# question about uniqueness of solution of initial value problem

consider the IVP $$\dot{y}=\sqrt{|y|},y(x_0)=y_0$$ then it has two solutions namely y=0 and y= \begin{cases} \frac{x^2}{4}, & \text{if $x\ge0$} \\ -\frac{x^2}{4}, & \text{if $x\lt0$} \\ \end{cases} but on solving i found that it has solution $y=0$ and y=\begin{cases} \frac{x^2}{4}, & \text{if $y\ge0$} \\ -\frac{x^2}{4}, & \text{if $y\lt0$} \\ \end{cases} i found it as $|y|=y$ if $y \ge 0$ and $|y|=-y$ if $y<0$ and doing integration can someone tell me where I am wrong how the case become at $x$ not on $y$. $\bf EDIT$if we changes the initial condition as $y(0)=0$ then how to solve the problem?the actuall problem i wanted to ask is with these initials condition

• Are you asking which is correct? That is, using $x \ge 0$ or $y \ge 0$, for example. Regards Oct 4, 2013 at 5:49
• yes i want to ask the same
– abc
Oct 4, 2013 at 6:07
– abc
Oct 4, 2013 at 6:14

Your solutions need to be functions of $x_0$ and $y(x_0)$.

Since $\dot{y}(x) \ge 0$, it is clear that $y$ is non-decreasing for $x \ge x_0$.

First I will deal with $y(x_0) >0$. Then it is clear that $\dot{y}(x) > 0$, hence $y(x) \ge y(x_0)$ for all $x \ge x_0$.

Then there is a unique solution $y(x) = (\frac{1}{2}(x-x_0)+ \sqrt{y(x_0)})^2$.

If $y(x_0) = 0$, two solutions are $y(x) = 0$ and $y(x) = (\frac{1}{2}(x-x_0))^2$.

If $y(x_0) <0$, then the solution is $y(x) = -(\frac{1}{2}(x_0-x)+ \sqrt{-y(x_0)})^2$, and this is valid for $x \in [x_0, x_0+2 \sqrt{-y(x_0)})$. For $x \ge x_0+2 \sqrt{-y(x_0)}$, the two solutions described above apply (with appropriate changes for $x_0$, of course). The solutions is unique on $[x_0, x_0+2 \sqrt{-y(x_0)})$.

– abc
Oct 4, 2013 at 6:36
• I have given all possible answers for all $x_0$ and all $y(x_0)$. If $x_0 = 0$ and $y(x_0) = 0$, then the two solutions are $y(x) = 0$ and $y(x) = \frac{1}{4} x^2$, for $x \ge 0$. Oct 4, 2013 at 6:40
• thanks sir is then y(x)=-$\frac{x^2}\4$.
– abc
Oct 4, 2013 at 6:45
• Was that a rhetorical question? Oct 4, 2013 at 6:47
• merriam-webster.com/dictionary/rhetorical Oct 4, 2013 at 6:50

\begin{align} {\rm d}\sqrt{\left\vert y\right\vert\,} &= {1 \over 2\sqrt{\left\vert y\right\vert\,}}\,\,{\rm sgn}\left(y\right){\rm d}y \\[3mm] {{\rm d}y \over \sqrt{\left\vert y\right\vert\,}} &= 2{\rm sgn}\left(y\right){\rm d}\sqrt{\left\vert y\right\vert\,} = 2{\rm d}\left[{\rm sgn}\left(y\right)\sqrt{\left\vert y\right\vert\,}\right] - 2\sqrt{\left\vert y\right\vert\,}\,\left[2\delta\left(y\right)\right] = 2{\rm d}\left[{\rm sgn}\left(y\right)\sqrt{\left\vert y\right\vert\,}\right] \end{align}
$$2{\rm sgn}\left(y\right)\sqrt{\left\vert y\right\vert\,} - 2{\rm sgn}\left(y_{0}\right)\sqrt{\left\vert y_{0}\right\vert\,} = x - x_{0}$$ $$\sqrt{\left\vert y\right\vert\,} = {\rm sgn}\left(y\right)\left[% {\rm sgn}\left(y_{0}\right)\,\sqrt{\left\vert y_{0}\right\vert\,} + {1 \over 2}\,\left(x - x_{0}\right) \right]$$

$$\left\vert y\right\vert = \left[% {\sqrt{\left\vert y_{0}\right\vert\,} + {1 \over 2}\,\rm sgn}\left(y_{0}\right)\left(x - x_{0}\right) \right]^{2}$$

$$\begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% y \color{#000000}{\ =\ } {\rm sgn}\left(y\right)\left[% {\sqrt{\left\vert y_{0}\right\vert\,} + {1 \over 2}\,\rm sgn}\left(y_{0}\right)\left(x - x_{0}\right) \right]^{2} \quad} \\ \\ \hline \end{array}$$