The conditions for the solution of the problem:
to exist and to be unique is that and Lipschitz continuous for . Then, the solution of the problem is given by:
Note that for the function to be integrable in , it must contain finitely many discontinuities so remains continuous in whilst, otherwise, may not, in general.
As an example, consider the problem:
Its solution is then given by:
which is continuous but not differentiable at .