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I would like to solve the differential equation given by $$ y' = \sqrt{|y|},\qquad y(0) = 0 $$ This is equivalent, if we suppose that $y > 0$, to $$ \frac{dy}{dt} = y^{1/2} \text{ if and only if } y^{-1/2} dy = dt $$ so it should be: $$ 2 y^{1/2} = t + c \implies y = \frac{(t+c)^2}{4} $$ As a test I have checked that $$ y' = \frac{t+c}{2} = \sqrt{|y|} = \sqrt{y} $$ However, I would like to know how to obtain other solutions, just as: $$ y_{\alpha,\beta}(t) = \begin{cases} -(t-\alpha)^2 / 4 & t < \alpha,\\ 0 & \alpha \leq t \leq \beta,\\ (t-\beta)^2/4 & t > \beta \end{cases} $$ for any $\alpha < 0 < \beta$, real numbers. I mean, I don't know how would you find out every solution to this differential equation. Thanks in advance.

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    $\begingroup$ What are $\alpha$ and $\beta$? Do you have boundary conditions for this problem? $\endgroup$
    – Vlad
    Jun 9, 2015 at 21:21
  • $\begingroup$ @Vlad I've edited the question $\endgroup$
    – user55268
    Jun 9, 2015 at 21:26
  • $\begingroup$ @AlbertT. What Vlad asked you is the role of $\alpha$ and $\beta$ in your problem. The fact that $\alpha < \beta$ is just an hypotesis you had, not the "meaning" of $\alpha$ and $\beta$. $\endgroup$ Jun 9, 2015 at 21:28
  • $\begingroup$ @the_candyman For every two $\alpha,\beta \in \mathbb R$, such that $\alpha < \beta$, $y$ defined as above satisfies the equation, so there are infinite solutions $\endgroup$
    – user55268
    Jun 9, 2015 at 21:29

3 Answers 3

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$\newcommand{\Strut}{\vphantom{(}}$When performing separation of variables, you can't re-write your ODE as $y^{-1/2}\, dy = dt$ in a neighborhood of $t_{0}$ if $y(t_{0}) = 0$.

What you might do instead is:

  1. Observe that $y(t) = 0$ is a solution in an arbitrary interval.

  2. If $y(t_{0}) = y_{0} > 0$, separate variables in a neighborhood of $t_{0}$ on which $y$ is positive: $$ t - t_{0} = \int_{t_{0}}^{t} y^{-1/2}\, dy = 2\left(\sqrt{y(t)} - \sqrt{y_{0}\Strut}\right), $$ so $y(t) = \frac{1}{4}\bigl(t - t_{0} + 2\sqrt{y_{0}\Strut}\bigr)^{2}$.

  3. If $y(t_{0}) = y_{0} < 0$, separate variables in a neighborhood of $t_{0}$ on which $y$ is negative: $$ t - t_{0} = \int_{t_{0}}^{t} (-y)^{-1/2}\, dy = -2\left(\sqrt{-y(t)} - \sqrt{-y_{0}\Strut}\right), $$ so $y(t) = -\frac{1}{4}\bigl(t - t_{0} + 2\sqrt{-y_{0}\Strut}\bigr)^{2}$.

  4. Observe that all three solutions have $y' = 0$ when $y = 0$ (as required by the ODE), so piecing together formulas over abutting intervals gives continuously-differentiable solutions.

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    $\begingroup$ Incidentally, every solution $y$ is non-decreasing (obvious from the ODE, or can be read off the integrated equations in 2 and 3). The quadratics as given come with implicit "fine print": The solutions are the "right half" ($t > t_{0} - 2\sqrt{y_{0}\Strut}$) of the quadratic in 2, or the "left half" ($t < t_{0} - 2\sqrt{-y_{0}\Strut}$) in 3. Particularly, the function $y_{\alpha,\beta}$ in the question appears not to be a solution; you'd need a minus sign in the portion where $t < \alpha$. $\endgroup$ Jun 10, 2015 at 1:53
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$$y=0$$ is a trivial solution, and as $\sqrt{|y|}\ge0$, any solution must be growing. More precisely, negative growing, then zero for a while, then positive growing.

If $y>0$, we can write

$$\frac{y'}{2\sqrt y}=\frac12$$ and

$$\sqrt y=\frac{x-x_+}2$$ for $x>x_+$ or $$y=\frac{(x-x_+)^2}4.$$

Similarly, if $y<0$,

$$y=-\frac{(x_--x)^2}4$$ for $x<x_-$.

Note that the negative and positive sections are optional. In case they both exist, we must have $$x_-\le x_+,$$ and with the given initial condition,

$$x_-\le0\le x_+$$

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Since you have the initial condition $y(0)=0$, we should now focus on solving the equation $y'=\sqrt{|y|}$ on $\mathbb{R}\setminus\{0\}$. For $y\lt0$, $|y|=-y$, and for $y\gt0$, $|y|=y$. Hence for $y\lt0$, $y=-(-y)=-\sqrt{-y}^2=-\sqrt{|y|}^2$, and for $y\gt0$, $y=|y|=\sqrt{|y|}^2$. Hence $$-2\sqrt{-y}\left[\sqrt{-y}\right]'=\left[-\sqrt{-y}^2\right]'=\sqrt{-y},\,y\lt0$$ $$2\sqrt{y}\left[\sqrt{y}\right]'=\left[\sqrt{y}^2\right]'=\sqrt{y},\,y\gt0,$$ leaving you with $$\left[\sqrt{-y}\right]'=-\frac12,\,y\lt0$$ $$\left[\sqrt{y}\right]'=\frac12,\,y\gt0$$ which implies $$\sqrt{-y(t)}=A-\frac12t,\,y\lt0$$ $$\sqrt{y(t)}=B+\frac12t,\,y\gt0.$$ If we want the equation to be satisfied on $\mathbb{R}$ rather than just $\mathbb{R}\setminus\{0\}$, then the condition $y(0)=0$ forces $A=B=0$. This causes somewhat of a problem. In fact, the two equations that were just derived imply this much: $A\geq\frac12t$ and $B\geq-\frac12t$, which for a choice of constants $A,B$, this constrains the domain of $y$ for any individual case rather significantly. However, what we have is that $\sqrt{-y(t)}=-\frac12t,\,y\leq0$ implies $t\leq0$, and $\sqrt{y(t)}=\frac12t,\,y\geq0$ implies $t\geq0$. In fact, we have that $y(t)=-\frac{t^2}{4},\,t\leq0$, $y(t)=\frac{t^2}{4},\,t\geq0$ is the nontrivial solution of the equation in $\mathbb{R}$. To put it succinctly, $$y(t)=\frac{t|t|}{4}.$$ It is easy now to verify, $$y'(t)=\frac{|t|}{2}=\sqrt{|y|}$$ for every $t\in\mathbb{R}$.

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  • $\begingroup$ Why the downvote? $\endgroup$
    – Angel
    Jan 14 at 17:24

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