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In my research I need to show that the set

$$M := \{X \in \mathbb{R}^4,X≥0\}$$

where

$$X(t)=(x_1(t),x_2(t),x_3(t),x_4(t))^T$$

is positively invariant with respect the following system of fractional ordinary differential equation

$$D^{\alpha}(x(t))=f(t,x(t))$$

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 $f(t,M) \in M$. $\endgroup$ – Ehsan Mottaghi Jan 27 '15 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 ($\alpha=1$). $\endgroup$ – Denis Serre Jan 27 '15 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 $M$ $\endgroup$ – Ehsan Mottaghi Jan 29 '15 at 7:09

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