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In my research I need to show that the set $$M \mathrel{:=} \{X \in \mathbb{R}^4,X\ge0\}$$ where $$X(t)=(x_1(t),x_2(t),x_3(t),x_4(t))^T$$ is positively invariant with respect to the following system of fractional ordinary differential equations $$D^{\alpha}(x(t))=f(t,x(t))$$ with initial non-negative condition $x(0)=x_0$, where $f$ is nonlinear and continuous.

My question is: how do I show that $M$ is positively invariant with respect to the system given? Any ideas, references are appreciated.

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Let $\alpha\in(0,1]$. Rewrite the equation in an integral form $$x(t)=x_0+\frac1{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}f(s,x(s))ds.$$ The classical method to prove the existence of a solution $x(t)$ is to define a sequence $(x^n)_{n\ge0}$ by $x^0\equiv x_0$ and $$x^{n+1}(t)=x_0+\frac1{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}f(s,x^n(s))ds,$$ and to prove that it converges to some $x$.

If your assumption on $f$ is that $f(t,M)\subset M$, then every $x^n$ takes values in $M$, and therefore the limit $x$ takes values in $M$.

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  • $\begingroup$ Thanks for your consideration, How do i proved that $x_n$ converges to $x$. Plus, i can not assume $f(t,M) \in M$. $\endgroup$ Commented Jan 27, 2015 at 13:46
  • $\begingroup$ If $f(t,M)$ is not included in $M$, there is no hope that $M$ be positively invariant. You can convince yourself by considering the ODE case ($\alpha=1$). $\endgroup$ Commented Jan 27, 2015 at 13:54
  • $\begingroup$ Thanks for your hint, if i show that $D^{\alpha}(x_i)|_{x_i=0} \geq 0$ (for i=1,2,3,4), then according "Generalized mean value theorem", can i conclude $f(t,x(t))$ on $M$ is non-decreasing and solution of the above system remain in $M$ $\endgroup$ Commented Jan 29, 2015 at 7:09
  • $\begingroup$ @DennisSerre: Actually, one does need to assume $f(t,M) \subseteq M$ in the ODE case; for (first order) ODEs, positive invariance of $M$ is characterised by the (considerably weaker) property of *quasi-positivity" of $f$ (but I don't know whether it's also the right condition for the fractional case). $\endgroup$ Commented Jan 20, 2021 at 5:23
  • $\begingroup$ Ouch - now I justed noted that the post is over five years old. Sorry for that, it's still early in the morning where I live, so I guess I'm not completely awake, yet. $\endgroup$ Commented Jan 20, 2021 at 5:24

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