Please help me to solve the following problem:
Assume we have an ODE $f$ such that Picard–Lindelöf theorem holds. Let $g_k(t)$ is $k$-th Picard approximation. Prove or disprove the following:
- If there exists $k$ such that $g_k(t)=g_{k+1}(t)$ then $g_k(t)$ is solution for $f$.
- If for any $k$ it is true that $g_{2k}(t)=g_{0}(t)$ and $g_{2k+1}(t)=g_{1}(t)$ then $g_{i}(t)=g_{j}(t)$ for any $i$, $j$.
My attempt:
I think it is true, but I can not prove this. Intuitively I feel that every Picard iteration give a function that is closer to solution. As long as two of them are the same, there is not way to get closer to solution, so we found one. But I do not see how to make it precise, so I can not be sure it is true. But my vague idea looks reasonable.
From Picard–Lindelöf theorem we know that there exists unique solution $h=\lim g_k(k)$ of our ODE with given initial conditions. Since any subsequence of a convergent sequence has the same limit and since our even and odd subsequences are constant we conclude that statement is true. I feel ok with my proof, but can you please validate it?
Thanks a lot for your help!
Let $P$ denote the Picard operator.
If $g_{k+1} = Pg_k = g_k$, then $g_k$ is a fixed point of the Picard operator, hence it is a solution of the related Cauchy problem.
Since the sequence of iterates converges, we must have $g_0 = g_1$ and $g_k = g_0$ for every $k$.