# To show that this IVP does/does not have a unique solution

Given the IVP $y'=2√y, y(0)=0$. I find $y=x^2$ as the only solution to the problem. But I am asked to show that it does not have a unique solution. Is there a mistake or am I just wrong? Is there a sufficient condition for the existence of more than one solution to an IVP?

• did you check to make sure you didnt divide by zero when you separated variables? Dec 4, 2017 at 17:29
• Oh. There's the trivial solution isn't there? Dec 4, 2017 at 17:32

This is a standard type of example problem to illustrate the difference of the Cauchy-Peano and Picard-Lindelöf theorems. You get any of $$y(x)=\begin{cases}0,&x<c,\\(x-c)^2,&x\ge c,\end{cases}$$ as valid solutions to this IVP.

• I understand now that the trivial solution is also a solution. But could you elaborate on the first line of your answer please? Dec 4, 2017 at 17:43
• probably worth showing how you actually come up with this family Dec 4, 2017 at 17:44
• I meant the part about Cauchy-Peano and Picard-Lindelof existence theorems. Dec 4, 2017 at 17:56
• @HritRoy the idea is that the Lipschitz condition is what gives you uniqueness, here you have continuity alone, this gives you the existence of the family Dec 4, 2017 at 18:16

When you separate variables, recall that either $y=0$, which is easy to see is a solution, or $$\int\frac{\mathrm dy}{\sqrt{y}}=2\int \mathrm dt\implies \sqrt{y}=t+c\\ \implies y=t^2+2ct+c$$ and $c=0$ thanks to BC, you have, as you found $$y=t^2$$

• sigh, why the downvote Dec 4, 2017 at 20:17
• I felt this answer was inadequate, as it does not address solutions other than the trivial one. Dec 4, 2017 at 22:37
• I acknowledge your answer is still correct, but it was too late to undo the vote Dec 4, 2017 at 22:42
• @Dylan How could this be inadequate? The question is "to show that this IVP does/does not have a unique solution." Unique means one. Dec 4, 2017 at 23:19
• If it's any consolation, I upvoted :) Dec 4, 2017 at 23:55