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I have the following analogue of Picard's theorem for Banach space valued ode's:

Let $O$ be an open subset of a Banach space $B$ and let $F$ be a nonlinear operator satisfying the following criteria

i) F maps $O$ into $B$

ii) F is locally Lipschitz continuous

Then for any $X_0 \in O$ there exists $T>0$ such that the ODE \begin{equation} \frac{dX}{dt}=F(X) ,\quad X|_{t=0}=X_0 \end{equation} has a unique solution $X \in C^1((-T,T),O)$.

So suppose $B \subset C(\mathbb{R}^n, \mathbb{R}^n)$ be a Banach space with some norm $\| \cdot \|$. Also suppose I have some open $O \subset B$ and nonlinear $F$ as in the above theorem.

My question is the following. Assuming we have For $s,t \in \mathbb{R}$ such that $0 \leq s \leq t \leq t+s < T$, can we conclude (perhaps by uniqueness in the theorem) that \begin{equation} X(t) \circ X(s) = X(t+s) \quad \text{?} \end{equation}

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1 Answer 1

No. Consider $$ \frac{\mathrm{d}X}{\mathrm{d}t} = \begin{pmatrix}0&1\\1&0\end{pmatrix}, $$ with the initial condition $X(0)=I$, the identity. The solution to this problem is $$ X(t)=\begin{pmatrix}1&t\\t&1\end{pmatrix}, $$ but $$ X(t)X(s)=\begin{pmatrix}1+st&s+t\\s+t&1+st\end{pmatrix}, \qquad X(t+s)=\begin{pmatrix}1&s+t\\s+t&1\end{pmatrix}. $$ Note: I assume you know that this is not the usual setting to talk about flows associated to ODEs, and you intentionally took $B$ to be a space of mappings. In the usual setting you would have the mapping sending initial conditions in $B$ to the solution at time $t$. Of course, you have the flow property in that setting.

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