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So I have a differential equation $$\frac{\mathrm{d}y}{\mathrm{d}t}= 2 \sqrt{|y|}$$ with $y(0)=0$. So, I found that the solution to this ODE is $y=(t+c)^2$ and $y=-(t+c)^2$. Doesn't this Differential equation contradict the Uniqueness theorem?

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Let $c \ge 0$. Here is a family of solutions all satisfying $y_c(0) = 0$: $y_c (t) = \begin{cases} 0 , & t < c \\ (t-c)^2, & t \ge c \end{cases}$.

The usual proof of uniqueness depends on the right hand side being Lipschitz, which fails here.

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    I see your point. Okay. What if I ask this: for any t y(t)= 0 is also a solution to the Differential equation. and y=t^2 is also a non-zero solution. Does it make sense to ask whether the presence of these two solutions contradict uniqueness theorem? – 2017-02-06
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    @Avery: No, because the hypotheses of the relevant theorems are not satisfied, so the theorems do not apply. – 2017-02-06
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Firstly, we have the stipulation that $y(0) = 0$, and so we obtain $c=0$ in both cases. So your question is really "why are $y=\pm t^2$ both solutions?".

Well, the simple answer is that they're not both solutions. Let $y=t^2$. Then $$\mathrm{LHS} = \dfrac{\mathrm{d}y}{\mathrm{d}t} = 2t$$ while $$\mathrm{RHS} = 2 \sqrt{|y|} = 2 |t|$$

As an aside, for completeness: the uniqueness theorem I know for first-order ODEs in general is:

Let $\phi: \mathbb{R}^2 \to \mathbb{R}$ be continuous, and Lipschitz in the second variable. Let $y_0, t_0 \in \mathbb{R}$. Then there is a ball around $t_0$ such that there is a unique solution $y$ to the DE $$y'(t) = \phi(t, y(t)); y(t_0) = y_0$$ in that ball.

This theorem doesn't apply to your ODE because $\phi: (t, y) \mapsto 2 \sqrt{|y|}$ is not Lipschitz in the variable $y$.

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    So do you mean to say that they are solutions for their individual cases? y=t^2 is a solution only for y>=0 and y = -t^2 is a solution for y<0? – 2017-02-06
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    Why is the theorem pathetic??? – 2017-02-06
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    @Avery The theorem I quoted doesn't say anything about global properties like "$y=t^2$ is a solution for $y \geq 0$". That theorem can only hope to apply to small regions. – 2017-02-06
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    (I've got rid of the first uniqueness theorem I quoted, because I'm becoming increasingly unsure that it's true. I'm pretty sure the later theorem is true because I copied it verbatim from my second-year analysis notes.) – 2017-02-06
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    @copper.hat I called it pathetic because it's so local, basically. It usually takes work to stitch that theorem's unique local solutions into unique global ones. But I've removed the word "pathetic" anyway :) – 2017-02-06
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    Well, global is out of fashion at the moment... – 2017-02-06
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    I like the theorem, I think it is powerful (if local). Plus the proof uses contraction maps, another favourite :-). – 2017-02-06
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    I agree that the proof is lovely, and the theorem gave me a great sense of accomplishment when we did it in lectures. It's just quite a bit of work for something that doesn't justify anything the applied people do. – 2017-02-06
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    It is the basis of relaxation techniques for solving ODEs... – 2017-02-06