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I have to verify the applicability of Picard-Lindelöf theorem for the Cauchy problems associated with the ODE $$x'(t) = |\sin{x(t)}|(t-e^{x(t)}).$$

In order to answer I have to verify that:

  1. $f(x,t)=|\sin{x(t)}|(t-e^{x(t)})$ is $\mathcal{C}(\mathbb{R^2})$

  2. $f(x,t)$ is locally Lipschitz continuous

The first condition is verified. For the second condition I tried to calculate the first order partial derivatives $\frac{\partial f(x,t)}{\partial x}$. If they are continuous, the condition is verified. $$f(x,t)=\cases{\sin{x(t)}(t-e^{x(t)})\quad \sin{x(t)}\ge0\\-\sin{x(t)}(t-e^{x(t)})\quad \sin{x(t)}<0}$$ $$f'(x,t)=\cases{\cos{x(t)}(t-e^{x(t)})-\sin{x(t)}e^{x(t)}\quad \sin{x(t)}\ge0\\-\cos{x(t)}(t-e^{x(t)})+\sin{x(t)}e^{x(t)}\quad \sin{x(t)}<0}$$ but the derivative is not continuous for $\sin{x(t)}=0$ (i.e $\cos{x(t)}=1$ and $x=0+2k\pi$) because $$t\neq-t,$$ so I cannot say that there is an unique solution for the Cauchy problems associated.

Is my approach correct?

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  • $\begingroup$ $f$ is just a function of two numbers, $f(x,t)=|\sin(x)|(e-e^x)$. What do you know about the arithmetic and composition of locally Lipschitz functions? $\endgroup$ – LutzL Jun 30 '18 at 13:52
  • $\begingroup$ Composition of locally Lipschitz functions is locally Lipschitz and I know that $(t-e^{x(t)})$ is locally Lipschitz and also $|\sin{x(t)}|$ is locally Lipschitz because its derivative with respect to x is $\pm\cos{x(t)}$. So I can conclude that $f(x,t)$ is locally Lipschitz and a unique solution is locally guaranteed. Is it right? $\endgroup$ – Phi_24 Jun 30 '18 at 14:07
  • $\begingroup$ Yes, that you can do. There is no $x(t)$ in the function $f$, it appears only in the differential equation $\dot x(t)=f(x(t),t)$. $\endgroup$ – LutzL Jun 30 '18 at 14:46

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