MHB Maximal Existence Interval for DE with Continuous Derivative

onie mti
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I am given that a DE with the form x' = f(x) is defined on the interval (c,b) where f has continuous derivative on its domain
How do i show that if f(p) = f(q) = 0 and x(t) is between p and q then the maximal interval of existence of x is (-∞, ∞)
 
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onie mti said:
I am given that a DE with the form x' = f(x) is defined on the interval (c,b) where f has continuous derivative on its domain
How do i show that if f(p) = f(q) = 0 and x(t) is between p and q then the maximal interval of existence of x is (-∞, ∞)

Lets suppose that in p < x < q is f(x)>0 and no other zero of f(x) exists outside this interval. In thi case x=p is a repulsive fixed point, i.e. all the solution of the DE with the only exception of the constant solution x=p will diverge from p, and x=q is an attractive fixed point, i.e. all the solution of the DE with the only exception of the constant solution x=p will converge to q. That means that any solution of the DE with initial condition $x(t_{0})= x_{0},\ p < x_{0} < q$ will converge to q. Furthemore f(x) is function of the x alone, so that a if x(t) is solution of the DE, then $x(t + \tau),\ \tau \in \mathbb{R}$ is also solution of the DE. All that is sufficient to prove the hypothesis...

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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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