Finding value of $a$ to satisfy Ordinary Differential Equation

Question

I am very confused by this problem. I think it is probably quite easy and I must be missing something simple. Any help is appreciated!

Suppose that $y(t) \in C^{1}[0,+\infty)$ (meaning that $y:[0,+\infty) \to \mathbb{R}$ is continuously differentiable) satisfies

$$y'(t)=2\sqrt{|y(t)|}$$ for $t>0$ and $$y(0)=0.$$

Give a detailed proof that then there is $a \in [0, +\infty]$ such that $$y(t)= \begin{cases} 0 \text{ if } 0 \leq t \leq t\\ (t-a)^{2} \text{ if } t \geq a. \end{cases}$$ Hint: How to determine a?

Answer 1

$$\frac{dy}{dt}=2\sqrt{|y|}$$ $$\frac{dy}{\sqrt{|y|}}=2.dt$$ $$2\sqrt{y}=2t+C$$ $$|y|=(t+a)^2 $$

Hope you can handle the rest?

Answer 2

Since $y'(t)\geq 0$ for every $t > 0$, the function $y$ is monotone non-decreasing. Let $$ a := \sup\{t\geq 0:\ y(t) = 0\}. $$ If $a = +\infty$ then $y(t) = 0$ for every $t\geq 0$, so that $y$ is a solution of the problem and we are done.

If $a<+\infty$, then $y(t) = 0$ for every $t\in [0,a]$ whereas $y(t) > 0$ for every $t > a$. Solving the differential equation in the region $(a, +\infty)$ you get that $y$ must be of the form $y(t) = (t-C)^2$, $t>C$. In order to have a continuous function in $[0,+\infty)$ the only possibility is to choose $C=a$, obtaining $$ y(t) = \begin{cases} 0, &\text{if}\ t\in [0,a],\\ (t-a)^2, &\text{if} \ t > a. \end{cases} $$ Now, you can check that this function is of class $C^1$ (you need only to chech that it is differentiable at $t = a$ and that its derivative is continuous).