If there is local uniqueness to a solution to an IVP (such as implied Picard–Lindelöf theorem), and if we restrict ourselves to solutions over intervals, then there is global uniqueness of solutions.
Theorem. Uniqueness over intervalsEdit
Theorem If solutions over intervals coincide at a single point then they are the same
- and are solutions to an IVP
- and are locally unique solutions (by the Picard–Lindelöf theorem for example)
- the domains of and are both intervals (which contain , otherwise the initial condition makes no sense)
- and coincide inside their common domain of definition:
- If ,
is also a solution, with domain . This notation is unambiguous because of the above hypothesis.
Example. y'=y in various interval domainsEdit
Take the IVP . Therefore .
Then and satisfy all the hypothesis of the theorem
- both are solutions to the IVP
- both are locally unique because satisfies the Picard–Lindelöf theorem in all of its domain ( and therefore is also locally Lipschitz)
- and are both intervals and respectively.
Then we observe all of our conclusions:
- Inside ,
- If we fix the intervals and , then and are the only solutions with exactly those domains
is also a solution to the IVP with domain .
Remark. Different domains, different functionsEdit
Remark Different domains mean completely different functions.
Remember from set theory that a function simply a set of ordered pairs. For example
are two functions so that and , and , and . Note that they coincide in the intersection of their domains:
However they are not equal. Remember that two sets are equal iff the have exactly the same elements, which is obviously not the case for and since but . Therefore and are two completely different sets, and therefore two completely different solutions.
The same goes for two functions such as
Many times the domain of a function is implicit, and we forget about it, usually taking the largest possible. But sometimes taking the largest possible domain may not be appropriate. For example when solving differential equations, taking a domain that is too large (and not an interval) may not lead to uniqueness, which is undesirable. In those cases it is necessary to specify very well what domain we are talking about.
Counter-example. Not an interval.Edit
Take the IVP . Look at the infinite family solutions
which are each determined by any value of a ( =(y(3) ).
Those solutions satisfy all the conditions of the theorem, except that their common domain is not an interval. Then we observe that all the conclusions fail for
- Inside ,
- Both have the same domain, but , but
All of this happens because the uniqueness of the initial condition cannot propagate from to the other side of the domain .