# Definition of unique solution of initial value problem

I am confused about the definition of unique solution of ODE. I wonder which of following definitions ((1) or (2)) is correct?

Given an ODE $y'=f(x,y)$ and the initial condition $y(0) = 1, y'(0) = 2$ .
If $y=f(x)$ (defined on (-3,3) for example) satisfies the ODE and the initial condition then $y$ is said to be the unique solution on (-3,3) if

1. For any function z(x) that satisfies the ODE and the initial condition on $(-3,3)$, we have $z(x) = y(x) \quad \forall x \in (-3,3)$ .
2. For any function z(x) that satisfies the ODE and the initial condition on $(a,b) \subset (-3,3)$ and $0 \in (a,b)$, we have $z(x) = y(x) \quad \forall x \in (a,b)$ .

I mean if a solution $y_1$ if said to be unique on an interval $I$, is there any chance that there exists another solution $y_2$ defined on a subset $B$ of $(a,b)$ but does not coincide with $y_1$ on $B$? I think they are equivalent and tried to prove it by extending $y_2$ on $I$ but I am not sure if one can do that.

## 1 Answer

You are on the right track.

One can show that the maximal interval only depends on the initial condition, no matter if the solutions are unique or not! In other words, all solutions corresponding to a given initial condition have the same maximal interval.

So, even if we decided to take your number $2$ as the definition of uniqueness, the truth is that we could always extend the solution from $(a,b)$ to the larger interval, thus making $1$ and $2$ equivalent. Summing up, one can take your number $1$ as the definition in case $(-3,3)$ is not necessarily the maximal interval:

For some initial condition $x(t_0)=x_0$, the solution is said to be unique on the interval $(a,b)$ containing $t_0$ (and $(a,b)$ may not be the maximal interval!) if any other solution on the same interval coincides.