In the previous lecture we defined the Young’s integral when and with . The integral path has then a bounded -variation. Now, if is a Lipschitz map, then the integral, is only defined when , that is for . With this in mind, it is apparent that Young’s integration should be useful to solve differential equations driven by continuous paths with bounded -variation for . If , then the Young’s integral is of no help and the rough paths theory later explained is the correct one.
The basic existence and uniqueness result is the following. Throughout this lecture, we assume that .
Theorem: Let and let be a Lipschitz continuous map, that is there exists a constant such that for every ,
For every , there is a unique solution to the differential equation:
Proof: The proof is of course based again of the fixed point theorem. Let and consider the map going from the space into itself, which is defined by
By using basic estimates on the Young’s integrals, we deduce that
If is small enough, then , which means that is a contraction of the Banach space endowed with the norm .
The fixed point of , let us say , is the unique solution to the differential equation:
By considering then a subdivision
such that , we obtain a unique solution to the differential equation:
As for the bounded variation case, the solution of a Young's differential equation is a function of the initial condition,
Proposition: Let and let be a Lipschitz continuous map. Let be the flow of the equation
Then for every , the map is and the Jacobian is the unique solution of the matrix linear equation
As we already mentioned it before, solutions of Young’s differential equations are continuous with respect to the driving path in the -variation topology
Theorem: Let and let be a Lipschitz and bounded continuous map such that for every ,
Let be the solution of the differential equation:
If converges to in -variation, then converges in -variation to the solution of the differential equation:
Proof: Let . We have
Thus, if is such that , we obtain
In the very same way, provided , we get
Let us fix and pick a sequence such that . Since , for with big enough, we have
We deduce that for ,
For with and big enough, we have