MHB Prove Local Uniqueness of DE Solutions on Interval

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if a function ls locally lip then considering this diff eq x'(t)= f(x(t) where now x and y are solutions of the DE on some interval J
and x(s)=y(s) for some s in J. then how can I prove that there exists a positive number delta such that x=y on (s-delta, s+delta)∩ J
 
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The proof of the Picard–Lindelöf theorem converts the original IVP
\begin{align*}
x'(t)&=f(x(t))\\
x(t_0)&=x_0
\end{align*}
into an integral equation
\[
x(t)=x_0+\int_{t_0}^tf(x(s))\,ds.\qquad{(*)}
\]
Define an operator $P(x)(t)=x_0+\int_{t_0}^tf(x(s))\,ds$, so (*) becomes
\[
x(t)=P(x).
\]
Thus, $x(t)$ is a fixpoint of $P$ iff $x(t)$ is a solution to the original IVP. The proof shows that there exists a $\delta$ such that $P$ is a contraction on $C[t_0-\delta,t_0+\delta]$ and thus has a unique fixpoint. Therefore, the solution to the IVP is also unique.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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