# Differential Equation $u'(t) = \sqrt{|u(t)|}$

I got the equation (exercise of an old exam) $$u'(t) = \sqrt{|u(t)|} \quad ; \qquad u(t_0) = u_0$$ with $u(t) \in \mathbb R$. Then I have to say on which intervals $\mathcal I$ solutions exist and if they are unique.

What I got so far:

Because $f(t,v) = \sqrt{|v|}$ is not differentiable in $0$ I concluded that if $u_0 \neq 0$ there are locally unique solutions because then $f(t,v)$ is continuously differentiable around $(t_0,u_0)$.

If $u_o > 0$ then we get the solution $$\left ( \frac {t+C}{2} \right )^2$$ with $C \in \mathbb R$. If $t_0 = 0$ then we get $C = \pm 2 \sqrt{u_0}$ which leads to the conclusion that there are two solutions which fulfill the conditions and pass trough $u_0$ in $t_0$. So the solution is not unique in $(0,u_0)$ ?! I am confused :D

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Since $\frac{\mathrm{d}}{\mathrm{d}t}u=\sqrt u$, we get $$\int\frac{\mathrm{d}u}{\sqrt{|u|}}=\int\mathrm{d}t\tag{1}$$ so that $$2\,\mathrm{sgn}(u)\sqrt{|u|}=t+C\tag{2}$$ Since $u'\ge0$, the solution is: \begin{align} u &=\mathrm{sgn}(t+C)\frac14(t+C)^2\\ &=\frac14|t+C|(t+C)\tag{3} \end{align} Given $(t_0,u_0)$, using $(2)$, we must have $$C=2\,\mathrm{sgn}(u_0)\sqrt{|u_0|}-t_0\tag{4}$$
But for $t_0 = 0$ we get $C = \pm 2 \sqrt{u_0}$ so that $C$ is not unique and thus there are two solutions which pass through $(t_0,u_0)$? (For $u_0 > 0$) – Epsilon Jan 18 '13 at 13:59
I got a corollary in my text which says (according to Picard) : Assume $f$ is cont. diff. on an open set $\mathcal E$ around $(u_0,t_0)$ Then the problem $u'(t) = f(u(t)), u(t_0) = u_0$ has a unique solution on $[t_0-\delta,t_0+\delta]$ for some $\delta > 0$.--- This should guarantee uniqueness if $u_0 \neq 0$ around $t_0$. Because if $u_0 \neq 0$ then $\sqrt{|v|}$ is con. diff. arund $v$ for some interval. – Epsilon Jan 18 '13 at 14:11
@André: I have fixed my answer to account for the fact that $u'\ge0$. The answer is now indeed unique. – robjohn Jan 18 '13 at 14:15