Hello interested helpers,
my actual problem is that I want to define a solution operator to a "nice" initial value problem, say $$ \begin{cases} \dot y = f(y),\\ y(0) = y_0, \end{cases} $$ where $f$ is (locally) Lipschitz-continous. This initial value problem has a unique solution on some maximal interval of existence. This is where my problem lies. I want to define a solution operator $$ S\colon Y\times \mathrm{Lip}(Y,Y)\supset U \to \mathcal C(0,T; Y) $$ that maps the initial value $y_0$ and the right hand side $f$ to the solution of the problem. However, if a blow-up occures before the time $T$ this operator is not well defined. To restrict the domain is a bit too bothersome in my case since it might destroy the properties of U.
What I'd like to do, is to introduce a point $\infty$ and extend the solution with infinity after the blow-up. Then the space $\bar Y:= Y \cup\{\infty\}$ can be identified with a sphere in $Y\times \mathbb R$. This should be a manifold with inherited topology from the plane $Y$ onto the sphere via sterografic projection, it has a notion of Cauchy-sequences and their convergence and one can define a curve integral, at least I imagine, that I can.
My imagination reaches its end, however, at the space $\mathcal C(0,T; \bar Y)$. It is not a linear space, due the lack of an additive inverse. If I want to apply Picard-Lindelöfs theorem, I need at least a complete metric space. Therefore, I consider $$d_p: \bar Y \times \bar Y \to \mathbb R \cup \{\infty\}, d_p(y_1,y_2) = \begin{cases}0, & y_1 = y_2;\\ \Vert y_1-y_2\Vert_Y,& y_1,y_2 \in Y;\\ \infty, & \text{otherwise} \end{cases}$$ and $$d:\mathcal C(0,T; \bar Y)\times \mathcal C(0,T; \bar Y)\to\mathbb R \cup \{\infty\}, d(f,g)=\sup_{t\in [0,T]} d_p(f(t),g(t)).$$ Of course, those are not metrics anymore, but they induce the same topology as we had on the plane $Y$ before and can be used to define pointwise continuity and Lipschitz-continuity.
Has anyone seen this approach before or can tell me, why this is not done anywhere? Where do I hit the wall?
Thanks for reading (and helping).