TITLE: If an IVP does not enjoy uniqueness, then it possesses infinitely many solutions
QUESTION [5 upvotes]: I am trying to prove that when an IVP in ODEs does not enjoy uniqueness, then it has infinitely many different solutions. I know that when the Lipschitz condition is satisfied, then there is a unique solution.  
I guess, I am trying to show that if the IVP possesses two solutions, then we can create infinitely many solutions using these two. However, I am slightly lost as where to start the proof.

REPLY [2 votes]: You can start like this. Let the IVP be 
\begin{eqnarray} L(D)y=f,y(0)=y_0. \end{eqnarray}
Let $y_1,y_2$ be two different solutions; namely, $y_1,y_2$ satisfy
$$ L(D)y_i=f, y_i(0)=y_0, i=1,2. $$
Now let $y_\alpha=(1-\alpha)y_1+\alpha y_2$ ($\alpha$ is a constant) and you can check that $y_\alpha$ is a solution of the IVP for any $\alpha$. Hence your argument is proved.