Suppose that we have a piecewise, differential equation. Say
$\frac{dx}{dt} = \begin{cases} x \sin \frac{1}{x} & x \neq 0 \\ 0 & x = 0\end{cases}$
$x(0) = 0$
I like to ask if the differential equation has a unique solution. I have tried two different methods, but none of them can seem to tell me anything. Later, a friend told me about his theorem, but I am unsure if his theorem is true or not.
Theorem 1 Lipschitz
$|f(x) – f(y)| \leq L |x – y|$,
It doesn't seem to satisfy the Lipschitz condition when $x^{-1} = \frac{\pi}{2} + 2 n \pi$ and $y^{-1} = \frac{3\pi}{2} + 2 n \pi$ because I end up getting $L$ to be $2(2n+1)$ and when $n \to \infty$, $L \to \infty$. So I don't think I can apply the theorem because it doesn't fit the first part of the theorem.
Theorem 2 Continuity
If $f(t,y)$ and $\frac{\partial f}{\partial y}$ are continuous functions in some rectangle $α<t<β$, $γ<y<δ$ containing the point $(t_o,y_o)$, then there is a unique solution to the IVP in some interval $t_o – h < t < t_o+h$ that is contained in $α<t<β$.
$\frac{\partial f}{\partial y}$ is not continuous so I don't think I can apply the theorem because it also doesn't fit the first part of the theorem.
Theorem 3 Absolute Asymptotic Condition Number (unsure of this theorem)
If $f(x)$ has a constant as an absolute asymptotic condition number, then the differential equation has a unique solution. It seems to be saying that if
$\underset{x \to 0}{\lim} \sup \frac{|f(x) – f(0)|}{|x – 0|} < \infty$
then it has a unique solution.
I calculated the absolute asymptotic condition number to be 1, which is a constant. I believe that it is also saying that it is asymptotically stable. Furthermore, since it satisfies the theorem, then I must say the differential equation has a unique solution.
So does this differential equation actually have a unique solution?
If I am wrong saying that, since I can't use the first part of the theorem, I cannot use the theorem, then let me know.
Please and thank you.
Best Answer
If $x(0)\in \left[\frac1{(n+1)\pi},\frac1{n\pi}\right]$ then go some step below, set $a=\frac1{2n\pi}$ and recognize that on $x\in[a,\infty)$ the right side is smooth, so the existence and uniqueness theorem applies.
But as $x_L(t)=\frac1{(n+1)\pi}$ and $x_U(t)=\frac1{n\pi}$ are constant solutions, they bound the solution $x(t)$ which thus exists for all times.
By exclusion, the solution $x(t)=0$ is also unique, no branching to non-zero values is possible.
Your counter-example for the Lipschitz condition is not correctly constructed, you probably wanted $x^{-1} = \frac{\pi}{2} + 2 n \pi$ and $y^{-1} = \frac{3\pi}{2} + 2 n \pi$ so that the sine factor has values $\pm 1$. But then $$ |f(x)-f(y)|=x+y=\frac{2\pi+4n\pi}{(\frac{\pi}{2} + 2 n\pi)(\frac{3\pi}{2} + 2 n\pi)}\sim\frac1n $$ which is still asymptotically larger than $$x-y=\frac{\pi}{(\frac{\pi}{2} + 2 n\pi)(\frac{3\pi}{2} + 2 n\pi)}\sim\frac1{n^2},$$ so that no Lipschitz constant exists at $x=0$.