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?
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2$\begingroup$ did you check to make sure you didnt divide by zero when you separated variables? $\endgroup$– operatorerrorDec 4, 2017 at 17:29
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$\begingroup$ Oh. There's the trivial solution isn't there? $\endgroup$– Not EulerDec 4, 2017 at 17:32
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2 Answers
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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.
answered Dec 4, 2017 at 17:32
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$\begingroup$ I understand now that the trivial solution is also a solution. But could you elaborate on the first line of your answer please? $\endgroup$ Dec 4, 2017 at 17:43
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1$\begingroup$ probably worth showing how you actually come up with this family $\endgroup$ Dec 4, 2017 at 17:44
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$\begingroup$ I meant the part about Cauchy-Peano and Picard-Lindelof existence theorems. $\endgroup$ Dec 4, 2017 at 17:56
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1$\begingroup$ @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 $\endgroup$ Dec 4, 2017 at 18:16
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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 $$ y=t^2 $$
answered Dec 4, 2017 at 17:33
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$\begingroup$ I felt this answer was inadequate, as it does not address solutions other than the trivial one. $\endgroup$– DylanDec 4, 2017 at 22:37
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$\begingroup$ I acknowledge your answer is still correct, but it was too late to undo the vote $\endgroup$– DylanDec 4, 2017 at 22:42
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$\begingroup$ @Dylan How could this be inadequate? The question is "to show that this IVP does/does not have a unique solution." Unique means one. $\endgroup$ Dec 4, 2017 at 23:19
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