Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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}

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.