Ordinary Differential Equations/Global uniqueness of solution over interval

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 intervals
Theorem If solutions over intervals coincide at a single point then they are the same

Hypothesis


 * 1) $$y_1$$ and $$y_2$$ are solutions to an IVP
 * 2) $$y_1$$ and $$y_2$$ are locally unique solutions (by the Picard–Lindelöf theorem for example)
 * 3) the domains of $$y_1$$ and $$y_2$$ are both intervals (which contain $$x_0$$, otherwise the initial condition makes no sense)

Conclusion


 * 1) $$y_1$$ and $$y_2$$ coincide inside their common domain of definition: $$ \forall x \in Dom(y_1) \cap Dom(y_2) \, y_1(x) = y_2(x)$$
 * 2) If $$Dom(y_1) = Dom(y_2)$$, $$y_1 = y_2$$
 * 3) $$ y=\begin{cases}y_1 & x \in Dom(y_1) \\y_2 & x \in Dom(y_2)\end{cases}$$

is also a solution, with domain $$Dom(y_1) \cup Dom(y_2)$$. This notation is unambiguous because of the above hypothesis.

Example. y'=y in various interval domains
Example

Take the IVP $$y'=y,\, y(0)=1$$. Therefore $$F(x,y)=y$$.

$$ \begin{align}y_1 : \, & ]-2,2[\\ & x \mapsto e^x \end{align} $$

$$ \begin{align}y_2 : \, & ]-1,3[ \\ & x \mapsto e^x \end{align} $$

$$ \begin{align}y_3 : \, & ]-2,3[\\ & x \mapsto e^x \end{align} $$

Then $$y_1$$ and $$y_2$$ satisfy all the hypothesis of the theorem


 * 1) both are solutions to the IVP
 * 2) both are locally unique because $$F(x,y)$$satisfies the Picard–Lindelöf theorem in all of its domain ( $$F \in \mathbb{C}^1$$ and therefore is also locally Lipschitz)
 * 3) $$Dom(y_1)$$ and $$Dom(y_2)$$ are both intervals $$]-2,2[$$ and $$]-1,3[$$ respectively.

Then we observe all of our conclusions:


 * 1) Inside $$Dom(y_1) \cap Dom(y_2) = ]-1,2[ $$, $$y_1 = y_2$$
 * 2) If we fix the intervals $$]-2,2[$$ and $$]-1,3[$$, then $$y_1$$ and $$y_2$$ are the only solutions with exactly those domains
 * 3) $$ y=\begin{cases}y_1 & x \in Dom(y_1) \\y_2 & x \in Dom(y_2)\end{cases} = y_3$$

is also a solution to the IVP with domain $$]-2,3[$$.

Remark. Different domains, different functions
Remark Different domains mean completely different functions.

Remember from set theory that a function simply a set of ordered pairs. For example

$$ f_1 = \{(0,1)\} $$

$$ f_2 = \{(0,1),(1,2)\} $$

are two functions so that $$Dom(f_1)={0}$$ and $$f(0)=1$$, and $$Dom(f_2)={0,1}$$, $$f(0)=1$$ and $$f(1)=2$$. Note that they coincide in the intersection of their domains: $$f_1(0)=f_2(0)=1$$

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 $$f_1$$ and $$f_2$$ since $$(1,2) \in f_2$$ but $$(1,2) \notin f_1$$. Therefore $$f_1$$ and $$f_2$$ are two completely different sets, and therefore two completely different solutions.

The same goes for two functions such as

$$ \begin{align}y_1 : \, & ]-2,2[ \, \to \, ]e^{-2},e^{2}[ \\ & x \mapsto e^x \end{align} $$

$$ \begin{align}y_2 : \, & ]-1,3[ \, \to \, ]e^{-1},e^{3}[ \\ & x \mapsto e^x \end{align} $$

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.
Counter example

Take the IVP $$y'=y,\, y(0)=1$$. Look at the infinite family solutions

$$ \begin{align}y_a : & Dom(y_a) = ]-1,1[ \,t \cup \, ]2,4[ \\ & x \mapsto \begin{cases}e^x & x \in ]-1,1[ \\ae^{(x-3)} & ]2,4[\end{cases}\end{align} $$



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 $$]-1,1[ \, \cup \, ]2,4[$$ is not an interval. Then we observe that all the conclusions fail for $$a \neq a'$$


 * 1) Inside $$ ]2,4[ \subset Dom(y_1) \cap Dom(y_2)$$, $$y_a \neq y_{a'}$$
 * 2) Both have the same domain, but $$Dom(y_a) = Dom(y_{a'}) = ]-1,1[ \, \cup \, ]2,4[$$, but $$y_a \neq y_{a'}$$

All of this happens because the uniqueness of the initial condition cannot propagate from $$x_0 = 0$$ to the other side of the domain $$]2,4[$$.