_k \frac{k^{j-1}}{(j-1)!}$$
following holds:
The process has consistency order $p$ iff $C_j=0$ for $j=1,\dots, p$ and $\rho(1)=0$
Now I am asked to conclude, that
The process is for $f\in C^1$ consistent iff $\rho(1)=0$ and $\rho'(1)=\sigma(1)$
Note, that $\rho$ and $\sigma$ are the first and second characteristic polynomials, with
$$\rho(x)=\sum_{j=0}^m a_jx^j~~\text{ and }~~\sigma(x)=\sum_{j=0}^m b_j x^j$$
My problem is, that I am not exactly sure, what to do, since $f\in C^1$. Do I have to show, that the process is consistent of order 1?
Help is appreciated.
Thanks in advance.
If $f$ is $C^1$, then any solution $y$ is $C^2$. Insert the exact solution into the method error equation, and then apply Taylor polynomials of the available degree \begin{align} e(t,h)&=\sum a_ky(t+kh)-h\sum b_k y'(t+kh) \\ &=\sum a_k(y(t)+y'(t)kh+O(h^2))-h\sum b_k(y'(t)+O(h)) \\ &=ρ(1)y(t)+ρ'(1)y'(t)h-σ(1)y'(t)h+O(h^2) \end{align} so that for $ρ(1)=0$ and $ρ'(1)=σ(1)$ you get $e(t,h)=O(h^2)$. Which means, by one definition, that the method is consistent.
You could also check directly another definition, that the method is exact for the ODE $y'(x)=f(x,y(x))=1$ with solution $y(x)=x+c$.