# Prove that an initial value problem has more than 1 solution

let $$y' = \sin{\sqrt{|y|}}$$, I need to prove that there is more than one solution with the initial value $$y(0) = 0$$, and I was given guidance to use $$y = z^2$$, reach $$z'=f(x, z)$$ and use the existence and uniqueness theorem to find a solution for $$z'=f(x, z)$$, with it I'm suppose to prove the existence of a second solution.

the first solution is easy to find $$y \equiv 0$$ is a solution. But I'm not sure how to find the second even with the guidence, here is what I tried:

Using the guidance I say $$y = z^2 \Longrightarrow y' = 2z \cdot z' \Longrightarrow 2z\cdot z' = \sin{\sqrt{|z^2|}} = \sin{z} \Longrightarrow z' = \frac{\sin{z}}{2z}$$

Because $$\frac{\sin{z}}{2z}$$ is continuous and $$\frac{\partial}{dz}\frac{\sin{z}}{2z}$$ is also continuous when $$z\neq 0$$ from the existence and uniqueness theorem I get that there is a solution $$u(x)$$ for $$z' = \frac{\sin{z}}{2z}$$ and a point $$(x_0, z_0)$$ ($$z_0 \neq 0$$), so I create a new function $$v(x)$$ where $$v(0) = 0$$ and $$v(x) = u^2(x)$$ for $$x\neq 0$$, $$v$$ is a solution for $$y' = \sin{\sqrt{|y|}}$$, and it upholds the initial value. Because it doesn't uphold the existence and uniqueness theorem 's requirements there is no contradictions with having 2 solution to the same initial value problem

Though my solution seems good to me I feel like something is missing, but I can't put a finger on it

EDIT:

I'm was told to work with $$y = z^2$$, I understand that the solution to the IVP should look like $$v(x) = \left\{ \begin{array}{ll} 0 & \mbox{if } x \leq 0 \\ u^2(x) & \mbox{if } x > 0 \end{array} \right.$$ where $$u(x)$$ is a solution for $$z' = \frac{\sin{z}}{2z}$$. $$u(x)\not\equiv 0$$ because $$0$$ isn't a solution for $$z$$, so $$v(x) \not\equiv 0$$ but I don't know how I can prove that $$v(x)$$ is derivable

• There is one solution that "immediately moves as you go in either direction", which is the one that you get from separating variables. It is implicitly described by $\int_0^y \frac{d\tilde{y}}{\sin(\sqrt{\tilde{y}})} = x$. There are other solutions that first budge from $y=0$ at a different time, or (as you noticed) never budge at all. This is easier to write down to develop intuition with $y'=\sqrt{y}$ instead, but the behavior is really the same near $y=0$.
– Ian
Oct 30, 2021 at 18:28
• Without trying it I am not sure whether this suggestion to play with $y=z^2$ helps you any.
– Ian
Oct 30, 2021 at 18:30
• @Ian so If I understand correctly, my answer is good but there are simpler ways to find another solution? Oct 30, 2021 at 20:42
• Now that I think about it, $z = \frac{\pi}{2}$ is a solution, thus $y = \frac{\pi^2}{4}$ is also a solution for ODE, so if I define a new function $v(x)$, such that $v(0) = 0$ and $v(x)=\frac{\pi^2}{4}$ for $x \neq 0$, $v(x)$ is a solution for the initial value problem and it doesn't contradict the existence and uniqueness theorem because $v(x)$ is not continuous in any area around $(0, 0)$ Oct 31, 2021 at 10:03
• A solution to an IVP is is differentiable by definition and thus is also continuous by definition. So no, jumping between squares of zeros of $\sin$ doesn't give you the non-uniqueness. It is really exactly the same as the situation with $y'=|y|^{1/2}$ even though that only has one zero.
– Ian
Oct 31, 2021 at 13:25

Based on $$z'=\frac{\sin(|z|)}{2z}$$ for $$z\ne 0$$, you can consider the ODEs $$z'=\pm f(z)~~\text{ where }~~f(z)=\begin{cases}\frac{\sin(z)}{2z},&z\ne 0\\\frac12,&z=0\end{cases}$$ As you observed, $$f(z)$$ has a nice power series expansion, so satisfies the conditions of the existence-and-uniqueness theorem.
Solutions of $$z'=f(z)$$ give solutions of the original ODE whenever $$z\ge0$$, and those of $$z'=-f(z)$$ whenever $$z\le0$$. Both variants give a solution each for $$x\ge 0$$ that passes through $$z(0)=0$$.