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Let $z(t)$ be a differentiable function for all $t≥0$. I want to solve the inequality: $$z′(t)+1≥0$$ for all $t≥0$.

where $z′(t)$ is the derivative of $z(t)$.

DER
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    Any constant, in fact any linear function with slope $≥-1$ will work. Aren't there any more conditions? The problem seems quite open. – rewritten Dec 23 '13 at 09:17
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    I cant see an unique solution every $z(t)=t^k$ will give you a solution – Eli Elizirov Dec 23 '13 at 09:21
  • @rewritten: We can add the fact that $z(t)$ is bounded. – DER Dec 23 '13 at 09:26
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    Then $\sin (ax+b)$ will also do for any $b$ and $|a|\leq 1$. And you can easily imagine sums of such functions with varying $|a_n|$ so that $\sum_n |a_n| \leq 1$. – Jean-Claude Arbaut Dec 23 '13 at 09:31
  • @arbautjc: I ask if there is a general form of solutions. – DER Dec 23 '13 at 09:38
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    The most general I can see for now is $f(x)=K+\sum_n \lambda_n \sin(a_nx+b_n)$ with $\sum_n |\lambda_n a_n| \leq 1$, with a possibly infinte sum (provided it converges to a differentiable function). If $a_n=n$, $b_n=0$ and suitably decreasing $\lambda_n$, you get Fourier series (you can add a $\cos (nx)$ part of course), so maybe there is something to look at here. – Jean-Claude Arbaut Dec 23 '13 at 09:41
  • @arbautjc: What if $z(t)$ is a component of a 3D chaotic system of the form $$x′=y$$ $$y′=-x-yz$$ $$z′=y²-1$$ – DER Dec 23 '13 at 09:45
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    Not easy to generalize. You may find classes of solutions, but I'm not sure you can find a characterization of all solutions that will differ from the mere definition in the question. Just a silly opinion, I'm not an expert ;-) Also, are you sure the solution of your system is bounded? – Jean-Claude Arbaut Dec 23 '13 at 09:50
  • Yes, at least numerically. And there is a conjecture claiming this result. – DER Dec 23 '13 at 09:55

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