Ordinary Differential Equations/Maximum domain of solution

< Ordinary Differential Equations

Even if a differential equation satisfies the Picard–Lindelöf theorem, it may still not have a solution for all of that is uniquely determined by a single initial condition. Here we study the maximum interval to which a solution may be extended.


Uniqueness of solution over intervalsEdit

Theorem Local uniqueness implies global uniqueness over intervals


  • y is a solution to an IVP
  • y is locally unique (by the Picard–Lindelöf theorem for example)
  • the domain of y is an interval (which contains  , otherwise the initial condition makes no sense)

y(x) is the only solution to that IVP of the which has that domain.

Maximal solutionEdit


Definition Maximal solution and maximal domain of solution

The maximal solution   of an IVP which is locally unique is the solution such that

  • has an interval as domain (which contains  , otherwise the initial condition makes no sense)
  • is not a restriction of any other solution whose domain is a larger interval

  is called the maximal domain of solution and is denoted  


Theorem Every IVP has an unique maximal solution

Theorem If there is a solution   with domain   then it is the maximal solution.

We must verify that:

  •   is an interval. Obviously true.
  •   is not a restriction of any other solution. Obvious because the domain of   is already all of  

Behavior at boundaryEdit

Theorem Behavior at the boundary of a maximal solution


The domain of a maximal solution is smaller than infinity ( )

At  one or two of the following happen:

  • explosion in finite time:  
  • y falls out of the domain of F:  

Sufficient condition for Dom(y) = REdit

Theorem Growth at most linear implies  


F grows at most linearly with y


The domain of the maximal solution is all of  


y'=y,\, y(0)=1Edit

Example   has an infinity of solutions

F(x,y)=y , which is   in both x and y, and therefore satisfies the Picard–Lindelöf theorem on all of its domain, so that solutions are locally unique.

All of the following are solutions to this IVP:




 Different domains mean completely different solutions

A function is a set of ordered pairs {(x,f(x))}, with  . If the domains are different, the sets are completely different, and therefore the solutions are completely different too.

 ,  and   are the only solutions on their respective domains

This is a conclusion of the theorem on the uniqueness on intervals, because their domains are all intervals.

If the domain is not an interval there is generally no uniqueness

For the fixed domain of  


which is clearly not an interval, then for any value y(3) we choose we have a different solution.

Therefore, even if the domain is fixed but not an interval and there is local uniqueness, there may not be global uniqueness. This is the major reason why we restrict ourselves to a maximal domain that is an interval, even if there may be larger domains with non-unique solutions.

To determine the solution uniquely the initial value at 0 is not enough. We would have to set a value on   such as y(3). This happens because the initial condition at 0 is separated from the interval  . We need to set the a value inside of  , such as y(3) in order to fully determine the solution.

  is the maximal solution  

Since the domain is all of  , it must be the maximal solution by the proposition above.

We could immediately see that   without solving it because the growth of F is linear, as stated on the theorem of sublinear growth.

We note also that as stated on the definition, any solution that has a domain that is an interval such as   and   are restrictions of  .

Solutions which are not defined on intervals may not be a restriction of the the maximal solution  

  is not a a restriction of   unless  . However,   is not unique, and in unacceptable in a physical situation since there is no causality between the two intervals of its domain. So it is not very serious if it is not a restriction of  , which is generally the 'best' one.

y' = y^2Edit


 , which is   and therefore Lipschitz continuous, satisfying the Picard–Lindelöf theorem.

  • the maximal domain may depend on the initial condition
  • the maximal domain may be different than  
  • at the borders of the maximal domain the function may go to infinity

The maximal solutions are


which clearly depend on the initial condition, and are not all of   unless for the trivial solution.

At   the maximal solution tends to infinity.

Extending the domain other side of the singularity leads to non uniqueness

Fix  . One might want to extend the domain of   to  . But then there is no more uniqueness since for any   the following is a solution


This is no surprise since   is not an interval.

  above are not maximal solutions

Actually any solution that is an interval is a restriction of those solutions. But this is not a maximal solution since its domain is not an interval.

We wouldn't like those to be maximal solutions since they would not be unique.



 , which is   and therefore Lipschitz continuous  , satisfying the Picard–Lindelöf theorem. If  , f is undefined, so that


The maximum solution is

 At the border the solution may fall out of the domain of F

The solution is only defined for  , because if   then  , which is not in the domain of F. The maximal solution ends at those points, which is one of the two possibilities for finite domains. The solution does not explode in this case.