We just read Picard existence and uniqueness theorem for initial value problems. Please someone give us one example to show that IVP may not have a solution, and second example to show that if solution exists then it may not be unique

  • $\begingroup$ Are you more than one :D $\endgroup$ Apr 4, 2017 at 7:52
  • $\begingroup$ Hint: Picard (basically) works when the $F$ is Lipschitz (in $y'=F(x,y)$) - play around with some standard non-Lipschitz $F$ and see what happens. $\endgroup$
    – πr8
    Apr 4, 2017 at 7:54
  • $\begingroup$ dear Jaideep by "we" I mean all my classmates $\endgroup$
    – Tadde
    Apr 4, 2017 at 7:55
  • $\begingroup$ math.stackexchange.com/questions/757245/… $\endgroup$
    – Abelois
    Apr 4, 2017 at 8:45

1 Answer 1


No solution: $$ \dot x =\begin{cases}-1&x\ge 0\\1&x<0\end{cases} $$ with $x(0)=0$. In some weakened generalized context, $x\equiv0$ could be called a solution, but since $\dot x=0$ it does not satisfy this ODE in the strong sense.

Note that the Peano theorem guarantees local solutions if the right side is continuous, which covers all the "nice" cases.

More than one solution: The classical example is $$ \dot x=2\sqrt{|x|} $$ with $x(0)=0$ which has solutions $$ x(t)=\begin{cases}0&0\le t<c\\\pm(t-c)^2&t\ge c\end{cases} $$ for any $c>0$ including $c=\infty$.


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