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I was thinking about a ODE problem recently when I was reading about dynamical system. In school we used to solve the ODE problem $\frac{dx}{dt}=\sqrt{1-x^2}, x=0, t=0$ as $x=\sin(t),$ which will have the graph enter image description here

Now in dynamical system we can see that the fixed points are $\pm 1$ so specifically we can observe if the solution hits $1$ or $-1$ it should not increase or decrease from there. Specifically if we draw the phase diagram we can conclude that the solution passing through $(0,0)$ should look like

enter image description here

and it seems reasonable. So I am surprised that we were taught wrong for many days. Isn't it? Or, am I making any mistake?

Ri-Li
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  • Related: https://stackoverflow.com/questions/74165228/notimplementederror-initial-conditions-produced-too-many-solutions-for-constant – Galen Oct 22 '22 at 16:26

1 Answers1

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Since $\frac{dx}{dt}=\sqrt{1-x^2}\geq 0$, the solution should be increasing (not necessarily strictly)!

The global solution to the Cauchy problem $$\begin{cases}\frac{dx}{dt}=\sqrt{1-x^2}\\ x(0)=0 \end{cases}$$ is $$x(t)=\begin{cases} \sin(t) & t\in [-\pi/2,\pi/2],\\ 1 & t\geq \pi/2,\\ -1 & t\leq -\pi/2. \end{cases}$$

Robert Z
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  • Ahh! I can see why you are saying so. The uniqueness of the global solution. Yeah, it was strange that I didn't realise this at school. Thanks a lot. – Ri-Li Oct 22 '22 at 07:09