I should prove, that the following RK method:
$$x_{k+1} = x_k + \frac{h}{6} (K_1 + 2 K_2 + 2 K_3 + K_4)$$
for:
$$K_1 = f(t_k, x_k)$$
$$K_2 = f(t_k + \frac{1}{2} h, x_k + \frac{1}{2} h K_1)$$
$$K_3 = f(t_k + \frac{1}{2} h, x_k + \frac{1}{2} h K_2)$$
$$K_4 = f(t_k + h, x_k + h K_3)$$
has a unique solution for any given ODE:
$$x'(t) = \phi (t) \quad x(t_0) = x_0$$
where $\phi \in \pi^3$
What I've found out is that the corresponding matrix for this RK method is equal to:
$$\begin{pmatrix}c&A\\ &b^T\end{pmatrix} = \begin{pmatrix}0&0&0&0&0\\ \frac{1}{2}&\frac{1}{2}&0&0&0\\ \frac{1}{2}&0&\frac{1}{2}&0&0\\ 1&0&0&1&0\\ &\frac{1}{6}&\frac{1}{3}&\frac{1}{3}&\frac{1}{6}\end{pmatrix}$$
Which means, that:
$$A = \begin{pmatrix}0&0&0&0\\ \frac{1}{2}&0&0&0\\ 0&\frac{1}{2}&0&0\\ 0&0&1&0\end{pmatrix}$$
This tells us that this RK method is an explicit method.
And I was wondering: Is this enough of a proof that this numerical method has exactly one unique solution?
Because what we've been doing was:
$$x_{k+1} = x_k + h \Psi_f (h, t_k, x_k, x_{k+1}) = F(x_{k+1})$$
And to show whether $F(\cdot)$ is Lipschitz or not (if yes, then we'd have found the interval on which our method does indeed have a single unique solution)
But the problem here is, that my RK method does not have any $x_{k+1}$ involved, which means that this approach cannot be applied here